0
$\begingroup$

Let $k \in {L^2}((0,4) \times (0,1))$, $g \in {L^2}(0,1)$.

We consider the following first kind Fredholm equation $$\int\limits_0^4 {k(s,x)f(s)ds=g(x), x\in(0,1).} $$ Where $f$ is the unknown. How can I prove that there exists $g$ in $L^2(0,1)$ such that the above equation doesn't have solution. I thought about the compactness of Hilbert-Schmidt operator, but I think that as the kernel $k$ is not defined on a square, we cannot apply that. Any suggestions? Thanks

$\endgroup$
  • 2
    $\begingroup$ Can a compact operator which is not of finite rank ever be surjective? $\endgroup$ – MisterRiemann Dec 30 '18 at 16:56
  • $\begingroup$ No, this is what I wrote. But I'm not sure about that because the kernel is not defined in a square $\endgroup$ – Gustave Dec 30 '18 at 17:02
  • 2
    $\begingroup$ Does that affect the compactness of your operator in any way? (In fact, one can prove that any integral operator on $\mathscr{L}^2(\mathbb{R})$ with a square-integrable kernel over $\mathbb{R}^2$ forms a Hilbert-Schmidt operator, and is therefore compact.) $\endgroup$ – MisterRiemann Dec 30 '18 at 17:09
  • $\begingroup$ Thank you @MisterRiemann. $\endgroup$ – Gustave Dec 30 '18 at 17:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.