Consider a Hilbert Space $H$ such that $\sum_{n=0}^\infty \|x_n-y_n\| < 1.$ Show that if $z\perp y_n$ with $\forall n>0$ then $z=0$

Question

Let $H$ be a separable Hilbert space and $(x_n)_{n\in \mathbb{N}}$ a complete orthonormal set. If $(y_n)_{n \in \mathbb{N}}$ is a sequence such that $$\sum_{n=0}^\infty \|x_n-y_n\| < 1.$$ Show that if $z\perp y_n$ with $\forall n>0$ then $z=0$

Hi, can you help me with this exercise, I think in resolve it with Parseval's identity.

Answer 1

\begin{align*} \|z\|&=\left(\sum\left|\left<z,x_{n}\right>\right|^{2}\right)^{1/2}\\ &=\left(\sum\left|\left<z,x_{n}-y_{n}\right>+\left<z,y_{n}\right>\right|^{2}\right)^{1/2}\\ &=\left(\sum\left|\left<z,x_{n}-y_{n}\right>\right|^{2}\right)^{1/2}\\ &\leq\sum\left|\left<z,x_{n}-y_{n}\right>\right|\\ &\leq\sum\|z\|\|x_{n}-y_{n}\|\\ &=\|z\|\sum\|x_{n}-y_{n}\|, \end{align*} if $z\ne 0$, then we get a contradiction.