Sobolev regularity required for $|\nabla u|^2 \in H^1(\Omega)$

Question

We have $n=2$ and $\Omega=\{x \in \mathbb{R}^2\| ||x||_2<1 \}$. I need $|\nabla u|^2 \in H^1(\Omega)$. So i would choose $u \in W^{2,4}(\Omega)$. But since we have $n=2$ I thought that a lower regularization of $u$ is enough. Probably with Sobolev embedding theorem? Any ideas?

Answer 1

First we verify that $u\in W^{2,4}$ is sufficient. $|\nabla u |^2\in L^2$ requires $\nabla u\in L^4$ so this means $u\in W^{1,4}$. Then we compute $$ \partial_i |\nabla u|^2 = 2(\partial_i \nabla u)\cdot\nabla u$$ and we need this to be in $L^2$. Since the strongest constraint so far is $\nabla u\in L^4$ we can use Holder with $ \frac 12 = \frac14 + \frac14$to bound $$ \| \nabla |\nabla u|^2\|_{L^2} \le 2 \| \nabla^2 u\|_{L^4} \|\nabla u\|_{L^4}$$ A posteriori, Sobolev embedding tells us that $\nabla u\in W^{1,4}\subset C^{0,\gamma}\subset L^\infty$, which hints that we can use different Holder exponents. In fact, if we assume $\nabla^2 u\in L^p$ for some $p>n=2$ then we get $\nabla u\in L^\infty$. In which case, we can use Holder with $\frac12 = \frac12 + \frac1\infty$ (and the fact that $\Omega$ is bounded) to see $$ \| \nabla |\nabla u|^2\|_{L^2} \le 2 \| \nabla^2 u\|_{L^2} \|\nabla u\|_{L^\infty} \le C \| \nabla^2 u\|_{L^p} \|\nabla u\|_{L^\infty}.$$

For $p=n=2$, it only holds that $\|\nabla u\|_{L^q}\le C_q \|u\|_{H^2}$ for all $q<\infty$, which is not enough, since to use $\|\nabla u\|_{L^q}$ with $q<\infty$ in the Holder estimation, we would need to use $\|\nabla^2 u\|_{L^r}$ with $r>2$.

It follows that the proof also fails when we try $p<2$ (because $L^2 \subset L^p$ on bounded domains).

Summary: $u\in W^{2,p}$ for any $p>2$ is sufficient, as this implies $|\nabla u|^2 \in L^\infty\subset L^2$ and $\nabla|\nabla u|^2\in L^2$.