I've been reading Lawson's book, Spin Geometry recently. In this book, a pseudo-differential operator is defined as a linear map on Schwartz space $P\colon \mathcal{S} \longrightarrow \mathcal{S} $ by

$$ Pu(x) = (2\pi)^{-n/2} \int e^{-i<x,\xi>} p(x,\xi) \hat{u}(\xi) d\xi $$

with $p(x,\xi)$ satisfying the usual assumptions.

In theorem 3.9. of chapter III, it states that for all open $U \subset \mathbb{R}^n$, if $u$ is in the domain of $P$ (Sobolev space $L^2_s$ for some $s$), then $$ u|_{U} \in C^\infty \Longrightarrow Pu|_{U} \in C^\infty.$$

He uses a Workhorse Theorem here. The theorem states if we have a smooth matrix-valued function $a(x,y,\xi)$ with compact $x$- and $y$-support plus some conditions which is similar to $p(x,\xi)$, then the operator $K\colon \mathcal{S} \longrightarrow \mathcal{S} $ defined by

$$ Ku(x) = (2\pi)^{-n} \int e^{i<x-y,\xi>} a(x,y,\xi) {u}(y) dy d\xi.$$ is a pseudo-differential operator. Its asymptotic development can be written explicitly modulo a smooth operator. And we can thus know that $K$ is a smoothing operator if $a(x,y,\xi)$ vanishes on a neighborhood of the diagonal of $(x,y)$.

However, in the proof of preserving smoothness, he takes $a(x,y,\xi) = \chi_1(x)p(x,\xi) (1- \chi_2(y) ) $ where $\chi_1, \chi_2$ are smooth functions with compact support such $\operatorname{supp} \chi_1$ $\subset$ $\operatorname{supp}\chi_2$ and $\chi_1(x)=1$ near a fixed point, $\chi_2(x)=1$ near $\operatorname{supp}\chi_1$. So he actually cannot apply the Workhorse Theorem to this $a(x,y,\xi)$ because it is not compact $y$-support.

My question is whether this statement is correct. Since other books such as Gilkey's Invariance Theory: The Heat Equation and the Atiyah-Singer Index Theorem, and Booss' Topology and Analysis, the Atiyah-Singer Index Formula and Gauge-Theoretical Physics consider only $u$ with compact support. If it is correct, how the proof can be fixed? If it is not, then is there a easy counterexample?

  • 5
    $\begingroup$ Yes. Smmothness is a local condition. If $A$ is $\psi$do on $\mathbb{R}^n$ and $u$ is smooth, then $Au$ is smooth iff $\phi Au$ is smooth for any compactly supported smooth function $\phi$. The operator $u\mapsto \phi Au$ is properly supported and thus maps smooth functions to smmooth functions; see Prop.2.3.1 of these notes www3.nd.edu/~lnicolae/Pseudo.pdf $\endgroup$ – Liviu Nicolaescu Jan 4 '16 at 12:41
  • 1
    $\begingroup$ $U$ needs to be open. $\endgroup$ – Deane Yang Jan 4 '16 at 13:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.