Why $\|u_k(0)\|_{H^2(\Omega)}^2\leq C\|\Delta u_k(0)\|_{L_{2}(\Omega)}^2 $? Let $u_k(t)$ the Galerkin approximations satisfying 
$$(u_k'(t),w_i) + B[u_k(t),w_i;t]=(f(t),w_i),\;\;\;i=1,\ldots,k $$
constructed form $\{w_i\}_{i\in\mathbb{N}}$ that is a complete collection of smooth eigenfunctions for $-\Delta$ on $H^1_0(\Omega).$ In particular $\Delta u_k(t)=0$ on $\partial\Omega.$ Then
$$\|u_k(0)\|_{H^2(\Omega)}^2\leq C\|\Delta u_k(0)\|_{L_{2}(\Omega)}^2\;\;\;\;(1) $$
then 
$$\|\Delta u_k(0)\|_{L_{2}(\Omega)}^2=C(u_k(0),\Delta^2u_k(0))\;\;\;(2) $$
This is a part of the proof from regularity theory for linear evolutions equations on Evans book, page 384 on second edition. Can anyone give me a hint why $(1)$ and $(2)$ it is true?
Edit:
$$u_k(t):=\sum_{i=1}^kd_i(t)w_i(x),$$
where $d_i(t)$ sufficiently smooth functions. 
 A: The first equation is due to elliptic regularity for the problem 
$$
\begin{cases}
-\Delta u = f & \text{in } \Omega \\
u = 0 & \text{on }\partial \Omega.
\end{cases}
$$
This says that if $f \in H^k(\Omega)$ and $\partial \Omega$ is sufficiently regular, for instance $C^{2+k}$ will do the job, and $u \in H^1_0(\Omega)$ is a weak solution to the above problem, then 
$$
\Vert u\Vert_{H^{2+k}} \le C \Vert f \Vert_{H^k}
$$
for a constant $C = C(k,\Omega)>0$.  The details of this estimate are too much for me to reproduce here, so since you seem to be reading Evans I will recommend you look in Chapter 6 there.  He works out this estimate in full detail.  Re-interpreting this for your problem, we have for $k=0$ that $H^0 = L^2$, and so 
$$
\Vert u_k\Vert_{H^2} \le C \Vert \Delta u_k \Vert_{L^2},
$$
which is your first bound.
The second comes from the fact that if $\partial \Omega$ is $C^4$, then each of the eigenfunctions, which satisfy 
$$
\begin{cases}
-\Delta w_i = \lambda_i w_i & \text{in } \Omega \\
w_i = 0 & \text{on }\partial \Omega,
\end{cases}
$$
will belong to $H^4$ since 
$$
\Vert w_i\Vert_{H^4} \le C \Vert  \lambda_i w_i \Vert_{H^2} \le C \lambda_i^2 \Vert   w_i \Vert_{L^2} = C\lambda_i^2 < \infty
$$
due to the normalization $\Vert w_i \Vert_{L^2}=1$.  As a result $\Delta^2 w_i$ is well-defined, as is $\Delta^2 u_k$ for each $k$.  Now the rest follows by using the fact that the eigenfunctions are $L^2$ orthogonal:
$$
\Vert \Delta u_k \Vert_{L^2}^2 = \int_\Omega \Delta u_k \Delta u_k =  \sum_{i,j} d_i d_j \int_\Omega \Delta w_i \Delta w_j =  \sum_{i,j} d_i d_j \lambda_i \lambda_j\int_\Omega  w_i w_j =  \sum_i d_i^2 \lambda_i^2 \int_\Omega w_i w_i = \sum_i d_i^2  \int_\Omega w_i (\lambda_i^2 w_i) =  \sum_i d_i^2  \int_\Omega w_i \Delta^2 w_i =   \int_\Omega u_k \Delta^2 u_k = (u_k, \Delta^2 u_k)_{L^2}.
$$
This is the second identity.
