# Complete orthonormal and countable set in a Hilbert space

$\{x_n\}$ is a complete orthonormal countable set in a Hilbert space and $\{y_n\}$ is a countable set. It is known that $\underset {n\ge 1} \sum ||x_n-y_n||<1$. I need to show that $\{y_n\}$ is also complete (not necessarily orthonormal).

I tried to use Parseval's identity and another facts which are equivalent to say that an orthonormal system is complete, but no results.

Suppose $\langle x,y_n\rangle = 0$ for all $n\in \mathbb{N}$, then $$\|x\|^2 = \sum_{n=1}^{\infty} |\langle x,x_n\rangle|^2 = \sum_{n=1}^{\infty} |\langle x,x_n\rangle - \langle x,y_n\rangle|^2 \leq \sum_{n=1}^{\infty} \|x\|^2\|x_n - y_n\|^2$$ If $x\neq 0$, then this implies $$\sum_{n=1}^{\infty} \|x_n-y_n\|^2 \geq 1$$ But since $\sum_{n=1}^{\infty} \|x_n -y_n\| < 1$, $\|x_n - y_n\| < 1$ for all $n\in \mathbb{N}$, whence $$\sum_{n=1}^{\infty} \|x_n-y_n\|^2 \leq \sum_{n=1}^{\infty}\|x_n-y_n\| < 1$$ This is a contradiction, unless $x=0$. Hence, $\{y_n\}$ is complete.
Let $A$ be the linear map on the span of $\{x_n\}$ such that $A(x_n)=y_n$. The hypothesis and the triangle inequality imply that $A$ is bounded, so it extends uniquely to a bounded operator on the whole space. If $\|\cdot\|$ is the operator norm and $\|\cdot\|_2$ is the Hilbert-Schmidt norm, then $\|I-A\|^2\leq \|I-A\|_2^2=\sum\limits_{n=1}^\infty\|x_n-y_n\|^2<\sum\limits_{n=1}^\infty\|x_n-y_n\|<1$, which implies that $A$ is invertible with inverse $\sum\limits_{k=0}^\infty (I-A)^k$. In particular, $A$ is surjective, and because the span of $\{y_n\}$ is dense in the image of $A$, it is dense in the whole space.