For $(l^2,\|\cdot\|_2)$ and $e_n=(0,0,.,1,0,.)$ and a bounded linear functional $\Phi$ find $p\geq 1$ where $\sum_{n=1}^\infty |b_n|^p$ converges?

Question

For $(l^2,\|\cdot\|_2)$ and $e_n=(0,0,...,1,0,...)$ and a bounded linear functional $\Phi$ find a value of $p\geq 1$ where $\sum_{n=1}^\infty |b_n|^p$ converges for $b_n=\Phi(e_n)$?

Ok so since $\Phi$ is bounded I know $|\Phi(e_n)|\leq C\|e_n\|_2 = C$ so:

$$\sum_{n=1}^\infty |b_n|^p=\sum_{n=1}^\infty |\Phi(e_n)|^p \leq \sum_{n=1}^\infty |C|^p$$

but what does this prove? Or have I made a mistake?

Answer 1

$p=2$ does the job. To see this, represent the bounded linear form $\Phi$ as $\Phi(x)=\langle x_0,x \rangle$. Bessel's equality gives the wanted convergence. Indeed, we have $$\sum_{n=1} ^\infty|\langle x_0,e_n\rangle|^2=\lVert x_0\rVert^2 $$ and $\langle x_0,e_n\rangle =\Phi(e_n)$. In general, we cannot hope for a better $p$, that is, $p\lt 2$. Indeed, take $1/2\lt\alpha\leqslant 1/p$ and $x_0:=\sum_n n^{-\alpha}e_n$.