# Maximum Modulus Principle

Let $$u(z)$$ be a real-valued harmonic function on a domain $$D$$ such that $$u(z) \leq M$$ for all $$z \in D$$. If $$u\left(z_{0}\right)=M$$ for some $$z_{0} \in D$$, then $$u(z)=M$$ for all $$z \in D$$.

The idea of the proof is to use the mean value property to show that the set of points for which $$u(z)=M$$ is open. Indeed, suppose $$u\left(z_{1}\right)=M$$, and express the mean value equality (4.1) in the form \begin{align*} 0=\int_{0}^{2 \pi}\left[u\left(z_{1}\right)-u\left(z_{1}+r e^{i \theta}\right)\right] \frac{d \theta}{2 \pi}, \quad 0 Since the integrand is nonnegative $$(\geq 0)$$ and continuous, the integral (5.1) can be zero only if the integrand is zero. Thus $$u\left(z_{1}+r e^{i \theta}\right)=u\left(z_{1}\right)=M$$ for $$0 \leq \theta \leq 2 \pi$$ and $$0<r<\rho$$, and the set $$\{u(z)=M\}$$ contains a disk centered at each of its points, hence is open. Now, the set $$\{u(z)<M\}$$ is also open, since $$u(z)$$ is continuous. Since $$D$$ is a domain, one of these two sets is empty and the other coincides with all of $$D$$. (See Section II.1.) In other words, either $$u(z)<M$$ for all $$z \in D$$, or $$u(z)=M$$ for all $$z \in D$$, and this proves the theorem.

Let $$h$$ be a bounded complex-valued harmonic function on a domain $$D$$. If $$|h(z)| \leq M$$ for all $$z \in D$$, and $$\left|h\left(z_{0}\right)\right|=M$$ for some $$z_{0} \in D$$, then $$h(z)$$ is constant on $$D$$.

This can be derived easily from the real version of the strict maximum principle. We replace $$h(z)$$ by $$\lambda h(z)$$ for an appropriate unimodular constant $$\lambda$$, and we can assume that $$h\left(z_{0}\right)=M$$. Let $$u(z)=\operatorname{Re} h(z)$$. Then $$u(z)$$ is a harmonic function on $$D$$ that attains its maximum at $$z_{0}$$. By the strict maximum principle for real-valued harmonic functions, $$u(z)=M$$ for all $$z \in D$$. Since $$|h(z)| \leq M$$ and $$\operatorname{Re} h(z)=M$$, we must have $$\operatorname{Im} h(z)=0$$ for $$z \in D$$. Hence $$h(z)$$ is constant on $$D$$.

Let $$h(z)$$ be a complex-valued harmonic function on a bounded domain $$D$$ such that $$h(z)$$ extends continuously to the boundary $$\partial D$$ of $$D .$$ If $$|h(z)| \leq M$$ for all $$z \in \partial D$$, then $$|h(z)| \leq M$$ for all $$z \in D$$.

Use the extreme value theorem: a continuous function on a compact space attains a maximum absolute value. But then if $$h$$ attains its maximum modulus, by the strict maximum principle above, $$h$$ is constant.

If $$f$$ is nonzero on $$\Omega$$, then $$f$$ attains a minimum on $${{\partial}}\Omega$$. This follows from applying the MMP to $$1/f$$.

## Exercises

Show that if $$f$$ is holomorphic on $${\mathbb{D}}\setminus\left\{{0}\right\}$$ and $${\left\lvert {f(z)} \right\rvert} \leq \log\qty{1\over {\left\lvert {z} \right\rvert}}$$, then $$f\equiv 0$$.

• Claim: $$f$$ has a removable singularity at $$z_0=0$$.
• Thus $$f$$ extends to some holomorphic $$F$$ defined on $${\mathbb{D}}$$.
• By continuity, $${\left\lvert {F} \right\rvert}$$ satisfies the same inequality as $$f$$.
• Now $$\lim_{{\left\lvert {z} \right\rvert} \to 1}{\left\lvert {F(z)} \right\rvert}\leq \lim_{{\left\lvert {z} \right\rvert}\to 1} \log\qty{1\over {\left\lvert {z} \right\rvert}} = 0$$, so by the MMP on $${\mathbb{D}}$$, $$F\equiv 0$$ on $${\mathbb{D}}\setminus\left\{{0}\right\}$$.