# Lipschitz Condition implies Uniform Continuity

## Theorem

Let $\left({M_1, d_1}\right)$ and $\left({M_2, d_2}\right)$ 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:

$d_1 \left({x, y}\right) \le 0 d_2 \left({g \left({x}\right), g \left({y}\right)}\right)$ 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 $M_1$.

$\blacksquare$