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Eigenvalues for a bounded operator

  1. Jan 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Let [itex]C[/itex] be the composition operator on the Hilbert space [itex]L_{2}(\mathbb{R})[/itex] with the usual inner product. Let [itex]f\in L_{2}(\mathbb{R})[/itex], then [itex]C[/itex] is defined by

    [itex](Cf)(x) = f(2x-1)[/itex], [itex]\hspace{9pt}x\in\mathbb{R}[/itex]

    give a demonstration, which shows that [itex]C[/itex] does not have any eigenvalues.

    2. Relevant equations
    [itex]C[/itex] is a unitary operator.

    Let [itex]\mathcal{F}[/itex] denote the Fourier Transformation on [itex]L_{2}(\mathbb{R})[/itex], then
    [itex](\mathcal{F}C\mathcal{F}^{\ast}f)(p) =
    \frac{\exp\big(-i\tfrac{1}{2}p\big)}{\sqrt{2}}\hspace{1pt}f\big(\tfrac{1}{2}p\big)
    [/itex]

    3. The attempt at a solution
    Direct application of the eigenvalue equation of course yields Schröders equation, that is
    [itex]f(2x-1) = \lambda f(x)[/itex]

    I don't have a slightest idea on how to proceed from here.


    Any good suggestions are more than welcome!
     
  2. jcsd
  3. Jan 20, 2015 #2

    RUber

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    If C were to have any eigenvalues, they would apply to any choice of f in your space.
    Consdider a function that is has the property f(0)=0 and f(-1) ##\neq## 0.
    Such a function clearly exists in your space.
     
  4. Jan 20, 2015 #3
    Thanks for your reply.
    I still don't see how this implies that [itex]C[/itex] doesn't have any eigenvalues ?

    I mean, does i not only show that there doesn't exist an eigenfunction having these properties?
     
  5. Jan 20, 2015 #4
    Is this a valid argument ?
    Assume that [itex]f:\mathbb{R}\rightarrow\mathbb{R}[/itex] is non vanishing almost everywhere and [itex]\lambda[/itex] is an eigenvalue of [itex]C[/itex], then
    [itex]f(2x-1) = \lambda f(x)[/itex]
    [itex]2f'(2x-1) = \lambda f'(x)[/itex]

    For x=1 the first of the equations yields : [itex]\hspace{3pt} f(1) = \lambda f(1) \hspace{12pt}\Rightarrow\hspace{12pt} \lambda=1[/itex].
    While the second of the equations yields : [itex]\hspace{3pt} 2f'(1) = \lambda f'(1) \hspace{12pt}\Rightarrow\hspace{12pt} \lambda=2[/itex].
    This is obvious a contradiction, therefore [itex]\lambda[/itex] can't be a eigenvalue.

    Of course [itex] f [/itex] must behave properly around [itex] x=1 [/itex], but this is guaranteed by [itex] f \in L_{2}(\mathbb{R}) [/itex].
     
  6. Jan 20, 2015 #5

    Ray Vickson

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    No, the eigenvalue equation ##f(2x-1) = \lambda f(x) \; \forall x## need apply only to an eigenfunction ##f = f_{\lambda}##.
     
  7. Jan 21, 2015 #6

    RUber

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    Thanks, Ray. I must have been operating without caffeine when I posted that.

    Wuberdall, your contradiction argument looks reasonable to me.
     
  8. Jan 21, 2015 #7

    LCKurtz

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    But ##f\in \mathbb L^2## doesn't give you that ##f## is differentiable.
     
  9. Jan 21, 2015 #8

    RUber

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    LCKurtz makes a good point. The goal should be to contradict your assumptions.
    As it stands, 1 is the only potential eigenvalue, as long as f(x) is defined at x=1.
    So if f(x) is defined at ##x = 1+ \delta##, ##f(1+ \delta) = f(1+ 2\delta)=f(1+2^k \delta) ## for any delta and integer k .
    You may be able to show that this sort of function is either the zero function (trivial) or not in ##L^2##.
     
  10. Jan 21, 2015 #9

    Ray Vickson

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    More generally: if ##f \in C^{\infty}## near ##x = 1## then we have ##2^n f^{(n)}(2x-1) = \lambda f^{(n)}(x), n = 0,1,2,\ldots##, where ##f^{(n)}## denotes the ##n##th derivative. Thus ##2^n f^{(n)}(1) = \lambda f^{(n)}(1), n = 0,1,2, \ldots##.
    (1) If ##\lambda = 0## that would imply that ##f^{(n) = 0, n = 0,1,2, \ldots##, meaning that ##f## would be non-analytic (but still ##C^{\infty}##. That means that ##f## would not equal its own Taylor series. There are functions like that, and our ##f## would have to be one of them.
    (2) If ##\lambda \neq 0## then either (i) ##f(0) = 0##; or (ii) ##\lambda = 1##. In case (i) we have ##2 f'(1) = \lambda f'(1), 4 f''(1) = \lambda f''(1), \ldots##, and this allows ##\lambda = 2, f^{(n)}(1) = 0## for ##n \geq 2## or ##\lambda = 4, f^{(n)}(1)=0## for ##n = 1,3,4,\ldots#, etc. Again, ##f## would not be analytic
    The equation ##f(1) = \lambda f(1)## implies ##\lambda = 1## only in the case ##f(1) \neq 0##; otherwise, if ##f(1) = 0## there is no restriction on ##\lambda##.
     
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