 Solution

Use normalized Fourier-Walsh transform coefficients $\hat{g}(\xi )=2^{-n}{\sum }_{x}g(x)(-1{)}^{x\cdot \xi }$. The Hamming weight $|\xi |$ counts the nonzero coordinates. Norms on the input cube use uniform probability, and a frequency integral here is counting over $\xi \in {\mathbb{F}}_2^n$. A common alternative normalization multiplies both sides of the requested comparison by the same factor and does not change it.

Beckner's inequality states that for $1<p\le q<\infty$ and $0\le \rho \le \sqrt{(p-1)/(q-1)}$, the noise operator on the Boolean hypercube, with coefficients $\widehat{T_{\rho }g}(\xi )={\rho }^{|\xi |}\hat{g}(\xi )$, satisfies $\Vert T_{\rho }g{\Vert }_q\le \Vert g{\Vert }_p$. In particular choose $q=2$, $p=4/3$, $\rho =1/\sqrt{3}$.

Let $g=1_A$ and $\alpha =|A|/2^n$. Then $\Vert g{\Vert }_{4/3}^2={\alpha }^{3/2}$. On frequencies of weight at most two, ${\rho }^{2|\xi |}\ge 1/9$, so Beckner's inequality and Parseval's identity give the low-degree Fourier mass of a sparse Boolean set estimateThe indicator function is real and the cube characters are real, so the displayed squares equal the squared absolute values. Total Fourier weight is $\sum \hat{g}(\xi {)}^2=\Vert g{\Vert }_2^2=\alpha$. If $2^n>2003$, then $0<\alpha \le 1/2003$ and $9\sqrt{\alpha }\le 9/\sqrt{2003}<1/2$, since $2003>18^2$. ThereforeThis establishes the strict comparison for every power $N=2^n>2003$, not merely for an unspecified very large threshold. The original PDF supplies the density condition and Beckner request, both missing from the converted TeX.