FALSE: every additive $f : \mathbb{R} \to \mathbb{R}$ is of the form $x \mapsto cx$ for a single real $c$

FALSE. Every additive $f : \mathbb{R} \to \mathbb{R}$ (Cauchy's functional equation $f(x+y) = f(x) + f(y)$, and the additive functions $\mathbb{R} \to \mathbb{R}$) is of the form $x \mapsto c\,x$ for a single real $c$.

What is true is the $\mathbb{Q}$-linear part of it, $f(qx) = q f(x)$ for rational $q$ (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 $q$), and the conditional statements of 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 $\mathbb{R}^{2}$, each of which adds a regularity hypothesis. The claim above asserts the conclusion with no hypothesis at all, and it is false.

The refutation assumes the Axiom of Choice (The Axiom of Choice), which it uses through Assuming the Axiom of Choice, $\mathbb{R}$ has a Hamel basis over $\mathbb{Q}$: there is $B \subseteq \mathbb{R}$ such that every real is a finite $\mathbb{Q}$-linear combination of elements of $B$ in exactly one way, and each basis vector carries a well-defined $\mathbb{Q}$-linear coefficient map and hence through Zorn's lemma. The hypothesis is carried explicitly in the Facts below and in every step that needs it. It is an axiom already adopted in this library, so the refutation is a refutation and not a conditional one; what it does not settle is whether a counterexample exists without choice, and nothing here bears on that question.

• What the witness looks like, by the regularity theorem. Since $f$ is additive and not of the form $x \mapsto cx$, the contrapositive of each clause of 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 $\mathbb{R}^{2}$ applies: $f$ is continuous at no point of $\mathbb{R}$, is bounded neither above nor below on any nondegenerate interval, is monotone on no nondegenerate interval, is of constant sign on none, and its graph is dense in $\mathbb{R}^{2}$. The companion page states and uses exactly this in full.

• The cost is the Axiom of Choice, and only that. The construction uses no other principle, and AC is an axiom this library has adopted, so nothing here is conditional in the sense of resting on unproved material. It is worth being precise about what is not claimed: it is not claimed that no explicit non-linear additive function can be written down, only that this one is produced by a proof that exhibits nothing.

• Every hypothesis that would rescue the claim is already recorded. Adding any single one of the six conditions of 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 $\mathbb{R}^{2}$ turns the false statement into a theorem. That is the reason the false statement is worth stating: the failure is not marginal, and yet it is repaired by an extremely weak hypothesis, as little as continuity at one single point.