Conformal to a scalar-flat complete asymptotically flat metric I'm studying a proof in Lee's Geometric Relativity shown below. The first half of the proof was just fine (hard though) with me, but after seeing his argument for $0<u<1$, I was completely lost. I mean, just because $u$ cannot be identically equal to $1$ doesn't mean $0<u<1$ everywhere. Yeah, Lee did mention the maximum principle and the fact that $u\to 1$ at infinity on each asymptotically flat chart, but what would his point be really? How could $u$ not exceed $1$ at some point? I need some assistance, please. Thank you.



Edit 1. The reason why I can't deny the possibility of $u\geq 1$ is that there exists, by definition, a bounded set $K$ that is the complement of the union of all ends. On this $K$, $u$ may assume any values if no further constraint is imposed. Even if we work on the AF coordinate neighborhood, $u$ may still assume values no less than $1$.
Edit 2. I'm sorry. It seems like I will have to first know why $L_g$ is an isomorphism if I want to deal with this very question that also involves the maximum principle. Now I'm trying to know why $w$ attains a non-negative maximum or a non-positive minimum.
Edit 3. For future reference, let me enclose the definition of an AF manifold:



Edit 4. I just found that the author assumes all manifolds are connected throughout his book. Then, since the bounded set $K$ is closed, the Hopf-Rinow theorem tells us that $K$ is compact being a subset of a complete Riemannian manifold.
 A: TLDR
Recall strong maximum principle from Evans. If $U$ is a connected domain in $\mathbb{R}^n$ and $u$ is a subsolution (with usual regularity assumptions) to an 2nd order elliptic operator $L$ on $U$, either $u$ is constant or no interior point in $U$ can reach $\text{max } u$. Theorem A.2 from Lee's appendix tells us essentially the same thing holds when replacing $U$ with an arbitrary Riemannian manifold.
Since $L_g$ is the scalar curvature PDE, it is a 2nd order elliptic PDE, so we may use maximum principle on $M$. Since $u|_{\partial M}\leq 1$, so by strong max, $u<1$ everywhere in the interior of $M$.
Edit: Here is an attempt to go through the entire proof, since the bounty asks for it.
Before starting this proof, let me recall the weak and strong maximum principles (paraphrased from Evans) for 2nd order elliptic operators $L$ over a connected, bounded domain $U\subset \mathbb{R}^n$. Lee tells us strong max holds upon replacing $U$ with $M$ in A.2.

Let $L$ be a 2nd order elliptic operator with smooth coefficients nonnegative $c$, namely $Lu:=\sum_{i,j} a^{ij}u_{ij}+b^iu_i+cu$, where $u_i,u_{ij}$ denotes taking one partial and two partials respectively. We take smooth coefficient functions $a^{ij},b^i, c$. Suppose $L_u=0$.
Weak: For some $x_o\in \partial U$, we have $u(x_o)=\max_{\overline{U}} u$.
Strong: Suppose for some interior point $x\in U^\circ$ we have $u(x)=\max_{\overline{U}} u$, then $u\equiv \max_{\overline{U}} u$.

Let $(M,g)$ be an AF manifold with nonnegative scalar curvature. We seek a conformally flat metric $\tilde{g}$ which amounts to finding a positive function $u$ such that
$$\tilde{g}=u^{\frac{4}{n-2}}g$$
and $u\to 1$ as $|x|\to \infty$. With this choice, this is equivalent to solving the elliptic 2nd order PDE
$$L_g u:=-\frac{4(n-1)}{n-2}\Delta_g u+ R_g u=0$$
such that $u\to 1$ as $|x|\to \infty$. Equivalently, making the change of variable $v=u-1$, this is equivalent to solving the elliptic 2nd order PDE
$$-\frac{4(n-1)}{n-2}\Delta_g v+ R_g v=-R_g$$
such that $v\to 0$ as $|x|\to \infty$. The specific asymptotic behavior of $R_g$ tells us that we can think of $L_g$ as an operator between the Sobolev spaces
$$L_g:W^{2,p}_{-q}(M)\to L^p_{-q-2}(M)=W^{0,p}_{q-2}(M)$$
since we ''lose two derivatives", and I guess the asymptotic behavior is also affected somehow. Our goal now is to show that $L_g$ is an isomorphism after putting specific constraints on $p,q$. In this case, we find that $v:=L_g^{-1}(-R_g)$.
Because we've transformed our problem to be $L_g v= -R_g$, this means $-R_g\in L^p_{-q-2}(M)$ which puts a constraint on $q$. Since we are also asking that $v\to 0$ as $|x|\to\infty$, this puts another constraint on $q$. If we choose $p>\frac{n}{2}$, weighted Sobolev embedding tells us $W^{2,p}_{q}\subset C^0_{-q}$. We need to put one more constraint on $q$ to ensure $L_g$ is surjective. To summarize the constraints:

*

*$q<\text{ decay rate of g}$ since $-R_g\in L^p_{-q-2}$.

*$q>0$ to ensure $v\to 0$ as $|x|\to\infty$.

*$p>\frac{n}{2}$ for weighted Sobolev embedding (I know Lee mentions he needs a constraint on $q$ also, but the statement of $A.25$ does not seem to need any constraints on the asymptotic decay).

*$q<n-2$ for surjectivity of the Laplacian.

With these constraints in place, we've successfully imitated the finite dimensional case where surjectivity is equivalent to injectivity. Therefore, suppose $L_gw=0$ and we aim to show $w\equiv 0$. By the above constraints, $w\to 0$ at infinity, and $R_g\geq 0$ and elliptic regularity of $L_g$ implies $w$ is smooth enough to satisfy a version of the strong maximum principle, which tells us $w\equiv 0$ everywhere. Thus, $L_g$ is an isomorphism, and such a $v$ exists.
Since $u=1+v$, it is smooth since $v$ is, so it also satisfies a strong maximum principle. Since $u\to 1$ as $|x|\to\infty$ and $R_g$ is nontrivial, $u$ is nonconstant and by strong maximum, $u<1$ everywhere on $M=\overline{M}-\partial M$. Since we asked that $u$ is positive, this tells us $0<u<1$, so indeed it is a conformal factor.
