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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!

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    $\begingroup$ consider posting this on MO if nothing happens once the bounty expires, it is maybe quite technical $\endgroup$
    – FShrike
    Feb 19 at 18:29

1 Answer 1

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At the beginning of the section where Schrohe states Proposition 4.1 (top of page 14 in your link), Schrohe cites Nest, R. & Schrohe, E., Dixmier’s trace for boundary value problems, Manuscripta Mathematica, 96(2), 203–218 (1998), available at https://doi.org/10.1007/s002290050062 (but paywalled), which contains this (at page 209):

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.

Here

  • The reference [7] is Grubb, G., Functional Calculus of Pseudodifferential Boundary Problems [mis-cited in the text as "Functional Calculus for Boundary Value Problems"], Progress in Mathematics, vol. 65. Birkhäuser, Boston (2d ed. 1996), available at https://doi.org/10.1007/978-1-4612-0769-6 (but paywalled), and
  • "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).
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