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Let $\{e_n\}$ be an orthonormal sequence in an inner product space E. Then I'm trying to show the following inequality:

$$\sum_1^\infty| \langle x, e_n \rangle \langle y, e_n \rangle | \leq ||x||\cdot ||y||$$

for any $x,y \in E$. This is Bessel's inequality if $x = y$, so I'm trying to modify the proof of Bessel's inequality to show this result, but the proof of Bessel's inequality is by the Pythagorean formula which only involves one indeterminate vector in $E$. Any thoughts?

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    $\begingroup$ Expand $x=\sum_n \langle x,e_n \rangle e_n$ and $y=\sum_n \langle y,e_n \rangle e_n$... $\endgroup$ – Siminore Jun 24 '14 at 13:00
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    $\begingroup$ The left hand side cries for an application of the Cauchy-Schwarz inequality. $\endgroup$ – Daniel Fischer Jun 24 '14 at 13:00
  • $\begingroup$ You mean Cauchy-Schwarz on the factors $\langle x, e_n \rangle$ and $\langle y, e_n \rangle $, right? $\endgroup$ – Nick Jun 24 '14 at 13:01
  • $\begingroup$ @Siminore: With a twist, yes: see my answer. $\endgroup$ – Han de Bruijn Jun 25 '16 at 13:41
  • $\begingroup$ @DanielFischer: You're quite right: see my answer. $\endgroup$ – Han de Bruijn Jun 25 '16 at 13:41
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Here is a simpler proof: $$ \sum_{n=1}^\infty| \langle x, e_n \rangle \langle y, e_n \rangle | \le \sum_{n=1}^\infty| \langle x, e_n \rangle|\cdot | \langle y, e_n \rangle | \\ \le \left(\sum_{n=1}^\infty| \langle x, e_n \rangle|^2\right)^{1/2}\left(\sum_{n=1}^\infty | \langle y, e_n \rangle |^2\right)^{1/2} \le \|x\| \cdot \|y\|. $$ First inequality is Cauchy-Schwarz, second inequality is Cauchy-Schwarz in $l^2$, then Bessel's inequality.

It is worth noting that this proof does not use the assumption that $(e_n)$ is a complete orthonormal sequence, i.e., that $x=\sum_{n=1}^\infty \langle x,e_n\rangle e_n$ holds.

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According to Daniel Fischer: The left hand side cries for an application of the Cauchy-Schwarz inequality. And according to siminore: $$ x=\sum_n x_n e_n = \sum_n \langle x,e_n \rangle e_n \quad ; \quad y=\sum_n y_n e_n = \sum_n \langle y,e_n \rangle e_n $$ But we give it a twist: $$ x'=\sum_n |x_n| e_n = \sum_n \left|\langle x,e_n \rangle\right| e_n \quad ; \quad y'=\sum_n |y_n| e_n = \sum_n \left|\langle y,e_n \rangle\right| e_n $$ Then actually it is the Cauchy-Schwarz inequality: $$ \langle x',y' \rangle^2 \le \langle x',x' \rangle \langle y',y' \rangle \\ \langle \; \sum_i \left|\langle x,e_i \rangle\right| e_i \; , \; \sum_j \left|\langle y,e_j \rangle\right| e_j \; \rangle^2 \le \sum_n |x_n|^2 \cdot \sum_n |y_n|^2 \\ \left(\sum_i \sum_j \left|\langle x,e_i \rangle\right| \left| \langle y,e_j \rangle\right| \langle e_i , e_j \rangle \right)^2 \le ||x||^2 \cdot ||y||^2 $$ The Kronecker delta comes into play here: $$ \delta_{ij} = \langle e_i , e_j \rangle = \begin{cases} 1 & \text{for } i = j \\ 0 & \text{for } i \ne j \end{cases} $$ Resulting in: $$ \left( \sum_n \left|\langle x, e_n \rangle \langle y, e_n \rangle\right| \right)^2 \leq \left(||x||\cdot ||y||\right)^2 \\ $$ Now take the square root of this and you're done.

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