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In this part of the book "Elliptic PDE" of Qing Han & Fanghua Lin, the Leray–Schauder existence theorem is applied to prove the existence of $C^{2, \alpha}(\bar{\Omega})$ solution.

For $\beta \in(0,1)$, we consider the Banach space $X=C^{1, \beta}(\bar{\Omega})$ where $\Omega$ is a $C^{2, \alpha}$, bounded domain in $\mathbb{R}^{n}$. Let $L$ be an operator given by $$ L u=a^{i j}(x, u, \nabla u) u_{x_{i} x_{j}}+b(x, u, \nabla u) . $$ We assume that $L$ is elliptic in $\bar{\Omega}$; i.e., $\left(a^{i j}(x, \zeta, p)\right)$ is positive definite for all $(x, \zeta, p) \in \bar{\Omega} \times \mathbb{R} \times \mathbb{R}^{n}$. We also assume, for some $\alpha \in(0,1)$, that $a^{i j}, b \in$ $C^{\alpha}\left(\Omega \times \mathbb{R} \times \mathbb{R}^{n}\right) .$ Let $\phi \in C^{2, \alpha}(\partial \Omega) .$

For all $v \in C^{1, \beta}(\bar{\Omega})=X$, we let $u=T v$ be the unique solution in $C^{2, \alpha \beta}(\bar{\Omega})$ of the linear Dirichlet problem $$ \quad \begin{cases}a^{i j}(x, v, D v) u_{x_{i} x_{j}}+b(x, v, D v)=0 & \text { in } \Omega, \\ \left.u\right|_{\partial \Omega}=\phi & \text { on } \partial \Omega .\end{cases}$$ We note that the solvability of $L u=0$ in $\Omega$ with $u=\phi$ on $\partial \Omega$ in the space $C^{2, \alpha}(\bar{\Omega})$ is equivalent to the solvability of $T u=u$ in $X$. Let $$ L_{\sigma} u=a^{i j}(x, u, D u) u_{x_{i} x_{j}}+\sigma b(x, u, \nabla u) . $$ Then $u=\sigma T u$ in $X$ is the same as $L_{\sigma} u=0$ in $\Omega$ and $u=\sigma \phi$ on $\partial \Omega$. As a consequence of the Leray-Schauder theorem, we have the following:

Theorem $6.23$ Let $\Omega, \phi$, and $L$ be as above. If, for some $\beta>0$, there is a constant $M$ independent of $u$ and $\sigma$ such that every $C^{2, \alpha}(\bar{\Omega})$-solution of the Dirichlet problem $$ \begin{cases}L_{\sigma} u=0 & \text { in } \Omega \\ u=\sigma \phi & \text { on } \partial \Omega\end{cases} $$ satisfies $$ \|u\|_{C^{1, \beta}(\bar{\Omega})}<M, $$ then it follows that the Dirichlet problem $L u=0$ in $\Omega$ with $u=\phi$ on $\partial \Omega$ is solvable in $C^{2, \alpha}(\bar{\Omega})$.

The proof is short.

Proof: From the preceding discussion, it suffices to verify that $T$ is continuous and compact. Again, this is simply a consequence of the Schauder estimates. We note that $C^{2, \alpha \beta}(\bar{\Omega})$ is precompact in $C^{1, \beta}(\bar{\Omega})$.

My question is: Why the solvability of $L u=0$ in $\Omega$ with $u=\phi$ on $\partial \Omega$ in the space $C^{2, \alpha}(\bar{\Omega})$ is equivalent to the solvability of $T u=u$ in $C^{1, \beta}(\bar{\Omega})$.

First I define an operator$$\begin{aligned} T: C^{1, \beta}(\bar{\Omega}) \times[0,1] & \rightarrow C^{2, \alpha \beta}(\bar{\Omega}) \subset C^{1, \beta}(\bar{\Omega}) \\(v, \sigma) & \mapsto u, \end{aligned}$$ where $u=T(v, \sigma)$ is the unique solution of the linear elliptic Dirichlet problem $$ \left\{\begin{aligned} L_{\sigma} u=0 & \text { in } \Omega \\ u=\sigma \varphi & \text { on } \partial \Omega \end{aligned}\right\} . $$ The existence of a unique $C^{2, \alpha \beta}(\bar{\Omega})$ solution is guaranteed by the linear theory.

After checking that $T$ satisfies the condition of the Leray-Schauder fixed point theorem:

(Leray-Schauder fixed point theorem). Let $\mathcal{B}$ be a Banach space and $T: \mathcal{B} \times[0,1] \rightarrow \mathcal{B}$ a compact map such that

  • $T(x, 0)=0$ for each $x \in \mathcal{B}:$
  • there exists a constant $M>0$ such that for each pair $(x, \sigma) \in \mathcal{B} \times[0,1]$ which satisfies $x=T(x, \sigma)$, we have $$ \|x\|<M $$ Then $x$ is a fixed point of the map $T_{1}: \mathcal{B} \rightarrow \mathcal{B}$ given by $T_{1} y=T(y, 1), y \in \mathcal{B} .$

Then it's enough to show that the operator $T$ satisfies the hypotheses of Leray-Schauder fixed point theorem. It then follows that $T_{1}$ has a fixed point $u \in C^{1, \beta}(\bar{\Omega})$ and this is a $C^{2, \alpha}(\bar{\Omega})$.

But why this fixed point $u \in C^{1, \beta}(\bar{\Omega})$ is a $C^{2, \alpha}(\bar{\Omega})$ solution?

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  • $\begingroup$ because $Tu\in C^{2,\alpha}$ by construction? $\endgroup$
    – username
    Dec 13, 2021 at 10:26

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