$C(S^1)$, as a Banach algebra, does not have a single generator 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. 
 A: 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,
