A non-zero function satisfying $g(x+y) = g(x)g(y)$ must be positive everywhere

Question

Let $g: \mathbf R \to \mathbf R$ be a function which is not identically zero and which satisfies the equation $$ g(x+y)=g(x)g(y) \quad\text{for all } x,y \in \mathbf{R}. $$ Show that $g(x)\gt0$ for all $x \in \mathbf{R}$.

Answer 1

We have $g(x) = g(\tfrac{x}{2} + \tfrac{x}{2}) = g(\tfrac{x}{2})^2 \geq 0$ for all $x \in \mathbf{R}$.

Suppose we have $g(x_0) = 0$ for some $x_0 \in \mathbf{R}$. Then $g(x_0+y) = g(x_0)g(y) = 0$ for all $y \in \mathbf{R}$, hence $g$ must be identically zero. Since you assume that's not the case, there can't be any such $x_0$, thus $g(x) \gt 0$ for all $x \in \mathbf{R}$.