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If harmonic functions converges in the distribution sense to a distribution. Then can we prove that the functions are actually converges uniformly to a function on every compact set. And the limit function is actually also harmonic? I'm totally stuck by this exercise and I have achieved nothing. So please someone help me with this problem, thanks....

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I believe this is just an application of Weyl's Lemma, with multiplying by bump functions on the compact subsets of convergence. – Alex Youcis Dec 10 '12 at 13:57
Can you tell me what does the lemma state? Thanks a lot. – lee Dec 10 '12 at 14:16
Thanks a lot, but how to use this lemma to show that the sequence will converge uniformly in a compact set? – lee Dec 10 '12 at 14:53
I have not thought about that as much--but once you do that it seems feasible to use Weyl's lemma to prove that its harmonic. The only thing, which I assume you would have thought of, is Harnack's Principle--but I don't know how to apply that directly here. – Alex Youcis Dec 10 '12 at 14:55
up vote 4 down vote accepted

Convergence in the sense of distributions has a very nice property: the limit of derivatives is the derivative of the limit. Precisely, if $f_n\to f$ as distributions, then for every test function $\phi$ we have $\langle f_n,\nabla \phi\rangle \to \langle f,\nabla \phi\rangle$ (since $\nabla \phi$ is just a bunch of test functions), which by definition of distributional derivative means $\nabla f_n\to \nabla \phi$.

In particular, $f_n\to f$ implies $\Delta f_n\to \Delta f$ in the sense of distributions. But $\Delta f_n\equiv 0$, hence $\Delta f\equiv 0$. Weyl's lemma tells us that both $f$ and $f_n$ are harmonic functions.

It remains to upgrade the convergence to locally uniform. Harmonic functions have another nice property: $f(a)=\int f\varphi$ for any test function of the form $\varphi(x)=\psi(|x-a|)$ such that $\int \varphi =1$. (Proof is a sub-exercise.) By distributional convergence, $\int f_n\varphi\to \int f \varphi$, hence $f_n(a)\to f(a)$. This is pointwise convergence; I leave it for you to upgrade it to locally uniform.

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Is it because when constrained the domain to a compact set, the test function space is a Frechet space. And use the resonance theorem to make the convergence locally uniform? – lee Dec 19 '12 at 12:27
@lee That is an interesting idea; please post the details if it works. My personal inclination was to forget distributions and instead prove a lemma saying "if pointwise limit of harmonic functions is harmonic, then the convergence is locally uniform". But then the proof escaped me... – user53153 Dec 19 '12 at 17:17
By Banach-Steinhaus theorem, they are equicontinuous, then the locally uniformness follows immediately when we apply the Ascoli's theorem. – lee Dec 20 '12 at 4:55
At least when holomorphic functions are involved, it seems that pointwise convergence to a holomorphic function is not enough to deduce locally uniform convergence and that a similar example works also for harmonic functions. – levap Dec 22 '12 at 11:31
@levap Thanks for the reference. Indeed, taking real or imaginary part of a holomorphic counterexample we get a harmonic one. Interesting. – user53153 Dec 22 '12 at 18:30

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