TITLE: Let $f:\mathbb{R}\to(0,\infty)$ be a differentiable function. For all $x\in\mathbb{R}$ $f'(x)=f(f(x)).$ Then show that such function does not exists
QUESTION [6 upvotes]: What i have done is very small.
$$f'(x)=f(f(x))\implies f(f'(x))=f(f(f(x)))$$Now $$f(f(f(x)))=f'(f(x))$$Hence$$f(f'(x))=f'(f(x))$$Now i am blank. What to do for the proof

REPLY [11 votes]: Note that under the hypothesis $f$ is increasing. So, $f(f(x))>f(0)$ for all $x\in\Bbb R$. So, $f'(x)$ has a lower bound which is $f(0)$.
Hence $f(x)<f(0)+xf(0)=(1+x)f(0)$ for all $x<0$. So, for $x\leq -1$ we have $f(x)\leq 0$, contradiction.