Subharmonic functions in the punctured disk

Question

I want to prove the following (exercise from Ahlfors' text):

If $\Omega$ is the punctured disk $0<|z|<1$ and if $f$ is given by $f(\zeta)=0$ for $|\zeta|=1$, $f(0)=1$, show that all functions $v \in \mathfrak B(f)$ are $\leq0$ in $\Omega$.

Here $\mathfrak B(f)$ is the class of all functions $v$ which satisfy:

(a) $v$ is subharmonic in $\Omega$.

(b) $\limsup_{z \to \zeta} v(z) \leq f(\zeta)$ for all $\zeta \in \partial \Omega$.

After quite some time with no success, a short google search gave me this , which has some hints, but I need to get the full book in order to understand the references.

Any help on solving this problem will be highly appreciated!

Thanks!

Answer 1

Given an arbitrary $v \in \mathfrak{B}(f)$, consider the functions

$$u_\varepsilon(z) = v(z) + \varepsilon\cdot \log \lvert z\rvert$$

for $\varepsilon > 0$. Since $\log \lvert z\rvert$ is harmonic in the punctured disk, $u_\varepsilon$ is subharmonic. Since $\limsup\limits_{z\to 0} v(z) < \infty$, we have $\lim\limits_{z\to 0} u_\varepsilon(z) = -\infty$. Also, $\log \lvert z\rvert \equiv 0$ on the unit circle.

By the maximum principle⁽¹⁾, $u_\varepsilon \leqslant 0$ in the punctured disk. Let $\varepsilon \to 0$.

⁽¹⁾ If we had $u_\varepsilon(z_0) = c > 0$ for some $z_0 \in \Omega$, then $u_\varepsilon$ would attain a global maximum in $\Omega$, hence be constant ($\equiv -\infty$):

Since $\limsup_{z\to \zeta} \leqslant 0$ for all $\zeta\in \partial \Omega$, each $\zeta\in\partial\Omega$ has an open neighbourhood $U_\zeta$ with $u_\varepsilon \leqslant c/2$ on $U_\zeta$.

$$K = \Omega \setminus \bigcup_{\zeta\in\partial\Omega} U_\zeta$$

is then a compact subset of $\Omega$ containing $z_0$, in particular non-empty. The upper semicontinuous (or even continuous) function $u_\varepsilon\lvert_K$ attains its maximum in $K$, say at $z_1$. Then $u_\varepsilon$ attains its global maximum in $\Omega$ in the point $z_1$.