IMPROVED BOUNDS IN THE SCALED ENFLO TYPE INEQUALITY FOR BANACH SPACES

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1 IMPROVED BOUNDS IN THE SCALED ENFLO TYPE INEQUALITY FOR BANACH SPACES OHAD GILADI AND ASSAF NAOR Abstract. It is shown that if (, ) is a Banach sace with Rademacher tye 1 then for every n N there exists an even integer m n 2 1/ log n such that for every f :, [ ] f ( E x,ε x + m ) 2 ε n ] f(x) m E x [ f(x + e j ) f(x), where the exectation is with resect to uniformly chosen x and ε { 1, 1} n. This imroves a bounds of m n 3 2/ that was obtained in [7]. The roof is based on an augmentation of the smoothing and aroximation scheme, which was imlicit in [7]. 1. Introduction A Banach sace (, ) is said to have Rademacher tye 1 if there is a constant T < such that for every n N and for every x 1, x 2,..., x n, [ ] E ε ε j x j T x j, (1) where the exectation in (1) is taken with resect to the uniform robability measure on { 1, 1} n. By considering the case of the real line, we necessarily have 2. The smallest ossible T for which (1) holds is denoted by T (). The notion of Rademacher tye is clearly a linear notion, as inequality (1) involves random linear combinations of vectors in. A Banach sace (, ) is said to be finitely reresentable in a Banach sace (Y, Y ) if there exists a constant D < such that for every finite dimensional subsace E of there exists a subsace F of Y and a ma T : E F with T T 1 D. A classical theorem of Ribe (see [11] and also [2]) states that if two Banach saces and Y are uniformly homeomorhic, then is finitely reresentable in Y and vice versa. This theorem motivated what is now known as the Ribe rogram : finding concrete metric characterizations of local roerties of Banach saces (a roerty is said to be local if it deends only on finitely many vectors). In articular, Ribe s theorem suggests that the notion of Rademacher tye has a urely metric characterization. Finding a concrete metric characterization of Rademacher tye is a long standing roblem that goes back to the work of Enflo. Following Enflo, we say that a metric sace (M, d M ) has Enflo tye > 0 if there exists a constant T < such that for 2010 Mathematics Subject Classification. 46B07,46B20,51F99. O. G. was artially suorted by NSF grant CCF A. N. was suorted in art by NSF grants CCF and CCF , BSF grant , and the Packard Foundation. 1

2 every n N and every f : { 1, 1} n M, [ ] E ε d M (f(ε), f( ε)) T E ε [ d M (f(ε 1,..., ε j 1, ε j, ε j+1..., ε n ), f(ε 1,..., ε j 1, ε j, ε j+1,..., ε n )) ]. (2) In [4], Enflo asked whether for the class of Banach saces, Enflo tye is equivalent to Rademacher tye. Clearly, Enflo tye imlies Rademacher : simly aly inequality (2) to the function f(ε) = n ε jx j, and inequality (1) is obtained. In the other direction, Pisier roved (see [10, Ch. 7]) that if a Banach sace has Rademacher tye then it has Enflo tye for every <. The question of whether Rademacher tye imlies Enflo tye remains an interesting oen roblem. Naor and Schechtman showed [9] that the answer is ositive for the class of UMD Banach saces. We refer to the work of Gromov [6, Sec. 9.1] and Bourgain, Milman and Wolfson [3] for earlier results related to the notion of non-linear tye, and to the work of Ball [1] for an imortant variant of (2) known as Markov tye. Motivated by their work on metric cotye [8], Mendel and Naor defined in [7] the notion of scaled Enflo tye. A metric sace (M, d M ) is said to have scaled Enflo tye > 0 with constant θ < if for every n N there exists an m 2N such that for every f : M, [ ] E x,ε d M (f (x + m ) ) 2 ε, f(x) θ m E x [d ] M (f(x + e j ), f(x)). (3) In (3) and in what follows, {e j } n denotes the standard basis of. In [7] the following theorem was roved: Theorem 1.1. A Banach sace has Rademacher tye [1, 2] if and only if it has scaled Enflo tye. The notion of scaled Enflo tye thus gives a urely metric characterization of Rademacher tye. While the value of m is imlicit in Theorem 1.1, it does lay a crucial role. Note that by choosing m = 2 in (3), the original Enflo tye inequality (2) is obtained. Therefore, finding the smallest m for which inequality (3) holds is a question of great interest. In [7] it was shown that if a Banach sace has Rademacher tye [1, 2] then it has scaled Enflo tye with m n 3 2/. Motivated by the recent rogress in [5], we obtain the following imroved bound on m: Theorem 1.2. Assume that a Banach sace has Rademacher tye [1, 2]. Then for every n N, there exists m 4N with m n 2 1/ log n such that for every f :, [ ] f ( E x,ε x + m ) 2 ε [ ] f(x) m E x f(x + e j ) f(x). (4) The outline of the roof of Theorem 1.2 is as follows: we begin by describing a general smoothing and aroximation scheme. This scheme allows us to relace f in inequality (3) by a smoothed version of f and then use inequality (1). This is discussed in detail in Section 2. Once we have the smoothing and aroximation scheme, the roof of Theorem 1.2 is straightforward. This is done in Section 3. Nevertheless, the smoothing and aroximation 2

3 scheme relies on two technical lemmas. These lemmas are roved in Section 4. The logarithmic factor in Theorem 1.2 aears due to an additional comlication that does not arise in [5] (where there is no such logarithmic term); this is overcome here via an alication of Pisier s inequality [10]. Notation. We use, to indicate that an inequality holds true with an imlied absolute constant. Also, we use, to indicate that the imlied constant deends on and T () and, if the constant deends on only. Also, µ will denote the uniform robability measure on and τ will denote the uniform robability measure on { 1, 1} n. Finally, [n] will denote the set {1, 2,..., n}. 2. The smoothing and aroximation scheme for the case of tye Following [5], we investigate the aroach which is imlicit in [7]. Given f : and a robability measure ν on, let f ν(x) = f(x y)dν(y). For a Banach sace with Rademacher tye, suose that we are given a robability measure ν that satisfies the following two roerties: (A) Aroximation roerty: f ν(x) f(x) dµ(x) A (S) Smoothing roerty: { 1,1} n f ν(x + ε) f ν(x ε) dτ(ε)dµ(x) S f(x + e j ) f(x) dµ(x). (5) f(x + e j ) f(x) dµ(x). (6) The goal is to deduce inequality (3) from inequality (1). It is known that (3) holds for linear functions x n x jv j (this statement is not comletely accurate but is sufficient to give an intuition). So, the idea is to first relace f by a smoothed version of it which locally linear on average. The way to measure the smoothness of the function is given by the smoothing roerty (6). On the other hand, the smoothed version of f has to be close enough to f itself, since the final goal is to rove inequality (3) for f. This is measured by the aroximation roerty (5). To obtain inequality (3) from (5) and (6), choose m 4N and note that by the triangle inequality and convexity, f (x + m ) 2 ε f(x) ( 3 1 f ν x + m ) 2 ε f ν(x) ( f ν x + m ) 2 ε f (x + m ) 2 ε f ν(x) f(x). (7) 3

4 Also, by the triangle inequality and Hölder s inequality (remembering that m is divisible by 4), f ν (x + m ) 2 ε f ν(x) m/4 f ν(x + 2tε) f ν(x + 2(t 1)ε) ( m ) m/4 1 f ν(x + 2tε) f ν(x + 2(t 1)ε) 4. (8) Integrating (7) over while using (8) and the translation invariance of µ, f (x + m ) Z 2 ε f(x) dµ(x) 3 f ν(x) f(x) dµ(x) n m + m f ν(x + ε) f ν(x ε) dµ(x). (9) Using the aroximation roerty (5) and the smoothing roerty (6), we get f (x + m ) 2 ε f(x) dµ(x) (A + m S ) f(x + e j ) f(x) dµ(x). (10) If we could find a smoothing and aroximation scheme for which S 1 and A m, then (10) would imly f (x + m ) 2 ε f(x) dµ(x) m S f(x + e j ) f(x) dµ(x), (11) which is recisely the desired scaled Enflo tye inequality (3). Thus, the goal is to come u with a smoothing and aroximation scheme with S 1 and A as small as ossible. In [7] it was shown that one can choose and A n 3 2/. Here we show that we can in fact choose A n 2 1/ log n. Remark 2.1. In the context of metric cotye, it was shown in [5] that the scheme of smoothing and aroximation necessarily has limitations. Secifically, it was shown that using such a scheme imlies that m is bigger than some function of n. However, in the context of tye, this is no longer the case. If we assume that the notions of Rademacher tye and Enflo tye are in fact the same for the class of Banach saces, then the smoothing roerty (6) should hold for f itself in which case the aroximation roerty (5) becomes trivial. 3. Proof of Theorem 1.2 Let f :, and fix an odd integer 0 < k < m/2. We follow the notations in [5] and define the following family of averaging oerators. For f :, k < m/2 an odd integer 4

5 and B [n], let where B f(x) def = 1 f(x + y)dµ(y), (12) µ(l B ) L B def = {y : i / B, y i = 0; i [n] y i is even; d (0, y) < k}. As we mentioned in Section 2, the roof of inequality (3) will follow once we have the smoothing and aroximation roerties. The aroximation roerty is given by the following lemma, which was already roved in [7]. Lemma 3.1 (Lemma 2.2 in [7]). For every Banach sace and for every f :, [n] f(x) f(x) dµ(x) (k 1) n 1 f(x + e j ) f(x) dµ(x). The smoothing roerty is the new ingredient in this note. Lemma 3.2. Assume that is a Banach sace with Rademacher tye. Then for every k n log n we have for every f :, [n] f(x + ε) [n] f(x ε) dτ(ε)dµ(x) { 1,1} n f(x + e j ) f(x) dµ(x). The roof of Theorem 1.2 now follows immediately. Proof of Theorem 1.2. As we saw in Section 2, we must have S 1 and A m. By Lemma 3.1 and Lemma 3.2 the smoothing and aroximation roerties hold with A = (k 1)n 1 1/ and S = 1, assuming that k n log n. This imlies A n 2 1/ log n and therefore m n 2 1/ log n. 4. Proof of Lemma 3.1 and Lemma 3.2 Lemma 3.1 was already roved in [7]. However, we reeat its roof for the sake of comleteness in Subsection 4.2. The roof of Lemma 3.2 is resented in Subsection Proof of Lemma 3.2. We recall some notation from [5]: for ε { 1, 1} n and B [n], let ε B be the restriction of ε to the coordinates of B. Also, for ε, ε { 1, 1} n, let ε, ε = n ε jε j. Fix x and ε { 1, 1} n. Let R i,l f(x, ε) def = [ [n]\s f(x + δ S k + ε [n]\s ) [n]\s f(x + δ S k ε [n]\s ) ]. (13) S [n] S =i δ { 1,1} S δ S,ε S =i 2l Also, recall from [8] the following averaging oerators: E j f(x) def 1 = f(x + y)dµ(y), (14) µ(s(j, k)) 5

6 where S(j, k) def = {y : y j is even; l [n] {j}, y l is odd; d (0, y) k}. The following identity follows immediately from Lemma 3.8 in [5]: i k n i 1 ε j [E j f(x + e j ) E j f(x e j )] = h i,l (k + 1) R i,lf(x, ε), (15) n 1 where {h i,l } li are scalars satisfying i=0 h 0,0 = 1, (16) (i l)!l! h i,l. (17) 2 i Note that the first term on the right hand side of (15) equals ( ) n 1 ( ) n 1 k k [ R 0,0f(x, ε) = [n]f(x + ε) [n]f(x ε) ]. k + 1 k + 1 Using identity (15), the triangle inequality and the fact that ( ) k+1 n 1 k 1, [n] f(x + ε) [n] f(x ε) i The triangle inequality imlies i h i,l k R i,lf(x, ε) i ε j [E j f(x + e j ) E j f(x e j )] ( i h i,l k i + h i,l R i,l f(x, ε) k R i,lf(x, ε) i Convexity of the function t t and Hölder s inequality imly ( ) ( i h i,l i R k i i,l f(x, ε) = 2 (i+1) 2 i+1 h i,l R k i i,l f(x, ε). (18) ). (19) ( i 2 (i+1) 2 i+1 h i,l R k i i,l f(x, ε) i 2 (i+1)( 1) (i + 1) 1 h i,l R k i i,l f(x, ε). (20) Therefore, combining (18) and (20), we get [n] f(x + ε) [n] f(x ε) ε j [E j f(x + e j ) E j f(x e j )] + i 2 (i+1)( 1) (i + 1) 1 h i,l ) ) k i R i,l f(x, ε). (21) The next goal is to estimate the average of R i,l f(x, ε) over x Zn m and ε { 1, 1} n. 6

7 Lemma 4.1. Assume that is a Banach sace with Rademacher tye. 0 l i n, Then for all { 1,1} n R i,l f(x, ε) dτ(ε)dµ(x) (log n) ( n i ) ( ) i l f(x + e j ) f(x) dµ(x). (22) Proof. First, note that on the right hand side of (13) there are ( i l) terms for which δs, ε S = i 2l and there are ( n i) sets S of size i. Therefore, alying the triangle inequality and then Hölder s inequality, ( ) 1 ( ) 1 n i R i,l f(x, ε) i l [n]\s f(x + δ S k + ε [n]\s ) [n]\s f(x + δ S k ε [n]\s ). (23) S [n] S =i δ { 1,1} S δ S,ε S =i 2l We would like to integrate inequality (23) over x and ε { 1, 1} n. First, note that by the invariance of µ we have [n]\s f(x + δ S k + ε [n]\s ) [n]\s f(x + δ S k ε [n]\s ) dµ(x) = [n]\s f(x + ε [n]\s ) [n]\s f(x ε [n]\s ) dµ(x). (24) Since [n]\s is a convolution with a robability measure we also have [n]\s f(x + ε [n]\s ) [n]\s f(x ε [n]\s ) dµ(x) Thus, integrating (23) over x while using (24) and (25), R i,l f(x, ε) dµ(x) 2 ( n i f(x + ε[n]\s ) f(x ε [n]\s ) dµ(x). (25) ) 1 ( ) i f(x + ε [n]\s ) f(x ε [n]\s ) dµ(x). (26) l S [n] S =i 7

8 Recall Pisier s inequality (see [10, Ch. 7]): for every g : { 1, 1} n, g(ε) gdτ dτ(ε) { 1,1} { 1,1} n n (e log n) [ ( ε ) j g ε (j) g(ε) ] dτ(ε )dτ(ε), { 1,1} n { 1,1} n where ε (j) def = (ε 1,..., ε j 1, ε j, ε j+1,..., ε n ). For a fixed x and S [n], define g x : { 1, 1} n to be g x (ε) def = f(x + ε [n]\s ) f(x ε [n]\s ). Clearly, gdτ = 0. { 1,1} n Alying Pisier s inequality to g x thus imlies, g x (ε) dτ(ε) { 1,1} n (e log n) [ ( ε ) j gx ε (j) g x (ε) ] dτ(ε )dτ(ε). (27) { 1,1} n { 1,1} n Alying the Rademacher tye roerty of, we get [ ( ε ) j gx ε (j) g x (ε) ] dτ(ε ) { 1,1} n Now, by the definition of g x it follows immediately that ( g x (ε) g ) [ ( )] x ε (j) = f(x + ε [n]\s ) f x + ε (j) [n]\s + ( g ) x ε (j) g x (ε). (28) [ ( )] f(x ε [n]\s ) f x ε (j) [n]\s (in the case j S we let ε (j) [n]\s = ε [n]\s, in which case the difference is zero). Thus, using convexity and the translation invariance of µ, gx (ε (j) ) g x (ε) dµ(x) 2 f(x + e j ) f(x e j ) dµ(x) 4 Integrating (27) over x and using (28) and (29), we get f(x + ε[n]\s ) f(x ε [n]\s ) dτ(ε)dµ(x) { 1,1} n Plugging (30) into (26), we get { 1,1} n R i,l f(x, ε) dτ(ε)dµ(x) (log n) f(x + e j ) f(x) dµ(x). (29) ( ) ( ) n i (log n) i l 8 f(x + e j ) f(x) dµ(x). (30) f(x + e j ) f(x) dµ(x).

9 The roof of Lemma 4.1 is therefore comlete. We are now in a osition to rove Lemma 3.2. Proof of Lemma 3.2. Integrating inequality (21) over x and ε { 1, 1} n, we get [n] f(x + ε) [n] f(x ε) dτ(ε)dµ(x) { 1,1} n ε j [E j f(x + e j ) E j f(x e j )] dτ(ε)dµ(x) + { 1,1} n i 2 (i+1)( 1) (i + 1) 1 h i,l k i R i,l f(x, ε) dτ(ε)dµ(x). (31) { 1,1} n Alying the Rademacher tye inequality to the vectors {E j f(x + e j ) E j f(x e j )} n, ε j [E j f(x + e j ) E j f(x e j )] dτ(ε) { 1,1} n E j f(x + e j ) E j f(x e j ). (32) Integrating (32) over x and using convexity and the fact the E j is a convolution with a robability measure imlies ε { 1,1} j [E j f(x + e j ) E j f(x e j )] dτ(ε)dµ(x) n f(x + e j ) f(x) dµ(x). (33) It remains to bound the second term in (31). For that, fix 1 i n and 0 l i. Using Lemma 4.1 and the estimate (17), each term in the sum on the right hand side of (31) can be bounded as follows: 2 (i+1)( 1) (i + 1) 1 h i,l k i 2 i( 1) (i + 1) 1 ( (i l)!l! 2 i k i Now, 2 i( 1) (i + 1) 1 ( (i l)!l! 2 i k i R i,l f(x, ε) dτ(ε)dµ(x) { 1,1} ) n ( ) ( ) n i (log n) f(x + e j ) f(x) i l dµ(x). ) ( n i ) ( i l ) (log n) = 9 (i + 1) 1 2 i (i + 1) 1 2 i ( n k ( n! (n i)!k i ) (log n) ) i (log n). (34)

10 Thus, (34) becomes 2 (i+1)( 1) (i + 1) 1 h i,l k i (i + 1) 1 2 i ( n k Plugging (35) and (33) into (31) imlies R i,l f(x, ε) dτ(ε)dµ(x) { 1,1} n ) i (log n) [n] f(x + ε) [n] f(x ε) dτ(ε)dµ(x) { 1,1} n (i + 1) [1 ( n ) i + (log n) ] 2 i k f(x + e j ) f(x) dµ(x). (35) f(x + e j ) f(x) dµ(x). (36) We always have (i+1) 2 i 1. Thus, if we choose k n log n, Lemma 3.2 follows from (36) Proof of Lemma 3.1. Given z let z def = n z j, where z j def = min{z j, m z j }. For z j 0, we let sgn(z j ) = 1 if z j = z j and sgn(z j ) = 1 otherwise. By the triangle inequality, for every x, z we have f(x + z) f(x) z j ( ) j 1 f x + z t e t + l sgn(z j ) e j l=1 ( ) j 1 f x + z t e t + (l 1) sgn(z j ) e j. (37) On the right hand side of (37) we have at most z non-zero elements and therefore Hölder s inequality imlies f(x + z) f(x) z 1 z j ( ) j 1 f x + z t e t + l sgn(z j ) e j l=1 ( ) j 1 f x + z t e t + (l 1) sgn(z j ) e j. (38) Integrating (38) over x and using the translation invariance of µ, we get f(x + z) f(x) dµ(x) z 1 z j l=1 = z 1 z j 10 f(x + sgn(z j )e j ) f(x) dµ(x) f(x + e j ) f(x) dµ(x). (39)

11 Since k is odd, we have ( k, k) n (2Z) n = k n, and thus we get by (12) [n] f(x) = 1 f(x + y). (40) k n y ( k,k) n (2Z) n Using (40) and convexity imlies [n] f(x) f(x) 1 dµ(x) k n Combining (39) and (41) we get [n] f(x) f(x) dµ(x) z ( k,k) n (2Z) n f(x + z) f(x) dµ(x). (41) 1 k n z ( k,k) n (2Z) n (k 1) n 1 The roof is therefore comlete. z 1 z j f(x + e j ) f(x) dµ(x) f(x + e j ) f(x) dµ(x). References [1] K. Ball. Markov chains, Riesz transforms and Lischitz mas. Geom. Funct. Anal., 2(2): , [2] Y. Benyamini and J. Lindenstrauss. Geometric nonlinear functional analysis. Vol. 1, volume 48 of American Mathematical Society Colloquium Publications. American Mathematical Society, Providence, RI, [3] J. Bourgain, V. Milman, and H. Wolfson. On tye of metric saces. Trans. Amer. Math. Soc., 294(1): , [4] P. Enflo. On infinite-dimensional toological grous. In Séminaire sur la Géométrie des Esaces de Banach ( ), ages Ex. No , 11. École Polytech., Palaiseau, [5] O. Giladi, M. Mendel, and A. Naor. Imroved bounds in the metric cotye inequality for Banach saces. Prerint, Available at htt://arxiv.org/abs/ [6] M. Gromov. Filling Riemannian manifolds. J. Differential Geom., 18(1):1 147, [7] M. Mendel and A. Naor. Scaled Enflo tye is equivalent to Rademacher tye. Bull. Lond. Math. Soc., 39(3): , [8] M. Mendel and A. Naor. Metric cotye. Ann. of Math. (2), 168(1): , [9] A. Naor and G. Schechtman. Remarks on non linear tye and Pisier s inequality. J. Reine Angew. Math., 552: , [10] G. Pisier. Probabilistic methods in the geometry of Banach saces. In Probability and analysis (Varenna, 1985), volume 1206 of Lecture Notes in Math., ages Sringer, Berlin, [11] M. Ribe. On uniformly homeomorhic normed saces. Ark. Mat., 14(2): , Courant Institute, New York University address: giladi@cims.nyu.edu Courant Institute, New York University address: naor@cims.nyu.edu 11