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The whole theorem looks like-

Let, $f_n:\Omega\to\Bbb{C}$ be a sequence of holomorphic functions where $\Omega$ is open set. Suppose, $\{f_n\}$ converges uniformly to a function $f$ on every compact subset. Then $f$ is holomorphic on $\Omega$ and $\{f_n'\}$ converges uniformly to a function $f'$ on every compact subset.

I have proved the holomorphicity of $f$ using Morera's theorem. Now I try to prove the second part.
Let $D=D_\delta(z_0)$ be an open disc in $\Omega$ such that $\overline {D}\subset\Omega$
Let, $F_n(z)=f_n(z)-f(z)\ \forall z\in \Omega$, $F_n$ is holomorphic on $\Omega$.
Note that since $\overline D$ is compact, $\operatorname{sup}_{z\in\overline D}|F_n(z)|$ exists.
Choose $\varepsilon>0$, then $\exists K\in\Bbb{N}$ such that $|f_n(z)-f(z)|<\delta\varepsilon\ \forall n\ge K, \ \forall z\in \overline D$ (as $\overline D$ is compact) thus, $\operatorname{sup}_{z\in\overline D}|F_n(z)|<\delta\varepsilon.$
Now we apply Cauchy integral formula, $F_n'(z)={1\over 2\pi i}\int_{\partial D}{F_n(\zeta)\over (\zeta-z_0)^2}d\zeta\implies |F_n'(z)|\le{1\over 2\pi}\operatorname{sup}_{z\in\partial D}|F_n(z)| 2\pi\delta{1\over \delta^2}\le\varepsilon\ \forall z\in \overline D\ \forall n\ge K$.
Thus, $|f_n'(z)-f'(z)|<\varepsilon\ \forall z\in \overline D\ \forall n\ge K$.
Hence, $f_n'$ coverges uniformly to $f'$ on every compact disc inside $\Omega$.
Now, how to generalize the proof for any arbitrary compact disc? Thanks for assistance in advance.

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For an arbitrary compact $K\subset\Omega$, we may find finitely many disks $D_1,\cdots,D_n$ which cover $K$ and whose closures are contained in $\Omega$. We may apply your proof separately to each disk $D_i$, giving us indices $n_i$ for which $\sup_{D_i}|f'_n-f'|<\epsilon \forall n\geq n_i$. Taking the maximum $N:= \max_i n_i$, we obtain that $\sup_{K}|f'_n-f'|\leq \max_i \sup_{D_i}|f'_n-f'|<\epsilon \forall n\geq N$ giving us locally uniform convergence as required.

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  • $\begingroup$ Ok, it's easy, got it. $\endgroup$
    – MathBS
    Apr 13, 2019 at 8:03

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