Valeriu Soltan Lectures on Convex Sets

Valeriu Soltan Lectures on Convex Sets

i

ii

Lectures on convex sets

Preface

As any other well-established mathematical discipline, convex geometry is divided into various subfields, like generalized convexity, finite-dimensional Banach spaces, asymptotic theory of convex bodies, combinatorial convexity, mixed volumes, etc. Among all these subfields, algebraic and topological properties of convex sets in the Euclidean space Rn have a place apart: they serve the whole convex geometry and are widely used in various mathematical disciplines. The role of these properties outside convex geometry is especially visible in convex analysis, optimization, operations research, and their applications. The first systematic treatment of algebraic and topological properties of convex sets in Rn was given in the seminal monograph of Rockafellar [184], published in 1970. Since then, the list of chapters on convex sets used in convex analysis became almost canonical (see, e. g., the books of Florenzano and Le Van [87], G¨ uler [107], Hiriart-Urruty and Lemaréchal [114], Stoer and Witzgall [210], and some others). The topics usually include general properties of convex sets and convex hulls, cones and conic hulls, polyhedral sets, the extreme structure and separation properties of convex sets. Various monographs on convex geometry provide another source for the study of the above properties of convex sets, predominantly for the case of convex bodies (see, for instance, Gardner [93], Gruber [102], Schneider [191], and many others). Despite the presence of a large number of monographs on convex geometry, there are quite few books on this topic, suitable for the undergraduate study. Examples could be the books of Lay [147] and Webster [222] on convex sets, most recent of them published in 1994. The present book provides a systematic treatment of the above topics, which concern arbitrary convex sets in Rn , possibly non-closed or uniii

iv


bounded. An essential part of the text is adapted from various research articles or is written ad hoc by the author for the purpose of completeness. Comments on further related topics and results, as well as some bibliographic notes, are given at the end of each chapter. A selection of exercises (with solutions given at the end of the book) provides additional results to the main text. This book grew up out of various courses on geometry and convexity, taught by the author at George Mason University. The audience usually consisted of students interested in all fields of mathematics and operations research. The text can be used for a one-semester undergraduate or an entry-level graduate course on convex sets, or as a supplementary book for any course on convex geometry or convex analysis. Also, the book may be viewed as a source for independent study of the subject, suitable to non-geometers. Prerequisites include proof-based undergraduate courses on linear algebra, analysis, and elementary topology. The text is divided into ten chapters. The introductory Chapter 0 gives a brief account of necessary prerequisites. It is assumed that the reader starts the book from Chapter 1 and uses the first one as a reference material. Chapter 1 deals with the affine structure of the Euclidean space Rn . It describes various properties of planes, affine spans, and affine transformations. Assuming that students have little knowledge of n-dimensional geometry, this chapter develops the topic as a natural extension of linear algebra and sets up the methods of proof for the entire book. The next three chapters are devoted to basic constructions concerning algebraic and topological properties of convex sets (see Chapter 2) and corresponding properties of convex and conic hulls (Chapters 3 and 4). Chapters 5–8 deal with general properties of convex sets: recession and normal cones, support and separation, the extreme and exposed structures. The concluding Chapter 9 demonstrates how the results from previous chapters can be applied to the study of polyhedral sets, which are important in many applications. Further material on polyhedral sets and polytopes can be found in the books of Gr¨ unbaum [104], Brøndsted [41], and Ziegler [228]. For comments, information, and corrections to this book, please contact the author at [email protected]. V.S.

Notation

|a| sgn (a) R [a, b] [a, b) (a, b) [a, ∞) (−∞, a) X ∪Y X ∩Y X \Y ∅ card X f :U →V f −1 g ◦f f (X) f −1 (Y ) rng f Rn X +Y X ⊕Y span X null f x·y X⊥

absolute value of a scalar a sign value of a scalar a set of real numbers (scalars) closed interval in R semi-open interval in R open interval in R closed halfline in R open halfline in R union of sets X and Y intersection of sets X and Y set difference of sets X and Y empty set cardinality of a finite set X mapping from a set U into a set V inverse of a mapping f composition of mappings f and g f -image of a set X inverse f -image of a set Y range of a mapping f n-dimensional coordinate space (Minkowski) sum of sets X and Y direct sum of sets X and Y span of a set X null space of a linear transformation f dot product of vectors x and y orthogonal complement of a set X

v

1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 4 4 5 5 5

vi

kxk kx − yk Bρ (c) Uρ (c) Sρ (c) Bρ (X) int X cl X δ(X, Y ) sub L dim L hx, yi [x, yi (x, yi [x, y] [x, y) (x, y) aff X sub X dim X rint X rbd X conv X convr X ap C cones X rec X lin X pK (x) NK (z) nor K bar K FK (X) ext K extr K GK (X) exp K expr K


norm of a vector x distance between points x and y closed ball with center c and radius ρ open ball with center c and radius ρ sphere with center c and radius ρ ρ-neighborhood of a set X interior of a set X closure of a set X inf -distance between sets X and Y characteristic subspace of a plane L dimension of a plane L line through points x and y closed halfline through y with endpoint x open halfline through y with endpoint x closed segment with endpoints x and y semi-open segment with endpoints x and y open segment with endpoints x and y affine span of a set X characteristic subspace of a set X dimension of a set X relative interior of a set X relative boundary of a set X convex hull of a set X union of convex hulls of r-pointed subsets of X apex set of a cone C conic hull with apex s of a set X recession cone of a set X linearity space of a set X metric projection of point x on convex set K normal set of K at point z ∈ K normal cone of a convex set K barrier cone of a convex set K extreme face of K generated by a set X ⊂ K set of extreme points of K set of r-extreme points of K exposed face of K generated by a set X ⊂ K set of exposed points of K set of r-exposed points of K

6 6 7 7 7 7 7 8 9 12 13 25 25 25 26 26 26 26 42 55 78 104 117 122 153 161 187 197 208 209 211 212 260 267 278 298 299 310

Contents

Preface

iii

Notation

v

0.

Prerequisites

1

1.

The affine structure of Rn

11

1.1 1.2 1.3

11 37 57

2.

3.

Convex sets

71

2.1 2.2 2.3

71 78 93 117

Algebraic properties of convex hulls . . . . . . . . . . . . . 117 Topological properties of convex hulls . . . . . . . . . . . . 128

Convex cones and conic hulls 4.1 4.2

5.

Algebraic properties of convex sets . . . . . . . . . . . . . Relative interior of convex sets . . . . . . . . . . . . . . . Closure and relative boundary of convex sets . . . . . . .

Convex hulls 3.1 3.2

4.

Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Affine spans . . . . . . . . . . . . . . . . . . . . . . . . . . Affine transformations . . . . . . . . . . . . . . . . . . . .

145

Convex cones . . . . . . . . . . . . . . . . . . . . . . . . . 145 Conic hulls . . . . . . . . . . . . . . . . . . . . . . . . . . 161

Recession and normal directions 5.1 5.2

183

Recession cones . . . . . . . . . . . . . . . . . . . . . . . . 183 Lineality spaces . . . . . . . . . . . . . . . . . . . . . . . . 197 vii

viii


5.3 6.

Support and separation properties 6.1 6.2 6.3

7.

9.

287

Exposed faces . . . . . . . . . . . . . . . . . . . . . . . . . 287 Exposed representations . . . . . . . . . . . . . . . . . . . 299

Polyhedra 9.1 9.2

255

Extreme faces . . . . . . . . . . . . . . . . . . . . . . . . . 255 Extreme representations . . . . . . . . . . . . . . . . . . . 267

The exposed structure of convex sets 8.1 8.2

227

Bounds, supports, and asymptotes . . . . . . . . . . . . . 227 Containing and supporting halfspaces . . . . . . . . . . . . 235 Separation by hyperplanes and slabs . . . . . . . . . . . . 241

The extreme structure of convex sets 7.1 7.2

8.

Normal and barrier cones . . . . . . . . . . . . . . . . . . 206

319

Intersections of halfspaces . . . . . . . . . . . . . . . . . . 319 Properties of polyhedra . . . . . . . . . . . . . . . . . . . 328

Solutions to Exercises

345

Bibliography

387

Author Index

399

Subject Index

401

Chapter 0

Prerequisites

Basic Set Theory and Notation In what follows, R stands for the set of real numbers, called scalars. Scalars a and b, with a < b, determine the closed interval [a, b], two semi-open intervals [a, b) and (a, b], and open interval (a, b) of the number line R. Similarly, [a, ∞), (a, ∞), (−∞, a], and (−∞, a) are the halflines of R determined by a. The absolute value and sign value of a scalar a are denoted |a| and sgn (a), respectively. The symbol marks the end of a proof or of a statement without proof. Given a universal set U , the union, intersection, and set difference of sets X and Y from U are denoted X ∪ Y , X ∩ Y , and X \ Y , respectively. The symbol ∅ stands for the empty set. In a standard way, x ∈ X means that an element x of U belongs to X, and Y ⊂ X shows that Y is a subset of X (possibly, X = Y ). In what follows, a subset Y of X is called proper if ∅ 6= Y 6= X. A set X ⊂ U is often described by its indexed elements, like X = {xα }, or by a given condition P (x) on U , as X = {x ∈ U : P (x)}. We distinguish finite, denumerable, countable (finite or denumerable), and uncountable sets. The number of elements in a finite set X is denoted card X. If F = {Xα : α ∈ A} is an indexed family of subsets of a universal set U , then De Morgan’s Laws state that U \ ∪ Xα = ∩ (U \ Xα ) and U \ ∩ Xα = ∪ (U \ Xα ). α

α

α

α

The family F is called disjoint if ∩ Xα = ∅, and is called pairwise disjoint α provided Xβ ∩ Xγ = ∅ for any distinct indices β, γ ∈ A. We say that F is nested if Xβ ⊂ Xγ or Xγ ⊂ Xβ whenever β, γ ∈ A. To simplify the arguments, we considered only nonempty families of sets. Assuming the axiom of choice, Zorn’s lemma implies that a family F of 1

2


subsets of a given set U contains a maximal element provided that every totally ordered subfamily of F has an upper bound in F. A mapping f from a set U into a set V (notation f : U → V ) assigns to every element x ∈ U a unique element f (x) ∈ V . The mapping f : U → V is onto if every element v ∈ V is an f -image of an element u ∈ U ; it is oneto-one if f (x) 6= f (y) whenever x 6= y; finally, f is invertible (equivalently, is a bijection) if it is both onto and one-to-one. A bijection f : U → V has the inverse mapping f −1 : V → U , defined by the identities f −1 (f (x)) = x and f (f −1 (y)) = y for all x ∈ U and y ∈ V. For a mapping f : U → V (not necessarily invertible), the image of a set X ⊂ U and the inverse image of a set Y ⊂ V are defined by and f −1 (Y ) = {x : f (x) ∈ Y }.

f (X) = {f (x) : x ∈ X}

Obviously, f (∅) = ∅ and f −1 (∅) = ∅. Furthermore, X ⊂ f −1 (f (X)),

f −1 (Y ) = f −1 (Y ∩ rng f ),

f (f −1 (Y )) = Y ∩ rng f.

The set f (U ), denoted rng f , is called the range of f . For a family {Xα } of subsets of U , one has f (∪ Xα ) = ∪ f (Xα ) α

α

and f (∩ Xα ) ⊂ ∩ f (Xα ), α

α

with equality if f is one-to-one. Similarly, f −1 (∪ Yα ) = ∪ f −1 (Yα ) α

α

and f −1 (∩ Yα ) = ∩ f −1 (Yα ) α

α

for a family {Yα } of subsets of V . If X1 , X2 ⊂ U and Y1 , Y2 ⊂ V , then f (X1 ) \ f (X2 ) ⊂ f (X1 \ X2 )

and f −1 (Y1 ) \ f −1 (Y2 ) = f −1 (Y1 \ Y2 ),

with f (X1 ) \ f (X2 ) = f (X1 \ X2 ) provided f is one-to-one. The composition g ◦ f of mappings f : U → V and g : V → W is defined by (g ◦ f )(x) = g(f (x)). The Vector Space Rn Throughout this book we deal with elements and subsets of the n-dimensional vector space Rn , n > 1, consisting of all n-tuples x = (x1 , . . . , xn ), where x1 , . . . , xn are scalars, named the coordinates of x. The elements of Rn are called vectors, or points. We make no difference between these two notions, and their use should support the reader’s intuition. To distinguish similarly looking elements, 0 will stand for the number zero, and o for the origin (zero vector) of Rn .

Prerequisites

3

Given vectors x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ) and a scalar λ, the sum x + y and the product λx are defined, respectively, as x + y = (x1 + y1 , . . . , xn + yn )

and λx = (λx1 , . . . , λxn ).

These operations can be expanded to the case of nonempty sets X and Y in Rn : X + Y = {x + y : x ∈ X, y ∈ Y }

and λX = {λx : x ∈ X}.

The set X + Y is often called the Minkowski sum of sets X and Y , and λX is the scalar multiple of X. Put X + Y = ∅ if at least one of the sets X and Y is empty; also, let λ ∅ = ∅ for all scalars λ. For sets X, Y, Z ⊂ Rn and scalars λ, µ, one has X + Y = Y + X,

(X + Y ) + Z = X + (Y + Z),

(X ∪ Y ) + Z = (X + Z) ∪ (Y + Z), (X ∩ Y ) + Z ⊂ (X + Z) ∩ (Y + Z), (λ + µ)X ⊂ λX + µX.

λ(X + Y ) = λX + λY,

If X consists of a single vector x, we use x + Y instead of {x} + Y and say that x + Y is a translate of Y . We write −x and −X for (−1)x and (−1)X, respectively. X +Y X @ @ Y H x HH

@ @ *H Y x + yH

HH H

HH H q o

Fig. 0.1

H * y

Y

QQ Q λX Q Q Q * Q XQQ q o

The sum X + Y and a scalar multiple λX.

A nonempty set S ⊂ Rn is called a subspace if it is closed under vector addition and scalar multiplication: x+y ∈ S and λx ∈ S whenever x, y ∈ S and λ ∈ R. The zero subspace {o} and the whole space Rn are trivial subspaces. Subspaces S and T of Rn are called independent provided S ∩ T = {o} (possibly, S = {o} or T = {o}); equivalently, every vector x in S + T is uniquely expressible as x = y + z, where y ∈ S and z ∈ T . The

4


sum of independent subspaces S and T is called direct and is denoted S ⊕T . A similar notation, X ⊕ Y , is used for the sum of nonempty subsets X and Y of independent subspaces S and T , respectively. If a subspace P of Rn is the sum of independent subspaces S and T , then we say that S and T are complementary in P . In particular, S and T are called complementary if S ⊕ T = Rn . The intersection of all subspaces containing a given set X ⊂ Rn is called the span of X (clearly, the span of ∅ is {o}). The span of a nonempty set X ⊂ Rn can be described in terms of linear combinations: span X = {λ1 x1 + · · · + λk xk : k > 1, λ1 , . . . , λk ∈ R, x1 , . . . , xk ∈ X}. For nonempty sets X and Y in Rn , one has span (X + Y ) ⊂ span (X ∪ Y ) = span X + span Y. Furthermore, span (X + Y ) = span X + span Y if o ∈ X ∩ Y . A set {x1 , . . . , xr } of vectors in Rn is called linearly dependent if there are scalars λ1 , . . . , λr , not all zero, such that λ1 x1 +· · ·+λr xr = o; otherwise {x1 , . . . , xr } is linearly independent. Each linearly independent subset of Rn contains at most n vectors. A basis for a subspace S ⊂ Rn is a linearly independent set whose span is S (the set ∅ is linearly independent and forms a basis for {o}). The vectors e1 = (1, 0, . . . , 0), e2 = (0, 1, . . . , 0), . . . , en = (0, 0, . . . , 1) form the standard basis for Rn . The (unique) number of vectors in a basis for a subspace S is called the dimension of S and is denoted dim S. A subspace S ⊂ Rn of dimension n − 1 is called a hypersubspace. For subspaces S and T of Rn , one has dim (S + T ) = dim S + dim T − dim (S ∩ T ). Furthermore, if a basis b1 , . . . , br for S ∩ T is completed into bases b 1 , . . . , b r , c1 , . . . , cp

and b1 , . . . , br , e1 , . . . , eq

for S and T , respectively, then the combined list b1 , . . . , br , c1 , . . . , cp , e1 , . . . , eq is a basis for S + T . A mapping f : Rn → Rm is called a linear transformation provided f (x + y) = f (x) + f (y)

and f (λx) = λf (x)

Prerequisites

5

whenever x, y ∈ Rn and λ ∈ R. A linear transformation f : Rn → R is called a linear functional. Any linear transformation f preserves linear combinations of finitely many vectors, and f (X + Y ) = f (X) + f (Y )

and f (λX) = λf (X)

for any sets X, Y ⊂ Rn and a scalar λ. The subspaces null f = {x ∈ Rn : f (x) = o}

and

rng f = {f (x) : x ∈ Rn }

are called, respectively, the null space and the range of f . One has dim (null f ) + dim (rng f ) = n. Furthermore, the following statements hold. 1. If {c1 , . . . , cr } is a linearly independent set in Rn and f is one-toone, then the set {f (c1 ), . . . , f (cr )} is linearly independent. 2. If {a1 , . . . , ar } is a linearly independent set in Rm and vectors c1 , . . . , cr ∈ Rn satisfy the conditions f (ci ) = ai , 1 6 i 6 r, then the set {c1 , . . . , cr } is linearly independent. For a linearly independent set {c1 , . . . , cr } in Rn and vectors a1 , . . . , ar in Rm , there is a linear transformation f : Rn → Rm such that f (ci ) = ai for all 1 6 i 6 r. If, additionally, r = n, then f is uniquely determined. Furthermore, if n = m = r and {a1 , . . . , an } is a basis for Rn , then f is invertible. If S and T are complementary subspaces of Rn and x = y + z is a (unique) expression of a vector x ∈ Rn as the sum of vectors y ∈ S and z ∈ T , then the linear transformation f : Rn → Rn defined by f (x) = y is called the linear projection on S along T . The Euclidean Structure of Rn The dot product x·y of vectors x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ) is defined by x·y = x1 y1 + · · · + xn yn . Vectors x and y are called orthogonal provided x·y = 0 (possibly, x = o or y = o). Nonempty sets X and Y in Rn are orthogonal if x·y = 0 whenever x ∈ X and y ∈ Y . The orthogonal complement of a nonempty set X ⊂ Rn is defined by X ⊥ = {y ∈ Rn : x·y = 0 for all x ∈ X}. The set X ⊥ is a subspace of Rn ; moreover, (X ⊥ )⊥ = span X

and

span X ⊕ X ⊥ = Rn .

6


If S and T are subspaces such that S ⊂ T , then the subspace S ⊥ ∩ T is the orthogonal complement of S within T . If S ⊂ Rn is a subspace, then S ⊕ S ⊥ = Rn , which gives dim S + dim S ⊥ = n. For subspaces S1 and S2 , one has (S1 + S2 )⊥ = S1⊥ ∩ S2⊥

and

(S1 ∩ S2 )⊥ = S1⊥ + S2⊥ .

Any real-valued linear functional ϕ(x) on Rn is expressible in the form ϕ(x) = x·c, where c ∈ Rn is a suitable vector. The norm, or length, of a vector x = (x1 , . . . , xn ) is defined by q √ kxk = x·x = x21 + · · · + x2n . Basic properties of the norm: 1. kxk > 0, with kxk = 0 if and only if x = o. 2. kλxk = |λ| kxk. 3. |x·y| 6 kxk kyk, with |x·y| = kxk kyk if and only if x and y are linearly dependent (Cauchy-Schwarz inequality). 4. kx + yk 6 kxk + kyk, with kx + yk = kxk + kyk if and only if one of x and y is a positive scalar multiple of the other. A unit vector is a vector of unit norm. A set X ⊂ Rn of pairwise orthogonal unit vectors is called orthonormal. For a pair of points x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ), the norm p kx − yk = (x1 − y1 )2 + · · · + (xn − yn )2 is called the distance between x and y. Clearly, 5. kx − yk > 0, with kx − yk = 0 if and only if x = y. 6. kλx − λyk = |λ| kx − yk. 7. kx − yk 6 kx − zk + kz − yk whenever x, y, z ∈ Rn . The linear projection p of Rn on a subspace S along its orthogonal complement S ⊥ is called the orthogonal projection on S. For any vectors x ∈ Rn and y ∈ S, one has kx − yk2 = kx − p(x)k2 + kp(x) − yk2 Consequently, p(x) is the unique nearest to x point in S and x − p(x) is orthogonal to S. If, additionally, dim S = n − 1, then S = {x ∈ Rn : x·c = 0}, where c is a certain nonzero vector, and the orthogonal projection p on S can be written as x·c c for all x ∈ Rn . p(x) = x − kck2

Prerequisites

7

Elementary Topology in Rn Given a point c ∈ Rn and a scalar ρ > 0, the sets Bρ (c) = {x ∈ Rn : kx − ck 6 ρ}, Uρ (c) = {x ∈ Rn : kx − ck < ρ}, Sρ (c) = {x ∈ Rn : kx − ck = ρ} are called, respectively, the closed ball, open ball, and sphere with center c and radius ρ. The sets B = B1 (o) and U = U1 (o) are called, respectively, the closed and open unit balls of Rn . Remark. To simplify certain geometric arguments, we consider neighborhoods of sets and points in terms of closed balls. Because of the inclusions Uδ (c) ⊂ Bγ (c) ⊂ Uρ (c) whenever 0 < δ < γ < ρ, our approach is topologically equivalent to the standard one. The ρ-neighborhood of a set X ⊂ Rn , denoted Bρ (X), is defined as Bρ (X) = ρB + X = ∪ (Bρ (x) : x ∈ X). $

'

X &

Bρ (X) %

A point x ∈ Rn is called interior for a set X ⊂ Rn provided there is a scalar ρ > 0 such that Bρ (x) ⊂ X. A set Y ⊂ Rn is said to be open if every point y ∈ Y is interior for Y . The union of interior points of a given set X, denoted int X, is called the interior of X (put int ∅ = ∅). It is easy to see that int X is the largest open set contained in X; furthermore, int X ⊂ X,

int (int X) = int X,

int X ⊂ int Y if X ⊂ Y.

n

For sets X1 , . . . , Xr in R , one has int (X1 ∩ · · · ∩ Xr ) = int X1 ∩ · · · ∩ int Xr . The union of a family of open sets and the intersection of finitely many open sets are open sets. Moreover, int X + Y ⊂ int (X + Y ) for sets X and Y in Rn . If a set X ⊂ Rn is covered with a family F of open sets, then F contains a countable subfamily of sets whose union also covers X (Lindel¨ of’s theorem).

8


We say that an infinite sequence of points x1 , x2 , . . . converges to a point x (and write lim xi = x, or xi → x) if lim kxi − xk = 0. With i→∞

i→∞

(i)

x = (x1 , . . . , xn ) and xi = (x1 , . . . , xn(i) ), i > 1, one has lim xi = x if and only if

i→∞

(i)

lim xj = xj for all 1 6 j 6 n.

i→∞

Clearly, kxi k → kxk provided xi → x. A point x ∈ Rn is called a closure point of a nonempty set X ⊂ Rn if every ball Bρ (x), ρ > 0, meets X. All closure points of X form the closure of X, denoted cl X, which can be expressed as cl X = ∩ (ρB + X : ρ > 0), where B is the closed unit ball (let cl ∅ = ∅). Given sets X and Y in Rn , X ⊂ cl X,

cl (cl X) = cl X,

cl X ⊂ cl Y if X ⊂ Y.

For sets X1 , . . . , Xr in Rn , one has cl (X1 ∪ · · · ∪ Xr ) = cl X1 ∪ · · · ∪ cl Xr . A set X ⊂ Rn is called closed if cl X = X. Equivalently, X is closed if and only if its complement Rn \ X is open. The intersection of a family of closed sets and the union of finitely many closed sets are closed sets. A subset Y of a set X is called dense in X provided X ⊂ cl Y . Every set X in Rn has a countable dense subset. The boundary of a set X ⊂ Rn , denoted bd X, is the set of all points x ∈ Rn such that every ball Bρ (x), ρ > 0, meets both X and Rn \ X. Clearly, X is closed if and only if bd X ⊂ X. Furthermore, bd X 6= ∅ if and only if X is a proper subset of Rn (that is, ∅ 6= X 6= Rn ). A set X ⊂ Rn is called bounded if it lies in a ball of positive radius. A nonempty set X ⊂ Rn is bounded if and only if every infinite sequence of points from X contains a convergent subsequence. The diameter of a bounded set X is defined by diam X = sup{ku − vk : u, v ∈ X}. A set X ⊂ Rn is called compact if every family of open sets in Rn whose union covers X contains a finite subfamily that also covers X. Criteria for the compactness of a nonempty set X in Rn : 1. X is bounded and closed. 2. Every infinite sequence of points from X contains a subsequence converging to a point in X (the Heine-Borel theorem).

Prerequisites

9

If X ⊂ Rn is a nonempty bounded set and ρ > 0, then there is a finite set {x1 , . . . , xr } ⊂ X satisfying the condition X ⊂ Bρ (x1 ) ∪ · · · ∪ Bρ (xr ). The inf -distance between nonempty sets X and Y in Rn is defined by δ(X, Y ) = inf {kx − yk : x ∈ X, y ∈ Y }. We say that X and Y are strongly disjoint provided δ(X, Y ) > 0. A mapping f : Rn → Rm is called continuous at a point x0 ∈ Rn if for any ε > 0 there is a δ = δ(ε) > 0 such that kf (x) − f (x0 )k 6 ε whenever kx − x0 k 6 δ. The mapping f is said to be continuous on Rn if it is continuous at every point of Rn ; the latter occurs if and only if any of the following equivalent conditions is satisfied: 1. f (cl X) ⊂ cl f (X) for every set X ⊂ Rn . 2. cl f −1 (Y ) ⊂ f −1 (cl Y ) for every set Y ⊂ Rm . 3. The inverse image f −1 (Y ) of every closed set Y ⊂ Rm is a closed set. 4. f −1 (int Y ) ⊂ int f −1 (Y ) for every set Y ⊂ Rm . 5. The inverse image f −1 (Y ) of every open set Y ⊂ Rm is an open set. If a continuous mapping f : Rn → Rm is one-to-one, or if a set X ⊂ Rn is bounded, then f (cl X) = cl f (X). In particular, f (X) is compact if X is compact. A continuous real-valued function f on Rn attains its maximum and minimum values on every nonempty compact subset of Rn . A countable union of closed sets is called an Fσ -set. A set X ⊂ Rn is an Fσ -set if and only if it is a countable union of a nested family of compact sets. Countable unions and finite intersections of Fσ -sets are Fσ -sets. A countable intersection of open sets is called a Gδ -set. A nonempty set X ⊂ Rn is called connected if it cannot be partitioned into two nonempty subsets such that each of them does not meet the closure of the other. Exercises for Chapter 0 Exercise 0.1. Let f : Rn → Rm be a linear transformation and S be a subspace of Rn complementary to null f . Show that the mapping h : S → rng f defined by h(x) = f (x) is an invertible linear transformation and that the inverse image f −1 (x) of every vector x ∈ rng f equals h−1 (x) + null f . Exercise 0.2. Let f : Rn → Rm be a linear transformation and L and M

10


be subspace in Rn and Rm , respectively. Show that dim f (L) = dim L − dim (L ∩ null f ), dim f −1 (M ) = dim (M ∩ rng f ) + dim (null f ). Exercise 0.3. Show that a linear transformation f : Rn → Rn is a linear projection if and only if f 2 = f . Exercise 0.4. Given a vector c ∈ Rn , show that the linear functional ϕ(x) = x·c is continuous on Rn . Exercise 0.5. Show that kx − yk − kz − yk 6 kx − zk for any points x, y, z in Rn . Conclude from here that for a given point c ∈ Rn , the distance function δc (x) = kx − ck is continuous on Rn . Exercise 0.6. Show that every linear transformation f : Rn → Rm is continuous on Rn . Exercise 0.7. Let {b1 , . . . , br } ⊂ Rn be a linearly independent set. Show (i) (i) that an infinite sequence xi = λ1 b1 + · · · + λr br , i > 1, converges to (i) the vector x = λ1 b1 + · · · + λr br if and only if limi→∞ λj = λj for all 1 6 j 6 r. Exercise 0.8. Let X and Y be nonempty set in Rn such that at least one of them is bounded. Show the existence of points x ∈ cl X and y ∈ cl Y with the property kx − yk = δ(X, Y ). Exercise 0.9. Given nonempty sets X and Y in Rn , show the equivalence of the following conditions: (1) δ(X, Y ) > 0, (2) o ∈ / cl (X − Y ), (3) there is a scalar ρ > 0 such that Bρ (X) ∩ Bρ (Y ) = ∅. Exercise 0.10. Show that cl X + cl Y ⊂ cl (X + Y ) for sets X and Y in Rn . Furthermore, cl X + cl Y = cl (X + Y ) provided any of the following conditions is satisfied: (1) at least one of X and Y is bounded, (2) X and Y lie in independent subspaces. Exercise 0.11. Let X and Y be Fσ -sets in Rn . Show that the sum X + Y is an Fσ -set. Also, for a linear transformation f : Rn → Rm , the set f (X) is an Fσ -set.

Chapter 1

The Affine Structure of Rn

1.1

Planes

Definition and Basic Properties Definition 1.1. A nonempty set L ⊂ Rn is called a plane provided it is a translate of a subspace: L = a + S (= {a + x : x ∈ S}), where a is a point in Rn and S is a subspace of Rn . The empty set is defined to be a plane. A plane L ⊂ Rn is called proper if ∅ 6= L 6= Rn .

Fig. 1.1

a 6

L=a+S

ro

S

A subspace S and a plane L.

For example, planes in R3 are the empty set, singletons, lines, usual planes, and R3 itself. Synonyms for “plane”, used by various authors, are “affine set”, “affine subspace”, “affine variety”, and “flat.” Ways to describe a nonempty plane L = a + S derive from those for the subspace S. Thus, if S is the span of vectors b1 , . . . , br , then L can be expressed parametrically: L = a + span {b1 , . . . , br } = {a + λ1 b1 + · · · + λr br : λ1 , . . . , λr ∈ R}. 11

12


Similarly, if S is the solution set of a system of homogenous linear equations ai1 x1 + · · · + ain xn = 0,

1 6 i 6 m,

then L is the solution set of a suitable system ai1 x1 + · · · + ain xn = bi ,

1 6 i 6 m.

One more way to describe planes in terms of affine spans of points is studied in Section 1.2. Theorem 1.2. Any nonempty plane L ⊂ Rn is a translate of a unique subspace S of Rn , given by S = L − L (= {x − y : x, y ∈ L}).

(1.1)

n

For a vector c ∈ R , the equality L = c + S holds if and only if c ∈ L. Furthermore, L = X + S and S = L − X for a nonempty subset X of L. Consequently, L is a subspace if and only if o ∈ L. Proof. By the definition, L can be expressed as L = a + S, where a ∈ Rn and S is a subspace of Rn . To prove (1.1), choose points x and y in L. Then x = a + x0 and y = a + y0 for certain points x0 , y0 ∈ S. This argument gives x − y = (a + x0 ) − (a + y0 ) = x0 − y0 ∈ S. Hence L − L = {x − y : x, y ∈ L} ⊂ S. On the other hand, every point x ∈ S can be written as x = (a + x) − (a + o) ∈ L − L, which implies the opposite inclusion S ⊂ L − L. The equality S = L − L shows the desired uniqueness of S. If L = c + S for a point c ∈ Rn , then c = c + o ∈ c + S = L. Conversely, if c ∈ L, then c = a + c0 , where c0 ∈ S. Since c0 + S = S, we have c + S = (a + c0 ) + S = a + (c0 + S) = a + S = L. Now, assume that X is a nonempty subset of L. By the above proved, X + S = ∪ (x + S : x ∈ X) = ∪ (L : x ∈ X) = L. Similarly, L − X = ∪ (L − x : x ∈ X) = ∪ (S : x ∈ X) = S. In view of Theorem 1.2, we introduce the following definition. Definition 1.3. For a nonempty plane L ⊂ Rn , the subspace S = L − L is called the characteristic subspace of L and is denoted sub L.


13

Theorem 1.4. The following statements hold. (1) If F = {Lα } is a family of planes in Rn , then the intersection ∩ Lα α is a plane. (2) If L1 , . . . , Lr are planes in Rn and µ1 , . . . , µr are scalars, then the sum µ1 L1 + · · · + µr Lr is a plane. Proof. (1) Suppose that ∩ Lα 6= ∅ and choose a point a ∈ ∩ Lα . Accordα α ing to Theorem 1.2, every plane Lα ∈ F can be written as Lα = a + Sα , where Sα is a subspace. Since ∩ Sα is a subspace, the equality α

∩ Lα = ∩ (a + Sα ) = a + ∩ Sα α

α

α

shows that ∩ Lα is a plane as a translate of ∩ Sα . α

α

(2) If at least one of the planes L1 , . . . , Lr is empty, then the whole sum µ1 L1 + · · · + µr Lr is empty. Suppose that all L1 , . . . , Lr are nonempty. Let Li = ai + Si , where ai ∈ Rn and Si is a subspace, 1 6 i 6 r. Then µ1 L1 + · · · + µr Lr = (µ1 a1 + · · · + µr ar ) + (µ1 S1 + · · · + µr Sr ). Hence the sum µ1 L1 + · · · + µr Lr is a plane as a translate of the subspace µ1 S1 + · · · + µr Sr . Additional results on algebraic operations with planes are given in Corollary 1.89 and Exercise 1.2. Dimension of Planes Theorem 1.2 gives a base to the definition below. Definition 1.5. If L ⊂ Rn is a nonempty plane and S is its characteristic subspace, then the dimension of L, denoted dim L, is defined by dim L = dim S. If L = ∅, then we let dim L = −1. Planes of dimension 0 and 1 are called points and lines, respectively. Theorem 1.6. The following statements take place. (1) If L ⊂ Rn is a plane, then for a point c ∈ Rn and a scalar µ 6= 0, the plane c + µL is a translate of L and has the same dimension as L. (2) If planes L1 and L2 in Rn satisfy the inclusion L1 ⊂ L2 , then dim L1 6 dim L2 , with dim L1 = dim L2 if and only if L1 = L2 .

14


Proof. Since both statements are obvious for the case of empty planes, we assume that all planes involved are nonempty. (1) Let L = a + S, where a ∈ Rn and S is a subspace. From the equalities c + µL = c + µ(a + S) = c + µa + S = c + (µ − 1)a + L if follows that c + µL is a translate of L and of S. Therefore, dim (c + µL) = dim S = dim L. (2) Choose a point b ∈ L1 . By Theorem 1.2, L1 = b+S1 and L2 = b+S2 for certain subspaces S1 and S2 . Theorem 1.2 shows that S1 = L1 − b ⊂ L2 − b = S2 . From linear algebra we know that dim S1 6 dim S2 , with dim S1 = dim S2 if and only if S1 = S2 . Hence dim L1 = dim S1 6 dim S2 = dim L2 , and dim L1 = dim L2 if and only if L1 = b + S1 = b + S2 = L2 . Theorem 1.7. For nonempty planes L1 and L2 in Rn , one has dim (L1 + L2 ) = dim L1 + dim L2 − dim (sub L1 ∩ sub L2 ). If, additionally, L1 ∩ L2 6= ∅, then dim (L1 + L2 ) = dim L1 + dim L2 − dim (L1 ∩ L2 ). Proof. Let Si = sub Li , i = 1, 2. By Theorem 1.2, we can write Li = ci +Si for suitable points ci ∈ Li , i = 1, 2. Then L1 + L2 = c1 + c2 + (S1 + S2 ). This argument and the equality dim (S1 + S2 ) = dim S1 + dim S2 − dim (S1 ∩ S2 ) give dim (L1 + L2 ) = dim L1 + dim L2 − dim (S1 ∩ S2 ). Suppose that L1 ∩ L2 6= ∅ and choose a point c ∈ L1 ∩ L2 . By Theorem 1.2, Li = c + Si , i = 1, 2. Consequently, L1 + L2 = 2c + (S1 + S2 )

and L1 ∩ L2 = c + S1 ∩ S2 .

Therefore, dim (L1 + L2 ) = dim (S1 + S2 ) = dim L1 + dim L2 − dim (L1 ∩ L2 ). Theorem 1.8. If planes L1 and L2 lie within a nonempty plane L ⊂ Rn such that L1 + L2 is a translate of L, then L1 ∩ L2 6= ∅.


15

Proof. Translating all three planes L, L1 , and L2 on the same suitable vector, we may assume that L is a subspace. Then L1 + L2 ⊂ L, which gives L1 + L2 = L due to the assumption on L1 + L2 . Clearly, both planes L1 and L2 are nonempty. Let Li = ai + Si , where ai ∈ Rn and Si is a subspace, i = 1, 2. Since ai ∈ Li ⊂ L, i = 1, 2 (see Theorem 1.2), one has a1 + a2 ∈ L. Therefore, S1 + S2 = (L1 + L2 ) − (a1 + a2 ) = L − (a1 + a2 ) = L. With r = dim (S1 ∩ S2 ), p = dim S1 − r, and q = dim S2 − r, a dimension argument yields p + q + r = dim S1 + dim S2 − dim (S1 ∩ S2 ) = dim (S1 + S2 ) = dim L. Choose a basis b1 , . . . , br for S1 ∩ S2 and complete it into bases b 1 , . . . , b r , c1 , . . . , cp

and b1 , . . . , br , e1 , . . . , eq

for S1 and S2 . The combined list b1 , . . . , br , c1 , . . . , cp , e1 , . . . , eq is a basis for L (see page 4). Therefore, we can express a1 and a2 as linear combinations a1 = ξ1 b1 + · · · + ξr br + ξr+1 c1 + · · · + ξp+r cp + ξp+r+1 e1 + · · · + ξq eq , a2 = η1 b1 + · · · + ηr br + ηr+1 c1 + · · · + ηp+r cp + ηp+r+1 e1 + · · · + ηq eq . Let u0 = ηr+1 c1 + · · · + ηp+r cp + ξp+r+1 e1 + · · · + ξn eq , u1 = −(ξ1 b1 + · · · + ξr br ) + (ηr+1 − ξr+1 )c1 + · · · + (ηp+r − ξp+r )cp , u2 = −(η1 b1 + · · · + ηr br ) + (ξp+r+1 − ηp+r+1 )e1 + · · · + (ξq − ηq )eq . Obviously, u1 ∈ S1 , u2 ∈ S2 , and u0 = a1 + u1 = a2 + u2 . Furthermore, u1 + S1 = S1 and u2 + S2 = S2 . Hence Li = ai + Si = ai + (ui + Si ) = (ai + ui ) + Si = u0 + Si , i = 1, 2. Then u0 ∈ L1 ∩ L2 according to Theorem 1.2. Independent and Complementary Planes We recall that subspaces S and T are independent provided S ∩ T = {o}. If, additionally, S + T = Rn , then S and T are complementary. Definition 1.9. Nonempty planes L1 and L2 in Rn are called independent (respectively, complementary) if their characteristic subspaces are independent (respectively, complementary).

16


S2 rc S1

ro

L1 = a1 + S1

L2 = a2 + S2

Fig. 1.2

Complementary planes L1 and L2 .

Theorem 1.10. Nonempty planes L1 and L2 in Rn are independent if and only if dim (L1 + L2 ) = dim L1 + dim L2 . Furthermore, L1 and L2 are complementary if and only if dim (L1 + L2 ) = dim L1 + dim L2 = n. Proof. Let Li = ai + Si , where ai ∈ Rn and Si is a subspace, i = 1, 2. Then L1 + L2 = (a1 + a2 ) + (S1 + S2 ), which yields dim (L1 + L2 ) = dim (S1 + S2 ). From linear algebra we know that S1 and S2 are independent if and only if dim (S1 + S2 ) = dim S1 + dim S2 . Furthermore, S1 and S2 are complementary if and only if dim S1 + dim S2 = n. Now, both statements of the theorem follows from the definitions and the equalities dim Li = dim Si , i = 1, 2. Theorem 1.11. For nonempty planes L1 and L2 in Rn , the following conditions are equivalent. (1) L1 and L2 are complementary. (2) L1 + L2 = Rn and L1 ∩ L2 is a singleton. (3) There are complementary subspaces S1 and S2 and a point c ∈ Rn such that L1 = c + S1 and L2 = c + S2 . (4) There is a point c ∈ L1 ∩L2 such that every point x ∈ Rn is uniquely expressible as x = x1 + x2 − c, where x1 ∈ L1 and x2 ∈ L2 . Proof. Let Li = ai + Si , where ai ∈ Rn and Si is a subspace, i = 1, 2. (1) ⇒ (2). If L1 and L2 are complementary, then the subspaces S1 and S2 are complementary. Therefore, L1 + L2 = (a1 + a2 ) + (S1 + S2 ) = (a1 + a2 ) + Rn = Rn .


17

By Theorem 1.8, L1 ∩ L2 6= ∅ and dim (L1 ∩ L2 ) = dim L1 + dim L2 − n = dim S1 + dim S2 − n = 0. The latter equality shows that L1 ∩ L2 is a singleton. (2) ⇒ (3). Denote by c the unique point of L1 ∩ L2 . Then L1 = c + S1 and L2 = c + S2 (see Theorem 1.2). Furthermore, S1 + S2 = (L1 + L2 ) − 2c = Rn − 2c = Rn , {c} = L1 ∩ L2 = (c + S1 ) ∩ (c + S2 ) = c + S1 ∩ S2 . Hence S1 ∩ S2 = {o}, and S1 and S2 are complementary. (3) ⇒ (4). Clearly, the point c satisfying condition (3) belongs to L1 ∩ L2 . Since the subspaces S1 and S2 are complementary, for every point x ∈ Rn , the point x − c is uniquely expressible as the sum z1 + z2 , where z1 ∈ S1 and z2 ∈ S2 . Consequently, the representation x = x1 + x2 − c, where x1 = z1 + c and x2 = z2 + c, is unique. (4) ⇒ (1). Let S1 = L1 − c and S2 = L2 − c. Condition (4) shows that every vector u ∈ Rn is uniquely expressible as u = u1 + u2 , where u1 ∈ S1 and u2 ∈ S2 . Therefore, the subspaces S1 and S2 are complementary, and Definition 1.9 shows that the planes L1 and L2 are complementary. Parallel Planes Definition 1.12. Nonempty planes L1 and L2 in Rn are called parallel if one of them contains a translate of the other.

L1 L2 Fig. 1.3

L1 L2

Parallel planes L1 and L2 .

Remark. Parallelism of planes defined above is not an equivalence relation on the family of all planes in Rn . For example, the coordinate x- and y-axes of R3 are not parallel, while both are parallel to the coordinate xy-plane. Obviously, any singleton is parallel to each nonempty plane.

18


Theorem 1.13. Suppose that nonempty planes L1 and L2 in Rn satisfy the inequality dim L1 6 dim L2 . If S1 and S2 are the characteristic subspaces of L1 and L2 , respectively, then L1 and L2 are parallel if and only if S1 ⊂ S2 . Consequently, if dim L1 = dim L2 , then L1 and L2 are parallel if and only if they are translates of each other. Proof. Let Li = ai + Si for a certain point ai ∈ Rn , i = 1, 2. Suppose first that L1 and L2 are parallel. From Theorem 1.6 and the assumption dim L1 6 dim L2 it follows that namely L2 contains a translate of L1 . Choose a vector c ∈ Rn satisfying the condition c + L1 ⊂ L2 . Put b = c + a1 . Then b = c + a1 ∈ c + a1 + S1 = c + L1 ⊂ L2 . Thus L2 = b + S2 according to Theorem 1.2. Therefore, S1 = L1 − a1 = (c + L1 ) − (c + a1 ) ⊂ L2 − b = S2 . Conversely, if S1 ⊂ S2 , then (a2 − a1 ) + L1 = (a2 − a1 ) + (a1 + S1 ) = a2 + S1 ⊂ a2 + S2 = L2 , implying that L1 and L2 are parallel. XXX L1 XX X

Fig. 1.4

L01 = L1 + sub L2

L 2

L02 = L2 + sub L1

Parallel planes L01 and L02 through planes L1 and L2

Theorem 1.14. If L1 and L2 are nonempty planes in Rn , then L01 = L1 + sub L2

and

L02 = sub L1 + L2

are parallel planes containing L1 and L2 , respectively, and having the same dimension dim (L1 + L2 ). Furthermore, L01 and L02 are disjoint if and only if L1 and L2 are disjoint. Proof. Choose points ai ∈ Li , i = 1, 2. According to Theorem 1.2, one has Li = ai + Si , where Si = sub Li , i = 1, 2. Hence both sets L01 = L1 + S2 = a1 + (S1 + S2 ), L02 = S1 + L2 = a2 + (S1 + S2 )

(1.2)


19

are parallel planes as translates of the same subspace S1 +S2 . Consequently, they have the same dimension dim (S1 + S2 ) = dim (L1 + L2 ). Clearly, L1 ⊂ L1 + S2 = L01

and L2 ⊂ S1 + L2 = L02 .

If L01 and L02 are disjoint, then so are L1 and L2 . Suppose that L01 ∩L02 6= ∅ and choose a point c ∈ L01 ∩ L02 . From Theorem 1.2 and the above argument it follows that L01 = c + S1 + S2 = L02 . Comparing the latter equalities with (1.2), we conclude from the same theorem that a1 − c ∈ L01 − L01 = S1 + S2 , a2 − c ∈ L02 − L02 = S1 + S2 . Therefore, a1 + a2 − 2c ∈ S1 + S2 . Furthermore, Li − c ⊂ Li + Sj − c = L0i − c = S1 + S2 ,

i 6= j,

and (L1 − c) + (L2 − c) = (a1 + S1 − c) + (a2 + S2 − c) = (a1 + a2 − 2c) + (S1 + S2 ) = S1 + S2 . Finally, Theorem 1.8 shows that (L1 − c) ∩ (L2 − c) 6= ∅. Consequently, L1 ∩ L2 = (L1 − c) ∩ (L2 − c) + c 6= ∅. In elementary geometry, two lines in the plane (a line and a plane, or a pair of planes in the 3-space) are defined to be parallel provided they do not meet. Theorem 1.15 below relates Definition 1.12 with this approach. Theorem 1.15. Let L1 and L2 be nonempty planes in Rn such that dim L1 6 dim L2 . Then L1 and L2 are parallel if and only if the following two conditions are satisfied. (1) The union L1 ∪ L2 lies within a plane of dimension dim L2 + 1. (2) Either L1 ∩ L2 = ∅ or L1 ⊂ L2 . Proof. Let Li = ai + Si , where ai ∈ Rn and Si is a subspace, i = 1, 2. Suppose first that L1 and L2 are parallel. From Theorem 1.6 and the assumption dim L1 6 dim L2 it follows that L2 contains a translate of L1 . Therefore, S1 ⊂ S2 by Theorem 1.13. Choose a point c ∈ Rn satisfying the inclusion c + L1 ⊂ L2 , and put b = c + a1 . Then b = c + a1 ∈ c + a1 + S1 = c + L1 ⊂ L2 . Thus L2 = b + S2 according to Theorem 1.2. Let S = span {c ∪ S2 } and L = b + S. Clearly, S1 − c ⊂ S2 − c ⊂ S and dim L = dim S 6 dim S2 + 1 = dim L2 + 1

20


This argument yields the inclusions L1 = a1 + S1 = b + (S1 − c) ⊂ b + S = L, L2 = b + S2 ⊂ b + S = L. If L1 ∩ L2 6= ∅ and a is a point in L1 ∩ L2 , then Theorem 1.2 gives L1 = a + S1 ⊂ a + S2 = L2 . Conversely, let L1 and L2 satisfy conditions (1) and (2) of the theorem. Since the case L1 ⊂ L2 implies the parallelism of L1 and L2 , we may assume that L1 ∩ L2 = ∅. Choose a plane L ⊂ Rn of dimension dim L2 + 1 or less, which contains L1 ∪L2 . Denote by S the characteristic subspace of L. Then S1 ∪ S2 ⊂ S according to Theorem 1.13. Suppose that L1 and L2 are not parallel. Then S1 6⊂ S2 by the same Theorem 1.13. Put m = dim S2 . Choose a vector b0 ∈ S1 \ S2 and a basis b1 , . . . , bm for S2 . Then b0 , b1 , . . . , bm is a basis for S because of the condition dim S 6 m + 1. Since both a1 and a2 belong to L, their difference a1 − a2 is in L − L = S. Hence a1 − a2 can be expressed as a linear combination a1 − a2 = λ0 b0 + λ1 b1 + · · · + λm bm . Put c = λ1 b1 + · · · + λm bm . Then c ∈ S2 and a1 − λ0 b0 = a2 + c. Furthermore, a1 − λ0 b0 ∈ a1 + S1 = L1

and a2 + c ∈ a2 + S2 = L2 ,

contrary to the assumption L1 ∩ L2 = ∅. Hence L1 and L2 are parallel. Hyperplanes Definition 1.16. A hyperplane in Rn is a plane of dimension n − 1. Theorem 1.17. A set H ⊂ Rn is a hyperplane if and only if there is a nonzero vector c ∈ Rn and a scalar γ such that H = {x ∈ Rn : x·c = γ}.

(1.3)

Furthermore, this representation of H is unique up to a common nonzero scalar multiple of c and γ. Proof. A set H ⊂ Rn is a hyperplane if and only if H = a + S, where a ∈ Rn and S is a subspace of dimension n − 1. From linear algebra we know (see page 6) that S can be written as S = {u ∈ Rn : u·c = 0}, c 6= o,

(1.4)


21

and this expression is unique up to a nonzero scalar multiple of c. Put γ = a·c. Then H is a hyperplane if and only if H = {a + u ∈ Rn : u·c = 0} = {x ∈ Rn : (x − a)·c = 0} = {x ∈ Rn : x·c = γ}. Assume now that H has another representation H = {x ∈ Rn : x·c0 = γ 0 } for a nonzero vector c0 ∈ Rn and a scalar γ 0 . Then a·c0 = γ 0 because of a ∈ H. From S = H − a, we obtain S = {x − a ∈ Rn : x·c0 = γ 0 } = {u ∈ Rn : (u + a)·c0 = γ 0 } = {u ∈ Rn : u·c0 = 0}. A comparison of the latter equality with (1.4) yields that c0 = λc for a certain scalar λ 6= 0. Finally, γ 0 = a·c0 = a·(λc) = λγ. Corollary 1.18. If a hyperplane H ⊂ Rn is given by (1.3), then a set H 0 ⊂ Rn is a translate of H if and only if H 0 can be expressed as H 0 = {x ∈ Rn : x·c = γ 0 }.

(1.5)

Consequently, a set S ⊂ Rn is the characteristic subspace of H if and only if S = {x ∈ Rn : x·c = 0}.

(1.6)

Proof. A set H 0 is a translate of H if and only H 0 = b + H for a certain vector b ∈ Rn . With γ 0 = γ + b·c, we have H 0 = {b + x ∈ Rn : x·c = γ} = {u ∈ Rn : (u − b)·c = γ} = {u ∈ Rn : u·c = γ 0 }. The second statement of the corollary is obvious. Q Q Q Q S QQ

r

Q

oQ Q

Fig. 1.5

Q Q Q

Q H Q Q Q

Q

Normal vectors to a hyperplane H.

22


If a hyperplane H = a + S is given by (1.3), then (1.6) shows that c is orthogonal to the subspace S. In this regard, all vectors of the form λc, λ 6= 0, are called normal to H (see example on page 211). For the next theorem, we recall that S ⊥ denotes the orthogonal complement of a subspace S ⊂ Rn (see page 5). Theorem 1.19. For a hyperplane H ⊂ Rn and a proper plane L ⊂ Rn , the following statements hold. (1) H and L are parallel if and only if either H ∩ L = ∅ or L ⊂ H. (2) H and L are parallel if and only if H + L is a translate of H. (3) H and L are parallel if and only if there is a unique hyperplane H 0 ⊂ Rn containing L and parallel to H. (4) If H is given by (1.3), then H and L are parallel if and only if c belongs to the subspace (sub L)⊥ . (5) H and L are not parallel if and only if ∅ 6= H ∩ L 6= L. (6) H and L are not parallel if and only if H + L = Rn . (7) H and L are not parallel if and only if H ∩L is a plane of dimension dim L − 1. Proof. In what follows, let H = a + S and L = b + T , where a and b are points, and S and T are subspaces. (1) Assume first that H and L are parallel. Then T ⊂ S according to Theorem 1.13. If H ∩ L 6= ∅ and c is a point in H ∩ L, then L = c + T ⊂ c + S = H (see Theorem 1.2). Conversely, suppose that H ∩ L = ∅ or L ⊂ H. Since the case L ⊂ H implies the parallelism of L and H, we may assume that H ∩L = ∅. Assume for a moment that H and L are not parallel. Then T 6⊂ S by Theorem 1.13. Choose a vector b1 ∈ T \ S and a basis b2 , . . . , bn for S. Then b1 , . . . , bn is a basis for Rn . Hence b − a can be expressed as a linear combination b − a = λ1 b1 + · · · + λn bn . Put c = λ2 b2 + · · · + λn bn . Then c ∈ S and b − λ1 b1 = a + c. Furthermore, b − λ1 b1 ∈ b + T = L and a + c ∈ a + S = H, contrary to the assumption H ∩ L = ∅. Hence H and L are parallel. (2) By Theorem 1.13, H and L are parallel if and only if T ⊂ S, which is equivalent to the equality T + S = S. Consequently, H and L are parallel if and only if H + L = (a + b) + (S + T ) = (a + b) + S = b + (a + S) = b + H.


23

(3) Assume that H and L are parallel. Since dim L 6 dim H, Theorem 1.6 implies that namely H contains a translate of L. If c + L ⊂ H for a certain point c ∈ Rn , then the hyperplane G = H − c contains L and is parallel to H. For the uniqueness of G, choose a hyperplane G0 containing L and parallel to H. According to Theorem 1.13, G0 is a translate of G, and Theorem 1.2 shows that G − b and G0 − b are identical subspaces. Consequently, G = G0 . Conversely, suppose that H 0 is a unique hyperplane containing L and parallel to H. Then H 0 is a translate of H (see Theorem 1.13). Expressing H 0 as H 0 = u+H for a certain point u ∈ Rn , we obtain L−u ⊂ H 0 −u = H. Hence L is parallel to H. (4) By Theorem 1.13, H and L are parallel if and only if T ⊂ S, which is equivalent to S ⊥ ⊂ T ⊥ . Since S ⊥ = span {c} and T = sub L, the inclusion S ⊥ ⊂ T ⊥ is equivalent to c ∈ (sub L)⊥ . Statement (5) is the contrapositive of (1). (6) According to Theorem 1.13, H and L are not parallel if and only if T 6⊂ S. Since dim S = n − 1, the latter is equivalent to S + T = Rn , or, to H + L = (a + b) + (S + T ) = Rn . (7) Assume first that H and L are not parallel. Then H + L = Rn by statement (6) above, and Theorem 1.8 gives dim (H ∩ L) = dim H + dim L − n = dim L − 1. Conversely, if H ∩ L is a plane of dimension dim L − 1, then ∅ 6= H ∩ L 6= L, and statement (5) above shows that H and L are not parallel. Theorem 1.20. Every proper plane L ⊂ Rn of dimension m can be expressed as the intersection of n − m hyperplanes. Furthermore, if L is the intersection of a certain family H of hyperplanes, then H contains a subfamily of n − m members whose intersection is L.

ppp ppp p pp

Fig. 1.6

pppp

pp p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p

A line as the intersection of two planes in R3 .

24


Proof. Let L = a + S, where a ∈ Rn and S is a subspace of dimension m. From linear algebra we know that S can be expressed as the intersection of certain (n − 1)-dimensional subspaces S1 , . . . , Sn−m ⊂ Rn . The sets Fi = a + Si , 1 6 i 6 n − m, are hyperplanes, and L = a + S = a + S1 ∩ · · · ∩ Sn−m = (a + S1 ) ∩ · · · ∩ (a + Sn−m ) = F1 ∩ · · · ∩ Fn−m . Suppose that L is the intersection of a family H = {Hα } of hyperplanes. Choose a point c ∈ L. According to Theorem 1.2, all sets L − c and Hα − c are subspaces. Clearly, L − c = ∩ Hα − c = ∩ (Hα − c), α

dim (L − c) = m,

α

dim (Hα − c) = n − 1 for all Hα ∈ H.

From linear algebra we know that the family {Hα − c : Hα ∈ H} contains n − m members, say H1 − c, . . . , Hn−m − c, whose intersection is L − c. Consequently, L = c + (L − c) = c + (H1 − c) ∩ · · · ∩ (Hn−m − c) = H1 ∩ · · · ∩ Hn−m . The following result complements Theorem 1.20. Theorem 1.21. The intersection of r, 1 6 r 6 n, hyperplanes in Rn of the form Hi = {x ∈ Rn : x·ci = γi }, 1 6 i 6 r, is an (n − r)-dimensional plane if and only if the set {c1 , . . . , cr } is linearly independent. Proof. The intersection H1 ∩ · · · ∩ Hr is the solution set of the system of r linear equations in x1 , . . . , xn : x·ci = x1 c1i + · · · + xn cni = γi ,

1 6 i 6 r.

This solution set is a translate of an (n − r)-dimensional subspace if and only if the r × n matrix (cij ) has rank r, which occurs if and only if the set {c1 , . . . , cr } is linearly independent. Corollary 1.22. Every plane L in Rn is a closed set.


25

Proof. The statement is trivial if L = ∅ or L = Rn . Assume that L is a proper plane. Suppose first that L is a hyperplane. Then L can be expressed by (1.3). Because the linear functional ϕ(x) = x·c is continuous on Rn (see Exercise 0.4), the hyperplane L is closed as the inverse image ϕ−1 (γ) of the singleton {γ}. If dim L 6 n − 2, then L is the intersection of n − m hyperplanes, say H1 , . . . , Hn−m (see Theorem 1.20). Since each hyperplane Hi , 1 6 i 6 n − m, is a closed set, their intersection also is closed. Lines, Halflines, and Segments According to Definitions 1.1 and 1.5, a line in Rn is a translate of a onedimensional subspace. In what follows, we use another description of lines. (1 − λ)x + λy r x

λ60 Fig. 1.7

06λ61

r y

λ>1

The line through distinct points x and y.

Definition 1.23. The line through distinct points x and y in Rn , denoted hx, yi, is defined by hx, yi = {(1 − λ)x + λy : λ ∈ R}.

(1.7)

Remark. For distinct points x, y ∈ Rn , the mapping λ 7→ (1 − λ)x + λy is a bijection from R on the line hx, yi. Rewriting (1.7) as hx, yi = x + {λ(y − x) : λ ∈ R}, we see that hx, yi is a translate of the 1-dimensional subspace span {y −x}. Definition 1.24. For distinct points x and y in Rn , the closed halflines [x, yi and hx, y], and the open halflines (x, yi and hx, y) are defined, respectively, as [x, yi = {(1 − λ)x + λy : λ > 0},

(x, yi = {(1 − λ)x + λy : λ > 0},

hx, y] = {(1 − λ)x + λy : λ 6 1},

hx, y) = {(1 − λ)x + λy : λ < 1}.

Definition 1.25. For distinct points x and y in Rn , the closed segment [x, y], the semi-open segments [x, y) and (x, y], and the open segment

26


(x, y) are defined, respectively, as [x, y] = {(1 − λ)x + λy : 0 6 λ 6 1}, [x, y) = {(1 − λ)x + λy : 0 6 λ < 1}, (x, y] = {(1 − λ)x + λy : 0 < λ 6 1}, (x, y) = {(1 − λ)x + λy : 0 < λ < 1}. For every point x ∈ Rn , let [x, x] = {x} and [x, x) = (x, x] = (x, x) = ∅. Clearly, [x, yi = hy, x], (x, yi = hy, x), [x, y] = [y, x], [x, y) = (y, x]. Lemma 1.26. For distinct points x and y in Rn , the following statements hold. (1) If u and v are distinct points in hx, yi, then hu, vi = hx, yi. (2) If u ∈ (x, yi, then [x, ui = [x, yi. (3) If v ∈ [x, y] and w ∈ hx, yi, then one of the inclusions v ∈ [x, w] and v ∈ [y, w] holds. Proof. (1) Since u, v ∈ hx, yi, we can write u = (1 − λ)x + λy

and v = (1 − µ)x + µy

(1.8)

for certain scalars λ and µ. Clearly, λ 6= µ because of u 6= v. To prove the inclusion hu, vi ⊂ hx, yi, choose a point w ∈ hu, vi. Then w = (1 − ξ)u + ξv for a scalar ξ. With ν = λ − λξ + µξ, one has w = (1 − ξ)((1 − λ)x + λy) + ξ((1 − µ)x + µy) = (1 − λ + λξ − µξ)x + (λ − λξ + µξ)y = (1 − ν)x + νy ∈ hx, yi. Conversely, let z ∈ hx, yi. Then z = (1 − η)x + ηy for a scalar η. Solving (1.8) for x and y, x=

λ µ u− v µ−λ µ−λ

and y =

1−λ 1−µ v− u, µ−λ µ−λ

and letting α = (η − λ)/(µ − λ), we obtain µ 1−λ λ 1−µ z = (1 − η) u− v +η v− u µ−λ µ−λ µ−λ µ−λ η − λ η−λ = 1− u+ v = (1 − α)u + αv ∈ hu, vi. µ−λ µ−λ


27

(2) Let u = (1 − λ)x + λy, where λ > 0. To show the inclusion [x, ui ⊂ [x, yi, choose a point v ∈ [x, ui. Then v = (1 − η)x + ηu for a scalar η > 0. Because λη > 0 and v = (1 − η)x + η((1 − λ)x + λy) = (1 − λη)x + ληy, it follows that v ∈ [x, yi. Conversely, let z ∈ [x, yi. Then z = (1−µ)x+µy for a certain scalar µ > 0. Since y = (1 − λ−1 )x + λ−1 u and µλ−1 > 0, one has z = (1 − µ)x + µ((1 − λ−1 )x + λ−1 u) = (1 − µλ−1 )x + µλ−1 u ∈ [x, ui. (3) Let v = (1−λ)x+λy and w = (1−µ)x+µy for scalars λ ∈ [0, 1] and µ ∈ R. Assume first that λ 6 µ. Leaving aside the case λ = µ = 0 (which corresponds to v = w = x), we suppose that µ > 0. Since 0 6 λ/µ 6 1, the equality λ λ λ λ x+ (1 − µ)x + µy = 1 − x+ w v = 1− µ µ µ µ gives v ∈ [x, w]. Suppose now that λ > µ. Excluding the case λ = µ = 1 (which corresponds to v = w = y), we suppose that µ < 1. Then 0 6 (λ−µ)/(1−µ) 6 1, and the equality λ−µ λ − µ 1 − λ λ−µ y = 1− y v= (1 − µ)x + µy + w+ 1−µ 1−µ 1−µ 1−µ yields the inclusion v ∈ [y, w]. Lemma 1.27. For any points x, y, and z in Rn and points u ∈ [x, y] and v ∈ [x, z], the segments [u, z] and [v, y] have a point in common. Similarly, if q ∈ [u, z], then q ∈ [w, y] for a certain point w ∈ [x, z]. z r P P

v r

w r

PP pP r PP r P r q u r y

x r

Proof. Let u = (1 − λ)x + λy and v = (1 − µ)x + µz, where λ, µ ∈ [0, 1]. Excluding the case λ = µ = 1 (which corresponds to u = y and v = z), we

28


assume that λµ < 1. Put p=

λ(1 − µ) (1 − λ)µ (1 − λ)(1 − µ) x+ y+ z, 1 − λµ 1 − λµ 1 − λµ 1−µ 1−λ ξ= , η= . 1 − λµ 1 − λµ

Obviously, ξ, η ∈ [0, 1], and the equalities 1−µ 1−µ p= 1− z+ (1 − λ)x + λy = (1 − ξ)z + ξu, 1 − λµ 1 − λµ 1−λ 1−λ p= 1− y+ (1 − µ)x + µz = (1 − η)y + ηv 1 − λµ 1 − λµ imply the inclusion p ∈ [u, z] ∩ [v, y]. Similarly, given a point q ∈ [u, z], we can write q = (1 − γ)z + γu for a certain scalar γ ∈ [0, 1]. Leaving aside the case λ = γ = 1 (corresponding to q = u = y), we assume that λγ < 1. Let ζ = 1 − λγ,

ν=

1−γ , 1 − λγ

w = (1 − ν)x + νz.

Then ζ, ν ∈ [0, 1] and w ∈ [x, z]. Finally, the equalities q = (1 − γ)z + γ u = (1 − γ)z + γ((1 − λ)x + λ y) (1 − λ)γ 1−γ = λγ y + (1 − λγ) x+ z 1 − λγ 1 − λγ = (1 − ζ)y + ζ((1 − ν)x + ν z) = (1 − ζ)y + ζw show that q ∈ [w, y]. Halfspaces and Slabs Definition 1.28. If a hyperplane H ⊂ Rn is given by (1.3), then the sets V1 = {x ∈ Rn : x·c 6 γ}

and V2 = {x ∈ Rn : x·c > γ}

(1.9)

are called the opposite closed halfspaces determined by H, and the sets W1 = {x ∈ Rn : x·c < γ}

and W2 = {x ∈ Rn : x·c > γ}

(1.10)

are the opposite open halfspaces determined by H. The vectors λc, λ > 0, are called normal to both V1 and W1 , and the vectors λc, λ < 0, are normal to both V2 and W2 .


29

Topological properties of halfspaces are described in the next theorem. Theorem 1.29. If H ⊂ Rn is a hyperplane expressed by (1.3), then the halfspaces V1 and V2 given by (1.9) are closed sets, and the halfspaces W1 and W2 given by (1.10) are open sets. Furthermore, Vi = cl Wi ,

Wi = int Vi ,

H = bd Vi = bd Wi ,

i = 1, 2.

(1.11)

Proof. The linear functional ϕ(x) = x · c is continuous on Rn (see Exercise 0.4). Therefore, the halfspaces (1.9) are closed sets as ϕ−1 -images of the closed halflines (−∞, γ] and [γ, ∞) of R. Similarly, the halfspaces (1.10) are open sets as ϕ−1 -images of the respective open halflines of R. For the equalities (1.11), it suffices to consider the case i = 1. From the inclusion W1 ⊂ V1 it follows that W1 = int W1 ⊂ int V1 . Hence bd V1 = V1 \ int V1 ⊂ V1 \ W1 = H. For the opposite inclusion, choose a point u ∈ H and a scalar ρ > 0. Let ρ ρ c and v2 = u + c. v1 = u − kck kck Since kv1 − uk = kv2 − uk = ρ, both points v1 and v2 belong to the closed ball Bρ (u). Furthermore, from the inequalities ρ v1 ·c = (u − c)·c = γ − ρkck < γ kck and v2 ·c = (u +

ρ c)·c = γ + ρkck > γ kck

it follows that v1 ∈ V1 and v2 ∈ W2 . Hence Bρ (u) meets both sets V1 and Rn \ V1 , implying that u ∈ bd V1 . Consequently, bd V1 = H

and

int V1 = V1 \ bd V1 = V1 \ H = W1 .

Similarly, the inclusion W1 ⊂ V1 gives cl W1 ⊂ V1 . Therefore, bd W1 = cl W1 \ W1 ⊂ V1 \ W1 = H. To show the opposite inclusion, choose a point u ∈ H. By the above argument, every point u ∈ H belongs to bd W1 . Therefore, bd W1 = H

and V1 = W1 ∪ H = W1 ∪ bd W1 = cl W1 .

Remark. There are obvious relations between closed and open halfspaces: (a) a closed halfspace {x ∈ Rn : x·c 6 γ} is the intersection of denumerably many open halfspaces {x ∈ Rn : x·c < γ + 1/k}, k > 1,

30


(b) an open halfspace {x ∈ Rn : x·c < γ} is the union of denumerably many closed halfspaces {x ∈ Rn : x·c 6 γ − 1/k}, k > 1. The next result complements Theorem 1.19 Theorem 1.30. If H ⊂ Rn is a hyperplane and L ⊂ Rn is a proper plane, then the following conditions are equivalent. (1) H and L are parallel. (2) Either L ⊂ H, or L lies in an open halfspace determined by H. Proof. Let H = a + S and L = b + T , where a and b are points, and S and T are subspaces. Assume that H is given by (1.3). Then Corollary 1.18 shows that S is expressed by (1.6). (1) ⇒ (2). If H and L are parallel, then, according to Theorem 1.19, either H ∩ L = ∅ or L ⊂ H. Theorem 1.13 yields the inclusion T ⊂ S. Let H ∩ L = ∅. Then b ∈ / H, since otherwise L = b + T ⊂ b + S = H (see Theorem 1.2). Put β = b·c. Clearly, β 6= γ because of b ∈ / H. Suppose that β < γ (the case β > γ is similar). We state that L lies in the open halfspace W = {x ∈ Rn : x · c < γ}. Indeed, for a point x ∈ L, we can write x = b + x0 , where x0 ∈ T , which gives x·c = b·c + x0 ·c = β + 0 < γ. (2) ⇒ (1). Since the case L ⊂ H is obvious, we may suppose that L lies in an open halfspace determined by H. Then H ∩L = ∅, and Theorem 1.19 shows that H and L are parallel. The theorem below plays an important role in separation properties of convex sets. Theorem 1.31. For a hyperplane H ⊂ Rn and distinct points y, z ∈ Rn , the following statements hold. (1) If both y and z belong to H, then the line hy, zi lies in H. (2) If y ∈ H and z belongs to an open halfspace determined by H, then the open halfline (y, zi lies in this halfspace. (3) If both y and z belong to an open halfspace determined by H, then at least one of the halflines [y, zi, hy, z] lies in this halfspace. (4) If y and z belong, respectively, to distinct open halfspaces determined by H, then hy, zi meets H at a single point, which belongs to the open segment (y, z).


31

Proof. Suppose that H is given by (1.3). Any point u ∈ hy, zi can be written as u = (1 − λ)y + λz, and this expression implies u·c = ((1 − λ)y + λz)·c = (1 − λ)y·c + λ z·c.

(1.12)

(1) If y, z ∈ H and u ∈ hy, zi, then y ·c = z ·c = γ, and (1.12) gives u·c = γ. Hence hy, zi ⊂ H. (2) Assume that y ∈ H and z belongs to the halfspace W = {x ∈ Rn : x·c < γ}. Then y·c = γ and z·c < γ. If u ∈ (y, zi, then λ > 0, and (1.12) gives u·c = y·c + λ(z·c − y·c) < y·c = γ. Therefore, (y, zi ⊂ W . (3) Suppose that y and z belong to the same open halfspace W determined by H. As above, we assume that W = {x ∈ Rn : x · c < γ}. If z·c 6 y·c and u ∈ [y, zi, then λ > 0, and (1.12) gives u·c = y·c + λ(z·c − y·c) 6 y·c < γ. Hence [y, zi ⊂ W . If y·c 6 z·c and u ∈ hy, z], then λ 6 1, and (1.12) gives u·c = y·c + λ(z·c − y·c) 6 y·c + (z·c − y·c) = z·c < γ. Hence hy, z] ⊂ W . (4) Suppose that y and z belong, respectively, to distinct open halfspaces determined by H. Without loss of generality, we assume that y·c < γ < z·c. Put µ=

γ − y·c . z·c − y·c

(1.13)

Since 0 < µ < 1, the point u = (1 − µ)y + µz belongs to (y, z). Combining (1.12) and (1.13), we obtain u·c = (1 − µ)y·c + µz·c γ − y·c γ − y·c = 1− y·c + z·c = γ. z·c − y·c z·c − y·c Hence u ∈ H. By statement (2) above, the open halflines (u, yi and (u, zi lie, respectively, in distinct open halfspaces determined by H. So, u is the unique point from hy, zi which belongs to H. Corollary 1.32. Let y and z be distinct points in a closed halfspace V determined by a hyperplane H ⊂ Rn . If (y, z) ∩ H 6= ∅, then [y, z] ⊂ H.

32


Proof. From Theorem 1.31 it follows that [y, z] ⊂ V . Assume for a moment that at least one of the points y and z belongs to V \ H. According to Theorem 1.29, the set W = V \ H is an open halfspace determined by H. Theorem 1.31 shows that the open segment (y, z) lies in W , contrary to the assumption (y, z) ∩ H 6= ∅. Hence both points y and z belong to H, and the same Theorem 1.31 implies the inclusion [y, z] ⊂ H. Definition 1.33. Let H and H 0 be distinct parallel hyperplanes in Rn , given, respectively, by (1.3) and (1.5), where γ < γ 0 . The set M = {x ∈ Rn : γ 6 x·c 6 γ 0 }

(1.14)

is called the closed slab between H and H 0 . Similarly, the set N = {x ∈ Rn : γ < x·c < γ 0 }

(1.15)

is called the open slab between H and H 0 . Furthermore, the sets {x ∈ Rn : x·c 6 γ}

and {x ∈ Rn : γ 0 6 x·c}

are the opposite closed halfspaces determined by M , and the sets {x ∈ Rn : x·c < γ}

and {x ∈ Rn : γ 0 < x·c}

are the opposite open halfspaces determined by M . Theorem 1.29 implies that if M and N are slabs defined by (1.14) and (1.15), respectively, then M = cl N,

N = int M,

and

bd M = bd N = H ∪ H 0 .

Theorem 1.34. For a closed slab M ⊂ Rn and distinct points y, z ∈ Rn , the following statements take place. (1) If both y and z belong to M , then the segment [y, z] lies in M . (2) If y ∈ int M and z ∈ Rn \ M , then [y, z] meets bd M at exactly one point. (3) If y and z belong, respectively, to opposite open halfspaces determined by M , then [y, z] meets bd M at exactly two points. (4) If the halfline [y, zi lies in M (respectively, in int M ), then the whole line hy, zi lies in M (respectively, in int M ). Proof. Assume that M is given by (1.14). Then M can be expressed as the intersection of closed halfspaces {x ∈ Rn : x·c > γ}

and {x ∈ Rn : x·c 6 γ 0 }.


33

Statements (1)–(3) follow from Theorem 1.31. Hence it remains to prove statement (4). For this, choose a point u ∈ [y, zi. Then u = (1 − λ)y + λz, where λ > 0. The inclusion [y, zi ⊂ M implies that the inequalities γ 6 u·c = (1 − λ)y·c + λ z·c = y·c + λ(z·c − y·c) 6 γ 0

(1.16)

hold for all λ > 0. Clearly, this is possible only if y·c = z·c. Let β = y · c. Then γ 6 β 6 γ 0 . If v is a point in hy, zi, then v = (1 − µ)y + µz for a certain scalar µ. So, v·c = (1 − µ)y·c + µz·c = β, implying that the entire line hy, zi lies in M . Similarly, if [y, zi ⊂ int M , then both inequalities in (1.16) are strict, which yields γ < β < γ 0 . Consequently, hy, zi ⊂ int M . Halfplanes and Plane Slabs As we will see below, given a plane L ⊂ Rn of positive dimension m, its subplanes of dimension m−1 have properties similar to those of hyperplanes in Rn . Theorem 1.35. For a plane L ⊂ Rn of positive dimension m and a proper subplane M of L, the following statements hold. (1) dim M = m − 1 if and only if there is a hyperplane H ⊂ Rn such that M = H ∩ L. (2) dim M = m − 1 if and only if there is a vector c in Rn \ (sub L)⊥ such that M = {x ∈ L : x·c = γ} for a suitable scalar γ. (3) If dim M = m − 1 and c is a vector in (Rn \ (sub L)⊥ ) ∩ (sub M )⊥ , then M = {x ∈ L : x·c = γ} for a suitable scalar γ. Proof. The “if” parts in statements (1) and (2) follow from Theorems 1.17 and 1.19, while the “only if” parts in these statements follow from statement (3) of the theorem. So, it remains to prove statement (3). (3) Let dim M = m − 1 and c be a vector in (Rn \ (sub L)⊥ ) ∩ (sub M )⊥ . Choose a point u ∈ M . Then L = u+S and M = u+T , where S and T are the characteristic subspaces of L and M , respectively (see Theorem 1.2). Clearly, T ⊂ S, dim S = m, dim T = m − 1, and c ∈ (Rn \ S ⊥ ) ∩ T ⊥ . Consider the hypersubspace P = {x ∈ Rn : x·c = 0}. Then T ⊂ {c}⊥ = P , implying the inclusion T ⊂ P ∩ S. Furthermore, S 6⊂ P because c ∈ / S⊥.

34


Theorem 1.19 shows that dim (P ∩ S) = m − 1. Hence T = P ∩ S. Since the set H = u + P is a hyperplane, and since M = u + T = u + P ∩ S = (u + P ) ∩ (u + S) = H ∩ L, we obtain a desire expression of M . Finally, with γ = u·c, one has M = H ∩ L = {x ∈ L : x·c = γ}.

V1 D1

Fig. 1.8

p p p p p p pH p p p p p p p p pp pp pp pp p p p ppp

V2 D2

L

Closed halfplanes D1 and D2 of the plane L.

Definition 1.36. Let L ⊂ Rn be a plane of positive dimension (possibly, L = Rn ). A closed halfplane of L is the intersection of L and a closed halfspace V of Rn satisfying the condition ∅ 6= L ∩ V 6= L. Similarly, an open halfplane of L is the intersection of L and an open halfspace W of Rn such that ∅ 6= L ∩ W 6= L. Combining Theorems 1.19 and 1.30, we obtain two corollaries. Corollary 1.37. Let V (respectively, W ) be a closed (respectively, open) halfspace determined by a hyperplane H ⊂ Rn . For a proper plane L ⊂ Rn of positive dimension, the following conditions are equivalent. (1) (2) (3) (4)

∅ 6= L ∩ H 6= L. ∅ 6= L ∩ V 6= L (respectively, ∅ 6= L ∩ W 6= L). L meets both open halfspaces of Rn determined by H. No normal vector of V (respectively, of W ) belongs to the characteristic subspace (sub L)⊥ .

Corollary 1.38. Let L ⊂ Rn be a plane of positive dimension. A subset of L is a closed halfplane of L if and only if it can be expressed as D1 = {x ∈ L : x·c 6 γ}

or

D2 = {x ∈ L : x·c > γ},

(1.17)

⊥

where c ∈ / (sub L) and γ ∈ R. Similarly, a subset of L is an open halfplane of L if and only if it can be expressed as E1 = {x ∈ L : x·c < γ}

or

E2 = {x ∈ L : x·c > γ},

(1.18)


35

where c ∈ / (sub L)⊥ and γ ∈ R. Consequently, if Vi and Wi are halfspaces given by (1.9) and (1.10), respectively, then Di = Vi ∩ L

and

Ei = Wi ∩ L = int Vi ∩ L,

i = 1, 2.

The next corollary, which follows from Theorems 1.19 and 1.29, describes topological properties of halfplanes. Corollary 1.39. Let L ⊂ Rn be a plane of positive dimension m. A set E ⊂ L is an open halfplane of L if and only if there is a subplane M ⊂ L of dimension m − 1 such that E is one of the two components of L \ M . A set D ⊂ L is a closed halfplane of L if and only if there is a subplane M ⊂ L of dimension m − 1 such that D = M ∪ E, where E is an open halfplane of L determined by M . Remark. Any closed halfplane of a proper plane L ⊂ Rn is a closed set as the intersection L and a suitable closed halfspace (see Corollary 1.22 and Theorem 1.29). On the contrary, no open halfplane E of L is an open set if dim L 6 n − 1. As we will see later (Definition 2.13), open halfplanes are relatively open sets. One more corollary follows from Theorems 1.19 and 1.30. Corollary 1.40. Let L ⊂ Rn be a plane of positive dimension m and M be a subplane of L of dimension m − 1. If N is a proper subplane of L, then M and N are parallel if and only if either N ⊂ M , or N lies in an open halfplane of L determined by M . A set of useful properties of halfplanes follows from Theorem 1.31 and Corollary 1.32. Corollary 1.41. Let L ⊂ Rn be a plane of positive dimension m and M be a subplane of L of dimension m − 1. For distinct points y, z ∈ L, the following statements hold. (1) If both y and z belong to M , then the line hy, zi lies in M . (2) If y ∈ M and z belongs to an open halfplane of L determined by M , then the open halfline (y, zi lies in this halfplane. (3) If both y and z belong to an open halfplane of L determined by M , then at least one of the halflines [y, zi, hy, z] lies in this halfplane. (4) If y and z belong, respectively, to distinct open halfplanes of L determined by M , then hy, zi meets M at a single point, which belongs to the open segment (y, z).

36


(5) If both y and z belong to a closed halfplane of L determined by M such that (y, z) ∩ M 6= ∅, then [y, z] ⊂ M . An algebraic description of halfplanes is given by the following theorem. Theorem 1.42. Let L ⊂ Rn be a plane of positive dimension m. A set X ⊂ Rn is a closed (respectively, open) halfplane of L if and only if X can be expressed as X = h + M , where M is a subplane of L of dimension m − 1 and h is a closed (respectively, open) halfline with endpoint o, which lies in sub L and is not parallel to M . Proof. Assume first that X is a closed halfplane of L. By Corollary 1.38, X can be expressed as X = {x ∈ L : x·c 6 γ},

c∈ / (sub L)⊥ ,

γ ∈ R.

Let H = {x ∈ Rn : x·c = γ}

and M = {x ∈ L : x·c = γ}.

According to Theorem 1.17, H is a hyperplane. Clearly, M = H ∩ L, and Theorem 1.19 implies that M is a plane of dimension m−1. Choose a point a ∈ M and let X 0 = X − a,

S = L − a,

T = M − a.

Theorem 1.2 shows that S and T are subspaces of dimension m and m − 1, respectively. One can write T = {x ∈ S : x·c = 0}

and X 0 = {x ∈ S : x·c 6 0}.

If b1 , . . . , bm−1 is a basis for T and bm ∈ X 0 \ T , then b1 , . . . , bm is a basis for S. Clearly, bi ·c = 0 for all 1 6 i 6 m − 1, and bm ·c < 0. Since every vector x ∈ S can be written as x = ξ1 b1 + · · · + ξm bm , one has X 0 = {x ∈ S : x·c 6 0} = {ξ1 b1 + · · · + ξm bm : (ξm bm )·c 6 0} = {ξ1 b1 + · · · + ξm bm : ξm > 0}. Let h = {λbm : λ > 0} be the closed halfline generated by bm . Obviously, h ⊂ S = sub L such that h is not parallel to M . From the argument above it follows that a point x ∈ S belongs to X 0 if and only if x = y + z, where y ∈ h and z ∈ T . Consequently, X 0 = h + M , and X = a + h + T = h + M. Conversely, if X = h + M , then the above argument shows that X is a closed halfplane of L determined by the subplane M . The case of open halfplanes is similar.


37

Various descriptions of halfplanes and their properties are given in Theorem 1.71, Lemma 9.17, and Exercises 9.3 and 9.4. Definition 1.43. Let L ⊂ Rn be a plane of positive dimension. A closed slab of L is the intersection of L and a closed slab M of Rn such that ∅ 6= L ∩ M 6= L. Similarly, an open slab of L is the intersection of L and an open slab N of Rn such that ∅ 6= L ∩ N 6= L. M H H0 p p p p p p p p p p p p p p p pp p p p p p p p p p p p p p p p pp ppp ppp p pp F p p pp p pp p p p pp p p p

Fig. 1.9

L

A closed slab F of the plane L.

Definition 1.33 and Corollary 1.38 imply the following description of plane slabs. Corollary 1.44. Let L ⊂ Rn be a plane of positive dimension. F ⊂ L is a closed slab of L if and only if it can be expressed as F = {x ∈ L : γ 6 x·c 6 γ 0 }, c ∈ / (sub L)⊥ , γ < γ 0 . A set G ⊂ L is an open slab of L if and only if it can be expressed G = {x ∈ L : γ < x·c < γ 0 }, c ∈ / (sub L)⊥ , γ < γ 0 .

1.2

A set (1.19) as (1.20)

Affine Spans

Affine Combinations of Points and Planes Definition 1.45. An affine combination of points x1 , . . . , xr in Rn is a linear combination λ1 x1 + · · · + λr xr , where λ1 + · · · + λr = 1. For example, the affine combinations of distinct points x and y in Rn fulfill the line hx, yi. The next two theorems show that affine combinations provide a suitable characteristic property of planes. Theorem 1.46. A nonempty set X ⊂ Rn is a plane if and only if it contains all points (1 − λ)x + λy whenever x, y ∈ X and λ ∈ R.

38


Proof. Suppose first that X is a plane and express it as X = a + S for a certain a ∈ Rn and a subspace S. Choose points x, y ∈ X and a scalar λ. Then x = a + x0 and y = a + y0 , where x0 , y0 ∈ S. Consequently, (1 − λ)x + λy = a + ((1 − λ)x0 + λy0 ) ∈ a + S = X. Conversely, assume that (1 − λ)x + λy ∈ X whenever x, y ∈ X and λ ∈ R. Choose a point c ∈ X and let T = X − c. Clearly, o = c − c ∈ T . We state that (1 − λ)x + λy ∈ T for all x, y ∈ T and λ ∈ R. Indeed, expressing x and y as x = x0 − c and y = y 0 − c, with x0 , y 0 ∈ X, we obtain (1 − λ)x + λy = ((1 − λ)x0 + λy 0 ) − c ∈ X − c = T. Based on this argument, for points x, y ∈ T and a scalar λ, one has λx = (1 − λ)o + λx ∈ T,

x + y = 2((1 − 21 )x + 12 y) ∈ T.

Hence T is a subspace, which shows that X = c + T is a plane. Remark. We can reformulate Theorem 1.46, saying that a set X ⊂ Rn with two or more points is a plane if and only if with every pair of distinct points x, y ∈ X it contains the whole line hx, yi. The next result expands Theorem 1.46 to the case of affine combinations of finitely many points. Theorem 1.47. A nonempty set X ⊂ Rn is a plane if and only if it contains all affine combinations of points from X. Proof. If X contains all affine combinations of its points, then λ1 x1 + λ2 x2 ∈ X whenever x1 , x2 ∈ X and λ1 + λ2 = 1. By Theorem 1.46, X is a plane. Conversely, let X be a plane. By induction on k, we are going to prove that X contains all affine combinations of k points from X. The case k = 1 is obvious, and the case k = 2 is proved in Theorem 1.46. Assuming that the statement holds for all positive integers k 6 r − 1 (r > 2), choose an affine combination y = λ1 x1 + · · · + λr xr of points x1 , . . . , xr ∈ X. One of the scalars λ1 , . . . , λr is not 1, since otherwise λ1 + · · · + λr = r > 2. Let, for example, λ1 6= 1. By the induction hypothesis, the affine combination λ2 λr z= x2 + · · · + xr 1 − λ1 1 − λ1 belongs to X. Finally, the equality y = λ1 x1 + (1 − λ1 )z and Theorem 1.46 imply the inclusion y ∈ X.


39

Theorem 1.48. A nonempty set X ⊂ Rn is a plane if and only if any of the following conditions holds. (1) µ1 X + · · · + µr X = (µ1 + · · · + µr )X for every choice of scalars µ1 , . . . , µr , where µ1 + · · · + µr 6= 0. (2) µ1 X + · · · + µr X = (µ1 + · · · + µr − 1)c + X for every choice of c ∈ X and scalars µ1 , . . . , µr , where µ1 + · · · + µr 6= 0. Proof. (1) Suppose X is a plane. Since the inclusion (µ1 + · · · + µr )X ⊂ µ1 X + · · · + µr X holds for every set X ⊂ Rn and every choice of scalars µ1 , . . . , µr , it remains to prove that µ1 X + · · · + µr X ⊂ (µ1 + · · · + µr )X provided X is a plane and µ1 + · · · + µr 6= 0. For this, put µ = µ1 + · · · + µr

and λi = µi /µ, 1 6 i 6 r.

Then λ1 + · · · + λr = 1. According to Theorem 1.47, λ1 X + · · · + λr X = {λ1 x1 + · · · + λr xr : x1 , . . . , xr ∈ X} ⊂ X. Multiplying both sides of this inclusion by µ, one has (µ1 + · · · + µr )X ⊂ µ1 X + · · · + µr X. Conversely, let X satisfy condition (1) of the exercise. Choose points x, y ∈ X and a scalar λ. Then (1 − λ)x + λy ∈ (1 − λ)X + λX = X, and Theorem 1.46 implies that X is a plane. (2) If X is a plane, then, according to Theorem 1.2, the set S = X − c is a subspace. This argument gives µ1 X + · · · + µr X = (µ1 + · · · + µr )c + (µ1 + · · · + µr )S = (µ1 + · · · + µr )c + S = (µ1 + · · · + µr − 1)c + X. Conversely, let X satisfy condition (2) of the exercise. Choose points x, y ∈ X and a scalar λ. Then (1 − λ)x + λy ∈ (1 − λ)X + λX = (1 − λ + λ − 1)c + X = X, and Theorem 1.46 shows that X is a plane.

40


Affine Spans Definition 1.49. For a given set X ⊂ Rn , the intersection of all planes containing X is called the affine span of X and denoted aff X. Theorem 1.4 implies that the affine span of any set X ⊂ Rn exists and is the smallest plane containing X. Furthermore, aff ∅ = ∅. Example. The affine span of a point x ∈ Rn is the singleton {x}, and the affine span of two distinct points x, y ∈ Rn is the line hx, yi. Example. If L ⊂ Rn is a plane of positive dimension, then the affine span of a halfplane or a slab of L coincides with L (see Corollary 1.80). Some elementary properties of affine spans are described below. Theorem 1.50. For sets X and Y in Rn , the following statements hold. (1) (2) (3) (4) (5) (6)

X ⊂ aff X, with X = aff X if and only if X is a plane. aff (aff X) = aff X. aff X ⊂ span X = aff ({o} ∪ X). aff X = span X if and only if o ∈ aff X. aff X ⊂ aff Y if X ⊂ Y . aff X = aff Y if X ⊂ Y ⊂ aff X.

Furthermore, if {Xα } is a family of sets in Rn , then the statements below are true. (7) aff (∩ Xα ) ⊂ ∩ aff Xα . α

α

(8) ∪ aff Xα ⊂ aff (∪ Xα ). α

α

(9) ∪ aff Xα = aff (∪ Xα ) if the family {Xα } is nested. α

α

(10) aff (∪ aff Xα ) = aff (∪ Xα ). α

α

Proof. Let P(X) denote the family of all planes containing a given set X ⊂ Rn . The proofs of statements (1)–(10) derive from the following simple arguments. (a) (b) (c) (d)

aff X is the smallest element in P(X). P(X) is exactly the family of all planes containing aff X. P({o} ∪ X) is exactly the family of all subspaces containing X. If X ⊂ Y , then P(Y ) ⊂ P(X).


41

The next theorem gives an important description of affine spans in terms of affine combinations of points. Theorem 1.51. The affine span of a nonempty set X ⊂ Rn is the collection of all affine combinations of points from X: aff X = {λ1 x1 + · · · + λk xk : k > 1, x1 , . . . , xk ∈ X, λ1 + · · · + λk = 1}. Proof. Since aff X is a plane containing X, Theorem 1.47 yields that the set L = {λ1 x1 + · · · + λk xk : k > 1, x1 , . . . , xk ∈ X, λ1 + · · · + λk = 1} lies in aff X. We state that L also is a plane. For this, choosing points x, y ∈ L and expressing them as affine combinations x = γ1 x 1 + · · · + γp x p

and y = µ1 y1 + · · · + µq yq

of certain points x1 , . . . , xp , y1 , . . . , yq from X, we see that (1 − λ)x + λy also is an affine combination of these points for every choice of λ ∈ R: (1 − λ)x + λy = (1 − λ)γ1 x1 + · · · + (1 − λ)γp xp + λµ1 y1 + · · · + λµq yq . Hence (1 − λ)x + λy ∈ L, and Theorem 1.46 shows that L is a plane. Since X ⊂ L (every point x ∈ X can be written as 1x), the inclusions X ⊂ L ⊂ aff X and Theorem 1.50 give aff X = L. Corollary 1.52. For points x1 , . . . , xr ∈ Rn , one has aff {x1 , . . . , xr } = {λ1 x1 + · · · + λr xr : λ1 + · · · + λr = 1}. Proof. By Theorem 1.51, a point x ∈ Rn belongs to aff {x1 , . . . , xr } if and only if it can be written as an affine combination x = λi1 xi1 + · · · + λit xit of certain points xi1 , . . . , xit from {x1 , . . . , xr }. If any of these points, say xip and xiq , coincide, then we replace λip xip + λiq xiq with (λip + λiq )xip . Since such a replacement can be performed at most finitely many times, one may assume that all xi1 , . . . , xit are pairwise distinct. Finally, for every index i ∈ {1, . . . , r} \ {i1 , . . . , it }, we add 0xi to the right-hand side of x = λi1 xi1 + · · · + λit xit to express x as an affine combination of all points x1 , . . . , xr . Theorem 1.51 allows to establish the following important algebraic property of affine spans. Theorem 1.53. If X1 , . . . , Xr are sets in Rn and µ1 , . . . , µr are scalars, then aff (µ1 X1 + · · · + µr Xr ) = µ1 aff X1 + · · · + µr aff Xr .

42


Proof. Excluding the trivial case when at least one of the sets X1 , . . . , Xr is empty, we assume that all these sets are nonempty. An induction argument shows that the proof can be reduced to the case r = 2 (if r = 1, then we can write µ1 X1 = µ1 X1 + µ2 o). Since the set µ1 aff X1 + µ2 aff X2 is a plane (see Theorem 1.4), the obvious inclusion µ1 X1 + µ2 X2 ⊂ µ1 aff X1 + µ2 aff X2 and Theorem 1.50 give aff (µ1 X1 + µ2 X2 ) ⊂ µ1 aff X1 + µ2 aff X2 . For the opposite inclusion, choose an x ∈ µ1 aff X1 + µ2 aff X2 . Then x = µ1 x1 + µ1 x2 for certain points x1 ∈ aff X1 and x2 ∈ aff X2 . By Theorem 1.51, x1 and x2 can be written as affine combinations x1 = λ1 u1 + · · · + λp up

and x2 = γ1 v1 + · · · + γq vq ,

where u1 , . . . , up ∈ X1 and v1 , . . . , vq ∈ X2 . Because µ1 ui + µ2 vj ∈ µ1 X1 + µ2 X2

for all

16i6p

and

1 6 j 6 q,

the equalities x = µ1 x1 + µ2 x2 =

p P q P i=1 j=1

λi γj (µ1 ui + µ2 vj ),

p P q P

λ i γj = 1

i=1 j=1

show that x is an affine combination of points from µ1 X1 +µ2 X2 . Therefore, Theorem 1.51 gives x ∈ aff (µ1 X1 + µ2 X2 ). A sharper version of Theorem 1.53 for the case of large sums is considered in Exercise 1.7. Definition 1.54. For a nonempty set X ⊂ Rn , the subspace sub X = aff X − aff X is called the characteristic subspace of X. Corollary 1.55. For a nonempty set X ⊂ Rn and a point c ∈ aff X, the characteristic subspace of X can be expressed in the following ways. (1) sub X = aff X − aff X = aff (X − X) = span (X − X). (2) sub X = aff X − c = aff (X − c) = span (X − c). (3) sub X = {λ1 x1 + · · · + λk xk : x1 , . . . , xk ∈ X, λ1 + · · · + λk = 0}. Proof. Statements (1) and (2) immediately follow from a combination of Theorems 1.50 and 1.53. (3) Choose a point z ∈ X. If x ∈ sub X, then x ∈ aff X − z, and Theorem 1.51 shows that x can be expressed as x = λ1 x1 + · · · + λm xm + (−1)z, x1 , . . . , xm ∈ X, λ1 + · · · + λm = 1.


43

Therefore, x belongs to the set S(X) = {λ1 x1 + · · · + λk xk : k > 1, x1 , . . . , xk ∈ X, λ1 + · · · + λk = 0}. Conversely, if x ∈ S(X), then x = µ1 x1 + · · · + µk xk for suitable x1 , . . . , xk ∈ X and scalars µ1 , . . . , µk whose sum is 0. Then x + z ∈ aff X, implying that x = (x + z) − z ∈ aff X − z = sub X. Affine Span and Union of Sets For the sake of simplicity, we consider the affine spans of two sets in Rn . Theorem 1.56. For nonempty sets X and Y in Rn , the following statements hold. (1) aff (X ∪ Y ) = (aff X + sub Y ) ∪ (sub X + aff Y ) ∪ L(X, Y ), where L(X, Y ) = {λx + µy : x ∈ aff X, y ∈ aff Y, λ + µ = 1}. (2) If aff X ∩ aff Y 6= ∅ and c ∈ aff X ∩ aff Y , then aff (X ∪ Y ) = L(X, Y ), aff (X ∪ Y ) = aff (X + Y ) − c = aff (X + Y ) − X ∩ Y.

(1.21) (1.22)

(3) If aff X ∩ aff Y = ∅, c ∈ aff X, e ∈ aff Y , and l is the line through c and e, then aff (X ∪ Y ) = aff X + aff Y + l − c − e.

(1.23)

Proof. (1) By Theorem 1.51, a point z ∈ Rn belongs to aff (X ∪ Y ) if and only if it can be expressed as an affine combination z = λ1 x1 + · · · + λp xp + µ1 y1 + · · · + µq yq , where x1 . . . , xp ∈ X and y1 , . . . , yq ∈ Y . If one of the sums λ = λ1 + · · · + λp

and µ = µ1 + · · · + µq

equals 0, then x belongs to the respective set aff Y +sub X or aff X +sub Y . Suppose that both λ and µ are distinct from 0. Then x0 = λ−1 (λ1 x1 + · · · + λp xp ) ∈ aff X, y 0 = µ−1 (µ1 y1 + · · · + µq yq ) ∈ aff Y, implying the inclusion z = λx0 + µy 0 ∈ L(X, Y ).

44


(2) First, we will prove the equalities (1.22). For this, we observe that o ∈ (aff X − c) ∪ (aff Y − c). Therefore, Theorem 1.50 and the properties of spans (see page 4) give aff ((X − c) ∪ (Y − c)) = span ((X − c) ∪ (Y − c)) = span (X − c) + span (Y − c). Using Theorem 1.53 twice, we obtain aff (X ∪ Y ) = aff ((X − c) ∪ (Y − c) + c) = aff ((X − c) ∪ (Y − c)) + c = span (X − c) + span (Y − c) + c = aff (X − c) + aff (Y − c) + c = aff X + aff Y − c = aff (X + Y ) − c. The second part of (1.22), aff (X ∪ Y ) − c = aff (X + Y ) − X ∩ Y, follows from Theorem 1.2. Now, we will prove the equality (1.21). Theorem 1.51 shows that the set L(X, Y ) lies in aff (aff X ∪ aff Y ). Since aff (aff X ∪ aff Y ) = aff (X ∪ Y ) (see Theorem 1.50), one has L(X, Y ) ⊂ aff (X ∪ Y ). For the opposite inclusion, choose points z ∈ aff (X ∪ Y ) and c ∈ X ∩ Y . By the above argument, z − c ∈ aff (X ∪ Y ) − c = span (X − c) + span (Y − c). Therefore, z − c can be written as z − c = u + v, where u ∈ span (X − c) and v ∈ span (Y − c). Because 2u ∈ span (X − c) and 2v ∈ span (Y − c), it follows that 2u + c ∈ span (X − c) + c = aff X, 2v + c ∈ span (Y − c) + c = aff Y. Finally, z = 21 (2u + c) + 12 (2v + c) ∈ L(X, Y ). (3) Since every point x ∈ l can be written as x = (1 − λ)c + λe, Theorem 1.51 shows that l ⊂ aff (aff X ∪ aff Y ) = aff (X ∪ Y ). Hence X ∪Y ⊂ X ∪Y ∪l ⊂ aff (X ∪Y ), and Theorem 1.50 gives aff (X ∪Y ) = aff (X ∪ Y ∪ l). Since e ∈ aff Y ∩ l

and c ∈ aff X ∩ aff (Y ∪ l),


45

a combination of Theorem 1.50 and statement (2) above gives aff (X ∪ Y ) = aff (X ∪ Y ∪ l) = aff (X ∪ aff (Y ∪ l)) = aff X + aff (Y ∪ l) − c = aff X + aff Y + l − c − e. Corollary 1.57. For nonempty sets X and Y in Rn , the following statements hold. (1) If aff X ∩ aff Y = 6 ∅, then dim (X ∪ Y ) = dim (X + Y ). (2) If aff X ∩ aff Y = ∅, then dim (X ∪ Y ) = dim (X + Y ) + 1. Proof. (1) In terms of statement (2) of Theorem 1.56, one has dim (X ∪ Y ) = dim (aff (X ∪ Y )) = dim (aff (X + Y ) − c) = dim (aff (X + Y )) = dim (X + Y ). (2) Similarly, based on statement (2) of Theorem 1.56, we obtain first dim (X + Y ) = dim (aff (X + Y )) 6 dim (aff X + aff Y + l) = dim (X ∪ Y ) 6 dim (aff (X + Y )) + 1

(1.24)

= dim (X + Y ) + 1. Next, we state that l 6⊂ L ≡ aff X + aff Y − c − e. Indeed, assuming the contrary, one has {c, e} ⊂ L. The set L is a subspace as the sum of subspaces aff X − c and aff Y − e. Therefore, c − e ∈ L, and we can write c − e = α(x − c) + β(y − e),

where

x ∈ aff X, y ∈ aff Y, α, β ∈ R.

Thus c − α(x − c) = e + β(y − e). The left-hand side of the latter equality belongs to aff X, and the right-hand side belongs to aff Y , contrary to the assumption aff X ∩ aff Y = ∅. Summing up, l 6⊂ L. Consequently, dim (X + Y ) = dim (aff X + aff Y ) = dim L < dim (aff X + aff Y + l). Combined with (1.24), this gives dim (X ∪ Y ) = dim (X + Y ) + 1. Affinely Independent Sets Definition 1.58. A set of points x1 , . . . , xr in Rn , with r > 2, is called affinely dependent if there are scalars ν1 , . . . , νr , not all zero, such that ν1 x1 + · · · + νr xr = o

and ν1 + · · · + νr = 0.

(1.25)

The set {x1 , . . . , xr } is called affinely independent if it is not affinely dependent. The empty set ∅ and every singleton {x} are assumed to be affinely independent.

46


Occasionally, we will write “affinely independent points x1 , . . . , xr ” instead of “affinely independent set {x1 , . . . , xr }.” The next result refines Theorem 1.51. Theorem 1.59. The affine span of a nonempty set X ⊂ Rn is the collection of all affine combinations of affinely independent points from X. Proof. By Theorem 1.51, a point x ∈ Rn belongs to aff X if and only if is expressible as an affine combination x = λ1 x1 + · · · + λk xk of certain points x1 , . . . , xk ∈ X. We are going to show that if {x1 , . . . , xk } is affinely dependent, then x can be written as an affine combination of k − 1 of fewer points from {x1 , . . . , xk }. So, let {x1 , . . . , xk } be affinely dependent. Then there are scalars ν1 , . . . , νk , not all zero, such that ν1 x1 + · · · + νk xk = o

and ν1 + · · · + νk = 0.

Without loss of generality, we assume that ν1 6= 0. Then νk ν2 x1 = − x2 − · · · − xk . ν1 ν1 Hence x can be written as an affine combination νk ν2 x2 + · · · + λk − λ1 xk , x = λ1 x1 + · · · + λk xk = λ2 − λ1 ν1 ν1 which involves only points x2 , . . . , xk . Consecutively repeating this argument, we obtain a desired expression of x. Theorem 1.60. For points x1 , . . . , xr ∈ Rn , the following conditions are equivalent. (1) The set {x1 , . . . , xr } is affinely independent. (2) None of x1 , . . . , xr is an affine combination of the others. (3) For every index i = 1, . . . , r, the set {x1 − xi , . . . , xi−1 − xi , xi+1 − xi , . . . , xr − xi }

(1.26)

is linearly independent (it is empty if r = 1). (4) There is an index i ∈ {1, . . . , r} such that the set (1.26) is linearly independent. (5) The plane aff {x1 , . . . , xr } has dimension r − 1. (6) Every point x ∈ aff {x1 , . . . , xr } is uniquely expressible as an affine combination of x1 , . . . , xr .


47

Proof. Since the case r = 1 is obvious, we assume that r > 2. (1) ⇒ (2). Assume for a moment the existence of a point xi , i ∈ {1, . . . , r}, which is an affine combination of x1 , . . . , xi−1 , xi+1 , . . . , xr : xi = λ1 x1 + · · · + λi−1 xi−1 + λi+1 xi+1 · · · + λr xr , λ1 + · · · + λi−1 + λi+1 + · · · + λr = 1. Rewriting these equalities as λ1 x1 + · · · + λi−1 xi−1 + (−1)xi + λi+1 xi+1 · · · + λr xr = 0, λ1 + · · · + λi−1 + (−1) + λi+1 + · · · + λr = 0, we conclude that the set {x1 , . . . , xr } is affinely dependent, in contradiction with (1). (2) ⇒ (3). Suppose the existence of an index i ∈ {1, . . . , r} such that the set (1.26) is linearly dependent. Then one of these vectors, say x1 − xi , is a linear combination of the others: x1 − xi = λ2 (x2 − xi ) + · · · + λi−1 (xi−1 − xi ) + λi+1 (xi+1 − xi ) + · · · + λr (xr − xi ). With λi = 1 − λ2 − · · · − λi−1 − λi+1 − · · · − λr , the equality x1 = λ2 x2 + · · · + λr xr shows that x1 is an affine combination of x2 , . . . , xr , contradicting (2). Since (3) implies (4), we show next that (4) ⇒ (5). For this, consider the (r − 1)-dimensional subspace S = span {x1 − xi , . . . , xi−1 − xi , xi+1 − xi , . . . , xr − xi }.

(1.27)

A combination of Theorems 1.50 and 1.53 gives S = aff {x1 − xi , . . . , xi−1 − xi , o, xi+1 − xi , . . . , xr − xi } = aff {x1 , . . . , xr } − xi . Hence the plane aff {x1 , . . . , xr } = xi + S has dimension r − 1. (5) ⇒ (6). Choose a point x ∈ aff {x1 , . . . , xr }. By Corollary 1.52, x can be written as an affine combination x = λ1 x1 + · · · + λr xr . To show its uniqueness, choose an affine combination x = µ1 x1 + · · · + µr xr . Consequently, the vector y = x − xi can be written as y = λ1 (x1 − xi ) + · · · + λi−1 (xi−1 − xi ) + λi+1 (xi+1 − xi ) + · · · + λr (xr − xi ), y = µ1 (x1 − xi ) + · · · + µi−1 (xi−1 − xi ) + µi+1 (xi+1 − xi ) + · · · + µr (xr − xi ).

48


The subspace (1.27) is (r −1)-dimensional as a translate of aff {x1 , . . . , xr }. Hence the set (1.26) is linearly independent, which implies that the above expressions for y are identical: λj = µj for all j 6= i. Finally, P P λi = 1 − λj = 1 − µj = µi . j6=i

j6=i

(6) ⇒ (1). Assume for a moment that the set {x1 , . . . , xr } is affinely dependent. Then there are scalars ν1 , . . . , νr , not all zero, such that (1.25) holds. Choose a point x ∈ aff {x1 , . . . , xr }. By Corollary 1.52, x can be expressed as an affine combination x = λ1 x1 + · · · + λr xr . Then x = (λ1 + ν1 )x1 + · · · + (λr + νr )xr is another expression of x as an affine combination of x1 , . . . , xr , contrary to (6). Since every m-dimensional subspace of Rn contains at most m linearly independent vectors, Theorem 1.60 implies the following corollary. Corollary 1.61. An affinely independent set X ⊂ Rn which lies within a plane of dimension m consists of m + 1 or fewer points. Theorem 1.62. A set X = {x1 , . . . , xr } ⊂ Rn is affinely independent if and only if it does not contain disjoint subsets whose affine spans meet. Proof. We will prove the contrapositive statement: the set X is affinely dependent if and only if it contains disjoint subsets whose affine spans meet. Let X be affinely dependent. This means the existence of scalars ν1 , . . . , νr , not all zero, such that ν1 x1 + · · · + νr xr = o,

ν1 + · · · + νr = 0.

Put I = {i : νi > 0} and J = {i : νi < 0}. Then both sets I and J are nonempty and partition the set {1, . . . , r}. Furthermore, P P νi xi = (−νj )xj . (1.28) i∈I

P

Let ν = i∈I νi . Clearly, ν > 0 and of (1.28) by ν, we obtain

i∈J

P

i∈J (−νj )

= ν. Dividing both parts

P (−νj ) P νi xi = xj . ν ν i∈J i∈I

(1.29)

As easily seen, left-hand part of (1.29) is an affine combination of points from the set Y = {xi : i ∈ I}; similarly, right-hand side of (1.29) is an affine


49

combination of points from the set Z = {xj : j ∈ J}. Hence Y ∩ Z = ∅ and aff Y ∩ aff Z 6= ∅. Conversely, let X = Y ∪Z be a partition of X such that aff Y ∩aff Z 6= ∅. Choose a point x ∈ aff Y ∩ aff Z. By Theorem 1.51, x can be expressed as affine combinations, x = λ1 y1 + · · · + λk yk = µ1 z1 + · · · + µm zm , of certain points y1 , . . . , yk ∈ Y and z1 , . . . , zm ∈ Z. Definition 1.58 and the equalities λ1 y1 + · · · + λk yk − µ1 z1 − · · · − µm zm = o, λ1 + · · · + λk − µ1 − · · · − µm = 0 show that the set {y1 , . . . , yk , z1 , . . . , zm } is affinely dependent. Hence X is affinely dependent. Corollary 1.63. Let a set {x1 , . . . , xr } ⊂ Rn be affinely independent and xr+1 be a point in Rn . Then the set {x1 , . . . , xr+1 } is affinely independent if and only if xr+1 does not belong to the plane aff {x1 , . . . , xr }. Affine Bases Definition 1.64. We say that points c1 , . . . , cr ∈ Rn form an affine basis for a nonempty plane L ⊂ Rn provided the set {c1 , . . . , cr } is affinely independent and L = aff {c1 , . . . , cr }. The empty set ∅ is assumed to be the affine basis for the empty plane. r c 3 c3 − c1 r

o

r

Fig. 1.10

-

c1

r

c2

c2 − c1

Affine basis c1 , c2 , c3 and basis c2 − c1 , c3 − c1 for R2 .

Theorem 1.65. Every plane L ⊂ Rn has an affine basis. Furthermore, if dim L = m, then each affine basis for L consists of m + 1 points.

50


Proof. Since the case L = ∅ is obvious, we assume that L is nonempty. Let m = dim L. The case m = 0 is clear (L is a singleton, {c}, and c is its affine basis). Let m > 1. Choose a point a ∈ L and consider the m-dimensional subspace S = L − a. If b1 , . . . , bm is a basis for S, then, according to Theorem 1.50, S = span {b1 , . . . , bm } = aff {o, b1 , . . . , bm }. Therefore, L = a + S = a + aff {o, b1 , . . . , bm } = aff {a, a + b1 , . . . , a + bm } (see Theorem 1.53). Finally, Theorem 1.60 shows that a, a + b1 , . . . , a + bm is an affine basis for L. The same theorem implies that each affine basis for L consists of m + 1 points. Theorem 1.60 implies one more corollary. Corollary 1.66. For a nonempty plane L ⊂ Rn of dimension m and points c1 , . . . , cm+1 ∈ L, the following conditions are equivalent. (1) (2) (3) (4)

The points c1 , . . . , cm+1 form an affine basis for L. The set {c1 , . . . , cm+1 } is affinely independent. L = aff {c1 , . . . , cm+1 }. For every index i = 1, . . . , m + 1, the vectors c1 − ci , . . . , ci−1 − ci , ci+1 − ci , . . . , cm+1 − ci

(1.30)

form a basis for the subspace sub L (this set of vectors is empty if m = 1). (5) There is an index i ∈ {1, . . . , m + 1} such that the vectors (1.30) form a basis for the subspace sub L. (6) Every point x ∈ L is uniquely expressible as an affine combination of c1 , . . . , cm+1 . Corollary 1.67. If L ⊂ Rn is a plane of dimension m, −1 6 m 6 n − 1, and c is a point in Rn \ L, then aff (c ∪ L) is a plane of dimension m + 1. Proof. Since the case L = ∅ is obvious, we let L 6= ∅. By Theorem 1.65, L has an affine basis c1 , . . . , cm+1 , and Corollary 1.63 shows that the set {c, c1 , . . . , cm+1 } is affinely independent. Theorem 1.50 gives aff (c ∪ L) = aff (c ∪ aff {c1 , . . . , cm+1 }) = aff {c, c1 , . . . , cm+1 }. Finally, Theorem 1.60 implies that dim (aff (c ∪ L)) = m + 1.


51

The next result refines Theorem 1.65 by putting a restriction on the choice of an affine basis. Theorem 1.68. If Y is an affinely independent subset of a set X ⊂ Rn , then there is an affine basis Z for aff X such that Y ⊂ Z ⊂ X. Proof. The case X = ∅ is obvious; so, we assume that X 6= ∅. Choose a point c ∈ Y if Y 6= ∅ or any c ∈ X if Y = ∅. Consider the set Y 0 = (Y − c) \ {o}. Clearly, Y 0 ⊂ X − c. By Theorem 1.60, the set Y 0 is linearly independent (if card Y 6 1, then Y 0 = ∅ is linearly independent by the definition). From linear algebra we know that Y 0 can be expanded to a basis U for the subspace span (X − c). Furthermore, span (X − c) = aff (X − c) because of o ∈ X − c (see Theorem 1.50). Finally, let Z = {c} ∪ (c + U ). Then Y ⊂ Z ⊂ X and, according to Corollary 1.66, Z is an affine basis for the plane c + span (X − c) = c + aff (X − c) = aff X. Corollary 1.69. If L is a plane in Rn , then every affinely independent subset of L can be completed into an affine basis for L. ppp

pp ppp

pp

p ppp ppp ppppp pp pp rpp p c1 ppp p p ppppp ppp pp pp ppp ppp

ppp

ppp

r x = 12 c1 − 21 c2 + 1c3

ppp pp pp pp pcp 2 p p p p p p p p p pppprp p p p p p p p p p p p p p p p p p p p p p p p p ppppprpppp pcp p3p p p p p p ppp p ppp ppp p ppp Fig. 1.11

Affine coordinates of x relative to c1 , c2 , c3 .

Theorem 1.60 gives a base to the following definition. Definition 1.70. Let {c1 , . . . , cr } ⊂ Rn be an affinely independent set and x be a point in aff {c1 , . . . , cr }. If x is (uniquely) expressed as an affine combination x = λ1 c1 + · · · + λr cr ,

λ1 + · · · + λr = 1,

then λ1 , . . . , λr are called the affine coordinates of x relative to c1 , . . . , cr . The next theorem shows that affine coordinates are useful in descriptions of halfplanes (see Definition 1.36).

52


Theorem 1.71. Let L ⊂ Rn be a plane of positive dimension m. A set D ⊂ Rn is a closed halfplane of L if and only if there is an affine basis c1 , . . . , cm+1 for L such that D = {λ1 c1 + · · · + λm+1 cm+1 : λ1 + · · · + λm+1 = 1, λm+1 > 0}. (1.31) Similarly, a set E ⊂ Rn is an open halfplane of L if and only if there is an affine basis c1 , . . . , cm+1 for L such that E = {λ1 c1 + · · · + λm+1 cm+1 : λ1 + · · · + λm+1 = 1, λm+1 > 0}. (1.32)

rc1 r c2

D = {λ1 c1 + λ2 c2 + λ3 c3 :

r c3

λ1 + λ2 + λ3 = 1, λ3 > 0}

Fig. 1.12

A closed halfplane of R2 .

Proof. Assume first that D is a closed halfplane of L. By Corollary 1.38, D can be expressed as D = {x ∈ L : x·c 6 γ},

c∈ / (sub L)⊥ ,

γ ∈ R.

Let H = {x ∈ Rn : x·c = γ} and M = {x ∈ L : x·c = γ}. According to Theorem 1.17, H is a hyperplane. Clearly, M = H ∩ L, and Theorem 1.19 shows that M is a plane of dimension m − 1. Choose a point a ∈ M and let D0 = D − a, S = L − a, T = M − a. Theorem 1.2 shows that S and T are subspaces of dimension m and m − 1, respectively. One can write T = {x ∈ L0 : x·c = 0}

and D0 = {x ∈ L0 : x·c 6 0}.

If b1 , . . . , bm−1 is a basis for T and bm ∈ D0 \ T , then b1 , . . . , bm is a basis for S. Clearly, bi ·c = 0 for all 1 6 i 6 m − 1, and bm ·c < 0. Since every vector x ∈ S can be written as x = ξ1 b1 + · · · + ξm bm , one has D0 = {x ∈ S : x·c 6 0} = {ξ1 b1 + · · · + ξm bm : (ξm bm )·c 6 0} = {ξ1 b1 + · · · + ξm bm : ξm > 0}. Furthermore, with c1 = a, ci+1 = a + bi , and λ1 = 1 − ξ1 − · · · − ξm , λi+1 = ξi , 1 6 i 6 m,


53

the halfplane D can be described as D = a + D0 = a + {ξ1 b1 + · · · + ξm bm : ξm > 0} = {λ1 a + ξ1 (a + b1 ) + · · · + ξm (a + bm ) : ξm > 0} = {λ1 c1 + · · · + λm+1 cm+1 : λ1 + · · · + λm+1 = 1, λm+1 > 0}. Repeating the above argument in the converse order, we obtain that every set of the form (1.31) is a closed halfplane of L. The case of an open halfplane is similar. Affine Bases and Operations with Planes Corollary 1.72. Let L1 = a1 + S1 and L2 = a2 + S2 be nonempty planes in Rn , expressed as translates of subspaces S1 and S2 , respectively. If a basis b1 , . . . , br for S1 ∩ S2 is completed into bases b1 , . . . , br , c1 , . . . , cp

and

b1 , . . . , br , e1 , . . . , eq

(1.33)

for S1 and S2 , respectively, then {(a1 + a2 ), b01 , . . . , b0r , c01 , . . . , c0p , e01 , . . . , e0q }

(1.34)

is an affine basis for L1 + L2 , where b0i = (a1 + a2 ) + bi , c0i = (a1 + a2 ) + ci , e0i = (a1 + a2 ) + ei for all suitable indices i. Proof. From linear algebra we know (see page 4), that the combined list b1 , . . . , br , c1 , . . . , cp , e1 , . . . , eq is a basis for the subspace S1 + S2 . Corollary 1.66 implies that the set (1.34) is an affine basis for the plane L1 + L2 = (a1 + a2 ) + (S1 + S2 ). Corollary 1.73. Let L1 = a1 + S1 and L2 = a2 + S2 be nonempty planes in Rn , expressed as translates of subspaces S1 and S2 , respectively. If a basis b1 , . . . , br for S1 ∩ S2 is completed into bases b1 , . . . , br , c1 , . . . , cp

and

b1 , . . . , br , e1 , . . . , eq

(1.35)

for S1 and S2 , respectively, then the following statements take place. (1) If L1 and L2 are disjoint, then {a1 , a2 , b01 , . . . , b0r , c01 , . . . , c0p , e01 , . . . , e0q } is an affine basis for aff (L1 ∪ L2 ), where b0i = a1 + bi , c0i = a1 + ci , e0i = a2 + ei for all suitable indices i.

(1.36)

54


(2) If L1 ∩ L2 6= ∅ and a is a point in L1 ∩ L2 , then {a, b01 , . . . , b0r , c01 , . . . , c0p , e01 , . . . , e0q }

(1.37)

is an affine basis for aff (L1 ∪ L2 ), where b0i = a + bi , c0i = a + ci , e0i = a + ei for all suitable indices i. Proof. (1) According to Corollary 1.66, the sets X1 = {a1 , b01 , . . . , b0r , c01 , . . . , c0p }

and X2 = {a2 , b01 , . . . , b0r , e01 , . . . , e0q }

are affine bases for L1 and L2 , respectively. In particular, L1 = aff X1 and L2 = aff X2 , and Theorem 1.50 gives aff (L1 ∪ L2 ) = aff (aff X1 ∪ aff X2 ) ⊂ aff (aff (X1 ∪ X2 )) = aff (X1 ∪ X2 ) ⊂ aff (L1 ∪ L2 ). Hence aff (L1 ∪ L2 ) = aff (X1 ∪ X2 ). According to Corollary 1.57, dim (L1 ∪ L2 ) = dim (L1 + L2 ) + 1 = p + q + r + 1. Let Y2 = {a2 , e01 , . . . , e0q }. It is easy to see that X1 ∪ Y2 is an affinely independent set of cardinality p + q + r + 2. Corollary 1.66 implies that X1 ∪ Y2 is an affine basis for aff (L1 ∪ L2 ). (2) By Theorem 1.2, L1 = a + S1 and L2 = a + S2 . Repeating the above argument, with a1 = a2 = a and p + q + r instead of p + q + r + 1, we obtain the desired conclusion. Affine bases provide a suitable description of complementary planes, as shown below. Theorem 1.74. Nonempty planes L1 and L2 in Rn are complementary if and only if there is an affine basis c1 , . . . , cn+1 for Rn such that L1 = aff {c1 , . . . , cm+1 }

and

L2 = aff {cm+1 , . . . , cn+1 },

(1.38)

where m = dim L1 . Proof. Assume first that the planes L1 and L2 are complementary. By Theorem 1.11, L1 + L2 = Rn and the set L1 ∩ L2 is a singleton. Let cm+1 denote the unique point from L1 ∩ L2 . Corollary 1.69 shows that {cm+1 } can be completed into affine bases c1 , . . . , cm+1 and cm+1 , . . . , cn+1 for L1 and L2 , respectively. Then both equalities (1.38) hold, and a combination


55

of Theorem 1.56 and Corollary 1.73 implies that c1 , . . . , cn+1 is an affine basis for aff (L1 ∪ L2 ) = L1 + L2 − cm+1 = Rn − cm+1 = Rn . Conversely, suppose the existence of an affine basis c1 , . . . , cn+1 for Rn such that both equalities (1.38) hold. Combining Theorems 1.50 and 1.56, we obtain L1 + L2 = aff (L1 ∪ L2 ) + cm+1 = aff (aff {c1 , . . . , cm+1 } ∪ aff {cm+1 , . . . , cn+1 }) + cm+1 = aff {c1 , . . . , cn+1 } + cm+1 = Rn . Since dim L1 = m and dim L2 = n − m (see Theorem 1.60), one has dim (L1 + L2 ) = n = dim L1 + dim L2 . By Theorem 1.10, the planes L1 and L2 are complementary. Dimension of Sets Definition 1.75. The dimension of a set X ⊂ Rn , denoted dim X, is defined by dim X = dim (aff X). In particular, dim ∅ = −1. Theorems 1.59, 1.60, and 1.68 imply one more corollary. Corollary 1.76. The dimension of a set X ⊂ Rn equals the maximum number of affinely independent points in X minus 1. If m = dim X, then the following statements hold. (1) X contains an affinely independent subset Y of m + 1 points such that aff Y = aff X. (2) Every affinely independent set of m + 1 points from X is an affine basis for aff X. (3) aff X is the set of all affine combinations of m + 1 or fewer affinely independent points from X. Corollary 1.77. If L ⊂ Rn is a plane of dimension m and X is subset of L, then aff X = L if and only if dim X = m. Proof. If aff X = L, then dim X = m according to the definition. Conversely, let dim X = m. The inclusion X ⊂ L shows that aff X ⊂ L (see Theorem 1.50). Since the planes aff X and L have the same dimension, Theorem 1.6 yields the equality aff X = L.

56


Corollary 1.78. If X ⊂ Rn is a set of dimension m, −1 6 m 6 n − 1, and c is a point in Rn \ aff X, then dim ({c} ∪ X) = m + 1. Proof. Let L = aff X. Then dim L = m, and, according to Theorem 1.50, aff ({c} ∪ X) = aff ({c} ∪ aff X) = aff ({c} ∪ L). Because dim (aff ({c} ∪ L)) = m + 1 (see Corollary 1.67), we conclude that dim ({c} ∪ X) = m + 1. Theorem 1.79. If L ⊂ Rn is a nonempty m-dimensional plane and Bρ (c) ⊂ Rn is the closed ball of radius ρ > 0 centered at a point c ∈ L, then aff (Bρ (c) ∩ L) = L

and

dim (Bρ (c) ∩ L) = m.

Proof. Since the case m = 0 is trivial, we assume that m > 1. Consider the subspace S = L − c and the ball Bρ (o) = Bρ (c) − c. Because dim S = m, one can choose in Bρ (o) ∩ S a basis b1 , . . . , bm for S. Put ci = bi + c, 1 6 i 6 m. By Corollary 1.66, the set {c, c1 , . . . , cm } is an affine basis for L. Furthermore, ci ∈ Bρ (c) for all 1 6 i 6 m because of kci − ck = kbi k 6 ρ. From the inclusions {c, c1 , . . . , cm } ⊂ Bρ (c) ∩ L ⊂ L = aff {c, c1 , . . . , cm } and Theorem 1.50 it follows that aff (Bρ (c) ∩ L) = L. Finally, Definition 1.75 shows that dim (Bρ (c) ∩ L) = dim L = m. Remark. A section of a ball Bρ (c) by an m-dimensional plane L ⊂ Rn through c is often called an m-dimensional ball (see Exercise 1.3 for a characterization of planes in terms of m-balls). Corollary 1.80. If L ⊂ Rn is a plane of positive dimension m and F is a halfplane of L (respectively, G is a slab of L), then aff F = aff G = L. Consequently, dim F = dim G = m. Proof. Suppose that F is a closed halfplane of L (the case of an open halfplane is similar). By the definition, F = L ∩ V , where V is a closed halfspace of Rn such that ∅ 6= L ∩ V 6= L. According to Corollary 1.37, the set L ∩ int V is nonempty. Choose a point c ∈ L ∩ int V . Since the set int V is open, there is a scalar ρ > 0 such that the ball Bρ (c) lies in V . Therefore, Bρ (c) ∩ L ⊂ V ∩ L = F . A combination of Theorems 1.50 and 1.79 shows that L = aff (Bρ (c) ∩ L) ⊂ aff F ⊂ L. Hence aff F = L, which gives dim F = m. The proof for G is similar.


1.3

57

Affine Transformations

Definition and Basic Properties Definition 1.81. A mapping f : Rn → Rm is called an affine transformation provided it can be expressed as f (x) = a + g(x),

x ∈ Rn ,

where a is a point in Rm and g : Rn → Rm is a linear transformation. An affine functional on Rn is an affine transformation f : Rn → R. Theorem 1.82. The following statements take place. (1) Any affine transformation f : Rn → Rm is uniquely expressible as f (x) = a + g(x), where a ∈ Rm and g : Rn → Rm is a linear transformation. (2) If an affine transformation f : Rn → Rn is invertible, then its inverse also is an affine transformation. (3) If fi : Rn → Rm , i = 1, 2, are affine transformations, then every linear combination µ1 f1 + µ2 f2 is an affine transformation. (4) If f1 : Rn → Rm and f2 : Rm → Rk are affine transformations, then their composition f2 ◦ f1 : Rn → Rk is an affine transformation. Proof. (1) Let f (x) = a1 + g1 (x) = a2 + g2 (x), where a1 , a2 ∈ Rm and both g1 and g2 are linear transformations. Then a1 = f (o) = a2 , and g1 (x) = f (x) − a1 = f (x) − a2 = g2 (x)

for all

x ∈ Rn .

(2) Let f (x) = a + g(x), where a ∈ Rn and g : Rn → Rn is a linear transformation. Clearly, g is invertible, and from linear algebra we know that g −1 is a linear transformation. Consider the affine transformation h(x) = −g −1 (a) + g −1 (x). The identities f (h(x)) = a + g(−g −1 (a) + g −1 (x)) = x, h(f (x)) = −g −1 (a) + g −1 (a + g(x)) = x show that h = f −1 . (3) If fi (x) = ai + gi (x), where ai ∈ Rn and gi : Rn → Rm is a linear transformation, i = 1, 2, then (µ1 f1 + µ2 f2 )(x) = (µ1 a1 + µ2 a2 ) + (µ1 g1 + µ2 g2 )(x). Since µ1 g1 + µ2 g2 is a linear transformation, µ1 f1 + µ2 f2 is an affine one.

58


(4) Let f1 and f2 be expressed as f1 (x) = a1 + g1 (x)

and f2 (x) = a2 + g2 (x),

where a1 ∈ Rm , a2 ∈ Rk and g1 and g2 are respective linear transformations. With a = a2 + g2 (a1 ), we have (f2 ◦ f1 )(x) = f2 (f1 (x)) = a2 + g2 (a1 + g1 (x)) = a2 + g2 (a1 ) + g2 (g1 (x)) = a + (g2 ◦ g1 )(x). Since g2 ◦ g1 is a linear transformation, f2 ◦ f1 is an affine one. Theorem 1.83. For an affine transformation f : Rn → Rm , the following statements hold. (1) If {c1 , . . . , cr } is an affinely independent set in Rn and f is oneto-one, then the set {f (c1 ), . . . , f (cr )} is affinely independent. (2) If {a1 , . . . , ar } is an affinely independent set in Rm and points c1 , . . . , cr ∈ Rn satisfy the conditions f (ci ) = ai , 1 6 i 6 r, then the set {c1 , . . . , cr } is affinely independent. Proof. Let f (x) = a + g(x), where a ∈ Rm , and g : Rn → Rm is a linear transformation. Since both statements (1) and (2) are obvious when r = 1, we may assume that r > 2. (1) By Theorem 1.60, the set {c2 − c1 , . . . , cr − c1 } is linearly independent. Because g is one-to-one, the set {g(c2 − c1 ), . . . , g(cr − c1 )} also is linearly independent (see page 5). The equalities f (ci ) − f (c1 ) = g(ci ) − g(c1 ) = g(ci − c1 ),

2 6 i 6 r,

and Theorem 1.60 show that {f (c1 ), . . . , f (cr )} is affinely independent. (2) As above, the set {a2 − a1 , . . . , ar − a1 } is linearly independent, and the equalities ci − c1 = g(ai − a1 ),

2 6 i 6 r,

show that the set {c2 −c1 , . . . , cr −c1 } also is linearly independent. Finally, Theorem 1.60 implies that {c1 , . . . , cr } is affinely independent. Theorem 1.84. If f : Rn → Rm is an affine transformation and X is a set in Rn , then dim X > dim f (X), with dim X = dim f (X) provided f is one-to-one.


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Proof. Let k = dim f (X). Since the case X = ∅ is obvious, we assume that k > 0. Choose in f (X) an affinely independent set {a1 , . . . , ak+1 } (see Corollary 1.76). If c1 , . . . , ck+1 are points in X satisfying the conditions f (ci ) = ai , 1 6 i 6 k + 1, then Theorem 1.83 shows that the set {c1 , . . . , ck+1 } is affinely independent. By the same corollary, dim X > k = dim f (X). Assume now that f is one-to-one, and let m = dim X. As above, X contains m + 1 affinely independent points c1 , . . . , cm+1 . By Theorem 1.83, the set {f (c1 ), . . . , f (cm+1 )} is affinely independent. Corollary 1.76 implies that dim f (X) > m = dim X, and the above argument gives dim f (X) = dim X. Theorem 1.85. For an affinely independent set {c1 , . . . , cr } ⊂ Rn and points a1 , . . . , ar ∈ Rm , there is an affine transformation f : Rn → Rm such that f (ci ) = ai , 1 6 i 6 r. If, additionally, r = n + 1, then f is uniquely determined. Furthermore, if r = n+1 = m+1 and {a1 , . . . , an+1 } is an affine basis for Rn , then f is invertible. Proof. Since the case r = 1 is obvious (let, for instance, f (x) = a1 + o(x), where o : Rn → Rm is the zero linear transformation), we may assume that r > 2. By Theorem 1.60, the set {c2 − c1 , . . . , cr − c1 } is linearly independent. Choose a linear transformation g : Rn → Rm (see page 5) such that g(ci − c1 ) = ai − a1 ,

2 6 i 6 r.

Put f (x) = a1 − g(c1 ) + g(x). Clearly, f is an affine transformation, and f (ci ) = a1 + g(ci − c1 ) = a1 + (ai − a1 ) = ai ,

1 6 i 6 r.

The transformation g is uniquely determined if r = n+1. Consequently, f is uniquely determined if r = n + 1. Finally, if r = n + 1 = m + 1 and {a1 , . . . , an+1 } is an affine basis for Rn , then the set {a2 −a1 , . . . , an+1 −a1 } is a basis for Rn (see Corollary 1.66). In this case, g is invertible, and so if f. Affine Transformations and Affine Combinations Theorem 1.86. For a mapping f : Rn → Rm , the following conditions are equivalent. (1) f is an affine transformation.

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(2) f ((1 − λ)x + λy) = (1 − λ)f (x) + λf (y) whenever x, y ∈ Rn and λ ∈ R. (3) f (λx + µy) = (1 − λ − µ)f (o) + λf (x) + µf (y) whenever x, y ∈ Rn and λ, µ ∈ R. Proof. (1) ⇒ (3). Let f be an affine transformation, expressed as f (x) = a + g(x), where a ∈ Rm and g : Rn → Rm is a linear transformation. Then f (o) = a. For points x, y ∈ Rn and scalars λ, µ, one has f (λx + µy) = a + g(λx + µy) = (1 − λ − µ)a + λ(a + g(x)) + µ(a + g(y)) = (1 − λ − µ)f (o) + λf (x) + µf (y). The statement (3) ⇒ (2) is obvious; so, it remains to show that (2) implies (1). Assuming (2), we state that the mapping g(x) = f (x) − f (o) is a linear transformation. Indeed, for every x ∈ Rn and λ ∈ R, g(λx) = f ((1 − λ)o + λx) − f (o) = (1 − λ)f (o) + λf (x) − f (o) = λ(f (x) − f (o)) = λg(x). Using the above property for λ = 2, we obtain g(x + y) = 2g( 21 x + 21 y) = 2f ((1 − 12 )x + 12 y) − 2f (o) = 2((1 − 21 )f (x) + 12 f (y)) − 2f (o) = (f (x) − f (o)) + (f (y) − f (o)) = g(x) + g(y). Hence g(x) is a linear transformation, and f (x) = f (o) + g(x) is an affine one. Theorem 1.87. A mapping f : Rn → Rm is an affine transformation if and only if it preserves affine combinations of points: f (λ1 x1 + · · · + λk xk ) = λ1 f (x1 ) + · · · + λk f (xk ) for all k > 1, x1 , . . . , xk ∈ Rn , and λ1 + · · · + λk = 1. Proof. If a mapping f : Rn → Rm preserves affine combinations of any two points, then f is an affine transformation (see Theorem 1.86). Conversely,


61

let f : Rn → Rm be an affine transformation. By induction on k, we will show that f preserves affine combinations of any k points from Rn . The case k = 1 is trivial, and the case k = 2 is proved in Theorem 1.86. Assuming that the induction hypothesis holds for all positive integers k 6 r − 1 (r > 2), choose a affine combination λ1 x1 + · · · + λr xr ∈ Rn . At least one of λ1 , . . . , λr is distinct from 1, since otherwise λ1 + · · · + λr = r > 2. Let, for example, λ1 6= 1. Consider the affine combination z=

λ2 λr x2 + · · · + xr . 1 − λ1 1 − λ1

By the induction hypothesis, f (z) =

λ2 λr f (x2 ) + · · · + f (xr ). 1 − λ1 1 − λ1

Hence f (λ1 x1 + · · · + λr xr ) = f (λ1 x1 + (1 − λ1 )z) = λ1 f (x1 ) + (1 − λ1 )f (z) = λ1 f (x1 ) + · · · + λr f (xr ). Theorem 1.88. For an affine transformation f : Rn → Rm and sets X ⊂ Rn and Y ⊂ Rm , one has aff f (X) = f (aff X), aff f

−1

(Y ) = f

−1

(aff (Y ∩ rng f )).

(1.39) (1.40)

Proof. Excluding the trivial cases X = ∅ and Y ∩ rng f = ∅, we assume that both sets X and Y ∩ rng f are nonempty. Consequently, f −1 (Y ) 6= ∅. For the equality (1.39), choose a point x ∈ aff f (X). By Theorem 1.51, x can be written as an affine combination x = λ1 x1 + · · · + λp xp ,

where

x1 , . . . , xp ∈ f (X).

Let z1 , . . . , zp be points in X such that f (zi ) = xi for all 1 6 i 6 p. Put z = λ1 z1 + · · · + λp zp . Then z ∈ aff X by the same theorem, and x = λ1 x1 + · · · + λp xp = λ1 f (z1 ) + · · · + λp f (zp ) = f (z) according to Theorem 1.87. Hence x = f (z) ∈ f (aff X), which proves the inclusion aff f (X) ⊂ f (aff X). Conversely, let x ∈ aff X. Similarly to the above, x can be expressed as an affine combination x = µ1 x1 + · · · + µq xq ,

where

x1 , . . . , xq ∈ X.

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Then f (x) = µ1 f (x1 ) + · · · + µq f (xq ) according to Theorem 1.87, and Theorem 1.51 gives f (x) ∈ aff f (X). Hence f (aff X) ⊂ aff f (X). In particular, rng f = f (Rn ) = f (aff Rn ) = aff f (Rn ) = aff (rng f ), which shows that rng f is a plane. It remains to prove (1.40). Letting X = f −1 (Y ) in (1.39), one has f (aff f −1 (Y )) = aff f (f −1 (Y )) = aff (Y ∩ rng f ). Hence aff f −1 (Y ) ⊂ f −1 (f (aff f −1 (Y ))) = f −1 (aff (Y ∩ rng f )). For the opposite inclusion, let x ∈ f −1 (aff (Y ∩ rng f )). Then f (x) belongs to aff (Y ∩ rng f ), and Theorem 1.51 implies that f (x) can be written as an affine combination f (x) = γ1 x1 + · · · + γr xr ,

where

x1 , . . . , xr ∈ Y ∩ rng f.

−1

Choose points z1 , . . . , zr ∈ f (Y ) such that f (zi ) = xi , 1 6 i 6 r, and put z = γ1 z1 + · · · + γr zr . Then z ∈ aff f −1 (Y ) and f (z) = γ1 f (z1 ) + · · · + γr f (zr ) = γ1 x1 + · · · + γr xr = f (x). Let ui = zi + (x − z), 1 6 i 6 r. Since ui is an affine combination, one has f (ui ) = f (zi ) + f (x) − f (z) = f (zi ) = xi . Hence ui ∈ f −1 (xi ) ⊂ f −1 (Y ), 1 6 i 6 r. From the equalities x = ui − zi + z,

1 6 i 6 r,

it follows that x can be written as an affine combination of u1 , . . . , ur : x = (γ1 + · · · + γr )x = γ1 (u1 − z1 + z) + · · · + γr (ur − zr + z) = (γ1 u1 + · · · + γr ur ) − (γ1 z1 + · · · + γr zr ) + (γ1 + · · · + γr )z = γ1 u1 + · · · + γr ur . Therefore, x ∈ aff f −1 (Y ). Summing up, f −1 (aff (Y ∩ rng f )) ⊂ aff f −1 (Y ). Remark. The equality (1.40) implies the inclusion aff f −1 (Y ) ⊂ f −1 (aff Y ), which may be proper. Indeed, let f : R2 → R2 be the orthogonal projection on the x-axis of R2 and Y = {(0, 1), (0, −1)}. Then aff f −1 (Y ) = f −1 (Y ) = ∅, while aff Y is the y-axis and f −1 (aff Y ) = aff Y .


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Corollary 1.89. If f : Rn → Rm is an affine transformation and L ⊂ Rn and M ⊂ Rm are planes, then both sets f (L) and f −1 (M ) also are planes. In particular, the set rng f is a plane in Rm . Proof. By Theorem 1.88, f (L) = f (aff L) = aff f (L), which shows that f (L) is a plane. The set N = M ∩ rng f is a plane as the intersection of the planes M and rng f (see Theorem 1.4), and the same Theorem 1.88 gives f −1 (M ) = f −1 (N ) = f −1 (aff N ) = aff f −1 (N ) = aff f −1 (M ). Hence f −1 (M ) is a plane. Theorem 1.90. For an affine transformation f : Rn → Rm and nonempty planes L1 and L2 in Rn , the following statements hold. (1) If L1 and L2 are parallel, then the planes f (L1 ) and f (L2 ) are parallel. (2) If L1 and L2 are independent and f is one-to-one, then the planes f (L1 ) and f (L2 ) are independent. (3) If L1 and L2 are complementary, n = m, and f is invertible, then the planes f (L1 ) and f (L2 ) are complementary. Proof. Let f (x) = a + g(x), where a ∈ Rm and g : Rn → Rm is a linear transformation. (1) Let Li = ci + Si , where ci ∈ Rn and Si is a subspace of Rn , i = 1, 2. Then f (Li ) = a + g(ci ) + g(Si ), i = 1, 2. Without loss of generality, we may suppose that dim L1 6 dim L2 . From Theorem 1.6 it follows that namely L2 contains a translate of L1 , and Theorem 1.13 yields the inclusion S1 ⊂ S2 . Because g(S1 ) ⊂ g(S2 ), the same lemma implies that the planes f (L1 ) and f (L2 ) are parallel. (2) Suppose that L1 and L2 are independent, and f is one-to-one. From Theorem 1.86 it follows that f (L1 + L2 ) = {f (x1 + x2 ) : x1 ∈ L1 , x2 ∈ L2 } = {f (x1 ) + f (x2 ) − a : x1 ∈ L1 , x2 ∈ L2 } = f (L1 ) + f (L2 ) − a, and Theorem 1.6 gives dim (f (L1 ) + f (L2 )) = dim f (L1 + L2 ) = dim(L1 + L2 ). Since f is one-to-one, a combination of Theorems 1.10 and 1.84 gives dim (f (L1 ) + f (L2 )) = dim f (L1 + L2 ) = dim(L1 + L2 ) = dim L1 + dim L2 = dim f (L1 ) + dim f (L2 ).

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Theorem 1.10 shows that the planes f (L1 ) and f (L2 ) are independent. (3) Let L1 and L2 be complementary, n = m, and f be invertible. By statement (2) above, dim (f (L1 ) + f (L2 )) = dim f (L1 ) + dim f (L2 ) = dim L1 + dim L2 = n = m. Consequently, the planes f (L1 ) and f (L2 ) are complementary in Rm , as follows from Theorem 1.10. Homotheties and Projections Homotheties and projections make important classes of affine transformations used in the study of convex sets. Definition 1.91. Given a point a ∈ Rn and a scalar µ 6= 0, the affine transformation f (x) = a + µx,

x ∈ Rn ,

is called a homothety with ratio µ. Furthermore, f is a positive homothety if µ > 0 (respectively, a negative homothety if µ < 0). Among all homotheties, those with µ = 1 are the simplest ones. An affine transformation f (x) = a + x is called the translate on a vector a.

@

@ @

@ -@ a+x

@ x Fig. 1.13

The translate on a vector a.

A homothety f (x) = a + µx, µ 6= 1, can be written as f (x) = c + µ(x − c),

where

c = a/(1 − µ)

is the center of f . Furthermore, a homothety with center is called a contraction if 0 < |µ| < 1 (respectively, an expansion if |µ| > 1). Theorem 1.92. The following statements hold. (1) Every homothety f (x) = a + µx is invertible, and its inverse is a homothety: f −1 (x) = −µ−1 a + µ−1 x.


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(2) The composition of homotheties f (x) = a + µx and h(x) = c + γx again is a homothety: (f ◦ h)(x) = (a + µc) + µγ x. Proof. (1) With g(x) = −µ−1 a + µ−1 x, we have (f ◦ g)(x) = f (−µ−1 a + µ−1 x) = a + µ(−µ−1 a + µ−1 x) = x, (g ◦ f )(x) = g(a + µx) = −µ1 a + µ−1 (a + µx) = x. Hence f is invertible, and f −1 (x) = −µ−1 a + µ−1 x. (2) (f ◦ h)(x) = f (c + γx) = a + µ(c + γx) = (a + µc) + µγx, implying that (f ◦ h) is a homothety. @ @ @ @ `X @ X XXX c XXX XXX x z X c + µ(x − c) Fig. 1.14

A positive homothety with center c.

A characteristic property of homotheties is given in Exercise 1.8. Definition 1.93. Let L and N be complementary planes in Rn , expressed as translates of complementary subspaces S and T , respectively. Let g denote the linear projection on S along T (see page 5). Given a point c ∈ L, the affine transformation f (x) = c + g(x − c) is called the (affine) projection on L along N . If, additionally, T is the orthogonal complement of S, then f is called the orthogonal projection on L.

r x

N

L Fig. 1.15

r

? f (x)

The projection on a plane L along a complementary plane N .

Remark. The projection f in Definition 1.93 does not depend on the choice of c in L. Indeed, let f1 (x) = c1 + g(x − c1 ) for another point c1 ∈ L.

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Then c1 − c ∈ S (see Theorem 1.2). Therefore, g(c1 − c) = c1 − c because g is a linear projection on S. Consequently, c1 − g(c1 ) = c − g(c) due to the linearity of g. Hence f1 (x) = c1 + g(x − c1 ) = (c1 − g(c1 )) + g(x) = (c − g(c)) + g(x) = c + g(x − c) = f (x). Theorem 1.94. Let f be the projection on a plane L ⊂ Rn along a complementary plane N ⊂ Rn , and let c1 , . . . , cn+1 be an affine basis for Rn such that L = aff {c1 , . . . , cm+1 } and N = aff {cm+1 , . . . , cn+1 }, where m = dim L. If a point x ∈ Rn is written as an affine combination x = λ1 c1 + · · · + λn+1 cn+1 , then f (x) = λ1 c1 + · · · + λm cm + (λm+1 + · · · + λn+1 ) cm+1 . Proof. Since both cases m = 0 and m = n are obvious, we may assume that 1 6 m 6 n − 1. Put bi = ci − cm+1 , 1 6 i 6 m,

bi = ci+1 − cm+1 , m + 1 6 i 6 n.

The subspaces S = L − cm+1 and T = N − cm+1 are complementary, and b1 , . . . , bm and bm+1 , . . . , bn are, respectively, their bases (see Corollary 1.66). According to Definition 1.93 and the above remark, one can write f (x) = cm+1 + g(x − cm+1 ), where g is the linear projection on S along T . From the equality x − cm+1 = λ1 b1 + · · · + λm bm + λm+2 bm+1 + · · · + λn+1 bn , it follows that g(x − cm+1 ) = λ1 b1 + · · · + λm bm . Therefore, f (x) = g(x − cm+1 ) + cm+1 = λ1 b1 + · · · + λm bm + cm+1 = λ1 c1 + · · · + λm cm + (1 − λ1 − · · · − λm ) cm+1 = λ1 c1 + · · · + λm cm + (λm+1 + · · · + λn+1 ) cm+1 . Theorem 1.95. An affine transformation f : Rn → Rn is a projection if and only if f 2 = f . Proof. Let f be a projection; that is f (x) = c + g(x − c), where g : Rn → Rn is a linear projection and c ∈ Rn . Since g 2 = g (see Exercise 0.3), one has f 2 (x) = f (f (x)) = f (c + g(x − c)) = c + g(c + g(x − c) − c) = c + g(g(x − c)) = c + g(x − c) = f (x).


67

Conversely, let f : Rn → Rn be an affine transformation with the property f 2 = f . Put e = f (o) and g(x) = f (x) − e. Then g(x) is a linear transformation such that g(e) = f (e) − e = f (f (o)) − f (o) = o. Furthermore, g 2 (x) = g(g(x)) = g(f (x) − e) = g(f (x)) = f (f (x)) − e = f (x) − e = g(x). Hence g is a linear projection (see Exercise 0.3), and f (x) = e + g(x) = e + g(x − e) is a projection. Theorem 1.96. Let L ⊂ Rn be a nonempty plane and f be the orthogonal projection of Rn on L. For any points u ∈ Rn and v ∈ L, one has ku − vk2 = ku − f (u)k2 + kf (u) − vk2 . Consequently, f (u) is the nearest to u point in L, and v = f (u) if and only if u − v ∈ (sub L)⊥ . Proof. Let S be the characteristic subspace of L (see Theorem 1.2). Then f is the projection on L along the subspace T = S ⊥ (see remark on page 65). By Theorem 1.11, the set L ∩ T is a singleton, say {c}. Theorem 1.2 shows that L = c + S. If p is the orthogonal projection on S, then p(c) = o, and f (x) = c + p(x − c) = c + p(x) for all x ∈ Rn . Let v = c + w, where w ∈ S. From linear algebra we know that k(u − c) − wk2 = k(u − c) − p(u − c)k2 + kp(u − c) − wk2 . Combining this equality with p(u − c) = p(u), we obtain ku − vk2 = k(u − c) − wk2 = k(u − c) − p(u)k2 + kp(u) − wk2 = ku − f (u)k2 + kf (u) − vk2 . Finally, since both vectors u − p(u) and c are orthogonal to S, their difference, u − f (u), also is orthogonal to S = L − L. Theorem 1.97. The orthogonal projection f of Rn on a hyperplane H = {x ∈ Rn : x·c = γ} ⊂ Rn is given by γ − u·c f (u) = u + c, u ∈ Rn . kck2 Furthermore, f (u) is the nearest to u point in H. Consequently, δ(u, H) = ku − f (u)k = |γ − u·c|/kck.

68


Proof. Let S = {x ∈ Rn : x · c = 0} be the characteristic subspace of H (see Corollary 1.18). From linear algebra we know that the orthogonal projection on S is given by u·c c p(u) = u − kck2 (see page 6). Since the point a = (γ/kck2 )c belongs to H, Definition 1.93 shows that the orthogonal projection on H is expressed as γ γ f (u) = a + p(u − a) = c + p(u − c) kck2 kck2 γ γ γ 2 c + (u − c) − (u − c)·c/kck c = kck2 kck2 kck2 γ − u·c =u+ c. kck2 From Theorem 1.96 it follows that f (u) is the nearest to u point in H. Therefore, γ − u·c |γ − u·c| δ(u, H) = ku − f (u)k = . kck = kck2 kck Theorem 1.98. If L1 and L2 are nonempty disjoint planes in Rn , then δ(L1 , L2 ) > 0. Proof. Put Li = ci + Si , where Si is the characteristic subspace of Li , i = 1, 2. Let L02 = L2 − c1 and L = S1 + L02 . Then S1 ∩ L02 = ∅, since otherwise L1 ∩ L2 = (c1 + S1 ) ∩ (c1 + L02 ) = c1 + S1 ∩ L02 6= ∅, contrary to the assumption L1 ∩ L2 = ∅. We state that S1 ∩ L = ∅. Indeed, assume for a moment the existence of a point x ∈ S1 ∩ L. Then x = y + z, where y ∈ S1 and z ∈ L02 . Consequently, z = x − y ∈ S1 , which is impossible because of S1 ∩ L02 = ∅. Denote by a the orthogonal projection of o on L. Then a is the nearest to o point in L (see Theorem 1.96). We state that δ(S1 , L) = kak. Indeed, for any points u ∈ S1 and v ∈ L, one has ku − vk = ko − (v − u)k > kak due to v − u ∈ L − S1 = L. Finally, the inclusion L02 ⊂ L implies that δ(L1 , L2 ) = δ(L1 − c1 , L2 − c1 ) = δ(S1 , L02 ) > δ(S1 , L) = kak > 0. Remark. Theorem 1.98 can be sharpened by showing the existence of points u1 ∈ L1 and u2 ∈ L2 with the property ku1 − u2 k = δ(L1 , L2 ).


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Exercises for Chapter 1 Exercise 1.1. Show that a proper plane L ⊂ Rn is not a hyperplane if and only if its complement Rn \ L is connected. Exercise 1.2. Let F = {Lα } be a countable family of planes in Rn . Show that the set L = ∪ Lα is a plane if and only if one of the planes from F α contains all the others. Exercise 1.3. Show that a nonempty set X ⊂ Rn of dimension m is a plane if and only if there is a scalar ρ > 0 satisfying the condition: for every point c ∈ X, the set Bρ (c) ∩ X is an m-dimensional ball (see page 56). If, additionally, X is closed, then the value of ρ may depend on c ∈ X. Exercise 1.4. Let H ⊂ Rn be a hyperplane, a and b be points in Rn \ H, and h be a halfline with endpoint a, which meets H. Show that the halfline h0 = (b − a) + h meets H if and only if a and b belong to the same open halfspace determined by H. Exercise 1.5. Let F = {Fα } be a family of halfplanes of a nonempty plane L ⊂ Rn of positive dimension (each Fα may be closed or open). Show that the following statements hold. (1) If the union U = ∪ Fα is a proper subset of L, then U is a halfplane α of L if and only if F is nested. (2) If the intersection Z = ∩ Fα is nonempty, then Z is a halfplane of α L if and only if F is nested. Exercise 1.6. Let X and Y be nonempty sets in Rn such that a translate of Y lies in aff X. Show that for any point y ∈ Y , aff (X + Y ) = aff X + Y = aff X + y. Exercise 1.7. Let X1 , . . . , Xr ⊂ Rn be nonempty sets in Rn , µ1 , . . . , µr scalars, and ai ∈ Xi , 1 6 i 6 r. Let also r > m = dim (µ1 X1 + · · · + µr Xr ). Show the existence of an index set I ⊂ {1, . . . , r}, card I 6 m, satisfying the condition P P aff (µ1 X1 + · · · + µr Xr ) = µi ai + µi aff Xi . i∈I /

i∈I

Exercise 1.8. Show that a mapping f : Rn → Rn , n > 2, is a homothety if and only if it takes every line in Rn onto a parallel line.

70


Exercise 1.9. Show that for points x, y, z ∈ Rn and a scalar λ ∈ [0, 1], k(1 − λ)x + λy − zk 6 (1 − λ)kx − zk + λky − zk. Furthermore, if x 6= y and 0 < λ < 1, then the equality k(1 − λ)x + λy − zk = (1 − λ)kx − zk + λky − zk holds if and only if z ∈ hx, yi such that either x ∈ [z, y] or y ∈ [x, z]. Notes for Chapter 1 The affine structure of Rn . A substantial part of Chapter 1 is scattered through various books on linear algebra and affine geometry and can be viewed as a body of “common knowledge.” Affine spans. Given a nonempty set X ⊂ Rn , let aff r X denote the collection of all affine combinations of r points from X: aff r X = {λ1 x1 + · · · + λi xi : x1 , . . . , xi ∈ X, λ1 + · · · + λr = 1}. In particular, aff 2 X is the union of points and lines through distinct points of X. From Corollary 1.76 it follows that aff m+1 X = aff X for every m-dimensional set X in Rn . It is easy to see that aff r (aff s X) ⊂ aff rs X, and, unlike convex hulls (see Theorem 3.10), this inclusion may be proper. For example, if X consists of four affinely independent points x, y, z, u in R3 , then aff 4 X \ aff 2 (aff 2 X) is the four-point set 1 {−x 2

+ y + z + u, x − y + z + u, x + y − z + u, x + y + z − u}.

Given a set X ⊂ Rn and integers r1 , . . . , rk > 2, Klee [129] studies properties of sets aff r1 (. . . (aff rk X) . . . ). Collineations. One of the fundamental theorems of affine geometry states that an invertible mapping f : Rn → Rn is an affine transformation if and only if it takes every three collinear points x, y, z ∈ Rn onto collinear points, f (x), f (y), f (z) (see, e. g., Veblen and Whitehead [220]). Lenz [149] showed that the statement still holds if “invertible” is replaced with “one-to-one”, while Chubarev and Pinelis [57] succeeded to replace the condition “invertible” with that of “onto.” Frenkel [88, p. 91] (see also Aboubabdullah [2, Article 5]) proved that a mapping f : Rn → Rm , where n > 2 and m > 2, is an affine transformation provided its range is at least 2-dimensional and f (aff {x, z}) = aff {f (x), f (z)} for all x, z ∈ Rn .

Chapter 2

Convex Sets

2.1

Algebraic Properties of Convex Sets

Definition and Basic Properties Definition 2.1. A nonempty set X ⊂ Rn is called convex provided it contains all points (1 − λ)x + λy whenever x, y ∈ X and λ ∈ [0, 1]. The empty set ∅ is assumed to be convex. r Q Q Q

r r Fig. 2.1

Q

Qr

Convex and nonconvex sets.

We can reformulate Definition 2.1, saying that a nonempty set X in Rn is convex if and only if it contains all segments [x, y] with endpoints x, y ∈ X. The condition λ ∈ [0, 1] in Definition 2.1 can be weakened provided the set X is closed or open (see Exercises 2.2 and 2.3). Example. Every singleton in Rn is a convex set. Convex subsets of a line l ⊂ Rn are the empty set, all singletons, segments and halflines of all kinds (see Definitions 1.24 and 1.25), and the whole line l. Example. Every plane L in Rn is a convex set. Indeed, the case L = ∅ is trivial. If L 6= ∅ and x and y are points in L, then [x, y] ⊂ L according to Theorem 1.46. 71

72


Example. Every halfplane of a plane L ⊂ Rn of positive dimension is a convex set. Indeed, if D is a closed halfplane of L, then, by Corollary 1.38, it can be expressed as D = {x ∈ L : x·c 6 γ},

c∈ / (sub L)⊥ ,

γ ∈ R.

Choose points x, y ∈ D and a scalar λ ∈ [0, 1]. Then (1 − λ)x + λy ∈ L because L is convex, and the inequality ((1 − λ)x+ λ y)·c = (1 − λ)x·c + λy·c 6 (1 − λ)γ + λγ = γ shows that (1 − λ)x + λy ∈ D. Similarly, every open halfplane of L is a convex set. Example. Every slab of a plane L ⊂ Rn of positive dimension is a convex set. Let, for instance, F be a closed slab of L. According to Corollary 1.44, F can be described as F = {x ∈ L : γ 6 x·c 6 γ 0 },

c∈ / (sub L)⊥ ,

γ < γ0.

If x, y ∈ F and λ ∈ [0, 1], then (1 − λ)x + λy ∈ L because L is convex, and the inequalities γ 6 ((1 − λ)x+ λ y)·c 6 γ 0 show that (1 − λ)x + λy ∈ F . In a similar way, every open slab of L is a convex set. Example. Every closed ball Bρ (c) ⊂ Rn is a convex set. Indeed, for points x, y ∈ Bρ (c) and a scalar λ ∈ [0, 1], one has k((1 − λ)x + λy) − ck = k(1 − λ)(x − c) + λ(y − c)k 6 (1 − λ)kx − ck + λky − ck 6 (1 − λ)ρ + λρ = ρ. Hence (1 − λ)x + λy ∈ Bρ (c), implying the convexity of Bρ (c). Definition 2.2. A convex combination of points x1 , . . . , xr in Rn is a linear combination λ1 x1 + · · · + λr xr , where λ1 , . . . , λr > 0 and λ1 + · · · + λr = 1. If, additionally, all scalars λ1 , . . . , λr are positive, then λ1 x1 + · · · + λr xr is called a positive convex combination. For example, all convex combinations of distinct points x and y in Rn fulfil the segment [x, y] (see Definition 1.25), while all positive convex combinations of x and y fulfill the open segment (x, y). Theorem 2.3. A nonempty set X ⊂ Rn is convex if and only if it contains all convex combinations (equivalently, all positive convex combinations) of points from X.

Convex sets

73

Proof. If X contains all positive convex combinations of its points, then, in particular, (1−λ)x+λy ∈ X whenever x, y ∈ X and 0 < λ < 1. Because [x, y] = {x} ∪ (x, y) ∪ {y}, the segment [x, y] lies in X for any choice of x, y ∈ X. Hence X is convex. Conversely, let X be convex. By induction on k, we are going to prove that X contains all convex combinations of k points from X. The case k = 1 is trivial (every point x ∈ X can be written as 1x), and the case k = 2 holds by the definition. Assuming that the statement is true for all positive integers k 6 r−1 (r > 2), choose a convex combination x = λ1 x1 + · · · + λr xr of points x1 , . . . , xr ∈ X. If λ1 = 1, then λ2 = · · · = λr = 0, and x = 1x1 ∈ X. Suppose that λ1 < 1. By the induction hypothesis, the convex combination λr λ2 x2 + · · · + xr y= 1 − λ1 1 − λ1 belongs to X. Since x can be written as x = λ1 x1 +(1−λ1 )y, the inclusion x ∈ X follows from the convexity of X. See Theorem 2.42 for an extension of Theorem 2.3 to the case of converging convex series. Theorem 2.4. A set X ⊂ Rn is convex if and only if µ1 X + · · · + µr X = (µ1 + · · · + µr )X for every choice of r > 2 and scalars µ1 , . . . , µr > 0.

(2.1)

Proof. Let X be convex. The inclusion (µ1 + · · · + µr )X ⊂ µ1 X + · · · + µr X holds for every set X ⊂ Rn and every choice of scalars µ1 , . . . , µr . Hence it remains to prove that µ1 X + · · · + µr X ⊂ (µ1 + · · · + µr )X provided µ1 , . . . , µr > 0. Since the statement is trivial when X = ∅, we may assume that X is nonempty. Excluding one more trivial case µ1 = · · · = µr = 0, we suppose that µ = µ1 + · · · + µr > 0. Put λi = µi /µ, 1 6 i 6 r. Then λ1 , . . . , λr > 0 and λ1 + · · · + λr = 1. According to Theorem 2.3, λ1 X + · · · + λr X = {λ1 x1 + · · · + λr xr : x1 , . . . , xr ∈ X} ⊂ X. Multiplying both sides of the latter inclusion by µ, we obtain µ1 X + · · · + µr X ⊂ (µ1 + · · · + µr )X. Conversely, let a nonempty set X ⊂ Rn satisfy the condition (2.1) for every choice of r > 2 and scalars µ1 , . . . , µr > 0. If x, y ∈ X and λ ∈ [0, 1], then (1 − λ)x + λy ∈ (1 − λ)X + λX = X, implying the convexity of X.

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Simplices Definition 2.5. Let {x1 , . . . , xr+1 }, r > 0, be an affinely independent set in Rn . The r-simplex ∆ = ∆(x1 , . . . , xr+1 ) with vertices x1 , . . . , xr+1 is defined as ∆ = {λ1 x1 + · · · + λr+1 xr+1 : λ1 , . . . , λr+1 > 0, λ1 + · · · + λr+1 = 1}. Equivalently, ∆ is the set of all convex combinations of x1 , . . . , xr+1 .

r x1

r x1

r x2

r x3 A A A A Ar r x1 x2

Fig. 2.2

r x3

CJ

CJ

C J

C J rX

Jr x1 XXXC r x2 x4

Simplices in R3 .

Theorem 2.6. Every r-simplex ∆(x1 , . . . , xr+1 ) ⊂ Rn is a convex set of dimension r, and aff ∆ = aff {x1 , . . . , xr+1 }. Proof. If x and y are points in ∆ = ∆(x1 , . . . , xr+1 ), then x = γ1 x1 + · · · + γr+1 xr+1 ,

γ1 + · · · + γr+1 = 1,

y = µ1 x1 + · · · + µr+1 xr+1 ,

µ1 + · · · + µr+1 = 1,

for certain nonnegative scalars γ1 , . . . , γr+1 and µ1 , . . . , µr+1 . Choose a scalar λ ∈ [0, 1], and let αi = (1 − λ)γi + λµi , 1 6 i 6 r + 1. Clearly, α1 , . . . , αr+1 > 0, and the equalities (1 − λ)x + λy = α1 x1 + · · · + αr+1 xr+1 ,

α1 + · · · + αr+1 = 1

show that (1 − λ)x + λy ∈ ∆. Hence ∆ is a convex set. Since convex combinations are particular cases of affine combinations, Corollary 1.52 implies the inclusions {x1 , . . . , xr+1 } ⊂ ∆ ⊂ aff {x1 , . . . , xr+1 }, and Theorem 1.50 gives aff ∆ = aff {x1 , . . . , xr+1 }. Furthermore, from Definition 1.75 and Theorem 1.60 one has dim ∆ = dim (aff ∆) = dim (aff {x1 , . . . , xr+1 }) = r.

Convex sets

75

Combining Theorems 2.3 and 2.6 and Corollary 1.76, we obtain one more corollary. Corollary 2.7. For a nonempty convex set K ⊂ Rn , the following statements hold. (1) If {x1 , . . . , xr+1 } is an affinely independent subset of K, then the r-simplex ∆(x1 , . . . , xr+1 ) lies in K. (2) The dimension of K equals the maximum dimension of a simplex contained in K. (3) aff K = aff ∆ for every simplex ∆ ⊂ K whose dimension equals dim K. Algebra of Convex Sets Theorem 2.8. If F = {Kα } is a family of convex sets in Rn , then the following statements hold. (1) The intersection M = ∩ Kα is a convex set. α

(2) If the family F is nested, then the union N = ∪ Kα is a convex α set. Proof. (1) Since the statement is obvious when M is empty, we may suppose that M 6= ∅. If x, y ∈ M , then x, y ∈ Kα for every Kα ∈ F, which gives [x, y] ⊂ Kα by the convexity of Kα . Hence [x, y] ⊂ ∩ Kα = M , and α M is convex. (2) Excluding the trivial case N = ∅, we assume that N is nonempty. Let x and y be points in N . Then x ∈ Kγ and y ∈ Kµ for certain sets Kγ , Kµ ∈ F. By the hypothesis, one of the sets Kγ , Kµ contains the other. Let, for example, Kγ ⊂ Kµ . In this case, x, y ∈ Kµ , and [x, y] ⊂ Kµ by the convexity of Kµ . Thus [x, y] ⊂ Kµ ⊂ ∪ Kα = N , and N is a convex α set. Theorem 2.9. If K1 , . . . , Kr are convex sets in Rn and µ1 , . . . , µr are scalars, then the sum µ1 K1 + · · · + µr Kr is a convex set. Proof. Without loss of generality, we may suppose that all sets K1 , . . . , Kr are nonempty. An induction argument shows that the proof can be reduced to the case r = 2 (if r = 1, then we can write µ1 K1 = µ1 K1 + µ2 o). Choose points x and y in µ1 K1 + µ2 K2 and a scalar λ ∈ [0, 1]. Then x = µ1 x1 + µ2 x2 and y = µ1 y1 + µ2 y2 for certain points x1 , y1 ∈ K1 and

76


x2 , y2 ∈ K2 . By a convexity argument, (1 − λ)xi + λyi ∈ Ki , i = 1, 2. Finally, the inclusion (1 − λ)x + λy = µ1 ((1 − λ)x1 + λy1 ) + µ2 ((1 − λ)x2 + λy2 ) ∈ µ1 K1 + µ2 K2 shows that µ1 K1 + µ2 K2 is a convex set. Theorem 2.10. For a convex set K ⊂ Rn and a scalar µ ∈ (0, 1), there is a set X ⊂ Rn satisfying the condition µK = ∩ (z + K : z ∈ X). Proof. Since the case K = ∅ is obvious (let X = ∅), we assume that K is nonempty. It suffices to show that for every u ∈ Rn \ µK there is a point z ∈ Rn such that µK ⊂ z + K and u ∈ / z + K. The union of all such points z will form the desired set X. We consider apart the following cases. 1. Suppose that u+µ(x−u) ∈ µK for all x ∈ µK. Put z = (1−µ−1 )u. Then the implications x ∈ µK ⇒ u + µ(x − u) ∈ µK ⇒ x ∈ (1 − µ−1 )u + K = z + K, u∈ / µK ⇒ µu ∈ / (µ − 1)u + µK ⇒ u ∈ / (1 − µ−1 )u + K = z + K prove the statement. 2. Assume the existence of a point v ∈ µK such that u+µ(v−u) ∈ / µK. Since u + µ(v − u) ∈ [u, v], the convexity of µK shows that u + η(v − u) ∈ / µK for all η ∈ (0, µ). Hence there is a positive integer r such that u + (1 − µ)r (v − u) ∈ / µK. Assuming that r is the smallest integer with respect to the above property, one has w = u + (1 − µ)r−1 (v − u) ∈ µK. Let z = (1 − µ−1 )w. If x ∈ µK, then (1 − µ)w + µx ∈ µK, implying x ∈ µ−1 ((µ − 1)w + µK) = (1 − µ−1 )w + K = z + K. On the other hand, from µu + (1 − µ)w = u + (1 − µ)(w − u) = u + (1 − µ)r (v − u) ∈ / µK it follows that u∈ / µ−1 ((µ − 1)w + µK) = (1 − µ−1 )w + K = z + K. Exercises 2.5 and 2.6 give additional results on algebraic properties of convex sets. The next theorem shows that affine spans of convex sets can be describes in terms of affine combinations of two points (compare with Theorem 1.51 and Corollary 1.76). Theorem 2.11. The affine span of a nonempty convex set K ⊂ Rn can be expressed as aff K = {(1 − λ)x + λy : x, y ∈ K, λ ∈ R}.

Convex sets

77

Proof. Because aff K is a plane containing K, Theorem 1.46 shows that the set L = {(1 − λ)x + λy : x, y ∈ K, λ ∈ R} lies in aff K. For the opposite inclusion, choose a point x ∈ aff K. By Theorem 1.51, x is an affine combination, x = λ1 x1 + · · · + λr xr , of certain points x1 , . . . , xr ∈ K. Omitting zero terms of the form 0xi , we may assume that all scalars λ1 , . . . , λr are distinct from zero. Since λ1 + · · · + λr = 1, at least one of the scalars λ1 , . . . , λr is positive. We suppose that all positive scalars among λ1 , . . . , λr are given by λ1 , . . . , λp , where 1 6 p 6 r. If p = r, then x is a convex combination of x1 , . . . , xr , and x ∈ K ⊂ aff K according to Theorem 2.3. Assume that p 6 r − 1 and let µ = λ1 + · · · + λp . Clearly, µ > 1. Put ( λi /µ if 1 6 i 6 p, µi = λi /(1 − µ) if p + 1 6 i 6 r. Obviously, u = µ1 x1 + · · · + µp xp

and v = µp+1 xp+1 + · · · + µr xr

are convex combinations of points from K. Therefore, u, v ∈ K according to Theorem 2.3. Since x can be written as x = (1 − µ)v + µu, it follows that x ∈ L. Summing up, aff K ⊂ L. A sharper version of Theorem 2.11 is given in Theorem 2.30. Theorem 2.12. If f : Rn → Rm is an affine transformation and K ⊂ Rn and M ⊂ Rm are convex sets, then both sets f (K) and f −1 (M ) are convex. Proof. Excluding the trivial cases K = ∅ and M ∩ rng f = ∅, we assume that both sets K and M ∩rng f are nonempty. Consequently, f −1 (M ) 6= ∅. Choose points x, y ∈ f (K) and a scalar λ ∈ [0, 1]. Let x0 and y0 be points in K satisfying the conditions f (x0 ) = x and f (y0 ) = y. Then (1 − λ)x0 + λy0 ∈ K because K is convex. By Theorem 1.86, (1 − λ)x + λy = (1 − λ)f (x0 ) + λf (y0 ) = f ((1 − λ)x0 + λy0 ) ∈ f (K), which shows the convexity of f (K). Similarly, if x, y ∈ f −1 (M ) and λ ∈ [0, 1], then f (x), f (y) ∈ M , and (1 − λ)f (x) + λf (y) ∈ M because of the convexity of M . Theorem 1.86 results in f ((1 − λ)x + λy) = (1 − λ)f (x) + λf (y) ∈ M ∩ rng f.

78


Hence (1 − λ)x + λy ∈ f −1 (M ∩ rng f ) = f −1 (M ), and f −1 (M ) is convex. See Exercise 2.12 and references on page 114 for a family of mappings which preserve convexity. 2.2

Relative Interior of Convex Sets

Definition and Basic Properties Definition 2.13. A point c ∈ Rn is called relatively interior for a set X ⊂ Rn provided there is a scalar ρ > 0 such that Bρ (c) ∩ aff X ⊂ X. The set of relatively interior points of X is called the relative interior of X and denoted rint X (we let rint ∅ = ∅). A set X ⊂ Rn is said to be relatively open if rint X = X. Example. The relative interior of a plane L ⊂ Rn is the plane itself; in particular, rint {c} = {c} for every point c ∈ Rn . Example. If L ⊂ Rn is a plane of positive dimension and D is a closed halfplane of L given by D = {x ∈ L : x·c 6 γ},

c∈ / (sub L)⊥ ,

γ∈R

(see Corollary 1.38), then rint D coincides with the open halfplane E = {x ∈ L : x·c < γ},

c∈ / (sub L)⊥ ,

γ ∈ R.

Indeed, if D is given by D = V ∩ L, where V = {x ∈ Rn : x·c 6 γ} is the closed halfspace, Corollary 2.27 below shows that rint D = int V ∩ L = E. Similarly, the relative interior of a closed slab of L, expressed by F = {x ∈ L : γ 6 x·c 6 γ 0 },

c∈ / (sub L)⊥ ,

γ < γ0,

is the respective open slab (see Corollary 1.44). Remark. The difference between the relative interior and topological interior in Rn can be illustrated by the following example. The relative interior of a segment [x, y] in R2 is the open segment (x, y), while the topological interior of [x, y] is empty (indeed, no ball Bρ (z) ⊂ R2 centered at a point z ∈ [x, y] can lie in [x, y]).

Convex sets

79

'$ Bρ (z) r

x

r r z y &%

Theorem 2.14. Let X ⊂ Rn be a set with nonempty relative interior. For any point c ∈ rint X, there is a scalar δ > 0 such that Bδ (c) ∩ aff X ⊂ rint X. Furthermore, if 0 < ε 6 δ, then aff (Bε (c) ∩ rint X) = aff (rint X) = aff X.

(2.2)

Proof. Choose a scalar ρ > 0 satisfying the condition Bρ (c) ∩ aff X ⊂ X and put δ = ρ/2. Let x be a point in Bδ (c) ∩ aff X. We first observe that Bδ (x) ⊂ Bρ (c). Indeed, if y ∈ Bδ (x), then ky − ck 6 ky − xk + kx − ck 6 ρ/2 + ρ/2 = ρ, implying the inclusion y ∈ Bρ (c). Consequently, Bδ (x) ∩ aff X ⊂ Bρ (c) ∩ aff X ⊂ X. Hence x ∈ rint X, and Bδ (c) ∩ aff X ⊂ rint X. If 0 < ε 6 δ, then Bε (c) ∩ aff X ⊂ Bδ (c) ∩ aff X ⊂ rint X. Now, a combination of Theorems 1.50 and 1.79 gives aff X = aff (Bε (c) ∩ aff X) ⊂ aff (rint X) ⊂ aff X. Therefore, the equalities (2.2) hold. Theorem 2.15. For sets X and Y in Rn , the following statements hold. (1) (2) (3) (4) (5) (6) (7)

If int X 6= ∅, then dim X = n and int X = rint X. int X ⊂ rint X ⊂ X. If X ⊂ Y and aff X = aff Y , then rint X ⊂ rint Y . rint (rint X) = rint X. If rint X ⊂ Y ⊂ X, then rint X = rint Y . If a translate of Y lies in aff X, then rint X + Y ⊂ rint (X + Y ). rint (X + v) = rint X + v for every point v ∈ Rn .

Proof. (1) Choose a point u ∈ int X and a scalar ρ > 0 such that Bρ (u) ⊂ X. Then aff X = Rn because of Rn = aff Bρ (u) ⊂ aff X ⊂ Rn . Hence dim X = n. In the latter case, int X = rint X. (2) If int X = ∅, then the inclusion int X ⊂ rint X is obvious. If int X 6= ∅, then statement (2) follows from (1).

80


To exclude trivial cases, we suppose below that both sets rint X and rint Y are nonempty. (3) Choose a point x ∈ rint X and a scalar ρ > 0 satisfying the condition Bρ (x) ∩ aff X ⊂ X. Since aff X = aff Y , we have Bρ (x) ∩ aff Y = Bρ (x) ∩ aff X ⊂ X ⊂ Y. Hence x ∈ rint Y . (4) Because of rint X ⊂ X and aff (rint X) = aff X (see Theorem 2.14), statement (3) implies that rint (rint X) ⊂ rint X. For the opposite inclusion, choose a point x ∈ rint X. By the same theorem, there is a scalar δ > 0 such that Bδ (x) ∩ aff X ⊂ rint X. Thus Bδ (x) ∩ aff (rint X) = Bδ (x) ∩ aff X ⊂ rint X. Therefore, x ∈ rint (rint X). (5) We first observe that aff X = aff (rint X) = aff Y . Indeed, a combination of Theorems 1.50 and 2.14 gives aff X = aff (rint X) ⊂ aff Y ⊂ aff X. Now, statements (3) and (4) imply that rint X = rint (rint X) ⊂ rint Y ⊂ rint X. Summing up, rint X = rint Y . (6) Let z ∈ rint X + Y . Then z = x + y, where x ∈ rint X and y ∈ Y . Choose a scalar ρ > 0 satisfying the condition Bρ (x) ∩ aff X ⊂ X. Because aff (X + Y ) = aff X + Y = aff X + y (see Exercise 1.6) and Bρ (z) = Bρ (x) + y, one has Bρ (z) ∩ aff (X + Y ) = (Bρ (x) + y) ∩ (aff X + y) = Bρ (x) ∩ aff X + y ⊂ X + y ⊂ X + Y. Therefore, z ∈ rint (X + Y ), and rint X + Y ⊂ rint (X + Y ). (7) By statement (6) above, it suffices to show that rint (X + v) ⊂ rint X + v. Let u ∈ rint (X + v). Then u ∈ X + v, whence we can write u = x + v for a certain point x ∈ X. Choose a scalar δ > 0 such that Bδ (u) ∩ aff (X + v) ⊂ X + v. Consequently, Bδ (x) ∩ aff X = (Bδ (u) − v) ∩ (aff X + v − v) = Bδ (u) ∩ aff (X + v) − v ⊂ (X + v) − v = X. Hence x ∈ rint X and u = x + v ∈ rint X + v.

Convex sets

81

Remark. The converse of statement (1) in Theorem 2.15 does not hold. Indeed, if X consists of n + 1 affinely independent points, then dim X = n (see Corollary 1.76), while int X = rint X = ∅. Nevertheless, it becomes true provided X is convex (see Corollary 2.18). Furthermore, the condition aff X = aff Y is essential in statement (3). For example, if X is a triangle in R2 and Y one of its sides, then Y ⊂ X, while rint X and rint Y are disjoint. Finally, the inclusion rint X + Y ⊂ rint (X + Y ) in statement (6) may be proper. Indeed, let X = [0, 1] ∪ Q and Y = [0, 1], where Q is the set of all rational numbers in R. Then rint X = (0, 1) and rint X + Y = (0, 2) 6= R = X + Y = rint (X + Y ). Relative Interior and Simplices We will need the following auxiliary statement. Theorem 2.16. Let {c1 , . . . , cr } be an affinely independent set in Rn and x, x1 , x2 , . . . be an infinite sequence of points in aff {c1 , . . . , cr }: x = λ1 c1 + · · · + λr cr , λ1 + · · · + λr = 1, (i)

(i)

(i) xi = λ1 c1 + · · · + λ(i) r cr , λ1 + · · · + λr = 1, i > 1. (i)

Then lim xi = x if and only if lim λj = λj for all 1 6 j 6 r. i→∞

i→∞

Proof. Since the case r = 1 is obvious, we assume that r > 2. By Theorem 1.60, the vectors b2 = c2 −c1 , . . . , br = cr −c1 are linearly independent. Put y = x − c1 and yi = xi − c1 , i > 2. Then y = λ2 b2 + · · · + λr br

(i)

and yi = λ2 b2 + · · · + λ(i) r br , i > 2.

Since kx − xi k = ky − yi k, one has xi → x if and only if yi → y. Further(i) more, yi → y if and only if λj → λj for all 2 6 j 6 r (see Exercise 0.7). In the latter case, (i)

(i)

lim λ1 = lim (1 − λ2 − · · · − λr(i) ) = 1 − λ2 − · · · − λr = λ1 .

i→∞

i→∞

Theorem 2.17. The relative interior of an r-simplex ∆ = ∆(x1 , . . . , xr+1 ) in Rn is the set of all positive convex combinations of x1 , . . . , xr+1 : rint ∆ = {λ1 x1 + · · · + λr+1 xr+1 : λ1 , . . . , λr+1 > 0, λ1 + · · · + λr+1 = 1}.

82


Proof. We state first that the set M = {λ1 x1 + · · · + λr+1 xr+1 : λ1 , . . . , λr+1 > 0, λ1 + · · · + λr+1 = 1} lies in rint ∆. For this, choose a point x = λ1 x1 + · · · + λr+1 xr+1 ∈ M and put ε = min {λ1 , . . . , λr+1 }. Theorem 2.16 yields the existence of a scalar ρ > 0 with the following property: for every point u ∈ Bρ (x) ∩ aff ∆, its affine coordinates µ1 , . . . , µr+1 in the expression u = µ1 x1 + · · · + µr+1 xr+1 satisfy the inequalities |λi − µi | 6 ε, 1 6 i 6 r + 1. Consequently, µi > λi − ε > 0 for all i = 1, . . . , r + 1, which gives the inclusion u ∈ ∆. Hence Bρ (x)∩ aff ∆ ⊂ ∆, and x ∈ rint ∆. Conversely, let x = λ1 x1 + · · · + λr+1 xr+1 ∈ rint ∆. Then all scalars λ1 , . . . , λr+1 are nonnegative. Assume for a moment that at least one of them, say λ1 , equals zero. Choose a scalar ρ > 0, and put −1 1 γ = ρ kx1 k + kx2 k + · · · + kxr+1 k r and γ γ v = −γx1 + λ2 + x2 + · · · + λr+1 + xr+1 . r r Clearly, v is an affine combination of x1 , . . . , xr+1 . Therefore, v ∈ aff {x1 , . . . , xr+1 } = aff ∆ according to Theorem 2.6. Furthermore, v ∈ Bρ (x) because of γ γ kv − xk 6 γkx1 k + kx2 k + · · · + kxr+1 k = ρ. r r On the other hand, v ∈ / ∆ since its first affine coordinate, −γ, is negative. Summing up, v ∈ (Bρ (x) ∩ aff ∆) \ ∆, and Definition 2.13 shows that x∈ / rint ∆, contrary to the assumption. Hence all scalars λ1 , . . . , λr+1 are positive, which gives x ∈ M . Corollary 2.18. If K ⊂ Rn is a nonempty convex set, then rint K 6= ∅. Furthermore, rint ∆ ⊂ rint K for every simplex ∆ ⊂ K whose dimension equals dim K. Proof. Let m = dim K. By Corollary 2.7, K contains an m-simplex ∆0 such that aff ∆0 = aff K. A combination of Theorems 2.15 and 2.17 implies that ∅ 6= rint ∆0 ⊂ rint K. Similarly, if ∆ is an m-simplex in K, then aff ∆ = aff K according to Corollary 2.7, and Theorem 2.15 gives the inclusion rint ∆ ⊂ rint K.

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83

Theorem 2.19. Let K ⊂ Rn be a nonempty convex set of dimension m. A point x ∈ Rn belongs to rint K if and only if there is an m-simplex ∆ ⊂ K with the property x ∈ rint ∆. Proof. If x belongs to the relative interior of an m-simplex ∆ ⊂ K, then x ∈ rint K according to Corollary 2.18. Conversely, let x ∈ rint K. Since the case m = 0 is obvious (then K is a singleton, {u}, and ∆ = ∆(u) is the desired 0-simplex), we assume that m > 1. Choose a scalar ρ > 0 satisfying the inclusion Bρ (x) ∩ aff K ⊂ K. Because of dim (aff K) = dim K = m, the set S = aff K − x is an mdimensional subspace (see Theorem 1.2). Let b1 , . . . , bm be a basis for S such that kbi k = ρ/m, 1 6 i 6 m. Put bm+1 = −(b1 + · · · + bm ). Then bm+1 ∈ S and kbm+1 k 6 m (ρ/m) = ρ. Consequently, x + {b1 , . . . , bm+1 } ⊂ x + S = aff K. Let xi = x + bi , 1 6 i 6 m + 1. From kxm+1 − xk 6 ρ and kxi − xk = kbi k = ρ/m, 1 6 i 6 m, it follows that {x1 , . . . , xm+1 } ⊂ Bρ (x) ∩ aff K ⊂ K. By Corollary 1.63, the set {b1 , . . . , bm+1 } is affinely independent. Hence, its translate {x1 , . . . , xm+1 } also is affinely independent, as follows from Theorem 1.83. By Corollary 2.7, the m-simplex ∆(x1 , . . . , xm+1 ) lies in K. Because of x= =

1 1 m+1 ((m + 1)x + o) = m+1 ((m + 1)x + b1 + · · · + bm+1 ) 1 1 1 m+1 ((x + b1 ) + · · · + (x + bm+1 )) = m+1 x1 + · · · + m+1 xm+1 ,

Theorem 2.17 shows that x belongs to rint ∆(x1 , . . . , xm+1 ). The next result refines Theorem 2.19 (see also Exercise 2.7). Theorem 2.20. If K ⊂ Rn is a convex set of positive dimension m and u is a point in K, then a point x ∈ K \ {u} belongs to rint K if and only if K contains an m-simplex ∆ with vertex u such that x ∈ rint ∆. Proof. We observe that the basis b1 , . . . , bm for S in the proof of Theorem 2.19 was chosen arbitrarily. We modify the argument of that proof by letting ρ b1 = δ(u − x), where δ = . mku − xk As above, kb1 k = ρ/m and x=

1 m+1 x1

+ ··· +

1 m+1 xm+1 ,

(2.3)

84


where x1 = x + b1 = (1 − δ)x + δu. Using this value of x1 in (2.3), we obtain x=

1−δ m+1 x

+

δ m+1 u

+

1 m+1 x2

+ ··· +

1 m+1 xm+1 .

Solving the latter equation for x, we express it as a positive convex combination x=

δ m+δ u

+

1 m+δ x2

+ ··· +

1 m+δ xm+1 .

Finally, combining Corollary 2.7 and Theorem 2.17, one has ∆(u, x2 , . . . , xm+1 ) ⊂ K

and x ∈ rint ∆(u, x2 , . . . , xm+1 ).

Geometric Properties of the Relative Interior Theorem 2.21. If K ⊂ Rn is a nonempty convex set, x ∈ rint K, and y ∈ K, then (1 − λ)x + λy ∈ rint K for all 0 < λ < 1. Consequently, the open segment (x, y) lies in rint K. Proof. The case x = y is trivial, we assume that x 6= y. Let z = (1 − λ)x + λy for a certain scalar 0 < λ < 1. Choose a scalar ρ > 0 satisfying the condition Bρ (x) ∩ aff K ⊂ K. We state that B(1−λ)ρ (z) ∩ aff K ⊂ K (see the figure below). Bρ (x) '$ ``` B(1−λ)ρ (z) ` `` ``` r `r r x z y &% λ 1 u − 1−λ y Indeed, let u be a point in B(1−λ)ρ (z) ∩ aff K. Put v = 1−λ (so that u = (1 − λ)v + λy). Then v ∈ aff K as an affine combination of λ 1 z − 1−λ y, one the points u, y ∈ aff K (see Theorem 1.46). Since x = 1−λ has 1 1 kx − vk = ku − zk 6 (1 − λ)ρ = ρ. 1−λ 1−λ

Hence v ∈ Bρ (x)∩ aff K ⊂ K, and u = (1−λ)v +λy ∈ K by the convexity of K. Thus B(1−λ)ρ (z) ∩ aff K ⊂ K, which gives z ∈ rint K. Corollary 2.22. If K ⊂ Rn is a convex set, then its relative interior is a convex set.

Convex sets

85

Proof. The statement is trivial when K = ∅. If K 6= ∅, then the convexity of rint K follows from Theorem 2.21. The next theorem gives an important algebraic description of relatively interior points of a convex set. Theorem 2.23. For a nonempty convex set K ⊂ Rn and a point x ∈ Rn , the following conditions are equivalent. (1) x ∈ rint K. (2) For any point y ∈ aff K, there is a scalar 0 < λ < 1 such that (1 − λ)x + λy ∈ K. (3) For any point y ∈ aff K, there is a scalar γ > 1 satisfying the inclusion γx + (1 − γ)y ∈ K. (4) For any point y ∈ K, there is a scalar γ > 1 with the property γx + (1 − γ)y ∈ K. (5) There is a point y ∈ rint K and a scalar γ > 1 satisfying the inclusion γx + (1 − γ)y ∈ K. Proof. (1) ⇒ (2). Choose a scalar ρ > 0 such that Bρ (x) ∩ aff K ⊂ K. Let y ∈ aff K. Excluding the trivial case x = y, we suppose that x 6= y. Let a scalar δ satisfy the inequalities 0 < δ < min {ρ, kx − yk}. The line l = hx, yi lies in aff K (see Theorem 1.46) and meets the sphere Sδ (x) at two distinct points, u and v, as depicted below. Hence u, v ∈ Bδ (x) ∩ aff K ⊂ Bρ (x) ∩ aff K ⊂ K. By Lemma 1.26, one of the points u, v, say v, belongs to the open halfline (x, yi. Then v = (1 − λ)x + λy for a certain scalar λ > 0. From λkx − yk = kx − vk = δ < kx − yk it follows that 0 < λ < 1. '$ Bδ (x) r r r r u x v y &% (2) ⇒ (3). Choose a point y ∈ aff K, and let y 0 = 2x − y. Then y 0 belongs to aff K as an affine combination of x, y ∈ aff K (see Theorem 1.46). By condition (2), there is a scalar 0 < λ < 1 such that (1 − λ)x + λy 0 ∈ K. Put γ = 1 + λ. Then γ > 1 and γx + (1 − γ)y = (1 + λ)x + (−λ)(2x − y 0 ) = (1 − λ)x + λy 0 ∈ K.

86


Since (3) ⇒ (4) ⇒ (5), it remains to show that (5) ⇒ (1). Choose a point y ∈ rint K (this is possible since rint K 6= ∅ according to Corollary 2.18). Since the case x = y is trivial, we assume that x 6= y. By the assumption, there is a scalar γ > 1 such that the point z = γx + (1 − γ)y belongs to K. Because x can be expressed as x = (1−γ −1 )y +γ −1 z, where 0 < γ −1 < 1, one has x ∈ (y, z) ⊂ rint K according to Theorem 2.21. Theorem 2.23 implies the corollary below. Corollary 2.24. Let K ⊂ Rn be a convex set of positive dimension. A point x ∈ Rn belongs to rint K if and only if any of the following conditions is satisfied. (1) For any y ∈ aff K \ {x}, there is a point u ∈ (x, y) ∩ K. (2) For any y ∈ aff K \ {x}, there is a point u ∈ K \ {x} such that x ∈ (u, y). (3) For any y ∈ K \ {x}, there is a point z ∈ K \ {x} such that x ∈ (y, z). (4) There are points y ∈ rint K \ {x} and z ∈ K \ {x} such that x ∈ (y, z). Remark. The condition dim K > 0 in Corollary 2.24 has a purpose to exclude from consideration empty intervals (x, x). Also, the convexity of K is essential in Theorem 2.23 and Corollary 2.24. Indeed, if K is the union of two circles and a closed segment centered at x, as depicted below, then x satisfies conditions (1)–(4) of the corollary, but is not relatively interior for K. r '$ '$ rx &% r&% Theorem 2.25. Let K ⊂ Rn be a nonempty convex set, and c be a point in rint K. Then rint K = M1 ∪ M2 ∪ · · · , where Mi is the homothetic copy of K with center c given by Mi = c + (1 −

1 i+1 )(K

− c)

i > 1.

Proof. Choose a point y ∈ rint K. By Theorem 2.23, there is a scalar γ > 1 such that the point z = (1 − γ)c + γy belongs to K. Obviously,

Convex sets

87

y = (1 − γ −1 )c + γ −1 z and 0 < γ −1 < 1. Choose an integer i > 1 such 1 1 1 that γ −1 6 1 − i+1 and put u = i+1 c + (1 − i+1 )z. Then y ∈ (c, u) and 1 u = c + (1 − i+1 )(z − c) ∈ Mi . Since the set Mi is convex (see Theorem 2.9), we have y ∈ (c, u) ⊂ Mi . This argument proves the inclusion rint K ⊂ M1 ∪ M2 ∪ · · · . Conversely, if y ∈ Mi , then 1 1 1 )(z − c) = i+1 c + (1 − i+1 )z y = c + (1 − i+1 for a certain point z ∈ K. Therefore, y ∈ rint K according to Theorem 2.21. Hence M1 ∪ M2 ∪ · · · ⊂ rint K. Relative Interior and Algebra of Convex Sets Theorem 2.26. Let F = {Kα } be a family of convex sets in Rn whose relative interiors have a point in common. Then rint (∩ Kα ) ⊂ ∩ rint Kα . α

α

If, additionally, the family F is finite, then rint (∩ Kα ) = ∩ rint Kα . α

α

Proof. Since ∅ 6= ∩ rint Kα ⊂ ∩ Kα , Corollary 2.18 shows that rint (∩ Kα ) α

α

α

is nonempty. Choose points x ∈ rint (∩ Kα ) and y ∈ ∩ rint Kα . Then α α y ∈ ∩ Kα , and, by Theorem 2.23, there is a scalar γ > 1 such that the α

point z = γx + (1 − γ)y belongs to ∩ Kα . Since 0 < γ −1 < 1, Theorem 2.21 α gives x = (1 − γ −1 )y + γ −1 z ∈ rint Kα for all Kα ∈ F. Therefore, x ∈ ∩ rint Kα . α

Suppose now that F is finite, and let F = {K1 , . . . , Kr }. By the above proved, it remains to verify the inclusion rint K1 ∩ · · · ∩ rint Kr ⊂ rint (K1 ∩ · · · ∩ Kr ). For this, choose points x ∈ rint K1 ∩ · · · ∩ rint Kr and y ∈ K1 ∩ · · · ∩ Kr . By Theorem 2.23, there are scalars γi > 1 such that zi = γi x + (1 − γi )y ∈ Ki , 1 6 i 6 r. Put γ = min {γ1 , . . . , γr }. Then γ > 1 and γx + (1 − γ)y ∈ [y, z1 ] ∩ · · · ∩ [y, zr ] ⊂ K1 ∩ · · · ∩ Kr by the convexity of K1 , . . . , Kr . Therefore, Theorem 2.23 shows that x belongs to rint (K1 ∩ · · · ∩ Kr ).

88


Remark. The inclusion rint (∩ Kα ) ⊂ ∩ rint Kα in Theorem 2.26 may be α

α

proper if the family {Kα } is infinite. Indeed, consider the closed intervals Ki = [0, 1 + 1i ], i > 1, on the real line R. Then rint (∩ Ki ) = rint [0, 1] = (0, 1) 6= (0, 1] = ∩ (0, 1 + 1i ) = ∩ rint Ki . i

i

i

n

Since rint L = L for every plane L ⊂ R , Theorem 2.26 implies the following corollary. Corollary 2.27. If a plane L ⊂ Rn meets the relative interior of a convex set K ⊂ Rn , then rint (K ∩ L) = rint K ∩ L. Theorem 2.28. Let F = {Kα } be a nested family of convex sets in Rn . Then the set ∪ rint Kα is convex and α

rint (∪ Kα ) ⊂ ∪ rint Kα . α

(2.4)

α

If, additionally, aff Kγ = aff (∪ Kα ) for every Kγ ∈ F, then α

rint (∪ Kα ) = ∪ rint Kα . α

(2.5)

α

Proof. First, we are going to show that the set ∪ rint Kα is convex. Exα cluding the trivial case ∪ Kα = ∅, we suppose that ∪ Kα is nonempty. Then α α ∪ rint Kα 6= ∅ according to Corollary 2.18. Let x, y be points in ∪ rint Kα . α α Then x ∈ rint Kβ and y ∈ rint Kδ for certain sets Kβ and Kδ from F. Assuming, for instance, that Kβ ⊂ Kδ , we obtain x ∈ Kδ . Consequently, Theorem 2.21 gives (x, y) ⊂ rint Kδ . Thus [x, y] = {x} ∪ (x, y) ∪ {y} ⊂ rint Kβ ∪ rint Kδ ⊂ ∪ rint Kα , α

which shows the convexity of ∪ rint Kα . α

To prove the inclusion (2.4), let x ∈ rint (∪ Kα ). Since the family F α

is nested, the family {aff Kα } of planes also is nested (see Theorem 1.50). By a dimension argument, the family {aff Kα } contains at most finitely many distinct planes. Hence there is a set Kδ ∈ F such that aff Kδ = aff (∪ Kα ). Choose a point y ∈ rint Kδ . Because the set ∪Kα is convex α

α

(see Theorem 2.8) and y ∈ ∪Kα , Theorem 2.23 shows the existence of a α

scalar γ > 1 such that the point z = γx + (1 − γ)y belongs to ∪ Kα . α Therefore, z ∈ Kβ for a certain Kβ ∈ F. If Kβ ⊂ Kδ , then z ∈ Kδ , and the equality x = (1 − γ −1 )y + γ −1 z,

0 < γ −1 < 1,

(2.6)

Convex sets

89

together with Theorem 2.21, implies x ∈ rint Kδ ⊂ ∪ rint Kα . α Assume that Kδ ⊂ Kβ . By the choice of Kδ , Theorem 1.50 gives aff Kδ = aff Kβ = aff (∪ Kα ). Hence y ∈ rint Kδ ⊂ rint Kβ because of Theα

orem 2.15. As above, (2.6) and Theorem 2.21 give x ∈ rint Kβ ⊂ ∪ rint Kα . α

Summing up, the inclusion (2.4) holds. Finally, if aff Kγ = aff (∪ Kα ) for every Kγ ∈ F, then Theorem 2.15 α

implies that rint Kγ ⊂ rint (∪ Kα ) for all Kγ ∈ F. Hence the opposite to α

(2.4) inclusion holds. Theorem 2.29. If K1 , . . . , Kr are convex sets in Rn and µ1 , . . . , µr are scalars, then rint (µ1 K1 + · · · + µr Kr ) = µ1 rint K1 + · · · + µr rint Kr . Proof. Without loss of generality, we may assume that all sets K1 , . . . , Kr are nonempty. An induction argument shows that the proof can be reduced to the case r = 2 (for r = 1, we can write µ1 K1 = µ1 K1 + µ2 o). By Corollary 2.18, all three sets rint K1 , rint K2 , and rint (µ1 K1 + µ2 K2 ) are nonempty. Let x ∈ rint (µ1 K1 + µ2 K2 ). Choose points yi ∈ rint Ki , i = 1, 2, and put y = µ1 y1 + µ2 y2 . By Theorem 2.23, there is a scalar γ > 1 such that z = γx + (1 − γ)y ∈ µ1 K1 + µ2 K2 . Hence z can be written as z = µ1 z1 + µ2 z2 , where zi ∈ Ki , i = 1, 2. Put xi = (1 − γ −1 )yi + γ −1 zi , Since 0 < γ

−1

i = 1, 2.

< 1, Theorem 2.21 gives xi ∈ rint Ki , i = 1, 2. Therefore,

x = (1 − γ −1 )y + γ −1 z = µ1 x1 + µ2 x2 ∈ µ1 rint K1 + µ2 rint K2 . Conversely, choose points x ∈ µ1 rint K1 + µ2 rint K2

and y ∈ µ1 K1 + µ2 K2 .

Then x = µ1 x1 + µ2 x2 and y = µ1 y1 + µ2 y2 for certain points xi ∈ rint Ki and yi ∈ Ki , i = 1, 2. By Theorem 2.23, there are scalars γi > 1 such that ui = γi xi + (1 − γi )yi ∈ Ki , i = 1, 2. Let γ = min {γ1 , γ2 }. Then γ > 1 and γ/γi ∈ [0, 1], i = 1, 2. Furthermore, 1 1 vi = γxi + (1 − γ)yi = γ ui + (1 − )yi + (1 − γ)yi γi γi γ γ = ui + 1 − yi ∈ [ui , yi ] ⊂ Ki , i = 1, 2. γi γi

90


Thus γx + (1 − γ)y = µ1 v1 + µ2 v2 ∈ µ1 K1 + µ2 K2 . Finally, Theorem 2.23 and (2.7) give x ∈ rint (µ1 K1 + µ2 K2 ).

(2.7)

A sharper version of Theorem 2.29 for the case of large sums is considered in Exercise 2.9. Relative Interior and the Affine Structure The following result can be viewed as a refinement of Theorem 2.11. Theorem 2.30. If K ⊂ Rn is a nonempty convex set and x ∈ rint K, then aff K = {(1 − λ)x + λy : y ∈ rint K, λ ∈ R}. Proof. Since aff K is a plane containing K, Theorem 1.46 implies that the set L = {(1 − λ)x + λy : y ∈ rint K, λ ∈ R} lies in aff K. Conversely, let u ∈ aff K. By Theorem 2.23, there is a scalar λ ∈ (0, 1) for which the point z = (1−λ)x+λu belongs to K. Theorem 2.21 shows that the point y = 12 (x + z) lies in rint K. Hence u = (1 − λ−1 )x + λ−1 z = (1 − λ−1 )x + λ−1 (2y − x) = (1 − 2λ−1 )x + 2λ−1 y ∈ L. Theorem 2.31. If K1 , . . . , Kr are convex sets in Rn whose relative interiors have a point in common, then aff (K1 ∩ · · · ∩ Kr ) = aff K1 ∩ · · · ∩ aff Kr . Proof. From Theorem 1.50 it follows that aff (K1 ∩ · · · ∩ Kr ) ⊂ aff K1 ∩ · · · ∩ aff Kr . For the opposite inclusion, choose points x ∈ aff K1 ∩ · · · ∩ aff Kr and y ∈ rint K1 ∩ · · · ∩ rint Kr . By Theorem 2.23, there are scalars γi > 1 with the property zi = (1 − γi )x + γi y ∈ Ki , 1 6 i 6 r. Put γ = min {γ1 , . . . , γr } and u = (1 − γ)x + γy. Obviously, γ > 1 and, by a convexity argument, u ∈ [y, z1 ] ∩ · · · ∩ [y, zr ] ⊂ K1 ∩ · · · ∩ Kr . γ 1 Since x can be expressed as an affine combination x = γ−1 y − γ−1 u of points y and u from K1 ∩ · · · ∩ Kr , Theorem 1.51 implies the inclusion x ∈ aff (K1 ∩ · · · ∩ Kr ).

Convex sets

91

Theorem 2.32. If K1 and K2 are convex sets in Rn whose relative interiors meet, then dim (K1 ∪ K2 ) = dim (K1 + K2 ) = dim K1 + dim K2 − dim (K1 ∩ K2 ). Proof. Because aff (K1 ∪ K2 ) is a translate of aff (K1 + K2 ) (see Theorem 1.56), one has dim (K1 ∪ K2 ) = dim (aff (K1 ∪ K2 )) = dim (aff (K1 + K2 )) = dim (K1 + K2 ). By Theorems 1.53 and 2.31, one has aff (K1 + K2 ) = aff K1 + aff K2 , aff (K1 ∩ K2 ) = aff K1 ∩ aff K2 . Finally, Theorem 1.7 gives dim (K1 + K2 ) = dim (aff (K1 + K2 )) = dim (aff K1 + aff K2 ) = dim (aff K1 ) + dim (aff K2 ) − dim (aff K1 ∩ aff K2 ) = dim (aff K1 ) + dim (aff K2 ) − dim (aff (K1 ∩ K2 )) = dim K1 + dim K2 − dim (K1 ∩ K2 ). The following corollary is important for the study of extreme faces of convex sets. Corollary 2.33. If a plane L ⊂ Rn of dimension m meets the relative interior of a convex set K ⊂ Rn , then dim K 6 n − m + dim (K ∩ L). In particular, if L is a hyperplane not containing K, then dim K = dim (K ∩ L) + 1. Proof. Since rint L = L, Theorem 2.32 gives dim K = dim (K ∪ L) − dim L + dim (K ∩ L) 6 n − m + dim (K ∩ L).

(2.8)

In particular, if L is a hyperplane not containing K, then dim (K ∪ L) = n, and (2.8) becomes as dim K = n − (n − 1) + dim (K ∩ L) = dim (K ∩ L) + 1.

92


Remark. The condition L ∩ rint K 6= ∅ is essential in Corollary 2.33. Indeed, if K is the closed circle in R2 and L a line supporting K, then dim K = 2 2 − 1 + 0. Theorem 2.34. Let f : Rn → Rm be an affine transformation. For convex sets K ⊂ Rn and M ⊂ Rm , one has rint f (K) = f (rint K), rint f

−1

(M ) = f

−1

(rint (M ∩ rng f )).

(2.9) (2.10)

If, additionally, rint M ∩ rng f 6= ∅, then rint f −1 (M ) = f −1 (rint M ). Proof. Excluding the trivial cases K = ∅ and M ∩ rng f = ∅, we suppose that both sets K and M ∩rng f are nonempty. Consequently, f −1 (M ) 6= ∅. According to Theorem 2.12, the sets f (K) and f −1 (M ) are convex. For the equality (2.9), choose points x ∈ rint f (K) and y0 ∈ rint K. Then y = f (y0 ) ∈ f (rint K) ⊂ f (K). By Theorem 2.23, there is a scalar γ > 1 such that u = γx + (1 − γ)y ∈ f (K). Let a point u0 ∈ K satisfy the condition f (u0 ) = u. Put x0 = (1 − γ −1 )y0 + γ −1 u0 . Since 0 < γ −1 < 1, Theorem 2.21 yields x0 ∈ rint K. By the affinity of f , one has x = (1 − γ −1 )y + γ −1 u = f ((1 − γ −1 )y0 + γ −1 u0 ) = f (x0 ) ∈ f (rint K). Conversely, let x ∈ f (rint K) and y ∈ f (K). Then x = f (x0 ) and y = f (y0 ) for certain points x0 ∈ rint K and y0 ∈ K. By Theorem 2.23, there is a scalar γ > 1 such that u0 = γx0 + (1 − γ)y0 ∈ K. Consequently, γx + (1 − γ)y = γf (x0 ) + (1 − γ)f (y0 ) = f (u0 ) ∈ f (K). Hence x ∈ rint f (K) according to the same theorem. To prove the equality (2.10), let x ∈ rint f −1 (M ) = rint f −1 (M ∩ rng f ). Choose points z ∈ M ∩ rng f and z0 ∈ f −1 (M ∩ rng f ), with f (z0 ) = z. As above, there is a scalar γ > 1 such that γx + (1 − γ)z0 ∈ f −1 (M ∩ rng f ). Therefore, γf (x) + (1 − γ)z ∈ M ∩ rng f , implying f (x) ∈ rint (M ∩ rng f ). Hence x ∈ f −1 (rint (M ∩ rng f )). Conversely, let x ∈ f −1 (rint (M ∩rng f )) and z ∈ f −1 (M ∩rng f ). Then f (x) ∈ rint (M ∩ rng f )

and f (z) ∈ M ∩ rng f.

Convex sets

93

Theorem 2.23 gives the existence of a scalar γ > 1 with the property f (γx + (1 − γ)z) = γf (x) + (1 − γ)f (z) ∈ M ∩ rng f. Hence γx + (1 − γ)z ∈ f −1 (M ∩ rng f ), which gives x ∈ rint f −1 (M ∩ rng f ) = rint f −1 (M ). Now, assume that rint M ∩ rng f 6= ∅. According to Corollary 1.89, the set rng f is a plane, and Corollary 2.27 implies that rint M ∩ rng f = rint (M ∩ rng f ). Therefore, (2.9) shows that rint f −1 (M ) = f −1 (rint (M ∩ rng f )) = f −1 (rint M ∩ rng f ) = f −1 (rint M ).

2.3

Closure and Relative Boundary

Closure and Relative Interior Theorem 2.35. For a convex set K ⊂ Rn , its closure cl K is a convex set with the same affine span and the same dimension as K: aff (cl K) = aff K

and

dim (cl K) = dim K.

Proof. The statement is obvious when K = ∅. So, we assume that K 6= ∅. Because the closed unit ball B of Rn is convex (see example on page 72), Theorem 2.9 implies that every neighborhood Bρ (K) = ρB + K, ρ > 0, of K is convex. Theorem 2.8 shows that cl K is convex as the intersection of convex sets Bρ (K), ρ > 0 (see page 8). By Theorem 1.50, the inclusion K ⊂ cl K yields that aff K ⊂ aff (cl K). On the other hand, from K ⊂ aff K and the closedness of aff K (see Corollary 1.22) it follows that cl K ⊂ aff K. Hence aff (cl K) ⊂ aff K. Finally, the equality aff (cl K) = aff K gives dim (cl K) = dim K. The next result is a sharper version of Theorem 2.21. Theorem 2.36. If K ⊂ Rn is a nonempty convex set, x ∈ rint K, and y ∈ cl K, then (1 − λ)x + λy ∈ rint K for all 0 < λ < 1. Consequently, the open segment (x, y) lies in rint K.

94


Proof. Since the case x = y is trivial, we assume that x 6= y. Choose a point z = (1−λ)x+λy, where 0 < λ < 1. By Theorem 2.14, there is a scalar δ > 0 such that Bδ (x) ∩ aff K ⊂ rint K. Because y ∈ cl K ⊂ aff K, the segment [x, y] lies in aff K. Let a point u ∈ K be such that ku−yk 6 1−λ λ δ, λ 1 and put v = 1−λ z − 1−λ u (whence z = (1 − λ)v + λu, as depicted below). From x=

1 λ z− y 1−λ 1−λ

and kx − vk =

λ ku − yk 6 δ 1−λ

it follows that v ∈ Bδ (x). Furthermore, v ∈ aff K because v is an affine combination of the points u, z ∈ aff K (see Theorem 1.46). Hence v ∈ Bρ (x) ∩ aff K ⊂ rint K. By Theorem 2.21, z ∈ rint K. Bδ (x)

'$

ru xr r (((r ( ( ( y z r v((( ( &%

Theorem 2.37. Let K ⊂ Rn be a nonempty convex set, and c be a point in rint K. Then cl K = N1 ∩ N2 ∩ · · · , where Ni is the homothetic copy of K with center c given by Ni = c + (1 +

1 i+1 )(K

− c),

i > 1.

Proof. Let y ∈ cl K. Since the case y = c is obvious, we assume that y 6= c. Then [c, y) ⊂ rint K ⊂ K by Theorem 2.36. For every integer i > 1, there is a point z ∈ [c, y) ⊂ K so close to y that y ∈ [c, zi ), where 1 )(z − c). Because zi ∈ Ni and Ni is a convex set (see zi = c + (1 + i+1 Theorem 2.9), we have y ∈ Ni . Hence y ∈ N1 ∩ N2 ∩ · · · , which gives the inclusion K ⊂ N1 ∩ N2 ∩ · · · . Conversely, let y ∈ N1 ∩ N2 ∩ · · · . For every integer i > 1, there is 1 )(zi − c). Expressing zi as a point zi ∈ K such that y = c + (1 + i+1 i i zi = (1 − i+1 )c + i+1 y, we see that zi → y when i → ∞. Hence y ∈ cl K, implying the inclusion N1 ∩ N2 ∩ · · · ⊂ K. A relation between homothetic copies of a convex set and its neighborhoods is given in Theorem 2.56. Theorem 2.38. For a convex set K ⊂ Rn , one has cl (rint K) = cl K

and

rint (cl K) = rint K.

Convex sets

95

Proof. Without loss of generality we may assume that K 6= ∅. Because the inclusion cl (rint K) ⊂ cl K is obvious, it suffices to prove the opposite one. Let x ∈ cl K and y ∈ rint K. By Theorem 2.36, the open segment (x, y) lies in rint K. Since the case x = y is trivial (then x ∈ rint K ⊂ cl (rint K)), we suppose that x 6= y. Put zλ = (1 − λ)x + λy, 0 < λ < 1. Clearly, zλ ∈ (x, y) ⊂ rint K and zλ → x as λ → 0. Hence x ∈ cl (rint K). Because aff K = aff (cl K) (see Theorem 2.35), the inclusion K ⊂ cl K and Theorem 2.15 give rint K ⊂ rint (cl K). Conversely, choose points x ∈ rint (cl K) and z ∈ rint K. Then u = γx + (1 − γ)z ∈ cl K for a certain γ > 1 (see Theorem 2.23). Therefore, x = (1 − γ −1 )z + γ −1 u ∈ rint K according to Theorem 2.36. Hence rint (cl K) ⊂ rint K. Corollary 2.39. If K1 and K2 are convex sets in Rn , then cl K1 = cl K2 if and only if rint K1 = rint K2 . Corollary 2.40. For a convex set K ⊂ Rn , the following conditions are equivalent. (1) K is a plane. (2) rint K is a plane. (3) cl K is a plane. Proof. (1) ⇔ (2). If K is a plane, then rint K = K, which shows that rint K also is a plane. Conversely, if rint K is a plane, then rint K ⊂ K ⊂ aff K = aff (rint K) = rint K according to Theorem 2.14. Hence K = rint K, and K is a plane. (1) ⇔ (3). If K is a plane, then cl K = K (see Corollary 1.22), which shows that cl K is a plane. Conversely, assume that cl K is a plane. Then Theorem 2.38 gives rint K = rint (cl K) = cl K, implying that rint K is a plane. By the above proved, K is a plane. The next theorem describes some local properties of convex sets. Theorem 2.41. If K ⊂ Rn is a nonempty convex set and c is a point in cl K, then Bρ (c) ∩ rint K 6= ∅ for every closed ball Bρ (c) ⊂ Rn , ρ > 0. Furthermore, aff (Bρ (c) ∩ rint K) = aff (Bρ (c) ∩ K) = aff (Bρ (c) ∩ cl K) = aff K.

(2.11)

96


Proof. First, we state that Bρ (c) ∩ rint K 6= ∅. Since the case K = {c} is obvious, we may assume that K has positive dimension. Then rint K also has positive dimension (see Theorem 2.14), implying that rint K 6= {c}. Choose points u ∈ rint K \ {c} and z ∈ (u, c) such that kz − ck < ρ. Then z ∈ rint K according to Theorem 2.36. By Theorem 2.14, there is a δ > 0 such that Bδ (z) ∩ aff K ⊂ rint K. Let 0 < ε < min{δ, ρ − kz − ck}. For every point x ∈ Bε (z), one has kx − ck 6 kx − zk + kz − ck 6 ε + kz − ck < ρ, implying the inclusion x ∈ Bρ (c). Consequently, Bε (z) ∩ rint K ⊂ Bρ (c) ∩ rint K. A combination of Theorem 2.14 (with z instead of c), and Theorems 1.50 and 2.35 gives aff K = aff (Bε (z) ∩ rint K) ⊂ aff (Bρ (c) ∩ rint K) ⊂ aff (Bρ (c) ∩ K) ⊂ aff (Bρ (c) ∩ cl K) ⊂ aff (cl K) = aff K. Summing up, the equalities (2.11) hold. Theorem 2.42. A nonempty convex set K ⊂ Rn contains every convergent series ∞ ∞ P P z= λi xi , where xi ∈ K, λi = 1, and λi > 0 for all i > 1. (2.12) i=1

i=1

Proof. Choose a point z of the form (2.12). Excluding zero multiples, we assume that all scalars λi are positive. Let X = {x1 , x2 , . . . } and L = aff X. Put r = dim L. Then X contains r + 1 affinely independent points (see Corollary 1.76). Without loss of generality, we may assume that these are x1 , . . . , xr+1 . So, L = aff {x1 , . . . , xr+1 }. Consider the convex set M = K ∩ L. We first state that z ∈ cl M . For this, we express z as ∞ k k P P P (k) z= λi xi = lim λi xi = lim γk ηi xi k→∞ i=1

i=1

where γk =

Pk

j=1 λj

and

zk =

(k) ηi

k→∞

i=1

= λi /γk for all 1 6 i 6 k. Clearly, every

(k) η1 x1

(k)

+ · · · + ηk xk ,

k > 1,

is a positive convex combination of points x1 , . . . , xk ∈ M . Hence all z1 , z2 , . . . belong to M (see Theorem 2.3). Since limk→∞ γk = 1, one has lim γk zk γk z k = lim = lim zk ∈ cl M. z = lim γk zk = k→∞ k→∞ γk k→∞ k→∞ lim γk k→∞

Convex sets

97

To prove the inclusion z ∈ M , we write z=

∞ P

λi xi =

i=1

r+1 P i=1

= γr+1

r+1 P i=1

∞ P

λi xi +

λi xi

i=r+2

(r+1)

ηi

xi + (1 − γr+1 )

∞ P

δi xi ,

i=r+2

(r+1) ηi

where γr+1 and are defined as above, and δi = λi /(1 − γr+1 ) for all i > r + 2. Then z = γr+1 u + (1 − γr+1 )v, where u=

r+1 P i=1

(r+1)

ηi

xi ,

∞ P

v=

δi xi ,

0 < γr+1 < 1.

i=r+2

By the above argument, v ∈ cl M . We state that u ∈ rint M . Indeed, (r+1) since all scalars ηi are positive, Theorem 2.17 shows that u belongs to the relative interior of the simplex ∆ = ∆(x1 , . . . , xk+1 ). By the choice of x1 , . . . , xr+1 , we have aff ∆ = L = aff M . Now, Theorem 2.15 implies u ∈ rint ∆ ⊂ rint M . Finally, from the inclusions u ∈ rint M , v ∈ cl M and Theorem 2.36 it follows that the point z ∈ (u, v) belongs to rint M . Hence z ∈ M ⊂ K. Closure and Algebra of Convex Sets Theorem 2.43. If F = {Kα } is a family of convex sets in Rn , then cl (∩ Kα ) ⊂ ∩ cl Kα . α

(2.13)

α

Furthermore, cl (∩ Kα ) = ∩ cl Kα provided ∩ rint Kα 6= ∅. α

α

α

Proof. The convexity of ∩ cl Kα follows from Theorems 2.8 and 2.35. We α

observe that the inclusion (2.13) holds for any family {Kα } of sets, not necessarily convex. We state that ∩ cl Kα ⊂ cl (∩ Kα ) provided ∩ rint Kα 6= ∅. For this, α α α choose points x ∈ ∩ cl Kα and y ∈ ∩ rint Kα . By Theorem 2.36, one has α

α

(1 − λ)x + λy ∈ ∩ rint Kα α

for all

0 < λ < 1.

Therefore, x = lim ((1 − λ)x + λy) ∈ cl (∩ rint Kα ) ⊂ cl (∩ Kα ). λ→0

α

α

Remark. The second statement of Theorem 2.43 fails if ∩ rint Kα = ∅. α Indeed, let H be a hyperplane in Rn and W be an open halfspace determined by H. Then ∅ = cl (H ∩ W ) 6= cl H ∩ cl W = H.

98


Since rint L = cl L = L for every plane L ⊂ Rn , Theorem 2.43 implies the following corollary. Corollary 2.44. If a plane L ⊂ Rn meets the relative interior of a convex set K ⊂ Rn , then cl (K ∩ L) = cl K ∩ L. Corollary 2.45. If F = {Kα } is a nested family of convex sets in Rn , then ∪ cl Kα is a convex set satisfying the inclusion α

∪ cl Kα ⊂ cl (∪ Kα ). α

(2.14)

α

Proof. The convexity of ∪ cl Kα follows from Theorems 2.8 and 2.35, and α

the inclusion (2.14) holds for any family {Kα } of sets, not necessarily convex. We will need the following corollary, which is a particular cases of a more general statement (see Exercise 0.10). Corollary 2.46. For convex sets K1 , K2 ⊂ Rn , the sum cl K1 + cl K2 is a convex set satisfying the inclusion cl K1 + cl K2 ⊂ cl (K1 + K2 ).

(2.15)

If at least one of the sets K1 , K2 is bounded or the planes aff K1 and aff K2 are independent, then the sum cl K1 + cl K2 is a closed set, and cl K1 + cl K2 = cl (K1 + K2 ).

(2.16)

Proof. The convexity of cl K1 + cl K2 follows from Theorems 2.9 and 2.35, and the remaining part of the statement easily derives from the properties of closures (see Exercise 0.10). y

6 K1 K2

Fig. 2.3

r

-

x

The sum of closed convex sets may be nonclosed.

Convex sets

99

Remark. The inclusion (2.15) in Corollary 2.46 may be proper if both convex sets K1 and K2 are unbounded. Indeed, let K1 = {(x, y) : x > 0, xy > 1}

and K2 = {(x, 0) : x 6 0}.

Then both K1 and K2 are closed, while their sum K1 +K2 = {(x, y) : y > 0} is an open halfplane (see the picture above). Theorem 2.47. If K and M are convex sets and X is a nonempty bounded set in Rn such that K + X = M + X, then cl K = cl M . Proof. Without loss of generality, we may assume that both sets K and M are nonempty. By a symmetry argument, it suffices to prove the inclusion K ⊂ cl M . Choose points u ∈ K and x0 ∈ X. Since u + x0 ∈ K + X = M + X, there are points v1 ∈ M and x1 ∈ X such that u + x0 = v1 + x1 . Similarly, we select recursively points vi ∈ M and xi ∈ X such that u + xi−1 = vi + xi

for all

i > 1.

Adding the first r of these equalities, one has ru + x0 = v1 + · · · + vr + xr ,

r > 1.

Equivalently, u − 1r (v1 + · · · + vr ) = 1r (xr − x0 ),

r > 1.

By Theorem 2.3, wr = 1r (v1 + · · · + vr ) ∈ M for all r > 1. If δ = diam X, then kx0 − xr k 6 δ, and ku − wr k = 1r kx0 − xr k 6 δ/r → 0

when r → ∞.

Hence u = limr→∞ wr ∈ cl M , which gives the inclusion K ⊂ cl M . A similar to Theorem 2.47 statement is considered in Theorem 8.11. Theorem 2.48. Let f : Rn → Rm be an affine transformation and K ⊂ Rn and M ⊂ Rm be convex sets. Then f (cl K) is a convex set and f (cl K) ⊂ cl f (K), cl f

−1

(M ) = f

−1

(cl (M ∩ rng f )).

Furthermore, (1) f (cl K) = cl f (K) if f is one-to-one or K is bounded, (2) cl f −1 (M ) = f −1 (cl M ) provided rint M ∩ rng f 6= ∅.

(2.17) (2.18)

100


Proof. Theorems 2.12 and 2.35 imply the convexity of f (cl K). The inclusions f (cl X) ⊂ cl f (X)

and

cl f −1 (Y ) ⊂ f −1 (cl (Y ∩ rng f ))

hold for any continuous function f : Rn → Rm and sets X ⊂ Rn and Y ⊂ Rm (see page 9). So, it remains to prove the opposite inclusion f −1 (cl (M ∩ rng f )) ⊂ cl f −1 (M ). Excluding the trivial case M ∩ rng f = ∅, we assume that M ∩ rng f is nonempty. Choose points x ∈ f −1 (cl (M ∩ rng f ))

and y ∈ rint (M ∩ rng f ).

Then f (x) ∈ cl (M ∩ rng f ). Let u ∈ f −1 (y) and vλ = (1 − λ)u + λx,

zλ = (1 − λ)y + λf (x),

0 < λ < 1.

Theorem 2.36 shows that zλ ∈ rint (M ∩ rng f ) for all 0 < λ < 1. Since f (vλ ) = (1 − λ)f (u) + λf (x) = zλ , Theorem 2.34 gives vλ ∈ f −1 (zλ ) ⊂ f −1 (rint (M ∩ rng f )) = rint f −1 (M ). Now, from Theorem 2.38 it follows x = lim vλ ∈ cl (rint f −1 (M )) = cl f −1 (M ). λ→1

Statement (1) of the theorem holds for any continuous function f and a set X ⊂ Rn (see page 9). For statement (2), we first observe that rng f is a plane (see Corollary 1.89). Then Corollary 2.44 gives cl M ∩ rng f = cl (M ∩ rng f ). Therefore, (2.18) implies that cl f −1 (M ) = f −1 (cl (M ∩ rng f )) = f −1 (cl M ∩ rng f ) = f −1 (cl M ). Remark. The inclusion f (cl K) ⊂ cl f (K) in Theorem 2.48 can be proper if f is not one-to-one and K is unbounded. Indeed, let f be the orthogonal projection of R2 on the x-axis, and K = {(x, y) : x > 0, xy > 1}. Then K is a closed convex set, while f (K) is the open halfline {(x, 0) : x > 0}.

Convex sets

101

Closedness of Affine Images Given a convex set K ⊂ Rn and an affine transformation f : Rn → Rm , the closedness condition f (cl K) = cl f (K) plays an important role in convex analysis. Theorem 2.52 below shows that asymptotic planes can provide a useful tool in formulating such a condition for the case of any set X ⊂ Rn . We recall that the inf -distance between nonempty sets X and Y in Rn is defined by δ(X, Y ) = inf {kx − yk : x ∈ X, y ∈ Y }. Definition 2.49. A nonempty plane L ⊂ Rn is called asymptotic to a nonempty set X ⊂ Rn (or an asymptote of X) provided cl X ∩ L = ∅ and δ(X, L) = 0.

X L Fig. 2.4

An asymptotic plane L to a set X.

The next two lemmas show a relation of asymptotes with closedness of sums. Lemma 2.50. Let X ⊂ Rn be a nonempty set and S ⊂ Rn be a subspace. A translate L of S is an asymptote of X if and only if L ⊂ cl (X + S) \ (cl X + S). Proof. Let a translate L = c + S be an asymptote of X. We first state that L ∩ (cl X + S) = ∅. Indeed, assume for a moment the existence of a point z ∈ L ∩ (cl X + S). Then z = c + u = x + v, where x ∈ cl X and u, v ∈ S. Consequently, x = c + (u − v) ∈ c + S = L, contrary to the assumption cl X ∩ L = ∅. For the inclusion L ⊂ cl (X + S), choose a point z ∈ L. Then z = c + u, where u ∈ S. The condition δ(X, L) = 0 means the existence of a sequence

102


x1 , x2 , . . . of points from X such that limi→∞ δ(xi , L) = 0. Denote by zi the orthogonal projection of xi on L. Clearly, δ(xi , L) = kxi − zi k (see Theorem 1.96). Because of zi ∈ L = c + S, we can write zi = c + ui for a certain ui ∈ S. Since xi + (z − zi ) = xi + (u − ui ) ∈ X + S, and since k(xi + z − zi ) − zk = kxi − zi k = δ(xi , L) → 0

as i → ∞,

one has z ∈ cl (X + S). Hence L ⊂ cl (X + S). Conversely, suppose that L ⊂ cl (X + S) \ (cl X + S). From L + S = L, one has (cl X ∩ L) + S ⊂ (cl X + S) ∩ (L + S) = (cl X + S) ∩ L = ∅. Therefore, cl X ∩ L = ∅. Now, let z ∈ L. The inclusion z ∈ L ⊂ cl (X + S) implies the existence of an infinite sequence z1 , z2 , . . . of points from X + S such that limi→∞ kzi − zk = 0. Expressing zi and z as zi = x + ui and z = c + u, where xi ∈ X and ui , u ∈ S, we obtain, because of c + u − ui ∈ c + S = L, that δ(xi , L) 6 kxi − (c + u − ui )k = kzi − zk → 0

as i → ∞.

Hence δ(X, L) = 0, and L is an asymptote of X. Lemma 2.51. If X ⊂ Rn is a nonempty set and S ⊂ Rn is a subspace, then a certain translate of S is an asymptote of X if and only if the sum cl X + S is not closed. Proof. First, we observe that cl (X + S) ⊂ cl (cl X + S) ⊂ cl (cl (X + S)) = cl (X + S), which gives the equality cl (X + S) = cl (cl X + S). If a certain translate L = c + S of S is an asymptote of X, then, by Lemma 2.50, L ⊂ cl (X + S) \ (cl X + S) = cl (cl X + S) \ (cl X + S), implying that the sum cl X + S is not closed. Conversely, suppose that the sum cl X + S is not closed. Choose a point a ∈ cl (cl X + S) \ (cl X + S) = cl (X + S) \ (cl X + S)

Convex sets

103

and consider the plane M = a + S. We state that (cl X + S) ∩ M = ∅. Indeed, assuming the existence of a point z ∈ (cl X +S)∩M and expressing z as z = x+u = a+v, where x ∈ cl X and u, v ∈ S, we obtain the inclusion a = x + (u − v) ∈ cl X + S, which is impossible. Next, we state that M ⊂ cl (X + S). Indeed, let e ∈ M = a + S. Then e = a + u for a certain point u ∈ S. Because of a ∈ cl (X + S), there is a sequence of points a1 , a2 , . . . in X + S such that limi→∞ kai − ak = 0. Let ei = ai + u. Then ei ∈ (X + S) + S = X + S, lim kei − ek = lim kai − ak = 0.

i→∞

i→∞

Hence e ∈ cl (X + S), and M ⊂ cl (X + S). By Lemma 2.50, M is an asymptote of X. Theorem 2.52. Let f : Rn → Rm be an affine transformation expressed as f (x) = a + g(x), where a ∈ Rm and g : Rn → Rm is a linear transformation. For every set X ⊂ Rn , one has f (cl X) ⊂ cl f (X). Furthermore, the following conditions are equivalent. (1) f (cl X) = cl f (X). (2) The sum cl X + null g is closed. (3) No translate of the subspace null g is an asymptote of X. Proof. Because of f (cl X) = a + g(cl X), we may assume that f is a linear transformation; that is, f = g. Without loss of generality, we may consider that X 6= ∅. For any set X ⊂ Rn , the inclusion f (cl X) ⊂ cl f (X) because f is continuous (see Exercise 0.6). (1) ⇒ (2). Condition (1) implies that the set f (cl X) is closed. Choose a subspace S ⊂ Rn complementary to null f and denote by Y the projection of cl X on S along null f . Then cl X + null f = Y + null f , which gives f (cl X) = f (Y ). Consequently, the set f (Y ) is closed. The mapping h : S → rng f defined by h(x) = f (x) is an invertible linear transformation (see Exercise 0.1). Clearly, h(Y ) = f (Y ). Furthermore, the set Y = h−1 (h(Y )) is closed by the continuity of h−1 : rng f → S and closedness of h(Y ). Since the subspaces S and null f are complementary, the sum Y + null f = cl X + null f is closed according to Theorem 2.48. (2) ⇒ (1). Repeating the above argument in the converse order, we obtain that the set f (cl X) is closed. Therefore, cl f (X) ⊂ cl f (cl X) = f (cl X) ⊂ cl f (X),

104


implying that f (cl X) = cl f (X). The equivalence of conditions (2) and (3) follows from Lemma 2.51. Relative Boundary Definition 2.53. The relative boundary of a set X ⊂ Rn , denoted rbd X, is defined as rbd X = cl X \ rint X. Example. If L ⊂ Rn is a plane, then rbd L = cl L \ rint L = L \ L = ∅ (see Corollary 1.22 and example on page 78). Example. If L ⊂ Rn is a plane of positive dimension and D is a closed halfplane of L given by D = {x ∈ L : x·c 6 γ},

c∈ / (sub L)⊥ ,

γ∈R

(see Corollary 1.38), then rbd D is the (m − 1)-dimensional plane H ∩ L, where H is the hyperplane expressed by (1.3) (see Corollary 2.62 below). Similarly, the relative boundary of a closed slab of L, given by F = {x ∈ L : γ 6 x·c 6 γ 0 },

c∈ / (sub L)⊥ ,

γ < γ0,

coincides with the union of two parallel (m − 1)-dimensional planes H ∩ L and H 0 ∩ L, where H and H 0 are parallel hyperplanes, given by (1.3) and (1.4), respectively. Theorem 2.54. The relative boundary of a set X ⊂ Rn is a closed set. Furthermore, if X is convex, then rbd X = rbd (cl X) = rbd (rint X). Proof. Since the case rbd X = ∅ is obvious, we assume that rbd X is nonempty. Let x1 , x2 , . . . be a sequence of points in rbd X converging to a point x ∈ Rn . Then x ∈ cl X because of rbd X ⊂ cl X. We state that no set Bρ (x) ∩ aff X, ρ > 0, lies in X. Indeed, choose an index i large enough to satisfy the inequality kx − xi k 6 ρ/2. Since xi ∈ rbd X, there is a point yi ∈ Bρ/2 (xi ) ∩ (aff X \ X). The inequalities kx − yi k 6 kx − xi k + kxi − yi k 6 ρ give yi ∈ Bρ (x). Therefore, Bρ (x) ∩ aff X 6⊂ X, and x ∈ / rint X. Summing up, x ∈ cl X \ rint X = rbd X, which confirms the closedness of rbd X.

Convex sets

105

Assume that the set X is convex. Theorems 2.15 and 2.38 yield rbd (cl X) = cl (cl X) \ rint (cl X) = cl X \ rint X = rbd X, rbd (rint X) = cl (rint X) \ rint (rint X) = cl X \ rint X = rbd X. The next result describes an important property of the relative boundary of a convex set (see also the complementary Theorem 5.1 concerning the inclusion [x, zi ⊂ rint K). Theorem 2.55. Let K ⊂ Rn be a nonempty convex set, x ∈ rint K, and z ∈ aff K \ {x}. Then either [x, zi ⊂ rint K, or the halfline [x, zi contains a unique point u ∈ rbd K such that [x, u) ⊂ rint K and [x, zi \ [x, u] ⊂ aff K \ cl K.

K r x

r u

r z

-

Proof. Suppose that [x, zi 6⊂ rint K and choose a point v ∈ [x, zi \ rint K. We observe first that no point from [x, vi \ [x, v] belongs to cl K. Indeed, assuming the existence of a point y ∈ ([x, vi \ [x, v]) ∩ cl K, we would obtain the inclusion v ∈ (x, y) ⊂ rint K (see Theorem 2.36). Hence the set cl K ∩[x, zi is bounded, and, by the convexity of cl K, it is a closed segment, say [x, u]. We state that u ∈ rbd K. For if u belonged to rint K, then, according to Corollary 2.24, there would be a point y ∈ K ∩[x, zi satisfying the inclusion u ∈ (x, y), in contradiction with cl K ∩ [x, zi = [x, u]. Assume for a moment the existence of another point u0 ∈ [x, zi which belongs to rbd K. Then either u0 ∈ (x, u) or u ∈ (x, u0 ), and Theorem 2.36 shows that one of the points u, u0 belongs to rint K, which is impossible. Theorem 2.55 is used in establishing the following relation between homothetic copies of a convex set and its neighborhoods. Theorem 2.56. Let K ⊂ Rn be a nonempty convex set, c ∈ rint K, and ρ > 0 such that Bρ (c) ∩ aff K ⊂ K. For any scalar γ > 1, the homothetic copy Kγ = c + γ(K − c) contains the set B(γ−1)ρ (K) ∩ aff K, where B(γ−1)ρ (K) means the (γ − 1)ρ-neighborhood of K.

106


Proof. Choose a point y ∈ B(γ−1)ρ (K) ∩ aff K. The case y ∈ K is obvious because of the inclusion K ⊂ Kγ for all γ > 1. Let y ∈ / K. Denote by u the nearest to y point in cl K. Then ky − uk 6 (γ − 1)ρ by the choice of y. If u ∈ [c, y], then kc − yk/kc − uk = (kc − uk + ku − yk)/kc − uk 6 (ρ + (γ − 1)ρ))/ρ = γ, implying the inclusion y ∈ Kγ . Assume that u ∈ / [c, y] and denote by L the 2-dimensional plane containing u, c, y. By Theorem 2.55, [c, y] meets rbd K at a single point, z. Denote by l the line in L which is tangent to the circle Bρ (c) ∩ L and meets [u, y] at a point u0 . Let v be the nearest to y point in l, and denote by w the point at which l supports Bρ (c) ∩ L. Clearly, kv − yk 6 ku0 − yk 6 ky − uk 6 (γ − 1)ρ, which gives kc − yk/kc − zk = kw − vk/kw − zk = (kc − wk + ky − vk)/kc − wk 6 (ρ + (γ − 1)ρ)/ρ = γ, again implying the inclusion y ∈ Kγ . Corollary 2.57. The relative boundary of a convex set K ⊂ Rn is empty if and only if K is a plane. Proof. If K is a plane, then rbd K = ∅ (see example on page 104). Suppose that K is not a plane. Then K 6= aff K and K 6= ∅. Choose points x ∈ rint K (see Corollary 2.18) and z ∈ aff K \ K. By Theorem 2.55, there is a point u ∈ [x, zi ∩ rbd K. Hence rbd K 6= ∅. Theorem 2.58. If K ⊂ Rn is a convex set distinct from a plane, then the following conditions are equivalent. (1) rbd K is a plane. (2) rbd K is a convex set. (3) There is an open halfplane E of aff K such that K is the union of E and a convex subset F of rbd E (possibly, F = ∅ or F = rbd E). Proof. (1) ⇒ (2). This part is trivial because every plane is a convex set. (2) ⇒ (1). According to Corollary 2.57, rbd K 6= ∅. If rbd K is a singleton, then, obviously, it is a plane. Suppose that rbd K is not a singleton

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and choose distinct points x, y ∈ rbd K. We state that hx, yi lies in rbd K. Assume for a moment the existence of a point z ∈ hx, yi \ rbd K. Since rbd K is convex, we have [x, y] ⊂ rbd K. So, z ∈ hx, yi \ [x, y]. Without loss of generality, we may suppose that y ∈ (x, z), as depicted below. Then z ∈ Rn \ cl K, because otherwise z ∈ cl K \ rbd K = rint K, implying the inclusion y ∈ (x, z) ⊂ rint K (see Theorem 2.36), which contradicts the assumption y ∈ rbd K. x rP P

y z r r PP Pr w PPP r v r u

Choose a point u ∈ rint K. By Theorem 2.55, there is a point v ∈ (u, z) which belongs to rbd K. By the assumption, [x, v] ⊂ rbd K. On the other hand, the segment [u, y] meets [x, v] at some point w (see Lemma 1.27 and the picture below). Theorem 2.36 shows that w ∈ rint K, in contradiction with w ∈ [x, v] ⊂ rbd K. Hence z ∈ rbd K, and hx, yi ⊂ rbd K. Finally, Theorem 1.46 shows that rbd K is a plane. (3) ⇒ (1). According to Corollary 1.38, the closure of E is a closed halfplane D of aff K such that D \ E is a plane. Since E ⊂ K ⊂ D, one has D = cl E ⊂ cl K ⊂ D, implying that rbd K = cl K \ rint K = D \ E. (1) ⇒ (3). Put L = aff K. Then cl K 6= L by Corollary 2.40. Let z ∈ L \ cl K and N = aff ({z} ∪ rbd K). Obviously, N ⊂ L. We state that cl K ⊂ N . Indeed, suppose cl K 6⊂ N . Then rint K 6⊂ N by Theorem 2.38. Choose a point y ∈ rint K \ N and consider the line l = hy, zi. We observe that l∩N = {z} (otherwise, l∩N would contain another point w, and y ∈ l = hw, zi ⊂ N by Lemma 1.26 and Theorem 1.46). On the other hand, the open segment (y, z) should contain a point of rbd K (see Theorem 2.55) which is distinct from z. The obtained contradiction proves the inclusion cl K ⊂ N . So, L = aff K ⊂ N , which gives L = N . Corollary 1.67 shows that dim (rbd K) = dim L − 1. The set L \ rbd K is the union of two open halfplanes of L, say E1 and E2 , as stated in Corollary 1.39. Without loss of generality, we may assume that z ∈ E1 . Since [u, z] ⊂ E1 for every point u ∈ E1 , and since each segment [x, z], with x ∈ rint K, should contain a point of rbd K (see Corollary 1.41), one has rint K ⊂ E2 . So,

108


K ⊂ cl K ⊂ cl E2 , which shows that K = F ∪ E2 , where F = K ∩ rbd E2 is a convex subset of rbd E2 . Theorem 2.59. If F and K are nonempty convex sets in Rn , with F ⊂ cl K, then either F ⊂ rbd K or rint F ⊂ rint K. Furthermore, if F ⊂ rbd K, then aff F ∩ cl K ⊂ rbd K and dim F 6 dim K − 1. Proof. Suppose F 6⊂ rbd K. We observe first that rint F ∩ rint K 6= ∅, since otherwise rint F ⊂ rbd K, and a combination of Theorems 2.38 and 2.54 would give F ⊂ cl (rint F ) ⊂ rbd K. Let z ∈ rint F ∩ rint K. Choose a point x ∈ rint F . Then y = γx + (1 − γ)z ∈ F for a certain scalar γ > 1 (see Theorem 2.23). Since y ∈ F ⊂ cl K, Theorem 2.36 shows that the point x = (1 − γ −1 )z + γ −1 y belongs to rint K. Summing up, rint F ⊂ rint K. Now, let F ⊂ rbd K. Assume for a moment that aff F ∩ cl K 6⊂ rbd K. Then there is a point y ∈ aff F ∩ rint K. Let x ∈ rint F . By Theorem 2.23, a certain point z = (1 − λ)x + λy, 0 < λ < 1, belongs to F . On the other hand, z ∈ rint K because of x ∈ cl K (see Theorem 2.36), contradicting the assumption F ⊂ rbd K. Hence aff F ∩ cl K ⊂ rbd K. Finally, suppose that dim F = dim K. Then dim (aff F ) = dim (aff K), which, combined with the inclusion aff F ⊂ aff K, gives aff F = aff K (see Theorem 1.6). Therefore, rint F ⊂ rint K according to Theorem 2.15, in contradiction with rint F ⊂ F ⊂ rbd K. Hence dim F 6 dim K − 1. Corollary 2.60. If a hyperplane H ⊂ Rn meets a nonempty convex set K ⊂ Rn such that K 6⊂ H, then dim (H ∩ K) 6 dim K − 1. Proof. Clearly, the set F = H ∩ K is convex. If F ⊂ rbd K, then the inequality dim F 6 dim K − 1 follows from Theorem 2.59. Suppose that F 6⊂ rbd K. By the same theorem, rint F ⊂ rint K. Hence H meets rint K, and Corollary 2.33 gives dim (H ∩ K) = dim K − 1. Relative Boundary and Algebra of Convex Sets Theorem 2.61. Let F = {Kα } be a family of convex sets whose relative interiors have a point in common. If K = ∩ Kα , then α

∪ (rbd K ∩ rbd Kα ) = ∪ (cl K ∩ rbd Kα ) ⊂ rbd K. α

α

If, additionally, the family F is finite, then ∪ (rbd K ∩ rbd Kα ) = ∪ (cl K ∩ rbd Kα ) = rbd K. α

α

(2.19)

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Proof. By Theorem 2.26, rint K ⊂ ∩ rint Kα . Hence α

rint K ∩ rbd Kα = ∅ for all

Kα ∈ F,

implying that rbd K ∩ rbd Kα = (cl K \ rint K) ∩ rbd Kα = cl K ∩ rbd Kα . Hence ∪ (rbd K ∩ rbd Kα ) = ∪ (cl K ∩ rbd Kα ). α

α

Furthermore, since cl K ⊂ cl Kα for all Kα ∈ F, Theorem 2.26 gives ∪ (cl K ∩ rbd Kα ) = ∪ (cl K \ rint Kα ) = cl K \ (∩ rint Kα ) α

α

α

⊂ cl K \ rint (∩ Kα ) = cl K \ rint K = rbd K. α

If F is finite, then rint (∩ Kα ) = ∩ rint Kα (see Theorem 2.26), and α

α

(2.19) holds. Corollary 2.62. If a plane L ⊂ Rn meets the relative interior of a convex set K ⊂ Rn , then rbd (K ∩ L) = rbd K ∩ L. Furthermore, any side of the latter equality is empty if and only if K ∩ L is a plane which lies in rint K. Proof. Since rint L = cl L = L, Corollaries 2.27 and 2.44 give rbd (K ∩ L) = cl (K ∩ L) \ rint (K ∩ L) = (cl K ∩ L) \ (rint K ∩ L) = (cl K \ rint K) ∩ L = rbd K ∩ L. Furthermore, if K ∩L is a plane which lies in rint K, then rbd K ∩L = ∅. Conversely, assume that the set rbd K ∩ L is empty. First, we observe that L ⊂ rint K. Indeed, assume for a moment the existence of a point v ∈ L \ rint K. By the assumption, there is a point u ∈ rint K ∩ L. Clearly, u 6= v. Theorem 2.55 shows the existence of a point w ∈ rbd K ∩ [u, vi. Since w ∈ [u, vi ⊂ hu, vi ⊂ L (see Theorem 1.46), one has rbd K ∩ L 6= ∅, contrary to the assumption. Next, suppose that K ∩ L is not a plane. Then rbd (K ∩ L) 6= ∅ according to Corollary 2.57. Hence rbd K ∩ L 6= ∅ by the above proved, again contradicting the assumption. Theorem 2.63. For an affine transformation f : Rn → Rm and convex sets K ⊂ Rn and M ⊂ Rm , the following statements hold. (1) If K is bounded, then rbd f (K) ⊂ f (rbd K).

110


(2) If f is one-to-one, then rbd f (K) = f (rbd K). (3) rbd f −1 (M ) = f −1 (rbd (M ∩ rng f )). (4) rbd f −1 (M ) = f −1 (rbd M ) provided rint M ∩ rng f 6= ∅. Proof. (1) If K is bounded, then Theorems 2.34 and 2.48 give rbd f (K) = cl f (K) \ rint f (K) = f (cl K) \ f (rint K) ⊂ f (cl K \ rint K) = f (rbd K). (2) Similarly, if f is one-to-one, then rbd f (K) = cl f (K) \ rint f (K) = f (cl K) \ f (rint K) = f (cl K \ rint K) = f (rbd K). (3) Again by Theorems 2.34 and 2.48, rbd f −1 (M ) = cl f −1 (M ) \ rint f −1 (M ) = f −1 (cl (M ∩ rng f )) \ f −1 (rint (M ∩ rng f )) = f −1 (cl (M ∩ rng f ) \ rint (M ∩ rng f )) = f −1 (rbd (M ∩ rng f )). (4) Since the set rng f is a plane (see Corollary 1.89), statement (3) above and Corollary 2.62 give rbd f −1 (M ) = f −1 (rbd (M ∩ rng f )) = f −1 (rbd M ∩ rng f ) = f −1 (rbd M ). Remark. The inclusion rbd f (K) ⊂ f (rbd K) in statement (1) of Theorem 2.63 does not hold if K is unbounded. Indeed, if f is the orthogonal projection of R2 on the x-axis and K = {(x, y) : xy > 1, x > 0}, then rbd f (K) = {o} 6⊂ {(x, 0) : x > 0} = f (rbd K). Exercises for Chapter 2 Exercise 2.1. Show the convexity of the following solid quadrics: solid ellipsoid, solid elliptic paraboloid, and solid elliptic hyperboloid, given, respectively, as Γ = {(x1 , . . . , xn ) ∈ Rn : a1 x21 + · · · + an x2n 6 1}, Φ = {(x1 , . . . , xn ) ∈ Rn : a1 x21 + · · · + an−1 x2n−1 6 xn }, Ω = {(x1 , . . . , xn ) ∈ Rn : a1 x21 + · · · + an−1 x2n−1 + 1 6 an x2n , xn > 0}, where a1 , . . . , an are positive scalars.

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Exercise 2.2. Show that a nonempty closed set X ⊂ Rn is convex if and only if for any pair of points x, y ∈ X there is a scalar 0 < λ < 1, possibly depending on x and y, such that (1 − λ)x + λy ∈ X. Exercise 2.3. Given a scalar 0 < λ0 < 1, show that a relatively open nonempty set X ⊂ Rn is convex if and only if (1 − λ0 )x + λ0 y ∈ X whenever x, y ∈ X. Exercise 2.4. Show that a nonempty convex set K ⊂ Rn is closed (respectively, relatively open) if and only if for every line l ⊂ aff K, the set K ∩ l is closed (respectively, relatively open). Exercise 2.5. (Grzybowski and Urba´ nski [106]) Show that for every convex set K ⊂ Rn and scalars λ, µ, one has λK + µK = (λ + µ)K +

|λ| + |µ| − |λ + µ| (K − K). 2

Exercise 2.6. (Grzybowski and Urba´ nski [106]) Let K, M ⊂ Rn be compact convex sets, and λ, µ, γ, δ be scalars, with λK + µM = γK + δM . Show that K=

δ−µ λ−γ M if λ 6= γ, or M = K if δ 6= µ. λ−γ δ−µ

Exercise 2.7. Let K ⊂ Rn be a nonempty convex set of dimension m, and ∆(x1 , . . . , xm+1 ) be an m-simplex in aff K. Show that a point z ∈ Rn belongs to rint K is and only if there is an m-simplex ∆(z1 , . . . , zm+1 ) ⊂ K positively homothetic to the simplex ∆(x1 , . . . , xm+1 ) such that z = 1 m+1 (z1 + · · · + zm+1 ). Exercise 2.8. Let K1 and K2 be convex sets in Rn such that a certain translate of K2 lies in aff K1 . Show that rint (K1 + K2 ) = rint K1 + K2 . Exercise 2.9. Let K1 , . . . , Kr be nonempty convex sets in Rn and µ1 , . . . , µr be scalars. Let also r > m = dim (µ1 K1 + · · · + µr Kr ). Show the existence of an index set I ⊂ {1, . . . , r}, card I 6 m, satisfying the condition P P rint (µ1 K1 + · · · + µr Kr ) = µi rint Ki + µi Ki . i∈I

i∈I /

Exercise 2.10. Show that for r-simplices ∆ and ∆0 in Rn , there is an invertible affine transformation f : Rn → Rn such that f (∆) = ∆0 .

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Exercise 2.11. (Starshaped set) Given a nonempty set X ⊂ Rn show that the intersection of all maximal convex subsets of X coincides with the set of points c ∈ X satisfying the property: [c, x] ⊂ X for all x ∈ X. Exercise 2.12. A projective transformation h : Rn → Rm is defined by h(x) = (a + g(x))/(α + x · c), where a ∈ Rm , g : Rn → Rm is a linear transformation, α is a scalar, and c is a nonzero vector in Rn . Show that the image h(K) of a convex set K lying in Rn \ {x ∈ Rn : α + x·c = 0} is a convex set. Notes for Chapter 2 Convex geometry. The first ideas on convexity date back to ancient Greece. For example, Archimedes (in his book “On the sphere and cylinder”, see [3, p. 404]) defines a line concave in the same direction by the following property: “... if every two points on it are taken either, all straight lines connecting the points fall on the same side of the line, or some fall on one and the same side while others fall on itself, but none on the other side.” Based on results of Euler, Cauchy, Schal¨ afly, and Steiner, convex geometry became an independent branch of mathematics around 1900, with notable contributions of Brunn [43, 44] and Minkowski [157, 158]. A comprehensive survey of convex geometry prior to 1934 is summarized in the well-known book of Bonnesen and Fenchel [29]. There are some articles on history of convex geometry, written by Fenchel [85] and Gruber [100, 101]. Intersections and unions of convex sets. Klee [134] posed the problem to describe those sets in Rn which are the intersections of decreasing infinite sequences of pairwise distinct convex sets. The problem is solved in [134] for dimensions two and three. In particular, a convex set in the plane is not the intersection of an infinite decreasing sequence of pairwise distinct convex sets if and only if it has the form S ∪ P ∪ Q, where S is an open convex m-gonal region (possibly, unbounded), P is the union of m open segments or halflines properly contained in the respective open sides of S, and every point of Q is a vertex of S which is an end point of two segments or halflines forming P . A similar problem on the union of convex sets is much easier: the convex sets in Rn which are not the unions of infinite increasing sequences of pairwise distinct convex sets are exactly the convex polytopes. Borovikov [32] showed that the intersection of a nested sequence of simplices in Rn also is a simplex; this statement was generalized by Eggleston, Gr¨ unbaum, and Klee [78] for the case of Hausdorff vector space. In this regard, Gruber [98] posed the problem to describe the families of closed convex sets in Rn which are closed under the intersections of nested sequences of their elements; he also proved in [98, 99] that the following two families have the above property: the family consisting of direct sums of simplices, simplicial cones and subspaces, and the family consisting of direct sums of parallelotopes, simplicial cones and subspaces

Convex sets

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(see also Lawrence [145]). Theorem 2.10 is proved by Jamison [120]. Topological properties of convex sets. Various elementary properties of relative interior and relative boundary of a convex set in n dimensions can be found in Steinitz [207]. Theorem 2.47 is attributed to R˚ adstr¨ om [178]. Klee [126] proved that for a proper subset X of Rn , the following conditions are equivalent: (a) X is a convex Fσ -set, (b) X is an orthogonal projection of a closed convex set, (c) X is an orthogonal projection of a convex Fσ -set. See also Yaksubaev [225] for condition (b) above, and Bromek and Kaniewski [36] for similar results for the case of convex cones. Answering a question of Klee [123], Larman [140] (for n = 3) and Preiss [175] (for all n > 3) showed that every convex Borel set in Rn can be obtained, starting from compact convex sets, by iteration of countable increasing unions or decreasing intersections. (We recall that a Borel set is a set in a topological space that can be formed from closed sets through the operations of countable union, countable intersection, and relative complement.) Theorem 2.42 is independently proved by Rubin and Wesler [188], Blackwell and Girshick [26, pp. 48–49], Cook and Webster [61]. Local convexity. A variety of sufficient conditions for a given set X ⊂ Rn to be convex are summarized in the surveys of Burago and Zalgaller [48] and ManiLevitska [152]. For instance, a closed connected set X ⊂ Rn is convex if and only if it is locally convex (see Nakajima [163] for n 6 3 and Tietze [215] for all n > 2). Similarly, if a connected nonempty set X ⊂ Rn is locally convex, then its relative interior is nonempty and convex (see Tamura [214]). We recall that a set X ⊂ Rn is called locally convex if for any point x ∈ X there is a ball Bρ (x) ⊂ Rn such that Bρ (x) ∩ X is convex (compare with Theorem 2.41). Universal convex sets. Mazur (see Mauldin [154, p. 111–112]) posed the problem on the existence of a convex body K ⊂ R3 symmetric about the origin of R3 such that every planar centrally symmetric convex disc is affinely equivalent to the intersection of K with a certain 2-dimensional subspace. This problem was solved in the negative by Bessaga [23] and Gr¨ unbaum [103] (see Klee [127] for further references). Grz¸a´slewicz [105] showed the existence of a compact convex set K ⊂ Rn+2 such that every compact convex subset of the unit ball of Rn can be obtained as the intersection of K with a certain n-dimensional plane of Rn+2 . Starshaped sets. A subset X of a vector space E is said to be starshaped with respect to a point c ∈ X if the segment [x, c] belongs to X whenever x ∈ X. The kernel of X consists of all those points with respect to which X is starshaped. Brunn [46] proved that the kernel of a compact starshaped set X ⊂ Rn is a compact convex set. The statement of Exercise 2.11 is proved by Toranzos [216] (see also Smith [193]). Obviously, a set X ⊂ E is convex if and only if X is identical with its kernel.

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The kernel of a star-shaped set X ⊂ Rn is called proper if its closure is different from the closure of X. L. Fejes T´ oth asked (see [173]) for a characterization of those convex sets in the plane which can be realized as proper kernels. Post [173] proved the following statements: (a) a strictly convex set K ⊂ R2 with regular boundary is a proper kernel if and only if bd K contains an arc C such that the set C \ K is countable (that is finite or denumerable), (b) a closed convex set K ⊂ R2 is a proper kernel if and only if it is neither a halfplane nor a slab between two parallel lines. Klee [131] established a much stronger result, by proving that a closed convex set K in a separable Banach space is a proper kernel of a starshaped set if and only if K contains no hyperplane. A convex set K ⊂ Rn containing a hyperplane is a proper kernel of a starshaped set if and only if the relative boundary of K contains a line which does not meet K (see Breen [35]). Convex combinations and distance. A point x in a normed vector space E is a convex combination, x = λ1 x1 + · · · + λr xr , of points x1 , . . . , xr ∈ E if and only if kx − zk 6 λ1 kx1 − zk + · · · + λr kxr − zk for all z ∈ E (see Bilyeu [24] for r = 2 and Wolfe [224] for r > 2). Convexity-preserving mappings. A mapping f : Rn → Rm is called convexity-preserving if the f -images of all convex sets in Rn are convex sets in Rm . As shown in Theorem 2.12, affine transformations are convexity-preserving. The converse statement is not true: for example, the mapping f in Rn , defined by f (x1 , x2 , . . . , xn ) = (x31 , 0, . . . , 0), is convexity-preserving but not affine. Walsh [221] proved that a one-to-one convexity-preserving mapping of R2 or 3 R into itself is an affine transformation. This result was expanded by Meyer and Kay [156], who showed that a one-to-one and convexity-preserving mapping f of a real vector space V , dim V > 2, into another real vector space W is an affine transformation. Kuz0 minyh [139] observed that the one-to-one condition here cannot be dropped, by constructing a convexity-preserving mapping R2 → R2 which is onto and discontinuous at every point of R2 . Positively answering Alexandrov’s question, Kuz0 minyh [139] proved that a one-to-one mapping of a real vector space L, dim L > 2, into itself is an affine transformation if it maps every segment onto a convex set. Shaidenko-K¨ unzi [192] (correcting a proposition from [139]) established the following result: If K ⊂ Rn is a convex body which has at least n tangent hyperplanes in general position, and if H is the family of all (positive or negative) homothetic copies of K, then a bijection f : Rn → Rn which maps every convex body M ∈ H onto a convex set is an affine transformation. Ariyawansa, Davidon, and McKennon [4] showed that if D is an open connected subset of Rn , n > 2, and f : D → Rm is a continuous one-to-one convexitypreserving mapping, then there is a projective transformation h(x), expressed as h(x) = g(x)/ϕ(x), where g : Rn → Rm is an affine transformation and ϕ is a nonzero affine functional on Rn , such that h|D = f (see also Exercise 2.12).

Convex sets

115

Homothetic convex sets. Rogers [185] showed that convex bodies K1 and K2 in Rn are positively homothetic if and only if all orthogonal projections of K1 and K2 on 2-dimensional planes are positively homothetic (see [199] for further references on this topic). Soltan [202] refined this statement by proving that compact (respectively, closed) convex sets K1 and K2 in Rn are homothetic provided for any given integer m, 2 6 m 6 n − 1 (respectively, 3 6 m 6 n − 1), the orthogonal projections of K1 and K2 on every m-dimensional plane of Rn are homothetic, where the homothety ratio may depend on the projection plane. Less attention in the literature is given to similar results involving planar sections of convex bodies, since they are viewed as dual forms of orthogonal projections with respect to polarity. Nevertheless, Rogers [185] (for the case when p1 ∈ int K1 and p2 ∈ int K2 ) and later Burton [49] proved of the following statement: if K1 , K2 ⊂ Rn are convex bodies and p1 , p2 ∈ Rn are some points such that for any pair of parallel 2-dimensional planes L1 , L2 through p1 , p2 , respectively, the intersections K1 ∩ L1 and K2 ∩ L2 are both empty or positively homothetic, then K1 and K2 are positively homothetic. Moreover, as proved by Burton and Mani [50], if no homothety is taking p1 to p2 transforms K1 onto K2 , then K1 and K2 are homothetic ellipsoids. For the case of unbounded convex sets, partial results in this direction are obtained by Soltan [199, 201].

116


Chapter 3

Convex Hulls

3.1

Algebraic Properties of Convex Hulls

Definition and Basic Properties Definition 3.1. For a given set X ⊂ Rn , the intersection of all convex sets containing X is called the convex hull of X and denoted conv X.

JJ J J

X Fig. 3.1

Q Q Q

conv X

The convex hull of a set X.

Theorem 2.8 shows that the convex hull of any set X ⊂ Rn exists and is the smallest convex set containing X. Furthermore, conv ∅ = ∅. Example. Let L ⊂ Rn be a plane of positive dimension m and F = {x ∈ L : γ 6 x·c 6 γ 0 } ⊂ Rn ,

c∈ / (sub L)⊥ ,

γ < γ0,

be a closed slab of L, as represented in Corollary 1.44. Then F is the convex hull of its relative boundary (see Theorem 3.15 for a more general statement), which is the union of parallel (m − 1)-dimensional planes M = {x ∈ L : x·c = γ}

and M 0 = {x ∈ L : x·c = γ 0 }.

Example. Every closed ball Bρ (c) ⊂ Rn is the convex hull of its boundary sphere Sρ (c) (see Exercise 3.2). 117

118


Elementary properties of convex hulls are given in the theorem below. Theorem 3.2. For sets X and Y in Rn , the following statements hold. (1) (2) (3) (4) (5)

X ⊂ conv X, with X = conv X if and only if X is convex. conv (conv X) = conv X. conv X ⊂ aff X and aff (conv X) = aff X. conv X ⊂ conv Y if X ⊂ Y . conv X = conv Y if X ⊂ Y ⊂ conv X.

Furthermore, if {Xα } is a family of sets in Rn , then the statements below are true. (6) conv (∩ Xα ) ⊂ ∩ conv Xα . α

α

(7) ∪ conv Xα ⊂ conv (∪ Xα ). α

α

(8) ∪ conv Xα = conv (∪ Xα ) if the family {Xα } is nested. α

α

(9) conv (∪ conv Xα ) = conv (∪ Xα ). α

α

Proof. Let K(X) denote the family of all convex sets containing a given set X ⊂ Rn . The proofs of statements (1)–(9) derive from the following simple arguments. (a) (b) (c) (d)

conv X is the smallest element in K(X). K(X) is exactly the family of all convex sets containing conv X. If X ⊂ Y , then K(Y ) ⊂ K(X). The union of a nested family of convex sets is a convex set (see Theorem 2.8).

The next theorem gives an important description of convex hulls in terms of convex combinations of points. Theorem 3.3. The convex hull of a nonempty set X ⊂ Rn is the collection of all convex combinations of points from X: conv X = {λ1 x1 + · · · + λk xk : k > 1, x1 , . . . , xk ∈ X, λ1 , . . . , λk > 0, λ1 + · · · + λk = 1}. Equivalently, conv X is the collection of all positive convex combinations of points from X. Proof. According to Theorem 2.8, conv X is a convex set containing X, and Theorem 2.3 implies that the set M = {λ1 x1 + · · · + λk xk : k > 1, x1 , . . . , xk ∈ X, λ1 , . . . , λk > 0, λ1 + · · · + λk = 1}

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lies in conv X. We state that M is a convex set. Indeed, choosing points x, y ∈ M and expressing them as convex combinations x = γ1 x 1 + · · · + γp x p

and y = µ1 y1 + · · · + µq yq

of certain points x1 , . . . , xp , y1 , . . . , yq from X, we see that (1 − λ)x + λy is a convex combination of these points for every choice of λ ∈ [0, 1]: (1 − λ)x + λy = (1 − λ)γ1 x1 + · · · + (1 − λ)γp xp + λµ1 y1 + · · · + λµq yq . Hence (1 − λ)x + λy ∈ M . Since X ⊂ M (every point x ∈ X can be written as 1x), the inclusions X ⊂ M ⊂ conv X and Theorem 3.2 give conv X = M . The second statement follows from the first one. Indeed, if a point x ∈ conv X is written as a convex combination x = λ1 x1 + · · · + λk xk of points x1 , . . . , xk ∈ X, then, eliminating all terms of the form 0xi , we obtain a positive convex combination. Corollary 3.4. For points x1 , . . . , xr ∈ Rn , one has conv {x1 , . . . , xr } = {λ1 x1 + · · · + λr xr : λ1 , . . . , λr > 0, λ1 + · · · + λr = 1}. Proof. By Theorem 3.3, a point x ∈ Rn belongs to conv {x1 , . . . , xr } if and only if it can be written as a convex combination x = λi1 xi1 + · · · + λit xit of certain points xi1 , . . . , xit from {x1 , . . . , xr }. If any of these points, say xip and xiq , coincide, then we replace λip xip + λiq xiq with (λip + λiq )xip . Since such a replacement can be performed at most finitely many times, one may assume that all points xi1 , . . . , xit are pairwise distinct. Finally, for every index i ∈ {1, . . . , r} \ {i1 , . . . , it }, we add 0xi to the right-hand side of x = λi1 xi1 + · · · + λit xit to express x as a convex combination of all points x1 , . . . , xr . A combination of Definition 2.5 and Corollary 3.4 shows that every rsimplex ∆(x1 , . . . , xr+1 ) ⊂ Rn is the convex hull of the set {x1 , . . . , xr+1 }. Theorem 3.5. The convex hull of a nonempty set X ⊂ Rn is the collection of all convex combinations (equivalently, of all positive convex combinations) of affinely independent points from X. Proof. By Theorem 3.3, a point x ∈ Rn belongs to conv X if and only if it can be written as a convex combination x = λ1 x1 + · · · + λk xk of certain

120


points x1 , . . . , xk from X. Elimination all terms of the form 0xi , we assume that the scalars λ1 , . . . , λk are positive. We are going to show that x can be written as a convex combination of k − 1 of fewer points from the set {x1 , . . . , xk } provided x1 , . . . , xk are affinely dependent. So, let the set {x1 , . . . , xk } be affinely dependent. By the definition, there are scalars ν1 , . . . , νk , not all zero, such that ν1 x1 + · · · + νk xk = o

and ν1 + · · · + νk = 0.

Clearly, at least one of the scalar ν1 , . . . , νk is positive. Put t = min {λi /νi : νi > 0, 1 6 i 6 k}. Then λi − tνi > 0 for all 1 6 i 6 k, and λi − tνi = 0 for at least one index i ∈ {1, . . . , k}. Furthermore, (λ1 − tν1 ) + · · · + (λk − tνk ) = 1. Hence x = (λ1 − tν1 )x1 + · · · + (λk − tνk )xk is a convex combination of k − 1 or fewer points from X. Consecutively repeating this argument, we obtain a required expression of x. Remark. Although Theorem 3.5 looks similar to Theorems 1.59, convex hulls do not allow, in general, the existence of finite “convex bases” (compare with Theorem 1.68). For example, every point x in the convex hull of the unit circumference C = {(x, y) : x2 + y 2 = 1} of R2 is a convex combination of at most two points from C, while no finite subset X of C satisfies the condition conv X = conv C. The next result is a slight refinement of the well known Carathéodory’s theorem. Theorem 3.6. For an m-dimensional set X ⊂ Rn , the following statements hold. (1) conv X is the collection of all positive convex combinations of m+1 or fewer affinely independent points from X, (2) conv X is the collection of all convex combinations of exactly m+1 affinely independent points from X, (3) conv X is the union of all m-simplices with vertices in X. Proof. (1) Since dim (aff X) = dim X = m, Corollary 1.61 shows that every affinely independent subset of X contains m + 1 or fewer points. Thus the statement follows from Theorem 3.5. (2) By statement (1) above, a point x ∈ Rn belongs to conv X if and only x is a convex combination x = λ1 x1 + · · · + λk xk of m + 1 or fewer

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affinely independent points x1 , . . . , xk ∈ X. Theorem 1.68 shows that the set {x1 , . . . , xk } can be expanded to an affinely independent subset {x1 , . . . , xm+1 } of X. Then x = λ1 x1 + · · · + λk xk + 0xk+1 + · · · + 0xm+1 is a desired expression of x. From Corollary 3.4 it follows that statement (3) is an equivalent form of statement (2). An extension of Theorem 3.6 to the case of finite subsets of conv X is considered in Exercise 3.6. As shown below, one point in the set X from Theorem 3.5 can be chosen in advance. Theorem 3.7. Given a nonempty set X ⊂ Rn and a point x0 ∈ X, the convex hull of X is the collection of all convex combinations of affinely independent subsets of X each containing x0 . Proof. By Theorem 3.5, a point x ∈ Rn belongs to conv X if and only if can be expressed as a positive convex combination x = λ1 x1 + · · · + λk xk of affinely independent points x1 , . . . , xk ∈ X. Without loss of generality, we suppose that the number k is this representation is minimum possible. If x0 ∈ Rn \ aff {x1 , . . . , xk }, then {x0 , x1 , . . . , xk } is affinely independent (see Corollary 1.63), and x = 0x0 +λ1 x1 +· · ·+λk xk is a desired expression. Let x0 ∈ aff {x1 , . . . , xk }. By Theorem 1.60, x0 can be uniquely written as an affine combination x0 = µ1 x1 + · · · + µk xk . Leaving aside the trivial case x = x0 (then x = 1x0 is the convex combination of x0 ), we assume that x 6= x0 . Then at least one of the scalars λ1 − µ1 , . . . , λk − µk is not zero, and the equality (λ1 − µ1 ) + · · · + (λk − µk ) = 0 shows that at least one of them is negative. Since all scalars µi + t(λi − µi ), 1 6 i 6 k, are positive for t = 1, a continuity argument shows the existence of a t0 > 1 such that µi + t0 (λi − µi ) > 0

for all

16i6k

and µi + t0 (λi − µi ) = 0 for at least one index i ∈ {1, . . . , k}. Therefore, the expression z = (1 − t0 )x0 + t0 x = (µ1 + t0 (λ1 − µ1 ))x1 + · · · + (µk + t0 (λk − µk ))xk is a convex combination of k − 1 or fewer points from {x1 , . . . , xk }. Let, for example, z = η1 x1 + · · · + ηk−1 xk−1 , η1 , . . . , ηk−1 > 0, η1 + · · · + ηk−1 = 1.

122


Then −1 −1 −1 −1 x = (1 − t−1 0 )x0 + t0 z = (1 − t0 )x0 + t0 η1 x1 + · · · + t0 ηk−1 xk−1

is a convex combination of x0 , x1 , . . . , xk−1 (as depicted below for k = 3).

r x0

r x1 J r J rz x J J rh x3 hhhhhh J h Jr

x2

Finally, we observe that the set {x0 , x1 , . . . , xk−1 } is affinely independent. Indeed, if {x0 , x1 , . . . , xk−1 } were affinely dependent, then, by Theorem 3.5, x would be a positive convex combination of k − 1 or fewer points from {x0 , x1 , . . . , xk−1 }, contradicting the assumption on minimality of k. The following statement is a refinement of Theorem 3.6. Corollary 3.8. Let X ⊂ Rn be a nonempty set of dimension m, and x0 be a point in X. Then conv X coincides with the union of all m-simplices ∆(x0 , x1 , . . . , xm ), where x1 , . . . , xm ∈ X. Proof. According to Theorem 3.2, conv X contains every m-simplex with vertices x0 , x1 , . . . , xm ∈ X. Conversely, let x ∈ conv X. By Theorem 3.7, x can be written as a convex combination x = λ0 x0 + λ1 x1 + · · · + λk xk of affinely independent points x0 , x1 , . . . , xk from X. Theorem 1.68 shows that {x0 , x1 , . . . , xk } can be expanded to an affinely independent subset {x0 , x1 , . . . , xm } of X. Rewriting x as x = λ0 x0 + λ1 x1 + · · · + λk xk + 0xk+1 + · · · + 0xm , we obtain the inclusion x ∈ ∆(x0 , x1 , . . . , xm ). Iterative Construction of Convex Hulls Definition 3.9. For a nonempty set X ⊂ Rn and a positive integer r, denote by convr X the collection of all convex combination of r or fewer points from X: convr X = ∪ ( conv {x1 , . . . , xk } : k 6 r, x1 , . . . , xk ∈ X).

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123

Example. If X = {x1 , x2 , x3 , x4 } is an affinely independent set in R3 , then conv2 X is the union of all edges of the simplex ∆ = ∆(x1 , x2 , x3 , x4 ), and conv2 (conv2 X) = ∆. Indeed, every point x ∈ ∆ belongs to a closed segment with endpoints on [x1 , x2 ] and [x3 , x4 ], respectively, as depicted below. r x3

CJ

C Jr C J

x rC J Jr r

XX r C XXXCCr x4 x1 x2 A combination of Theorems 3.2 and 3.6 shows that X = conv1 X ⊂ conv2 X ⊂ · · · ⊂ convm+1 X = conv X

(3.1)

n

for any nonempty set X ⊂ R of dimension m. Theorem 3.10. For a nonempty set X ⊂ Rn and positive integers r and s, one has convr (convs X) = convrs X. Proof. Let z ∈ convr (convs X). Then z is a convex combination, z = λ1 y1 + · · · + λp yp , where y1 , . . . , yp ∈ convs X, p 6 r. Similarly, any yi , 1 6 i 6 p, can be written as a convex combination (i)

(i)

(i)

(i) (i) yi = µ1 x1 + · · · + µ(i) mi xmi , where x1 , . . . , xmi ∈ X, mi 6 s.

So, z is a convex combination of at most rs points from X: (1)

(1)

(1) z = λ1 µ1 x1 + · · · + λ1 µ(1) m1 xm1 + . . . (p)

(p)

(p) + λp µ1 x1 + · · · + λp µ(p) mp xmp ,

implying that z ∈ convrs X. Hence convr (convs X) ⊂ convrs X. Conversely, let z ∈ convrs X. Then z can be expressed as a convex combination z = η1 x1 + · · · + ηq xq of certain points x1 , . . . , xq ∈ X, q 6 rs. We may assume that q = rs (otherwise choose any points xq+1 , . . . , xrs in X and add 0xq+1 + · · · + 0xrs to η1 x1 + · · · + ηq xq ). Let λi = η(i−1)s+1 + η(i−1)s+2 + · · · + ηis ,

1 6 i 6 r.

For every λi = 6 0, 1 6 i 6 r, consider the convex combination η(i−1)s+1 η(i−1)s+2 ηis yi = x(i−1)s+1 + x(i−1)s+2 + · · · + xis . λi λi λi

124


Then yi ∈ convs X. Also, choose any yi ∈ convs X if λi = 0, 1 6 i 6 r. Since z = λ1 y1 + · · · + λr yr is a convex combination of y1 , . . . , yr , one has z ∈ convr (convs X). Hence convrs X ⊂ convr (convs X). A combination of (3.1) and Theorem 3.10 gives the following result. Corollary 3.11. If X ⊂ Rn is a nonempty set of dimension m and r1 , . . . , rk are positive integers satisfying the inequality r1 · · · rk > m + 1, then convr1 (convr2 (. . . (convrk X) . . . )) = conv X. In particular, conv2 (conv2 (. . . (conv2 X) . . . )) = conv X, where the operation conv2 is consequently applied dlog2 (m + 1)e times. Algebra of Convex Hulls Theorem 3.12. For sets X1 , . . . , Xr ⊂ Rn and scalars µ1 , . . . , µr , one has conv (µ1 X1 + · · · + µr Xr ) = µ1 conv X1 + · · · + µr conv Xr . Proof. Excluding the trivial case when at least one of the sets X1 , . . . , Xr is empty, we assume that all these sets are nonempty. An induction argument shows that the proof can be reduced to the case r = 2 (if r = 1, then we can write µ1 X1 = µ1 X1 + µ2 o). Since the set µ1 conv X1 + µ2 conv X2 is convex (see Theorem 2.9), the obvious inclusion µ1 X1 + µ2 X2 ⊂ µ1 conv X1 + µ2 conv X2 , and Theorem 3.2 give conv (µ1 X1 + µ2 X2 ) ⊂ µ1 conv X1 + µ2 conv X2 . For the opposite inclusion, choose an x ∈ µ1 conv X1 +µ2 conv X2 . Then x = µ1 x1 + µ1 x2 for certain points x1 ∈ conv X1 and x2 ∈ conv X2 . By Theorem 3.3, x1 and x2 can be written as convex combinations x1 = λ1 u1 + · · · + λp up

and x2 = γ1 v1 + · · · + γq vq

of certain points u1 , . . . , up ∈ X1 and v1 , . . . , vq ∈ X2 . Because µ1 ui + µ2 vj ∈ µ1 X1 + µ2 X2

for all

1 6 i 6 p and 1 6 j 6 q,

the equalities x = µ1 x1 + µ2 x2 =

p P q P i=1 j=1

λi γj (µ1 ui + µ2 vj ),

p P q P

λi γj = 1,

i=1 j=1

show that x is a convex combination of points from µ1 X1 + µ2 X2 . Theorem 3.3 gives x ∈ conv (µ1 X1 + µ2 X2 ).

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125

A sharper version of Theorem 3.12 for the case of large sums is considered in Exercise 3.9. The next theorem describes a “decomposition” of the convex hull of the union of sets (compare with Theorem 3.10). Theorem 3.13. For nonempty sets X1 , . . . , Xr in Rn , one has conv (X1 ∪ · · · ∪ Xr ) = {λ1 x1 + · · · + λr xr : x1 ∈ conv X1 , . . . , xr ∈ conv Xr , λ1 , . . . , λr > 0, λ1 + · · · + λr = 1}. Proof. Theorem 3.2 implies the inclusion conv X1 ∪ · · · ∪ conv Xr ⊂ conv (X1 ∪ · · · ∪ Xr ). This argument and Theorem 3.3 show that the set M = {λ1 x1 + · · · + λr xr : x1 ∈ conv X1 , . . . , xr ∈ conv Xr , λ1 , . . . , λr > 0, λ1 + · · · + λr = 1} lies in conv (X1 ∪ · · · ∪ Xr ). Conversely, let x ∈ conv (X1 ∪ · · · ∪ Xr ). By the same Theorem 3.3, x is expressible as a convex combination x = µ1 z1 + · · · + µk zk ,

where

z1 , . . . , zk ∈ X1 ∪ · · · ∪ Xr .

Renumbering the terms µ1 z1 , . . . , µk zk , we assume that {z1 , . . . , zk } is partitioned into subsets Y1 , . . . , Yr (some of them may be empty) such that Yi = {zpi−1 +1 , . . . , zpi } ⊂ Xi , 1 6 i 6 r, where p0 = 0, pr = k. Let λi = µpi−1 +1 + · · · + µpi , 1 6 i 6 r. Denote by I the set of all indices i ∈ {1, . . . , r} satisfying the condition: either Yi is empty or λi = 0. Let J = {1, . . . , r} \ I. For every i ∈ I, choose a point xi ∈ Xi , and for every i ∈ J, let µp +1 µp xi = i−1 zpi−1 +1 + · · · + i zpi . λi λi Then every xi is a convex combination of points from Xi , i ∈ J, and Theorem 3.3 shows that xi ∈ conv Xi for all 1 6 i 6 r. Finally, x = λ1 x1 + · · · + λr xr is a desired convex combination. Corollary 3.14. For nonempty convex sets K1 , . . . , Kr in Rn , one has conv (K1 ∪ · · · ∪ Kr ) = {λ1 x1 + · · · + λr xr : x1 ∈ K1 , . . . , xr ∈ Kr , λ1 , . . . , λr > 0, λ1 + · · · + λr = 1}. Theorem 3.15. For nonempty planes L1 and L2 in Rn , the following statements hold.

126


(1) If L1 ∩ L2 6= ∅, then conv (L1 ∪ L2 ) = conv2 (L1 ∪ L2 ) = aff (L1 ∪ L2 ). (2) If L1 ∩ L2 = ∅ and G is the open slab of aff (L1 ∪ L2 ) determined by the parallel planes L01 = L1 + sub L2 and L02 = L2 + sub L1 , then conv (L1 ∪ L2 ) = conv2 (L1 ∪ L2 ) = L1 ∪ G ∪ L2 . Proof. (1) Let c ∈ L1 ∩ L2 . A combination of Theorem 1.56 and Corollary 3.14 shows that it suffices to prove the equality conv2 (L1 ∪ L2 ) = L1 + L2 − c. From Theorem 3.12 it follows that the latter equality can be rewritten as conv2 (S1 ∪ S2 ) = S1 + S2 , where S1 = L1 − c and S2 = L2 − c. Clearly, both S1 and S2 are subspaces. Because the subspace S1 + S2 is a convex set which contains S1 ∪ S2 , one has conv2 (S1 ∪ S2 ) ⊂ S1 + S2 . For the opposite inclusion, choose a point u ∈ S1 +S2 . Then u = u1 +u2 , where u1 ∈ S1 and u2 ∈ S2 . Consequently, 2u1 ∈ S1 and 2u2 ∈ S2 , which gives u = 12 (2u1 + 2u2 ) ∈ [2u1 , 2u2 ] ⊂ conv2 (S1 ∪ S2 ). (2) As above, Corollary 3.14 shows that it suffices to prove the equality conv2 (L1 ∪ L2 ) = L1 ∪ G ∪ L2 . Since all the sets L1 , G, L2 are convex, and since L1 and L2 belong to the opposite components of rbd G, we easily conclude that the set L1 ∪ G ∪ L2 is convex. Consequently, conv2 (L1 ∪ L2 ) ⊂ L1 ∪ G ∪ L2 . Conversely, since L1 ∪ L2 ⊂ conv2 (L1 ∪ L2 ), it suffices to prove the inclusion G ⊂ conv2 (L1 ∪ L2 ). For this, choose a point x ∈ G. Statement (1) of Theorem 1.56 implies that x ∈ hc1 , c2 i for suitable points c1 ∈ L1 and c2 ∈ L2 . Since G lies between L01 and L02 , we conclude that, in fact, x ∈ (c1 , c2 ) ⊂ conv2 (L1 ∪ L2 ). Theorem 3.16. For an affine transformation f : Rn → Rm and sets X ⊂ Rn and Y ⊂ Rm , one has conv f (X) = f (conv X), conv f

−1

(Y ) = f

−1

(conv (Y ∩ rng f )).

(3.2) (3.3)

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127

Proof. Excluding the trivial cases X = ∅ and Y ∩ rng f = ∅, we assume that both sets X and Y ∩ rng f are nonempty. Consequently, f −1 (Y ) 6= ∅. For the equality (3.2), choose a point x ∈ conv f (X). By Theorem 3.3, x can be written as a convex combination x = λ1 x1 + · · · + λp xp ,

x1 , . . . , xp ∈ f (X).

where

Let z1 , . . . , zp be points in X such that f (zi ) = xi for all 1 6 i 6 p. Put z = λ1 z1 + · · · + λp zp . Then z ∈ conv X by same theorem, and it follows that x = λ1 x1 + · · · + λp xp = λ1 f (z1 ) + · · · + λp f (zp ) = f (z) according to Theorem 1.87. Hence x = f (z) ∈ f (conv X), which proves the inclusion conv f (X) ⊂ f (conv X). Conversely, let x ∈ conv X. Similarly to the above, x can be written as a convex combination x = µ1 x1 + · · · + µq xq ,

where

x1 , . . . , xq ∈ X.

Then f (x) = µ1 f (x1 ) + · · · + µq f (xq ) according to Theorem 1.87, and Theorem 3.3 gives f (x) ∈ conv f (X). Hence f (conv X) ⊂ conv f (X). It remains to prove (3.3). Letting X = f −1 (Y ) in (3.2), one has f (conv f −1 (Y )) = conv f (f −1 (Y )) = conv (Y ∩ rng f ). Hence conv f −1 (Y ) ⊂ f −1 (f (conv f −1 (Y ))) = f −1 (conv (Y ∩ rng f )). For the opposite inclusion, let x ∈ f −1 (conv (Y ∩ rng f )). Then f (x) belongs to conv (Y ∩rng f ), and Theorem 3.3 implies that f (x) can be written as a convex combination f (x) = γ1 x1 + · · · + γr xr ,

where

x1 , . . . , xr ∈ Y ∩ rng f.

Choose points z1 , . . . , zr ∈ f −1 (Y ) such that f (zi ) = xi , 1 6 i 6 r, and put z = γ1 z1 + · · · + γr zr . Then z ∈ conv f −1 (Y ) and f (z) = γ1 f (z1 ) + · · · + γr f (zr ) = γ1 x1 + · · · + γr xr = f (x). Let ui = zi + (x − z), 1 6 i 6 r. Since ui is an affine combination, we have f (ui ) = f (zi ) + f (x) − f (z) = f (zi ) = xi . Hence ui ∈ f

−1

(xi ) ⊂ f −1 (Y ), 1 6 i 6 r. From the equalities x = ui − zi + z,

1 6 i 6 r,

128


it follows that x can be written as a convex combination of u1 , . . . , ur : x = (γ1 + · · · + γr )x = γ1 (u1 − z1 + z) + · · · + γr (ur − zr + z) = (γ1 u1 + · · · + γr ur ) − (γ1 z1 + · · · + γr zr ) + (γ1 + · · · + γr )z = γ1 u1 + · · · + γr ur . Therefore, x ∈ conv f −1 (Y ). Summing up, f −1 (conv (Y ∩ rng f )) ⊂ conv f −1 (Y ). Remark. The equality (3.3) implies the inclusion conv f −1 (Y ) ⊂ f −1 (conv Y ), which may be proper. Indeed, let f : R2 → R2 be the orthogonal projection on the x-axis of R2 and Y = {(0, 1), (0, −1)}. Then conv f −1 (Y ) = f −1 (Y ) = ∅, while conv Y = {(0, y) : −1 6 y 6 1} and f −1 (conv Y ) is the y-axis.

3.2

Topological Properties of Convex Hulls

Convex Hulls and Closure Theorem 3.17. For a set X ⊂ Rn , one has conv (cl X) ⊂ cl (conv X). Furthermore, if X bounded, then conv X also is bounded and conv (cl X) = cl (conv X). Consequently, conv X is compact provided X is compact. Proof. Excluding the trivial case X = ∅, we assume that X is nonempty. Since X ⊂ conv X, one has cl X ⊂ cl (conv X). By Theorem 2.35, cl (conv X) is a convex set. Hence conv (cl X) ⊂ cl (conv X). Let X be bounded, and Bρ (c) ⊂ Rn be a ball containing X. Since Bρ (c) is a convex set (see example on page 72), Theorem 3.2 implies the inclusion conv X ⊂ Bρ (c). Hence conv X is a bounded set. For the equality conv (cl X) = cl (conv X), it remains to show, by the above proved, that cl (conv X) ⊂ conv (cl X). Let x ∈ cl (conv X). Then x = limi→∞ xi for a sequence of points x1 , x2 , . . . from conv X. Put m = dim X. According to Theorem 3.6, every xi can be written as a convex combination (i) (i)

(i)

(i)

xi = λ1 y1 + · · · + λm+1 ym+1 ,

(i)

(i)

y1 , . . . , ym+1 ∈ X,

i > 1.

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129

(1)

(2)

Consider the m + 1 sequences of points yk , yk , . . . from X and m + (1) (2) 1 sequences of scalars λk , λk , . . . from [0, 1], 1 6 k 6 m + 1. By a compactness argument, there is an infinite sequence of indices i1 , i2 , . . . such that all 2m + 2 subsequences (i )

(i )

yk 1 , yk 2 , . . . ,

(i )

(i )

λk 1 , λk 2 , . . . ,

1 6 k 6 m + 1,

converge. Put (i )

yk = lim yk s s→∞

(i )

and λk = lim λk s , s→∞

1 6 k 6 m + 1.

Then y1 , . . . , ym+1 ∈ cl X, λ1 , . . . , λm+1 > 0, λ1 + · · · + λm+1 = 1. Furthermore, (i ) (i )

(i )

(i )

s s x = lim xis = lim (λ1 s y1 s + · · · + λm+1 ym+1 )

s→∞

s→∞

= λ1 y1 + · · · + λm+1 ym+1 , which shows that x is a convex combination of y1 , . . . , ym+1 . Hence x belongs to conv (cl X), and the inclusion cl (conv X) ⊂ conv (cl X) is proved. If X is compact, then the equality conv X = conv (cl X) = cl (conv X) implies the closedness of conv X. Since conv X is bounded, it also is compact. Remark. If a set X ⊂ Rn is closed but not compact, then conv X may be nonclosed. For example, the set X = {(x, y) : y = 1/|x|} ⊂ R2 is closed, while its convex hull is the open halfplane {(x, y) : y > 0}. The following particular result gives a closedness condition for the convex hull of a pair of planes. Theorem 3.18. For nonempty planes L1 and L2 in Rn , the set conv (L1 ∪ L2 ) is closed if and only if any of the following two conditions holds: (a) L1 ∩ L2 6= ∅,

(b) L1 is a translate of L2 .

Proof. If L1 ∩ L2 6= ∅, then, by Theorem 3.15, conv (L1 ∪ L2 ) coincides with the plane aff (L1 ∪ L2 ), which is a closed set (see Corollary 1.22). If L1 and a translate of L2 , then, by the same theorem, conv (L1 ∪ L2 ) is the closed slab of aff (L1 ∪ L2 ) bounded by L1 and L2 , which also is a closed set (see Corollary 1.44).

130


Conversely, let conv (L1 ∪ L2 ) be a closed set. If L1 ∩ L2 = ∅, then, by Theorem 3.15, conv (L1 ∪ L2 ) = L1 ∪ G ∪ L2 , where G is the open slab of aff (L1 ∪ L2 ) bounded by the parallel planes L01 = L1 + sub L2 and L02 = L2 + sub L1 . Since the closure of conv (L1 ∪ L2 ) equals L01 ∪ G ∪ L02 , one has L1 = L01 and L2 = L2 . The latter is possible if and only if sub L1 = sub L2 , that is, when L1 is a translate of L2 . Convex Hulls and Relative Interior Theorem 3.19. For a set X in Rn , one has conv (rint X) ⊂ rint (conv X). Furthermore, the set conv X is relatively open provided X is relatively open. Proof. By Theorem 3.2, aff X = aff (conv X). The inclusion X ⊂ conv X and Theorem 2.15 imply that rint X ⊂ rint (conv X). Since rint (conv X) is convex (see Corollary 2.22), one has conv (rint X) ⊂ rint (conv X). If X is relatively open (that is, X = rint X), then, by the above proved, conv X = conv (rint X) ⊂ rint (conv X) ⊂ conv X. Hence rint (conv X) = conv X, and conv X is relatively open. Theorem 3.20. Let X ⊂ Rn be a nonempty set. A point x ∈ Rn belongs to rint (conv X) if and only if x is a positive convex combination of finitely many points x1 , . . . , xk ∈ X such that aff {x1 , . . . , xk } = aff X. Proof. Put m = dim X. Assume first that x ∈ rint (conv X). According to Corollary 1.76, X contains m + 1 affinely independent points y1 , . . . , ym+1 with the property aff X = aff {y1 , . . . , ym+1 }. Let ∆ = ∆(y1 , . . . , ym+1 )

and y =

1 m+1 (y1

+ · · · + ym+1 ).

A combination of Theorems 2.15 and 2.17 gives y ∈ rint ∆ ⊂ rint (conv X). Since the case x = y is obvious, we assume that x and y are distinct. Theorem 2.23 shows the existence of a scalar γ > 1 such that the point z = γx + (1 − γ)y belongs to conv X. By Theorem 3.3, z can be written as a convex combination z = λ1 z1 +· · ·+λp zp of certain points z1 , . . . , zp ∈ X. Excluding zero scalars, we suppose that all λ1 , . . . , λp are positive. Then x = (1 − γ −1 )y + γ −1 z =

1 − γ −1 1 − γ −1 y1 + · · · + ym+1 + γ −1 λ1 z1 + · · · + γ −1 λp zp m+1 m+1

(3.4)

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131

is a positive convex combination of y1 , . . . , ym+1 , z1 , . . . , zp . Theorem 1.50 and the inclusions {y1 , . . . , ym+1 } ⊂ {y1 , . . . , ym+1 , z1 , . . . , zp } ⊂ aff X = aff {y1 , . . . , ym+1 } show that the affine span of these points equals aff X. Hence (3.4) gives a desired expression of x. Conversely, let x = λ1 x1 + · · · + λk xk be a positive convex combination of points x1 , . . . , xk ∈ X satisfying the condition aff {x1 , . . . , xk } = aff X. According to Theorem 1.68, the set {x1 , . . . , xk } contains m + 1 affinely independent points whose affine span equals aff X. Without loss of generality, we may suppose that these are exactly x1 , . . . , xm+1 . If k = m + 1, then conv {x1 , . . . , xm+1 } is the m-simplex ∆ = ∆(x1 , . . . , xm+1 ), and a combination of Theorems 2.17 and 2.15 gives x ∈ rint ∆ ⊂ rint (conv X). Let k > m + 2, and put λ = λ1 + · · · + λm+1 . Clearly, 0 < λ < 1, and λm+1 λm+2 λk λ1 x1 + · · · + xm+1 , z = xm+2 + · · · + xk y= λ λ 1−λ 1−λ are convex combinations of points from X. Hence y, z ∈ conv X according to Theorem 3.3. Since y ∈ rint ∆ ⊂ rint (conv X), Theorem 2.21 implies the inclusion x = λy + (1 − λ)z ∈ rint (conv X). Remark. Unlike Theorem 3.5, the points x1 , . . . , xk from Theorem 3.20 cannot be chosen affinely independent. Indeed, let X = {x1 , x2 , x3 , x4 } be the set of points in R2 depicted below. Clearly, o ∈ int (conv X) and X is affinely dependent. At the same time, no proper subset Y of X satisfies the condition o ∈ int (conv Y ). x3 r Q Q Q Qr r r x1 Q x4 o Q Q Q r x2 The next corollary can be viewed as a generalization of Theorem 2.17. Corollary 3.21. For a finite set X = {x1 , . . . , xr } in Rn , one has rint (conv X) = {λ1 x1 + · · · + λr xr : λ1 , . . . , λr > 0, λ1 + · · · + λr = 1}.

132


Proof. Theorem 3.20 shows that every positive convex combination of points x1 , . . . , xr belongs to rint (conv X). Conversely, choose a point x in rint (conv X). Since the point y = 1r (x1 + · · · + xr ) is a convex combination of x1 , . . . , xr and whence belongs to conv X, there is a scalar γ > 1 such that z = γx + (1 − γ)y ∈ conv X (see Theorem 2.23). By Corollary 3.4, z can be written as a convex combination z = µ1 x1 + · · · + µr xr . Consequently, x = (1 − γ −1 )y + γ −1 z 1 − γ −1 1 − γ −1 = + γ −1 µ1 x1 + · · · + + γ −1 µr xr r r is a positive convex combination of x1 , . . . , xr . Algebra of Convex Hulls and Relative Interior A combination of Theorems 2.29 and 3.12 implies the following statement. Corollary 3.22. For sets X1 , . . . , Xr ⊂ Rn and scalars µ1 , . . . , µr , rint (conv (µ1 X1 + · · · + µr Xr )) = µ1 rint (conv X1 ) + · · · + µr rint (conv Xr ). The result below complements Theorem 3.13. Theorem 3.23. For nonempty sets X1 , . . . , Xr ⊂ Rn , the relative interior of conv (X1 ∪ · · · ∪ Xr ) is the collection of all positive convex combinations of the form λ1 x1 + · · · + λr xr , where xi ∈ rint (conv Xi ) for all 1 6 i 6 r. (3.5) Consequently, rint (conv (X1 ∪ · · · ∪ Xr )) ⊂ conv (rint (conv X1 ) ∪ · · · ∪ rint (conv Xr )). Proof. Denote by M the set of all points of the form (3.5). Then the statement can be rewritten as M = rint (conv (X1 ∪ · · · ∪ Xr )).

(3.6)

First, we establish the convexity of M . Indeed, let x = γx1 + · · · + γxr

and y = µ1 y1 + · · · + µr yr

be points from M , expressed in the form (3.5). For any 0 < λ < 1, the point (1 − λ)x + λy = η1 z1 + · · · + ηr zr is a positive convex combination, where λµi (1 − λ)γi xi + yi , 1 6 i 6 r. ηi = (1 − λ)γi + λµi and zi = ηi ηi

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133

By Theorem 2.21, zi ∈ (xi , yi ) ⊂ rint (conv Xi ),

1 6 i 6 r,

which shows that (1 − λ)x + λy ∈ M . Summing up, M is a convex set. Next, we will prove the inclusion M ⊂ rint (conv (X1 ∪ · · · ∪ Xr )). For this, choose a point x ∈ M , expressed in the form (3.5). By Theorem 3.20, (i) (i) (i) (i) each xi is a positive convex combination xi = µ1 y1 + · · · + µpi ypi , where (i)

(i)

y1 , . . . , yp(i) ∈ Xi and aff {y1 , . . . , yp(i) } = aff Xi , 1 6 i 6 r. i i Clearly, (1) (1)

(r) (r)

(1) (r) (r) x = λ1 µ1 y1 + · · · + λ1 µ(1) p1 yp1 + · · · + λr µ1 y1 + · · · + λr µpr ypr

is a positive convex combination of all points from the set (1)

(r)

Y = {y1 , . . . , yp(1) , . . . , y1 , . . . , yp(r) }. 1 r Furthermore, Theorems 1.50 and 3.2 give (1)

(r)

aff Y = aff (aff {y1 , . . . , yp(1) } ∪ · · · ∪ aff {y1 , . . . , yp(r) }) 1 r = aff (aff X1 ∪ · · · ∪ aff Xr ) = aff (X1 ∪ · · · ∪ Xr ) = aff (conv (X1 ∪ · · · ∪ Xr )). Therefore, x ∈ rint (conv (X1 ∪ · · · ∪ Xr )) according to Theorem 3.20. Next, we state that conv (X1 ∪ · · · ∪ Xr ) ⊂ cl M . For this, let x ∈ conv (X1 ∪ · · · ∪ Xr ). To prove the inclusion x ∈ cl M , it suffices to show that for every ρ > 0 there is a point z ∈ M such that kx − zk < ρ. By Theorem 3.13, x can be expressed as a convex combination x = γ1 x 1 + · · · + γr x r ,

where

x1 ∈ conv X1 , . . . , xr ∈ conv Xr .

Renumbering the sets X1 , . . . , Xr , we assume that all positive scalars γi are placed before all zero scalars: γ1 , . . . , γ s > 0

and γs+1 = · · · = γr = 0,

1 6 s 6 r.

Consider first the case s = r. Since conv Xi ⊂ cl (rint (conv Xi )) due to Theorem 2.38, there are points zi ∈ rint (conv Xi ) satisfying the conditions kxi − zi k 6 ρ for all 1 6 i 6 r. Let z = γ1 z1 + · · · + γr zr . Then z ∈ M and kx − zk 6 γ1 kx1 − z1 k + · · · + γr kxr − zr k 6 ρ. Now, assume that s < r and put µ = ρ2 ,

δ = max {kx1 k, . . . , kxr k},

ρ 0 < ε < min γ1 , . . . , γs , 4δs .

134


As above, choose points zi ∈ rint (conv Xi ) such that kxi − zi k 6 µ for all 1 6 i 6 r. Then kzi k 6 kzi − xi k + kxi k 6 µ + δ, 1 6 i 6 r. Consider the positive convex combination z = (γ1 − ε)z1 + · · · + (γs − ε)zs +

εs εs zs+1 + · · · + zr . r−s r−s

Clearly, z ∈ M and kx − zk 6 kγ1 x1 − (γ1 − ε)z1 k + · · · + kγs xs − (γs − ε)zs k εs εs + kzs+1 k + · · · + kzr k r−s r−s = k(γ1 − ε)(x1 − z1 ) + εx1 k + · · · + k(γs − ε)(xs − zs ) + εxs k εs + kzs+1 k + · · · + kzr k r−s 6 (γ1 − ε)kx1 − z1 k + · · · + (γs − ε)kxs − zs k εs kzs+1 k + · · · + kzr k + ε kx1 k + · · · + kxs k + r−s εs 6 (γ1 + · · · + γs − εs)µ + εsδ + (r − s)(δ + µ) r−s ρ ρ = µ + 2εδs < + 2 δs = ρ. 2 4δs Summing up, x ∈ cl M , and conv (X1 ∪ · · · ∪ Xr ) ⊂ cl M . Finally, Theorem 2.38 gives M ⊂ rint (conv (X1 ∪ · · · ∪ Xr )) ⊂ rint (cl M ) = rint M ⊂ M, implying (3.6). Theorem 3.24. For convex sets K1 , . . . , Kr ⊂ Rn , one has rint (conv (K1 ∪ · · · ∪ Kr )) ⊂ conv (rint K1 ∪ · · · ∪ rint Kr ).

(3.7)

If the relative interiors of K1 , . . . , Kr have a point in common, then rint (conv (K1 ∪ · · · ∪ Kr )) = conv (rint K1 ∪ · · · ∪ rint Kr ).

(3.8)

Proof. The inclusion (3.7) immediately follows from Theorem 3.23. Assume that the relative interiors of K1 , . . . , Kr have a point c in common. Due to (3.7), it suffices to prove that conv (rint K1 ∪ · · · ∪ rint Kr ) ⊂ rint (conv (K1 ∪ · · · ∪ Kr )). First, we state that c ∈ rint (conv (K1 ∪ · · · ∪ Kr )). Indeed, let x be any point in conv (K1 ∪ · · · ∪ Kr ). By Theorem 3.13, x is a convex combination,

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135

x = λ1 x1 + · · · + λr xr , of some points xi ∈ Ki , 1 6 i 6 r. Theorem 2.23 shows the existence of scalars γi > 1 such that the point γi c + (1 − γi )xi belongs to Ki , 1 6 i 6 r. Replacing each γi with γ = min{γ1 , . . . , γr }, a convexity argument implies the inclusions yi = γc + (1 − γ)xi ∈ Ki for all 1 6 i 6 r. Finally, with y = λ1 y1 + · · · + λr yr , we obtain the equality y = γc + (1 − γ)x. Since 0 < γ −1 < 1, Theorem 2.23 gives c = (1 − γ −1 )x + γ −1 y ∈ rint (conv (K1 ∪ · · · ∪ Kr )). Next, choose any point z ∈ conv (rint K1 ∪ · · · ∪ rint Kr ). As above, z is a convex combination, z = µ1 z1 + · · · + µr zr , of some points zi ∈ rint Ki , 1 6 i 6 r. By Theorem 2.23, there are scalars ηi > 1 such that the point ui = ηi zi + (1 − ηi )c belongs to Ki , 1 6 i 6 r. As above, we may suppose that η1 = · · · = ηr = η. Let u = µ1 u1 + · · · + µr ur . Then u ∈ conv (K1 ∪ · · · ∪ Kr ), and Theorem 2.23 gives z = (1 − η −1 )c + η −1 u ∈ rint (conv (K1 ∪ · · · ∪ Kr )). Remark. The inclusion (3.7) may be proper. Indeed, if K1 = {(0, y) : 0 6 y 6 1} and K2 = {(1, y) : 0 6 y 6 1}, then rint (conv (K1 ∪ K2 )) = {(x, y) : 0 < x < 1, 0 < y < 1}, conv (rint K1 ∪ rint K2 ) = {(x, y) : 0 < x < 1, 0 6 y 6 1}. The next corollary follows from Theorems 2.34 and 3.16. Corollary 3.25. For an affine transformation f : Rn → Rm and sets X ⊂ Rn and Y ⊂ Rm , one has rint (conv f (X)) = f (rint (conv X)), rint (conv f −1 (Y )) = f −1 (rint (conv (Y ∩ rng f ))).

Steinitz’s Theorem If X ⊂ Rn is a nonempty set, then a combination of Theorem 3.20 and Corollary 3.21 shows that a point x ∈ Rn belongs to rint (conv X) if and only if there is a finite subset Y of X such that x ∈ rint (conv Y ) and aff Y = aff X. Theorem 3.26, which is attributed to Steinitz, sharpens this statement by establishing bounds on the cardinality of Y . The picture below, with X = {x1 , x2 , x3 , x4 , x5 } and x ∈ int (conv X), illustrates two possible cases described in this theorem for m = n = 2 (where Y = {x1 , x2 , x4 } corresponds to the first case, and Y = {x1 , x2 , x3 , x4 } does to the second one).

136


xr 3 r x4 !!

! !

! r ! r

r x x@ x5 2

@ r @

x1

xr 3 @ r x@ 2

@ r @r x x4

r x5

@ @r x1

Theorem 3.26. Let X ⊂ Rn be a set of positive dimension m. A point x ∈ Rn belongs to rint (conv X) if and only if there is a subset Y of X such that x ∈ rint (conv Y ),

m + 1 6 card Y 6 2m,

aff Y = aff X.

(3.9)

Furthermore, if Y is a minimal subset of X satisfying (3.9), then either card Y 6 2m − 1 or Y consists of 2m points collinear in pairs with x. Proof. If a subset Y of X satisfies the conditions (3.9), then Theorem 3.20 and Corollary 3.21 imply the inclusion x ∈ rint (conv X). Conversely, let x ∈ rint (conv X). By the same argument, there is a finite subset Z ⊂ X with the properties x ∈ rint (conv Z) and aff Z = aff X. We are going to choose in Z a subset Y satisfying (3.9). Because the cases m 6 1 are obvious, we suppose that m > 2. Denote by F the family of planes in Rn , each of dimension m − 1 or less, such that every plane L ∈ F contains x and is the affine span of points from {x}∪Z. Since the set Z is finite, the family F also is finite. Choose in aff X a line l through x which does not lie in any plane L ∈ F. According to Theorem 3.17, conv Z is a compact set, and from Theorem 2.55 it follows that the intersection of l and conv Z is a segment [y, y 0 ], where y, y 0 ∈ rbd (conv Z) and x ∈ (y, y 0 ). So, x = (1 − η)y + ηy 0 for a scalar η ∈ (0, 1). By Theorem 3.6, y is a positive convex combination, y = λ1 x1 + · · · + λk xk , of m + 1 or fewer affinely independent points x1 , . . . , xk ∈ Z. According to Corollary 3.21, y ∈ rint (conv {x1 , . . . , xk }). We state that k = m and x ∈ / aff {x1 , . . . , xm }. Indeed, assume for a moment that k = m + 1. Then dim (aff {x1 , . . . , xm+1 }) = m according to Corollary 1.66. Since aff {x1 , . . . , xm+1 } ⊂ aff Z

and

dim (aff Z) = m,

Theorem 1.6 implies that aff {x1 , . . . , xm+1 } = aff Z. Consequently, Theorem 2.15 gives y ∈ rint (conv {x1 , . . . , xm+1 }) ⊂ rint (conv Z), in contradiction with y ∈ rbd (conv Z). Hence k 6 m.

Convex hulls

137

If k 6 m − 1 or x belonged to aff {x1 , . . . , xk }, then the plane L = aff {x, x1 , . . . , xk } would be a member of F, and l = hx, yi ⊂ L, contrary to the choice of l. Summing up, y = λ1 x1 + · · · + λm xm , where the set {x, x1 , . . . , xm } is affinely independent (see Corollary 1.63). Similarly, y 0 is a positive convex combination, y 0 = µ1 x01 + · · · + µm x0m , of points x01 , . . . , x0m ∈ Z such that the set {x, x01 , . . . , x0m } is affinely independent. Put Y = {x1 , . . . , xm , x01 , . . . , x0m }. Clearly, card Y 6 2m and aff Y = aff X. Since x is expressible as a positive convex combination x = (1 − η)y + ηy 0 = (1 − η)λ1 x1 + · · · + (1 − η)λm xm + ηµ1 x01 + · · · + ηµm x0m , Theorem 3.20 implies that x ∈ rint (conv Y ). Finally, card Y > m + 1 by Corollary 1.76. Now, suppose that Y is a minimal subset of X satisfying the conditions (3.9), with card Y = 2m. Then, with the notation above, y and y 0 belong, respectively, to the relative interiors of the simplices ∆ = ∆(x1 , . . . , xm )

and

∆0 = ∆(x01 , . . . , x0m ).

We observe that both ∆ and ∆0 lie in rbd (conv Z). Indeed, assume for a moment that ∆ contains a point u ∈ rint (conv Z). According to Theorem 2.23, there is a scalar γ > 1 satisfying the condition v = γy + (1 − γ)u ∈ ∆ ⊂ conv Z, and Theorem 2.21 gives y = (1 − γ −1 )u + γ −1 v ∈ rint (conv Z), in contradiction with y ∈ rbd (conv Z). The above argument holds for every choice of y in rint ∆ and the corresponding point y 0 in rint ∆0 . Clearly, y → y 0 is a one-to-one correspondence between rint ∆ and rint ∆0 . Allowing some scalars λi , µj from y = λ1 x1 + · · · + λm xm

and y 0 = µ1 x01 + · · · + µm x0m ,

take zero values, we expand this correspondence to a bijection f : ∆ → ∆0 (see Theorem 2.16). We state that f ({x1 , . . . , xm }) ⊂ {x01 , . . . , x0m }.

(3.10)

Indeed, assume for a moment that f (x1 ) ∈ / {x01 , . . . , x0m }, and express f (x1 ) 0 as a convex combination f (x1 ) = α1 x1 + · · · + αm x0m , where at least two scalars, say α1 and α2 , are not 0. Put z1 = (α1 + α2 )x01 + α3 x03 + · · · + αm x0m , z2 = (α1 + α2 )x02 + α3 x03 + · · · + αm x0m .

138


Then z1 and z2 are distinct points in ∆0 . Hence their pre-images u1 = f −1 (z1 ) and u2 = f −1 (z2 ) are distinct points in ∆. Because of α1 α2 f (x1 ) = z1 + z2 ∈ (z1 , z2 ), α1 + α2 α1 + α2 the point x1 should belong to (u1 , u2 ), as depicted below. u1 r PP PPr r r xPPP f (x1 ) PP x1 PP r r z u2 P 1 z PP r 2 PP

Let x1 = (1 − λ)u1 + λu2 , where 0 < λ < 1. Since both u1 and u2 belong to ∆, they can be expressed as convex combinations u1 = β1 x1 + · · · + βm xm

and u2 = γ1 x1 + · · · + γm xm .

Consequently, x1 = (1 − λ)u1 + λu2 = ((1 − λ)β1 + λγ1 )x1 + · · · + ((1 − λ)βm + λγm )xm . Because x1 = 1x1 + 0x2 + · · · + 0xm is the only way to express x1 as an affine combination of points x1 , . . . , xm (see Theorem 1.60), one has (1 − λ)β1 + λγ1 = 1,

(1 − λ)βi + λγi = 0,

2 6 i 6 m.

(3.11)

Solving (3.11) for βi and γi , we obtain β1 = γ1 = 1 and βi = γi = 0 for all 2 6 i 6 m. Thus u1 = u2 , contrary to the assumption. So, the inclusion (3.10) holds. This argument implies that the set Y can be partitioned into pairs {xi , x0j }, each of them collinear with x. An extension of Theorem 3.26 to the case of finite (or even compact) subsets of rint (conv X) is considered in Exercises 3.7 and 3.8. The following lemma will be of use in the next two theorems. Lemma 3.27. Let {x1 , . . . , xm } ⊂ Rn be a linearly independent set, and let y = λ1 x1 + · · · + λm xm , with exactly r nonzero scalars among λ1 , . . . , λm , 0 6 r 6 m. For negative scalars α1 , . . . , αm , there is a subset Y of the set Z = {x1 , α1 x1 , . . . , xm , αm xm } satisfying the conditions o ∈ rint (conv ({y} ∪ Y )),

card Y = 2m − r,

dim ({y} ∪ Y ) = m.

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139

Proof. Since the case r = 0 is trivial (put Y = Z), we suppose that r > 1. Renumbering the vectors x1 , . . . , xm , we assume that namely the scalars λ1 , . . . , λr are distinct from zero. Replacing, if necessary, each term λi xi with (αi λi )(αi−1 xi ) (and, respectively, xi with αi−1 xi ), one can suppose that all scalars λ1 , . . . , λr are negative. Suppose first that r = m. Dividing both parts of the equality o = y − λ1 x1 − · · · − λm xm by the positive scalar γ = 1 − λ1 − · · · − λm , we express o as a positive convex combination of the points y, x1 , . . . , xm . So, we can put Y = {x1 , . . . , xm }. Assume now that r 6 m − 1 and let x0i = αi xi , r + 1 6 i 6 m. Adding the equalities o = y − λ1 x1 − · · · − λr xr ,

o = −αi xi + x0i ,

r + 1 6 i 6 m,

we obtain o = y − (λ1 x1 + · · · + λr xr + αr+1 xr+1 + · · · + αm xm ) + x0r+1 + · · · + x0m . Dividing both parts of the latter equality by the positive scalar γ = m − r + 1 − (λ1 + · · · + λr + αr+1 + · · · + αm ), we express o as a positive convex combination of the points y, x1 , . . . , xm , x0r+1 , . . . , x0m . Put Y = {x1 , . . . , xm , x0r+1 , . . . , x0m }. From Theorem 1.60 it follows that the set {o, x1 , . . . , xm } is affinely independent. Therefore, Theorem 3.20 gives o ∈ rint (conv ({y} ∪ Y ))

and

dim ({y} ∪ Y ) = m.

The next statement complements Theorem 3.26. Theorem 3.28. For a set X ⊂ Rn of positive dimension m, the following conditions are equivalent. (1) There is a point x ∈ rint (conv X) such that every m-dimensional subset Y ⊂ X with the property x ∈ rint (conv Y ) has 2m or more points. (2) There is a point x ∈ Rn and lines l1 , . . . , lm through x such that: (a) l1 , . . . , lm are independent, (b) X ⊂ l1 ∪ · · · ∪ lm , (c) each of l1 , . . . , lm contains points of X on both sides of x.

140


Proof. (1) ⇒ (2). Let a point x ∈ Rn satisfy condition (1). Translating X on −x, we assume that x = o. By Theorem 3.26, X contains a subset Z of 2m points which can be partitioned into pairs {xi , x0i } such that o ∈ (xi , x0i ) and the lines li = hxi , x0i i are independent, 1 6 i 6 m. It remains to show that X ⊂ l1 ∪ · · · ∪ lm . For this, choose a point y ∈ X. Because the case y = o is obvious, we may assume that y 6= o. Since {x1 , . . . , xm } is a basis for span X = aff X, the vector y is uniquely expressed as a linear combination y = λ1 x1 + · · · + λm xm . Let r be the number of nonzero scalars among λ1 , . . . , λm . Clearly, r > 1 because of y 6= o. By Lemma 3.27, there is a subset Y of Z such that o ∈ rint (conv ({y} ∪ Y )),

card Y = 2m − r,

dim ({y} ∪ Y ) = m.

By the assumption, 2m 6 card ({y} ∪ Y ). Therefore, 2m − 1 6 card Y = 2m − r, which gives r 6 1. Hence r = 1 and y is a scalar multiple of a vector xi , i ∈ {1, . . . , m}; that is, y ∈ li . Summing up, X ⊂ l1 ∪ · · · ∪ lm . (2) ⇒ (1). Let a point x ∈ Rn and lines l1 , . . . , lm through x satisfy conditions (a)–(c). Choose an m-dimensional subset Y of X \ {x} with the property z ∈ rint (conv Y ), and consider the partition Y = Y1 ∪ · · · ∪ Ym ,

where

Yi = Y ∩ li ,

1 6 i 6 m.

If at least one of the sets Y1 , . . . , Ym were empty, then Y would lie in the sum of fewer than m lines l1 , . . . , lm , which is impossible because of dim Y = m. Assume for a moment that at least one of the sets Y1 , . . . , Ym , say Y1 , lies on one side of x on the line l1 . Denote by D the closed halfplane of aff (l1 ∪ · · · ∪ lm ) which is determined by the plane L = aff (l2 ∪ · · · ∪ lm ) and contains Y1 . Clearly, x ∈ L, Y ⊂ D, and Y 6⊂ L. By Theorem 2.59, rint (conv Y ) ⊂ rint D, implying that rint (conv Y ) ∩ L = ∅. The last is in contradiction with x ∈ rint (conv Y ). Hence every set Yi contains a pair of points which lie on opposite sides of x. Summing up, Y contains at least 2m points. The following proposition is analogous to Theorem 3.7 Theorem 3.29. Let X ⊂ Rn be a set of positive dimension m, and let x0 ∈ X. A point x ∈ Rn belongs to rint (conv X) if and only if there is a subset Y ⊂ X such that x ∈ rint (conv ({x0 } ∪ Y )),

m + 1 6 card Y 6 2m − 1,

aff ({x0 } ∪ Y ) = aff X.

(3.12)

Convex hulls

141

Proof. If a subset Y of X satisfies the conditions (3.12), then a combination of Theorem 3.20 and Corollary 3.21 implies the inclusion x ∈ rint (conv X). Conversely, let x ∈ rint (conv X). By Theorem 3.26, X contains a subset Z such that x ∈ rint (conv Z),

m + 1 6 card Z 6 2m,

aff Z = aff X.

Without loss of generality, we assume that Z is minimal. If card Z 6 2m − 1, then we put Y = Z \ {x0 }. Consequently, m 6 card Y 6 2m − 1

and

aff ({x0 } ∪ Y ) = aff Z = aff X

according to Theorem 1.50. Furthermore, Theorem 2.15 gives x ∈ rint (conv Z) ⊂ rint (conv ({x0 } ∪ Y )). Assume that card Z = 2m. Translating X on −x, we assume that x = o. By Theorem 3.26, Z consists of m disjoint pairs {xi , x0i } such that o ∈ (xi , x0i ) and the lines li = hxi , x0i i, 1 6 i 6 m, are independent. Excluding the trivial case x0 = o, we suppose that x0 6= o. Since {x1 , . . . , xm } is a basis for span X, the vector x0 is uniquely expressed as a linear combination x0 = λ1 x1 + · · · + λm xm . Let r be the number of nonzero scalars among λ1 , . . . , λm . Clearly, r > 1 because of x0 6= o. By Lemma 3.27, there is a subset Y of Z such that o ∈ rint (conv ({x0 } ∪ Y )), card Y = 2m − r, aff ({x0 } ∪ Y ) = aff X. Furthermore, Theorem 1.50 shows that card Y = 2m − r 6 2m − 1

and

aff ({x0 } ∪ Y ) = aff X.

By Corollary 1.76, one has m + 1 6 card ({x0 } ∪ Y ). Therefore, m 6 card Y . Exercises for Chapter 3 Exercise 3.1. Let ∆ = ∆(x1 , . . . , xr+1 ) be an r-simplex in Rn . For every proper subset X of {x1 , . . . , xr+1 }, denote by ∆1 and ∆2 the simplices with the vertex sets X and Y = {x1 , . . . , xr+1 } \ X, respectively. Show that ∆ = conv2 (∆1 ∪ ∆2 ) = ∪ ([u1 , u2 ] : u1 ∈ ∆1 , u2 ∈ ∆2 ). Exercise 3.2. Show that every closed ball Bρ (c) ⊂ Rn is the convex hull of its boundary sphere Sρ (c). Furthermore, Bρ (c) = conv2 Sρ (c).

142


Exercise 3.3. Let Q = {(x1 , . . . , xn ) : 0 6 xi 6 1, i = 1, . . . , n} be the unit cube of Rn and X be the set of points (x1 , . . . , xn ) in Q with all irrational coordinates x1 , . . . , xn . Show that conv X = conv2 X = int Q. Exercise 3.4. Let a convex set K ⊂ Rn and points x, y ∈ Rn \ K satisfy the conditions x ∈ conv (y ∪ K) and y ∈ conv (x ∪ K). Show that x = y. Exercise 3.5. (Radon [177]) Show that a set X ⊂ Rn is affinely independent if and only if it does not contain disjoint subsets whose convex hulls meet. Exercise 3.6. (Reay [179, Lemma 4.1]) Let X ⊂ Rn be a set of positive dimension m and Y be a subset of conv X of cardinality r > 2. Show the existence of a set Z ⊂ X of cardinality rm such that Y ⊂ conv Z. Exercise 3.7. (Reay [179, Lemma 4.6]) Let X ⊂ Rn be a set of positive dimension m and Y be a subset of rint (conv X) of cardinality r > 2. Show the existence of a set Z ⊂ X of cardinality (r +1)m such that Y ⊂ rint (conv Z) and aff Z = aff X. Exercise 3.8. Let X ⊂ Rn and Y be a compact subset of rint (conv X). Show the existence of a finite subset Z of X such that Y ⊂ rint (conv Z) and aff Z = aff X. Exercise 3.9. Let X1 , . . . , Xr be nonempty sets in Rn , µ1 , . . . , µr ∈ R, and x ∈ conv (µ1 X1 + · · · + µr Xr ). Let r > m = dim (µ1 X1 + · · · + µr Xr ). Show the existence of an index set I = I(x) ⊂ {1, . . . , r}, card I 6 m, and points zi ∈ Xi , i ∈ {1, . . . , r} \ I, such that P P x∈ µi zi + µi conv Xi . i∈I /

i∈I

Notes for Chapter 3 Carath´ eodory’s theorem and its variations. The concept of convex hull is originated by Carathéodory [51, 52], who defined it as the smallest convex set containing a compact set in Rn . Statement (2) of Theorem 3.6 was proved in [52, p. 200] for the case of compact sets X ⊂ Rn , using the support properties of convex sets and induction on n. The extension of Theorem 3.6 to arbitrary sets in Rn , as well as Theorems 3.3 and 3.5, are attributed to Steinitz [205, Part I, § 10]. Corollary 3.8 is proved by Reay [179, Lemma 4.1]. Cook [60] established the following analogue of Carathéodory’s theorem for the case of convex combinations with linear constraints. Baker [19] showed that if a point x belongs to the convex

Convex hulls

143

hull of a finite set X ⊂ Rn such that card X = k and dim X = m, then there are k − m or more m-simplices with vertices in X that contain x. Fenchel [82] (for n = 3) and Stoelinga [209, Theorem 25] (for all n > 3) proved that every point in the convex hull of an n-dimensional connected set can be expressed as a convex combination of n points from X. Bunt [47, p. 23] (see also Kramer [138]) further observed that connectedness of X can be replaced by the weaker condition “has n or fewer components.” Hanner and R˚ adstr¨ om [111] defined a set M ⊂ Rn as convexly connected if there is no hyperplane H ⊂ Rn such that H ∩ M = ∅ and M has points in both open halfspaces determined by H. They proved that if a set X in Rn is compact and has n or fewer convexly connected components, then every point x ∈ X can be expressed as a convex combination of n points from X. Similar results on convex hulls of special curves in Rn are given by Derry [72], Danielyan and Movsisyan [64]. Tverberg [218] proved that if X is the union of convex sets in Rn , all meeting a fixed hyperplane, then every point x ∈ conv X is a convex combination of n points from X. See Danzer, Gr¨ unbaum, and Klee [66], Eckhoff [77], and Reay [179] for additional references on Carathéodory-type theorems. Iterative construction of convex hulls. The property of convr described in Theorem 3.10 was discovered by Lepin [150] and independently by Klee (see [30]). A particular case r1 = · · · = rk = 2 and m = n of Corollary 3.11, which implies the inequality k 6 dlog2 (n + 1)e, is well-known due to Brunn [45] (also Abe, Kubota, and Yoneguchi [1], Bonnesen and Fenchel [29, p. 10], Danzer, Gr¨ unbaum, and Klee [66, pp. 116–117], and Straszewicz [212, pp. 29–36]). Topological properties of convex hulls. Theorem 3.18 is due to Bair [17]. Theorem 3.26 is proved by Steinitz [206, § 20] in terms of positive hulls. By using positive convex combinations of points, this theorem was proved by Robinson [183] for the case of convex polytopes (see also Blumenthal [27, pp. 197–198]); other proofs can be found in Bonnice and Klee [30], Gustin [108], Peterson [170], and Valentine [219, pp. 41–42]. Theorem 3.28 is due to Reay [179, p. 14]. Exercise 3.7 is a weaker version of a result of Reay [179, Lemma 4.6]: If X ⊂ Rn is a set of positive dimension m, and Y ⊂ rint (conv X) is a set of cardinality r > 2, then there is a set Z ⊂ X of cardinality rm such that Y ⊂ rint (conv Z) and aff Z = aff X. Given a set X ⊂ Rn , denote by intr X the union of relative interiors of all r-simplices contained in X; equivalently, a point x belongs to intr X if and only if there is an r-dimensional plane L ⊂ Rn such that x ∈ rint (L ∩ X). Bonnice and Klee [30] proved the following result, which becomes Carathéodory’s theorem when r = 0 and Steinitz’s theorem when r = n: For a set X ⊂ Rn , an integer r (0 6 r 6 n), and a point x ∈ intr (conv X), there is a subset Y ⊂ X such that x ∈ intr (conv Y ) and card Y 6 max {2r, n + 1}. A simplified proof of this assertion is given by Reay [180]. Furthermore, Bonnice and Reay [31] showed that if m is the highest dimension of a simplex with vertices in X that contains x in its relative interior, then r + 1 + b(r − 1)/mc 6 card Y 6 max {2r − m + 1, min {2m + 1, n + 2}}

144


for r 6 m, and r + 1 6 card Y 6 m + 1 for m 6 r − 1. Convex hull and large sums. Exercise 3.9 is a refinement of a lemma of Shapley and Folkman (see Starr [203]), which states that for every point x in conv (X1 + · · · + Xr ), where X1 , . . . , Xr are compact subset of Rn and r > n + 1, there is an index set I ⊂ {1, . . . , r} such that card I 6 n and x∈

P i∈I

conv Xi +

P

Xi .

i∈I /

Their proof was gradually simplified and expanded to the case of arbitrary sets by Artstein [6], Cassels [53], Howe [117], Starr [204], and Zhou [227]. Lawrence and Soltan [146] gave the following refinement of this statement: If X1 , . . . , Xr are nonempty sets in Rn , then for every point x ∈ conv (X1 + · · · + Xr ), there is an index set I ⊂ {1, . . . , r} with card I 6 n and nonempty subsets Yi ⊂ Xi , 1 6 i 6 r, such that x∈

P i∈I

conv Yi +

P i∈I /

Yi ,

P

card Yi 6 n + card I,

card Yi = 1, i ∈ / I.

i∈I

Shapley-Folkman’s lemma is used in an asymptotic approximation of the set conv (X1 + · · · + Xr ) by the sum of compact sets X1 , . . . , Xr ⊂ Rn , with applications in mathematical economics (see, e.g., Arrow and Hahn [5, pp. 392–400], Ekeland and Temam [80, Apendix 1], Ichiishi [119, pp. 24–25], Schneider [191, Capter 3] for additional references and results). For the case r = n + 1 and X1 = · · · = Xr = X, Shapley-Folkman’s lemma gives X + n conv X = conv X + n conv X (see also Borwein and O’Brien [33]). This equality is closely related to the function c(X) = inf {λ > 0 : X + λ conv X is convex}, studied by Schneider [189, 191]. Convex closure. The correspondence h : X → conv X can be considered as a closure mapping on the family of all subsets of Rn , since it has the following obvious properties: X ⊂ hX, hhX = hX, hX ⊂ hY if X ⊂ Y . Koenen [137] proved that hchchX = chchX, where cZ means the complement Rn \ Z of a set Z. Therefore, applying the operations h and c in any order to a given set X ⊂ Rn , one can obtain 10 or fewer distinct sets. Similar problems on semigroups generated by the mappings h, c, i = chc, and f = h∩hc are studied by Soltan [195, § 7]: if m(ϕ, η, . . . ) denotes the number of elements in the semigroup generated by mappings ϕ, η, . . . on the family of all subsets of Rn , then m(c, f ) = 6, m(f, i) = 7, m(h, i) = 5, m(f, h) = 5, m(f, h, i) = 12, m(c, f, h) = 22.

Bibliography

[1] Y. Abe, T. Kubota, H. Yoneguchi, Some properties of a set of points in Euclidean space, K¯ odai Math. Sem. Rept. 2 (1950), 117–119. [2] D. Abouabdillah, Contributions en géométrie. Monographie de cinq articles ´ de recherche en géométrie, Ecole Normale Supérieure de Rabat, 2001. [3] M. J. Adler, C. Fadiman, P. W. Goetz (eds.), Great books of the western world. No. 11 (Euclid, Archimedes, Nicomachus), Encyclopaedia Britannica, Chicago, 1990. [4] K. A. Ariyawansa, W. C. Davidon, K. D. McKennon, A characterization of convexity-preserving maps from a subset of a vector space into another vector space, J. London Math. Soc. 64 (2001), 179–190. [5] K. J. Arrow, F. H. Hahn, General competitive analysis, Holden-Day, San Francisco, CA, 1971. [6] Z. Artstein, Discrete and continuous bang-bang and facial spaces or: look for the extreme points, SIAM Rev. 22 (1980), 172–185. [7] E. Asplund, A k-extreme point is the limit of k-exposed points, Israel J. Math. 1 (1963), 161–162. [8] A. Auslender, M. Teboulle, Asymptotic cones and functions in optimization and variational inequalities, Springer, New York, 2003. [9] J. Bair, Sur les partitions convexes d’un espace vectoriel, Bull. Soc. Roy. Sci. Liège 42 (1973), 23–30. [10] J. Bair, Partitions ternaires proprement convexes d’un espace vectoriel, Bull. Soc. Roy. Sci. Liège 42 (1973), 95–103. [11] J. Bair, Sur la nature de l’attenance des éléments d’une partition convexe, Bull. Soc. Roy. Sc. Liège 42 (1973), 402–407. [12] J. Bair, Quelques compléments sur les partitions convexes finies d’un espace vectoriel, Bull. Soc. Roy. Sci. Liège 43 (1974), 20–25. [13] J. Bair, Une mise au point sur la décomposition des convexes, Bull. Soc. Roy. Sci. Liège 44 (1975), 698–705. [14] J. Bair, Une étude des sommans d’un polyedre convexe, Bull. Soc. Roy. Sci. Liège 45 (1976), 307–311. ` propos des points extrêmes de la somme de deux ensembles con[15] J. Bair, A vexes, Bull. Soc. Roy. Sci. Liège 48 (1979), 262–264.

387

388


[16] J. Bair, Some characterizations of the asymptotic cone and the lineality space of a convex set, Math. Operationsforsch. Statist. Ser. Optim. 12 (1981), 173–176. [17] J. Bair, Critères de fermeture pour l’enveloppe convexe d’une réunion, Bull. Soc. Roy. Sci. Liège 52 (1983), 72–79. ´ [18] J. Bair, R. Fourneau, Etude géometrique des espaces vectoriels: une introduction, Lecture Notes in Math. Vol. 489, Springer, Berlin, 1975. [19] M. J. C. Baker, Covering a polygon with triangles: a Carathéodory-type theorem, J. Austral. Math. Soc. Ser. A 28 (1979), 229–234. [20] M. L. Balinski, On the graph structure of convex polyhedra in n-space, Pacific J. Math. 11 (1961), 431–434. [21] R. G. Batson, Necessary and sufficient condition for boundedness of extreme points of unbounded convex sets, J. Math. Anal. Appl. 130 (1988), 365–374. [22] S. K. Berberian, Compact convex sets in innner product spaces, Amer. Math. Monthly 74 (1967), 702–705. [23] C. Bessaga, A note on universal Banach spaces of a finite dimension, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 6 (1958), 97–101. [24] R. G. Bilyeu, Metric definition of the linear structure, Proc. Amer. Math. Soc. 25 (1970), 205–206. [25] G. Bj¨ orck, The set of extreme points of a compact convex set, Arkiv Math. 3 (1958), 463–468. [26] D. Blackwell, M. A. Girshick, Theory of games and statistical decisions, Willey & Sons, New York, 1954. [27] L. M. Blumenthal, Theory and applications of distance geometry, Clarendon Press, Oxford, 1953. ¨ [28] G. Bol, Uber Eik¨ orper mit Vieleckschatten, Math. Z. 48 (1942), 227–246. [29] T. Bonnesen, W. Fenchel, Theorie der konvexen K¨ orper, Springer, Berlin, 1934. English translation: Theory of convex bodies, BCS Associates, Moscow, ID, 1987. [30] W. E. Bonnice, V. L. Klee, The generation of convex hulls, Math. Ann. 152 (1963), 1–29. [31] W. E. Bonnice, J. R. Reay, Relative interiors of convex hulls, Proc. Amer. Math. Soc. 20 (1969), 246–250. [32] V. A. Borovikov, On the intersection of a sequence of simplices, (Russian) Uspehi Matem. Nauk 7 (1952), 179–180. [33] J. M. Borwein, R. C. O’Brien, Cancellation characterizizes convexity, Nanta Math. 11 (1978), 100–102. [34] N. Bourbaki, Topological vector spaces. Chapters 1–5. Elements of Mathematics, Springer, Berlin, 1987. [35] M. Breen, Admissible kernels for starshaped sets, Proc. Amer. Math. Soc. 82 (1981), 622–628. [36] T. Bromek, J. Kaniewski, Linear images and sums of closed cones in Euclidean spaces, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 24 (1976), 231–238. [37] E. M. Bronshte˘ın, Extremal boundaries of finite-dimensional convex compacta, (Russian) Optimizatsiya No. 26 (1981), 119–128.

Bibliography

389

[38] E. M. Bronshte˘ın, One-dimensional extremal boundaries of three-dimensional compact convex sets, (Russian) Optimizatsiya, No. 51 (1992), 34–46. [39] E. M. Bronshte˘ın, On convex compacta with given extreme points, Russian Acad. Sci. Dokl. Math. 49 (1994), 497–500. [40] E. M. Bronshte˘ın, On compact convex sets with given extreme points, (Russian) Sibirsk. Mat. Zh. 36 (1995), 20–27. [41] A. Brøndsted, An introduction to convex polytopes, Springer, New York, 1983. [42] A. L. Brown, Chebyshev sets and facial systems of convex sets in finite-dimensional spaces, Proc. London Math. Soc. 41 (1980), 297–339. ¨ [43] H. Brunn, Uber Ovale und Eifl¨ achen, Inaugural Dissertation, Straub, M¨ unchen, 1887. ¨ [44] H. Brunn, Uber Kurven ohne Wendepunkte, Habilitationsschrift, Ackermann, M¨ unchen, 1889. ¨ [45] H. Brunn, Uber das durch eine belibige endliche Figur bestimmte Eigebilde, In: Ludwig Boltzmann Festschrift, pp. 94–104, Barth, Leipzig, 1904. ¨ [46] H. Brunn, Uber Kerneigebiete, Math. Ann. 73 (1913), 436–440. [47] L. N. H. Bunt, Bijdrage tot de theorie der convexe puntverzamelingen, (Dutch) Proefschrifft Groningen, Noord-Hollandsche Uitgevers Maatschappij, Amsterdam, 1934. [48] Yu. D. Burago, V. A. Zalgaller, Sufficient conditions for convexity, J. Soviet Math. 16 (1978), 395–434. [49] G. R. Burton, Sections of convex bodies, J. London Math. Soc. 12 (1976) 331–336. [50] G. R. Burton, P. Mani, A characterization of the ellipsoid in terms of concurrent sections, Comment. Math. Helv. 53 (1978) 485–507. ¨ [51] C. Carathéodory, Uber den Variabilit¨ atsbereich der Koefficienten von Potenzreihen, die gegebene Werte nicht annehmen, Math. Ann. 64 (1907), 95–115. ¨ [52] C. Carathéodory, Uber den Variabilit¨ atsbereich der Fourierschen Konstanten von positiven harmonischen Funktionen, Rend. Circ. Mat. Palermo 32 (1911), 193–217. [53] J. W. S. Cassels, Measures of the non-convexity of sets and the ShapleyFolkman-Starr theorem, Math. Proc. Cambridge Philos. Soc. 78 (1975), 433–436. ˇ [54] S. N. Cernikov, Theorems on separation of convex polyhedral sets, Soviet Math. Dokl. 2 (1961), 838–840. [55] G. Choquet, Unicité des représentations intégrales au moyen de points extrémaux dans les cˆ ones convexes réticulés, C. R. Acad. Sci. Paris 243 (1956), 555–557. [56] G. Choquet, H. H. Corson, V. L. Klee, Exposed points of convex sets, Pacif. J. Math. 17 (1966), 33–43. [57] A. Chubarev, I. Pinelis, Fundamental theorem of geometry without the 1-to1 assumption, Proc. Amer. Math. Soc. 127 (1999), 2735–2744. [58] J. B. Collier, On the set of extreme points of a convex body, Proc. Amer. Math. Soc. 47 (1975), 184–186.

390


[59] J. B. Collier, On the facial structure of a convex body, Proc. Amer. Math. Soc. 61 (1976), 367–370. [60] W. D. Cook, Carathéodory’s theorem with linear constraints, Canad. Math. Bull. 17 (1974), 189–191. [61] W. D. Cook, R. J. Webster, Carathéodory’s theorem, Canad. Math. Bull. 15 (1972), 293. [62] H. H. Corson, A compact convex set in E 3 whose exposed points are of the first category, Proc. Amer. Math. Soc. 16 (1965), 1015–1021. [63] L. Dalla, On the measure of the one-skeleton of the sum of convex compact sets, J. Austral. Math. Soc. Ser. A. 42 (1987), 385–389. [64] E. A. Danielyan, G. S. Movsisyan, Sharpening the Carathéodory-Riesz theorem, (Russian) Akad. Nauk Armyan. SSR Dokl. 89 (1989), 105–108. [65] E. A. Danielyan, G. S. Movsisyan, K. R. Tatalyan, Generalization of the Carathéodory theorem, (Russian) Akad. Nauk Armenii Dokl. 92 (1991), 69–75. [66] L. Danzer, B. Gr¨ unbaum, V. L. Klee, Helly’s theorem and its relatives, In: Convexity. Proc. Sympos. Pure Math. Vol. VII, pp. 101–179, Amer. Math. Soc., Providence, RI, 1963. [67] C. Davis, Theory of positive linear dependence, Amer. J. Math. 76 (1954), 733–746. [68] C. Davis, Remarks on a previous paper, Michigan Math. J. 2 (1954), 23–25. [69] A. Dax, The distance between two convex sets, Linear Algebra Appl. 416 (2006), 184–213. [70] M. De Wilde, Some properties of the exposed points of finite dimensional convex sets, J. Math. Anal. Appl. 99 (1984), 257–264. [71] G. Debs, Applications affines ouvertes et convexes compacts stables, Bull. Sci. Math. 102 (1978), 401–414. [72] D. Derry, Convex hulls of simple space curves, Canad. J. Math. 8 (1956), 383–388. [73] R. Deumlich, K.-H. Elster, R. Nehse, Recent results on separation of convex sets, Math. Operationsforsch. Statist. Ser. Optim. 9 (1978), 273–296. [74] J. Dieudonné, Sur la séparation des ensembles convexes, Math. Ann. 163 (1966), 1–3. [75] M. Dragomirescu, A strong separation theorem with a side-condition, Rev. Roumaine Math. Pures Appl. 39 (1994), 27–35. [76] L. E. Dubins, On extreme points of convex sets, J. Math. Anal. Appl. 5 (1962), 237–244. [77] J. Eckhoff, Helly, Radon, and Carathéodory type theorems, In: Handbook of convex geometry. Vol. A, pp. 389–448, North-Holland, Amsterdam, 1993. [78] H. G. Eggleston, B. Gr¨ unbaum, V. L. Klee, Some semicontinuity theorems for convex polytopes and cell-complexes, Comment. Math. Helv. 39 (1964), 165–188. [79] H. G. Eggleston, Intersection of convex sets, J. London Math. Soc. 5 (1972), 753–754. [80] I. Ekeland, R. Temam, Analyse convexe et problèmes variationnels, Gauthier-Villars, Montreal, 1974. [81] G. V. Epifanov, Universality of sections of cubes, (Russian) Mat. Zametki

Bibliography

391

2 (1967), 93–95. ¨ [82] W. Fenchel, Uber Kr¨ ummung und Windung geschlossener Raumkurven, Math. Ann. 101 (1929), 238–252. [83] W. Fenchel, A remark on convex sets and polarity, Comm. Sém. Math. Univ. Lund. Tome Suppl. (1952), 82–89. [84] W. Fenchel, Convex cones, sets, and functions, Mimeographed lecture notes. Spring Term 1951, Princeton University, 1953. [85] W. Fenchel, Convexity through the ages, In: Convexity and its applications, pp. 120–130, Birkh¨ auser, Basel, 1983. [86] T. Fischer, Strong unicity and alternation for linear optimization, J. Optim. Theory Appl. 69 (1991), 251–267. [87] M. Florenzano, C. Le Van, Finite dimensional convexity and optimization, Springer, Berlin, 2001. [88] J. Frenkel, Géométrie pour l’élève-professur, Hermann, Paris, 1973. ¨ [89] M. Fujiwara, Uber den Mittelk¨ orper zweier konvexer K¨ orper, Sci. Rep. Tˆ ohoku Univ. 5 (1916), 275–283. [90] D. Gale, Linear combination of vectors with non-negative coefficients (solution of problem 4395), Amer. Math. Monthly 59 (1952), 46–47. [91] S. Gallivan, R. J. Gardner, A counterexample to a “simplex algorithm” for convex bodies, Geom. Dedicata 11 (1981), 475–488. [92] S. Gallivan, D. G. Larman, Further results on increasing paths in the oneskeleton of a convex body, Geom. Dedicata 11 (1981), 19–29. [93] R. J. Gardner, Geometric tomography, Cambridge University Press, New York, 1995. [94] M. A. Goberna, V. Jornet, M. M. L. Rodr´ıguez, On the characterization of some families of closed convex sets, Beitr¨ age Algebra Geom. 43 (2002), 153–169. [95] M. A. Goberna, V. Jornet, M. M. L. Rodr´ıguez, On linear systems containing strict inequalities, Linear Algebra Appl. 360 (2003), 151–171. [96] M. A. Goberna, M. M. L. Rodr´ıguez, Analyzing linear systems containing strict inequalities via evenly convex hulls, European J. Oper. Res. 169 (2006), 1079–1095. [97] P. Goossens, Hyperbolic sets and asymptotes, J. Math. Anal. Appl. 116 (1986), 604–618. ¨ [98] P. M. Gruber, Uber den Durchscnitt einer abnehmenden Folge von Parallelepipeden, Elem. Math. 32 (1977), 13–15. [99] P. M. Gruber, Durchsnitte und Vereinigungen monotoner Folgen spezieller konvexer K¨ orper, Abh. Math. Sem. Univ. Hamburg 49 (1979), 189–197. [100] P. M. Gruber, Zur Geschichte der Konvexgeometrie und der Geometrie der Zahlen, In: Ein Jahrhundert Mathematik 1890–1990, pp. 421–455, Deutsche Mathematiker Vereinigung, Freiburg, 1990. [101] P. M. Gruber, History of convexity, In: Handbook of convex geometry. Vol. A, pp. 1–15, North-Holland, Amsterdam, 1993. [102] P. M. Gruber, Convex and discrete geometry, Springer, Berlin, 2007. [103] B. Gr¨ unbaum, On a problem of S. Mazur, Bull. Res. Council Israel. Sect. F 7 (1957/1958), 133–135.

392


[104] B. Gr¨ unbaum, Convex polytopes, Springer, New York, 2003. [105] R. Grz¸a´slewicz, A universal convex set in Euclidean space, Colloq. Math. 45 (1981), 41–44. [106] J. Grzybowski, R. Urba´ nski, Affine straight lines in family of bounded closed convex sets, Rend. Circ. Mat. Palermo 53 (2004), 225–230. [107] O. G¨ uler, Foundations of optimization, Springer, New York, 2010. [108] W. Gustin, On the interior of the convex hull of a Euclidean set, Bull. Amer. Math. Soc. 53 (1947), 299–301. [109] P. C. Hammer, Maximal convex sets, Duke Math. J. 22 (1955), 103–106. [110] P. C. Hammer, Semispaces and the topology of convexity, In: Convexity. Proc. Sympos. Pure Math. Vol. VII, pp. 305–316, Amer. Math. Soc., Providence, RI, 1963. [111] O. Hanner, H. R˚ adstr¨ om, A generalization of a theorem of Fenchel, Proc. Amer. Math. Soc. 2 (1951), 589–593. [112] W. Hansen, V. L. Klee, Interesection theorems for positive sets, Proc. Amer. Math. Soc. 22 (1969), 450–457. [113] Y. He, J. Sun, Minimum recession-compatible subsets of closed convex sets, J. Global Optim. 52 (2012), 253–263. [114] J.-B. Hiriart-Urruty, C. Lemaréchal, Fundamentals of convex analysis, Springer, Berlin, 2001. [115] A. J. Hoffman, On the covering of polyhedra by polyhedra, Proc. Amer. Math. Soc. 23 (1969), 123–126. [116] P. Holick´ y, M. Laczkovich, Descriptive properties of the set of exposed points of compact convex sets in R3 , Proc. Amer. Math. Soc. 132 (2004), 3345– 3347. [117] R. Howe, On the tendency toward convexity of the vector sum of sets, Cowles Foundation Discussion Papers, No. 538, Yale University, CN, 1979. [118] T. Husain, I. Tweddle, On the extreme points of the sum of two convex sets, Math. Ann. 188 (1970), 113–122. [119] T. Ichiishi, Game theory for economic analysis, Academic Press, New York, 1983. [120] R. E. Jamison, Contractions of convex sets, Proc. Amer. Math. Soc. 62 (1976), 129–130. [121] B. Jessen, Zwei Z¨ atze u ¨ber Konvexe Punktmengen, (Danish) Mat. Didsskr. B (1940), 66–70. [122] A. Kirsch, Konvexe Figuren als Durchschnitte abz¨ ahlbar vieler Halbr¨ aume, Arch. Math. (Basel) 18 (1967), 313–319. [123] V. L. Klee, Convex sets in linear spaces, Duke Math. J. 18 (1951), 443–466. [124] V. L. Klee, Extremal structure of convex sets, Arch. Math. 8 (1957), 234– 240. [125] V. L. Klee, Extremal structure of convex sets. II, Math. Z. 69 (1958), 90– 104. [126] V. L. Klee, Some characterizations of convex polyhedra, Acta Math. 102 (1959), 79–107. [127] V. L. Klee, Polyhedral sections of convex bodies, Acta Math. 105 (1960), 243–267.

Bibliography

393

[128] V. L. Klee, Asymptotes and projections of convex sets, Math. Scand. 8 (1960), 356–362. [129] V. L. Klee, The generation of affine hulls, Acta Sci. Math. (Szeged) 24 (1963), 60–81. [130] V. L. Klee, On a theorem of Dubins, J. Math. Anal. Appl. 7 (1963), 425–427. [131] V. L. Klee, A theorem on convex kernels, Mathematika 12 (1965), 89–93. [132] V. L. Klee, Asymptotes of convex bodies, Math. Scand. 20 (1967), 89–90. [133] V. L. Klee, Separation and support properties of convex sets–a survey, In: Control theory and the calculus of variations, pp. 235–303, Academic Press, New York, 1969. [134] V. L. Klee, Unions of increasing and intersections of decreasing sequences of convex sets, Israel J. Math. 12 (1972), 70–78. [135] V. L. Klee, E. Maluta, C. Zanco, Basic properties of evenly convex sets, J. Convex Analysis 14 (2007), 137–148. [136] V. L. Klee, M. Martin, Semicontinuity of the face-function of a convex set, Comment. Math. Helv. 46 (1971), 1–12. [137] W. Koenen, The Kuratowski closure problem in the topology of convexity, Amer. Math. Monthly 73 (1966), 704–708. ¨ [138] H. Kramer, Uber einen Satz von Fenchel, Mathematica (Cluj) 9 (1967), 281–283. [139] A. V. Kuz0 minyh, Affineness of convex-invariant mappings, Siberian Math. J. 16 (1975), 918–922. [140] D. G. Larman, The convex borel sets in R3 are convexly generated, J. London Math. Soc. 4 (1971), 5–14. [141] D. G. Larman, The skeleta of convex bodies, in: Convex and fractal geometry, pp. 57–62, Polish Acad. Sci. Inst. Math., Warsaw, 2009. [142] D. G. Larman, C. A. Rogers, Paths in the one-skeleton of a convex body, Mathematika 17 (1970), 293–314. [143] D. G. Larman, C. A. Rogers, Increasing paths on the one-skeleton of a convex body and the directions of segments on the boundary of a convex body, Proc. London Math. Soc. 23 (1971), 683–694. [144] M. Lassak, Convex half-spaces, Fund. Math. 120 (1984), 7–13. [145] J. Lawrence, Intersections of descending sequences of affinely equivalent convex bodies, Beitr. Algebra Geom. 54 (2013), 669–676. [146] J. Lawrence, V. Soltan, Carathéodory-type results for the sums and unions of convex sets, Rocky Mountain J. Math. 43 (2013), 1675–1688. [147] S. R. Lay, Convex sets and their applications, John Wiley & Sons, New York, 1982. [148] K. Leichtweiß, Konvexe Mengen, Springer, Berlin, 1980. [149] H. Lenz, Einige Anwendungen der projektiven Geometrie auf Fragen der Fl¨ achentheorie, Math. Nachr. 18 (1958), 346–359. [150] A. A. Lepin, On a method of forming the convex hull, (Russian) Uˇcen. Zap. Belorussk. Univ. Ser. Mat. (1959), No. 1, 27–30. ¨ [151] K. Lommatzsch, Uber Extremalmengen und St¨ utzhyperebenen konvexer Mengen, Wiss. Z. Humboldt-Univ. Berlin. Math.-naturwiss. R. 30 (1981), 411–413.

394


[152] P. Mani-Levitska, Characterizations of convex sets, In: Handbook of convex geometry. Vol. A, pp. 19–41, North-Holland, Amsterdam, 1993. [153] J.-E. Mart´ınez-Legaz, I. Singer, The structure of hemispaces in Rn , Linear Algebra Appl. 110 (1988), 117–179. [154] R. D. Mauldin (ed.), The Scottish book, Birk¨ auser, Boston, 1981. [155] R. L. McKinney, Positive bases for linear spaces, Trans. Amer. Math. Soc. 103 (1962), 131–148. [156] W. Meyer, D. C. Kay, A convexity structure admits but one real linearization of dimension greater than one, J. London Math. Soc. 7 (1973), 124–130. [157] H. Minkowski, Geometrie der Zahlen, Teubner, Leipzig, 1896. [158] H. Minkowski, Gesammelte Abhandlungen. Bd 2, Teubner, Leipzig, 1911. [159] H. Mirkil, New characterizations of polyhedral cones, Canad. J. Math. 9 (1957), 1–4. [160] J.-J. Moreau, Décomposition orthogonale d’un espace hilbertien selon deux cˆ ones mutuellement polaires, C. R. Acad. Sci. Paris 255 (1962), 238–240. [161] T. Motzkin, Beitr¨ age zur Theorie der linearen Ungleichungen, Ph.D. Thesis, Basel, 1933 (Completed in 1936, Azriel Printers, Jerusalem). English translation: Contributions to the theory of linear inequalities, Selected Papers, pp. 1–80, Birkh¨ auser, Boston, 1983. [162] T. Motzkin, Sur quelques propriétés caractéristiques des ensembles convexes, Rend. Reale Accad. Lincei, Classe Sci. Fis., Mat. Nat. 21 (1935), 562–567. ¨ [163] S. Nakajima, Uber konvexe Kurven und Fl¨ achen, Tˆ ohoku Math. J. 29 (1928), 227–230. [164] H. Naumann, Beliebige konvexe Polytope als Schnitte und Projektionen h¨ oherdimensionaler W¨ urfel, Simplizes und Masspolytope, Math. Z. 65 (1956), 91–103. [165] J. W. Nieuwenhuis, About separation by cones, J. Optim. Theory Appl. 41 (1983), 473–479. [166] G. H. Orland, On a class of operators, Proc. Amer. Math. Soc. 15 (1964), 75–79. [167] Z. P´ ales, Separation theorems for convex sets and convex functions with invariance properties, Lecture Notes in Econom. and Math. Systems, Vol. 502, pp. 279–293, Springer, Berlin, 2001. [168] S. Papadopoulou, On the geometry of stable compact convex sets, Math. Ann. 229 (1977), 193–200. [169] S. Papadopoulou, Stabile konvexe Mengen, Jahresber. Deutsch. Math.Verein. 84 (1982), 92–106. [170] B. B. Peterson, The geometry of Radon’s theorem, Amer. Math. Monthly 79 (1972) 949–963. [171] E.-A. Pforr, Zum Bewis des Carathéodory-Theorems f¨ ur Kegel, Wiss. Z. Techn. Univ. Dresden 27 (1978), 931–933. [172] R. R. Phelps, Lectures on Choquet’s theorem, Van Nostrand, Princeton, 1966. [173] K. A. Post, Star extension of plane convex sets, Indag. Math. 26 (1964), 330–338.

Bibliography

395

[174] W. Pranger, Extreme points of convex sets, Math. Ann. 205 (1973), 299– 302. [175] D. Preiss, The convex generation of convex Borel sets in Banach spaces, Mathematika 20 (1973), 1–3. [176] G. B. Price, On the extreme points of convex sets, Duke Math. J. 3 (1937), 56–67. [177] J. Radon, Mengen konvexen K¨ orper, die einen gemeinamen punkt enhalten, Math. Ann. 83 (1921), 113–115. [178] H. R˚ adstr¨ om, An embedding theorem for spaces of convex sets, Proc. Amer. Math. Soc. 3 (1952), 165–169. [179] J. R. Reay, Generalizations of a theorem of Catathéodory, Mem. Amer. Math. Soc. No. 54 (1965). [180] J. R. Reay, A new proof of the Bonnice-Klee theorem, Proc. Amer. Math. Soc. 16 (1965), 585–587. [181] J. R. Reay, Unique minimal representations with positive bases, Amer. Math. Monthly 73 (1966), 253–261. [182] H. B. Reiter, N. M. Stavrakas, On the compactness of the hyperspace of faces, Pacific J. Math. 73 (1977), 193–196. [183] C. V. Robinson, Spherical theorems of Helly type and congruence indices of spherical caps, Amer. J. Math. 64 (1942), 260–272. [184] R. T. Rockafellar, Convex analysis, Princeton Universty Press, Princeton, NJ, 1970. [185] C. A. Rogers, Sections and projections of convex bodies, Portugal. Math. 24 (1965), 99–103. [186] C. A. Rogers, G. C. Shephard, The difference body of a convex body, Arch. Math. (Basel) 8 (1957), 220–233. [187] A. K. Roy, Facial structure of the sum of two compact convex sets, Math. Ann. 197 (1972), 189–196. [188] H. Rubin, O. Wesler, A note on convexity in euclidean n-space, Proc. Amer. Math. Soc. 9 (1958), 522–523. [189] R. Schneider, A measure of convexity for compact sets, Pacific J. Math. 58 (1975), 617–626. [190] R. Schneider, Boundary structure and curvature of convex bodies, In: Contributions to geometry, pp. 13–59, Birkh¨ auser, Basel, 1979. [191] R. Schneider, Convex bodies: the Brunn-Minkowski theory, Second expanded edition, Cambridge University Press, Cambridge, 2014. [192] A. V. Sha˘ıdenko-K¨ unzi, On mappings preserving convexity, Amer. Math. Soc. Transl. Ser. 2, Vol. 163, pp. 155–163, Amer. Math. Soc., Providence, RI, 1995. [193] C. R. Smith, A characterization of star-shaped sets, Amer. Math. Monthly 75 (1968), 368. [194] P. Soltan, A. Vasiloi, Certain characterizations of a convex cone, (Russian) Mat. Issled. 6 (1971), 155–164. [195] V. Soltan, Introduction to the axiomatic theory of convexity, (Russian) S ¸ tiint¸a, Chi¸sin˘ au, 1984. [196] V. Soltan, A characterization of homothetic simplices, Discrete Comput.

396


Geom. 22 (1999), 193–200. [197] V. Soltan, Choquet simplexes in finite dimension–a survey, Expo. Math. 22 (2004), 301–315. [198] V. Soltan Affine diameters of convex-bodies—a survey, Expo. Math. 23 (2005), 47–63. [199] V. Soltan, Sections and projections of homothetic convex bodies, J. Geom. 84 (2005), 152–163. [200] V. Soltan, Addition and subtraction of homothety classes of convex sets, Beitr¨ age Algebra Geom. 47 (2006), 351–361. [201] V. Soltan, Convex solids with homothetic sections through given points, J. Convex Anal. 16 (2009), 473–486. [202] V. Soltan, Convex sets with homothetic projections, Beitr¨ age Algebra Geom. 51 (2010), 237–249. [203] R. M. Starr, Quasi-equilibria in markets with non-convex preferences, Econometrica 37 (1969), 25–38. [204] R. M. Starr, Approximation of points of convex hull of a sum of sets by points of the sum: an elementary approach, J. Econom. Theory 25 (1981), 314–317. [205] E. Steinitz, Bedingt konvergente Reihen und konvexe Systeme. I–IV, J. Reine Angew. Math. 143 (1913), 128–175. [206] E. Steinitz, Bedingt konvergente Reihen und konvexe Systeme. V, J. Reine Angew. Math. 144 (1914), 1–40. [207] E. Steinitz, Bedingt konvergente Reihen und konvexe Systeme. VI–VII, J. Reine Angew. Math. 146 (1916), 1–52. [208] E. Steinitz, Polyeder und Raumeinteilungen, in: Encyklop¨ adie der Mathematischen Wissenschaften, Bande III, 1. Teil, IIIAB 12, pp. 1–139, Teubner Verlag, Leipzig, 1922. [209] T. G. D. Stoelinga, Convexe Puntverzemelingen, (Dutch) Proefschrifft Groningen, Amsterdam, Noordhollandsche Uitgeversmaatschappij, 1932. [210] J. Stoer, C. Witzgall, Convexity and optimization in finite dimensions. I, Springer, Berlin, 1970. [211] J. J. Stoker, Unbounded convex sets, Amer. J. Math. 62 (1940), 165–179. [212] S. Straszewicz, Beitr¨ age zur Theorie der konvexen Punktmengen, Inaugural Dissertation, Meier, Z¨ urich, 1914. [213] C. H. Sung, B. S. Tam, On the cone of a finite dimensional compact convex set at a point, Linear Algebra Appl. 90 (1987), 47–55. [214] T. Tamura, Remarks on the convexity of connected sets, J. Gakugei Tokushima Univ. 3 (1953), 24–27. ¨ [215] H. Tietze, Uber Konvexheit im kleinen und im grossen und u ¨ber gewisse den Punkten einer Menge zugeordnete Dimensionszahlen, Math. Z. 28 (1928), 697–707. [216] F. A. Toranzos, Radial functions of convex and star-shaped bodies, Amer. Math. Monthly 74 (1967), 278–280. [217] J. W. Tukey, Some notes on the separation of convex sets, Portugal. Math. 3 (1942), 95–102. [218] H. Tverberg, A separation property of plane convex sets, Math. Scand. 45

Bibliography

397

(1979), 255–260. [219] F. A. Valentine, Convex sets, McGraw-Hill, New York, 1964. [220] O. Veblen, J. H. C. Whitehead, The foundations of differential geometry, Cambridge University Press, Cambridge, 1932. [221] J. L. Walsh, On the transformation of convex point sets, Ann. Math. 22 (1921), 262–266. [222] R. J. Webster, Convexity, Oxford University Press, Oxford, 1994. [223] H. Weyl, Elementare Theorie der konvexen Polyeder, Comment. Math. Helv. 7 (1935), 290–306. English translation: The elementary theory of convex polyhedra, In: Contributions to the theory of games. Vol. I, pp. 3– 18, Princeton University Press, Princeton, NJ, 1950. [224] D. Wolfe, Metric inequalities and convexity, Proc. Amer. Math. Soc. 40 (1973), 559–562. [225] K. D. Yaksubaev, Construction of a linear closed prototype of convex sets of type Fσ , (Russian) Dokl. Akad. Nauk UzSSR No. 12 (1983), 4–6. [226] T. Zamfirescu, Minkowski’s theorem for arbitrary convex sets, European J. Combin. 29 (2008), 1956–1958. [227] L. Zhou, A simple proof of the Shapley-Folkman theorem, Econom. Theory 3 (1993), 371–372. [228] G. M. Ziegler, Lectures on polytopes, Springer, New York, 1995. [229] V. Zizler, Note on separation of convex sets, Czechoslovak Math. J. 21 (1971), 340–343.