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I am trying to prove the $(\ell^2)^{\ast}$ is isometrically isomorphic to $\ell^2.$ For this I considered a linear map $\phi : \ell^2 \longrightarrow (\ell^2)^{\ast}$ defined by $\phi (x) = \widehat {x},$ where $\widehat {x} (y) = \left \langle x, y \right \rangle.$ I have shown that $\phi$ is a linear isometry. So the only thing which is left to show is that the map is surjective. For that I note that if $x = \sum\limits_{n \geq 1} x_n e_n \in \ell^2$ then for $\tau \in V^{\ast}$ we have $\tau (x) = \sum\limits_{n \geq 1} x_n \tau (e_n).$ Now we define a sequence $y$ by $y_n = \overline {\tau (e_n)}.$ If we can show that $y \in \ell^2$ then we have $\tau = \widehat {y} = \phi (y)$ and hence the surjection of $\phi$ follows. This leads me to ask the following question $:$

Given $T \in (\ell^2)^{\ast}$ is it always true that the sequence $\{T(e_n)\}_{n \geq 1}$ is square summable i.e. $\sum\limits_{n \geq 1} | T (e_n)|^2 \lt \infty\ $? If so, how do I prove it?

Any help in this regard would be warmly appreciated. Thanks for your time.

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Let $x_n=\overline {Te_n}$ for $n \leq N$ and $0$ for $n>N$. Then $\tau(x_n)=\sum\limits_{k=1}^{N}|Te_n))|^{2}$. Since $\tau$ is a bounded linear functional we get $\sum\limits_{k=1}^{N}|Te_n|^{2} \leq \|\tau\| \|(x_n)\|\leq \|\tau\| \sqrt {\sum\limits_{k=1}^{N}|Te_n)|^{2}}$. Can you finish?

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  • $\begingroup$ Actually I am considering the sequences to be complex. $\endgroup$
    – RKC
    Apr 18, 2023 at 7:38
  • $\begingroup$ Got it on my own. I have just saw your answer while I am writing my answer. Thanks though. $\endgroup$
    – RKC
    Apr 18, 2023 at 7:40
  • $\begingroup$ Idea $:$ I am thinking about taking sequences $x^N \in \ell^2$ such that $x_n =\overline {T(e_n)}$ for all $n \leq N$ and $0$ otherwise. Then using the fact that $|T(x^N)| \leq K \|x^N\|.$ $\endgroup$
    – RKC
    Apr 18, 2023 at 7:40
  • $\begingroup$ Actually using Cauchy-Schwarz inequality (more generally using Hölder's inequality for $p = q = 2$) we have $\|T\|_{\text {op}} = \sum\limits_{n = 1}^{\infty} |T(e_n)|^2.$ $\endgroup$
    – RKC
    Apr 18, 2023 at 7:43
  • $\begingroup$ $:$ I didn't downvote you. In fact, I haven't downvoted anybody in this site so far till I've been to this site. So please don't blame me. As I have said I have solved it on my own. I have added my ideas in the above comment which coincides with that of your's and so I refrained from answering my own question. Extremely sorry to know that you were getting downvoted after posting a valid answer. $\endgroup$
    – RKC
    Apr 18, 2023 at 7:45

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