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!

  • $\begingroup$ In fact I have deleted my comment because Daniel Fisher has answered your question I think. $\endgroup$ – user119228 Mar 6 '14 at 21:54
  • $\begingroup$ Copy that - I'm just reading his response. I'll delete my comments also. Thank you for your response. $\endgroup$ – Sam T Mar 6 '14 at 21:55
  • $\begingroup$ Look here, perhaps it's helpful too : math.stackexchange.com/questions/53035/… $\endgroup$ – user119228 Mar 6 '14 at 21:59
  • $\begingroup$ You are welcome! $\endgroup$ – user119228 Mar 6 '14 at 22:08

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 $J$.

  • $\begingroup$ 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 $\endgroup$ – Sam T Mar 6 '14 at 22:06
  • $\begingroup$ 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? $\endgroup$ – Sam T Mar 6 '14 at 22:06
  • $\begingroup$ 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!) $\endgroup$ – Sam T Mar 6 '14 at 22:07
  • $\begingroup$ 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. $\endgroup$ – Steven Gubkin Mar 6 '14 at 22:08
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    $\begingroup$ The sets with the property you desire are exactly characterized as the compact $J$ with empty interior, and $\mathbb{C}\\J$ connected. $\endgroup$ – Steven Gubkin Mar 6 '14 at 22:12

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