# Right shift operator is not compact but is a limit of finite rank operator?

by user 123456   Last Updated January 12, 2018 20:20 PM

I'm doing a preparation for an exam, and I have a doubt concerning the right shift operator, for example in $\ell^2$, $S_d : \ell^2 \to \ell^2$ such that $(x_1,x_2,\ldots) \mapsto (0,x_1,\ldots)$ is standard to show tha $S_d$ is not compact (using the spectral values for example). But if we define $S_{dn} : \ell^2 \to \ell^2$ such that $(x_1,x_2,\ldots,x_n,\ldots) \mapsto (0,x_1,\ldots,x_n,0,0,\ldots)$ then $S_{dn} \to S_d$ and $S_{dn}$ is of finite rank, then $S_d$ should be a compact operator.

I know that this affirmation is false, but I can't find why.

Thank you for any help.

Tags :