 Solution

The Cauchy functional equation first gives $g(0)=0$ and $g(-x)=-g(x)$. If $h\to 0$, then continuity at the given point $z$ givesso $g$ is continuous at zero. For every $x$,so $g$ is continuous everywhere.

Put $c=g(1)$. Additivity gives $g(n)=nc$ for integers $n$ and then $g(m/n)=(m/n)c$ for rational numbers. For any real $x$, choose rationals $q_j\to x$; continuity gives

For the multiplicative equation, $h(0)=h(0{)}^2$. If $h(0)=0$, then $h(x)=h(x)h(0)=0$ for every $x$. Otherwise $h(0)=1$. In that caseand $h(x)$ cannot vanish because $1=h(0)=h(x)h(-x)$. Hence $h$ is everywhere positive. The continuous function $g=\log h$ satisfies the Cauchy functional equation, so $g(x)=cx$. Therefore the complete family is