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Fall 2014 - Problem 12

subharmonic functions

Let $\Omega = \set{z \in \C \mid \abs{z} > 1}$ . Suppose $\func{u}{\cl{\Omega}}{\R}$ is bounded and continuous on $\cl{\Omega}$ and that it is subharmonic on $\Omega$ . Prove the following: if $u\p{z} \leq 0$ for all $\abs{z} = 1$ then $u\p{z} \leq 0$ for all $z \in \Omega$ .

Solution.

Let $\epsilon > 0$ and consider $v_\epsilon\p{z} = u\p{z} + \epsilon\log\,\abs{\frac{1}{z}}$ , which is subharmonic as a sum of two subharmonic functions ( $\log{\abs{\frac{1}{z}}}$ is harmonic away from $0$ ). Observe that because $u$ is bounded by, say, $M > 0$ ,

v_\epsilon\p{z} \leq M + \epsilon\log\,\abs{\frac{1}{z}} \xrightarrow{\abs{z}\to\infty} -\infty.

Thus, let $R > 0$ be big enough so that $v_\epsilon\p{z} \leq 0$ on $\abs{z} \geq R$ . Notice also that $v_\epsilon\p{z} = u\p{z} \leq 0$ on $\abs{z} = 1$ , so because $v_\epsilon$ is subharmonic on the annulus $\set{1 < \abs{z} < R}$ , the maximum principle tells us that $v_\epsilon\p{z} \leq 0$ on all of $\Omega$ . Since $\epsilon$ was arbitrary, sending $\epsilon \to 0$ , we see that $u\p{z} \leq 0$ on all of $\Omega$ , which completes the proof.