# Fourier transform on the discrete cube

Notation: identify an element of $\{-1,1\}^n$ with the set $S \subseteq \{1, \ldots, n\}$ on which it takes the value $-1$.

The following is an asymptotic question. "Close to one" means "more than $r_n$" and "away from $\frac{n}{2}$" means "outside the interval $[\frac{n}{2} - k_n\sqrt{n}, \frac{n}{2} + k_n\sqrt{n}]$", for some $r_n$ and $k_n$ which respectively increase to one and infinity as $n \to \infty$.

Conjecture: for any subset of $\{-1,1\}^n$ there is a complex valued function with $l^2$ norm equal to 1, supported either on that subset or on its complement, whose Fourier transform has $l^2$ norm close to one on $\{S: |S|$ is away from $\frac{n}{2}\}$.

A positive answer would have very interesting consequences. It would mean that a single subspace of $l^2(\{-1,1\}^n)$ whose dimension is small compared to the whole space is close to every subspace spanned by standard basis vectors or its complement.

I'm not familiar with the literature on Fourier analysis on the discrete cube. Is there anything there that would help settle this question?

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Nice question, Nik. Do you know what happen for a random subset of {-1,1}^n? –  Gil Kalai Jan 1 '13 at 12:12
@Gil: I don't know. Does a random subset approximately contain a face of dimension greater than $n/2$? I think the characteristic function of such a face has transform supported on small $S$ ... –  Nik Weaver Jan 1 '13 at 18:56
In a separate thread (mathoverflow.net/questions/117912/faces-in-the-discrete-cube) Anthony Quas informs me that typical subsets do not contain faces that big. So that idea doesn't work. –  Nik Weaver Jan 3 '13 at 1:10