The map $\varphi : E \to \mathbb R \cup \{+\infty\}$ is proper convex l.s.c. if and only if $\varphi = \varphi^{**}$ This thread is meant to record a question that I feel interesting during my self-study. I'm very happy to receive your suggestion and comments. See: SE blog: Answer own Question and MSE meta: Answer own Question.

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) $\varphi = \varphi^{**}$.


 A: 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.
