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Let $(f_{n})_{n}$ be a sequence of holomorphic functions on a domain D which satisfies the following conditions: there exists some $z_{0}$ in D such that $f_{n}(z_{0})$ converges and the sequence of $(f_{n}')_{n}$ of their derivatives converges uniformly on compact subsets of D.Then the sequence $(f_{n})_{n}$ itself is also locally uniformly convergent. I try to prove the statement above by using the identity theorem for analytic functions but I couldn't. Thank you for help.

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In any dimension, uniform convergence of derivatives and convergence in a point of functions implies uniform convergence of functions (See Baby Rudin). In the analytic case, uniform convergence of analytic functions implies analytic limit using the theorem of Morera.

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  • $\begingroup$ I understand the real case but complex case.In your last sentence you do not mention the pointwise convergence of the sequence itself in some point and also you express the opposite direction of the implication we need! Thanks.. $\endgroup$ – user135582 Apr 11 '14 at 12:01
  • $\begingroup$ In your case: the real case implies that $(f_n)_{n\in\Bbb N}$ converges uniformly; now, the complex case implies analytic limit. $\endgroup$ – Martín-Blas Pérez Pinilla Apr 12 '14 at 13:48
  • $\begingroup$ Could you explain more precisely, please? $\endgroup$ – user135582 Apr 12 '14 at 19:14
  • $\begingroup$ Open subsets of $\Bbb R^n$ are connected if and only if they are path-connected. Suppose that your result fails in some $z_1$. Take a path from $z_0$ to $z_1$ and a connected compact $K\subset D$ s.t. path$\subset\mathring{K}$. $\endgroup$ – Martín-Blas Pérez Pinilla Apr 13 '14 at 18:08
  • $\begingroup$ Yes, the uniform convergence of derivatives plus the "anchor" at $z_0$ implies the uniform convergence of functions. $\endgroup$ – Martín-Blas Pérez Pinilla Apr 13 '14 at 21:52

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