# Do powers of the shift operator applied to a non-zero vector always yield a total set?

Let $$S$$ be the (say, left) shift operator on $$\ell^2(\mathbb{Z})$$. For a non-zero vector $$x \in \ell^2(\mathbb{Z})$$, consider the set $$X = \{ S^n v \mid n \in \mathbb{Z} \}.$$ Is this always a total set, i.e., is its span dense in $$\ell^2(\mathbb{Z})$$?

• What if $v$ is the indicator function of $\{1,...,100\}$ (i.e. it has value 1 on these numbers, zero everywhere else). Can you approximate the indicator function of $\{1\}$ arbitrarily well? Aug 4 at 9:03
• The shift operator is unitarily equivalent to multiplication by z on L^2(S^1), for which there are vectors that are not cyclic (e.g. characteristic function of an interval). Aug 4 at 9:04
• Thanks for the observation! Can you maybe post this as an answer so that I can accept? Aug 4 at 10:57
• Sure I can do that! Aug 4 at 11:19
• @MatthiasLudewig It may also be worth pointing out that the property that you ask about will still fail if you replace S by any other unitary operator U (on a Hilbert space of dimension greater than 1). The commutant {U,U*}' will contain a non-trivial projection; if it did not then {U,U*}'' would be all of B(H), but the von Neumann algebra {U,U*}'' generated by U is abelian. Any vector in the range of such a projection gives a counterexample. Aug 5 at 0:49

Such sets are not always total. The shift operator $$S$$ is unitarily equivalent to multiplication by $$z$$ on $$L^2(S^1)$$. From this perspective you can see vectors for which the set you write is not total, for example the characteristic function of an interval.
• The unitary equivalence between $L^2(S^1)$ and $\ell^2(\mathbb{Z})$ that translates the multiplication by $z$ operator into $S$ is the operation that associates to a function on $S^1$ (equivalently, a periodic function on $\mathbb{R}$) the coefficients of its Fourier series. Aug 4 at 13:04
• Taking $v$ to be the characteristic function of a (proper, closed) subinterval $I$ of $S^1$, the functions you get from repeated multiplication by $z$ and taking linear combinations will all have support in $I$, hence cannot be dense in $L^2(S^1)$. Aug 4 at 13:06