Is it possible to have $f(x)f(y) = g(x)+g(y)$?

Question

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).

Answer 1

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

$$c = \frac{2d}{c} \implies c^2 = 2d.$$

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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.

Answer 2

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").