18
$\begingroup$

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?

  • 2
    Do You mean additivity instead of linearity? – 2017-02-06
  • 0
    @PeterMelech Sorry, yes, I mean additive, I correct it – 2017-02-06
  • 2
    https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation – 2017-02-06
  • 0
    See [Do there exist functions satisfying $f(x+y)=f(x)+f(y)$ that aren't linear?](http://math.stackexchange.com/q/385586) or some other related posts which can be found [here](http://math.stackexchange.com/questions/423492/overview-of-basic-facts-about-cauchy-functional-equation). – 2017-02-06
  • 2
    For a kind of "converse", see http://math.stackexchange.com/questions/192396/are-all-multiplicative-functions-additive – 2017-02-07

2 Answers 2

15

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 $j \neq i$.

  • 0
    If the Hamel basis link of this response doesn't work, try the following link instead: http://mathworld.wolfram.com/search/?q=Hamel+Basis – 2017-02-07
  • 0
    @Stephan: thanks. I've fixed the link in the answer (I hope). – 2017-02-07
8

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
    Alternatively, I believe you can directly prove linearity from additivity and measurability, which is a pretty striking result in itself. – 2017-02-07