2
$\begingroup$

Let $\Omega$ be a domain in $\mathbb{C}$ and $z_0 \in \Omega$. Further, let $f_n$ and $f$ be holomorphic functions on $\Omega \setminus \{z_0\}$ such that $z_0$ is a simple pole for each $f_n$ and at most a simple pole (i.e. a simple pole or a removable singularity) for $f$. If we assume that $f_n \to f$ uniformly on any compact subset of $\Omega \setminus \{z_0\}$, is there anything we can say about the behaviour of the Residues $\operatorname{Res}_{z_0}(f_n)$ under taking the limit?

$\endgroup$
4
$\begingroup$

The Residue Theorem is usually used to evaluate integrals in terms of residues, but here you'll use it to write a residue as an integral (let's say around a circle around $z_0$ within $\Omega$). The circle is a compact set, so $f_n$ converges uniformly to $f$ on it. Therefore...

$\endgroup$
2
$\begingroup$

Let$$g_n(z)=\begin{cases}zf_n(z)&\text{ if }z\neq z_0\\\lim_{z\to z0}zf_n(z)&\text{ if }z=z_0.\end{cases}$$Then $(g_n)_{n\in\mathbb N}$ is a sequence of holomorphic functions that converges to$$\begin{array}{rccc}g\colon&\Omega\cup\{z_0\}&\longrightarrow&\mathbb C\\&z&\mapsto&\begin{cases}zf(z)&\text{ if }z\neq z_0\\\lim_{z\to z0}zf(z)&\text{ if }z=z_0\end{cases}\end{array}$$uniformly on any compact subset of $\Omega\cup\{z_0\}$. In particular, $g(z_0)=\lim_{n\in\mathbb N}g_n(z_0)$. But this means that$$\operatorname{Res}_{z_0}(f)=\lim_{n\in\mathbb N}\operatorname{Res}_{z_0}(f_n).$$

$\endgroup$
  • $\begingroup$ How do you know that $g_n(z_0) \to g(z_0)$? $\endgroup$ – Adayah Jul 31 '17 at 8:39
  • $\begingroup$ @Adayah Take a circle centered at $z_0$ and contained in $\Omega$. Then the restriction of the $f_n$' s to that circle converge to the restriction of $f$. A standard theorem from Complex Analysis says that then $(f_n)_{n\in\mathbb N}$ converges unifomly to $f$ in the closed disk whose boundary is that circle. $\endgroup$ – José Carlos Santos Jul 31 '17 at 8:49
  • $\begingroup$ I see, thanks. $\mbox{}$ $\endgroup$ – Adayah Jul 31 '17 at 9:26
  • $\begingroup$ Thanks for your answer. Unfortunately I can only accept one, had to ask the die ;-) $\endgroup$ – agb Jul 31 '17 at 18:22
  • $\begingroup$ No problem. I upvoted Robert Israel's answer myself. $\endgroup$ – José Carlos Santos Jul 31 '17 at 18:31
0
$\begingroup$

I can't quite comment yet, so I'll leave you my hint here: Recall the integral definition of the residue and the dominated convergence theorem. I'm sure you can take it from there :)

$\endgroup$

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.