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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.

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.

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.

Irrationality of $\sqrt{2}$ makes the expression $q+r\sqrt{2}$ unique for rational $q,r$: equality of two such expressions with different $r$ would express $\sqrt{2}$ as a rational number. Therefore the rule selecting the coefficient $r$ defines a function on $A$. Its domain is a vector subspace over $\mathbb{Q}$, and direct coefficient addition and rational scaling show it is $\mathbb{Q}$-linear, hence weakly linear. In particularUse Zorn's lemma on the nonempty family of weakly linear extensions of this particular function. Chain unions remain extensions by (a); hence there is a maximal extension $h$. Any further extension in $\mathcal{F}$ would also contain the original function, so this $h$ is maximal in $\mathcal{F}$. Part (b) therefore gives $\mathrm{dom}h=\mathbb{R}$.

A real-linear map $\mathbb{R}\to \mathbb{R}$ is multiplication by its value at $1$. Were $h$ real-linear, it would satisfycontradicting the inherited value $h(\sqrt{2})=1$. Thus a weakly linear function on all of $\mathbb{R}$ need not be real-linear. Its rational homogeneity of an additive function is insufficient for real homogeneity. This particular extension is also discontinuous: rational numbers approaching $\sqrt{2}$ have value zero, while the value at their limit is one.

A rational-linear map is a linear map when its domain and codomain are regarded as vector spaces over the rational numbers. An additive map on a rational vector subspace of the real numbers is automatically rational-linear: integer homogeneity follows by repeated addition, and division by a positive integer gives rational homogeneity of an additive function.

A weakly linear partial function on the real numbers has a domain closed under rational scalar multiplication and addition, and it preserves addition on that domain. A nonempty domain is a vector space over the rational numbers, and additivity implies rational homogeneity of an additive function by integer multiplication and division. The empty function also satisfies the defining conditions vacuously. This terminology does not assert homogeneity under arbitrary real scalars.