The Fenchel conjugate of a proper closed convex function is proper closed convex

Choose $x_0\in \mathrm{dom}f$ by [F3]. The affine minorant [F2] gives $⟨a,x⟩-f(x)\le -b$ for every $x$ with finite $f(x)$, so $f^{\ast }(a)\le -b<+\mathrm{\infty }$. The term at $x_0$ gives $f^{\ast }(a)\ge ⟨a,x_0⟩-f(x_0)>-\mathrm{\infty }$. Thus $f^{\ast }$ is proper.

If $n=0$, [F4] gives the singleton ${\mathbb{R}}^0=\{0\}$ and properness makes $f(0)$ finite, so $f^{\ast }(0)=-f(0)$ is proper, closed and convex directly. Hence assume $n\ge 1$. For each $x\in \mathrm{dom}f$ and $s,s^{\mathrm{\prime }}\in {\mathbb{R}}^n$, [F5] gives $\mathrm{\mid }⟨s,x⟩-⟨s^{\mathrm{\prime }},x⟩\mathrm{\mid }\le \parallel s-s^{\mathrm{\prime }}{\parallel }_2\parallel x{\parallel }_2=d_2(s,s^{\mathrm{\prime }})\parallel x{\parallel }_2$. Thus $s\mapsto ⟨s,x⟩-f(x)$ is finite continuous affine. A pointwise supremum of these affine functions is convex: apply each affine function to $(1-\theta )s_0+\theta s_1$, bound by $(1-\theta )f^{\ast }(s_0)+\theta f^{\ast }(s_1)$, then take the supremum.

The epigraph of $f^{\ast }$ is the intersection, over $x\in \mathrm{dom}f$, of the closed half-spaces $\{(s,t):t\ge ⟨s,x⟩-f(x)\}\mathrm{.}$ Each half-space is closed because its defining affine function is continuous by step 1.2, and any intersection of closed sets is closed. Hence $f^{\ast }$ is closed in the epigraph convention of [F3].