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I am trying to solve this old qual problem:

Suppose $\{u_n\}_{n=1}^\infty$ is a sequence of functions harmonic on an open set $U \subset \mathbb{C}$ and uniformly bounded by 1. Suppose there is a function $u : U \to \mathbb{R}$ such that $u_n \to u$ pointwise. Show $u$ is harmonic on $U$.

I would like to solve this using only elementary complex analysis. My idea is this:

Pick a point $z \in U$ and a disk $D_r(z) \subset U$. Since the disk is simply connected, each $u_n$ is the real part of some holomorphic $f_n$. Now, we would like to show that $f_n$ converges uniformly on compact subsets...

I haven't gotten any further than this.

Will this approach work? Other ideas which use complex analysis are appreciated.

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As you indicate, we can assume that $U$ is a disk, since being harmonic is a local property, and so there exists a sequence of holomorphic functions $f_n$ with $u_n = \Re f_n$. In order to elevate pointwise convergence to locally uniform convergence you need something like boundedness, and one common trick to pass from a bounded real part to a bounded function is to exponentiate, i.e., consider $g_n = \exp f_n$. Then $(g_n)$ is a sequence of holomorphic functions with $|g_n| = \exp u_n \le e$. By Montel's theorem every uniformly bounded sequence of holomorphic functions is normal, i.e., there exists a locally uniformly convergent subsequence $g_{n_k} \to g$, where $g$ is holomorphic. Then $|g_{n_k}| = \exp u_n \to \exp u$ pointwise, so $|g| = \exp u$, and $u = \log |g|$ is harmonic.

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  • $\begingroup$ Thanks--this is the sort of solution I was looking for. One question: how do we know that $\log|g|$ is harmonic? I think it is because, since we are working for $z$ in a small neighborhood, we can position the branch cut of $\log$ to make it holomorphic on the image of $g$. Thus, $\log g$ is holomorphic, so its real part is harmonic. Is this correct? $\endgroup$ Jan 7 '15 at 23:52
  • $\begingroup$ This is one way to argue this, correct. $\endgroup$ Jan 8 '15 at 6:01
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This answer does not use complex analysis per se, but a basic property of the Lebesgue integral, the dominated convergence theorem. Sorry! I thought you might find it useful nonetheless.

To show that $u$ is harmonic, choose a point $z_0\in U$ and observe that each $u_n$ satisfies the mean-value property at $z_0$: $$ u_n(z_0) = \frac{1}{2\pi r}\int_{|z-z_0|=r} u_n(z), \text{ for all sufficiently small $r>0$.} $$ Since each $u_n$ is dominated in modulus by $1$, the dominated convergence theorem allows us to let $n$ approach infinity and interchange limit with integral. Then $$ u(z_0) = \frac{1}{2\pi r}\int_{|z-z_0|=r} u(z), \text{ for all sufficiently small $r>0$.} $$

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