Can I write the fractional Laplacian
$$(-\Delta)^{\alpha/2} u(x) : = c\_{\alpha,d} 
 \mathrm{P.V.}\int\_{\mathbb{R}^2} \frac{u(x) - u(y)}{|x-y|^{d+\alpha}}dy$$
in the divergence form 
$$(-\Delta)^{\alpha/2} u(x) = \nabla \cdot J(x)$$
for some function $J$? Or something else?



Divergence form of the fractional Laplacian
**The answer is yes, under some regularity assumptions for $u$.**


I will show two such representations:


1. Since $\Delta=\operatorname{div}\nabla=\nabla\cdot\nabla$ we have:



> 
> $$ (-\Delta)^{\alpha/2}u=(-\Delta)(-\Delta)^{(\alpha-2)/2}u=
> \nabla\cdot(-\nabla((-\Delta)^{(\alpha-2)/2}u). $$
> 
> 
> 


2. Another representation uses Riesz transforms.
With the Fourier transform defined by 
$$
\hat{u}(\xi) = \int\_{\mathbb{R}^n} u(x)e^{-2\pi i x\cdot\xi}\, dx
$$
we can express the derivatives and the fractional Laplacian by
$$
\frac{\partial u}{\partial x\_j} = (2\pi i\xi\_j\hat{u})^\vee,
\qquad
(-\Delta)^{\alpha/2}u=((4\pi^2|\xi|^2)^{\alpha/2}\hat{u})^\vee.
$$
The Riesz transforms are defined by 
$$
R\_j u = \left(-\frac{i\xi\_j}{|\xi|}\hat{u}\right)^\vee
\quad
\text{for $j=1,2,\ldots,n$,}
$$
and the vector valued Riesz transform is
$$
Ru=\langle R\_1 u,\ldots,R\_n u\rangle
$$
Then it is easy to check that



> 
> $$ (-\Delta)^{\alpha/2}u =
>  \nabla\cdot(R(-\Delta)^{\frac{\alpha-1}{2}}u). $$
> 
> 
> 


Indeed, by taking the Fourier transform this equality reduces to
$$
(4\pi^2|\xi|^2)^{\alpha/2}=
\sum\_{j=1}^n (2\pi i\xi\_j)\left(-\frac{i\xi\_j}{|\xi|}\right)(4\pi^2|\xi|^2)^{(\alpha-1)/2}.
$$


The Riesz transform can be expressed as the singular integral operator.


