# showing something odd for an operator

• Jan 15th 2012, 10:34 AM
FalexF
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!
• Jan 15th 2012, 11:30 AM
Opalg
Re: showing something odd for an operator
Quote:

Originally Posted by FalexF
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.
• Jan 15th 2012, 01:29 PM
FalexF
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?
• Jan 16th 2012, 07:51 AM
FalexF
Re: showing something odd for an operator
Quote:

Originally Posted by FalexF
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 $\omega<1$...
• Jan 16th 2012, 12:11 PM
Opalg
Re: showing something odd for an operator
Quote:

Originally Posted by FalexF
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.