If $(e_n)_{n\geq 1}$ is total in $H$ and $\sum \| e_n - f_n \| < 1$, prove $(f_n)$ is total.

Question

Sorry about the title. Not enough space for me.

Proposition

If $E = (e_n)$ and $F = (f_n)$ are orthonormal sequences in a real Hilbert space $H$ that satisfy

$$\sum_{n=1}^\infty \|e_n - f_n\| < 1 \tag{A}$$

and $(e_n)$ is total, then $(f_n)$ is total.

Discussion

One thing we can immediately see is that there's a subsequence such that $e_{n_k} \longrightarrow f_{n_k}$ as $n_k \longrightarrow \infty$. That means, we can say stuff like: for $x \in H$ and any $\epsilon > 0$ there exists a sufficiently large $k$ such that

$$|\langle x, e_{n_k}\rangle - \langle x, f_{n_k}\rangle| < \epsilon$$ $$\text{or I think that}$$ $$\| \langle x, e_{n_k}\rangle e_{n_k} - \langle x, f_{n_k}\rangle f_{n_k}\| < \epsilon$$

I don't know if it's easy to apply that fact to prove the proposition. Or maybe we could assume there's some rearrangement of a geometric series that dominates each term to help us out.

My main thinking, however, is that since $(e_n)$ is total, every $x \in H$ can be written as linear combination of the elements of $(e_n)$. Then, seeking a contradiction, if we suppose that $(f_n)$ is not total, then there exists some $x\in H$ not in $\overline{span\{F\}}$ such that $x \perp y \in \overline{span\{F\}}$. Then we take the inner product

$$\langle x, y \rangle = 0$$

But that procedure is sort of leading me in circles. Should I represent $y$ as

$$y = \sum_{n=1}^\infty \langle x, f_n \rangle f_n \tag{B}$$

which I believe is the minimizing vector for x mentioned in the paragraph above?

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Any hints for the proof would be great. Thank you.

Answer 1

Assume $\{ e_n \}$ is total. Then $f=\sum_n \langle f,e_n\rangle e_n$. Consider $Lf=\sum_n \langle f,e_n\rangle f_n$. In particular $$ \|(I-L)f\|^2=\|\sum_n \langle f,e_n\rangle(e_n-f_n)\| \\ \le \sum_n\|f\|\|e_n-f_n\|, $$ which implies that $\|I-L\| \le \sum_n \|e_n-f_n\| < 1$. Therefore $L$ is invertible and, hence, is surjective. So the closed linear span of $\{ f_n \}$ must be all of $H$, which is what you want to prove.

Answer 2

It suffices to prove that $F^{\perp}=\left\{0\right\}$. Now suppose for the sake of obtaining a contradiction that there exists a non-zero vector $x\in F^{\perp}$. Using the fact that $(e_{n})_{n}$ is total, we can expand $$x=\sum_{n} \langle x,e_n\rangle e_n =\sum_{n} \, \langle x, e_n-f_n\rangle e_{n}$$ Applying Parseval, we get $$\lvert\lvert x \rvert \rvert^{2} = \sum_{n} \, \lvert \langle x, e_n-f_n\rangle \rvert^{2} \leq \left(\sum_{n} \,\lvert \langle x, e_n-f_n\rangle \rvert\ \right)^{2} \leq \lvert\lvert x\rvert\rvert^{2} \left(\sum_{n}\lvert \lvert e_n - f_n\rvert \rvert \right)^{2} < \lvert \lvert x \rvert \rvert^{2}$$ Hence we conclude that $F^{\perp}=\left\{0\right\}$, implying that $F$ is an orthonormal sequence which spans $H$, i.e $F$ is total.