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. Prove that $f$ is holomorphic too
Saw the following solution: 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 $n$, $$\left|\int_{\partial\Delta} f(z)dz\right|=\left|\int_{\Delta} (f(z)-f_n(z))dz\right|\leq\mathrm{length}({\partial\Delta})\sup_{z\in\partial\Delta}|f(z)-f_n(z)|$$
By letting $n\rightarrow\infty$, you find that $\displaystyle\int_{\partial\Delta} f(z)dz=0$.
By Morera's theorem, $f$ is holomorphic.
(*) My question - Couldn't understand the following equality: $$\left|\int_{\partial\Delta} f(z)dz\right|=\left|\int_{\Delta} (f(z)-f_n(z))dz\right|$$
There should be $\partial \Delta$ instead of $\Delta$. Leaving aside that typo, the point is that $$\int_{\partial \Delta} f(z)\,dz=\int_{\partial \Delta} f(z)-f_n(z)\,dz+\underbrace{\int_{\partial \Delta} f_n(z)\,dz}_{(A)}$$
And $(A)=0$ because $f_n$ is holomorphic on a contractible open set that contains $\Delta$. Namely, since $\Delta$ is convex and compact and $U\supseteq \Delta$ is open, there is a convex open set $V$ such that $\Delta\subseteq V\subseteq U$.