Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi,

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 $fAg \in \mbox{Op}S_{0,0}^0$.

Thanks

share|improve this question
    
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

1 Answer 1

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 $v\not=0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.