# Why is this resolvent trace-class?

I am trying to read Richard Froese's paper Asymptotic distribution of resonances in one dimension and I cannot follow the logic of Proposition 7.3. He defines a cutoff $$\chi$$ to be multiplication by the indicator function of $$[-1, 1]$$. He then defines an operator $$\mathbf R(\lambda) = \chi (\Delta - \lambda)^{-1} \chi$$, where $$(\Delta - \lambda)^{-1}$$ is the resolvent function of the Laplacian. It is not too hard to show that this is a meromorphic family of operators with just a simple pole at $$0$$, but he claims that it is actually trace-class, essentially because if $$\operatorname{Im} \lambda > 0$$ then we have $$\mathbf R(\lambda) = (\chi(\partial + \lambda)^{-1})((\partial - \lambda)^{-1}\chi),$$ and the two factors are Hilbert-Schmidt. But I do not know why that is true.

Intuitively I want to diagonalize $$\partial \pm \lambda$$, but this doesn't quite work because the eigenvalues of those operators cluster at $$0$$, so their resolvents cannot be Hilbert-Schmidt. So we have to use that the operator is cut off by $$\chi$$, but this doesn't quite help either, because we get an ugly commutator which gives delta functions. I thought about trying to compute the integral kernels of the claimed Hilbert-Schmidt operators, but again the annoying $$\chi$$ gets in the way.

More generally, if $$f,g\in L^2$$, then the operator $$A\colon\phi\mapsto \mathcal{F}^{-1}f\mathcal{F}(g\phi)$$ is Hilbert-Schmidt. This can be seen by computing its kernel (if you want to be rigorous, you should first take $$f,g,\phi$$ from the Schwartz space and then approximate to guarantee convergence of all the integrals): $$A\phi(x)=\int e^{2\pi i p x}f(p)\widehat{g\phi}(p)\,dp=\int\int e^{2\pi i p (x-y)}f(p)g(y)\phi(y)\,dy\,dp=\int \check f(x-y)g(y)\phi(y)\,dy.$$ Clearly, the kernel given by $$k(x,y)=\check f(x-y)g(y)$$ is in $$L^2$$.
In particular, if $$f(p)=(p-\lambda)^{-1}$$ and $$g=\chi$$, this shows that $$(-iD-\lambda)^{-1}\chi$$ is Hilbert-Schmidt. The other factor is essentially the adjoint of this operator, hence also Hilbert-Schmidt.