Abstract 1 Introduction 2 Preliminaries 3 A Gaussian Variant 4 Linearity Testing Requires Pairwise Independence 5 Queries vs. Bias Tradeoff 6 Putting Everything Together 7 Analysis of the Linearity Test References

Biased Linearity Testing in the 1% Regime

Subhash Khot ORCID Courant Institute of Mathematical Sciences, New York University, NY, USA Kunal Mittal ORCID Department of Computer Science, Princeton University, NJ, USA
Abstract

We study linearity testing over the p-biased hypercube ({0,1}n,μpn) in the 1% regime. For a distribution ν supported over {x{0,1}k:i=1kxi=0 (mod 2)}, with marginal distribution μp in each coordinate, the corresponding k-query linearity test Lin(ν) proceeds as follows: Given query access to a function f:{0,1}n{1,1}, sample (x1,,xk)νn, query f on x1,,xk, and accept if and only if i[k]f(xi)=1.

Building on the work of Bhangale, Khot, and Minzer (STOC ’23), we show, for 0<p12, that if k1+1p, then there exists a distribution ν such that the test Lin(ν) works in the 1% regime; that is, any function f:{0,1}n{1,1} passing the test Lin(ν) with probability 12+ϵ, for some constant ϵ>0, satisfies Prxμpn[f(x)=g(x)]12+δ, for some linear function g, and a constant δ=δ(ϵ)>0.

Conversely, we show that if k<1+1p, then no such test Lin(ν) works in the 1% regime. Our key observation is that the linearity test Lin(ν) works if and only if the distribution ν satisfies a certain pairwise independence property.

Keywords and phrases:
Linearity test, 1% regime, p-biased
Funding:
Subhash Khot: Research supported by NSF Award CCF-1422159, NSF Award CCF-2130816, and the Simons Investigator Award.
Kunal Mittal: Research supported by NSF Award CCF-2007462, and the Simons Investigator Award.
Copyright and License:
[Uncaptioned image] © Subhash Khot and Kunal Mittal; licensed under Creative Commons License CC-BY 4.0
2012 ACM Subject Classification:
Theory of computation Computational complexity and cryptography
Related Version:
Full Version: https://arxiv.org/abs/2502.01900
Acknowledgements:
We thank Amey Bhangale, Yang P. Liu, and Dor Minzer for discussions that helped this project. Amey and Dor politely declined to be co-authors.
Editors:
Srikanth Srinivasan

1 Introduction

A function f:{0,1}n{1,1} is said to be linear over 𝔽2111by identifying the range 𝔽2 with {1,1}, under the map b(1)b, if there exists a set S[n], such that f(x)=iS(1)xi; this function is denoted by χS. The classical linearity testing problem, asks, given query access222the algorithm is allowed to ask/query the value of f(x) at any x{0,1}n to a function f:{0,1}n{1,1}, to distinguish between the following two cases333the algorithm is allowed to answer arbitrarily for functions f which violate both the conditions:

  1. 1.

    f is a linear function.

  2. 2.

    f is far from being linear; that is, for every linear function χS, the functions f and χS disagree on many points.

Linearity testing was first studied by Blum, Luby, and Rubinfeld, who gave a very simple 3-query test for this problem [15]. This test, known as the BLR test, proceeds in the following manner: Sample x,y{0,1}n uniformly and independently; query f at x,y, and xy, and accept if and only if f(xy)=f(x)f(y). Observe that this test accepts all linear functions with probability 1. Blum, Luby and Rubinfeld proved that any function f passing this test with high probability (1δ, for some small δ>0), must agree with some linear function χS on most (at least 1O(δ) fraction of) points in {0,1}n. This result, with the acceptance/agreement probability close to 1, is known as the 99%-regime of the test.

It was shown later [4, 23] that the above result extends to the 1% regime as well; more precisely, for every δ[0,1], and f:{0,1}n{1,1} such that

𝔼x,y{0,1}n[f(x)f(y)f(xy)]δ,

there exists S[n] such that 𝔼x{0,1}n[f(x)χS(x)]δ.

The above test is of fundamental importance in theoretical computer science, and has several applications; for example, it is one of the ingredients in the proof of the celebrated PCP theorem [19, 3, 2]. Furthermore, the analysis of the BLR test by Bellare et al. [4] is one of the early uses of Fourier analysis over the boolean hypercube, an area which now plays a crucial role in many diverse subfields of mathematics and computer science, like complexity theory, harness of approximation, learning theory, coding theory, social choice theory, etc. [28].

In this work, we are interested in the problem of linearity testing over the p-biased hypercube. For p(0,1), we denote by μp the p-biased distribution on {0,1}, which assigns probability p to 1, and 1p to 0. The p-biased hypercube refers the set {0,1}n, with the n-fold product measure μpn. Linearity testing, in this p-biased setting, asks to distinguish between linear functions, and functions which are far (with respect to the p-biased measure) from being linear.

The 99% regime of this problem is well-understood [24, 17], and a simple 4-query test works in this case (see Example 4 below). The question for the 1% regime turns out to be significantly more challenging for any p1/2, and was wide open until a recent work of Bhangale, Khot and Minzer [11] made significant progress. In particular, for every p(13,23), they give a 4-query test that works in the 1% regime.

Building upon the work of Bhangale, Khot and Minzer, we consider a very general class of tests, where, very roughly, some k queries x1,,xk{0,1}n, satisfying i[k]xi=0 (mod 2) are chosen, and the test accepts f:{0,1}n{1,1} if i[k]f(xi)=1. We shall require the following definitions:

Definition 1 (Class of Distributions).

For k,p(0,1), we define 𝒟(p,k) to be the class of all distributions ν on {0,1}k having μp as the marginal distribution on each coordinate i[k], and such that supp(ν){x{0,1}k:i=1kxi=0 (mod 2)}. We say that such a distribution ν has full even-weight support, if the above inclusion is an equality.

For a distribution ν𝒟(p,k), we say that i[k] is a pairwise independent coordinate, if for each j[k],ji, it holds that 𝔼Xν[XiXj]=p2. We say that ν is pairwise independent, if all its coordinates are pairwise independent.

Definition 2 (Class of Linearity Tests).

For a distribution ν𝒟(p,k), we define a corresponding linearity test, denoted by Lin(ν), as follows. Given query access to a function f:{0,1}n{1,1}: Sample444here, by x=(x1,,xk)νn, we mean that for each j[n], sample (x1(j),,xk(j))ν independently (also see Section 2 for notation). x=(x1,,xk)νn, and accept if and only if f(x1)f(x2)f(xk)=1.

Note that every linear function passes such a test with probability 1555When k is even, affine functions of the form ±χS also pass the test with probability 1. In this work, we shall ignore the distinction between these functions and linear functions.. More strongly, each query in νn having marginal distribution μpn ensures that functions that are close to linear (with respect to p-biased measure) are also accepted with high probability; in the property testing literature, such tests are called tolerant. Furthermore, this is a very general class of linearity tests, containing many of the mentioned previously tests, as demonstrated by the following examples:

Example 3.

The BLR test uses ν to be uniform over {x{0,1}3:x1+x2+x3=0 (mod 2)}.

Example 4.

The 4-query p-biased test of [17] (for the 99% regime) uses a distribution ν over {0,1}4 of the following form: With probability p0, set all coordinates to 0; with probability p1, set all coordinates to 1; and with probability 1p0p1, sample uniformly from the set {x{0,1}4:x1+x2+x3+x4=0 (mod 2)}. Note that each coordinate has bias p1+12(1p0p1), and p0,p1 are chosen so that this equals p.

In this work, we analyze the precise conditions under which tests in Defintion 2 work for linearity testing, in the 1% regime. Our main result (proven in Section 6) is the following:

Theorem 5.

Let p(0,1).

  1. 1.

    For every integer k>1+1min{p,1p}, there exists a distribution ν𝒟(p,k), such that the test Lin(ν) is a k-query linearity test over the p-biased hypercube, for the 1% regime.

    That is, for every ϵ>0, there exists a δ>0, such that for every large n, and every function f:{0,1}n[1,1] satisfying

    |𝔼(X1,,Xk)νn[i[k]f(Xi)]|ϵ,

    there exists a set S[n], such that |𝔼Xμpn[f(X)χS(X)]|δ.

  2. 2.

    The above point also holds for all integers k3 with p=1k1, and for all even integers k4 with p=11k1.

  3. 3.

    Conversely, for every positive integer k<1+1min{p,1p}, and every distribution ν𝒟(p,k), the test Lin(ν) fails in the 1% regime.

    That is, there exists a constant α>0, such that for every large n, there exists a function f:{0,1}n{1,1} satisfying

    |𝔼(X1,,Xk)νn[i[k]f(Xi)]|α,

    and such that for every S[n], it holds that |𝔼Xμpn[f(X)χS(X)]|on(1).

 Remark 6.

Note that the above theorem does not discuss the case when k5 is an odd integer, and p=11k1. This case is very interesting and is discussed in more detail in Section 6.1. Informally speaking, the test corresponding to the “natural” distribution ν𝒟(p,k), in this case, ensures correlation with a character of /(k1), and not a linear function χS (that is, a character of /2). In Section 6.1, we also present an alternative test to get around this.

Next, we shall describe the main technical results we prove along the way to prove Theorem 5. We start by stating (a generalized version of) the main linearity testing result of Bhangale, Khot and Minzer [11]:

Theorem 7 (General version proved later as Theorem 33).

Let k3 be a positive integer, and let p(0,1),ϵ(0,1] be constants, and let ν𝒟(p,k) be a distribution with full even-weight support (see Definition 1). Then, there exist constants δ>0,d (possibly depending on k,p,ϵ,ν), such that for every large enough n, the following is true:

Let f:{0,1}n[1,1] be a function such that

|𝔼(X1,,Xk)νn[i=1kf(Xi)]|ϵ.

Then, there exists a set S[n], and a polynomial g:{0,1}n of degree at most d and with 2-norm 𝔼Xμpn[g(X)2]1, such that

|𝔼Xμpn[f(X)χS(X)g(X)]|δ.

Moreover, if the distribution ν has some pairwise independent coordinate, then we may assume g1; that is, f correlates with a linear function χS.

We remark that Bhangale, Khot and Minzer only consider the case k=4, and only show g1 in the case that all coordinates of ν are pairwise independent. However, their proofs extend to the more general setting of Theorem 7; we give an outline of this proof in Section 7. Furthermore, we note that we are able to analyze the linearity test for a class of distributions which is much larger than the class of full even-weight support distributions; these distributions, in some sense, contain the BLR test, and are formally defined in Section 7.

In the above work, the authors ask whether the conclusion g1 can be obtained without assumption that ν has a pairwise independent coordinate. We show this is not possible, and in fact the assumption that ν has a pairwise independent coordinate is necessary.

Theorem 8 (Restated and proved later as Theorem 24).

Let k,p(0,1), and let ν𝒟(p,k) be a distribution having no pairwise independent coordinate (see Definition 1).

Then, there exists a constant α>0, such that for every large enough n, there exists a function f:{0,1}n[1,1] such that

  1. 1.

    |𝔼Xνn[i=1kf(Xi)]|α.

  2. 2.

    For every S[n], it holds that |𝔼Xμpn[f(X)χS(X)]|on(1).

Moreover, if the distribution ν is such that η:=maxi,j[k],ijPrXν[Xi=Xj]<1 (that is, no two coordinates are almost surely equal), the above holds for a function f with range {1,1}.

 Remark 9.
  1. 1.

    The assumption η<1 in the second part of the Theorem 8 is necessary. For example, if the ith and jth coordinates of ν are equal, then, for functions f with range {1,1}, the terms f(Xi) and f(Xj) cancel out in the product 𝔼Xνn[i=1kf(Xi)]. In particular, the test is equivalent to the (k2)-query test with coordinates i,j removed from ν, and this new distribution may possibly satisfy the conditions of Theorem 7.

  2. 2.

    The function f we construct in Theorem 8 does not correlate well with any linear function, although, as possibly required by Theorem 7, it does correlate well with some constant degree function.

  3. 3.

    The above theorem, answers in the negative a question of [11], who ask if

    |𝔼(X,Y,Z,W)νn[g1(X)g2(Y)g3(Z)g4(W)]|=on(1)

    for distributions ν𝒟(p,4) with full even-weight support, and g1,,g4:{0,1}n bounded, noise stable, and resilient functions.

  4. 4.

    It is an easy check that the distribution ν from Example 4 cannot have a pairwise independent coordinate, unless p=1/2. This shows that for p12, simple tests that work in the 99% regime fail to work in the 1% regime.

  5. 5.

    Recall that every ν𝒟(p,k) satisfies iXi=0 (mod 2) almost surely, for Xν. We never use this in the proof of the above theorem, and the conclusion holds without it.

Very roughly speaking, in the proof of the above theorem we first construct a counter-example function in Gaussian space which “passes the test” with decent probability, while having zero expectation; this function is then converted to a boolean function using the Central Limit Theorem and a rounding procedure. Along the way, we prove a simple characterization for a random vector to have an independent coordinate, which we believe to be of independent interest, and is stated as follows:

Proposition 10 (Restated formally and proved later as Proposition 19).

Let X=(X1,,Xk) be a k-dimensional multivariate Gaussian random vector, such that for each i[k], the marginal is Xi𝒩(0,1). Then, the following are equivalent:

  1. 1.

    For every “nice” function f: satisfying 𝔼Z𝒩(0,1)[f(Z)]=0, it holds that 𝔼[f(X1)f(X2)f(Xk)]=0.

  2. 2.

    There exists i[k] such that Xi is independent of (X1,,Xi1,Xi+1,,Xk).

Finally, to use the above theorems (Theorem 7 and Theorem 8), we analyze the tradeoff between the number of queries k and the bias p, such that a distribution ν𝒟(p,k) with some pairwise independent coordinate exists. In particular, we prove the following (restated and proved later as Proposition 25 and Proposition 27):

Proposition 11.

Let k,p(0,1). Then, there exists a distribution ν𝒟(p,k) with some pairwise independent coordinate if and only if k1+1min{p,1p}.

We note that the above generalizes the parameter setting for both the BLR test, corresponding to p=12,k=3, and the case of p(13,23),k=4 considered in [11].

1.1 Related work

The problem of linearity testing has been extensively studied, starting with the work of Blum, Luby and Rubinfeld [15], who gave a test for the uniform distribution, in the 99% regime. The analysis of their test was later extended to the 1% regime [4, 23]. Tests for linearity have been also been studied in the low-randomness regime, and in the setting of non-abelian groups [6, 5, 29].

For the p-biased case, in the 99% regime, Halevy and Kushilevitz [20] gave a 3-query linearity test, that only uses random samples from the p-biased distribution! However, the test is not tolerant, makes queries that are not distributed according to μpn, and hence may reject functions that are very close to linear (with respect to the p-biased measure). Tolerant testers were analyzed later [24, 17]. More strongly, the work of Dinur, Filmus and Harsha [17] gives 2d-query tolerant tester for p-biased testing of degree d functions over 𝔽2, a problem which has been well studied over the uniform distribution [1, 14].

As a part of their work on approximability of satisfiable constraint satisfaction problems [9, 10, 11, 12, 13], Bhangale, Khot and Minzer study the p-biased version of linearity testing, in the 1% regime. As mentioned before, they give a 4-query test for p(13,23).

David, Dinur, Goldberg, Kindler and Shinkar [16] study linearity testing on the k-slice (vectors of hamming-weight k), denoted by Lk,n, of the n-dimensional boolean hypercube, for even integers k. They show that if f:{0,1}n{1,1} is such that f(xy)=f(x)f(y) with probability 1ϵ over x,y,xy (conditioned on all lying in Lk,n), then f agrees with a linear function on 1δ fraction of Lk,n, where δ=δ(ϵ)0 as ϵ0. In a recent work, Kalai, Lifshitz, Minzer and Ziegler [22] prove a similar result for the n/2-slice, in the 1% regime.

1.2 Organization of the paper

We start by presenting some preliminaries in Section 2. In Section 3, we prove a variant of Theorem 8 over the Gaussian distribution, which then is used in Section 4 to prove Theorem 8. In Section 5, we analyze the tradeoff between the bias p and the number of queries k, for the existence of a valid linearity test. Combining all results, we prove Theorem 5 in Section 6. In Section 7, we outline of the proof of Theorem 7.

2 Preliminaries

We use exp to denote the exponential function, given by exp(x)=ex for x.

Let ={1,2,} be the set of natural numbers. For each n, we use [n] to denote the set {1,2,,n}. For non-negative functions f,g:, we say that f(n)=on(g(n)) if limnf(n)g(n)=0.

For a probability distribution ν on 𝒳, we use supp(ν) to denote its support. For n, we use νn to denote the n-fold product distribution on 𝒳n. In particular, we shall be interested in the case when 𝒳k for some k. In this case, for vectors xkn, we shall use subscripts for indices in [k] and superscripts for indices in [n]; that is, for each i[k],j[n], we use xi(j) to denote the (i,j)th coordinate of x. Further, for each i[k], we use xi to denote the vector (xi(1),,xi(n))n, and similarly for each j[n], we use x(j) to denote the vector (x1(j),,xk(j))k.

For k, let Sk denote the group of all permutations on [k]. For each πSk,xk, we use xπ to denote (xπ(1),,xπ(k))k. With this notation, we define the symmetrization of functions over k:

Definition 12.

For any function f:k, we define its symmetrization as the function Sym(f):k, given by Sym(f)(x)=πSkf(xπ).

We shall use the following facts from probability theory:

Fact 13 (Chebyshev’s Inequality; see [18] for reference).

Let X be a random variable such that 𝔼[X2]<. Then, for any any a>0,

Pr[|X𝔼[X]|a]Var[X]a2.
Fact 14 (Hoeffding’s Inequality [21]).

Let X1,,Xn be independent random variables such that aiXibi almost surely, and let S=i=1nXi. Then, for all t>0,

Pr[|S𝔼[S]|t]2exp(2t2i=1n(biai)2).
Theorem 15 (Multivariate Central Limit Theorem; see [18] for reference).

Let X(1),X(2), be k-valued i.i.d. random vectors, with mean zero, and finite a covariance matrix Σk×k given by Σi,j=𝔼[Xi(1)Xj(1)]. If Sn=1ni=1nX(i), then, Sn𝒟Z as n, for Z𝒩(0,Σ). That is, for every bounded continuous function H:k,

limn𝔼[H(Sn)]=𝔼Z𝒩(0,Σ)[H(Z)].

We shall also use the following fact about zeros of polynomials:

Lemma 16.

Let p1,,pr:k be non-zero polynomials. Then, there exists yk such that for each permutation πSk, and each i[r], it holds that pi(yπ)0.

Proof Sketch.

The zero-set of any non-zero polynomial has measure zero, with respect to the Lebesgue measure on k. Hence, by sub-additivity, the set of points yk violating the statement of the lemma is of measure zero as well.

Next, we give some basic results about the probabilist’s Hermite polynomials. The reader is referred to Chapter 11 in [28] for more details.

Definition 17.

The Hermite polynomials (Hj)j0 are univariate polynomials, with Hj a monic polynomial of degree j, satisfying the power series expression

exp(txt22)=j=01j!Hj(x)tj,for t,x.

Note that the series above is absolutely convergent, with j=01j!|Hj(x)||t|jexp(|t||x|+t22) for each t,x.

Lemma 18.

Let k and s1,s2,,sk0, and let Σk×k be a positive semi-definite matrix such that Σi,i=1 for each i. For V=ΣI, it holds that

𝔼X𝒩(0,Σ)[Hs1(X1)Hsk(Xk)]=s1!sk!d!2d[(tVt)d:t1s1tksk]

where s1++sk=2d, and [(tVt)d:t1s1tksk] denotes the coefficient of t1s1tksk in the polynomial (tVt)d. Also, the above expectation is zero when s1++sk is odd.

Proof.

Recall that the moment generating function of a multivariate Gaussian distribution is given by

𝔼X𝒩(0,Σ)[exp(t1X1+tkXk)]=exp(12tΣt),

for each tk. Multiplying the above by exp(12tt), and plugging in the power series in Definition 17, we get for each tk that

d=01d!2d(tVt)d =exp(12tVt)
=𝔼X𝒩(0,Σ)[exp((t1X1t122)++(tkXktk22))]
=𝔼X𝒩(0,Σ)[(s1=01s1!Hs1(X1)t1s1)(sk=01sk!Hsk(Xk)tksk)]
=s1,,sk0t1s1tksks1!sk!𝔼X𝒩(0,Σ)[Hs1(X1)Hsk(Xk)].

Note that since the power series in Definition 17 is absolutely convergent, all steps above of interchanging limits and expectations are valid by the dominated convergence theorem. Finally, comparing coefficients, we have the desired result.

3 A Gaussian Variant

The first step towards proving Theorem 8 is to prove a Gaussian variant, stated below:

Proposition 19.

Let k, and let Σk×k be a symmetric positive semi-definite matrix such that:

  1. 1.

    For each i[k], it holds that Σi,i=1.

  2. 2.

    The matrix V=ΣI has no row/column as all zeros.

Then, there exists a Lipschitz continuous function f:[1,1] such that:

𝔼X𝒩(0,1)[f(X)]=0,and|𝔼X𝒩(0,Σ)[i[k]f(Xi)]|>0.

3.1 Symmetric Powers of Polynomials

Before we prove the above proposition, we first prove a lemma about (symmetrization of) powers of multivariate polynomials. We show that if a polynomial q(x1,,xk) depends on all the variables x1,,xk, then the symmetric power Sym(qd), for some integer d (see Definition 12), contains a monomial divisible by x1x2xk.

Lemma 20.

Let k, and let q:k be a polynomial such that for each i[k], the polynomial i=iq is not identically zero. Then, there exists some d, and positive integers s1,,sk such that the coefficient of x1s1x2s2xksk in the polynomial Sym(qd) is non-zero.

We start by proving the following lemma about derivates of powers of q.

Lemma 21.

Let k, and let q:k be a polynomial. For each i[k], let i=iq.

Then, for every s=(s1,,sk)0k with |s|=i[k]si, there exist polynomials p0,,p|s|, with p|s|=i[k]isi, such that for each d|s|, it holds that

1s12s2ksk(qd)=qd|s|(i=0|s|dipi).

Proof.

The proof is by induction on |s|. For the base case, if |s|=0, we have s=(0,0,,0), and p0=1 satisfies the statement of the lemma.

For the inductive step, consider any s=(s1,,sk)0k with |s|=i[k]si>0. Without loss of generality, by symmetry, we can assume that s1>0. By the inductive hypothesis applied to (s11,s2,sk), we have the existence of polynomials p0,,p|s|1, with p|s|1=1s11i=2kisi, and such that for each d|s|1, we have

1s112s2ksk(qd)=qd|s|+1(i=0|s|1dipi).

Now, if d|s|, differentiating the above with respect to x1, we get

1s12s2ksk(qd) =qd|s|((d|s|+1)1i=0|s|1dipi)+qd|s|+1(i=0|s|1di1(pi))
=qd|s|(i=1|s|di1pi1+i=0|s|1di((|s|+1)1pi+q1pi))
=qd|s|(i=0|s|dip~i),

where the polynomials p~1,,p~|s| do not depend on d, and are such that p~|s|=p|s|11=i[k]isi, as desired.

With the above lemma in hand, next we shall consider the symmetrization operation applied to derivatives of powers of q.

Lemma 22.

Let k, and let q:k be a polynomial such that for each i[k], the polynomial i=iq is not identically zero.

Then, for each large enough even integer d, the polynomial Sym(1222k2(qd)) is not identically zero.

Proof.

By applying Lemma 21 on s=(2,2,,2), we have the existence of polynomials p0,,p2k, with p2k=i[k]i2, such that for each d2k, it holds that 1222k2(qd)=qd2k(i=02kdipi).

By Lemma 16, let yk be such that y (and its permutations) don’t lie in the zero set of any of the polynomials 1,,k,q. We define

A=minπSk[i[k]i(yπ)2]>0,B=max0i2k1,πSk|pi(yπ)|0.

Then, for any even integer dmax{2k,4kBA}, it holds that

Sym(1222k2(qd))(y) =πSkq(yπ)d2k(d2ki[k]i(yπ)2+i=02k1dipi(yπ))
πSkq(yπ)d2k(d2kAi=02k1diB)
(πSkq(yπ)d2k)(d2kA2kd2k1B)
(πSkq(yπ)d2k)d2kA2>0.

Hence, for even integers dmax{2k,4kBA}, the polynomial Sym(1222k2(qd)) is not identically zero.

Finally, we prove the main lemma of this section.

Proof of Lemma 20.

Let k, and let q:k be a polynomial such that for each i[k], the polynomial i=iq is not identically zero. It suffices to prove that for some d, the polynomial 1222k2(Sym(qd)) is not identically zero, since then the coefficient of some monomial divisible by x12x22xk2 is non-zero.

For each polynomial p:k, and each πSk, we shall use pπ to denote the polynomial given by pπ(x)=p(xπ). Then, for all s1,,sk0, we have that 1s12s2ksk(pπ)=(π1(1)s1π1(2)s2π1(k)sk(p))π.

By the above, we have that for each d,

1222k2(Sym(qd)) =1222k2(πSkqπd)
=πSk1222k2(qπd)
=πSk(π1(1)2π1(2)2π1(k)2(qd))π
=πSk(1222k2(qd))π
=Sym(1222k2(qd)).

Now, the result follows from Lemma 22.

3.2 Proving the Gaussian Variant

We start by proving a slight variant of Proposition 19, where we allow f to be an arbitrary (possibly unbounded) polynomial.

Lemma 23.

Let k, and let Σk×k be a symmetric positive semi-definite matrix such that:

  1. 1.

    For each i[k], it holds that Σi,i=1.

  2. 2.

    The matrix V=ΣI has no row/column as all zeros.

Then, there exists a polynomial f: such that 𝔼X𝒩(0,1)[f(X)]=0, and

|𝔼X𝒩(0,Σ)[i[k]f(Xi)]|>0.

Proof.

For s=(s1,,sk)k and α=(α1,,αk)k, let fs,α: be the polynomial defined by fs,α(x)=α1Hs1(x)++αkHsk(x), where the polynomials Hsi are Hermite polynomials (see Definition 17). Observe that since s1,,sk1, we have 𝔼X𝒩(0,1)[f(X)]=0; this follows from the orthogonality of the Hermite polynomials.

Suppose, for the sake of contradiction, that for every sk,αk, it holds that

𝔼X𝒩(0,Σ)[i[k]fs,α(Xi)]=𝔼X𝒩(0,Σ)[i[k]j[k]αjHsj(Xi)]=0.

Observe that for every sk, the above expression can be written as a multivariate polynomial in α1,,αk. If the polynomial vanishes for all αk, the coefficient of α1α2αk must be zero; that is,

πSk𝔼X𝒩(0,Σ)[i[k]Hsπ(i)(Xi)]=0.

Now, applying Lemma 18, we get that for each d, and each s1,,sk1 with s1++sk=2d,

πSk[(tVt)d:t1sπ(1)tksπ(k)] =πSk[(tπVtπ)d:t1sktksk]
=[Sym((tVt)d):t1sktksk]=0.

Note that the assumption that V has no zero row/column implies that for every i[k], the polynomial i(tVt) is not identically zero. By Lemma 20, this is a contradiction.

With the above, we now prove Proposition 19 via a standard truncation argument.

Proof of Proposition 19.

By Lemma 23, we know that there exists a polynomial f: such that 𝔼X𝒩(0,1)[f(X)]=0, and |𝔼X𝒩(0,Σ)[i[k]f(Xi)]|>0.

For each integer M, we define the truncated function fM:[M,M] by

fM(x)=f(x)𝟙|f(x)|M+M𝟙f(x)>MM𝟙f(x)<M.

Also, let gM:[2M,2M], be given by gM(x)=fM(x)𝔼X𝒩(0,1)[fM(X)]. Observe that

  1. 1.

    For every M, it holds that 𝔼X𝒩(0,1)[gM(X)]=0.

  2. 2.

    For every M, the function gM is bounded and Lipschitz continuous.

  3. 3.

    For every x, fM(x)f(x) as M. Further, since |fM(x)||f(x)| for each x,M, by the dominated convergence theorem, we have 𝔼X𝒩(0,1)[fM(x)]𝔼X𝒩(0,1)[f(x)]=0 as M. This implies that for each x, gM(x)f(x) as M.

    Also, for each x,M, we have |gM(x)||f(x)|+𝔼X𝒩(0,1)[|f(X)|]. Hence, by the dominated convergence theorem, we have that 𝔼X𝒩(0,Σ)[i[k]gM(Xi)]𝔼X𝒩(0,Σ)[i[k]f(Xi)]0 as M.

By the above, for some large enough M, the function 12MgM:[1,1] satisfies the desired properties.

4 Linearity Testing Requires Pairwise Independence

In this section, we prove Theorem 8, which is restated below.

Theorem 24.

Let k,p(0,1), and let ν𝒟(p,k) be a distribution having no pairwise independent coordinate (see Definition 1). Then, there exists a constant α>0, such that for every large enough n, there exists a function f:{0,1}n[1,1] such that

  1. 1.

    |𝔼Xνn[i=1kf(Xi)]|α.

  2. 2.

    For every S[n], it holds that |𝔼Xμpn[f(X)χS(X)]|on(1).

Moreover, if the distribution ν is such that η:=maxi,j[k],ijPrXν[Xi=Xj]<1 (that is, no two coordinates are almost surely equal), the above holds for a function f with range {1,1}.

The remainder of this section is devoted to the proof of Theorem 24. In Section 4.1, we prove the first part of the theorem, dealing with functions with range [1,1]. Then, in Section 4.2, we show how to round to functions with range {1,1}.

4.1 Function with Range [𝟏,𝟏]

Let k,p(0,1), and let ν𝒟(p,k) be a distribution having no pairwise independent coordinate. Let Σk×k be the (normalized) covariance matrix corresponding to the distribution ν, given by, Σi,j=𝔼Xν[(Xip)(Xjp)pp2]. Observe that the matrix Σ satisfies the conditions of Proposition 19, and hence there exists a function h:[1,1] such that

  1. 1.

    𝔼Z𝒩(0,1)[h(Z)]=0

  2. 2.

    The function H:k[1,1] given by H(x)=i[k]h(xi) is such that

    α:=12|𝔼Z𝒩(0,Σ)[H(Z)]|>0.
  3. 3.

    The function h is K-Lipschitz for some K>0; in particular, both h and H are bounded continuous functions.

Consider any large n. We define f:{0,1}n[1,1] by

f(x)=h(1nj=1nx(j)ppp2),

The function f satisfies the two properties in the theorem statement, as follows:

  • Let Xνn, and let Y=(Y1,,Yk) be a {0,1}k-valued random vector, defined as Yi=1nj=1nXi(j)ppp2.

    Let F:{0,1}kn[1,1] be given by F(x)=i[k]f(xi). Since H is continuous and bounded, we have by the Multivariate CLT (Theorem 15) that

    |𝔼[F(X)]𝔼Z𝒩(0,Σ)[H(Z)]|=|𝔼[H(Y)]𝔼Z𝒩(0,Σ)[H(Z)]|on(1).

    Hence, for large n, we get |𝔼Xνn[i=1kf(Xi)]|2αon(1)α, as desired.

  • Consider any subset S[n], and let TS be any subset of size |T|=min{n1/4,|S|}. Let f~:{0,1}n[1,1] be defined by f~(X)=h(1n|T|j[n]Tx(j)ppp2); note that this function only depends on the coordinates of x outside the set T. Further, for each x{0,1}n, by the Lipschitz bound on h, we get

    |f(x)f~(x)| K|1nj=1nx(j)ppp21n|T|j[n]Tx(j)ppp2|
    Kpp2(|T|n+(n|T|)|1n|T|1n|)
    Kpp2(|T|n+n|T|n|T|n)
    Kpp22|T|n=on(1),

    where we used that (1t)1/21+t for each t[0,1/2].

    Now, for Xμpn, we have

    |𝔼X[f(X)χS(X)]| |𝔼X[f~(X)χS(X)]|+on(1)
    =|𝔼X[f~(X)χST(X)]𝔼X[χT(X)]|+on(1)
    =|𝔼X[f~(X)χST(X)]||12p||T|+on(1).

    If |S|n1/4, then |12p||T|=on(1). Otherwise, we have that S=T, and by the Central Limit Theorem (see Theorem 15) , the first term in the above product equals

    |𝔼X[f~(X)]|=|𝔼X[f~(X)]𝔼Z𝒩(0,1)[h(Z)]|=on(1).

4.2 Rounding to a Function with Range {𝟏,𝟏}

Now, we shall prove the second part of Theorem 24.

Let k,p(0,1), and let ν𝒟(p,k) be a distribution having no pairwise independent coordinate. Further suppose that the distribution ν is such that

η:=maxi,j[k],ijPrXν[Xi=Xj]<1.

Let α>0 be as obtained in Section 4.1. Consider any large n, and let f:{0,1}n[1,1] be the function obtained in Section 4.1.

Let g:{0,1}n{1,1} be a random function, defined as g(x)={1,w.p.1+f(x)21,w.p.1f(x)2, independently for each x{0,1}n. Observe that this satisfies 𝔼g[g(x)]=f(x) for each x{0,1}n. We will show that the function g satisfies the two desired properties with probability 1on(1), and hence by the probabilistic method, this guarantees the existence of a non-random g as desired. This is done as follows:

  1. 1.

    Let F,G:{0,1}kn[1,1] be defined as F(x)=i[k]f(xi) and G(x)=i[k]g(xi). Let X,Yνn be independent (of each other and of g) and let E be the event that X1,,Xk,Y1,,Yk are all distinct. Then, by a union bound, we have that Pr[E¯]2(k2)ηn+k2(p2+(1p)2)n=on(1), and hence

    |𝔼g𝔼Xνn[G(X)]𝔼Xνn[F(X)]| Pr[E¯]+|𝔼g𝔼X,Y[G(X)𝟙E]𝔼X[F(X)]|
    Pr[E¯]+|𝔼X,Y[F(X)𝟙E]𝔼X[F(X)]|
    2Pr[E¯]=on(1).

    Similarly, we have

    |𝔼g[𝔼X[G(X)]]2[𝔼X[F(X)]]2| =|𝔼g𝔼X,Y[G(X)G(Y)]𝔼X,Y[F(X)F(Y)]|
    2Pr[E¯]=on(1).

    Letting β=|𝔼X[F(X)]|α, we get Varg[𝔼X[G(X)]]β2+on(1)(βon(1))2=on(1). Hence, by Chebyshev’s inequality (Fact 13), we have |𝔼X[G(X)]|α2 with probability 1on(1).

  2. 2.

    Fix S[n]. Let Xμpn, and let W=𝔼X[χS(X)g(X)]=x{0,1}nPr[X=x]χS(x)g(x). Observe that W is a sum of 2n independent and bounded random variables, and such that 𝔼g[W]=𝔼X[χS(X)f(X)]. For q=max{p,1p}<1, it holds that x(2Pr[X=x])24qnxPr[X=x]=4qn, and by Hoeffding’s inequality (Fact 14), we have for each t>0 that

    Pr[|W𝔼[W]|t]2exp(2t24qn).

    Let t=qn/4. Then, with probability at least 1on(2n), it holds that |W|=|𝔼X[χS(X)g(X)]||𝔼X[χS(X)f(X)]|+qn/4=on(1).

    Now, a union bound over S[n] shows that with probability 1on(1), the above holds for every S[n]. ∎

5 Queries vs. Bias Tradeoff

In this section, we analyze the relation between p (the bias) and k (the number of queries) for the existence of a distribution ν𝒟(p,k) with some pairwise independent coordinate, and with full even-weight support (see Definition 1).

5.1 Query Lower Bound

We prove a lower bound on k in terms of the p, as follows:

Proposition 25.

Let k,p(0,1), and let ν𝒟(p,k) be a distribution that has some pairwise independent coordinate. Then, it holds that k3 and 1k1p11k1.

Proof.

Let Xν, and let i[k] be a pairwise independent coordinate under ν.

For Z=jiXj, we have by linearity of expectation, that 𝔼[XiZ]=(k1)p2. On the other hand, observe that if Xi=1, then Z=1 (mod 2) and so Z1. Hence,

p=𝔼[Xi1]𝔼[XiZ]=(k1)p2,

and we have (k1)p1; in particular, this shows k3.

For the upper bound on p, we consider the following cases:

  • k is odd: In this case, if Xi=1, then Z=1 (mod 2) and so Zk2. Hence,

    (k1)p2=𝔼[XiZ]𝔼[Xi(k2)]=p(k2),

    and we have (k1)p(k2), as desired.

  • k is even: In this case, observe that the distribution of the random variable (1X1,,1Xk) also satisfies the hypothesis of the proposition, with p replaced by 1p. Hence, the above proof gives us (k1)(1p)1, as desired.

 Remark 26.

The proof of Proposition 25 also shows that for k>3 and p{1k1,11k1}, any distribution satisfying the assumptions of Proposition 25 cannot have full even-weight support. This is because if p{1k1,11k1}, in all cases in the above proof, the random variable Z must be constant under some value of Xi (either Xi=0 or Xi=1); this cannot be the case for a distribution with full even-weight support when k>3.

5.2 Query Upper Bound

In this subsection, we shall prove the following proposition.

Proposition 27.

Let k3 be a positive integer, and let p[1k1,11k1] (note that this interval is non-empty for k3).

Then, there exists a permutation-invariant666we say that a distribution ν over {0,1}k is permutation-invariant, if for X=(X1,,Xk)ν, and any permutation π:[k][k], the distribution of (Xπ(1),,Xπ(k)) is the same as ν. and pairwise independent distribution ν(k,p)𝒟(p,k) (see Definition 1). Furthermore, if k=3 or if p{1k1,11k1}, then there exists such a distribution with full even-weight support.

The proof involves various cases, considered below in Lemma 28 and Lemma 29.

Lemma 28.

Let k4 be a positive integer, and let p[1k1,2k1)(12k1,11k1] (note that this interval is contained in [1k1,11k1] for k4). Then, there exists a pairwise independent distribution ν(k,p)𝒟(p,k).

Moreover, if p{1k1,11k1}, then there exists such a distribution with full even-weight support.

Proof.

Let k4 be a positive integer, and let p[1k1,2k1)(12k1,11k1]. Let s=k2; we shall exhibit a vector q=(q0,q1,,qs)[0,1]s+1 satisfying:

i=0s(k2i)qi=1,i=1s(k12i1)qi=p,i=1s(k22i2)qi=p2.

The distribution ν(p,k) is then defined by assigning probability {q|x|/2,|x|=0 (mod 2)0,|x|=1 (mod 2) to the point x{0,1}k, where |x|=i=1kxi. Note that the above properties correspond to ν(k,p) being a valid probability distribution supported on even-hamming-weight vectors, having marginals μp, and pairwise independent coordinates. Furthermore, the distribution ν(p,k) has full even-weight support if and only if each qi(0,1).

The vector q is defined as follows in different cases (for brevity, we omit the verification of the above properties):

  1. 1.

    k5 is odd, p[1k1,2k1): Let q0=1+kp22k2p2(k1), q1=(k2)p(k1)p2(k1)(k3), q(k1)/2=(k1)p2p(k1)(k3), and zero otherwise.

  2. 2.

    k5 is odd, 1p[1k1,2k1): Let q0=1+kp2k3k(2k5)p(k1)(k3), q(k3)/2=3(k2)p3(k1)p2(k1)(k2)(k3), q(k1)/2=(k1)p2(k4)p2(k1), and zero otherwise.

  3. 3.

    k4 is even, p[1k1,2k1): Let q0=(k1)p2(k+1)p+22, q1=pp2k2, qk/2=(k1)p2pk2, and zero otherwise.

  4. 4.

    k4 is even, 1p[1k1,2k1): In this case, we define ν(k,p) to be the distribution obtained by flipping each coordinate of ν(k,1p).

Next, we show that if p{1k1,11k1}, then such a distribution ν(p,k) with full even-weight support exists. We only need to do this for the first three cases, as the procedure described in the fourth case preserves the property of full even-weight support.

The same argument applies in all cases, and we present it for the first case: that is when k5 is odd, and p(1k1,2k1). We observe if p1k1, each of the probabilities q0,q1,q(k1)/2 above lie in the interval (0,1). Now, consider the equations

i=0s(k2i)q~i=0,i=1s(k12i1)q~i=0,i=1s(k22i2)q~i=0.

In these equations, the variables q~0,q~1,q~(k1)/2 are linearly independent, and hence, there exists a vector q~s+1 satisfying these equations, which has all coordinates equal to 1, other than possibly q~0,q~1,q~(k1)/2. Then, for some small δ>0, the vector q+δq~ has all coordinates in (0,1), and satisfies the required properties.

Lemma 29.

Let k6 be a positive integer, and let p[2k1,12k1]{12} (note that this interval is non-empty for k6). There, there exists a pairwise independent distribution ν(k,p)𝒟(p,k) with full even-weight support.

Proof.

Let k6 be a positive integer, and let p[2k1,12k1],p12. That is, for q=min{p,1p}<12, we have k1+2q. Let be the smallest odd integer satisfying >1+1q>3. Note that this satisfies 43+1q<1+2qk, and we have q(11,21).

By Lemma 28, there exist pairwise independent distributions ν(,p) and ν(,1p), with full even-weight support. Let ν~0=ν(,p), and let ν~1 be the distribution obtained by flipping each coordinate of ν(,1p). Since is odd, for each b{0,1}, it holds that ν~b has pairwise independent coordinates, each with marginal μp, and such that supp(ν~b)={x{0,1}k:i=1kxi=b (mod 2)}. Finally, we define Xν(k,p) via the following random process: Let (X+1,,Xk)μp(k), and with Z=i=+1kXi (mod 2), we let (X1,,X)ν~Z. It is an easy check that this distribution satisfies the required properties.

Finally, we prove Proposition 27.

Proof of Proposition 27.

Note that it suffices to find such a distribution that is not necessarily permutation invariant, since averaging the distribution over all permutations preserves pairwise independence and full even-weight support.

If p=1/2, for any k3, we let ν(k,p) be the uniform distribution on the set {x{0,1}k:i=1kxi=0 (mod 2)}.

Now, for k=3, it must hold that p=1/2, in which case ν(k,p) is as above. For k=4 or k=5, and p1/2, it must hold that p[1k1,2k1)(12k1,11k1], and the result follows from Lemma 28. For k6 and p12, the result follows from Lemma 28 and Lemma 29.

6 Putting Everything Together

We are now ready to prove our main result.

Proof of Theorem 5.

Let p(0,1).

  1. 1.

    Consider any positive integer k>1+1min{p,1p}3 (or k=3 with p=12). By Proposition 27, there exists a pairwise independent distribution ν𝒟(p,k) with full even-weight support. The result now follows by Theorem 33.

  2. 2.

    Suppose that k3 with p=1k1, or k4 is even with p=11k1. In these cases, we observe that the distribution ν𝒟(p,k) constructed in Lemma 28 is pairwise independent, and contains BLR (see Definition 32):

    1. (a)

      If k3,p=1k1, the distribution ν contains all vectors in {0,1}k of hamming-weights 0 and 2 in its support. In this case, Definition 32 is satisfied with b~=0 and z~ as the all-zeros vector.

    2. (b)

      If k4 is even, and p=11k1, the distribution ν contains all vectors in {0,1}k of hamming-weights k2 and k in its support. In this case, Definition 32 is satisfied with b~=1 and z~ as the all-ones vector.

    The result now follows by Theorem 33.

  3. 3.

    Suppose that k<1+1min{p,1p} is a positive integer, and let ν𝒟(p,k). We perform the following operation on the distribution ν: if i,j[k],ij are such that PrXν[Xi=Xj]=1, we remove coordinates i,j from ν, and repeat until no such pairs remain.

    Finally, we are left with a distribution ν~ on k~k coordinates. We consider the following two cases:

    1. (a)

      Suppose that k~=0. In this case, for every n, and every f:{0,1}n{1,1}, it holds that 𝔼Xνn[i=1kf(Xi)]=1, since the k terms in the product cancel out in pairs. Hence, it suffices to show the existence of a function f:{0,1}n{1,1} satisfying |𝔼Xμpn[f(X)χS(X)]|on(1) for every S[n]. Note that a (uniformly) random function f:{0,1}n{1,1} satisfies this with high probability, by an argument similar to the one at the end of Section 4.2 (a random function can be thought of as rounding the constant zero function as in Section 4.2).

    2. (b)

      Now, suppose that k~0. Then, it holds that ν~𝒟(p,k~), and by Proposition 25, we have that ν~ has no pairwise independent coordinate. Now, by Theorem 24 there exists a constant α>0, such that for every large n, there exists a function f:{0,1}n{1,1} such that

      |𝔼(X1,,Xk)νn[i[k]f(Xi)]|=|𝔼(X1,,Xk~)ν~n[i[k~]f(Xi)]|α,

      and such that |𝔼Xμpn[f(X)χS(X)]|on(1) for every S[n].

6.1 A Corner Case

In the above proof, we leave the case of odd k5 and p=11k1. This turns out to be very interesting, and we discuss it next. For the remainder of this section, we fix such a k and p.

In this case, the pairwise independent distribution ν𝒟(p,k) constructed in Lemma 28, is supported on vectors of hamming weights 0 and k1 (and does not contain BLR as in Definition 32). In particular, for every xsupp(ν), it holds that i=1kxi=0 (mod k1). For this reason, as we show next, the best we can expect from the test Lin(ν), is to guarantee correlation with a character over /(k1), and this is indeed true.

Definition 30 (Characters over /(k1)).

Let ω be a primitive (k1)th root of unity. For every 0rk2, we define the function ϕr:{0,1} as ϕr(x)=ωrx.

For every n, and every integers 0r(1),,r(n)k2, we define the product character ϕr(1),,r(n):{0,1}n by ϕr(1),,r(n)(x)=j=1nϕr(j)(x(j))=ωj=1nr(j)x(j).

Now, consider the test Lin(ν). Observe that any character f=ϕr(1),,r(n) passes this test with probability 1:

𝔼Xνn[i[k]f(Xi)]=j=1n𝔼Yν[i[k]ϕr(j)(Yi)]=j=1n𝔼Yν[ωr(j)(i[k]Yi)]=1.

Next, we claim that characters explain the success of Lin(ν) for any function f:

Theorem 31.

For every constant ϵ>0, there exists a constant δ>0 such that for every large enough n, the following is true:

Let f:{0,1}n[1,1] be a function such that |𝔼Xνn[i=1kf(Xi)]|ϵ. Then, there exist integers 0r(1),,r(n)k2, such that

|𝔼Xμpn[f(X)ϕr(1),,r(n)(X)]|δ.

Proof.

The result follows from the work of Bhangale, Khot, Liu and Minzer [7, 8], and we omit the details. Very roughly speaking, the proof follows a similar strategy as in Section 7: first show that f has good correlation with a character under random restrictions; then, use this to show that f has good correlation with character times a low-degree function; finally, use that ν is pairwise independent to get rid of the low-degree function.

Finally, we present an alternative solution to deal with this corner case of odd k5 and p=11k1. Instead of the test Lin(ν), we can perform the following test:

Let f:{0,1}n[1,1], and let ν𝒟(1p,k)=𝒟(1k1,k) be the pairwise independent distribution from Lemma 28.

  1. 1.

    Sample X=(X1,,Xk)νn.

  2. 2.

    Let X be the vector obtained by negating each of the kn coordinates of X.

  3. 3.

    Query f on X1,,Xk and accept if and only if i[k]f(Xi)=1.

Each query Xi of the above test is distributed according to μpn, and the analysis of the test simply follows from the analysis for Lin(ν) in Theorem 5. The drawback here, though, is that the test does not accept all linear functions with probability 1, but only functions of the form (1)|S|χS, for S[n].

7 Analysis of the Linearity Test

In this section, we shall state and prove a generalized version of Theorem 7. The proof follows the work of Bhangale, Khot and Minzer [11], and hence we only give a rough outline (skipping many of the technical points), pointing out the places where the proof differs from the above work. We start with the following definition:

Definition 32.

Let k3,p(0,1), and let ν𝒟(p,k) be a distribution. We say that ν contains BLR, if there exists some b~{0,1},z~{0,1}k3, such that

{(x1,x2,x1x2b~,z~):x1,x2{0,1}}supp(ν){0,1}k.

Furthermore, for technical reasons, we shall also require that

span𝔽2(supp(ν))={x{0,1}k:i=1kxi=0 (mod 2)}.

Observe that any ν with full even-weight support contains BLR (with b~=0, and z~ the all-zeros vector). With this, we state the following generalization of Theorem 7:

Theorem 33.

Let k3 be a positive integer, and let p(0,1),ϵ(0,1] be constants, and let ν𝒟(p,k) be a distribution containing BLR (see Definition 32). Then, there exist constants δ>0,d (possibly depending on k,p,ϵ,ν), such that for every large enough n, the following is true:

Let f:{0,1}n[1,1] be a function such that

|𝔼(X1,,Xk)νn[i=1kf(Xi)]|ϵ.

Then, there exists a set S[n], and a polynomial g:{0,1}n of degree at most d and with 2-norm 𝔼Xμpn[g(X)2]1, such that

|𝔼Xμpn[f(X)χS(X)g(X)]|δ.

Moreover, if the distribution ν has some pairwise independent coordinate, then we may assume g1; that is, f correlates with a linear function χS.

The remainder of this section is devoted to the proof of the above theorem. Let k3 be an integer, and let p(0,1),ϵ(0,1] be constants, and let ν𝒟(p,k) be a distribution containing BLR (see Definition 32). Also, let f:{0,1}n[1,1] be a function such that

|𝔼X=(X1,,Xk)νn[i=1kf(Xi)]|ϵ. (1)

Step 1: Large Fourier Coefficient under Random Restriction

We note that the proof of this step is where we differ from [11].

Since the distribution ν𝒟(p,k) contains BLR, we can write ν=(1β)ν+βμ, for some small constant 0<β<12min{p,1p}, some distribution ν over {0,1}k, and with μ the uniform distribution over {(x1,x2,x1x2b~,z~):x1,x2{0,1}}, where b~,z~ are as in Definition 32. Using this, we can describe choosing Xνn as the following two step process. First choose a set I[n], denoted I1β[n], by choosing iI with probability 1β, independently for each i[n]. Then, choose ZνI and YμI¯, and set X=(Y,Z).

With the above, we can prove that the function f satisfies the property of having a large fourier coefficient under random restrictions; the reader is referred to [28] for an introduction to Fourier analysis over the hypercube.

Lemma 34.

With δ=ϵ/2, it holds that

PrI1β[n],ZνI[S[n]I:|fIZ1^(S)|δ]δ.

Here, fIZ1 refers to the restriction of the function f, with the variables in I set to Z1.

Proof.

By Equation 1, we have

ϵ |𝔼X=(X1,,Xk)νn[i=1kf(Xi)]|
=|𝔼I1β[n],ZνI𝔼YμI¯[i=1kfIZi(Yi)]|
𝔼I1β[n],ZνI|𝔼YμI¯[i=1kfIZi(Yi)]|

Observe that in the above expression, the random variables Y4,,Yk are constants (determined by z~). Now, using a (classical) Fourier analytic argument to analyze the BLR linearity test over the uniform distribution (see Chapter 1 of [28]), we get

ϵ 𝔼I1β[n],ZνI|𝔼YμI¯[i=13fIZi(Yi)]|
=𝔼I1β[n],ZνI|SI¯fIZ1^(S)fIZ2^(S)fIZ3^(S)(1)b~|S||
𝔼I1β[n],ZνI[maxSI¯|fIZ1^(S)|]
PrI1β[n],ZνI[SI¯:|fIZ1^(S)|ϵ/2]+ϵ/2.

Step 2: Direct Product Test

Using Theorem 1.1 in [11], by Lemma 34 we get the existence of constants d,δ>0, a set S[n], and a polynomial g:{0,1}n of degree at most d, and with 2-norm 𝔼Xμpn[g(X)2]1, such that

|𝔼Xμpn[f(X)χS(X)g(X)]|δ.

This proves the first part of Theorem 7. It remains to show that if ν has some pairwise independent coordinate, it is possible to remove the function g in the above expression.

Step 3: List Decoding

This step follows Section 4.2 and Section 4.3 in [11].

Using an iterative list-decoding process,777we remark that this is a subtle argument using different degree parameters for the different polynomials. we can find a constant r, and functions χS1,,χSr, and constant degree polynomials g1,,gr, such that it is possible to “replace” f by i[r]χSigi in Equation 1 (and lose at most some constant factor in ϵ). Now, this implies that for some constant ϵ>0, and some indices j1,,jk[r], we have

|𝔼(X1,,Xk)νn[i=1kχSji(Xi)gji(Xi)]|ϵ. (2)

We remark that for the next step, some extra structure on Sji’s is needed, and ensuring that it holds requires the condition on span𝔽2(supp(ν)) in Definition 32.

Step 4: Invariance Principle Argument

This step follows Section 4.4, Section 4.5, and Section 4.6 in [11].

Assume, for the sake of contradiction, that f is not correlated well with any χS; that is, 𝔼Xμpn[f(X)χS(X)]on(1) for each S[n]. Using this, it can be shown, roughly, that for each i[k], the expectation 𝔼Xμpn[χSji(X)gji(X)]on(1); note that for this conclusion to hold, we might have to modify Sji’s and gji’s, however it is possible to do so while maintaining Equation 2 – roughly, this follows by showing Sji’s are close to each other.

Now, by an invariance principle argument [27, 25, 26], very roughly,888technically speaking, this requires two extra steps: first, noise is applied to ensure function boundedness and distribution connectivity; second, a regularity lemma is used to make low-degree influences small. it is possible to replace the expectation in Equation 2 over (X1,,Xk)νn, by an expectation over (Z1,,Zk)𝒩(0,Σ)n, where Σk×k is the (normalized) covariance matrix of ν. Finally, we use that some coordinate Xi is pairwise independent of each Xi, for ii. Since the Gaussian distribution is determined by its covariance matrix, this implies that Zi is mutually independent of (Zi)ii. We have

ϵ |𝔼X=(X1,,Xk)νn[i=1kχSji(Xi)gji(Xi)]|
|𝔼Z=(Z1,,Zk)𝒩(0,Σ)n[i=1kχSji(Zi)gji(Zi)]|
|𝔼Zi𝒩(0,1)n[χSji(Zi)gji(Zi)]||𝔼Z[i[k],iiχSji(Zi)gji(Zi)]|
|𝔼Xiμpn[χSji(Xi)gji(Xi)]||𝔼Z[i[k],iiχSji(Zi)gji(Zi)]|
on(1),

which is a contradiction. ∎

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