Are the polynomial functions on $S^1$ dense in $C(S^1,ℂ)$? 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.
 A: 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. 
A: Assuming "polynomial" means "polynomial in $z$", i.e. of the form $\sum_{k=0}^n a_k z^k$, then no, they are not dense, and the function $\bar{z}$ is not in their uniform closure.
The high-level reason is that polynomials are holomorphic on the unit disk, holomorphic functions are closed under uniform convergence, and $\bar{z}$ is not holomorphic.
For a direct proof, consider the linear functional $I(f) = \int_{S^1} f(z)\,dz = \int_0^{2\pi} f(e^{i\theta}) i e^{i\theta}\,d\theta$.  This is clearly continuous with respect to the uniform norm since
$$|I(f)| \le \int_0^{2\pi} |f(e^{i\theta})| |ie^{i\theta}|\,d\theta \le 2 \pi \|f\|_{\infty}.$$ 
But for any polynomial $f$, you can check that $I(f)=0$.   Hence by continuity, $I(f)=0$ for any $f$ in the closure of the polynomials.  On the other hand, for $f(z)=\bar{z}$, $I(f) = 2\pi i$, so $f(z)=\bar{z}$ is not in the closure.
(This proof is easy to discover when you know the Cauchy integral theorem, which says that any holomorphic function has $\int_{S^1} f(z)\,dz = 0$, or indeed the same replacing $S^1$ with any nice closed curve.)
