Show that the distance of $x^ast$ to $C$ is the maximum of distances from $x^ast$ to closed half-spaces...












2












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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







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$endgroup$

















    2












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






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      share|cite|improve this question













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      edited Dec 17 '18 at 1:20







      lntls

















      asked Nov 26 '18 at 8:47









      lntlslntls

      666




      666






















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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.






          share|cite|improve this answer









          $endgroup$









          • 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













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          1 Answer
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          1 Answer
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          active

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          active

          oldest

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          2












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$









          • 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


















          2












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$









          • 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
















          2












          2








          2





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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
















          • 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




















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