# Why is $\det(\nabla u)\in L^{p/2}(\Omega)?$

Let $\Omega\subseteq \mathbb{R}^2$ be open, bounded. Let $p\ge 2$ Consider $u\in W^{1,p}(\Omega ;\mathbb{R}^2)$, it's weak derivation $\nabla u$ is in $L^p(\Omega; \mathbb{R}^{2\times 2})$.

How to prove that $\det(\nabla u)\in L^{p/2}(\Omega)?$

For that we have to show that $\int_\Omega |\det(\nabla u)|^{p/2}dx<\infty$ and my guess is to use Hölder's inequality. But I don't know how to do this in detail.

Thank you.

If you calculate $\det$ explicitly, it has the form $$ab-cd$$ where $a,b,c,d \in L^p(\Omega)$. Then you have to use that the multiplication of two $L^p(\Omega)$ functions is in $L^{\frac p2}(\Omega)$.
• thank you. I wasn't aware of that the multiplication of two $L^p$ functions is in $L^{p/2}$. – user421621 Jul 4 '17 at 11:09