Entropy versus influence for complex functions of modulus one
aa r X i v : . [ m a t h . C O ] N ov Entropy versus influence for complex functionsof modulus one ∗ Gideon Schechtman † Abstract
We present an example of a function f from {− , } n to the unitsphere in C with influence bounded by 1 and entropy of | ˆ f | larger than log n . We also present an example of a function f from {− , } n to R with L norm 1, L ∞ norm bounded by √
2, influence bounded by 1and entropy of ˆ f larger than log n . We denote by ε i : {− , } n → {− , } the projection onto the i -s coordi-nate: ε i ( δ , . . . , δ n ) = δ i . For a subset A of [ n ] := { , . . . , n } we denote W A = Q i ∈ A ε i , W A : {− , } n → {− , } . The W A -s are the characters ofthe Cantor group {− , } n (with coordintewise multiplication) and form anorthonormal basis in L of the Cantor group equipped with the normalizedcounting measure. In most of this note we shall be concerned with functionsfrom {− , } n into the real numbers, R , but later on we shall also considerfunctions into the complex plane, C . These can also be considered as a cou-ple of real functions. Each such function f : {− , } n → C has a uniqueexpansion f = X A ⊆ [ n ] ˆ f ( A ) W A , ∗ AMS subject classification:42C10 (42C05, 05A20) Key words: Influence, Fourier–Entropy, Rudin–Shapiro polynomials † Supported in part by the Israel Science Foundation. f ( A ) ∈ C are given byˆ f ( A ) = E ( f W A ) = 2 − n X ε ∈{− , } n f ( ε ) W A ( ε ) . Note that if f : {− , } n → R , then ˆ f ( A ) ∈ R for every A ⊆ [ n ]. Theorthonormality of the W A -s implies that k f k := 2 − n X ε ∈{− , } n | f ( ε ) | = X A ⊆ [ n ] | ˆ f ( A ) | . Define the influence of a function f : {− , } n → C by I ( f ) = X A ⊆ [ n ] | ˆ f ( A ) | | A | (1)where for A ⊆ [ n ], | A | denotes the cardinality of A . This object, especiallyfor boolean functions, is a deeply studied one and quite influential (but thisis not the reason for the name...) in several directions. We refer to [O] forsome information. A recent paper dealing with the subject is [KKLMS].The entropy of the sequence | ˆ f ( A ) | is given by H ( | ˆ f ( A ) | ) = − X A ⊆ [ n ] | ˆ f ( A ) | log | ˆ f ( A ) | . (2)(0 log 0 := 0). The base of the log does not really matter here (as long as it isconsistent throughout the paper, so the log in the statements of the results isthe same as the one here). For concreteness we take the log to base 2. Notethat if f has L norm 1 then the sequence {| ˆ f ( A ) | } A ⊆ [ n ] sums up to 1 andthus this is the usual definition of entropy of this probability distribution,but we shall use this notation and term also for non normalized functions.The entropy influence conjecture of Friedgut and Kalai [FK] is that forsome absolute constant K , for all n and all boolean functions f : {− , } n →{− , } H ( | ˆ f ( A ) | ) ≤ KI ( f ) . For the significance of this conjecture we refer to the original paper [FK], andto Kalai’s blog [K] (embedded in Tao’s blog) which report on all significantresults concerning the conjecture. [KKLMS] establishes a weaker version of2he conjecture. Its introduction is also a good source of information on theproblem.As is well known the conjecture fails if we replace boolean functions withgeneral real functions (say, normalized to have L norm 1): f ( ε , . . . , ε n ) = n − / P ni =1 ε i gives such an example.In this note we show that the analogous version of the conjecture for twoclasses of ( L normalized) functions which resemble boolean functions fail aswell.The first class is the set of well bounded real functions on {− , } n . Thesecond is the complex functions on {− , } n which have modulus 1. Thesecond example solves a question raised by Gady Kozma some time ago (see[K], comment from April 2, 2011). More specifically, we prove the followingtwo theorems: Theorem 1.
For each n = 1 , , . . . there is a function f : {− , } n → R with k f k = 1 ≤ k f k ∞ ≤ √ , I ( f ) < , and H ( ˆ f ) > nn + 1 log n. Theorem 2.
For each n = 1 , , . . . there is a function f : {− , } n → { z ∈ C ; | z | = 1 } with I ( f ) < , and H ( | ˆ f | ) > nn + 1 log n. Actually, Theorem 1 (with a somewhat different lower bound for H ( | ˆ f | )but still of order log n ) follows from Theorem 2 but we prefer to give an easyindependent proof.The innovative contribution of this note is in the idea of the examples.The proofs are elementary, easy and self contained. We hope that somevariation of the examples will serve related purposes.After the first version of this note was posted, Joe Neeman sent me asimpler proof of Theorem 1 (with different constants). See the remark atthe end of this note. Also, an anonymous referee wrote that theorem 2 wasknown to experts but not published. No hint was given for the construction.Consequently, this note will not be further sent for publication (at leastin the near future). 3 The main results
Define P ≡ Q ≡ { a n } ∞ n =1 ⊂ (0 , P n , Q n : {− , } n → R inductively by P n +1 = P n + ε n +1 a n +1 Q n , Q n +1 = ε n +1 a n +1 P n − Q n , n = 0 , , . . . (3)The definition is inspired by that of the Rudin–Shapiro polynomials. Actu-ally, for the constant sequence a n = 1 for all n , P n and Q n are Rudin–Shapiropolynomials. These are functions R n : {− , } n → R all having all Fourier–Walsh coefficients of absolute value one (and in particular k R n k = 2 n/ ) and k R n k ∞ ≤ √ k R n k . These of course are no good for our purposes as 2 − n/ R n has influence equals to n/ n . Proposition 1.
For all n = 1 , , . . . , | P n | + | Q n | is a constant function, | P n | + | Q n | ≡ n Y i =1 (1 + a i ) , (4) k P n k = k Q n k = n Y i =1 (1 + a i ) / , (5) n Y i =1 (1 + a i ) / ≤ k P n k ∞ , k Q n k ∞ ≤ √ n Y i =1 (1 + a i ) / . (6) For each A ⊆ [ n ] , ˆ P n ( A ) = ˆ Q n ( A ) = Y i ∈ A a i . (7) Proof:
By Induction, starting from n = 0. (4) holds for n = 0 (or checkfor n = 1, if the case n = 0 bothers you). Assume it holds for n then by (3), | P n +1 | + | Q n +1 | = (1 + a n +1 )( | P n | + | Q n | ) ≡ n +1 Y i =1 (1 + a i ) . (5) holds for n = 0. Assume it holds for n then the orthogonality of P n and ε n +1 Q n implies k P n +1 k = k P n k + a n +1 k Q n k = k Q n k + a n +1 k P n k = k Q n +1 k . Proposition 2.
For each n = 1 , , . . .I ( P n ) = I ( Q n ) = n X i =1 a i n Y j = i (1 + a j ) . (8) Proof:
Note first that for each two functions
R, S : {− , } n → R and a ∈ R , if we put T = R + aε n +1 S , i.e., T ( ε , . . . , ε n +1 ) = R ( ε , . . . , ε n ) + aε n +1 S ( ε , . . . , ε n ), then I ( T ) = I ( R ) + a ( I ( S ) + k S k ) . (9)We now prove (8) by induction. I ( P ) = I ( Q ) = 0 (and I ( P ) = I ( Q ) = a ). Assume (8) then by (9) and (5), I ( P n +1 ) = I ( P n ) + a n +1 ( I ( Q n ) + n Y i =1 (1 + a i ))= n X i =1 a i Y ≤ j ≤ n,j = i (1 + a j ) + a n +1 n X i =1 a i Y ≤ j ≤ n,j = i (1 + a j ) + a n +1 n Y i =1 (1 + a i )= n X i =1 a i Y ≤ j ≤ n +1 ,j = i (1 + a j ) + a n +1 n Y i =1 (1 + a i ) = n +1 X i =1 a i Y ≤ j ≤ n +1 ,j = i (1 + a j ) . (10)The proof for I ( Q n +1 ) is almost identical. Remark 1.
It follows that putting K = P ni =1 a i , I ( P n ) = I ( Q n ) < Ke K . Proposition 3.
For each n = 1 , , . . .H ( ˆ P n ) = H ( ˆ Q n ) = − n X i =1 Y ≤ j ≤ n,j = i (1 + a j ) ! a i log a i . (11)5 roof: By the last assertion of Proposition 1 H ( ˆ P n ) = − X A ⊆ [ n ] Y j ∈ A a j ! log Y i ∈ A a i ! = − X A ⊆ [ n ] Y j ∈ A a j ! X i ∈ A log a i = − n X i =1 (log a i ) X A,i ∈ A Y j ∈ A a j = − n X i =1 a i (log a i ) X A,i ∈ A Y j ∈ A \{ i } a j = − n X i =1 a i (log a i ) Y j = i (1 + a j ) . Theorem 1.
For each n = 1 , , . . . there is a function f : {− , } n → R with k f k = 1 ≤ k f k ∞ ≤ √ , I ( f ) < , and H ( ˆ f ) > nn + 1 log n. Proof:
Note that for any g : {− , } n → R and a ∈ R , I ( ag ) = a I ( g ) and H ( c ag ) = a H (ˆ g ) − ( a log a ) k g k . (12)Put f = P n k P n k = n Y i =1 (1 + a i ) − / P n . (The last equation is by (5).) Then k f k = 1 and k f k ∞ ≤ √ I ( f ) = n X i =1 a i a i , and H ( ˆ f ) = − n X i =1 a i a i log a i − log n Y i =1 (1 + a i ) − > −
11 + max ≤ i ≤ n a i n X i =1 a i log a i . a i = 1 / √ n, i = 1 , . . . , n , gives I ( f ) < H ( ˆ f ) >
11 + n − log n. Remark 2.
The functions produced above have expectations different fromzero. It is easy to rectify this by looking instead at ε n +1 f : {− , } n +1 → R .The function g = ε n +1 f has the same L , L ∞ and H (ˆ g ) as f and I ( g ) = I ( f ) + 1 < S the unit sphere in R : S = { z ∈ C ; | z | = 1 } Theorem 2.
For each n = 1 , , . . . there is a function f : {− , } n → S with I ( f ) < , and H ( | ˆ f | ) > nn + 1 log n. Proof:
Complex functions satisfy a similar scaling property to (12): Forany g : {− , } n → C and a ∈ C , I ( ag ) = | a | I ( g ) and H ( | c ag | ) = | a | H ( | ˆ g | ) − ( | a | log | a | ) k g k . (13)Also, the absolute values of the Fourier–Walsh coefficients are preserved un-der taking conjugates and in particular, H ( | ˆ¯ g | ) = H ( | ˆ g | ) . (14)Put f n = 1 √ Q ni =1 (1 + a i ) / ( P n + ıQ n ) . By (4) f n : {− , } → S . By (13) and (8), I ( f n ) = n X i =1 a i a i . (15)To evaluate H ( | ˆ f n | ) notice first that, for each of the four functions F n = P n + ıQ n , F n = P n − ıQ n , F n = Q n + ıP n and F n = Q n − ıP n , each of the7our sets {| c F kn ( A ) | } A ⊆ [ n ] , k = 1 , , ,
4, is equal to { Q i ∈ A a i } A ⊆ [ n ] . Moreprecisely, for each A ⊆ [ n ] and k = 1 , , , | c F kn ( A ) | = √ Y i ∈ A | a i | . This is easily proved by induction on n : Assuming the assertion for n thenby (3), F n +1 = P n − ıQ n + ε n +1 a n +1 ( Q n + ıP n ) = F n + ε n +1 a n +1 F n . Let A ⊆ [ n + 1]. If A ⊆ [ n ] then [ F n +1 ( A ) = c F n ( A )and by the induction hypothesis (for F n ), | [ F n +1 ( A ) | = √ Q i ∈ A | a i | . If A [ n ] then n + 1 ∈ A and [ F n +1 ( A ) = a n +1 c F n ( A \ { n + 1 } )and by the induction hypothesis (for F n ), | [ F n +1 ( A ) | = | a n +1 |√ Q i ∈ A \{ n +1 } | a i | = √ Q i ∈ A | a i | . This proves the needed assertion for F n . The other three casesare proved similarly.It follows from (the proof of) Proposition 3 that H (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) \ P n + ıQ n √ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = − n X i =1 Y ≤ j ≤ n,j = i (1 + a j ) ! a i log a i . Now, just as in the proof of Theorem 1, using (13), we get the bound H ( | ˆ f n | ) > −
11 + max ≤ i ≤ n a i n X i =1 a i log a i . The choice a i = 1 / √ n, i = 1 , . . . , n , gives I ( f n ) < H ( | ˆ f n | ) > nn + 1 log n. emark 3. By taking a i -s with higher P a i one can get examples of func-tions of the same kind ( L norm 1, L ∞ norm at most √ R or into S ) with influence going to infinity with n and entropy of a larger orderof magnitude than the influence. This works as long as the influence is ofsmaller order of magnitude than its maximal order, n . More precisely, for1 < a < n let a i = √ a/ √ n , i = 1 , . . . , n . Then the proofs of Theorems1 and 2 give functions f : {− , } n → R and g : {− , } n → S with k f k = k g k = 1 , k f k ∞ ≤ √ a/ < I ( f ) , I ( g ) < a and H ( ˆ f ) , H ( | ˆ g | ) > a n − log a ) . So the ratio between the Fourier entropy and the influence tends to infinitywith n whenever a = o ( n ) (and is of order log n whenever a < n c for some c < Remark 4.
Joe Neeman pointed out to me that the following is also anexample of a well bounded real function with bounded influence and entropyof order log n : f ( ε , . . . , ε n ) = (cid:16) √ n P ni =1 ε i (cid:17) , (cid:12)(cid:12) √ n P ni =1 ε i (cid:12)(cid:12) ≤ CC, √ n P ni =1 ε i ≥ C, − C, √ n P ni =1 ε i ≤ − C. (Stricktly speaking, one needs to normalize f to have L norm exactly 1. Thisdoes not cause any complication.) References [FK] Friedgut, Ehud; Kalai, Gil. Every monotone graph property has a sharpthreshold. Proc. Amer. Math. Soc. 124 (1996), no. 10, 2993–3002[K] Kalai, Gil. https://terrytao.wordpress.com/2007/08/16/gil-kalai-the-entropyinfluence-conjecture[KKLMS] Kelman, Esty; Kindler, Guy; Lifshitz, Noam; Minzer, Dor;Safra, Muli. Towards a Proof of the Fourier–Entropy Conjecture?https://arxiv.org/abs/1911.105799O] O’Donnell, Ryan. Analysis of Boolean functions. Cambridge UniversityPress, New York, 2014.G. SchechtmanDepartment of MathematicsWeizmann Institute of ScienceRehovot, Israel [email protected]@weizmann.ac.il