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I have been studying the book Some Nonlinear Problems In Riemannian Geometry - Thierry Aubin. On page $46$ he begins the proof of the Sobolev imbedding theorem to manifolds. The proof is divided in two steps, when $d=dist(p,q)\geq\delta$ and when $d<\delta$. In the second step ($d<\delta$), he considers a ball $\tilde{B}=B(O,d/2)$ such that $p,q\in\tilde{B}$ and defines the function $h(x)=f(\mbox{exp}_{O}(x))$ in $\tilde{B}$. After some manipulations, he uses the theorem 1.53 ( to obtain the inequality $$ \left(\int_{\tilde{B}}|\nabla_{E}h(x)|^{q}dE\right)^{1/q}\leq\frac{\sinh(a\delta)}{a\delta}\left(\frac{\pi}{2}\right)^{(n-1)q}\|\nabla f\|_{q} $$ where $\nabla_{E}$ is the Euclidean gradient. (Suppose that the assumptions of the theorem are satisfied and try to apply the theorem to obtain the inequality.)

Theorem 1.53: Let $M$ be a Riemannian manifold whose sectional curvature $K$ satisfies the bounds $-a^2\leq K\leq b^2$, the Ricci curvature being greater than $a'=(n - 1)\alpha^2$. Let $S_{P}(r_o)$ be a ball of $M$ with center $P$ and radius $r_o< \delta p$ the injectivity radius at P. Consider $(S_{P}(r_o), \exp_{P}^{-1})$, a normal geodesic coordinate system. Denote the coordinates of a point $Q = (r,\theta)\in[0, r_o]\times S_{n-1}(1)$, locally by $\theta = \{\theta^{i}\}$, ($i=1, 2,. . ., n - 1$). The metric tensor $g$ can be expressed by $$ ds^2=(dr)^2+r^2g_{\theta^{i}\theta^{j}}(r,\theta)d\theta^{i}d\theta^{j} $$ For convenience let $g_{\theta\theta}$ be one of the components $g_{\theta^{i}\theta^{i}}$ and $|g|=\mbox{det}((g_{\theta^{i}\theta^{j}}))$. Then $g_{\theta\theta}$ and $|g|$ satisfy the following inequalities: $$ \frac{\partial}{\partial r}\sqrt{g_{\theta\theta}(r,\theta)}\geq\frac{\partial}{\partial r}\log[\sin(br)/r], \ g_{\theta\theta}(r,\theta)\geq[\sin(br)/br]^2 \ \ (\mbox{if} \ br<\pi) $$ $$ \frac{\partial}{\partial r}\sqrt{g_{\theta\theta}(r,\theta)}\leq\frac{\partial}{\partial r}\log[\sin(ar)/r], \ g_{\theta\theta}(r,\theta)\leq[\sin(ar)/ar]^2 $$ $$ \frac{\partial}{\partial r}\sqrt{|g(r,\theta)|}\leq(n-1)\frac{\partial}{\partial r}\log[\sin(\alpha r)/r]\leq-a'r/3, \sqrt{|g(r,\theta)|}\leq\left[\frac{\sin(\alpha r)}{\alpha r}\right]^{n-1} $$ $$ \frac{\partial}{\partial r}\sqrt{|g(r,\theta)|}\geq(n-1)\frac{\partial}{\partial r}\log[\sin(br)/r], \sqrt{|g(r,\theta)|}\geq\left[\frac{\sin(\alpha r)}{br}\right]^{n-1} \ (\mbox{if} \ br<\pi) $$

I don't understand how to use this theorem to obtain the inequality above. Has anyone studied this demonstration of this book? Thank you.

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    $\begingroup$ You're more likely to get an answer if your question is self-contained instead of citing page and theorem numbers from a book that we might not own. $\endgroup$ – Deane Yang Jul 12 '14 at 1:54
  • $\begingroup$ I cited the book because he explain that this is a consequence of a theorem. This inequality is a step of a proof, which he uses another result to obtain it. If I had wrote here, would be confusing. The reference is necessary to understand it. $\endgroup$ – José Carlos Jul 13 '14 at 2:31
  • $\begingroup$ There's nothing wrong with citing the book, but your question assumes that we know what Theorem 1.53, $\nabla_E$, $dE$, $a$, $\delta$ are. But if we don't have the book, we don't. All you need to do is say what these are. $\endgroup$ – Deane Yang Jul 13 '14 at 5:29
  • $\begingroup$ I will write the theorem 1.53 here and will try to explain the notations. However, we need a lot of definitions and notations of this book to understand my doubt. Unfortunatly, the own notation is confusing. Anyway, I will write it later. Thank you. $\endgroup$ – José Carlos Jul 13 '14 at 16:11
  • $\begingroup$ I think the definition of each constant (for example, $-a$ is presumably the lower bound of the sectional curvature and $\delta$ the diameter of $\tilde{B}$) plus the statement of Theorem 1.53 will be enough. To be honest, I can probably figure out what everything is but you really should state your question so that anyone can read and understand the question and answer without having to consult the book. $\endgroup$ – Deane Yang Jul 13 '14 at 19:35

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