Hi all,
I need the solution of the following exercise:
Show that the integral equation g(s)=$\displaystyle \lambda$$\displaystyle \int^{\pi}_{0} (\sin s \sin2t) g(t) dt$ has no eigenvalues
After one integration wrt s from 0 to pi, you get:
$\displaystyle 2* \lambda \int^{\pi}_{0} ( \sin2t) g(t) dt = \int^{\pi}_{0} g(t)dt$
so you get:
$\displaystyle \int_{0}^{\pi} (2\lambda \sin(2t) -1) g(t) dt=0$
g is different than zero, otherwise it wouldn't be an eigenfunction of this operator.
So 2 lambda *sin(2t)-1 should be zero almost everywhere in the interval [0,pi], there isn't such lambda cause lambda is constant and sin(2t) varies along this interval.
Up till here it's fineg is different than zero, otherwise it wouldn't be an eigenfunction of this operator.
This is not true, that an integral is zero doesn't mean that the function is zero unless you know that said function is non-positive or non-negative. Since we know nothing about g, you can't concludeSo 2 lambda *sin(2t)-1 should be zero almost everywhere in the interval [0,pi]
For the problem I only get that if $\displaystyle |\lambda | <\frac{1}{\pi}$ then there is no eigenvector. I don't know how to deal with the other cases.
Is the integral zero for all functions g? If so, then the fundamental theorem of the calculus of variations kicks in, and your conclusion is sound. Otherwise not. I'm just not entirely sure of the logic here, whether you get to say the integral is zero for all choices of the function g.
I don't think this is the case, since he is trying to solve an eigenvalue problem so assuming the identity holds for all your domain is equivalent to asserting that your operator is a multiple of the identity, which clearly has no eigenvalues.
If you're working on $\displaystyle C[0,\pi ]$ (or some interval) then maybe you could appeal to the compactness of your operator (Fredholm alternative comes to mind, but as far as I know it's only valid in Hilbert spaces, I'm not too knowledgeable in Banach algebras though so maybe there you could find something).