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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.

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1 Answer 1

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$\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.

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  • $\begingroup$ Is it the Holder's inequality? I was worried whether it can be applied to an uncountable set. $\endgroup$
    – Tab1e
    Commented Dec 6, 2019 at 5:57
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    $\begingroup$ @UnbelieveTable An uncoutable sum of positive numbers is defined the supremum of finite sums. So if you write down the Cauchy - Schwartz /Holder's inequality for finite sums you can take suprema on both sides to get it for any uncoutable sum also. $\endgroup$ Commented Dec 6, 2019 at 6:00

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