I'm trying to prove this well-known approximation of proper l.s.c. convex function. Could you have a check on my attempt?
Let $(X, | \cdot|)$ be a normed space and $f:X \to \mathbb R \cup \{+\infty\}$ convex proper lower semi-continuous. Then there is an increasing sequence of Lipchitz-continuous convex functions $(f_n)$ with $f_n:X \to \mathbb R$ such that $f_n \nearrow f$ pointwise.
My attempt: We need the following lemma.
Lemma: Any convex proper lower semi-continuous function is bounded from below by a continuous affine function.
We define $f_n:X \to \mathbb R \cup \{\pm\infty\}$ by $$ f_n (x) := \inf_{y\in X} \{ f(y) + n|x-y| \} \quad \forall x\in X. $$
Clearly, $f_n$ is convex and $f_n \le f_{n+1} \le f$. We have $$ \begin{align} f_n (x) &= \inf_{y\in X}\{f(y)+n |x-y|\} \\ &\le \inf_{y\in X}\{f(y)+n |x-z| + n |y-z|\} \\ &= f_n(z)+n |x-z|. \end{align} $$ By symmetry, we get $|f_n(x)-f_n(z)|\ \le n |x-z|$. Thus $f_n$ is $n$-Lipschitz continuous.
Because $f$ is proper, there is $a \in X$ such that $f(a) < +\infty$. Then $f_n (x) \le f(a) + n|a-x| <+\infty$ for all $x\in X$. By our Lemma, there is a linear continuous functional $g$ and $\alpha \in \mathbb R$ such that $g +\alpha \le f$. Then $$ \begin{align} f(y) + n|x-y| &\ge g(y)+n|x-y| +\alpha \\ &\ge -\|g\||y| + n|x-y|+\alpha \\ &\ge (n-\|g\|) |y| -n |x| +\alpha. \end{align} $$
If $n \ge \|g\|$, then $f_n (x) \ge-n|x| +\alpha > -\infty$ for all $x\in X$. Clearly, $(f_n)$ is increasing. WLOG, we re-index $(f_n)$ such that $f_n(x) \in \mathbb R$ for all $n \in \mathbb N$ and $x \in X$.
Let's prove that $f_n (x) \to f(x)$.
- Fix $x \in \operatorname{dom} f$. We pick $x_n \in X$ such that $$ f(x_n)+n|x-x_n|<f_n(x)+\frac{1}{n}. $$
First, $f_n(x) \le f(x)$. Second, $f(x_n) \ge g(x_n)+\alpha \ge -\|g\||x_n|+\alpha$, so $$ n|x-x_n| < f(x)+1/n+\|g\||x_n|-\alpha. $$ This implies $|x_n-x| \to 0$. Hence $$ \lim_n f_n(x) \ge \lim_n f(x_n) \ge f(x). $$
- Fix $x \not\in \operatorname{dom} f$. Then $f(x) = +\infty$. We claim that $\lim_n f_n(x) = +\infty$. Assume the contrary that $(f_n(x))$ is bounded by $C <+\infty$. We pick $x_n \in X$ such that $$ f(x_n)+n|x-x_n|<f_n(x)+\frac{1}{n}. $$
Then $$ f(x_n)+n|x-x_n| \le C+1/n. $$
Just as above, we get $|x-x_n| \to 0$. Then $f(x) \le \liminf_n f(x_n) \le C$, which is a contradiction. This completes the proof.
