# Exercise 3.4.16 Introduction to Real Analysis by Jiri Lebl

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.

New contributor
popping900 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• What does $S$ represent ? – Suzet 2 days ago
• S is just a subset of the reals – popping900 2 days ago

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 $$|f(x)-f(y)| \le g(|x-y|) < \epsilon.$$