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Cartesian product of reflexive spaces is reflexive
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$begingroup$
Given $(E,||_E),(F,||_F)$ reflexive normed vector spaces.
I have to prove that also $(Etimes F,||_{Etimes F})$ is reflexive where $||_{Etimes F}$ is the product norm.
What I know is that $(Etimes F)'$ is algebrically and topologically isomorphic to $E'times F'$.
functional-analysis normed-spaces
edited Sep 25 '16 at 2:45
Pedro
10.9k23475
asked Jun 12 '14 at 14:03
avati91avati91
1,0071024
$endgroup$
add a comment |
$begingroup$
Given $(E,||_E),(F,||_F)$ reflexive normed vector spaces.
I have to prove that also $(Etimes F,||_{Etimes F})$ is reflexive where $||_{Etimes F}$ is the product norm.
What I know is that $(Etimes F)'$ is algebrically and topologically isomorphic to $E'times F'$.
functional-analysis normed-spaces
edited Sep 25 '16 at 2:45
Pedro
10.9k23475
asked Jun 12 '14 at 14:03
avati91avati91
1,0071024
$endgroup$
$begingroup$
search over the site this question has already been answered
$endgroup$
– Norbert
Jun 12 '14 at 17:55
$begingroup$
Can you post the link of that question? I can't find it
$endgroup$
– avati91
Jun 12 '14 at 21:40
$begingroup$
@Well, I must confess I've not found it, but you should look at this post.
$endgroup$
– Norbert
Jun 12 '14 at 22:13
add a comment |
$begingroup$
Given $(E,||_E),(F,||_F)$ reflexive normed vector spaces.
I have to prove that also $(Etimes F,||_{Etimes F})$ is reflexive where $||_{Etimes F}$ is the product norm.
What I know is that $(Etimes F)'$ is algebrically and topologically isomorphic to $E'times F'$.
functional-analysis normed-spaces
edited Sep 25 '16 at 2:45
Pedro
10.9k23475
asked Jun 12 '14 at 14:03
avati91avati91
1,0071024
$endgroup$
Given $(E,||_E),(F,||_F)$ reflexive normed vector spaces.
I have to prove that also $(Etimes F,||_{Etimes F})$ is reflexive where $||_{Etimes F}$ is the product norm.
What I know is that $(Etimes F)'$ is algebrically and topologically isomorphic to $E'times F'$.
functional-analysis normed-spaces
functional-analysis normed-spaces
edited Sep 25 '16 at 2:45
Pedro
10.9k23475
asked Jun 12 '14 at 14:03
avati91avati91
1,0071024
edited Sep 25 '16 at 2:45
Pedro
10.9k23475
asked Jun 12 '14 at 14:03
avati91avati91
1,0071024
edited Sep 25 '16 at 2:45
Pedro
10.9k23475
edited Sep 25 '16 at 2:45
Pedro
10.9k23475
edited Sep 25 '16 at 2:45
Pedro
10.9k23475
10.9k23475
asked Jun 12 '14 at 14:03
avati91avati91
1,0071024
asked Jun 12 '14 at 14:03
avati91avati91
1,0071024
asked Jun 12 '14 at 14:03
avati91avati91
1,0071024
1,0071024
$begingroup$
search over the site this question has already been answered
$endgroup$
– Norbert
Jun 12 '14 at 17:55
$begingroup$
Can you post the link of that question? I can't find it
$endgroup$
– avati91
Jun 12 '14 at 21:40
$begingroup$
@Well, I must confess I've not found it, but you should look at this post.
$endgroup$
– Norbert
Jun 12 '14 at 22:13
add a comment |
$begingroup$
search over the site this question has already been answered
$endgroup$
– Norbert
Jun 12 '14 at 17:55
$begingroup$
Can you post the link of that question? I can't find it
$endgroup$
– avati91
Jun 12 '14 at 21:40
$begingroup$
@Well, I must confess I've not found it, but you should look at this post.
$endgroup$
– Norbert
Jun 12 '14 at 22:13
$begingroup$
search over the site this question has already been answered
$endgroup$
– Norbert
Jun 12 '14 at 17:55
$begingroup$
search over the site this question has already been answered
$endgroup$
– Norbert
Jun 12 '14 at 17:55
$begingroup$
Can you post the link of that question? I can't find it
$endgroup$
– avati91
Jun 12 '14 at 21:40
$begingroup$
Can you post the link of that question? I can't find it
$endgroup$
– avati91
Jun 12 '14 at 21:40
$begingroup$
@Well, I must confess I've not found it, but you should look at this post.
$endgroup$
– Norbert
Jun 12 '14 at 22:13
$begingroup$
@Well, I must confess I've not found it, but you should look at this post.
$endgroup$
– Norbert
Jun 12 '14 at 22:13
add a comment |
1 Answer
1
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$begingroup$
Let $J:(E'times F')to (Etimes F)'$ be defined by $(J(e,f))(x,y)=e(x)+f(y)$. We know that
$$J(E'times F')=(Etimes F)'.tag{$*$}$$
Take a bounded sequence $(x_n,y_n)$ in $Etimes F$. Then $(x_n)$ is bounded in $E$ and $(y_n)$ is bounded in $F$. Since $E$ and $F$ are reflexive, there are subsequences $(x_{n_k})$, $(y_{n_k})$ and vectors $xin E$, $yin F$ such that
$$x_{n_k}rightharpoonup xtext{ in }E,quad y_{n_k}rightharpoonup ytext{ in }F$$
and thus
$$(J(e,f))(x_{n_k},y_{n_k})=e(x_{n_k})+f(y_{n_k})to e(x)+f(y)=(J(e,f))(x,y),quadforall;(e,f)in E'times F'.$$
It follows from $(*)$ that
$$g(x_{n_k},y_{n_k})to g(x,y),quad forall; gin (Etimes F)'$$
and thus
$$(x_{n_k},y_{n_k})rightharpoonup (x,y)text{ in }Etimes F.$$
This shows that every bounded sequence in $Etimes F$ admits a weakly convergent subsequence. So, by the Eberlein-Šmulian theorem, $Etimes F$ is reflexive.
edited Apr 13 '17 at 12:20
Community♦
1
answered Sep 25 '16 at 2:45
PedroPedro
10.9k23475
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Let $J:(E'times F')to (Etimes F)'$ be defined by $(J(e,f))(x,y)=e(x)+f(y)$. We know that
$$J(E'times F')=(Etimes F)'.tag{$*$}$$
Take a bounded sequence $(x_n,y_n)$ in $Etimes F$. Then $(x_n)$ is bounded in $E$ and $(y_n)$ is bounded in $F$. Since $E$ and $F$ are reflexive, there are subsequences $(x_{n_k})$, $(y_{n_k})$ and vectors $xin E$, $yin F$ such that
$$x_{n_k}rightharpoonup xtext{ in }E,quad y_{n_k}rightharpoonup ytext{ in }F$$
and thus
$$(J(e,f))(x_{n_k},y_{n_k})=e(x_{n_k})+f(y_{n_k})to e(x)+f(y)=(J(e,f))(x,y),quadforall;(e,f)in E'times F'.$$
It follows from $(*)$ that
$$g(x_{n_k},y_{n_k})to g(x,y),quad forall; gin (Etimes F)'$$
and thus
$$(x_{n_k},y_{n_k})rightharpoonup (x,y)text{ in }Etimes F.$$
This shows that every bounded sequence in $Etimes F$ admits a weakly convergent subsequence. So, by the Eberlein-Šmulian theorem, $Etimes F$ is reflexive.
edited Apr 13 '17 at 12:20
Community♦
1
answered Sep 25 '16 at 2:45
PedroPedro
10.9k23475
$endgroup$
add a comment |
$begingroup$
Let $J:(E'times F')to (Etimes F)'$ be defined by $(J(e,f))(x,y)=e(x)+f(y)$. We know that
$$J(E'times F')=(Etimes F)'.tag{$*$}$$
Take a bounded sequence $(x_n,y_n)$ in $Etimes F$. Then $(x_n)$ is bounded in $E$ and $(y_n)$ is bounded in $F$. Since $E$ and $F$ are reflexive, there are subsequences $(x_{n_k})$, $(y_{n_k})$ and vectors $xin E$, $yin F$ such that
$$x_{n_k}rightharpoonup xtext{ in }E,quad y_{n_k}rightharpoonup ytext{ in }F$$
and thus
$$(J(e,f))(x_{n_k},y_{n_k})=e(x_{n_k})+f(y_{n_k})to e(x)+f(y)=(J(e,f))(x,y),quadforall;(e,f)in E'times F'.$$
It follows from $(*)$ that
$$g(x_{n_k},y_{n_k})to g(x,y),quad forall; gin (Etimes F)'$$
and thus
$$(x_{n_k},y_{n_k})rightharpoonup (x,y)text{ in }Etimes F.$$
This shows that every bounded sequence in $Etimes F$ admits a weakly convergent subsequence. So, by the Eberlein-Šmulian theorem, $Etimes F$ is reflexive.
edited Apr 13 '17 at 12:20
Community♦
1
answered Sep 25 '16 at 2:45
PedroPedro
10.9k23475
$endgroup$
add a comment |
$begingroup$
Let $J:(E'times F')to (Etimes F)'$ be defined by $(J(e,f))(x,y)=e(x)+f(y)$. We know that
$$J(E'times F')=(Etimes F)'.tag{$*$}$$
Take a bounded sequence $(x_n,y_n)$ in $Etimes F$. Then $(x_n)$ is bounded in $E$ and $(y_n)$ is bounded in $F$. Since $E$ and $F$ are reflexive, there are subsequences $(x_{n_k})$, $(y_{n_k})$ and vectors $xin E$, $yin F$ such that
$$x_{n_k}rightharpoonup xtext{ in }E,quad y_{n_k}rightharpoonup ytext{ in }F$$
and thus
$$(J(e,f))(x_{n_k},y_{n_k})=e(x_{n_k})+f(y_{n_k})to e(x)+f(y)=(J(e,f))(x,y),quadforall;(e,f)in E'times F'.$$
It follows from $(*)$ that
$$g(x_{n_k},y_{n_k})to g(x,y),quad forall; gin (Etimes F)'$$
and thus
$$(x_{n_k},y_{n_k})rightharpoonup (x,y)text{ in }Etimes F.$$
This shows that every bounded sequence in $Etimes F$ admits a weakly convergent subsequence. So, by the Eberlein-Šmulian theorem, $Etimes F$ is reflexive.
edited Apr 13 '17 at 12:20
Community♦
1
answered Sep 25 '16 at 2:45
PedroPedro
10.9k23475
$endgroup$
Let $J:(E'times F')to (Etimes F)'$ be defined by $(J(e,f))(x,y)=e(x)+f(y)$. We know that
$$J(E'times F')=(Etimes F)'.tag{$*$}$$
Take a bounded sequence $(x_n,y_n)$ in $Etimes F$. Then $(x_n)$ is bounded in $E$ and $(y_n)$ is bounded in $F$. Since $E$ and $F$ are reflexive, there are subsequences $(x_{n_k})$, $(y_{n_k})$ and vectors $xin E$, $yin F$ such that
$$x_{n_k}rightharpoonup xtext{ in }E,quad y_{n_k}rightharpoonup ytext{ in }F$$
and thus
$$(J(e,f))(x_{n_k},y_{n_k})=e(x_{n_k})+f(y_{n_k})to e(x)+f(y)=(J(e,f))(x,y),quadforall;(e,f)in E'times F'.$$
It follows from $(*)$ that
$$g(x_{n_k},y_{n_k})to g(x,y),quad forall; gin (Etimes F)'$$
and thus
$$(x_{n_k},y_{n_k})rightharpoonup (x,y)text{ in }Etimes F.$$
This shows that every bounded sequence in $Etimes F$ admits a weakly convergent subsequence. So, by the Eberlein-Šmulian theorem, $Etimes F$ is reflexive.
edited Apr 13 '17 at 12:20
Community♦
1
answered Sep 25 '16 at 2:45
PedroPedro
10.9k23475
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Sep 25 '16 at 2:45
PedroPedro
10.9k23475
answered Sep 25 '16 at 2:45
PedroPedro
10.9k23475
answered Sep 25 '16 at 2:45
PedroPedro
10.9k23475
10.9k23475
add a comment |
add a comment |
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$begingroup$
search over the site this question has already been answered
$endgroup$
– Norbert
Jun 12 '14 at 17:55
$begingroup$
Can you post the link of that question? I can't find it
$endgroup$
– avati91
Jun 12 '14 at 21:40
$begingroup$
@Well, I must confess I've not found it, but you should look at this post.
$endgroup$
– Norbert
Jun 12 '14 at 22:13