# Functional Analysis – Density of Bounded Lipschitz Continuous Functions

functional-analysislipschitz-functionsmetric-spaces

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.

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 $$\|f-f_n\|_\infty \le \varepsilon$$.