Let $\Omega \subset R^n$ a open set and $p >1$. Let $K \subset \Omega$ a compact set. Define $$A:= \{ u \in C^{\infty}_{0} (\Omega) ; \ u \geq 1 \text{ on } K\}$$

$$B:= \{ u \in C^{\infty}_{0} (\Omega) ; \ u \geq 1 \text{ on } \partial K\}$$

Show that

$$\inf_{u \in A} \int_{\Omega} |\nabla u |^p \ dx = \inf_{u \in B} \int_{\Omega} |\nabla u |^p \ dx$$

A good time ago i tried to prove this and i only proved the obvious inequality (first inf $\geq$ second inf)...

Someone can give me a hint ?

• Equivalence of two definitions of capacity would be a far better title. Think of this question being in the archive alongside hundreds of questions about infima and suprema in real analysis. – 40 votes Jul 16 '13 at 21:00

Given $u\in B$, observe that $u$ is Lipschitz on an open set containing $K$. So is $\tilde u=\max(u,1)$, and this truncation does not increase the norm of gradient. Next, let $v=1$ on the interior of $K$, and $v= \tilde u$ otherwise. The Lipschitz condition survives. Therefore, $v\in W^{1,p}_0(\Omega)$. Let $\psi$ be a compactly supported smooth function with small $W^{1,p}$ norm such that $\psi >0$ on $K$. Mollifying $v+\psi$, we get a function $w\in C_0^\infty$ such that $w>1$ on $K$ and $\|w-\tilde u\|_{W^{1,p}}$ is small. Since $w\in A$, and $\int|\nabla w|^p$ is not much larger than $\int|\nabla u|^p$, the statement follows.