Spectrum of $(x_1,x_2,x_3, )\mapsto(0,x_1,2^{-1}x_2, )$

1. Sep 29, 2014

DavideGenoa

Dear friends, I would like to find the spectrum of the linear operator $A\in\mathscr{L}(\ell_2,\ell_2)$, which I have verified to be compact and eigenvalueless, defined by

$A(x_1,x_2,x_3,...,x_n,...)=(0,x_1,\frac{1}{2}x_2,...,\frac{1}{n-1}x_{n-1},...)$
but my book does not give examples of how to do so. Could anybody help me in finding its (continuous) spectrum, i.e. the set $\sigma(A)=\{\lambda\in\mathbb{C}\quad|\quad\nexists B\in\mathscr{L}(\ell_2,\ell_2):B=(A-\lambda I)^{-1}\}$?
$\infty$ thanks!

2. Sep 30, 2014

DavideGenoa

I am very inexpert, but I se that, for any $(y_1,y_2,...)=A(x_1,y_2,...)$ I think we can calculate $(x_1,x_2,...)$ observing that $x_1=-\lambda^{-1}y_1$ and $\forall k\geq 2\quad x_k=-\lambda^{-1}(y_k-(n-1)^{-1}x_{k-1})$, therefore an application inverting the image $A(\ell_2)$ exists for any $\lambda\ne 0$.
I also see that for any $(y_1,y_2,...)\in\ell_2$ we can recursively chose a sequence $z_1=-\lambda^{-1}y_1$, ..., $z_n=-\lambda^{-1}(y_n-(n-1)^{-1}z_{n-1})$, which I intuitively guess to belong to $\ell_2$, such that $A(z_1,z_2,...)=(y_1,y_2,...)$.
Therefore, applying Banach bounded inverse theorem I would say that for any $\lambda \ne 0$ the inverse $(A-\lambda I)^{-1}$ would be continuous and defined on $\ell_2$.
Am I saying stupid things?
Anyhow, I cannot prove that $(z_1,z_2,...)\in\ell_2$...:(
$\infty$ thanks for any answer!!!

Last edited: Sep 30, 2014
3. Sep 30, 2014

DavideGenoa

EDIT: Read $B=A-\lambda I$ instead of $A$ ofr every occurrence in the preceding post.

4. Oct 1, 2014

5. Oct 1, 2014

WWGD

Davide: why don't you set it up and we can work it together; I need to review the material myself.

6. Oct 2, 2014

DavideGenoa

Thank you so much for your answer! I rewrite the second post, correctly, I hope, this time (it is not possible to edit after some time one has posted, apparently):

I would say that, for any $(y_1,y_2,...)=(A-\lambda I)(x_1,y_2,...)$, i.e. for any $(y_1,y_2,...)$ in the image of $A-\lambda I$, we can calculate $(x_1,x_2,...)$ observing that $x_1=-\lambda^{-1}y_1$ and $\forall k\geq 2\quad x_k=-\lambda^{-1}(y_k-(n-1)^{-1}x_{k-1})$, therefore an application inverting the image $(A-\lambda I)(\ell_2)$ exists for any $\lambda\ne 0$.
I also see that for any $(y_1,y_2,...)\in\ell_2$ we can recursively chose a sequence $z_1=-\lambda^{-1}y_1$, ..., $z_n=-\lambda^{-1}(y_n-(n-1)^{-1}z_{n-1})$, which I intuitively guess to belong to $\ell_2$, such that $(A-\lambda I)(z_1,z_2,...)=(y_1,y_2,...)$.
Therefore, applying Banach bounded inverse theorem I would say that for any $\lambda \ne 0$ the inverse $(A-\lambda I)^{-1}$ would be continuous and defined on $\ell_2$, so that $\sigma(A)=\{0\}$. Am I saying stupid things?

If my calculations are correct I would say that $z_n=-\frac{1}{\lambda}(y_n+\sum_{k=1}^{n-1}\frac{(-1)^{n+k}}{(n-1)...k} y_k)$, but I cannot prove that $(z_1,z_2,...)\in\ell_2$...

Thank you a lot again!

P.S.: In math.stackexchange I find no answer about our $A$...

Last edited: Oct 2, 2014
7. Oct 2, 2014

Hawkeye18

The easiest way is to use the Gelfand formula for the spectral radius $\rho(A)$ of $A$, $$\rho(A) = \lim_{n\to\infty} \|A^n\|^{1/n}.$$ You can easily see that $\|A^n\|= 1/n!$, so the spectral radius is 0. And just to remind, $$\rho(A) = \max\{|\lambda|:\lambda\in\sigma(A)\}.$$

8. Oct 2, 2014

DavideGenoa

...and I have verified that $(n!)^{1/n}\xrightarrow{n\to+\infty}+\infty$ using the Puiseux series, which I happen to have studied in discrete mathematics.
I've just studied a chapter about Banach algebras and know the formula for the spectral radius: how stupid am I!
I heartily thank you!

Last edited: Oct 2, 2014
9. Oct 2, 2014

Hawkeye18

The limit also can be computed using an elementary estimate $$n! \ge \left( \frac{n}{2}\right)^{\lfloor n/2\rfloor} \ge \left( \frac{n}{2}\right)^{ n/2 -1} ,$$ where $\lfloor x\rfloor$ is the integer part of $x$. I guess that it is easier than the proof you got using the Puiseux series.

10. Oct 2, 2014

DavideGenoa

Very interesting. How can we see the first inequality? Thanks $\to+\infty$!;)

11. Oct 2, 2014

Hawkeye18

At least half of the numbers $1,2, \ldots, n$ are $\ge n/2$. So in fact even better estimate $$n!\ge\left(\frac{n}{2}\right)^{n/2}$$ holds.
I was too lazy to count, so I just estimated that there surely will be $\lfloor n/2\rfloor$ such numbers. But if you look carefully, you see that if $n$ is even then $n/2+1$ numbers are $\ge n/2$, and if $n$ is odd then $n/2+1/2$.

12. Oct 3, 2014

DavideGenoa

Beautiful! Thank you very much!