Finding an orthonormal basis from an existing one in a Hilbert space

Question

Suppose we are given a separable Hilbert space $H$ with countable orthonormal basis $\{e_n\}$. Suppose we are given an orthonormal set $\{f_n\}$ such that $\sum\|e_n-f_n\| < 1$. How do we prove that $\{f_n\}$ is in fact an orthonormal basis?

Answer 1

Recall that an orthonormal sequence in a Hilbert space is an orthonormal basis if and only if it is complete, that is $x\perp f_n$ for all $n$ if and only if $x=0$.

So, suppose that $x\perp f_n$ for all $n$ but $x\neq 0$. Then,

$$ \vert \langle x,e_n\rangle\vert = \vert \langle x,e_n-f_n+f_n\rangle\vert \leq \vert \langle x,e_n-f_n\rangle\vert+\vert \langle x,f_n\rangle\vert=\vert\langle x,e_n-f_n\rangle\vert $$ where the last equality follows since $\langle x,f_n\rangle = 0$ for all $n$ by assumption. Squaring both sides and using CSB inequality, we obtain

$$ \vert \langle x,e_n\rangle\vert^2\leq \vert\langle x,e_n-f_n\rangle\vert^2\leq \|x\|^2\|e_n-f_n\|^2 $$ Summing both sides, we obtain

$$ \sum_{n=1}^\infty\vert \langle x,e_n\rangle\vert^2\leq \|x\|^2\sum_{n=1}^\infty\|e_n-f_n\|^2 $$ Now, your assumption that $\sum\|e_n-f_n\|<1$ implies that the sum of the squares is also strictly less than one. Thus if $x\perp f_n$ for all $n$, we have shown that

$$ \sum_{n=1}^\infty\vert\langle x,e_n\rangle\vert^2 < \|x\|^2, $$which is a contradiction of the Pythagorean theorem/Parseval's identity. Thus $x=0$, and $\{f_n\}$ is complete.