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?

$\begingroup$ You mean the span is dense? And you mean $x_2$, $x_3$, etc. are zero? Is Halmos: Hilbert space problem book, problem 160 related to your question? $\endgroup$ – András Bátkai Feb 19 '16 at 7:22

$\begingroup$ If the $x_2,x_3$ etc, are zero, the answer is clear. So I don't assume this to holds. $\endgroup$ – user1688 Feb 19 '16 at 7:33

$\begingroup$ Intuitively yes, because you can recursively approximate the basis vectors: $xx_2\cdot Sx + (x_2^2x_3)S^2x+ \ldots$. But I am travelling now, cannot think about it seriously. Sorry. I hope this works out. $\endgroup$ – András Bátkai Feb 19 '16 at 8:40
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 $\logh$ belongs to $L^1$ of the unit circle, and the inequality \begin{equation} \logh(0)\leq \frac{1}{2\pi} \int_0^{2\pi} \logh(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.

1$\begingroup$ Nice. I had some vague memory that Hardy spaces were involved but I didn't remember just how. $\endgroup$ – Nik Weaver Feb 19 '16 at 17:22

1$\begingroup$ Hm, I do not understand something here. At first, if $h(z)=1z$, then $h$ is cyclic, since $1(z+z^2+\dots+z^n)/n=:h(z)p_n(z)$ tends to 1 in $H^2$. But $1/h\notin H^2$. $\endgroup$ – Fedor Petrov Feb 20 '16 at 17:18

$\begingroup$ At second, if $h$ does not has zeroes inside the unit disc, $\log h$ is locally harmonic function (it is a real part of analytic function $\log h$ which is locally well defined), hence globally harmonic in the unit disc, hence it coincides with its average over a circle, i.e. your inequality is always an equality. $\endgroup$ – Fedor Petrov Feb 20 '16 at 17:21

$\begingroup$ Of course $1/h$ need not belong to $H^2$ when $h$ is outer, my edit was too hasty and I have removed it. However the inequality is not always an equality for nonvanishing functions; in particular it is a strict inequality whenever $h$ has a singular inner factor...the simplest example is $h(z)=exp((z+1)/(z1))$. $\endgroup$ – Mike Jury Feb 20 '16 at 17:38

2$\begingroup$ OK, I see the source of confusion, I had a superfluous $r$ inside the integral...the integral is meant to be taken over the unit circle. $\endgroup$ – Mike Jury Feb 20 '16 at 17:49
Not always. Let $(e_n)$ be the standard basis of $l^2$ and take $x = e_1  2e_2$. For any $k$ the vector $S^kx$ is orthogonal to the sequence $(2^{n})$, so $x$ is not cyclic.
You ask for a handy criterion that $\overline{\rm span}(x, Sx, S^2x, \ldots)$ equals $l^2$. I don't know, but note that this subspace is invariant for $S$, so it is necessary and sufficient that it contain the vector $e_1$. So a sufficient condition for cyclicity is that the series mentioned by András in the comments should converge. I guess that is probably pretty close to a necessary condition, too, because you need to use a series close to this one in order to get close to $e_1$.
I suppose my conjecture is that $x$ is cyclic if and only if András's series converges weakly. You can see that weak convergence might be necessary by taking $x = e_1  e_2$. This vector is cyclic because anything that is orthogonal to $S^kx$ for all $k$ must be a constant sequence and hence, if it is in $l^2$, the zero sequence. But the series you get by trying to solve $\sum a_k S^kx = e_1$ is $x + Sx + S^2 x + \cdots$, which only converges weakly.
Consider $x=(x_0,x_1,\dots)\in \ell^2$ as a function $F_x:=\sum_{k=0}^{\infty} x_k z^k\in L^2({\mathbb S}^1)$ (this is normpreserving). Then shift operator multiples functions by $z$. If $x$ is so that $F_x(a)=0$ for some complex $a$, $a<1$, then also $a^k F_x(a)=F_{S^k x}(a)=0$, and since $$ f(z)\rightarrow f(a)=\frac1{2\pi i}\int \frac{f(z)dz}{za} $$ is a continuous functional on $L^2$ restricted to our space (well, not just our, but Hardy space), we see that orbit of $x$ under $S$ lies in a proper closed subspace. In Nik Weaver's example function is $12z$ and $a=1/2$.

2$\begingroup$ I am starting to suspect that you are more than one person. You know too many things ... $\endgroup$ – Nik Weaver Feb 20 '16 at 17:56

$\begingroup$ @NikWeaver have you ever been in St. Petersburg? Every busker on Nevsky knows Hardy spaces better than me. $\endgroup$ – Fedor Petrov Feb 23 '16 at 11:23