Question

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.

Answer 1

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

Answer 2

There is a paper by A. Eduardo Gatto containing $L^p$- estimates for a fractional derivative of a product.

Answer 3

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]（ http://www.jstor.org/stable/10.2307/25098514.）

Answer 4

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.