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Let the linear operator $T:l^2\rightarrow l^2$ be defined by $y=Tx$ where $x=\{\xi_j\}$, $y=\{\eta_j\}$, and $\eta_j = \alpha_j \xi_j$, where $\{\alpha_j\}$ is a dense sequence in $[0,1]$. Does $y=Tx\in l^2, \forall x\in l^2$, imply that $\{\alpha_j\}\in l^2$?

In Kreyszig's "Introductory Functional Analysis with Applications", he defines such an operator in Sec. 7.3, Problem 4 and asks about the spectrum.

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    I probably missed something, but if we take any bounded sequence and we define $T$ in this way, we have $Tx\in\ell^2$ whenever $x\in\ell^2$.2012-12-15

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No. As Davide said, every bounded sequence $(\alpha_j)$ determines a bounded operator on $\ell^2$ in this way. The assumption that each $\alpha_j$ is in $[0,1]$ guarantees that $T\in B(\ell^2)$. Furthermore, the assumption that $\{\alpha_j\}$ is dense in $[0,1]$ guarantees that $(\alpha_j)$ is not in $\ell^2$.

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    Yes. Corrected the original. Thank you.2012-12-15