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A sequence of complex functions $\{f_n\}$ such that $f_i:D\rightarrow\mathbb C$ for all $i\in\mathbb N$, is locally uniformly convergent to $f$ if for all $P\in D$, exists a neighborhood $U$ of $P$ such that $\{f_{n} \big|U\cap D\}$ is uniformly convergent to the function $f|U\cap D$. Now it is clear that

Uniform convergence $\Rightarrow$ local uniform convergence

but i can't find a counterexample that disprove the converse implication. Can someone help me?

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up vote 0 down vote accepted

z^n, see Remmert "Funktionentheorie 1" ISBN 3-540-55233-2, Page 74

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It is usually best to at least lightly expand on what one might find in that page of the book. Saying «z^n», while suggestive, does not an answer make! – Mariano Suárez-Alvarez May 7 '12 at 1:35

Consider $f_n(z) = z^n$ on the unit disc.

Added later: In fact, locally uniform convergence without uniform convergence is the norm rather than the exception. Consider for example a holomorphic function $f$ on the unit disc such that its Maclaurin series has radius of convergence equal to $1$. Then the Maclaurin series converges locally uniformly to $f$ on the unit disc, and unless the coefficients $a_n$ in the series tend to zero relatively quickly (a little faster than $1/n$), the convergence won't be uniform on the whole disc.

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