Thread: proof involving lipschitz and uniformly continuity. is it correct?

1. proof involving lipschitz and uniformly continuity. is it correct?

Let f: D-R satisfy the Lipschitz condition: there exists an L such that for all x,y$\displaystyle \in$D |f(x)-f(y)|<=L|x-y|

Prove that f is uniformly continuous on D

Here is what I have, is this correct?

Let $\displaystyle \epsilon$>0. Choose $\displaystyle \delta$=$\displaystyle \epsilon$/L. Then by the definition of uniform continuity, if |x-y|<$\displaystyle \delta$=$\displaystyle \epsilon$/L, then |f(x)-f(y)|<L$\displaystyle \epsilon$/L=$\displaystyle \epsilon$.

2. Your logic is correct but your writing is misleading. When you say "by the definition of uniform continuity" you are signalling that you're about to use uniform continuity as a hypothesis. But instead you use Lipschitz continuity---which is what you should do!

A proper write-up would look something like this...

Let $\displaystyle f\to R$ satisfy a Lipschitz condition with constant $\displaystyle L$, and let $\displaystyle \epsilon>0$. Then there is $\displaystyle \delta:=\epsilon/L>0$ such that for any $\displaystyle x,y\in D$, if $\displaystyle |x-y|<\delta$ then $\displaystyle |f(x)-f(y)|<L|x-y|=L\epsilon/L=\epsilon$. Hence $\displaystyle f$ is uniformly continuous.