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

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}?
• Welcome to Math.SE! I have tried to improve the readability of your question by improving the $\rm \LaTeX$ code. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened. Commented Jan 27, 2021 at 10:14
• Thanks for your edit. Commented Jan 27, 2021 at 10:39
• What is $\Sigma$? Commented Jan 30, 2021 at 4:45
• $\Simga$ are the sets that the elliptic equation doesn't have a unique weak solution. You can see it in Evans. Commented Jan 30, 2021 at 6:02
• @CameronWilliams in addition $\Sigma$ is called the real spectrum of the uniformly elliptic operator $L$. The above theorem is on page 324 of the second edition of Evan's PDEs book. Commented Jan 30, 2021 at 6:30

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, $$v=u$$.