# Cauchy's functional equation

Cauchy's functional equation is the functional equation:

${\displaystyle f(x+y)=f(x)+f(y).\ }$

A function ${\displaystyle f}$ that solves this equation is called an additive function. Over the rational numbers, it can be shown using elementary algebra that there is a single family of solutions, namely ${\displaystyle f:x\mapsto cx}$ for any rational constant ${\displaystyle c.}$ Over the real numbers, the family of linear maps ${\displaystyle f:x\mapsto cx,}$ now with ${\displaystyle c}$ an arbitrary real constant, is likewise a family of solutions; however there can exist other solutions not of this form that are extremely complicated. However, any of a number of regularity conditions, some of them quite weak, will preclude the existence of these pathological solutions. For example, an additive function ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ is linear if:

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

On the other hand, if no further conditions are imposed on ${\displaystyle f,}$ 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 ${\displaystyle c}$ such that ${\displaystyle f(cx)\neq cf(x)}$ 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 3D to higher dimensions.[2]

This equation is sometimes referred to as Cauchy's additive functional equation to distinguish it from Cauchy's exponential functional equation ${\displaystyle f(x+y)=f(x)f(y),}$ Cauchy's logarithmic functional equation ${\displaystyle f(xy)=f(x)+f(y),}$ and Cauchy's multiplicative functional equation ${\displaystyle f(xy)=f(x)f(y).}$

## Solutions over the rational numbers

A simple argument, involving only elementary algebra, demonstrates that the set of additive maps ${\displaystyle f:\mathbb {Q} \to \mathbb {Q} }$ is identical to the set of linear maps from ${\displaystyle \mathbb {Q} }$ to ${\displaystyle \mathbb {Q} .}$

Theorem: Let ${\displaystyle f:\mathbb {Q} \to \mathbb {Q} }$ be an additive function. Then ${\displaystyle f}$ is linear.

Proof: We want to prove that any solution ${\displaystyle f:\mathbb {Q} \to \mathbb {Q} }$ to Cauchy's functional equation, ${\displaystyle f(x+y)=f(x)+f(y),}$ takes the form ${\displaystyle f(q)=cq}$ for some ${\displaystyle c\in \mathbb {Q} .}$ It is convenient to consider the cases ${\displaystyle q=0,\ q>0,\ q<0.}$

Case I: (${\displaystyle q=0}$)

By setting ${\displaystyle y=0}$ we conclude that

${\displaystyle f(x)=f(x)+f(0),\,\ \ x\in \mathbb {Q} }$
${\displaystyle \implies f(0)=0.}$

Case II: (${\displaystyle q>0}$)

By repeated application of Cauchy's equation to ${\displaystyle f(x+x+\cdots +x)=f(nx),}$ we obtain

${\displaystyle nf(x)=f(nx),\quad n\in \mathbb {N} ,\ x\in \mathbb {Q} .\qquad (*)}$

Substitution of ${\displaystyle x}$ by ${\displaystyle {\frac {x}{n}}}$ in ${\displaystyle (*),}$ and multiplication of the result by ${\displaystyle {\frac {m}{n}},}$ where ${\displaystyle m\in \mathbb {N} ,}$ yields

${\displaystyle mf\!\left({\frac {x}{n}}\right)={\frac {m}{n}}f(x),\quad m,n\in \mathbb {N} ,\ x\in \mathbb {Q} .\qquad (**)}$

Application of ${\displaystyle (*)}$ to the left-hand side of ${\displaystyle (**)}$ then affords

${\displaystyle f\!\left({\frac {m}{n}}x\right)={\frac {m}{n}}f(x),\quad m,n\in \mathbb {N} ,\ x\in \mathbb {Q} }$
${\displaystyle \implies f(qx)=qf(x),\quad q,x\in \mathbb {Q} ,\ q>0}$
${\displaystyle \implies f(q)=qf(1)=cq,\quad q\in \mathbb {Q} ^{+},}$

where ${\displaystyle c=f(1)\in \mathbb {Q} }$ is an arbitrary rational constant.

Case III: (${\displaystyle q<0}$)

Setting ${\displaystyle y=-x}$ in the functional equation and recalling that ${\displaystyle f(0)=0,}$ we obtain

${\displaystyle f(-x)=-f(x),\,\ \ x\in \mathbb {Q} .}$

Combining this with the conclusion drawn for the positive rational numbers (Case II) gives

${\displaystyle f(q)=-f(-q)=-{\big (}c(-q){\big )}=cq,\,\ \ q\in \mathbb {Q} ^{-}.}$

Considered together, the three cases above allow us to conclude that the complete solutions of Cauchy's functional equation over the rational numbers are given by:

${\displaystyle f:\mathbb {Q} \to \mathbb {Q} ,\ q\mapsto cq,\quad c=f(1)\in \mathbb {Q} .\quad \blacksquare }$

## Properties of nonlinear solutions over the real numbers

We prove below that any other solutions must be highly pathological functions. In particular, it is shown that any other solution must have the property that its graph ${\displaystyle y=f(x)}$ is dense in ${\displaystyle \mathbb {R} ^{2},}$ that is, 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 ${\displaystyle f(q)=q}$ for all ${\displaystyle q\in \mathbb {Q} ,}$ and ${\displaystyle f(\alpha )\neq \alpha }$ for some ${\displaystyle \alpha \in \mathbb {R} .}$ Put

${\displaystyle f(\alpha )=\alpha +\delta ,\ \delta \neq 0.}$

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

Put ${\displaystyle \beta ={\frac {y-x}{\delta }}}$ and choose a non-zero rational number ${\displaystyle 0\neq b\in \mathbb {Q} }$ close to ${\displaystyle \beta }$ with

${\displaystyle |\beta -b|<{\frac {r}{2|\delta |}}\ .}$

Then choose a rational number ${\displaystyle a\in \mathbb {Q} }$ close to ${\displaystyle \alpha }$ with

${\displaystyle |\alpha -a|<{\frac {r}{2|b|}}\ .}$

Now put:

${\displaystyle X=x+b(\alpha -a)}$
${\displaystyle Y=f(X).}$

Then using the functional equation, we get:

{\displaystyle {\begin{alignedat}{9}Y&=f(x+b(\alpha -a))~&&\\&=f(x+b\alpha -ba)~&&\\&=x+bf(\alpha )-bf(a)~&&\\&=y-\delta \beta +bf(\alpha )-ba~&&{\text{ since }}f(a)=a{\text{ and }}x=y-\beta \delta \\&=y-\delta \beta +b(\alpha +\delta )-ba~&&{\text{ since }}f(\alpha )=\alpha +\delta \\&=y+b(\alpha -a)-\delta (\beta -b)~&&\\\end{alignedat}}}

Because of our choices above, the point ${\displaystyle (X,Y)}$ is inside the disk.

## Existence of nonlinear solutions over the real numbers

The linearity proof given above also applies to ${\displaystyle f:\alpha \mathbb {Q} \to \mathbb {R} ,}$ where ${\displaystyle \alpha \mathbb {Q} }$ is a scaled copy of the rationals. This shows that only linear solutions are permitted when the domain of ${\displaystyle f}$ is restricted to such sets. Thus, in general, we have ${\displaystyle f(\alpha q)=f(\alpha )q}$ for all ${\displaystyle \alpha \in \mathbb {R} }$ and ${\displaystyle q\in \mathbb {Q} .}$ However, as we will demonstrate below, highly pathological solutions can be found for functions ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ based on these linear solutions, by viewing the reals as a vector space over the field of rational numbers. Note, however, that this method is nonconstructive, relying as it does on the existence of a (Hamel) basis for any vector space, a statement proved using Zorn's lemma. (In fact, the existence of a basis for every vector space is logically equivalent to the axiom of choice.)

To show that solutions other than the ones defined by ${\displaystyle f(x)=f(1)x}$ exist, we first note that because every vector space has a basis, there is a basis for ${\displaystyle \mathbb {R} }$ over the field ${\displaystyle \mathbb {Q} ,}$ i.e. a set ${\displaystyle {\mathcal {B}}\subset \mathbb {R} }$ with the property that any ${\displaystyle x\in \mathbb {R} }$ can be expressed uniquely as ${\textstyle x=\sum _{i\in I}{\lambda _{i}x_{i}},}$ where ${\displaystyle \{x_{i}\}_{i\in I}}$ is a finite subset of ${\displaystyle {\mathcal {B}},}$ and each ${\displaystyle \lambda _{i}}$ is in ${\displaystyle \mathbb {Q} .}$ We note that because no explicit basis for ${\displaystyle \mathbb {R} }$ over ${\displaystyle \mathbb {Q} }$ can be written down, the pathological solutions defined below likewise cannot be expressed explicitly.

As argued above, the restriction of ${\displaystyle f}$ to ${\displaystyle x_{i}\mathbb {Q} }$ must be a linear map for each ${\displaystyle x_{i}\in {\mathcal {B}}.}$ Moreover, because ${\displaystyle x_{i}q\mapsto f(x_{i})q}$ for ${\displaystyle q\in \mathbb {Q} ,}$ it is clear that ${\displaystyle f(x_{i}) \over x_{i}}$ is the constant of proportionality. In other words, ${\displaystyle f:x_{i}\mathbb {Q} \to \mathbb {R} }$ is the map ${\displaystyle \xi \mapsto [f(x_{i})/x_{i}]\xi .}$ Since any ${\displaystyle x\in \mathbb {R} }$ can be expressed as a unique (finite) linear combination of the ${\displaystyle x_{i}}$s, and ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ is additive, ${\displaystyle f(x)}$ is well-defined for all ${\displaystyle x\in \mathbb {R} }$ and is given by:

${\displaystyle f(x)=f{\Big (}\sum _{i\in I}\lambda _{i}x_{i}{\Big )}=\sum _{i\in I}f(x_{i}\lambda _{i})=\sum _{i\in I}f(x_{i})\lambda _{i}.}$

It is easy to check that ${\displaystyle f}$ is a solution to Cauchy's functional equation given a definition of ${\displaystyle f}$ on the basis elements, ${\displaystyle f:{\mathcal {B}}\to \mathbb {R} .}$ Moreover, it is clear that every solution is of this form. In particular, the solutions of the functional equation are linear if and only if ${\displaystyle f(x_{i}) \over x_{i}}$ is constant over all ${\displaystyle x_{i}\in {\mathcal {B}}.}$ Thus, in a sense, despite the inability to exhibit a nonlinear solution, "most" (in the sense of cardinality[3]) solutions to the Cauchy functional equation are actually nonlinear and pathological.

3. ^ It can easily be shown that ${\displaystyle \mathrm {card} ({\mathcal {B}})={\mathfrak {c}}}$; thus there are ${\displaystyle {\mathfrak {c}}^{\mathfrak {c}}=2^{\mathfrak {c}}}$ functions ${\displaystyle f:{\mathcal {B}}\to \mathbb {R} ,}$ each of which could be extended to a unique solution of the functional equation. On the other hand, there are only ${\displaystyle {\mathfrak {c}}}$ solutions that are linear.