# Differentiable functions satisfying $f'(f(x))=f(f'(x))$

I am wondering whether or not there is a reasonable characterization of differentiable functions $f: \mathbb{R}\to \mathbb{R}$ such that $f'(f(x))=f(f'(x))$ for each $x\in\mathbb{R}$. (Or, if you like the composition sign, $f'\circ f=f\circ f'$).

I could only come up with trivial examples of such functions: $f(x)=0$ and $f(x)=e^{x}$.

This reminds me of a recent Putnam problem (2010), which asked whether or not there exists a strictly increasing function $f:\mathbb{R}\to\mathbb{R}$ satisfying $f'(x)=f(f(x))$. (The answer is: No).

Note: I see that a question of similar type has been asked here.

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$f(x) = x$ is also a solution since $f'(x) = 1$. – Cameron Williams Jun 2 '13 at 2:58
@CameronWilliams: Right. :) – Prism Jun 2 '13 at 3:06
Actually there is a larger class of functions than just $f(x) = x$ amongst monomial solutions. Let $n\in\mathbb{N}$, then $f(x) = n^{-n+1}x^n$ is a solution to the functional equation. – Cameron Williams Jun 2 '13 at 3:10
As for functions of the form $ax^2 + b$, the only nontrivial solution is $a = \frac{1}{2}, b = 1$. As for functions of the form $ax^2 + bx + c$ (with $c\neq 0$), there are no solutions. There are also no solutions for $ax^n + b$ when $n>2$. I think some binomial theorem or (abstract) algebraic arguments can be used to state what kinds of restrictions are needed for polynomial functions. – Cameron Williams Jun 2 '13 at 3:47
$f(x) = \frac{1}{x}$ is another solution. In fact, we can adapt the solution from before to get that $f(x) = (-1)^{n+1}n^{n+1}x^{-n}$ is also a solution. – Cameron Williams Jun 2 '13 at 4:07

This question seems like it will not have a simple solution.

As others have pointed out, there are a couple of interesting solutions based on , such as $f(x) = 0$ or $f(x) = e^x$ or $f(x) = (-1)^{n+1}n^{n+1}x^{-n}$ or $f(x) = 1/x$ (Though technically the last one isn't a function from $\mathbb{R}$ to $\mathbb{R}$).

However, what hasn't been done so far in the comments is to characterize the solutions by looking at the various values that $f \circ f'$ can take.

Therefore let's define $g = f \circ f' = f' \circ f$. Now the reason that makes this problem seem like it may not have a simple solution, is the fact that there are already a whole bunch of solutions just for the case $g = 0$, not all of which are smooth.

For $g = 0$, a trivial solution would be the function $f = 0$. However, if we take a function $h$ with compact support whose derivative is bounded, we can consider the function $f(x) := h(x - T)$ for some $T \in \mathbb{R}$. If we set $T$ large enough, $f$ takes on zero for all values of $f'$, satisfying $f \circ f' = 0$. Similarly, we know that $f$ is bounded given that it is continuous, leading us to conclude that $f' \circ f = 0$ for large enough T. This means that by taking any differentiable function with compact support and a bounded derivative, we can shift it far enough to the right (or left) to obtain a function that satisfies the equation of the question just for the special case of $g=0$. If I'm not mistaken, there are even functions in $C^1(\mathbb R)\setminus C^2(\mathbb{R})$ with finite support and a bounded derivative, meaning that we have some quite "not-nice" solutions.amongst the set of solutions.

Of course it isn't as easy to find solutions for other values of $g$ as it is for $g = 0$, but I think that the fact that this special case already has such a wide variety of solutions highlights the fact that this problem may be quite difficult to solve.

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