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

• Apr 9th 2011, 09:35 AM
alice8675309
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 $\in$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 $\epsilon$>0. Choose $\delta$= $\epsilon$/L. Then by the definition of uniform continuity, if |x-y|< $\delta$= $\epsilon$/L, then |f(x)-f(y)|<L $\epsilon$/L= $\epsilon$.
• Apr 9th 2011, 10:23 AM
hatsoff
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 $f:D\to R$ satisfy a Lipschitz condition with constant $L$, and let $\epsilon>0$. Then there is $\delta:=\epsilon/L>0$ such that for any $x,y\in D$, if $|x-y|<\delta$ then $|f(x)-f(y)|. Hence $f$ is uniformly continuous.