Condenser capacity

Remarks

When $\Omega$ is bounded and its boundary and that of $K$ are regular, the infimum is attained in a larger class by the potential $u$ of the condenser. This is the harmonic function on $\Omega \setminus K$ with $u = 1$ on $\partial K$ and $u = 0$ on $\partial \Omega$.

Example: in $\R^3$ let $0 < r < R$, $K = \bar B_r$ and $\Omega = B_R$. The potential is $u(x) = (1/\abs{x} - 1/R) / (1/r - 1/R)$ for $r \le \abs{x} \le R$, and $\abs{\nabla u} = \abs{x}^{-2} / (1/r - 1/R)$. Integration over spheres gives $4 \pi \int_r^R s^{-2} \, ds / (1/r - 1/R)^2 = 4 \pi / (1/r - 1/R)$. So $\operatorname{cap}(\bar B_r, B_R) = 4 \pi r R / (R - r)$. In the plane the same computation with $u(x) = \log(R/\abs{x}) / \log(R/r)$ gives $\operatorname{cap}(\bar B_r, B_R) = 2 \pi / \log(R/r)$.

Non-example: if $K$ is compact but not contained in $\Omega$, the pair is not a condenser. No $u \in C_c^\infty(\Omega)$ can be at least $1$ at a point of $K$ outside $\Omega$.

Some authors normalize by a constant, such as $1/(4\pi)$ in $\R^3$ or $1/(2\pi)$ in the plane, or define the capacity with $\abs{\nabla u}^n$, which makes it invariant under conformal maps.