Upper bound for the Dvoretzky dimension in Milman-Schechtman theorem
aa r X i v : . [ m a t h . F A ] D ec Upper bound for the Dvoretzky dimension inMilman-Schechtman theorem
Han Huang, Feng Wei
Abstract
For a symmetric convex body K ⊂ R n , the Dvoretzky dimension k ( K ) is thelargest dimension for which a random central section of K is almost spherical. ADvoretzky-type theorem proved by V. D. Milman in 1971 provides a lower bound for k ( K ) in terms of the average M ( K ) and the maximum b ( K ) of the norm generatedby K over the Euclidean unit sphere. Later, V. D. Milman and G. Schechtmanobtained a matching upper bound for k ( K ) in the case when M ( K ) b ( K ) > c ( log( n ) n ) .In this paper, we will give an elementary proof of the upper bound in Milman-Schechtman theorem which does not require any restriction on M ( K ) and b ( K ). Given a symmetric convex body K in R n , we have a corresponding norm k x k K = inf { r > , x ∈ rK } . Let | · | denote the Euclidean norm, ν n denote thenormalized Haar measure on the Euclidean sphere, S n − , and ν n,k denote thenormalized Haar measure on the Grassmannian manifold Gr n,k . Let M = M ( K ) := R S n − k x k K dν n and b = b ( K ) := sup {k x k K , x ∈ S n − } be the mean andthe maximum of the norm over the unit sphere.In 1971, V. D. Milman proved the following Dvoretzky-type theorem [3]: Theorem 1.
Let K be a symmetric convex body in R n . Assume that k x k K ≤ b | x | forall x ∈ R n . For any ǫ ∈ (0 , , there is k ≥ C ǫ ( M/b ) n such that ν n,k { F ∈ G n,k : (1 − ε ) M < k · k K ∩ F < (1 + ε ) M } > − exp( − ˜ ck ) where ˜ c > is a universal constant, C ǫ > is a constant only depending on ǫ . The quantity C ǫ was of the order ǫ log − ( ǫ ) in the original proof of V. D. Milman. Itwas improved to the order of ǫ by Y. Gordon [2] and later, with a simpler argument,by G. Schechtman [6].In 1997, V. D. Milman and G. Schechtman [5] found that the bound on k appearing inTheorem 1 is essentially optimal. More precisely, they proved the following theorem.1 heorem A. (Milman–Schechtman, see e.g., section 5.3 in [1]). Let K be a symmetricconvex body in R n . For ǫ ∈ (0 , , define k ( K ) to be the largest dimension k such that ν n,k (cid:0) { F ∈ G n,k : ∀ x ∈ S n − ∩ F , (1 − ε ) M < k x k K < (1 + ε ) M } (cid:1) > p n,k = nn + k . Then, ˜ C ǫ n ( M/b ) ≥ k ( K ) ≥ ¯ C ǫ n ( M/b ) when Mb > c ( log( n ) n ) for some universal constant c , where k · k F denotes the normcorresponding to the convex body K ∩ F in F , and ˜ C ǫ , ¯ C ǫ > are constants dependingonly on ǫ . Because the Dvoretzky-Milman theorem cannot guarantee the lower bound with small Mb for p n,k = nn + k , the original proof required an assumption that Mb > c ( log( n ) n ) forsome c . In [1, p. 197], S. Artstein-Avidan, A. A. Giannopoulos, and V. D. Milmanaddressed it as an open question whether one can prove the same result when p n,k is aconstant, such as . When p n,k = , the lower estimate on k ( K ) is a direct result ofDvoretzky-Milman theorem [3], but the upper bound was unknown. In this paper, weare going to give upper bound estimate with p n,k = , our main result is the followingtheorem: Theorem B.
Let K be a symmetric convex body in R n . Fix a constant ǫ ∈ (0 , , let k ( K ) be the largest dimension such that ν n,k { F ∈ G n,k : (1 − ε ) M < k · k K ∩ F < (1 + ε ) M } > . Then, Cn ( M/b ) ≥ k ( X ) ≥ ¯ C ǫ n ( M/b ) where C > is a universal constant and ¯ C ǫ > is a constant depending only on ǫ . In the next section, we will provide a proof of Theorem B with no restriction on Mb . Infact, from the proof, one can see that can be replaced by any c ∈ (0 ,
1) or1 − exp( − ˜ ck ), which is the probability appearing in Milman-Dvoretzky theorem. Let P k be the orthogonal projection from S n − to some fixed k -dimensional subspace,and | · | be the Euclidean norm. The upper estimate is related to the distribution of | P k ( x ) | , where x is uniformly distributed on S n − .Recall the concentration inequality for Lipschitz functions on the sphere (see, e.g., [4]): heorem 2 (Measure Concentration on S n − ) . Let f : S n − → R be a Lipschitzcontinuous function with Lipschitz constant b . Then, for every t > , ν n ( { x ∈ S n − : | f ( x ) − E ( f ) | ≥ bt } ) ≤ − c t n ) where c > is a universal constant. Theorem 2 implies the following elementary lemma.
Lemma 3.
Fix any c > , let P k be an orthogonal projection from R n to somesubspace R k . If t > c √ n and ν n ( { x ∈ S n − : | P k ( x ) | < t } ) > , then k < c t n , where c > is a constant depending only on c .Proof. | P k ( x ) | is a 1-Lipschitz function on S n − with E | P k ( x ) | about q kn . If we wantthe measure of { x : | P k ( x ) | < t } to be greater than 1 /
2, then measure concentrationwill force E | P k | to be bounded by the size of t , which means k < c t n for someuniversal constant c . Since t n > c , we may and shall assume k is bigger than someabsolute constant in our proof, then adjust c .To make it precise, we will first give a lower bound on E | P k | . By Theorem 2, ν n ( || P k ( x ) | − E | P k ( x ) || > t ) ≤ − c tn ) . Thus, E | P k | − ( E | P k | ) = E ( | P k | ( x ) − E | P k | ) < Z ∞ ν n ( || P k ( x ) | − E | P k ( x ) || > t ) dt ≤ Z ∞ − c tn ) dt = 4 c n . With E | P k | = E P ki =1 | x i | = kn , we get E ( | P k | ) > q kn − c n . If we assume that k > c ,then we have E ( | P k | ) > r kn . Assuming k > t n , we have E ( | P k | ) − t > r kn − t ≥ r kn > . Applying Theorem 2 again, we obtain ν n ( | P k | < t ) < ν n ( || P k | − E | P k || > E ( | P k | ) − t ) ≤ − c ( E ( | P k | ) − t ) n ) ≤ − c ( 12 r kn ) n ) ≤ − c k ) ≤ − < , which proves our result by contradiction. heorem 4. Let K be a convex body with inradius b . For ǫ ∈ (0 , , let k be the largestinteger such that ν n,k { F ∈ G n,k : (1 − ε ) M < k · k K ∩ F < (1 + ε ) M } > . Then k < Cn ( Mb ) where C is an absolute constant.Proof. We may assume k e k K = b , then K ⊂ S = { x ∈ R n : | x | < b } , thus k x k K ≥ k x k S = b |h x, e i| . This implies { V ∈ G n,k : ∀ x ∈ V ∩ S n − , (1 − ǫ ) M < k x k K < (1 + ǫ ) M }⊂ { V ∈ G n,k : ∀ x ∈ V ∩ S n − , k x k S < (1 + ǫ ) M } = { V ∈ G n,k : sup x ∈ V ∩ S n − h x, e i < (1 + ǫ ) Mb } = { V ∈ G n,k : | P V ( e ) | < (1 + ǫ ) Mb } (1)where P V is the orthogonal projection from R n to V . If V is uniformly distributed on G n,k and x is uniformly distributed on S n − , then | P V ( x ) | and | P V ( e ) | areequi-distributed for any fixed k -dimensional subspace V . Therefore, ν n,k ( { V ∈ G n,k : | P V ( e ) | < (1 + ǫ ) Mb } ) = ν n ( { x ∈ S n − : | P V ( x ) | < (1 + ǫ ) Mb } ) . As shown in the Remark 5.2.2(iii) of [1, p. 164], the ratio Mb has a lower bound c ′ √ n .Setting c = c ′ and t = (1 + ǫ ) Mb , it is easy to see that if ν n,k { F ∈ G n,k : (1 − ε ) M < k · k K ∩ F < (1 + ε ) M } > , then k ≤ c (1 + ǫ ) (cid:0) Mb (cid:1) n < Cn ( Mb ) by Lemma 3 and (1).Now we can prove Theorem B as a corollary of Theorem 4 and Theorem 1: Proof of Theorem B.
Theorem 1 shows that if C ǫ ( M/b ) n > log(2)˜ c , then there is k ≥ C ǫ ( M/b ) n such that ν n,k { F ∈ G n,k : (1 − ε ) M < k · k F < (1 + ε ) M } > − exp( − ˜ ck ) > . Otherwise, (
M/b ) n < log(2)˜ cC ǫ . Therefore, k ( K ) ≥ min { ˜ cC ǫ log(2) , C ǫ } ( M/b ) n . Combining itwith Theorem 4, we get C ( Mb ) n ≥ k ( K ) ≥ min { ˜ cC ǫ log(2) , C ǫ } ( M/b ) n. emark. (1) It is worth noticing that the number plays no special role in our proof.Thus, if we define the Dvoretzky dimension to be the largest dimension such that ν n,k { F ∈ G n,k : (1 − ε ) M < k · k K ∩ F < (1 + ε ) M } > c for some c ∈ (0 , k ( K ) ∼ ( Mb ) n . Similarly, if we fix ǫ and replace by 1 − exp( − ˜ ck ), then the lowerbound of k ( K ) is the one from Theorem 1. For k bigger than some absolute constant,we have 1 − exp( − ˜ ck ) > . Thus, the upper bound is still of order (cid:0) Mb (cid:1) n . Therefore,we can replace by 1 − exp( − ˜ ck ) in Theorem A. With this probability choice, it alsoshows Theorem 1 provides an optimal k depending on M, b .(2) Usually, we are only interested in ǫ ∈ (0 , C ǫ = o ǫ (1). It is anatural question to ask if we could improve the upper bound from a universal constant C to o ǫ (1). Unfortunately, it is not possible due to the following observation. Let K = conv( B n , Re ) ◦ . By passing from the intersection on K to the projection of K ◦ ,one can show that k ( K ) does not exceed the maximum dimension k such that ν n ( P k ( Rx ) < ǫ ) > . Choosing R = p nl , we get n ( Mb ) ∼ l and k ( X ) ∼ l byTheorem 2 and a similar argument to that of Lemma 3. This example shows that nomatter what Mb is, one can not improve the upper bound in Theorem A from anabsolute constant C to o ǫ (1). We want to thank our advisor Professor Mark Rudelson for his advise and encouragementon solving this problem. And thank both Professor Mark Rudelson and Professor VitaliMilman for encouraging us to organize our result as this paper.
References [1] S. Artstein-Avidan, A. A. Giannopoulos, and V. D. Milman, Asymptotic geometricanalysis. Part I. Mathematical Surveys and Monographs, 202. American Mathemat-ical Society, Providence, RI, 2015.[2] Y. Gordon, Some inequalities for Gaussian processes and applications. Israel J. Math.50 (1985), 265-289.[3] V. D. Milman, New proof of the theorem of A. Dvoretzky on sections of convexbodies, Funct. Anal. Appl. 5(4) (1971), 288-295.[4] V. D. Milman, G. Schechtman, Asymptotic theory of finite-dimensional normedspaces. With an appendix by M. Gromov. Lecture Notes in Mathematics, 1200.Springer-Verlag, Berlin, 1986.5] V. D. Milman and G. Schechtman, Global versus Local asymptotic theories of finitedimensional normed spaces, Duke Math. Journal 90 (1997), 73-93.[6] G. Schechtman, A remark concerning the dependence on ǫǫ