Suppose $f: S \rightarrow \mathbb{R}$ and $g :[0, \infty) \rightarrow [0,\infty)$ are functions, $g$ is continuous at $0$, $g(0)=0$, and whenever x and y are in S we have $|f(x)-f(y)| \leq g(|x-y|)$. Prove $f$ is uniformly continuous.
I am basically stuck on not knowing what to do with $g(|x-y|)$ and how to use the fact that g is continuous at $0$.
Any help is appreciated thanks.
Best Answer
Let $ \epsilon >0.$ Since $g$ is continuous at $0$ and $g(0)=0$, there is $ \delta >0$ such that $g(t) < \epsilon$ for $0 \le t < \delta$.
Now let $x,y \in S$ with $|x-y| < \delta.$ Then we get $|f(x)-f(y)| \le g(|x-y|) < \epsilon.$