Let $\{e_n\}_{n=1}^{\infty}$ be an orthonormal basis in $H$. Show that, given $a_n\in\mathbb{C}$, $a_n\neq0$ such that $\{|a_n|\}_{n=1}^{\infty}$ is decreasing, the set $\{f_n=e_n-a_ne_{n+1}\}_{n=1}^{\infty}$ is complete.
Attempt of the book. Set $x=\sum_{n=1}^{\infty}x_ne_n$ such that $0=(x, f_n)=x_n-\overline{a_n}x_{n+1}$ we have $|x_{n+1}|^2=\frac{|x_n|^2}{|a_n|^2}$ that implies that $\{|x_n|\}_n$ is non-decreasing. Hence $x_1=0$ and $x_n=0$ so that $x=0$.
I do not understand why $|x_{n+1}|^2=\frac{|x_n|^2}{|a_n|^2}$ implies that $\{|x_n|\}_n$ is non-decreasing and the conclusion. I turn out that $\{|x_n|\}_n$ is decreasing.
Is the attempt of the book correct?
Thank You
The equality $$|x_{n+1}|=\frac{|x_n|}{|a_{n}|}$$ cannot decide if $\{|x_n|\}$ is increasing or decreasing. For instance, if $|a_n|>1$ for all $n$, then $\{|x_n|\}$ is decreasing; if $|a_n|<1$ for all $n$, then $\{|x_n|\}$ is increasing.
One can see that $$\tag1 |x_n|=\frac{x_1}{|a_{n-1}a_{n-2}\cdots a_1|}. $$ It is easy to choose $\{a_n\}$ so that $(1)$ defines either a sequence in $\ell^2$ or not (for instance, $a_n=2+1/2$ or $2-1/n$); so the proof does not conclude as stated.
If $\{a_n\}$ produces, via $(1)$, a sequence in $\ell^2$, then the set $\{f_n\}$ is not total. For example, if $|a_n|\geq\delta$ for some fixed $\delta>1$. Indeed, in this case $$ x_n=\frac1{a_{n-1}\cdots a_1} $$ defines an $x$ such that $\langle x,f_n\rangle=0$ for all $n$.
The way I see the proof working is if one prescribes that $|a_n|\leq1$ for all $n$. In such case, the sequence $\{|x_n|\}$ will be indeed non-decreasing, giving the desired contradiction.