Inspired by this question I wondered whether there are any "notable" functions $f,g$ on (or on some subset $\Omega$ of) $\mathbb R$ or $\mathbb C$ that satisfy
$$f(x)f(y) = g(x) + g(y) \:\forall x,y \in \Omega$$
By "notable" I mean nontrivial solutions (for example $f(x) = c, g(x) = \frac{c^2}{2}$ for some $c$ and all $x$ would be trivial, or also if you chose e.g. a one element domain $|\Omega|=1$), that are sufficiently well behaved (e.g. continuous or even differentiable).
Suppose that $0 \in \Omega$ and that $f(0) = c, g(0) = d$. We then have
$$cf(x) = g(x) + d$$
for all $x \in \Omega$. If $c = 0$, then $g(x)$ is also constant $0$.
Otherwise, we have $$f(x) = \frac{g(x) + d}{c}.$$ Note that $x = 0$ gives us the equation
$$c = \frac{2d}{c} \implies c^2 = 2d.$$
Looking at the diagonal case $x = y$, we have
$$f(x)^2 = 2g(x)$$
which reduces us to
$$f(x) = \frac{\frac 1 2 f(x)^2 + \frac{f(0)^2}{2}}{f(0)} = \frac 1 {2f(0)} (f(x)^2 + f(0)^2).$$
There aren't so many solutions to this.
To be even more explicit, we have a quadratic equation
$$(f(x) - f(0))^2 = f(x)^2 - 2f(0) f(x) + f(0)^2 = 0$$
which implies that $f$ is constant.
Notice that deriving both side ($dx$):
$$f'(x)f(y)=g'(x)$$
And now both sides $dy$:
$$f'(x)f'(y)=0$$
So $f'(x)=0 \Rightarrow f(x)=K $
And:
$$K^2=g(x)+g(y)$$
That has trivially as only solution $g(x)=\frac{K^2}{2}$ . These are the only solution if $f$ and $g$ are both derivable(you asked for "notable function").
