
> 
> Given a mapping in the Sobolev space $f\in W^{2,n}\_{\rm loc}(\mathbb{R}^n,\mathbb{R}^n)$ I would like to know what is the
>  Sobolev regularity of the Jacobian $J\_f=\operatorname{det} Df$.
> 
> 
> 


It is well known and easy to prove that if $u,v\in W^{1,p}\cap L^\infty(\mathbb{R}^n)$, then $uv\in W^{1,p}\cap L^\infty$. Indeed, product of a bounded and an $L^p$ function is in $L^p$ and the same argument applies to the derivatives $\partial\_i(uv)=(\partial\_i u)v+v\partial\_i\in L^p$. Now if $u\in W^{1,n}$ than $u$ has very high integrability (Trudinger's inequality) so if $u,v\in W^{1,n}$ (no longer bounded), then $uv\in W^{1,n}$ must belong to some Orlicz-Sobolev space slightly larger than $W^{1,n}$. Thus my question is:



> 
> Let $u\_1,\ldots,u\_n \in W^{1,n}(B^n(0,1))$. Find an optimal (or close
>  to optimal) Orlicz-Sobolev space $W^{1,P}$ for some Young function $P$
>  such that $u\_1\cdot\ldots\cdot u\_n\in W^{1,P}$.
> 
> 
> 


In fact I would like to know if one can find $P$ so that it satisfies the so called divergence condition:
$$
\int\_0^1 \frac{P(t)}{t^{n+1}}\, dt =\infty.
$$
is satisfied.



> 
> Since the derivatives of $f\in W^{2,n}(\mathbb{R}^n,\mathbb{R}^n)$
>  belong to $W^{1,n}$ such a result will imply that $J\_f=\det Df\in
>  W^{1,P}.$
> 
> 
> 



Regularity of the Jacobian of a $W^{2,n}$ Sobolev mapping
Here is an answer from
[**Andrea Cianchi**](https://mathscinet.ams.org/mathscinet/search/author.html?mrauthid=260742):


A form of Hölder's inequality in Orlicz spaces asserts that, if $f\_1\in L^{A\_1},\ldots,f\_n\in L^{A\_n}$, and $B$ is such that
$$
A\_1^{-1}(t)\cdots A\_n^{-1}(t)\leq cB^{-1}(t)
\quad
\text{for $t\geq 0$},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
$$
for some constant $c$, then $f\_1 f\_2\cdots f\_n\in L^B$ and
$$
\Vert f\_1 f\_2\cdots f\_n\Vert\_{L^B}\leq C\Vert f\_1\Vert\_{L^{A\_1}}\cdots\Vert f\_n\Vert\_{L^{A\_n}},
$$
for some constant $C$. If the domain has finite measure, then (1) is only required for sufficiently large $t$.


Now it $u\_1,\ldots,u\_n\in W^{1,n}$, then $u\_i\in\exp L^{n'}$ for every $i$ (Trudinger's inequality). In view of the condition (1), with $A\_i(t)=t^n$ and $A\_j(t)=e^{t^{n'}}$ for $j\neq i$, the product rule yields that
$$
\nabla(u\_1\cdots u\_n)\in L^P
$$
if
$$
t^{1/n}(\log t)^{1/n'}\cdots(\log t)^{1/n'}\leq cP^{-1}(t)
$$
for large $t$ (if the domain has finite measure), where $(\log t)^{1/n'}$ appears ($n-1$)-times. Thus $P$ has to fulfill
$$
t^{1/n}(\log t)^{\frac{(n-1)^2}{n}}\leq cP^{-1}(t)
$$
so the best possible choice of $P$ is
$$
P(t)=t^n(\log t)^{-(n-1)^2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
$$
for large $t$. The divergence condition is only satisfied for $n=2$.


Therefore we have:



> 
> If $f\in W^{2,n}\_{\rm loc}(\mathbb{R}^n,\mathbb{R}^n)$, then $J\_f=\det
>  Df\in W^{1,P}\_{\rm loc}$, where $P$ is given by (2).
> 
> 
> 


