Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose you have a sequence $f_n$ of analytic functions on an open set $\Omega$, which converges uniformly on compact subsets of $\Omega$. Can you conclude that $f_n$ converges uniformly on the whole open $\Omega$?

share|improve this question
add comment

2 Answers 2

up vote 3 down vote accepted

No: if $\Omega:=\{z,|z|<1\}$, and $f_n(z):=z^n$, this sequence converges uniformly to $0$ on compact sets (because such a set is contained in $B(0,r),r<1$) but not on $\Omega$ as $f_n(1-n^{—1)}\to e$.

share|improve this answer
    
and the simple convergence? if we have uniform convergence on compacts, do we have convergence on $\Omega$? –  Federica Maggioni Jan 5 '13 at 13:03
    
Yes, we have simple/pointwise convergence as $\{z\}$ is compact for all $z\in\Omega$. –  Davide Giraudo Jan 5 '13 at 13:11
add comment

Another example: the series $\displaystyle\sum \frac{z^k}{k!}$ converges uniformly on compact subsets of $\Omega$, but not on the whole $\mathbb{C}$, as $$ e^{n} - \sum_{k=0}^{n-1}\frac{n^k}{k!} \geq \frac{n^n}{n!} $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.