span density of shift operator Let $\mathcal{H}$ be a Hilbert space with complete orthonormal basis $\{e_n | n\in \mathbb{N}\}$. Let $R$ be the shift operator on $\mathcal{H}$ defined by $R(e_n)=e_{n+1}$, and extended by linearity and continuity. Show that there exists no $x\in \mathcal{H}$, such that $\text{span}\{R^{2n}(x)|n\geq 0\}$ is dense in $\mathcal{H}$.
Assuming the contrary, we get a $x\in \mathcal{H}$ such that $R^{2n}(x)$ is spanned over $\mathcal{H}$ for some $n\geq 0$, i.e. $R^{2n}(x)=c_1e_1+c_2e_2+\dots +c_ne_n$ for all $c_i\in \mathbb{R}$. Taking $R$ operated on both sides, $R^{2n+1}(x)=c_1R(e_1)+c_2R(e_2)+\dots +c_nR(e_n)=c_1e_2+c_2e_3+\dots +c_ne_{n+1}$, since $R$ is extended by linearity. Proceeding as above, after $n$-times we get, $R^{3n}(x)=c_1e_{n+1}+c_2e_{n+2}+\dots +c_ne_{2n}$. But from here, I'm unable to figure out what to do, specially how to use the fact that $R$ is extended by continuity, and all $e_i$'s are orthonormal. How can I bring the contradiction by the above process?
 A: Let $x \in H, x \ne 0$ and let $n \in \mathbb{N}$ be the smallest number such that $\langle x, e_{2n-1}\rangle \ne 0$.
Consider $y = \langle x, e_{2n}\rangle e_{2n-1}- \langle x, e_{2n-1}\rangle e_{2n}$. Clearly $y \ne 0$.
For $k = 0$ we have $\langle y, x\rangle = 0$ and for $k \ge 1$ we have $R^{2k}(x) \in \operatorname{span}\{e_j\}_{j \ge 2n+1}$ so again $\langle y, R^{2k}(x)\rangle = 0$.
Therefore $y \perp \{R^{2k}(x) : k \ge 0\}$ so it cannot be dense in $H$.
A: I have no idea why you bring the power $2n+1$ and $3n$; you only care about the even powers here. 
If you write $x=\sum_j x_je_j$ and you consider a linear combination $z=\sum_{n=0}^k c_nR^{2n}x$, you have 
$$
z=\sum_{n=0}^k c_n\sum_j x_j e_{j+2n}=\sum_j\sum_{n=0}^kc_nx_je_{j+2n}.
$$
Let us simply consider what happens with the coefficients of $e_1$ and $e_2$. Those guys can only appear when $n=0$. So the series begins with 
$$
z=c_0x_1e_1+c_0x_2e_2+\cdots
$$
For any $y=y_1e_1+y_2e_2+\cdots$, we have $$\|y-z\|^2\geq|y_1-c_0x_1|^2+|y_2-c_0x_2|^2.$$ If for instance we take $y_1=x_1$, $x_2=-x_2$, then 
$$
\|y-z\|^2\geq|1-c_0|^2|x_1|^2+|1+c_0|^2|x_2|^2.
$$
Let $r=\min\{|x_1|,|x_2|\}$. If $r>0$, we have 
$$
\|y-z\|^2\geq r\,(|1-c_0|^2+|1+c_0|^2)=2r(1+|c_0|^2)\geq 2r,
$$
and the distance cannot be arbitrarily small. If $r=0$, say $x_1=0$, then $\|e_1-z\|^2\geq1$; and similarly, if $x_2=0$, then $e_2-z\|^2\geq1$. 
Note that the above estimates work for any choice of $c_0$, so it works for any linear combination of $x,R^2x,R^4x,\ldots$
