Show minimum distance to a convex set is a convex function.
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
asked Dec 2 '18 at 23:16
Saeed
651310
add a comment |
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
asked Dec 2 '18 at 23:16
Saeed
651310
Hint: you can prove this for any jointly convex function $f(x,z)$, not just for $f(x,z) = ||x-z||$.
– LinAlg
Dec 2 '18 at 23:44
add a comment |
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
asked Dec 2 '18 at 23:16
Saeed
651310
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
asked Dec 2 '18 at 23:16
Saeed
651310
asked Dec 2 '18 at 23:16
Saeed
651310
asked Dec 2 '18 at 23:16
Saeed
651310
asked Dec 2 '18 at 23:16
Saeed
651310
asked Dec 2 '18 at 23:16
Saeed
651310
651310
Hint: you can prove this for any jointly convex function $f(x,z)$, not just for $f(x,z) = ||x-z||$.
– LinAlg
Dec 2 '18 at 23:44
add a comment |
Hint: you can prove this for any jointly convex function $f(x,z)$, not just for $f(x,z) = ||x-z||$.
– LinAlg
Dec 2 '18 at 23:44
Hint: you can prove this for any jointly convex function $f(x,z)$, not just for $f(x,z) = ||x-z||$.
– LinAlg
Dec 2 '18 at 23:44
Hint: you can prove this for any jointly convex function $f(x,z)$, not just for $f(x,z) = ||x-z||$.
– LinAlg
Dec 2 '18 at 23:44
add a comment |
1 Answer
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Hint: try starting from the opposite direction, consider the subadditivity of the infimum and see if you can show it that way
answered Dec 3 '18 at 2:51
svail
12
Your statement is not an answer, please write it as a comment and delete your answer.
– Saeed
Dec 3 '18 at 4:17
add a comment |
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Hint: try starting from the opposite direction, consider the subadditivity of the infimum and see if you can show it that way
answered Dec 3 '18 at 2:51
svail
12
Your statement is not an answer, please write it as a comment and delete your answer.
– Saeed
Dec 3 '18 at 4:17
add a comment |
Hint: try starting from the opposite direction, consider the subadditivity of the infimum and see if you can show it that way
answered Dec 3 '18 at 2:51
svail
12
Your statement is not an answer, please write it as a comment and delete your answer.
– Saeed
Dec 3 '18 at 4:17
add a comment |
Hint: try starting from the opposite direction, consider the subadditivity of the infimum and see if you can show it that way
answered Dec 3 '18 at 2:51
svail
12
Hint: try starting from the opposite direction, consider the subadditivity of the infimum and see if you can show it that way
answered Dec 3 '18 at 2:51
svail
12
answered Dec 3 '18 at 2:51
svail
12
answered Dec 3 '18 at 2:51
svail
12
answered Dec 3 '18 at 2:51
svail
12
12
Your statement is not an answer, please write it as a comment and delete your answer.
– Saeed
Dec 3 '18 at 4:17
add a comment |
Your statement is not an answer, please write it as a comment and delete your answer.
– Saeed
Dec 3 '18 at 4:17
Your statement is not an answer, please write it as a comment and delete your answer.
– Saeed
Dec 3 '18 at 4:17
Your statement is not an answer, please write it as a comment and delete your answer.
– Saeed
Dec 3 '18 at 4:17
add a comment |
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Hint: you can prove this for any jointly convex function $f(x,z)$, not just for $f(x,z) = ||x-z||$.
– LinAlg
Dec 2 '18 at 23:44