Weak convergence (in $H_0^1$) in the proof of the bounded inverse theorem in Evans's PDE book The following is the bounded inverse theorem for the elliptic equations in Evans's Partial Differential Equations (Chapter 6):



Here $\Sigma$ is the (real) spectrum of the elliptic operator $L$, where


The proof essential uses weak convergence in $H_0^1(U)$:



Here are my questions:


*

*How is (30) used in passing the limit (in subsequence) in the first read box? Here "in the weak sense" means for every $v\in H_0^1(U)$, 
$$
\int_U\sum a^{ij}\cdot(u_k)_{x_i}v_{x_j}+\sum b^i\cdot(u_k)_{x_i}v+cu_kv=\lambda (u_k,v)+(f_k,v)
$$ 
where $(\cdot,\cdot)$ is the $L^2$ inner product. Convergence of the term $(f_k,v)$ term is trivial by Cauchy-Schwartz. How to give the convergence in other terms? Relabeling the subsequence as $u_k$, weak convergence in $H_0^1(U)$ means for each $v\in H_0^1(U)$:
$$
(\nabla u_k,\nabla v)+(u_k,v)\to(\nabla u,\nabla v)+(u,v).
$$

*Why is $L^2$ convergence in (30) needed?
 A: First Question
You know that
$$u_k\rightharpoonup u\text{ in } H_0^1(U).$$
Using this you want to pass to the limit in
$$\int_U\sum a^{ij}\cdot(u_k)_{x_i}v_{x_j}+\sum b^i\cdot(u_k)_{x_i}v+cu_kv\;dx=\lambda (u_k,v)+(f_k,v)$$
to obtain
$$\int_U\sum a^{ij} u_{x_i}v_{x_j}+\sum b^i u_{x_i}v+cuv\;dx=\lambda (u,v)+(f,v).\tag{1}$$
You know how to pass the limit in the last term. So, let us consider the other terms.
Note that, for each fixed $i$ and $j$, the maps
\begin{align}
H_0^1 (U)\ni w &\mapsto a^{ij} w_{x_i}\in L^2(U)\\
H_0^1 (U)\ni w &\mapsto b^{i} w_{x_i}\in L^2(U)\\
H_0^1 (U)\ni w &\mapsto cw\in L^2(U)\\
H_0^1 (U)\ni w &\mapsto \lambda w\in L^2(U)
\end{align}
are linear and continuous and thus they preserve weak convergence. So, 
\begin{align}a^{ij} (u_k)_{x_i}&\rightharpoonup a^{ij}u_{x_i} \text{ in } L^2(U)\\
b^{i} (u_k)_{x_i}&\rightharpoonup b^iu_{x_i}\text{ in }L^2(U)\\
cu_k&\rightharpoonup cu\text{ in } L^2(U)\\
\lambda u_k&\rightharpoonup\lambda u\text{ in } L^2(U)\end{align}
for each $i$ and each $j$, which imply
\begin{align}
\int_U\sum a^{ij}\cdot(u_k)_{x_i}v_{x_j}=\sum \int_Ua^{ij}(u_k)_{x_i}v_{x_j}&\longrightarrow \sum \int_Ua^{ij}u_{x_i}v_{x_j}=\int_U\sum a^{ij}u_{x_i}v_{x_j}\\
\int_U\sum b^{i}\cdot(u_k)_{x_i}v=\sum \int_Ub^{i}(u_k)_{x_i}v&\longrightarrow \sum \int_Ub^{i}u_{x_i}v=\int_U\sum b^{i}u_{x_i}v\\
\int_Ucu_kv&\longrightarrow\int_Ucuv\\
\lambda(u_k,v)=\int_U\lambda u_kv&\longrightarrow\int_U\lambda uv=\lambda(u,v)
\end{align}
Adding  you get $(1)$.
Second Question
We know that $\|u_k\|_{L^2}=1$ for all $k$. Due to the strong convergence
$$u_{k_j}\longrightarrow u\text{ in } L^2(U)$$
we conclude that $\|u\|_{L^2}=1$ (which is used to get the contradiction).
A: What Evans is really doing here is using the fact that $\nabla u_k$ converges weakly in $L^2$.  The assumptions on $A$ guarantee that $A \nabla v \in L^2$ whenever $v \in H^1$, and so 
$$
\int A \nabla u_k \cdot \nabla v = \int \nabla u_k \cdot A \nabla v \to \int \nabla u \cdot A \nabla v = \int A \nabla u \cdot \nabla v.
$$
Similarly, 
$$
\int b \cdot \nabla u_k v \to \int b \cdot \nabla u v
$$
due to the fact that $b v \in L^2$ whenever $v \in H^1$.
The strong convergence $u_k \to u$ in $L^2$ is overkill for the convergence in the weak solution since weak convergence is enough to get 
$$
\int \lambda u_k v \to \int \lambda u v, 
$$
but he has the strong convergence, so he uses it.
