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?

  • 1
    $\begingroup$ You probably also want $K$ and $\overline{\Omega}$ to be disjoint, otherwise it could be that $d(K,\partial\Omega)=0$ $\endgroup$
    – Lorago
    Sep 21, 2023 at 16:12

1 Answer 1


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?


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