Perturbation property of Orthonormal basis for Hilbert space Suppose $\{e_i\}_{i\in I}$ is an orthonormal basis for some Hilbert space. $\{f_i\}_{i\in I}$ is an orthonormal family with the property $\sum^\infty_{i\in I} \|e_i-f_i\|^2<\infty$, show that $\{f_i\}_{i\in I}$ is also orthonormal basis. 
I can think of for a $x$ with $(x,f_i)=0$ for all $i$, prove $x=0$. However, I don't know how to continue to the next step.
 A: Let $H$ denote the given Hilbert space. As noted by copper.hat, one may assume $I=\mathbb{N}$. First observe that
$$
\sum_{i=1}^\infty \lvert 1 - \langle e_i,f_i\rangle\rvert^2 + 2\sum_{i=1}^\infty \sum_{j=i+1}^\infty \lvert \langle f_j,e_i\rangle\rvert^2 = \sum _{i = N + 1} ^{\infty} \lVert e _i - \sum_{j=1}^\infty \langle e_j,f_i\rangle e_j \rVert ^{2} = \sum _{i = 1} ^{\infty} \lVert e _i - f _i \rVert ^{2}.
$$
Pick $N\in\mathbb{N}$ such that
\begin{align*}
\sum _{i = N + 1} ^{\infty} \lVert e _i - f _i \rVert ^{2} &< 1/2, \\
\sum_{i=1}^{N} \sum_{j=N+1}^\infty \lvert \langle f_j,e_i\rangle\rvert^2 &< 1/2.
\end{align*}
Define $E_N=\overline{\mbox{span}}(e_{N+1},e_{N+2},\ldots)$ and $F_N=\overline{\mbox{span}}(f_{N+1},f_{N+2},\ldots)$. For a closed subspace $A\subseteq H$, let $P_A$ denote the orthogonal projection onto $A$ in $H$. Then we find
\begin{align*}
\lVert P_{E_N} - P_{F_N} \rVert_{\text{HS}}^2 & = \sum_{i=1}^N \lVert P_{F_N}e_i \rVert^2 + \sum_{i=N+1}^\infty \lVert e_i - P_{F_N}e_i \rVert^2 \\
& = \sum_{i=1}^N \sum_{j=N+1}^\infty \lvert\langle f_j,e_i\rangle \rvert^2 + \sum_{i=N+1}^\infty \lVert e_i - P_{F_N}e_i \rVert^2 \\
& \leqslant \sum_{i=1}^N \sum_{j=N+1}^\infty \lvert\langle f_j,e_i\rangle \rvert^2 + \sum_{i=N+1}^\infty \lVert e_i - f_i \rVert^2 < 1.
\end{align*}
Thus, we have
$$
\lVert P_{E_N^\perp} - P_{F_N^\perp} \rVert_{\text{HS}}^2 = \lVert P_{E_N} - P_{F_N} \rVert_{\text{HS}}^2 < 1.
$$
But this implies that $\dim F_N^\perp = \dim E_N^\perp = N$, and therefore $F_N^\perp = \mbox{span}(f_1,\ldots,f_n)$. Which was what we wanted!
