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Metrizability of a subset in the weak topology.
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Let $X$ be a Banach space (not reflexive). It is well-known that $(X,w)$, which is $X$ with its weak topology, is not metrizable if $X$ is infinite dimensional.
I want to know under which condition can a subset $Ssubset X$ be given a metric that is compatible with the weak topology $(S,w)$?
If $X$ is reflexive then, if I recall correctly, the norm-boundedness of $S$ is enough. However, I am dealing with a non-reflexive space $X=W^1,1(Omega)$ so the previous criterion is not applicable. What if I assume that $S$ is norm-compact? Would that be enough?
functional-analysis pde banach-spaces sobolev-spaces
asked Jul 16 at 11:40
BigbearZzz
5,70311344
 |Â
show 7 more comments
up vote
2
down vote
favorite
Let $X$ be a Banach space (not reflexive). It is well-known that $(X,w)$, which is $X$ with its weak topology, is not metrizable if $X$ is infinite dimensional.
I want to know under which condition can a subset $Ssubset X$ be given a metric that is compatible with the weak topology $(S,w)$?
If $X$ is reflexive then, if I recall correctly, the norm-boundedness of $S$ is enough. However, I am dealing with a non-reflexive space $X=W^1,1(Omega)$ so the previous criterion is not applicable. What if I assume that $S$ is norm-compact? Would that be enough?
functional-analysis pde banach-spaces sobolev-spaces
asked Jul 16 at 11:40
BigbearZzz
5,70311344
1
Norm-boundedness is not enough, even if $X$ is reflexive. You also need a countable separating set in $X'$. For that, separability of $X'$ suffices, but is not a necessary condition, separability of $X$ is enough. For non-reflexive $X$, the same argument yields metrisability of relatively weakly compact subsets, if $X'$ contains a countable separating set.
– Daniel Fischer♦
Jul 16 at 12:01
2
For a norm-compact $S$, the answer is always "yes, it's metrisable in the weak topology". For the restriction of the weak topology is a coarser Hausdorff topology on $S$ than the restriction of the norm topology, but that is compact, hence every coarser Hausdorff topology on $S$ is identical to it. So on norm-compact subsets, the weak topology and the norm topology coincide.
– Daniel Fischer♦
Jul 16 at 12:09
@DanielFischer Thank you, my first reaction to this problem was to look up theorems regarding metrizability like the one Rhys suggested below. It didn't occur to be that standard topological argument is enough to solve it.
– BigbearZzz
Jul 16 at 12:16
@DanielFischer May I ask 1 more related question? I suppose that if I only require $S$ to be precompact (in norm) then it would inherit the metric (which is equivalent to the weak topology on $S$) from its norm closure right? If I denote the said metric space by $(S,d)$, then would its metric completion coincide with its norm-closure in $X$.?
– BigbearZzz
Jul 16 at 13:50
The weak topology and the norm topology coincide on every subset of a compact set. But the completion depends not only on the induced topology, it depends on the used metric. If you use the restriction of a metric on $overlineS$ (which induces the topology), then $overlineS$ is a completion; if you use a different metric, the completion can be different.
– Daniel Fischer♦
Jul 16 at 13:58
 |Â
show 7 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $X$ be a Banach space (not reflexive). It is well-known that $(X,w)$, which is $X$ with its weak topology, is not metrizable if $X$ is infinite dimensional.
I want to know under which condition can a subset $Ssubset X$ be given a metric that is compatible with the weak topology $(S,w)$?
If $X$ is reflexive then, if I recall correctly, the norm-boundedness of $S$ is enough. However, I am dealing with a non-reflexive space $X=W^1,1(Omega)$ so the previous criterion is not applicable. What if I assume that $S$ is norm-compact? Would that be enough?
functional-analysis pde banach-spaces sobolev-spaces
asked Jul 16 at 11:40
BigbearZzz
5,70311344
Let $X$ be a Banach space (not reflexive). It is well-known that $(X,w)$, which is $X$ with its weak topology, is not metrizable if $X$ is infinite dimensional.
I want to know under which condition can a subset $Ssubset X$ be given a metric that is compatible with the weak topology $(S,w)$?
If $X$ is reflexive then, if I recall correctly, the norm-boundedness of $S$ is enough. However, I am dealing with a non-reflexive space $X=W^1,1(Omega)$ so the previous criterion is not applicable. What if I assume that $S$ is norm-compact? Would that be enough?
functional-analysis pde banach-spaces sobolev-spaces
asked Jul 16 at 11:40
BigbearZzz
5,70311344
asked Jul 16 at 11:40
BigbearZzz
5,70311344
asked Jul 16 at 11:40
BigbearZzz
5,70311344
asked Jul 16 at 11:40
BigbearZzz
5,70311344
5,70311344
1
Norm-boundedness is not enough, even if $X$ is reflexive. You also need a countable separating set in $X'$. For that, separability of $X'$ suffices, but is not a necessary condition, separability of $X$ is enough. For non-reflexive $X$, the same argument yields metrisability of relatively weakly compact subsets, if $X'$ contains a countable separating set.
– Daniel Fischer♦
Jul 16 at 12:01
2
For a norm-compact $S$, the answer is always "yes, it's metrisable in the weak topology". For the restriction of the weak topology is a coarser Hausdorff topology on $S$ than the restriction of the norm topology, but that is compact, hence every coarser Hausdorff topology on $S$ is identical to it. So on norm-compact subsets, the weak topology and the norm topology coincide.
– Daniel Fischer♦
Jul 16 at 12:09
@DanielFischer Thank you, my first reaction to this problem was to look up theorems regarding metrizability like the one Rhys suggested below. It didn't occur to be that standard topological argument is enough to solve it.
– BigbearZzz
Jul 16 at 12:16
@DanielFischer May I ask 1 more related question? I suppose that if I only require $S$ to be precompact (in norm) then it would inherit the metric (which is equivalent to the weak topology on $S$) from its norm closure right? If I denote the said metric space by $(S,d)$, then would its metric completion coincide with its norm-closure in $X$.?
– BigbearZzz
Jul 16 at 13:50
The weak topology and the norm topology coincide on every subset of a compact set. But the completion depends not only on the induced topology, it depends on the used metric. If you use the restriction of a metric on $overlineS$ (which induces the topology), then $overlineS$ is a completion; if you use a different metric, the completion can be different.
– Daniel Fischer♦
Jul 16 at 13:58
 |Â
show 7 more comments
1
Norm-boundedness is not enough, even if $X$ is reflexive. You also need a countable separating set in $X'$. For that, separability of $X'$ suffices, but is not a necessary condition, separability of $X$ is enough. For non-reflexive $X$, the same argument yields metrisability of relatively weakly compact subsets, if $X'$ contains a countable separating set.
– Daniel Fischer♦
Jul 16 at 12:01
2
For a norm-compact $S$, the answer is always "yes, it's metrisable in the weak topology". For the restriction of the weak topology is a coarser Hausdorff topology on $S$ than the restriction of the norm topology, but that is compact, hence every coarser Hausdorff topology on $S$ is identical to it. So on norm-compact subsets, the weak topology and the norm topology coincide.
– Daniel Fischer♦
Jul 16 at 12:09
@DanielFischer Thank you, my first reaction to this problem was to look up theorems regarding metrizability like the one Rhys suggested below. It didn't occur to be that standard topological argument is enough to solve it.
– BigbearZzz
Jul 16 at 12:16
@DanielFischer May I ask 1 more related question? I suppose that if I only require $S$ to be precompact (in norm) then it would inherit the metric (which is equivalent to the weak topology on $S$) from its norm closure right? If I denote the said metric space by $(S,d)$, then would its metric completion coincide with its norm-closure in $X$.?
– BigbearZzz
Jul 16 at 13:50
The weak topology and the norm topology coincide on every subset of a compact set. But the completion depends not only on the induced topology, it depends on the used metric. If you use the restriction of a metric on $overlineS$ (which induces the topology), then $overlineS$ is a completion; if you use a different metric, the completion can be different.
– Daniel Fischer♦
Jul 16 at 13:58
1
1
Norm-boundedness is not enough, even if $X$ is reflexive. You also need a countable separating set in $X'$. For that, separability of $X'$ suffices, but is not a necessary condition, separability of $X$ is enough. For non-reflexive $X$, the same argument yields metrisability of relatively weakly compact subsets, if $X'$ contains a countable separating set.
– Daniel Fischer♦
Jul 16 at 12:01
Norm-boundedness is not enough, even if $X$ is reflexive. You also need a countable separating set in $X'$. For that, separability of $X'$ suffices, but is not a necessary condition, separability of $X$ is enough. For non-reflexive $X$, the same argument yields metrisability of relatively weakly compact subsets, if $X'$ contains a countable separating set.
– Daniel Fischer♦
Jul 16 at 12:01
2
2
For a norm-compact $S$, the answer is always "yes, it's metrisable in the weak topology". For the restriction of the weak topology is a coarser Hausdorff topology on $S$ than the restriction of the norm topology, but that is compact, hence every coarser Hausdorff topology on $S$ is identical to it. So on norm-compact subsets, the weak topology and the norm topology coincide.
– Daniel Fischer♦
Jul 16 at 12:09
For a norm-compact $S$, the answer is always "yes, it's metrisable in the weak topology". For the restriction of the weak topology is a coarser Hausdorff topology on $S$ than the restriction of the norm topology, but that is compact, hence every coarser Hausdorff topology on $S$ is identical to it. So on norm-compact subsets, the weak topology and the norm topology coincide.
– Daniel Fischer♦
Jul 16 at 12:09
@DanielFischer Thank you, my first reaction to this problem was to look up theorems regarding metrizability like the one Rhys suggested below. It didn't occur to be that standard topological argument is enough to solve it.
– BigbearZzz
Jul 16 at 12:16
@DanielFischer Thank you, my first reaction to this problem was to look up theorems regarding metrizability like the one Rhys suggested below. It didn't occur to be that standard topological argument is enough to solve it.
– BigbearZzz
Jul 16 at 12:16
@DanielFischer May I ask 1 more related question? I suppose that if I only require $S$ to be precompact (in norm) then it would inherit the metric (which is equivalent to the weak topology on $S$) from its norm closure right? If I denote the said metric space by $(S,d)$, then would its metric completion coincide with its norm-closure in $X$.?
– BigbearZzz
Jul 16 at 13:50
@DanielFischer May I ask 1 more related question? I suppose that if I only require $S$ to be precompact (in norm) then it would inherit the metric (which is equivalent to the weak topology on $S$) from its norm closure right? If I denote the said metric space by $(S,d)$, then would its metric completion coincide with its norm-closure in $X$.?
– BigbearZzz
Jul 16 at 13:50
The weak topology and the norm topology coincide on every subset of a compact set. But the completion depends not only on the induced topology, it depends on the used metric. If you use the restriction of a metric on $overlineS$ (which induces the topology), then $overlineS$ is a completion; if you use a different metric, the completion can be different.
– Daniel Fischer♦
Jul 16 at 13:58
The weak topology and the norm topology coincide on every subset of a compact set. But the completion depends not only on the induced topology, it depends on the used metric. If you use the restriction of a metric on $overlineS$ (which induces the topology), then $overlineS$ is a completion; if you use a different metric, the completion can be different.
– Daniel Fischer♦
Jul 16 at 13:58
 |Â
show 7 more comments
1 Answer
1
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oldest
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up vote
2
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Reflexivity isn't the right kind of condition for metrizability of the weak topology. Instead we have the following result.
Lemma: Let $X$ be a normed space. The relative weak topology on the unit ball $B_X$ of $X$ is metrizable if and only if $X^*$ is separable.
In particular, if $X$ is reflexive but has non-separable dual then we can conclude that your first claim about boundedness of $S$ being enough in reflexive spaces is false.
It also follows from the lemma that no set with non-empty interior (in the norm topology) can be weakly metrizable if $X^*$ is not separable. If such a set were metrizable, we could fit a homeomorphic image of $B_X$ into it and then, since a subspace of a metrizable space is metrizable, we would have that $B_X$ is metrizable.
In the particular case $X = W^1,1(Omega)$ we have that $X^* = W^1,infty(Omega)$ is not separable and so $B_X$ (and hence any set with non-empty interior in the norm topology) is not metrizable for the weak topology.
answered Jul 16 at 12:00
Rhys Steele
5,6101828
Thanks, this is probably the theorem that I remember (minus the separability part). I am particularly interested in the case where $S$ is norm-compact, do you happen to know any result related to this case?
– BigbearZzz
Jul 16 at 12:07
1
@BigbearZzz I think Daniel Fischer's latest comment answers your question
– Rhys Steele
Jul 16 at 12:10
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Reflexivity isn't the right kind of condition for metrizability of the weak topology. Instead we have the following result.
Lemma: Let $X$ be a normed space. The relative weak topology on the unit ball $B_X$ of $X$ is metrizable if and only if $X^*$ is separable.
In particular, if $X$ is reflexive but has non-separable dual then we can conclude that your first claim about boundedness of $S$ being enough in reflexive spaces is false.
It also follows from the lemma that no set with non-empty interior (in the norm topology) can be weakly metrizable if $X^*$ is not separable. If such a set were metrizable, we could fit a homeomorphic image of $B_X$ into it and then, since a subspace of a metrizable space is metrizable, we would have that $B_X$ is metrizable.
In the particular case $X = W^1,1(Omega)$ we have that $X^* = W^1,infty(Omega)$ is not separable and so $B_X$ (and hence any set with non-empty interior in the norm topology) is not metrizable for the weak topology.
answered Jul 16 at 12:00
Rhys Steele
5,6101828
Thanks, this is probably the theorem that I remember (minus the separability part). I am particularly interested in the case where $S$ is norm-compact, do you happen to know any result related to this case?
– BigbearZzz
Jul 16 at 12:07
1
@BigbearZzz I think Daniel Fischer's latest comment answers your question
– Rhys Steele
Jul 16 at 12:10
add a comment |Â
up vote
2
down vote
Reflexivity isn't the right kind of condition for metrizability of the weak topology. Instead we have the following result.
Lemma: Let $X$ be a normed space. The relative weak topology on the unit ball $B_X$ of $X$ is metrizable if and only if $X^*$ is separable.
In particular, if $X$ is reflexive but has non-separable dual then we can conclude that your first claim about boundedness of $S$ being enough in reflexive spaces is false.
It also follows from the lemma that no set with non-empty interior (in the norm topology) can be weakly metrizable if $X^*$ is not separable. If such a set were metrizable, we could fit a homeomorphic image of $B_X$ into it and then, since a subspace of a metrizable space is metrizable, we would have that $B_X$ is metrizable.
In the particular case $X = W^1,1(Omega)$ we have that $X^* = W^1,infty(Omega)$ is not separable and so $B_X$ (and hence any set with non-empty interior in the norm topology) is not metrizable for the weak topology.
answered Jul 16 at 12:00
Rhys Steele
5,6101828
Thanks, this is probably the theorem that I remember (minus the separability part). I am particularly interested in the case where $S$ is norm-compact, do you happen to know any result related to this case?
– BigbearZzz
Jul 16 at 12:07
1
@BigbearZzz I think Daniel Fischer's latest comment answers your question
– Rhys Steele
Jul 16 at 12:10
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Reflexivity isn't the right kind of condition for metrizability of the weak topology. Instead we have the following result.
Lemma: Let $X$ be a normed space. The relative weak topology on the unit ball $B_X$ of $X$ is metrizable if and only if $X^*$ is separable.
In particular, if $X$ is reflexive but has non-separable dual then we can conclude that your first claim about boundedness of $S$ being enough in reflexive spaces is false.
It also follows from the lemma that no set with non-empty interior (in the norm topology) can be weakly metrizable if $X^*$ is not separable. If such a set were metrizable, we could fit a homeomorphic image of $B_X$ into it and then, since a subspace of a metrizable space is metrizable, we would have that $B_X$ is metrizable.
In the particular case $X = W^1,1(Omega)$ we have that $X^* = W^1,infty(Omega)$ is not separable and so $B_X$ (and hence any set with non-empty interior in the norm topology) is not metrizable for the weak topology.
answered Jul 16 at 12:00
Rhys Steele
5,6101828
Reflexivity isn't the right kind of condition for metrizability of the weak topology. Instead we have the following result.
Lemma: Let $X$ be a normed space. The relative weak topology on the unit ball $B_X$ of $X$ is metrizable if and only if $X^*$ is separable.
In particular, if $X$ is reflexive but has non-separable dual then we can conclude that your first claim about boundedness of $S$ being enough in reflexive spaces is false.
It also follows from the lemma that no set with non-empty interior (in the norm topology) can be weakly metrizable if $X^*$ is not separable. If such a set were metrizable, we could fit a homeomorphic image of $B_X$ into it and then, since a subspace of a metrizable space is metrizable, we would have that $B_X$ is metrizable.
In the particular case $X = W^1,1(Omega)$ we have that $X^* = W^1,infty(Omega)$ is not separable and so $B_X$ (and hence any set with non-empty interior in the norm topology) is not metrizable for the weak topology.
answered Jul 16 at 12:00
Rhys Steele
5,6101828
answered Jul 16 at 12:00
Rhys Steele
5,6101828
answered Jul 16 at 12:00
Rhys Steele
5,6101828
answered Jul 16 at 12:00
Rhys Steele
5,6101828
5,6101828
Thanks, this is probably the theorem that I remember (minus the separability part). I am particularly interested in the case where $S$ is norm-compact, do you happen to know any result related to this case?
– BigbearZzz
Jul 16 at 12:07
1
@BigbearZzz I think Daniel Fischer's latest comment answers your question
– Rhys Steele
Jul 16 at 12:10
add a comment |Â
Thanks, this is probably the theorem that I remember (minus the separability part). I am particularly interested in the case where $S$ is norm-compact, do you happen to know any result related to this case?
– BigbearZzz
Jul 16 at 12:07
1
@BigbearZzz I think Daniel Fischer's latest comment answers your question
– Rhys Steele
Jul 16 at 12:10
Thanks, this is probably the theorem that I remember (minus the separability part). I am particularly interested in the case where $S$ is norm-compact, do you happen to know any result related to this case?
– BigbearZzz
Jul 16 at 12:07
Thanks, this is probably the theorem that I remember (minus the separability part). I am particularly interested in the case where $S$ is norm-compact, do you happen to know any result related to this case?
– BigbearZzz
Jul 16 at 12:07
1
1
@BigbearZzz I think Daniel Fischer's latest comment answers your question
– Rhys Steele
Jul 16 at 12:10
@BigbearZzz I think Daniel Fischer's latest comment answers your question
– Rhys Steele
Jul 16 at 12:10
add a comment |Â
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1
Norm-boundedness is not enough, even if $X$ is reflexive. You also need a countable separating set in $X'$. For that, separability of $X'$ suffices, but is not a necessary condition, separability of $X$ is enough. For non-reflexive $X$, the same argument yields metrisability of relatively weakly compact subsets, if $X'$ contains a countable separating set.
– Daniel Fischer♦
Jul 16 at 12:01
2
For a norm-compact $S$, the answer is always "yes, it's metrisable in the weak topology". For the restriction of the weak topology is a coarser Hausdorff topology on $S$ than the restriction of the norm topology, but that is compact, hence every coarser Hausdorff topology on $S$ is identical to it. So on norm-compact subsets, the weak topology and the norm topology coincide.
– Daniel Fischer♦
Jul 16 at 12:09
@DanielFischer Thank you, my first reaction to this problem was to look up theorems regarding metrizability like the one Rhys suggested below. It didn't occur to be that standard topological argument is enough to solve it.
– BigbearZzz
Jul 16 at 12:16
@DanielFischer May I ask 1 more related question? I suppose that if I only require $S$ to be precompact (in norm) then it would inherit the metric (which is equivalent to the weak topology on $S$) from its norm closure right? If I denote the said metric space by $(S,d)$, then would its metric completion coincide with its norm-closure in $X$.?
– BigbearZzz
Jul 16 at 13:50
The weak topology and the norm topology coincide on every subset of a compact set. But the completion depends not only on the induced topology, it depends on the used metric. If you use the restriction of a metric on $overlineS$ (which induces the topology), then $overlineS$ is a completion; if you use a different metric, the completion can be different.
– Daniel Fischer♦
Jul 16 at 13:58