# proof of Bessel inequality

Let

 $r_{n}=x-\sum_{k=1}^{n}\left\langle x,e_{k}\right\rangle\cdot e_{k}.$

Then for $j=1,\ldots,n$,

 $\displaystyle\left\langle r_{n},e_{j}\right\rangle$ $\displaystyle=\left\langle x,e_{j}\right\rangle-\sum_{k=1}^{n}\left\langle% \left\langle x,e_{k}\right\rangle\cdot e_{k},e_{j}\right\rangle$ (1) $\displaystyle=\left\langle x,e_{j}\right\rangle-\left\langle x,e_{j}\right% \rangle\left\langle e_{j},e_{j}\right\rangle=0$ (2)

so $e_{1},\ldots,e_{n},r_{n}$ is an orthogonal series.

Computing norms, we see that

 $\left\|x\right\|^{2}=\left\|r_{n}+\sum_{k=1}^{n}\left\langle x,e_{k}\right% \rangle\cdot e_{k}\right\|^{2}=\left\|r_{n}\right\|^{2}+\sum_{k=1}^{n}\left|% \left\langle x,e_{k}\right\rangle\right|^{2}\geq\sum_{k=1}^{n}\left|\left% \langle x,e_{k}\right\rangle\right|^{2}.$

So the series

 $\sum_{k=1}^{\infty}\left|\left\langle x,e_{k}\right\rangle\right|^{2}$

converges and is bounded by $\left\|x\right\|^{2}$, as required.

Title proof of Bessel inequality ProofOfBesselInequality 2013-03-22 12:46:41 2013-03-22 12:46:41 ariels (338) ariels (338) 4 ariels (338) Proof msc 46C05