- Prove the maximum principle for harmonic functions, that is:
If u is a non-constant real-valued harmonic function in a region \(\Omega ,\) , then u cannot attain a maximum (or a minimum) in Ω.
Suppose that Ω is a region with compact closure \(\overline { { \Omega } } .\) If u is harmonic in \(\Omega\) and continuous in \({ \overline { { \Omega } } } ,\) then
\[ \sup _ {z \in \Omega} | u (z) | \leq \sup _ {z \in \overline {{\Omega}} - \Omega} | u (z) |. \]
[Hint: To prove the first part, assume that u attains a local maximum at \(z _ { \mathrm { 0 } }\) . Let f be holomorphic near \(z _ { 0 }\) with \(u = { \mathrm { R e } } ( f )\) , and show that \(f\) is not open. The second part follows directly from the first.]
Solution
Let \(u\) be harmonic and suppose it has a local maximum at \(z_0\). On a small disc \(D\) about \(z_0\), choose a harmonic conjugate \(v\), so that \[ F=u+iv \] is holomorphic on \(D\). Then \[ |e^{F(z)}|=e^{u(z)}\le e^{u(z_0)}=|e^{F(z_0)}| \] near \(z_0\). Thus the holomorphic function \(e^F\) attains a local maximum of its modulus, so by the maximum modulus principle it is constant. Hence \(F\) is constant and therefore \(u\) is constant on \(D\). Now suppose \(u\) attains its maximum value \(M\) over \(\Omega\) at \(z_0\), and let \(A\) be the set of points of \(\Omega\) near which \(u\equiv M\). The argument just given shows that every point where \(u=M\) lies in \(A\); so \(A\) is open, nonempty, and closed in \(\Omega\) because \(u\) is continuous. Since \(\Omega\) is connected, \(A=\Omega\) and \(u\) is constant. Thus a nonconstant harmonic function cannot attain an interior maximum. Applying this to \(-u\) gives the corresponding statement for minima.
Since \(\overline\Omega\) is compact and \(u\) is continuous there, \(|u|\) attains a maximum at some point of \(\overline\Omega\). If the maximum were attained at an interior point and were strictly larger than the boundary maximum, then either \(u\) or \(-u\) would attain an interior maximum, forcing \(u\) to be constant by part (a). In that constant case the same value occurs on the boundary. Therefore in all cases \[ \sup_{z\in\Omega}|u(z)|\le \sup_{z\in\overline\Omega\setminus\Omega}|u(z)|. \]