Non-trivial faces of the closed convex hull of a non-convex closed set with connected complement
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I'm trying to prove or disprove a problem, but I'm struggling to make headway. Any help is appreciated.
Suppose $X$ is a Hilbert Space, and $C subseteq X$ is closed, bounded, non-convex, and $X setminus C$ is connected. Does the boundary of $operatornameconv C$ necessarily contain a line segment?
The following are my thoughts on the problem:
The connectedness and openness of the complement gives us an open subset of the boundary of $operatornameconv C$ that doesn't intersect $C$. The Bishop-Phelps theorem tells us that support points are dense in the boundary of convex sets, so there must exist support points of $overlineoperatornameconv C$ that aren't contained in $C$.
If we choose one such point, by translation, we can say it is $0$ without loss of generality. If this point is not extreme in $overlineoperatornameconv C$, then we are done. However, such points can be extreme (e.g. take $C$ to be the standard orthonormal basis in $l^2$, in which case $0$ is such a point).
I decided to look at the tangent cone of $overlineoperatornameconv C$ from the point $0$. I then picked a support point $x$ of this cone, other than $0$, but close enough that we could guarantee that it would not be in $C$. The supporting hyperplane must support the set $overlineoperatornameconv C$ at $0$. I'm hoping to show that $x in overlineoperatornameconv C$.
This is where I get stuck. If the functional that supports at $x$ achieves its maximum on $C$, then I get to the result, but this may not be the case.
Again, any help is appreciated. Thanks in advance.
functional-analysis convex-analysis hilbert-spaces convex-hulls
asked Oct 29 '17 at 13:16
Theo Bendit
12.7k1944
 |Â
show 4 more comments
up vote
6
down vote
favorite
I'm trying to prove or disprove a problem, but I'm struggling to make headway. Any help is appreciated.
Suppose $X$ is a Hilbert Space, and $C subseteq X$ is closed, bounded, non-convex, and $X setminus C$ is connected. Does the boundary of $operatornameconv C$ necessarily contain a line segment?
The following are my thoughts on the problem:
The connectedness and openness of the complement gives us an open subset of the boundary of $operatornameconv C$ that doesn't intersect $C$. The Bishop-Phelps theorem tells us that support points are dense in the boundary of convex sets, so there must exist support points of $overlineoperatornameconv C$ that aren't contained in $C$.
If we choose one such point, by translation, we can say it is $0$ without loss of generality. If this point is not extreme in $overlineoperatornameconv C$, then we are done. However, such points can be extreme (e.g. take $C$ to be the standard orthonormal basis in $l^2$, in which case $0$ is such a point).
I decided to look at the tangent cone of $overlineoperatornameconv C$ from the point $0$. I then picked a support point $x$ of this cone, other than $0$, but close enough that we could guarantee that it would not be in $C$. The supporting hyperplane must support the set $overlineoperatornameconv C$ at $0$. I'm hoping to show that $x in overlineoperatornameconv C$.
This is where I get stuck. If the functional that supports at $x$ achieves its maximum on $C$, then I get to the result, but this may not be the case.
Again, any help is appreciated. Thanks in advance.
functional-analysis convex-analysis hilbert-spaces convex-hulls
asked Oct 29 '17 at 13:16
Theo Bendit
12.7k1944
@Desire: Ah yes, that's right. Boundedness is necessary. I'll edit to include that.
– Theo Bendit
Oct 29 '17 at 15:56
While I don't have a solution to this problem, I've found something of a work-around, so I'm less invested in the solution, but it'd be nice to validate (or invalidate) my intuition.
– Theo Bendit
Oct 29 '17 at 15:58
You said : "The connectedness and openness of the complement gives us an open subset of the boundary of $bartextconv(C) $ that doesn't intersect $C.$ " ?! what if $C$ just contains two points.?
– Red shoes
Oct 31 '17 at 0:34
@Redshoes Then $overlineoperatornameconvC$ is a line segment. The open line segment is open in the boundary (in the subspace topology), and doesn't intersect $C$.
– Theo Bendit
Oct 31 '17 at 0:46
OK .. I misunderstood it.
– Red shoes
Oct 31 '17 at 1:03
 |Â
show 4 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I'm trying to prove or disprove a problem, but I'm struggling to make headway. Any help is appreciated.
Suppose $X$ is a Hilbert Space, and $C subseteq X$ is closed, bounded, non-convex, and $X setminus C$ is connected. Does the boundary of $operatornameconv C$ necessarily contain a line segment?
The following are my thoughts on the problem:
The connectedness and openness of the complement gives us an open subset of the boundary of $operatornameconv C$ that doesn't intersect $C$. The Bishop-Phelps theorem tells us that support points are dense in the boundary of convex sets, so there must exist support points of $overlineoperatornameconv C$ that aren't contained in $C$.
If we choose one such point, by translation, we can say it is $0$ without loss of generality. If this point is not extreme in $overlineoperatornameconv C$, then we are done. However, such points can be extreme (e.g. take $C$ to be the standard orthonormal basis in $l^2$, in which case $0$ is such a point).
I decided to look at the tangent cone of $overlineoperatornameconv C$ from the point $0$. I then picked a support point $x$ of this cone, other than $0$, but close enough that we could guarantee that it would not be in $C$. The supporting hyperplane must support the set $overlineoperatornameconv C$ at $0$. I'm hoping to show that $x in overlineoperatornameconv C$.
This is where I get stuck. If the functional that supports at $x$ achieves its maximum on $C$, then I get to the result, but this may not be the case.
Again, any help is appreciated. Thanks in advance.
functional-analysis convex-analysis hilbert-spaces convex-hulls
asked Oct 29 '17 at 13:16
Theo Bendit
12.7k1944
I'm trying to prove or disprove a problem, but I'm struggling to make headway. Any help is appreciated.
Suppose $X$ is a Hilbert Space, and $C subseteq X$ is closed, bounded, non-convex, and $X setminus C$ is connected. Does the boundary of $operatornameconv C$ necessarily contain a line segment?
The following are my thoughts on the problem:
The connectedness and openness of the complement gives us an open subset of the boundary of $operatornameconv C$ that doesn't intersect $C$. The Bishop-Phelps theorem tells us that support points are dense in the boundary of convex sets, so there must exist support points of $overlineoperatornameconv C$ that aren't contained in $C$.
If we choose one such point, by translation, we can say it is $0$ without loss of generality. If this point is not extreme in $overlineoperatornameconv C$, then we are done. However, such points can be extreme (e.g. take $C$ to be the standard orthonormal basis in $l^2$, in which case $0$ is such a point).
I decided to look at the tangent cone of $overlineoperatornameconv C$ from the point $0$. I then picked a support point $x$ of this cone, other than $0$, but close enough that we could guarantee that it would not be in $C$. The supporting hyperplane must support the set $overlineoperatornameconv C$ at $0$. I'm hoping to show that $x in overlineoperatornameconv C$.
This is where I get stuck. If the functional that supports at $x$ achieves its maximum on $C$, then I get to the result, but this may not be the case.
Again, any help is appreciated. Thanks in advance.
functional-analysis convex-analysis hilbert-spaces convex-hulls
asked Oct 29 '17 at 13:16
Theo Bendit
12.7k1944
edited Aug 29 at 5:53
asked Oct 29 '17 at 13:16
Theo Bendit
12.7k1944
asked Oct 29 '17 at 13:16
Theo Bendit
12.7k1944
asked Oct 29 '17 at 13:16
Theo Bendit
12.7k1944
12.7k1944
@Desire: Ah yes, that's right. Boundedness is necessary. I'll edit to include that.
– Theo Bendit
Oct 29 '17 at 15:56
While I don't have a solution to this problem, I've found something of a work-around, so I'm less invested in the solution, but it'd be nice to validate (or invalidate) my intuition.
– Theo Bendit
Oct 29 '17 at 15:58
You said : "The connectedness and openness of the complement gives us an open subset of the boundary of $bartextconv(C) $ that doesn't intersect $C.$ " ?! what if $C$ just contains two points.?
– Red shoes
Oct 31 '17 at 0:34
@Redshoes Then $overlineoperatornameconvC$ is a line segment. The open line segment is open in the boundary (in the subspace topology), and doesn't intersect $C$.
– Theo Bendit
Oct 31 '17 at 0:46
OK .. I misunderstood it.
– Red shoes
Oct 31 '17 at 1:03
 |Â
show 4 more comments
@Desire: Ah yes, that's right. Boundedness is necessary. I'll edit to include that.
– Theo Bendit
Oct 29 '17 at 15:56
While I don't have a solution to this problem, I've found something of a work-around, so I'm less invested in the solution, but it'd be nice to validate (or invalidate) my intuition.
– Theo Bendit
Oct 29 '17 at 15:58
You said : "The connectedness and openness of the complement gives us an open subset of the boundary of $bartextconv(C) $ that doesn't intersect $C.$ " ?! what if $C$ just contains two points.?
– Red shoes
Oct 31 '17 at 0:34
@Redshoes Then $overlineoperatornameconvC$ is a line segment. The open line segment is open in the boundary (in the subspace topology), and doesn't intersect $C$.
– Theo Bendit
Oct 31 '17 at 0:46
OK .. I misunderstood it.
– Red shoes
Oct 31 '17 at 1:03
@Desire: Ah yes, that's right. Boundedness is necessary. I'll edit to include that.
– Theo Bendit
Oct 29 '17 at 15:56
@Desire: Ah yes, that's right. Boundedness is necessary. I'll edit to include that.
– Theo Bendit
Oct 29 '17 at 15:56
While I don't have a solution to this problem, I've found something of a work-around, so I'm less invested in the solution, but it'd be nice to validate (or invalidate) my intuition.
– Theo Bendit
Oct 29 '17 at 15:58
While I don't have a solution to this problem, I've found something of a work-around, so I'm less invested in the solution, but it'd be nice to validate (or invalidate) my intuition.
– Theo Bendit
Oct 29 '17 at 15:58
You said : "The connectedness and openness of the complement gives us an open subset of the boundary of $bartextconv(C) $ that doesn't intersect $C.$ " ?! what if $C$ just contains two points.?
– Red shoes
Oct 31 '17 at 0:34
You said : "The connectedness and openness of the complement gives us an open subset of the boundary of $bartextconv(C) $ that doesn't intersect $C.$ " ?! what if $C$ just contains two points.?
– Red shoes
Oct 31 '17 at 0:34
@Redshoes Then $overlineoperatornameconvC$ is a line segment. The open line segment is open in the boundary (in the subspace topology), and doesn't intersect $C$.
– Theo Bendit
Oct 31 '17 at 0:46
@Redshoes Then $overlineoperatornameconvC$ is a line segment. The open line segment is open in the boundary (in the subspace topology), and doesn't intersect $C$.
– Theo Bendit
Oct 31 '17 at 0:46
OK .. I misunderstood it.
– Red shoes
Oct 31 '17 at 1:03
OK .. I misunderstood it.
– Red shoes
Oct 31 '17 at 1:03
 |Â
show 4 more comments
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@Desire: Ah yes, that's right. Boundedness is necessary. I'll edit to include that.
– Theo Bendit
Oct 29 '17 at 15:56
While I don't have a solution to this problem, I've found something of a work-around, so I'm less invested in the solution, but it'd be nice to validate (or invalidate) my intuition.
– Theo Bendit
Oct 29 '17 at 15:58
You said : "The connectedness and openness of the complement gives us an open subset of the boundary of $bartextconv(C) $ that doesn't intersect $C.$ " ?! what if $C$ just contains two points.?
– Red shoes
Oct 31 '17 at 0:34
@Redshoes Then $overlineoperatornameconvC$ is a line segment. The open line segment is open in the boundary (in the subspace topology), and doesn't intersect $C$.
– Theo Bendit
Oct 31 '17 at 0:46
OK .. I misunderstood it.
– Red shoes
Oct 31 '17 at 1:03