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

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 $x \in \mathbf{R}$.

• Look at $g(x)=g(x/2+x/2)$. – Brian M. Scott May 5 '12 at 7:41
• $g(x) = g(x/2)^2 \geq 0$. Suppose $g(x) = 0$ then $g(y+x) = 0$ for all $y$. – t.b. May 5 '12 at 7:42
• thanks. but how can one that g(x)<0 as wel? to complete the proof? – Sikhanyiso May 5 '12 at 8:00
• I don't understand: the whole point is that $g(x)$ is never negative. – Brian M. Scott May 5 '12 at 8:02
• sorry, i ment since we've already shown that g(x) is never zero, how can one show that g(x) is also never nevative? – Sikhanyiso May 5 '12 at 8:08

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 $x \in \mathbf{R}$.