Push Forward Measures And Concentration Phenomena

Transcription

1 Push Forward Measures And Concentration Phenomena Joint work with C. Hugo Jiménez and Raffael Villa Márton Naszódi Dept. of Geometry, Eötvös University, Budapest

2 Notations All measures regular Borel measures, not supported on a singleton. (X, d, µ): a metric probability space. ε-expansion: For A X, A ε = {x X : d(a, x) < ε} = {x X : a A : d(a, x) < ε}. Concentration function: For ε > 0, { α µ (ε) = sup 1 µ(a ε ) : A X measurable, µ(a) 1 }. 2 m f R is a median of a random variable f : X R if µ({x X : f (x) m f }), µ({x X : f (x) m f }) 1/2. Push Forward Measures And Concentration Phenomena 2 / 19

3 Motivation Problem Transfer a well concentrated measure from one finite dim. Banach space to another. (UC) Uniformly convex space (CM) Concentration of measure (CD) Concentration of distance (AE) Almost equilateral set of large cardinality Swanepoel Is there an almost equilateral set of exponential size in any normed space? Is it true that for any ε there is a c > 0 : N(ε) e cn? Push Forward Measures And Concentration Phenomena 3 / 19

4 Motivation (UC) (CM) for normalized Lebesgue measure λ [Gromov V. Milman, 83] α λ (ε) 2e 2nδ(ε), with the modulus of convexity δ(ε) = 1 sup{ x + y /2 : x = y = 1, x y = ε}. Push Forward Measures And Concentration Phenomena 4 / 19

5 Motivation (UC) (CM) for normalized Lebesgue measure λ [Gromov V. Milman, 83] α λ (ε) 2e 2nδ(ε), with the modulus of convexity δ(ε) = 1 sup{ x + y /2 : x = y = 1, x y = ε}. (CM) (CD) for λ [Arias-de-Reyna Ball Villa, 98] If α λ (ε) e nφ(ε), for some increasing function φ, then a [ 1 2, 2] : { (λ λ) (x, y) : x y [ a(1 ε), a(1 + ε) ]} 1 4e nφ(ε/6). Push Forward Measures And Concentration Phenomena 4 / 19

6 Motivation (UC) (CM) for normalized Lebesgue measure λ [Gromov V. Milman, 83] α λ (ε) 2e 2nδ(ε), with the modulus of convexity δ(ε) = 1 sup{ x + y /2 : x = y = 1, x y = ε}. (CM) (CD) for λ [Arias-de-Reyna Ball Villa, 98] If α λ (ε) e nφ(ε), for some increasing function φ, then a [ 1 2, 2] : { (λ λ) (x, y) : x y [ a(1 ε), a(1 + ε) ]} 1 4e nφ(ε/6). (CD) (AE) for any µ. (CM) (AE) for any µ. Push Forward Measures And Concentration Phenomena 4 / 19

7 Motivation Families of spaces: X n (UC) Uniformly convex spaces: δ n (ε) δ(ε). (CM) Concentration of measure for λ n α λn (ε) Ce nφ(ε). (CD) Concentration of distance for λ n λ n λ n {(x, y) : x y [a(1 ε), a(1 + ε)]} 1 Ce nφ(ε). Push Forward Measures And Concentration Phenomena 5 / 19

8 Motivation Families of spaces: X n (UC) Uniformly convex spaces: δ n (ε) δ(ε). (CM) Concentration of measure for λ n α λn (ε) Ce nφ(ε). (CD) Concentration of distance for λ n λ n λ n {(x, y) : x y [a(1 ε), a(1 + ε)]} 1 Ce nφ(ε). Neither of the three are preserved by isomorphisms. Compare l n+1 2 and (l n 2 l 1 ). Push Forward Measures And Concentration Phenomena 5 / 19

9 Motivation Families of spaces: X n (UC) Uniformly convex spaces: δ n (ε) δ(ε). (CM) Concentration of measure for λ n α λn (ε) Ce nφ(ε). (CD) Concentration of distance for λ n λ n λ n {(x, y) : x y [a(1 ε), a(1 + ε)]} 1 Ce nφ(ε). Neither of the three are preserved by isomorphisms. Compare l n+1 2 and (l n 2 l 1 ). (AE) Almost equilateral set of large cardinality Is it preserved under isomorphism? N n Ce nφ(ε) Push Forward Measures And Concentration Phenomena 5 / 19

10 Push-forward measure Definition (X, µ) measure spaces, Y a measurable space, φ : X Y a measurable map. The push-forward of µ by φ is ν where for any measurable set A Y. ν(a) := φ (µ)(a) := µ(φ 1 (A)) The Lipschitz bound For any (X, d X, µ) metric prob. space, and (Y, d Y ) metric space, if φ : X Y is λ-lipschitz then α ν (ε) α µ (ε/λ). Push Forward Measures And Concentration Phenomena 6 / 19

11 Concentration of λ and BM distance Banach Mazur distance K, L R n o-symmetric convex bodies. d BM (K, L) = inf{λ > 0 : K T (L) λk, T GL(R n )}. Central projection of K to L The Lipschitz bound π : R n R n ; π(x) = x K x L x If K and L are in Banach Mazur position and d BM (K, L) = λ then π is λ-lipschitz, and hence α ν (ε) α µ (ε/λ) for ν = π (µ) for any prob. measure µ on K. Push Forward Measures And Concentration Phenomena 7 / 19

12 A Positive Result The Lipshcitz bound If K and L are in Banach Mazur position and d BM (K, L) = λ then π is λ-lipschitz, and hence α ν (ε) α µ (ε/λ) for ν = π (µ) for any prob. measure µ on K. Push Forward Measures And Concentration Phenomena 8 / 19

13 A Positive Result The Lipshcitz bound If K and L are in Banach Mazur position and d BM (K, L) = λ then π is λ-lipschitz, and hence α ν (ε) α µ (ε/λ) for ν = π (µ) for any prob. measure µ on K. Theorem 1 K, L R n o-symmetric convex bodies, L K λl, µ a prob. measure on R n and π(x) = x x L x K the central projection. ν := π (µ). Then ( ε m L α (L,ν) (ε) 16α (K,µ) 14 λm K where m K := median µ (. K : R n R) and m L := median µ (. L : R n R). Note that. K. L, so m K m L. ), Push Forward Measures And Concentration Phenomena 8 / 19

14 A Distance-like Quantity: β m K := median µ (. K : R n R) and m L := median µ (. L : R n R). If supp(µ) K then supp(ν) L and m K = 1. If µ is the normalized Lebesgue measure restricted to K then m K 1. Push Forward Measures And Concentration Phenomena 9 / 19

15 A Distance-like Quantity: β m K := median µ (. K : R n R) and m L := median µ (. L : R n R). If supp(µ) K then supp(ν) L and m K = 1. If µ is the normalized Lebesgue measure restricted to K then m K 1. A natural quantity to study: { β((k, µ), L) = inf λ m } K : TL K λtl, T GL(R n ) m TL Push Forward Measures And Concentration Phenomena 9 / 19

16 A Distance-like Quantity: β m K := median µ (. K : R n R) and m L := median µ (. L : R n R). If supp(µ) K then supp(ν) L and m K = 1. If µ is the normalized Lebesgue measure restricted to K then m K 1. A natural quantity to study: { β((k, µ), L) = inf λ m } K : TL K λtl, T GL(R n ) m TL β((k, µ), L) d BM (K, L). Push Forward Measures And Concentration Phenomena 9 / 19

17 A Distance-like Quantity: β m K := median µ (. K : R n R) and m L := median µ (. L : R n R). If supp(µ) K then supp(ν) L and m K = 1. If µ is the normalized Lebesgue measure restricted to K then m K 1. A natural quantity to study: { β((k, µ), L) = inf λ m } K : TL K λtl, T GL(R n ) m TL β((k, µ), L) d BM (K, L). So, Theorem( 1 replaces ) the obvious bound α ν (ε) α ε µ d BM by ( ) α ν (ε) 16α ε µ 14β. Push Forward Measures And Concentration Phenomena 9 / 19

18 More on β β is not new. Let K be the Euclidean ball with the normalized Lebesgue measure, and k(l, ε) denote the dimension of an ε-almost Euclidean section of L. Milman s Theorem: k(l, ε) c(ε) n β 2. β can be far below d BM : while d BM (B n 2, Bn 1 ) = n, we have β(b n 2, Bn 1 ) c. Push Forward Measures And Concentration Phenomena 10 / 19

19 More on β β is not new. Let K be the Euclidean ball with the normalized Lebesgue measure, and k(l, ε) denote the dimension of an ε-almost Euclidean section of L. Milman s Theorem: k(l, ε) c(ε) n β 2. β can be far below d BM : while d BM (B n 2, Bn 1 ) = n, we have β(b n 2, Bn 1 ) c. We can replace the medians in the definition of β by means to obtain β. If µ is a log-concave measure, then β and β are equivalent. Push Forward Measures And Concentration Phenomena 10 / 19

20 More on β β is not new. Let K be the Euclidean ball with the normalized Lebesgue measure, and k(l, ε) denote the dimension of an ε-almost Euclidean section of L. Milman s Theorem: k(l, ε) c(ε) n β 2. β can be far below d BM : while d BM (B n 2, Bn 1 ) = n, we have β(b n 2, Bn 1 ) c. We can replace the medians in the definition of β by means to obtain β. If µ is a log-concave measure, then β and β are equivalent. For any o-symmetric convex body L, n β(b 2 n, L) C log n. Push Forward Measures And Concentration Phenomena 10 / 19

21 Examples: l p spaces Let 1 p < 2, consider B n 2 with λ. Then β(b 2 n, Bp n ) b p, where b p depends only on p. Better than the Lipschitz bound. Let 2 p <, consider B2 n with λ. Then β(b n 2, B n p ) C p n p implies α (B n p,ν)(ε) C 3 exp{ c 3 ε 2 n 2/p }, Not better than the Lipschitz bound. Push Forward Measures And Concentration Phenomena 11 / 19

22 A Negative Result B n has no concentration w.r.t. the Lebesgue measure: α λ (ε) 1/2 ε. β(b 2 n, Bn ) n log n. This won t yield a well-concentrated measure on B. n Push Forward Measures And Concentration Phenomena 12 / 19

23 A Negative Result B n has no concentration w.r.t. the Lebesgue measure: α λ (ε) 1/2 ε. β(b 2 n, Bn ) n log n. This won t yield a well-concentrated measure on B. n Theorem 2 For any ν o-symmetric prob. measure on B n, α ν (ε) 1 2n (1 ν(εbn )). Push Forward Measures And Concentration Phenomena 12 / 19

24 A Stronger Negative Result Definition T : X Y linear map between two normed spaces is a d-embedding, if a x X Tx Y b x X with b/a d. Theorem 3 Let T : (R n,. K ) l N be a d-embedding, µ an o-symmetric prob. measure on (R n,. K ). Then for any 0 < ε < 1/d, if α µ (ε) > 0 then N 1 µ(dεk). 2α µ (ε) Applying this to T = id l n yields Theorem 2. Push Forward Measures And Concentration Phenomena 13 / 19

25 Combining the results Theorem 4 K R n a convex body, µ an o-symmetric probability measure on (R n,. K ). Assume that for some 0 < s, and (R n,. L ) d l N. Then α (K,µ) (ε) Ce cεsn, β((k, µ), L) m ( K 14d cn log(64cn) ) 1/s. Push Forward Measures And Concentration Phenomena 14 / 19

26 Proof of Theorem 3 Theorem 3 Let T : (R n,. K ) l N be a d-embedding, µ an o-symmetric prob. measure on K. Then for any 0 < ε < 1/d, if α µ (ε) > 0 then N 1 µ(dεk). 2α µ (ε) T : X l N, assume that 1 d x Tx x. f i : X R the i th coordinate of T. Let A i := {x R n : f i (x) ε}. Push Forward Measures And Concentration Phenomena 15 / 19

27 Proof of Theorem 3 Theorem 3 Let T : (R n,. K ) l N be a d-embedding, µ an o-symmetric prob. measure on K. Then for any 0 < ε < 1/d, if α µ (ε) > 0 then N 1 µ(dεk). 2α µ (ε) T : X l N, assume that 1 d x Tx x. f i : X R the i th coordinate of T. Let A i := {x R n : f i (x) ε}. {x R n : f i (x) ε} {x R n : f i (x) 0} +εk. }{{} of measure 1/2 Thus, µ(a C i ) 2α µ (ε), and hence ( N ) µ A i 1 2Nα µ (ε). i=1 Push Forward Measures And Concentration Phenomena 15 / 19

28 Proof of Theorem 3 Theorem 3 Let T : (R n,. K ) l N be a d-embedding, µ an o-symmetric prob. measure on K. Then for any 0 < ε < 1/d, if α µ (ε) > 0 then N 1 µ(dεk). 2α µ (ε) T : X l N, assume that 1 d x Tx x. f i : X R the i th coordinate of T. Let A i := {x R n : f i (x) ε}. {x R n : f i (x) ε} {x R n : f i (x) 0} +εk. }{{} of measure 1/2 Thus, µ(a C i ) 2α µ (ε), and hence ( N ) µ A i 1 2Nα µ (ε). i=1 Let r (0, 1) be such that µ(rk) < 1 2Nα µ (ε). We have N i=1 A i rk. Push Forward Measures And Concentration Phenomena 15 / 19

29 Proof of Theorem 3 Let r (0, 1) be such that µ(rk) < 1 2Nα µ (ε). We have N i=1 A i rk. x N A i with x K r. i=1 r < x K d Tx dε. Push Forward Measures And Concentration Phenomena 16 / 19

30 Proof of Theorem 3 Let r (0, 1) be such that µ(rk) < 1 2Nα µ (ε). We have N i=1 A i rk. x N A i with x K r. i=1 r < x K d Tx dε. So we have: if dε < r then µ(rk) 1 2Nα µ (ε). µ(dεk) 1 2Nα µ (ε). Push Forward Measures And Concentration Phenomena 16 / 19

31 Proof of Theorem 1 Theorem 1 L K λl, µ a prob. measure on (R n,. K ). π(x) = central projection, ν := π (µ). Then ) m L, ( ε α (L,ν) (ε) 16α (K,µ) 14 λm K x x L x K the where m K := median µ (. K : R n R) and m L := median µ (. L : R n R). Let A R n with 1/2 ν(a) = µ(π 1 (A)). Goal: µ(π 1 (A L ε)) is large. Let δ := ε 7m K. G L := {x R n : (1 δ)m L < x L < (1 + δ)m L } G K := {x R n : (1 δ)m K < x K < (1 + δ)m K }. Push Forward Measures And Concentration Phenomena 17 / 19

32 Proof of Theorem 1 G L := {x K : (1 δ)m L < x L < (1 + δ)m L }, G K := {x K : (1 δ)m K < x K < (1 + δ)m K }, J := π 1 (A) G L G K. L K λl =. K. L λ. K = m K m L λm K. By the Lipschitz bound µ(j) 1/2 2α (K,µ) (δm L /λ) 2α (K,µ) (δm K ) 1/2 4α (K,µ) (δm L /λ). By computation: J K δm L λ π 1 ( A L ε ) (1) Push Forward Measures And Concentration Phenomena 18 / 19

33 Proof of Theorem 1 ν J K δm L λ π 1 ( A L ε (( ) c ) ( (( ) c )) A L ε = µ π 1 A L ε ) (( ) c ) µ J K δm L λ... (1) Push Forward Measures And Concentration Phenomena 19 / 19

34 Proof of Theorem 1 ν J K δm L λ π 1 ( A L ε (( ) c ) ( (( ) c )) A L ε = µ π 1 A L ε ) (( ) c ) µ J K δm L λ... (1) Lemma (Ledoux): µ(j)µ(f ) 4α (X,d,µ) (dist(j, F )/2).... 4α (K,µ)( δm L 2λ ) 4α (K,µ) ( δm L 2λ ) µ(j) 1/2 4α (K,µ) ( δm L λ ). Hence, ε > 0 such that 16α (K,µ) (εm L /(7λm K )) 1, (( ) c ) ( ) ν A L εml ε 16α (K,µ). 14λm K And so, α (L,ν) (ε) 16α (K,µ) ( εml 14λm K ). Push Forward Measures And Concentration Phenomena 19 / 19