linear algebra - A real function which is additive but not homogenous

From the theory of linear mappings, we know linear maps over a vector space satisfy two properties:

Additivity: $$f(v+w)=f(v)+f(w)$$

Homogeneity: $$f(\alpha v)=\alpha f(v)$$

which $\alpha\in \mathbb{F}$ is a scalar in the field which the vector space is defined on, and neither of these conditions implies the other one. If $f$ is defined over the complex numbers, $f:\mathbb{C}\longrightarrow \mathbb{C}$, then finding a mapping which is additive but not homogenous is simple; for example, $f(c)=c^*$. But can any one present an example on the reals, $f:\mathbb{R}\longrightarrow \mathbb{R}$, which is additive but not homogenous?

Answer

If $f : \Bbb{R} \to \Bbb{R}$ is additive, then you can show that $f(\alpha v) = \alpha f(v)$ for any $\alpha \in \Bbb{Q}$ (so $f$ is a linear transformation when $\Bbb{R}$ is viewed as a vector space over $\Bbb{Q}$). As $\Bbb{Q}$ is dense in $\Bbb{R}$, it follows that an additive function that is not homogeneous must be discontinuous. To construct non-trivial discontinuous functions on $\Bbb{R}$ with nice algebraic properties, you usually need to resort to the existence of a basis for $\Bbb{R}$ viewed as a vector space over $\Bbb{Q}$. Such a basis is called a Hamel basis. Given a Hamel basis $B = \{x_i \mid i \in I\}$ for $\Bbb{R}$ (where $I$ is some necessarily uncountable index set), you can easily define a function that is additive but not homogeneous, e.g., pick a basis element $x_i$ and define $f$ such that $f(x_i) = 1$ and $f(x_j) = 0$ for $j \neq i$.