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Finite dimensional subspace is weak star closed
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I want to show the weak star closed convex hull of a finite set of points is contained in the linear span of those points.
It's enough to show that any finite dimensional subspace $V$ of a Banach space $Z$ is weak star closed in $Z$. Since $V$ is finite dimensional, it is a closed subspace of $Z$ in the norm topology. How can I show that it is also weak star closed? Also, it this result true for arbitrary subspaces?
functional-analysis
asked Aug 6 at 21:49
Tom Chalmer
271212
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up vote
1
down vote
favorite
I want to show the weak star closed convex hull of a finite set of points is contained in the linear span of those points.
It's enough to show that any finite dimensional subspace $V$ of a Banach space $Z$ is weak star closed in $Z$. Since $V$ is finite dimensional, it is a closed subspace of $Z$ in the norm topology. How can I show that it is also weak star closed? Also, it this result true for arbitrary subspaces?
functional-analysis
asked Aug 6 at 21:49
Tom Chalmer
271212
Is $Z$ given as the dual space of another Banach space? Otherwise what do you mean by weak star?
– Robert Israel
Aug 6 at 22:48
Yes, if $Z=X^*$ for some Banach space $X$
– Tom Chalmer
Aug 7 at 21:27
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to show the weak star closed convex hull of a finite set of points is contained in the linear span of those points.
It's enough to show that any finite dimensional subspace $V$ of a Banach space $Z$ is weak star closed in $Z$. Since $V$ is finite dimensional, it is a closed subspace of $Z$ in the norm topology. How can I show that it is also weak star closed? Also, it this result true for arbitrary subspaces?
functional-analysis
asked Aug 6 at 21:49
Tom Chalmer
271212
I want to show the weak star closed convex hull of a finite set of points is contained in the linear span of those points.
It's enough to show that any finite dimensional subspace $V$ of a Banach space $Z$ is weak star closed in $Z$. Since $V$ is finite dimensional, it is a closed subspace of $Z$ in the norm topology. How can I show that it is also weak star closed? Also, it this result true for arbitrary subspaces?
functional-analysis
asked Aug 6 at 21:49
Tom Chalmer
271212
asked Aug 6 at 21:49
Tom Chalmer
271212
asked Aug 6 at 21:49
Tom Chalmer
271212
asked Aug 6 at 21:49
Tom Chalmer
271212
271212
Is $Z$ given as the dual space of another Banach space? Otherwise what do you mean by weak star?
– Robert Israel
Aug 6 at 22:48
Yes, if $Z=X^*$ for some Banach space $X$
– Tom Chalmer
Aug 7 at 21:27
add a comment |Â
Is $Z$ given as the dual space of another Banach space? Otherwise what do you mean by weak star?
– Robert Israel
Aug 6 at 22:48
Yes, if $Z=X^*$ for some Banach space $X$
– Tom Chalmer
Aug 7 at 21:27
Is $Z$ given as the dual space of another Banach space? Otherwise what do you mean by weak star?
– Robert Israel
Aug 6 at 22:48
Is $Z$ given as the dual space of another Banach space? Otherwise what do you mean by weak star?
– Robert Israel
Aug 6 at 22:48
Yes, if $Z=X^*$ for some Banach space $X$
– Tom Chalmer
Aug 7 at 21:27
Yes, if $Z=X^*$ for some Banach space $X$
– Tom Chalmer
Aug 7 at 21:27
add a comment |Â
2 Answers
2
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2
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All finite diemensional subspaces of Hausdorff topological vector spaces are closed. This is theorem 1.21 in Rudin's Functional Analysis.
answered Aug 7 at 6:58
Jochen
6,320722
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up vote
1
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This is true not only for arbitrary subspaces of Banach spaces, but in fact for arbitrary subspaces of locally convex spaces. For any locally convex space $X$, the weak and original closures of any convex set $E subset X$ are the same (see, for instance, Theorem 3.12 here). Since Banach spaces are locally convex and subspaces of any space are convex, this shows that the weak and original closures of any subspace of a Banach space are the same.
answered Aug 6 at 22:03
Alex Nolte
2,1281419
1
If the OP meant weak-* (presumably relative to a Banach space $X$ such that $Z = X^*$), this is not relevant.
– Robert Israel
Aug 6 at 22:50
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
All finite diemensional subspaces of Hausdorff topological vector spaces are closed. This is theorem 1.21 in Rudin's Functional Analysis.
answered Aug 7 at 6:58
Jochen
6,320722
add a comment |Â
up vote
2
down vote
All finite diemensional subspaces of Hausdorff topological vector spaces are closed. This is theorem 1.21 in Rudin's Functional Analysis.
answered Aug 7 at 6:58
Jochen
6,320722
add a comment |Â
up vote
2
down vote
up vote
2
down vote
All finite diemensional subspaces of Hausdorff topological vector spaces are closed. This is theorem 1.21 in Rudin's Functional Analysis.
answered Aug 7 at 6:58
Jochen
6,320722
All finite diemensional subspaces of Hausdorff topological vector spaces are closed. This is theorem 1.21 in Rudin's Functional Analysis.
answered Aug 7 at 6:58
Jochen
6,320722
answered Aug 7 at 6:58
Jochen
6,320722
answered Aug 7 at 6:58
Jochen
6,320722
answered Aug 7 at 6:58
Jochen
6,320722
6,320722
add a comment |Â
add a comment |Â
up vote
1
down vote
This is true not only for arbitrary subspaces of Banach spaces, but in fact for arbitrary subspaces of locally convex spaces. For any locally convex space $X$, the weak and original closures of any convex set $E subset X$ are the same (see, for instance, Theorem 3.12 here). Since Banach spaces are locally convex and subspaces of any space are convex, this shows that the weak and original closures of any subspace of a Banach space are the same.
answered Aug 6 at 22:03
Alex Nolte
2,1281419
1
If the OP meant weak-* (presumably relative to a Banach space $X$ such that $Z = X^*$), this is not relevant.
– Robert Israel
Aug 6 at 22:50
add a comment |Â
up vote
1
down vote
This is true not only for arbitrary subspaces of Banach spaces, but in fact for arbitrary subspaces of locally convex spaces. For any locally convex space $X$, the weak and original closures of any convex set $E subset X$ are the same (see, for instance, Theorem 3.12 here). Since Banach spaces are locally convex and subspaces of any space are convex, this shows that the weak and original closures of any subspace of a Banach space are the same.
answered Aug 6 at 22:03
Alex Nolte
2,1281419
1
If the OP meant weak-* (presumably relative to a Banach space $X$ such that $Z = X^*$), this is not relevant.
– Robert Israel
Aug 6 at 22:50
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is true not only for arbitrary subspaces of Banach spaces, but in fact for arbitrary subspaces of locally convex spaces. For any locally convex space $X$, the weak and original closures of any convex set $E subset X$ are the same (see, for instance, Theorem 3.12 here). Since Banach spaces are locally convex and subspaces of any space are convex, this shows that the weak and original closures of any subspace of a Banach space are the same.
answered Aug 6 at 22:03
Alex Nolte
2,1281419
This is true not only for arbitrary subspaces of Banach spaces, but in fact for arbitrary subspaces of locally convex spaces. For any locally convex space $X$, the weak and original closures of any convex set $E subset X$ are the same (see, for instance, Theorem 3.12 here). Since Banach spaces are locally convex and subspaces of any space are convex, this shows that the weak and original closures of any subspace of a Banach space are the same.
answered Aug 6 at 22:03
Alex Nolte
2,1281419
answered Aug 6 at 22:03
Alex Nolte
2,1281419
answered Aug 6 at 22:03
Alex Nolte
2,1281419
answered Aug 6 at 22:03
Alex Nolte
2,1281419
2,1281419
1
If the OP meant weak-* (presumably relative to a Banach space $X$ such that $Z = X^*$), this is not relevant.
– Robert Israel
Aug 6 at 22:50
add a comment |Â
1
If the OP meant weak-* (presumably relative to a Banach space $X$ such that $Z = X^*$), this is not relevant.
– Robert Israel
Aug 6 at 22:50
1
1
If the OP meant weak-* (presumably relative to a Banach space $X$ such that $Z = X^*$), this is not relevant.
– Robert Israel
Aug 6 at 22:50
If the OP meant weak-* (presumably relative to a Banach space $X$ such that $Z = X^*$), this is not relevant.
– Robert Israel
Aug 6 at 22:50
add a comment |Â
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Is $Z$ given as the dual space of another Banach space? Otherwise what do you mean by weak star?
– Robert Israel
Aug 6 at 22:48
Yes, if $Z=X^*$ for some Banach space $X$
– Tom Chalmer
Aug 7 at 21:27