 Solution

For $0<a<x<b$, put $S={\tau }_a\wedge {\tau }_b$. The process $\varphi (\Vert B_{t\wedge S}\Vert )$, where $\varphi (r)=r^{2-d}$, is a bounded martingale by part (a). Also $S<\infty$ almost surely: before exit the endpoint $B_t$ lies inside the radius-$b$ ball, whose Gaussian probability tends to zero as $t\to \infty$. Path continuity makes the terminal radius either $a$ or $b$. If $p={\mathbb{P}}_{\bar{x}}({\tau }_a<{\tau }_b)$, the dominated convergence theorem applied to this bounded stopped martingale givesSolving yieldsThe denominator is nonzero because $\varphi$ is strictly decreasing for $d\ge 3$.

As $b\uparrow \infty$, the events $\{{\tau }_a<{\tau }_b\}$ increase to $\{{\tau }_a<\infty \}$. In one direction membership already implies a finite hit of $a$; in the other direction a continuous path up to its finite hit of $a$ has a finite maximum radius, so it lies in the event for every sufficiently large $b$. Continuity from below of probability and $\varphi (b)\to 0$ now give