Functional equation

A functional equation specifies a function through an identity involving values of that function at several arguments.

A continuous real function satisfying $g(x+y)=g(x)g(y)$ is either identically zero or $g(x)=e^{cx}$ for a real constant $c$. If it is nonzero somewhere, $g(0)=1$ and $g(x)g(-x)=1$, so it never vanishes. The identity $g(x)=g(x/2{)}^2$ then makes it everywhere positive. Its real logarithm is a continuous additive function, hence linear by additive function continuous at one point is linear. Exponentiating proves the classification, and each listed function satisfies the equation.

Cauchy's functional equation asks for functions $f:\mathbb{R}\to \mathbb{R}$ satisfyingThe identity must hold for every pair of real numbers, and a solution is an additive function. It says that combining two inputs before applying $f$ gives the same result as combining their outputs.^[ref]

For example, $f(x)=3x$ satisfies the identity because $3(x+y)=3x+3y$. At $x=2/3$ and $y=5/6$ the two sides giveFor positive widths, a rectangle of height three makes this visible: cutting its width into $x$ and $y$ partitions its area into $3x$ and $3y$. The same identity also holds for negative inputs, where $3x$ is a signed output rather than an ordinary area.

Every function $f(x)=cx$ with a fixed real number $c$ solves Cauchy's functional equation. The surprising question is whether these are all the solutions. Rational homogeneity of an additive function forces $f(q)=qf(1)$ for every rational number $q$, but reaching an irrational number requires an additional argument. Continuity at just one point or being a bounded function on an interval of positive length supplies that argument; without such a condition, discontinuous additive functions exist.^[ref]

This makes Cauchy's functional equation a useful test of the boundary between algebra and analysis: a rule about addition fixes all rational multiples, while a tiny amount of continuity fixes the entire real line. For example, the continuous exponential functional equation reduces to an additive function after taking a logarithm, so this result explains why its nonzero solutions are exponential functions.

An additive function on an additive semigroup satisfies $f(x+y)=f(x)+f(y)$ whenever $x$, $y$, and $x+y$ belong to its domain.

For real numbers, an additive function is a group homomorphism from $(\mathbb{R},+)$ to itself. It preserves scalar multiplication by every rational number, so it is a linear map when domain and codomain are viewed as vector spaces over the rational numbers. Being a linear map over the real numbers additionally requires $f(tx)=tf(x)$ for every real number $t$. On the real line this stronger condition is equivalent to $f(x)=xf(1)$. Additivity alone does not imply it.

For every additive function $f:\mathbb{R}\to \mathbb{R}$, every real number $x$, and every rational number $q$, one has $f(qx)=qf(x)$. This is the scalar multiplication part of being a linear map over the rational numbers.

First $f(0)=f(0)+f(0)$ gives $f(0)=0$, and $f(x)+f(-x)=f(0)$ gives $f(-x)=-f(x)$. Repeated addition gives $f(nx)=nf(x)$ for each positive integer $n$, and the negative case follows from the sign identity. If $q=m/n$ with $m\in \mathbb{Z}$ and $n>0$, thenNo continuity was used. In particular $f(1)=3$ fixes $f(-2/3)=-2$, $f(5/6)=5/2$, and every other rational value, even for a discontinuous additive function.

A real additive function $f:\mathbb{R}\to \mathbb{R}$ that is continuous at one point $x_0$ is continuous everywhere and has form $f(x)=xf(1)$. Indeed $f(0)=0$, and $f(h)=f(x_0+h)-f(x_0)\to 0$ as $h\to 0$. Then $f(x+h)-f(x)=f(h)\to 0$ for every $x$. Additivity gives $f(q)=qf(1)$ for every rational number; continuity and the density of the rational numbers extend this identity to every real argument.

The step from the rational numbers to the real numbers can be made explicit. Put $c=f(1)$. For an arbitrary real number $x$, the density of the rational numbers supplies a sequence of rational numbers $q_n\to x$. Because continuity at the original point already implied continuity everywhere,This is where continuity enters: knowing all the rational values alone does not justify the first equality.

For an original numerical example, use the floor function to define the rational lower approximants $q_n=\lfloor 10^n\sqrt{2}\rfloor /10^n$. They obey $0\le \sqrt{2}-q_n<10^{-n}$. When $f(1)=3$ and $f$ is continuous at one point, rational homogeneity of an additive function and the preceding argument giveThe error bound here follows from the resulting linear map, rather than from a quantitative bound on the initially assumed continuity.

Let $f:\mathbb{R}\to \mathbb{R}$ be an additive function. Suppose $|f(x)|\le M$ for all $x$ in a nondegenerate interval $(a,b)$, meaning $a<b$. Then $f$ is continuous everywhere and $f(x)=xf(1)$. A finite bound on even one such interval is enough; a bound at one isolated point is not.

Here is a proof outline that turns a finite bound into arbitrarily small outputs. Set $z=(a+b)/2$ and $\delta =(b-a)/2$. If $|h|<\delta$, then both $z$ and $z+h$ lie in $(a,b)$, so Cauchy's functional equation givesGiven $\epsilon >0$, choose a positive integer $N$ with $2M/N<\epsilon$. For $|h|<\delta /N$, the argument $Nh$ lies in $(-\delta ,\delta )$, and rational homogeneity of an additive function givesThis proves continuity at zero. Now additive function continuous at one point is linear completes the proof. The key mechanism is that repeated addition magnifies a tiny input without escaping the bounded region, forcing the original output to be small.

Every Lebesgue-measurable additive function $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is continuous. Small-value preimages are measurable; one of their integer dilates has positive measure, so the Steinhaus theorem puts a neighbourhood of zero in their difference set. Additivity then makes $f$ small on that neighbourhood.

If $f:[0,\infty )\to \mathbb{R}$ is continuous and is an additive function, then $f(t)=tf(1)$. Additivity proves this first for nonnegative rational numbers; because $f$ is continuous, the identity extends to every nonnegative real number.

Assuming the axiom of choice, Cauchy's functional equation has solutions that are not linear maps over the real numbers. Regard $\mathbb{R}$ as a vector space over the rational numbers. Because $\sqrt{2}$ is an irrational number, $1$ and $\sqrt{2}$ are linearly independent over $\mathbb{Q}$. Zorn's lemma extends them to a Hamel basis $B$; assigning arbitrary real outputs to its basis vectors and extending by finite rational linear combinations gives an additive function.^[ref]

For a particular choice, assign $g(1)=3$ and $g(b)=0$ for every $b\in B\setminus \{1\}$. Every real number has a unique finite linear combination $x={\sum }_{b\in F}{q}_{b}b$ with $F\subset B$ finite and $q_b\in \mathbb{Q}$. Definewhere the coefficient $q_1$ is zero if $1\notin F$. The uniqueness of the Hamel basis representation makes this definition unambiguous, and adding the coefficients proves $g(x+y)=g(x)+g(y)$. On the rational two-dimensional vector subspace spanned by $1$ and $\sqrt{2}$, the rule is particularly simple:For instance, $u=2/3+\sqrt{2}$ and $v=5/6-2\sqrt{2}$ give $g(u)=2$, $g(v)=5/2$, and $g(u+v)=9/2$. Thus the map still preserves addition, even though $g(\sqrt{2})=0\ne 3\sqrt{2}$ rules out being a linear map over the real numbers.

The rational approximants $q_n$ from additive function continuous at one point is linear expose the failure of continuity: $q_n\to \sqrt{2}$ but $g(q_n)=3q_n\to 3\sqrt{2}\ne g(\sqrt{2})$. Moreover $h_n=\sqrt{2}-q_n\to 0$ while $g(h_n)=-3q_n\to -3\sqrt{2}$. The map is continuous nowhere, since continuity at any point would imply additive function continuous at one point is linear. It is unbounded on every interval of positive length by additive function bounded on an interval is linear.

The axiom of choice is used to obtain the Hamel basis; the coefficient rule does not provide an explicit way to find the representation of an arbitrary real number. A plot sampled only at rational numbers would show exactly the same values as $f(x)=3x$, so that plot cannot reveal this discontinuous additive function.