A function $f:D\rightarrow \mathbb{R}$ is said to be a Lipschitz function provided that there is a nonnegative number $C$ such that
$|f(u)-f(v)|\le C|u-v|$ for all $u,v\in D$.
Show that a Lipschitz function satisfies the $\epsilon$-$\delta$ criterion on $D$.
In order for this to be true, for $\epsilon>0$, we must find a $\delta>0$ such that
$|f(x)-f(x_0)|<\epsilon$ $\hspace{10pt}$ if $\hspace{10pt}$ $|x-x_0|<\delta$.
If we let $u=x$ and $v=x_0$, then we have
$|f(x)-f(x_0)|\le C|x-x_0|$ for all $x\in D$, but I'm not sure where to go after this. Any suggestions?
