The maximum modulus principle from the Cauchy integral formula

We prove the strong local form of the maximum modulus principle. Let \(G\) be a region, let \(f\in H(G)\), and suppose that \(|f|\) has a local maximum at \(z_0\in G\). We show that \(f\) is constant.

Choose \(r>0\) so small that \(\overline{D(z_0,r)}\subset G\) and \[ |f(z)|\le |f(z_0)| \qquad (z\in \overline{D(z_0,r)}). \] If \(f(z_0)=0\), then \(f\equiv0\) on this disk and hence on \(G\) by the identity theorem. Thus assume \(M:=|f(z_0)|>0\), and multiply \(f\) by the unimodular constant \(\overline{f(z_0)}/M\). This does not change its modulus, so we may assume \(f(z_0)=M>0\).

For every \(0<\rho<r\), Cauchy’s integral formula gives the mean-value identity \[ f(z_0)=\frac1{2\pi}\int_0^{2\pi} f(z_0+\rho e^{it})\,dt. \] Taking real parts, \[ M=\frac1{2\pi}\int_0^{2\pi} \Re f(z_0+\rho e^{it})\,dt. \] But for every \(t\), \[ \Re f(z_0+\rho e^{it}) \le |f(z_0+\rho e^{it})|\le M. \] The integrand is continuous and bounded above by \(M\), while its average is exactly \(M\); therefore it is identically \(M\). Hence on the circle \(|z-z_0|=\rho\) we have both \(\Re f(z)=M\) and \(|f(z)|\le M\). This forces \(f(z)=M\) at every point of that circle.

Thus \(f-M\) has infinitely many zeros with an accumulation point in \(G\), so the identity theorem gives \(f\equiv M\) on \(G\). Consequently a nonconstant analytic function cannot attain a local maximum of its modulus at an interior point. In particular, on any bounded domain where \(f\) is continuous up to the boundary, the maximum of \(|f|\) is attained on the boundary.