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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. –  user31373 Jul 28 '12 at 19:18
    
@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 ? –  Theorem Jul 28 '12 at 19:36
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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. –  user31373 Jul 28 '12 at 20:33
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2 Answers

See theorem 2.6 in the page 35 here and the comment below.

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You can see this book

D.H. Armitage, S.J. Gardiner, Classical Potential Theory, Springer, London, 2000.

proposition 1.5.11. This will exactly answer your question. This is a link of this book in booksgoogle

Enjoy.

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Can you quote the proposition? We can't see it in booksgoogle. –  Davide Giraudo Nov 3 '12 at 11:06
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