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

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

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 $f(qx) = q\,f(x)$ for every rational $q$ and every real $x$; in particular $f(q) = q\,f(1)$ at every rational $q$.