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

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

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

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

    Many Thanks!
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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 $\displaystyle Vg(x)=\int_0^{1}g(s)ds$ for $\displaystyle g\in L^2([0,1])$.
    Are you sure about that? In that definition, $\displaystyle Vg(x)$ is independent of x. I think it is more likely that it should be $\displaystyle Vg(x)=\int_x^{1}g(s)\,ds.$ 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,

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

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

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

    This will now tend to infinity as long as our fixed $\displaystyle \omega>0$. Correct? Hmmm... I am however not using the fact that $\displaystyle \omega<1$...
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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: $\displaystyle (distance\{\lambda, spectrum(T)\})^{-1} \leq |(\lambda I-T)^{-1}|$.

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

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

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

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

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

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

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