Capacitary-Distance Hardy Inequality

Abstract

Let $n\ge3$, $\Omega\subset\mathbb R^n$ 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_\alpha(x) := \inf\left\{ r>0: \operatorname{cap}(\overline{F\cap B(x,r)}) \ge \alpha\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 $u\in C_{\rm{c}}^\infty(\Omega)$, \begin{align*} \int_\Omega \frac{|u(x)|^2}{d_\alpha(x)^2}\,d x \le \frac{C_n}{\alpha^{2}} \int_\Omega|\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.