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Fall 2013 - Problem 4

subharmonic functions

Let $u$ be a non-negative continuous function on $\cl{\D} \setminus \set{0} = \set{z \in \C \mid 0 < \abs{z} \leq 1}$ that is subharmonic on $\D \setminus \set{0}$ . Suppose that $\res{u}{\partial\D} \equiv 0$ and

\lim_{r\to0^+} \frac{1}{r^2\log\,\p{1/r}} \int_{\set{0 < \abs{z} < r}} u\p{z} \,\diff\lambda\p{z} = 0,

where integration is with respect to Lebesgue measure $\lambda$ on $\C$ . Show that then $u \equiv 0$ .

Solution.

Notice that if $a < b$ , then

\log{a} \leq \log{b} \implies \frac{1}{\log\,\p{1/a}} \leq \frac{1}{\log\,\p{1/b}}.

Let $R \in \p{0, 1}$ so that $\cl{B\p{0, R}} \subseteq \D$ . Thus, given any $z \in B\p{0, \frac{R}{2}}$ , we have $\cl{B\p{z, \frac{R}{4}}} \subseteq B\p{0, R}$ and so non-negativity of $u$ and the sub-mean value property give

\begin{aligned} \frac{u\p{z}}{\log\,\p{1/\abs{z}}} &\leq \frac{1}{\log\,\p{1/\abs{z}}} \frac{1}{\pi \p{R/4}^2} \int_{B\p{z,\frac{R}{4}}} u\p{w} \,\diff\lambda\p{w} \\ &\leq \frac{16}{\pi R^2\log\,\p{1/R}} \int_{B\p{0, R}} u\p{w} \,\diff\lambda\p{w} \xrightarrow{R\to0^+} 0, \end{aligned}

by assumption. Set $f\p{z} = \log\frac{1}{\abs{z}}$ and for $\epsilon > 0$ , let $u_\epsilon\p{z} = u\p{z} - \epsilon f\p{z}$ which is subharmonic on $\D \setminus \set{0}$ since $u$ is subharmonic and $f$ is harmonic.

u_\epsilon\p{z} = f\p{z}\p{\frac{u\p{z}}{f\p{z}} - \epsilon} \xrightarrow{\abs{z}\to0} -\infty

by our previous calculation. Thus, for each $\epsilon$ , there exists $r_\epsilon > 0$ such that $u_\epsilon \leq 0$ on $\cl{B\p{0, r_\epsilon}} \setminus \set{0}$ . By the maximum principle, $u_\epsilon$ attains its maximum on $\partial\D \cup \partial B\p{0, r_\epsilon}$ . Since $u_\epsilon\p{z} = 0$ for $z \in \partial\D$ and $u_\epsilon\p{z} \leq 0$ for $z \in \partial B\p{0, r_\epsilon}$ , it follows that $u_\epsilon \leq 0$ on all of $\D \setminus \set{0}$ .

Letting $\epsilon \to 0$ , we see that $u\p{z} \leq 0$ on all of $\D \setminus \set{0}$ , and so $u \equiv 0$ , since $u\p{z} \geq 0$ .