# How to show that the sequence $\{T (e_n)\}_{n \geq 1}$ is square summable?

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.

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?

• Actually I am considering the sequences to be complex.
– RKC
Apr 18, 2023 at 7:38
• Got it on my own. I have just saw your answer while I am writing my answer. Thanks though.
– RKC
Apr 18, 2023 at 7:40
• 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 $|T(x^N)| \leq K \|x^N\|.$
– RKC
Apr 18, 2023 at 7:40
• Actually using Cauchy-Schwarz inequality (more generally using Hölder's inequality for $p = q = 2$) we have $\|T\|_{\text {op}} = \sum\limits_{n = 1}^{\infty} |T(e_n)|^2.$
– RKC
Apr 18, 2023 at 7:43
• $:$ 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.
– RKC
Apr 18, 2023 at 7:45