Fractional Leibniz formula

Question

Let $T=(-\Delta)^{1/2}$. Can we have estimates, similar to the one below $$ \| T^{\alpha}(fg)-(T^{\alpha}f)g-f(T^{\alpha}g) \|_p \leq \|T^{\alpha-1}f\|_p \|T^{\alpha-1}g\|_p, $$
hold in $L^p$, where $\alpha>0$ and $p>1$. If such a fractional Leibniz formula holds, can we then estimate a fractional integration by parts as well?

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

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.

Answer 4

I think an estimate very similar to what you have written appears in http://www.ams.org/mathscinet-getitem?mr=1211741

But the scaling is a little off in your estimate. You should have something like

$$\| \vert\nabla\vert^\alpha(fg)-f\vert\nabla\vert^\alpha g-g\vert\nabla\vert^\alpha f\|_p \lesssim \| \vert\nabla\vert^{\alpha_1}f\|_{p_1}\|\vert\nabla\vert^{\alpha_2}g\|_{p_2}$$

where $1<p,p_1,p_2<\infty,\quad$ $\frac{1}{p}=\tfrac{1}{p_1}+\frac{1}{p_2},\quad 0<\alpha,\alpha_1,\alpha_2<1,\quad\alpha=\alpha_1+\alpha_2.$

Answer 5

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 6

This

T. J. Osler, Fractional Derivatives and Leibniz Rule, Amer. Math. Monthly, Vol. 78, No. 6 (Jun. - Jul., 1971), pp. 645-649,

could be helpful.