Family of Lipschitz Continuous Functions with same Lipschitz Constant is Uniformly Equicontinuous

Theorem
Let $\struct {X, d}$ and $\struct {Y, d'}$ be metric spaces.

Let $\map {\mathcal C} {X, Y}$ be the set of continuous functions $X \to Y$.

Let $\mathcal F \subset \map {\mathcal C} {X, Y}$ be a set of Lipschiz continuous functions all with Lipschitz constant $M \ge 0$.

Then $\mathcal F$ is uniformly equicontinuous.

Proof
Note that for each $f \in \mathcal F$, we have:


 * $\map {d'} {\map f x, \map f y} \le M \map d {x, y}$

for each $x, y \in X$.

Note that if $M = 0$, we have:


 * $\map {d'} {\map f x, \map f y} = 0$

for each $f \in \mathcal F$ and $x, y \in X$.

That is:


 * $\map {d'} {\map f x, \map f y} < \epsilon$

for each $\epsilon > 0$, $f \in \mathcal F$ and $x, y \in X$.

So $\mathcal F$ is uniformly equicontinuous in this case.

Suppose now that $M > 0$.

Note that whenever:


 * $\ds \map d {x, y} < \frac \epsilon M$

we have:

for each $f \in \mathcal F$.

So $\mathcal F$ is uniformly equicontinuous.