Chapter 9: Basic of Hypercontractivity

Transcription

1 Analysis of Boolean Functions Prof. Ryan O Donnell Chapter 9: Basic of Hypercontractivity Date: May 6, 2017 Student: Chi-Ning Chou Index Problem Progress 1 Exercise 9.3 (Tightness of Bonami Lemma) 2/2 2 Exercise 9.6 (Bonami Lemma implies (2,4)-Hypercontractivity) 4/4 3 Exercise 9.9 (Basic properties of (p, q, ρ)-hypercontractivity) 2/2 4 Exercise 9.10 (Basic properties of (p, q, ρ)-hypercontractivity) 2/2 5 Exercise 9.11 (Basic properties of (p, q, ρ)-hypercontractivity) 3/3 6 Exercise 9.14 ((p, 2)-Bonami Lemma) 2/2 7 Exercise 9.15 (Generality of Theorem 9.22) 1/1 8 Exercise 9.16 ((2, q)-hypercontractivity by induction of derivative) 1/1 9 Exercise 9.18 (Sharpen Level-1 Inequality) 2/2 10 Exercise 9.19 (Level-k inequalities) 2/2 11 Exercise 9.20 (KKL fails for function with range [ 1, 1]) 1/1 14 Exercise 9.29 (General-variance KKL) 1/1 15 Exercise 9.30 (State-of-the-art constant of KKL) 2/3 16 Exercise 9.31 (Delicate version of Theorem 9.28) 1/1 Table 1: Problem Set 6: Progress 1

2 Problem 1. Exercise 9.3 (Tightness of Bonami Lemma) (a) Let A := {S 1 S 2 S 7 ( E[f 4 ] = E x { 1,1} n[ = = [n] ) k }, we have 3, k 3,..., k 3,n 2k x T 1 x T 2 x T 3 x T 4 ] (1) T 1,...,T 4 [n], T 1 = = T 4 =k T 1,...,T 4 [n], T 1 = = T 4 =k S 1 S 7 A S 1 S 7 A E x { 1,1} n[x T 1 x T 2 x T 3 x T 4 ] (2) E x { 1,1} n[x S 1 S 2 S 3 x S 4 S 5 S 6 x S 1 S 3 S 5 x S 2 S 4 S 6 ] (3) E x { 1,1} n[(x i [6]S i ) 2 ] (4) ( ) n = A = k 3, k 3,..., k 3, n 2k. (5) Also, E[f 2 ] = E x { 1,1} n[ = = T 1,T 2 [n], T 1 = T 2 =k T [n], T =k T 1,T 2 [n], T 1 = T 2 =k 1 = x T 1 x T 2 ] (6) E x { 1,1} n[x T 1 x T 2 ] (7) ( ) n. (8) k As a result, we have E[f 4 ] ( n ) 1 = k 3, k 3,..., k 3,n 2k E[f 2 ] 2 ( n k) 2. (9) (b) Expand the formula, we have ( n ) k 3,..., k 3 (,n 2k n k ) 2 = (n) 2k (k!) 2 (n) 2 k ( k, (10) 3!)6 where (n) k = n (n 1) (n k + 1). By Stirling s formula and taking n, we have (k!) 2 (n) 2k lim n (n) 2 k Plug-in (a), we have f 4 Ω(k 1/2 ) 3 k f 2. ( k 3!)6 = Θ(k 2 9 k ), (11) = 1. (12) 2

3 Problem 2. Exercise 9.6 (Bonami Lemma implies (2,4)-Hypercontractivity) (a) By triangle inequality, Bonami lemma, and the fact that f =k 2 f 2 for any k N, we have T (1 δ)/ 3 f 4 ( Bonami lemma) = T (1 δ)/ 3 f =k 4 (13) k=0 k 3 T(1 δ)/ 3 f =k 2 (14) k=0 (1 δ) k f =k 2 (15) k=0 (1 δ) k f 2 1 δ f 2. (16) k=0 Note that k=0 f =k 2 2 = f 2 2 by Parseval s equality. (b) Here, we show that for any h : { 1, 1} n R, g d p = g d p for any p R + and d N +. g d p p = g(x (1) ) p g(x (2) ) p g(x (d) ) p (17) x (1),x (2),...,x (d) { 1,1} n d ( ) = (18) i=1 (c) Apply (a) on T (1 δ)/ 3 f d we get x (i) { 1,1} n g(x (i) ) p = g dp p. (19) T (1 δ)/ 3 f d 4 = T (1 δ)/ 3 f d 4 (20) 1 δ f d 2 = 1 δ f d 2. (21) That is, T (1 δ)/ 3 f 4 ( 1 δ )1/d f 2. When d, we have T (1 δ)/ 3 f 4 f 2. (d) Take δ 0 +, we have T 1/ 3 f 4 f 2, i.e., the (2,4)-Hypercontractivity Theorem. 3

4 Problem 3. Exercise 9.9 (Basic properties of (p, q, ρ)-hypercontractivity) (a) For any a, b R, Thus, cx is (p, q, ρ)-hypercontractive. a + ρb(cx) q = a + ρbcx q (22) a + bcx p (23) = a + b(cx) p. (24) (b) Take a = 0 and b = 1, by the (p, q, ρ)-hypercontractivity of X, we have Thus, ρ X p X q. ρ X q = ρx q (25) X p. (26) 4

5 Problem 4. Exercise 9.10 (Basic properties of (p, q, ρ)-hypercontractivity) (a) By Taylor expansion to the first order, we have 1 + bx r 1 + be[x] + O(b 2 ). (27) As X is (p, q, ρ)-hypercontractive, we have 1 + ρɛx q 1 + ɛx p. Combine with (27), we have 1 + ρɛe[x] + O(ρ 2 ɛ 2 ) 1 + ɛe[x] + O(ɛ 2 ). (28) By letting ɛ 0 +, as 0 < ρ < 1, there exists ɛ 0 such that As ρ <, the above inequality holds only if E[X] = 0. ρɛ 0 E[X] ɛ 0 E[X]. (29) (b) To Taylor expand 1 + bx r, let s compute d 2 d(bx i ) bx r as follows. d 1 + bx r = b 1 + bx 1 r r (1 + bx i ) r 1, (30) dx i d 2 dx 2 i 1 + bx r = (1 r)b bx 1 2r r Thus, Taylor expand 1 + bx r to the second order, we have (1 + bx i ) 2r 1 + b(r 1) 1 + b 2 X 1 r r (1 + bx i ) r 2. (31) 1 + bx r = 1 + be[x] + b2 (r 1) E[X 2 ] + O(b 3 ) (32) 2 From (a), we know that E[X] = 0 so by the (p, q, ρ)-hypercontractivity of X, we have 1 + ρ2 ɛ 2 (q 1) 2 Similarly, when ɛ 0 +, there exists ɛ 0 > 0 such that That is, ρ p 1 q 1. E[X 2 ] + O(ɛ 3 ρ 3 ) 1 + ɛ2 (p 1) E[X 2 ] + O(ɛ 3 ). (33) 2 ρ 2 ɛ 2 0(q 1)E[X 2 ] ɛ 2 0(p 1)E[X 2 ]. (34) 5

6 Problem 5. Exercise 9.11 (Basic properties of (p, q, ρ)-hypercontractivity) (a) For any a, b R, Thus, X is (q, q, 0)-hypercontractive. (b) WLOG, consider arbitrary a R, a + 0 bx q = a (35) = E[a + bx] (36) E[ a + bx ] (37) = a + bx 1 (38) a + bx q. (39) a + ρx q = (1 ρ)a + ρ(a + X) q (40) Thus, X is (q, q, ρ)-hypercontractive for any 0 ρ 1. (1 ρ)a + ρ a + X q (41) (1 ρ) a + X q + ρ a + X q (42) = a + X q. (43) (c) By Exercise 9.10 (a), E[X] = 0. Let 0 ρ ρ 1. By Exercise 9.11 (b), X is (q, q, ρ /ρ)- hypercontractive. That is, for any a, b R, a + ρ bx q a + ρbx q (44) a + bx p, (45) where the first inequality is due to the (q, q, ρ /ρ)-hypercontractivity and the second inequality is due to the (p, q, ρ)-hypercontractive. Thus, if X is (p, q, ρ)-hypercontractive, then X is also (p, q, ρ )-hypercontractive for any 0 ρ ρ. 6

7 Problem 6. Exercise 9.14 ((p, 2)-Bonami Lemma) (a) For any 1 p 2, take q = By Holder s inequality, we have p p 1 2. By Theorem 9.21 (the (2, q)-bonami Lemma), we have f q q 1 k f 2. (46) f 2 2 f p f q. (47) Combine the above two equations, we have the (p, 2)-Bonami Lemma as follows. f 2 k 1 f p. (48) p 1 (b) It suffices to show that exp(2 4 p ) > p 1 for any 1 p < 2. First, observe that when p = 2, the two sides are both equal to 1. Next, differentiate both sides and get By first order Taylor approximation, d dp LHS = 4 p 2 exp(2 4 ), p (49) d RHS = 1. dp (50) e 2 p p 1 = 2 p. (51) Plug-in (49) and (50), we have d p LHS d dprhs and thus the inequality holds. 7

8 Problem 7. Exercise 9.15 (Generality of Theorem 9.22) Similarly to the proof of Theorem 9.22, take 1 p 2 such that 1 2 = θ p + 1 θ 2+ɛ θ and ɛ > 0. By the interpolation version of Holder s inequality, we have for some f 2 f θ p f 1 θ 2+ɛ (52) f θ p 1 + ɛ k(1 θ) f 1 θ 2. (53) That is, Compute that Let ɛ 0 +, we have f 2 (1 + ɛ) 1 θ 2θ k f p. (54) 1 θ 2θ θ = ɛp 2(2 + ɛ p), (55) = ( 2 p 1)1 ɛ + 1 p 1 2. (56) f 2 e k( 2 p 1) f p. (57) 8

9 Problem 8. Exercise 9.16 ((2, q)-hypercontractivity by induction of derivative) Let s prove the (2, q)-hypercontractivity theorem using induction of derivative. Recall that f = e + x n d. That is, by the linearity of T, we can rewrite T f q as follows. As (1/ q 1) 1 and x n { 1, 1}, we have T f 2 q = T e + (1/ q 1)x n T d 2 q (58) = E x [E xn [ T e + (1/ q 1)x n T d q ]] 2/q (59) ( = E x [E xn [ T e + (1/ q 1)x n T d 2) q/2 ]] 2/q (60) E xn [ T e + (1/ q 1)x n T d 2 ] = (T e) q 1 (T d)2 (T e) 2 + (T d) 2. (61) Since h(t) = t q/2 is convex when q 2, we have E xn [ ( T e + (1/ q 1)x n T d 2) q/2 ( ] E xn [ T e + (1/ q/2. q 1x n T d) ]) 2 (62) Combine (61) and (62), we have E x [E xn [ T e + (1/ ( q 1)x n T d q ]] 2/q E x [ (T e) 2 + (T d) 2) q/2 ] 2/q = (T e) 2 + (T d) 2 q/2. (63) By the triangle inequality of q/2, we have (T e) 2 + (T d) 2 q/2 (T e) 2 q/2 + (T d) 2 q/2 = T e 2 q + T d 2 q. (64) By induction hypothesis, T e 2 q/2 e 2 q and T d 2 2 d 2 2. Thus, we conclude that T f 2 q e d 2 2 = f 2 2. (65) 9

10 Problem 9. Exercise 9.18 (Sharpen Level-1 Inequality) (a) Combine the fact that E[f] = α and the expansion of stability of f, for any 0 < ρ 1, we have Stab ρ [f] = n ρ k W k [f] (66) k=0 Also, by the Small-Set Expansion theorem, Thus, W 1 [f] 1 ρ (α 2 1 ρ α 2 ). (b) Reformulate the RHS in (a), we have ρw 1 [f] + E[f] 2 = ρw 1 [f] + α 2. (67) Stab ρ [f] α 2 1 ρ. (68) 1 ρ (α 2 1 ρ α 2 ) = α 2 1 2ρ (α 1 ρ 1) ρ (69) = α 2 1 2ρ ln α (1 ρ 1 ρ + O(ρ2 ) 1) (70) = α 2 ( 2 ln(1/α) + O(ρ)). 1 ρ (71) Let ρ 0 and combine with (a), we have W 1 [f] 2α 2 ln(1/α). 10

11 Problem 10. Exercise 9.19 (Level-k inequalities) (a) Plug-in k ɛ = 2e 1 ɛ ln(1/α), we have (b) By second order Taylor expansion, we have W k 2e [f] ( 2 ln(1/α)(1 ɛ) ln(1/α))2(1 ɛ) ln(1/α) α 2 (72) e = ( 1 ɛ )2(1 ɛ) ln(1/α) α 2 (73) = α 2 2(1 ɛ)(1 ln(1 ɛ)) (74) = α 2ɛ 2(1 ɛ) ln(1/(1 ɛ)). (75) ln(1 ɛ) ɛ + ɛ2 2. (76) Thus, α 2ɛ 2(1 ɛ) ln(1/(1 ɛ)) = α 2ɛ+2(1 ɛ) ln((1 ɛ)) (77) α 2ɛ+2(1 ɛ)( ɛ+ɛ2 /2) (78) = α 2ɛ2 ɛ 2 = α ɛ2. (79) 11

12 Problem 11. Exercise 9.20 (KKL fails for function with range [ 1, 1]) i [n] Consider f(x) = trunc( x i n ) and let f (x) = Var[f] P[ i [n] x i n. Observe that i [n] x i n > 1] = Ω(1). (80) Also, for any i [n], As Inf i [f ] = E[x i ][(D i f ) 2 ] = 1 n, we have As a result, the statement of KKL doesn t hold on f. Inf i [f] Inf i [f ]. (81) MaxInf[f] MaxInf[f ] = 1 n. (82) 12

13 Problem 12. Exercise 9.24 (Hamming ball achieves Small-Set Expansion stability bound) By Small-Set Expansion theorem, we have an upper bound for the ρ-stability of indicator function. Concretely, for any A { 1, 1} n, E[1 A ] = α, we have Stab ρ [1 A ] α 2 1+ρ. (83) In this exercise, we are going to show that the Hamming ball indicator function will asymptotically achieve the bound by proving that Λ ρ (Φ 1 (t)) = Θ(µ 2 1+ρ ). (84) (a) 13

14 Problem 13. Exercise 9.25 (a) 14

15 Problem 14. Exercise 9.29 (General-variance KKL) It suffices to prove that S: S 1 ( 1 3 ) S f(s) 2 Var[f] 3 I[f]/ Var[f]. Recall that Var[f] = S: S 1 f(s) 2. Thus, by the concavity of h(t) = (1/3) t, we have S: S 1 f(s) 2 Var[f] (1 3 ) S ( 1 f(s) 2 3 ) S: S 1 Var[f] S (85) = 3 I[f]/ Var[f]. (86) Note that replace 1 3 with 0 < ρ < 1, we have ρ S f(s) 2 Var[f] ρ I[f]/ Var[f]. (87) S: S 1 15

16 Problem 15. Exercise 9.30 (State-of-the-art constant of KKL) (a) From Exercise 9.29, we have S: S 1 ρ S f(s) 2 Var[f] ρĩ[f]. (88) Note that the LHS is upper bounded by 1 ρ I(ρ) [f] because I (ρ) [f] = S S ρ S 1 f(s) 2 (89) 1 ρ S: S 1 Also, for each i [n], by Corollary 9,25, we have Thus, we have I (ρ) [f] = i [n] ρ S f(s) 2. (90) I (ρ) i [f] I i [f] 2 1+ρ. (91) I (ρ) i [f] I i [f] 2 1+ρ MaxInf[f] 1 ρ 1+ρ I[f]. (92) i [n] As a result, from (88), (90), and (92), we have MaxInf[f] 1 ρ 1+ρ 1 ρ Ĩ[f] 1 ρĩ[f]. (93) For any 0 < δ < 1, take ρ = 1 δ 1+δ 1 ρ (0, 1) such that δ = 1+ρ. We have MaxInf[f] ( 1 + δ 1 δ ) 1 δ ( 1 Ĩ[f] ) 1 δ ( 1 δ 1 + δ ) 1 δ Ĩ[f]. (94) Now, consider how small the base of Ĩ[f] can be. ( 1 + δ 1 δ )1/δ > ( /δ )1/δ. (95) Note that the RHS is a non-decreasing function. Thus, it suffices to lower bound the case when δ 0. lim (1 + 2 δ 0 1/δ )1/δ = e 2. (96) As a result, we have for any C > e 2 MaxInf[f] Ω(C Ĩ[f] ). (97) 16

17 (b) Let s lower bound ( 1 δ 1+δ )1/δ for 0 < δ < 1/2. By the definition of e, we have ( 1 δ ( 1 + δ )1/δ = (1 2δ 1 + δ )1+δ/δ) 1/(1+δ) (98) ( e 2) 1/(1+δ). (99) When 0 < δ < 1/2, by second order Taylor expansion, we have Combine (99) and (100), we have δ = (1 + δ) 1 1 δ + δ δ2 2. (100) Now, take δ = 1, by (94), (96), and (101), we have 2Ĩ[f]1/3 ( 1 δ 1 + δ )1/δ e 2 δ2. (101) MaxInf[f] ( 1 + δ 1 δ ) 1 δ ( 1 Ĩ[f] ) 1 δ ( 1 δ 1 + δ ) 1 δ Ĩ[f] (102) e 2 ( 1 Ĩ[f] )2Ĩ[f]1/3 exp( 2Ĩ[f]) exp( Ĩ[f]1/3 ). (103) 4 17

18 ρ k S / F Problem 16. Exercise 9.31 (Delicate version of Theorem 9.28) Similarly, we take the parameters in Theorem 9.28 in the following setting. For any η > 0, 0 ɛ < 1, and k 0, define τ = ɛ1+η I[f] 1+η C(η) k, J = {j [n] : I j [f] τ}, so J I[f]2+η ɛ 1+η C(η)k. (104) Also, define F = {S : S J} {S : S > k}. (105) Now, consider 0 < ρ < 1 chosen later according to η. The idea is to upper bound i/ J I(ρ) i [f] with the error and lower bound it with the Fourier mass not in F. By Corollary 9.25 i/ J I (ρ) i [f] I[f] 2 1+ρ i/ J (106) max i[f] 1 ρ 1+ρ I i [f] i/ J (107) τ 1 ρ 1+ρ I[f] = i/ J I[f] ɛ(1+η) 1 ρ 1+ρ 1 ρ (1+η) 1+ρ 1 ρ C(η) k 1+ρ. (108) 1 As to the lower bound, we have I (ρ) i [f] = i/ J i/ J = S S / F S: i S ρ S 1 f(s) 2 S J ρ S 1 f(s) 2 ρ S 1 f(s) 2 (109) (110) (111) f(s) 2. (112) Now, take ρ = η 2+η = C(η)1/2, we have 1 + η = 1+ρ 1 ρ C(η) k/2 S / F f(s) 2 i/ J. As a result, (108) and (112) become I (ρ) i [f] C(η) k/(1+η) ɛ (113) That is, S / F f(s) 2 C(η) k( η ) ɛ. (114) 18

19 Problem 17. (a) 19

20 Problem 18. (a) 20

21 Problem 19. (a) 21

22 Problem 20. (a) 22