# Existence of a strictly increasing transformation between two functions [closed]

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?

## closed as off-topic by Frpzzd, max_zorn, Alexander Gruber♦Jan 30 at 1:12

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• We usually put $\forall x$ in front of whatever statement it applies to. Not when using words, but when using symbols. – 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 $$\frac{3}{2}\pi and $$f'(x)>0$$ if $$\frac{1}{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.

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

• Why should $m$ not exist in this case? It can be defined differently for $x \leq 0$ and $x>0$. – Bertrand Jan 29 at 15:30
• @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. – 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 $$m \circ g = f$$.