Let $N$ be a real vector space of finite dimension and $M$ be the dual vector space.
We shall fix a convex body $P\subseteq M$.
The following classes are introduced in [BB13].
Definition 21.11.41.2AEBD4E3830A We define the set $\mathcal{P}(N,P)$ as the set of proper convex functions $F\in \mathrm{Conv}(N)$ such that $F\preceq \operatorname{Supp}_ P$.
We define the set $\mathcal{E}^{\infty }(N,P)$ as the set of closed convex functions $F\in \mathrm{Conv}(N)$ such that $F\sim \operatorname{Supp}_ P$.
We define the set $\mathcal{E}(N,P)$ as follows: Suppose that $\operatorname{Int}P=\varnothing $. Then $\mathcal{E}(N,P)\coloneqq \mathcal{P}(N,P)$; otherwise, let
\[ \mathcal{E}(N,P)=\left\{ F\in \mathcal{P}(N,P) : P=\overline{\nabla F(N)} \right\} . \]
We define the set $\mathcal{E}^1(N,P)$ as the subset of $\mathcal{E}(N,P)$ consisting of $F\in \mathcal{E}(N,P)$ with
\[ \int _ P F^*\, \mathrm{d}\operatorname{vol}<\infty , \]
where $\mathrm{d}\operatorname{vol}$ is any Lebesgue measure on $M$.
Observe that for any $F\in \mathcal{P}(N,P)$, we have $\operatorname{Dom}F=N$ and $F$ is necessarily closed.
Proposition 21.11.42.A63267E9EA69 We have
\[ \mathcal{E}^{\infty }(N,P)\subseteq \mathcal{E}^{1}(N,P)\subseteq \mathcal{E}(N,P) \subseteq \mathcal{P}(N,P). \]
Proof.
When $\operatorname{Int}P=\varnothing $, the assertion is clear. We assume that $\operatorname{Int}P\neq \varnothing $. The second inclusion follows from the definition. We only handle the first inequality. Take $F\in \mathcal{E}^{\infty }(N,P)$. By definition, $F\sim \operatorname{Supp}_ P$ and hence $F^*\sim \chi _ P$. It follows that $P=\operatorname{Dom}F^*$.
By proposition 21.11.38, we already know that
\[ \nabla F(N)\subseteq P=\operatorname{Dom}F^*. \]
On the other hand, by proposition 21.11.34, we have
\[ \operatorname{Int}P\subseteq \nabla F(N). \]
So it follows that
\[ P=\overline{\nabla F(N)}. \]
It is clear that $F^*\sim \chi _ P$ is integrable.
$\square$
Proposition 21.11.43.281F97D00D5D For any $F\in \mathcal{E}^{\infty }(N,P)$, we have $F^*|_{M\setminus P}\equiv \infty $ and $F^*$ is bounded on $P$.
Proof.
From $F\sim \operatorname{Supp}_ P$, we take the Legendre transform to get $F^*\sim \operatorname{Supp}_ P^*=\chi _ P$, where we applied example 21.11.25.
$\square$
Definition 21.11.44.0E804C9E9F89 We endow the topology of pointwise convergence on $\mathcal{P}(N,P)$.
Note that this topology coincides with the compact-open topology. See [Theorem 10.8, Roc70].
Proposition 21.11.45.C0908AA2E3B2 Let $F\in \mathcal{P}(N,P)$. Then there is a decreasing sequence $F_ j\in \mathcal{E}^{\infty }(N,P)\cap C^{\infty }(N)$ converging to $F$.
Proof. Apply the proved decreasing smooth approximation with fixed growth body. It gives smooth convex $F_j$ with bounded difference from the support function of $P$, decreasing pointwise, in fact locally uniformly, to $F$. The bounded-difference condition is precisely membership in $\mathcal E^\infty(N,P)$, and $\mathcal P(N,P)$ is the same upper-growth class in both formulations. That construction preserves the common growth body $P$, including when $P$ has empty interior; therefore the empty-interior case requires no full-dimensional approximation of $P$.
See [Lemma 2.2, BB13].
We observe that the point $0\in N$ plays a special role since it does in the definition of the support function.
Proposition 21.11.46.93CDDD09D6B0 For any $F\in \mathcal{P}(N,P)$, we have
\[ \max _{N}(F-\operatorname{Supp}_ P)=F(0). \]
Proof.
It follows from (21.11.39) that
\[ \sup _{N}(F-\operatorname{Supp}_ P)\leq F(0). \]
The equality is clearly obtained at $0\in N$.
$\square$
Lemma 21.11.47.53E69530968E Let $P'\subseteq M$ be another convex body. Then for any $F\in \mathcal{P}(N,P)$ and $F'\in \mathcal{P}(N,P')$, we have
\[ F+F'\in \mathcal{P}(N,P+P'). \]
Moreover, let $F$ and $F'$ be real-valued convex functions on $N$. Assume that $\operatorname{Dom}F^*$ is bounded. Then
\[ \overline{\nabla F(N)}+\overline{\nabla F'(N)}=\overline{\nabla (F+F')(N)}. \]
In particular, if $F\in \mathcal{P}(N,P)$, $F'\in \mathcal{P}(N,P')$ and
\[ \overline{\nabla F(N)}=P,\quad \overline{\nabla F'(N)}=P', \]
then $F+F'\in \mathcal{E}(N,P+P')$.
Proof.
The former assertion follows immediately from the observation
\[ \operatorname{Supp}_{P+P'}=\operatorname{Supp}_ P+\operatorname{Supp}_{P'}. \]
We prove the gradient-image equality. In view of proposition 21.11.34, this means
B39039E9725D
21.11.48
\begin{equation} \overline{\operatorname{Dom}(F+F')^*}=\overline{\operatorname{Dom}F^*}+\overline{\operatorname{Dom}F'{}^*}. \end{equation}
It follows from proposition 21.11.37 that
\[ \operatorname{Dom}(F+F')^*=\operatorname{Dom}F^*+\operatorname{Dom}F'{}^*. \]
Since $\overline{\operatorname{Dom}F^*}$ is compact, (21.11.48) follows.
Finally, under the assumptions in the last assertion, the former assertion gives $F+F'\in \mathcal{P}(N,P+P')$. If $\operatorname{Int}(P+P')=\varnothing $, this is exactly the definition of $\mathcal{E}(N,P+P')$. Otherwise, the gradient-image equality gives
\[ \overline{\nabla (F+F')(N)}=P+P', \]
so again $F+F'\in \mathcal{E}(N,P+P')$.
$\square$
Pome
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