# The map $\varphi : E \to \mathbb R \cup \{+\infty\}$ is proper convex l.s.c. if and only if $\varphi = \varphi^{**}$

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Let $$(E, |\cdot|)$$ be a normed space and $$\varphi : E \to \mathbb R \cup \{+\infty\}$$ proper. The convex conjugates $$\varphi^*:E^* \to \mathbb R \cup \{+\infty\}$$ and $$\varphi^{**}:E \to \mathbb R \cup \{+\infty\}$$ of $$\varphi$$ are defined as $$\varphi^* (f) := \sup_{x \in X} [f(x) -\varphi (x)] \quad \forall f \in E^*\\ \varphi^{**} (x) := \sup_{f \in X^*} [f(x) -\varphi^* (f)] \quad \forall x \in E.$$ Then $$\varphi^*, \varphi^{**}$$ are convex l.s.c.

Theorem: Let $$\varphi : E \to \mathbb R \cup \{+\infty\}$$ be proper. Then the following statements are equivalent.

• (i) $$\varphi$$ is convex l.s.c.
• (ii) $$\varphi = \psi^*$$ for some proper function $$\psi:E^* \to \mathbb R \cup \{+\infty\}$$.
• (iii) $$\varphi = \varphi^{**}$$.

Clearly, (iii) implies (ii), and (ii) implies (i). Let's prove (i) implies (iii).

• First, we assume $$\varphi \ge 0$$.

We have $$\varphi^* (f) \ge f(x) - \varphi (x)$$ and thus $$\varphi (x) \ge f(x) - \varphi^* (f)$$ for all $$f \in E^*$$. Then $$\varphi (x) \ge \varphi^{**} (x)$$. Next we prove $$\varphi (x) \le \varphi^{**} (x)$$. Assume the contrary that there is $$a \in X$$ such that $$\varphi (a) > \varphi^{**} (a)$$. Then $$\varphi^{**} (a) \in \mathbb R$$.

Because $$\varphi$$ is convex l.s.c., $$\operatorname{epi} \varphi$$ is closed convex. We apply Hahn-Banach theorem to strictly separate $$\operatorname{epi} \varphi$$ and $$\{(a, \varphi^{**} (a))\}$$. There exist $$f \in E^*$$ and $$k,\alpha \in \mathbb R$$ such that $$f(x) + k \lambda > \alpha > f(a) + k \varphi^{**} (a) \quad \forall (x, \lambda) \in \operatorname{epi} \varphi.$$

It follows that $$k \ge 0$$. Fix $$\varepsilon >0$$. Because $$\varphi \ge 0$$, we get $$f(x) + (k + \varepsilon) \varphi (x) > \alpha \quad \forall x \in \operatorname{dom} \varphi.$$

This implies $$\frac{-f(x)}{k + \varepsilon} - \varphi (x)< \frac{-\alpha}{k + \varepsilon} \quad \forall x \in \operatorname{dom} \varphi.$$

So $$\varphi^* \left ( \frac{-f}{k + \varepsilon} \right ) \le \frac{-\alpha}{k + \varepsilon}.$$

Then $$\varphi^{**} (a) \ge \frac{-f(a)}{k + \varepsilon} - \varphi^* \left ( \frac{-f}{k + \varepsilon} \right ) \ge \frac{-f(a)}{k + \varepsilon} + \frac{\alpha}{k + \varepsilon} = \frac{-f(a) +\alpha}{k + \varepsilon}.$$

Taking the limit $$\varepsilon \to 0^+$$, we get $$\varphi^{**} (a) \ge \frac{-f(a) +\alpha}{k}$$, which is a contradiction.

• Now we consider the general case.

Lemma: Every proper l.s.c. convex function $$f:E \to \mathbb R \cup \{+\infty\}$$ on a normed space $$X$$ is bounded below by an affine continuous function.

By our Lemma, we get $$\operatorname{dom} \varphi^* \neq \emptyset$$. Fix $$f_0 \in \operatorname{dom} \varphi^*$$. We define $$\overline \varphi: E \to \mathbb R \cup \{+\infty\}$$ by $$\overline \varphi (x) := \varphi (x) - f_0 (x) + \varphi^* (f_0).$$

Then $$\overline \varphi$$ is proper l.s.c. convex and $$\overline \varphi \ge 0$$. By above result, we get $$\overline \varphi = \overline \varphi^{**}.$$

We have \begin{align} \overline \varphi^{**} (x) &= \sup_{f \in E^*} [f(x) - \overline{\varphi}^* (f)] \\ &= \sup_{f \in E^*} [f(x) - [\sup_{y \in E} [f(y) - \overline{\varphi} (y)]]] \\ &= \sup_{f \in E^*} [f(x) - [\sup_{y \in E} [f(y) - [\varphi (y) - f_0 (y) + \varphi^* (f_0)]]]] \\ &= \varphi^* (f_0) +\sup_{f \in E^*} [f(x) - [\sup_{y \in E} [(f+f_0)(y) - \varphi (y)]]] \\ &= \varphi^* (f_0) -f_0 (x)+\sup_{f \in E^*} [(f+f_0)(x) - [\sup_{y \in E} [(f+f_0)(y) - \varphi (y)]]] \\ &= \varphi^* (f_0) -f_0 (x)+\sup_{f \in E^*} [f(x) - [\sup_{y \in E} [f(y) - \varphi (y)]]] \\ &= \varphi^* (f_0) -f_0 (x)+\sup_{f \in E^*} [f(x) - \varphi^*(y)] \\ &= \varphi^* (f_0) -f_0 (x)+\varphi^{**} (x). \end{align}

It follows that $$\varphi(x) = \varphi^{**} (x)$$. This completes the proof.