Lipschitz functions are uniformly continuous

Definition reminder. $f$ is uniformly continuous if for every $\varepsilon>0$ there is $\delta>0$ such that for all $x,y$ with $|x-y|<\delta$, $|f(x)-f(y)|<\varepsilon$. The difference with ordinary continuity is that $\delta$ must work for all points at once.

Proof. Assume $K>0$ (the case $K=0$ is trivial since $f$ is constant). Let $\varepsilon>0$ and set $\delta=\frac\varepsilon K$. If $|x-y|<\delta$, then
$$|f(x)-f(y)|\le K|x-y|<K\delta=\varepsilon.$$

Examples.
- $x\mapsto\sin x$ is $1$-Lipschitz on $\mathbb{R}$, by the mean value theorem since $|\cos|\le1$.
- $x\mapsto|x|$ is $1$-Lipschitz by the reverse triangle inequality, although it is not differentiable at $0$.

Remark. On an interval, a differentiable function with bounded derivative $|f'|\le K$ is $K$-Lipschitz (mean value theorem). That is the most common way to prove uniform continuity in practice.