Firstly, I would like to define the notion of a sub solution and a supersolution and enunciate the Schauder's fixed point theorem.
Consider the non-linear elliptic problem
$$(P) \begin{cases} \begin{align*} L u &= f(x,u), x \in \Omega,\\ u &= 0, x \in \partial \Omega, \end{align*} \end{cases}$$
where $\Omega \subset \mathbb{R}^N$ is a bounded domain and $L$ is an operator uniformly elliptic of second order of the form
$$L := - \sum_{i,j=1}^N \frac{\partial}{\partial x_j} \left( a_{ij}(x) \frac{\partial}{\partial x_i} \right) + b_i(x) \frac{\partial}{\partial x_i} + c(x), x \in \Omega,$$
where $a_{ij}, b_i, c \in C^{\alpha}(\overline{\Omega})$.
$\textbf{Definition 1.}$ We say that $\underline{u} \in C^2(\Omega) \cap C(\overline{\Omega})$ is a $\textbf{subsolution}$ of $(P)$ if
$$\begin{cases} \begin{align*} L \underline{u} &\leq f(x,\underline{u}), x \in \Omega,\\ \underline{u} &\leq 0, x \in \partial \Omega. \end{align*} \end{cases}$$
$\textbf{Definition 2.}$ We say that $\overline{u} \in C^2(\Omega) \cap C(\overline{\Omega})$ is a $\textbf{supersolution}$ of $(P)$ if
$$\begin{cases} \begin{align*} L \overline{u} &\geq f(x,\overline{u}), x \in \Omega,\\ \overline{u} &\geq 0, x \in \partial \Omega. \end{align*} \end{cases}$$
$\textbf{Theorem 1 (Schauder's fixed point theorem).}$ Let $E$ be a Banach space and $M \subset E$ a non-empty set, convex and closed. If $T: M \rightarrow M$ is a continuous and compact map, then $T$ has a fixed point.
It is used to prove that
$\textbf{Theorem 2.}$ Suppose that $f \in C^1(\overline{\Omega} \times \mathbb{R})$ and that $(P)$ has a pair of sub-super solution ordened, then exists a solution $u \in C^2(\overline{\Omega})$ that verifies $\underline{u} \leq u \leq \overline{u}$.
$\textbf{Proof.}$ Define $$I = [\underline{u},\overline{u}] := \{ u \in L^2(\Omega) ; \underline{u} \leq u \leq \overline{u} \}$$ and the solution operator $T: I \rightarrow I$ that, for each $w \in I$, is associated to the function $T(w) := u$, where $u$ is the unique solution of the problem $$(P_1) \begin{cases} \begin{align*} L u + K u &= f(x,w) + K w, x \in \Omega,\\ u &= 0, x \in \partial \Omega, \end{align*} \end{cases}$$ where $K > 0$ is a constant such that the map $t \mapsto f(x,t) + K t$ is crescent when restrict to the compact $$\overline{\Omega} \times [\min_{x \in \overline{\Omega}} \underline{u}(x), \max_{x \in \overline{\Omega}} \overline{u}(x)].$$ We are going to show that $T$ is well-defined, i.e., $T(I) \subset I$. Indeed, for each $u \in T(I)$, then exists $w \in I$ such that $T(w) = u$, where $u$ is a solution of the problem $(P_1)$. We are going to show that $w \in I$, i.e., $\underline{u} \leq w \leq \overline{u}$. We show that $\underline{u} \leq w$, the other inequality is analogous. Using that $\underline{u}$ is a subsolution of $(P)$, we obtain $$\begin{cases} \begin{align*} L (u - \underline{u}) + K ( u - \underline{u}) &\geq f(x,w) + K w - \left( f(x,\underline{u}) + K \underline{u} \right) \geq 0, x \in \Omega,\\ u - \underline{u} &\geq 0, x \in \partial \Omega, \end{align*} \end{cases}$$ where the last inequality occurs because $t \mapsto f(x,t) + K t$ is crescent and $\underline{u} \leq w$. As the operator $L + K$ satisfies the maximum principle, follows that $$ u - \underline{u} \geq 0.$$ Observe that $T$ is continuous and compact due to compact Sobolev embbedings. Furthermore, $I$ is non-empty, convex and closed.
My doubts arises here:
- Why $T$ is continuous and compact?
- Why $I$ is closed?
Thanks in advance!