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.

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    $\begingroup$ The general problem you are thinking about is the spectrum of an integral operator Those search terms might find you something useful. $\endgroup$ Jun 9, 2019 at 16:45
  • $\begingroup$ Hi, thanks for the indication :) $\endgroup$
    – Azur
    Jun 9, 2019 at 17:11
  • $\begingroup$ up. I am still interested if anyone should have any sources to tackle down this problem. :) $\endgroup$
    – Azur
    Jun 17, 2019 at 0:46
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    $\begingroup$ Texts on homogeneous Fredholm equations on the 2nd kind? $\endgroup$ Jun 17, 2019 at 16:39
  • 1
    $\begingroup$ Mikhlin. $\endgroup$ Jun 17, 2019 at 17:00

1 Answer 1


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.

  • 1
    $\begingroup$ Thanks for your answer and sorry about the delay. In this case though, the operator I'm interested in is not a Volterra operator, because the bounds of the integral are not dependent on $x$. But Fredholm integrals do seem to be the way to go to prove this particular transform has no non-zero eigenvalues. Thank you very much for all the links you provided, I am starting to read them right about now. $\endgroup$
    – Azur
    Jun 27, 2019 at 10:11
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    $\begingroup$ I'm sorry, you actually wrote that you had found a proof for the volterra operator yet I posted it again, so I'm even more thankful you decided to accept this answer/comment anyway. I think maybe trying to extend the proof is a possibility, if you ever know how to prove it be sure to post it and if I know one will do the same and delete this post. $\endgroup$
    – Dabed
    Jun 27, 2019 at 23:58
  • $\begingroup$ (1) I don't see how that link shows that this operator has no eigenvalues. (3) that's "singular integral operator", or "singular integral", not ""singular operator"! Again I don't see where in that pdf this operator is treated... $\endgroup$ Jun 29, 2019 at 17:57
  • $\begingroup$ Yes, my confusion has been already pointed out to me by Arthur, edited to singular integral of the Volterra operator as suggested, thanks $\endgroup$
    – Dabed
    Jun 30, 2019 at 2:19
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    $\begingroup$ It is no problem haha. I tried to extend the proof that a Volterra operator has no eigenvalues, and thought I almost got it, but it turned out there was a missing hypothesis and it didn't work, it was worth trying though. And I did accept your answer cause it was well-documented, and got as close to the answer I sought as I could in three weeks' time, so I appreciate it :). Thanks for the links $\endgroup$
    – Azur
    Jul 1, 2019 at 9:37

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