Analysis Problem of the Day 67

Today’s problem appeared as Problem 12 on UCLA’s Fall 2024 Analysis Qual:

Problem 12. Let $U \subseteq \mathbb{C}$ be a bounded open set and $u: U \to \mathbb{R}$ be a continuous subharmonic function.

a) Show that if $M$ is such that

$\limsup_{z \to z_0} u(z) \leq M$

for all $z_0 \in \partial U,$ then $u(z) \leq M$ for all $z \in U.$

b) Show that if the above condition holds for all except possibly finitely many points $z_0 \in \partial U,$ then the conclusion still holds.

Solution: a) Suppose $u(a)>v(a)$ for some $a \in U$ and $v \equiv M$ harmonic. Then, the set $\{b \in U: (u-v)(b)>0\}$ is nonempty. Since $u-v$ is upper-semicontinuous, it achieves its maximum at some point $p \in U$ (which is not on the boundary since $\limsup_{z \to \partial U} (u-v)(z) \leq 0$ ) and the set of points $A$ where the maximum is achieved is closed in $U.$ Moreover, since $u-v$ is subharmonic, by the sub-mean value property, it follows that around $a \in A,$ $u-v$ has to be locally constant, i.e. the set of points where the maximum is achieved is open. It follows that $u-v>0$ is constant on each connected component of $U,$ which contradicts the fact that $\limsup_{z \to z_0} u(z) \leq M.$

b) The trick in this part is inspired by the trick used to prove that a subharmonic function on $\mathbb{C}$ that is bounded above is constant. Namely, one takes the subharmonic function $h(z)=\log |z|,$ and for points $z_1,...,z_n \in \partial U$ for which the inequality in (a) does not hold, one considers

$g_\epsilon(z):=u(z) + \epsilon \sum_{i=1}^n h(\epsilon(z-z_i)).$

As a sum of subharmonic functions, $g_\epsilon$ is subharmonic. Moreover, since $u$ is bounded from above, it follows that $\limsup_{z \to \partial U} g_\epsilon(z) \leq M,$ i.e. $g_\epsilon$ satisfies the hypotheses in (a). Thus, $g_\epsilon \leq M$ on $U.$ But now, notice that for $\epsilon < \frac{1}{2 \text{diam}(U)},$ $h(\epsilon(z-z_i))<0$ on $U.$ Moreover, $\lim_{\epsilon \to 0}\epsilon h(\epsilon(z-z_i)) =0.$ It follows that $u=\sup_\epsilon g_\epsilon,$ and since $g_\epsilon \leq M$ everywhere, one obtains $u \leq M$ on $U.$

Leave a Reply Cancel reply