# Regarding Right Shift Operator

Let $H$ be a Hilbert space with complete orthonormal basis $\{e_n$ : $n$ $\in$ $\mathbb{N}$}. Let $R$ be the right shift operator on $H$ given by $R(e_n)$ = $e_{n+1}$ and extend it by linearity and continuity. Show that there is no $x$ $\in$ $H$ such that $X$ = $span$ $\{R^{2k}x$ : $k$ $\in$ $\mathbb{N}$} is dense in $H$. I have used the definition of $R$ and the Fourier expansion formula to get $X$ = $span$ {$\sum_{n=1}^{\infty} <x,e_n>e_{n+2k}$: $k$ $\in$ $\mathbb{N}$}, but I cannot conclude that this set is not dense in $X$ for a fixed $x$ $\in$ $H$. Thanks for any help.

Think of elements of $H$ as infinite "vectors" $(x_1,x_2,x_3,\ldots)$. Call this $x$, then the $R^{2k}x$ are $(x_1,x_2,x_3,\ldots)$, $(0,0,x_1,x_2,x_3,\ldots)$, $(0,0,0,0,x_1,x_2,x_3,\ldots)$. All of these are orthogonal to $(-x_2,x_1,0,0,\ldots)$.
OK, this doesn't work (but is easily fixed) if $x_1=x_2=0$.
• $(-x_2,x_1,0,\ldots)$ is orthogonal to the (closure of the) span of the $T^{2k}x$. May 4 '17 at 5:16