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Let $S$ be the bilateral shift on $\ell^2(\mathbb{Z})$ and let $T = S + S^*$. I want to show that there is no cyclic vector for the representation of $T$ on $\ell^2(\mathbb{Z})$ i.e. $\forall x\in \ell^2(\mathbb{Z})$ the set $P_x = \{f(x) : f\mbox{ is a polynomial in }T\}$ is not dense in $\ell^2(\mathbb{Z})$.

I tried to start simple, showing that if $x = \delta_n$ for $n \in \mathbb{Z}$ then $P_x$ isn't dense. This worked out, but I was exploiting a symmetry that doesn't exist when considering finite linear combinations of such elements. I'm pretty sure this idea isn't going to work out.

Then I tried considering translating this operator to an operator on $L^2(S^1)$ via the Fourier transform. The corresponding operator $\hat{T}$ is simply a multiplication operator, the multiplying function being $g(t) = 2\cos(t)$. This seemed more promising initially, but I still don't see how to approach the problem outside of explicitly constructing an element $y_x$ s.t. $y_x \notin \bar{P}_x$ for arbitrary $x$.

Any hints or pointers in the right direction would be much appreciated.

Edit: Working in the $L^2(S^1)$ setting, the problem amounts to showing that $\forall f\in L^2(S^1)$ $\exists h \in L^2(S^1)$ such that $h$ cannot be approximated in $L^2$ norm by functions of the form $(\sum_{n=0}^m c_n cos^n(t))f(t).$ This seems like it should be clear, but I have been unable to prove it for general $f$.

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    Nevermind, I've figured it out. I figured I'd mention it in the comments instead of editing since that would bring it back to the front page.2011-11-17
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    The system has bounced the question back to the front page anyway (Jan 12 2012) so why not type up your solution here? That way the question isn't left "hanging".2012-01-13
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    Bounced again (Feb 11 2012)2012-02-12
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    To stop the cycle of bouncing, I upvoted the incorrect answer: user19255 gets +1 for effort.2012-08-10
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    That's probably not wise. We don't want to give credence to incorrect answers.2013-02-17

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