Reference request: Where is this functional equation found?

Question

$$ g\left(\frac{x+y}{1+xy}\right) = g(x)g(y). $$ One solution is $$ g(x) = \frac{1+x}{1-x}. $$ Another is $$ g(x) = \sqrt\frac{1+x}{1-x}. $$ Any other power of the first solution is also a solution, but for the moment I suspect the second one may be in some reasonable sense the natural one.

Is there a standard name of this functional equation?

Is there an extensive literature?

Does it occur naturally in some particular field?

Are there interesting results? What are they?

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\gg}{{\rm g}}$ $\ds{\gg\pars{x + y \over 1 + xy} = \gg\pars{x}\gg\pars{y}:\ {\large ?}}$

\begin{align} \gg'\pars{x + y \over 1 + xy}\,{1 - y^{2} \over \pars{1 + xy}^{2}} &= \gg'\pars{x}\gg\pars{y} \\[3mm] \gg'\pars{x + y \over 1 + xy}\,{1 - x^{2} \over \pars{1 + xy}^{2}} &= \gg\pars{x}\gg'\pars{y} \end{align} which leads to $\ds{{1 - y^{2} \over 1 - x^{2}} = {\gg'\pars{x}\gg\pars{y} \over \gg\pars{x}\gg'\pars{y}}\quad\imp\quad \pars{1 - x^{2}}\,{\gg'\pars{x} \over \gg\pars{x}} = \pars{1 - y^{2}}\,{\gg'\pars{y} \over \gg\pars{y}} = C}$. $C$ is a constant. Then, \begin{align} &\ln\pars{\verts{g}} =\int{\dd\gg \over g} = \half\,C\int{\dd x \over 1 - x^{2}} + \overbrace{D}^{\ds{\mbox{constant}}} = \half\,C\int\pars{{1 \over x - 1} - {1 \over 1 + x}}\,\dd x + D \\[3mm]&= \half\,C\ln\pars{\verts{x - 1 \over x + 1}} + D \end{align}

$$ \ln\pars{\verts{\gg\pars{x}}} = \half\,C\ln\pars{\verts{x - 1 \over x + 1}} + D\,, \qquad C, D: \mbox{constants} $$ $$ \verts{\gg\pars{x}} = A\verts{x - 1 \over x + 1}^{C/2}\,,\qquad A:\mbox{constant} $$

1. $\gg\pars{0} = 0$ leads to the trivial solution $\gg\pars{x} = 0\,,\ \forall\ x$.
2. $\gg\pars{0} = 1$ yields $\ds{\verts{\gg\pars{x}} = \verts{x - 1 \over x + 1}^{C/2}}$.
3. What else ?.