# Showing a sequence is in $\ell_2$

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?

• Hint: what does an element of the dual look like? Also, Parseval’s identity...
– J.G
Jul 9 '19 at 21:54
• presumably it is $\phi^*(e_k)^2$ in your final expression as you are taking the L2 norm here on your L2 sequence. Jul 9 '19 at 21:59
• And also yes follow the hint to under why this expression would converge to a finite number Jul 9 '19 at 22:00
• 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 $\lambda \le \lVert \phi \rVert \cdot \lambda^{1/2}$. Jul 9 '19 at 22:01
• @DanielSchepler Thanks, that should do it Jul 9 '19 at 23:04