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 $f$.
I think that I have a partial answer.
Let $a = (a_i)_{i \in \mathbb{Z}}$ be cyclic vector for $T$
Claim: this forces $a$ to vanish either on all odd or all even integers.
Define two new vectors $b_1, b_2$ by the rule $$ (b_k)_i = (-1)^{i+k}a_i.$$
Notice that we must have at least one $b_k \neq a$. Otherwise $a = 0$.
Sub-Claim: We have $<b_k,T^n a> = 0$ for all integers $n > 0$ and for $k=1,2$.
Proof: Just going through by hand and checking that a bunch of stuff cancels.
Now $b_k$ is orthogonal to all of $\{f(T)a: f \in \mathbb{C}[x], f(0)=0\}$ by linearity. If $Q = \{f(T)a: f \in \mathbb{C}[x]\}$ is dense in $\ell^2(\mathbb{Z})$, then we must be able to approximate $b_k$ with elements of $Q$. Thus we must be able to approximate $b_k$ with scalar multiples of $a$.
It's not too hard to see that this forces $a$ to vanish either on all even or all odd integers. For if $a$ has a nonzero entry $a_{2k+1}$ and a nonzero entry $a_{2j}$, then if $(\lambda_m a)_{m=1}^{\infty}$ is a sequence of multiples of $a$ converging to $b_1$, then we must have $\lambda_m \to -1$ and $\lambda_m \to 1$ by division by the entry $a_{2k+1}$ or $a_{2j}$ respectively.
Thus we have shown that if $a$ is a cyclic vector, then it vanishes either on all even or all odd integers.
I think that you can keep on replicating this procedure to force all the entries of $a$ to be $0$, but I don't feel like churning out the rest of the details.