The question is the following
Let $S= \{e_\alpha:\alpha \in \mathcal{A}\}$ be an orthonormal set in an inner space $X$. Show that for any $x,y \in X$
$$
\sum_{\alpha \in \mathcal{A}} |\langle e_\alpha, x\rangle \langle e_\alpha,y \rangle| \leq \|x\| \cdot \|y\|
$$
My attempt:
I tried to follow from the proof of Bessel's inequality, and here is what I did:
Let $S = \{e_\alpha:\alpha \in \mathcal{A}\}$ be an orthonormal set in an inner product space $X$. For any $x,y \in X$, take a finite subset $\mathcal{A}' \subset \mathcal{A}$ and compute the following:
\begin{align*}
&\left\langle \left(x-\sum_{\alpha \in \mathcal{A}'} \langle e_\alpha x\rangle e_\alpha\right),\ \left(y-\sum_{\beta \in \mathcal{A}'} \langle e_\alpha y\rangle e_\alpha\right) \right\rangle \\
=\ &\langle x,y\rangle-\sum_{\beta \in \mathcal{A}'} \langle e_\beta, y\rangle \langle x, e_\beta \rangle -\sum_{\alpha \in \mathcal{A}'} \overline{\langle e_\alpha x\rangle} \langle e_\alpha, y \rangle + \sum_{\alpha,\beta \in \mathcal{A}'}\overline{\langle e_\alpha x\rangle} \langle e_\beta, y \rangle \langle e_\alpha, e_\beta \rangle \\
=\ &\langle x,y\rangle-\sum_{\alpha \in \mathcal{A}'} \langle e_\alpha, y\rangle \langle x, e_\alpha \rangle
\end{align*}
Therefore
$$
\sum_{\alpha \in \mathcal{A}'} \langle e_\alpha, y\rangle \langle x, e_\alpha \rangle = \langle x,y\rangle-\left\langle \left(x-\sum_{\alpha \in \mathcal{A}'} \langle e_\alpha x\rangle e_\alpha\right),\ \left(y-\sum_{\beta \in \mathcal{A}'} \langle e_\alpha y\rangle e_\alpha\right) \right\rangle
$$
Here I wanted to apply the Cauchy-Schwarz inequality:
$$|\langle x,y\rangle| \leq \|x\|\cdot \|y\|$$
but I don't know how to apply the absolute value. Or is there any simpler way to do it? Any help is appreciated.
Best Answer
$\sum_{\alpha \in \mathcal A} |\langle e_{\alpha} , x \rangle| |\langle e_{\alpha}, y \rangle| \leq \sqrt {\sum_{\alpha \in \mathcal A}|\langle e_{\alpha} , x \rangle|^{2}} \sqrt {\sum_{\alpha \in \mathcal A}|\langle e_{\alpha} , y \rangle|^{2}} $ from which the result follows.