Intersection of two conic hulls is equal to the conic hull of the intersection
$begingroup$
since the above conjecture is wrong in general, I would like to know (and maybe prove) that the following Statement holds:
Let $A,B$ be two closed, convex sets in $mathbb{R}^n$ such that $Acap B=text{bd}(A)cap text{bd}(B)neqemptyset$ (where $text{bd}$ denotes the boundary).
Then $text{cone}(Acap B)=text{cone}(A)cap text{cone}(B)$.
(cone is the conic hull operator)
So far I found out that $text{cone}(Acap B)subseteqtext{cone}(A)cap text{cone}(B)$ is clear. So lets consider an element $xintext{cone}(A)cap text{cone}(B)$. So there exisits $ain A$, $bin B$, $lambda,mu>0$ such that $x=lambda a$ and $x=mu b$.
Hence $ain Acap text{cone}(B)$ and $bin Bcap text{cone}(A)$.
All I need to show the conjecture is to show that $ain Acap B$ or $bin Bcap A$ since this would prove $xintext{cone}(Acap B)$.
If $text{int}(A)captext{int}(B)neqemptyset$ this is not possible I guess, so what about $text{int}(A)captext{int}(B)=emptyset$ (int denotes the interior). Do you have an idea?
Edit: The statement should follow immediately if we assume the conic representation of $x$ to be unique. This should be true if $A,B$ are simplices for instance.
convex-cone
asked Nov 28 '18 at 13:31
KluyvertKluyvert
84
$endgroup$
add a comment |
$begingroup$
since the above conjecture is wrong in general, I would like to know (and maybe prove) that the following Statement holds:
Let $A,B$ be two closed, convex sets in $mathbb{R}^n$ such that $Acap B=text{bd}(A)cap text{bd}(B)neqemptyset$ (where $text{bd}$ denotes the boundary).
Then $text{cone}(Acap B)=text{cone}(A)cap text{cone}(B)$.
(cone is the conic hull operator)
So far I found out that $text{cone}(Acap B)subseteqtext{cone}(A)cap text{cone}(B)$ is clear. So lets consider an element $xintext{cone}(A)cap text{cone}(B)$. So there exisits $ain A$, $bin B$, $lambda,mu>0$ such that $x=lambda a$ and $x=mu b$.
Hence $ain Acap text{cone}(B)$ and $bin Bcap text{cone}(A)$.
All I need to show the conjecture is to show that $ain Acap B$ or $bin Bcap A$ since this would prove $xintext{cone}(Acap B)$.
If $text{int}(A)captext{int}(B)neqemptyset$ this is not possible I guess, so what about $text{int}(A)captext{int}(B)=emptyset$ (int denotes the interior). Do you have an idea?
Edit: The statement should follow immediately if we assume the conic representation of $x$ to be unique. This should be true if $A,B$ are simplices for instance.
convex-cone
asked Nov 28 '18 at 13:31
KluyvertKluyvert
84
$endgroup$
add a comment |
$begingroup$
since the above conjecture is wrong in general, I would like to know (and maybe prove) that the following Statement holds:
Let $A,B$ be two closed, convex sets in $mathbb{R}^n$ such that $Acap B=text{bd}(A)cap text{bd}(B)neqemptyset$ (where $text{bd}$ denotes the boundary).
Then $text{cone}(Acap B)=text{cone}(A)cap text{cone}(B)$.
(cone is the conic hull operator)
So far I found out that $text{cone}(Acap B)subseteqtext{cone}(A)cap text{cone}(B)$ is clear. So lets consider an element $xintext{cone}(A)cap text{cone}(B)$. So there exisits $ain A$, $bin B$, $lambda,mu>0$ such that $x=lambda a$ and $x=mu b$.
Hence $ain Acap text{cone}(B)$ and $bin Bcap text{cone}(A)$.
All I need to show the conjecture is to show that $ain Acap B$ or $bin Bcap A$ since this would prove $xintext{cone}(Acap B)$.
If $text{int}(A)captext{int}(B)neqemptyset$ this is not possible I guess, so what about $text{int}(A)captext{int}(B)=emptyset$ (int denotes the interior). Do you have an idea?
Edit: The statement should follow immediately if we assume the conic representation of $x$ to be unique. This should be true if $A,B$ are simplices for instance.
convex-cone
asked Nov 28 '18 at 13:31
KluyvertKluyvert
84
$endgroup$
since the above conjecture is wrong in general, I would like to know (and maybe prove) that the following Statement holds:
Let $A,B$ be two closed, convex sets in $mathbb{R}^n$ such that $Acap B=text{bd}(A)cap text{bd}(B)neqemptyset$ (where $text{bd}$ denotes the boundary).
Then $text{cone}(Acap B)=text{cone}(A)cap text{cone}(B)$.
(cone is the conic hull operator)
So far I found out that $text{cone}(Acap B)subseteqtext{cone}(A)cap text{cone}(B)$ is clear. So lets consider an element $xintext{cone}(A)cap text{cone}(B)$. So there exisits $ain A$, $bin B$, $lambda,mu>0$ such that $x=lambda a$ and $x=mu b$.
Hence $ain Acap text{cone}(B)$ and $bin Bcap text{cone}(A)$.
All I need to show the conjecture is to show that $ain Acap B$ or $bin Bcap A$ since this would prove $xintext{cone}(Acap B)$.
If $text{int}(A)captext{int}(B)neqemptyset$ this is not possible I guess, so what about $text{int}(A)captext{int}(B)=emptyset$ (int denotes the interior). Do you have an idea?
Edit: The statement should follow immediately if we assume the conic representation of $x$ to be unique. This should be true if $A,B$ are simplices for instance.
convex-cone
convex-cone
asked Nov 28 '18 at 13:31
KluyvertKluyvert
84
asked Nov 28 '18 at 13:31
KluyvertKluyvert
84
edited Nov 29 '18 at 10:15
Kluyvert
asked Nov 28 '18 at 13:31
KluyvertKluyvert
84
asked Nov 28 '18 at 13:31
KluyvertKluyvert
84
asked Nov 28 '18 at 13:31
KluyvertKluyvert
84
84
add a comment |
add a comment |
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