Sequence of 1-Lipschitz functions pointwise converge

Question

I'm asked to prove the following:

Let $(X, d_x)$ and $(Y, d_y)$ be two metric spaces. Let $D \subset X$ be dense. Show the following: If $f_1, f_2, \ldots$ is a sequence of $1$-Lipschitz functions so that there is another $1$-Lipschitz function, g, where $\lim f_i(d) = g(d)$ for all $d \in D$ then $f_1, f_2, ...$ converge pointwise to $g$.

This is my first time encountering Lipschitz functions, but I do know of the Arzela-Ascoli Theorem, which seems relevant. My version of the theorem is as follows:

For a sequence $f_1$, $f_2$, ... of continuous functions X to Y, if:

1. $X$ is $\sigma$-compact
2. $Y$ is complete
3. The sequence of functions is point-wise equicontinuous
4. The closure of the sequence of functions is totally bounded

Then there exists $n_1,n_2, ...$ such that $f_{n_1}, f_{n_1}, ...$ converge pointwise to a function g.

From Wikipedia, I know that 1-Lipschitz functions are uniformly continuous, so the sequence of functions as a whole is equicontinuous, so condition (3) of my theorem is met. I think (1) is met because we are told that $X$ has a dense subset D, but am unsure of that. And I don't know how to prove (2) and (4) so that I can use my theorem. Any help appreciated!

Answer 1

To establish pointwise convergence, we need to show that for every $x \in X$ and for any $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for all $i \geq N$, $d_Y(f_i(x), g(x)) < \epsilon$.

Since $D$ is dense in $X$, for any $x \in X$ and for any $\epsilon > 0$, there exists a sequence $\{d_n\}_{n \in \mathbb{N}} \subset D$ converging to $x$ in the sense that $\lim_{n \to \infty} d_X(d_n, x) = 0$.

Given that $g$ is $1$-Lipschitz, for any $d_n \in D$ and $x \in X$, we have $$ d_Y(g(d_n), g(x)) \leq d_X(d_n, x). $$ As $n \to \infty$, this implies $g(d_n)$ converges to $g(x)$, i.e., $\lim_{n \to \infty} g(d_n) = g(x)$.

For each $f_i$, also being $1$-Lipschitz, we analogously have $$ d_Y(f_i(d_n), f_i(x)) \leq d_X(d_n, x). $$ Thus, as $n \to \infty$, $f_i(d_n)$ converges to $f_i(x)$ for each $i$.

Consider $\epsilon > 0$. Given $\lim_{i \to \infty} f_i(d) = g(d)$ for all $d \in D$, there exists $N \in \mathbb{N}$ such that for all $i \geq N$ and for sufficiently large $n$,

$$|f_i(d_n) - g(d_n)| < \frac{\epsilon}{3}.$$

By the continuity of $g$, there exists $M \in \mathbb{N}$ such that for all $n \geq M$, $|g(d_n) - g(x)| < \frac{\epsilon}{3}$.

Hence, for $i \geq N$ and $n$ sufficiently large, we can apply the triangle inequality as follows: $$ d_Y(f_i(x), g(x)) \leq d_Y(f_i(x), f_i(d_n)) + d_Y(f_i(d_n), g(d_n)) + d_Y(g(d_n), g(x)). $$ This ensures that for $i \geq N$, $d_Y(f_i(x), g(x)) < \epsilon$, demonstrating the pointwise convergence of the sequence $\{f_i\}_{i \in \mathbb{N}}$ to $g$ on $X$.