$(e_{n})$ orthonormal basis, $(f_{n})$ orthonormal sequence such that $\sum\left\|e_{n}-f_{n}\right\|^{2}<\infty$ Then $(f_{n})$ is orthonormal basis.

Let $H$ be a hilbert space, $(e_{n})$ a orthonormal basis of $H$, and, $(f_{n})$ a orthonormal sequence on $H$ such that $$\sum_{n=1}^{\infty}\left\|e_{n}-f_{n}\right\|^{2}<\infty. \tag{I}$$ Show that $(f_{n})$ is also a orthonormal basis.

Remark: My idea was the following:

First I show the following fact:

Fact 1: Let $H$ be a Hilbert space, $(x_{n})_{n=1}^{\infty}$ a orthonormal basis of $H$, and let $(y_{n})_{n=1}^{\infty}$ be a sequence in $H$ such that $$\sum_{n=1}^{\infty}\left\|x_{n}-y_{n}\right\|^{2}<1. \tag{II}$$ Then, if $z\bot y_{n}$ for all $n\in\mathbb{N}$, then $z=0$.

Let $f\bot f_{n}$ for all $n\leq 1$, then by (I) there exists $m$ such that $$\sum_{n=m+1}^{\infty}\left\|e_{n}-f_{n}\right\|^{2}<1. \tag{III}$$ Therefore, by Fact 1 we have that $\left\{e_{1},\ldots,e_{m},f_{m+1},f_{m+2},\ldots\right\}$ is total in $H$.

I need to show that $\left\{f,f_{1},\ldots,f_{m}\right\}$ is linearly dependent.

$\def\span{\operatorname{span}}$

This solution uses ideas from Demophilus and DisintegratingByParts , who removed his/her nearly complete solution for some unknown reason.

Take $m \in \mathbb{N}$ such that $\sum_{n=m+1}^\infty \|e_n-f_n\|^2 = c <1$ and take $V = \overline{\text{span}\{e_{m+1},e_{m+2}, \ldots\}}$. Let $W = \span\{e_1, \dots, e_m\} = V^\perp$. Let $V' = \overline{\text{span}\{f_{m+1},f_{m+2}, \ldots\}}$.

There is an isometry $A'$ from $V$ to $V'$ determined by $e_j \mapsto f_j$ for $j \ge m+1$. Define $A : H = W \oplus V \to H$ by $A( w + v) = w + A'(v)$. The range of $A$ is $W + V'$. Then $$ (A - I)(w + v) = A'(v) - v = \sum_{ j = m+1}^\infty \langle v, e_j\rangle (f_j - e_j). $$

$$ \begin{aligned} || (A - I) (w + v)|| &\le \sum_{ j = m+1}^\infty | \langle v, e_j\rangle| \ \ || f_j - e_j|| \\ & \le \left( \sum_{ j = m+1}^\infty | \langle v, e_j\rangle|^2\right ) ^{1/2} \left( \sum_{ j = m+1}^\infty || f_j - e_j ||^2\right )^{1/2} \\ & = c^{1/2} ||v|| \le c^{1/2} ||w + v||, \end{aligned} $$ where the very last estimate uses that $w$ and $v$ are orthogonal, so $||v + w||^2 = ||v||^2 + ||w||^2$. Thus $|| A - I ||\le c^{1/2} < 1$, and $A$ is invertible. (See the proof sketch below.) In particular $A$ is surjective so $W+ V' = H$. Check that $W \cap V' = (0)$, so $H = W \oplus V'$ (not an orthogonal direct sum).

In particular $H/V'$ is $m$--dimensional, so $(V')^\perp$ is $m$ dimensional. But $\{f_1,\ldots, f_m\}$ is an orthonormal subset of $(V')^\perp$ of cardinality $m$, so $\{f_1,\ldots, f_m\}$ is an orthonormal basis of $(V')^\perp$. Therefore $\{f_j : 1 \le j < \infty\}$ is an orthonormal basis of $H$.

Note: In any Banach algebra with identity $I$ (and $||I|| = 1$) if $x$ is an element satisfying $||x|| < 1$, then $I - x$ is invertible with inverse $\sum_{j = 0}^\infty x^j$. Apply this in our context, noting $A = I - (I - A)$.