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Which topology is finer in dual spaces?
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Let $V$ and $H$ be Banach spaces and suppose that $$Vsubset H.$$
Which topology (either from $V$ or $H$) is finer? If topology induced by $H$ is finer, then is topology in $H^*$ finer than $V^*$? Can we ommit the assumption that $V$ and $H$ are Banach spaces and obtain the same relations?
I think that we should distinguish two cases. The first one is that $V$ has topology inherited from $H$, which means that it consists of all subsets from $V$ that has nonempty intersection with open sets from $H$. Then, topology in $V$ is at least as big as topology from $H$. Hence, topology in $H^*$ is finer than topology in $V^*$.
What about the second case - we have two topologies induced by different norms? Can we somehow compare those topologies? For instance $V=H^1(mathbbR)$ and $H=L^2(mathbbR)$.
general-topology functional-analysis
asked Aug 2 at 10:10
zorro47
495412
add a comment |Â
up vote
0
down vote
favorite
Let $V$ and $H$ be Banach spaces and suppose that $$Vsubset H.$$
Which topology (either from $V$ or $H$) is finer? If topology induced by $H$ is finer, then is topology in $H^*$ finer than $V^*$? Can we ommit the assumption that $V$ and $H$ are Banach spaces and obtain the same relations?
I think that we should distinguish two cases. The first one is that $V$ has topology inherited from $H$, which means that it consists of all subsets from $V$ that has nonempty intersection with open sets from $H$. Then, topology in $V$ is at least as big as topology from $H$. Hence, topology in $H^*$ is finer than topology in $V^*$.
What about the second case - we have two topologies induced by different norms? Can we somehow compare those topologies? For instance $V=H^1(mathbbR)$ and $H=L^2(mathbbR)$.
general-topology functional-analysis
asked Aug 2 at 10:10
zorro47
495412
Could you provide some context and tell us what you have done?
– Kavi Rama Murthy
Aug 2 at 10:19
I'm trying to understand $"H^*subset V^*$ and I found math.stackexchange.com/questions/655718/… But, I don't know much about topology.
– zorro47
Aug 2 at 10:24
In the post you have quoted it is shown that we cannot consider $H^*$ as a subspace of $V^*$ in general. So where is the question of comparing topologies on $H^*$?
– Kavi Rama Murthy
Aug 2 at 10:27
Exactly, but if $V$ is dense in $H$ than the restriction map is injective and hence, we may in some sense think of "$H^*=V^*$". But still, we write $H^*subset V^*$. According to the post, the inclusion may be in common with topologies.
– zorro47
Aug 2 at 10:31
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $V$ and $H$ be Banach spaces and suppose that $$Vsubset H.$$
Which topology (either from $V$ or $H$) is finer? If topology induced by $H$ is finer, then is topology in $H^*$ finer than $V^*$? Can we ommit the assumption that $V$ and $H$ are Banach spaces and obtain the same relations?
I think that we should distinguish two cases. The first one is that $V$ has topology inherited from $H$, which means that it consists of all subsets from $V$ that has nonempty intersection with open sets from $H$. Then, topology in $V$ is at least as big as topology from $H$. Hence, topology in $H^*$ is finer than topology in $V^*$.
What about the second case - we have two topologies induced by different norms? Can we somehow compare those topologies? For instance $V=H^1(mathbbR)$ and $H=L^2(mathbbR)$.
general-topology functional-analysis
asked Aug 2 at 10:10
zorro47
495412
Let $V$ and $H$ be Banach spaces and suppose that $$Vsubset H.$$
Which topology (either from $V$ or $H$) is finer? If topology induced by $H$ is finer, then is topology in $H^*$ finer than $V^*$? Can we ommit the assumption that $V$ and $H$ are Banach spaces and obtain the same relations?
I think that we should distinguish two cases. The first one is that $V$ has topology inherited from $H$, which means that it consists of all subsets from $V$ that has nonempty intersection with open sets from $H$. Then, topology in $V$ is at least as big as topology from $H$. Hence, topology in $H^*$ is finer than topology in $V^*$.
What about the second case - we have two topologies induced by different norms? Can we somehow compare those topologies? For instance $V=H^1(mathbbR)$ and $H=L^2(mathbbR)$.
general-topology functional-analysis
asked Aug 2 at 10:10
zorro47
495412
edited Aug 2 at 10:58
asked Aug 2 at 10:10
zorro47
495412
asked Aug 2 at 10:10
zorro47
495412
asked Aug 2 at 10:10
zorro47
495412
495412
Could you provide some context and tell us what you have done?
– Kavi Rama Murthy
Aug 2 at 10:19
I'm trying to understand $"H^*subset V^*$ and I found math.stackexchange.com/questions/655718/… But, I don't know much about topology.
– zorro47
Aug 2 at 10:24
In the post you have quoted it is shown that we cannot consider $H^*$ as a subspace of $V^*$ in general. So where is the question of comparing topologies on $H^*$?
– Kavi Rama Murthy
Aug 2 at 10:27
Exactly, but if $V$ is dense in $H$ than the restriction map is injective and hence, we may in some sense think of "$H^*=V^*$". But still, we write $H^*subset V^*$. According to the post, the inclusion may be in common with topologies.
– zorro47
Aug 2 at 10:31
add a comment |Â
Could you provide some context and tell us what you have done?
– Kavi Rama Murthy
Aug 2 at 10:19
I'm trying to understand $"H^*subset V^*$ and I found math.stackexchange.com/questions/655718/… But, I don't know much about topology.
– zorro47
Aug 2 at 10:24
In the post you have quoted it is shown that we cannot consider $H^*$ as a subspace of $V^*$ in general. So where is the question of comparing topologies on $H^*$?
– Kavi Rama Murthy
Aug 2 at 10:27
Exactly, but if $V$ is dense in $H$ than the restriction map is injective and hence, we may in some sense think of "$H^*=V^*$". But still, we write $H^*subset V^*$. According to the post, the inclusion may be in common with topologies.
– zorro47
Aug 2 at 10:31
Could you provide some context and tell us what you have done?
– Kavi Rama Murthy
Aug 2 at 10:19
Could you provide some context and tell us what you have done?
– Kavi Rama Murthy
Aug 2 at 10:19
I'm trying to understand $"H^*subset V^*$ and I found math.stackexchange.com/questions/655718/… But, I don't know much about topology.
– zorro47
Aug 2 at 10:24
I'm trying to understand $"H^*subset V^*$ and I found math.stackexchange.com/questions/655718/… But, I don't know much about topology.
– zorro47
Aug 2 at 10:24
In the post you have quoted it is shown that we cannot consider $H^*$ as a subspace of $V^*$ in general. So where is the question of comparing topologies on $H^*$?
– Kavi Rama Murthy
Aug 2 at 10:27
In the post you have quoted it is shown that we cannot consider $H^*$ as a subspace of $V^*$ in general. So where is the question of comparing topologies on $H^*$?
– Kavi Rama Murthy
Aug 2 at 10:27
Exactly, but if $V$ is dense in $H$ than the restriction map is injective and hence, we may in some sense think of "$H^*=V^*$". But still, we write $H^*subset V^*$. According to the post, the inclusion may be in common with topologies.
– zorro47
Aug 2 at 10:31
Exactly, but if $V$ is dense in $H$ than the restriction map is injective and hence, we may in some sense think of "$H^*=V^*$". But still, we write $H^*subset V^*$. According to the post, the inclusion may be in common with topologies.
– zorro47
Aug 2 at 10:31
add a comment |Â
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Could you provide some context and tell us what you have done?
– Kavi Rama Murthy
Aug 2 at 10:19
I'm trying to understand $"H^*subset V^*$ and I found math.stackexchange.com/questions/655718/… But, I don't know much about topology.
– zorro47
Aug 2 at 10:24
In the post you have quoted it is shown that we cannot consider $H^*$ as a subspace of $V^*$ in general. So where is the question of comparing topologies on $H^*$?
– Kavi Rama Murthy
Aug 2 at 10:27
Exactly, but if $V$ is dense in $H$ than the restriction map is injective and hence, we may in some sense think of "$H^*=V^*$". But still, we write $H^*subset V^*$. According to the post, the inclusion may be in common with topologies.
– zorro47
Aug 2 at 10:31