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Math Help - showing something odd for an operator

  1. #1
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    Post showing something odd for an operator

    Hi,

    How do I do the following:

    Define Vg(x)=\int_0^{1}g(s)ds for g\in L^2([0,1]). Show that for all \omega \in (0,1) the following holds:

    \limsup_{\lambda \rightarrow 0}e^{-\frac{\omega}{|\lambda|}}\|(\lambda I-V)^{-1})\|= \infty.

    (The hint given is that one should compute the resolvent of V)

    So far I have been able to show that the given operator V is compact and has no eigenvalues. This then gives me the spectrum of V, \sigma(V)=\{0\}. From this, since the spectrum is the complement of the resolvent, i get that the resolvment(V)=\mathbb{C}\setminus \{0\}. What now?

    Many Thanks!
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  2. #2
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    Re: showing something odd for an operator

    Quote Originally Posted by FalexF View Post
    Hi,

    How do I do the following:

    Define Vg(x)=\int_0^{1}g(s)ds for g\in L^2([0,1]).
    Are you sure about that? In that definition, Vg(x) is independent of x. I think it is more likely that it should be Vg(x)=\int_x^{1}g(s)\,ds. In that case, V is the Volterra integral operator. You could try looking online (or better still, in a library) for information about it.
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    Re: showing something odd for an operator

    Oh yes! It's supposed to go from 0 to x. And I actually (by accident) sumbled upon the real name (Volterra) a couple of hours ago. So, I've been googeling like a mad man since then. And I've worked with it before (that's why I knew that it's compact) but I've had no luck when trying to solve the above. Any ideas?
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    Re: showing something odd for an operator

    Quote Originally Posted by FalexF View Post
    Hi,

    \limsup_{\lambda \rightarrow 0}e^{-\frac{\omega}{|\lambda|}}\|(\lambda I-V)^{-1})\|= \infty.
    Now, I'm thinking that this is obvious. You see I have a thm that says: (distance\{\lambda, spectrum(T)\})^{-1} \leq |(\lambda I-T)^{-1}|.

    Now if i apply that to the above I should get the follwoing:

    e^{-\frac{\omega}{|\lambda|}}\|(\lambda I-V)^{-1})\|> \frac{e^{-\frac{\omega}{|\lambda|}}}{|\lambda|}

    This will now tend to infinity as long as our fixed \omega>0. Correct? Hmmm... I am however not using the fact that \omega<1...
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  5. #5
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    Re: showing something odd for an operator

    Quote Originally Posted by FalexF View Post
    Now, I'm thinking that this is obvious. You see I have a thm that says: (distance\{\lambda, spectrum(T)\})^{-1} \leq |(\lambda I-T)^{-1}|.

    Now if I apply that to the above I should get the following:

    e^{-\frac{\omega}{|\lambda|}}\|(\lambda I-V)^{-1})\|> \frac{e^{-\frac{\omega}{|\lambda|}}}{|\lambda|}

    This will now tend to infinity as long as our fixed \omega>0. Correct? Hmmm... I am however not using the fact that \omega<1...
    No, it will not tend to infinity, because when |\lambda| is very small, \omega/|\lambda| will be large, and so e^{-\omega/|\lambda|} will be very small.

    It looks to me as though you should follow the hint in the question and compute the resolvent (\lambda I-V)^{-1}. Do this in a totally nonrigorous way using methods of Differential Equations 101.

    If f = (\lambda I-V)^{-1}g then (\lambda I-V)f = g. That is, \lambda f(x) - \int_0^x f(t)\,dt = g(x). Assume that f and g are both differentiable. Also assume (for convenience later on) that f(0) = 0. Differentiate, to get \lambda f'(x) - f(x) = g'(x). Solve that differential equation by the Diff. Eq. 101 method of introducing an integrating factor -e^{-x/\lambda}, so as to write the equation in the form \tfrac d{dx}\bigl(\lambda e^{-x/\lambda}f(x)\bigr) = e^{-x/\lambda}g'(x). Integrate, getting

    \boxed{f(x) = \frac{e^{x/\lambda}}{\lambda}\int_0^xe^{-t/\lambda}g'(t)\,dt}

    (the lower limit 0 for the integral ensures that f(0) = 0, which is why I made that assumption earlier).

    I reckon that is what the hint wanted you to do. Why is that useful? Well, we want to show that \|(\lambda I-V)^{-1}\| can be very large when |\lambda| is small. So we want to find f and g as in the boxed equation such that \|f\| is very much larger than \|g\|. I'm not sure quite how to achieve that. The best I can do is to try taking g(t) = t. The L^2([0,1])-norm of g is then given by \|g\|^2 = \int_0^1t^2\,dt = 1/3. But g'(t) = 1 and so f(x) = \frac{e^{x/\lambda}}{\lambda}\int_0^xe^{-t/\lambda}\,dt = e^{x/\lambda} - 1. If you then work out \|f\|, it comes to something of the order of  e^{1/\lambda}/\sqrt\lambda, which looks promising. But I don't see where that \omega in the problem comes from.
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