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












0














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










  • 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
    – Ross
    Dec 11 '18 at 22:41












  • 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.
    – Henno Brandsma
    Dec 12 '18 at 4:51


















0














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
























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










  • 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
    – Ross
    Dec 11 '18 at 22:41












  • 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.
    – Henno Brandsma
    Dec 12 '18 at 4:51
















0












0








0







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















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

























asked Dec 11 '18 at 22:09









Ross

1137




1137












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










  • 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
    – Ross
    Dec 11 '18 at 22:41












  • 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.
    – Henno Brandsma
    Dec 12 '18 at 4:51




















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










  • 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
    – Ross
    Dec 11 '18 at 22:41












  • 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.
    – Henno Brandsma
    Dec 12 '18 at 4:51


















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




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












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
– Ross
Dec 11 '18 at 22:41






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
– Ross
Dec 11 '18 at 22:41














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.
– Henno Brandsma
Dec 12 '18 at 4:51






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.
– Henno Brandsma
Dec 12 '18 at 4:51












2 Answers
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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





























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    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

      votes









      2














      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


























        2














        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
























          2












          2








          2






          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












          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 Murthy

          51.3k31855




          51.3k31855























              0














              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


























                0














                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
























                  0












                  0








                  0






                  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












                  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









                  Mindlack

                  1,70717




                  1,70717






























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