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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.

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gt is very appropriate. Have you tried Hirsch's textbook "Differential Topology"? This looks like the relative smooth approximation theorem. –  Ryan Budney Aug 11 '13 at 1:28
    
@RyanBudney It does look like relative smooth approximation. I couldn't see how to add Lipschitz closeness to the proof though. –  Danny Brown Aug 11 '13 at 4:19
    
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? –  Ryan Budney Aug 11 '13 at 4:57
    
...yep. Thanks. Sorry for such a silly question. –  Danny Brown Aug 11 '13 at 5:13
    
You're welcome. It's helpful to think through some basic examples like this when contemplating these kinds of questions. –  Ryan Budney Aug 11 '13 at 5:23

1 Answer 1

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.

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Apparently, this didn't help at all... the question is back on the front page! –  Vidit Nanda Sep 10 '13 at 17:49

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