# 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 $$\max_{z\in \overline{G}}{\left\lvert {1/f(z)} \right\rvert} = \max_{z\in {{\partial}}G} {\left\lvert {1/f(z)} \right\rvert}$$.

# Exercises

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 $${\mathbb{D}}$$.

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