TITLE: Are the polynomial functions on $S^1$ dense in $C(S^1,ℂ)$?
QUESTION [12 upvotes]: A friend of mine came up with this problem: Let $S^1$ be the unit circle in $ℂ$ and $P$ the space of polynomial functions $S^1 → ℂ$ (with complex coefficients). Is $P$ dense in $C(S^1,ℂ)$?
Stone–Weierstraß is not applicable because $P$ is not closed under complex conjugation. We’re wondering if complex conjugation on $S^1$ (= inverting) is a uniform limit of polynomials. We suspect not, but don’t know how to prove it.

REPLY [2 votes]: Ok (almost) every periodic function $g$ on $[0,1]$ can be written in the form
$$
g(x) = \sum_{n=-\infty}^\infty c_n e^{2\pi i n x}
$$
You can switch from function $f : S^1 \subset \mathbb{C} \rightarrow \mathbb{C}$ to the periodic function on $[0,1]$ via simple transformation: $e^{2\pi i \theta}$.
$$
g(\theta) = f(e^{2\pi i \theta})
$$
This transformation can be seen in some sense as bijection between periodic functions and functions defined on $S^1$. (You would need to exactly specify those function spaces to be true.)
But let's have a look what you get when you transform polynomial.
$$
f(z) = \sum_{n=0}^N a_n z^n
$$
$$
g(\theta) = f(e^{2\pi i \theta}) = \sum_{n=0}^N a_n e^{2\pi i n \theta}
$$
Uupss you are missing those negative terms $e^{-2 \pi i n \theta}$. So you can't get every periodic function on $[0,1]$.

From above you can see that "polynomials" of form $\sum_{n=-N}^n a_n z^n$ are dense in $C(S^1,\mathbb{C})$ and they are holomorphic in $\mathbb{C} \setminus \{0\}$, which is interesting because $\overline z$ is not holomorphic and $\overline z$ was exactly missing in previous. This seams to me as very special to the $S^1$ because $\frac1{z} = \frac{\overline z}{z\overline z} = \overline z $ only on the $S^1$.
Could anyone expand on the thing that harmonic polynomials form basis of $L^2(S^1)$ and real part of holomorphic function is harmonic I guess it is connected to the above but right now I do not have the power to think any more.