Two orthonormal sets in a Hilbert space. One is complete, the other must be complete. Given two orthonormal sets $\{e_k\}_{k=1,2\ldots}$, $\{e'_k\}_{k=1,2\ldots}$ in a Hilbert space $H$, which satisfy
$$
\sum_{k=1}^\infty \|e_k-e'_k\|^2 < 1. \tag{*}
$$
Prove that if $\{e_k\}_{k=1,2\ldots}$ is complete, $\{e'_k\}_{k=1,2\ldots}$ must be complete. Here by complete, we mean $\overline{span\{e_k\}}=H$.
I tried to prove that it if there is a non-zero $x \perp \overline{span\{e'_k\}}$, i.e.
$$\forall k,\quad
\langle x,e'_k \rangle = \sum_{j=1}^\infty \langle x,e_j \rangle\langle e_j, e'_k \rangle = 0,
$$
then $(*)$ would not hold. But I have not reached any useful results in that direction yet.
 A: Assuming $\{ e_n \}$ and $\{ e_n'\}$ are orthonormal, define
$$
              Lx = \sum_{n=1}^{\infty}(x,e_n)e_n'
$$
The operator $L$ is linear and satisfies
$$
             \|Lx\|^2=\sum_{n=1}^{\infty}|(x,e_n)|^2 \le \|x\|^2.
$$
Now assume that $\{e_n\}$ is a complete orthonormal basis, and assume that $d=\left(\sum_{n=1}^{\infty}\|e_n-e_n'\|^2\right)^{1/2} < 1$. Then $x=\sum_{n=1}^{\infty}(x,e_n)e_n$ and
\begin{align}
       \|(I-L)x\|^2&=\|\sum_{n=1}^{\infty}(x,e_n)(e_n-e_n')\|^2 \\
  &\le \left(\sum_{n=1}^{\infty}|(x,e_n)|\|e_n-e_n'\|\right)^{2} \\
   & \le\sum_{n=1}^{\infty}|(x,e_n)|^2\sum_{n=1}^{\infty}\|e_n-e_n'\|^2 \\
   & = \|x\|^2\sum_{n=1}^{\infty}\|e_n-e_n'\|^2 = d^2\|x\|^2.
\end{align}
Because $d < 1$ by assumption, then $\|I-L\| \le d < 1$. It follows that
$$
        L = I-(I-L)
$$
is invertible. Therefore $L$ is surjective, which implies that the closure of the linear space spanned by $\{ e_n'\}_{n=1}^{\infty}$ is $H$. Hence, $\{ e_n'\}_{n=1}^{\infty}$ is a complete orthonormal basis.
Note: I do not know if this is the standard proof for your problem, but I like the proof, and I found on this site. Unfortunately, I cannot find it again. If this is not the standard proof, it should be. :)
