Family of Lipschitz Continuous Functions with same Lipschitz Constant is Uniformly Equicontinuous
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Theorem
Let $\struct {X, d}$ and $\struct {Y, d'}$ be metric spaces.
Let $\map \CC {X, Y}$ be the set of continuous functions $X \to Y$.
Let $\FF \subset \map \CC {X, Y}$ be a set of Lipschitz continuous functions all with Lipschitz constant $M \ge 0$.
Then $\FF$ is uniformly equicontinuous.
Proof
Note that for each $f \in \FF$, 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 $f \in \FF$ and $x, y \in X$.
That is:
- $\map {d'} {\map f x, \map f y} < \epsilon$
for each $\epsilon > 0$, $f \in \FF$ and $x, y \in X$.
So $\FF$ is uniformly equicontinuous in this case.
Suppose now that $M > 0$.
Note that whenever:
- $\ds \map d {x, y} < \frac \epsilon M$
we have:
\(\ds \map {d'} {\map f x, \map f y}\) | \(\le\) | \(\ds M \map d {x, y}\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds M \times \frac \epsilon M\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \epsilon\) |
for each $f \in \FF$.
So $\FF$ is uniformly equicontinuous.
$\blacksquare$