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Suppose $U$ is a bounded open subset of the plane and $f$ is continuous on the closure of $U$ and holomorphic on $U$. Show that if $C \geq 0$ and $\begin{vmatrix} f(z) \end{vmatrix} \leq C$ for all $z$ on the boundary of $U$, then $\begin{vmatrix} f(z) \end{vmatrix} \leq C$ for $z \in U$.

This problem seems to scream Max Modulus Principle, but I am not sure how to use it here. I am not sure how, a non-constant function cannot attain its maximum is even applicable here.

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Here would be my main argument, leaving the details and punchline for you:

Since $f$ is continuous on $\bar{U} = U \cup \partial U$, there must be a point $z_0 \in \bar{U}$ such that $|f(z_0)| = \max_{z \in \bar{U}}|f(z)|$ (since the continuous image of a compact set will be compact). If $z_0 \in U$, then $|f(z)| \leq |f(z_0)|$ for all $z \in U$. By MMP, this implies that $f$ is constant on $U$. Otherwise, $z_0 \in \partial U$.

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  • $\begingroup$ Thanks, so the closure of $U$ is compact since it is closed and bounded? That is why you mention the image of a compact set is compact? $\endgroup$ Sep 23, 2016 at 2:37
  • $\begingroup$ Correct, since $\bar{U}$ is closed and bounded (and we're working in the Euclidean space $\mathbb{R}^2$ topologically), it is compact, and since $z \mapsto |f(z)|$ is continuous, then $f$ must map $\bar{U}$ to a compact subset of $\mathbb{R}$ and therefore achieve its maximum. $\endgroup$
    – Tom
    Sep 23, 2016 at 2:38
  • $\begingroup$ Great, thanks for the clarity and hints. $\endgroup$ Sep 23, 2016 at 2:40

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