Prove that $\|e_n\|_{n\geq1}$ is bounded. Let $(e_n)_{n\geq1}$ a sequence in $\ell^2$ such that for all $x\in \ell^2$, we have :
$$\sum_{n=1}^\infty |\langle x,e_n\rangle|^2<\infty$$
Prove that $||e_n||_{n\geq1}$ is bounded. By Cauchy-Schwarz
$$\sum_{n=1}^\infty |\langle x,e_n\rangle|^2 \leq \sum_{n=1}^\infty | x|^2 |e_n|^2\leq \|x\|_{\ell^2}^2\|e_n\|_{\ell^2}^2$$
I'm stuck here. Any help please?
 A: Let $f_n : \ell^2 \to \ell^2$ be given by
$$ f_n (x) = (\langle x, e_1\rangle,\langle x, e_2\rangle, \cdots, \langle x, e_n\rangle, 0,0,\cdots).$$
Since
$$\sup _{n\in \mathbb N} \|f_n (x)\| = \sup_{n\in \mathbb N} \sqrt {\sum_{k=1}^n |\langle x, e_k\rangle|^2} = \sqrt {\sum_{k=1}^\infty|\langle x, e_k\rangle|^2} < \infty,$$
the uniform boundedness principle implies
$$\sup_{n\in \mathbb N} \|f_n\| < \infty.$$
Since
$$\|f_n\| \ge \|f_n (e_n/\|e_n\|)\| = \sqrt {\sum_{k=1}^n |\langle e_k ,e_n\rangle|^2} \ge \|e_n\|, $$
The sequence of norms $\{\|e_n\|\}$ is also uniformly bounded.
A: One has to prove that $\sum_n |(x,e_n)|^2 <= C$ when $||x|| <=1$ since if this is the case then for $x=e_n/||e_n||$ we get $||e_n||^2 <= C$. If this is not the case then for any k there is $||x_k|| <=1$ such that $$\sum_n |(x_k,e_n)|^2 >k$$ so for some large $n_k$ $$\sum_1^{n_k}|(x_k,e_n)|^2 > k$$ Now, $x_k$ as bounded admit a weakly convergent subsequent. Say the whole $x_k$ converges weakly to $x_0$. Then one can go with limit in the sum on the left to get $$\sum_1^{n_k}|(x_0,e_n)|^2 > k$$ Hence $$\sum_n |(x_0,e_n)|^2 = \infty$$
