3
$\begingroup$

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 $\Omega$?

$\endgroup$
3
$\begingroup$

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 $f_n(1-n^{—1)}\to e$.

$\endgroup$
  • $\begingroup$ and the simple convergence? if we have uniform convergence on compacts, do we have convergence on $\Omega$? $\endgroup$ – Federica Maggioni Jan 5 '13 at 13:03
  • $\begingroup$ Yes, we have simple/pointwise convergence as $\{z\}$ is compact for all $z\in\Omega$. $\endgroup$ – Davide Giraudo Jan 5 '13 at 13:11
1
$\begingroup$

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 $$ e^{n} - \sum_{k=0}^{n-1}\frac{n^k}{k!} \geq \frac{n^n}{n!} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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