that $\B(\oy,\delta)\cap\aff(F(\ox))\subset F(\ox)$ and so verifies that $\oy\in\ri(F(\ox))$.

To prove next the opposite inclusion in , fix $\ox\in\ri(\dom(F))$ and $\oy\in\ri(F(\ox))$. Arguing by contradiction, suppose that $(\ox,\oy)\notin\ri(\gph(F))$ and then find by Theorem \[pst1\] such a pair $(u,v)\in\R^m\times\R^n$ that $$\label{1}
\la u,x\ra+\la v,y\ra\le\la u,\ox\ra+\la v,\oy\ra\;\mbox{ whenever }\;y\in F(x).$$ In addition, it follows from the proper separation that there is $(x_0,y_0)\in\gph(F)$ satisfying $$\label{2}
\la u,x_0\ra+\la v,y_0\ra<\la u,\ox\ra+\la v,\oy\ra.$$ Letting $x=\ox$ in yields $\la v,y\ra\le\la v,\oy\ra$ for all $y\in F(\ox)$. Since $\ox\in\ri(\dom(F))$ and $x_0\in \dom(F)$, we deduce from Theorem \[pts2\](a) that there exists $\tilde{x}\in\dom(F)$ such that $\ox=tx_0+(1-t)\tilde{x}$ for some $t\in(0,1)$. Choose $\tilde{y}\in F(\tilde{x})$ and consider the convex combination $$y^\prime:=ty_0+(1-t)\tilde{y},$$ where $y^\prime\in F(\ox)$ since $\gph(F)$ is convex. Since $\tilde{x}\in\dom(F)$ we use and to get $$\begin{aligned}
&\la u,\tilde{x}\ra+\la v,\tilde{y}\ra\le\la u,\ox\ra+\la v,\oy\ra,\\
&\la u,x_0\ra+\la v,y_0\ra<\la u,\ox\ra+\la v,\oy\ra.\end{aligned}$$ Multiplying the first inequality above by $1-t$ and the second one by $t$, and then adding them together gives us the condition $$\la u,\ox\ra+\la v,y^\prime\ra<\la u,\ox\ra+\la v,\oy\ra,$$ which yields $\la v,y^\prime\ra<\la v,\oy\ra$. Thus the sets $\{\oy\}$ and $F(\ox)$ are properly separated. Applying Theorem \[pst1\] tells us that $\oy\notin\ri(F(\ox))$, a contradiction that verifies $(\ox,\oy)\in\ri(\gph(F))$. $\h$

As a direct consequence of Theorem \[TheoRoc1\], we get the following useful result on calculating relative 