# Proof of Taylor's Schwartz Kernel Estimate of Pseudodifferential Operators

I am interested in the proof of the following result which gives an estimate on the Schwartz Kernel of a $\Psi$DO. There is one aspect the proof that is not clear to me which I would like to ask the mathoverflow community about:

The theorem

Suppose that $q(x,D)$ is a pseudodifferential operator acting on distributions in $\mathbb{R}^n$ with symbol $q$ in $S^{-s}_{1,0}$ and define

$\displaystyle \tilde \Phi(x,z) = \int q(x,\xi) e^{ix\cdot\xi}\,\text{d}\xi\,.$.

In his book "Pseudodifferential Operators" from 1981, Michael E. Taylor claims in Proposition 3.1 of Chapter XII that for $|\xi| \leq C$ and $s < n$, we have the estimate

$\displaystyle |\tilde \Phi(x,z)| \leq C |z|^{-n + s}\,.$

Note: $C$ is a generic constant, so the meaning of $C$ may change from one line to the next.

The beginning of the proof and my question

Firstly, Taylor observes that it suffices to consider symbols that do not depend on $x$. Furthermore, he observes that $q \in S^{-s}_{1,0}$ implies that the family of functions $q_r(\xi) = r^sq(r\xi)$ where $r$ runs from $1$ to $\infty$ forms a bounded set in $C^\infty(1 \leq |\xi| \leq 2)$. I assume he means here that all the derivatives are uniformly bounded. What I do not understand is the following part:

Taylor claims that the above facts imply that $q$ can be written as

$\displaystyle q(\xi) = q_0(\xi) + \int_{0}^\infty p_\tau(e^{-\tau}\xi)\,\text{d}\tau$

where $q_0(\xi) \in C^\infty_c$ and $e^{s\tau}p_\tau(\xi)$ is bounded in the Schwartz space $\mathcal{S}(\mathbb{R}^n)$ for $0 \leq \tau < \infty$. Unfortunately, I do not see how this follows.

The rest of the proof

I give a sketch of the rest of the proof here for completeness sake:

Firstly, we have

$\displaystyle \hat q(z) = \hat q_0(z) + \int_{0}^\infty e^{n\tau} \hat p_\tau(e^{-\tau}z)\,\text{d}\tau$

We observe that $e^{s\tau} \hat p_\tau(z)$ is also bounded in $\mathcal{S}$, which leads to the estimates

$e^{s\tau} |\hat p_\tau(z)| \leq C_N(1 + |z|)^{-N}$

with the constants independent of $\tau$. Then, we can estimate

$\displaystyle |\hat q(z)| \leq C + C_N |z|^{s-n}\int_0^\infty e^\tau|z|^{n-s}(1 + |e^\tau z|)^{-n}\,\text{d}\tau$.

Using a transformation of the form $\tau \mapsto \tau + \log(|z)$, one can bound the last integral and finish the proof.

• Maybe there is some typo here: in the definition of $\tilde\Phi$, the right hand side is independent of $z$. – Fan Zheng Jul 8 '18 at 23:21