# Uniform limit of holomorphic functions

Let $\{f_n\}$ be a sequence of holomorphic functions defined in a generic domain $D \subset \mathbb{C}$. Suppose that there exist $f$ such that $f_n \to f$ uniformly.

My question is: is it true that $f$ is holomorphic too?

For a function $f: D \to \mathbb{C}$ to be holomorphic, by Morera's theorem it's enough that for every triangle $\Delta \subset D$, we have $\int_{\partial \Delta} f = 0$. Now, if $f_n \to f$ uniformly, it's easy to show that $\int_{\partial \Delta} f_n \to \int_{\partial \Delta} f$, but as $f_n$ are holomorphic, $\int_{\partial \Delta} f_n = 0$, so the limit is of course also 0.

Note that it's actually enough to assume almost uniform convergence, i.e. uniform convergence on all compact subsets $K \subset D$.

• The first sentence is wrong. Consider, e.g. $f(0) = 1$ and $f(z) = 0$ for each $z \not = 0$. You should first note that $f$, being the uniform limit of continuous functions is continuous. Jun 19, 2019 at 2:30
• Your $f$ is obviously not continuous, so it cannot be a uniform limit of continuous functions. Jun 20, 2019 at 5:40
• I think you should reread my comment. The first sentence of your solution is wrong (it is stated generally). Jun 20, 2019 at 6:03
• I assume it is obvious to the reader, just like you just assumed that it’s obvious that a value at a single point doesn’t affect integrals. Jun 21, 2019 at 15:57
• I don't think the "just like" is valid. You are assuming that the reader will make an assumption (i.e. that $f$ is continuous), while I am assuming the reader will know something is true. I really think you should edit your answer to add the word "continuous" before "function". Jun 21, 2019 at 16:44

You only need that $$f_n\to f$$ uniformly on every compact subsets of $$D$$. It's a well-known fact that $$f$$ is then continuous. The idea is to use Morera's theorem.

Let $$\Delta\subset D$$ be a closed triangle. Since each $$f_n$$ is holomorphic, by Cauchy's theorem, you have $$\displaystyle\int_{\partial\Delta} f_n(z)dz=0$$ for all $$n$$.

$$\partial\Delta$$ is a compact subset of $$D$$, so you know that $$f_n\to f$$ uniformly on $$\partial\Delta$$.

So you get, for all $$n$$, $$\left|\int_{\partial\Delta} f(z)dz\right|=\left|\int_{\partial\Delta} (f(z)-f_n(z))dz\right|\leq\mathrm{length}({\partial\Delta})\sup_{z\in\partial\Delta}|f(z)-f_n(z)|$$

By letting $$n\rightarrow\infty$$, you find that $$\displaystyle\int_{\partial\Delta} f(z)dz=0$$.

By Morera's theorem, $$f$$ is holomorphic.

You've already seen an approach using Morera's theorem from the other excellent answers. For a slightly more concrete demonstration of why $f$ is complex differentiable, you can use the fact that every $f_n$ satisfies Cauchy's integral formula, so by uniform convergence $f$ also satisfies Cauchy's integral formula. This allows you to differentiate $f$ by differentiating the integral.

I just want to add a small precision :

• (If $f(z)$ is piecewise continuous) and $$\int_\gamma f(z) dz = 0$$ for every closed $C^1$ contour $\gamma \subset U$ some open, then $f(z)$ is holomorphic on $U$.

• But if $U$ is closed, say $U = \{ \ |z| \le 1 \ \}$ then we only get that $f(z)$ is holomorphic on $U \setminus \partial U$.

Try with $$f(z) = \sum_{n=2}^\infty \frac{z^n}{n (n-1)}$$

then $\displaystyle\sum_{n=2}^\infty \left|\frac{1}{n (n-1)} \right|$ converges hence $\displaystyle\sum_{n=2}^\infty \frac{z^n}{n (n-1)}$ converges uniformly on $|z| \le 1$, but $$f'(z) = \sum_{n=1}^\infty \frac{z^n}{n} = -\log(1-z)$$ has a singularity at $z=1$ : $\quad f(z)$ is holomorphic only on $|z| < 1$ and not on $|z| \le 1$.