Existence of a strictly increasing transformation between two functions

Question

Assume $f$ and $g$ are two differentiable functions defined on a compact interval $X \subseteq \mathbb{R}$ mapping into $\mathbb{R}$ . I want to proof or disproof the following statement

$ \forall x \in X: \operatorname{sign}(f'(x))=\operatorname{sign}(g'(x))\;\; \implies \exists \;\; m: \mathbb{R} \to \mathbb{R}$, strictly increasing s.t. $f=m \circ g$

My attempts raised the elementary question which conditions on arbitrary $f,g$ are in general sufficient for the existence of an $m$ such that $f=m \circ g$.

Any suggestions?

Answer 1

Counter-example: Let $f(x)=\frac{1}{3}x^{3}-\pi x^{2}+\frac{3}{4}\pi^{2} x$ and let $g(x)=\sin(x)$ on the interval [0,2\pi]. Note that $f'(x)=x^{2}-2\pi x+\frac{3}{4}\pi^{2}=(x-\frac{1}{2}\pi)(x-\frac{3}{2}\pi)$ so $f'(x)<0$ if $0\leq x<\frac{1}{2}\pi$ and $\frac{3}{2}\pi<x\leq 2\pi$ and $f'(x)>0$ if $\frac{1}{2}\pi<x<\frac{3}{2}\pi$ which corresponds with $g'(x)$.

Now suppose a transformation function exists, then $0=f(0)=m(g(0))=m(0)$ and $\frac{1}{12}\pi^{3}=f(\pi)=m(g(\pi))=m(0)$. This is a contradiction.

I believe it should work if $f$ and $g$ are both strictly increasing.

Answer 2

Let $I=[-{\pi \over 3} , {\pi \over 3}]$.

Let $f = \cos$, and let $g(x) = \begin{cases} -x^2,& x \le 0 \\ -2x^2, & \text{otherwise}\end{cases}$.

The sign of derivative condition is satisfied, but $f$ is even and $g$ is not.

Answer 3

To summarize this discussion, now that the initial claim has been disproved, it may be interesting to compare it with a case which works. May be this one:

Assume that $f$ and $g$ are one to one, defined on an interval $X \subseteq \mathbb{R}$ mapping into $\mathbb{R}$ . Then

$ \forall x \in X, \exists \;\; m: \mathbb{R} \to \mathbb{R}$ s.t. $f=m \circ g$

Proof. If $g$ is one to one, then $g^{-1}$ exists and $m \equiv f \circ g^{-1}$ satisfies $m \circ g = f$.