Is it possible to extend Urysohn's lemma to $C^1(X)$?

Question

So what I mean precisely: take $(\mathbb{R}, \mathcal{T}_{eucl})$ can we find a function $f$ for every closed $A, B \subset \mathbb{R}$ such that $A \subset f^{-1}(\{0\})$ and $B \subset f^{-1}(\{1\})$ and $f'$ is continious?

I think this makes sense because we can find a continious function $f$ such that this is true, and then redefine $f$ in a small neighboorhood around a point where $f'$ is not continious such that it is smooth, but I can't proof it, and I can't find mentoin anywhere on the internet, so I'm wondering whether this is actually true.

And if $X$ is a differential manifold then a standard partition of unity type of argument should give the result. — Daniel Schepler, Jan 15 '18 at 18:14
@MoisheCohen you're right, I should have said $\mathbb{R}$, I edited the question. — user388557, Jan 15 '18 at 18:15
see https://math.stackexchange.com/questions/198748/c-infty-version-of-urysohn-lemma-in-bbb-rn (I got there with the very first result of googling "smooth urysohn") — Mariano Suárez-Álvarez, Jan 15 '18 at 18:39

score 2 · Accepted Answer · answered Jan 15 '18 at 18:26

2

In the smooth situation oe can be much more precise than Unrysohn's lemma: see the Whitney extension theorem.

answered Jan 15 '18 at 18:26

Mariano Suárez-Álvarez

139,300

Is it possible to extend Urysohn's lemma to $C^1(X)$?

1 Answers1