Is it possible to estimate $| u |^2_{H^1}$ by $|u|_{H^2}$ for bounded functions?

Question

Let $u\in [L^\infty(\Omega)]^m \cap [H^2(\Omega)]^m$ be a vector valued function with bounded $\Omega \subset \mathbb{R}^n$. Moreover, let $\|u\|_{L^\infty} \leq 1$. Is it possible to bound the square of $L^2$ norm of $\nabla u$ from above by the $L^2$ norms of second order weak derivatives?

I know some inequalities of Landau-Kolmogorov type, however, they only allow to bound $|u|_{H^1}^2$ by $(|u|_{H^2}+\|u\|_{L^2})^2$. Is it possible to strengthen the inequality to either $$|u|_{H^1}^2\leq C(|u|_{H^2}+\|u\|_{L^2})$$ or $$|u|_{H^1}^2\leq C(1+|u|_{H^2})\|u\|_{L^2}?$$ In fact, any right hand side that grows only linearly with $|u|_{H^2}$ and does not contain derivatives of order other than 0 or 2 would help.

As a motivation/plausibility argument, take the basic building block of Fourier series, $\sin(kx)$ - its $H^1$ seminorm grows as $k$ and $H^2$ seminorm grows as $k^2$, suggesting $|\cdot|_{H^1}^2 \leq c|\cdot|_{H^2}$.

Thanks!

Answer 1

You can derive such an inequality from the Gagliardo-Niremberg inequality.

Answer 2

You can not have an inequality of the form $$ |u|_{H^1}^2\le C(|u|_{H^2}+\|u\|_{L^2}) $$ Assume that this is true and replace $u$ by $Mu$, where $M>0$ constant. Then the left hand side is ${\mathcal O}(M^2)$ while the right-hand side is ${\mathcal O}(M).$

In your argument, try instead $M\sin kx$, for $k,M$ large.