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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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    $\begingroup$ 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 $x\in\ell^2$. $\endgroup$ – Davide Giraudo Dec 15 '12 at 21:47
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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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  • $\begingroup$ Ah yes thank you. So $||y||^2 = \sum_{j=1}^\infty | \alpha_j \xi_j |^2 \leq \sup_j | \alpha_j |^2 \sum |\xi_j|^2| \leq ||x||^2$. And since $(\alpha_j)$ is dense in $[0,1]$ we can find a subsequence of $(\alpha_j)$ converging to $1$, so that $(\alpha_j)$ cannot be square-summable. $\endgroup$ – CCL Dec 15 '12 at 23:36
  • $\begingroup$ CCL: Yes, except you mean $\|y\|^2$ and $\|x\|^2$. And thanks for the edit. I don't know how $\alpha$ became $\eta$. $\endgroup$ – Jonas Meyer Dec 15 '12 at 23:37
  • $\begingroup$ Yes. Corrected the original. Thank you. $\endgroup$ – CCL Dec 15 '12 at 23:39

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