Show minimum distance to a convex set is a convex function.












1












$begingroup$


Show that



$$
g(x)=inf_{z in C}|x-z|
$$
where $g:mathbb{R}^n rightarrow mathbb{R}$, $C$ is a convex set in $mathbb{R}^n$ (nor close neither bounded), and $|cdot|$ is a norm on $mathbb{R}^n$.
Let $x,y$ be in $mathbb{R}^n$. We need to show that



$$
g(lambda x +(1-lambda)y) leq lambda g(x)+ (1-lambda)g(y) tag{1}
$$



I tried the following:



$$
|lambda x +(1-lambda)y-z| leq lambda| x -z| + (1-lambda)| y-z| ,, forall {z in C}
$$
Since



$$
g(lambda x +(1-lambda)y)=inf_{z in C}|lambda x +(1-lambda)y-z| leq |lambda x +(1-lambda)y-z| ,, forall {z in C}
$$



So



$$
g(lambda x +(1-lambda)y)=inf_{z in C}|lambda x +(1-lambda)y-z| leq lambda| x -z| + (1-lambda)| y-z| ,, forall {z in C}
$$



I do not know how to handle the right hand side and apply infimum in a right way because the following is not correct in general



$$
inf_{z in C}|lambda x +(1-lambda)y-z| nleq lambda inf_{z in C} | x -z| + (1-lambda) inf_{z in C} | y-z|
$$



Or maybe my initial way to prove the convexity is wrong. Can you complete my proof or show the claim using another way?






convex-analysis convex-optimization convex-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hint: you can prove this for any jointly convex function $f(x,z)$, not just for $f(x,z) = ||x-z||$.
    $endgroup$
    – LinAlg
    Dec 2 '18 at 23:44
















1












$begingroup$


Show that



$$
g(x)=inf_{z in C}|x-z|
$$
where $g:mathbb{R}^n rightarrow mathbb{R}$, $C$ is a convex set in $mathbb{R}^n$ (nor close neither bounded), and $|cdot|$ is a norm on $mathbb{R}^n$.
Let $x,y$ be in $mathbb{R}^n$. We need to show that



$$
g(lambda x +(1-lambda)y) leq lambda g(x)+ (1-lambda)g(y) tag{1}
$$



I tried the following:



$$
|lambda x +(1-lambda)y-z| leq lambda| x -z| + (1-lambda)| y-z| ,, forall {z in C}
$$
Since



$$
g(lambda x +(1-lambda)y)=inf_{z in C}|lambda x +(1-lambda)y-z| leq |lambda x +(1-lambda)y-z| ,, forall {z in C}
$$



So



$$
g(lambda x +(1-lambda)y)=inf_{z in C}|lambda x +(1-lambda)y-z| leq lambda| x -z| + (1-lambda)| y-z| ,, forall {z in C}
$$



I do not know how to handle the right hand side and apply infimum in a right way because the following is not correct in general



$$
inf_{z in C}|lambda x +(1-lambda)y-z| nleq lambda inf_{z in C} | x -z| + (1-lambda) inf_{z in C} | y-z|
$$



Or maybe my initial way to prove the convexity is wrong. Can you complete my proof or show the claim using another way?






convex-analysis convex-optimization convex-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hint: you can prove this for any jointly convex function $f(x,z)$, not just for $f(x,z) = ||x-z||$.
    $endgroup$
    – LinAlg
    Dec 2 '18 at 23:44














1












1








1





$begingroup$


Show that



$$
g(x)=inf_{z in C}|x-z|
$$
where $g:mathbb{R}^n rightarrow mathbb{R}$, $C$ is a convex set in $mathbb{R}^n$ (nor close neither bounded), and $|cdot|$ is a norm on $mathbb{R}^n$.
Let $x,y$ be in $mathbb{R}^n$. We need to show that



$$
g(lambda x +(1-lambda)y) leq lambda g(x)+ (1-lambda)g(y) tag{1}
$$



I tried the following:



$$
|lambda x +(1-lambda)y-z| leq lambda| x -z| + (1-lambda)| y-z| ,, forall {z in C}
$$
Since



$$
g(lambda x +(1-lambda)y)=inf_{z in C}|lambda x +(1-lambda)y-z| leq |lambda x +(1-lambda)y-z| ,, forall {z in C}
$$



So



$$
g(lambda x +(1-lambda)y)=inf_{z in C}|lambda x +(1-lambda)y-z| leq lambda| x -z| + (1-lambda)| y-z| ,, forall {z in C}
$$



I do not know how to handle the right hand side and apply infimum in a right way because the following is not correct in general



$$
inf_{z in C}|lambda x +(1-lambda)y-z| nleq lambda inf_{z in C} | x -z| + (1-lambda) inf_{z in C} | y-z|
$$



Or maybe my initial way to prove the convexity is wrong. Can you complete my proof or show the claim using another way?






convex-analysis convex-optimization convex-geometry







share|cite|improve this question









$endgroup$




Show that



$$
g(x)=inf_{z in C}|x-z|
$$
where $g:mathbb{R}^n rightarrow mathbb{R}$, $C$ is a convex set in $mathbb{R}^n$ (nor close neither bounded), and $|cdot|$ is a norm on $mathbb{R}^n$.
Let $x,y$ be in $mathbb{R}^n$. We need to show that



$$
g(lambda x +(1-lambda)y) leq lambda g(x)+ (1-lambda)g(y) tag{1}
$$



I tried the following:



$$
|lambda x +(1-lambda)y-z| leq lambda| x -z| + (1-lambda)| y-z| ,, forall {z in C}
$$
Since



$$
g(lambda x +(1-lambda)y)=inf_{z in C}|lambda x +(1-lambda)y-z| leq |lambda x +(1-lambda)y-z| ,, forall {z in C}
$$



So



$$
g(lambda x +(1-lambda)y)=inf_{z in C}|lambda x +(1-lambda)y-z| leq lambda| x -z| + (1-lambda)| y-z| ,, forall {z in C}
$$



I do not know how to handle the right hand side and apply infimum in a right way because the following is not correct in general



$$
inf_{z in C}|lambda x +(1-lambda)y-z| nleq lambda inf_{z in C} | x -z| + (1-lambda) inf_{z in C} | y-z|
$$



Or maybe my initial way to prove the convexity is wrong. Can you complete my proof or show the claim using another way?






convex-analysis convex-optimization convex-geometry




convex-analysis convex-optimization convex-geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 '18 at 23:16









SaeedSaeed

928310




928310












  • $begingroup$
    Hint: you can prove this for any jointly convex function $f(x,z)$, not just for $f(x,z) = ||x-z||$.
    $endgroup$
    – LinAlg
    Dec 2 '18 at 23:44


















  • $begingroup$
    Hint: you can prove this for any jointly convex function $f(x,z)$, not just for $f(x,z) = ||x-z||$.
    $endgroup$
    – LinAlg
    Dec 2 '18 at 23:44
















$begingroup$
Hint: you can prove this for any jointly convex function $f(x,z)$, not just for $f(x,z) = ||x-z||$.
$endgroup$
– LinAlg
Dec 2 '18 at 23:44




$begingroup$
Hint: you can prove this for any jointly convex function $f(x,z)$, not just for $f(x,z) = ||x-z||$.
$endgroup$
– LinAlg
Dec 2 '18 at 23:44










1 Answer
1






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oldest

votes


















-1












$begingroup$

Hint: try starting from the opposite direction, consider the subadditivity of the infimum and see if you can show it that way






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your statement is not an answer, please write it as a comment and delete your answer.
    $endgroup$
    – Saeed
    Dec 3 '18 at 4:17











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1 Answer
1






active

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votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









-1












$begingroup$

Hint: try starting from the opposite direction, consider the subadditivity of the infimum and see if you can show it that way






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your statement is not an answer, please write it as a comment and delete your answer.
    $endgroup$
    – Saeed
    Dec 3 '18 at 4:17
















-1












$begingroup$

Hint: try starting from the opposite direction, consider the subadditivity of the infimum and see if you can show it that way






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your statement is not an answer, please write it as a comment and delete your answer.
    $endgroup$
    – Saeed
    Dec 3 '18 at 4:17














-1












-1








-1





$begingroup$

Hint: try starting from the opposite direction, consider the subadditivity of the infimum and see if you can show it that way






share|cite|improve this answer









$endgroup$



Hint: try starting from the opposite direction, consider the subadditivity of the infimum and see if you can show it that way







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 3 '18 at 2:51









svailsvail

12




12












  • $begingroup$
    Your statement is not an answer, please write it as a comment and delete your answer.
    $endgroup$
    – Saeed
    Dec 3 '18 at 4:17


















  • $begingroup$
    Your statement is not an answer, please write it as a comment and delete your answer.
    $endgroup$
    – Saeed
    Dec 3 '18 at 4:17
















$begingroup$
Your statement is not an answer, please write it as a comment and delete your answer.
$endgroup$
– Saeed
Dec 3 '18 at 4:17




$begingroup$
Your statement is not an answer, please write it as a comment and delete your answer.
$endgroup$
– Saeed
Dec 3 '18 at 4:17


















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