Polarity and Separation of Cones Valeriu Soltan Department of Mathematical Sciences, George Mason University 4400 University Drive, Fairfax, VA 22030, USA,
[email protected]
Abstract Given a closed convex cone C ⊂ Rn and its polar cone C ◦ , properties of the set C ∩(−C ◦ ) are studied. In particular, we solve a problem of Stoker concerning nonemptiness of rint C ∩ (−rint C ◦ ). Based on these properties, new results on separation of C and C ◦ by hyperplanes are established. MSC: 90C25, 52A20, 15A39 Key words: Cone, convex, polar, hyperplane, separation
1. Introduction We recall that a nonempty set C in the n-dimensional Euclidean space Rn is a cone if λx ∈ C whenever λ > 0 and x ∈ C. (Obviously, this definition implies that the origin o of Rn belongs to C, although a stronger condition λ > 0 can be beneficial; see, e. g., [6]). The cone C is called convex if it is a convex set. In a standard way, the (negative) polar cone C ◦ of C is defined by C ◦ = {e ∈ Rn : x·e 6 0 for all x ∈ C}, where x·e means the dot (scalar) product of vectors x and e. Despite a wide usage of polar cones C and C ◦ in various mathematical disciplines, not much is known about the sets D = C ∩ (−C ◦ )
and E = rint C ∩ (−rint C ◦ ),
where rint C denotes the relative interior of C (we observe that the set −C ◦ is often called the dual cone of C). Blumenthal [1] and Dines [2] showed that the existence of positive solutions of certain systems of homogenous linear inequalities can be geometrically formulated as the property D 6= {o} of a suitable convex cone C ⊂ Rn . In this regard Gaddum [4], using a simple argument, proved that D 6= {o} if and only if the cone C is not a subspace. Independently, Stoker [9] asked whether the set E is nonempty and gave a partial affirmative answer by proving that rint C ∩ (−C ◦ ) 6= ∅ for the case when C is a closed convex cone in R3 , distinct from a subspace. Another motivation for the study of the sets D and E comes from separation theory. For instance, an assertion of Klee [5] on the existence of a hyperplane specially separating closed convex cones C1 and C2 in Rn Preprint submitted to Linear Algebra and Its Applications
November 1, 2017
satisfying the condition C1 ∩ C2 = {o} can be equivalently reformulated as rint C1◦ ∩ (−rint C2◦ ) 6= ∅ (see Theorem 4.2 below). In this paper, we determine the dimensions of the sets D and E and prove that rint D = E, as shown in Theorems 3.1 and 3.2. These results allow us to answer affirmatively Stoker’s question on the nonemptiness of E (see Corollary 3.1). The concluding Section 4 contains new assertions on separation of arbitrary convex cones C1 and C2 in Rn (which refine a result of Klee [5]), and, in particular, on separation of cones C and C ◦ (see Theorems 4.1–4.3 and their corollaries) 2. Notation, Terminology, and Preliminaries We follow standard notation and terminology of finite-dimensional convex analysis (see, for instance, the books [7] and [8]). In particular, cl F , dim F , rbd F , rint F , and span F stand, respectively, for the closure, dimension, relative boundary, relative interior, and span of a convex set F ⊂ Rn . For a simplicity of language, we will be dealing with closed convex cones in Rn . Indeed, the obtained results can be easily rewritten for the case of any convex cones, based on the equalities rint C = rint (cl C) and C ◦ = (cl C)◦ . Given a closed convex cone C ⊂ Rn , the set lin C = C ∩ (−C) is called the lineality space of C, and C is called pointed provided lin C = {o}. It is known that lin C is the largest subspace contained in C and C = C + lin C (see, e. g., [8], Theorems 4.14 and 4.15). Obviously, C 6= lin C if and only if C is not a subspace. Equivalently, the number s(C) = dim C − dim (lin C)
(1)
is positive if and only if C is not a subspace. We will need some known results about cones (given below as auxiliary propositions without proof). The first one holds for any closed convex sets; its first part is due to Fenchel [3, p. 44] (also see Rockafellar [7, p. 73]), while the second part can be found in [8, Corollary 5.30]. Proposition 2.1. Let C ⊂ Rn be a closed convex cone, and let L ⊂ Rn be the orthogonal complement of lin C in Rn . With S = L ∩ span C, the cone C can be expressed as the direct sum C = (C ∩ S) + lin C, (2) where C ∩ S is a closed pointed convex cone such that span (C ∩ S) = S. Furthermore, rint C = (rint C ∩ S) + lin C = rint (C ∩ S) + lin C.
(3)
The next two propositions are due to Rockafellar [7] (see, respectively, Theorem 6.5 and Corollary 9.1.2). Proposition 2.2. If closed convex cones C1 and C2 in Rn satisfy the condition rint C1 ∩ rint C2 6= ∅, then rint (C1 ∩ C2 ) = rint C1 ∩ rint C2 . 2
Proposition 2.3. If closed convex cones C1 and C2 in Rn satisfy the condition C1 ∩ C2 = lin C1 ∩ lin C2 , then the cone C1 + C2 is closed. Consequently, if both cones C1 and C2 are pointed and C1 ∩ C2 = {o}, then the cone C1 + C2 is closed and pointed. One more proposition collects some known properties of polar cones (here X ⊥ denotes the orthogonal complement of a nonempty set X ⊂ Rn ). Proposition 2.4. For a closed convex cone C ⊂ Rn , the following assertions hold. C ◦ is a closed convex cone such that C ∩ C ◦ = {o} and (C ◦ )◦ = C. C ◦ is a subspace if and only if C is a subspace. lin C ∈ rbd C (equivalently, lin C ◦ ∈ rbd C ◦ ) if and only if C is not a subspace. span C = (lin C ◦ )⊥ and span C ◦ = (lin C)⊥ . Consequently, lin C ◦ and lin C are orthogonal subspaces, and C ◦ ∩ lin C = C ∩ lin C ◦ = {o}. 5) dim C ◦ = n if and only if C is pointed. 6) Assuming C is not a subspace, a nonzero vector e ∈ Rn belongs to rint C ◦ if and only if x·e < 0 for all x ∈ C \ lin C (see, e. g., Fenchel [3, page 18]).
1) 2) 3) 4)
Also, we will need the following known identities regarding closed convex cones C1 and C2 in Rn : (C1 + C2 )◦ = C1◦ ∩ C2◦
and
(C1 ∩ C2 )◦ = cl (C1◦ + C2◦ ).
(4)
3. Cone Polarity Our approach is based on considering the subspace S from Proposition 2.1, which we will call the orthogonal complement of lin C within span C. Since span C is the sum of orthogonal subspaces lin C and S, the dimension of S equals the number s(C), defined by (1). Lemma 3.1 below shows that S may be considered as a common “axis” in decomposition of C and C ◦ . Lemma 3.1. Let C ⊂ Rn be a closed convex cone, and S be the orthogonal complement of lin C within span C. Then S is the orthogonal complement of lin C ◦ within span C ◦ . Consequently, dim S = dim C ◦ − dim (lin C ◦ ) = s(C ◦ ). (5) Furthermore, C ◦ = (C ◦ ∩ S) + lin C ◦ ,
(6)
◦
and the conclusions of Proposition 2.1 hold for C . Proof. Proposition 2.4 and the choice of S imply that Rn is the sum of pairwise orthogonal subspaces span C, lin C ◦ , and S: Rn = span C + lin C ◦ = lin C + S + lin C ◦ .
(7)
By the same proposition, Rn = span C ◦ + lin C. 3
(8)
Comparing (7) and (8), we see that span C ◦ is the sum of orthogonal subspaces lin C ◦ and S: span C ◦ = lin C ◦ + S. Hence S is the orthogonal complement of lin C ◦ within span C ◦ . The remaining part of the lemma follows from Proposition 2.1. The example below illustrates Lemma 3.1. Example 3.1. For the 2-dimensional closed convex cone C = {(x, y, 0) : x > 0} in R3 , its polar cone is another 2-dimensional closed convex cone, given by C ◦ = {(x, 0, z) : x 6 0}. The lineality spaces lin C and lin C ◦ are, respectively, the y- and z-coordinate axes of R3 , and the subspace S is the x-axis, confirming the compositions (2) and (6). Lemma 3.2. Let C ⊂ Rn be a closed convex cone, and S be the orthogonal complement of lin C within span C. Then (C ∩ S)◦ = C ◦ + lin C
and
(C ◦ ∩ S)◦ = C + lin C ◦ .
Proof. Since S ◦ = S ⊥ , the equalities (4) and (7) give (C ∩ S)◦ = cl (C ◦ + S ◦ ) = cl (C ◦ + S ⊥ ) = cl (C ◦ + lin C + lin C ◦ ). Because C ◦ ∩ lin C = {o}, Proposition 2.3 shows that the convex cone C ◦ + lin C is closed. Based on this argument and the equality C ◦ + lin C ◦ = C ◦ , we have cl (C ◦ + lin C + lin C ◦ ) = cl (C ◦ + lin C) = C ◦ + lin C. Similarly, (C ◦ ∩ S)◦ = cl ((C ◦ )◦ + S ◦ ) = cl (C + S ⊥ ) = cl (C + lin C + lin C ◦ ) = cl (C + lin C ◦ ) = C + lin C ◦ . Theorem 3.1. Let C ⊂ Rn be a closed convex cone which is not a subspace. Also, let S be the orthogonal complement of lin C within span C. Then the set D = C ∩ (−C ◦ ) is a closed convex pointed cone of positive dimension s(C), with span D = S
and
D◦ = C ◦ − C.
Proof. According to Proposition 2.1 and Lemma 3.1, C = (C ∩ S) + lin C
and C ◦ = (C ◦ ∩ S) + lin C ◦ .
This argument and the equalities (7) give D = C ∩ (−C ◦ ) = ((C ∩ S) + lin C) ∩ ((−C ◦ ∩ S) + lin C ◦ ) = (C ∩ S) ∩ (−C ◦ ∩ S) = D ∩ S. 4
Hence D is a closed convex pointed cone as the intersection of closed convex pointed cones C ∩ S and −C ◦ ∩ S (see again Proposition 2.1 and Lemma 3.1). Furthermore, since (C ∩ S) ∩ (C ◦ ∩ S) = (C ∩ C ◦ ) ∩ S = {o} ∩ S = {o}, Proposition 2.3 implies that the set B = (C ∩ S) + (−C ◦ ∩ S) also is a closed convex pointed cone. Consequently, dim B ◦ = n, as follows from Proposition 2.4. On the other hand, a combination of the equalities (2), (4), (6), (7), and Lemma 3.2 gives B ◦ = ((C ∩ S) + (−C ◦ ∩ S))◦ = (C ∩ S)◦ ∩ (−C ◦ ∩ S)◦ = (C ◦ + lin C) ∩ (−C + lin C ◦ ) = ((C ◦ ∩ S) + lin C ◦ + lin C) ∩ (−(C ∩ S) + lin C + lin C ◦ ) = ((C ◦ ∩ S) + S ⊥ ) ∩ (−(C ∩ S) + S ⊥ ) = (C ◦ ∩ S) ∩ −(C ∩ S)) + S ⊥ = (D ∩ S) + S ⊥ . Therefore, dim D = dim (D ∩ S) = dim B ◦ − dim S ⊥ = n − dim S ⊥ = dim S = s(C). Combined with the inclusion D ⊂ S, the last equality implies that span D = S. A similar argument gives D◦ = (C ∩ (−C ◦ ))◦ = cl (C ◦ + (−C ◦ )◦ ) = cl (C ◦ + (−C)) = C ◦ − C, where the last equality follows from C ∩ C ◦ = {o} and Proposition 2.3. Theorem 3.2. Under the assumptions of Theorem 3.1, the following assertions hold. 1) The relative interior, rint D, of the cone D = C ∩ (−C ◦ ) is a convex set of positive dimension s(C). 2) rint D = E = rint (C ∩ S) ∩ (−rint (C ◦ ∩ S)). Proof. 1) This part immediately follows from Theorem 3.1 and the fact that the relative interior of a nonempty convex set F ⊂ Rn is nonempty and has the same dimension as F (see, e. g., [7, Section 6]). 2) According to Proposition 2.1, Lemma 3.1, and Theorem 3.1, span D = span (C ∩ S) = span (C ◦ ∩ S) = S. From the definition of relative interior, one can easily obtain the following elementary fact: if C1 and C2 are convex cones in Rn such that C1 ⊂ C2 and span C1 = span C2 , then rint C1 ⊂ rint C2 . Consequently, the equality D = (C ∩ S) ∩ (−C ◦ ∩ S) 5
implies the inclusion ∅ 6= rint D ⊂ rint (C ∩ S) ∩ rint (−C ◦ ∩ S), and Proposition 2.2 gives rint D = rint (C ∩ S) ∩ (−rint (C ◦ ∩ S)). Finally, combining Propositions 2.1, 2.4, and Lemma 3.1, we obtain E = rint C ∩ (−rint C ◦ ) = (rint (C ∩ S) + lin C) ∩ (−rint (C ◦ ∩ S) + lin C ◦ ) = rint (C ∩ S) ∩ (−rint (C ◦ ∩ S)) = rint D. Theorem 3.2 gives an affirmative answer to the question of Stoker (see the introduction). Corollary 3.1. If C ⊂ Rn is a closed convex cone, then the set E = rint C ∩ (−rint C ◦ ) is nonempty. Proof. If C is not a subspace, then the assertion immediately follows from Theorem 3.2. If C is a subspace, then rint C = C and C ◦ = C ⊥ is a subspace orthogonal to C. Therefore, rint C ◦ = C ◦ , which gives E = rint C ∩ (−rint C ◦ ) = C ∩ C ◦ = C ∩ C ⊥ = {o} = 6 ∅. 4. Separation of Convex Cones If a hyperplane H ⊂ Rn separates closed convex cones C1 and C2 , then, due to the inclusion o ∈ C1 ∩ C2 , it has to be of the form H = {x ∈ Rn : x·e = 0}
(9)
for some nonzero vector e. Following Rockafellar [7, p. 95], a hyperplane of the form (9) properly separates C1 and C2 provided C1 and C2 lie in distinct closed halfspaces of Rn determined by H such that C1 ∪ C2 6⊂ H. As proved in [7, Theorem 11.3], nonzero convex cones C1 and C2 are properly separated by a hyperplane if and only if rint C1 ∩ rint C2 = ∅. The theorem below shows that polar cones poses a stronger form of proper separation. Theorem 4.1. Let C ⊂ Rn be a closed convex cone distinct from a subspace. Also, let D = C ∩ (−C ◦ ). The following assertions hold. 1) There is a hyperplane H ⊂ Rn separating C and C ◦ . 2) Every hyperplane H ⊂ Rn separating C and C ◦ satisfies the conditions C 6⊂ H
and
C ◦ 6⊂ H.
(10)
3) A hyperplane of the form (9) separates C and C ◦ if and only if e ∈ D ∪ (−D) for some nonzero vector e. 6
Proof. 1) The above criterion of proper separation and the inclusions rint C ∩ rint C ◦ ⊂ (C \ lin C) ∩ (C ◦ \ lin C ◦ ) ⊂ (C \ {o}) ∩ (C ◦ \ {o}) = (C ∩ C ◦ ) \ {o} = {o} \ {o} = ∅ (which follow from assertions 1) and 3) of Proposition 2.4) show the existence of at least one hyperplane H of the form (9) properly separating C and C ◦ . 2) Let a hyperplane H ⊂ Rn separate C and C ◦ . Assume, for contradiction, that one of the cones C and C ◦ , say C, is contained in H. Then span C ⊂ H. Consider the line l = span {e}. By Proposition 2.4, l = H ⊥ ⊂ (span C)⊥ = lin C ◦ . Since l meets both open halfspaces determined by H, the hyperplane H cannot support C ◦ , contrary to the choice of H. 3) Suppose a hyperplane of the form (9) separates C and C ◦ . Then C and C ◦ are included into the opposite closed halfspaces V1 = {x ∈ Rn : x·e 6 0}
and V2 = {x ∈ Rn : x·(−e) 6 0}.
If, for instance, C ⊂ V1 and C ◦ ⊂ V2 , then e ∈ C ◦ ∩ (−C ◦ )◦ = C ◦ ∩ (−C) = −D. Similarly, e ∈ D if C ⊂ V2 and C ◦ ⊂ V1 . The converse assertion is obvious. We will need the following lemma. Lemma 4.1. If a hyperplane H ⊂ Rn supports a closed convex cone C ⊂ Rn , then lin C ⊂ H and H = {x ∈ Rn : x·e = 0} for a suitable vector e 6= o. Proof. Let H = {x ∈ Rn : x·e = γ}, where e 6= o and γ ∈ R. Denote by V a closed halfspace determined by H and containing C. Without loss of generality, we may assume that V = {x ∈ Rn : x·e 6 γ}. Since o ∈ C ⊂ V , one has 0 = o·e 6 γ. By the hypothesis, H ∩ C 6= ∅. Choose any point u ∈ H ∩ C. Then u·e = γ. Since λu ∈ C ⊂ V for all λ > 0, one has λγ = (λu)·e 6 γ whenever λ > 0. The latter is possible only if γ = 0. So, H = {x ∈ Rn : x·e = 0}. For the inclusion lin C ⊂ H, choose any point v ∈ lin C. Because lin C is a subspace, one has µv ∈ lin C ⊂ V whenever µ ∈ R. Thus (µv)·e 6 0 for all µ ∈ R, which is possible only if v·e = 0. Hence v ∈ H. Suppose that a hyperplane H of the form (9) separates convex cones C1 and C2 . By Lemma 4.1, both subspaces lin C1 and lin C2 lie H. Consequently, lin C1 ⊂ C1 ∩ H
and
lin C2 ⊂ C2 ∩ H.
In this regard, we will say that H sharply separates C1 and C2 provided H separates them and C1 ∩ H = lin C1 and C2 ∩ H = lin C2 . (11) A result of Klee [5, Theorem 2.7], formulated here for the case of Rn , states that closed convex cones C1 and C2 in Rn satisfying the condition C1 ∩ C2 = {o} are sharply separated by a hyperplane. The theorem below refines this assertion. 7
Theorem 4.2. Let C1 and C2 be closed convex cones in Rn , each distinct from a subspace. The following conditions are equivalent. 1) C1 and C2 are sharply separated by a hyperplane. 2) C1 ∩ C2 = lin C1 ∩ lin C2 . 3) The set F = rint C1◦ ∩ (−rint C2◦ ) has positive dimension. Proof. 1) ⇔ 2). Suppose C1 and C2 are sharply separated by a hyperplane H ⊂ Rn such that C1 and C2 lie, respectively in the closed halfspaces V1 and V2 determined by H. Then, by (11), C1 ∩ C2 = (C1 ∩ V1 ) ∩ (C2 ∩ V2 ) = (C1 ∩ C2 ) ∩ (V1 ∩ V2 ) = (C1 ∩ C2 ) ∩ H = (C1 ∩ H) ∩ (C2 ∩ H) = lin C1 ∩ lin C2 . Conversely, suppose that C1 ∩ C2 = lin C1 ∩ lin C2 . Put M = lin C1 ∩ lin C2 ,
N = M ⊥,
Ci0 = Ci ∩ N,
i = 1, 2.
Clearly, M is a nontrivial subspace of Rn . From Proposition 2.1 we easily conclude that Ci = Ci0 + M
and
lin Ci0 = lin Ci ∩ N,
i = 1, 2.
Furthermore, C10 ∩ C20 = (C1 ∩ C2 ) ∩ N = (lin C1 ∩ lin C2 ) ∩ N = M ∩ N = {o}. By the above assertion of Klee, used now for the case of vector space N , there is a subspace G ⊂ N of dimension dim N − 1 sharply separating C10 and C20 . It is easy to see that the subspace H = G + M is a hyperplane in Rn sharply separating C1 and C2 . 1) ⇔ 3). Suppose C1 and C2 are sharply separated by a hyperplane of the form (9) such that C1 ⊂ V1 = {x ∈ Rn : x·e 6 0}
and C2 ⊂ V2 = {x ∈ Rn : x·(−e) 6 0}.
Combining assertion 6) of Proposition 2.4 and the equalities (11), we obtain the inclusions e ∈ rint C1◦ and −e ∈ rint C2◦ . Thus e ∈ F . Similarly, −e ∈ F if C1 ⊂ V2 and C2 ⊂ V1 . Summing up, e ∈ F ∪ (−F ), which shows that F has positive dimension. Conversely, if F has positive dimension and e is a nonzero vector in F ∪ (−F ), then the above argument implies that the hyperplane (9) sharply separates C1 and C2 . Remark 4.1. From the proof of Theorem 4.2 it follows that a hyperplane of the form (9) sharply separates closed convex cones C1 and C2 if and only if e ∈ F ∪ (−F ) for some nonzero vector e. A combination of Theorems 3.2 and 4.2 gives the corollary below. Corollary 4.1. Let C ⊂ Rn be a closed convex cone which is not a subspace. Also, let D = C ∩ (−C ◦ ). The following assertions hold. 1) There is a hyperplane sharply separating C and C ◦ . 8
2) A hyperplane of the form (9) sharply separates C and C ◦ if and only if e ∈ rint D ∪ (−rint D) for some nonzero vector e. The next result describes conditions for the uniqueness of a hyperplane which sharply separates given cones. Theorem 4.3. Let C1 and C2 be closed convex cones in Rn , neither being a subspace, which are separated by a hyperplane. The following conditions are equivalent. 1) There is a unique hyperplane sharply separating C1 and C2 . 2) C1 ∩ C2 = lin C1 ∩ lin C2 and the subspace lin C1 + lin C2 is a hyperplane. 3) The set F = rint C1◦ ∩ (−rint C2◦ ) is one-dimensional. Proof. 1) ⇔ 2). Let H be a unique hyperplane of the form (9) sharply separating C1 and C2 . Without loss of generality, we may assume that kek = 1. Then C1 ∩ C2 = lin C1 ∩ lin C2 according to Theorem 4.2. Since both subspaces lin C1 and lin C2 lie in H, their sum M = lin C1 + lin C2 also lies in H. Thus dim M 6 n − 1. Assume, for contradiction, that dim M < n − 1. Then the orthogonal complement of M is at least two-dimensional. Clearly, e ∈ M ⊥ . By Proposition 2.1, we can write Ci = (Ci ∩ Si ) + lin Ci , where Si is the orthogonal complement of lin Ci within span Ci , i = 1, 2. The condition Ci ∩ H = lin Ci obviously implies that (Ci ∩ Si ) ∩ H = {o}, i = 1, 2. Let Pi denote the intersection of Ci ∩ Si with the unit sphere of Rn , i = 1, 2. Clearly, Pi 6= ∅ because Ci ∩ Si is not a subspace, i = 1, 2. The above argument shows that Pi ∩ H = ∅, i = 1, 2. Since both sets P1 and P2 are nonempty compacts, there is a scalar ε > 0 such that, for any unit vector e0 ∈ Rn satisfying the condition ke − e0 k < ε, the hyperplane H 0 = {x ∈ Rn : x·e0 = 0} is disjoint from P1 ∪ P2 . Because Ci ∩ Si = {λy : λ > 0, y ∈ Pi },
i = 1, 2,
we have (Ci ∩ Si ) ∩ H 0 = {o},
i = 1, 2,
whenever ke − e0 k < ε. Finally, choose e0 in M ⊥ such that e and e0 are not collinear (this is possible due to dim M ⊥ > 2) and ke − e0 k < ε. Then M ⊂ H0
and Ci ∩ H 0 = lin Ci ,
i = 1, 2.
Consequently, H 0 sharply separates C1 and C2 , contrary to condition 1). Thus M is a hyperplane. Conversely, if condition 2) is satisfied, then, by Theorem 4.2, there is a hyperplane H sharply separating C1 and C2 . Furthermore, the inclusion lin C1 + lin C2 ⊂ H and the assumption dim (lin C1 + lin C2 ) = n − 1 show that lin C1 + lin C2 = H, implying the uniqueness of H. 9
1) ⇔ 3). By Theorem 4.2 and Remark 4.1, the set F has positive dimension and every nonzero vector e ∈ F ∪ (−F ) determines the hyperplane of the form (9) which sharply separates C1 and C2 . Assuming condition 1), suppose for a moment that dim F > 2 and choose in F non-collinear vectors e1 and e2 . Then the hyperplanes H1 = {x ∈ Rn : x·e1 = 0}
and H2 = {x ∈ Rn : x·e2 = 0}
are distinct and both sharply separate C1 and C2 , contrary to 1). Hence 1) ⇒ 3). Similarly, 3) ⇒ 1). Corollary 4.2. For a closed convex cone C ⊂ Rn , which is not a subspace, the following conditions are equivalent. 1) There is a unique hyperplane sharply separating C and C ◦ . 2) The subspace lin C + lin C ◦ is a hyperplane. 3) C is a closed halfplane of the form C = L + h, where L is a subspace, with 0 6 dim L 6 n − 1, and h is a closed halfline with endpoint o, orthogonal to L. Proof. By Proposition 2.4, C ∩ C ◦ = lin C ∩ lin C ◦ = {o}. This argument and Theorem 4.3 show that it suffices to prove the equivalence of conditions 2) and 3). Assume first that lin C + lin C ◦ is a hyperplane and let L = lin C. Denote by S the orthogonal complement of L within span C. Then (7) implies the equality dim S = 1. Consequently, the set h = C ∩ S is a closed halfline with endpoint o, orthogonal to L. This argument and Proposition 2.1 immediately imply that C = lin C + (C ∩ S) = L + h. Conversely, suppose that C is the sum of a subspace L and a closed halfline h with endpoint o, orthogonal to L. Obviously, lin C = L and
span C = L + (h ∪ −h).
If S is the one-dimensional subspace containing h, then S is the orthogonal complement of lin C within span C. A combination of Propositions 2.1 and 2.4 shows that dim (lin C ◦ ) = dim (span C)⊥ = n − dim (span C) = n − dim (lin C) − 1. Since the subspaces lin C and lin C ◦ are orthogonal, their sum lin C + lin C ◦ is a hyperplane. Acknowledgment. The author thanks the anonymous referees for helpful suggestions on the original manuscript.
10
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