The closednees in Moreau - Rockafellar Theorem.

Question

One says that $x\mapsto \langle x^*,x\rangle +\alpha\;$ is an affine minorant of $f: \; X\to \overline{\mathbb R}\;$ if $\;\langle x^*,x\rangle +\alpha \leq f(x)\;$ for all $x\in X$.

The Moreau - Rockafellar Theorem stated that: If $f$ is a proper, l.s.c and convex function then $$f^{**}=f.$$ If the l.s.c assumption is violated then $f$ could not have any affine minorant and the Theorem could fail.

I would like to construct a counterexample to verify this. Anyone can help me? Thank you very much.

Answer 1

What about $f : [-1,+1] \to \Bbb R$ with $$f=0 ~~\text{on} ~(-1,1) ,~~~f(-1)=f(1) = 1$$

Answer 2

Denote by $c_0$ the space of sequences converging to $0$, i.e.

$$c_0 := \{x=(x_n)_{n \in \mathbb{N}}; \lim_{n \to \infty} x_n=0\},$$

endowed with the norm

$$\|x\| := \sup_{n \in \mathbb{N}} |x_n|,$$

and set

$$c_c := \{x=(x_n)_{n \in \mathbb{N}}; \exists k \in \mathbb{N}: x_n = 0 \, \text{for all} \, n \geq k\}.$$

Then the function $f: c_0 \to \mathbb{R}$,

$$f(x) := \begin{cases} \sum_{i=1}^n x_i & \|x\| \leq 1, x \in c_c, \\ \infty, & \text{otherwise}, \end{cases}$$

defines a convex proper function. Since

$$\inf_{\|x\| \leq 1, x \in c_c} f(x) =- \infty$$

we conclude that $f$ does not have an affine minorant. (Note that an affine function is bounded on any bounded non-empty set.)