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Aug 11, 2013 at 17:22 answer added Vidit Nanda timeline score: 1
Aug 11, 2013 at 5:23 comment added Ryan Budney You're welcome. It's helpful to think through some basic examples like this when contemplating these kinds of questions.
Aug 11, 2013 at 5:13 comment added Danny Brown ...yep. Thanks. Sorry for such a silly question.
Aug 11, 2013 at 4:57 comment added Ryan Budney 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?
Aug 11, 2013 at 4:53 history edited Danny Brown CC BY-SA 3.0
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Aug 11, 2013 at 4:19 comment added Danny Brown @RyanBudney It does look like relative smooth approximation. I couldn't see how to add Lipschitz closeness to the proof though.
Aug 11, 2013 at 1:28 comment added Ryan Budney gt is very appropriate. Have you tried Hirsch's textbook "Differential Topology"? This looks like the relative smooth approximation theorem.
Aug 10, 2013 at 22:12 history edited Danny Brown CC BY-SA 3.0
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Aug 9, 2013 at 23:31 history edited Danny Brown CC BY-SA 3.0
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Aug 9, 2013 at 23:29 review First posts
Aug 9, 2013 at 23:35
Aug 9, 2013 at 23:10 history asked Danny Brown CC BY-SA 3.0