Let $H$ be a Hilbert space, $\{e_n\}_{1}^{\infty}$ be a Hilbert basis and $(u_n)_{1}^{\infty}$ an orthonormal set such that $\sum_{i=1}^{\infty}||e_n-u_n||^2$ converges then $\{u_n\}$ is also a Hilbert basis.
I tried this: there exists $n$ such that $\sum_{i=n}^{\infty}||e_i-u_i||^2 < 1$, $X=A+A^{\perp}$, where $A=span\{e_i\}_{i}^{n-1}$, let $f \in A^{\perp}$ such that $<f,u_i>=0 ,\forall i\geq n$.
$||f||^2 = \sum_{i=n}^{\infty}|<f,e_i>|^2 = \sum_{i=n}^{\infty}|<f,e_i -u_i + u_i>| \\ = \sum_{i=n}^{\infty}|<f,e_i -u_i>|^2 \leq \sum_{i=n}^{\infty}||f||^2||e_i-u_i||^2 < ||f||^2$
So we can see that $A^{\perp} \cap (span\{u_i\}_{i=n}^{\infty})^{\perp} = 0$, but I dont know where to go from here, because I can't seem to see that these $u_i$ will be in $A^{\perp}$ to conclude that they form an hilbertian base for it. I also know that if $N=(span\{u_i\}_{i=n}^{\infty})$ , since dim$A \in \mathbb{N}$ and dim$N$>dim$A$ we have that $A^{\perp} \cap N \neq 0$ but this doenst tell me that its a subspace of it .
Any help is aprecciated Thanks in advance,
Try defining $$ Af = \sum_{n}\langle f,e_n\rangle u_n. $$ Let $C=\sum_{n}\|u_n-e_n\|^2$, and assume that $C < 1$. Then $$ Af-f = \sum_{n}\langle f,e_n\rangle u_n-\sum_{n}\langle f,e_n\rangle e_n \\ = \sum_{n}\langle f,e_n\rangle(u_n-e_n) . $$ By the Cauchy-Schwarz inequality, $$ \|Af-f\|^2 \le \sum_{n}|\langle f,e_n\rangle|^2 \sum_{n}\|u_n-e_n\|^2 = C\|f\|^2 $$ Therefore $\|A-I\| \le \sqrt{C} < 1$, which is enough to prove that $A$ is invertible. In particular $A$ is surjective, which guarantees that $\{ u_n \}$ is a spanning orthonormal set and, hence, is a complete orthonormal set.