I wanted to prove the estimate:
$$ \| \Delta u \|_{L_2(\Omega)} \leq C \| u \|_{H^2(\Omega)} \ \ \forall \ u \in H^2_0(\Omega) $$ Where $ C > 0$ is a constant not depending on $ u $.
On the internet (http://www.math.ucsb.edu/~grigoryan/246B/lecs/246B_ch5.pdf) i've found the identity:
$$ \int\limits_\Omega \left( \sum_{i=1}^d \frac{\partial^2 u}{\partial^2 x_i}(x) \right)^2 dx \ = \ \sum_{i,j = 1}^d \int\limits_\Omega \left( \frac{\partial^2 u}{\partial x_i \partial x_j}(x) \right)^2 dx $$
With this the estimate would be easy, but how can this be true? I mean the mixed derivatives come out of nowhere.
I would be glad for some help. Thank you!
On $H_0^2(\Omega)$, consider the two bilinear forms
$$\beta_1(u,v) = \sum_{i,j} \int_\Omega \frac{\partial^2u}{\partial x_i^2}(x)\frac{\partial^2v}{\partial x_j^2}(x)\,dx$$
and
$$\beta_2(u,v) = \sum_{i,j} \int_\Omega \frac{\partial^2u}{\partial x_i\partial x_j}(x) \frac{\partial^2v}{\partial x_i\partial x_j}(x)\,dx.$$
Both are continuous. On the subspace $C_c^\infty(\Omega)$, we have
$$\begin{align} \int_\Omega \frac{\partial^2u}{\partial x_i\partial x_j}(x)\frac{\partial^2 v}{\partial x_i\partial x_j}(x)\,dx &= - \int_\Omega \frac{\partial^3u}{\partial x_i^2\partial x_j}(x)\frac{\partial v}{\partial x_j}(x)\,dx\\ &= \int_\Omega \frac{\partial^2 u}{\partial x_i^2}(x)\frac{\partial^2v}{\partial x_j^2}(x)\,dx \end{align}$$
by partial integration. So $\beta_1 = \beta_2$ on $C_c^\infty(\Omega)\times C_c^\infty(\Omega)$. By continuity, you have $\beta_1 = \beta_2$ on $H_0^2(\Omega)\times H_0^2(\Omega)$, since $C_c^\infty(\Omega)$ is dense.