What holomorphic functions are limits of polynomials?

Question

Let $\Omega$ be a connected open set in the complex plane. What is the closure of the polynomials in $\mathcal{H}(\Omega)$ the set of holomorphic functions on $\Omega$? The topology is the usual compact convergence topology. Take, for instance, an annulus such as $D(r,R)$, the set of all complex $z$ such that $r<|z|< R$, you cannot recover the function $z\mapsto \frac{1}{z}$ because of the residue at $0$, so what holomorphic functions are limits of polynomials?

Answer 1

Let $\Sigma\supset\Omega$ be the union of $\Omega$ and all bounded components of ${\mathbb C}\setminus \Omega$. The algebra you get is the algebra of all holomorphic functions on $\Sigma$.

First, every $f\in{\cal H}(\Sigma)$ is a locally uniform limit of polynomials as a consequence of Runge's Theorem, see Corollary 1.15 in John Conway's "Functions of one complex variable I", 2nd Ed.

Second, if $p_n$ is a sequence of polynomials converging locally uniformly on $\Omega$ and if $K$ is a bounded component of ${\mathbb C}\setminus \Omega$, then $p_n$ also converges uniformly on a neighborhood of $K$ by Cauchy's integral Theorem, as there exists a path around $K$ with winding number 1. (Take the positively oriented boundary of an $\varepsilon$-neighborhood of $K$, where $\varepsilon$ is chosen so small that it does not hit any other components.)

Answer 2

let's call Runge an open subset $\Omega \subset \mathbb C $ such that polynomials are dense in $\mathcal H(\Omega)$ . A hole of $\Omega$ is a compact connected component of $ \mathbb C \setminus \Omega $. We then have the equivalent statements, for the open subset $\Omega \subset \mathbb C$ (not assumed connected).

a) $\Omega$ is Runge
b) $\Omega$ has no hole
c) Every connected component of $\Omega$ is simply connected

In the general case, when these equivalent conditions are not fulfilled, Runge's theorem says that if you choose one point in each hole of $\Omega$, then the rational functions with poles only in these points are dense in $\mathcal H(\Omega)$. Beware that you can have a non-denumerable set of holes : take for $\Omega$ the complement of a Cantor set in $\mathbb R \subset \mathbb C$.

In a related vein, Mergelyan's (difficult) theorem says that if you take a compact subset $K \subset \mathcal H (\Omega)$ with connected complement in $\mathbb C$, then any continuous function on $K$ which is holomorphic in the interior of $K$ can be uniformly approximated by polynomials.

Bibliography: Remmert's book is probably the best reference for this question ( and many others...) The equivalence of the statements quoted above is proved in Chapter 13, section 2.

Mergelyan's theorem is not in Remmert's book but is proved on page 386 of Rudin's well known Real and complex analysis, of which you can find a review here

Remark: These results are somewhat astonishing. Take $\Omega=\mathbb C \setminus \{x\in \mathbb R| x \leq -1\}$ and for $f \;$ the holomorphic branch of the logarithm $f(z)=log (1+z) $ which is zero at the origin. Its Taylor series $\sum_{k=0}^{\infty} (-1)^k \frac {z^k}{k}$ diverges for $|z|\gt 1$ and the partial sums of the series are polynomials which definitely don't converge to $f$. However, since $\Omega$ has no holes, there does exist some sequence of polynomials converging to $f$ uniformly on compact subsets of $\Omega$.

Answer 3

In a simply connected open set, any holomorphic function is a locally uniform limit of polynomials : by the conformal mapping theorem, one needs only check this is true in the disc.

EDIT: The argument of the conformal mapping theorem is false, but it remains true that In a simply connected open set, any holomorphic function is a locally uniform limit of polynomials, by

If the set is not simply connected, consider $\Omega_0$ its simply connected hull.

A holomorphic function $f$ on $\Omega$ extends to a holomorphic function if, and only if, for any polynomial $P$ the integral of $P.f$ over any contour is zero. Its extension $f_0$ at $z \in \Omega_0 - \Omega$ is then given by $f_0(z) = \frac{1}{2i\pi}\oint_\gamma \frac {f(\zeta)}{\zeta - z}dz, $ where $\gamma$ is any simple contour in $\Omega$ around $z$.

Such a function is therefore a locally uniform limit of polynomials in $\Omega_0$, and the condition is also necessary. This condition is equivalent to being the restriction of a certain holomorphic function on $\Omega_0$.

The proof of Runge's theorem is constructive and therefore probably tells you more in that case.