The space of bounded Lipschitz continuous functions is dense in that of bounded uniformly continuous functions

Question

In proving that condition (2) implies condition (3), i.e.,

Let $S$ be a metric space. A sequence of Borel probability measures $P_1, P_2, \ldots$ on $S$ is said to converge weakly to a Borel probability measure $P$ (denoted $P_{n} \Rightarrow P$) if any of the following equivalent conditions is true (here $\mathrm{E}_{n}$ denotes expectation w.r.t. $P_{n}$, while $\mathrm{E}$ denotes expectation w.r.t. $P$):

• $\mathrm{E}_{n}[f] \rightarrow \mathrm{E}[f]$ for all bounded, continuous functions $f$;
• $\mathrm{E}_{n}[f] \rightarrow \mathrm{E}[f]$ for all bounded, uniformly continuous functions $f$;
• $\mathrm{E}_{n}[f] \rightarrow \mathrm{E}[f]$ for all bounded, Lipschitz continuous functions $f$;
• $\lim \sup \mathrm{E}_{n}[f] \leq \mathrm{E}[f]$ for every upper semi-continuous function $f$ bounded from above;
• $\lim \inf _{n}[f] \geq \mathrm{E}[f]$ for every lower semi-continuous function $f$ bounded from below;
• $\lim \sup P_{n}(C) \leq P(C)$ for all closed sets $C$ of space $S$;
• $\liminf P_{n}(U) \geq P(U)$ for all open sets $U$ of space $S$;
• $\lim P_{n}(A)=P(A)$ for all continuity sets $A$ of measure $P$.

I come up with below result

Let $(X,d)$ be a metric space. Then the space of bounded Lipschitz continuous functions is dense in the space of bounded uniformly continuous functions.

I have found a proof here but my proof seems much more simpler. Could you confirm if I made some subtle mistakes?

I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.

Answer 1

Let $f$: $X$ $\rightarrow$ $\mathbb{R}$ be uniformly continuous and bounded. Let $$ f_n(x) := \inf_{y\in X}\{f(y)+n\text{d}(x,y)\}. $$

Let $\alpha >0$ such that $|f(x)| \le \alpha$ for all $x\in X$, so $|f(x)-f(y)| \le 2 \alpha$ for all $x,y\in X$.

It's clear that $-\alpha \le f_n \le f$ is bounded and $n$-Lipschitz continuous. Let's prove that $f_n \to f$ uniformly. Fix $\varepsilon>0$, there is $\delta>0$ such that $d(x,y)<\delta$ implies $|f(x)-f(y)|<\varepsilon$. Then for all $x\in X$, we have

$$ \begin{align} 0\le f(x) - f_n(x) &= f(x)- \inf \{ f(y)+n\text{d}(x,y) \mid y\in X \} \\ &= \sup \{ f(x)-f(y)-n\text{d}(x,y) \mid y\in X \} \\ &= \sup \{ f(x)-f(y)-n\text{d}(x,y) \mid y\in X \text{ s.t. } d(x,y) \le 2\alpha/n \}. \end{align} $$

If $n$ such that $2\alpha/n <\delta$ or equivalently $n>2\alpha/\delta$. Then $$ f(x) - f_n(x) \le \sup \{ \varepsilon-n\text{d}(x,y) \mid y\in X \text{ s.t. } d(x,y) \le 2\alpha/n \} \le \varepsilon \quad \forall x\in X. $$

It follows that $\|f-f_n\|_\infty \le \varepsilon$.