Necessity of compactness of manifolds $M,N$ for smooth approximation of $W^{1,p}(M,N)$.

I'm currently studying the Sobolev space $W^{1,p}(M,N)$ between manifolds $M,N$. One result by Schoen & Uhlenbeck is existence of approximation through $C^\infty(M,N)$-functions, if $M$ and $N$ are smooth and compact and $p \geq \dim M$.

A raw sketch of the result goes as follows. We use the embedding (by Whitney/Nash) of the target manifold $N$ into a Euclidean space. (Here we use smoothness.) Now we consider the smooth convolution approximation which is in $C^\infty(M, \mathbb{R}^k)$. A Poincaré inequality will show that the approximating sequence lies in some tubular neighbourhood of $N$ (here we use $p \geq \dim M$) and thus can be smoothly projected for the projected approximation to be in $C^\infty(M,N)$.

Where is compactness of the manifolds needed? Is it something in the proof? Or is something without compactness not welldefined at all?

I'd appreciate any help.

Note: Crossposted to Mathoverflow.

• One has that $C^\infty(\mathbb R^n; \mathbb R)$ is not embedded in the usual Sobolev space $W^{1,p}(\mathbb R^n; \mathbb R)$. One considers $C^\infty_c$, the class of compactly-supported smooth functions, which is dense in the closed subspace $W^{1,p}_0(\mathbb R^n; \mathbb R)$. All those subtleties disappear on compact manifolds. Maybe that's the only reason why one consider compact manifolds only? (Unfortunately I cannot answer this question myself, I don't know anything about those geometric analysis tools you mention) – Giuseppe Negro Sep 22 '16 at 16:07
• I would be skeptical of the bit where you say "the convolution lies in a tubular neighborhood" if N is not compact. If M is not compact convolution itself is dangerous, and Giuseppe's comment is also relevant. – user98602 Sep 22 '16 at 16:09
• @GiuseppeNegro So what you are saying is that in the euclidean case one either has only approximation on a bounded subset of $\mathbb{R}^n$ or if on $\mathbb{R}^n$ one can only approximate the Sobolev space $W^{1,p}(M,N)$ with $C^\infty_c$ functions. Now if $M$ is not compact, the convolution approximation itself might not work, thus one only considers compact manifolds. Is that your idea? – Nhat Sep 22 '16 at 16:24
• @MikeMiller To be honest, i have a hard time imagining a tubular neighbourhood for some non-compact manifold, however the neighbourhood does exist even for non-compact manifolds. What exactly are you skeptic about? – Nhat Sep 22 '16 at 16:26
• @JohnMa The point of my comment was to see the picture, rather than to provide an example. – user98602 Sep 22 '16 at 19:19