Sobolev inequality for fractional derivative Let $f:\mathbb R^n \to \mathbb C$ be a nice function. We define the gradient of $f$ as 
$$\nabla f(x)= (\frac{\partial f(x)}{\partial x_1},..., \frac{\partial f(x)}{\partial x_n})$$ for $x=(x_1,...,x_n)\in \mathbb R^n.$

My Question is: How to define the fractional gradient(derivative)? Dose it make sense to talk of $|\nabla  |^{\alpha} f     \    \    (\alpha \in \mathbb R)$? 

We have the Sobolev embedding:  $\|u\|_{L^{p*}} \leq C \|\nabla u \|_{L^p}$ for $p<n$ and $p*=\frac{np}{n-p}.$

Can we use Sobolev embedding  to show: $\|u\|_{L^{\frac{2n^2(n+2)}{(n+4)(n-2)^2}}} \leq C_1 \||\nabla|^{4/n+2} u \|_{L^{\frac{2n^2(n+2)}{n^3-4n+16}}} (n\geq 3)$?

 A: Yes, it does make sense to talk about "fractional derivatives". The spaces you
are looking for are either fractional Sobolev spaces or Besov spaces. In this
case instead of gradients you have difference quotients.
For fractional Sobolev spaces $W^{s,p}(\mathbb{R}^{n})$ the subcritical case
is $s>0$, $1\leq p<n/s$. For you $s=\frac{4}{n+2}$ so you need
$$
p=\frac{2n^{2}(n+2)}{n^{3}-4n+16}<n/s=\frac{n(n+2)}{4}%
$$
which reduces to $8n<n^{3}-4n+16$, Solution is: $\left(  -4,2\right)
\cup\left(  2,\infty\right)  $ which holds for $n>2$. The critical exponent
becomes%
\begin{align*}
p_{s}^{\ast} &  :=\frac{np}{n-sp}=\frac{n\frac{2n^{2}(n+2)}{n^{3}-4n+16}%
}{n-\frac{4}{n+2}\frac{2n^{2}(n+2)}{n^{3}-4n+16}}\\
&  =\frac{2n^{2}(n+2)}{(n-2)^{2}(n+4)},
\end{align*}
as you wanted. The inequality would actually give you%
$$
\Vert u\Vert_{L^{p_{s}^{\ast}}(\mathbb{R}^{n})}\leq c\Vert u\Vert
_{W^{s,p}(\mathbb{R}^{n})},$$
namely on the right-hand side the full norm and not the seminorm.
For more info on fractional Sobolev spaces, take a look at Adams and Fournier
"Sobolev spaces", Burenkov "Sobolev spaces", Leoni "A First Course in Sobolev
Spaces", Mayza "Sobolev spaces" and then there are really a lot of books on
Besov spaces, Triebel is the bible but it is difficult to read. Take a look at
wikipedia for that.
