I came across this question when reading the proof of Lemma 6.19 p246 in the book A Course in Complex Analysis and Riemann Surfaces by Schlag. In the proof, they make the following reasoning.

Take $V\subset \mathbb{C}$ open and $\{u_n \}_n \subset C^{\infty}(V)$ a sequence of harmonic functions on $V$ which converges in the sense of $L^1_{loc}$ to $u_{\infty}$, meaning that for any compact $K\subset V$, we have $$ \iint_K |u_n(z)-u_{\infty}(z)|dz \rightarrow 0 $$ as $n\rightarrow \infty$. Now, by the mean value property (MVP), we have for each disk $D(z_0,r)$: $$u_n(z_0)= \frac{1}{\pi r^2}\iint_{D(z_0,r)} u_n(z)dz$$ which implies that $\{u_n \}_n$ is a Cauchy sequence in $C(V)$ and therefore converges uniformly on compact subsets of $V$ to $u_{\infty}$.

My questions are:

1) What do they mean with '$\{u_n \}_n$ is a Cauchy sequence in $C(V)$'? I am pretty sure that this means $\sup_{z\in V}|u_n(z)-u_m(z)|\rightarrow 0$ as $n,m\rightarrow \infty$ but I just want to make sure this interpretation is correct.

2) How does this alternative way of writing the MVP imply the uniform convergence on compact subsets of $V$ to $u_{\infty}$? I can only show that this implies pointwise convergence, but this is not as strong as their claim.

Any help would be very much appreciated.


I'll answer question 2, which will hopefully clarify question 1 a bit. (I don't have the book so I am not sure about what they use as the notational convention.)

Let $K\subset V$ be compact. I claim first that there exists $K' \subset V$ and an open set $W \subset K'$ such that $K \subset W$.

This follows by noting that $K$ has positive distance to $V^C$ which is closed. So let $r = \frac12 \mathrm{dist}(K, V^C)$ and you can set $W = \cup_{p\in K} D(p,r)$ and $K'$ the closure of $W$.

Then for any $p\in K$, using the same $r$ as in the above construction, we have that using the mean value property

$$ |u_n(p) - u_\infty(p)| \leq \frac{1}{\pi r^2} \iint_{D(p,r)} | u_n(z) - u_\infty(z)| ~\mathrm{d}z \leq \frac{1}{\pi r^2} \iint_{K'} |u_n(z) - u_\infty(z)| ~\mathrm{d}z $$

Note that on the far right the domain of integration is $K'$ independent of $p\in K$. Hence the convergence of $u_n\to u_\infty$ is uniform on $K$. (But note that the "$\epsilon$" in the convergence is determined by the $L^1$ convergence on a somewhat bigger $K'$, not just $K$ itself.)


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