eigenvalues of homogeneous integral equation of second kind, with singular kernel There is a homogeneous integral equation of second kind with a singular kernel(non-symmetric). The equation has the form:
$\int_{a}^{b} k(x,t)Γ(t)dt =λΓ(x).$
It's 2-norm  is infinity, $||k(x,t)||_2 \rightarrow\infty$. Then, all the eigenvalues are zero, correct or not?
btw: is there any way to justify the range of eigenvalues of such problems?
Thx!
 A: Generally we need the kernel to have a finite norm so that it's operation on $\Gamma$  doesn't blow up in some direction. Also, the non-symmetry seems suspicious, since Hermitian and anti-Hermitian operators have nice properties (real eigenvalues and therefore a basis). If you can think of continuous kernels $k(x,t)$ to have a countably infinite basis, the representations in this basis are through a countably infinite set of eigenvalues. The 2-norm condition seems to imply that the modulus squared of these eigenvalues diverges (i.e. not in $L^2(a,b)$). If the kernel is just a finite basis, one of the eigenvalues must be infinite. Spectral theory and resolvent set applies here.
If I understood your question correctly, then only one of your potentially countably infinite set of eigenvalues needs to be infinite. Or you can have any divergent sequence of complex numbers (whose 2-norm blows up) and then blow up your $\Gamma$  The kernel with zero eigenvalues is also not that interesting. 
I hope this helps.
