TITLE: Functional equation: $f(2x)=f(x)+f^{-1}(x)$
QUESTION [6 upvotes]: I have been working to find solutions to the functional equation
$$f(2x)=f(x)+f^{-1}(x)$$
$$f:\mathbb R^+\to \mathbb R$$
So far I have found the trivial solution
$$f(x)=x$$
and, by mere luck, I stumbled upon the solution
$$f(x)=\ln(e^x-1)$$
But I don't know how to go after this problem strategically without using "guess and check". Can anybody find any other solutions, or show me how I might find the second solution that I mentioned analytically, without just getting lucky and happening across it?
Thanks!

REPLY [4 votes]: As requested by Gregory, here is the proof that the only solutions analytic in a neighbourhood of $0$ are $x$ and $-x$.
Rearranging and applying $f$ to both sides, we get the functional equation
$$ f(f(2x) - f(x)) = x \tag{1}$$
Taking $x=0$ (assuming this is in the domain) we get $f(0) = 0$.
If $f$ is differentiable at $0$, differentiating both sides of (1) at $x=0$ gives
$f'(0)^2 = 1$, thus $f'(0)= \pm 1$.
Now suppose $f$ is a solution of (1) that is analytic in a neighbourhood of $0$ with $f'(0) = 1$.  If $f(x)$ is not $x$, there exist some integer $m > 1$ and constant $a \ne 0$ such that
$f(x) = x + a x^m + O(x^{m+1})$.  Then the left side of (1) is is
$f(x + (2^m-1) a x^m) = x + 2^m a x^m + O(x^{m+1})$
Similarly for the case $f'(0)=-1$.