# Solving a function equation

I would like to solve $$f(x)=g(x)\cdot g(-x)$$ The goal is to express $$g(x)$$ in terms of $$f(x)$$. Any idea how to solve this?

• There isn't a way to do this in general. Oct 13 at 23:53
• Yes, there is no way unless g(-x) is the inverse of g(x) or equal to g(x) or anything similar. Oct 13 at 23:55
• There is a solution only when $f \geq 0$ and $f(-x)=f(x)$. In this case $g(x)=\sqrt {f(x)}$ is the unique solution. Oct 14 at 0:01
• You gave a rule to get $f$ from $g$. This can be understood as mapping $g \mapsto f$. You ask for a mapping $f \mapsto g$ (and it should be in some sense well behaved that you can write it down). However, your $g \mapsto f$ isn't even injective... Oct 14 at 0:14

Assume $$g$$ is an arbitrary function $$g:\mathbb{R} \to \mathbb{R}$$ and $$h^+$$ is an arbitrary function $$h^+:\mathbb{R^+} \to \mathbb{R}.$$ Define $$f:\mathbb{R} \to \mathbb{R}$$ as $$f(x):=g(x)g(-x)$$. Now define $$h:\mathbb{R} \to \mathbb{R}$$ by $$h(x)=\begin{cases}h^+(x), &x>0\\g(0), &x=0\\ \frac{f(x)}{h^+(-x)},&x<0\end{cases}$$ We have $$f(x)=h(x)h(-x)$$, so it satisfies the same equation as $$g$$, but $$h(x)$$ is for positive $$x$$ an arbitrary function, so $$g$$ cannot be determined by $$f.$$