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What kinds of function $f: \mathbb{R} \to \mathbb{R}$ can be written as some function of itself? I.e. $f'(x) = g(f(x))$ for some function $g$?

If $f$ is given, can $g$ be solved in terms of the symbol $f$ (not in terms of specific $f$), if $g$ exists?

My question is related to part 3 of my another question, which asks about when the variance can be represented as a function of mean, both as functions of a distribution parameter, and in particular, when the variance is the derivative of the mean.

Thanks!

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Injective functions and constant ones can but there must be others. –  xavierm02 Jun 30 '13 at 14:42
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$f(x)=\exp(x)$ and $g(x)=x$. Do you have any additional constraints on $f$ and $g$? –  deoxygerbe Jun 30 '13 at 14:43
    
@deoxygerbe: No additional constraints –  Tim Jun 30 '13 at 14:46
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If $f$ is given (smooth and injective) then $g=f'\circ f^{-1}$. –  Did Jun 30 '13 at 14:58
    
@deoxygerbe He wasn't asking for examples. –  Git Gud Jun 30 '13 at 15:13
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2 Answers

up vote 4 down vote accepted

A necessary and sufficient condition is that $[f(x_1)=f(x_2)\implies f'(x_1)=f'(x_2)]$.

When this condition is met, one can define $g$ as follows:

  • If $t$ is not in $f(\mathbb R)$, then $g(t)=0$.
  • If $t$ is in $f(\mathbb R)$, then $g(t)=f'(x)$ for any $x$ such that $t=f(x)$.

The condition above is what is needed for the second part of this definition to be independent of the choice of $x$.

Thus, strictly monotone (smooth) functions $f$ are allright but $f=\cos$ is not.

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I thought I would somehow find a way to prove $S=L \cup I$ but looking at Did's answer, I start doubting so I'll just post that for the time being and edit if I find something.


Let $S=\left\{f \in \mathcal C^1\left(\Bbb R, \Bbb R\right) \mid \exists g \in \mathcal F\left(\Bbb R,\Bbb R\right), \forall x \in \Bbb R, f'(x) = g(f(x))\right\}$

Let $A=\left\{x \mapsto ax+b \mid a,b \in \Bbb R\right\}$

Let $I=\left\{f\in C^1\left(\Bbb R, \Bbb R\right)\mid \forall x,y \in \Bbb R, f(x)=f(y) \implies x = y\right\}$


Let $f\in A$

$\forall x \in \Bbb R,f'(x)=k=k\Bbb 1(x)$

So we can take $g=k\Bbb 1$

So $f \in S$

$\boxed{A \subset S}$


Let $f\in I$

We can find $h:f\left(\Bbb R\right) \to \Bbb R$ so that

$\forall x \in \Bbb R, h(f(x))=x$

Then, we have that $f'(x)=f'(h(f(x))$ so we can take $g = f'\circ h$

So $f \in S$

$\boxed{I\subset S}$

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