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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.

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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.

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  • $\begingroup$ Thank you, this is just what I needed! $\endgroup$ Dec 18, 2019 at 23:29

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