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Suppose that $\varphi$ is an element of the dual of $\ell_2$. Then I need to show that the sequence $$(\varphi(e_1),\varphi(e_2), \varphi(e_3), \cdots)$$ is in $\ell_2$ (where $\{e_k\}$ is of course the standard basis for $\ell_2$).

So far, I've found that $$ \begin{align*} \sum_{k = 1}^N |\varphi(e_k)|^2 &= \sum_{k = 1}^N \varphi(e_k) \varphi^*(e_k) \\ &= \varphi\left(\sum_{k = 1}^N e_k\varphi^*(e_k) \right) \end{align*} $$ This seems promising, but the best I can seem to do in order to bound the right-hand side is $$ \begin{align*} \varphi\left(\sum_{k = 1}^N e_k\varphi^*(e_k) \right) &\leq \Vert \varphi \Vert \cdot \left\Vert \sum_{k = 1}^N e_k \varphi^*(e_k) \right\Vert \\ &\leq \Vert \varphi \Vert \cdot \left( \sum_{k = 1}^N \varphi^*(e_k) \right)^{1/2} \end{align*} $$ which appears to go $\infty$ as $N \to \infty$. How can I find a better bound?

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    $\begingroup$ Hint: what does an element of the dual look like? Also, Parseval’s identity... $\endgroup$ – J.G Jul 9 at 21:54
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    $\begingroup$ presumably it is $\phi^*(e_k)^2$ in your final expression as you are taking the L2 norm here on your L2 sequence. $\endgroup$ – George Dewhirst Jul 9 at 21:59
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    $\begingroup$ And also yes follow the hint to under why this expression would converge to a finite number $\endgroup$ – George Dewhirst Jul 9 at 22:00
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    $\begingroup$ Umm, the $\ell_2$-norm of $\sum_{k=1}^N e_k \overline{\phi(e_k)}$ is $\left[ \sum_{k=1}^N |\phi(e_k)|^2 \right]^{1/2}$. So, with $\lambda = \sum_{k=1}^N |\phi(e_k)|^2$ you have $\lambda \le \lVert \phi \rVert \cdot \lambda^{1/2}$. $\endgroup$ – Daniel Schepler Jul 9 at 22:01
  • $\begingroup$ @DanielSchepler Thanks, that should do it $\endgroup$ – jacer21 Jul 9 at 23:04

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