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I am looking for the proof of the following Theorem : Does anyone know where i can find out ?

If $\Omega$ is open and connected and $u_k$ be uniformly bounded sequence of harmonic functions . There exists a subsequence that converges uniformly to a harmonic function $u:\Omega \to \mathbb R$ on any compact subset of $\Omega$.

In case you think that its not hard, i look forward to hints as well.

I hope the statement is true . Thanks.

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    You can add a constant to make them all $\ge 1$, and apply Harnack's inequality to get equicontinuity on compact subsets... then Arzela-Ascoli.2012-07-28
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    @LeonidKovalev : Sir, i was thinking of using derivative bound of harmonic equation but before using it how do i set up so that i can use it ?2012-07-28
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    That works too. If $z\in\Omega$, then there exists $r>0$ such that the disk $D(z,r)$ is contained in $\Omega$, and you have an upper bound on $|\nabla u_k|$ in $D(z,r/2)$. Hence the family is locally uniformly Lipschitz, which implies equicontinuity on compact subsets as well.2012-07-28

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