Proposition 21.4.2D6E7C0B06FAD

Proof. If $F\leq h_ P+C$, conjugation reverses the inequality and gives $F^*\geq h_ P^*-C=\chi _ P-C$ by theorem 21.3.2, so $\operatorname {dom}F^*\subset P$. Every subgradient of $F$ lies in $\operatorname {dom}F^*$ by the same theorem, so the second condition implies the third. If all slopes of $F$ lie in $P$ and $p\in \partial F(x)$, the subgradient inequality at $x$, evaluated at $0$, gives

Thus $F\leq h_ P+F(0)$, which proves the first condition and, since $h_ P(0)=0$, the identity $\max _ x(F(x)-h_ P(x))=F(0)$. If $p_ x\in \partial F(x)$ and $p_ y\in \partial F(y)$, the subgradient inequalities give $\langle p_ y,y-x\rangle \geq F(y)-F(x)\geq \langle p_ x,y-x\rangle $, so $|F(y)-F(x)|\leq \max _{p\in P}|p|\, |y-x|$.

Conjugation reverses inequalities, and $F^{**}=F$ because the finite convex function $F$ is continuous. Hence $h_ P-C\leq F\leq h_ P+C$ holds if and only if $\chi _ P-C\leq F^*\leq \chi _ P+C$, that is, if and only if $F^*=+\infty $ outside $P$ and $|F^*|\leq C$ on $P$. In that case $K_ F=P$, so $F\in \mathcal E(N_{\mathbb R},P)$ (by the convention if $P$ has empty interior), and $\int _ PF^*\, \mathrm dp\leq C\operatorname {vol}(P)<\infty $; thus $\mathcal E^\infty \subset \mathcal E^1$. The inclusions $\mathcal E^1\subset \mathcal E\subset \mathcal P$ hold by definition. If $P$ has empty interior and $n>0$, then $P$ lies in a proper affine subspace, because its relative interior is nonempty and open in its affine span (lemma 21.2.3(4)); hence $\operatorname {vol}(P)=0$, every integral over $P$ vanishes, and $\mathcal E^1=\mathcal E=\mathcal P$.

Now let $F,G$ be finite convex functions with $K_ F$ compact. By proposition 21.3.3(4), $(F+G)^*(p)$ is finite exactly when $p=q+r$ with $q\in \operatorname {dom}F^*$ and $r\in \operatorname {dom}G^*$, so

It remains to pass to closures. Since addition $M_{\mathbb R}\times M_{\mathbb R}\to M_{\mathbb R}$ is continuous, $K_ F+K_ G$ lies in the closure of $\operatorname {dom}F^*+\operatorname {dom}G^*$. Conversely, if $q_ k+r_ k\to s$ with $q_ k\in \operatorname {dom}F^*$ and $r_ k\in \operatorname {dom}G^*$, then a subsequence of $(q_ k)$ converges to some $q\in K_ F$, because $K_ F$ is compact, and along it $r_ k\to s-q\in K_ G$; hence $s\in K_ F+K_ G$. This proves $K_{F+G}=K_ F+K_ G$. Finally, if $K_ F=P$ and $K_ G=P'$, then $F\preceq h_ P$ and $G\preceq h_{P'}$ by the first part, and $h_ P+h_{P'}=h_{P+P'}$; hence $F+G\in \mathcal P(N_{\mathbb R},P+P')$ and $K_{F+G}=P+P'$, so that $F+G\in \mathcal E(N_{\mathbb R},P+P')$. $\square$