Leibniz rule for Pseudo-differential operators of negative order - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T04:45:55Z http://mathoverflow.net/feeds/question/102450 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102450/leibniz-rule-for-pseudo-differential-operators-of-negative-order Leibniz rule for Pseudo-differential operators of negative order David 2012-07-17T14:16:00Z 2012-07-17T17:03:06Z <p>Does anyone know of some good references for a fractional Leibniz rule for pseudo-differential operators of negative order? As a specific example, I would like to compute $\partial_{x}^{-1}(uv)$, assuming $\xi \neq 0$ in phase space and mean-zero $u,v$.</p> http://mathoverflow.net/questions/102450/leibniz-rule-for-pseudo-differential-operators-of-negative-order/102468#102468 Answer by Bazin for Leibniz rule for Pseudo-differential operators of negative order Bazin 2012-07-17T17:03:06Z 2012-07-17T17:03:06Z <p>Let $P$ be a pseudodifferential operator with symbol $p(x,\xi)$ belonging to $S^m_{1,0}$ and let $a(x)$ be a $C^\infty$ function. Then the operator $Pa$ defined by $Pa u=P(au)$ is a pseudodifferential operator of order $m$ with symbol $q(x,\xi)$ such that $$q-\sum_{\vert \alpha\vert\le N} \frac{1}{i^{\vert \alpha\vert}\alpha !}(\partial_\xi^\alpha p)(x,\xi) (\partial_x^\alpha a)(x)\in S^{m-N-1}_{1,0},$$ which provides an asymptotic expansion of the symbol of the operator Pa, following simply from the composition formula.</p>