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Let $(X,\langle ,\rangle)$ a euclidean space and $(e_n)\in X^\infty$ a orthonormal succession. Show that $\forall x,y \in X$

$$\sum_{k=1}^{\infty}|\langle x,e_k\rangle\langle y,e_k\rangle| \leq \|x\|\|y\|$$

Solution: Let $u_k=(\langle x,e_k\rangle)^{\frac12}$ and $v_k=(\langle y,e_k\rangle)^{\frac12}$, now by C-B-S (Cauchy-Bunyakovsky-Schwarz Inequality) we have,

$$\sum_{k=1}^{\infty }u_k^2 v_k^2=\|u_k\|^2\|v_k\|^2$$

where

$$\sum_{k=1}^{\infty}|\langle x,e_k\rangle\langle y,e_k\rangle| \leq \langle x,e_k\rangle\langle y,e_k\rangle\leq \|x\|\|y\|$$

then

$$\sum_{k=1}^{\infty}|\langle x,e_k\rangle\langle y,e_k\rangle| \leq \|x\|\|y\|$$

this is correct?

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For any $x$, the sequence $x_n = \langle x,e_n \rangle$ is in the classic Hilbert space $\ell^2$, and $\sum_{n}|\langle x,e_n \rangle|^2 \le \|x\|^2$ follows from Bessel's inequality. Using Cauchy-Schwarz for $\ell^2$ and Bessel's inequality gives $$ \sum_n |\langle x,e_n\rangle\langle e_n,y\rangle| \le \left(\sum_n |\langle x,e_n\rangle|^2\right)^{1/2}\left(\sum_n |\langle e_n,y\rangle|^2\right)^{1/2} \le \|x\|\|y\|. $$ Your statement after "where" doesn't add up.

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  • $\begingroup$ I'm using Cauchy-Bunyakovsky-Schwarz Inequality $| \langle x, e_k \rangle| \leq \| x\| \|e_k \|$, but $\| e_k \| =1 $. This right? $\endgroup$ – Sergio Gonzales Feb 6 '17 at 2:41

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