Show that equality for support function of two compact convex sets implies that two sets are equal.











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Let $Ssubseteq mathbb{R}^n$.



The support function of set $S$ is defined as the following



$$
sigma_S(x)=sup_{y in S} x^Ty
$$

where $x in mathbb{R}^n$.



Let $F$ and $G$ be two compact convex sets in $mathbb{R}^n$ such that $
sigma_F(x)=sigma_G(x)$
.



Show that $F=G$.



Hint: use appropriate separation theorem.










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    down vote

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    Let $Ssubseteq mathbb{R}^n$.



    The support function of set $S$ is defined as the following



    $$
    sigma_S(x)=sup_{y in S} x^Ty
    $$

    where $x in mathbb{R}^n$.



    Let $F$ and $G$ be two compact convex sets in $mathbb{R}^n$ such that $
    sigma_F(x)=sigma_G(x)$
    .



    Show that $F=G$.



    Hint: use appropriate separation theorem.










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $Ssubseteq mathbb{R}^n$.



      The support function of set $S$ is defined as the following



      $$
      sigma_S(x)=sup_{y in S} x^Ty
      $$

      where $x in mathbb{R}^n$.



      Let $F$ and $G$ be two compact convex sets in $mathbb{R}^n$ such that $
      sigma_F(x)=sigma_G(x)$
      .



      Show that $F=G$.



      Hint: use appropriate separation theorem.










      share|cite|improve this question













      Let $Ssubseteq mathbb{R}^n$.



      The support function of set $S$ is defined as the following



      $$
      sigma_S(x)=sup_{y in S} x^Ty
      $$

      where $x in mathbb{R}^n$.



      Let $F$ and $G$ be two compact convex sets in $mathbb{R}^n$ such that $
      sigma_F(x)=sigma_G(x)$
      .



      Show that $F=G$.



      Hint: use appropriate separation theorem.







      convex-analysis






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      asked Nov 24 at 20:27









      Sepide

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      1227






















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          This is a straightforward application of Hahn Banach separation theorem: if there is point $u$ in $F$ which is not in $G$ then (we can separate $u$ from $G$ in the sense) there exists a vector $y$ and a real number $r$ such that $y^{T}x <r<y^{T} u $ for all $xin G$. Hence $sigma_G(y) leq r<y^{T} u leq sigma_F(y)$ so $sigma_F(y)neq sigma_G(y)$.






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            This is a straightforward application of Hahn Banach separation theorem: if there is point $u$ in $F$ which is not in $G$ then (we can separate $u$ from $G$ in the sense) there exists a vector $y$ and a real number $r$ such that $y^{T}x <r<y^{T} u $ for all $xin G$. Hence $sigma_G(y) leq r<y^{T} u leq sigma_F(y)$ so $sigma_F(y)neq sigma_G(y)$.






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              This is a straightforward application of Hahn Banach separation theorem: if there is point $u$ in $F$ which is not in $G$ then (we can separate $u$ from $G$ in the sense) there exists a vector $y$ and a real number $r$ such that $y^{T}x <r<y^{T} u $ for all $xin G$. Hence $sigma_G(y) leq r<y^{T} u leq sigma_F(y)$ so $sigma_F(y)neq sigma_G(y)$.






              share|cite|improve this answer























                up vote
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                up vote
                1
                down vote









                This is a straightforward application of Hahn Banach separation theorem: if there is point $u$ in $F$ which is not in $G$ then (we can separate $u$ from $G$ in the sense) there exists a vector $y$ and a real number $r$ such that $y^{T}x <r<y^{T} u $ for all $xin G$. Hence $sigma_G(y) leq r<y^{T} u leq sigma_F(y)$ so $sigma_F(y)neq sigma_G(y)$.






                share|cite|improve this answer












                This is a straightforward application of Hahn Banach separation theorem: if there is point $u$ in $F$ which is not in $G$ then (we can separate $u$ from $G$ in the sense) there exists a vector $y$ and a real number $r$ such that $y^{T}x <r<y^{T} u $ for all $xin G$. Hence $sigma_G(y) leq r<y^{T} u leq sigma_F(y)$ so $sigma_F(y)neq sigma_G(y)$.







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                share|cite|improve this answer










                answered Nov 24 at 23:49









                Kavi Rama Murthy

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                44.2k31852






























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