# Rudin Functional Analysis Chapter 6 Problem 11

Suppose $\Omega$ is open in $\mathbb R^2$, and $\{f_i\}$ is a sequence of harmonic functions in $\Omega$ that coverges in the distribution sense to $\Lambda\in D'(\Omega)$ i.e. $$\Lambda(\phi)=\lim_{i\rightarrow \infty}\int_\Omega f_i(x)\phi(x)\ dx\ \ \ \ \ \ \ \ \ \ \ \ \forall\ \phi\in D(\Omega)$$Prove that $\{f_i\}$ converges uniformly on every compact subset of $\Omega$ and that $\Lambda$ is a harmonic function.

It was easy to see that if each of the $\{f_i\}$ is harmonic then $\Lambda$ is harmonic. But I am not able to get that how to prove $\{f_i\}$ converges uniformly on compact subset of $\Omega$.

Any type of help will be appreciated. Thanks in advance.

• @reuns Are your $f_i$ harmonic functions? – Pozz Sep 10 '17 at 17:35
• @Pozz yes but it didn't converge. I tried an answer – reuns Sep 10 '17 at 17:51

If $T_n \to T$ in $D'(\Omega)$ then for any $\phi \in C^\infty_c$, $T_n \ast \phi \to T\ast \phi$ in $C^\infty(\Omega)$. If the $T_n$ are harmonic functions then so are $T_n \ast \phi$ and $T \ast \phi$.
By the maximum principle or the mean value property for harmonic functions, from a local bound for $T_n\ast \phi$ and $(T_n-T_n(a)) \ast \phi$ we obtain a local bound for $T_n-T_n(a)$. And those bounds transfer to $T \ast \phi$ so that $\lim_{\epsilon\to 0} T \ast \phi_\epsilon$ converges locally uniformly (with $\phi_\epsilon(x) = \frac{1}{\epsilon} \phi(x/\epsilon)$) and the limit $T$ is harmonic and $T_n \to T$ locally uniformly.
• @bunny it follows from $\partial_{xx}h+\partial_{yy}h = 0$ that (real and imaginary part of) $h$ has no local maximum – reuns Sep 10 '17 at 18:38
• how are u defining $T_n(a)$ since $T_n$ is a distribution – bunny Sep 11 '17 at 5:16
• @bunny $T_n$ is assumed to be an harmonic (and hence real analytic) function.. Did you downvote ? – reuns Sep 12 '17 at 0:25