Prove that $\forall x,y \in M \; \sum^{\infty}_{n=1} \lvert \langle x,u_n\rangle\langle y,u_n\rangle\rvert \leq \lVert x\rVert\lVert y\rVert$

Question

Let $\{u_n \}^\infty_{n=1}$ an orthornormal set in an inner product space $\langle M,\langle \cdot,\cdot\rangle\rangle$

Prove that $$\forall x,y \in M \; \sum^{\infty}_{n=1} \lvert \langle x,u_n\rangle\langle y,u_n\rangle\rvert \leq \lVert x\rVert\lVert y\rVert \rvert$$

Hint: use Holder's inequality in $\ell^2$ . i.e $\forall x= (x_1,x_2,\dots ) \in \ell^2$, $\forall y =(y_1,y_2,\dots) \in \ell^2 $

$$ \sum^{\infty}_{n=1} \lvert x_n y_n\rvert\leq \left(\sum^{\infty}_{n=1} \lvert x _n \rvert\right)^{1/2} \left(\sum^{\infty}_{n=1} \lvert y_n\rvert ^2\right)^{1/2} $$

---

Attempt 1 Using the inequality right away

$$\begin{aligned} \sum^{\infty}_{n=1} \langle x,u_n\rangle\langle y,u_n\rangle &\leq (\sum\lvert\langle x,u_n\rangle\rvert^2)^{1/2} (\sum\lvert\langle y,u_n\rangle\rvert ^2)^{1/2} \\& = (\sum(\langle x,u_n\rangle \overline{\langle x,u_n\rangle}))^{1/2} (\sum (\langle y,u_n\rangle\overline{\langle y,u_n\rangle})^{1/2} \\ &=(\sum(\langle x,u_n\rangle \langle u_n,x\rangle))^{1/2} (\sum (\langle y,u_n\rangle\langle u_n,y\rangle)^{1/2} \\& \ \ \vdots \qquad (\textrm{Missing steps}) \\&=\sqrt{\langle x,x\rangle}\cdot\sqrt{\langle y,y\rangle} \\&=\lVert x\rVert \lVert y\rVert \end{aligned} $$

Answer 1

Recall Bessel's inequality:

Let $\{u_n\}$ be an orthonormal sequence in an inner product space $M$. Then for any $x\in M$ we have $$ \sum_{n=1}^\infty\left|\langle x,u_n\rangle\right|^2\leq\|x\|^2 $$

First applying Hölder's inequality, then Bessel's inequality, we have \begin{align*} \sum_{n=1}^\infty \left|\langle x,u_n\rangle\langle y,u_n\rangle\right| &\leq \left(\sum_{n=1}^\infty \left|\langle x,u_n\rangle\right|^2 \right)^{1/2}\left(\sum_{n=1}^\infty \left|\langle y,u_n\rangle\right|^2 \right)^{1/2} \\ &\leq \left(\|x\|^2 \right)^{1/2}\left(\|y\|^2 \right)^{1/2} \\ &=\|x\|\ \|y\|. \end{align*}