ESI
The Erwin Schr¨ odinger International Institute for Mathematical Physics
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Extremal Problems and Isotropic Positions of Convex Bodies
A.A. Giannopoulos V.D. Milman
Vienna, Preprint ESI 688 (1999)
Supported by Federal Ministry of Science and Transport, Austria Available via anonymous ftp or gopher from FTP.ESI.AC.AT or via WWW, URL: http://www.esi.ac.at
April 2, 1999
EXTREMAL PROBLEMS AND ISOTROPIC POSITIONS OF CONVEX BODIES A.A. Giannopoulos and V.D. Milman1
Abstract Let K be a convex body in Rn and let Wi (K), i = 1, . . . , n − 1 be its quermassintegrals. We study minimization problems of the form min{Wi (T K) | T ∈ SLn } and show that bodies which appear as solutions of such problems satisfy isotropic conditions or even admit an isotropic characterization for appropriate measures. This shows that several well known positions of convex bodies which play an important role in the local theory may be described in terms of classical convexity as isotropic ones. We provide new applications of this point of view for the minimal mean width position.
1. Introduction Given a convex body K in Rn we consider the family {T K | T ∈ SLn } of its positions. One of the main problems in the asymptotic theory of finite dimensional normed spaces is introducing the right position of the unit ball KX of a space X. There exist many well-known positions which have been introduced and used for different purposes in this theory: John’s position, the `-position, M -positions are among them (see [MSch1], [Pi2] and [TJ] for a description and important applications). Because of the isomorphic nature of the results of the asymptotic theory, an isomorphic point of view dominates the study of these special positions as well. Even the definition of some of them (the M -position is such an example) is done in isomorphic form. The purpose of this paper is to discuss the possibility of an isometric approach to these questions. The standard isotropic position of a convex body provides a good example for our point of view: Let K be a convex body in Rn with centroid at the origin and volume equal to one. We say that K is in isotropic position if Z hx, θi2 dx = L2K K
1
The second named author was supported in part by the Israel Science Foundation founded by the Academy of Sciences and Humanities. Part of this work was done while both authors were visiting the Erwin Schroedinger International Institute for Mathematical Physics in Vienna.
1
for every θ ∈ S n−1 . It is not hard to see that every body K of volume one has a position which is isotropic. Moreover, this position is uniquely determined up to an orthogonal transformation. Therefore, LK is an affine invariant which is called the isotropic constant of K. The isotropic position is well studied and has several connections with classical convexity problems (see [MP1]). In particular, the question if LK ≤ c for some absolute positive constant and every body K is a major open problem. The starting point of our present discussion is the following remark: R R Fact I: A body K is isotropic if and only if K |x|2 dx ≤ T K |x|2 dx for every T ∈ SLn , where | · | is the standard Euclidean norm. The proof of the “if” part is given by a simple variational argument: If T ∈ L(Rn , Rn ) and ε > 0 is small enough, then (I + εT )/[det(I + εT )]1/n is volume preserving, therefore Z Z |x|2 dx. |x + εT x|2 dx ≥ [det(I + εT )]2/n (1) K
K
Writting |x + εT x|2 = |x|2 + 2εhx, T xi + O(ε2 ) and [det(I + εT )]2/n = 1 + 2ε trT n + O(ε2 ), and letting ε → 0+ we get Z Z trT (2) hx, T xidx ≥ |x|2 dx, n K K and replacing T by −T we see that there must be equality in (2) for every T ∈ L(Rn , Rn ). This in turn implies that K is isotropic. R Starting with the functional T → f (T K) = T K |x|2 dx on SLn we saw that its minimum is achieved on some isotropic position (for the Lebesgue measure on K). In this paper we show that this is a general scheme which produces isometric descriptions for many classical positions of the theory. As a second example, we mention the minimal surface area position: Let K be a convex body, and write ∂(K) for its surface area. We say that K has minimal surface area if ∂(K) ≤ ∂(T K) for every T ∈ SLn . A characterization of the minimal surface area position was given by Petty ([Pe], see also [GP]): Fact II: A convex body K has minimal surface area if and only if Z ∂(K) (3) hu, θi2 σK (du) = n S n−1 for every θ ∈ S n−1 , where σK is the area measure of K.
Recall that the area measure σK of K is defined on S n−1 by
(4)
σK (A) = ν({x ∈ bd(K) : the outer normal to K at x is in A}),
2
where ν is the (n − 1)-dimensional surface measure on K. The key point for the proof of the fact is the observation that Z −1 ∗ |T x|σK (dx). (5) ∂((T ) K) = S n−1
Then, we employ a variational argument identical to the one used for Fact I. One can also check that the minimal surface position is unique up to orthogonal transformations (see [GP] for the details). In view of the above result we give the following definition: Definition. A Borel measure µ on S n−1 will be called isotropic if Z µ(S n−1 ) (6) hu, θi2 µ(du) = n S n−1 for every θ ∈ S n−1 . In this terminology, a body K has minimal surface area if and only if its area measure is isotropic: The minimum of the functional T → ∂(T K), T ∈ SLn is again achieved on an isotropic position (for the appropriate measure on the sphere). Surface area is one of the quermassintegrals Wi (K) of the body K (see Section 2 for notation and definitions). We consider the minimization problems (7)
min{Wi (T K) | T ∈ SLn } ,
i = 1, . . . , n − 1.
In every case, a necessary condition for the minimal position is that the corresponding mixed area measure Sn−i (K, ·) should be isotropic (see Section 4). In particular, in Section 3 we find a necessary and sufficient condition for the minimal mean width position: a body K has minimal mean width if and only if Z hK (u)hu, θi2 σ(du) S n−1
n−1
does not depend on θ ∈ S , where hK is the support function of K and σ is the rotationally invariant probability measure on the sphere. In the symmetric case, using a classical estimate of Pisier [Pi1] (after work of Lewis [L] and Figiel and Tomczak-Jaegermann [FT]) we see that isotropicity of the measure hK dσ implies the inequality ¶1/n µ Z |K| . (8) hK (u)σ(du) ≤ c log d(XK , `n2 ) |Dn | S n−1 In Section 5 we see the maximal volume ellipsoid position (John’s position) as a solution of the problem (9)
min{kT : `n2 → XK k | T ∈ SLn }.
3
Using the same general method we give a simple proof of John’s theorem in its full strength. In our present setting, John’s representation of the identity may be interpreted as an isotropic condition: a symmetric body K is in John’s position if and only if there is an isotropic measure supported by its contact points with the inscribed ball. Finally, in Section 6 we show that M -position may also be described in an isometric way. If |K| = |Dn |, we study the problem (10)
min{|T K + Dn | | T ∈ SLn }
and show that if K is a solution, then K + Dn must have minimal surface area. In view of Petty’s result, this opens the possibility of an isotropic M -position. K. Ball [Ba1,2,3] realized that John’s representation of the identity could be combined with the Brascamp-Lieb inequality. This led him to sharp bounds for the volume ratio, the volume of the central sections of the cube, and an exact reverse isoperimetric inequality. The reverse Brascamp-Lieb inequality [Bar] has been recently applied for an estimate of the volume of the central sections of the difference body of a non-symmetric body [Ru]. Petty’s isotropic description of the minimal surface area position (combined with the Brascamp-Lieb inequality) leads to sharp inequalities for the volume of the projection body and its polar in terms of the minimal surface parameter [GP]. All these results show that the general isotropic point of view we propose in this paper might help towards a new understanding of several isomorphic results of the theory. 2. Definitions and Preliminaries We first recall some facts about mixed volumes and mixed area measures. For detailed proofs we refer the reader to [Sch]. 2.1. Let Kn denote the set of all non-empty, compact convex subsets of Rn . We may view Kn as a convex cone under Minkowski addition and multiplication by nonnegative real numbers. Minkowski’s theorem (and the definition of the mixed volumes) asserts that if K1 , . . . , Km ∈ Kn , m ∈ N, then the volume of t1 K1 + . . . + tm Km is a homogeneous polynomial of degree n in ti > 0. That is, X (1) |t1 K1 + . . . + tm Km | = V (Ki1 , . . . , Kin )ti1 . . . tin , 1≤i1 ,...,in ≤m
where the coefficients V (Ki1 , . . . , Kin ) are chosen to be invariant under permutations of their arguments. The coefficient V (K1 , . . . , Kn ) is called the mixed volume of K1 , . . . , Kn . 2.2. Steiner’s formula may be seen as a special case of Minkowski’s theorem. The volume of K + tDn , t > 0, can be expanded as a polynomial in t: n µ ¶ X n Wi (K)ti , (2) |K + tDn | = i i=0
4
where Wi (K) = V (K; n − i, Dn ; i) is the i-th quermassintegral of K. Here and elsewhere we use the notation L; j for L, . . . , L j-times. The quermassintegrals inherit properties of mixed volumes: they are monotone, continuous with respect to the Hausdorff metric, and homogeneous of degree n − i. 2.3. The mixed area measures were introduced by Alexandrov [Al1,2] and may be viewed as a local generalization of the mixed volumes. For any (n − 1)tuple C = K1 , . . . , Kn−1 ∈ Kn , the Riesz representation theorem guarantees the existence of a Borel measure S(C, ·) on the unit sphere S n−1 such that Z 1 hL (u)dS(C, u) (3) V (L, K1 , . . . , Kn−1 ) = n S n−1 for every L ∈ Kn , where hL is the support function of L. The local analogue of Minkowski’s theorem is (4)
Sn−1 (
m X
X
ti Ki , ω) =
i=1
S(Ki1 , . . . , Kin−1 , ω)ti1 . . . tin−1
1≤i1 ,...,in ≤m
for all Borel ω ⊆ S n−1 , ti > 0, Ki ∈ Kn , m ∈ N (see below for the definition of Sn−1 ). The j-th area measure of K is defined by Sj (K, ·) = S(K; j, Dn ; n − j − 1, ·), j = 0, 1, . . . , n − 1. It follows that the quermassintegrals of K can be represented by Z 1 hK (u)dSn−i−1 (K, u) , i = 0, 1, . . . , n − 1 (5) Wi (K) = n S n−1 or, alternatively, 1 Wi (K) = n
(6)
Z
dSn−i (K, u) ,
i = 1, . . . , n.
S n−1
2.4. Let Ki ∈ Kn and assume for simplicity that hKi is twice continuously differentiable. Then, the mixed area measure of K1 , . . . , Kn−1 has a continuous density s(K1 , . . . , Kn−1 , ·) with respect to the Lebesgue measure on S n−1 , the mixed discriminant of the second differentials of hKi . We write sj (K, u) for s(K; j, Dn ; n − j − 1, u). It follows that (7)Z Z S n−1
hK1 (u)s(K2 , K3 , . . . , Kn , u)du =
S n−1
hK2 (u)s(K1 , K3 , . . . , Kn , u)du.
In particular, for i = 1, . . . , n − 1 we have Z Z 1 1 sn−i (K, u)du = hK (u)sn−i−1 (K, u)du. (8) Wi (K) = n S n−1 n S n−1
5
2.5. Let f be a real function on Rn \{o}. We write fˆ for the restriction of f to S n−1 . If F is defined on S n−1 , the radial extension f of F to Rn \{o} is given by f (x) = F (x/|x|). If F is a twice differentiable function on S n−1 , we define ˆ ) and ∇o F = (∇f ˆ ), ∆o F = (∆f
(9)
where f is the radial extension of F . The operator ∆o is usually called the Laplace-Beltrami operator, while ∇o is referred to as the gradient. As a consequence of Green’s formula we have Z Z Z (10) F ∆o G = G∆o F = − (∇o F ) · (∇o G). S n−1
S n−1
S n−1
For more details we refer the reader to [Gr]. 2.6. If K is an origin symmetric convex body in Rn , then K induces a norm k · kK on Rn in a natural way. We shall write XK for the normed space with unit ball K, and KX for the unit ball of X. The polar body of K is defined by kxkK ◦ = maxy∈K |hx, yi| = hK (x), and will be denoted by K ◦ . We consider the average Z kxkK σ(dx) (11) M (K) = S n−1
of the norm k · kK on S n−1 , and define M ∗ (K) = M (K ◦ ). If K and L are bodies in Rn , their multiplicative distance d(K, L) is defined by (12)
d(K, L) = inf{ab : a, b > 0, K ⊆ bL, L ⊆ aK}.
The Banach-Mazur distance between XK and XL is (13)
d(XK , XL ) = inf{d(K, T L) | T ∈ GLn }.
Whenever we write (1/a)|x| ≤ kxkK ≤ b|x|, we assume that a, b are the smallest positive numbers for which this inequality holds true for every x ∈ Rn . In particular, we then have d(K, Dn ) = ab. Finally, we denote by Gn,k the Grassmannian of all k-dimensional subspaces of Rn , equipped with the Haar probability measure νn,k . We write |K| for the volume of K, and ωn for the volume of the Euclidean unit ball. The letters c, c0 , C etc. are reserved for absolute positive constants. 3. Minimal Mean Width Let K be a convex body in Rn (without loss of generality we may assume that o ∈ intK). The mean width w(K) of K is the quantity Z (1) w(K) = 2 hK (u)σ(du). S n−1
6
This is equal to 2M ∗ (K) in the symmetric case. From 2.3 we see that Z Z 1 hK (u)du = ωn (2) Wn−1 (K) = hK (u)σ(du), n S n−1 S n−1 hence, (3)
w(K) =
2Wn−1 (K) . ωn
We say that K has minimal mean width if w(K) ≤ w(T K) for every T ∈ SLn . This notion was heavily used in the literature under a different name: K has minimal mean width if and only if the `-ellipsoid of K ◦ is a multiple of Dn [FT]. Our purpose is to find necessary and sufficient conditions for a body K to have minimal mean width. We assume for simplicity that hK is twice continuously differentiable (we then say that K is smooth enough). Theorem 3.1: A smooth enough convex body K in Rn has minimal mean width if and only if Z trT w(K) h∇hK (u), T uiσ(du) = (4) 2 n S n−1 for every T ∈ L(Rn , Rn ). Moreover, this minimal mean width position is unique up to an orthogonal transformation. Proof: Assume first that K has minimal mean width. Let T ∈ L(Rn , Rn ) and ε > 0 be small enough. Then (I + εT )∗ /[det(I + εT )]1/n is volume preserving, and this means that Z Z 1/n hK (u)σ(du). hK (u + εT u)σ(du) ≥ [det(I + εT )] (5) S n−1
S n−1
Since hK (u + εT u) = hK (u) + εh∇hK (u), T ui + O(ε2 ) and [det(I + εT )]1/n = 2 + 1 + ε trT n + O(ε ), letting ε → 0 we obtain Z trT w(K). (6) 2 h∇hK (u), T uiσ(du) ≥ n S n−1 Replacing T by −T in (6) we see that there must be equality in (4) for every T ∈ L(Rn , Rn ). Conversely, assume that (4) is satisfied and let T ∈ SLn . Up to an orthogonal transformation we may assume that T ∗ is symmetric positive-definite. Then, Z Z (7) w(T K) = 2 hT K (u)σ(du) = 2 hK (T ∗ u)σ(du). S n−1
S n−1
7
It is a known fact that ∇hK (u) is the unique point on the boundary of K at which u is the outer normal to K (see [Sch], pp.40). In particular, ∇hK (u) ∈ K, which implies h∇hK (u), zi ≤ hK (z)
(8)
for every z ∈ Rn . Therefore, by (7),(8) and (4) we get Z trT ∗ (9) w(T K) ≥ 2 h∇hK (u), T ∗ uiσ(du) = w(K) ≥ w(K). n S n−1 This shows that K has minimal mean width. Moreover, we can have equality in (9) only if T is the identity. This proves uniqueness of the minimal mean width position up to U ∈ O(n). 2
Consider the measure νK on S n−1 with density hK with respect to σ. We shall prove that a smooth enough convex body K has minimal mean width if and only if νK is isotropic. Lemma 3.2: Let K be a smooth enough convex body in Rn . We define Z h∇hK (u), θihu, θiσ(du) , θ ∈ S n−1 . (10) IK (θ) = S n−1
Then, (11)
w(K) + IK (θ) = (n + 1) 2
Z
hK (u)hu, θi2 σ(du) S n−1
for every θ ∈ S n−1 .
Proof: Let θ ∈ S n−1 , and consider the function f (x) = hx, θi2 /2. A direct computation shows that (∇◦ fˆ)(u) = hu, θiθ − hu, θi2 u
(12) and
(∆◦ fˆ)(u) = 1 − nhu, θi2 .
(13)
Since hK is positively homogeneous of degree 1, we have (∇◦ hˆK )(u) = ∇hK (u) − hK (u)u and hK (u) = h∇hK (u), ui, u ∈ S n−1 . Taking into account (12) we obtain (14)
h(∇◦ fˆ)(u), (∇◦ hˆK )(u)i = h∇hK (u), θihu, θi − hK (u)hu, θi2 .
Integrating on the sphere and using Green’s formula (see 2.5), we have Z Z hK (u)(∆◦ fˆ)(u)σ(du), (15) IK (θ) − hK (u)hu, θi2 σ(du) = − S n−1
S n−1
8
which is equal to −
w(K) +n 2
Z
hK (u)hu, θi2 σ(du) S n−1
by (13). This proves (11). 2 Theorem 3.3: A smooth enough convex body K has minimal mean width if and only if Z w(K) hK (u)hu, θi2 σ(du) = (16) 2n S n−1 for every θ ∈ S n−1 (equivalently, if νK is isotropic).
Proof: It is not hard to check that (4) is true for every T ∈ L(Rn , Rn ) if and only if (17)
IK (θ) =
w(K) 2n
for every θ ∈ S n−1 . The result now follows from Theorem 3.1 and Lemma 3.2. 2 The fact that (4) and (16) are linear in K has the following immediate consequence: Corollary 3.4: Let K1 and K2 be smooth enough convex bodies in Rn . (i) If K1 and K2 have minimal mean width, then their Minkowski sum K1 +K2 has also minimal mean width. (ii) If K1 and K1 + K2 have minimal mean width, then K2 has also minimal mean width. Proof: Obvious from Theorem 3.1 or 3.3, since hK1 +K2 = hK1 + hK2 and w(K1 + K2 ) = w(K1 ) + w(K2 ). 2 In the symmetric case, it is a well-known fact [L], [FT], [Pi1] that if W has minimal mean width then M (W )M ∗ (W ) ≤ c log d(XW , `n2 ). As an application of this estimate and of Corollary 3.4 we obtain: Theorem 3.5: Let k · k be a norm on Rn and assume that its unit ball K has the property M (K) ≤ M (T K) for every T ∈ SLn . Then, for every λ ∈ (0, 1), there exists a [(1 − λ)n]-dimensional section K ∩ E of K such that ¶ µ b 2b √ (18) d(K ∩ E, Dn ∩ E) ≤ c √ log , M λ M λ where c > 0 is an absolute constant. Proof: Without loss of generality we may assume that M (K) = 1 and λ < 1/2. Let t0 be the smallest integer t for which log(t) (bM ∗ (K)) ≤ 2 (where log(t) denotes
9
the t-th iterated logarithm). The Low M ∗ -estimate [M1], [PT], [Go] implies that, for some absolute constant δ > 0, p δ λ/2t0 |x| (19) kxk ≥ M ∗ (K) for all x ∈ E0 or x ∈ E0⊥ , where E0 is in a subset L0 of Gn,[(1−2−t0 λ)n] of measure greater than p(λ, n, t0 ) = 1 − c1 exp(−c2 2−t0 λn), and c1 , c2 > 0 are absolute constants. Consider the orthogonal transformation U = U (E0 ) = PE0 − PE0⊥ , E0 ∈ L0 . Then, p δ λ/2t0 kxk + kU xk |x| (20) ≥√ 2 2M ∗ (K) ◦
∗
◦
K for all x ∈ Rn . Define a new body K1 = K1 (E0 ) by K1◦ = K +U . Then, by 2 ◦ Corollary 3.4, K1 has minimal mean width equal to M (K1 ) = 1. It follows that Ã√ ! t0 +1 M ∗ (K)b 2 √ (21) M ∗ (K1 ) = M (K1 )M ∗ (K1 ) ≤ c log . δ λ
Observe that kxkK = kxkK1 (E0 ) on E0 , for every E0 ∈ L0 . We now iterate this step: assume that Li ⊂ Gn,[(1−2−t0 +i λ)n] , Ei ∈ Li , and Ki+1 (E0 , . . . , Ei ), i = 0, . . . , s − 1 have been defined and satisfy the following: (i) (Ki+1 )◦ has minimal mean width, and M (Ki+1 ) = 1. √ √ (ii) M ∗ (Ki+1 ) ≤ c log( 2t0 −i+1 M ∗ (Ki )b/δ λ).
(iii) kxkKi+1 = kxkKi = . . . = kxk, for all x ∈ Fi = E0 ∩ . . . ∩ Ei .
We apply the Low M ∗ -estimate to Ks , and find Ls ⊂ Gn,[(1−2−t0 +s λ)n] with measure p(λ, n, t0 − s) such that p δ λ/2t0 −s |x| (22) b|x| ≥ kxkKs ≥ M ∗ (Ks ) ◦ on Es and on Es⊥ , for every Es ∈ Ls . If Es ∈ Ls , we define Ks+1 by Ks+1 = ◦ ∗ ◦ Ks +U (Es )Ks . Then, 2
(23)
b|x| ≥ kxkKs+1
p δ λ/2t0 −s √ ≥ |x| 2M ∗ (Ks )
on Rn , and (24)
kxkKs+1 = kxkKs = . . . = kxk
10
for every x ∈ Fs = E0 ∩ . . . ∩ Es . This means that p (25) d(K ∩ Fs , Dn ∩ Fs ) ≤ δ 2t0 −s+1 /λbM ∗ (Ks ).
We stop the procedure when s = t0 . Note that if (E0 , E1 , . . . , Et0 ) is a sequence as above, we have dimFt0 ≥ (1 − λ)n. Also, since each (Ks )◦ has minimal mean width, exactly as in (21) we get ! Ã√ 2t0 −s+1 M ∗ (Ks )b ∗ √ , (26) M (Ks+1 ) ≤ c log δ λ
and this implies that M ∗ (Kt0 ) ≤ C log
³
√b λ
´
. By (25), we have
b d(K ∩ Ft0 , Dn ∩ Ft0 ) ≤ c √ log λ
µ ¶ b . 2 λ
Theorem 3.5 should be compared to an analogous result for the M -position: In [MSch2] it is proved that if K is in M -position of orderP α > 1/2, and if there exist t t orthogonal transformations U1 , . . . , Ut such that 1t i=1 Ui K ◦ is c-equivalent to a ball, then for every λ ∈ (0, 1) there exists a subspace F ∈ Gn,(1−λ)n such that d(K ∩ F, Dn ∩ F ) ≤ C(t, λ, c). We can now show that the same is true for the minimal mean width position: Corollary 3.6: Let k · k be a norm on Rn and assume that its unit ball K has the property M (K) ≤ M (T K) for every T ∈ SLn . Assume further that for some t orthogonal transformations U1 , . . . , Ut and for some 0 < r, C < ∞, t
(27)
r|x| ≤
1X kUi xk ≤ Cr|x| t i=1
for all x ∈ Rn . Then, for every λ ∈ (0, 1), there exists a [(1 − λ)n]-dimensional section K ∩ E of K such that √ µ √ ¶ t 2C t √ (28) d(K ∩ E, Dn ∩ E) ≤ cC √ log . λ λ Proof: Lemma 2.1 from [MSch2] and (27) imply that √ (29) b(K) = max kxk ≤ Cr t. x∈S n−1
Since M (K) ≥ r, we have (30)
√ b(K) ≤ C t. M (K)
11
The result is now a consequence of Theorem 3.5. 2 An inspection of the argument we used for Theorem 3.5 shows that the statement holds true for a random [(1 − λ)n]-dimensional section of K. This allows a “global” reformulation of Corollary 3.6: Corollary 3.7: With the same hypotheses as in Corollary 3.6, there exists one orthogonal transformation U such that, for some r 0 > 0, √ √ (31) r0 |x| ≤ kxk + kU xk ≤ r 0 C t log(2C t)|x| for all x ∈ Rn . 2
n/10
The example of X = `1 cannot hold in general.
9n/10
⊕ `∞
from [MSch2] shows that such a statement
Let t(K) be the smallest integer t for which there exist orthogonal transformations U1 , . . . , Ut such that t
1X M (K) |x| ≤ kUi xk ≤ 2M (K)|x| 2 t i=1 for all x ∈ Rn . In [MSch2] it is shown that t(K) ' (b/M (K))2 . We will prove below an “isomorphic” version of this fact for bodies in `-position.PWe fix s ∈ s {2, . . . , t(K)} and ask how close to Euclidean can a norm kxks = 1s i=1 kUi xk, Ui ∈ O(n) be. More precisely, let gK (s) be the smallest A > 0 for which there exist r > 0, m ≤ s, and U1 , . . . , Um ∈ O(n) satisfying m
r|x| ≤
1 X kUi xk ≤ rA|x| , m i=1
x ∈ Rn .
From Lemma √ (see also the proof of Corollary 3.6), we must have √ 2.1 in [MSch2] b(K) ≤ rA m ≤ M (K)A s. This shows that p (32) gK (s) ≥ c t(K)/s.
We shall show that if K ◦ has minimal mean width (if K has minimal M ), then this estimate is sharp: Theorem 3.8: Let k · k be a norm on Rn such that its unit ball K satisfies M (K) ≤ M (T K), T ∈ SLn . Then, r r ¶ µ 2t(K) t(K) t(K) (33) c1 , ≤ gK (s) ≤ c2 log s s s where c1 , c2 > 0 are absolute constants.
12
Proof: Let s ∈ {2, . . . , t(K)}, and set b = b(K), M = M (K). Following the proof of Theorem 2 in [BLM], one can check that there exist s1 = [s/2] and U1 , . . . , Us1 ∈ O(n) such that (34)
kxks1
s1 b b 1 X kUi xk ≤ c √ |x| ≤ c0 √ |x| := s1 i=1 s1 s
for all x ∈ Rn . Let K1 be the unit ball of k·ks1 and set b1 = b(K1 ), M1 = M (K1 ). Since M1 = M , (34) implies that (35)
t(K1 ) ≤ c00 t(K)/s.
Observe that K1 has minimal M , therefore we can apply Corollary 3.7 with C = 4 and t = t(K1 ) to find r > 0 and V ∈ O(n) such that p (36) r|x| ≤ kxks1 + kV xks1 ≤ c000 r t(K1 ) log(2t(K1 ))|x|
for all x ∈ Rn . Setting Us1 +i = Ui V , i = 1, . . . , s1 , and taking into account (35) we conclude the proof. 2
Remark. Let k(K) be the largest integer k for which a random k-dimensional central section of K is 4-equivalent to Euclidean. In [Msch2] it is proved that 1 C n ≤ t(K)k(K) ≤ Cn, where C > 0 is an absolute constant. Having this duality in mind, one may view Theorem 3.8 as a global analogue (for bodies with minimal M ) of the isomorphic version of Dvoretzky’s theorem proved in [MSch3] (see also [GGM]): There exists a constant c > 0 such that, for every k ≥ c log n p every n-dimensional space K has a k-dimensional subspace F with d(F, `k2 ) ≤ c k/ log(n/k).
Let us also mention the following common property of the M -position and the minimal mean width position: If both a/M ∗ and b/M are bounded by some constant C, then the space is f (C)-isomorphic to `n2 . This is proved in [MSch2] for the M -position, and follows from Pisier’s inequality (37)
M M ∗ ≤ c log(ab) ≤ c log(C 2 M M ∗ ) n/2
n/2
for the minimal mean width position. The space X = `1 ⊕ `∞ shows that the position of the unit ball is crucial for this statement as well. We close this section with a variation of the minimal mean width position: Consider a symmetric convex body K in Rn , and the problem of minimizing M (T K)M ∗ (T K) over all T ∈ SLn . Repeating the procedure of Theorems 3.1 and 3.3 we obtain the following condition for the minimum position: Theorem 3.9: Let K be a symmetric convex body in Rn , and assume that M (K)M ∗ (K) ≤ M (T K)M ∗ (T K) for every T ∈ SLn . Then, Z Z 1 1 (38) kukK hu, θi2 σ(du) = ∗ kukK ◦ hu, θi2 σ(du), M S n−1 M S n−1
13
for every θ ∈ S n−1 . Proof: Without loss of generality we may assume that K is smooth enough. Let R ∈ L(Rn , Rn ) and ε >P0 be small enough, and write T −1 = I + εR. Then, ∞ T ∗ = (I + εR∗ )−1 = I + k=1 (−1)k εk (Rk )∗ , and our assumption about K takes the form Z Z (39) M (K)M ∗ (K) ≤ ku+εRukK σ(du) ku−εR∗ ukK ◦ σ(du)+O(ε2 ), S n−1
which implies (40) µ Z MM∗ ≤ M + ε
S n−1
h∇hK ◦ (u), Rui
S n−1
¶µ
M∗ − ε
Z
S n−1
¶ h∇hK (u), R∗ ui +O(ε2 ).
Letting ε → 0+ and replacing R by −R, we have Z Z 1 1 h∇hK (u), R∗ uiσ(du) h∇hK ◦ (u), Ruiσ(du) = ∗ (41) M S n−1 M S n−1 for every R ∈ GLn . Using (40) with Rθ (x) = hx, θiθ, θ ∈ S n−1 , we get Z Z 1 1 h∇hK (u), θihu, θiσ(du) (42) h∇hK ◦ (u), θihu, θiσ(du) = ∗ M S n−1 M S n−1 for every θ ∈ S n−1 . Taking into account Lemma 3.2, we conclude the proof. 2 We do not know if (38) implies the minimality condition of Theorem 3.9 (nevertheless, we find (38) quite appealing, since it demonstrates once again the deep relation between a body and its polar). 4. Quermassintegrals and Volume Preserving Transformations We say that a convex body K minimizes Wi if Wi (K) ≤ Wi (T K) for every volume preserving linear transformation T . Since nW1 (K) = ∂(K), a body K minimizes W1 if and only if it has minimal surface area. Also, since 2Wn−1 (K) = ωn w(K), a body K minimizes Wn−1 if and only if it has minimal mean width. Our purpose is to find necessary and sufficient conditions for a convex body K to minimize Wi , i = 1, . . . , n − 1. We first show that such a body is a solution of a much more general problem: Proposition 4.1: Let i = 1, . . . , n − 1, and assume that the convex body K minimizes Wi . Then, (1)
V (T1 K, . . . , Tn−i K, Dn ; i) ≥ Wi (K)
for any T1 , . . . , Tn−i ∈ SLn .
14
Proof: We have Wi (Tj K) ≥ Wi (K), j = 1, . . . , n − i. As a consequence of the Alexandrov-Fenchel inequality we see that (2)
1
1
V (T1 K, . . . , Tn−i K, Dn ; i) ≥ Wi (T1 K) n−i . . . Wi (Tn−i K) n−i ,
and this proves our claim. 2 The arguments we used for the surface area and the mean width apply to every quermassintegral and provide necessary conditions for the minimal position: Proposition 4.2: Assume that K is smooth enough and minimizes Wi . Then, Z (3) h∇hK (u), RuidSn−i−1 (K, u) = [trR]Wi (K) S n−1
for any R ∈ L(Rn , Rn ).
Proof: Let T ∈ L(Rn , Rn ) and ε > 0 be small enough. Then, (I + εT )/[det(I + εT )]1/n is volume preserving. Therefore, (4)
[det(I + εT )]
n−i n
Wi (K) ≤ V ((I + εT )K; n − i, Dn ; i).
Since (I + εT )K ⊆ K + εT K, using the monotonicity of the mixed volumes we get (5)
[det(I + εT )]
n−i n
Wi (K) ≤ V (K + εT K; n − i, Dn ; i).
n−i
2 We have [det(I + εT )] n = 1 + ε n−i n trT + O(ε ), and linearity of the mixed volumes with respect to its arguments shows that V (K + εT K; n − i, Dn ; i) = Wi (K) + (n − i)εV (T K, K; n − i − 1, Dn ; i) + O(ε2 ). Letting ε → 0+ we see that Z trT 1 (6) Wi (K) ≤ V (T K, K; n − i − 1, Dn ; i) = hT K (u)dSn−i−1 (K, u). n n S n−1
Now, let R ∈ L(Rn , Rn ) and set T ∗ = I + εR where ε > 0. Since hT K (u) = hK (T ∗ u) = hK (u + εRu), we get Z 1 trR Wi (K) ≤ hK (u + εRu)dSn−i−1 (K, u). (7) Wi (K) + ε n n S n−1 But, hK (u + εRu) = hK (u) + εh∇hK (u), Rui + O(ε2 ), so letting ε → 0+ and using (2.5), we have Z 1 trR Wi (K) ≤ (8) h∇hK (u), RuidSn−i−1 (K, u). n n S n−1 Replacing R by −R we get the reverse inequality, therefore Z (9) [trR]Wi (K) = h∇hK (u), RuidSn−i−1 (K, u) S n−1
15
for every R ∈ L(Rn , Rn ). 2
Proposition 4.3: Let i = 1, . . . , n − 1. If a convex body K in Rn minimizes Wi , then Sn−i (K, ·) is isotropic. Proof: Assume that K minimizes Wi . For every U ∈ SLn we have Wi (U K) = V (K; n − i, U −1 Dn ; i) ≥ Wi (K).
(10)
Let T ∈ L(Rn , Rn ) and ε > 0 be small enough. Then, U −1 = (I + εT )/[det(I + εT )]1/n is volume preserving, therefore V (K; n − i, Dn + εT Dn ; i) ≥ [det(I + εT )]i/n Wi (K).
(11)
2 Observe that the right hand side is Wi (K) + iεtrT n Wi (K) + O(ε ), while the left 2 hand side is Wi (K) + iεV (K; n − i, Dn ; i − 1, T Dn ) + O(ε ). Letting ε → 0+ and taking into account (2.3) we get Z trT 1 Wi (K) hT Dn (u)dSn−i (K, u) ≥ (12) n S n−1 n
for every T ∈ L(Rn , Rn ). Let R ∈ L(Rn , Rn ) and set T ∗ = I + εR. We have hT Dn (u) = |T ∗ u| = |u + εRu| = 1 + εhu, Rui + O(ε2 ), so (12) becomes Z (13) {1 + εhu, Rui + O(ε2 )}dSn−i (K, u) ≥ nWi (K) + ε[trR]Wi (K). S n−1
Letting ε → 0+ , using (2.6) and replacing R by −R we conclude that Z (14) hu, RuidSn−i (K, u) = [trR]Wi (K) S n−1
for every R ∈ L(Rn , Rn ). This shows that Sn−i (K, ·) is isotropic. 2 In order to proceed we need to introduce some terminology and notation. If A is a selfadjoint linear transformation of Rn , we denote by sj (A) the j-th elementary symmetric function sj (λ1 , . . . , λn ) of the eigenvalues λ1 , . . . , λn of A: X (15) sj (A) = λk1 . . . λ kj . 1≤k1 0. We can find xε such that kxε kK = |xε | = 1 and (7)
kxε + εT xε kK ≥
tr(I + εT ) trT =1+ε . n n
We write kxε + εT xε kK = 1 + εh∇kxε kK , T xε i + O(ε2 ), and from (7) we get + such that xεm → x ∈ h∇kxε kK , T xε i ≥ trT n + O(ε). Choosing again εm → 0 n−1 S , we see that x is a contact point of K and Dn which satisfies (8)
h∇kxkK , T xi ≥
20
trT . n
Moreover, since ∇kxkK is the point on the boundary of K ◦ at which the outer unit normal is parallel to x and x is a contact point of K and Dn , we must have ∇kxkK = x. This proves the theorem. 2 ¿From Theorem 5.1 we can easily recover all the well-known properties of the maximal volume ellipsoid: √ Theorem 5.2: Let Dn be the maximal volume ellipsoid of K. Then, K ⊂ nDn .
Proof: Let x ∈ Rn and consider the map T y = hy, xix. By Theorem 5.1, we can find a contact point z of K and Dn such that hz, T zi ≥
(9)
trT |x|2 = . n n
But, hz, T zi = hz, xi2 ≤ kzk2K ◦ kxk2K = kxk2K .
(10)
√ nkxkK . This is equivalent to the assertion of the theorem. 2 √ Theorem 5.2 provides the estimate d(X, `n2 ) ≤ n for the Banach-Mazur distance from an arbitrary n-dimensional normed space to `n2 . From Theorem 5.1 we can also deduce the Dvoretzky-Rogers lemma: Therefore, |x| ≤
Theorem 5.3: Let Dn be the maximal volume ellipsoid of K. There exist pairwise orthogonal vectors y1 , . . . , yn in Rn such that µ
n−i+1 n
¶1/2
≤ kyi kK ≤ |yi | = 1 ,
i = 1, . . . , n.
Proof: We define the yi ’s inductively. The first vector y1 can be any of the contact points of K and Dn . Assume that y1 , . . . , yi−1 have been defined. Let Fi = span{y1 , . . . , yi−1 }. Then, tr(PFi⊥ ) = n − i + 1, and by Theorem 5.1 there exists a contact point xi such that (11)
|PFi⊥ xi |2 = hxi , PFi⊥ xi i ≥
It follows that kPFi xi k ≤ |PFi xi | ≤ Then, (12)
p
n−i+1 . n
(i − 1)/n. We set yi = PFi⊥ xi /|PFi⊥ xi |.
1 = |yi | ≥ kyi kK ≥ hxi , yi i = |PFi⊥ xi | ≥
µ
n−i+1 n
¶1/2
. 2
Note that the argument shows that for every k-dimensional subspace F there exists a contact point x of K and Dn such that |PF x|2 = hx, PF xi ≥ k/n.
21
Finally, a separation argument and Theorem 5.1 give us John’s representation of the identity: Theorem 5.4: Let Dn be the maximal volume ellipsoid of K. There exist contact points x1 , . . . , xm of K and Dn and positive real numbers λ1 , . . . , λm such that I=
m X i=1
λi x i ⊗ x i .
Proof: Consider the convex hull C of all operators x⊗x, where x is a contact point of K and Dn . One can easily see that the assertion of the theorem is equivalent to I/n ∈ C. If this is not true, there exists T ∈ L(Rn , Rn ) such that hT, I/ni > hx ⊗ x, T i
(13)
for every contact point x. But, hT, I/ni = trT /n and hx ⊗ x, T i = hx, T xi. Therefore, (13) would contradict Theorem 5.1. 2 Theorem 5.4 implies that m X
(14)
i=1
λi hxi , θi2 = 1
n−1
for every θ ∈ S . In our terminology, the measure µ on S n−1 that gives mass λi to the point xi , i = 1, . . . , m, is isotropic. In this sense, John’s position is an isotropic position. Conversely, following [Ba4] we have: Proposition 5.5: Let K be a symmetric convex body in Rn which contains the Euclidean unit ball Dn . Assume that there exists an isotropic Borel measure µ on S n−1 which is supported by the contact points of K and Dn . Then, Dn is the maximal volume ellipsoid of K. Proof: Let kµk = µ(S n−1 ) and A ⊂ S n−1 be the support of µ. Define (15)
L = {y ∈ Rn : |hx, yi| ≤ 1, x ∈ A}.
Since K ⊆ L, it clearly suffices to prove that Dn is the maximal volume ellipsoid of L. Let (16)
E = {y ∈ Rn :
n X j=1
αj−2 hy, vj i2 ≤ 1},
where {vj } is an orthonormal basis of Rn and αj > 0. Assume that E ⊆ L. For every x ∈ A we have −1/2 n n X X 2 2 αj hx, vj i αj hx, vj ivj ∈ E ⊆ L, (17) y(x) = j=1
j=1
22
hence, |hx, y(x)i| ≤ 1 gives n X
(18)
j=1
αj2 hx, vj i2 ≤ 1 ,
x ∈ A.
Our hypotheses imply that n X
n X
n αj αj = kµk j=1 j=1
(19)
=
n kµk
Z
n X
S n−1 j=1
Z
S n−1
hx, vj i2 µ(dx)
αj hx, vj i2 µ(dx).
Using (18) and the Cauchy-Schwarz inequality we see that
(20)
n X j=1
1/2 1/2 n n X X αj2 hx, vj i2 hx, vj i2 αj hx, vj i2 ≤ ≤1 j=1
j=1
for every x ∈ A. Then, (19) becomes (21)
n X j=1
αj ≤ n.
Q By the arithmetic-geometric means inequality we get αj ≤ 1. That is |E| ≤ |Dn |. Moreover, we can have equality only if all αj ’s are equal to 1, which shows that Dn is the unique maximal volume ellipsoid of L. 2 Theorem 5.4 and Proposition 5.5 provide the following characterization of John’s position: “Let K be a symmetric convex body in Rn which contains the Euclidean unit ball Dn . Then, Dn is the maximal volume ellipsoid of K if and only if there exists an isotropic measure µ supported by the contact points of K and Dn .” Let us discuss one more problem of the same nature: Let K be a symmetric convex body in Rn and k · k be the corresponding norm. Assume that (1/a)|x| ≤ kxk ≤ b|x| for every x ∈ Rn . It is clear that M (K)a(K) ≥ 1, and we are interested in (22)
min{M (T K) | T ∈ GLn , a(T K) = 1}.
The condition a(T K) = 1 means that T K ⊆ Dn but there exist contact points of T K and Dn . We then have the following condition for the minimum position:
23
Theorem 5.6: Let K be a symmetric convex body in Rn satisfying a(K) = 1 and M (K) ≤ M (T K) for every T ∈ GLn with a(T K) = 1. Then, for every θ ∈ S n−1 we can find contact points x1 , x2 of K and Dn such that Z n+1 (23) 1 + hx1 , θi2 ≤ kukK hu, θi2 σ(du) ≤ 1 + hx2 , θi2 . M S n−1 Proof: Let T ∈ L(Rn , Rn ) and ε > 0 be small enough. Then T1 := (minS n−1 kx + εT xk)(I + εT )−1 satisfies a(T1 K) = 1. Therefore, Z ku + εT ukσ(du) ≥ M (K) min kx + εT xk. (24) n−1 x∈S
S n−1
If we write ku + εT uk = kuk + εh∇hK ◦ (u), T ui + O(ε2 ), we see that Z minS n−1 kx + εT xk − 1 . (25) h∇hK ◦ (u), T uiσ(du) + O(ε) ≥ M (K) ε n−1 S Let xε be a point on S n−1 at which the minimum is attained. If x is a contact point of K and Dn , we must have 1 + εkT k ≥ kx + εT xk ≥ kxε + εT xε k ≥ kxε k − εkT k, where kT k := kT : `n2 → XK k. It follows that 1 ≤ kxε k ≤ 1 + 2εkT k.
(26)
Since xε ∈ S n−1 and k · k ≥ | · |, (25) takes the form Z |xε + εT xε | − 1 (27) h∇hK ◦ (u), T uiσ(du) + O(ε) ≥ M (K) ε S n−1 = M (K)[hxε , T xε i + O(ε)].
Now, we can find a sequence εm → 0 and a point x ∈ S n−1 such that xεm → x. Letting m → ∞ in (27), we obtain Z h∇hK ◦ (u), T uiσ(du) ≥ M (K)hx, T xi. (28) S n−1
Also, x ∈ S n−1 and using (26) we see that kxk = limm kxεm k = 1. That is, x is a contact point of K and Dn . Replacing T by −T we find another contact point x0 of K and Dn such that Z (29) h∇hK ◦ (u), T uiσ(du) ≤ M (K)hx0 , T x0 i. S n−1
Choosing Tθ (x) = hx, θiθ, θ ∈ S n−1 , and applying Lemma 3.2, we obtain (23). 2 The condition of the Theorem shows in a sense that the minimum position of the problem is rich in contact points with the circumscribed ball. The dual problem of maximizing M under the condition b = 1 has exactly the same answer.
24
6. Minimal Surface Area and M -position If K and L are convex bodies in Rn , we write N (K, L) for the covering number of K by L (that is, the minimum number of translates of L whose union covers K). If |K| = |Dn |, we say that K is in M -position (with parameter δ > 0) if (1)
N (K, Dn ) ≤ exp(δn).
One can then prove (see [MP2] for the non-symmetric case) that (2)
N (K, Dn ) · N (Dn , K) · N (K ◦ , Dn ) · N (Dn , K ◦ ) ≤ exp(δ1 n),
where δ1 = cδ, and c > 0 is an absolute constant. Moreover, condition (1) is equivalent to (3)
|K + Dn |1/n ≤ c|Dn |1/n .
This isomorphically defined position is the best representative of the affine class of a body in volume computations: this is mainly due to the fact that reverse Brunn-Minkowski inequalities hold for bodies in M -position [M2]. We define a function f : [0, +∞) → R by (4)
f (t) = min{|T K + tDn | | T ∈ SLn }.
For every t > 0 there exists a volume preserving Tt such that |Tt K + tDn | = f (t). It is clear that U Tt has the same property for every U ∈ O(n). By (3) we see that T1 K is in M -position. This suggests that M -position can be described as the solution of a minimum problem similar to the ones we discussed in the previous sections. We start with the following observation: Lemma 6.1: Let K be a convex body in Rn . Then, (5)
|K + tA1 Dn + sA2 Dn | ≥ min{|K + (t + s)A1 Dn |, |K + (t + s)A2 Dn |}
for every A1 , A2 ∈ GLn and t, s > 0. Proof: It is an immediate consequence of the Brunn-Minkowski inequality, since (6) s t (K + (t + s)A1 Dn ) + (K + (t + s)A2 Dn ) . 2 K + tA1 Dn + sA2 Dn ⊇ t+s t+s Theorem 6.2: Let K be a convex body in Rn . Assume that (7)
|K + tDn | = f (t)
for some t > 0. Then, K + tDn has minimal surface area.
25
Proof: Let T ∈ SLn . From Steiner’s formula we see that (8)
|T (K + (t − ε)Dn ) + εDn | − |T (K + (t − ε)Dn )| = nεW1 (T (K + (t − ε)Dn ) + O(ε2 ).
By the continuity of W1 with respect to the Hausdorff metric, (9)
∂(T (K + tDn )) = nW1 (T (K + tDn )) = n lim W1 (T (K + (t − ε)Dn )) ε→0+
|T (K + (t − ε)Dn ) + εDn | − |T (K + (t − ε)Dn | ε −1 |K + (t − ε)Dn + εT Dn | − |K + (t − ε)Dn | . = lim ε ε→0+ Since |K + tDn | = f (t), Lemma 6.1 implies that |K + (t − ε)Dn + εT −1 Dn | ≥ |K + tDn |. Hence, = lim+ ε→0
(10)
∂(T (K + tDn )) ≥ lim
ε→0+
|K + tDn | − |K + (t − ε)Dn | = ∂(K + tDn ). ε
This shows that K + tDn has minimal surface area. 2 Remark. It is not hard to show that f 0 (t) = ∂(Tt K + tDn )
(11)
for every t > 0. It follows that for every t > s > 0 we have Z t Z t ∂(Tt K + xDn )dx, ∂(Tx K + xDn )dx ≥ (12) s
s
with equality if s = 0. In the planar case, a convex body K has minimal perimeter (surface area) if and only if it has minimal mean width. Since |Tt K + tDn | = f (t), Theorem 6.2 shows that Tt K + tDn has minimal mean width and, using Corollary 3.4(ii) we see that Tt K has minimal mean width. Moreover, Tt is constant up to an orthogonal transformation. That is, the solution of Problem (4) is the minimal mean width position, independently of t > 0: Corollary 6.3: A convex body K in R2 satisfies |K + tDn | ≤ |T K + tDn | for every T ∈ SLn and every t > 0 if and only if it has minimal mean width. 2
It would be interesting to see if the minimal surface area position is an M position in higher dimensions. This would provide an isometric description of the M -position. Observe that, by Theorem 6.2, the limit of Tt K as t → 0+ is the minimal surface position and, by Steiner’s formula, the limit of Tt K as t → +∞ is the minimal mean width position.
26
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[Sch]
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[TJ]
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A.A. Giannopoulos: Department of Mathematics, University of Crete, Iraklion, Greece. E-mail:
[email protected] V.D. Milman: Department of Mathematics, Tel Aviv University, Tel Aviv, Israel. E-mail:
[email protected]
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