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Let $ (M,g) $ be a Riemannian manifold ($ g$ Riemannian metric) and let $ f: M \rightarrow R $ be a Lipschitz function (with respect to $ g $) with compact support. I want to study if it is possible to approximate $ f $ with smooth functions with compact support with respect to the norm of the Sobolev space $ H^p_1 $.

I know an argument that solves the problem when $ M $ is complete. The argument is the following:

Since $ f $ is a Lipschitz function with compact support then $ f \in H^p_1(M) $ (for every $ p \geq 1 $). Now, since $ M $ is complete the set of smooth functions with compact support is dense in $ H^p_1(M) $.

Now my question:

Is it possible to extend this result to non complete case? In detail if $ M $ is not complete, is it possible to approximate a lipschitz function with compact support (with respect to the norm of $ H^p_1 $) with smooth functions ($ C^\infty $) with compact support ?

Thank you

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  • $\begingroup$ perhaps you can use an exaustion of the manifold by compact balls, approximate inside each of these balls and usa some sort of "cantor's diagonal argument"... $\endgroup$ – matgaio Feb 8 '13 at 17:28
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Since everything happens on some compact set $K$, non-completeness is a non-issue. One way to get around it is to cook up a smooth function $u:M\to [1,\infty)$ such that $u=1$ in a neighborhood of $K$ and $u(x)\ge \epsilon \, \operatorname{dist}(x,\partial M)^{-1}$ for all $x$, where $\epsilon>0$. Then the conformally deformed manifold $(M,u^2ds^2)$ is complete, but the metric in a neighborhood of $K$ is the same as it was. (This might simultaneously resolve other issues with adapting the results of Fischer-Colbrie and Schoen to the noncomplete case).

Added later: I meant $\partial M$ as the metric boundary of $M$ (yes, the notation is misleading). Given any incomplete metric space $X$, we can identify it with a subset of the completion $\overline{X}$. Then it is natural to let $\partial X = \overline{X}\setminus X$, the metric boundary of $X$. The function $x\mapsto \operatorname{dist}(x,\partial X)$ is defined on $X$, and this is all we actually need from this construction: no need to ponder the nature of the elements of $\partial X$.

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  • $\begingroup$ I don't understand what you mean when you consider $ dist(x,\partial M) $. The manifolds that we are considering are manifolds without boundary (in the article of fischer-colbrie schoen i think that all manifolds are without boundaty; is it true?). $\endgroup$ – user55449 Feb 9 '13 at 11:07
  • $\begingroup$ Previously i've thought that our approximation problem is easy to extend if we consider extendable manifolds. So, i've thought that the problem remains with the class of non complete and non extendable manifolds. This is a my reflection on this problem. $\endgroup$ – user55449 Feb 9 '13 at 11:12
  • $\begingroup$ The articles 'The existence of complete riemannian metrics' (Nomizu-Ozeki) and 'Denseness of complete riemannian metrics' (Morrow) and the idea of your answer give a definitive answer to my question. Thanks $\endgroup$ – user55449 Feb 11 '13 at 10:57
  • $\begingroup$ @user55449 Sorry I lost track of your follow-up question; edited now to answer it. Glad to hear that you sorted this out meanwhile. $\endgroup$ – user53153 Feb 11 '13 at 13:51
  • $\begingroup$ i post another possible argument that should solve my question. $\endgroup$ – user55449 Feb 12 '13 at 16:02
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I think that also this argument can work: If $ (M,g) $ is a riemannian manifold (without any assumption about completeness) and $ f: M \rightarrow R $ is a lipschitz function with compact support $ K $ then $ f \in H^p_1(M) $ (for every $ p \geq 1 $ ). Then there exists a sequence of smooth function $ \zeta_n \in H^p_1 (M) $ (from definition of $ H^p_1(M) $ ) converging to $ f $ in $ H^P_1(M) $. Now let $ \phi $ be a bump function such that $ \phi=1 $ on $ K $. Then $ \phi \zeta_n $ is a sequence of smooth functions with compact support converging to $ f $ in $ H^p_1(M) $.

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