I have studied rudin's Real and complex analysis, and have a question on the proof that why the level set $E$ is compact. Could you give me some hint?
11.13 Theorem If a continuous function $u$ has the mean value property in an open set $\Omega$, then $u$ is harmonic in $\Omega$.
PROOF It is enough to prove this for real $u$. Fix $D(a; R) \subseteq \Omega$. The Poisson integral gives us a continuous function $h$ on $D(a; R)$ which is harmonic in $D(a; R)$ and which coincides with $u$ on the boundary of $D(a; R)$. Put $v=u- h$,and let $m=\sup\{v(z):z \in D(a;R)\}$. Assume $m >0$, and let $E$ be the set of $\forall z \in D(a; R)$, at which $v(z) = m$. Since $v = 0$ on the boundary of $D(a; R)$, $E$ is a compact subset of $D(a; R).$
The function $v$ is continuous, so $v^{-1} (m)$ is a closed set. Moreover since $v = 0$ at the boundary and $m > 0$, $v^{-1}(m)$ does not intersect $\partial D(a;R)$. Thus $E = v^{-1}(m) \cap D(a; R)=v^{-1}(m) \cap \overline{D(a;R)}$, and $E$ is closed. Since it's bounded, it's compact, too.