 Additive function bounded on an interval is linear

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.