Let $T=(\Delta)^{1/2}$.
Can we have estimates, similar to the one below
$$
\ T^{\alpha}(fg)(T^{\alpha}f)gf(T^{\alpha}g) \_p \leq \T^{\alpha1}f\_p \T^{\alpha1}g\_p,
$$
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?

$\begingroup$ By "lebniz fomular" you probably meant "Leibniz formula". $\endgroup$ – GH from MO Apr 24 '12 at 14:48
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 twosided derivative case, but it is easy to obtain as I did in the onesided case. See the paper Magin et al, On the fractional signals and systems, Signal Processing 91 (2011) 350–371
There is a paper by A. Eduardo Gatto containing $L^p$ estimates for a fractional derivative of a product.
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_{m2} $ pseudodifferential operator of order $m2$, $ T(au)=aTu+[T,a]u=aTu+Op(\frac{\partial t}{i\partial \xi}\cdot \frac{\partial a}{\partial x})u+R_{m2} u, $ so that $$ T(au)=aTu+[T,a]u=aTu+ \frac{\partial a}{\partial x}\cdot [T,x]u+S_{m2} u, $$ with $S_{m2} $ pseudodifferential operator of order $m2$. 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.

$\begingroup$ Right,thanks a lot,when $T=\nabla$,then $S_{m2}\equiv 0$. $\endgroup$ – user23078 Sep 17 '12 at 12:15
I think an estimate very similar to what you have written appears in http://www.ams.org/mathscinetgetitem?mr=1211741
But the scaling is a little off in your estimate. You should have something like
$$\ \vert\nabla\vert^\alpha(fg)f\vert\nabla\vert^\alpha gg\vert\nabla\vert^\alpha f\_p \lesssim \ \vert\nabla\vert^{\alpha_1}f\_{p_1}\\vert\nabla\vert^{\alpha_2}g\_{p_2}$$
where $1<p,p_1,p_2<\infty,\quad$ $\frac{1}{p}=\tfrac{1}{p_1}+\frac{1}{p_2},\quad 0<\alpha,\alpha_1,\alpha_2<1,\quad\alpha=\alpha_1+\alpha_2.$
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.）
This
could be helpful.