Show that equality for support function of two compact convex sets implies that two sets are equal.
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
asked Nov 24 at 20:27
Sepide
2808
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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.
convex-analysis
asked Nov 24 at 20:27
Sepide
2808
add a comment |
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
asked Nov 24 at 20:27
Sepide
2808
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
convex-analysis
asked Nov 24 at 20:27
Sepide
2808
asked Nov 24 at 20:27
Sepide
2808
asked Nov 24 at 20:27
Sepide
2808
asked Nov 24 at 20:27
Sepide
2808
asked Nov 24 at 20:27
Sepide
2808
2808
add a comment |
add a comment |
1 Answer
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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 $sigma_F(y)neq sigma_G(y)$.
answered Nov 24 at 23:49
Kavi Rama Murthy
48.5k31854
add a comment |
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1 Answer
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1 Answer
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oldest
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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 $sigma_F(y)neq sigma_G(y)$.
answered Nov 24 at 23:49
Kavi Rama Murthy
48.5k31854
add a comment |
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 $sigma_F(y)neq sigma_G(y)$.
answered Nov 24 at 23:49
Kavi Rama Murthy
48.5k31854
add a comment |
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 $sigma_F(y)neq sigma_G(y)$.
answered Nov 24 at 23:49
Kavi Rama Murthy
48.5k31854
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 $sigma_F(y)neq sigma_G(y)$.
answered Nov 24 at 23:49
Kavi Rama Murthy
48.5k31854
answered Nov 24 at 23:49
Kavi Rama Murthy
48.5k31854
answered Nov 24 at 23:49
Kavi Rama Murthy
48.5k31854
answered Nov 24 at 23:49
Kavi Rama Murthy
48.5k31854
48.5k31854
add a comment |
add a comment |
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