# Trace Class Operators On Manifolds With Boundary

Let $$X$$ be an $$n$$-dimensional manifold with nonempty boundary $$\partial X$$ and $$n\geq 2$$. Proposition 4.1 of this paper by Schrohe states that it is "not very difficult" to show:

Proposition: A bounded operator on $$L^2(X)$$ with range in $$H^{n+1}(X)$$ is trace class.

Tragically, I cannot figure out a proof. I've tried generalizing the approach outlined in chapter 8 of Roe's Elliptic Operators, Topology and Asymptotic Methods for the boundaryless case, I've thought about Mercer's Theorem but the setting seems quite different, and looked around in Hörmander's PDEs book III and I don't think it's in there. A hint, proof outline or reference would be greatly appreciated, thank you!

• consider posting this on MO if nothing happens once the bounty expires, it is maybe quite technical Feb 19 at 18:29

Proposition 2.1. A bounded operator on $$L^2(X)$$ with range in $$H^n(X)$$ is an element of $$\mathcal{L}^{1,\infty}(L^2(X))$$; if its range even is contained in $$H^{n+1}(X)$$ then it is trace class.
Proof: Let $$A$$ be bounded on $$L^2(X)$$ with range in $$H^n(X)$$. It is well-known that there is an invertible pseudodifferential operator $$R$$ of order $$−n$$ on $$M$$ such that $$R_{+}: L^2(X) \to H^n(X)$$ is an isomorphism with inverse equal to $$(R^{−1})_{+}$$, for a proof see e.g. [7, Theorem 3.2.14]. We then may write $$A = R_{+}(R^{−1})_{+} A$$. The composition $$(R^{−1})_{+} A$$ yields a bounded operator on $$L^2(X)$$, since $$(R^{−1})_{+}: H^n(X) \to L^2(X)$$ is bounded. On the other hand the singular values of $$R_{+}$$ on $$L^2(X)$$ can be estimated in terms of the singular values of $$R$$ on $$L^2(M)$$ and the norms of the operators $$e^{+}: L^2(X) \to L^2(M)$$ and $$r^+: L^2(M) \to L^2(X)$$. Since $$R$$ is of order $$−n$$, it is an element of $$\mathcal{L}^{1,\infty}(L^2(M))$$ according to Theorem 1.1. So we get the first assertion. The second statement is proven similarly, noting that operators of order $$−n−1$$ are trace class.
• "Theorem 1.1" is credited to Connes, and stated as follows: "Let $$P \in \Psi(M)$$ have order $$−n$$. Then $$P: L^2(M, E) \to L^2(M, E)$$ defines an element of $$\mathcal{L}^{1,\infty}(L^2(M,E))$$ and $$\operatorname{Tr}_{\omega}(P) = \frac{1}{(2 \pi)^n n} \operatorname{res} P$$, independent of $$\omega$$." For proof, the paper refers to Connes, A. The action functional in non-commutative geometry, Comm. Math. Phys. 117, 673–683 (1988), available at this link (no paywall).