Lecture 8: The Maximum Modulus Principle

In this lecture, we study two important properties of holomorphic functions: the mean value property and the Maximum Modulus Principle.

The mean value property

Let $\Omega\subseteq\mathbb C$ be an open set, and $f(z)$ be holomorphic in $\Omega$ . For any $\mathcal B_r(z_0)$ such that $\overline{\mathcal B_r(z_0)}\subset\Omega$ , we have Cauchy’s Integral Formula $\label{EQ:Mean Value Property 1} f(z_0) = \frac{1}{2 \pi i} \int_{\mathcal B_r(z_0)} \frac{f(z)}{z - z_0} dz = \frac{1}{2 \pi i} \int_{0}^{2 \pi} i r e^{i \theta} \frac{f(z_0 + r e^{i \theta})}{r e^{i \theta}} d\theta = \frac{1}{2 \pi} \int_{0}^{2 \pi} f(z_0+ r e^{i \theta})d \theta.$ This result says that the value of $f$ at $z_0$ equals the average of the value of the function on the circle of radius $r$ centered at $z_0$ . What is even more interesting is that this is valid for a circle of any radius, as long as $\mathcal B_r(z_0)$ is still inside $\Omega$ (so tha Cauchy’s Integral Formula can still hold).

This important mean value propoerty of holomorphic function can be explored a little further. Indeed, [EQ:Mean Value Property 1] allows us to obtain in a straighforward manner that $\iint_{\mathcal B_r(z_0)} f(z) dx dy=\int_0^{2\pi}\int_0^r f(z_0+ r' e^{i\theta})r' dr' d\theta=\pi r^2 f(z_0)\,.$ Therefore, we have $\label{EQ:Mean Value Property 2} f(z_0)=\dfrac{1}{\pi r^2} \iint_{\mathcal B_r(z_0)} f(z) dx dy\,.$ This result says that the value of a holomorphic function at a point $z_0$ equals the average of the function value in any disk centered at the point. We summarize it into the following theorem.

Theorem 1 (Mean Value Property). Let $\Omega\subseteq\mathbb C$ be an open set, and $f(z)$ be holomorphic in $\Omega$ . Then for any $\mathcal B_r(z_0)$ such that $\overline{\mathcal B_r(z_0)}\subset\Omega$ , we have [EQ:Mean Value Property 1] and [EQ:Mean Value Property 2].

The Maximum Modulus Principle

Holomorphic functions have the following maximum principle.

Theorem 2 (Maximum Modulus Principle). Let $\Omega\subseteq \mathbb C$ be a connected open set, and $f(z)$ a holomorphic function on $\Omega$ . Then
(i) Given $z_0\in\Omega$ , we have that $|f(z_0)|=\sup_{z\in\Omega}|f(z)| \implies f\ \mbox{is constant in $\Omega$}\,.$ (ii) If $\Omega$ is bounded and $f(z)\in C(\overline\Omega)$ , then $\sup_{z\in\overline\Omega}|f(z)|=\sup_{z\in\partial\Omega} |f(z)|\,.$

Proof. We first prove part (i). (a) Let $z_0\in\Omega$ be such that $|f(z_0)|=\sup_{z\in\Omega}|f(z)|$ . Denote $\mathcal O:=\{z'\in\Omega \mid f(z')=f(z_0)\}\,.$ Then clearly $z_0\in\mathcal O$ . Since $f$ is continuous ( $f$ is continuous iff preimages of closed subsets under $f$ are closed), we conclude that $\mathcal O$ is a closed subset of $\Omega$ and is nonempty.

(b) Next, if $z'_0\in\mathcal O$ , there is a disk of radius $r$ , $\overline{B_r(z_0')}\subset\Omega$ and $f(z_0')=\dfrac{1}{\pi r^2} \iint_{B_r(z_0')} f(z) dx dy\,.$ The fact that $|f(z_0')|\ge |f(z)|$ for all $z\in B_r(z_0')$ forces $f(z_0')= f(z), \ \ \ \forall z\in B_r(z_0')\,.$ Therefore $\mathcal O$ is an open subset of $\Omega$ .

(c) The fact that $\mathcal O$ is an open and nonempty closed subset of $\Omega$ . Using the assumption that $\Omega$ is connected, we conclude that $\mathcal O=\Omega$ .

To prove (ii), we use the fact that continuous functions achieve supremum on compact domains. By the result in part (i), if the supremum of $|f|$ is achieved in the interior, then $|f|$ is a constant. Continuity then implies that $|f|$ on $\partial\Omega$ takes the same constant. Therefore, the supremum of $|f|$ can only be achieved on the boundary $\partial\Omega$ . ◻

The Maximum Modulus Principle can be used to prove the following result.

Theorem 3 (Schwarz Lemma). Let $f(z)$ be analytic on the unit disk $\mathbb D:=\{z \in\mathbb C\mid |z|<1\}$ such that: (i) $f(0=0$ , and (ii) $|f(z)|\le 1$ on $\mathbb D$ . Then $|f(z)|\le |z|$ for all $z\in\mathbb D$ . Moreover, $|f'(0)|\le 1$ .

Proof. Let us define the function $g(z)$ as $g(z)=\left\{\begin{matrix}\frac{f(z)}{z}, & z\neq 0 \\ f'(0), & z=0\end{matrix}\right.$ Then $g(z)$ is analytic on $\mathbb D$ . For any $r<1$ , let $\mathbb D_r:=\{z\in\mathbb C\mid |z|\le r\}$ . By the Maximum Modulus Principle, $|g(z)|\le |g(z_r)|=\frac{|f(z_r)|}{|z_r|}\le \frac{1}{|z_r|}=\frac{1}{r}$ for any $z\in\mathbb D_r$ , where $z_r$ denotes a point on the boundary of $\mathbb D_r$ . We then have that $|g(z)|\le 1$ when $r\to 1$ . Therefore, $|g(z)|\le 1$ for any $z\in\mathbb D$ . This finishes the proof. ◻

Liouville’s Theorem

Liouville’s Theorem says that bounded entire functions (that is, functions that are defined and analytic on the whole complex plane $\mathbb C$ ) have to be constant.

Theorem 4 (Liouville’s Theorem). Let $f(z): \mathbb C\to \mathbb C$ be entire and bounded. Then $f$ is a constant.

Proof. For any $z\in\mathbb C$ , we have by Cauchy’s Derivative Formula that for any $R\in(0, \infty)$ , $f'(z)=\dfrac{1}{2\pi i}\int_{\partial B_R(z)} \dfrac{f(\xi)}{(\xi-z)^2}d\xi\,.$ We can then parameterize $B_R(z)$ by $\xi(t)=z+Re^{it}$ , $0\le t\le 2\pi$ to have $f'(z)=\dfrac{1}{2\pi i}\int_0^{2\pi} \dfrac{f(z+Re^{it})}{R^2e^{2it}} iRe^{it} dt = \dfrac{1}{2\pi R}\int_0^{2\pi} f(z+Re^{it}) e^{-it} dt\,.$ If $|f(z)|\le M$ for all $z\in\mathbb C$ , the above formula gives that $|f'(z)|\le \dfrac{M}{R}$ . Taking $R\to \infty$ will give us that $f'(z)=0, \ \ \forall z\in\mathbb C\,.$ Therefore $f$ is constant. The proof is complete. ◻

Guided review and additional examples

Learning goals

Apply the maximum modulus principle on bounded regions.
Distinguish maximum-modulus statements from minimum-modulus statements.
Derive Liouville’s theorem from Cauchy estimates.
Use boundary information to control an analytic function in the interior.

Concept connection: a nonconstant analytic function cannot peak inside

If $f$ is nonconstant and analytic on a connected open set, $|f|$ has no local maximum in the interior. For a bounded region and a function continuous on its closure, the global maximum is therefore found on the boundary. A minimum statement requires $f$ to be nonzero so that the principle can be applied to $1/f$ .

Worked example 1: maximize $|z^2+1|$ on the unit disk

The polynomial $f(z)=z^2+1$ is analytic, so its maximum modulus on $|z|\le1$ occurs on $|z|=1$ . There,

$|z^2+1|\le |z|^2+1=2.$

Equality occurs at $z=1$ and $z=-1$ , where $z^2=1$ . Thus the maximum is $2$ . Notice that the minimum is $0$ at $z=\pm i$ ; zeros prevent a positive minimum-modulus conclusion.

Worked example 2: why bounded entire functions are constant

Suppose $|f(z)|\le M$ on all of $\mathbb C$ . Cauchy’s estimate on a circle of radius $R$ about $z_0$ gives

$|f'(z_0)|\le\frac{M}{R}.$

Letting $R\to\infty$ yields $f'(z_0)=0$ . Since $z_0$ was arbitrary, $f$ is constant. This is Liouville’s theorem.

Check your understanding

May a nonconstant analytic function attain its maximum modulus at a boundary point?

Show the answer

Yes. The principle rules out an interior maximum for a nonconstant function; on a compact closure the maximum is expected on the boundary.