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The following is taken from Rudin's Principles of Mathematical Analysis; it is Theorem 7.17 with some extra assumptions made based on a remark at the end of the proof of the theorem.

Let $f_n,g$ be functions defined on an interval $[a,b]$. Assume that $f_n$ are differentiable with continuous derivatives. Assume also that $f_n' \rightarrow g$ uniformly and $f_n \rightarrow f$ pointwise in $[a,b]$ for some function $f$ (or more generally, assume that $f_n(x_o)$ is convergent for some $x_o \in [a,b]$). Then, $f_n \rightarrow f$ uniformly, $f$ is differentiable with continuous derivative and $f'=g$.

Does a similar result hold for holomorphic functions on an open set? Since Rudin speaks of intervals, should such a result be valid for connected or convex sets only, instead of open sets in general? Namely, if $\Omega$ is a domain and $f_n$ are holomorphic functions on $\Omega$, does uniform convergence of $f_n′$ on the whole domain (not uniform convergence on compacts subsets of the domain!) plus pointwise convergence of $f_n$, imply uniform convergence of $f_n$ on the whole domain?

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Yes. Is an easy consequence of Morera's theorem and uniform convergence on compact sets sets is sufficient. For your exact hypothesis a connected domain is required.

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  • $\begingroup$ Thanks for your response, i have written down a proof and indeed connectedness was used. Though i wouldn't say uniform convergence on compact subsets is sufficient. By assuming that $f_n' \rightarrow f'$ uniformly on compact subsets and $f_n(z_o) \rightarrow f(z_o)$, i can only show that $f_n \rightarrow f$ uniformly on compact subsets. How can i show uniform convergence with these assumptions? Is really the uniform convergence of the derivatives on the whole open set not needed? $\endgroup$ Jun 5 '17 at 21:06
  • $\begingroup$ @kleinmeinpouts, uniform convergence on the whole domain isn't required for proving that the limit is holomorphic. $\endgroup$ Jun 5 '17 at 21:55
  • $\begingroup$ Of course it is not required, uniform convergence on compact subsets implies that the limit is holomorphic. That's not what i want to prove. I showed the following: Let $\Omega$ be an open connected set and $z_o\in \Omega$. Let $f_n,f$ be holomorphic functions on $\Omega$, such that $f_n' \rightarrow f'$ uniformly on compact subsets of $\Omega$ and $f_n(z_o) \rightarrow f(z_o)$. Then, $f_n \rightarrow f$ uniformly on compact subsets of $\Omega$. $\endgroup$ Jun 5 '17 at 22:07
  • $\begingroup$ What i want to show is this: if instead of ''$f_n' \rightarrow f'$ uniformly on compact subsets of $\Omega$'', i assume that ''$f_n' \rightarrow f'$ uniformly on $\Omega$'', do i get that $f_n \rightarrow f$ uniformly? Essentially, my question is about the analogue of the theorem in Rudin's book in the complex variable case, as far as the uniform convergence is concerned, not the differentiability/holomorphicity part; that was assumed in the first place, maybe i did not state it correctly at first. I appreciate your answers though! $\endgroup$ Jun 5 '17 at 22:14
  • $\begingroup$ @kleinmeinpouts, uniform convergence on compacts of $f'_n$ plus pointwise convergence in a point of $f_n$ not implies uniform convergence of $f_n$ in the whole domain. Consider $f_n(z) = (z+1/n)^2$ in $\Bbb C$ of $f_n(z) = 1/(z+1/n)$ in $D(1,1)$. $\endgroup$ Jun 6 '17 at 7:12

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