# Is harmonicity preserved when taking limits (normal convergence) on the unit disk.

I'm reading Koosis's book on $H^p$ spaces and have a question. He is proving a $L^p$ version of the Dirichlet problem which states that if $F(t)$ is in $L^p$ on the unit circle then $$U_{r}(\theta)=F\ast P_{r}(\theta)$$ is harmonic in the open unit disk. He does this by noticing $$U_{r}(\theta)=\sum_{n=-\infty}^{\infty} A_n r^{|n|} e^{i n \theta}.$$ We get this from noting that the Poisson kernel has the form $$P_{r}(\theta)=\sum_{n=-\infty}^{\infty}r^{|n|} e^{i n\theta},$$ where the sum converges uniformly on compact subsets of the open unit disk. Since we have uniform convergence, we can interchange the summation and integral to get the form above.

My question is this: if the partial sums $$\sum_{n=-N}^{N}A_n r^{|n|}e^{i n \theta}$$ satisfy Laplace's equation, does the limit satisfy it as well? The limit function would have the mean value property since the partial sums do, but is this equivalent to harmonicity?

EDIT: I actually proved the result myself. The last summand, when split between summing $n\geq 0$ and $n<0$, is the sum of an analytic function and a conjugate analytic function. We know this because the partial sums converge uniformly on compact subsets, preserving holomorphicity. Thus, the real and imaginary parts are harmonic.

However, if anyone has anything interesting to add I'd love to hear it.

The idea of the proof is very simple: If $D$ is an arbitrary closed disk contained in the domain of a continuous function $u$ satisfying the mean-value property, then the function $\tilde{u}$ obtained by replacing $u$ in the interior of $D$ by its Poisson integral is harmonic in $D$ with the same boundary values as $u$. This shows that the function $v=u-\tilde{u}$ is a continuous function on $D$ satisfying the mean value property. However, this implies that $v$ attains both maximum and minimum on the boundary $\partial D$, so $v=0$ in $D$, which implies $u = \tilde{u}$ in $D$. Since the Poisson integral $\tilde{u}$ is real-analytic, this implies that $u$ is, too, and since $D$ was arbitrary, this is true for the whole domain of $u$. (Obviously, this argument works in any dimension with disks replaced by balls.)