Show that if $|f(z)| \leq M$ for $z \in \partial D$ for $z \in \mathbb{C}$ and M being a constant, then $|f(z)| \leq M$ for all $z \in D$

Question

We are given that $f$ is analytic in a bounded domain $D$ and continuous on the boundary $\partial D$. Also we can assume that $|f(z)| \leq M$ for $z \in \partial D$ for $z \in \mathbb{C}$.

I'm unsure what it means to be continuous on the boundary $\partial D$, but here is what I have thought up so far:

Since $f$ is analytic in the bounded domain $D$ we have that $f$ is continuously differentiable at some point $z \in D$. Since $f$ is continuously differentiable at $z$ we have that $f'$ is also continuous at $z$. As $f$ is continuous on the boundary $\partial D$ we have by the definition of $\epsilon - \delta$ continuity that for every $\epsilon >0$ there exists a $\delta >0$ such that $|z-z_0|<\delta \implies |f(z)-f(z_0)|<\epsilon$. We are also allowed to assume that $|f(z)| \leq M$ for $z \in \partial D$ for $z \in \mathbb{C}$. I'm not sure where to head with this problem, or if I have set up what was given correctly.

I think since $|f(z)-f(z_0)|<\epsilon$ we can rewrite this as $||f(z)|-|f(z_0)|| \leq \epsilon$ and so $|f(z)| \geq -\epsilon + |f(z_0)|$ where we then have that $M \geq -\epsilon +|f(z_0)|$. But, I don't think this leads in any particular direction.

Answer 1

This turned out a little more longwinded than I had hoped.

Suppose $f$ is analytic at $z^* \in D$. Since $D$ is open there is some $r>0$ such that $A=B(z^*,r) \subset \overline{B(z^*,r)} \subset D$. (This nitpick is just so that $f$ is defined on a set containing the closure of $A$.)

Let $R= \{ |f(z)| \}_{z \in A} $ and $\mu = \sup R$. ($|f|$ is bounded on the compact set $\overline{A}$ so $\mu$ is finite.)

If $\mu \in R$, then $|f|$ has a local $\max$ and hence $f$ is constant. Otherwise $|f(z)| < \mu$ for all $z \in A$.

If $f$ is constant then it is clear that $|f(z)| \le M$ for $z \in D$.

Hence we can assume that $f$ is non constant.

Let $\lambda = \sup_{z \in D} |f(z)|$. Then the above shows that $|f(z)| < \lambda$ for all $z \in D$ (show by contradiction).

Now note that $\lambda = \max_{z \in \overline{D}} |f(z)|$ and that the $\max$ is attained for some $z^* \in \overline{D}$ (using compactness).

From the second last sentence we see that $z^* \notin D$ and so $z^* \in \partial D$ and hence $\lambda \le M$.

Answer 2

“$f$ is analytic in a bounded domain $D$ and continuous on the boundary $\partial D$” is a somewhat imprecise formulation of the following:

• $D \subset \Bbb C$ is a bounded domain (i.e. a bounded, open, and connected set),
• $f: \overline D \to \Bbb C$ is a continuous function,
• the restriction of $f$ to $D$ is an analytic function.

In other words: $f$ is analytic in a bounded domain $D$ and has a continuous extension to $\overline D = D \cup \partial D$.

Then you can argue as follows: $|f|$ is continuous on the compact set $\overline D$, therefore it attains its maximum at some point $z_0 \in \overline D$. Now consider two cases:

• $z_0 \in \partial D$: Then $|f(z_0)| \le M$ according to the hypothesis, and we are done.
• $z_0 \in D$: Then $|f|$ has a local maximum at $z_0$. It follows from the maximum modulus principle that $f$ is constant, and we are done as well.

Remark: The condition “$f$ is analytic in a bounded domain $D$ and continuous on the boundary $\partial D$” is imprecise because it could be interpreted as

• The restriction of $f$ to $D$ is analytic, and
• the restriction of $f$ to $\partial D$ is continuous.

But then the statement is obviously wrong (take for example $D$ as the unit disk, $f(z) = z$ in $D$, and $f(z) = 0$ on $\partial D$).