Show that support function of a set $S$ and support function of the closure of that set $bar{S}$ 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 $x in mathbb{R}^n$.
Show that $sigma_S(x)=sigma_{bar{S}}(x) ,,,,forall ,,,x in mathbb{R}^n$.
If the set $S$ were closed then $S=bar{S}$ so we are done. For the case that $S neqbar{S}$ we need to prove the statement. My approach is to say that the closure of a set is the set of all supporting hyperplains (which I do not know how to prove!), then show that $sigma_S(x)$ takes its value for every $x$ when $y$ is at the boundary of $S$.
convex-analysis convex-optimization supremum-and-infimum convex-geometry convex-hulls
asked Nov 19 '18 at 4:56
Sepide
2918
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 $x in mathbb{R}^n$.
Show that $sigma_S(x)=sigma_{bar{S}}(x) ,,,,forall ,,,x in mathbb{R}^n$.
If the set $S$ were closed then $S=bar{S}$ so we are done. For the case that $S neqbar{S}$ we need to prove the statement. My approach is to say that the closure of a set is the set of all supporting hyperplains (which I do not know how to prove!), then show that $sigma_S(x)$ takes its value for every $x$ when $y$ is at the boundary of $S$.
convex-analysis convex-optimization supremum-and-infimum convex-geometry convex-hulls
asked Nov 19 '18 at 4:56
Sepide
2918
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 $x in mathbb{R}^n$.
Show that $sigma_S(x)=sigma_{bar{S}}(x) ,,,,forall ,,,x in mathbb{R}^n$.
If the set $S$ were closed then $S=bar{S}$ so we are done. For the case that $S neqbar{S}$ we need to prove the statement. My approach is to say that the closure of a set is the set of all supporting hyperplains (which I do not know how to prove!), then show that $sigma_S(x)$ takes its value for every $x$ when $y$ is at the boundary of $S$.
convex-analysis convex-optimization supremum-and-infimum convex-geometry convex-hulls
asked Nov 19 '18 at 4:56
Sepide
2918
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 $x in mathbb{R}^n$.
Show that $sigma_S(x)=sigma_{bar{S}}(x) ,,,,forall ,,,x in mathbb{R}^n$.
If the set $S$ were closed then $S=bar{S}$ so we are done. For the case that $S neqbar{S}$ we need to prove the statement. My approach is to say that the closure of a set is the set of all supporting hyperplains (which I do not know how to prove!), then show that $sigma_S(x)$ takes its value for every $x$ when $y$ is at the boundary of $S$.
convex-analysis convex-optimization supremum-and-infimum convex-geometry convex-hulls
convex-analysis convex-optimization supremum-and-infimum convex-geometry convex-hulls
asked Nov 19 '18 at 4:56
Sepide
2918
asked Nov 19 '18 at 4:56
Sepide
2918
asked Nov 19 '18 at 4:56
Sepide
2918
asked Nov 19 '18 at 4:56
Sepide
2918
asked Nov 19 '18 at 4:56
Sepide
2918
2918
add a comment |
add a comment |
1 Answer
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It is obvious that $sigma_S(x) leq sigma_{ overline {S}} (x)$. For the other inequality take $y in overline {S}$ and choose a sequence ${y_n} subset S$ such that $y_n to y$. Since $x^{T}y_n leq sigma_S(x)$ for all $n$ and $x^{T}y_n to x^{T}y$ we get $x^{T}y leq sigma_S(x)$. Taking supremum over $y$ we get $sigma_{ overline {S}} (x) leq sigma_S(x)$.
answered Nov 19 '18 at 5:20
Kavi Rama Murthy
50.3k31854
I need to show the first sentence. I understand that intuitively makes sense but how we can show it?
– Sepide
Nov 20 '18 at 3:50
If $A subset B$ then $sup A leq sup B$ by definition of supremum.
– Kavi Rama Murthy
Nov 20 '18 at 5:32
add a comment |
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1 Answer
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It is obvious that $sigma_S(x) leq sigma_{ overline {S}} (x)$. For the other inequality take $y in overline {S}$ and choose a sequence ${y_n} subset S$ such that $y_n to y$. Since $x^{T}y_n leq sigma_S(x)$ for all $n$ and $x^{T}y_n to x^{T}y$ we get $x^{T}y leq sigma_S(x)$. Taking supremum over $y$ we get $sigma_{ overline {S}} (x) leq sigma_S(x)$.
answered Nov 19 '18 at 5:20
Kavi Rama Murthy
50.3k31854
I need to show the first sentence. I understand that intuitively makes sense but how we can show it?
– Sepide
Nov 20 '18 at 3:50
If $A subset B$ then $sup A leq sup B$ by definition of supremum.
– Kavi Rama Murthy
Nov 20 '18 at 5:32
add a comment |
It is obvious that $sigma_S(x) leq sigma_{ overline {S}} (x)$. For the other inequality take $y in overline {S}$ and choose a sequence ${y_n} subset S$ such that $y_n to y$. Since $x^{T}y_n leq sigma_S(x)$ for all $n$ and $x^{T}y_n to x^{T}y$ we get $x^{T}y leq sigma_S(x)$. Taking supremum over $y$ we get $sigma_{ overline {S}} (x) leq sigma_S(x)$.
answered Nov 19 '18 at 5:20
Kavi Rama Murthy
50.3k31854
I need to show the first sentence. I understand that intuitively makes sense but how we can show it?
– Sepide
Nov 20 '18 at 3:50
If $A subset B$ then $sup A leq sup B$ by definition of supremum.
– Kavi Rama Murthy
Nov 20 '18 at 5:32
add a comment |
It is obvious that $sigma_S(x) leq sigma_{ overline {S}} (x)$. For the other inequality take $y in overline {S}$ and choose a sequence ${y_n} subset S$ such that $y_n to y$. Since $x^{T}y_n leq sigma_S(x)$ for all $n$ and $x^{T}y_n to x^{T}y$ we get $x^{T}y leq sigma_S(x)$. Taking supremum over $y$ we get $sigma_{ overline {S}} (x) leq sigma_S(x)$.
answered Nov 19 '18 at 5:20
Kavi Rama Murthy
50.3k31854
It is obvious that $sigma_S(x) leq sigma_{ overline {S}} (x)$. For the other inequality take $y in overline {S}$ and choose a sequence ${y_n} subset S$ such that $y_n to y$. Since $x^{T}y_n leq sigma_S(x)$ for all $n$ and $x^{T}y_n to x^{T}y$ we get $x^{T}y leq sigma_S(x)$. Taking supremum over $y$ we get $sigma_{ overline {S}} (x) leq sigma_S(x)$.
answered Nov 19 '18 at 5:20
Kavi Rama Murthy
50.3k31854
answered Nov 19 '18 at 5:20
Kavi Rama Murthy
50.3k31854
answered Nov 19 '18 at 5:20
Kavi Rama Murthy
50.3k31854
answered Nov 19 '18 at 5:20
Kavi Rama Murthy
50.3k31854
50.3k31854
I need to show the first sentence. I understand that intuitively makes sense but how we can show it?
– Sepide
Nov 20 '18 at 3:50
If $A subset B$ then $sup A leq sup B$ by definition of supremum.
– Kavi Rama Murthy
Nov 20 '18 at 5:32
add a comment |
I need to show the first sentence. I understand that intuitively makes sense but how we can show it?
– Sepide
Nov 20 '18 at 3:50
If $A subset B$ then $sup A leq sup B$ by definition of supremum.
– Kavi Rama Murthy
Nov 20 '18 at 5:32
I need to show the first sentence. I understand that intuitively makes sense but how we can show it?
– Sepide
Nov 20 '18 at 3:50
I need to show the first sentence. I understand that intuitively makes sense but how we can show it?
– Sepide
Nov 20 '18 at 3:50
If $A subset B$ then $sup A leq sup B$ by definition of supremum.
– Kavi Rama Murthy
Nov 20 '18 at 5:32
If $A subset B$ then $sup A leq sup B$ by definition of supremum.
– Kavi Rama Murthy
Nov 20 '18 at 5:32
add a comment |
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