# Particular series on Hilbert Space

Let $(H, \langle\cdot,\cdot\rangle)$ a Hilbert space and consider a sequence $\{x_n\}_{n\in\mathbb{N}}$ of $H$ such that: $$\langle x_n,x_m\rangle\ =\ \delta_{mn}\ =\ \left\{\begin{array}{ll}1, & n = m\\0, & n\neq m\end{array}\right.$$ Show that $$\sum_{n=1}^{\infty}|\langle x,x_n\rangle|^2\ \leq\ \|x\|^2,\ \forall x\in H.$$ Moreover, given a scalar sequence $\{\alpha_n\}_{n\in\mathbb{N}}$, show that the following are equivalent:

1. $\displaystyle\sum_{n=1}^{\infty}\alpha_nx_n$ converge in $H$.
2. $\displaystyle\sum_{n=1}^{\infty}|\alpha_n|^2 < +\infty$
• Is this homework? Have a look for Bessels inequality. – gerw Mar 28 '13 at 11:37
• Yes is a homework and I didn't know that this problem is about the Bessels inequality. Thanks – FASCH Mar 28 '13 at 11:59

Some hints: let $F_N$ be the subspace generated by $\{x_1,\dots,x_N\}$. It's a closed subspace of $H$. So we can consider the projection over this subspace. This gives that $\sum_{n=1}^N|\langle x,x_n\rangle|^2\leqslant \lVert x\rVert^2$.
• When I try to prove form b) to a), I have that $$\left\|\sum_{n=1}^{\infty}\alpha_nx_n\right\| = \left(\sum_{n=1}^{\infty}|\alpha_n|^2\right)^{1/2} < +\infty.$$ How I can justify that $\sum\limits_{n=1}^{\infty}\alpha_nx_n$ converge on $H$? – FASCH Mar 28 '13 at 16:13
• I proved that $$\left\|\sum_{k=n}^m\alpha_kx_k\right\| = \left(\sum_{k=n}^m|\alpha_k|^2\right)^{1/2},$$ then as $m>n\rightarrow \infty$, is correct to say that $\left\|\sum\limits_{k=n}^m\alpha_kx_k\right\| \rightarrow 0$ because $\sum\limits_{k=n}^m|\alpha|^2$ is bounded? – FASCH Mar 28 '13 at 18:24
• It's because the latest quantity you mention converges to $0$, not only because it is bounded. – Davide Giraudo Mar 28 '13 at 18:27