TITLE: Cyclic vectors for the shift operator
QUESTION [9 upvotes]: Let $S:\ell^2\to\ell^2$ be the shift operator
$$
S(x_1,x_2,\dots)=(0,x_1,x_2,\dots).
$$
Let $x\in \ell^2$ with $x_1=1$. Is $x$ cyclic for $S$? In other words, is the span of the vectors $x,Sx,S^2x,\dots$ dense in $\ell^2$?
If this does not hold for every $x$, is there a handy criterion to decide whether a given $x$ is cyclic or not?

REPLY [8 votes]: This is part of what the theory of Hardy spaces is for. There is a complete characterization of the cyclic vectors for the shift, though it applies to the Fourier transform of $x$ rather than the sequence $x=(x_0, x_1, \dots)$ itself. (For convenience I am starting the indexing at 0 rather than 1.) As in Fedor's answer, we can view the $x$ sequence as the sequence of Fourier coefficients of an $L^2$ function $h$ on the unit circle, which can be extended analytically to the disk $|z|<1$ by 
$$
h(z) = \sum_{n=0}^\infty x_n z^n.
$$
The function $h$ is said to belong to the Hardy space $H^2$, which sits as a closed subspace of $L^2$ on the circle. In this model, the right shift becomes the map $h(z)\to zh(z)$. It is then a fact that $\log|h|$ belongs to $L^1$ of the unit circle, and the inequality
\begin{equation}
\log|h(0)|\leq \frac{1}{2\pi} \int_0^{2\pi} \log|h(e^{i\theta})|
\,d\theta\end{equation}
always holds. Such an $h$ is called an outer function if equality holds (so in particular $x_0=h(0)\neq 0$ is necessary but not sufficient). Finally, it is a theorem that $h$ is cyclic for the shift if and only if $h$ is outer. All of these facts may be found e.g. in the book Banach Spaces of Analytic Functions by Kenneth Hoffman.