 Solution

For an extended-real function define its Fenchel conjugate by $f^{\ast }(p)={\sup }_x(⟨p,x⟩-f(x))$ and its biconjugate by $f^{\ast \ast }(x)={\sup }_p(⟨p,x⟩-f^{\ast }(p))$. The Fenchel-Moreau theorem states, in the standard proper-envelope setting,Here the right side is the largest lower semicontinuous convex function below $f$, equivalently the function whose epigraph is the closed convex hull of $\mathrm{epi}f$. It is enough to assume $f$ is proper and has an affine minorant, ensuring this envelope is proper. In particular, for a proper convex function that is lower semicontinuous, $\boxed{f^{\ast \ast }=f}$.

First, $⟨p,x⟩-f^{\ast }(p)\le f(x)$ by the definition of the convex conjugate. The biconjugate is a supremum of continuous affine functions, so it is convex, lower semicontinuous and no greater than $f$. Second, the best intercept for an affine minorant with slope $p$ is $-f^{\ast }(p)$: $⟨p,x⟩+a\le f(x)$ for all $x$ precisely when $a\le -f^{\ast }(p)$. Thus $f^{\ast \ast }$ is the supremum of all affine minorants.

To prove that no part of the closed convex envelope is missed, set $E=\mathrm{cl}\mathrm{conv}(\mathrm{epi}f)$. The half-space representation of a closed convex set from part (a), applied in ${\mathbb{R}}^{n+1}$, separates any $(x_0,t_0)\notin E$ from $E$ by an inequality $a\cdot x+bt\le d$. Because $E$ is upward closed, $b\le 0$. If $b<0$, division by $-b$ gives an affine minorant $\ell$ with $\ell (x_0)>t_0$.

A vertical separator has $b=0$. Let ${\ell }_0$ be an existing affine minorant. Combine $a\cdot x\le d$ with ${\ell }_0(x)\le t$ to obtainSince $a\cdot x_0-d>0$, a sufficiently small positive $\epsilon$ makes this affine function exceed $t_0$. Thus vertical half-spaces can be approximated by nonvertical epigraph supports. Every point below the envelope is excluded by an affine minorant, so the supremum of these minorants is exactly the envelope. This is the decisive use of part (a).

Properness and the minorant convention matter for unrestricted extended-real functions. For example, $f(x)=-x^2$ on $\mathbb{R}$ has no affine minorant; $f^{\ast }\equiv +\infty$ and $f^{\ast \ast }\equiv -\infty$. With the corresponding improper-envelope convention its closed convex envelope is also $-\infty$. The theorem should not silently describe such an envelope as proper. The identically $+\infty$ function is another degenerate case, handled separately by extended-real conventions.