# Cauchy's functional equation

Cauchy's functional equation is the functional equation

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

Solutions to this are called additive functions. Over the rational numbers, it can be shown using elementary algebra that there is a single family of solutions, namely f(x) = cx for any arbitrary rational number c. Over the real numbers, this is still a family of solutions; however there can exist other solutions that are extremely complicated. Further constraints on f sometimes preclude other solutions, for example:

• if f is continuous (proven by Cauchy in 1821). This condition was weakened in 1875 by Darboux who showed that it was only necessary for the function to be continuous at one point.
• if f is monotonic on any interval.
• if f is bounded on any interval.

On the other hand, if no further conditions are imposed on f, then (assuming the axiom of choice) there are infinitely many other functions that satisfy the equation. This was proved in 1905 by Georg Hamel using Hamel bases. Such functions are sometimes called Hamel functions.[1]

The fifth problem on Hilbert's list is a generalisation of this equation. Functions where there exists a real number $c$ such that $f(cx) \ne cf(x) \$ are known as Cauchy-Hamel functions and are used in Dehn-Hadwiger invariants which are used in the extension of Hilbert's third problem from 3-D to higher dimensions.[2]

## Proof of solution over rationals

We wish to prove that $f\left(q\right) = q f\left(1\right), q \in \mathbb{Q}$ is a solution to Cauchy's functional equation, $f(x+y) = f(x) + f(y)$.

Case 1: q=0

Set $x=1, y=0$.

$\Rightarrow f(0) = 0$.

Case 2: q>0

By repeated application of Cauchy's equation to $f\left(x + x + ... + x\right) = f\left(\alpha x\right)$:

$\alpha f\left(x\right) = f\left(\alpha x\right), \quad \alpha \in \mathbb{N^+}$

Replacing $x$ by $\frac{x}{\alpha}$, and multiplying by $\frac{\beta}{\alpha}$:

$\beta f\left(\frac{x}{\alpha}\right) = \frac{\beta}{\alpha} f\left(x\right), \quad \alpha \in \mathbb{N^+}$

By the first equation:

$f\left(\frac{\beta }{\alpha}x\right) = \frac{\beta}{\alpha} f\left(x\right), \quad \alpha, \beta \in \mathbb{N^+}$
$\Rightarrow f\left(qx\right) = q f\left(x\right), \quad q \in \mathbb{Q}, q > 0$
$\Rightarrow f\left(q\right) = q f\left(1\right), \quad q \in \mathbb{Q}, q > 0$.

Case 3: q<0

Set $y=-x$.

$\Rightarrow f(-x) = -f(x)$.

Combining this with the result from case 2:

$-f\left(q\right) = -q f\left(1\right), \quad q \in \mathbb{Q}, q > 0$
$\Rightarrow f\left(-q\right) = -q f\left(1\right), \quad q \in \mathbb{Q}, q > 0$

Replacing -q with q:

$f\left(q\right) = q f\left(1\right), \quad q \in \mathbb{Q}, q < 0. \; \blacksquare$

## Properties of other solutions

We prove below that any other solutions must be highly pathological functions. In particular, we show that any other solution must have the property that its graph $y = f(x)$ is dense in $\mathbb{R}^2$, i.e. that any disk in the plane (however small) contains a point from the graph. From this it is easy to prove the various conditions given in the introductory paragraph.

Suppose without loss of generality that $f(q) = q \ \forall q \in \mathbb{Q}$, and $f(\alpha) \neq \alpha$ for some $\alpha \in \mathbb{R}$.

Then put $f(\alpha) = \alpha + \delta, \delta \neq 0$.

We now show how to find a point in an arbitrary circle, centre $(x,y)$, radius $r$ where $x,y,r \in \mathbb{Q}, r > 0, x \neq y$.

Put $\beta = \frac{y - x}{\delta}$ and choose a rational number $b\neq 0$ close to $\beta$ with:

$\left| \beta - b \right| < \frac{r}{2 \left|\delta\right|}$

Then choose a rational number $a$ close to $\alpha$ with:

$\left| \alpha - a \right| < \frac{r}{2\left|b\right|}$

Now put:

$X = x + b (\alpha - a) \$
$Y = f(X) \$

Then using the functional equation, we get:

$Y = f(x + b (\alpha - a)) \$
$= x + b f(\alpha) - b f(a) \$
$= y - \delta \beta + b f(\alpha) - b f(a) \$
$= y - \delta \beta + b (\alpha + \delta) - b a \$
$= y + b (\alpha - a) - \delta (\beta - b) \$

Because of our choices above, the point $(X, Y)$ is inside the circle.

## Proof of the existence of other solutions

The linearity proof given above also applies to any set $\alpha \mathbb{Q}$, a scaled copy of the rationals. We can use this to find all solutions to the equation. Note that this method is highly non-constructive, relying as it does on the axiom of choice.

If we assume the axiom of choice, there is a basis for the reals over $\mathbb{Q}$ i.e. a set $A \sub \mathbb{R}$ such that for every real number $z$ there is a unique finite set $X = \left\{ x_1,\dots x_n \right\} \sub A$ and sequence $\left( \lambda_i \right)$ in $\mathbb{Q}$ such that:

$z= \sum_{i=1}^n { \lambda_i x_i }$

By the argument above, on each copy of the rationals, $x \mathbb{Q}, x \in A$, $f$ must coincide with a linear map, say with constant of proportionality g(x). In other words, f(y) = g(x)y for every y which is a rational multiple of x. Then by use of the decomposition above and repeated application of the functional equation, we can obtain the value of the function for any real number:

$f(z) = \sum_{i=1}^n { g(x_i) \lambda_i x_i }$

f(z) is a solution to the functional equation for any $g: A \rightarrow \mathbb{R}$, and every solution is of this form. f is linear if and only if g is constant.

## References

1. ^ Kuczma (2009), p.130
2. ^ V.G. Boltianskii (1978) "Hilbert's third problem", Halsted Press, Washington
• Kuczma, Marek (2009). An introduction to the theory of functional equations and inequalities. Cauchy's equation and Jensen's inequality. Basel: Birkhäuser. ISBN 9783764387495.