Intersection of two conic hulls is equal to the conic hull of the intersection












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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.










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$endgroup$

















    1












    $begingroup$


    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.










    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      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.










      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Nov 29 '18 at 10:15







      Kluyvert

















      asked Nov 28 '18 at 13:31









      KluyvertKluyvert

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