Is this bounded convex set in $mathbb{R}^n$ closed?
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Suppose we have a bounded set $C subset mathbb{R}^n$ that is convex and non-empty.
And suppose the family of linear functions $(f_{x})_{x in mathbb{R}^n}$ given by $f_{x}: C rightarrow mathbb{R}$, $ f_{x}(c) = x.c$ for $c in C$ attain their maximum and minimum in the set $C$.
Does this mean $C$ is closed (and hence compact) in $mathbb{R}^n$?
My idea: I think this does imply $C$ is closed but I am not sure how to write my argument "properly". For every vector $x in mathbb{R}^n$, there is a point in $C$ that is "furthest" in the direction of $x$. Then because we are in a convex set we can just "join up all these points" and our set is closed. But how do I write this formally.
Remark: Also is it true that if a linear function on a convex set attains its maximum/minimum it does so on the boundary?
convex-analysis convex-optimization compactness
asked Nov 14 at 16:25
Namch96
449314
add a comment |
up vote
2
down vote
favorite
Suppose we have a bounded set $C subset mathbb{R}^n$ that is convex and non-empty.
And suppose the family of linear functions $(f_{x})_{x in mathbb{R}^n}$ given by $f_{x}: C rightarrow mathbb{R}$, $ f_{x}(c) = x.c$ for $c in C$ attain their maximum and minimum in the set $C$.
Does this mean $C$ is closed (and hence compact) in $mathbb{R}^n$?
My idea: I think this does imply $C$ is closed but I am not sure how to write my argument "properly". For every vector $x in mathbb{R}^n$, there is a point in $C$ that is "furthest" in the direction of $x$. Then because we are in a convex set we can just "join up all these points" and our set is closed. But how do I write this formally.
Remark: Also is it true that if a linear function on a convex set attains its maximum/minimum it does so on the boundary?
convex-analysis convex-optimization compactness
asked Nov 14 at 16:25
Namch96
449314
Note that the argument used in this very interesting question math.stackexchange.com/questions/1434741/… may be of some help.....
– Namch96
Nov 14 at 16:25
Is $x.c$ the inner product between $x$ and $c$? The answer to the last question is yes due to the maximum principle.
– LinAlg
Nov 14 at 18:38
Yes, it is the inner product.
– Namch96
Nov 14 at 23:49
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose we have a bounded set $C subset mathbb{R}^n$ that is convex and non-empty.
And suppose the family of linear functions $(f_{x})_{x in mathbb{R}^n}$ given by $f_{x}: C rightarrow mathbb{R}$, $ f_{x}(c) = x.c$ for $c in C$ attain their maximum and minimum in the set $C$.
Does this mean $C$ is closed (and hence compact) in $mathbb{R}^n$?
My idea: I think this does imply $C$ is closed but I am not sure how to write my argument "properly". For every vector $x in mathbb{R}^n$, there is a point in $C$ that is "furthest" in the direction of $x$. Then because we are in a convex set we can just "join up all these points" and our set is closed. But how do I write this formally.
Remark: Also is it true that if a linear function on a convex set attains its maximum/minimum it does so on the boundary?
convex-analysis convex-optimization compactness
asked Nov 14 at 16:25
Namch96
449314
Suppose we have a bounded set $C subset mathbb{R}^n$ that is convex and non-empty.
And suppose the family of linear functions $(f_{x})_{x in mathbb{R}^n}$ given by $f_{x}: C rightarrow mathbb{R}$, $ f_{x}(c) = x.c$ for $c in C$ attain their maximum and minimum in the set $C$.
Does this mean $C$ is closed (and hence compact) in $mathbb{R}^n$?
My idea: I think this does imply $C$ is closed but I am not sure how to write my argument "properly". For every vector $x in mathbb{R}^n$, there is a point in $C$ that is "furthest" in the direction of $x$. Then because we are in a convex set we can just "join up all these points" and our set is closed. But how do I write this formally.
Remark: Also is it true that if a linear function on a convex set attains its maximum/minimum it does so on the boundary?
convex-analysis convex-optimization compactness
convex-analysis convex-optimization compactness
asked Nov 14 at 16:25
Namch96
449314
asked Nov 14 at 16:25
Namch96
449314
edited Nov 14 at 21:46
asked Nov 14 at 16:25
Namch96
449314
asked Nov 14 at 16:25
Namch96
449314
asked Nov 14 at 16:25
Namch96
449314
449314
Note that the argument used in this very interesting question math.stackexchange.com/questions/1434741/… may be of some help.....
– Namch96
Nov 14 at 16:25
Is $x.c$ the inner product between $x$ and $c$? The answer to the last question is yes due to the maximum principle.
– LinAlg
Nov 14 at 18:38
Yes, it is the inner product.
– Namch96
Nov 14 at 23:49
add a comment |
Note that the argument used in this very interesting question math.stackexchange.com/questions/1434741/… may be of some help.....
– Namch96
Nov 14 at 16:25
Is $x.c$ the inner product between $x$ and $c$? The answer to the last question is yes due to the maximum principle.
– LinAlg
Nov 14 at 18:38
Yes, it is the inner product.
– Namch96
Nov 14 at 23:49
Note that the argument used in this very interesting question math.stackexchange.com/questions/1434741/… may be of some help.....
– Namch96
Nov 14 at 16:25
Note that the argument used in this very interesting question math.stackexchange.com/questions/1434741/… may be of some help.....
– Namch96
Nov 14 at 16:25
Is $x.c$ the inner product between $x$ and $c$? The answer to the last question is yes due to the maximum principle.
– LinAlg
Nov 14 at 18:38
Is $x.c$ the inner product between $x$ and $c$? The answer to the last question is yes due to the maximum principle.
– LinAlg
Nov 14 at 18:38
Yes, it is the inner product.
– Namch96
Nov 14 at 23:49
Yes, it is the inner product.
– Namch96
Nov 14 at 23:49
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
I found a counterexample as I was trying to construct a proof. Let me explain my thought process.
Let $x'$ be on the boundary and assume $x'notin C$. Since $x'$ is on the boundary, it has a separating hyperplane. Let $c$ be perpendicular to that plane and consider the function $f(x)=c^Tx$. Since $x'$ is on the boundary, function values can get arbitrarily close to $c^Tx'$, but by construction, cannot exceed $c^Tx'$ (the separating hyperplane is an isocurve). Since $f$ attains its maximum on $C$, there is a point $x^* in C$ for which $c^Tx' = c^Tx^*$. So, $x^*$ and $x'$ share the same separating hyperplane.
The next step in my proof would be to construct a point on the other side of $x'$ that is in $C$ and reach a contradiction with convexity. This is where I had the aha moment.
Consider the following set, which is a closed rectangle and the first quadrant of a closed circle, minus the point where the rectangle meets the circle.
answered Nov 15 at 2:02
LinAlg
7,6241520
Very nice, thanks!
– Namch96
Nov 15 at 14:15
Yes, nice. I had gotten to the point of thinking about separating hyperplanes and started to doubt the claim but couldn't come up with a counterexample. Yours is excellent.
– JonathanZ
Nov 17 at 16:10
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I found a counterexample as I was trying to construct a proof. Let me explain my thought process.
Let $x'$ be on the boundary and assume $x'notin C$. Since $x'$ is on the boundary, it has a separating hyperplane. Let $c$ be perpendicular to that plane and consider the function $f(x)=c^Tx$. Since $x'$ is on the boundary, function values can get arbitrarily close to $c^Tx'$, but by construction, cannot exceed $c^Tx'$ (the separating hyperplane is an isocurve). Since $f$ attains its maximum on $C$, there is a point $x^* in C$ for which $c^Tx' = c^Tx^*$. So, $x^*$ and $x'$ share the same separating hyperplane.
The next step in my proof would be to construct a point on the other side of $x'$ that is in $C$ and reach a contradiction with convexity. This is where I had the aha moment.
Consider the following set, which is a closed rectangle and the first quadrant of a closed circle, minus the point where the rectangle meets the circle.
answered Nov 15 at 2:02
LinAlg
7,6241520
Very nice, thanks!
– Namch96
Nov 15 at 14:15
Yes, nice. I had gotten to the point of thinking about separating hyperplanes and started to doubt the claim but couldn't come up with a counterexample. Yours is excellent.
– JonathanZ
Nov 17 at 16:10
add a comment |
up vote
2
down vote
accepted
I found a counterexample as I was trying to construct a proof. Let me explain my thought process.
Let $x'$ be on the boundary and assume $x'notin C$. Since $x'$ is on the boundary, it has a separating hyperplane. Let $c$ be perpendicular to that plane and consider the function $f(x)=c^Tx$. Since $x'$ is on the boundary, function values can get arbitrarily close to $c^Tx'$, but by construction, cannot exceed $c^Tx'$ (the separating hyperplane is an isocurve). Since $f$ attains its maximum on $C$, there is a point $x^* in C$ for which $c^Tx' = c^Tx^*$. So, $x^*$ and $x'$ share the same separating hyperplane.
The next step in my proof would be to construct a point on the other side of $x'$ that is in $C$ and reach a contradiction with convexity. This is where I had the aha moment.
Consider the following set, which is a closed rectangle and the first quadrant of a closed circle, minus the point where the rectangle meets the circle.
answered Nov 15 at 2:02
LinAlg
7,6241520
Very nice, thanks!
– Namch96
Nov 15 at 14:15
Yes, nice. I had gotten to the point of thinking about separating hyperplanes and started to doubt the claim but couldn't come up with a counterexample. Yours is excellent.
– JonathanZ
Nov 17 at 16:10
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I found a counterexample as I was trying to construct a proof. Let me explain my thought process.
Let $x'$ be on the boundary and assume $x'notin C$. Since $x'$ is on the boundary, it has a separating hyperplane. Let $c$ be perpendicular to that plane and consider the function $f(x)=c^Tx$. Since $x'$ is on the boundary, function values can get arbitrarily close to $c^Tx'$, but by construction, cannot exceed $c^Tx'$ (the separating hyperplane is an isocurve). Since $f$ attains its maximum on $C$, there is a point $x^* in C$ for which $c^Tx' = c^Tx^*$. So, $x^*$ and $x'$ share the same separating hyperplane.
The next step in my proof would be to construct a point on the other side of $x'$ that is in $C$ and reach a contradiction with convexity. This is where I had the aha moment.
Consider the following set, which is a closed rectangle and the first quadrant of a closed circle, minus the point where the rectangle meets the circle.
answered Nov 15 at 2:02
LinAlg
7,6241520
I found a counterexample as I was trying to construct a proof. Let me explain my thought process.
Let $x'$ be on the boundary and assume $x'notin C$. Since $x'$ is on the boundary, it has a separating hyperplane. Let $c$ be perpendicular to that plane and consider the function $f(x)=c^Tx$. Since $x'$ is on the boundary, function values can get arbitrarily close to $c^Tx'$, but by construction, cannot exceed $c^Tx'$ (the separating hyperplane is an isocurve). Since $f$ attains its maximum on $C$, there is a point $x^* in C$ for which $c^Tx' = c^Tx^*$. So, $x^*$ and $x'$ share the same separating hyperplane.
The next step in my proof would be to construct a point on the other side of $x'$ that is in $C$ and reach a contradiction with convexity. This is where I had the aha moment.
Consider the following set, which is a closed rectangle and the first quadrant of a closed circle, minus the point where the rectangle meets the circle.
answered Nov 15 at 2:02
LinAlg
7,6241520
answered Nov 15 at 2:02
LinAlg
7,6241520
answered Nov 15 at 2:02
LinAlg
7,6241520
answered Nov 15 at 2:02
LinAlg
7,6241520
7,6241520
Very nice, thanks!
– Namch96
Nov 15 at 14:15
Yes, nice. I had gotten to the point of thinking about separating hyperplanes and started to doubt the claim but couldn't come up with a counterexample. Yours is excellent.
– JonathanZ
Nov 17 at 16:10
add a comment |
Very nice, thanks!
– Namch96
Nov 15 at 14:15
Yes, nice. I had gotten to the point of thinking about separating hyperplanes and started to doubt the claim but couldn't come up with a counterexample. Yours is excellent.
– JonathanZ
Nov 17 at 16:10
Very nice, thanks!
– Namch96
Nov 15 at 14:15
Very nice, thanks!
– Namch96
Nov 15 at 14:15
Yes, nice. I had gotten to the point of thinking about separating hyperplanes and started to doubt the claim but couldn't come up with a counterexample. Yours is excellent.
– JonathanZ
Nov 17 at 16:10
Yes, nice. I had gotten to the point of thinking about separating hyperplanes and started to doubt the claim but couldn't come up with a counterexample. Yours is excellent.
– JonathanZ
Nov 17 at 16:10
add a comment |
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Note that the argument used in this very interesting question math.stackexchange.com/questions/1434741/… may be of some help.....
– Namch96
Nov 14 at 16:25
Is $x.c$ the inner product between $x$ and $c$? The answer to the last question is yes due to the maximum principle.
– LinAlg
Nov 14 at 18:38
Yes, it is the inner product.
– Namch96
Nov 14 at 23:49