# ap.evaluation of pdes – Proof of Taylor’s Schwartz kernel appraise of pseudodifferential operators retort

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## ap.evaluation of pdes – Proof of Taylor’s Schwartz kernel appraise of pseudodifferential operators

I’m within the proof of the next end result which supplies an appraise on the Schwartz Kernel of a $$Psi$$assassinate. There is one side the proof that’s not limpid to me which I’d affection to query the Mathoverflow group about:

The theorem

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

$$displaystyle tilde Phi(x,z) = int q(x,xi) e^{izcdotxi},textual content{d}xi,.$$.

In his bespeak “Pseudodifferential Operators” from 1981, Michael E. Taylor claims in Lemma 3.1 of Chapter XII that for $$|xi| leq C$$ and $$s < n$$, we maintain the appraise

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

level to: $$C$$ is a common ceaseless, so the acceptation of $$C$$ might change from one line to the subsequent.

The beginning of the proof and my query

Firstly, Taylor observes that it suffices to assume symbols that assassinate not cipher on $$x$$. Furthermore, he observes that $$q in S^{-s}_{1,0}$$ implies that the household of features $$q_r(xi) = r^sq(rxi)$$ the place $$r$$ runs from $$1$$ to $$infty$$ types a bounded clique in $$C^infty(1 leq |xi| leq 2)$$. I occupy he means right here that every one the derivatives are uniformly bounded. What I assassinate not win is the next sever:

Taylor claims that the above details suggest that $$q$$ can breathe written as

$$displaystyle q(xi) = q_0(xi) + int_{0}^infty p_tau(e^{-tau}xi),textual content{d}tau$$

the place $$q_0(xi) in C^infty_c$$ and $$e^{stau}p_tau(xi)$$ is bounded within the Schwartz house $$mathcal{S}(mathbb{R}^n)$$ for $$0 leq tau < infty$$. Unfortunately, I assassinate not graze how this follows.

The leisure of the proof

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

Firstly, we maintain

$$displaystyle hat q(z) = hat q_0(z) + int_{0}^infty e^{ntau} hat p_tau(e^{-tau}z),textual content{d}tau$$

We retain that $$e^{stau} hat p_tau(z)$$ is just too bounded in $$mathcal{S}$$, which ends up in the estimates

$$e^{stau} |hat p_tau(z)| leq C_N(1 + |z|)^{-N}$$

with the constants unbiased of $$tau$$. Then, we are able to appraise

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

Using a change of the design $$tau mapsto tau + log(|z)$$, one can sure the ultimate integral and quick the proof.

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