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

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.

• What happens when you project $x$ onto the subspace of $H$ spanned by $a_{1}$, $a_{2}$, $\ldots$? – Brian Borchers Oct 31 '17 at 2:09
• This is Bessel's inequality – K.Power Oct 31 '17 at 2:09
• bessel inequality – Guy Fsone Oct 31 '17 at 2:11
• It's more or less just Pythagoras's Theorem. – Theo Bendit Oct 31 '17 at 2:24

If $\{\boldsymbol{\varphi}_1, \boldsymbol{\varphi}_2, ... \}$ be an orthonormal system in $\mathcal{H}$, then the Bessel's inequality is:
$$\sum_{j=1}^{\infty} |\langle \mathbf{x},\boldsymbol{\varphi}_j \rangle|^2 \leq \lVert \mathbf{x} \rVert^2 \quad \text{for every } \mathbf{x}\in \mathcal{H}.$$