高阶Hardy常数在晶格上的精确渐进行为
Sharp asymptotics for higher-order Hardy constants on lattices
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中文总结 AI 辅助
本文研究了晶格上高阶Hardy不等式最优常数的渐进行为，通过傅里叶变换和迭代方法推导出极限常数为$2^\ell$，揭示了高维情况下最优解的局部化特性。
中文摘要 AI 辅助
我们研究了晶格$\mathbb{Z}^d$上高阶Hardy不等式中的最优常数。对于每个固定的$\ell\in\mathbb{N}$，我们证明了最优常数$\mathcal{C}_\text{opt}^\ell(d)$在不等式$$\sum_{n\in\mathbb{Z}^d}|\Delta^{\ell/2}u(n)|^2\geq\mathcal{C}_\text{opt}^\ell(d)\sum_{n\in\mathbb{Z}^d}\frac{|u(n)|^2}{|n|^{2\ell}}$$中满足$$\lim_{d\rightarrow\infty}\frac{\mathcal{C}_\text{opt}^\ell(d)}{d^\ell}=2^\ell.$$证明基于将傅里叶变换转化为平滑紧致紧致 тор 上的一组奇异Hardy不等式，涉及权重$$\omega(x)^{-2\ell},\quad\omega(x)^2=\sum_{j=1}^d\sin^2\left(\frac{x_j}{2}\right),$$以及可接受函数的零平均条件。我们通过结合基态表示公式和加权积分Bochner恒等式，在迭代方案中建立了这些torus不等式。该方法得到显式常数，定义为递归顺序$\ell$，仅需在torus上使用经典的无权Poincaré不等式。极限常数$2^\ell$的出现尤为引人注目，因为它表明在高维情况下，最优解集中在晶格$\mathbb{Z}^d$中的单位球$\{n\in\mathbb{Z}^d:|n|=1\}$附近。
英文摘要
We study the optimal constants in higher-order Hardy inequalities on the lattice $\mathbb{Z}^d$. For each fixed $\ell \in \mathbb{N}$, we prove that the optimal constant $\mathcal{C}_\text{opt}^\ell(d)$ in $$ \sum_{n \in \mathbb{Z}^d} |Δ^{\ell/2}u(n)|^2 \geq \mathcal{C}_\text{opt}^\ell(d)\sum_{n \in \mathbb{Z}^d} \frac{|u(n)|^2}{|n|^{2\ell}}. $$ satisfies $$ \lim_{d\rightarrow\infty}\frac{\mathcal{C}_\text{opt}^\ell(d)}{d^\ell} =2^\ell. $$ The proof is based on a Fourier reduction to a family of singular Hardy inequalities on the flat torus, involving the weight \[ ω(x)^{-2\ell}, \qquad ω(x)^2=\sum_{j=1}^d\sin^2\left(\frac{x_j}{2}\right), \] and zero average condition on admissible functions. We establish these torus inequalities by combining a ground state representation formula with a weighted integrated Bochner identity in an iterative scheme. The method yields explicit constants, defined recursively in the order $\ell$, and requires only the classical unweighted Poincaré inequality on the torus. The appearance of the limiting constant $2^\ell$ is particularly striking, as it suggests that, in the high dimensional regime, the optimizers are localized near the unit sphere $\{n\in\mathbb{Z}^d:|n|=1\}$ in $\mathbb{Z}^d$.
