Lipschitz Condition implies Uniform Continuity

Theorem
Let $\struct {M_1, d_1}$ and $\struct {M_2, d_2}$ be metric spaces.

Let $g: M_1 \to M_2$ satisfy the Lipschitz condition.

Then $g$ is uniformly continuous on $M_1$.

Proof
Let $\epsilon > 0$, $x,y \in M_1$.

Let $K$ be a Lipschitz constant for $g$.

First, suppose that $K \le 0$.

Then:
 * $\map {d_1} {x, y} \le 0 \map {d_2} {\map g x, \map g y}$ by the Lipschitz condition on $g$.

So $\map {d_1} {x, y} \le 0 \implies \map {d_1} {x, y} = 0 \implies x = y$ for all $x$ and $y$ in $M_1$.

Thus $g$ is a constant function, which is uniformly continuous.

Next, suppose that $K > 0$.

Take $\delta = \epsilon / K$.

Then if $\map {d_1} {x, y} < \delta$, we have:
 * $K \map {d_1} {x, y} < \epsilon$

By the Lipschitz condition on $g$, we know that:
 * $\map {d_2} {\map g x, \map g y} \le K \map {d_1} {x, y}$

These last two statements together imply $\map {d_2} {\map g x, \map g y} < \epsilon$.

Thus, $g$ is uniformly continuous on $M_1$.