TITLE: Strictly increasing function with $f'(x) = f(f(x))$
QUESTION [5 upvotes]: Is there a strictly increasing function $f : \Bbb{R}\to\Bbb{R}$ such that $f'(x) = f(f(x))$ for all $x$?
I think the answer is no and my argument goes like this: If there were, $f'(x) = f(f(x))$ would imply that $f$ is linear on some interval $J$ and therefore has the form $f(x)=\alpha x$ for some real $\alpha>0$. But then $f(x)=f'(x)=\alpha$ for $x\in J$. Contradiction because $f$ is strictly increasing. 
I think my solution is too simple to be true. What is wrong with it?

REPLY [4 votes]: My quick proof to show there is no such surjective function:
Suppose $f'(x)=f(f(x))$ and $f$ is strictly increasing and surjective.  Since $f$ is strictly increasing it is invertible, and the surjectivity assumption implies $f^{-1}$ is defined on all of $\mathbb R$.
Then $0<f'(f^{-1}(x))=f(x)$ for all $x\in \mathbb R$, a contradiction.