Let $S^1$ be the unit circle in the complex plain and $C(S^1)$ be the continuous function space on $S^1$.$f\in C(S^1)$ is a generator means that $\{p(f) |\text{ p is a polynomial in z}\}$ is dense in $C(S^1)$.
I need to prove $C(S^1)$ does NOT have a generator.
$\forall f \in C(S^1)$, let $f=\sum_{n\in \mathbb{Z}}a_nz^n$ be the Fourier expansion of $f$.
If $\{f_k\}_{k=1}^{\infty}$ is convergent in $C(S^1)$,then $\{f_k\}_{k=1}^{\infty}$ is also convergent in $L^2(S^1)$.Hence $\{a_n^{(k)}\}_{k=1}^{\infty}$ is convergent,$\forall n \in \mathbb{Z}$.$a_n^{(k)}$ is the fourier coefficient at the n_th position of $f_k$.
It follows easily that functions like $f=\sum_{n\geq0}a_nz^n$ or $f=\sum_{n\lt0}a_nz^n$ can not be generators.
But I do not know how to prove functions which are not those kinds above can not be generator.
Any help would be appreciated.
Heh, good one.
If you were talking about the space of continuous real-valued functions on the circle it would be clear that there is no single generator, since a continuous real-valued function on $S^1$ cannot separate points; if $f(a)=f(b)$ then $p(f)(a)=p(f)(b)$ for all $p$ and hence the $p(f)$ are not dense.
For the space of complex-valued continuous functions, which is evidently what you're concerned with: Suppose $f$ is a generator. Arguing as above shows that $f$ must be injective on $S^1$. The Stone-Weierstrass theorem gives a clue what must go wrong: If $A$ is the uniform closure of the $p(f)$ then it must be impossible for $A$ to be self-adjoint (closed under complex conjugation), because if $f$ is injective and $A$ is self-adjoint then $A=C(S^1)$.
So we need to show that if $f\in C(S^1)$ is injective then there does not exist a sequence of polynomials $p_n$ so that $p_n(f)\to\overline f$ uniformly. This follows from some simple complex analysis!
Say $C=f(S^1)$. Saying $f$ is injective says $C$ is a simple closed curve in the plane. So it is the boundary of a bounded open set $V$. Suppose that $p_n(f)\to \overline f$ uniformly. Then $p_n(z)\to\overline z$ uniformly on $C$. The $p_n$ are holomorphic, hence they converge uniformly on $\overline V$ to a function $g$ continuous on the closure of $V$, holomorphic in $V$ and equal to $\overline z$ on the boundary.
But the function $\overline z$ is harmonic in $V$. So the maximum principle for harmonic functions shows that $g(z)=\overline z$ in $V$, contradicting the fact that $g$ is holomorphic,