Difference and functional equations Functional equations: foundations Proposition

Density of the graph of a discontinuous additive function

Builds on Additive function Graph of a function Dense set

Statement

Let $f \colon \R \to \R$ be an additive function, that is, $f(x + y) = f(x) + f(y)$ for all $x, y \in \R$ . Assume that there is no constant $c$ with $f(x) = cx$ for all $x \in \R$ . Then the graph $\set{(x, f(x)) : x \in \R}$ is a dense subset of the plane $\R^2$ .

Remarks

Additivity gives $f(qx) = q f(x)$ for all rational $q$ and real $x$ . Since $f$ is not linear, there are nonzero $x_1, x_2$ with $f(x_1)/x_1 \ne f(x_2)/x_2$ . Then the vectors $v_1 = (x_1, f(x_1))$ and $v_2 = (x_2, f(x_2))$ are linearly independent in $\R^2$ . The graph contains every point $q_1 v_1 + q_2 v_2$ with $q_1, q_2 \in \Q$ , because $f(q_1 x_1 + q_2 x_2) = q_1 f(x_1) + q_2 f(x_2)$ . These points are dense in $\R^2$ , since $v_1, v_2$ form a basis of $\R^2$ and $\Q$ is dense in $\R$ . A typical instance is the coordinate function of a Hamel basis.

Going deeper

For a student

This proposition explains why any mild regularity makes an additive function linear. A dense graph meets every small rectangle, so the function is unbounded above and below on every interval. Hence boundedness on one interval, as in Darboux's theorem, already rules out the non-linear case. The proof only uses that $f$ is $\Q$ -linear, which every additive function on $\R$ is.

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Additive function concept
Dense set concept
Graph of a function concept
Density of the graph of a discontinuous additive function statement

Tag	`MNAWWP9J1X`
URI	`m2://MNAWWP9J1X`
Kind	statement, proposition
Writing	W0 stub
MSC	39B22
Status	This copy of the site has no build data. Run `m2 status MNAWWP9J1X`.
About	Additive function
Uses	Additive function, Graph of a function, Dense set