 Solution

First derive the Brownian representation of half-plane capacity with the chosen normalization. SetThe height comparison from part (a), applied to $f_K=g_K^{-1}$, gives $h\ge 0$. On the boundary, $\mathrm{Im}{g}_K$ tends to zero; on $K$ the limiting value of $h$ is therefore $\mathrm{Im}z$, and on the real boundary it is zero. To see the asserted boundary-height limit without assuming a smooth hull, suppose a bounded sequence approaching the boundary had mapped images with imaginary part bounded away from zero. The inverse's expansion near infinity prevents these images from being unbounded, and an interior subsequential limit would map back to an interior point of $D$, a contradiction.

At infinity $h$ tends to zero by the Laurent series. The maximum principle for harmonic functions now bounds $h$ between zero and $R=\mathrm{rad}(K)$, since every point of $K$ has imaginary part at most $R$. Apply the optional sampling theorem to the bounded harmonic function $h(B_{t\wedge T})$. The exit time $T$ is almost surely finite because it is no larger than the first time the imaginary coordinate of planar Brownian motion reaches zero. Taking $t\to \infty$ givesThis uses the bounded harmonic difference, not optional sampling of the unbounded imaginary coordinate at an unbounded stopping time.

At $z=iy$, the expansion gives $yh(iy)\to a_K$. Sincewe obtainThus the quadratic half-plane capacity is controlled by a length times the linearly scaling harmonic capacity from infinity in the upper half-plane. The factor $1/\pi$ is essential if the alternative normalization $\mathrm{cap}(K)=\pi c(K)$ is used.