Sobolev space and norm

Question

Is it true that we have equivalence between these two norms?

$\|u\|_{H^{k}}^{2}\sim \|u\|_{L^{2}}^{2}+\sum_{\lvert \alpha\rvert=k} \|D^{\alpha}u\|_{L^{2}}^{2}$

Is it also true that we have this inequality and if it is true how can I prove it?

$\sum_{\lvert \alpha\rvert=2k} \|D^{\alpha}u\|_{L^{2}}^{2}\le \|\Delta ^{k}u\|_{L^{2}}^{2}$ where $\Delta=\sum_{i \le N}\dfrac{\partial^{2}}{\partial x_{i}^{2}}$

Here I am working on a bounded smooth domain $\Omega \subset \mathbf{R^{N}}$

Answer 1

For the first one, as a function $u \in H^2$ is zero on the boundary, you can integrate by parts to get that for all $i$, $$ \|\partial_iu\|_{L^2(\Omega)}^2 = \int_\Omega \partial_iu \cdot \partial_iu = -\int_\Omega u\partial_i^2u \leqslant \frac{1}{2}\|u\|_{L^2(\Omega)}^2 + \frac{1}{2}\|\partial_i^2u\|_{L^2(\Omega)}^2 $$ Using this on the different derivatives of $u \in H^k$, you can always bound the $L^2$ norm of $D^\alpha u$ in function of the $D^\beta u$ and the $D^\gamma u$ for $|\beta| = |\alpha| - 1$ and $|\gamma| = |\alpha| + 1$ so you bound $\|u\|_{H^k(\Omega)}$ in function of $\|u\|_{L^2(\Omega)} + \sum_{|\alpha| = k} \|D^\alpha u\|_{L^2}$.

Second point is obviously false because the right hand side is the same as the left hand side but with fewer terms so I guess you forgot a mutliplicative constante. When $u$ is smooth with compact support in $\Omega$, use the fact that for all $i,j$, $$ \|\partial_i\partial_ju\|_{L^2(\Omega)} = \left<\partial_i\partial_ju,\partial_i\partial_ju\right>_{L^2(\Omega)} = \left<\partial_i^2u,\partial_j^2u\right>_{L^2(\Omega)} \leqslant \frac{1}{2}\|\partial_i^2u\|_{L^2(\Omega)} + \frac{1}{2}\|\partial_j^2u\|_{L^2(\Omega)}, $$ because $\partial_i^* = -\partial_i$ for the scalar product in $L^2$. It should help you to get the wanted inequality.