Let $X$ be a non-compact complete Riemannian manifold and $P$ a first-order elliptic pseudodifferential operator on $X$. Let $Q$ be a parametrix for $P$, so that $PQ - 1 = T$ and $QP - 1 = R$ are lower order operators.

If $X$ were compact, then the operators $Q$, $T$, $PQ$ and $PT$ would all be bounded on $L^2(X)$, since they have non-positive order.

My question is: when $X$ is non-compact, is it possible to choose the parametrix $Q$ so that $Q$, $T$, $PQ$, $PT$ are still all bounded on $L^2(X)$?


Yes, at least if $Q(x,y)$ is locally integrable in $y$.

Given two integral kernels $Q(x,y)$ and $Q'(x,y)$ that agree on a neighborhood of the diagonal in $X \times X$, then if one is a parametrix, so is the other one. Hence, you can always restrict a parametrix to have support in an arbitrary neighborhood of the diagonal. Just multiply any given parametrix $Q(x,y)$ by a smooth function $\chi(x,y)$ that has sufficiently small support but is still identically $1$ on a smaller neighborhood of the diagonal.

Since $Q(x,y)$ is locally integrable, you can choose the neighborhood small enough so that $\int_X |Q(x,y)| dx < C$ and $\int_X |Q(x,y)| dy < C$. Then, by the Schur test, $Q$ is bounded on $L^2(X)$.

Each of the operators $T$ and $PT$ is also represented by a locally integrable integral kernel. Thus, you can make the exact same argument for each of them, possibly restricting the support of $Q$ even closer to the diagonal, so that $T$ and $PT$ both satisfy the Schur test.

Finally, by your definition $PQ = 1+T$, hence it is automatically bounded.

If $Q(x,y)$ is not locally integrable, the above argument may not work, since there might not exist a sufficiently small neighborhood of the diagonal that will make all the $y$-integrals bounded.

Update: However, take a look at Thm VI.2.1 of Stein's Harmonic Analysis (PUP, 1993). It seems to directly address the $L^2$ boundedness of (scalar) pseudodifferential operators of a certain class, that might cover your operators.

  • $\begingroup$ Thanks. Do you have a reference for order -1 $\implies$ kernel is locally integrable? Also, why is $\int_X |Q(x,y)| dy$ bounded, since a neighbourhood of the diagonal isn't compact? $\endgroup$ – geometricK Sep 6 '17 at 4:21
  • $\begingroup$ @ougoah: While the neighborhood of the diagonal itself is not compact, you can choose it as small as you like, in particular so that every $(x,-)$ slice through it is compact. About local integrability, I tried to find a quick argument, but the details turn out to be more subtle than the heuristic idea of looking at the identity $\widehat{|x|^{-1}} \sim |k|^{-n+1}$ in $n$-dimensions. I'll update the answer. $\endgroup$ – Igor Khavkine Sep 8 '17 at 1:23

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