Show distance to the half-space is equal to distance to the set
Let $C subseteq mathbb{R}^n$ be a closed convex set, and $x^* in C^c$ (not in $C$ and its closure).
Define the Euclidean distance from $x^*$ to $C$ as $d_C(x^*):=min_{z in C}|z -x^*|_2$.
Consider the closed half-space $H:={x in mathbb{R}^n: langle a,xrangle leq gamma }$, where $a=x^*-prod_C(x^*)$ and $gamma=langle a,prod_C(x^*)rangle$ where $prod_C(x^*)$ is projection of $x^*$ onto $C$.
Show that $d_H(x^*)=d_C(x^*)$.
Intuitively, it says that the hyperplane defining in this way is tangent to the set $C$ at point $prod_C(x^*)$.
Extension:
Show that the minimum distance from $x^*$ to $C$ is the maximum among the distance from $x^*$ to closed half-spaces $H$ containing $C$, i.e., $d_C(x^*)=max_{C subseteq H}d_H(x^*)$ where the max is taken over all closed half-spaces $H$'s containing $C$.
convex-optimization projection
asked Nov 25 at 21:06
Saeed
607110
add a comment |
Let $C subseteq mathbb{R}^n$ be a closed convex set, and $x^* in C^c$ (not in $C$ and its closure).
Define the Euclidean distance from $x^*$ to $C$ as $d_C(x^*):=min_{z in C}|z -x^*|_2$.
Consider the closed half-space $H:={x in mathbb{R}^n: langle a,xrangle leq gamma }$, where $a=x^*-prod_C(x^*)$ and $gamma=langle a,prod_C(x^*)rangle$ where $prod_C(x^*)$ is projection of $x^*$ onto $C$.
Show that $d_H(x^*)=d_C(x^*)$.
Intuitively, it says that the hyperplane defining in this way is tangent to the set $C$ at point $prod_C(x^*)$.
Extension:
Show that the minimum distance from $x^*$ to $C$ is the maximum among the distance from $x^*$ to closed half-spaces $H$ containing $C$, i.e., $d_C(x^*)=max_{C subseteq H}d_H(x^*)$ where the max is taken over all closed half-spaces $H$'s containing $C$.
convex-optimization projection
asked Nov 25 at 21:06
Saeed
607110
add a comment |
Let $C subseteq mathbb{R}^n$ be a closed convex set, and $x^* in C^c$ (not in $C$ and its closure).
Define the Euclidean distance from $x^*$ to $C$ as $d_C(x^*):=min_{z in C}|z -x^*|_2$.
Consider the closed half-space $H:={x in mathbb{R}^n: langle a,xrangle leq gamma }$, where $a=x^*-prod_C(x^*)$ and $gamma=langle a,prod_C(x^*)rangle$ where $prod_C(x^*)$ is projection of $x^*$ onto $C$.
Show that $d_H(x^*)=d_C(x^*)$.
Intuitively, it says that the hyperplane defining in this way is tangent to the set $C$ at point $prod_C(x^*)$.
Extension:
Show that the minimum distance from $x^*$ to $C$ is the maximum among the distance from $x^*$ to closed half-spaces $H$ containing $C$, i.e., $d_C(x^*)=max_{C subseteq H}d_H(x^*)$ where the max is taken over all closed half-spaces $H$'s containing $C$.
convex-optimization projection
asked Nov 25 at 21:06
Saeed
607110
Let $C subseteq mathbb{R}^n$ be a closed convex set, and $x^* in C^c$ (not in $C$ and its closure).
Define the Euclidean distance from $x^*$ to $C$ as $d_C(x^*):=min_{z in C}|z -x^*|_2$.
Consider the closed half-space $H:={x in mathbb{R}^n: langle a,xrangle leq gamma }$, where $a=x^*-prod_C(x^*)$ and $gamma=langle a,prod_C(x^*)rangle$ where $prod_C(x^*)$ is projection of $x^*$ onto $C$.
Show that $d_H(x^*)=d_C(x^*)$.
Intuitively, it says that the hyperplane defining in this way is tangent to the set $C$ at point $prod_C(x^*)$.
Extension:
Show that the minimum distance from $x^*$ to $C$ is the maximum among the distance from $x^*$ to closed half-spaces $H$ containing $C$, i.e., $d_C(x^*)=max_{C subseteq H}d_H(x^*)$ where the max is taken over all closed half-spaces $H$'s containing $C$.
convex-optimization projection
convex-optimization projection
asked Nov 25 at 21:06
Saeed
607110
asked Nov 25 at 21:06
Saeed
607110
edited Nov 26 at 5:55
asked Nov 25 at 21:06
Saeed
607110
asked Nov 25 at 21:06
Saeed
607110
asked Nov 25 at 21:06
Saeed
607110
607110
add a comment |
add a comment |
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Note that $Pi_C(x^ast) in Csubseteq H$, and $Pi_H(x^ast) in H$. Then by construction of $H$,
$$0 geq langle x^ast - Pi_C(x^ast),Pi_H(x^ast) - Pi_C(x^ast)rangle = langle Pi_C(x^ast) - x^ast, Pi_C(x^ast) - Pi_H(x^ast) rangle,$$
and by the variational inequality of Euclidean distance
$$0 geq langle x^ast - Pi_H(x^ast),Pi_C(x^ast) - Pi_H(x^ast)rangle.$$
So add the two inequalities together, and you should see that, in fact, $Pi_C(x^ast) = Pi_H(x^ast)$!
answered Nov 26 at 5:01
lntls
666
Oh boy. Perfect. Please keep answering. Can you look at the extension of the question?
– Saeed
Nov 26 at 5:50
Sure, but it seems that the community doesn't allow follow up questions to another question.(meta.stackoverflow.com/questions/296489/…) Do you mind creating a separate question with your extension, and linking to this question?
– lntls
Nov 26 at 7:32
I've provided a sketch that may be clear for the follow up here: math.stackexchange.com/questions/3014057/…. Please comment in the link provided if it is not clear.
– lntls
Nov 26 at 8:53
Can you help me for this question?
– Saeed
Nov 29 at 15:20
Could you help me to show this question: math.stackexchange.com/questions/3023525/…
– Saeed
Dec 3 at 2:27
add a comment |
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Note that $Pi_C(x^ast) in Csubseteq H$, and $Pi_H(x^ast) in H$. Then by construction of $H$,
$$0 geq langle x^ast - Pi_C(x^ast),Pi_H(x^ast) - Pi_C(x^ast)rangle = langle Pi_C(x^ast) - x^ast, Pi_C(x^ast) - Pi_H(x^ast) rangle,$$
and by the variational inequality of Euclidean distance
$$0 geq langle x^ast - Pi_H(x^ast),Pi_C(x^ast) - Pi_H(x^ast)rangle.$$
So add the two inequalities together, and you should see that, in fact, $Pi_C(x^ast) = Pi_H(x^ast)$!
answered Nov 26 at 5:01
lntls
666
Oh boy. Perfect. Please keep answering. Can you look at the extension of the question?
– Saeed
Nov 26 at 5:50
Sure, but it seems that the community doesn't allow follow up questions to another question.(meta.stackoverflow.com/questions/296489/…) Do you mind creating a separate question with your extension, and linking to this question?
– lntls
Nov 26 at 7:32
I've provided a sketch that may be clear for the follow up here: math.stackexchange.com/questions/3014057/…. Please comment in the link provided if it is not clear.
– lntls
Nov 26 at 8:53
Can you help me for this question?
– Saeed
Nov 29 at 15:20
Could you help me to show this question: math.stackexchange.com/questions/3023525/…
– Saeed
Dec 3 at 2:27
add a comment |
Note that $Pi_C(x^ast) in Csubseteq H$, and $Pi_H(x^ast) in H$. Then by construction of $H$,
$$0 geq langle x^ast - Pi_C(x^ast),Pi_H(x^ast) - Pi_C(x^ast)rangle = langle Pi_C(x^ast) - x^ast, Pi_C(x^ast) - Pi_H(x^ast) rangle,$$
and by the variational inequality of Euclidean distance
$$0 geq langle x^ast - Pi_H(x^ast),Pi_C(x^ast) - Pi_H(x^ast)rangle.$$
So add the two inequalities together, and you should see that, in fact, $Pi_C(x^ast) = Pi_H(x^ast)$!
answered Nov 26 at 5:01
lntls
666
Oh boy. Perfect. Please keep answering. Can you look at the extension of the question?
– Saeed
Nov 26 at 5:50
Sure, but it seems that the community doesn't allow follow up questions to another question.(meta.stackoverflow.com/questions/296489/…) Do you mind creating a separate question with your extension, and linking to this question?
– lntls
Nov 26 at 7:32
I've provided a sketch that may be clear for the follow up here: math.stackexchange.com/questions/3014057/…. Please comment in the link provided if it is not clear.
– lntls
Nov 26 at 8:53
Can you help me for this question?
– Saeed
Nov 29 at 15:20
Could you help me to show this question: math.stackexchange.com/questions/3023525/…
– Saeed
Dec 3 at 2:27
add a comment |
Note that $Pi_C(x^ast) in Csubseteq H$, and $Pi_H(x^ast) in H$. Then by construction of $H$,
$$0 geq langle x^ast - Pi_C(x^ast),Pi_H(x^ast) - Pi_C(x^ast)rangle = langle Pi_C(x^ast) - x^ast, Pi_C(x^ast) - Pi_H(x^ast) rangle,$$
and by the variational inequality of Euclidean distance
$$0 geq langle x^ast - Pi_H(x^ast),Pi_C(x^ast) - Pi_H(x^ast)rangle.$$
So add the two inequalities together, and you should see that, in fact, $Pi_C(x^ast) = Pi_H(x^ast)$!
answered Nov 26 at 5:01
lntls
666
Note that $Pi_C(x^ast) in Csubseteq H$, and $Pi_H(x^ast) in H$. Then by construction of $H$,
$$0 geq langle x^ast - Pi_C(x^ast),Pi_H(x^ast) - Pi_C(x^ast)rangle = langle Pi_C(x^ast) - x^ast, Pi_C(x^ast) - Pi_H(x^ast) rangle,$$
and by the variational inequality of Euclidean distance
$$0 geq langle x^ast - Pi_H(x^ast),Pi_C(x^ast) - Pi_H(x^ast)rangle.$$
So add the two inequalities together, and you should see that, in fact, $Pi_C(x^ast) = Pi_H(x^ast)$!
answered Nov 26 at 5:01
lntls
666
edited Nov 26 at 5:06
answered Nov 26 at 5:01
lntls
666
answered Nov 26 at 5:01
lntls
666
answered Nov 26 at 5:01
lntls
666
666
Oh boy. Perfect. Please keep answering. Can you look at the extension of the question?
– Saeed
Nov 26 at 5:50
Sure, but it seems that the community doesn't allow follow up questions to another question.(meta.stackoverflow.com/questions/296489/…) Do you mind creating a separate question with your extension, and linking to this question?
– lntls
Nov 26 at 7:32
I've provided a sketch that may be clear for the follow up here: math.stackexchange.com/questions/3014057/…. Please comment in the link provided if it is not clear.
– lntls
Nov 26 at 8:53
Can you help me for this question?
– Saeed
Nov 29 at 15:20
Could you help me to show this question: math.stackexchange.com/questions/3023525/…
– Saeed
Dec 3 at 2:27
add a comment |
Oh boy. Perfect. Please keep answering. Can you look at the extension of the question?
– Saeed
Nov 26 at 5:50
Sure, but it seems that the community doesn't allow follow up questions to another question.(meta.stackoverflow.com/questions/296489/…) Do you mind creating a separate question with your extension, and linking to this question?
– lntls
Nov 26 at 7:32
I've provided a sketch that may be clear for the follow up here: math.stackexchange.com/questions/3014057/…. Please comment in the link provided if it is not clear.
– lntls
Nov 26 at 8:53
Can you help me for this question?
– Saeed
Nov 29 at 15:20
Could you help me to show this question: math.stackexchange.com/questions/3023525/…
– Saeed
Dec 3 at 2:27
Oh boy. Perfect. Please keep answering. Can you look at the extension of the question?
– Saeed
Nov 26 at 5:50
Oh boy. Perfect. Please keep answering. Can you look at the extension of the question?
– Saeed
Nov 26 at 5:50
Sure, but it seems that the community doesn't allow follow up questions to another question.(meta.stackoverflow.com/questions/296489/…) Do you mind creating a separate question with your extension, and linking to this question?
– lntls
Nov 26 at 7:32
Sure, but it seems that the community doesn't allow follow up questions to another question.(meta.stackoverflow.com/questions/296489/…) Do you mind creating a separate question with your extension, and linking to this question?
– lntls
Nov 26 at 7:32
I've provided a sketch that may be clear for the follow up here: math.stackexchange.com/questions/3014057/…. Please comment in the link provided if it is not clear.
– lntls
Nov 26 at 8:53
I've provided a sketch that may be clear for the follow up here: math.stackexchange.com/questions/3014057/…. Please comment in the link provided if it is not clear.
– lntls
Nov 26 at 8:53
Can you help me for this question?
– Saeed
Nov 29 at 15:20
Can you help me for this question?
– Saeed
Nov 29 at 15:20
Could you help me to show this question: math.stackexchange.com/questions/3023525/…
– Saeed
Dec 3 at 2:27
Could you help me to show this question: math.stackexchange.com/questions/3023525/…
– Saeed
Dec 3 at 2:27
add a comment |
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