$\omega$ is a region, $f_n$ is holomorphic on $\omega$ for n=1,2,3,.., $u_{n}$ is the real part of $f_n$, $u_{n}$ converges uniformly on compact subsets of $\omega$, {$f_{n}(z)$} converges for atleast one $z \space \epsilon \space \omega$. Then how to prove that {$f_{n}$} converges uniformly on compact subsets of $\omega$? I think Harnack's theorem has to be used.
Uniform convergence on compact subsets
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complex-analysis
uniform-convergence
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0What is $\omega$? A connected open subset of $\mathbb C$? – 2012-12-13
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0@Jonas Meyer It's a region, i.e. open subset of $\mathbb C$. – 2012-12-13
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1aviress: Is it assumed that $\omega$ is connected, as is often part of the definition of "region"? Your comment leaves open the possibility that it is not connected, which would be problematic. Where is the problem from? – 2012-12-13
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0It's from Real and complex analysis, Walter Rudin chapter 11. The question just says that $\omega$ is a region. I believe the region is connected. – 2012-12-13
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0I hope so. The statement is false if the region is not connected. – 2012-12-13