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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.

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  • $\begingroup$ What does $S$ represent ? $\endgroup$ – Suzet 2 days ago
  • $\begingroup$ S is just a subset of the reals $\endgroup$ – popping900 2 days ago
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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 $|f(x)-f(y)| \le g(|x-y|) < \epsilon.$

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