I was trying to solve problem 7 in chapter 5(sobolev spaces) in evans. The first part requires to show that for all test functions $u$, we have $$\bigg(\int{|Du|^2}\bigg)^2 \leq C \int |u|^2 \cdot \int |D^2u|^2 $$
By integration by parts we have $\int |Du|^2 = -\int u\Delta u$ and then by cauchy ineqaulity we get the desired inequality with $\Delta u$ in place of $D^2u$ . My question is that is it obvious that $\int |\Delta u|^2 \leq \int |D^2u|^2$.
Doing an estimate like $$ (\sum_{i=1}^n \partial_i^2u)^2 \le n \sum_{i=1}^n (\partial_i^2u)^2 $$ gives you a constant that depends on space dimension. To get rid of this dependence, one has to integrate by parts again: $$ \int |\Delta u|^2 = \int (\sum_i \partial_i^2u)^2 = \int \sum_i\sum_j \partial_i^2u \cdot \partial_j^2 u = \int \sum_i\sum_j (\partial_i \partial_j u)^2 = \int |D^2u|^2. $$
Side note: integration by parts is used in the second to last equality, under the assumption of zero boundary condition. In general, we would have
$$ \int_U \partial_i^2 u \,\partial_j^2 u = \int_U (\partial_i \partial_j u)^2 + \int_{\partial U} \partial_i u \,(\partial_i \partial_j u \,\nu_j- \partial_j \partial_j u \,\nu_i) $$ where $\nu_i$ is the $i$-th component of boundary normal.