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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.

Richkent
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  • Conjugates are automatically convex and closed. In particular $f^{\star \star}$ must be closed. So if $f$ is not closed, it's impossible that $f^{\star \star} = f$. – littleO Apr 06 '14 at 03:04
  • Yes, I know that. I want to find an example to show that if $f$ is not l.s.c then $f$ could have no affine minorant. – Richkent Apr 06 '14 at 03:12

2 Answers2

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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.)

saz
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  • Thanks for your excellent example. I wonder why we don't find such a function in finite dimensional? – Richkent Apr 07 '14 at 03:29
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    @Richkent: A convex function on a finite dimensional space is continuous in the interiour of its domain. – anonymous Apr 28 '14 at 08:16
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What about $f : [-1,+1] \to \Bbb R$ with $$f=0 ~~\text{on} ~(-1,1) ,~~~f(-1)=f(1) = 1$$

Red shoes
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