1
$\begingroup$

Let $f$ be holomorphic function in an open set $\Omega$ in $\mathbb{C}$. Let $\{f_n\}$ be a sequence of holomorphic functions, converging uniformly to $f$ on $\Omega$.

For each $f_n$, let $\{{g_{n,m}}\}_{m=1}^{\infty}$ be a sequence of polynomials which converge uniformly to $f_n$ on $\Omega$.

Question: Does it follows that $f$ is a uniform limit on $\Omega$ of a sequence of polynomials? (More precisely, here do we need to assume $\Omega$ is compact set?)

Such arguments come in Runge's approximations theorem, which I didn't find rigorously explained in many (any) book. If one knows any elementary exposition on Runge's theorem, I would like to see it.)

$\endgroup$
  • $\begingroup$ I think you mean that $g_{n,m}\to f_{\color{red}n}$ uniformly as $m\to\infty$ for each $n$, right? $\endgroup$ – sranthrop Mar 12 '15 at 12:44
  • $\begingroup$ Oh yes! Thanks for notification. $\endgroup$ – Groups Mar 13 '15 at 6:21
0
$\begingroup$

Fix $\varepsilon>0$. For each $n\in\mathbb N$ there is some $m(n)\in\mathbb N$ such that $||g_{n,m(n)}-f_n||<\varepsilon/2$. Moreover, there is some $N\in\mathbb N$ such that $||f_n-f||<\varepsilon/2$ for each $n\geq N$. Finally, $||g_{n,m(n)}-f||\leq||g_{n,m(n)}-f_n||+||f_n-f||<\varepsilon$ for each $n\geq N$. Here, $||\cdot||$ is the sup-norm on $\Omega$. This shows, that the sequence $g_{n,m(n)}$ of polynomials converges uniformly to $f$ on $\Omega$.

$\endgroup$

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.