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$\newcommand{\R}{\mathbb{R}}$ This question is about Problem 3.16 in Ryan O'Donnell's Analysis of Boolean Functions book. The problem is stated as follows:

Let $f : \{-1,1\}^n\to\R$ and let $\epsilon>0$. Show that $f$ is $\epsilon$-concentrated on a collection $F\subseteq 2^{[n]}$ with $|F|\leq \hat{||}f\hat{||}_1^2/\epsilon$.

Here, $\hat{||}f\hat{||}_1 := \sum_{S\subseteq [n]} |\widehat{f}(S)|$, and $f$ being $\epsilon$-concentrated on $F$ means $\sum_{S\notin F} \widehat{f}(S)^2 \leq \epsilon$. I have tried several approaches for this problem, and all seem to run into the pitfall of relating $\hat{||}f\hat{||}_1^2$ to $||f||_2$ (equivalently $\hat{||}f\hat{||}_2$ by Parseval's) in any nontrivial way. The most fruitful approach was using a probabilistic method: randomly sampling $|\widehat{f}(S)|$ uniformly over $S$, bounding the variance of this random variable, and applying Chebyshev's. But since $\mathbf{E}[|\widehat{f}(S)|]$ might be large and Chebyshev's only dictates deviation from the expectation, I don't think this approach yields the desired concentration bound. Would appreciate any hints or insight!

Edit: I was eventually able to find a solution. As a hint, start by constructing a reasonable set $F$ which trivially satisfies $|F|\leq \hat{||}f\hat{||}_1^2/\epsilon$. To prove $\epsilon$-concentration, recall the fairly simple technique used to bound $\sum \widehat{g}(S)^3$ in the analysis for the BLR test (Section 1.6 in Analysis of Boolean Functions).

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2 Answers 2

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Hint: find a random variable $X=X(S)$ such that $\mathbf{E}[X]=\hat{||}f\hat{||}_1$ where the expectation is under the spectral sample. Use this to apply Markov to find an appropriate set with Fourier mass at least $1-\epsilon$, and then argue that the size of the set must be at most the claimed cardinality.

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  • $\begingroup$ Thanks for responding, just have a quick question. This approach seems to be assuming that the function is Boolean valued (so that total Fourier mass is equal to 1). Does scaling by $||f||_2^2$ give the same result for real-valued functions? $\endgroup$
    – Ash
    Commented Mar 25, 2023 at 15:45
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Here is a direct construction: $$F = \left\{S:|\hat{f}(S)|\ge \frac{\epsilon}{\hat{\|}f\hat{\|}_1}\right\}$$ And you can verify the two conditions.

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