Lipschitz Condition implies Uniform Continuity

Theorem
Suppose $$(M_1, d_1)$$ and $$(M_2, d_2)$$ are metric spaces.

Furthermore, suppose that $$g : M_1 \to M_2$$ satisfies the Lipschitz condition.

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

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

Let $$K$$ be $$g$$'s Lipschitz constant.


 * First suppose that $$K \le 0$$.

Then $$d_1 \left({x, y}\right) \le 0 d_2 \left({g \left({x}\right), g \left({y}\right)}\right)$$ by the Lipschitz condition on $$g$$.

So $$d_1 \left({x, y}\right) \le 0 \implies d_1 \left({x, y}\right) = 0 \implies x = y$$ 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 $$d_1 \left({x, y}\right) < \delta$$, we have:
 * $$K d_1 \left({x, y}\right) < \epsilon$$

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

These last two statements together imply $$d_2 \left({g \left({x}\right), g \left({y}\right)}\right) < \epsilon$$.

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