separation, of the important result by Rockafellar [@r Theorem 6.8] in finite dimensions (see also [@rw Proposition 2.43] for yet another proof), which can be formulated in the following way.

[**(Rockafellar’s theorem on relative interiors of convex graphs).**]{}\[TheoRoc1\] Let $F\colon\R^m\tto\R^n$ be a convex set-valued mapping. Then we have the representation $$\label{roc}
\ri\big(\gph(F)\big)=\big\{(x,y)\in\R^m\times\R^n\;\big|\;x\in\ri\big(\dom(F)\big),\;y\in\ri\big(F(x)\big)\big\}.$$

[**Proof.**]{} We first prove the inclusion “$\subset$" in . Picking $(\ox,\oy)\in\ri(\gph(F))$ and arguing by contradiction, suppose that $\ox\notin\ri(\dom (F))$. It follows from the relative interior descriptions of Theorem \[pst1\] that there exists $v\in\mathbb\R^m$ such that $$\la v,x\ra\le\la v,\ox\ra\;\mbox{ for all }\;x\in\dom(F)$$ together with a vector $x_0\in\dom(F)$ satisfying $$\la v,x_0\ra<\la v,\ox\ra.$$ Then for all $(x,y)\in\gph(F)$ we have $x\in\dom(F)$ and $$\la(v,0),(x,y)\ra=\la v,x\ra\le\la v,\ox\ra=\la(v,0),(\ox,\oy)\ra.$$ Taking now any fixed $y_0\in F(x_0)$ gives us $$\la(v,0),(x_0,y_0)\ra=\la v,x_0\ra<\la v,\ox\ra=\la(v,0),(\ox,\oy)\ra.$$ This shows that the sets $\gph(F)$ and $\{(\ox,\oy)\}$ are properly separated, and hence $(\ox,\oy)\notin\ri(\gph(F))$ by Theorem \[pst1\]. The obtained contradiction verifies that $\ox\in\ri(\dom(F))$.

To show now that $\oy\in\ri(F(\ox))$, we deduce from $(\ox,\oy)\in\ri(\gph(F))$ due to definition that there exists $\delta>0$ such that $$\big[\B(\ox;\delta)\times\B(\oy;\delta)\big]\cap\aff\big(\gph(F)\big)\subset\gph(F),$$ which yields in turn the inclusion $$\label{i1}
\big[\{\ox\}\times\B(\oy;\delta)\big]\cap\aff\big(\gph(F)\big)\subset\gph(F).$$ It follows furthermore from the definition of the affine hull that $$\{\ox\}\times\aff\big(F(\ox)\big)\subset\aff\big(\gph(F)\big),$$ which being combined with leads us to the relationships $$\begin{aligned}
\{\ox\}\times\big[(\B(\oy;\delta)\cap\aff\big(F(\ox)\big)\big]&=\big[\{\ox\}\times\B(\oy;\delta)\big]\cap\big[\{\ox\}\times\aff\big(F(\ox)\big)\big]\\
&\subset\big[\{\ox\}\times\B(\oy;\delta)\big]\times\aff\big(\gph(F)\big)\subset\gph(F).\end{aligned}$$ This shows 