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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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    $\begingroup$ 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. $\endgroup$
    – user31373
    Jul 28, 2012 at 19:18
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    $\begingroup$ @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 ? $\endgroup$
    – Theorem
    Jul 28, 2012 at 19:36
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    $\begingroup$ 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. $\endgroup$
    – user31373
    Jul 28, 2012 at 20:33

2 Answers 2

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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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    $\begingroup$ Can you quote the proposition? We can't see it in booksgoogle. $\endgroup$
    – Davide Giraudo
    Nov 3, 2012 at 11:06

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