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?


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  • $\begingroup$ We usually put $\forall x$ in front of whatever statement it applies to. Not when using words, but when using symbols. $\endgroup$ – Arthur Jan 29 at 14:31

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.

  • $\begingroup$ You're right, thank you. I've adjusted my example to ensure the signs are the same. $\endgroup$ – Floris Claassens Jan 29 at 14:59
  • $\begingroup$ Thanks a lot! I got the intuition why the statement cannot hold $\endgroup$ – StMa Jan 29 at 16:16

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.

  • $\begingroup$ Why should $m$ not exist in this case? It can be defined differently for $x \leq 0$ and $x>0$. $\endgroup$ – Bertrand Jan 29 at 15:30
  • 1
    $\begingroup$ @Bertrand: If such a function existed we would have $g(-1) = -1, g(1) = -2$. $f(-1) = m(-1) = f(1) = m(-2)$. Hence $m$ is not injective. $\endgroup$ – copper.hat Jan 29 at 15:55

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 $m \circ g = f$.


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