 Discontinuous additive function

Assuming the axiom of choice, Cauchy's functional equation has solutions that are not linear maps over the real numbers. Regard $\mathbb{R}$ as a vector space over the rational numbers. Because $\sqrt{2}$ is an irrational number, $1$ and $\sqrt{2}$ are linearly independent over $\mathbb{Q}$. Zorn's lemma extends them to a Hamel basis $B$; assigning arbitrary real outputs to its basis vectors and extending by finite rational linear combinations gives an additive function.^[ref]

For a particular choice, assign $g(1)=3$ and $g(b)=0$ for every $b\in B\setminus \{1\}$. Every real number has a unique finite linear combination $x={\sum }_{b\in F}{q}_{b}b$ with $F\subset B$ finite and $q_b\in \mathbb{Q}$. Definewhere the coefficient $q_1$ is zero if $1\notin F$. The uniqueness of the Hamel basis representation makes this definition unambiguous, and adding the coefficients proves $g(x+y)=g(x)+g(y)$. On the rational two-dimensional vector subspace spanned by $1$ and $\sqrt{2}$, the rule is particularly simple:For instance, $u=2/3+\sqrt{2}$ and $v=5/6-2\sqrt{2}$ give $g(u)=2$, $g(v)=5/2$, and $g(u+v)=9/2$. Thus the map still preserves addition, even though $g(\sqrt{2})=0\ne 3\sqrt{2}$ rules out being a linear map over the real numbers.

The rational approximants $q_n$ from additive function continuous at one point is linear expose the failure of continuity: $q_n\to \sqrt{2}$ but $g(q_n)=3q_n\to 3\sqrt{2}\ne g(\sqrt{2})$. Moreover $h_n=\sqrt{2}-q_n\to 0$ while $g(h_n)=-3q_n\to -3\sqrt{2}$. The map is continuous nowhere, since continuity at any point would imply additive function continuous at one point is linear. It is unbounded on every interval of positive length by additive function bounded on an interval is linear.

The axiom of choice is used to obtain the Hamel basis; the coefficient rule does not provide an explicit way to find the representation of an arbitrary real number. A plot sampled only at rational numbers would show exactly the same values as $f(x)=3x$, so that plot cannot reveal this discontinuous additive function.