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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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.