3
$\begingroup$

Suppose I have a triangulated smooth manifold, $\tau : |K| \rightarrow M$ (so that $\tau | _{\sigma}$ is smooth for each $\sigma \in K$), and a piecewise smooth map, $f: M \rightarrow \mathbb{R}^n$. Suppose further that this map is smooth (not just pw smooth) on the polyhedron of a subcomplex $L \subset K$ (feel free to assume its also a submanifold). My question is, can I approximate my $f$ with a smooth map $g$ which is also arbitrarily close to $f$ in the Lipschitz norm and with $g|_{\tau (|L|)}=f|_{\tau (|L|)}$? Here I assume K is sitting in some Euclidean space whose distance I use to define the Lipschitz norm. Please feel free to add hypotheses as needed. Edit: For starters I probably need M to be compact.

I have been browsing Hirsch's "smoothings of PL manifolds" but I haven't found anything about this particular question. Nonetheless, I suspect the answer is yes and that the argument is probably a fairly standard convolution argument so maybe this is really a reference request for the most natural general formulation of this question and where I can find the details of its proof.

I just added the geometric topology tag. If you feel this isn't a gt question please feel free to remove it.

$\endgroup$
5
  • $\begingroup$ gt is very appropriate. Have you tried Hirsch's textbook "Differential Topology"? This looks like the relative smooth approximation theorem. $\endgroup$ Aug 11, 2013 at 1:28
  • $\begingroup$ @RyanBudney It does look like relative smooth approximation. I couldn't see how to add Lipschitz closeness to the proof though. $\endgroup$ Aug 11, 2013 at 4:19
  • $\begingroup$ If I understand your question correctly, let $f : \mathbb R \to \mathbb R$ be the absolute value function. It is piecewise smooth for some triangulation of $\mathbb R$, and smooth on any subcomplex not containing the origin. But any smooth approximation to $f$ can not be close to $f$ in the Lipschitz norm. You could construct a version of this for compact manifolds, replace $\mathbb R$ with $[-1,1]$ for example. Does this answer your question? $\endgroup$ Aug 11, 2013 at 4:57
  • $\begingroup$ ...yep. Thanks. Sorry for such a silly question. $\endgroup$ Aug 11, 2013 at 5:13
  • $\begingroup$ You're welcome. It's helpful to think through some basic examples like this when contemplating these kinds of questions. $\endgroup$ Aug 11, 2013 at 5:23

1 Answer 1

1
$\begingroup$

Here is Ryan Budney's answer from the comments, I'm copying it here so that this question does not re-appear on the front page as unanswered.


Let $f:\mathbb{R}\to\mathbb{R}$ be the absolute value function. It is piecewise smooth for some triangulation of $\mathbb{R}$, and smooth on any subcomplex not containing the origin. But any smooth approximation to $f$ can not be close to $f$ in the Lipschitz norm. You could construct a version of this for compact manifolds, replace $\mathbb{R}$ with $[−1,1]$ for example.

$\endgroup$
1
  • $\begingroup$ Apparently, this didn't help at all... the question is back on the front page! $\endgroup$ Sep 10, 2013 at 17:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.