A non-zero function satisfying $g(x+y)=g(x)g(y)$ must be exponential function Let $g$ be a non-zero function satisfying $g(x+y)=g(x)g(y)$. Show that the function must be exponential function.
 A: First note that the zero function is a solution for the problem.
Now  we look for a non zero function: for $x$ s.t. $g(x)\neq 0$
$$g(x)=g(x+0)=g(x)g(0)$$
so we find $g(0)=1$
Let $g(1)=a$ and we have $g(1)=a=g(.5+.5)=g(.5)^2\geq0$.  We prove easily that $g(n)=a^n$ for $n\in\mathbb{N}$ and since
$$g(x-x)=g(0)=1=g(x)g(-x)$$
then $$g(-x)=\frac{1}{g(x)}$$
and then $$g(n)=a^n\quad\forall n\in\mathbb{Z}$$
moreover we have
$$g\left(\frac{p}{q}\right)=g\left(\frac{1}{q}\right)\times\cdots\times g\left(\frac{1}{q}\right)$$
product of $p$ factors
and since $g\left(\frac{q}{q}\right)=g(1)=a=g\left(\frac{1}{q}\right)\times\cdots\times g\left(\frac{1}{q}\right)=\left(g\left(\frac{1}{q}\right)\right)^q$ then $$g\left(\frac{1}{q}\right)=a^{\frac{1}{q}}\qquad\mathrm{and\,\, then}\qquad g\left(\frac{p}{q}\right)=a^{\frac{p}{q}}$$
hence we have proved that
$$\forall r\in\mathbb{Q}\quad g(r)=a^r$$
Finally if we we suppose that $g$ is continous and by density of $\mathbb{Q}$ in $\mathbb{R}$ we conclude
$$\forall x\in\mathbb{R}\quad g(x)=a^x$$
and don't forget to verify that this condition is sufficient.
A: Show that $g(rx) = g(x)^r$ for any $r \in \mathbb Q$. Then use continuity.
