Prove that a kernel operator has no eigenvalues Good evening!
I'm just popping here for a quick question. I'm just starting to work on kernel operators, from $L^2(\mathbb{R}_+)$ to itself, ie:
$f \mapsto \left(x \mapsto \displaystyle\int_{\mathbb{R}_+} K(x,y)f(y)dy\right)$, with $K$ a function from $\mathbb{R}^2$ to $\mathbb{R}$.
In the case where $K$ can be written as a finite sum of functions of $x$ and $y$ (ie $K(x,y) = \displaystyle\sum\limits_{i} u_i(x)v_i(y)$, with $u_i$ and $v_i$ real functions), it is fairly easy to find eigenvalues and eigenfunctions (or prove there are none). But what if $K$ cannot be written as such?
For example, I'm faced with the case $K(x,y) = \dfrac{1}{x+y}$. What to do then? I know this operator has no eigenfunctions, but I'm trying to prove it. I've been looking for solutions in books on the subject, mostly A Hilbert space problem book by Paul Helmos, but the closest I could find to it was the proof that a Volterra operator had no eigenfunctions.
So if anyone would have any idea where I could find resources on how to prove that.. maybe a book, or an article. I'd be grateful :)
Thanks, have a nice day.
 A: I haven't been formally introduced to operators myself but I have read about them on the wikipedia if that is something so this question got my attention and was waiting to see an answer, since the time already ended decided to look it up on the web and wanted to comment about what I found but needed space so this is not an answer per se and I'm new so didn't have the reputation to comment anyway but will delete this asap if there is some problem.
1)It seems your question in particular has been answered already on math.stackexchange
2)But interestingly it looks like that for the volterra operator you can define a notion of extended eigenvalue
3)Also it seems the Volterra operator corresponds to what is called a singular integral, and has been treated like so here
4)Finally this pages look like a short good treatment of the volterra operator that may be useful and in the comments has been mentioned the wikipedia page about fredholm operator, the linked terms Fredholm theory,fredholm kernel and fredholm determinant that you found there are also worth reading.

Try n° 2:
$\lambda f(x)=\int_{-\infty}^{\infty}K(x,y)f(x)dx=\lim_{\varepsilon \to 0^{+}} \int_{\mathbb{R} \setminus [-y- \varepsilon;-y+\varepsilon]}\dfrac{f(x)}{y+x}dx=\operatorname{p.v.}\int_{-\infty}^{\infty}\dfrac{f(x)}{y+x}dx=\operatorname{p.v.}\int_{\infty}^{-\infty}\dfrac{f(-x)}{y-x}d(-x)=\operatorname{p.v.}-\int_{\infty}^{-\infty}\dfrac{f(-x)}{y-x}dx=\operatorname{p.v.}\int_{-\infty}^{\infty}\dfrac{f(-x)}{y-x}dx=\pi H(f)(-x)$
$\therefore \lambda f(x)=\pi H(f)(-x)$
where H is the Hilbert transform and now  repeating what is done in this answer we take the Fourier transform and conclude
$\lambda\widehat{f}(\xi)=-i\pi \operatorname{sign}(\xi) \widehat{f}(-\xi)$
since $\lambda$ must be constant the only way is that $\widehat{f}(-\xi)=0$ and so $f(x)=0$ then $\lambda$ can't be an eigenvalue and therefore there are no eigenvalues.
