Cauchy's functional equation $f(x+y)=f(x)+f(y)$, and the additive functions $\mathbb{R}\to \mathbb{R}$

Let $\mathbb{R}$ be the complete ordered field (Complete ordered field (least-upper-bound property), Ordered field, Field). A function $f:\mathbb{R}\to \mathbb{R}$ is additive when it satisfies Cauchy's functional equation

$f(x+y)=f(x)+f(y)\text{for all }x,y\in \mathbb{R}\mathrm{.}$

Equivalently, $f$ is a homomorphism of the additive group of $\mathbb{R}$ into itself.

The linear maps are additive. For a fixed real $c$ the function $x\mapsto cx$ satisfies $c(x+y)=cx+cy$ by distributivity, so it is additive. Cauchy's question is whether these are the only additive functions, and the answer is a genuine dichotomy: with any one of a short list of regularity conditions the answer is yes (Six regularity conditions each force an additive $f:\mathbb{R}\to \mathbb{R}$ to be $x\mapsto f(1)x$: continuity at a single point, monotonicity on a nondegenerate interval, boundedness above on one, boundedness below on one, constancy of sign on one, and a graph that is not dense in ${\mathbb{R}}^2$), and without any of them it is no (FALSE: every additive $f:\mathbb{R}\to \mathbb{R}$ is of the form $x\mapsto cx$ for a single real $c$).

No continuity, no monotonicity and no measurability is part of the definition. The equation is purely algebraic, and every regularity hypothesis below is stated explicitly where it is used.

A first consequence, recorded here because it is used immediately. An additive $f$ satisfies $f(0)=0$: putting $x=y=0$ gives $f(0)=f(0)+f(0)$, and subtracting $f(0)$ gives $f(0)=0$. The remaining elementary consequences, including $f(-x)=-f(x)$ and $\mathbb{Q}$-homogeneity, are collected in An additive $f:\mathbb{R}\to \mathbb{R}$ satisfies $f(0)=0$, $f(-x)=-f(x)$ and $f(qx)=qf(x)$ for every rational $q$ and every real $x$; in particular $f(q)=qf(1)$ at every rational $q$.