Solution of the functional equation $f(x)+f\left(\frac 1 x\right) = f(x)f\left(\frac 1 x\right)$

Question

If $f(x)+f\left(\frac 1 x\right) = f(x)f\left(\frac 1 x\right)$, where $f$ is infinitely differentiable function, will only solution of this functional equations be $f(x)=x^n+1$ or $f(x)=-x^n+1$ for some natural number $n \in \mathbb{N}$. If yes, why?

Answer 1

I suppose that we work on $]0,+\infty[$. If we put $g(x)=f(x)-1$, we get $g(x)g(1/x)=1$. As $g$ is continuous, we must have ($g(x)>0$ for all $x$) or ($g(x)<0$ for all $x$). I leave to you the case $g(x)<0$. We get as only condition that $h(t)=\log g(\exp(t))$ verify $h(t)+h(-t)=0$ for all $t\in \mathbb{R}$, hence is odd. Hence for this case, all the functions $f(x)=1+\exp(h(\log x))$ with $h$ an odd function defined on $\mathbb{R}$, are the solutions.