Theorem 27.2.4 (Maximum Modulus Theorem).label Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open and connected, then
- (1)
For each $f \in H(U; E)$ and continuous seminorm $[\cdot]_{E}: E \to [0, \infty)$, if there exists $z_{0} \in U$ with
\[[f(z_{0})]_{E} = \sup_{z \in U}[f(z)]_{E}\]then $[f]_{E}$ is constant.
- (2)
For any $f \in H(U; \complex)$, if there exists $z_{0} \in U$ with
\[|f(z_{0})| = \sup_{z \in U}|f(z)|\]then $f$ is constant.
Proof. (1): Let $M = \sup_{z \in U}[f(z)]_{E}$. Since $[f]_{E}$ is continuous, $\bracsn{[f]_E = M}$ is closed. By Cauchy’s Integral Formula and Proposition 13.4.1, $\bracsn{[f]_E = M}$ is open. As $U$ is connected, $\bracsn{[f]_E = M}= U$.
(2): By (1), $|f|$ is constant. After rescaling, assume without loss of generality that $\ol f = 1/f$. In this case, for each $z \in U$,
and
so $Df = 0$ and $f$ is constant.$\square$
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