Prove the mapping $(x,y)mapsto x+y$ from $X times X to X$ is continuous when $X$ is given a weak topology.












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$begingroup$


Let X be a Banach space. Prove the mapping $(x,y)mapsto x+y$ from $X times X to X$ is continuous when $X$ is given a weak topology.



Could anybody give me some hints to start?






general-topology functional-analysis topological-vector-spaces product-space weak-topology







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$endgroup$












  • $begingroup$
    $X$ is any topological vector space? What is your definition of "a weak topology"?
    $endgroup$
    – Henno Brandsma
    Dec 11 '18 at 22:16










  • $begingroup$
    I am sorry, I forgot to write it: X is a Banach space and we define the weak topology as follows en.wikipedia.org/wiki/Weak_topology
    $endgroup$
    – Ross
    Dec 11 '18 at 22:41












  • $begingroup$
    Then it's basically trivial as addition commutes with linear functionals and $X$ has the weak topology wrt continuous functionals. Nets are the easiest way to see this. If these cannot be used, consider what the natural base for the weak topology on $X$ and its product on $X times X$ is.
    $endgroup$
    – Henno Brandsma
    Dec 12 '18 at 4:51


















0












$begingroup$


Let X be a Banach space. Prove the mapping $(x,y)mapsto x+y$ from $X times X to X$ is continuous when $X$ is given a weak topology.



Could anybody give me some hints to start?






general-topology functional-analysis topological-vector-spaces product-space weak-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    $X$ is any topological vector space? What is your definition of "a weak topology"?
    $endgroup$
    – Henno Brandsma
    Dec 11 '18 at 22:16










  • $begingroup$
    I am sorry, I forgot to write it: X is a Banach space and we define the weak topology as follows en.wikipedia.org/wiki/Weak_topology
    $endgroup$
    – Ross
    Dec 11 '18 at 22:41












  • $begingroup$
    Then it's basically trivial as addition commutes with linear functionals and $X$ has the weak topology wrt continuous functionals. Nets are the easiest way to see this. If these cannot be used, consider what the natural base for the weak topology on $X$ and its product on $X times X$ is.
    $endgroup$
    – Henno Brandsma
    Dec 12 '18 at 4:51
















0












0








0





$begingroup$


Let X be a Banach space. Prove the mapping $(x,y)mapsto x+y$ from $X times X to X$ is continuous when $X$ is given a weak topology.



Could anybody give me some hints to start?






general-topology functional-analysis topological-vector-spaces product-space weak-topology







share|cite|improve this question











$endgroup$




Let X be a Banach space. Prove the mapping $(x,y)mapsto x+y$ from $X times X to X$ is continuous when $X$ is given a weak topology.



Could anybody give me some hints to start?






general-topology functional-analysis topological-vector-spaces product-space weak-topology




general-topology functional-analysis topological-vector-spaces product-space weak-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 22:42







Ross

















asked Dec 11 '18 at 22:09









RossRoss

1237




1237












  • $begingroup$
    $X$ is any topological vector space? What is your definition of "a weak topology"?
    $endgroup$
    – Henno Brandsma
    Dec 11 '18 at 22:16










  • $begingroup$
    I am sorry, I forgot to write it: X is a Banach space and we define the weak topology as follows en.wikipedia.org/wiki/Weak_topology
    $endgroup$
    – Ross
    Dec 11 '18 at 22:41












  • $begingroup$
    Then it's basically trivial as addition commutes with linear functionals and $X$ has the weak topology wrt continuous functionals. Nets are the easiest way to see this. If these cannot be used, consider what the natural base for the weak topology on $X$ and its product on $X times X$ is.
    $endgroup$
    – Henno Brandsma
    Dec 12 '18 at 4:51




















  • $begingroup$
    $X$ is any topological vector space? What is your definition of "a weak topology"?
    $endgroup$
    – Henno Brandsma
    Dec 11 '18 at 22:16










  • $begingroup$
    I am sorry, I forgot to write it: X is a Banach space and we define the weak topology as follows en.wikipedia.org/wiki/Weak_topology
    $endgroup$
    – Ross
    Dec 11 '18 at 22:41












  • $begingroup$
    Then it's basically trivial as addition commutes with linear functionals and $X$ has the weak topology wrt continuous functionals. Nets are the easiest way to see this. If these cannot be used, consider what the natural base for the weak topology on $X$ and its product on $X times X$ is.
    $endgroup$
    – Henno Brandsma
    Dec 12 '18 at 4:51


















$begingroup$
$X$ is any topological vector space? What is your definition of "a weak topology"?
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:16




$begingroup$
$X$ is any topological vector space? What is your definition of "a weak topology"?
$endgroup$
– Henno Brandsma
Dec 11 '18 at 22:16












$begingroup$
I am sorry, I forgot to write it: X is a Banach space and we define the weak topology as follows en.wikipedia.org/wiki/Weak_topology
$endgroup$
– Ross
Dec 11 '18 at 22:41






$begingroup$
I am sorry, I forgot to write it: X is a Banach space and we define the weak topology as follows en.wikipedia.org/wiki/Weak_topology
$endgroup$
– Ross
Dec 11 '18 at 22:41














$begingroup$
Then it's basically trivial as addition commutes with linear functionals and $X$ has the weak topology wrt continuous functionals. Nets are the easiest way to see this. If these cannot be used, consider what the natural base for the weak topology on $X$ and its product on $X times X$ is.
$endgroup$
– Henno Brandsma
Dec 12 '18 at 4:51






$begingroup$
Then it's basically trivial as addition commutes with linear functionals and $X$ has the weak topology wrt continuous functionals. Nets are the easiest way to see this. If these cannot be used, consider what the natural base for the weak topology on $X$ and its product on $X times X$ is.
$endgroup$
– Henno Brandsma
Dec 12 '18 at 4:51












2 Answers
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$begingroup$

You have to say what topology you are using on $X times X$. Assuming that you are using the product topology here is a simple proof: if $(x_i,y_i) to (x,y)$ weakly and $f$ is a continuous linear functional then $x_i to x$ and $y_i to y$ weakly by definition of product topology. Hence $f(x_i) to f(x)$ and $f(y_i) to f(y)$ and $f(x_i+y_i)=f(x_i)+f(y_i) to f(x)+f(y)=f(x+y)$.



Remark: here $(x_i)$ and $(y_i)$ are nets. You can use sequences in the case of metrizable spaces but not in general.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You have to show that for every weakly open subset $U$ of $X$ containing $x_0+y_0$, there exists weakly open subsets $V$, $W$ of $X$ containing respectively $x_0$ and $y_0$, such that if $x in V$, $y in W$, $x+y in U$.



    I suggest that you prove it when $U$ is some ${y, phi(y) < a}$ where $a$ is a real number and $phi in X^*$. Then you can prove it when $U$ is a finite intersection of such subsets and from this address the general case.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      2












      $begingroup$

      You have to say what topology you are using on $X times X$. Assuming that you are using the product topology here is a simple proof: if $(x_i,y_i) to (x,y)$ weakly and $f$ is a continuous linear functional then $x_i to x$ and $y_i to y$ weakly by definition of product topology. Hence $f(x_i) to f(x)$ and $f(y_i) to f(y)$ and $f(x_i+y_i)=f(x_i)+f(y_i) to f(x)+f(y)=f(x+y)$.



      Remark: here $(x_i)$ and $(y_i)$ are nets. You can use sequences in the case of metrizable spaces but not in general.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        You have to say what topology you are using on $X times X$. Assuming that you are using the product topology here is a simple proof: if $(x_i,y_i) to (x,y)$ weakly and $f$ is a continuous linear functional then $x_i to x$ and $y_i to y$ weakly by definition of product topology. Hence $f(x_i) to f(x)$ and $f(y_i) to f(y)$ and $f(x_i+y_i)=f(x_i)+f(y_i) to f(x)+f(y)=f(x+y)$.



        Remark: here $(x_i)$ and $(y_i)$ are nets. You can use sequences in the case of metrizable spaces but not in general.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          You have to say what topology you are using on $X times X$. Assuming that you are using the product topology here is a simple proof: if $(x_i,y_i) to (x,y)$ weakly and $f$ is a continuous linear functional then $x_i to x$ and $y_i to y$ weakly by definition of product topology. Hence $f(x_i) to f(x)$ and $f(y_i) to f(y)$ and $f(x_i+y_i)=f(x_i)+f(y_i) to f(x)+f(y)=f(x+y)$.



          Remark: here $(x_i)$ and $(y_i)$ are nets. You can use sequences in the case of metrizable spaces but not in general.






          share|cite|improve this answer









          $endgroup$



          You have to say what topology you are using on $X times X$. Assuming that you are using the product topology here is a simple proof: if $(x_i,y_i) to (x,y)$ weakly and $f$ is a continuous linear functional then $x_i to x$ and $y_i to y$ weakly by definition of product topology. Hence $f(x_i) to f(x)$ and $f(y_i) to f(y)$ and $f(x_i+y_i)=f(x_i)+f(y_i) to f(x)+f(y)=f(x+y)$.



          Remark: here $(x_i)$ and $(y_i)$ are nets. You can use sequences in the case of metrizable spaces but not in general.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 23:30









          Kavi Rama MurthyKavi Rama Murthy

          61.8k42262




          61.8k42262























              0












              $begingroup$

              You have to show that for every weakly open subset $U$ of $X$ containing $x_0+y_0$, there exists weakly open subsets $V$, $W$ of $X$ containing respectively $x_0$ and $y_0$, such that if $x in V$, $y in W$, $x+y in U$.



              I suggest that you prove it when $U$ is some ${y, phi(y) < a}$ where $a$ is a real number and $phi in X^*$. Then you can prove it when $U$ is a finite intersection of such subsets and from this address the general case.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You have to show that for every weakly open subset $U$ of $X$ containing $x_0+y_0$, there exists weakly open subsets $V$, $W$ of $X$ containing respectively $x_0$ and $y_0$, such that if $x in V$, $y in W$, $x+y in U$.



                I suggest that you prove it when $U$ is some ${y, phi(y) < a}$ where $a$ is a real number and $phi in X^*$. Then you can prove it when $U$ is a finite intersection of such subsets and from this address the general case.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You have to show that for every weakly open subset $U$ of $X$ containing $x_0+y_0$, there exists weakly open subsets $V$, $W$ of $X$ containing respectively $x_0$ and $y_0$, such that if $x in V$, $y in W$, $x+y in U$.



                  I suggest that you prove it when $U$ is some ${y, phi(y) < a}$ where $a$ is a real number and $phi in X^*$. Then you can prove it when $U$ is a finite intersection of such subsets and from this address the general case.






                  share|cite|improve this answer









                  $endgroup$



                  You have to show that for every weakly open subset $U$ of $X$ containing $x_0+y_0$, there exists weakly open subsets $V$, $W$ of $X$ containing respectively $x_0$ and $y_0$, such that if $x in V$, $y in W$, $x+y in U$.



                  I suggest that you prove it when $U$ is some ${y, phi(y) < a}$ where $a$ is a real number and $phi in X^*$. Then you can prove it when $U$ is a finite intersection of such subsets and from this address the general case.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 11 '18 at 23:20









                  MindlackMindlack

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                  4,740210






























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