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Is there a function $f:\mathbb{R}\rightarrow (0,\infty)$, such that $f' = f\circ f$?

Apparently, I should assume by contradiction there is, and then it should imply that $f$ is increasing but I can't see the reason for that.

EDIT:

Now, we know that $f(0)$ is a lower bound for $f'(x).\forall x \in \mathbb{R}$.
The next claim is for $x<0.f(x) <f(0) + xf(0) = (1+x)f(0)$.

Why it the last claim true?

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    $\begingroup$ Look at the codomain of your function. $\endgroup$ – xavierm02 Feb 23 '14 at 19:21
  • $\begingroup$ @AlexR With your example $f\circ f= id \neq f'$, also your f is not defined for $x=0$ $\endgroup$ – user127.0.0.1 Feb 23 '14 at 19:21
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    $\begingroup$ Well, if $f'(t) = f(f(t))$, and $f: \mathbb{R} \to (0, \infty)$, then $f'(t) > 0$, so $f$ is increasing everywhere. $\endgroup$ – xyzzyz Feb 23 '14 at 19:24
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    $\begingroup$ @Max The obtained values of $p$ in your suggestion doesn't lead to positive real-valued functions. So, in a sense, it negates the possibility that $f$ can be a polynomial function :-) $\endgroup$ – Singhal Feb 24 '14 at 17:30
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    $\begingroup$ hmmm... i start to believe that there really is no such function. maybe this can be adopted to a local argument around zero using taylor to get a contradiction? (just a vague idea if you don't have anything to try, i had a long day and need a break from math puzzling so i will not try it) $\endgroup$ – Max Feb 24 '14 at 18:16
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To answer your specific question: try the fundamental theorem of calculus. Since $f$ is increasing, you know that $f\circ f$ is also increasing, and therefore $f'$ is as well. Therefore, fixing $x$, $f'(x)$ is larger than $f'(y)$ for any $y\lt x$ and since $f'(y) \lt f'(x)$, $\int_0^xf'(y)dy\lt\int_0^xf'(x)dy$ - but the former is $f(x)-f(0)$ and the latter is $xf'(x)$. This gives $f(x)-f(0)\lt xf'(x)$, or $f(x)\lt f(0)+xf'(x)\lt f(0)+xf(0)$, with the last inequality coming by the already-proved result on $f'(x)$.

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Observe $f'$ is strictly positive, and so $f$ is monotonically increasing. Thus $f'$ is a monotonically increasing function as well, so we can continuously extend $f$ and $f'$ to $-\infty$, and furthermore we must have $f'(-\infty) = 0$.

Then,

$$0 = \lim_{x \to -\infty} f'(x) = \lim_{x \to -\infty} f(f(x)) = f(\lim_{x \to -\infty} f(x)) $$

Since $f(x) > 0$ for all real $x$, we must have

$$\lim_{x \to -\infty} f(x) = -\infty $$

which is absurd.

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  • $\begingroup$ Can you elaborate on this? I don't quite understand. Why must $f'(-\infty) = 0$? Is it because $f$ is strictly nondecreasing? $\endgroup$ – Cameron Williams Feb 24 '14 at 17:50
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    $\begingroup$ @Cameron: Yes: differentiable + monotone + bounded implies the limit exists, and the limit of the derivative also exists and is $0$. $\endgroup$ – Hurkyl Feb 24 '14 at 17:52
  • $\begingroup$ Got it! Thanks. This is a very nice solution. $\endgroup$ – Cameron Williams Feb 24 '14 at 17:53
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    $\begingroup$ @Cameron: the composite of monotonic functions is monotonic: e.g. in the case of a nondecreasing function $g$ and nonincreasing function $h$, $$x \leq y \implies g(x) \leq g(y) \implies h(g(x)) \geq h(g(y)) $$ $\endgroup$ – Hurkyl Feb 24 '14 at 18:05
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    $\begingroup$ @Singhal: $f(\lim_{x \to -\infty} f(x)) = 0$. Therefore, $\lim_{x \to -\infty} f(x)$ cannot be any value $L$ for which $f(L) > 0$. There are surely other paths to the contradiction, but that's the one I used. $\endgroup$ – Hurkyl Feb 24 '14 at 22:03

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