I am working on solving for all complex-valued solutions $f: \mathbb{C} \rightarrow \mathbb{C}$ satisfying the following functional equation: $$f(x)\cdot f\left(\frac{1}{1-x}\right)=x, \; \forall x \neq 0,1.$$

I tried a linear solution $f(x)=A+Bx$ as a simple guess, which yields $(A+Bx)(A+B(\frac{1}{1-x}))=x \implies (A+Bx)(A+B-Ax)=x-x^2$. Since this holds $\forall x\in \mathbb{C}$ (other than 0 and 1), the coefficients of the polynomial on the left hand side must be the same as those of the corresponding terms in the polynomial on the right hand side. Equating $x^2$ terms $\implies -AB=-1$, equating $x$ terms $\implies (A+B)B-A^2=1$, and equating constants $\implies A(A+B)=0$. Calling these three equations (1), (2), and (3) respectively, (1) and (3) together imply that $A^2=-1 \implies A=\pm i \implies B=\mp i$ (from substituting $A$ into (1) or (3)). Thus there are two linear solutions, jointly given by $f(x)=\pm i \mp i x= \pm i (x-1)$. Are these the only solutions? If not, how can I obtain others? As an aside, I notice these values of $A$ and $B$ satisfy (2), although (2) was never used in obtaining them. Why is this the case?

  • $\begingroup$ Another class of well known functions are the Möbius transformations $z\mapsto \frac{az+b}{cz+d}$. Since we have Möbius trasnformations everywhere in this equation that might be worth checking out. $\endgroup$
    – flawr
    Sep 28 '16 at 22:38
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    $\begingroup$ Possible duplicate of Find $f$ if $ f(x)+f\left(\frac{1}{1-x}\right)=x $ $\endgroup$
    – user236182
    Sep 28 '16 at 22:45
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    $\begingroup$ That is not a duplicate, as there it is $f(x) + f(1/(1-x)) = x$ and here it is $f(x) \cdot f(1/(1-x)) = x$. $\endgroup$
    – TMM
    Sep 29 '16 at 0:42
  • $\begingroup$ Take a log(f), and that would make it an exact duplicate. $\endgroup$ Sep 29 '16 at 7:58

Let $\theta(x)$ be the function $x\mapsto \frac{1}{1-x}$. Notice that $$\theta(\theta(x)) = 1-\frac{1}{x}$$ $$\theta(\theta(\theta(x))) = x$$ Thus, $\theta$ has the remarkable property that $\theta\circ\theta\circ\theta$ is simply the identity function.

By taking the original functional equation and making repeated substitutions into the functional equation, we can obtain the following system of three equations: $$f(x)\cdot (f\circ \theta)(x)=x$$ $$(f\circ \theta)(x)\cdot (f\circ \theta\circ \theta)(x)=\theta(x)$$ $$(f\circ\theta\circ\theta)(x)\cdot(f\circ\theta\circ\theta\circ\theta)(x)= (f\circ\theta\circ\theta)(x)\cdot f(x) = (\theta\circ\theta)(x)$$ We may now solve this as a system of three equations with three unknowns (namely $f, f\circ\theta,$ and $f\circ\theta\circ\theta$). Solving yields the following: $$f(x)=\pm\sqrt{\frac{x\cdot(\theta\circ\theta)(x)}{\theta(x)}}=\pm i (x-1)$$ These are the solutions that you obtained. Since we obtained these solutions by solving the system of equations above without assuming that $f$ is linear, we know that these are the only two solutions.


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