Can left-shift be approximated by polynomials of right shift? Consider the left shift operator $L$ and the right shift $R$ on $l^2(\mathbb Z)$. Then both are unitary operators and inverse of each other. 
I have two question about them: 


*

*Is there $x\ne 0$ such that $Lx$ is in the closure of the span of $(R^nx)$?

*And can we show in addition that there numbers $(a_n)$ such that 
$$
Lx = \sum_{n=0}^\infty a_n R^n x \quad ?
$$


It is easy to see that the claim is false if $x$ is a one-sided sequence, i.e., there is $K$ such that ($x_k=0$ for  all $k<-K$) or ($x_k=0$ for  all $k>K$) holds. Now such sequences are dense in $l^2(\mathbb Z)$. 
But I was not able to show that no $x\ne0$ can be written as above. Is there a nice proof of this claim? Or is there an example of $x$, which works?
This question was inspired by this question and the example given there.

Edit: As commented by David C. Ullrich, the original version of the post contained two non-equivalent question. Both questions have been answered below. Thanks.
 A: Everything can be viewed in terms of $L^2[-\pi,\pi]$, where $Lf=e^{-i\theta}f$ and $Rf=e^{i\theta}f$. Then you want to know if there is $x\in L^2$ such that
$$
    e^{-i\theta}x(\theta) = \sum_{n=0}^{\infty}a_n e^{in\theta}x(\theta) \\
           x(\theta)\left(1-\sum_{n=0}^{\infty}a_n e^{i(n+1)\theta}\right)=0.
$$
To restrict the discussion, suppose $\sum_{n}|a_n|^2 < \infty$. Note that $1-\sum_n a_n e^{i(n+1)\theta}$ is the boundary function of a holomorphic function on the unit disk. This boundary function cannot vanish on a set of positive measure because that would force all of the power series coefficients to vanish, including the constant term $1$; in that case, $1-\sum_n a_ne^{i(n+1)\theta}$ would be non-zero almost everywhere, and that would force $x=0$ a.e.. The same type of argument works if $\sum_{n}|a_n|^p < \infty$ for some $1 \le p < \infty$. So it seems to me very unlikely that there would exist a sequence of coefficients $\{ a_n \}$.
A: Your two versions of the question are not equivalent (or at least they're not obviously equivalent, as you imply with that "i.e.".)
Yes, there does exist $x\ne 0$ such that $Lx$ is in the closed span of $(R^nx)$. (Which does not imply that $Lx=\sum a_nR^nx$, by any argument I see - if you believe it does imply that you should provide a proof.)
Say $K$ is a non-trivial closed arc on $\Bbb T$, with $K\ne\Bbb T$. Runge's theorem says that there exist (holomorphic) polynomials $P_n$ such that $P_n(z)\to1/z$ uniformly on $K$. If, say, $f$ is the characteristic function of $K$ this shows that $||e^{-it}f(t)-P_n(e^{it})f(t)||_2\to0$. If we say $$P_n(z)=\sum_{j=0}^{N_n}a_{j,n}z^j$$and $$x_j=\hat f(j)$$this says precisely that $$||Lx-\sum_{j=0}^{N_n}a_{j,n}R^jx||_2\to0.$$
