Solve for $f$: $$f(u)f(v)=f(u+v)(f(u)+f(v)-(q+q^{-1}))+1$$ where $q$ is a fixed complex value and $f$ is defined on $\mathbb R$.

We can assume $f(0)=q$ by symmetry. What I got is $f$ takes values in $\{q,q^{-1}\}$ and two relations $$f(u)=f(v)=q\Rightarrow f(u+v)=q,\ f(u+v)=f(v)^{-1}=q\Rightarrow f(u)=q.$$ So for example $$f(x)=\begin{cases}q\quad x\le 0\\q^{-1}\quad x>0\end{cases}$$ can be a solution. Also by those two relations if $f(x_0)$ is known then $f(kx_0)$ is known for $k\in\mathbb Q$.

Does this equation only give trivial solutions such as the one I have posted? Or is there any "horrible" solution?

  • $\begingroup$ By symmetry? Can you elaborate on the argument? I had to set up a quadratic equation and get that $q,1/q$ were the solutions before I could say that without loss of generality we could assume $f(0)=q$. $\endgroup$ – Aaron Sep 27 '17 at 13:43
  • $\begingroup$ It is the same as what you have said. There is no way to distinguish $q$ and $q^{-1}$. We can do a permutation $q\leftarrow q^{-1}$ to obtain the $f(0)=q^{-1}$ version. $\endgroup$ – erachang Sep 27 '17 at 14:07

It has some wild solutions. You can decompose $\Bbb R$ into two "wild" additively closed sets $X_1$ and $X_2$. Additively closed means that for $x,y\in X_i$ we also have $x+y\in X_i$. Define

$$f(x)=\begin{cases} q & \text{for }x\in X_1\\ q^{-1} & \text{for }x\in X_2\\ \end{cases}.$$

Your functional equation only describes dependencies between values from the same set.

  • $\begingroup$ Could you give such a wild example? $\endgroup$ – erachang Sep 27 '17 at 14:11
  • $\begingroup$ @erachang Follow the link. It needs the Axiom of Choice. $\endgroup$ – M. Winter Sep 27 '17 at 14:11

E.g. $v:=0$ : $\,\displaystyle f(u)f(0)=f(u)(f(u)+f(0)-(q+q^{-1}))+1 $

It follows $\,\displaystyle f(u)^2-(q+q^{-1})f(u)+1=(f(u)-q)(f(u)-q^{-1})=0 $

and therefore: $f(u)\in\{q,q^{-1}\}$


If $f(u)$ is real and therefore $\,q+q^{-1}=f(u)+f(u)^{-1}\,$ is real then it means that $\,|q|=1\,$ .

  • $\begingroup$ Not sure if OP meant $f:\Bbb R\to\Bbb R$ or $f:\Bbb R\to\Bbb C$ with "defined on $\Bbb R$". $\endgroup$ – M. Winter Sep 27 '17 at 13:33
  • $\begingroup$ $f$ is defined on $\mathbb R$. It doesn't mean that $f$ gives real values. $\endgroup$ – erachang Sep 27 '17 at 13:34
  • $\begingroup$ $q=\frac{1}{2}$ $\endgroup$ – erachang Sep 27 '17 at 13:35

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