= Additive function bounded on an interval is linear
{title2=$\sup_{a<x<b}|f(x)|<\infty\Longrightarrow f(x)=xf(1)$}
Let $f:\mathbb R\to\mathbb R$ be an <additive function>. Suppose $|f(x)|\leq M$ 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> gives
$$
|f(h)|=|f(z+h)-f(z)|\leq2M.
$$
Given $\varepsilon>0$, choose a positive <integer> $N$ with $2M/N<\varepsilon$. For $|h|<\delta/N$, the argument $Nh$ lies in $(-\delta,\delta)$, and <rational homogeneity of an additive function> gives
$$
|f(h)|=\frac{|f(Nh)|}{N}\leq\frac{2M}{N}<\varepsilon.
$$
This 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.
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