The Maximum-Modulus Theorem

# The Maximum-Modulus Theorem

 Theorem 1 (The Maximum-Modulus Theorem: Let $A \subseteq \mathbb{C}$ be open and connected and let $f : A \to \mathbb{C}$ be analytic on $A$. Then either $f$ is constant on $A$ or $\mid f(z) \mid$ does not obtain a maximum value on $A$.

The proof below is carried out by contradiction. We will assume that $f$ is not constant and satisfies the hypotheses in the theorem which will imply $f$ being constant - our contradiction.

• Proof: Assume that there exists a $z_0 \in A$ such that for all $z \in A$ we have that:
(1)
\begin{align} \quad \mid f(z) \mid \leq \mid f(z_0) \mid \end{align}
• In other words, assume that the $\mid f \mid$ attains a maximum value at $z_0$. Define a new set $B$ as the set of $z \in A$ for which $f(z) = f(z_0)$:
(2)
\begin{align} \quad B = \{ z \in A : f(z) = f(z_0) \} \end{align}
• Then $B$ is a nonempty set since $z_0 \in B$.
• We will first show that the set $B$ is closed. Let $(z_n)_{n=1}^{\infty}$ be any sequence of complex numbers in $B$ such that $\displaystyle{\lim_{n \to \infty} z_n = w}$. We must show that $w \in B$. Now since $z_n \in B$ for all $n \in \mathbb{N}$ we must have that $f(z_n) = f(z_0)$ for all $n\in \mathbb{N}$. Since $f$ is analytic on $A$ this means that $f$ is continuous on $A$ and so by the sequential criterion for continuity we have that:
(3)
\begin{align} \quad \lim_{n \to \infty} f(z_n) = f(w) \end{align}
• But then this means that $f(z_0) = f(w)$ and so $w \in B$. Therefore every sequence of complex numbers in $B$ converges to a complex number $w \in B$, so $B$ contains all of its accumulation points and hence $B$ is closed.
• We will now show that the set $B$ is also open. Let $z^* \in B$. Since $B \subseteq A$ and $A$ is open there exists an $r^* > 0$ such that $D(z^*, r^*) \subseteq A$. We claim that $D(z^*, r^*) \subseteq B$. Suppose note, i.e., suppose that there exists a $z^{**} \in D(z^*, r^*)$ such that $z^{**} \not \in B$. Since $z^{**} \not \in B$ this means that $f(z^{**}) \neq f(z_0)$ and so:
(4)
\begin{align} \quad \mid f(z^{**}) \mid < \mid f(z_0) \mid \end{align}
• Now since $f$ is continuous on $A$, $\mid f \mid$ is continuous on $A$ and so there exists an $r^{**} > 0$ such that $D(z^{**}, r^{**}) \subseteq D(z^*, r^*)$ and such that for all $z \in D(z^{**}, r^{**})$ we have that $\mid f(z) \mid < \mid f(z_0) \mid$.
• Let $\gamma$ be the circle centered at $z^*$ with radius $R = \mid z^* - z^{**} \mid$. Then this circle can be parameterized as $\gamma(t) = z^* + Re^{it}$ for $t \in [0, 2\pi]$.
• Let $\Gamma$ be the part of the arc of the circle contained in $D(z^{**}, r^{**})$.
• Note that the circle is a closed, piecewise smooth and positively oriented curve in $A$ and since $f$ is analytic on $A$ we can apply Cauchy's Integral Formula to get that:
(5)
\begin{align} \quad \mid f(z^*) \mid &= \biggr \lvert \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z - z^*} \biggr \rvert \\ & \leq \frac{1}{2\pi} \int_{\gamma} \frac{\mid f(z) \mid}{\mid z - z^* \mid} \: \mid dz \mid \\ & \leq \frac{1}{2\pi} \int_{\Gamma} \frac{\mid f(z) \mid}{\mid z - z^* \mid} \: \mid dz \mid + \frac{1}{2\pi} \int_{\gamma \setminus \Gamma} \frac{\mid f(z) \mid}{\mid z - z^* \mid} \: \mid dz \mid \\ & < \frac{1}{2\pi} \int_{\Gamma} \frac{\mid f(z_0) \mid}{\mid z - z^* \mid} \: \mid dz \mid + \int_{\gamma \setminus \Gamma} \frac{\mid f(z_0) \mid}{\mid z - z^* \mid} \: \mid dz \mid \\ & < \frac{1}{2\pi} \int_{\gamma} \frac{\mid f(z_0) \mid}{\mid z - z^* \mid} \: \mid dz \mid \\ & < \frac{\mid f(z_0) \mid}{2\pi} \int_{\gamma} \frac{1}{\mid z - z^* \mid} \: \mid dz \mid \\ \end{align}
• Note that the circle $\gamma$ has equation $\mid z - z^* \mid = R$. So:
(6)
\begin{align} \quad \mid f(z^*) \mid < \frac{\mid f(z_0) \mid}{2\pi R} \int_{\gamma} \mid dz \mid \end{align}
• The length of the curve $\gamma$ is $2\pi R$ though, so $\displaystyle{\int_{\gamma} \mid dz \mid = 2\pi R}$. So:
(7)
• Once again we note that $z^* \in B$ so $\mid f(z^*) \mid = \mid f(z_0) \mid$. Hence $\mid f(z_0) \mid < \mid f(z_0) \mid$ - a contradiction. So $B$ is also open.
• Therefore $B$ is a nonempty closed and open subset of $A$. Since $A$ is connected this means that $B = A$ (since if not then $B$ would be a nonempty proper closed and open subset of $A$ which would imply that $A$ is disconnected). But since $B = A$ and $f$ is constant on $B$ this means that $f$ is constant on $A$ - a contradiction.
• Therefore if $f$ is not constant on $A$ then $\mid f(z) \mid$ does not attain a maximum on $A$. $\blacksquare$