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A continuous real function satisfying $g(x+y)=g(x)g(y)$ is either identically zero or $g(x)=e^{cx}$ for a real constant $c$. If it is nonzero somewhere, $g(0)=1$ and $g(x)g(-x)=1$, so it never vanishes. The identity $g(x)=g(x/2{)}^2$ then makes it everywhere positive. Its real logarithm is a continuous additive function, hence linear by additive function continuous at one point is linear. Exponentiating proves the classification, and each listed function satisfies the equation.

For the additive function, setting both arguments to zero gives $f(0)=0$, and then $f(-x)=-f(x)$. Continuity at the given point $x_0$ impliesFor any real $x$, $f(x+h)-f(x)=f(h)$, so $f$ is continuous at every point. This proves the continuity part of additive function continuous at one point is linear.

Let $c=f(1)$. Repeated addition and negation give $f(m)=mc$ for every integer $m$. For integers $p$ and positive $q$,so $f(r)=cr$ for every rational number $r$. Given any real $x$, choose rational $r_n\to x$. Continuity and density of the rational numbers giveConversely each real multiple of $x$ satisfies the Cauchy functional equation.

For the multiplicative equation, assume first that $g$ is not identically zero and choose $x_1$ with $g(x_1)\ne 0$. Then $g(x_1)=g(x_1)g(0)$ forces $g(0)=1$. Nextmakes every value nonzero, andmakes every value positive. Thus $h(x)=\log g(x)$ is a continuous real additive function, since $h(x+y)=h(x)+h(y)$. The proved classification gives $h(x)=kx$ for a real constant $k$. This solves the continuous exponential functional equation:Both types satisfy the equation and continuity, and every nonzero solution is everywhere strictly positive.