Spring 2020 - Problem 12

Solution.

Let $\epsilon > 0$, and observe that if we set $u_\epsilon\p{z} = u\p{z} - \epsilon\log\,\abs{z}$, then $u_\epsilon$ is still subharmonic and

for $\abs{z} \geq R$ for some $R > 0$, by assumption. Thus, by the maximum principle, it follows that $u_\epsilon\p{z} \leq 0$ on $\abs{z} \leq R$, hence on all of $\C$. Sending $\epsilon \to 0$, we see that $u\p{z} \leq 0$ on $\C$. Thus,

so by the maximum principle applied to a large annular region, we see that $u_\epsilon\p{z}$ must attain its maximum in $\partial\D$. Thus, by the maximum principle applied to the unit disk,

since $u$ and $u_\epsilon$ agree on $\partial\D$. Hence, for $\abs{z} \geq 1$,

Sending $\epsilon \to 0$ again, we see that $u\p{z}$ must attain its maximum in $\cl{\D}$ by continuity, so by the maximum principle again, $u$ must be constant.