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Cartesian product of reflexive spaces is reflexive



Announcing the arrival of Valued Associate #679: Cesar Manara
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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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 25 '16 at 2:45









Pedro

10.9k23475




10.9k23475










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




















  • $begingroup$
    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












1 Answer
1






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oldest

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0












$begingroup$

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.






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    1 Answer
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    active

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    0












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 13 '17 at 12:20









        Community♦

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        1










        answered Sep 25 '16 at 2:45









        PedroPedro

        10.9k23475




        10.9k23475






























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