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?


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 $j \neq i$.

  • $\begingroup$ If the Hamel basis link of this response doesn't work, try the following link instead: mathworld.wolfram.com/search/?q=Hamel+Basis $\endgroup$ – Stephan Feb 7 '17 at 9:05
  • $\begingroup$ @Stephan: thanks. I've fixed the link in the answer (I hope). $\endgroup$ – Rob Arthan Feb 7 '17 at 22:40

Additive but not homogenous functions $f: \mathbb R\to\mathbb R$ have to be a little bit more complicated since one can show that those functions can't be measurable and therefore need the axiom of choice in some way to be constructed.

Consider $\mathbb R$ as a vectorspace over the field $\mathbb Q$ and select a basis $(r_i)_{i\in I}$. Call $(x,i)$ the coefficient of $r_i$ in the base representation of $x$ with respect to the base $(r_i)_{i\in I}$. Then $x\mapsto (x,i)$ is $\mathbb Q$-linear and therefore especially additive, but it is obviously not $\mathbb R$-homogenous because $(r_i,i) = 1$ and $0 = (r_j,i) = (\frac{r_j}{r_i}\cdot r_i,i)$ for $i\neq j$.

  • 3
    $\begingroup$ Alternatively, I believe you can directly prove linearity from additivity and measurability, which is a pretty striking result in itself. $\endgroup$ – R.. Feb 7 '17 at 4:01

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