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 $\ell^2(\mathbb{Z})$?
$\begingroup$
$\endgroup$
6
asked Aug 4, 2021 at 8:47
-
$\begingroup$ 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? $\endgroup$– M. WinterCommented Aug 4, 2021 at 9:03
-
3$\begingroup$ 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). $\endgroup$– James TenerCommented Aug 4, 2021 at 9:04
-
$\begingroup$ Thanks for the observation! Can you maybe post this as an answer so that I can accept? $\endgroup$– Matthias LudewigCommented Aug 4, 2021 at 10:57
-
$\begingroup$ Sure I can do that! $\endgroup$– James TenerCommented Aug 4, 2021 at 11:19
-
2$\begingroup$ @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. $\endgroup$– James TenerCommented Aug 5, 2021 at 0:49
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
4
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.
-
1$\begingroup$ @1.. Of what? Both statements are essentially obvious. $\endgroup$– abxCommented Aug 4, 2021 at 12:41
-
$\begingroup$ 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. $\endgroup$ Commented Aug 4, 2021 at 13:04
-
1$\begingroup$ 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 $L^2(S^1)$. $\endgroup$ Commented Aug 4, 2021 at 13:06
-