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I'm trying to prove this result. Could you verify if my attempt is fine?

Let $(X, d)$ be a metric space. Let $f:X \to \mathbb R \cup \{+\infty\}$ be a lower semi-continuous and bounded from below. Then there is an increasing sequence $(f_n)$ of $n$-Lipschitz continuous functions such that $f_n \nearrow f$ pointwise.

I post my proof separately as below answer. If other people post an answer, of course I will happily accept theirs. Otherwise, this allows me to subsequently remove this question from unanswered list.

Analyst
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1 Answers1

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WLOG, we assume $f$ is bounded from below by $0$. Let $$ f_n (x) := \inf_{y\in X}\{f(y)+n d(x,y)\} \quad \forall n \ge 1, x\in X. $$

Clearly, $0 \le f_n(x) \le f_{n+1} (x) \le f(x)$ for all $x\in X$. We have \begin{align} f_n (x) &= \inf_{y\in X}\{f(y)+n d(x,y)\} \\ &\le \inf_{y\in X}\{f(y)+n d(x,z) + n d(z,y)\} \\ &= f_n(z)+nd(x,z). \end{align}

By symmetry, we get $|f_n(x)-f_n(z)|\ \le n d(x,z)$. Thus $f_n$ is $n$-Lipschitz continuous. We have \begin{align} 0\le f(x) - f_n(x) &= f(x)- \inf \{ f(y)+nd(x,y) \mid y\in X \} \\ &= \sup \{ f(x)-f(y)-nd(x,y) \mid y\in X \} \\ &= \sup \{ f(x)- f(y)-nd(x,y) \mid y\in X \text{ s.t. } d(x,y) \le f(x)/n \}. \end{align}

Because $f$ is l.s.c., $f(x) \le \liminf_{y \to x} f(y) := \sup_{V \in \mathcal N(x)} \inf _{y \in V}$ with $\mathcal N(x)$ the collection of all neighborhoods of $x$. In our particular setting of metric space, $$ f(x) \le \lim_{n \to \infty} \inf \left \{ f(y) \,\middle\vert\, y\in B \left (x, \frac{f(x)+1}{n} \right) \right \}. $$

Fix $\varepsilon>0$. There is $N$ such that $$ \forall n \ge N, \forall y \in B(x, f(x)/n): f(y) \ge f(x)-\varepsilon. $$

It follows that $$ 0\le f(x) - f_n(x) \le \sup \{\varepsilon-nd(x,y) \mid y \in X\} \le \varepsilon \quad \forall n \ge N. $$

Analyst
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