Bessel-like inequality 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?
 A: 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.
A: 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.
