Show that the distance of $x^ast$ to $C$ is the maximum of distances from $x^ast$ to closed half-spaces...
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This Q&A is in response to the follow up question asked in here.
Let $Csubseteq mathbb{R}^n$ be a closed convex set, and $x^ast in C^c$. Define the Euclidean distance from $x^ast$ to $C$ by $d_C(x^ast) = ||x^ast - Pi_C(x^ast)||_2 = min_{zin C}||x^ast - z||_2$, where $Pi_C(x^ast)$ is the projection of $x^ast$ onto $C$.
Show that the distance from $x^ast$ to $C$ is the maximum among the distances from $x^ast$ to closed half-spaces $H$ containing $C$, i.e., $d_C(x^ast)=max_{Csubseteq H}d_H(x^ast)$, where the max is taken over all closed half-spaces $H$'s containing $C$.
linear-algebra convex-analysis convex-optimization
asked Nov 26 '18 at 8:47
lntlslntls
666
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add a comment |
$begingroup$
This Q&A is in response to the follow up question asked in here.
Let $Csubseteq mathbb{R}^n$ be a closed convex set, and $x^ast in C^c$. Define the Euclidean distance from $x^ast$ to $C$ by $d_C(x^ast) = ||x^ast - Pi_C(x^ast)||_2 = min_{zin C}||x^ast - z||_2$, where $Pi_C(x^ast)$ is the projection of $x^ast$ onto $C$.
Show that the distance from $x^ast$ to $C$ is the maximum among the distances from $x^ast$ to closed half-spaces $H$ containing $C$, i.e., $d_C(x^ast)=max_{Csubseteq H}d_H(x^ast)$, where the max is taken over all closed half-spaces $H$'s containing $C$.
linear-algebra convex-analysis convex-optimization
asked Nov 26 '18 at 8:47
lntlslntls
666
$endgroup$
add a comment |
$begingroup$
This Q&A is in response to the follow up question asked in here.
Let $Csubseteq mathbb{R}^n$ be a closed convex set, and $x^ast in C^c$. Define the Euclidean distance from $x^ast$ to $C$ by $d_C(x^ast) = ||x^ast - Pi_C(x^ast)||_2 = min_{zin C}||x^ast - z||_2$, where $Pi_C(x^ast)$ is the projection of $x^ast$ onto $C$.
Show that the distance from $x^ast$ to $C$ is the maximum among the distances from $x^ast$ to closed half-spaces $H$ containing $C$, i.e., $d_C(x^ast)=max_{Csubseteq H}d_H(x^ast)$, where the max is taken over all closed half-spaces $H$'s containing $C$.
linear-algebra convex-analysis convex-optimization
asked Nov 26 '18 at 8:47
lntlslntls
666
$endgroup$
This Q&A is in response to the follow up question asked in here.
Let $Csubseteq mathbb{R}^n$ be a closed convex set, and $x^ast in C^c$. Define the Euclidean distance from $x^ast$ to $C$ by $d_C(x^ast) = ||x^ast - Pi_C(x^ast)||_2 = min_{zin C}||x^ast - z||_2$, where $Pi_C(x^ast)$ is the projection of $x^ast$ onto $C$.
Show that the distance from $x^ast$ to $C$ is the maximum among the distances from $x^ast$ to closed half-spaces $H$ containing $C$, i.e., $d_C(x^ast)=max_{Csubseteq H}d_H(x^ast)$, where the max is taken over all closed half-spaces $H$'s containing $C$.
linear-algebra convex-analysis convex-optimization
linear-algebra convex-analysis convex-optimization
asked Nov 26 '18 at 8:47
lntlslntls
666
asked Nov 26 '18 at 8:47
lntlslntls
666
edited Dec 17 '18 at 1:20
lntls
asked Nov 26 '18 at 8:47
lntlslntls
666
asked Nov 26 '18 at 8:47
lntlslntls
666
asked Nov 26 '18 at 8:47
lntlslntls
666
666
add a comment |
add a comment |
1 Answer
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$begingroup$
The case $d_C(x^ast)geqmax_{Csubseteq H}d_H(x^ast)$ is trivial.
Note that it has been shown in the preceding question that $Csubseteq H_ast equiv overline{H}^-(a,gamma)$, where $H_ast$ is a closed half space, $a = x^ast - Pi_C(x^ast)$ and $gamma = langle x^ast, Pi_C(x^ast) rangle$, and in fact, $d_C(x^ast) = d_{H_ast}(x^ast)$. Suppose $d_C(x^ast)notleqmax_{Csubseteq H}d_H(x^ast)$. Then $d_C(x^ast)>max_{Csubseteq H}d_H(x^ast)$, and it's clear we have a contradiction.
answered Nov 26 '18 at 8:47
lntlslntls
666
$endgroup$
1
$begingroup$
Could you tell me why $d_C(x^ast)geqmax_{Csubseteq H}d_H(x^ast)$ is trivial?
$endgroup$
– Saeed
Nov 26 '18 at 12:51
$begingroup$
Do you agree that any closed half-space is convex? Then since for any closed half-space $H$ containing $C$, obviously $Csubseteq H$, even for the half-space $G$ containing $C$ such that $d_G(x^ast) = max_{Csubseteq H} d_H(x^ast)$. Then since $G$ is a closed convex set, $||x^ast - Pi_G(x^ast)||_2 = min_{zin G}||x^ast - z||_2 leq ||x^ast - y||_2$ for all $y in G$. Since $Csubseteq G$, $Pi_C(x^ast) in G$. So $Pi_C(x^ast)$ is one such $y$.
$endgroup$
– lntls
Nov 26 '18 at 18:21
add a comment |
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$begingroup$
The case $d_C(x^ast)geqmax_{Csubseteq H}d_H(x^ast)$ is trivial.
Note that it has been shown in the preceding question that $Csubseteq H_ast equiv overline{H}^-(a,gamma)$, where $H_ast$ is a closed half space, $a = x^ast - Pi_C(x^ast)$ and $gamma = langle x^ast, Pi_C(x^ast) rangle$, and in fact, $d_C(x^ast) = d_{H_ast}(x^ast)$. Suppose $d_C(x^ast)notleqmax_{Csubseteq H}d_H(x^ast)$. Then $d_C(x^ast)>max_{Csubseteq H}d_H(x^ast)$, and it's clear we have a contradiction.
answered Nov 26 '18 at 8:47
lntlslntls
666
$endgroup$
1
$begingroup$
Could you tell me why $d_C(x^ast)geqmax_{Csubseteq H}d_H(x^ast)$ is trivial?
$endgroup$
– Saeed
Nov 26 '18 at 12:51
$begingroup$
Do you agree that any closed half-space is convex? Then since for any closed half-space $H$ containing $C$, obviously $Csubseteq H$, even for the half-space $G$ containing $C$ such that $d_G(x^ast) = max_{Csubseteq H} d_H(x^ast)$. Then since $G$ is a closed convex set, $||x^ast - Pi_G(x^ast)||_2 = min_{zin G}||x^ast - z||_2 leq ||x^ast - y||_2$ for all $y in G$. Since $Csubseteq G$, $Pi_C(x^ast) in G$. So $Pi_C(x^ast)$ is one such $y$.
$endgroup$
– lntls
Nov 26 '18 at 18:21
add a comment |
$begingroup$
The case $d_C(x^ast)geqmax_{Csubseteq H}d_H(x^ast)$ is trivial.
Note that it has been shown in the preceding question that $Csubseteq H_ast equiv overline{H}^-(a,gamma)$, where $H_ast$ is a closed half space, $a = x^ast - Pi_C(x^ast)$ and $gamma = langle x^ast, Pi_C(x^ast) rangle$, and in fact, $d_C(x^ast) = d_{H_ast}(x^ast)$. Suppose $d_C(x^ast)notleqmax_{Csubseteq H}d_H(x^ast)$. Then $d_C(x^ast)>max_{Csubseteq H}d_H(x^ast)$, and it's clear we have a contradiction.
answered Nov 26 '18 at 8:47
lntlslntls
666
$endgroup$
1
$begingroup$
Could you tell me why $d_C(x^ast)geqmax_{Csubseteq H}d_H(x^ast)$ is trivial?
$endgroup$
– Saeed
Nov 26 '18 at 12:51
$begingroup$
Do you agree that any closed half-space is convex? Then since for any closed half-space $H$ containing $C$, obviously $Csubseteq H$, even for the half-space $G$ containing $C$ such that $d_G(x^ast) = max_{Csubseteq H} d_H(x^ast)$. Then since $G$ is a closed convex set, $||x^ast - Pi_G(x^ast)||_2 = min_{zin G}||x^ast - z||_2 leq ||x^ast - y||_2$ for all $y in G$. Since $Csubseteq G$, $Pi_C(x^ast) in G$. So $Pi_C(x^ast)$ is one such $y$.
$endgroup$
– lntls
Nov 26 '18 at 18:21
add a comment |
$begingroup$
The case $d_C(x^ast)geqmax_{Csubseteq H}d_H(x^ast)$ is trivial.
Note that it has been shown in the preceding question that $Csubseteq H_ast equiv overline{H}^-(a,gamma)$, where $H_ast$ is a closed half space, $a = x^ast - Pi_C(x^ast)$ and $gamma = langle x^ast, Pi_C(x^ast) rangle$, and in fact, $d_C(x^ast) = d_{H_ast}(x^ast)$. Suppose $d_C(x^ast)notleqmax_{Csubseteq H}d_H(x^ast)$. Then $d_C(x^ast)>max_{Csubseteq H}d_H(x^ast)$, and it's clear we have a contradiction.
answered Nov 26 '18 at 8:47
lntlslntls
666
$endgroup$
The case $d_C(x^ast)geqmax_{Csubseteq H}d_H(x^ast)$ is trivial.
Note that it has been shown in the preceding question that $Csubseteq H_ast equiv overline{H}^-(a,gamma)$, where $H_ast$ is a closed half space, $a = x^ast - Pi_C(x^ast)$ and $gamma = langle x^ast, Pi_C(x^ast) rangle$, and in fact, $d_C(x^ast) = d_{H_ast}(x^ast)$. Suppose $d_C(x^ast)notleqmax_{Csubseteq H}d_H(x^ast)$. Then $d_C(x^ast)>max_{Csubseteq H}d_H(x^ast)$, and it's clear we have a contradiction.
answered Nov 26 '18 at 8:47
lntlslntls
666
answered Nov 26 '18 at 8:47
lntlslntls
666
answered Nov 26 '18 at 8:47
lntlslntls
666
answered Nov 26 '18 at 8:47
lntlslntls
666
666
1
$begingroup$
Could you tell me why $d_C(x^ast)geqmax_{Csubseteq H}d_H(x^ast)$ is trivial?
$endgroup$
– Saeed
Nov 26 '18 at 12:51
$begingroup$
Do you agree that any closed half-space is convex? Then since for any closed half-space $H$ containing $C$, obviously $Csubseteq H$, even for the half-space $G$ containing $C$ such that $d_G(x^ast) = max_{Csubseteq H} d_H(x^ast)$. Then since $G$ is a closed convex set, $||x^ast - Pi_G(x^ast)||_2 = min_{zin G}||x^ast - z||_2 leq ||x^ast - y||_2$ for all $y in G$. Since $Csubseteq G$, $Pi_C(x^ast) in G$. So $Pi_C(x^ast)$ is one such $y$.
$endgroup$
– lntls
Nov 26 '18 at 18:21
add a comment |
1
$begingroup$
Could you tell me why $d_C(x^ast)geqmax_{Csubseteq H}d_H(x^ast)$ is trivial?
$endgroup$
– Saeed
Nov 26 '18 at 12:51
$begingroup$
Do you agree that any closed half-space is convex? Then since for any closed half-space $H$ containing $C$, obviously $Csubseteq H$, even for the half-space $G$ containing $C$ such that $d_G(x^ast) = max_{Csubseteq H} d_H(x^ast)$. Then since $G$ is a closed convex set, $||x^ast - Pi_G(x^ast)||_2 = min_{zin G}||x^ast - z||_2 leq ||x^ast - y||_2$ for all $y in G$. Since $Csubseteq G$, $Pi_C(x^ast) in G$. So $Pi_C(x^ast)$ is one such $y$.
$endgroup$
– lntls
Nov 26 '18 at 18:21
1
1
$begingroup$
Could you tell me why $d_C(x^ast)geqmax_{Csubseteq H}d_H(x^ast)$ is trivial?
$endgroup$
– Saeed
Nov 26 '18 at 12:51
$begingroup$
Could you tell me why $d_C(x^ast)geqmax_{Csubseteq H}d_H(x^ast)$ is trivial?
$endgroup$
– Saeed
Nov 26 '18 at 12:51
$begingroup$
Do you agree that any closed half-space is convex? Then since for any closed half-space $H$ containing $C$, obviously $Csubseteq H$, even for the half-space $G$ containing $C$ such that $d_G(x^ast) = max_{Csubseteq H} d_H(x^ast)$. Then since $G$ is a closed convex set, $||x^ast - Pi_G(x^ast)||_2 = min_{zin G}||x^ast - z||_2 leq ||x^ast - y||_2$ for all $y in G$. Since $Csubseteq G$, $Pi_C(x^ast) in G$. So $Pi_C(x^ast)$ is one such $y$.
$endgroup$
– lntls
Nov 26 '18 at 18:21
$begingroup$
Do you agree that any closed half-space is convex? Then since for any closed half-space $H$ containing $C$, obviously $Csubseteq H$, even for the half-space $G$ containing $C$ such that $d_G(x^ast) = max_{Csubseteq H} d_H(x^ast)$. Then since $G$ is a closed convex set, $||x^ast - Pi_G(x^ast)||_2 = min_{zin G}||x^ast - z||_2 leq ||x^ast - y||_2$ for all $y in G$. Since $Csubseteq G$, $Pi_C(x^ast) in G$. So $Pi_C(x^ast)$ is one such $y$.
$endgroup$
– lntls
Nov 26 '18 at 18:21
add a comment |
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