AI 中文总结
本文研究实数域上对称 Kahane--Salem--Zygmund 常数的精确阶，利用 Walsh 谱和 Gram 永久式刚性证明了下界 $\sqrt{2/e}\sqrt{m!}/m$ 与上界 $C_0\sqrt{m!}$，并给出截断 Hadamard 矩阵的矩形估计。
AI 中文摘要
Kahane--Salem--Zygmund 不等式提供了具有小上确界范数的单模 $m$ 线性形式。我们考虑实数标量域上的无维对称常数 $C_m^{\mathrm{sym}}$，其定义为：对于每个 $n$，存在某个实数对称单模 $m$ 线性形式 $A:(\ell_\infty^n)^m\to\mathbb R$ 满足 \\[ \\|A\\|\le C_m^{\mathrm{sym}}n^{(m+1)/2}. \\] 对于复数标量，Boas 在置换对称性下获得了阶至多为 $\sqrt{m\log m}\\,\sqrt{m!}$ 的上界；在实数标量域上，一个初等论证给出了更优的阶 $\sqrt m\\,\sqrt{m!}$。我们证明 \\[ C_m^{\mathrm{sym}}\ge \left(\sqrt{\frac2e}+o(1)\right)\frac{\sqrt{m!}}m, \\] 利用对角多项式的无平方 Walsh 谱。在相反方向上，我们建立 \\[ C_m^{\mathrm{sym}}\le C_0\sqrt{m!}, \\] 其中 $C_0$ 是绝对常数，从实数上界中去掉了因子 $\sqrt m$。该估计来自一个几何论证，其中 Gram 永久式的刚性将相关构型简化为由高斯宽度控制的集合。对于无限制的单模形式，我们还从截断 Hadamard 矩阵获得了定量矩形估计；在等维情形下，当 $m=o(n^{5/6})$ 时，归一化最小值至多为 $1+o(1)$。
英文摘要
The Kahane--Salem--Zygmund inequality provides unimodular $m$-linear forms with small supremum norm. We consider its dimension-free symmetric constant $C_m^{\mathrm{sym}}$ over the real scalar field, defined by the requirement that, for every $n$, some real symmetric unimodular $m$-linear form $A:(\ell_\infty^n)^m\to\mathbb R$ satisfy \[ \|A\|\le C_m^{\mathrm{sym}}n^{(m+1)/2}. \] For complex scalars, Boas obtained under permutation symmetry an upper bound of order at most $\sqrt{m\log m}\,\sqrt{m!}$; over the real scalar field, an elementary argument gives the sharper order $\sqrt m\,\sqrt{m!}$. We prove \[ C_m^{\mathrm{sym}}\ge \left(\sqrt{\frac2e}+o(1)\right)\frac{\sqrt{m!}}m, \] using the square-free Walsh spectrum of the diagonal polynomial. In the opposite direction, we establish \[ C_m^{\mathrm{sym}}\le C_0\sqrt{m!}, \] where $C_0$ is absolute, removing the factor $\sqrt m$ from the real upper bound. This estimate follows from a geometric argument in which rigidity for Gram permanents reduces the relevant configurations to sets controlled by Gaussian width. For unrestricted unimodular forms, we also obtain a quantitative rectangular estimate from truncated Hadamard matrices; in equal dimension, the normalized minimum is at most $1+o(1)$ whenever $m=o(n^{5/6})$.
