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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