Prove that maximum principle for harmonic functions, i.e.
If \(u\) is a non-constant real-valued harmonic function in a region \(\Omega\), then \(u\) can not attain a maximum or a minimum in \(\Omega\).
Suppose \(\Omega\) is a region with compact closure \(\bar \Omega\). If \(u\) is harmonic in \(\Omega\) and continuous in \(\bar \Omega\), then \[ \sup _{z \in \Omega}|u(z)| \leq \sup _{z \in \bar \Omega -\Omega}|u(z)| .\]
Hint: to prove (a), assume \(u\) attains a local maximum at \(z_0\). Let \(f\) be holomorphic near \(z_0\) with \(\Re(f) = u\), and show that \(f\) is not an open map. Then (a) implies (b).
Solution
For (a), suppose \(u\) attains a local maximum at \(z_0\in\Omega\). On a small disk \(D\Subset\Omega\) centered at \(z_0\), choose a harmonic conjugate \(v\), so \(F=u+iv\) is holomorphic on \(D\). If \(u(z)\le u(z_0)\) on \(D\), then \[ F(D)\subset\{w:\operatorname{Re}w\le u(z_0)\}, \] while \(F(z_0)\) lies on the boundary line of that half-plane. Thus \(F(D)\) is not open at \(F(z_0)\). By the open mapping theorem, \(F\) must be constant on \(D\), so \(u\) is constant on \(D\). Since \(u_x-i u_y\) is holomorphic on \(\Omega\), vanishing on the open set \(D\) forces it to vanish on the connected region \(\Omega\). Hence \(u\) is constant on \(\Omega\), contrary to the hypothesis. Therefore a nonconstant harmonic function has no interior local maximum. Applying the same argument to \(-u\) excludes an interior local minimum.
For (b), continuity on the compact set \(\overline\Omega\) implies that \(u\) attains its maximum and minimum there. By (a), unless \(u\) is constant, both extrema occur on \(\partial\Omega=\overline\Omega\setminus\Omega\). Thus \[ \sup_{z\in\Omega}|u(z)| \le \sup_{z\in\partial\Omega}|u(z)|. \] The same inequality is immediate when \(u\) is constant. Hence \[ \boxed{ \sup_{z\in\Omega}|u(z)| \le \sup_{z\in\overline\Omega-\Omega}|u(z)|.} \]