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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 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.

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.

For the additive function, setting both arguments to zero gives $f(0)=0$, and then $f(-x)=-f(x)$. Continuity at the given point $x_0$ impliesFor any real $x$, $f(x+h)-f(x)=f(h)$, so $f$ is continuous at every point. This proves the continuity part of additive function continuous at one point is linear.

Let $c=f(1)$. Repeated addition and negation give $f(m)=mc$ for every integer $m$. For integers $p$ and positive $q$,so $f(r)=cr$ for every rational number $r$. Given any real $x$, choose rational $r_n\to x$. Continuity and density of the rational numbers giveConversely each real multiple of $x$ satisfies the Cauchy functional equation.

For the multiplicative equation, assume first that $g$ is not identically zero and choose $x_1$ with $g(x_1)\ne 0$. Then $g(x_1)=g(x_1)g(0)$ forces $g(0)=1$. Nextmakes every value nonzero, andmakes every value positive. Thus $h(x)=\log g(x)$ is a continuous real additive function, since $h(x+y)=h(x)+h(y)$. The proved classification gives $h(x)=kx$ for a real constant $k$. This solves the continuous exponential functional equation:Both types satisfy the equation and continuity, and every nonzero solution is everywhere strictly positive.