An $m^{2.943}$ Bohnenblust--Hille Bound on the Boolean Cube

Abstract

Let $q_m=2m/(m+1)$ and put \[ \beta_0=\frac{3}{2}+\frac{1}{\log 2}=2.9426950408\ldots, \] where $\log$ is the natural logarithm. We give a proof scheme showing that, for every $\varepsilon>0$, there is $C_\varepsilon<\infty$ such that every complex-valued function $f:\{-1,1\}^n\to\C$ of Fourier degree at most $m$ satisfies \[ \left(\sum_{A\subseteq[n]}\abs{\wh f(A)}^{q_m}\right)^{1/q_m} \le C_\varepsilon m^{\beta_0+\varepsilon}\norm{f}_\infty. \] The improvement over the $m^9$ estimate of the earlier draft has two ingredients. The first three Fourier levels are estimated at the scales $m$, $m^{3/2}$, and $m^{5/3}$. These losses are encoded in the weight $M^{\mu_r/r}r^B\min\{r,L_M\}^8$, $L_M\asymp_B \log (M+1)$, with $\mu_r=\lceil3r/2\rceil$ for $r\ge2$. A parity-compatible central window handles $r<L_M$, while a shrinking balanced window handles $r\ge L_M\asymp_B\log(M+1)$, no parity is used in this regime. The leading high-degree loss--gain factor is $\ee\,2^{-B}$; careful uniform bounds close the bootstrap for every $B>1/\log2$. The formula for $\mu_r$ is explained below.