# Non existence of solution of a special first kind Fredholm integral equation

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

• Can a compact operator which is not of finite rank ever be surjective? – MisterRiemann Dec 30 '18 at 16:56
• No, this is what I wrote. But I'm not sure about that because the kernel is not defined in a square – Gustave Dec 30 '18 at 17:02
• 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.) – MisterRiemann Dec 30 '18 at 17:09
• Thank you @MisterRiemann. – Gustave Dec 30 '18 at 17:13