Show that $\sum_{n=1}^\infty |\langle x,a_n\rangle|^2 \le \|x\|^2$

Question

Let $H$ be a Hilbert space. Let $(a_1,a_2,...)$ be an orthonormal sequence in $H$. Let $x \in H$. Show that $$\sum_{n=1}^\infty |\langle x,a_n\rangle|^2 \le \|x\|^2$$

I was thinking about move $\|x\|$ form RHS to LHS to make $\displaystyle\bigg\|\frac{x}{\|x\|}\bigg\|=1$,but I don't think this is helpful. Can anyone give me a hint or idea of how to prove it since I don't know how to start it.

Answer 1

If $\{\boldsymbol{\varphi}_1, \boldsymbol{\varphi}_2, ... \}$ be an orthonormal system in $\mathcal{H}$, then the Bessel's inequality is:

$$\sum_{j=1}^{\infty} |\langle \mathbf{x},\boldsymbol{\varphi}_j \rangle|^2 \leq \lVert \mathbf{x} \rVert^2 \quad \text{for every } \mathbf{x}\in \mathcal{H}.$$

Proof: \begin{align*} 0 &\leq \|\mathbf{x} -\sum_{i=1}^{n}\langle \mathbf{x},\boldsymbol{\varphi}_j \rangle\boldsymbol{\varphi}_j \|^{2}\\ &= \Big\langle \mathbf{x} -\sum_{i=1}^{n}\langle \mathbf{x},\boldsymbol{\varphi}_j \rangle\varphi_j, \mathbf{x} -\sum_{i=1}^{n}\langle \mathbf{x},\boldsymbol{\varphi}_j \rangle\varphi_j \Big \rangle \\ &= \lVert \mathbf{x} \rVert^2 - \sum_{i=1}^{n} \langle \mathbf{x},\boldsymbol{\varphi}_j \rangle \langle \boldsymbol{\varphi}_j,\mathbf{x} \rangle - \sum_{i=1}^{n} \overline{\langle \mathbf{x},\boldsymbol{\varphi}_j \rangle} \langle \boldsymbol{\varphi}_j,\mathbf{x} \rangle +\sum_{i=1}^{n} \langle \boldsymbol{\varphi}_j,\mathbf{x} \rangle \overline{\langle \mathbf{x},\boldsymbol{\varphi}_j \rangle} \\ &= \lVert \mathbf{x} \rVert^2 - \sum_{i=1}^{n} | \langle \mathbf{x},\boldsymbol{\varphi}_j \rangle |^2, \quad \forall n. \end{align*}