Recently the following question was posed: does there exist a differentiable bijection $f:\mathbb R\to\mathbb R$ such that $f'=f^{-1}$? (Here, $f^{-1}$ is the inverse of $f$ with respect to composition of functions.) The answer, as it turns out, is negative, because such a bijection is monotonic, which implies that the derivative cannot be bijective.

There seem to be no such problems if we instead search for a bijection $f:(0,\infty)\to(0,\infty)$ such that $f'=f^{-1}$, so the following variation of the question seems interesting:

Does there exist a differentiable bijection $f:(0,\infty)\to(0,\infty)$ such that $f'=f^{-1}$?

(Edit: as mentioned at the end of this question, the question of existence has been resolved. Is the solution unique?)

Here's the first idea: if there is a function $f:(0,\infty)\to(0,\infty)$ such that $f'(x)=f^{-1}(x)$ holds for all $x\in(0,\infty)$, then $$f'(f(x))=f^{-1}(f(x))=x$$ also must hold for all $x$, since $f$ is bijective. This immediately reminds us of the chain rule, so we multiply by $f'(x)$ to yield $$f'(f(x))f'(x)=xf'(x)$$ which is equivalent to $$f'(f(x))f'(x)=xf'(x)+f(x)-f(x).$$ This implies (integrate from $1$ to $x$, for instance) that there is a primitive function $F$ of $f$ such that $$f(f(x)) = x f(x) - F(x)$$ holds for all $x\in(0,\infty)$, so we may try solving this equation instead. I don't know if this is any easier than the original problem, though. Maybe some kind of fixed point principle might work to show existence?

It would be very nice if it turns out that there is a nice characterization of such functions.

Edit: At the link pointed out by Christian Blatter in the comments there is an explicit solution ($f(x)=\frac{x^\phi}{\phi^{\phi-1}}$, where $\phi=\frac{1+\sqrt5}{2}$ is the golden ratio), so maybe it would also be interesting to know:

Is this solution unique?

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    $\begingroup$ There are various answers related to this question here: mathoverflow.net/questions/34052/function-satisfying-f-1-f/… $\endgroup$ Jan 15, 2013 at 20:54
  • $\begingroup$ @ChristianBlatter: Thanks, that looks useful. $\endgroup$
    – Dejan Govc
    Jan 15, 2013 at 21:03
  • $\begingroup$ Quick thoughts (hopefully not too wrong): if this is supposed to be a bijection, then this forces that as $x \to 0$ we have $f \to 0$ and also $f' \to 0$. From this we have that $f$ is monotone and $f'$ is also monotone (and actually very quickly growing, since $f'$ must be unbounded). But then we'd also have for $y$ big enough that $f(y) = z > y$ and so $f'(z) = y < z$, which is a contradiction with the quick growth of $f'$. So, there can be no bijection with this property. $\endgroup$
    – Marek
    Jan 15, 2013 at 21:15
  • $\begingroup$ Actually, my comment only shows that $f'$ is sublinear after certain point (which is consistent with the known solution). $\endgroup$
    – Marek
    Jan 16, 2013 at 11:29
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    $\begingroup$ I've come up with the following strategy (but it needs more work to carry out), let me know what you think: using the argument above, we can show that $f$ has a fixed point $f(x) = f'(x) = x$. Assume $f$ is analytic around this fixed point on some neighbourhood. Then all of the derivatives of $f$ are determined by $x$. (E.g. $f'' = (f^{-1})' = \dots$). Moreover, this analytic neighborhood extends all the way to the $(0, 2x)$ (and I think it actually determines $f$ completely on $(0, \infty))$. So this shows $f$ is unique up to choosing a fixed point. $\endgroup$
    – Marek
    Jan 16, 2013 at 12:03

1 Answer 1


I cannot find a general solution for this. But I may find one: The given DE eqn. is equivalent to $$ \frac{dg^{-1}}{dx} = g$$ setting $g = f^{-1}$. And this is in turn equivalent to $$ g^{-1}(x) = \int g(x)dx$$ or $$x = g(\int g(x)dx). $$ Try $g = ax^n$. Then you will find $$ x = a(\frac{a}{n+1}x^{n+1})^{n} = \frac{a^{n+1}}{(n+1)^{n}}x^{n(n+1)}.$$ Then by the root formula you will find two $n$ from the equation $n(n+1)=1$, and for these $n$ you also find suitable $a$.

And try if a linear combination of this solution works, i.e., if the solution space is linear.

If the above works, probably you can extend this idea even further: you man consider any function that is expressible as an infinite sum, i.e., the polynomials of $\infty$ power.

The above two processes, though they may indeed work, will not be that easy I guess.

  • $\begingroup$ Thanks, this is a very interesting method. Sadly, the second solution appears to be complex. I'll think about the infinite series case, though. (But it indeed doesn't seem easy.) $\endgroup$
    – Dejan Govc
    Jan 22, 2013 at 9:43

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