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?


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    $\begingroup$ I'm guessing the answer is no here, functional equations in general are difficult to handle, compound this with the inability to make any claims regarding symmetry (e.g. $x \to -x$) or $x = 0$ as a special guess (which is outside your domain). I'm guessing trial and error is the best you can hope for here. $\endgroup$
    – Gregory
    Sep 28, 2017 at 23:54
  • $\begingroup$ Another "trivial" solution (for $f: \mathbb R \to \mathbb R$) is $f(x) = -x$. $\endgroup$ Sep 29, 2017 at 0:01
  • $\begingroup$ @RobertIsrael Ah, good point. $\endgroup$ Sep 29, 2017 at 0:02
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    $\begingroup$ I can show that the only solutions that are analytic in a neighbourhood of $0$ are $x$ and $-x$. $\endgroup$ Sep 29, 2017 at 0:04
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    $\begingroup$ You might also try $f(x) = a \ln(e^{x/a}-1)$ for $a > 0$. $\endgroup$ Sep 29, 2017 at 0:37

2 Answers 2


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 $f'(0)=-1$.


Maybe plugging $x=f(x)$ can help. $f^{-1}(f(x))=x$

$f(2f(x))=f(f(x))+x$ $ \implies 2f^{'}(x)f^{'}(2f(x))=f^{'}(x)f^{'}(f(x))+1$

  • $\begingroup$ Probably a stupid attempt, but commenting on questions and suggesting something is not possible for me, commenting being disabled. $\endgroup$ Sep 29, 2017 at 1:02

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