Functional equation$ f(k-x)\cdot f(x)=f(k)$ with $k>0$

Question

The functional equation $f(k-x) f(x) = f(k)$ is satisfied by the exponential functions $e^{\lambda x}$.
I notice when $k=0$ that it also has the solutions $f(x) = \frac{-x+C}{x+C}$.

Does it have other "nice" solutions when $k > 0$?

Answer 1

Why, many. $f(x)=\dfrac{C}{k+C}\cdot\dfrac{k-x+C}{x+C}$ would do. Also, a product of two or more such functions with different $C$...

In fact, this equation is not much of an equation, in that it hardly restricts anything at all. Say, you define a function in a completely arbitrary way, continuous or not, on $\left[{k\over2},+\infty\right)$. If you want it to be continuous, just make sure that $f(k)=f^2\left({k\over2}\right)$. Then reconstruct the other half.