Half-plane capacity of a low rectangle
= Half-plane capacity of a low rectangle
For $R_r=[-r,r]\times(0,1]$ with $r\geq1$,
$$
\operatorname{hcap}(R_r)\leq Cr.
$$
Indeed, the imaginary part at the Brownian exit point is at most one and the probability of reaching the radius-$O(r)$ neighbourhood containing the rectangle from $iy$ is $O(r/y)$. Consequently $r^{-1}R_r=[-1,1]\times(0,r^{-1}]$ has capacity $O(r^{-1})$ although its diameter tends to two.