Find functions such that f(f(x))=f(x)e^x Are there monotonically increasing functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(f(x)) = e^x f(x)$?
 A: By monotonically increasing, do you mean strictly increasing or non-decreasing?
If you mean non-decreasing, then there is the trivial case $f(x)=0$. However, there may well be other cases that are harder to find.
If you mean strictly increasing, then there are no continuous solutions (there might still be discontinuous ones):
As proven by Carnahan, $f(0)=0$.
Now, consider the following limit:
$$\alpha=\lim_{x \to -\infty}{f(x)}$$
We shall consider three cases, $\alpha=0$, $-\infty<\alpha<0$, and $\alpha=-\infty$.
In the first case, one of three things occurs: it approaches from above, from below, or from equal to zero. All three of these contradict that $f$ is strictly increasing.
In the second case, we have:
$$f(\alpha)=\lim_{x \to -\infty}{f(f(x))}=\lim_{x \to -\infty}{{e^x}{f(x)}}=\lim_{x \to -\infty}{e^x} \times \lim_{x \to -\infty}{f(x)}=0 \times \alpha=0$$
So, $f(\alpha)=0=f(0)$ and since $\alpha<0$, we get a contradiction.
In the third case, 
$$-\infty=\lim_{x \to -\infty}{f(f(x))}=\lim_{x \to -\infty}{{e^x}{f(x)}}=\lim_{x \to -\infty}\frac{f(x)} {e^x}$$
Which means that it goes down to negative infinity at a greater than exponential rate.
Let $k=f(-1)$
Firstly, $k<0$ because $k=f(-1)<f(0)=0$.
Secondly, $f(k)=f(f(-1))={e^{-1}}{f(-1)}=\frac{k}{e}$
Thus, $f(k)>k$
Also, since it descends to negative infinity faster than exponentially, there must exist a finite $N$ such that $n<N \implies f(n)<n$
Thus, by assumed continuity of $f$, there must be a $N<p<k$ such that $f(p)=p$.
But, $p=f(p)=f(f(p))={e^p}{f(p)}=p{e^p}$
So, $p=p{e^p}$ which only has $p=0$ as a solution.
But, $p<k<0$. A contradiction.
