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Here is a theorem:

If $\{f_n\}_{n=1}^{\infty}$ is a sequence of holomorphic functions that converges uniformly to a function $f$ in every compact subset of $\Omega$, then $f$ is holomorphic in $\Omega$

Why do we use "compact subset" here?

The proof doesn't seem to use this condition.

Since $f_n$ is holomorphic, we have by Goursat's Theorem $$\int_Tf_n(z)\text{d}z=0$$ for any triangle $T$. and because $f_n \rightarrow f$ uniformly. Then by Morera's Theorem $$\left|\int_Tf(z)\text{dz}\right|=\left|\int_T (f(z)-f_n(z) ) \text{dz}\right|\leq\sup_{T}|f(z)-f_n(z)|\times length(T)\rightarrow 0$$ as $n\rightarrow \infty$ $$\Rightarrow$$ $$\int_Tf(z)\text{dz}=0$$ We can conclude that $f$ is holomorphic in $\Omega$.

It seems I don't use the compactness condition, so why does the theorem statement include compactness?

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Uniform convergence on compact sets is weaker than uniform convergence! –  Thomas Feb 1 '13 at 13:37
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Uniform convergence on compact subsets is a weaker hypothesis than uniform convergence. For example, $f_n(z) = z^n$ converges to zero uniformly on compact subsets of the open unit disk, but it does not converge uniformly to zero. –  Martin Feb 1 '13 at 13:37
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Hi: I worked on improving the grammar in your original post, and if you are interested, you can see exactly what changed in the edit history. I hope you find that I preserved your meaning, and that it's helpful. –  rschwieb Feb 1 '13 at 14:03
    
after you edit. it looks like more clear and exactly. thank you very much. –  Laura Feb 1 '13 at 15:05
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1 Answer 1

up vote 4 down vote accepted

Let me quote: "and because $f_n\to f$ uniformly" So you do use uniform convergence. However, you don't need uniform convergence on all of $\Omega$, we can get along with something weaker, namely uniform on any triangle. It is however more convenient to speak of all compact subsets and not just the triangles.

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