fractional Leibniz formula - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:37:46Z http://mathoverflow.net/feeds/question/95024 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95024/fractional-leibniz-formula fractional Leibniz formula Shanlin Huang 2012-04-24T14:18:19Z 2012-09-17T12:01:14Z <p>Let $T=(-\triangle)^{\frac{1}{2}}$,Can we have similar estimates below hold in $L^p$ ? $\| T^{\alpha}(fg)-(T^{\alpha}f)g-f(T^{\alpha}g) \|_p \leq \|T^{\alpha-1}f\|_p \|T^{\alpha-1}g\|_p$, where $\alpha>0$,p>1. If we really have such fractional Leibniz formula holds,we can then estimate the fractional integration by parts also. </p> http://mathoverflow.net/questions/95024/fractional-leibniz-formula/95028#95028 Answer by Manuel Ortigueira for fractional Leibniz formula Manuel Ortigueira 2012-04-24T15:03:52Z 2012-04-24T15:03:52Z <p>Definitively, no. The fractional derivative of a product verifies a generalised Leibniz formula that is defined by a series. I do not know any publication with it in the two-sided derivative case, but it is easy to obtain as I did in the one-sided case. See the paper Magin et al, On the fractional signals and systems, Signal Processing 91 (2011) 350–371</p> http://mathoverflow.net/questions/95024/fractional-leibniz-formula/95762#95762 Answer by Anatoly Kochubei for fractional Leibniz formula Anatoly Kochubei 2012-05-02T12:22:20Z 2012-05-02T12:22:20Z <p>There is a paper by <a href="http://www.sciencedirect.com/science/article/pii/S0022123601938364" rel="nofollow">A. Eduardo Gatto</a> containing $L^p$- estimates for a fractional derivative of a product.</p> http://mathoverflow.net/questions/95024/fractional-leibniz-formula/107309#107309 Answer by Shanlin Huang for fractional Leibniz formula Shanlin Huang 2012-09-16T09:20:45Z 2012-09-17T12:01:14Z <p>Denote by $D^{\alpha}=(-\triangle)^{\frac{\alpha}{2}}$,then we have $$\|D^{\alpha}(f\cdot g)\| \leq C(\|D^{\alpha+s}(f)\|_{p_1}\|D^{-s}(g)\|_{q_1}+\|D^{\alpha+t}(f)\|_{p_2}\|D^{-t}(f)\|_{q_2})$$ where $\alpha$,s,t are positive real numbers,and $\frac{1}{p}=\frac{1}{p_{i}}+\frac{1}{q_{i}}$ with $i=1,2$. The proof can be seen in [Exact smoothing properties of Schrödinger semigroups]（ <a href="http://www.jstor.org/stable/10.2307/25098514" rel="nofollow">http://www.jstor.org/stable/10.2307/25098514</a>.）</p> http://mathoverflow.net/questions/95024/fractional-leibniz-formula/107333#107333 Answer by Bazin for fractional Leibniz formula Bazin 2012-09-16T18:30:36Z 2012-09-16T18:30:36Z <p>Take a pseudodifferential operator $T$ of order $m$ with symbol $t(x,\xi)$ and $a=a(x)$ a smooth function with bounded derivatives of all orders (then $a$ is a symbol of order 0). Then with $R_{m-2}$ pseudodifferential operator of order $m-2$, $T(au)=aTu+[T,a]u=aTu+Op(\frac{\partial t}{i\partial \xi}\cdot \frac{\partial a}{\partial x})u+R_{m-2} u,$ so that $$T(au)=aTu+[T,a]u=aTu+ \frac{\partial a}{\partial x}\cdot [T,x]u+S_{m-2} u,$$ with $S_{m-2}$ pseudodifferential operator of order $m-2$. So somehow the two main terms are $$T(au)\equiv aTu+ \frac{\partial a}{\partial x}\cdot [T,x]u.$$ Note that for $T=\nabla_x$, you recover Leibniz formula.</p>