The following is exercise 11.8 in Rudin's Real and Complex Analysis:
Suppose $\Omega$ is a region, $f_n \in H(\Omega)$ for $n = 1, 2, 3, \ldots$. $u_n$ is the real part of $f_n$. $\{u_n\}$ converges uniformly on compact subsets of $\Omega$, and $\{f_n(z)\}$ converges for at least one $z \in \Omega$. Prove that then $\{f_n\}$ converges uniformly on compact subsets of $\Omega$.
My thoughts: By Harnack's theorem, the limit $u$ of $\{u_n\}$ is harmonic. Thus it's the real part of a holomorphic function $f$ on $\Omega$ defined up to an imaginary constant. Using the limit of $\{f_n(z)\}$, we can find this constant. Thus $f$ is well-defined.
What's left is to show that the imaginary part of $f_n$ converges to that of $f$. I suspect I need to use the Cauchy-Riemann equations for this, but I cannot apply the familiar uniform convergence theorems with partial derivatives. What should I do?
Note: A region is a connected open subset of the complex plain.
I'm not entirely sure about this answer but I'll give it a try and people can point out any mistakes.
Let $z_{0}=x_{0}+iy_{0}$ be the point in $\Omega$ where the sequence $(f_{n})$ converges. We write $f_{n}(z)=2u_{n}(\frac{z+\bar{z_{0}}}{2},\frac{z-\bar{z_{0}}}{2i})-u_{n}(x_{0},y_{0})+ic_{n}$ where $c_{n}\in\mathbb{R}$. Letting $z=z_{0}$, we see that $c_{n}=\mathrm{Im}f_{n}(z_{0})$. Then uniform convergence of $(u_{n})$ on compact subsets and convergence of $(f_{n}(z_{0}))$ imply uniform convergence of $(f_{n})$ on compact subsets.