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

  1. Sep 29, 2014 #1
    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. jcsd
  3. Sep 30, 2014 #2
    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
  4. Sep 30, 2014 #3
    EDIT: Read ##B=A-\lambda I## instead of ##A## ofr every occurrence in the preceding post.
     
  5. Oct 1, 2014 #4

    WWGD

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  6. Oct 1, 2014 #5

    WWGD

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    Davide: why don't you set it up and we can work it together; I need to review the material myself.
     
  7. Oct 2, 2014 #6
    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
  8. Oct 2, 2014 #7
    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, $$\rho(A) = \max\{|\lambda|:\lambda\in\sigma(A)\}.$$
     
  9. Oct 2, 2014 #8
    ...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
  10. Oct 2, 2014 #9
    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.
     
  11. Oct 2, 2014 #10
    Very interesting. How can we see the first inequality? Thanks ##\to+\infty##!;)
     
  12. Oct 2, 2014 #11
    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##.
     
  13. Oct 3, 2014 #12
    Beautiful! Thank you very much!
     
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