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I have a question which involves pdo. Let us consider a pseudodifferential operator $A:S(\mathbb{R}^d)\rightarrow S(\mathbb{R}^d) $ whose symbol $a(x,\xi)$ lives in the $S_{0,0}^0$ class : $$ \forall \alpha,\beta \in \mathbb{N}^d \qquad |\partial_x^{\alpha} \partial_\xi^{\beta} a(x,\xi) |\leq C(\alpha,\beta) $$ Let be $f$ and $g$ smooth real functions on $\mathbb{R}^d$ such that $$ \forall \alpha \in \mathbb{N}^d\qquad |\partial_x^\alpha f(x)|+ |\partial_x^\alpha g(x)|\leq C(\alpha) $$ If the supports of $f$ and $g$ are disjoint (one may assume that their distance is positive), is it true that for all $u\in S(\mathbb{R}^d)$ and $s',s\in \mathbb{R}$ one has $$ \left\vert \left\vert f A(gu)\right\vert\right\vert_{H^s} \leq C(s,s',f,g,A) \left\vert \left\vert u \right\vert\right\vert_{H^{s'}} $$ Notice that the case $s'\geq s$ is clear since $fAg \in \mbox{Op}S_{0,0}^0$.


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The symbol class does not allow asymptotic expansions. Therefore, it seems unlikely that standard pseudodifferential calculus arguments can be applied directly to prove the estimates in question. – Sönke Hansen May 10 '13 at 15:09

The answer is negative: take $f$ smooth compactly supported in $(-1/4,1/4)$ equal to 1 in $(-1/8,1/8)$, take $g(x) =f(x+1)$ so that $g$ is supported where

$-1/4<x+1<1/4,$ i.e. $-5/4<x<-3/4$

so that the supports of $f,g$ are disjoint. Now we consider $$ (f e^{-2i\pi D}g u)(x)=f(x) g(x-1) u(x-1)=f(x)^2 u(x-1) $$ which has the same $L^2$ norm as $f(x+1)^2 u(x)$. We note that the symbol $e^{i\xi}$ belongs to $S_{0,0}^0$. The estimate $$ \Vert f e^{-2i\pi D}g u)\Vert_{L^2}\le C\Vert u\Vert_{-\epsilon} $$ cannot hold if $\epsilon >0$: take $v$ supported in $(-9/8,-7/8)$ and $u(x)=e^{2i\pi x\lambda} v(x)$. If the previous estimate were true, we would have $$ \Vert v\Vert_{L^2}\le C \Vert v(x)e^{2i\pi x\lambda} \Vert_{-\epsilon}. $$ The rhs goes to 0 when $\lambda$ goes to infinity (for a fixed $\epsilon$ positive), making the estimate impossible when $v\not=0$.

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