Dual spaces of $ell^2$ of an arbitrary normed space
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Let $(V,|cdot|_V$ be a normed space, and consider $ell^2(V)$ to be the standard $l^2$ space using sequences in $V$ (instead of $mathbb{R})$. We can consider the dual space pair $(V,V')$. We know that $(ell^p,ell^q)$ are dual space pairs, when $(p,q)$ are conjugate pairs. What can we say about $ell^2(V')$, or $(ell^2(V))'$? Using the standard results of regular dual spaces, we know that these should be isometrically isomorphic. I would like to know why they are not.
real-analysis functional-analysis measure-theory
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user593295
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Let $(V,|cdot|_V$ be a normed space, and consider $ell^2(V)$ to be the standard $l^2$ space using sequences in $V$ (instead of $mathbb{R})$. We can consider the dual space pair $(V,V')$. We know that $(ell^p,ell^q)$ are dual space pairs, when $(p,q)$ are conjugate pairs. What can we say about $ell^2(V')$, or $(ell^2(V))'$? Using the standard results of regular dual spaces, we know that these should be isometrically isomorphic. I would like to know why they are not.
real-analysis functional-analysis measure-theory
asked yesterday
user593295
457
1
What do you mean by "Since we are now working in a normed space, rather a field."? Why do you think that these spaces are not isometrically isomorphic?
– gerw
yesterday
We don't really use a "dual space" of $mathbb{R}$ or $mathbb{C}$ that I know of. I guess that part doesn't matter for this question. As for the dual of $ell^2(V)$. I thought that is what it is, but I could not come up with/find a proof of it so I was skeptical. For instance $ell^2(V')$, is the space of all square summable linear functionals which is hard for me to work with.
– user593295
23 hours ago
1
You could try to mimic the standard proof and report, where it fails or where you have difficulties.
– daw
23 hours ago
1
@daw, I wouldn't think the standard proof for $(l^p,l^q)$ even apply. Note that the elements of $ell^2(X)'$ are the collection of all possible linear functionals of sequences that map into $R$. I don't have a good intuition for how this space looks.
– user593295
7 hours ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $(V,|cdot|_V$ be a normed space, and consider $ell^2(V)$ to be the standard $l^2$ space using sequences in $V$ (instead of $mathbb{R})$. We can consider the dual space pair $(V,V')$. We know that $(ell^p,ell^q)$ are dual space pairs, when $(p,q)$ are conjugate pairs. What can we say about $ell^2(V')$, or $(ell^2(V))'$? Using the standard results of regular dual spaces, we know that these should be isometrically isomorphic. I would like to know why they are not.
real-analysis functional-analysis measure-theory
asked yesterday
user593295
457
Let $(V,|cdot|_V$ be a normed space, and consider $ell^2(V)$ to be the standard $l^2$ space using sequences in $V$ (instead of $mathbb{R})$. We can consider the dual space pair $(V,V')$. We know that $(ell^p,ell^q)$ are dual space pairs, when $(p,q)$ are conjugate pairs. What can we say about $ell^2(V')$, or $(ell^2(V))'$? Using the standard results of regular dual spaces, we know that these should be isometrically isomorphic. I would like to know why they are not.
real-analysis functional-analysis measure-theory
real-analysis functional-analysis measure-theory
asked yesterday
user593295
457
asked yesterday
user593295
457
edited 23 hours ago
asked yesterday
user593295
457
asked yesterday
user593295
457
asked yesterday
user593295
457
457
1
What do you mean by "Since we are now working in a normed space, rather a field."? Why do you think that these spaces are not isometrically isomorphic?
– gerw
yesterday
We don't really use a "dual space" of $mathbb{R}$ or $mathbb{C}$ that I know of. I guess that part doesn't matter for this question. As for the dual of $ell^2(V)$. I thought that is what it is, but I could not come up with/find a proof of it so I was skeptical. For instance $ell^2(V')$, is the space of all square summable linear functionals which is hard for me to work with.
– user593295
23 hours ago
1
You could try to mimic the standard proof and report, where it fails or where you have difficulties.
– daw
23 hours ago
1
@daw, I wouldn't think the standard proof for $(l^p,l^q)$ even apply. Note that the elements of $ell^2(X)'$ are the collection of all possible linear functionals of sequences that map into $R$. I don't have a good intuition for how this space looks.
– user593295
7 hours ago
add a comment |
1
What do you mean by "Since we are now working in a normed space, rather a field."? Why do you think that these spaces are not isometrically isomorphic?
– gerw
yesterday
We don't really use a "dual space" of $mathbb{R}$ or $mathbb{C}$ that I know of. I guess that part doesn't matter for this question. As for the dual of $ell^2(V)$. I thought that is what it is, but I could not come up with/find a proof of it so I was skeptical. For instance $ell^2(V')$, is the space of all square summable linear functionals which is hard for me to work with.
– user593295
23 hours ago
1
You could try to mimic the standard proof and report, where it fails or where you have difficulties.
– daw
23 hours ago
1
@daw, I wouldn't think the standard proof for $(l^p,l^q)$ even apply. Note that the elements of $ell^2(X)'$ are the collection of all possible linear functionals of sequences that map into $R$. I don't have a good intuition for how this space looks.
– user593295
7 hours ago
1
1
What do you mean by "Since we are now working in a normed space, rather a field."? Why do you think that these spaces are not isometrically isomorphic?
– gerw
yesterday
What do you mean by "Since we are now working in a normed space, rather a field."? Why do you think that these spaces are not isometrically isomorphic?
– gerw
yesterday
We don't really use a "dual space" of $mathbb{R}$ or $mathbb{C}$ that I know of. I guess that part doesn't matter for this question. As for the dual of $ell^2(V)$. I thought that is what it is, but I could not come up with/find a proof of it so I was skeptical. For instance $ell^2(V')$, is the space of all square summable linear functionals which is hard for me to work with.
– user593295
23 hours ago
We don't really use a "dual space" of $mathbb{R}$ or $mathbb{C}$ that I know of. I guess that part doesn't matter for this question. As for the dual of $ell^2(V)$. I thought that is what it is, but I could not come up with/find a proof of it so I was skeptical. For instance $ell^2(V')$, is the space of all square summable linear functionals which is hard for me to work with.
– user593295
23 hours ago
1
1
You could try to mimic the standard proof and report, where it fails or where you have difficulties.
– daw
23 hours ago
You could try to mimic the standard proof and report, where it fails or where you have difficulties.
– daw
23 hours ago
1
1
@daw, I wouldn't think the standard proof for $(l^p,l^q)$ even apply. Note that the elements of $ell^2(X)'$ are the collection of all possible linear functionals of sequences that map into $R$. I don't have a good intuition for how this space looks.
– user593295
7 hours ago
@daw, I wouldn't think the standard proof for $(l^p,l^q)$ even apply. Note that the elements of $ell^2(X)'$ are the collection of all possible linear functionals of sequences that map into $R$. I don't have a good intuition for how this space looks.
– user593295
7 hours ago
add a comment |
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1
What do you mean by "Since we are now working in a normed space, rather a field."? Why do you think that these spaces are not isometrically isomorphic?
– gerw
yesterday
We don't really use a "dual space" of $mathbb{R}$ or $mathbb{C}$ that I know of. I guess that part doesn't matter for this question. As for the dual of $ell^2(V)$. I thought that is what it is, but I could not come up with/find a proof of it so I was skeptical. For instance $ell^2(V')$, is the space of all square summable linear functionals which is hard for me to work with.
– user593295
23 hours ago
1
You could try to mimic the standard proof and report, where it fails or where you have difficulties.
– daw
23 hours ago
1
@daw, I wouldn't think the standard proof for $(l^p,l^q)$ even apply. Note that the elements of $ell^2(X)'$ are the collection of all possible linear functionals of sequences that map into $R$. I don't have a good intuition for how this space looks.
– user593295
7 hours ago