0
$\begingroup$

Let $\Gamma \in C^{1,\lambda}$ be an oriented Jordan curve in complex plane $\mathbb{C} $, $\mathrm{R}(\Gamma)$ the set of all rational functions without poles on $\Gamma $. "$\mathrm{R}(\Gamma)$ is dense in $C(\Gamma)$" seems to be a classical result, where can we find it with detailed proof?

$\endgroup$

1 Answer 1

2
$\begingroup$

This is the main result in a paper by J.L. Walsh from 1927 (in german):

J.L. Walsh, Über die Entwicklung einer Funktion einer komplexen Veränderlichen nach Polynomen. Math. Ann. 96 (1927), no. 1, 437–450.

The precise statement is :

If the function $f(z)$ is continuous on a Jordan curve in whose interior the origin lies, then on that curve $f(z)$ can be approximated as closely as desired by a polynomial in $z$ and $1/z$.

A proof (in english) appears in Walsh's book, see Theorem 7 in Chapter 2, Section 2.4,

J.L. Walsh, Interpolation and approximation by rational functions in the complex domain. Fourth edition. American Mathematical Society Colloquium Publications, Vol. XX. American Mathematical Society, Providence, RI, 1965.

Here is how to derive the statement from Mergelyan's theorem :

Let $\Gamma$ be a Jordan curve with $0\in\text{Int}(\Gamma)$, $f$ continuous on $\Gamma$. Let $g$ be the conformal map from $\text{Int}(\Gamma)$ onto the open unit disk, with $g(0)=0$. The map $g$ can be continued to a homeomorphism from $\overline{\text{Int}(\gamma)}$ to $\overline{\mathbb D}$. The function $f\circ g^{-1}$ is continuous on $\partial{\mathbb D}$, hence there exists a trigonometric polynomial $T(w)=\sum_{n=-N}^{N}a_nw^{n}$, $w\in\partial{\mathbb D}$, such that $|f\circ g^{-1}(w)-T(w)|\leq\epsilon$, or equivalently, $|f(z)-\sum_{n=-N}^{N}a_n(g(z))^{n}|\leq\epsilon$. By Mergelyan's theorem, for $n\geq0$, each $g(z)^{n}$ can be approximated on $\Gamma$ by polynomials, and for $n<0$, $g(z)^{n} =(g(z)/z)^{n}.z^{n}$, where the first factor can also be approximated by polynomials.

$\endgroup$
6
  • $\begingroup$ Could you help give a website to access? $\endgroup$
    – Yidong Luo
    Commented Oct 7, 2023 at 7:35
  • $\begingroup$ If you know Mergelyan's theorem, then the statement also follows from it. $\endgroup$
    – user111
    Commented Oct 7, 2023 at 9:04
  • $\begingroup$ not familiar... $\endgroup$
    – Yidong Luo
    Commented Oct 7, 2023 at 9:21
  • $\begingroup$ Mergelyan's theorem is a classical result in complex analysis. I have edited to give the argument. $\endgroup$
    – user111
    Commented Oct 7, 2023 at 11:37
  • 2
    $\begingroup$ look in Rudin's book... $\endgroup$
    – user111
    Commented Oct 7, 2023 at 13:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .