Hilbert basis and Parseval's identity

Question

i am trying to do the following :

Let $H$ be a Hilbert space and $(e_n)_{n \in \mathbb{N}}$ a sequence of vectors of $H$ such that $\| e_n \|=1$ for all $n \in \mathbb{N}$. We suppose that $ \forall x\in H$,$\| x\|^2=\sum_{n=0}^{\infty}|\langle e_n,x\rangle|^2$

Show that $(e_n)_{n \in \mathbb{N}}$ is a Hilbert basis.

I know that using this i have to show that $B=Vect(e_n)_{n \in \mathbb{N}} $ is dense in $H$. I wanted to take $y\in H$ such that for all $n$, $y\perp e_n$. Then $\|y\|^2=0$ this implies $y=0$. So $B^{\perp}=\{0\}$.And i know that if $B$ is closed then $H=B \bigoplus B^{\perp} $. But i'm stuck here. Any hints ?

Thank you for your time.

Answer 1

Suppose that $\{ e_{k}\}_{k=1}^{\infty}$ is an orthonormal subset of a Hilbert space $\mathcal{H}$. Let $x\in\mathcal{H}$ be given. Then \begin{align} \left\|x-\sum_{n=1}^{k}\langle x,e_k\rangle e_k\right\|^2 &= \|x\|^2-2\Re\left\langle x,\sum_{n=1}^{k}\langle x,e_k\rangle e_k\right\rangle+\sum_{n=1}^{k}|\langle x,e_k\rangle|^2 \\ &=\|x\|^2-2\sum_{n=1}^{k}|\langle x,e_k\rangle|^2+\sum_{n=1}^{k}|\langle x,e_k\rangle|^2 \\ &=\|x\|^2-\sum_{n=1}^{k}|\langle x,e_k\rangle|^2. \end{align} From the above, it follows that $$ \lim_{k}\sum_{n=1}^{k}\langle x,e_k\rangle e_k = x\;\; \iff \sum_{n=1}^{\infty}|\langle x,e_k\rangle|^2=\|x\|^2. $$ Therefore, the Parseval equality holds for all $x$ with regard to the orthonormal set $\{e_k\}_{k=1}^{\infty}$ iff $\{ e_k\}_{k=1}^{\infty}$ is a Hilbert basis.