Let $(f_{n})_{n \in \mathbb{N}}$ be a sequence of holomorphic functions on the open unit disc $D$ in $\mathbb{C}$, and suppose that this sequence converges pointwise to a function $f$. By Osgood's theorem one can conclude then that there is an open and dense subset $V$ of the disc, so that the function $f$ is holomorphic there and that the convergence of the sequence is locally uniform on $V$. If we further suppose that the limit function $f$ is also holomorphic on the entire disc $D$, is it then possible to conclude that the sequence $(f_{n})_{n \in \mathbb{N}}$ converges locally uniformly to $f$ on $D$?


1 Answer 1


Let $K_n$ be the union of $\{ 0 \}$, the line segment $[2/n, 1]$ and the compact set $\{ z \in \mathbb{C} : |z| \le 1 \text{ and } \operatorname{dist}(z,\mathbb{R}_+) \ge 1/n \}$. Let $L_n = K_n \cup \{ 1/n \}$. (You should probably draw a picture...) Note that the complement of $L_n$ is connected.

Let $f_n$ be a holomorphic function on a (non-connected) neighborhood of $L_n$ such that $f_n = 0$ on $K_n$ and $f_n(1/n) = 1$. By Runge's theorem, there is a polynomial $p_n$ such that $|p_n - f_n| < 1/n$ on $L_n$. It's not hard to see that $p_n \to 0$ (pointwise) on the unit disc, but the convergence is not uniform on any neighborhood of $0$ since $p_n(1/n) \approx 1$. (In fact the convergence is not uniform near any point on the positive real axis.)

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    $\begingroup$ For a slightly simpler picture, take $K_n = \{r e^{i\theta}: 0 \le r \le 1, 2/n \le \theta \le 2 \pi$ and $L_n = K_n \cup \{e^{i/n}/n\}$, with $f_n(e^{i/n}/n) = 1$. $\endgroup$ Commented Feb 27, 2012 at 22:22

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