# Approximation of Lipschitz functions on Riemannian manifolds

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

• 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"... – matgaio Feb 8 '13 at 17:28

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 $\partial X$.
• 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?). – user55449 Feb 9 '13 at 11:07
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 $H^p_1(M)$.