# Inequality for holomorphic function on compact set

Exercise:

Let $$f$$ be holomorphic on an open region $$\Omega$$. Let $$|f(z)|\leq C$$ for all $$z$$. We define: $$d(z,\partial \Omega) = \inf(|z-c| \,\,\,|\,\,\ c \in \partial \Omega)$$ Now prove that on a compact set $$K$$, we have a bound for the derivative of $$f$$, which is only dependent on $$d(K,\partial \Omega) = \inf\{d(z,\partial \Omega) \,\,|\,\, z\in K\}$$ and $$C$$.

My attempt:

Using Cauchy's integral formula we get:

$$f'(z)=\frac 1 {2\pi i}\cdot \int_{B_R(z)} \frac{f(\xi)}{(\xi-z)^2}\,d\xi$$

where $$R= \frac 1 2 \cdot d(K,\partial \Omega)$$. Now, using the standard ring integral inequalities, we get the estimate:

$$f'(z) \leq \frac 1 {2\pi}\cdot 2 \pi R \cdot \frac C {R^2} = \frac C R$$

So our bound would be:

$$2 \cdot \frac C {d(K,\partial \Omega)}$$

I am confused because I seem to have found such a bound without using the compactness. And it seems too easy. However, I cannot find a mistake. Can someone tell me if my solution is correct?

• You probably also want $K$ and $\overline{\Omega}$ to be disjoint, otherwise it could be that $d(K,\partial\Omega)=0$ Sep 21, 2023 at 16:12

You used the fact that $$R$$ is positive and not $$0$$, but that's not true for a generic set $$K$$! Think about $$\Omega$$ being the unit disc and $$K$$ being the unit square to which you substract only the corners for example.
Compactness is what allows you to ensure that the distance between $$K$$ and $$\partial\Omega$$ is strictly positive, though it's of course not a necessary condition. You can find a proof of that for example here: How to show distance between boundary of an open set and a compact subset is positive?