# Simple Inequality for holomorphic functions

Let $g:D \longrightarrow \mathbb{C}$ be holomorphic on an open set $D$. Let $B$ be a disk with centre $c$ such that $\bar{B} \subset D$. Then $|g(c)| \leq \max_{z \in \partial B} |g(z)|$. Why is this statement true? Does it follow from Cauchy's integral formula?

• Okay I think I got it. Nevermind then May 29, 2017 at 17:46

This is a consequence of the maximum principle:

Proposition. Let $f\colon U\rightarrow\mathbb{C}$ be a holomorphic function on an open connected subset of $\mathbb{C}$ and let $z_0\in U$. If $|f|$ has a local maximum at $z_0$, then $f$ is constant.

Proof. This follows from the fact that non-constant holomorphic functions are open, which itself is derived from the fact that holomorphic functions satisfy the identity principle and the existence of a complex logarithm in $D(1,1)$. $\Box$

From there, one can derive the following:

Proposition. Let $f\colon U\rightarrow\mathbb{C}$ be a holomorphic function on a connected and bounded subset of $\mathbb{C}$ such that $f$ is continuous on $\overline{U}$, then $\sup\limits_{z\in U}|f(z)|\leqslant\sup\limits_{z\in\partial U}|f(z)|$.

Proof. Since $\overline{U}$ is compact and $|f|$ continuous on it, $|f|$ reaches a maximum on $\overline{U}$. If its maximum is reached in $U$, then using the above proposition, $f$ is constant and the desired equality holds. $\Box$

The last proposition is stronger than the result you aimed for.