TITLE: Is there a function such that $f' = f\circ f$?
QUESTION [10 upvotes]: 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?

REPLY [3 votes]: 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 $f'(x)$.