$u \in W^{2,2}(\Omega)$ implies $\Delta_p u \in L^q(\Omega)$ for some $q>1$?

Question

Let $\Omega$ be a bounded smooth domain in $\mathbb{R}^N$, and let $p \in (2, 2^*)$, where $2^* = \frac{2N}{N-2}$.

Take any $u \in W^{2,2}(\Omega)$. Is it true that $\text{div}(|\nabla u|^{p-2} \nabla u) \in L^{q}(\Omega)$ for some $q>1$?

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Evidently, if $p=2$, then $\Delta u \in L^2(\Omega)$, since $\|\Delta u\|_{L^2} \leq \|u\|_{W^{2,2}} < +\infty$.

The choice $p < 2^*$ is mostly motivated by the fact that $W^{2,2}(\Omega) \hookrightarrow W^{1,p}(\Omega)$ compactly.

Answer 1

It seems true. If you actually calculate the divergence, you obtain something bounded by $|D u|^{p-2} |D^2 u|$ times a constant. Since $|D u|^{p-2} \in L^{\frac{p}{p-2}}$ and $|D^2 u| \in L^2$, the product lies in $L^r$, where $$ \frac{1}{r} = \frac{p-2}{p} + \frac{1}{2} < \frac{1}{2n} + \frac{1}{2} \leq 1 \quad \text{for } p<2^*. $$ If needed, the bound on $r$ can be improved by noting that $|D u|^{p-2} \in L^{\frac{2^*}{p-2}}$.