Proposition 21.11.38.6FCC352452C3 Let $P\subseteq M$ be a convex body and $F\in \mathrm{Conv}^{\mathrm{prop}}(N)$. The following are equivalent:
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$F\preceq \operatorname{Supp}_ P$;
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$\operatorname{Dom}F=N$ and $F^*|_{M\setminus P}\equiv \infty $;
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$\operatorname{Dom}F=N$ and $\nabla F(N)\subseteq P$.
Moreover, under these conditions,
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21.11.39
\begin{equation} F(n)-\operatorname{Supp}_ P(n)\leq F(0),\quad \forall n\in N. \end{equation}
Proof.
1 $\implies $ 2. It is clear that $\operatorname{Dom}F=N$ since $\operatorname{Dom}\operatorname{Supp}_ P=N$. From $F\preceq \operatorname{Supp}_ P$ and example 21.11.25, we know that
\[ \chi _ P=\operatorname{Supp}_ P^*\preceq F^*. \]
So 2 follows.
2 $\implies $ 3. This follows from proposition 21.11.34.
3 $\implies $ 1. The function $t\mapsto F(tn)$ is convex on $[0,1]$, hence absolutely continuous and differentiable almost everywhere. For almost every $t\in (0,1)$,
\[ \frac{\mathrm d}{\mathrm dt}F(tn)=\partial _ n F(tn) \leq \operatorname{Supp}_ P(n) \]
by theorem 21.11.33 and the assumption $\nabla F(N)\subseteq P$. Integrating in $t$ gives
\[ F(n)-F(0)\leq \operatorname{Supp}_ P(n). \]
In particular, (21.11.39) also follows.
$\square$
Pome
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