Spring 2011 - Problem 8

Solution.

1. $f$ is upper-semicontinuous if $\set{z \mid f\p{z} < a}$ is open for any $a \in \R$.

2. An upper-semicontinuous function $\func{u}{\C}{\pco{-\infty, \infty}}$ is subharmonic if

   $$u\p{z} \leq \frac{1}{2\pi} \int_0^{2\pi} u\p{z + re^{i\theta}} \,\diff\theta$$

   for any $z \in \C$ and $r > 0$.

3. The first one is true as long as the supremum is upper-semicontinuous. Let $\set{u_\alpha}_\alpha$ be such a family and denote the pointwise supremum by $u$.

   Given $z \in \C$, there exists a sequence $\set{u_n}_n$ in this family such that $u_n\p{z} \to u\p{z}$. Moreover, if $M$ is a bound for the family, then $M + u_\alpha \geq 0$ for any $\alpha$, so we may apply Fatou's lemma:

   $$\begin{aligned} M + u\p{z} &= \limsup_{n\to\infty}\, \p{M + u_n\p{z}} \\ &\leq \limsup_{n\to\infty} \frac{1}{2\pi} \int_0^{2\pi} \p{M + u_n\p{z + re^{i\theta}}} \,\diff\theta \\ &\leq \frac{1}{2\pi} \int_0^{2\pi} \limsup_{n\to\infty}\, \p{M + u_n\p{z + re^{i\theta}}} \,\diff\theta \\ &\leq \frac{1}{2\pi} \int_0^{2\pi} M + u\p{z + re^{i\theta}} \,\diff\theta \\ &= M + \frac{1}{2\pi} \int_0^{2\pi} u\p{z + re^{i\theta}} \,\diff\theta \\ \implies u\p{z} &\leq \frac{1}{2\pi} \int_0^{2\pi} u\p{z + re^{i\theta}} \,\diff\theta, \end{aligned}$$

   so $u$ is subharmonic.

   The second one is false: consider $u_a\p{z} = \Re\p{e^{az}}$ indexed by $a \in \C$. Then $\inf_a u_a\p{z} = -\infty$ for $z \neq 0$, but $u_a\p{0} = 1$ for all $a \in \C$. Then if $u = \inf_a u_a$,

   $$1 = u\p{z} \geq -\infty = \frac{1}{2\pi} \int_0^{2\pi} u\p{z + re^{i\theta}} \,\diff\theta,$$

   so $u$ is not subharmonic.

4. Notice that because $\log{x}$ is continuous,

   $$\log\p{\norm{A\p{z}}} = \sup_{\norm{w} = 1}\, \log\p{\norm{A\p{z}w}} = \sup_{\norm{w} = \norm{\zeta} = 1}\, \log\,\abs{\inner{A\p{z}w, \zeta}}$$

   Since $A\p{z}$ is holomorphic, we see that $\inner{A\p{z}w, \zeta}$ is a composition of holomorphic functions, hence holomorphic. Thus, $z \mapsto \log\p{\norm{A\p{z}}}$ is the logarithm of the absolute value of an analytic function, hence subharmonic.