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Is it possible to show for $u:\Omega\subset\mathbb{R}^3\rightarrow \mathbb{R}^3$ that $$\||\nabla(\nabla\cdot u)|\|_2^2\leq C\||\Delta u|\|_2^2?$$

Here $\||f|\|_2$ is the norm in $(L^2(\Omega))^3$ and derivatives are taken in the sense of distributions.

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    $\begingroup$ Yes and this follows from the scalar case: $\|D^2 u\|_2 =\|\Delta u\|_2$ where $(D^2 u)$ is the Hessian matrix of $u$. Having this, $\|D^2_{ik}u_i\|_2^2 \leq \|\Delta u_i\|_2^2$ for $i=1,2,3$ which gives the inequality componentwise. $\endgroup$
    – Giorgio Metafune
    Commented Jan 24 at 20:52

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