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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. $\displaystyle\sum_{n=1}^{\infty}|\alpha_n|^2 < +\infty$
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  • $\begingroup$ Is this homework? Have a look for Bessels inequality. $\endgroup$ – gerw Mar 28 '13 at 11:37
  • $\begingroup$ Yes is a homework and I didn't know that this problem is about the Bessels inequality. Thanks $\endgroup$ – FASCH Mar 28 '13 at 11:59
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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 $\sum_{n=1}^N|\langle x,x_n\rangle|^2\leqslant \lVert x\rVert^2$.

For the equivalence, you can compute the norm of the partial sums.

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  • $\begingroup$ 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 $H$? $\endgroup$ – FASCH Mar 28 '13 at 16:13
  • $\begingroup$ Show that the sequence of partial sums is Cauchy. $\endgroup$ – Davide Giraudo Mar 28 '13 at 16:23
  • $\begingroup$ 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? $\endgroup$ – FASCH Mar 28 '13 at 18:24
  • $\begingroup$ It's because the latest quantity you mention converges to $0$, not only because it is bounded. $\endgroup$ – Davide Giraudo Mar 28 '13 at 18:27
  • $\begingroup$ I see that, but then, Where I did use the boundedness? $\endgroup$ – FASCH Mar 28 '13 at 18:28

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