FKN Theorem on the biased cube

Transcription

1 FKN Theorem on the biased cube Piotr Nayar Abstract In this note we consider Boolean functions defined on the discrete cube { γ, γ 1 } n equipped with a product probability measure µ n, where µ = βδ γ + αδ γ 1 and γ = α/β. This normalization ensures that the coordinate functions (x i ) i=1,...,n are orthonormal in L ({ γ, γ 1 } n, µ n ). We prove that if the spectrum of a Boolean function is concentrated on the first two Fourier levels, then the function is close to a certain function of one variable. Our theorem strengthens the non-symmetric FKN theorem due to Jendrej, Oleszkiewicz and Wojtaszczyk. Moreover, in the symmetric case α = β = 1 a [ 1, 1]-valued function defined on the discrete cube is close to a certain affine function, then it is also close to a [ 1, 1]-valued affine function. we prove that if 010 Mathematics Subject Classification. Primary 4C10; Secondary 60E15. Key words and phrases. Boolean functions, Walsh-Fourier expansion, FKN Theorem 1 Introduction and notation Let α, β > 0 with α + β = 1 and α (0, 1 ). We consider the discrete cube { γ, γ 1 } n equipped with the L structure given by the product probability measure µ n = µ n, where µ = βδ γ + αδ γ 1 and γ = α/β. For f, g : { γ, γ 1 } n R let us define the expectation Ef = f dµ n, the standard scalar product f, g = Efg and the induced norm f = f, f. We also Research partially supported by NCN Grant no. 011/01/N/ST1/

2 define the L p norm, f p = (E f p ) 1/p. Let [n] = {1,..., n}. For T [n] and x = (x 1,..., x n ) let w T (x) = i T x i and w 1. Note that we have Ex i = 0 and Ex i x j = δ ij. It follows that (w T ) T [n] is an orthonormal basis of L ({ γ, γ 1 } n, µ n ). Therefore, every function f : { γ, γ 1 } n R admits the unique expansion f = T [n] a T w T. The functions w T are sometimes called the Walsh-Fourier functions. If the function f is { 1, 1}-valued then it is called Boolean. The Fourier analysis of Boolean functions plays an important role in many areas of research, including learning theory, social choice, complexity theory and random graphs, see e.g. [O1] and [O]. One of the most important analytic tools in this theory is the so-called hypercontractive Bonami-Beckner- Gross inequality, see [Bo], [Be], [G1] and [G] for a survey on this topic. This inequality has been used in the celebrated papers by J. Kahn, G. Kalai and N. Linial, [KKL], and E. Friedgut, [F]. It can be stated as follows. Take α = β = 1 and q [1, ]. Then we have (q 1) T / a T w T a T w T (1) q T [n] T [n] for every choice of a T R. This inequality has been generalized in [Ol1] to the non-symmetric case. Namely, the following inequality holds true, c q (α, β) T a T w T a T w T, () q T [n] T [n] where c q (α, β) = αβ β q ( α q α q β q ). One can easily check that (1) is a special case of (), namely q 1 = lim ε 0 c q ( 1 ε, 1 + ε). Moreover, it is easy to see that c q(α, β) [0, 1]. In [FKN] the authors proved the following theorem, which is now called the FKN Theorem. Suppose α = β = 1 and we have a Boolean function f whose Fourier spectrum is concentrated on the first two levels, say T >1 a T < ε. Then f is Cε-close in the L norm to the constant function or to one of the functions ±x i. Here and in what follows C is a universal

3 constant that may vary from one line to another. The authors gave two proofs of this theorem. One of them contained an omission which was fixed by G. Kindler and S. Safra in their unpublished paper, [KS], see also [K]. The FKN Theorem was originally devised for applications in discrete combinatorics and social choice theory. It is useful in the proof of the robust version of the famous Arrow s theorem on Condorcet s voting paradox, see [A] and [KG]. It was also applied is theoretical computer science, e.g., it is useful in analyzing the Long Code Test in the proof of the PCP theorem by I. Dinur, [D]. Also the FKN Theorem in the biased case is worthy of attention, e.g., p-biased long code was used by I. Dinur and S. Safra in their PCP proof of NP-hardness of approximation of the Vertex Cover problem, see [DS]. In [JOW] the authors gave a proof of the following version of the FKN Theorem, Theorem 1 ([JOW], Theorem 5.3 and Theorem 5.8). Let f = T a T w T be the Walsh-Fourier expansion of a function f : { 1, 1} n { 1, 1} and ( ) 1/. let ρ = T >1 a T Then there exists B [n] with B 1 such that T 1,T B a T Cρ4 ln(/ρ) and a B 1 ρ Cρ 4 ln(/ρ). In particular, dist L (f, w B ) ρ + Cρ ln(/ρ). (3) Moreover, in the non-symmetric case, f : { γ, γ 1 } n { 1, 1}, there exists k [n] such that f (a + a {k} w {k} ) 8 ρ. The inequalities (3) is sharp, up to the universal constants. In the proof the inequality (1) has been used. However, in the non-symmetric case one can ask for a better bound involving bias parameter α. In this note we use inequality () to prove such an extension of the FKN Theorem. Namely, we have Theorem. Let f = T a T w T be the Walsh-Fourier expansion of a function ( ) 1/. f : { γ, γ 1 } n { 1, 1} and let ρ = T >1 a T Then there exists k [n] such that for ρ ln(e /ρ) < 3 α we have 10 e 4 f (a + a {k} w {k} ) ρ (4) and f sgn(a + a {k}w {k}) 4ρ. (5) 3

4 In this paper we use the { 1, 1}-valued function sgn(x) = I (,0) (x) + I [0, ) (x). Our proof of Theorem, which is given in the Section, is an application of the ideas used in the proof of Theorem 5.3 in [JOW]. Our inequality is closely related to the inequality of A. Rubinstein, see [R, Corollary 10]. Rubinstein s inequality states that for every function f : { γ, γ 1 } n { 1, 1} with T >1 a T = ρ we have f (a + a {k} w {k} ) Kρ, K = (6) (1 a )1/ However, our inequality (4) is a better bound in the regime ρ ln(e/ρ) < c 0 α. To see this consider the case when f 0 = sgn(a + a {k} w {k} ) is constant and equal to ε { 1, 1}. Then from (5) we have f ε 16ρ. It follows that 1 a = f Ef f ε 16ρ. Thus, the right hand side of (6) is greater than K/4, which gives no information. In the case when f 0 is not constant we have Ef 0 = 1 α. Thus, a 1 α = Ef Ef 0 E(f f 0 ) f f 0 4ρ. It follows that 1 a (1 a ) (α + 4ρ) 1α. Therefore, the right hand side in the Rubinstein bound is in this case Kρ/ 1α which is much greater than ρ when α 0. In the Section 3 we consider the case γ = 1 and we deal with the problem concerning [ 1, 1]-valued functions defined on the cube { 1, 1} n with uniform product probability measure. A function f : { 1, 1} n R is called affine if f(x) = a 0 + n i=1 a ix i, where a 0, a 1,..., a n R and x = (x 1,..., x n ). We will denote the set of all affine functions by A. Moreover, let A [ 1,1] A stands for the set of all affine functions satisfying f(x) 1 for every x { 1, 1} n. Note that f A [ 1,1] if and only if n i=0 a i 1. The function f(x) = x i will be denoted by r i, i = 1,..., n. Let us also notice that if f is [ 1, 1]-valued then a T = Ew T f E w T f 1. In [JOW] the authors gave the following example. Take g : { 1, 1} n R given by g(x) = s 1 n 1/ n i=1 x i. Note that g A. Define φ(x) = 1 (, 1) (x)+x1 [ 1,1] (x)+1 (1, ) (x) and take f = φ g. Clearly, f is [ 1, 1]- valued but may not be affine. The authors proved that lim n dist L (f, A) = O(e s /4 ) and lim n dist L (f, A [ 1,1] ) = Θ(s 1 ). Here we prove that this is the worst case as far as the dependence of these two quantities is concerned. Namely, we have the following theorem, which is the analogue of (3) in the case of [ 1, 1]-valued functions. 4

5 Theorem 3. Let us take f : { 1, 1} n [ 1, 1] and define ρ = dist L (f, A). Then dist L (f, A [ 1,1] ) 18. ln(1/ρ) Proof of Theorem In this section we give a proof of Theorem. We begin with a simple lemma. Lemma 1. Let 0 < α < β < 1 with α + β = 1 and let γ (0, 1]. Then α +γ β +γ β γ α γ γ γ α +γ β γ. Proof. Let x (0, 1) and µ 1. From the mean value theorem we have 1 x µ γ µ. Applying this with µ = and x = (α/β) γ yields an equivalent 1 x γ version of the statement. Proof of Theorem. Let k be given by Theorem 1. Let h = f (a +a {k} x k ) and h = f sgn(a + a {k} x k ). Moreover, let δ = h. It follows that δ 1. Note that for every u R and ε { 1, 1} we have u sgn(u) u ε. Therefore, ε sgn(u) ε u + u sgn(u) u ε. (7) It follows that h h. Thus, using the fact that h is {, 0, }-valued, we have P( h 0) = 1 4 h h = δ. Let us consider the expansion h = T ãt w T. Clearly, ã T = a T for T, {k}. Using () we obtain 4δ 4/q 4P( h 0) /q = h q = ã T w T c q (α, β) T ã T w T T q T = c q (α, β) T ã T c q (α, β) ã T, T T 1 where q [1, ]. Using Lemma 1 with γ = /q we obtain T 1, T,{k} ã T ã T T 1 4δ4/q c q (α, β) = 4δ4/q αβ α q β q β q α q 4δ4/q q 1 ( ) 1 α q. β 5

6 Take 1 = 1 1 [ 1, 1]. Note that q ln(e /δ) (α/β)1 /q α 1 /q α 1. It follows that ã T 4δ 4 α 1 e 4 ln(1/δ) ( ln(e /δ) ln e /δ ) 4e 4 δ 4 α 1 ln ( e /δ ). T 1, T,{k} From Theorem 1 we have ρ δ 8 ρ. Thus, 4e 4 δ 4 α 1 ln ( e /δ ) 8 e 4 α 1 δ ρ ln(e /ρ) 3 4 δ. Note that a + a {k} = 1 δ. We deduce 1 ρ = a T = a + a {k} + T 1 T 1, T,{k} Therefore, δ ρ. The inequality (5) follows from (7). ã T 1 δ δ = δ. Remark. The condition ρ ln(e /ρ) 1 9 e 4 α cannot be significantly improved. Indeed, if we take f : { γ, γ 1 } { 1, 1} given by f(x 1, x ) = (β βαx 1 )(β βαx ) 1, see the remark after the proof of Theorem 5.8 in [JOW], then we obtain ρ = αβ α and δ = β 3/ α 1/. Thus δ = ρβ ρ/. One can easily see that if we replace our assumption ρ ln(e /ρ) 1 α by 9 e 4 a slightly stronger condition, say ρ ln (e /ρ) α then we obtain δ ρ+o(ρ), which means that T 1, T,{k} a T = o(ρ ) and a + a {k} 1 ρ o(ρ ). 3 Proof of Theorem 3 We need the following lemma due to P. Hitczenko, S. Kwapień and K. Oleszkiewicz. Lemma. ([HK], Theorem 1 and [Ol], Theorem 1) Let a 1 a... a n 0 and let us take S : { 1, 1} n R given by S = n i=1 a ir i. Then for t 1 we have P ( S S ) > (8)

7 and We give a proof of Theorem 3. S t 1 ( ) 1/. t a i (9) 4 Proof of Theorem 3. Step 1. If f = T a T w T then dist L (f, A) = f S, where S = T 1 a T w T. For every u [ 1, 1] we have x u x φ(x) for all x R. Taking x = S and u = f we obtain E( S 1) + = S φ(s) S f ρ. For all g A [ 1,1] we have Therefore, i>t g f g S + S f g S + ρ. dist L (f, A [ 1,1] ) dist L (S, A [ 1,1] ) + ρ. (10) It suffices to prove that E( S 1) + ρ implies an appropriate bound on dist L (S, A [ 1,1] ), whenever S = a 0 + n i=1 a ir i, where a 0, a 1,..., a n R. Step. Suppose that for all n 1 we can prove that E( S 1) + ρ implies dist L (S, A [ 1,1] ) M for some M > 0, assuming that a 0 = 0. Then we can deal with the case a 0 0 as follows. Let us take S : { 1, 1} { 1, 1} n R given by S = a 0 x 0 + n i=1 a ix i. Clearly, E( S 1) + = E( S 1) + ρ. We can find a [ 1, 1]-valued function S 0 = b 0 x 0 + n i=1 b ix i such that S S 0 M. Take S0 = b 0 + n i=1 b ix i. Now it suffices to observe that the function S 0 is [ 1, 1]-valued and to notice that S S 0 = S S0. Step 3. Take S = n i=1 a ir i. Without loss of generality we can assume that 1 a 1 a... a n 0. Let τ = max{t 1 : t i=1 a i 1}. Clearly, τ 1. If f is already in A [ 1,1] then there is nothing to prove. Therefore we can assume that τ < n. We can also assume that ρ 1/3, since otherwise we have dist L (f, A [ 1,1] ) dist L (f, 0) = f 1 18 ln(1/ρ). Let A = { S 1 S t }. For t 1 we have E S t = E S t 1 A + E S t 1 A c E S t P(A) + 1 t E S t. 7

8 Since by the Khinchine inequality we have (E S t ) 1/t t 1 t 1 (E S t ) 1/t, we arrive at P ( S 1 ) S t (1 1 ) (E S t ) 1 t E S t 4 By the Chebyshev inequality we obtain (E S t ) 1 ( ) t t 1. E S t 4 t 1 P ( S 1 + ε) E( S 1) + ε ρ ε, (11) for all ε > 0. Let t 1 and assume that S t >. Take ε = 1 S t 1 > 0. We get ( ) t ( 1 t 1 P S 1 ) 4 t 1 S ρ t ( 1 S t 1). It follows that S t + 4ρ ( ) t/ t 1 t 1 which is also true in the case S t. From inequality (9) we obtain 1 ( t 4 i>t a i ) 1/ S t + 4ρ ( ) t/ t 1. (1) t 1 Step 4. We consider the case τ ln(1/ρ) 1. Let us now take ln 3 t = ln(1/ρ) > 1 and define S ln 3 1 = i ln 3 ln(1/ρ) a ir i. Notice that i ln 3 ln(1/ρ) a i i τ a i 1. Thus, S 1 A [ 1,1]. Moreover, since t, we have ρ ( ) t 1 t/ t 1 ρ3 t/ = 1 and therefore by (1) we have dist L (S, A [ 1,1] ) S S 1 = In this case (10) yields i> ln 3 ln(1/ρ) a i 1/ 4. ln(1/ρ) ln 3 dist L (f, A [ 1,1] ) ρ. ln(1/ρ) ln(1/ρ) ln 3 8

9 Step 5. We are to deal with the case τ < ln(1/ρ). Let us take S ln 3 = i τ+ a ir i. From inequality (8) we have ( P S ) a i + S 1 P ( S S τ+1 ) 1 1 τ ρ ln ln 3. i τ+1 Note that i τ+1 a i > 1. Therefore, from inequality (11) we obtain P ( It follows that S i τ+1 i τ+1 a i + S ) ρ (i τ+1 a i + S 1 ). a i + S 1 ln 1 0ρ ln 3. Take S 1 = τ i=1 a ir i +(1 (a a τ ))r τ+1. Clearly, S 1 A [ 1,1]. Moreover, S S 1 = ( (1 (a a τ ) a τ+1 ) + S ) 1/ Therefore, from (10) we have a a τ + a τ S ln 1 0ρ ln 3. dist L (f, A [ 1,1] ) ln ρ ln 3 + ρ. ln(1/ρ) Remark. If we perform our calculation with ln(.03) instead of ln 3 we will obtain the theorem with a constant 14.5 instead of 18. Acknowledgements I would like to thank Prof. Krzysztof Oleszkiewicz for his useful comments. I would also like to thank the anonymous referee for his comment regarding the statement of Lemma 1. 9

10 References [A] K. J. Arrow, A difficulty in the concept of social welfare, Journal of Political Economy 58(4), 1950, [Be] W. Beckner, Inequalities in Fourier analysis, Annals of Math. 10 (1975), [Bo] A. Bonami, Etude des coefficients Fourier des fonctiones de L p (G), Ann. Inst. Fourier 0 (1970), [D] I. Dinur, The PCP theorem by gap amplification, Journal of the ACM 54, no. 3 (007), article 1. [DS] I. Dinur, S. Safra, On the hardness of approximating minimum vertex cover, Annals of Mathematics, 16 (005), [F] E. Friedgut, Boolean functions with low average sensitivity depend on few coordinates, Combinatorica 18 (1998), [FKN] E. Friedgut, G. Kalai and A. Naor, Boolean functions whose Fourier transform is concentrated on the first two levels, Advances in Applied Mathematics 9, no. 3 (00), [G1] L. Gross, Logarithmic Sobolev inequalities, Amer. J. Math. 97 (1975), [G] L. Gross, Hypercontractivity, logarithmic sobolev inequalities and applications: a survey of surveys, preprint, 005. [HK] P. Hitczenko, S. Kwapień, On the Rademacher series, Probability in Banach spaces, 9 (Sandjberg, 1993), 31 36, Progr. Probab., 35, Birkhäuser Boston, Boston, MA, [JOW] J. Jendrej, K. Oleszkiewicz, J.O. Wojtaszczyk, On some extensions of the FKN theorem, preprint. [KKL] J. Kahn, G. Kalai and N. Linial, The influence of variables on Boolean functions, in Proc. 9-th Annual Symposium on Foundations of Computer Science, 1988,

11 [KG] G. Kalai, A Fourier-theoretic perspective on the Condorcet paradox and Arrow s theorem, Advances in Applied Mathematics, 9(3), 00, [K] G. Kindler, Property Testing, PCP and Juntas, PhD thesis, Tel Aviv University, 00. [KS] G. Kindler and S. Safra, Noise-resistant boolean functions are juntas, preprint, 00. [O1] R. O Donnell, Analysis of Boolean Functions, [O] R. O Donnell, Some topics in analysis of boolean functions, Proceedings of the 40th Annual ACM Symposium on Theory of Computing, 008, [Ol1] K. Oleszkiewicz, On a nonsymmetric version of the Khinchine- Kahane inequality, Proceedings of the Stochastic Inequalities Conference (Barcelona 00), Progress in Probab. 56 (003), [Ol] K. Oleszkiewicz, On the Stein property of Rademacher sequences, Prob. and Math. Stat., Vol. 16, Fasc. 1 (1996), [R] A. Rubinstein, Boolean functions whose Fourier transform is concentrated on pair-wise disjoint subsets of the inputs, Tel-Aviv University, master thesis, 01. Piotr Nayar, nayar@mimuw.edu.pl Institute of Mathematics, University of Warsaw, Banacha, Warszawa, Poland. 11