## 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,

- x is 0.
\(f(0) = f(0 + 0) = f(0) + f(0) = 2f(0)\).

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

- \(x\) 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 \(f(-x) = f(y) = -f(x)\).

- \(x\) 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 \(n\), \(f(n) = an\).

- \(x\) 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

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

- \(x\) 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

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

|Contact| |Front page| |Contents| |Algebra| |Did you know?|

Copyright © 1996-2018 Alexander Bogomolny71216442