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"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?

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    $\begingroup$ Could you perhaps provide a definition of $\text{rge}(A)$? $\endgroup$ – icurays1 May 17 '15 at 14:56
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    $\begingroup$ rge(A) is the range of map A. $\endgroup$ – user117890 May 17 '15 at 14:58
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    $\begingroup$ 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 $\text{ri conv rge}\,A\subset\text{ri rge} A \subset \text{ri conv rge}\,A$. $\endgroup$ – copper.hat May 19 '15 at 4:49
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    $\begingroup$ @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. $\endgroup$ – user117890 May 19 '15 at 11:46
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    $\begingroup$ 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. $\endgroup$ – copper.hat May 19 '15 at 15:57
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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.

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    $\begingroup$ @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. $\endgroup$ – copper.hat May 19 '15 at 4:55
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    $\begingroup$ @copper.hat My answer is correct. And the details you cannot explain for yourselves are a good exercise for both of you. It is also extremely annoying that you post the same proof as a comment before my answer. I assume that you finally understood the proof. (Yes the final explanation that you both should have found is that the affine hull the ri conv rge A equals the affine hull of (conv) rge A.) None of your comments told me anything new. If you want to argue about the validity of the proof I am happy to talk. Otherwise I will try to refrain from answering such posts. $\endgroup$ – Neutral Element May 19 '15 at 13:25
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    $\begingroup$ @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. $\endgroup$ – copper.hat May 19 '15 at 16:03
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    $\begingroup$ @copper.hat You put the same answer long after I posted my answer (timewise) but on the page it shows "before" my answer. I hope you understand now. As I explained before yes the relation does follow from applying ri while the rest of the details must be found by the person who posted it. An answer should not contain all details especially when the author of the post makes no effort towards solving it or takes an idea from a comment and writes another post. I don't know anything about your diapers or my high horses and as promised this is my last comment here. $\endgroup$ – Neutral Element May 19 '15 at 18:12
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    $\begingroup$ I didn't add any answer at all, so I have no idea what you are on about. You left out a crucial fact in your answer, how would someone who reads the answer have any idea that the same affine hull was involved? In fact, in that regard, it is very misleading, mitigated slightly by the fact that most don't deal with relative interiors. $\endgroup$ – copper.hat May 19 '15 at 18:19
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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

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

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