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Let $M$ be a complete, non-compact Riemannian manifold. Let $D$ be a bounded domain with smooth boundary $\partial D$ in $M$. What is the minimum requirement (about domain $D$, curvature,..) so that the Euclidean type $L^p$ Sobolev inequality holds on $D$?

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    $\begingroup$ There are many results about this. You might want to google "Sobolev inequality Riemannian manifold" and dig through the references cited in the papers that come up. For example, the book by Emmaunel Hebey is likely to have what you're looking for. springer.com/gp/book/9783540617228 $\endgroup$
    – Deane Yang
    Commented Nov 18, 2019 at 19:46
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    $\begingroup$ I suppose that by "bounded domain with smooth domain" you mean "bounded domain with smooth boundary"... If so, no other hypotheses besides the ones you've already put forward are needed. $\endgroup$ Commented Nov 18, 2019 at 21:29
  • $\begingroup$ Dear Ribeiro, yes my question has a typo." Let $M$ be a complete non-compact Riemannian manifold. Let $D$ be a bounded domain with smooth boundary $\partial D$ in $M$. What is the minimum requirement (about domain 𝐷, curvature,..) so that the Euclidean type 𝐿𝑝 Sobolev inequality holds on 𝐷?" You are saying that Euclidean type $L^p$ -Sobolev inequality holds on a bounded domain $D$ with smooth boundary $\partial D$ in $M$. Can you give me a reference please? Thanks for your answer. $\endgroup$
    – user1968
    Commented Nov 19, 2019 at 4:15
  • $\begingroup$ I took the liberty of correcting the typo. I'll answer your question below. $\endgroup$ Commented Nov 19, 2019 at 22:02

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Your hypotheses are enough. More precisely, if $(M,g)$ is a complete Riemannian manifold (be it compact or not), one is guaranteed that a closed and bounded subset of $M$ is compact by the Hopf-Rinow theorem. This is the only role of completeness. In that case, one has that $\overline D\doteq D\cup\partial D$ is a compact submanifold of $M$ with smooth boundary $\partial D$ and nonvoid interior $\mathring{D}\doteq D\smallsetminus\partial D$.

Compactness is important because it allows us to get the Sobolev inequalities without any hypotheses on curvature or injectivity radius (more precisely, such bounds are automatic on $\overline{D}$ by compactness) by just passing to a finite open cover of $\overline{D}$ by coordinate chart domains in $M$ by means of a subordinate partition of unity. In fact, the following form of the $L^p$ Sobolev inequality holds:

Theorem: Let $n=\dim M\geq 2$, $1\leq q<n$ and $\frac{1}{p}=\frac{1}{q}-\frac{1}{n}(>0)$. Then for all $u\in\mathscr{C}^\infty(\overline{D})$ we have that $$\|u\|_p\leq C_1\|\nabla u\|_q+C_0\|u\|_q$$ for some $C_0,C_1>0$.

In fact, it suffices to prove the Theorem for $q=1$, that is, $p=\frac{n}{n-1}$. All other $L^p$ Sobolev inequalities involving higher order derivatives and non-endpoint choices of $1\leq p,q<\infty$ follow by iterating the above inequality and applying Hölder's inequality, just like in the Euclidean case (edit: the Euclidean form of the Sobolev inequalities - that is, with $C_0=0$ - needs more work, see the end of the answer below)

If $u\in\mathscr{C}^\infty_c(\mathring{D})\subset\mathscr{C}^\infty(\overline{D})$, the proof is easily reducible to the Euclidean case by means of a partition of unity - more precisely, let $(U_1,\psi_1),\ldots,(U_m,\psi_m)$ be coordinate charts on $M$ adapted to $\partial D$ whenever $U_j\cap\partial D\neq\varnothing$ such that $\{U_1,\ldots,U_m\}$ is an open cover of $\overline{D}$, and let $\{f_1,\ldots,f_m\}$ be a partition of unity of $\overline{D}$ subordinate to this cover (i.e. $f_j\in\mathscr{C}^\infty_c(U_j)$, $0\leq f_j\leq 1$, $f_1(p)+\cdots+f_m(p)=1$ for all $p\in\overline{D}$). Then to get the above theorem is suffices to prove for all $j=1,\ldots,m$ the inequalities $$\|f_ju\|_p\leq K_j(\|f_j u\|_q+\|\nabla(f_j u)\|_q)\ ,\quad K_j>0\ ,$$ which on their turn clearly follow from the Euclidean $L^p$ Sobolev inequalities together with the fact that on $\mathrm{supp} f_j\subset U_j$ the components of $(\psi_j^{-1})^*g$ together with all their derivatives are bounded for each $j$. Indeed, since $|\nabla(f_j u)|\leq|\nabla u|+|\nabla f_j||u|$ ($|\nabla u|\doteq\sqrt{g(\nabla u,\nabla u)}$), we have that $$\|u\|_p\leq\sum^m_{j=1}K_j(\|f_j u\|_q+\|\nabla(f_j u)\|_q)\leq m\left(\sup_{1\leq j\leq m}K_j\right)\left[\|\nabla u\|_q+\left(1+\sup_{1\leq j\leq m}\|\nabla f_j\|_\infty\right)\|u\|_q\right]\ .$$ In fact, the above proof works for every $u\in\mathscr{C}^\infty_c(M)$ (the constants will depend on $\mathrm{supp}\,u$, of course). The details of the above argument may be found in the proofs of Theorem 2.20, pp. 44-45 and Theorem 2.30, pp. 50-53 of the excellent book by the late Thierry Aubin, Some Nonlinear Problems in Riemannian Geometry (Springer-Verlag, 1998), which also proves the $L^\infty$ version of the Sobolev inequalities and dedicates a considerable space to the determination of their optimal constants, a topic in which Aubin was a pioneer.

The general case $u\in\mathscr{C}^\infty(\overline{D})$ requires us to deal with the case that $\mathrm{supp}\,u\cap\partial D\neq\varnothing$. Unfortunately, Aubin's book does not provide all details on how to do this, which is usually done by appealing to a so-called smooth extension operator - that is, a linear map $\mathsf{K}:\mathscr{C}^\infty(\overline{D})\rightarrow\mathscr{C}^\infty_c(M)$ which is bounded with respect to all $L^p$ Sobolev norms, $1\leq p\leq\infty$ and such that $(\mathsf{K}u)|_{\overline{D}}=u$. It then suffices to apply the above proof to $\mathsf{K}u$ and finish the argument using the properties of $\mathsf{K}$.

Usually one requires a minimum of regularity from $\partial D$ in order to build such a $\mathsf{K}$ - typically one requires at least that $\partial D$ be Lipschitz (there are more specialized constructions for rougher boundaries, which I will not consider here). If $\partial D$ is smooth as you assumed, there is a particularly simple construction of such a $\mathsf{K}$ due to Seeley (Extensions of $C^\infty$ Functions Defined in a Half Space, Proc. Amer. Math. Soc. 15 (1963) 625-626). As the article's name suggests, Seeley actually did it for a closed half-space in $\mathbb{R}^n$ instead of $\overline{D}$, but this can be applied to $(\psi_j^{-1})^*(f_j u)$ for each $j$ such that $\mathrm{supp}\,f_j\cap\partial D\neq\varnothing$ while ensuring that the extension is still compactly supported in $\psi_j(U_j)\subset\mathbb{R}^n$ and hence can be pulled back to $U_j$. I do not recall right now a textbook which constructs Seeley's extension operator as I sketched above, but the missing steps are not difficult to fill if you use Seeley's original result as a black box. In any case, I strongly recommend having a look at Aubin's book.

Edit: to show that we can set $C_0=0$ in the above Theorem, one needs to apply to the case $q=1$, $p=\frac{n}{n-1}$ the following results, due to Federer, Fleming and Santalò:

Theorem 1 (Federer-Fleming): given an $n$-dimensional Riemannian manifold $(M,g)$, $n\geq 2$, set $$S_\nu(M)=\inf_{f\in\mathscr{C}^\infty_c(M)}\frac{\|\nabla f\|_1}{\|f\|_{\frac{\nu}{\nu-1}}}$$ and $$I_\nu(M)=\inf_D\frac{\mathrm{Area}(\partial D)}{\mathrm{Volume}(D)^{\frac{\nu-1}{\nu}}}\ ,$$ $\nu\in(1,\infty]$, where the last infimum is taken along all bounded subsets $D$ of $M$ with nonvoid interior and smooth boundary. Then $S_\nu(M)=I_\nu(M)$ for all $\nu$.

Theorem 2 (Santalò): Let $I_\nu$ as above. If $D$ is a nonvoid, bounded open subset of the complete Riemannian manifold $(M,g)$ with smooth boundary $\partial D$, then $I_\infty(D)>0$.

(for a proof of Theorem 1 see e.g. Theorem VIII.3.2, pp. 363-366 of the book by I. Chavel, Riemannian Geometry - A Modern Introduction, Second Edition, Cambridge University Press, 2006. As for a proof of Theorem 2, see e.g. Theorem VIII.3.5, pp. 369 of I. Chavel, loc. cit.)

Both Theorems together imply that $\|u\|_1\leq \frac{1}{I_\infty(D)}\|\nabla u\|_1$ for all $u\in\mathscr{C}^\infty_c(\mathring{D})$, hence yielding the Euclidean Sobolev inequality (with a possibly different constant from that of Euclidean space) in the case $q=1$, $p=\frac{n}{n-1}$. The remainder cases with $q>1$ follow from the Hölder inequality and by noticing that for $u$ smooth (in fact, even for $u$ locally Lipschitz) we have that $|\nabla u|=|\nabla|u||$ almost everywhere (see e.g. Lemma VIII.3.2, pp. 365-366 of I. Chavel, loc. cit.). Unfortunately the Euclidean Sobolev inequality no longer necessarily holds if $\mathrm{supp}\,u\cap\partial D\neq\varnothing$ (take e.g. $u\equiv$ a nonzero constant in $\overline{D}$) - the next best thing that survives is the so-called Sobolev-Poincaré inequality: if $u\in\mathscr{C}^\infty(\overline{D})$, then for $p,q$ as in the above Theorem there is some $C>0$ such that $$\|u-\overline{u}\|_p\leq C\|\nabla u\|_q\ ,\quad\overline{u}=\frac{\int_{\overline{D}} u d\mu_g}{\int_{\overline{D}}d\mu_g} \,$$ where $d\mu_g$ is the Riemannian volume element of $(M,g)$.

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  • $\begingroup$ Can we take $C_0=0$ in abobe theorem? Because I just need the Euclidean type : $||u||_q\le C_1||\nabla u||_p where $$ u\in C_0^{\infty}(D) $ $ and $$\frac{1}{p}=\frac{1}{q}-\frac{1}{n}.$ $\endgroup$
    – user1968
    Commented Nov 20, 2019 at 4:49
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    $\begingroup$ Hmm... I missed that when you said "Euclidean" you actually meant $C_0=0$. In this case the inequality only holds if the support of $u$ is contained in the interior of $D$ since nonzero constant functions in $\overline{D}$ clearly violate the Euclidean Sobolev inequality - I have adjusted my answer accordingly. $\endgroup$ Commented Nov 20, 2019 at 7:13

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