Capacitary-Distance Hardy Inequality

Abstract

Let $n\ge 3$, $\Omega \subset {\mathbb{R}}^n$ be an open set, $F:={\mathbb{R}}^n\setminus \Omega$, and $\alpha \in (0,\infty )$. For any $x\in \Omega$, we define the capacitary distance \begin{align*} d_α(x) := \inf\left\{ r>0: \operatorname{cap}(\overline{F\cap B(x,r)}) \ge α\operatorname{cap}(B(\mathbf0,r)) \right\}. \end{align*} In this article, we prove that there exists a positive constant $C_n$, depending only on $n$, such that, for any $\alpha \in (0,1]$ and any $u\in C_{\mathrm{c}}^{\infty }(\Omega )$, \begin{align*} \int_Ω\frac{|u(x)|^2}{d_α(x)^2}\,d x \le \frac{C_n}{α^{2}} \int_Ω|\nabla u(x)|^2\,d x. \end{align*} This gives an affirmative answer to Problem 8 of Maz'ya [25]. Moreover, this dependence on $\alpha$ is sharp: there exists a positive constant $c_n$, depending only on $n$, such that, for every $\alpha \in (0,1]$, we are able to construct a bounded connected domain ${\Omega }_{\alpha }$ on which the optimal constant in the above Hardy inequality is at least $\frac{c_n}{{\alpha }^2}$. The proof combines a variable-time semigroup estimate for the killed Brownian motion with finite-time exit estimates derived from capacity.

21 pages