I am looking at a proof regarding the uniform convergence of holomorphic functions. There is one step I do not understand.


We are given a sequence $(f_n)$ of holomorphic functions on an open set $U \subset \mathbb{C}$ such that for every compact subset $K$ of $U$ the sequence converges uniformely on $K$. Let $f$ be the limit of this sequence then $f$ is holomorphic.

(part of the) Proof:

Let $z_0 \in U$ and let $\bar{D}_R$ be a closed disc of radius $R$ centered at $z_0$ and contained in $U$. Then the sequence $(f_n)$ converges uniformely on $\bar{D}_R$. Let $\partial D_R$ be the boundary of $\bar{D}_R$ and consider the closed disc $\bar{D}_{R/2}$ centered at $z_0$ with radius $R/2$. Then for $z \in \bar{D}_{R/2}$ we have \begin{equation} f_n(z) = \frac{1}{2\pi i}\int_{\partial D_{R}}\frac{f_n(\zeta)}{\zeta-z}d\zeta \end{equation} and $|\zeta-z| \geq R/2$. Since $(f_n)$ converges uniformely ,for $|z-z_0| \leq R/2$, we have \begin{equation} f(z) = \frac{1}{2\pi i}\int_{\partial D_{R}}\frac{f(\zeta)}{\zeta-z}d\zeta \end{equation}


I do not understand why we are looking at the set $|z-z_0| \leq R/2$ when we take the limit in the integral. Hopefully someone can explain this to me. Much thanks in advance!

  • $\begingroup$ Is there a comma between "converges uniformly" and "for $|z-z_0|\le R/2$"? I ask this because what we really need is that $f_n$ converges uniformly on $\partial D_R$. $\endgroup$
    – robjohn
    Jun 3, 2017 at 8:34
  • $\begingroup$ Yes, you are right. I will change the text. Thank you! $\endgroup$
    – user421927
    Jun 3, 2017 at 9:53
  • $\begingroup$ Very related question math.stackexchange.com/questions/2307373/… $\endgroup$
    – reuns
    Jun 3, 2017 at 21:40
  • 1
    $\begingroup$ @kees1 You should not vandalize your own question. $\endgroup$
    – egreg
    Nov 18, 2017 at 11:48
  • $\begingroup$ Hiding your tracks? $\endgroup$
    – Did
    Nov 18, 2017 at 12:29

1 Answer 1


Because you want to use the fact that $\frac{f_n(\zeta)}{\zeta-z}$ converges uniformly to $\frac{f(\zeta)}{\zeta-z}$ (with respect to $\zeta$). You already know that $f_n$ converges uniformly to $f$. Since we are working on the disc centeres at $z$ with radius $\frac R2$, we know that$$\left|\frac{f_n(\zeta)}{\zeta-z}-\frac{f(\zeta)}{\zeta-z}\right|=\left|\frac{f_n(z)-f(z)}{\zeta-z}\right|\leqslant\frac2R\bigl|f_n(z)-f(z)\bigr|$$and this assures that $\frac{f_n(\zeta)}{\zeta-z}$ converges uniformly to $\frac{f(\zeta)}{\zeta-z}$.

Of course, you could use any number $R'\in(0,R)$ instead of $\frac R2$ and still reach the same conclusion.

  • $\begingroup$ Thank you very much for your clear explanation! $\endgroup$
    – user421927
    Jun 3, 2017 at 8:28

You must log in to answer this question.