## A continuous additive function is linear, i.e., has the form $f(x)=ax$. Discontinuous additive functions look dreadful.

To be more specific, I am going to discuss real valued functions of one real variable, i.e. $f: \mathbb{R}\rightarrow \mathbb{R}$, where $\mathbb{R}$ is, as usual, the set of all real numbers. Such functions are called additive provided the following condition holds:

(*)

For every two real $x_{1}$ and $x_{2}$, $f(x_{1} + x_{2}) = f(x_{1}) + f(x_{2})$.

Function $f: \mathbb{R}\rightarrow \mathbb{R}$ is called homogeneous if, for some $a$ and every $x\in\mathbb{R}$, $f(x)=ax$. Function both additive and homogeneous is called linear. A continuous additive function is necessarily linear as I am going to show below. However, this condition is too strong: additivity together with some weaker than continuity conditions still implies linearity. This will be shown on a separate page.

Thus, assuming that function f is continuous I plan to show that $f(x) = ax$ for some real $a$. Please note that if indeed $f(x) = ax$ then $a = f(1)$ which suggests a starting point for the proof. But first let me note that (*) contains an unknown which, as we are going to establish, is equal to $f(x) = ax.$ In other words, (*) serves as an example of a functional equation - an equation whose unknown is a function.

### Proof

The proof proceeds in several steps. Additivity is a strong condition which fixes values of the function at rational points. To start with,

1. x is 0.

$f(0) = f(0 + 0) = f(0) + f(0) = 2f(0)$.

Therefore $f(0) = 2f(0)$ and finally $f(0) = 0.$

2. $x$ is negative.

Let $x$ be negative, e.g., let $x + y = 0$, where $y$ is positive; so that $-x = y.$ Then

$0 = f(0) = f(x + y) = f(x) + f(y)$.

Therefore $f(-x) = f(y) = -f(x)$.

3. $x$ is an integer.

We have $f(2) = f(1 + 1) = f(1) + f(1) = 2f(1)$. By induction, assume $f(k - 1) = (k - 1)f(1)$. Then

$f(k) = f(1 + (k-1)) = f(1) + (k-1)f(1) = kf(1)$.

Let's denote $a = f(1)$. We have shown that for all integers $n$, $f(n) = an$.

4. $x$ is rational

First of all, for any integer $n\ne 0$, we have $1 = n/n$. Then, as before, $a = f(1) = f(n/n) = nf(1/n)$. Hence, $f(1/n) = a/n = a(1/n)$. For $p = m/n$ we similarly have

$f(p) = f(m/n) = mf(1/n) = m·a/n = a(m/n) = ap$.

5. $x$ is irrational

Any irrational number $r$ can be approximated by a sequence of rational numbers $\{p_{i}\}$. The closer $p_{i}$ is to $r$, the closer $ap_{i}$ is to $ar$. However, since $ap_{i} = f(p_{i})$ and, assuming $f$ continuous, we must necessarily get $f(r) = ar$.

Continuity of the function is quite essential as it's possible to show [Ref. 1, 2] that the graph of any discontinuous solution to (*) is dense in the plane $\mathbb{R}^2$. For the sake of reference, the graph of a function $f: \mathbb{R}\rightarrow\mathbb{R}$ is defined as a set of pairs $(x, y)$, i.e., elements of $\mathbb{R}^2$ such that $y = f(x)$. Formally, $graph(f) = \{(x, y)\in \mathbb{R}^2: y = f(x)\}$.

### Remark

In analytic geometry it is usual to designate as linear functions in the form $f(x)=ax+b$. It is well known that graphs of such function are straight lines, i.e., linear. ## References

1. B. R. Gelbaum and J. M. H. Olmsted, Counterexamples in Analysis, Holden-Day, 1964
2. B. R. Gelbaum and J. M. H. Olmsted, Theorems and Counterexamples in Mathematics, Springer-Verlag, 1990 