Proving that $\text{ri rge}\,A=\text{ri conv rge}\,A$

Question

"If $A:\mathbb R^n\rightrightarrows\mathbb R^n$ is maximal monotone,then $\text{ri rge}\,A$ is convex". This is a proposition in auslender's book about the asymptotic cones. We can prove that $$\text{ri conv rge}\,A\subset\text{rge} A$$ then author of book says "this relation shows that $\text{ri rge}\,A$ is convex", but I don't know how?

In answer below it is claimed that $\text{ri ri conv rge}A\subset\text{ri rge}A$, which is valid if $\text{ri conv rge}A$ and $\text{rge}A$ have the same affine hull, that is $$\text{aff ri conv rge}A=\text{aff rge}A$$ I'm well able to show that $$\text{aff ri conv rge}A\subset\text{aff rge}A$$ but I couldn't prove the converse relation yet, that is taking any $v\in\text{aff rge}A$, we have $$v=\sum_{i=1}^m\lambda_iv_i,\sum_{i=1}^m\lambda_i=1,v_i\in\text{rge}A$$ so $v$ is in $\text{aff ri conv rge}A$ if $v_i\in\text{ri conv rge}A$ that I can't verify it?

Answer 1

Actually, from your relation, just apply "ri" to get $$\rm{ri\,conv\,rge}A=\rm{ri}(\rm{ri\,conv\,rge}A)\subset \rm{ri\,rge}A\subset\rm{ri\,conv\,rge}A,$$ from which $\rm{ri\,rge}A=\rm{ri\,conv\,rge}A$ is convex.

Answer 2

Let $S=\text{conv rge}~A$, assume that $x\in\text{aff}~S$ and note that $\text{ri}~S\neq\emptyset$, so for $y\in\text{ri}~S$ and for sufficiently small $\epsilon>0$, $y+\epsilon(x-y)\in\text{ri}~S$, therefore

$$x=(1-\frac1\epsilon)y+\frac1\epsilon(y+\epsilon(x-y))\in\text{aff}\{y,y+\epsilon(x-y))\}\subset\text{aff ri}~S.$$