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

• Hint: take $\delta$ from uniform continuity (doesn't matter for which $\epsilon$) and use it repeatedly, moving from $x$ to $y$ in small steps. Count how many steps you need.
– user53153
Feb 23, 2013 at 4:28

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 $1 <1 + \frac{1}{M}|x-y| < 1 + \epsilon$

Thus

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

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)

• I don't see why $M\epsilon$ appeared instead of $\delta$ at the end of the first paragraph.
– user53153
Feb 23, 2013 at 5:07
• Correct me if I'm wrong, but if $g$ is uniformly continuous then $\delta$ just depends on $\epsilon$, so you can write $\delta = M\epsilon$ for some constant $M>0$, or is that not correct? Feb 23, 2013 at 5:11
• It is correct that $\delta$ depends only on $\epsilon$. However, writing $\delta=M\epsilon$ amounts to saying that $\delta$ depends linearly on $\epsilon$, which is not necessarily true. For example, the function $f(x)=\sqrt{x}$ is uniformly continuous on $[0,1]$. For this function the biggest $\delta$ we can take is $\delta=\epsilon^2$. Setting $\delta=M\epsilon$ will not work no matter what positive number $M$ you take.
– user53153
Feb 23, 2013 at 5:17
• Ah of course, do you think there's a simple way to change my solution to account for it? Feb 23, 2013 at 5:21
• No. It's best to start over. The first step is to give up the idea that some algebraic manipulations with $x,y,f(x),f(y),g(x),g(y)$ will yield the result. That's not going to happen. Other points, besides $x$ and $y$, must be considered. The comment I left under the question suggests what is to be done.
– user53153
Feb 23, 2013 at 5:24