6

The Weierstrass approximation theorem states that any continuous function $ f : I \rightarrow \Bbb R $ on a closed, bounded, connected subset $ I \subseteq \Bbb R $ can be uniformly approximated by polynomials.

Can any continuous function $ \phi : J \rightarrow \Bbb C $ on a closed, bounded, connected subset $ J \subseteq \Bbb C $ be uniformly approximated by polynomials?

What I mean is, for which subsets $ J \subseteq \Bbb C $ can all functions be approximated uniformly by polynomials.

This question is an example sheet question that I had (already supervised on $-$ non-examinable) but supervisor wasn't sure what the question meant exactly. There are some basic sets, such as closed, real intervals, that it clearly holds for, but others (such as closed unit ball) that it does not hold for. Is anyone able to shed any light on the answer. (Not just a few counter-examples, but some explanation as to why it does / does not hold on certain set (eg, because connected complement / similar).)

Thanks very much!

Sam OT
  • 6,675
  • In fact I have deleted my comment because Daniel Fisher has answered your question I think. –  Mar 06 '14 at 21:54
  • Copy that - I'm just reading his response. I'll delete my comments also. Thank you for your response. – Sam OT Mar 06 '14 at 21:55
  • Look here, perhaps it's helpful too : http://math.stackexchange.com/questions/53035/stone-weierstrass-theorem-in-mathbbc –  Mar 06 '14 at 21:59
  • You are welcome! –  Mar 06 '14 at 22:08

1 Answers1

11

It depends on what is meant by "polynomial".

If only $\sum c_n z^n$, then every function that is uniformly approximable by polynomials must be holomorphic on the interior of $J$.

Although that condition is trivially satisfied if $J$ has empty interior, that doesn't mean that for such $J$ every continuous function is the uniform limit of polynomials. For example the unit circle has empty interior, but a sequence of polynomials converging uniformly on the unit circle converges uniformly on the closed unit disk by the maximum principle, and thus if $f$ is a uniform limit of polynomials on the unit circle, then there is a holomorphic function $h$ on the unit disk that extends continuously to the unit circle, with boundary values $f$. In particular, we have

$$\int_{\lvert z\rvert = 1} f(z)\cdot z^n \,dz = 0\tag{1}$$

for all $n \geqslant 0$. (And, in this case, that condition is sufficient.)

That phenomenon generalises, if $J$ disconnects the plane, that is, if $\mathbb{C}\setminus J$ has at least two connected components, then the bounded components of the complement of $J$ impose restrictive conditions on the continuous functions that are uniform limits of polynomials similar to $(1)$.

Mergelyan's theorem asserts the converse, if $J$ is a compact subset of $\mathbb{C}$ with empty interior such that $\mathbb{C}\setminus J$ is connected, then every continuous function on $J$ can be uniformly approximated by polynomials (in $z$ only).

If "polynomial" means polynomial in $z$ and $\overline{z}$, or equivalently polynomial in $\operatorname{Re} z$ and $\operatorname{Im} z$, then the Weierstraß approximation theorem holds for all compact $J$.

Daniel Fischer
  • 206,697
  • Thanks for your detailed response.

    I understand most of what you have said, but I have only been doing complex analysis for 7 weeks, so don't 100% understand it all! =P

     – Sam OT Mar 06 '14 at 22:06
  • What I meant by a polynomial is a function $$ p(z) = \sum_{r=0}^{n} c_n z^n, $$ as opposed to a power series $$ P(z) = \sum_{r=0}^{\infty} c_n z^n. $$

    I realise that any holomorphic function can be expressed as a power series, but in real analysis any continuous function can be approximated uniformly by polynomials on a closed, bounded subset. This isn't the case for any closed, bounded subset in $ \Bbb C $, but I was wondering what conditions could be put on $J$ to make it true?

     – Sam OT Mar 06 '14 at 22:06
  • I see what you mean about extending $ |z| = 1 $ to $ |z| \le 1 $, but does this mean that any continuous $h$ in $ |z| \le 1 $ can be approximated by polys? (You have shown a sort of converse.)

    (Hopefully this makes sense!)

     – Sam OT Mar 06 '14 at 22:07
  • As Daniel says, if $J$ is compact, has empty interior, and $\mathbb{C}\J$ is connected, then every continuous function can be uniformly approximated by holomorphic polynomials. – Steven Gubkin Mar 06 '14 at 22:08
  • @SmileySam Any function on an set $J$ which can be uniformly approximated by polynomials on $J$ must be holomorphic on the interior of $J$. So choose any function which is continuous but not holomorphic on the disk, and that gives you function which cannot be uniformly approximated by polynomials. – Steven Gubkin Mar 06 '14 at 22:10
  • Ah yes, I see that, thank you.

    I imagine that this isn't a necessary condition, though? Is there some less restrictive condition?

    Am I right in thinking that saying that $J$ has empty non-interior means that I could form a bijection between it and the real line (eg with the boundary of the unit disc)?

     – Sam OT Mar 06 '14 at 22:10
  • In particular, the only domains which have the property you desire must have empty interior. – Steven Gubkin Mar 06 '14 at 22:10
  • Ah, I see. Thank you very much. Most appreciative! (Unfortunately, I can't give any upvotes/similar because this is in a comment, but still, thank you very much!) :) – Sam OT Mar 06 '14 at 22:12
  • 1
    The sets with the property you desire are exactly characterized as the compact $J$ with empty interior, and $\mathbb{C}\J$ connected. – Steven Gubkin Mar 06 '14 at 22:12
  • @SmileySam The comments exchange here came too fast for me, I lost track of what you understood or not. If you'd like me to elaborate on something, comment again, but please more slowly, so that I have a chance to keep up ;) – Daniel Fischer Mar 06 '14 at 22:14
  • Mergelyan's theorem does say the next best thing though:if $f$ is holomorphic on the interior of $J$ and continuous on $J$, then it can be uniformly approximated by complex polynomials. This is a reletively recent theorem for what it says: only proved in the 50s. The proof is very cool, I recommend reading Rudin. My research is on Mergelyan type properties in several complex variables. I love this stuff! – Steven Gubkin Mar 06 '14 at 22:15
  • @DanielFischer I understand now, thank you. Both yours and Steven's comments have been very helpful, thank you. – Sam OT Mar 06 '14 at 22:22
  • @StevenGubkin You say to "read Rudin" - is there a specific book that you mean? (At the moment I'm borrowing a Beardon and a Priestly book from my college's library because they're especially suited to my particular course, but I could try to get the Rudin one from the library also?) Also, I don't actually know what Mergelyan type properties are... how "far on" is the book? I'm a 2nd year at Cambridge - this is the first complex analysis course (after two real analysis courses) so I'm not exactly a master...! – Sam OT Mar 06 '14 at 22:23
  • Great, @SmileySam. By "Rudin", he means Rudin's "Real and Complex Analysis". A recommendation I fully endorse, it's a great book. It starts from the basics, but goes on rather far, at a fast pace, so it is probably a good idea to have another book besides it, to compare and see things from a (slightly) different angle. – Daniel Fischer Mar 06 '14 at 22:26
  • Thank you very much, both of you. I shall look out that book; I had heard of it before. – Sam OT Mar 06 '14 at 22:30