The Fourier Entropy Influence Conjecture for certain classes of Boolean functions

Transcription

1 The Fourier Entropy Influene Conjeture for ertain lasses of Boolean funtions Ryan O Donnell, John Wright, and Yuan Zhou Department of Computer Siene, Carnegie Mellon University Abstrat. In 996, Friedgut and Kalai made the Fourier Entropy Influene Conjeture: For every Boolean funtion f : {, } n {, } it holds that H[ f ] C I[f], where H[ f ] is the spetral entropy of f, I[f] is the total influene of f, and C is a universal onstant. In this work we verify the onjeture for symmetri funtions. More generally, we verify it for funtions with symmetry group S n S nd where d is onstant. We also verify the onjeture for funtions omputable by read-one deision trees. {odonnell,jswright,yuanzhou}@s.mu.edu. This researh performed while the first author was a member of the Shool of Mathematis, Institute for Advaned Study. Supported by NSF grants CCF and CCF , BSF grant 00877, and Sloan and Okawa fellowships.

2 Introdution The field of Fourier analysis of Boolean funtions f : {, } n {, } plays an important role in many areas of mathematis and omputer siene, inluding omplexity theory, learning theory, random graphs, soial hoie, inapproximability, arithmeti ombinatoris, oding theory, metri spaes, et. For a survey, see e.g. [7]. One of the most longstanding and important open problems in the field is the Fourier Entropy Influene (FEI Conjeture made by Friedgut and Kalai in 996 [6]: Fourier Entropy Influene (FEI Conjeture. C f, H[ f ] C I[f]. That is, f(s log f(s C f(s S. S [n] The quantity H[ f ] = f(s log f(s S [n] on the left is the spetral entropy or Fourier entropy of f. It ranges between 0 and n and measures how spread out f s Fourier spetrum is. The quantity I[f] = f(s S appearing on the right is the total influene or average sensitivity of f. It also ranges between 0 and n and measures how high up f s Fourier spetrum is. (For definitions of the terms used in this introdution, see Setion. The FEI Conjeture is superfiially similar to the well-known Logarithmi Sobolev Inequality [9] for the Boolean ube whih states that Ent[f ] I[f] holds for any f : {, } n R, where Ent[g] = E[g ln g] E[g] ln E[g]. However note that the FEI Conjeture requires f : {, } n {, } to be Boolean-valued, and it definitely fails for real-valued f.. Appliations of the onjeture Friedgut and Kalai s original motivation for the FEI Conjeture ame from the theory of random graphs. Suppose f represents a monotone graph property with Pr[f(G = ] = / when G G(v, / (here n = ( v. If we also onsider Pr[f(G = ] for G G(v, p, then the total influene I[f] equals the reiproal of the derivative of this quantity at p = /. Hene the property has sharp threshold if and only if I[f] is large. Friedgut and Kalai sought general onditions on f whih would fore I[f] to be large. They onjetured that having signifiant symmetry and hene, a spread-out Fourier spetrum was suh a property. The FEI Conjeture also easily implies the famed KKL Theorem []. To see this, first note that H[ f ] H [ f ] = min S {log f(s }, the min-entropy of f. Thus the FEI Conjeture is stritly stronger than the following: Fourier Min-Entropy Influene Conjeture C f f(s C I[f]. H [ f ] C I[f]. That is, S [n] suh that In partiular, for balaned f (i.e., E[f] = 0 = f( the above onjeture implies there is a nonempty S with f(s C I[f]. Sine Inf j [f] f(s for eah j S we onlude max{inf i [f]} C n max{inf i[f]} whene max{inf i [f]} Ω( C log n n, whih is KKL s onlusion. Indeed, by applying the above dedution just to the nonempty Fourier oeffiients it is straightforward to dedue I[f] C Var[f] log max i {Inf i [f]}, a strengthening of the KKL Theorem due to Talagrand [0]. We also remark that sine Inf i [f] = f({i} for monotone f, the KKL Theorem implies the Fourier Min-Entropy Influene Conjeture holds for monotone funtions. Finally, as emphasized by Klivans and oauthors [8, ], the FEI Conjeture is also important beause it implies a version of Mansour s Conjeture from 99.

3 Mansour s Conjeture [5]. ɛ > 0 K suh that if f : {, } n {, } is omputable by a DNF formula with m terms then by taking S to be the m K sets S for whih f(s is largest, S S f(s ɛ. In fat, Mansour onjetured more strongly that one may take K = O(log ɛ. It is well known [3] that if f is omputed by an m-term DNF then I[f] O(log m. Thus the Fourier Entropy Influene Conjeture would imply H[ f ] C O(log m, from whih it easily follows that one may take K = O(C/ɛ in Mansour s Conjeture. Mansour s Conjeture is important beause if it is true then the query algorithm of Gopalan, Kalai, and Klivans [7] would agnostially learn DNF formulas under the uniform distribution to any onstant auray in polynomial time. Establishing suh a result is a major open problem in omputational learning theory [8]. Further, suffiiently strong versions of Mansour s Conjeture would yield improved pseudorandom generators for DNF formulas; see, e.g., [, ] for more on this important open problem in pseudorandomness.. Prior work As far as we are aware, the result in [] showing that the FEI Conjeture holds for random DNFs is the only published progress on the FEI Conjeture sine it was posed. In this subsetion we ollet some observations related to the onjeture, all of whih were presumably known to Friedgut and Kalai and should be onsidered folklore. See also [] for additional reent disussion of the onjeture. The FEI Conjeture holds for the usual examples that arise in analysis of Boolean funtions Parities (for whih the onjeture is trivial, ANDs and ORs, Majority, Tribes [], and Inner- Produt-mod-. This may be established by diret alulation based on the known Fourier oeffiient formulas for these funtions (see [] for Majority and [6] for Tribes. By onsidering the AND and OR funtions it is easy to show that the onstant C must be at least. We an show that C = is neessary and suffiient for the Tribes funtions as well; smaller onstants suffie for Inner-Produt-mod- and Majority. The authors are also aware of an expliit family of funtions whih show the neessity of C 60/3.65. For a gross upper bound, it is not too hard to show that H[ f ] ( + log n I[f] + ; indeed, this will be shown in the ourse of the present paper. The FEI Conjeture tensorizes in the following sense: For f : {, } n {, } and M Z +, define f M {, } Mn {, } by f(x (,..., x (M = f(x ( f(x ( f(x (M. Then it s easy to hek that H[ f M ] = M H[ f ] and I[f] = M I[f]. This is of ourse onsistent with the FEI Conjeture; it also implies that the following weaker-looking onjeture is atually equivalent to FEI: for all f : {, } n {, }, H[ f ] C I[f] + o(n. To see the equivalene, given f : {, } n {, }, apply the above to f M, divide by M, and take the limit as M..3 Our results and approah In this work, we prove the FEI Conjeture for some lasses of Boolean funtions. Theorem. The FEI Conjeture holds for symmetri funtions, with C =.0. Although the lass of symmetri funtions is fairly small, there was sentiment that it might be a diffiult ase for FEI: for symmetri funtions, f(s = f(s whenever S = S and hene their Fourier spetrum is maximally spread out on eah level.

4 Our proof of Theorem uses the same high-level idea as in the well-known KKL Theorem []: namely, prove a ertain inequality for the disrete derivatives D i f of f, and then sum over i. In our ase, the key inequality we need for the derivatives is that they are very noise-sensitive: Theorem. Let g be a disrete derivative of a symmetri funtion f : {, } n+ {, }. Then / for all real n it holds that Stab [g] π n E[g ]. (For the notation used here, see Setion. Having established Theorem, it is not too hard to generalize it as follows: Theorem. The FEI Conjeture holds for d-part-symmetri funtions, with C =.0 + log d. A d-part-symmetri funtion is essentially a funtion with symmetry group of the form S n S nd. This theorem also generalizes the folklore fat that FEI holds (up to an additive onstant with C = O(log n, sine every funtion is n-part-symmetri. Finally, with an unrelated, diret indutive argument we show the following: Theorem 3. The FEI Conjeture holds for funtions omputable by read-one deision trees, with C =.88. Remark: In independent and onurrent work, the FEI Conjeture was verified for monotone symmetri funtions (a speial ase of Theorem and deision lists (a speial ase of Theorem 3 using different methods of proof []. Definitions and notation We use the notation N = {0,,,... }, Z + = N \ {0}, and [n] = {,,, n}. Throughout we write log for log ; for the natural logarithm we write ln. The expressions 0 log 0 and 0 log 0 are to be interpreted as 0.. Basis of Boolean Fourier analysis This paper is onerned with Boolean funtions f : {, } n R, espeially Boolean-valued funtions f : {, } n {, }. Every Boolean funtion f has a unique multilinear polynomial representation over R, f(x = f(s x i. i S S [n] This is known as the Fourier expansion of f, and the real numbers f(s are the Fourier oeffiients of f. We have the formula f(s = E[f(x i S x i]. (Here and throughout, expetation E[ ] is with respet to the uniform distribution of x on {, } n, unless otherwise speified. In partiular, f( = E[f]. An important basi fat about Fourier oeffiients is Parseval s identity: f(s S [n] = E[f(x ]. A onsequene of Parseval is that f(s S [n] = for Boolean-valued f. Thus the numbers f(s an be thought of as a probability distribution on the subsets of [n]. Given f : {, } n R and i [n], we define the disrete derivative D i f : {, } n R by D i f(x = f(x(i= f(x (i=, where x (i=b denotes (x,..., x i, b, x i+,..., x n. It holds that { 0 if i S, D i f(s = f(s {i} if i S;

5 i.e,. D i ats on the Fourier expansion as formal differentiation. The influene of i on f is Inf i [f] = E[(D i f ] = S i f(s. In the partiular ase that f is Boolean-valued, the derivative D i f is {, 0, }-valued and we have the ombinatorial interpretation Inf i [f] = Pr[f(x (i= f(x (i= ]. The total influene of f : {, } n R is I[f] = Inf i [f] = f(s S. i= For 0 ρ, we say that x, y {, } n are a pair of ρ-orrelated random strings if x is distributed uniformly randomly on {, } n and y is formed by setting y i = x i with probability + ρ, y i = x i with probability ρ, independently for eah i [n]. We may now define the noise stability of f at ρ and give its Fourier formula: Stab ρ [f] = S [n] E [f(xf(y] = x,y ρ-orrelated S [n] f(s ρ S. We often stratify the Fourier oeffiients into levels; the level of S is simply S. We define the weight of f at level k to be W k [f] = S =k f(s. Thus I[f] = W k [f] k, Stab ρ [f] = W k [f]ρ k. Finally, given a random variable or probability distribution we write H[ ] for its (binary Shannon entropy. Hene for f : {, } n {, }, H[ f ] = f(s S [n] log, alled the Fourier f(s entropy, or spetral entropy, of f. Thus the Fourier Entropy Influene Conjeture may be stated as follows: there is a universal onstant C suh that H[ f ] C I[f] holds for all Boolean-valued funtions f.. Some Boolean funtion lasses We will all a funtion f : {, } n R symmetri if it is invariant under any permutation of the oordinates [n]. Equivalently, f is symmetri if the value of f(x depends only the Hamming weight of x, defined to be #{i [n] : x i = }. In this ase we may identify f with the funtion f : {0,,..., n} R whose value at s equals f(x for any x of Hamming weight s. We generalize the notion to that of d-part-symmetri funtions, d Z +. We say the funtion f : {, } n R is d-part-symmetri if there is a partition [n] = V V V d suh that f is invariant under any permutation of the oordinates in any part V i. Equivalently, f is d-partsymmetri if its symmetry group is isomorphi to S n S nd for numbers n + + n d = n. Note that a symmetri funtion is -part-symmetri, and every funtion f : {, } n R is n- part-symmetri. We also generalize the notion of Fourier levels for d-part-symmetri funtions. Suppose f is d-part-symmetri with respet to the partition [n] = V V d, where V i = n i. Then f(s depends only on the numbers S V,..., S V d. We onsider all possible suh sequenes k {0,,..., n } {0,,..., n } {0,,..., n d },

6 and say that S [n] is at multi-level k if S V i = k i for eah i [d]. We also use the notation k = k + k + + k d, W k [f] = S at multi-level k f(s, so I[f] = k Wk [f] k. Finally, we reall the definition of deision trees. We may define the notion as follows. We say that f : {, } n {, } is omputable as a depth-0 deision tree if it is onstantly or. We indutively say that it is omputable as a depth-d deision tree if there is a oordinate i [n] suh that { f 0 (x if x i =, f(x = f (x if x i =, where f 0 and f are omputable by depth-(d deision trees. We further say that the deisiontree omputation is read-one if f 0 and f depend on disjoint sets of oordinates and are themselves indutively read-one. 3 Symmetri and d-part-symmetri funtions In this setion we prove Theorems and, establishing the FEI Conjeture for symmetri and O(-part-symmetri funtions. Although Theorem stritly generalizes Theorem, we prefer to prove Theorem separately and then generalize it afterward. When f : {, } n {, } is symmetri we have f(s = W k [f]/ ( n k whenever S = k. Hene H[ f ] = W k [f] log (n k W k [f] = ( n W k [f] log k Thus Theorem is an immediate onsequene of the following two theorems: + W k [f] log W k [f]. ( Theorem. Let f : {, } n {, } be a symmetri funtion. Then n Wk [f] log ( n k C I[f], where C = ln ( + e π 9.0. Theorem 5. Let f : {, } n {, } be any funtion, not neessarily symmetri. Then n Wk [f] log W k [f] 3 I[f]. We prove these theorems in the subsequent subsetions of the paper, following whih we give the extension to d-part-symmetri funtions. 3. Theorem : Derivatives of symmetri funtions are noise-sensitive We begin with an easy lemma. Lemma. Let p,..., p m be a nonnegative unimodal sequene; i.e., there exists k [m] suh that p,..., p k is a nondereasing sequene and p k,..., p m is a noninreasing sequene. Let g : [m] {, 0, } have the property that the sets g ( and g ( are interleaving. Then m i= p ig(i max{p i }.

7 Proof. It is without loss of generality to assume that g is never 0; for otherwise, we an restrit attention to the indies i i i l where g has value ±, noting that p i,..., p il is still a unimodal sequene. Now if m the result is trivial. Otherwise, hoose k [m ] suh that at least one of p k, p k+ equals max{p i }. By negating g if neessary we may assume that g(k =, g(k + =. We omplete the proof by showing p k+ m i= p ig(i p k. For the upper bound, unimodality implies p k+ p k+ p k+3 ; hene m i=k+ p ig(i = ( p k+ + p k+ + ( p k+3 + p k+ + 0 (regardless of whether this ends on p m or +p m. Similarly, we must have k i=0 p ig(i 0. Combining these, we obtain that m i= p ig(i p k as laimed. The proof of the lower bound m i= p ig(i p k+ is similar. We now show the key theorem stated in Setion.3 on the noise sensitivity of symmetri derivatives. Theorem 6. Let g be a disrete derivative of a symmetri funtion f : {, } n+ {, }. / Then for all real n it holds that Stab [g] π n E[g ]. Proof. Let (x, y be a ( n -orrelated pair of random strings in {, }n. We will show that E[g(y x] / π, independent of x. ( Sine g is {, 0, }-valued, it will follow from ( that Stab [g] = E[g(xg(y] / π E[ g(x ] = / π E[g ], n as required. To show ( we first observe that given x of Hamming weight s, the Hamming weight t of y is distributed as the sum of two independent binomial random variables, t Bin(s, n and t Bin(n s, n. Being the sum of independent Bernoulli random variables, it is well known [3] that t has a unimodal probability distribution. Sine g is a derivative of a symmetri Booleanvalued funtion, it is itself symmetri; hene we may identify it with a funtion g : {0,,..., n} {, 0, }. Further, beause f is {, }-valued, g must have the property that g ( and g ( are interleaving subsets of {0,,..., n}. Thus ( follows from Lemma assuming that max i {Pr[t = i]} / π. To show this, we may assume without loss of generality that s n/. Then max i {Pr[t = i]} max{pr[t = i]} i πs( n n π n n = / π where the seond inequality is the basi estimate max i {Pr[Bin(m, p] = i} whih πmp( p uses Stirling s formula (see [5, Ch. VII.3]. We an now give the proof of Theorem. Proof. (Theorem. Using ( n k ( en k k for all k n we have ( ( n W k [f] log W k [f]k log(e + W k [f]k log n k k = I[f] + ln k= and hene it suffies to show k= k= W k [f]k ln n k k= W k [f]k ln n k e π I[f]. (3

8 Let g be any derivative of f; say g = D n f. By symmetry of f, the right side of (3 is e π n E[g ]. As for the left side, for k [n] we have W k [g] = S [n ] S =k ĝ(s = S [n] S =k,s n f(s = k n Wk [f]. Hene the left side of (3 is n k= nwk [g] ln n k. Thus after dividing by n we see that (3 is equivalent to n W k n [g] ln k + e E[g ]. ( π Using the approximation ln m + γ + m m n n W k n [g] ln k + W k [g] j=k+ j = m j= j ln m + γ + m j= j j W k [g] exp( one may obtain j= j Stab [g], j where in the last step we used that exp( ( j k for all k j. We may now apply Theorem 6 with = n j to obtain W k [g] ln n k + exp( j= j π j n E[g ] e π E[g ], using n j= j n. Thus we have verified ( and ompleted the proof. 3. Theorem 5: Spetral level entropy versus influene In this setion we establish Theorem 5. We begin with a well-known fat about maximum entropy whih we prove for ompleteness: Proposition. Let K be a random variable supported on Z +. Then H[K] E[K]. Proof. For k Z + write p k = Pr[K = k], and let G be a Geometri( random variable. Then by the nonnegativity of binary relative entropy, 0 D(K G = p k log p k = H[P ] + (/ k k= p k log( k = E[K] H[K]. We will also need a simple orollary of the edge-isoperimetri inequality for {, } n. Proposition. Let f : {, } n {, } and write W 0 [f] = ɛ. Then I[f] ɛ log(/ɛ + ɛ. Proof. By negating f if neessary we may assume E[f] 0. We think of f as the indiator of a subset A {0, } n ; the density of this set is p = + E[f] = ɛ /. It follows from the edge-isoperimetri inequality [0, 9, ] that I[f], whih is n times the size of A s edge-boundary, is at least p log(/p. It is then elementary to hek that ( ɛ log( ɛ ɛ log(/ɛ+ɛ, as required. We an now establish Theorem 5. k=

9 Proof. (Theorem 5. Write W 0 [f] = ɛ. We may assume ɛ > 0 else the result is trivial. Then W k [f] log W k [f] W k [f] log + ɛ (as ( ɛ log W k [f] ɛ ln ɛ ɛ k 0 = ɛ k 0 ɛ k 0 W k [f] ɛ W k [f] ɛ ɛ log W k + ɛ log (/ɛ + ɛ [f] k + I[f] = 3 I[f], where the last inequality used Propositions and. 3.3 Theorem : extension to d-part-symmetri funtions We show now how to extend the proof of Theorem to obtain Theorem, the FEI Conjeture for d-part-symmetri funtions. Suppose then that f is d-part-symmetri with respet to the partition [n] = V V d, where V i = n i, and reall the multi-level index notation k from Setion.. Sine f(s = W k [f]/ d i= H[ f ] = ( ni k i whenever S is at multi-level k, we have d W k [f] log i= k ( ni k i + k W k [f] log W k [f], similarly to (. Sine I[f] = d W k [f] k = W k [f] k i k i= k we an prove Theorem by establishing the following two generalizations of Theorems and 5: Theorem 7. Let f : {, } V V d {, } be invariant under permutations of the oordinates in V i. Then k Wk [f] log ( n i k i C k Wk [f] k i, where C = ln ( + e π 9.0. Theorem 8. Let f : {, } V V d {, } be any funtion, not neessarily with any symmetries. Then k Wk [f] log (3 + log d I[f]. W k [f] We begin by proving Theorem 7. Proof. (Theorem 7. We assume i = d without loss of generality and write V d = V V d. For y {, } V d we define f y : {, } V d {, } by f y (z = f(y, z; the funtion f y is symmetri for eah y by assumption. Applying Theorem to eah f y and then taking expetations we obtain Now n d k =0 E y [ W k [f y ] ] = E y [ W k [f y ] ] log S V d S =k ( nd k C E y [ fy (S ] = S V d S =k n d k =0 E y [ W k [f y ] ] k. (5 T V d f(t S = k : k d =k W k [f] where the middle equality is an easy exerise using Parseval. Substituting this into (5 yields the desired inequality.

10 As for Theorem 8, its proof is essentially idential to that of Theorem 5, using the following generalization of Proposition 3: Proposition 3. Let K be a random variable supported on N d \ {0} and write L = K. Then H[K] ( + log d E[L]. Proof. Using the hain rule for entropy as well as Proposition, we have H[K] = H[L] + H[K L] E[L] + Pr[L = l] H[K L = l]. (6 Given L = l there are preisely ( l+d d possibilities for K; hene ( l + d H[K L = l] log = l log ( l+d d l log ( +d d = l log d. (7 d l The seond inequality here follows from the tehnial Lemma below. The proof is ompleted by substituting (7 into (6. Here we give the tehnial lemma used in the previous proof. Lemma. For eah N the funtion l log ( l+ is a dereasing funtion of l on Z+. Proof. We wish to show that for all l Z +, l ln ( l+ l+ ln ( l++ (l + ln ( l+ [ (l++ ( l+ l= ( l ln l++ / ( l+ ] l ( l+ ( l+ l+ ( l++ ( + l l+ l. This last inequality is easily shown by indution on l: if one multiplies both sides by ( + l+ one obtains ( l++ ( + l+ l+ whih exeeds ( + l+ l+ as required for the indution step. Proving the onjeture for read-one deision trees In this setion we prove Theorem 3, establishing the Fourier Entropy Influene Conjeture with onstant C.88 for funtions omputable by read-one deision trees. We begin with a tehnial lemma. Lemma 3. Let µ 0, µ [, ]. Define Then g(µ 0, µ = ( µ 0 +µ ( log ( µ 0 +µ + µ0 µ ( log µ µ 0 µ + 0 log µ µ 0 log. µ g(µ 0, µ C( µ 0µ (8 holds for some universal onstant C. The onstant C =.88 suffies. Proof. Note that g is unhanged when either of its arguments is negated. Sine the right side of (8 only dereases when µ 0 and µ have the same sign, we may assume both are nonnegative. By symmetry we may further assume 0 µ µ 0. Write µ = rµ 0 where r [0, ]. Simple manipulation shows that g(µ 0, µ = g(µ 0, rµ 0 = h(rµ 0,

11 where Thus it remains to show that h(r = r log r ( r log r + ( + r log( + r. h(rµ 0 C( rµ 0 ( rc h(rµ 0 C (9 holds for all µ 0, r [0, ] for some C. Assuming we take C, the right side of the seond inequality in (9 will be nonnegative; hene, it holds if and only it holds for µ 0 =. Putting µ 0 = into the first inequality in (9, we see that it suffies for C to be an upper bound on l(r = h(r r. Plainly, l is a ontinuous funtion on (0, with l(0+ = 0. It is also easy to hek that l( = 0. It follows that l is uniformly bounded on [0, ]; i.e., there is a universal C suh that l(r C for all r [0, ]. Numerially, one may hek that l(r.88. We may now prove Theorem 3 with C equal to the onstant from Lemma 3. Proof. (Theorem 3. The proof is by indution on the depth d of the deision tree. In the base ase, d = 0, the funtion f is onstant and so H[ f ] C I[f] holds trivially. For the general ase, assume without loss of generality that the root of the tree queries oordinate n. Let f 0 and f be the subfuntions omputed by the x n = and x n = subtrees, respetively. Sine the deision tree is read-one, the funtions f 0 and f depend on disjoint sets of the oordinates from [n ], say J 0 and J. In partiular, if we write µ i = E[f i ] = f i ( for i = 0,, then E[f 0 f ] = µ 0 µ. Also, f 0 and f are omputable by read-one deision trees of depth d ; hene we may later apply the indution hypothesis to them. Our task is now to prove H[ f ] C I[f] for f. For any Boolean funtion f whose subfuntions based on x n are f 0 and f, it holds that I[f] = Inf n [f] + I[f 0] + I[f ]. In our ase, Inf n [f] = Pr[f 0 f ] = E[f 0f ] = µ 0µ. Thus As for H[ f ], we have f(x = + x n I[f] = µ 0µ + I[f 0] + I[f ]. (0 f 0 (x + x n f (x = (f 0(x + f (x + x n(f 0 (x f (x. It follows that the Fourier oeffiients of f, whih are indexed by S J 0 J {n}, are given by (µ 0 + µ if S =, f(s = (µ 0 µ if S = {n}, f i (S if S J i, ( i f i (T if S = T {n}, T J i. Note that for eah S J i, we have the quantity f i (S ourring twie in the squared Fourier spetrum of f. We may therefore ompute H[ f ] = ( µ 0 +µ ( log ( µ 0 +µ + µ0 µ ( log µ 0 µ + i=0, =S J i f i (S log f i (S. (

12 We simplify the expression on the right in (: = i=0, =S J i f i (S log f i (S Substituting ( into ( yields i=0, = i=0, = µ i log µ i ( µ i log µ i i=0, µ i log µ i + S J i fi (S log f i (S + H[ f i ] + log + i=0, H[ f i ]. ( H[ f ] = g(µ 0, µ + H[ f 0 ] + H[ f ], (3 where g(µ 0, µ is the funtion from Lemma 3. Using this lemma along with the indution hypothesis on f 0 and f, we obtain from (3 that H[ f ] C( µ 0µ + C I[f 0] + C I[f ] = C I[f], where we used (0. The indution is omplete. Remark: The above proof tehnique an be extended to handle the more general lass of reursively read-one funtions (as defined in [8]. It seems that one an show C = 9 suffies for this lass; however, a formal proof would require proving a numerial lemma muh more diffiult than Lemma 3. Details will appear elsewhere. 5 Closing remarks As the general Fourier Entropy Influene Conjeture seems diffiult to resolve, one may try to prove it for additional interesting lasses of Boolean funtions. For example, Wan has suggested linear threshold funtions as a possibly tratable ase []. One may also try takling the Fourier Min- Entropy Influene Conjeture first (or the slightly stronger onsequene that S s.t. f(s C I[f]/ Var[f]. However this will already likely require stronger tools than the ones used in this paper. The reason is that as mentioned in Setion., this onjeture implies the KKL Theorem, and there is no known proof of KKL whih avoids the Hyperontrative or Logarithmi Sobolev Inequalities. Aknowledgments. The authors would like to thank Roo Servedio, Li-Yang Tan, and Andrew Wan for sharing their insights on the Entropy Influene Conjeture. Referenes. Mihael Ben-Or and Nathan Linial. Colletive oin flipping. In Silvio Miali, editor, Randomness and Computation. Aademi Press, New York, Arthur Bernstein. Maximally onneted arrays on the n-ube. SIAM Journal on Applied Mathematis, pages 85 89, Ravi Boppana. The average sensitivity of bounded-depth iruits. Information Proessing Letters, 63(5:57 6, Anindya De, Omid Etesami, Lua Trevisan, and Madhur Tulsiani. Improved pseudorandom generators for depth iruits. In Proeedings of the th Annual International Workshop on Randomized Tehniques in Computation, pages 50 57, 00.

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