Let us consider $\ell^p(\mathbb{Z})$. We know that the vector $e_1=(\dots,0,0,1,0,0,\dots)$ is a cyclic vector in sense that given the right shift operator $S:(\dots,x_0,x_1,x_2,\dots)\mapsto (\dots,x_{-1},x_0,x_1,\dots)$, $$\overline{\text{span}\{S^ne_1\}_{n\in\mathbb{Z}}}=\ell^2(\mathbb{Z}).$$

I wonder if the requirement for $n\in\mathbb{Z}$ is essential. I assume that we can find a vector $\vec{c}\in\ell^2(\mathbb{Z})$ such that

$$\overline{\text{span}\{S^n\vec c\}_{n\in\mathbb{N}}}=\ell^2(\mathbb{Z}).$$

I think that using $H^2(\mathbb{T})$ can be useful but I don't know how to use it. Am I right? Can you give an example of such vector?

  • $\begingroup$ what you ask is unclear but seems related to the en.wikipedia.org/wiki/Wiener's_tauberian_theorem $\endgroup$ – reuns Feb 15 '16 at 7:49
  • $\begingroup$ In the second displayed equation should $e_1$ be $c$? $\endgroup$ – Dirk Feb 15 '16 at 8:22
  • $\begingroup$ @user1952009 but his theorem are related to $L^2(\mathbb{R})$ while I look for theorem for $L^2(\mathbb{T})$. Dirk, I don't require that. $\endgroup$ – Michael Feb 15 '16 at 9:22
  • $\begingroup$ @Michael : there is also the $l^1,l^2$ version. and I didn't that only right shifts were allowed. then the right shifts of $c_n = a^n$ with $|a| < 1$ will span $l^2(\mathbb{N})$ and I'm quite sure it is impossible to span $l^2(\mathbb{\mathbb{Z}})$ only from right shifts $\endgroup$ – reuns Feb 15 '16 at 9:34
  • $\begingroup$ @user1952009 Maybe I'm wrong but I think right in math.stackexchange.com/questions/1647144/… the answer showed that $c_n=\frac{1}{|n+1|}$ is such a vector. However, I'm looking for a "hardy-space" explanation. $\endgroup$ – Michael Feb 15 '16 at 9:38

It's possible to prove the existence of such a vector using Hardy space methods, though the relevant space is $H^1(\mathbb T)$, not $H^2(\mathbb T)$. Taking Fourier transforms, we are looking for an $f\in L^2(\mathbb T)$ such that the space $$ M=\text{span}\{z^nf(z):n\geq 0\}$$ is dense in $L^2(\mathbb T)$. I claim that any $f\in L^2(\mathbb T)$ such that 1) $f$ is nonzero almost everywhere, and 2) $\log|f|\notin L^1(\mathbb T)$, does the job. Trivially, we must have $f(z)\neq 0$ for Lebesgue a.e. $z\in\mathbb T$. From now on consider only these $f$. If the associated $M$ is not dense, then there exists a nonzero $g\in L^2(\mathbb T)$ such that $$ \int_\mathbb{T} z^n f(z)\overline{g(z)}\, dm(z)=0 $$ for all $n\geq 0$. This means that the function $h=\overline{f}g$ is nonzero and belongs to $H^1(\mathbb T)$, and is hence log integrable, i.e. $\int_\mathbb{T} \log|h(z)|\, dm(z) >-\infty$. Since $f, g$ belong to $L^2$, they also belong to $L^1$, and therefore $\int \log|f|, \int\log |g| <+\infty$. So we must have $\int_\mathbb{T}\log|f(z)|\, dm(z)>-\infty$ also. Thus if $\int \log|f|=-\infty$, we see that $M$ is dense.

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  • $\begingroup$ Nice - I'd got down to the part of wanting to show that such an $h$ need not exists, but had forgotten how the rest of the proof went. $\endgroup$ – Yemon Choi Feb 15 '16 at 13:51

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