About $f(x)= f(\frac{1}{x})$

Question

Consider the equation

$$f(x)=f\left(\frac{1}{x}\right)$$

Where we want $f$ to be real-meromorphic.

Are all solutions $f$ of the form

$$f(x) = g\left(\frac{x}{1+x^2}\right)$$

Where $g$ is a continuous function ?

Answer 1

Note that $h\colon \Bbb R\setminus\{0\}\to \Bbb R\setminus (-2,2)$, $x\mapsto x+\frac1x$ is continuous and almost injective, in the sense that $h(x)=h(y)$ implies $y=x$ or $y=\frac1y$. Also, $h$ is never $0$. As a consequence, $f$ can be written as $f(x)=g(\frac 1{h(x)})$ for some $g\colon [-\frac12,\frac12]\setminus\{0\}$.

The function $g$ is also continuous. This is because if we restrict $h$ to $\Bbb R\setminus(-1,1)$ it becomes bijective with continuous inverse.

However, note that $g$ need only be continuous as function defined on $[-\frac12,\frac12]\setminus\{0\}$. Specifically, it is not guaranteed that a continuous extension to $[-\frac12,\frac12]$ exists. This exists only if $\lim_{x\to 0}g(x)$ exists, i.e., if $\lim_{x\to+\infty}f(x)=\lim_{x\to-\infty}f(x)$. If you demand that $f(x)=f(\frac1x)$ should also hold for $x=0$ in the sense that $\lim_{|x|\to\infty} f(x)=f(0)$, then this is the case.