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

"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?

• Could you perhaps provide a definition of $\text{rge}(A)$? – icurays1 May 17 '15 at 14:56
• rge(A) is the range of map A. – user117890 May 17 '15 at 14:58
• If $A,B$ have the same affine hull, and $A \subset B$, then $\operatorname{ri} A \subset \operatorname{ri} B$. Since $\text{ri conv rge}\,A\subset\text{rge} A \subset \text{conv rge}\,A$, we have $\text{ri conv rge}\,A\subset\text{ri rge} A \subset \text{ri conv rge}\,A$. – copper.hat May 19 '15 at 4:49
• @copper.hat I don't know from which property do you verify that $\text{ri conv rge}A\subset\text{ri rge}A$? Is it possible to please clarify that. – user117890 May 19 '15 at 11:46
• Sure. You have $\text{ri conv rge}\,A\subset\text{rge} A$ and both $\text{ri conv rge}\,A, \text{rge} A$ have the same affine hull, hence $\text{ri ri conv rge}\,A\subset\text{ri rge} A$, and since $\text{ri ri }A = \text{ri }A$, you have the desired result. – copper.hat May 19 '15 at 15:57

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.
• @NeutralElement: It doesn't follow that if $A \subset B$ that the relative interiors are also nested. If $A$, $B$ have the same affine hull, then it is true. – copper.hat May 19 '15 at 4:55
• @NeutralElement: Get off your high horse. I understood the proof while you were in diapers (I am guessing :-)). It is not true in general that if $A \subset B$ that you have $\operatorname{ri} A \subset \operatorname{ri} B$. That is exactly what I wrote two comments above. Your answer, which is correct (since the fact is true) does not follow from just applying $\operatorname{ri}$ to both side. It also depends on both sides having the same affine hull. Also, are you suggesting that no one should comment before you write an answer? I am sure you appreciate that this is a bit extreme. – copper.hat May 19 '15 at 16:03
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
$$x=(1-\frac1\epsilon)y+\frac1\epsilon(y+\epsilon(x-y))\in\text{aff}\{y,y+\epsilon(x-y))\}\subset\text{aff ri}~S.$$