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?



Compactness of integral operators
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.


