How many functions are there which are differentiable on $(0,\infty)$ and that satisfy the relation $f^{-1}=f'$?



Function satisfying $f^{-1} =f'$
Let $a=1+p>1$ be given. We shall construct a function $f$ of the required kind with $f(a)=a$ by means of an auxiliary function $h$, defined in the neighborhood of $t=0$ and coupled to $f$ via $x=h(t)$, $f(x)=h(a t)$, $f^{-1}(x)=h(t/a)$. The condition $f'=f^{-1}$ implies that $h$ satisfies the functional equation $$(\*)\quad h(t/a) h'(t)=a h'(at).$$ Writing $h(t)=a+\sum\_{k \ge 1} c\_k t^k$ we obtain from $(\*)$ a recursion formula for the $c\_k$, and one can show that $0< c\_r<1/p^{r-1}$ for all $r\ge 1$. This means that $h$ is in fact analytic for $|t|< p$, satisfies $(\*)$ and possesses an inverse $h^{-1}$ in the neighborhood of $t=0$. It follows that the function $f(x):=h(ah^{-1}(x))$ has the required properties.


