Weak convergence (in $H_0^1$) in the proof of the bounded inverse theorem in Evans's PDE book

Question

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?

Answer 1

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.

Answer 2

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).