I am trying understand the proof of the item $a$ of the theorem $2.34$ in Thierry Aubin's book. It is important clarify that theorem $2.33$ used in the proof of the theorem $2.34$ has the same statement of the theorem $2.34$ adding that $k \geq 0$ is an integer and changing $M_n$ to $\Omega$ a bounded open set of $\mathbb{R}^n$ with $C^1$-boundary or Lipschitzian boundary.
$\textbf{2.34 Theorem}$ The Kondrakov theorem, $2.33$, holds for the compact Riemannian manifolds $M_n$, and the compact Riemannian manifolds $\overline{W}_n$ with $C^1$ boundary. Namely, the following imbeddings are compact:
(a) $H_k^q(M_n) \subset L_p(M_n)$ and $H_k^q(\overline{W}_n) \subset L_p(\overline{W}_n)$, with $1 \geq \frac{1}{p} \geq \frac{1}{q} - \frac{k}{n} > 0$.
(b) $H_k^q(M_n) \subset C^{\alpha}(M_n)$ and $H_k^q(\overline{W}_n) \subset C^{\alpha}(\overline{W}_n)$, if $k - \alpha \geq \frac{n}{q}$, with $0 \leq \alpha < 1$.
$\textit{Proof.}$ Let $(\Omega_i,\varphi_i)$, ($i = 1,2,\cdots,N$) be a finite atlas of $M_n$ (respectively, $C^1$-atlas of $\overline{W}_n$), each $\Omega_i$ being homeomorphic either to a ball of $\mathbb{R}^n$ or to a half ball $D \subset \overline{E}$. We choose the atlas so that in each chart the metric tensor is bounded. Consider a $C^{\infty}$ partition of unity $\{ \alpha_i \}$ subordinate to the covering $\{ \Omega_i \}$. It is sufficient to prove the theorem in the special case $k = 1$ for the same reason as in the preceding proof, $2.33$.
a) Let $\{ f_m \}$ be a bounded sequence in $H_1^q$. Consider the functions defined on $B$ (or on $D$), $i$ being given:
$$h_m(x) = (\alpha_i f_m) \circ \varphi_i^{-1}(x)$$
Since the metric is bounded on $\Omega_i$, the set $\mathscr{A}_i$ of these functions is bounded in $H_1^q(\Omega)$ with $\Omega = B$ or $D$. (The boundary of $D$ is only Lipschitzian, but, since $\text{supp}(\alpha_i \circ \varphi_i^{-1})$ is included in $B$, we may consider a bounded open set $\Omega$ with smooth boundary which satisfies $D \subset \Omega \subset E$).
According to Theorem $2.33$, $\mathscr{A}_i$ is precompact. Thus there exists a subsequence which is a Cauchy sequence in $L_p$. Repeating this operation successively for $i = 1,2,\cdots,N$, we may select a subsequence $\{ \tilde{f}_m \}$ of the sequence $\{ f_m \}$, such that $\alpha_i \tilde{f}_m$ is a Cauchy sequence in $L_p$ for each $i$.
Thus $\{ \tilde{f}_m \}$ is a Cauchy sequence in $L_p$, since
$$|\tilde{f}_m - \tilde{f}_l| \leq \sum_{i=1}^N |\alpha_i \tilde{f}_m - \alpha_i \tilde{f}_l|.$$
$\textbf{My doubts:}$
We choose the atlas so that in each chart the metric tensor is bounded.
- Why isn't every metric tensor bounded for each atlas?
Every metric tensor is bounded for each atlas when $M$ is compact, see the answer for my second question below and the first link in this topic for details.
- How you can choose the atlas so that the metric tensor is bounded?
I was reading Nonlinear Analysis on Manifolds: Sobolev Spaces and Inequalities by Emmanuel Hebey and he gives more detail in this part of the proof, he stated that
Since $M$ is compact, $M$ can be covered by a finite number of charts $$(\Omega_s,\varphi_s)_{s=1,\cdots,N}$$ such that for any $s$ the components $g_{ij}^s$ of $g$ in $(\Omega_s,\varphi_s)$ satisfy $$\frac{1}{2} \delta_{ij} \leq g_{ij}^s \leq 2 \delta_{ij}$$ as bilinear forms.
where the $\delta_{ij}$ denote the components of a geodesic normal coordinates at $p \in M$, which follows from this result. This answers my question, but I would like to leave here an onbservation for the fact that $\mathscr{A}_i$ is bounded in $H_1^q(\Omega)$ in the proof for future reference for who read this post: we can assume that the inequalities above remain for each $p \in M$ once that the norm of tensors is invariant under change of coordinates as we can see in the comments of the answer in this topic, this invariance combined with the inequalities and the finite volume of $M$, give that $\mathscr{A}_i$ is bounded in $H_1^q(\Omega)$.
Consider the functions defined on $B$ (or on $D$), $i$ being given:
$$h_m(x) = (\alpha_i f_m) \circ \varphi_i^{-1}(x)$$
- Why the author define the sequence as $h_m(x) = (\alpha_i f_m) \circ \varphi_i^{-1}(x)$? I asking this because there can be exist $\alpha_j$ for $j \neq i$ such that the $\text{supp}(\alpha_j) \cap \varphi_i^{-1}(\Omega_i) \neq \emptyset$.
I think because a partition of unity is locally finite, then for each $p \in M_n$, there exists $l(p) \in \mathbb{N}$ and an open neighbourhood $V(p) \subset M_n$ such that $\text{supp}(\alpha_{k_j}) \cap V(p) \neq \emptyset$ for $j=1,\cdots,l(p)$. Suppose without loss of generality that $\{ \varphi_{\beta}^{-1}(\Omega_{\beta}) \} = \{ V(p) ; p \in M \}$, otherwise, we work with the atlas $\{ (\Omega_{\beta} \cap \varphi_{\beta}(V(p)), \varphi_{\beta}|_{\varphi_{\beta}^{-1}(\Omega_{\beta}) \cap V(p)}) ; p \in M \}$. Select a finite atlas $(\Omega_i,\varphi_i)$ ($i = 1,2,\cdots,N$) of $M_n$. Thus, for each $p \in M$, there exists $l(p) \in \mathbb{N}$ such that $\text{supp}(\alpha_{k_j}) \cap \varphi_i^{-1}(\Omega_i) \neq \emptyset$ for $j=1,\cdots,l(p)$. Choosing arbitrarly one of these $\alpha_{k_j}$ and fixing it, we can label as '$\alpha_i$' and define $h_m$.
$\Omega = B$ or $D$. (The boundary of $D$ is only Lipschitzian, but, since $\text{supp}(\alpha_i \circ \varphi_i^{-1})$ is included in $B$, we may consider a bounded open set $\Omega$ with smooth boundary which satisfies $D \subset \Omega \subset E$)
- What is the importance of this commentary in the proof?
$\alpha_i \tilde{f}_m$ is a Cauchy sequence in $L_p$ for each $i$.
- It is a Cauchy sequence in $L_p(M_n)$, but is it a Cauchy sequence in $L_p(M_n)$ if and only if its local representation is a Cauchy sequence in $L_p(\Omega_i)$?
I reviewed the definition of integral on smooth manifolds in Differential Forms and Applications by Do Carmo and the answer is positive for this question.
Thanks in advance!
