This question is related to [Monotone version of one-dimensional Whitney extension theorem](https://mathoverflow.net/q/215153/121665).
Let $m$ be a positive integer or $m=\infty$.



> 
> Suppose that $E\subset\mathbb{R}$ is a closed set and $f:E\to\mathbb{R}$ is a non-decreasing (strictly increasing) function such that there is a function $F\in C^m(\mathbb{R})$ that coincides with $f$ on $E$. Does it follow that we can choose $F$ to be non-decreasing (strictly increasing)?
> 
> 
> 


This looks like a reasonable conjecture, but I could not find it anywhere in the literature and I would like to know if this is a known fact.


**Edit.** This problem was not well thought. I have to think about it again and try to formulate it in a more reasonable way.



Whitney extension theorem preserving monotonicity
No also in the strictly increasing case.


Let $E = [0,1]$ and $f: x\mapsto x^2$ is strictly increasing, and is the restriction of a $C^\infty$ function. Any $C^2$ extension of this function must have $F'(0) = 0$ and $F''(0) = 2$, and so for some $\epsilon > 0$ must have $F'(-\epsilon) < 0$.




---


On the other hand, I believe the following refined conjecture is true:



> 
> If $f = F|\_E$ where $F\in C^m$, $m\geq 1$. Suppose $f$ is strictly increasing and $F'|\_E > 0$. Then $f$ admits a monotone extension.
> 
> 
> 


