Capacity of a disc and a segment

Example

Let $D_r$ be the closed disc of radius $r > 0$ and let $S_L$ be a closed segment of length $L > 0$ in $\C$. Their logarithmic capacities are $$\operatorname{cap}(D_r) = r, \qquad \operatorname{cap}(S_L) = L/4.$$

Check for the disc, with the Chebyshev constant. Translate the disc so that its centre is $0$; capacity does not change under translations. For a monic polynomial $p(z) = z^n + a_{n-1} z^{n-1} + \cdots + a_0$, Parseval's identity on the circle $\abs{z} = r$ gives $$\frac{1}{2\pi} \int_0^{2\pi} \abs{p(r e^{i\theta})}^2 \, d\theta = r^{2n} + \sum_{k < n} \abs{a_k}^2 r^{2k} \ge r^{2n}.$$ So $\max_{D_r} \abs{p} \ge r^n$, with equality for $p = z^n$. Hence the Chebyshev constant is $r$, and by the equality of capacity and Chebyshev constant the capacity is $r$.

Check for the segment. Move it by a rotation and a translation to $[-L/2, L/2]$, which does not change capacity. For $[-1, 1]$, the monic Chebyshev polynomial $2^{1-n} T_n$ has maximum $2^{1-n}$ on the interval, and no monic polynomial of degree $n$ has a smaller maximum there. So the Chebyshev constant of $[-1, 1]$ is $\lim_n (2^{1-n})^{1/n} = 1/2$. Capacity scales linearly, so the segment of length $L$ has capacity $(L/2) \cdot (1/2) = L/4$.

The two results show that capacity is not a measure of area. The segment has area zero but positive capacity $L/4$. The disc of diameter $L$ has capacity $L/2$.