A coercive Fenchel tilt attains its minimum and conjugate supremum

Since $f$ is proper, choose $x_0\in \mathrm{dom}f$ and put $c=g(x_0)\in \mathbb{R}$. By coercivity, first choose a radius $R_0>0$ such that $g(x)>c+1$ whenever $\mathrm{\parallel }x{\mathrm{\parallel }}_2\ge R_0$, then enlarge it to $R>\max \{R_0,\mathrm{\parallel }{x}_0{\mathrm{\parallel }}_2\}$; the same implication holds for this $R$. Let $K=\{x:\mathrm{\parallel }x{\mathrm{\parallel }}_2\le R\}$; it is nonempty, closed, bounded, and therefore compact by [L3].

The epigraph of $g$ is $T^{-1}(\mathrm{epi}f)$, because $g(x)\le r$ exactly when $f(x)\le r+⟨p,x⟩$. By [L1], $\mathrm{epi}f$ is closed. Fix $z\mathrm{\notin }{T}^{-1}(\mathrm{epi}f)$; openness of the complement of $\mathrm{epi}f$ gives $\epsilon >0$ such that the Euclidean ball of radius $\epsilon$ around $Tz$ misses $\mathrm{epi}f$. Set $\rho =\epsilon \mathrm{/}(K+1)>0$. If $\mathrm{\parallel }{z}^{\mathrm{\prime }}-z{\mathrm{\parallel }}_2<\rho$, [L2] gives $\mathrm{\parallel }T{z}^{\mathrm{\prime }}-Tz{\mathrm{\parallel }}_2\le K\mathrm{\parallel }{z}^{\mathrm{\prime }}-z{\mathrm{\parallel }}_2<\epsilon$, so $T{z}^{\mathrm{\prime }}$ also lies outside $\mathrm{epi}f$. Thus the complement of $T^{-1}(\mathrm{epi}f)$ is open and $\mathrm{epi}g$ is closed. By [L1], $g$ is lower semicontinuous.

For each real $r>0$, $U_r=\{x\in K:g(x)>-r\}$ is relatively open in $K$, since its complement is the weak sublevel $K\cap \{g\le -r\}$. These sets cover $K$ because $g(x)\in \mathbb{R}\cup \{+\mathrm{\infty }\}$ for every $x$ (properness excludes $-\mathrm{\infty }$ and the tilt is finite). Compactness gives a finite subcover, and [L6] supplies a largest real index $r_{\ast }$ among its finitely many indices. Since the $U_r$ are nested, $U_{r_{\ast }}=K$, so every finite value of $g$ on $K$ is greater than $-r_{\ast }$.

Let $S:=\{g(x):x\in K,g(x)<+\mathrm{\infty }\}$. It is nonempty because $x_0\in K$ and $g(x_0)=c$, and it is a real set bounded below by $-r_{\ast }$ by step 2.1. It therefore has a real infimum $m$. For every $\epsilon >0$, [L5] gives $s\in S$ with $s<m+\epsilon$; choosing $x\in K$ with $g(x)=s$ shows $F_{\epsilon }:=K\cap \{x:g(x)\le m+\epsilon \}$ is nonempty. It is closed in $K$ by [L1]. Any nonempty finite subfamily has nonempty intersection, because the member with least positive $\epsilon$ is contained in all the others; the empty subfamily has intersection $K\mathrm{\ne }\mathrm{\varnothing }$.

Applying [L4] to the family $(F_{\epsilon }{)}_{\epsilon >0}$ gives $x_{\ast }\in {\bigcap }_{\epsilon >0}{F}_{\epsilon }$. The bound $g(x_{\ast })\le m+1$ makes $g(x_{\ast })$ finite, so $g(x_{\ast })\in S$ and $m\le g(x_{\ast })$. If $g(x_{\ast })>m$, choose $\epsilon =(g(x_{\ast })-m)\mathrm{/}2>0$; then $x_{\ast }\in F_{\epsilon }$ would give $g(x_{\ast })\le m+\epsilon <g(x_{\ast })$, a contradiction. Thus $g(x_{\ast })=m$.

For $x\mathrm{\notin }K$, step 1.1 gives $g(x)>c+1\ge m+1$, because $x_0\in K$ implies $m\le c$. Thus $x_{\ast }$ is a global minimizer over ${\mathbb{R}}^d$, and its value $m$ is finite.

On $\mathrm{dom}f$, $-g(x)=⟨p,x⟩-f(x)$, and outside this domain $g(x)=+\mathrm{\infty }$. The global minimum $m$ is attained at $x_{\ast }\in \mathrm{dom}f$, so the supremum in the definition of $f^{\ast }(p)$ is finite, equals $-m$, and is attained at $x_{\ast }$. Therefore $f^{\ast }(p)=⟨p,x_{\ast }⟩-f(x_{\ast })$; [L7] yields $p\in \mathrm{\partial }f(x_{\ast })$.