Boundedness of the inverse (Evans) why the chosen subsequence converge in $L^2(U)$

Question

THEOREM 6 (Boundedness of the inverse). If $\lambda \notin \Sigma$, there exists a constant $C$ such that

$$ \|u\|_{L^{2}(U)} \leq C\|f\|_{L^{2}(U)} \tag{29} $$ whenever $f \in L^{2}(U)$ and $u \in H_{0}^{1}(U)$ is the unique weak solution of $$ \left\{\begin{array}{c1} L u=\lambda u+f & \text { in } U \\ u=0 & \text { on } \partial U \end{array}\right. $$ The constant $C$ depends only on $\lambda, U$ and the coefficients of $L .$ This constant will blow up if $\lambda$ approaches an eigenvalue.

Proof. If not, there would exist sequences $\left\{f_{k}\right\}_{k=1}^{\infty} \subset L^{2}(U)$ and $\left\{u_{k}\right\}_{k=1}^{\infty} \subset$ $H_{0}^{1}(U)$ such that $$ \left\{\begin{aligned} L u_{k} &=\lambda u_{k}+f_{k} & & \text { in } U \\ u_{k} &=0 & & \text { on } \partial U \end{aligned}\right. $$ in the weak sense, but $$ \left\|u_{k}\right\|_{L^{2}(U)}>k\left\|f_{k}\right\|_{L^{2}(U)} \quad(k=1, \ldots) $$ As we may with no loss suppose $\left\|u_{k}\right\|_{L^{2}(U)}=1,$ we see $f_{k} \rightarrow 0$ in $L^{2}(U)$. According to the usual energy estimates the sequence $\left\{u_{k}\right\}_{k=1}^{\infty}$ is bounded in $H_{0}^{1}(U) .$ Thus there exists a subsequence $\left\{u_{k_j}\right\}_{j=1}^{\infty}$ $\subset \left\{u_{k}\right\}_{k=1}^{\infty}$ such that $$ \left\{\begin{array}{c1} u_{k_j}&→u & & \text {weakly in } H_{0}^{1}(U) \\ u_{k_j} &→u & & \text {in } L^2(U)\\ \tag{30} \label{eq30} \end{array}\right. $$ (See §D.4 for weak convergence.)Then $u$ is a weak solution of $$ \left\{\begin{array}{c1} L u=\lambda u & \text { in } U \\ u=0 & \text { on } \partial U \end{array}\right. $$ since $λ \notin \Sigma, u \equiv 0 \dots$

1. Why could the subsequence $ u_{k_j} $ converge in $L^2(U)$ and why it converges to $u$? (in \eqref{eq30})
2. Why could we take the limit in \eqref{eq30}?

Answer 1

1. The reason is stated in the appendix D4. Let me give slightly more detail. By Banach-Alaoglu, a bounded sequence in a Banach space has a subsequence that converges in the weak* topology. Since $H^1_0$ is a (separable) Hilbert space (in particular reflexive), this subsequence, considered as elements in the dual of $H^1_0$, converge in the weak topology. Regarding why convergence is equal to $u$, see 3 below.

2. This is guaranteed by the Rellich-Kondrachov compactness theorem(see previous chapter): $L^2(\Omega)$ is compactly embedded into $H^1_0(\Omega)$. So norm-boundedness in $H^1_0$ implies strong convergence of a subsequence in $L^2$.

3. You first take a subsequence $u_{k_i}$ of $u_k$ such that $u_{k_i}$ is weakly convergent in $H^1_0$ to something, call the limit $u$. Now you take a subsequence $v_j=u_{k_{i_j}}$ of $u_{k_i}$ such that $u_{k_{i_j}}$ converges to something in $L^2$, say it is $v$. In particular it converges weakly to $v$ in $L^2$. But $L^2$ is a subset of the dual of $H^1_0$. So by uniqueness of weak limits, $v=u$.