$2$-capacity of a set in $\mathbb{R}^n$

Question

Let $B_r \subset \mathbb{R}^n$ denote the ball of radius $r$ centered at the origin. Consider any set $F \subset B_1$. For all sets $\Omega \subset \mathbb{R}^n$ such that $F \subset \Omega$, define $2$-capacity by

$$cap_2 (F, \Omega) = \inf_{\phi|_F = 1, \phi \in C^\infty_0(\Omega)}\int_\Omega |\nabla \phi|^2 dV$$

It is clear that $cap_2(F, B_2) \geq cap_2(F, B_3)$.

My question is, how do we prove that $cap_2(F, B_3) \geq C\; cap_2(F, B_2)$, with a constant $C$ independent of $F$?

Answer 1

Let $F$ be a diffeomorphism between $B_2$ and $B_3$ that is equal to the identity map on $B_1$. Such a diffeomorphism is easily constructed as a radial map $F(x)=g(|x|)x/|x|$ where $g(t) = t$ for $t\le 1$, and $g(2)=3$. Specific form doesn't matter, as long as both $F$ and $F^{-1}$ have uniformly bounded derivatives.

If $\phi$ is a function from the definition of $\operatorname{cap}_2(F,B_3)$, then $\phi\circ F$ qualifies for $\operatorname{cap}_2(F,B_2)$. The chain rule and the change of variable formula yield $$ \int_{B_2} |\nabla (\phi\circ F)|^2 \le C \int_{B_3} |\nabla \phi|^2 $$ where $C$ depends only on $F$. The claim follows.