# Approximation with complex polynomials on $S^1$ - can it be done?

Can one uniformly approximate a function 'similar' to identity on $S^1$ with complex polynomials? I mean a function like:
$f(z)=z \cdot (1+h \cdot \sin(m\cdot Arg(z)))$, for $|h| < 1,\ m \in \mathbb{N}$.

What I actually have is a function (actually I have a sequence of them but that is another story) like the following one (or a smooth version of it):
$f(z)=z \cdot (1+ h \cdot wave(Arg(z)))$, where
$wave(\phi)=[\phi^{-1} \in [\pi,k\pi]] \cdot \sin(\frac{1}{\phi})$, $[p]$ is $1$ if $p$ is true and $0$ otherwise and $k \in \mathbb{N_+}$.

Can anyone tell me if there are any theorems saying that this approximation can be done or not? It looks similarly to Fourier series. What actually needs to be done is approximating $wave(Arg(z))$. However I need $z^{-n}$ to use Fourier series theory and I have only polynomials. $z^{-n}$ can not be approximated on the whole $S^1$ with polynomials.

If I was able to extend the function to the whole 'unit+$\varepsilon$' disk in a holomorphic way I would be able to use Taylor series for this function, but I don't know if such an extending can be done, since the intuitive way with going orthogonally from the $f(S^1)$ when one goes orthogonally from the $S^1$ leads nowhere.

My question comes from a bigger problem and what I actually need to solve it is: make such an approximation $w$ of $f$, that:

1. $w(S^1) \subseteq{} \lbrace z \in \mathbb{C}: dist(z,f(S^1))< \delta \rbrace$
2. $|z-w(z)| < \varepsilon(h)$

where $\delta$ is some unknown constant depending on $f$. We may choose as $\varepsilon$ any function that is is continuous at zero and $\varepsilon(0)=0$.
This one looks more complicated but is weaker than the previous one.

If anyone could help me with that I would be really grateful.

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I believe the answer to your question is no.

Any continuous function on the unit circle can be approximated arbitrarily closely by rational functions that have no poles except at zero and infinity.

However, having a pole at zero is crucial. Indeed:

Proposition. A continuous function $f:S^1\to\mathbb{C}$ on the unit circle can be approximated arbitrarily closely by polynomials if and only if $f$ has an analytic extension to the unit disk.

Proof. If $p_n$ is a sequence of polynomials approximating $f$, then the maximum principle implies that $p_n$ is a uniform Cauchy sequence on the closed unit disk. Hence $p_n$ has a uniform limit, and this limit is analytic on the unit disk by the Weierstraß convergence theorem.

EDIT. I forgot to mention that the converse direction (any function continuous on the closed unit disk and holomorphic on the open unit disk can be approximated uniformly by polynomials) follows from Mergelyan's theorem.

Of course having an analytic extension to the unit disk is quite a special property, and your functions do not have it. The first function you mention is actually a rational function, since on $S^1$ we have $$\sin(m\cdot \arg(z) ) = \frac{z^m - z^{-m}}{2i}.$$ By the identity principle, this function, let's call it $F$, cannot have an analytic extension to the disk, so $z\cdot (1+h\cdot F)$ cannot have such an extension either.

(That the function cannot be approximated uniformly by polynomials is also easy to see in this case just using the argument principle.)

Your second function cannot have an analytic extension to the unit disk, since your function $\operatorname{wave}$ is zero on some intervals of the unit circle. That is impossible by the F. and M. Riesz theorem.

I hope this helps!

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Thank you very much! The wave does not have to be zero on some intervals, I just didn't want to create too complicated formulas. However the first function suggests that it's rather hard to approximate such a thing, it seems that we do need $z^{−n}$... –  savick01 Oct 30 '11 at 9:14
Of course, on the other hand there are plenty of functions that are close to the identity on the unit circle and do extend analytically to the unit disk (hence can be approximated by polynomials). –  mathstribble Oct 30 '11 at 21:12
There certainly are:) Anyway, I'm happy because even though you let me down about an easy way to approximate those functions, there were some facts new for me in your answer that inspired me to solve the original problem. –  savick01 Oct 30 '11 at 21:45
Great - I am glad I was of some help! –  mathstribble Oct 30 '11 at 22:41