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Let $P: (-a,a) \rightarrow \Psi_h^0(\mathbb{R}),$ be a pseudodifferential operator in Weyl quantization with $(-a,a) \ni z \mapsto P(z)$ depending smoothly on this parameter $z$. Note that this parameter is not the semiclassical parameter which is $h$.

Moreover, $P$ is supposed to be self-ajoint, of principal type, elliptic and satisfy $\partial_z p(z)\le -C<0$ bounded away by some constant. Here $p$ is the symbol of $P.$

Then, in the link it is claimed that we can write $P(z)= A^*(z)(P-z)A(z)$ where $P,A \in \Psi_h^0(\mathbb{R}):$

click me.

In other words, what is claimed here is that by an application of the implicit function theorem and some iteration we can transform any PDO that may depend non-linearly on $z$ into a form that depends on $z$ in the trivial way $P-z.$

What I understand so far is that the condition $\partial_zp(x,\xi,z)<0$ allows us to apply the implicit function theorem, i.e. we get that $$p(x,\xi,z)=g(x,\xi,z)(h(x,\xi)-z)$$

for some local functions $h,g.$

But I am not really sure what kind of iteration may be meant here? Which kind of symbolic iteration allows us to go from this equation to the identity above. Apparently $P$ is the symbol associated with $h$ but how do we construct $A$?

In other words, does anybody understand how those operators $A$ and $P$ are constructed then?

If you have any questions, please let me know.

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  • $\begingroup$ Can you find a squre root of $P(z)$ under the conditions given? If the answer is yes, then $Q(z)=P(z)^{1/2}$ should be able to be factorized into something similar to $E(I-R)$, with $R$ a smoothing operator, and the theorem then follows. $\endgroup$
    – Bombyx mori
    Commented Mar 6, 2017 at 2:40
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    $\begingroup$ I guess $g$ is positive and real, therefore $p(z) = g(z)^{1/2} (h - z) g(z)^{1/2}$ is the principal symbol of $A(z)^* (P - z) A(z)$ if $g(z)^{1/2}$ is the principal symbol of $A(z)$. Now use the usual iteration to get an error $O(h^\infty)$. $\endgroup$
    – mcd
    Commented Mar 6, 2017 at 10:30

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