Functional equation

A functional equation specifies a function through an identity involving values of that function at several arguments.

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.

An additive function on an additive semigroup satisfies $f(x+y)=f(x)+f(y)$ whenever $x$, $y$, and $x+y$ belong to its domain.

A real additive function $f:\mathbb{R}\to \mathbb{R}$ that is continuous at one point $x_0$ is continuous everywhere and has form $f(x)=xf(1)$. Indeed $f(0)=0$, and $f(h)=f(x_0+h)-f(x_0)\to 0$ as $h\to 0$. Then $f(x+h)-f(x)=f(h)\to 0$ for every $x$. Additivity gives $f(q)=qf(1)$ for every rational number; continuity and the density of the rational numbers extend this identity to every real argument.

Every Lebesgue-measurable additive function $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is continuous. Small-value preimages are measurable; one of their integer dilates has positive measure, so the Steinhaus theorem puts a neighbourhood of zero in their difference set. Additivity then makes $f$ small on that neighbourhood.

If $f:[0,\infty )\to \mathbb{R}$ is continuous and is an additive function, then $f(t)=tf(1)$. Additivity proves this first for nonnegative rational numbers; because $f$ is continuous, the identity extends to every nonnegative real number.