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

  • $\begingroup$ what do you mean by: "the set $\{f_n=e_n-a_n e_{n+1}: n\in\mathbb{N}\}$ is complete."? Complete in the sense of a metric space? $\endgroup$ – Nathanael Skrepek Feb 17 '18 at 14:58
  • $\begingroup$ @NathanaelSkrepek "complete" in the sense of "complete system in an Hilbert space" (if $(f_n, x)=0$, for every $x\in H$, then $x=0$). $\endgroup$ – Jeji Feb 17 '18 at 15:03
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    $\begingroup$ @Jeji: while I think it is clearly understood what you meant by "complete" in your context, I think that a more common word for it is "total". $\endgroup$ – Martin Argerami Feb 17 '18 at 15:08
  • $\begingroup$ @NathanaelSkrepek Yes, sorry, "complete" means "total". $\endgroup$ – Jeji Feb 17 '18 at 15:20

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.

  • $\begingroup$ Ok, I understand. How can I conclude that the set $\{f_n\}$ is then complete? $\endgroup$ – Jeji Feb 17 '18 at 15:02
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    $\begingroup$ You can't, in general. I have edited the answer a bit. $\endgroup$ – Martin Argerami Feb 17 '18 at 15:20
  • $\begingroup$ Now I understand. Indeed I conclude that the sequence of the absolute values of $x_n$ is non-decreasing only if $|a_n|\leq1$ but the exercise does not include this condition as hypotehesis. Probably it is implicit... $\endgroup$ – Jeji Feb 17 '18 at 15:22
  • $\begingroup$ One question. Why $|x_n|=\frac{1}{|a_{n-1}\cdot a_{n-2}\cdot\cdot\cdot a_1|}$? Should the numerator be $|x_1|$ instead of $1$? Thank You $\endgroup$ – Jeji Feb 17 '18 at 15:29
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    $\begingroup$ You are right about $x_1$. As for your last question, you have that the sequence of numbers $\{x_n\}$ is in $\ell^2$; if it were non-decreasing, you get a contradiction. $\endgroup$ – Martin Argerami Feb 17 '18 at 15:43

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