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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?

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By "lebniz fomular" you probably meant "Leibniz formula". – GH from MO Apr 24 '12 at 14:48
thanks it has been corrected – user23078 Apr 24 '12 at 14:58

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

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

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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.

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Right,thanks a lot,when $T=\nabla$,then $S_{m-2}\equiv 0$. – user23078 Sep 17 '12 at 12:15

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](

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I think an estimate very similar to what you have written appears in

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

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