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 $J_f=\det Df\in W^{1,P}.$


Here is an answer from Andrea Cianchi:

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.