Let $g:\mathbb{R}\to \mathbb{R}$ such that
(i) for all $y_{1},y_{2} \in (0, \infty)$, $g(y_{1}+y_{2})=g(y_{1})(y_{2})$
Suppose in addition that
(ii) there exists $y \in \mathbb{R}$ such that $g(y)\neq 0$
Prove that $g(0)=1$
I am supposed to use both conditions, however this is the only way I can figure out how to do it. Could anybody tell me if I am doing it correctly or in what step I am using the second condition?
Since $0 = 0 + 0$ we have $g(0)=g(0 + 0)$. By condition (i) we have $g(0+0)=g(0)g(0)$. So $g(0)=g(0)g(0)$. Then dividing both sides by $g(0)$, we have $g(0)=1$.
That seems to prove it, but I did not use the second condition. At least I don't think I did? Am I wrong?
Since $g(y)=g(0+y)=g(0)g(y)$, where $g(y)\not=0$, then $g(y)(1-g(0))=0 \implies g(0)=1.$