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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.

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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 $x_1=x_2=0$.

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  • $\begingroup$ I don't quite get your answer. Can you please explain a bit? $\endgroup$
    – Ester
    May 4 '17 at 5:14
  • $\begingroup$ $(-x_2,x_1,0,\ldots)$ is orthogonal to the (closure of the) span of the $T^{2k}x$. $\endgroup$ May 4 '17 at 5:16

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