# Given sequence of harmonic functions converges uniformly on compact subsets

Suppose that $u_n$ is a sequence of harmonic function on an open, connected subset $D \subset \mathbb{C}$ such that $u_n(z) \in (0, \infty)$ for all $z \in D$ and with $u_n(z_0) \to 0$ for some $z_0 \in D$. Show that $u_n \to 0$ uniformly on compact subset of $D$.

Thoughts so far: I'm new to harmonic functions, so I'm not sure where to begin. I know that locally each $u_n$ is the real part of a holomorphic function.

Context: I'm studying for a qual, so just a hint to get me going or any ideas to get started would be most helpful.

• Have you heard of Harnack's inequality? – Daniel Fischer Aug 25 '15 at 19:48

Consider first the case of a disk $B$ containing $z_0$. Then $u_n = \text{Re}(f_n)$ where $f_n$ is analytic on $B$. By Montel's theorem the $f_n$ form a normal family. What can you say about the limit of a subsequence that converges on $B$?
• Thank you for the response. My questions are: how do we know that $f_n$ is locally bounded? (or meets some other condition so we can apply Monte's theorem) And, based on my understanding of Montel's theorem, we would get that each subsequence has a further subsequence that converges to a holomorphic function uniformly on $B$. Do we want to show that these subsequences all converge to the same holomorphic function? – user19817 Aug 25 '15 at 19:45
• The real part is greater than $0$. All you need for Montel is that they omit two values (e.g. $0$ and $-1$). If the question is correct, the real parts all converge to $0$. We don't know what the imaginary parts do, but we can always add an imaginary constant so the imaginary part of $f_n(z_0)$ is, say, $0$. – Robert Israel Aug 25 '15 at 20:14