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Suppose $\Omega$ be a bounded region and $\{f_n\}_{n\in\mathbb N}$ a sequence of continuous functions on $\overline\Omega$ which are holomorphic in $\Omega$ and $f_n$ converges uniformly on the boundary of $\Omega$. Prove that $f_n$ converges uniformly on $\overline\Omega$.

How to prove this?

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  • $\begingroup$ but how? can you explain? $\endgroup$ – Topology Sep 23 '14 at 18:46
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According to the Maximum Principle for Holomorphic functions $$ \sup_{z\in\overline\Omega}\lvert\,f_m(z)-f_n(z)\rvert=\sup_{z\in\partial\Omega}\lvert\,f_m(z)-f_n(z)\rvert. $$ Hence, if $\{f_n\}$ converges uniformly on $\partial\Omega$, then it is uniformly Cauchy on $\partial\Omega$, i.e., $$ \lim_{m,n\to\infty}\sup_{z\in\partial\Omega}\lvert\,f_m(z)-f_n(z)\rvert=0, $$ and hence $\{f_n\}$ is uniformly Cauchy on $\overline\Omega$, and thus uniformly convergent on $\overline\Omega$.

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For a given $\epsilon >0$, there exists $N$ such that for any $n,m$ greater than $N$, $|f_n(z) - f_m(z)| \leq \epsilon$ on the boundary of $\omega$.

By the Maximum modulus principle, $|f_n(z) - f_m(z)|$ can only attain its maximal value on the boundary, so we get $|f_n(z) - f_m(z)| \leq \epsilon$ for all $z \in \bar\omega$

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