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.
• if f is Lebesgue measurable.

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)\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 3-D to higher dimensions.^[2]

Proof of solution over rationals

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

Case 1: $q=0$

Set $x=1,y=0$.

$\Rightarrow f(0)=0$.

Case 2: ${\displaystyle q>0}$

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:

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

$\Rightarrow f\left(qx\right)=qf\left(x\right),\quad q\in {\mathbb {Q}},q>0$

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

Case 3: ${\displaystyle q<0}$

Set $y=-x$.

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

Combining this with the result from case 2:

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

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

Replacing $-q$ with $q$:

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

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+bf(\alpha )-bf(a)\$

$=y-\delta \beta +bf(\alpha )-bf(a)\$

$=y-\delta \beta +b(\alpha +\delta )-ba\$

$=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, showing that the only solutions are linear when the domain of $f$ is restricted to such sets. However, as shown 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 theorem proved using Zorn's lemma.

Since every vector space has a basis, there is a basis for ${\displaystyle \mathbb {R} }$ over the field $\mathbb {Q}$, i.e. a set ${\displaystyle {\mathcal {B}}\subset \mathbb {R} }$ with the property that for any ${\displaystyle x\in \mathbb {R} }$, there is a unique finite set ${\displaystyle \{x_{i}\}_{i=1}^{n}\subset {\mathcal {B}}}$ and coefficients ${\displaystyle \left(\lambda _{i}\right)_{i=1}^{n}}$ with ${\displaystyle \lambda _{i}\in \mathbb {Q} }$ such that ${\textstyle x=\sum _{i=1}^{n}{\lambda _{i}x_{i}}}$. We note that because no explicit basis for ${\displaystyle \mathbb {R} }$ over $\mathbb {Q}$ can ever be constructed, none of the pathological functions below can be defined explicitly.

By the argument above, for each ${\displaystyle x_{i}\in {\mathcal {B}}}$, the restriction of $f$ to ${\displaystyle x_{i}\mathbb {Q} }$ (a scaled copy of the rationals) must be a linear map, with some constant of proportionality ${\displaystyle g(x_{i})}$. In other words, for ${\displaystyle \lambda _{i}x_{i}\in x_{i}\mathbb {Q} }$, $f$ is the map ${\displaystyle \lambda _{i}x_{i}\mapsto g(x_{i})\lambda _{i}x_{i}}$. Since any ${\displaystyle x\in \mathbb {R} }$ can be expressed as a unique (finite) linear combination of $x_{i}$ and $f$ is additive, the function is, by necessity, extended to all of ${\displaystyle \mathbb {R} }$ as

${\displaystyle f(x)=f{\Big (}\sum _{i=1}^{n}\lambda _{i}x_{i}{\Big )}=\sum _{i=1}^{n}f(\lambda _{i}x_{i})=\sum _{i=1}^{n}{g(x_{i})\lambda _{i}x_{i}}}$.

It is easy to check that $f$ is a solution to Cauchy's functional equation for any ${\displaystyle g:{\mathcal {B}}\rightarrow \mathbb {R} }$. 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 $g$ is a constant function. In this sense, despite the inability to explicitly exhibit a nonlinear solution, "most" solutions are actually nonlinear and pathological.

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.

External links

• Solution to the Cauchy Equation Rutgers University
• The Hunt for Addi(c)tive Monster
• Martin Sleziak and otherz (2013). "Overview of basic facts about Cauchy functional equation". Math.StackExchange. Retrieved 20 December 2015.