Fractional Leibniz formula 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? 
 A: 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
A: There is a paper by A. Eduardo Gatto containing $L^p$- estimates for a fractional derivative of a product.
A: 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.
A: 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 $1<p,p_1,p_2<\infty,\quad$ $\frac{1}{p}=\tfrac{1}{p_1}+\frac{1}{p_2},\quad 0<\alpha,\alpha_1,\alpha_2<1,\quad\alpha=\alpha_1+\alpha_2.$ 
A: 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.）
A: 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.
