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Taylor's PUP book on pseudodifferential operators in II.7 has an extension of the pseudodifferential version of Friedrichs' lemma to generalized Friedrichs' mollifiers $J_\epsilon$ on a compact manifold $M$, defined by the following three properties:

1) $J_\epsilon \in \Psi^{-\infty}(M)$ for each $\epsilon \in (0,1]$

2) $\{J_\epsilon: 0 < \epsilon \leq 1\}$ is a bounded subset of $\Psi^0(M)$.

3) $J_\epsilon \to u$ in $L^2(M)$ as $\epsilon \to 0$ for each $u \in L^2(M)$.

Here $\Psi^m$ is the space of pseudodifferential operators of order $m$. Then for $A \in \Psi^m$, the commutator $[A,J_\epsilon]$ is bounded in $\Psi^{m-1}$. Here the topology on $\Psi^m$ is that induced by the operator norms $H^{s} \to H^{s-m}$.

This is said to be a simple consequence of arguments in section 5, with which it seemingly bears little relation. There's a (somewhat tricky) proof of this in Treves' book for standard mollifiers, but it relies heavily on the fact that they are convolution operators. Any ideas on why this generalization should hold?

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  • $\begingroup$ After localization with a (finite) partition of unity, you can work in one single chart, where the mollifiers can be defined by convolution. Which step of this construction does not convince you? $\endgroup$
    – Piero D'Ancona
    Commented Jul 22, 2012 at 23:51
  • $\begingroup$ I must be missing something. Isn't $[A,B]$ bounded in $\Psi^{m-1}$ for any pseudodifferential operator $B$ of order $-1$ or less, because both $AB$ and $BA$ are? $\endgroup$
    – Deane Yang
    Commented Jul 23, 2012 at 1:11
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    $\begingroup$ There is no problem constructing these things; my favorite example is heat flow for time $\epsilon$. It is true that $[A,J_\epsilon] \in \Psi^{m-1}$ for each $\epsilon$. Actually, $[A,J_\epsilon] \in \Psi^{-\infty}$ since $J_\epsilon \in \Psi^{-\infty}$. The difficulty is in showing that $[A,J_\epsilon]$ is bounded uniformly as an operator from $H^{m-1}$ to $L^2$. To my mind there is no way a priori to bound the commutator in $\Psi^{m-1}$ in terms of the bounds for $J_\epsilon$ and $A$, despite the fact that $J_\epsilon \in \Psi^0$ uniformly. $\endgroup$
    – Boaz Haberman
    Commented Jul 23, 2012 at 3:14
  • $\begingroup$ Is it possible to do it as follows: Write $J_\epsilon = I + S_\epsilon$, where $S_\epsilon \rightarrow 0$. Then $[A,J_\epsilon] = [A,S_\epsilon]$. Now show the latter remains bounded in the appropriate norm. $\endgroup$
    – Deane Yang
    Commented Jul 23, 2012 at 22:28
  • $\begingroup$ It is true in the bornology sense defined in my answer. $\endgroup$
    – Bazin
    Commented Jul 24, 2012 at 20:13

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The topology on $\Psi^m$ is a ''bornology''. A sequence of symbols is converging if it is bounded in the Fréchet space $\Psi^m$ and if it converges in $C^\infty$. That bornology implies the strong convergence of operators. Then $J_\epsilon$ is bounded in $\Psi^0$ implies that the commutator $[A,J_\epsilon]$ is bounded in $\Psi^{m-1}$.

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  • $\begingroup$ I don't understand. Why should the map $[\cdot,\cdot]$ be continuous from $\Psi^0 \times \Psi^m$ to $\Psi^{m-1}$? $\endgroup$
    – Boaz Haberman
    Commented Jul 24, 2012 at 2:02
  • $\begingroup$ The commutator $[A_1,A_2]$ of operators $A_j\in \Psi^{m_j}$ indeed belongs to $\Psi^{m_1+m_2-1}$. This is a key classical -but nontrivial- point in the theory of pseudodifferential operators. Just to support that result, it is easy to check directly that the Poisson bracket {$a_1,a_2$} of symbols $a_j\in S^{m_j}_{1,0}$ belongs to $S^{m_1+m_2-1}_{1,0}$. $\endgroup$
    – Bazin
    Commented Jul 24, 2012 at 15:31
  • $\begingroup$ This is true, but that is just an algebraic fact. It doesn't tell us that the map $(A_1,A_2) \mapsto [A_1,A_2]$ is bounded in the appropriate sense. In fact I am pretty certain that it is not bounded in this sense. $\endgroup$
    – Boaz Haberman
    Commented Jul 24, 2012 at 19:24
  • $\begingroup$ This is true in the bornology sense defined in my answer. $\endgroup$
    – Bazin
    Commented Jul 24, 2012 at 20:15
  • $\begingroup$ Do you have a reference or explanation as to why that is the case? $\endgroup$
    – Boaz Haberman
    Commented Jul 24, 2012 at 21:48

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