Is there any non-constant function $f(x)$ satisfying $f(x) f(y) = f(x) + f(y)$?

Question

I am interested in the following functional equation:

$\begin{equation} f \left(x \right) f \left(y \right) = f \left(x \right) + f \left(y \right) \end{equation}$

In particular, I would like to know if there are any non-constant solutions for $f \left(x \right)$. Any help would be greatly appreciated, as I am completely new to functional equations.

Kind regards.

Answer 1

The desired property implies that, for every $x$, $$ (f(x))^2=2f(x).$$ Therefore, $$ f(x)(f(x)-2)=0,$$ so either $f(x)=0$ or $f(x)=2$. Suppose there are $x$ and $y$ such that $f(x)\neq f(y)$ (say, $f(x)=0$ and $f(y)=2$). Then, $$0=f(x)f(y)\neq f(x)+ f(y) = 2.$$ Therefore, $f$ must be constant.

Answer 2

If there exists $y$ with $f(y)\neq 1$ then we obtain $f(x)=\frac{f(y)}{f(y)-1}$ for every $x$, hence $f$ is constant.

Otherwise $f(y)=1$ for all $y$ and again, $f$ is constant.

Answer 3

Assuming differentiability, take the derivative with respect to $x$ of the property. $$f'(x)f(y) = f'(x)$$ More rigorously you define $h(x, y) = f(x)f(y)$ and then partially derive wrt $x$, but this works too. The resulting equation tells you that either $f(y) = 1$ for all $y$ (which, as someone has pointed out, is not possible), or $f'(x) = 0$ for all $x$. Hence $f$ must be constant.