Inverse of Sobolev interpolation inequality : $\lVert u \rVert_2 \lVert \Delta u \rVert_2 \leq C\lVert \nabla u \rVert_2^2$?

Question

If $u : \mathbb{T}^3 \to \mathbb{R}$ is a smooth function on the $3$-dimensional torus $\mathbb{T}^3$, I wonder it is possible to reverse the Sobolev interpolation inequality in the sense that

\begin{equation} \lVert u \rVert_2 \lVert \Delta u \rVert_2 \leq C\lVert \nabla u \rVert_2^2 \end{equation} for some constant $C>0$ independent of $u$. Here $\lVert \cdot \rVert_2$ is the usual $L^2$ norm with respect to the Lebesgue measure.

Could anyone please help me?

Answer 1

This inequality is incorrect for an essential reason. Assume it is true for smooth functions, then using an approximation by convolution we would conclude that it is true for Sobolev spaces and hence it would imply that functions in $W^{1,2}$ belong to $W^{2,2}$ so $W^{1,2}=W^{2,2}$. Then by an inductive argument we would conclude that $W^{1,2}=W^{2,\infty}$ and hence every $W^{1,2}$ is $C^\infty$ smooth by the Sobolev embedding.