TITLE: Uniformly continuous function acts almost like an Lipschitz function?
QUESTION [12 upvotes]: Can anyone help? 
Let $g:I \to \mathbb{R}$ be an uniformly continuous function, where $I$ is an interval. Prove that exists an constant $c$ that satisfies:
$$\lvert g(x)-g(y)\rvert < 1 + c \lvert x-y \rvert, \forall x,y \in I$$

REPLY [6 votes]: Apply the suggestion of @5pm. Take $\delta$ in the definition of uniformly continuity which corresponds to $\epsilon=1$. You can suppose $\delta<1$ (decreasing $\delta$ does not hurt). Given $x,y\in I$ let $x=x_0<x_1<\dots <x_n = y$ so that $|x_{i+1} - x_i|< \delta$. You can do this with $n = \lceil (x-y)/\delta \rceil$, so
$$
|f(x)-f(y)| \le \sum_{i=1}^n |f(x_{i})-f(x_{i-1})| \le n = \lceil (x-y)/\delta \rceil \le 1 + (x-y)/\delta.
$$
Let $c=1/\delta$ and you are done.