# Compactness of integral operators

We know that if an operator has $$L^2$$-kernel, then it is Hilbert-Schmidt. Is there a similar simple criterion to detect compact operators?

In particular, I'd like to know the following: Let $$f$$ be a Schwartz function on $${\mathbb R}^2$$ with $$\mathrm{supp}(f)\subset{\mathbb R}\times J$$ for some compact Interval $$J$$. Let $$k(x,y)=f(e^x,x-y)$$ Is the operator $$T:L^2({\mathbb R})\to L^2({\mathbb R})$$, $$T(\phi)(x)=\int_{\mathbb R}k(x,y)\phi(y)\,dy$$ a compact operator?

• If you take $f(x,y)=A(x)B(y)$ then $T\phi (x)=A(e^x) (B*\phi)(x)$ could be not in $L^2$ unless you assume more conditions on $A$. Jul 22 at 10:16
• @GiorgioMetafune: I don't see, why this function should not be in $L^2$. The convolution is in $L^2$ and you simply multiply with a bounded function, so it stays in $L^2$.
– Zero
Jul 22 at 12:49
• Sorry, you are right, I forgot the hypotheses on $f$. How do you see that $T$ is bounded in $L^2$? Jul 22 at 12:57
• @GiorgioMetafune: If we split $f$ into $f^+$ and $f^-$, then one can see that $T$ is the difference of two positive operators on $L^2$; positive operators are always bounded. Jul 22 at 17:21
• @Jochen Glueck Thanks, but at that time I was even doubting that $T$ acts on $L^2$. Jul 22 at 17:33

The answer is no, in general. Assume for example that $$J=[0,1]$$ so that $$|f| \leq C \chi_{\bf R \times [0,1]}$$. Then $$|T\phi(x)| \leq C \int_{x-1}^x |\phi(y)|\, dy \leq C\int_0^1 |\phi(x+y)|\, dy$$ and Minkowsky inequaility for integrals gives $$\|T\phi_2\| \leq C\|\phi\|_2$$. So boundedness follows without any smoothness assumption. Assume now that $$f=\chi_{[0,1]\times[0,1]}$$. Then $$T\phi(x)=\int_{x-1}^x \phi (y)\, dy$$ for $$x \leq e$$ and this is not a compact operator (consider $$\chi_{[-n-1,-n]}$$). To get a smooth counterexample it suffices to consider a smooth f with compact support which dominates $$\chi_{[0,1]\times[0,1]}$$. If $$S$$ is the corresponding integral operator, then $$S\phi \geq T\phi$$ for positive $$\phi$$ and $$S$$ is not compact.