Notes for seminar on volume ratio. These are notes for a few seminar talks on volume ratio and related notions delivered in fall 2009 and winter 2010.

Transcription

1 Notes for seminar on volume ratio These are notes for a few seminar talks on volume ratio and related notions delivered in fall 2009 and winter 200. Contents Introduction 2 2 Ball s recise uer bounds on volume ratio 3 3 Low volume ratio yields somewhat Euclidean sections 7 4 Digression: Convex bodies are siky 5 Alications of Szarek s theorem 2 6 Volume ratio of B n 5 7 Summary of arameters 7 8 Loose end: Control by ɛ-nets 8 9 Loose end: Construction of ɛ-nets 20 0 Comarison with Milman s theorem 2 References 25 Steven Taschuk 20 August 5 htt://

2 Introduction The volume ratio of a convex body K in R n is Nov 23 vr(k) = inf { /n vol(k) : E is an ellisoid and E K}. vol(e) It is easy to see that this quantity is affine invariant, meaning that for any invertible affine ma T we have vr(tk) = vr(k). Thus /n vol(kjohn ) vr(k) = vr(k John ) = vol(b n 2 ), where K John denotes an affine image of K which has B n 2 as its maximum volume ellisoid, as in John s theorem. In fact, slightly more is true than affine invariance: we have vr(k) d(k, L) vr(l). () Here d is one of the two natural generalizations of Banach Mazur distance to convex bodies that are not necessarily symmetric. The usual one is d(k, L) = inf {λ: λ > 0 and SL TK λsl for some invertible affine S, T}, but if we wish to allow the inner and outer L to be negative homothets, we use instead d(k, L) = inf { λ : SL TK λsl for some invertible affine S, T}. Clearly d(k, L) d(k, L); we have equality if one (or both) of K and L is symmetric. To rove (), consider any ellisoid E, any invertible affine mas S, T, and any real number λ such that Then vr(k) = vr(tk) E SL TK λsl. /n vol(tk) vol(e) /n vol(λsl) = λ vol(e) /n vol(sl). vol(e) As Sasha ointed out, this statement is slightly trickier than it aears. We d like to say that if L K λl (where λ > 0) and L is symmetric, then we can just relace λl by λl. But this argument assumes the centre of homothety (i.e., the origin wrt which we scale L to λl) is the same as the centre of symmetry (i.e., the origin for which L = L), and this is not art of the hyothesis. So we need to be a little more careful. But the idea is easy: if L is symmetric, then any negative homothet of L is also a ositive homothet (with the same absolute ratio, but with a different centre), so if K can be sandwiched between L and a negative homothet of some ratio then iso facto it can be sandwiched between L and a ositive homothet of the same ratio. Steven Taschuk 20 August 5 htt:// 2

3 Taking the infimum over all E such that E SL yields vr(k) λ vr(sl) = λ vr(l), and then taking the infimum over S, T, λ yields the desired result. An immediate corollary is that vr(k) d(k, B n 2 ) vr(b n 2 ) = d(k, B n 2 ) d(k, B n 2 ). From John s theorem we have well-known uer bounds on d(k, B n 2 ), yielding vr(k) n { n if K is symmetric, in general. (2) This estimate for symmetric K is retty good (as we will see, it s the right order), but the estimate for the general case is quite bad 2 ; the correct uer bound is again c n. 2 Ball s recise uer bounds on volume ratio Ball s Theorem 3 For any convex body K in R n, { vol(b n ) if K is symmetric, vol(k John ) vol(4 n ) in general. Here 4 n denotes the regular simlex in John s osition; it turns out that vol(4 n ) = nn/2 (n + ) (n+)/2 n! ce n. (3) (The asymtotic statement follows by Stirling s aroximation.) A corollary of Ball s theorem is that the cube B n has the highest volume ratio of all symmetric convex bodies, and the simlex 4 n has the highest volume ratio in general. (Another corollary is the reverse isoerimetric inequality; see [4].) To rove Ball s theorem we need a few facts about John s osition. John s Theorem 4 Every convex body K in R n contains a unique ellisoid of maximum volume. That ellisoid is B n 2 if and only if Bn 2 K and there exist contact oints (u i ) m bd K \ bd Bn 2 and ositive weights (c i) m such that 2 As Sasha mentioned, we can get the right order by comaring K to K \ K: as shown by Stein [7], for any K there exists a choice of origin so that vol(k \ K) 2 n vol(k), and it easily follows that vr(k) 2 vr(k \ K) 2 n. 3 The symmetric case aeared first in [], and the general case in [2]. A comlete treatment also aears in [4]. 4 The original aer of John [0] can be somewhat difficult to obtain through the usual methods; I have a coy if anyone is interested. Other roofs can be found in [3] and [8]. Steven Taschuk 20 August 5 htt:// 3

4 (i) c i u i u i = Id, and (ii) c i u i = 0. In condition (i), which is called John s decomosition of the identity, x y denotes the ma R n R given by x y(t) = hx, tiy. This ma is linear and has matrix yx T (taking x, y to be column vectors); note that tr(x y) = hx, yi. Unacking the definitions in condition (i) yields the equivalent statement 8x 2 R n : ci hu i, xiu i = x. Taking the inner roduct with x on both sides yields 8x 2 R n : ci hu i, xi 2 = x 2. (4) (In fact this statement is equivalent to condition (i).) This is one resect (among many) in which the vectors u i behave somewhat like an orthonormal basis. Since u i =, we have tr(u i u i ) =, so taking traces in condition (i) yields ci = n. (5) Also since u i =, the ma u i u i is the orthogonal rojection onto san {u i }. Since u i has the same san, the ma ( u i ) ( u i ) is the same rojection. If K is symmetric 5, then u i 2 bd K \ bd B n 2, that is, u i is also a contact oint; thus relacing c i u i u i with c i 2 u i u i + c i 2 ( u i) ( u i ) reserves condition (i) and makes condition (ii) automatic. Thus if K is symmetric then we can ignore condition (ii), or assume it freely, as we wish. Since u i 2 bd K, K has a suorting halfsace at u i. Since u i 2 bd B n 2 and B n 2 K, this halfsace also suorts Bn 2 at u i; but B n 2 only has one suorting halfsace at u i, and it is {x: hu i, xi }. Thus, defining ek = {x: (8i: hu i, xi )}, (6) we have K e K. When K is symmetric, we will instead define ek = {x: (8i: hu i, xi )}, (7) 5 An imortant detail: if K is symmetric then its maximum volume ellisoid has the same centre of symmetry. Indeed, if K = K and E K, then E K, and so conv(e [ E) K; if E 6= E then one can show that E can be stretched (in the direction joining the centres of E and E) to obtain an ellisoid of larger volume in K. Thus if K is symmetric then K John = K John. Steven Taschuk 20 August 5 htt:// 4

5 taking advantage again of the fact that in this case, u i and u i are both contact oints. The second major tool for Ball s theorem is the following inequality. Normalized Brascam Lieb Inequality 6 If (u i ) m are unit vectors and (c i) m are ositive real numbers such that c i u i u i = Id, then for any measurable nonnegative functions (f i ) m, R n fi (hu i, xi) ci dx R f i (t) dt ci. This is another examle of how the u i behave somewhat like an orthonormal basis; indeed, if the u i are an orthonormal basis then (taking all c i = ) we have equality. Now we can rove Ball s theorem. The symmetric case is quick: let K = K, let K be in John s osition, with (c i ) and (u i ) as in John s theorem, and e K as in (7). Then vol(k) vol(e K) = [x 2 K] e dx = [ hui, xi ] R ci dx n R n ci [ t ] dt = 2 ci = 2 c i = 2 n = vol(b n ), R making use of (5), the fact that B n = [, ] n, and the Iverson bracket notation, whereby { if P is true, [P] = 0 if P is false. If we try to reeat this roof in the general case, with e K defined by (6) instead of (7), we have [hu i, xi ] instead of [ hu i, xi ], and at the end we are confronted with an integral over (, ] instead of over [, ]. To remedy this, we introduce a weight function w(t); we wish to end u with ci [t ] w(t) dt, R where w(t) will be chosen as some function that decays to zero (as t ) fast enough that this integral is finite. In the revious ste, then, we should have R n [hui, xi ] ci w(hu i, xi) ci dx. 6 This name and formulation are due to Ball []; the original version of Brascam and Lieb is in [7]. Barthe ([5], [6]) gives a fairly simle roof of the inequality in this formulation, which is reeated in [4]. Steven Taschuk 20 August 5 htt:// 5

6 In order to introduce these weights at this oint, we desire that w(hui, xi) ci =, or equivalently, taking logs, ci ln w(hu i, xi) = 0. Since c i u i = 0, it is easy to see that we will obtain this condition by taking w(t) = e t. Carrying out this lan yields vol(k) e n, which as we see from (3) is asymtotically shar. The recise uer bound obtained by Ball requires another idea, which is motivated by the observation that the nicest way to resent an n-dimensional simlex is as the convex hull of the n + standard basis vectors in R n+, or equivalently, as a section of the ositive orthant of R n+. Likewise, we will relace our body K e with a cone in R n+ : the contact oints whose olar is K e will be relaced by a set of normal vectors for the cone; we will construct these normal vectors so that they give a decomosition of the identity in R n+, and aly the normalized Brascam Lieb inequality. (We will also use the idea of introducing an exonential weight to each factor.) So, define weights (d i ) m and unit vectors (v i) m in Rn+ by Nov 30 d i = n + n c i and v i = n + nui 2 R n R = R n+. As desired, these vectors, with these weights, give a decomosition of the identity as in John s theorem: di v i v T i = n + n c i nui nu T i n + " nui u T # i nui = nu T i = c i n In 0 0 T = I n+. They do not give a balanced configuration, since di v i = n + n c nui i n + = 0 n +. Define the cone C = {y 2 R n+ : (8i: hv i, yi 0)}. Steven Taschuk 20 August 5 htt:// 6

7 For y = (x, r) 2 R n R, we have y 2 C (8i: hv i, yi 0) (8i: h nu i, xi r 0) (8i: hu i, xi r 0 and x 2 r n ) r n e K (The condition r 0 is obtained because c i u i = 0, and so the values hu i, xi cannot all be negative.) By the normalized Brascam Lieb inequality, R n+ [hvi, yi 0] e dihv i,yi dy R On the other hand, [hvi, yi 0] e dihv i,yi dy R n+ = [y 2 C] e h div i,yi dy R n+ = [x 2 r nk] e e r n+ dx dr R n = = = r n n e r n+ dr vol(e K) [t 0] e t dt di =.! n t e t dt vol(e K) (t = r n + ) n(n + ) n + n! n n/2 (n + ) (n+)/2 vol( e K). Therefore as desired. vol(e K) n n/2 (n + ) (n+)/2 n! = vol(4 n ), 3 Low volume ratio yields somewhat Euclidean sections My treatment of this toic follows [5], chater Nicole exlained the history: in 977, Kašin [2] (whose name is rendered Kashin in some transliteration systems) showed, for K = B n, that there exists a decomosition Rn = F F? of the tye described in corollary 7; in 978, Szarek [8] gave a different roof, using the method shown here, but did not exlicitly isolate the notion of volume ratio; Szarek and Tomczak-Jaegermann [9] did that in 980, and stated exlicitly the result which I here call Szarek s theorem. Perhas it should be called the Kašin Szarek theorem, or the Szarek Tomczak theorem. Steven Taschuk 20 August 5 htt:// 7

8 Szarek s Theorem Let K be a convex body in R n with B n 2 K and vol(k) vol(b n 2 ) /n A. Then, with high robability, a random k-dimensional subsace F satisfies 8x 2 F: kxk K x (4πA) n/(n k) kxk K. Here F is random in the sense of the uniform (i.e., rotationally invariant) robability measure on the Grassmannian G n,k, and the statement holds with robability at least 2 n. The inequality kxk K x is immediate from the hyothesis that B n 2 K, so we need only rove the second inequality. Furthermore, by homogeneity it is enough to consider x 2 S n \ F. The lan of the roof is: (A) Since the volume of K is small, its radial function is small (i.e., k k K is large) at an average oint of S n. (B) Thus the norm is large at an average oint of an average section of S n. (C) Thus the norm is large at most oints of most sections of S n. (D) Thus the norm is large at all oints of most sections of S n. We need several lemmas, of whose roofs I will say very little. Lemma If f: R n R is measurable and nonnegative, then f(x) dx = f(rθ)r n dr dθ R n S n 0 = n vol(b n 2 ) f(rθ)r n dr dσ(θ) S n 0 where in the first line, dθ indicates Lebesgue measure on S n, and in the second, σ is the uniform (i.e., rotationally invariant) robability measure on S n. (The factor n vol(b n 2 ) = vol n (S n ) is the normalizing factor.) Lemma 2 For a starshaed body K in R n, vol(k) = vol(b n 2 ) S n kθk n dσ(θ). K Proof Aly lemma to the characteristic function of K. Steven Taschuk 20 August 5 htt:// 8

9 Lemma 2 exresses the volume of K as a kind of average of its norm, which is what we need for art (A). Lemma 3 For a nonnegative measurable function f: S n R, f(θ) dσ(θ) = f(θ) dσ F (θ) dµ(f), S n G n,k S n \F where σ is the uniform robability measure on S n, σ F is the uniform robability measure on S n \ F, and µ is the uniform robability measure on G n,k, the Grassmannian of k-dimensional subsaces of R n. Proof Define bσ(a) = G n,k σ F (A \ F) dµ(f), and show that bσ is a rotationally invariant robability measure on S n, whence bσ = σ. (The integral on the RHS is the integral wrt bσ.) Lemma 3 exresses the intuitively obvious equivalence between the notions of average oint on S n and average oint on average section of S n, as we need for art (B). Part (C) needs no lemmas, as to ass from average behaviour to behaviour in most laces we need merely invoke Markov s inequality. Lemma 4 For any x 0 2 S n and any δ 2 [0, 2], n δ σ(x 2 S n : x x 0 δ). π (See section 9 for a variant of this lemma.) In art (D) we wish to deduce the norm is large everywhere on the shere, knowing only that the norm is large for most oints, in the sense that the set of oints where the norm is small has small measure. We will deduce that the set of oints where the norm is small is small in terms of the metric (which is the role of lemma 4), and then use a Lischitz condition to deduce that the function cannot be very small anywhere. This argument is encasulated in the next and final lemma. Lemma 5 Let f: S n R be Lischitz and let t 2 [0, ( 2/π) n ]. If σ(f r) < t then, for every θ 2 S n, f(θ) > r πt /n. Proof Let δ = πt /n 2 [0, 2]. The hyothesis and lemma 4 show that {f r} contains no ca of radius δ. Thus every oint θ 2 S n is within δ of a oint ψ where f(ψ) > r, and so f(θ) f(ψ) δ > r πt /n. Now we can rove Szarek s theorem. Let K be as described therein. Then Dec Steven Taschuk 20 August 5 htt:// 9

10 A n vol(k) vol(b n 2 ) (hyothesis) = S n kθk n dσ(θ) (lemma 2; art (A)) K = G n,k S n \F kθk n dσ F (θ) dµ(f) (lemma 3; art (B)) K Let Ω 0 = {F 2 G n,k : dσ S n \F kθk n F (θ) < (2A) n }. By Markov s inequality, K µ(ω 0 ) = µ(ω 0 ) (2A) n G n,k S n \F kθk n dσ F (θ) dµ(f) K 2 n. (Ω 0 is the set of most sections described in art (C).) Let F 2 Ω 0. Then, for r > 0 to be chosen later, we have (by Markov s inequality again) σ F (θ: kθk K r) = σ F (θ: kθk n r ) K n r n S n \F kθk n dσ F (θ) K (2Ar) n (That comletes art (C).) As reviously noted, since B n 2 K we have k k K, and so kxkk kyk K kx ykk ky xk K x y, that is, k k K is Lischitz. (Note that since K is not assumed symmetric, we cannot use the usual other triangle inequality kxk kyk kx yk here.) So by lemma 5, rovided that r is later chosen so that 8θ 2 S n \ F: kθk K r π(2ar) n/k, (2Ar) n k 2. (8) π (Note that we aly lemma 5 to S n \ F, which is essentially S k, not to S n.) Finally, we choose r so that our estimate r π(2ar) n/k is ositive. The simlest way is to make the second term half of the first, that is, π(2ar) n/k = r 2. Thus we will take r n k = 2 π k (4A) n. Steven Taschuk 20 August 5 htt:// 0

11 k, To check that this value of r satisfies (8), note first that (2Ar) n = 2π so it suffices to check that r 2 r n k 2; and indeed, 2 =, so in fact π k (4A) n r 2 2. Thus we obtain the estimate kθk K r 2 (4πA) = θ. n/(n k) (4πA) n/(n k) The homogeneity of the norm yields the desired statement, comleting the roof of Szarek s theorem. r 4 Digression: Convex bodies are siky A key maneuver in the roof of Szarek s theorem is to aly Markov s inequality to the formula exressing volume as a kind of average of radius (lemma 2). Writing this idea down searately, we have: Proosition 6 If K is a star-shaed body in R n, then σ(s n \ rk) vol(rk) vol(b n 2 ). Proof σ(s n \ rk) = σ(s n \ r K) = σ(θ 2 S n : kθk K r ) = σ(θ 2 S n : kθk n r ) K n r n S n kθk n K dσ(θ) = r n vol(k) vol(b n 2 ) = vol(rk) vol(b n 2 ) Since σ is a robability measure, this uer bound is trivial when vol(rk) vol(b n 2 ), but when r decreases ast this oint, this uer bound decreases very raidly, indeed, like r n. So if K is even a little bit smaller (in volume) than the Euclidean ball, then it doesn t stick out very much (in terms of area on the shere); however, as we will see when we comute vr(b n ), a convex body with such volume can stick out quite far (in terms of distance from the origin). This is one reason that high-dimensional convex bodies should be drawn siky, 8 even though this makes the icture nonconvex: 8 Peter ointed out that siky ictures are good for showing measure henomena, but, as ever, we should be alert that they do not show metric henomena well. He also drew my attention to a brief discussion of exactly this kind of icture in [3], which refers to [9] for a similar exonential decay of volume, but using slabs instead of balls. Steven Taschuk 20 August 5 htt://

12 5 Alications of Szarek s theorem Corollary 7 Let n be even, and let K be a convex body in R n with B n 2 K and vol(k) vol(b n 2 ) /n A. Then there exists a subsace F of dimension n 2 such that 8x 2 F [ F? : kxk K x (4πA) 2 kxk K. Proof Say that a subsace is good if the stated inequality holds for vectors in that subsace. By Szarek s theorem, the set of good subsaces F has measure at least 2 > n 2. And since the ma F 7 F? reserves the measure on G n,k (indeed, one can check that the measure bµ(a) = µ(f? : F 2 A) is rotationally invariant, hence equal to µ), the set of subsaces F such that F? is good also has measure at least 2 > n 2. Therefore there exists a subsace F such that both F and F? are good. For examle, B n 2 nb n, and we will comute later that Jan 8 vol( nb n ) vol(b n 2 ) /n c, (9) where c is some constant (meaning in articular that it does not deend on n). Thus, for n = 2k, we obtain an orthogonal decomosition R n = F F? with dim F = k and 8x 2 F [ F? : kxk nb n x c 0 kxk nb n, Steven Taschuk 20 August 5 htt:// 2

13 that is, Another nice way to write this is 8x 2 F [ F? : kxk n x c 0 kxk. 8x 2 F [ F? : This is nice because, writing it asserts that x = n kxk n x c 0 n kxk. / xi = n n / kxk, 8x 2 F [ F? : x x 2 c 0 x, and these norms are just the usual L (µ) norms if we think of vectors in R n as functions {,..., n} R and take µ to be the uniform robability measure on {,..., n}. This motivates the notation L n for these normed saces (or for their unit balls). Corollary 8 Let n be even, and let K be a convex body in R n with B n 2 K and vol(k) vol(b n 2 ) /n A. Further assume that K = K. Then there exists a symmetric orthogonal ma Q such that 8x 2 R n : 2(4πA) 2 x 2 (kxk K + kqxk K ) kxk K kqxk K x. Proof The second inequality is obvious. The third follows from the facts that B n 2 K and Q is (will be) orthogonal. For the first inequality: let F be as in the revious corollary, let P be the orthogonal rojection onto F, let P 2 be the orthogonal rojection onto F?, and let Q = P P 2. (Q is reflection in F.) Qx F P x P 2 x x F? Qx Steven Taschuk 20 August 5 htt:// 3

14 We comute, first, that 2 (kxk K + kqxk K ) 2 kx + Qxk K = kp xk K (4πA) 2 P x. Using the fact that K = K, we have, similarly, Thus 2 (kxk K + kqxk K ) = 2 (kxk K + k Qxk K ) (4πA) 2 P 2x. 2 (kxk K + kqxk K ) (4πA) 2 ( P 2x P x ) 2(4πA) 2 x, as desired. In terms of unit balls, this corollary asserts that B n 2 K \ QK 2(K QK) 2(4πA) 2 B n 2, (0) where K L denotes the body whose norm is the sum of the norms of K and L. (It can be shown that this tye of addition is dual to Minkowski addition, in the sense that (K + L) = K L, and that we have the inclusions K L K \ L 2(K L) (K + L) conv(k [ L) K + L. 2 The third inclusion is tricky, but the others are straightforward. The last one assumes 0 2 K \ L.) In lecture 4 of [3], Ball roves a version of (0) directly. The method resembles the one used here, but the direct aroach yields several advantages: the result is roved for all n (not just even n); we obtain a better constant; the choice of Q is random and with high robability in all of O(n) (instead of just reflections in n 2 -dimensional subsaces); and we need not assume that K is symmetric. He also deduces Szarek s theorem (for even n) from this statement. The oint of these maneuvers is to show that the henomena described by Szarek s theorem (a section of the body is somewhat Euclidean) and (0) (combinations of the body and a coy of are somewhat Euclidean) are not merely arallel but, to some extent, mutually deducible. For more on this idea, but focussing on Dvoretzky s theorem, see [4]. The art of (0) that concerns K \ QK is intuitively reasonable from our Jan 25 revious discussion of sikiness : since the sikes of K occuy a small area on the shere, K and QK are not likely to have sikes in the same directions, and so intersecting them will cut off the sikes, leaving something close to B n 2. Steven Taschuk 20 August 5 htt:// 4

15 6 Volume ratio of B n As romised earlier, we will now comute the volume of B n (following the method used in [5], age ), and thereby show how the results of the revious section aly to such balls. Lemma 9 If K R n is star-shaed and 0 < <, then vol(k) = Γ( + n ) K dx. R n e kxk Proof e kxk R n K dx = R n kxk K e t dt dx = [kxk K t] e t dt dx = [kxk K t / ] e t dt dx (using > 0) = [t 0] [x 2 t / K] e t dt dx (using K star-shaed) = = 0 0 vol(t / K)e t dt t n/ e t dt vol(k) = Γ( + n ) vol(k) (Lemma 9 aears, with a slightly different roof, as Lemma 7 in [2], but I doubt this is its origin. 9 ) Proosition 0 If 0 < < then vol(b n ) = 2n Γ( + )n Γ( + n ). Proof First we comute that e kxk R n dx = e R xi dx = e x i dx n R n n n = e t dt = 2 n e dt t, 9 Sasha and Nicole susect this fact has been known for at least a hundred years. R 0 Steven Taschuk 20 August 5 htt:// 5

16 and then the substitution u = t yields n n = 2 n u e u du = 2 n Γ = 2 n Γ + n, 0 and the lemma yields the claim. This method relies on the rather clever lemma 9; it may be reassuring to know that the result for B n can also be obtained by the more obvious method of slicing. Indeed, the cross-section of B n obtained by fixing one coordinate is a scaled coy of B n ; thus vol(b n ) = vol(scaling factor B n ) dt, which with a little comutation leads to vol(b n ) vol(b n ) = 2 B, + n = = 2Γ( + n )Γ( + ) Γ( + n ), where B(s, t) is the beta integral. Induction on n then yields the desired formula. Examles. vol(b n ) = 2n n!, which can also be obtained by noting that Bn consists of 2n coies (one in each orthant) of a right-angled simlex, which is a cone whose height is and whose base is the analogous (n )-dimensional body. 2. vol(b n 2 ) = Γ ( 2 )n Γ (+ n 2 ). Since vol(b2 2 ) = π, this yields Γ( 2 ) = π. (Indeed, the method used here to comute vol(b n ) is a generalization of the classical method of comuting this value.) Thus vol(b n 2 ) = πn/2 Γ( + n 2 ). 3. The case = is not covered by roosition 0, but if we fix n and let, then by the continuity of measure, [ vol B n = lim n vol(bn ) = 2 n. Since >0 [ int B n B n Bn, >0 we thus have vol(b n ) = 2 n. (Of course we already knew this.) 0 0 Thanks to Niushan for ointing out that my original argument in this examle was wrong. Steven Taschuk 20 August 5 htt:// 6

17 4. If we fix and let n, then by Stirling s aroximation, whence vol(b n ) 2n Γ( + q )n n/, 2πn n e vol(b n ) /n 2Γ( + )(e)/ n / = c n /. Thus a natural normalization is n / B n, which is the unit ball of the L n norm mentioned on age 3. Now, recall the standard fact that for a robability measure µ, if 0 < q then k k L(µ) k k Lq(µ)). Alying this to the L n norms yields n /q B n q n/ B n. () (In fact we only care about, because we want convexity.) So if 2, then n /2 B n 2 n/ B n, with contact at the vertices of B n. Those vertices (after scaling) suort a decomosition of the identity as in John s theorem, so we conclude that n /2 B n 2 is the maximum volume ellisoid in n/ B n. Thus! /n vr(b n ) = vol(n/ B n ) vol(n /2 B n 2 ) Γ( + )(e)/ Γ( + if 2, 2 )(2e)/2 which establishes the romised result for B n (namely (9), on age 2). Again, note that the constant on the right does not deend on the dimension. On the other hand, if 2, then B n 2 is the maximum volume ellisoid of Bn, and so vol(b n /n vr(b n ) ) = vol(b n 2 ) Γ( + )(e)/ n/2 Γ( + if 2 2 )(2e)/2 n /, which is, alas, an unbounded function of n. Still, if <, this imroves the n estimate we obtained from John s theorem (see (2) on 3). 7 Summary of arameters K vol(k John ) vol(k John ) /n vr(k) d(k, B n 2 ) B n 2 n n n/2 /n! c/ n c n B n 2 π n/2 /Γ( + n 2 ) c/ n B n 2 n 2 c/ n n 4 n n n/2 (n + ) (n+)/2 /n! e c/ n n As Nicole ointed out, this also gives us an asymtotically shar lower bound on d(b n, B n 2 ). Indeed, for the range 2, we have d(b n, Bn 2 ) vr(bn ) cn/2 /, and B n 2 n /2 / B n 2. For the range 2, take duals. Bn Steven Taschuk 20 August 5 htt:// 7

18 In this table we can see that, although distance to the Euclidean ball and volume ratio are related (as in (2)), they may nevertheless be quite different: B n and B n both have extremal distance (for symmetric bodies) but their volume ratios are at the oosite ends of the scale. The table also shows again that asymmetric bodies can be much further from the ball, but asymmetry does not disturb volume ratio much. (Recall footnote 2, on age 3.) 8 Loose end: Control by ɛ-nets In the roof of Szarek s theorem, we used the following obvious fact about how controlling a (nice) function on an ɛ-net lets us control it on the whole shere: Observation Let Λ be an ɛ-net for S n and let f: S n R be c-lischitz. If α 2 R is such that 8θ 2 Λ: α f(θ), then 8θ 2 S n : α cɛ f(θ). (This observation, with c =, is imlicit in the roof of lemma 5.) When f is a norm (even an asymmetric one), there is a standard, more Feb owerful, roosition along the same lines. We need the following simle geometric roosition as a lemma. 2 Proosition Let K R n be closed and convex. Let A R n be bounded. Let 0 λ <. If A ( λ)k + λa, then A K. (We would erhas like to simly subtract λa from both sides (but this is not ossible with Minkowski addition), then factor out the common A on the left (but this is not ossible when A is not convex) and then cancel the ( λ).) For intuition, consider the case that A is a singleton. Then ( λ)k + λa is a smaller homothet of K, in centre A: 2 Actually, as Sasha ointed out, we can rove roosition 3 more directly and simly (but less geometrically). Such a roof is given in [3], Lemma 9.2. Steven Taschuk 20 August 5 htt:// 8

19 The roosition states that if A is in the smaller coy then we must be in the situation at right. Proof We rove the contraositive. Accordingly, suose A 6 K. Let x 2 A \ K, and searate x from K by a functional f, so that f(x) > su K f. then su ( λ)k+λa f = ( λ) su K f + λ su A f < su f A with strict inequality because λ < and su K f < f(x) su A f <. Therefore A 6 ( λ)k + λa. Corollary 2 If Λ is a ɛ-net for S n, where 0 ɛ <, then cl conv Λ B n 2 cl conv Λ. ɛ Proof The hyothesis asserts that Λ S n Λ + ɛb n 2. Taking closed convex hulls yields (using the facts that conv(a + B) = conv A + conv B and, for bounded sets, cl(a + B) = cl A + cl B) cl conv Λ B n 2 cl conv Λ + ɛbn 2 = ( ɛ) ɛ cl conv Λ + ɛbn 2 Alying roosition with K, A, λ := ɛ cl conv Λ, Bn 2, ɛ yields the desired result. Proosition 3 Let Λ be an ɛ-net for S n, where 0 ɛ <. Let K R n be a convex body with 0 2 int K. If α, β 2 R are such that 8θ 2 Λ: α kθk K β, then 8θ 2 S n : α ɛ ɛ β kθk K ɛ β. Proof By corollary 2, B n 2 ɛ Since kθk K β for θ 2 Λ, we have Λ βk, and so cl conv Λ βk. βk, which yields the desired uer estimate. It also β-lischitz, which yields the lower estimate. yields that k k K is ɛ A variant of this roosition is that if 8θ 2 Λ: δ kθk K + δ, then 8θ 2 S n δ 2ɛ : kθk K + δ ɛ ɛ. (This version is roven directly as Lemma 9.2 in [3].) Steven Taschuk 20 August 5 htt:// 9

20 9 Loose end: Construction of ɛ-nets In the roof of Szarek s theorem, we used a lower bound on the (relative) measure of a sherical ca, namely lemma 4, which asserted that ɛ n 0 ɛ 2 = σ(ɛ-ca), (2) π where σ is, as usual, the uniform robability measure on S n. As art of the argument for lemma 5, we deduced that any set smaller than this doesn t contain an ɛ-ca, and so its comlement is an ɛ-net. In other words, any big enough subset of the shere (secifically, any set with measure exceeding ɛ π n) is an ɛ-net, for ɛ 2 [0, 2]. This construction of an ɛ-net is good when you don t have much control over the set in question (as we didn t, in Szarek s theorem), but if you can choose the set yourself, there is a standard way to construct much smaller ɛ- nets (indeed, finite ones). First, a sketch of the main ideas: (a) a maximal ɛ-searated set is an ɛ-net, and since (b) any ɛ-searated set yields a acking of ɛ 2 -cas in Sn, we can (c) bound the number of oints of an ɛ-searated set by a volumetric argument. On the other hand, (d) any ɛ-net yields a cover of S n by ɛ-cas, which (e) by a similar volumetric argument yields a lower bound for the size of an ɛ-net. The resulting bounds are: σ(ɛ-ca) minimum number of oints in ɛ-net σ( ɛ 2 -ca). (Since an ɛ-ca is, for small ɛ, essentially the same as ɛb n 2, we exect these bounds to be, resectively, something like ɛ )n and 2 ɛ )n.) Now, the details behind the sketch above: (a) Let Λ be a maximal ɛ-searated subset of S n. Maximality means that adding any other oint of S n destroys the ɛ-searation. In other words, every other oint of S n is within ɛ of some oint in Λ. (b) For Λ to be ɛ-searated means Now, 8x, y 2 Λ: x 6= y = x y ɛ. x y ɛ x y /2 int ɛb n 2 x y /2 int ɛ 2 Bn 2 int ɛ 2 Bn 2 ( ) (x + int ɛ 2 Bn 2 ) \ (y + int ɛ 2 Bn 2 ) =? Thus Λ is ɛ-searated if and only if its oints are the centres of a acking of ɛ 2 -cas. (Note that ste ( ) uses the convexity and symmetry of Bn 2.) Steven Taschuk 20 August 5 htt:// 20

21 (c) Given a disjoint (or almost disjoint) collection of ɛ 2 -cas in Sn, we can comute = σ(s n ) σ(union of the ɛ 2 -cas) = (number of cas)σ( ɛ 2 -ca). (d) For Λ to be an ɛ-net for S n means that every oint of S n is in some ɛ-ca centred at a oint of Λ, which means that these cas cover S n. (e) Given a collection of ɛ-cas that cover S n, we have = σ(s n ) = σ(union of the ɛ-cas) (number of cas)σ(ɛ-ca). Now, comuting σ(ɛ-ca) is somewhat annoying. We can simlify our comutations considerably by noting that, by the argument in art (b) above, any ɛ-searated set of oints in S n yields a collection of almost disjoint coies of ɛ 2 Bn 2 with centres on Sn. Such balls are subsets of ( + ɛ 2 )Bn 2, so by the same volumetric argument as in art (c), any ɛ-searated set has at most vol(( + ɛ 2 )Bn 2 ) vol( ɛ 2 Bn 2 ) = ( + ɛ 2 )n ( ɛ = + 2 n 2 )n ɛ oints. If ɛ 2 [0, ], this is at most ( 3 ɛ )n, which is for many uroses close enough to otimal. Note that with such a net Λ we obtain σ(ɛ-ca) card(λ) 3 ɛ n (where card denotes cardinality) and so ɛ n 0 ɛ = σ(ɛ-ca), 3 which is very close to (2), that is, lemma 4. It is in fact ossible to rove Szarek s theorem using this estimate, and we get a slightly better constant. 0 Comarison with Milman s theorem Let K be a convex body in R n with B n 2 K. Milman s theorem asserts that Feb 8 For any ɛ > 0, most subsaces F of dimension c(ɛ)n(m(k)) 2 satisfy d(k \ F, B n 2 \ F) + ɛ. The M arameter is the average of the norm on the shere: M(K) = kθk K dσ(θ), S n Steven Taschuk 20 August 5 htt:// 2

22 where σ is, as usual, the uniform robability measure on S n. This arameter is not affine invariant, or even linear invariant, although we do have M(QK) = M(K) for orthogonal Q, and M(bK) = bm(k) for scalar b. In analogous language, Szarek s theorem asserts that For any k with k < n, most subsaces F of dimension k satisfy d(k \ F, B n 2 \ F) /n! n/(n k) vol(k) 4π vol(b n 2 ). Comaring these two theorems shows a trade-off between distance and dimension: if you want the section K \ F to be very close to Euclidean, then you can use Milman s theorem, and the dimension of F might not be large; on the other hand, if you want a high-dimensional section, then you can use Szarek s theorem, and the distance to the Euclidean ball might not be small. The arameters involved in the two theorems are related: /n vol(k) M(K) vol(b n 2 ) M(K ). (3) The uer inequality is called Urysohn s inequality; a roof aears in [5], age 6. 3 The lower inequality is easy: vol(k) vol(b n 2 ) = S n kθk n dσ(θ) (by lemma 2; see age 8) K = E(X n ) (with X: S n R, θ 7 kθk K ) (EX) n = M(K) n (Jensen s inequality: t 7 t n is convex) The lower inequality in (3) already yields good information about nearly- Euclidean sections of B n : recall from section 6 that Therefore vr(b n ) = vol( nb n ) vol(b n 2 ) /n c. c M( nb n ), so M( nb n ) is bounded (indeendently of n). Milman s theorem then yields that nb n (and hence Bn ) has ( + ɛ)-euclidean sections of dimension c(ɛ)n. 3 As Peter ointed out during the seminar, it also follows immediately by combining the lower inequality with the Blaschke Santaló inequality. Steven Taschuk 20 August 5 htt:// 22

23 This aroach fails for B n because vr(b n ) c n, so we only obtain M(B n ) c n, which yields only that B n has (+ɛ)-euclidean sections of dimension c(ɛ). But this is trivial, since if we are content with sections of constant dimension, we can just take one-dimensional sections, which are line segments and therefore exactly Euclidean. 4 Now, it turns out that, for symmetric K, r log n M(K John ) M(B n ) c n. (4) This yields that every symmetric convex body has ( + ɛ)-euclidean sections of dimension c(ɛ) log n, which is (Milman s imrovement of) Dvoretzky s theorem. Traditionally one deduces Dvoretzky s theorem from Milman s theorem not by roving exactly that M(K John ) M(B n ), but instead by first roving the Dvoretzky Rogers lemma, which shows that we can trade half the dimensions of our section for some resemblance to the cube (in the new, lower-dimensional section). This aroach resembles (4) in sirit. The last toic of this seminar is to rove the first inequality in (4) by exloiting much of what we have roved already. (The ossibility of using this method is the concluding remark of [3]. 5 ) Proosition 4 If f: R n R is measurable and ositively homogeneous (that is, f(λx) = λf(x) when λ 0), then R n f(x) dγ n (x) = 2 Γ( n+ 2 ) Γ( n 2 ) f(θ) dσ(θ), S n where γ n is the standard gaussian robability measure on R n, which has density e x 2 /2 /(2π) n/2, and σ is the uniform robability measure on S n. Proof In brief, R n f(x) dγ n (x) = f( x /2 (2π) n/2 x ) x e x 2 dx R n = n vol(bn 2 ) (2π) n/2 S n 0 f(θ)r n e r2 /2 dr dσ(θ) (by lemma ) = n vol(bn 2 ) (2π) n/2 2(n )/2 Γ( n+ 2 ) S n f(θ) dσ(θ) 4 As Sasha ointed out in seminar, the c(ɛ) that arises this way is robably less than anyway, so the result is esecially useless. 5 Udate: This roof was given exlicitly in [6], Proosition 4.. Steven Taschuk 20 August 5 htt:// 23

24 Substituting the value of vol(b n 2 ) (see age 6) and simlifying yields the desired result. (Incidentally, note that we can comute the normalizing constant (2π) n/2 for the gaussian measure by invoking lemma 9; see age 5.) In articular, roosition 4 and Stirling s aroximation yield M(K) = Γ( n 2 ) 2 Γ( n+ 2 ) kxk K dγ n (x) kxk K dγ n (x). R n R n n For examle, if X is a standard gaussian variable in R n, then E X = x dγ n (x) nm(b n 2 ) = n. R n Proosition 5 If K is a symmetric convex body in R n, then for all r 2 R, γ n (rk John ) γ n (rb n ). (This roosition resembles the symmetric case of Ball s theorem (see age 3), but for gaussian measure; we use a similar technique to rove it.) Proof Let (u i ) m and (c i) m Then be as in John s theorem. We may assume r 0. γ n (rk) γ n ({x 2 R n : (8i: hx, u i i r)}) = Corollary 6 Rn [8i: hx, u i i r] e x 2 /2 (2π) n/2 dx cihx,u ii 2 /2 Rn = [8i: hx, u i i r] e (2π) dx c i/2! m 2 ci = [ hx, u i i r] e hx,uii /2 dx R (2π) n /2 i= m! ci /2 [ t r] e t2 dt (Brascam Lieb; see age 5) (2π) /2 i= R = γ ([ r, r]) n = γ n (rb n ) If K is a symmetric convex body in R n, then M(K John ) M(B n ). Steven Taschuk 20 August 5 htt:// 24

25 Proof Indeed, for any symmetric convex body L, M(L) = = = Γ( n 2 ) 2 Γ( n+ Γ( n 2 ) 2 Γ( n+ 2 ) 2 ) 0 Γ( n 2 ) 2 Γ( n+ 2 ) R n kxk L dγ n (x) (by roosition 4) 0 γ n (x: kxk L t) dt ( γ n (tl)) dt (distribution formula) Since γ n (tl) is maximal for L = B n by roosition 5, it follows that M(L) is minimal for L = B n. References [] Keith Ball. Volumes of sections of cubes and related roblems. In Joram Lindenstrauss and Vitali Milman, editors, Israel Seminar on Geometrical Asects of Functional Analysis (987/88), volume 376 of Lecture Notes in Mathematics, ages Sringer, 990. (Cited on ages 3 and 5.) [2] Keith Ball. Volume ratios and a reverse isoerimetric inequality. J. London Math. Soc., 44:35 359, 99. doi:0.2/jlms/s (Cited on ages 3 and 5.) [3] Keith Ball. An elementary introduction to modern convex geometry. In Silvio Levy, editor, Flavors of Geometry, volume 3 of Mathematical Sciences Research Institute Publications, ages 58. Cambridge UP, Cambridge, 997. htt:// df. (Cited on ages 3, 4, 8, 9, and 23.) [4] Keith Ball. Convex geometry and functional analysis. In Johnson and Lindenstrauss [], ages (Cited on ages 3 and 5.) [5] Franck Barthe. Inégalités de Brascam Lieb et convexité. C. R. Acad. Sci. I, 324: , 997. In French. (Cited on age 5.) [6] Franck Barthe. On a reverse form of the Brascam Lieb inequality. Invent. Math., 34:335 36, 998. (Cited on age 5.) [7] Herm Jan Brascam and Elliot H. Lieb. Best constants in Young s inequality, its converse, and its generalization to more than three functions. Adv. Math., 20:5 73, 976. (Cited on age 5.) [8] Aostolos Giannooulos and Vitali Milman. Euclidean structure in finite dimensional normed saces. In Johnson and Lindenstrauss [], ages (Cited on age 3.) Steven Taschuk 20 August 5 htt:// 25

26 [9] Mikhail Gromov and Vitali Milman. Brunn theorem and a concentration of volume henomena for symmetric convex bodies. In Israel Seminar on Geometrical Asects of Functional Analysis (983/84), ages V. V.2. Tel Aviv UP, Tel Aviv, 984. (Cited on age.) [0] Fritz John. Extremum roblems with inequalities as subsidiary conditions. In K. O. Friedrichs, O. E. Neugebauer, and J. J. Stoker, editors, Studies and Essays Presented to R. Courant on his 60th birthday, January 8, 948, ages Interscience, New York, 948. (Cited on age 3.) [] William B. Johnson and Joram Lindenstrauss, editors. Handbook of the Geometry of Banach Saces, volume. North-Holland, Amsterdam, 200. (Cited on age 25.) [2] B. S. Kašin. Diameters of some finite-dimensional sets and classes of smooth functions. Math. USSR Izv., ():37 333, 977. doi:0.070/ IM977v0n02ABEH0079. (Cited on age 7.) [3] Vitali Milman. Surrising geometric henomena in high dimensional convexity theory. In Proceedings of ECM2, volume 2, ages 73 9, Budaest, 996. Birkhäuser. htt:// milman/les/survey4.s. (Cited on age.) [4] Vitali Milman and Gideon Schechtman. Global versus local asymtotic theories of finite-dimensional normed saces. Duke Math. J., 90():73 93, 997. (Cited on age 4.) [5] Gilles Pisier. The Volume of Convex Bodies and Banach Sace Geometry. Number 94 in Cambridge Tracts in Mathematics. Cambridge UP, 989. (Cited on ages 7, 5, and 22.) [6] Gideon Schechtman and Michael Schmuckenschläger. A concentration inequality for harmonic measures on the shere. In Joram Lindenstrauss and Vitali Milman, editors, Geometric Asects of Functional Analysis: Israel Seminar (GAFA) , number 77 in Oerator Theory: Advances and Alications, ages Birkhäuser Verlag, 995. (Cited on age 23.) [7] S. Stein. The symmetry function in a convex body. Pac. J. Math., 6:45 48, 956. (Cited on age 3.) [8] Stanisław J. Szarek. On Kashin s almost Euclidean orthogonal decomosition of l n. Bull. Acad. Polon. Sci. Sér. Math., 26:69 694, 978. (Cited on age 7.) [9] Stanisław J. Szarek and Nicole Tomczak-Jaegermann. On nearly Euclidean decomositions of some classes of Banach saces. Comos. Math., 40(3): , 980. (Cited on age 7.) Steven Taschuk 20 August 5 htt:// 26