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
 A: 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 $\partial X$.
A: 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) $.   
