Mixing
Closed cones and exposed faces
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I was wondering if a closed cone $C$ in a Banach space $X$ of dimension at least two always has an exposed face, that is, a face $F$ such that $F=Ccapkerphi$ for some positive $phiin X^*setminus{0}$.
Thanks a lot for your help!
functional-analysis convex-cone dual-cone
asked Nov 6 '17 at 7:21
Mark RoelandsMark Roelands
554
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add a comment |
$begingroup$
I was wondering if a closed cone $C$ in a Banach space $X$ of dimension at least two always has an exposed face, that is, a face $F$ such that $F=Ccapkerphi$ for some positive $phiin X^*setminus{0}$.
Thanks a lot for your help!
functional-analysis convex-cone dual-cone
asked Nov 6 '17 at 7:21
Mark RoelandsMark Roelands
554
$endgroup$
$begingroup$
What is a positive functional? $phi$ is linear....
$endgroup$
– daw
Nov 6 '17 at 7:23
$begingroup$
$phi$ is linear, continuous, and $phi(C)subseteq [0,infty)$.
$endgroup$
– Mark Roelands
Nov 6 '17 at 7:24
add a comment |
$begingroup$
I was wondering if a closed cone $C$ in a Banach space $X$ of dimension at least two always has an exposed face, that is, a face $F$ such that $F=Ccapkerphi$ for some positive $phiin X^*setminus{0}$.
Thanks a lot for your help!
functional-analysis convex-cone dual-cone
asked Nov 6 '17 at 7:21
Mark RoelandsMark Roelands
554
$endgroup$
I was wondering if a closed cone $C$ in a Banach space $X$ of dimension at least two always has an exposed face, that is, a face $F$ such that $F=Ccapkerphi$ for some positive $phiin X^*setminus{0}$.
Thanks a lot for your help!
functional-analysis convex-cone dual-cone
functional-analysis convex-cone dual-cone
asked Nov 6 '17 at 7:21
Mark RoelandsMark Roelands
554
asked Nov 6 '17 at 7:21
Mark RoelandsMark Roelands
554
asked Nov 6 '17 at 7:21
Mark RoelandsMark Roelands
554
asked Nov 6 '17 at 7:21
Mark RoelandsMark Roelands
554
asked Nov 6 '17 at 7:21
Mark RoelandsMark Roelands
554
554
$begingroup$
What is a positive functional? $phi$ is linear....
$endgroup$
– daw
Nov 6 '17 at 7:23
$begingroup$
$phi$ is linear, continuous, and $phi(C)subseteq [0,infty)$.
$endgroup$
– Mark Roelands
Nov 6 '17 at 7:24
add a comment |
$begingroup$
What is a positive functional? $phi$ is linear....
$endgroup$
– daw
Nov 6 '17 at 7:23
$begingroup$
$phi$ is linear, continuous, and $phi(C)subseteq [0,infty)$.
$endgroup$
– Mark Roelands
Nov 6 '17 at 7:24
$begingroup$
What is a positive functional? $phi$ is linear....
$endgroup$
– daw
Nov 6 '17 at 7:23
$begingroup$
What is a positive functional? $phi$ is linear....
$endgroup$
– daw
Nov 6 '17 at 7:23
$begingroup$
$phi$ is linear, continuous, and $phi(C)subseteq [0,infty)$.
$endgroup$
– Mark Roelands
Nov 6 '17 at 7:24
$begingroup$
$phi$ is linear, continuous, and $phi(C)subseteq [0,infty)$.
$endgroup$
– Mark Roelands
Nov 6 '17 at 7:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I was trying to find some information on exposed faces and found this problem, so in the category better late then never: The answer is yes, in fact, every element in the boundary lies in an exposed face.
Let $xinpartial C$. By the Hahn-Banach separation theorem there is a continuous linear functional $varphi$ and a $tinmathbb{R}$ such that
$$varphi(x)leq t<varphi(y)qquad yin C^{circ}.$$
Now let $yin C^{circ}$. Note that $lim_{nrightarrowinfty}varphi(frac{1}{n}y)=0$ so $tleq0$. Furthermore suppose that $varphi(y)<0$, then for $n$ large enough $varphi(ny)<tleq0$, which is a contradiction, so $varphi$ is a positive linear functional.
From this it follows that $emptysetneq F=text{ker}(varphi)cap Csubsetpartial C$. Finally let $x,yin C$ such that $lambda x+(1-lambda)yin F$, then $varphi(lambda x+(1-lambda)y)=0$ and as $varphi(x),varphi(y)geq0$ we find that $varphi(x)=varphi(y)=0$, so $F$ is a face.
answered Jan 29 at 11:04
Floris ClaassensFloris Claassens
1,21527
$endgroup$
$begingroup$
@ Floris Claassens. Thank you for your response! Yes, I understand why this would be true in case the cone in question has an interior. However, in general it is unclear to me why such an exposed face would exist.
$endgroup$
– Mark Roelands
Jan 30 at 16:24
add a comment |
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$begingroup$
I was trying to find some information on exposed faces and found this problem, so in the category better late then never: The answer is yes, in fact, every element in the boundary lies in an exposed face.
Let $xinpartial C$. By the Hahn-Banach separation theorem there is a continuous linear functional $varphi$ and a $tinmathbb{R}$ such that
$$varphi(x)leq t<varphi(y)qquad yin C^{circ}.$$
Now let $yin C^{circ}$. Note that $lim_{nrightarrowinfty}varphi(frac{1}{n}y)=0$ so $tleq0$. Furthermore suppose that $varphi(y)<0$, then for $n$ large enough $varphi(ny)<tleq0$, which is a contradiction, so $varphi$ is a positive linear functional.
From this it follows that $emptysetneq F=text{ker}(varphi)cap Csubsetpartial C$. Finally let $x,yin C$ such that $lambda x+(1-lambda)yin F$, then $varphi(lambda x+(1-lambda)y)=0$ and as $varphi(x),varphi(y)geq0$ we find that $varphi(x)=varphi(y)=0$, so $F$ is a face.
answered Jan 29 at 11:04
Floris ClaassensFloris Claassens
1,21527
$endgroup$
$begingroup$
@ Floris Claassens. Thank you for your response! Yes, I understand why this would be true in case the cone in question has an interior. However, in general it is unclear to me why such an exposed face would exist.
$endgroup$
– Mark Roelands
Jan 30 at 16:24
add a comment |
$begingroup$
I was trying to find some information on exposed faces and found this problem, so in the category better late then never: The answer is yes, in fact, every element in the boundary lies in an exposed face.
Let $xinpartial C$. By the Hahn-Banach separation theorem there is a continuous linear functional $varphi$ and a $tinmathbb{R}$ such that
$$varphi(x)leq t<varphi(y)qquad yin C^{circ}.$$
Now let $yin C^{circ}$. Note that $lim_{nrightarrowinfty}varphi(frac{1}{n}y)=0$ so $tleq0$. Furthermore suppose that $varphi(y)<0$, then for $n$ large enough $varphi(ny)<tleq0$, which is a contradiction, so $varphi$ is a positive linear functional.
From this it follows that $emptysetneq F=text{ker}(varphi)cap Csubsetpartial C$. Finally let $x,yin C$ such that $lambda x+(1-lambda)yin F$, then $varphi(lambda x+(1-lambda)y)=0$ and as $varphi(x),varphi(y)geq0$ we find that $varphi(x)=varphi(y)=0$, so $F$ is a face.
answered Jan 29 at 11:04
Floris ClaassensFloris Claassens
1,21527
$endgroup$
$begingroup$
@ Floris Claassens. Thank you for your response! Yes, I understand why this would be true in case the cone in question has an interior. However, in general it is unclear to me why such an exposed face would exist.
$endgroup$
– Mark Roelands
Jan 30 at 16:24
add a comment |
$begingroup$
I was trying to find some information on exposed faces and found this problem, so in the category better late then never: The answer is yes, in fact, every element in the boundary lies in an exposed face.
Let $xinpartial C$. By the Hahn-Banach separation theorem there is a continuous linear functional $varphi$ and a $tinmathbb{R}$ such that
$$varphi(x)leq t<varphi(y)qquad yin C^{circ}.$$
Now let $yin C^{circ}$. Note that $lim_{nrightarrowinfty}varphi(frac{1}{n}y)=0$ so $tleq0$. Furthermore suppose that $varphi(y)<0$, then for $n$ large enough $varphi(ny)<tleq0$, which is a contradiction, so $varphi$ is a positive linear functional.
From this it follows that $emptysetneq F=text{ker}(varphi)cap Csubsetpartial C$. Finally let $x,yin C$ such that $lambda x+(1-lambda)yin F$, then $varphi(lambda x+(1-lambda)y)=0$ and as $varphi(x),varphi(y)geq0$ we find that $varphi(x)=varphi(y)=0$, so $F$ is a face.
answered Jan 29 at 11:04
Floris ClaassensFloris Claassens
1,21527
$endgroup$
I was trying to find some information on exposed faces and found this problem, so in the category better late then never: The answer is yes, in fact, every element in the boundary lies in an exposed face.
Let $xinpartial C$. By the Hahn-Banach separation theorem there is a continuous linear functional $varphi$ and a $tinmathbb{R}$ such that
$$varphi(x)leq t<varphi(y)qquad yin C^{circ}.$$
Now let $yin C^{circ}$. Note that $lim_{nrightarrowinfty}varphi(frac{1}{n}y)=0$ so $tleq0$. Furthermore suppose that $varphi(y)<0$, then for $n$ large enough $varphi(ny)<tleq0$, which is a contradiction, so $varphi$ is a positive linear functional.
From this it follows that $emptysetneq F=text{ker}(varphi)cap Csubsetpartial C$. Finally let $x,yin C$ such that $lambda x+(1-lambda)yin F$, then $varphi(lambda x+(1-lambda)y)=0$ and as $varphi(x),varphi(y)geq0$ we find that $varphi(x)=varphi(y)=0$, so $F$ is a face.
answered Jan 29 at 11:04
Floris ClaassensFloris Claassens
1,21527
answered Jan 29 at 11:04
Floris ClaassensFloris Claassens
1,21527
answered Jan 29 at 11:04
Floris ClaassensFloris Claassens
1,21527
answered Jan 29 at 11:04
Floris ClaassensFloris Claassens
1,21527
1,21527
$begingroup$
@ Floris Claassens. Thank you for your response! Yes, I understand why this would be true in case the cone in question has an interior. However, in general it is unclear to me why such an exposed face would exist.
$endgroup$
– Mark Roelands
Jan 30 at 16:24
add a comment |
$begingroup$
@ Floris Claassens. Thank you for your response! Yes, I understand why this would be true in case the cone in question has an interior. However, in general it is unclear to me why such an exposed face would exist.
$endgroup$
– Mark Roelands
Jan 30 at 16:24
$begingroup$
@ Floris Claassens. Thank you for your response! Yes, I understand why this would be true in case the cone in question has an interior. However, in general it is unclear to me why such an exposed face would exist.
$endgroup$
– Mark Roelands
Jan 30 at 16:24
$begingroup$
@ Floris Claassens. Thank you for your response! Yes, I understand why this would be true in case the cone in question has an interior. However, in general it is unclear to me why such an exposed face would exist.
$endgroup$
– Mark Roelands
Jan 30 at 16:24
add a comment |
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$begingroup$
What is a positive functional? $phi$ is linear....
$endgroup$
– daw
Nov 6 '17 at 7:23
$begingroup$
$phi$ is linear, continuous, and $phi(C)subseteq [0,infty)$.
$endgroup$
– Mark Roelands
Nov 6 '17 at 7:24