Maximum modulus principle

Suppose \(f\) is holomorphic on \(\Omega\). If \(f\) has a relative maximum at \(z_0\in\Omega\), then \(f\) is constant in a neighborhood of \(z_0\). If \(\Omega\) is a bounded connected domain with \(f\) continuous on \(\Omega\) and \({{\partial}}\Omega\), then either \(f\) is constant or \(M \coloneqq\max_{z\in \Omega}{\left\lvert {f(z)} \right\rvert}\) is only attained by some \(z\in {{\partial}}\Omega\).

Use that \(z\mapsto {\left\lvert {z} \right\rvert}\) is an open map away from \(z=0\). If \(f\) is holomorphic, by the open mapping theorem it is an open map. If \(f\) attains a maximum at an interior point \(z_0\), then there is some neighborhood \(U\ni z_0\) where \({\left\lvert {f(U)} \right\rvert}\) is open in \({\mathbf{R}}\) – but such an interval contains values larger than \({\left\lvert {f(z_0)} \right\rvert}\), contradicting maximality at \(z_0\).

Let \(z_0\in\Omega\) and pick any \(R\) such that \({\mathbb{D}}_R(z_0) \subseteq \Omega\). We have \begin{align*} f(z_0) &= {1\over 2\pi i}\oint_{{\left\lvert {\xi-z_0} \right\rvert} = R} {f(\xi) \over \xi-z_0} \,dz\\ &= {1\over 2\pi i}\int_0^{2\pi} {f(Re^{it} + z_0) \over Re^{it} } iRe^{it} \,dt\\ &= {1\over 2\pi} \int_0^{2\pi } f(Re^{it} + z_0) \,dt ,\end{align*} so \begin{align*} {\left\lvert {f(z_0)} \right\rvert} \leq {1\over 2\pi}\int_0^{2\pi }{\left\lvert {f(Re^{it} + z_0 )} \right\rvert} \,dt\leq \max_{t \in [0, 2\pi]} {\left\lvert {f(Re^{it} + z_0) } \right\rvert} .\end{align*} Setting \(z_R\) to be the point \(Re^{it} + z_0\) that maximizes this last term, we have \(f(z_0) \leq f(z_R)\). Since this holds for all \(R\), this implies \(f(z_0) = f(z_R)\) for every \(R\), making \(f\) constant on \({\mathbb{D}}_R(z_0)\). By the identity principle \(f\) is constant on \(\Omega\).

Suppose \(f\) is holomorphic and nonvanishing on \(\Omega\). If any interior point \(z_0\in \Omega^\circ\) is a relative minimum for \(f\), then \(f\) is constant. If \(f\) is nonconstant, then the minimum must occur on \({{\partial}}\Omega\).

Suppose \(f\neq 0\) on \(G\). If \(f(z) = 0\) for some \(z\in {{\partial}}G\), we’re done, so suppose \(f\neq 0\) on \(\overline{G}\). Then \(1/f\) is holomorphic on \(G\) and continuous on \(\overline{G}\), so \(\max_{z\in \overline{G}}{\left\lvert {1/f(z)} \right\rvert} = \max_{z\in {{\partial}}G} {\left\lvert {1/f(z)} \right\rvert}\).


Suppose \(f\) is continuous and nonzero on \(\overline{{\mathbb{D}}}\) and holomorphic on \({\mathbb{D}}\). Show that if \({\left\lvert {f(z)} \right\rvert} = {\left\lvert {z} \right\rvert}\) for all \({\left\lvert {z} \right\rvert} = 1\) then \(f\) is constant.


If \(f\) has no zeros in \({\mathbb{D}}\), apply the MMP to \(1/f\) to get \({\left\lvert {f} \right\rvert} = 1\) on all of \({\mathbb{D}}\). By Cauchy-Riemann (or the open mapping theorem), if \({\left\lvert {f} \right\rvert}\) is constant, \(f\) is constant.

Let \(f: \Omega\to {\mathbf{C}}\) be holomorphic and suppose there is a \(z_0 \in \Omega\) with \({\left\lvert {f(z_0)} \right\rvert}\leq {\left\lvert {f(z)} \right\rvert}\) for all \(z\in \Omega\). Show that either \(f(a) = 0\) or \(f\) is constant.


Suppose \(f(z_0)\neq 0\), then the inequality forces there to be no zeros in \(\Omega\). So \(g(z) \coloneqq 1/f(z)\) is nonzero and holomorphic on \(\Omega\) and \({\left\lvert {g(z)} \right\rvert}\leq {\left\lvert {1\over f(z_0)} \right\rvert} \coloneqq{\left\lvert {g(z_0)} \right\rvert}\). Since \(z_0\in \Omega\), the MMP forces \(g\) to be constant, and thus so is \(f\).

Show that if \(\Re(f(z)) \geq 0\) for all \(z\in {\mathbf{C}}\), then \(f\) is constant.


Define \(g(z) \coloneqq e^{-f(z)}\), then \begin{align*} {\left\lvert {g(z)} \right\rvert} = e^{-\Re(f(z))} \leq e^0 = 1 .\end{align*} Since \(g\) is entire and bounded, \(g\) is constant and thus so is \(f\).

Show that if \(f\) is constant on any closed disk \(r\overline{{\mathbb{D}}}\) for \(r<1\), then \(f\) is constant on \({\mathbb{D}}\).


By MMP applied to \(r\overline{{\mathbb{D}}}\), since \(f\) achieves its maximum on the interior \(r{\mathbb{D}}\), \(f\) is necessarily constant.

Suppose \(f\) is entire and \(f^{(n)}\) is bounded on \({\mathbf{C}}\). Show that \(f\) is a polynomial of degree at most \(n\).


By Liouville or MMP, \(f^{(n)}\) is bounded and entire and thus constant. Integrating a constant \(n\) times yields a polynomial of degree t most \(n\).

Find all harmonic functions \(u:{\mathbb{D}}\to {\mathbf{C}}\) such that \(u(1/2) = 2\) and \({\left\lvert {u(z)} \right\rvert}\geq 2\) for all \({\left\lvert {z} \right\rvert} \leq 1\).


Since \(1/2 \in \left\{{{\left\lvert {z} \right\rvert}\leq 1}\right\}\) is an interior point of \({\mathbb{D}}\), by the MMP if \(u\) is nonconstant then \(u(1/2) = 2\) can not be a relative maximum for \(u\) on \({\mathbb{D}}\).