MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
If I'm not mistaken, it is the space of entire functions. – S. Carnahan Mar 25 '11 at 7:19
do you mean the restriction of entire functions to $\Omega$? In my understanding, an entire function is a holomorphic function defined on the whole complex plane. There are more than just the restrictions of those. take $\frac{1}{z}$ on the unit disc centered at $1$. – Olivier Bégassat Mar 25 '11 at 7:23
If I recall correctly, the algebra you get will only depend on the polynomial convex hull of $\Omega$, which is the union of $\Omega$ with all bounded components of ${\bf C} \setminus \Omega$. I am not sure if a more precise answer or characterization depends on the boundary of $\Omega$ – Yemon Choi Mar 25 '11 at 7:30
do you then get the restrictions of holomorphic functions on the polynomial convex hull of $\Omega$, or are there more functions? Is this hull always simply connected? – Olivier Bégassat Mar 25 '11 at 7:38
What if you take $\Omega$ with no holes, like $\lbrace z\in \mathbb{C}~|z|<1\rbrace\setminus\mathbb{R}^{-}$? – Olivier Bégassat Mar 25 '11 at 7:44
up vote 12 down vote accepted

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.)

share|cite|improve this answer
You don't say why the limit of your sequence of polynomials is also holomorphic inside $K.$ – Cédric Bounya Mar 25 '11 at 9:35
A locally uniform limit of holomorphic functions is holomorphic by a Theorem of Weierstrass. – Anton Mar 25 '11 at 9:37
Perhaps I am missing something; but since $K$ is compact, the convergence is uniform on $K$ and so the limit is holomorphic inside K (or inside any compact subset of $\Sigma$) by Morera... – Yemon Choi Mar 25 '11 at 18:16
The convergence is uniform on some neighborhood of K. So the limit is holomorphic in some neighborhood of K. – Anton Mar 25 '11 at 18:58
Anton: sure, I just wanted to suggest that there is nothing complicated about why the limit is holomorphic; I think Morera (rather than any normal family argument) will do – Yemon Choi Mar 25 '11 at 20:02

Dear Olivier, 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 $\Omega$.

share|cite|improve this answer

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.

share|cite|improve this answer
the conformal mapping theorem doesn't map polynomials to polynomials, because polynomials are bonuded on the unit disc, but for instance unbounded on the upper half plane while the two domains are biholomorphic, so I think the first part of the argument fails. I thought of using it too, but it seems not to work... – Olivier Bégassat Mar 25 '11 at 9:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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