Cyclic vector for the left regular representation 
Question : Let $f\in L^2(\mathbb S^1)$ be such that for every $n\in\mathbb Z$,
  $\hat{f}(n)\neq 0$. Prove that $f$ is a cyclic vector for the left
  regular representation $L$ on $L^2(\mathbb S^1)$.

Definition : Let $\pi$ be a unitary representation of a group $G$. An element $u$ in the representation space $H$ is called a cyclic vector if the set $\{\pi(g)u\mid g \in G\}$ (topologically) spans $H$. That is, the space of all finite linear combinations of the elements in $\{\pi(g)u\mid g \in G\}$ is dense in $H$.
The representation takes $g\in\mathbb S^1$ to $L(g)\in\text{GL}(L^2(\mathbb S^1))$, where $(L(g)f)(x)=f(g^{-1}x)$ for all $g\in \mathbb S^1,f\in L^2(\mathbb S^1) $.
I know that $L$ is unitary. So I can write $\langle L(g)f,h\rangle=\langle f,L(g^{-1})h\rangle$, where $\langle\cdot,\cdot\rangle$ is the usual inner product on $L^2(\mathbb S^1)\, \left(\langle f,h\rangle=\dfrac1{2\pi}\displaystyle\int_0^{2\pi}f(x)\overline{h(x)}\,\mathrm dx\right)$.

Let $M_f$ denote the closed subspace generated by the set $\{L(g)f\mid g \in \mathbb S^1\}$. So I need to prove that $M_f=L^2(\mathbb S^1)$. 
Assume the contrary, that $M_f\subsetneq L^2(\mathbb S^1)$. Then I can write $L^2(\mathbb S^1)=M_f\oplus M_f^{\perp}$. My idea is to show that there exists some $\chi_n\in M_f^{\perp}$ for some $n\in\mathbb Z$, where $\chi_n$ is a function on $\mathbb S^1$ defined by $\chi_n(z)=z^n$. In that case, I have $\langle f,\chi_n\rangle =0$. But it is given that $\hat{f}(n)=\langle f,\chi_n\rangle\neq 0$ for all $n\in\mathbb Z$. Hence we arrive at a contradicton.
I'm having trouble proving the existence of a $\chi_n$ in $M_f^{\perp}$. I know that $\{\chi_n\}_{-\infty}^{\infty}$ is an orthonormal basis and both $M_f$ and $M_f^{\perp}$ are $L\,$- invariant subspaces. But still, I can't seem to find a proper argument to show the existence.
Hints are appreciated. Thank you.  
 A: You probably know that the Pontryagin dual of $\mathbb S^1$ is $\mathbb Z$, and that by the Fourier inversion theorem you have
$$f = \sum _{n \in \mathbb Z} \hat f (n) \ \chi _n$$
where I am using your notations and $\hat f$ is the Fourier transform of $f$.
It follows immediately that
$$L_w f = \sum _{n \in \mathbb Z} \hat f (n) \ w^{-n} \ \chi _n$$
for every $w \in \mathbb S^1$.
Let $g = \sum _{n \in \mathbb Z} c_n \ \chi_n \in M_f ^\perp$, with $c_n \in \mathbb C \ \forall n$ and the series being convergent in the $L^2$ sense. We are going to show that $c_n = 0 \ \forall n$.
If $g \perp M_f$ then, in particular, $g \perp L_w f$ for all $w \in \mathbb S^1$, which means that (using $\langle \chi_m, \chi_n \rangle = \delta_{m,n}$)
$$0 = \langle g, L_w f \rangle = \sum _{n \in \mathbb Z} c_n \ \overline {\hat f (n)} \ w^n = \left( \sum _{n \in \mathbb Z} c_n \ \overline {\hat f (n)} \ \chi_n \right) (w) \ ,$$
whence, by dropping the argument $w$, we deduce that
$$0 = \sum _{n \in \mathbb Z} c_n \ \overline {\hat f (n)} \ \chi_n $$
in $L^2$. Since the characters $\chi_n$ are linearly independent, we deduce that $c_n  \ \overline {\hat f (n)} = 0 \ \forall n$ and since $\overline {\hat f (n)} \ne 0 \ \forall n$ by hypothesis, we are left with the only possibility $c_n = 0 \ \forall n$, whence $g=0$.
We conclude that $M_f ^\perp = 0$, whence $L^2 = M_f$.

I am too lazy to check it, but it seems to me that the proof can be adapted with only minimal notational changes to any arbitrary compact Abelian $G$.
A: First, my specialization is physics and electrical engineering, so I find some of your notation hard to understand.  I am also a relatively new user.  You seem to be describing a function $f(z): S^1 \mapsto C$, Where $z=e^{i \theta}$ is the unit circle.  You want to prove that by performing rotations on $f(z)$, $(L_ \theta f)(z) = f(e^{-i \theta}z)$, one can generate a set of functions which form a basis set in $L^2(S^1)$.  First note $(L_\theta \chi_m)(z)=e^{-im \theta}z^m$  Consider $(\int_0^{2 \pi } e^{in \theta} (L_\theta \chi_m)d\theta)(z)=2\pi \delta_{m,n}z^n$.  Thus
$(\int_0^{2 \pi } e^{in \theta} (L_\theta f)d\theta)(z)=2\pi \hat f(n)z^n$.
I believe that it can be proved by approximating the integral by a sum, that one can approximate $z^n$ to arbitrary accuracy, in the $L^2$ sense.  Certainly, for any $M$, one can find a finite set of rotations $\{\theta_i\}$ and constants $\{c_i\}$ such that
$(\sum_i c_i e^{in \theta_i} (L_{\theta_i} \chi_m))(z)=2\pi \delta_{m,n}z^n$,
for any $m,n < M$.  
