Whitney extension theorem preserving monotonicity

Question

This question is related to Monotone version of one-dimensional Whitney extension theorem. 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.

Answer 1

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.

Answer 2

No in the non decreasing case. Take $E=[0,1]\cup[2,3]$, and $f:x\to x$ on $[0,1]$ and $f:x\to x-1$ on $[2,3]$. This function is non decreasing. Its only non decreasing extension on $[0,3]$ is the function $$ f_p = \begin{cases} x & \textrm{ for } x<1 \\ 1 & \textrm{ for } 1\leq x\leq2 \\ x-1 & \textrm{ for } x>2 \end{cases} $$ which is only $C^{0,1}$. However there exists many $C^\infty$ functions on $\mathbb{R}$ which is equal to $f_p$ on $[0,1]\cup[2,3]$ (you can choose any value you wish in $\frac52$ for example).