Uniformly continuous function acts almost like an Lipschitz function?

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:
$$\lvert g(x)-g(y)\rvert < 1 + c \lvert x-y \rvert, \forall x,y \in I$$

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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.

From the definition of uniform continuity you have that $|x-y|<\delta$ implies $|g(x) – g(y)| < \epsilon$ and specifically $\delta$ can only depend on $\epsilon$ and must be independent of $x$ and $y$. So you would have that $|x-y|< M\epsilon$ where $M \in \mathbb{R}$ for all $x,y \in I$ implies $|g(x) – g(y)| < \epsilon$.

From here I’m not entirely sure where to go, I’m wondering if you can assume $\epsilon <1$ so you get $1 + \frac{1}{M}|x-y| < 1 + \epsilon$ but since, from the definition of continuity $\epsilon > 0$ you get $1 <1 + \frac{1}{M}|x-y| < 1 + \epsilon$


$|g(x) – g(y)| < 1 + \frac{1}{M}|x-y|$

Where $\frac{1}{M}$ is your constant.

There’s far more qualified people on here though (I’m just nervous about posting it, in case it takes you in a wrong direction)