$x=(x_n)$ satisfy $sum x_k y_k lt infty$ for all $yin c_0$












-1












$begingroup$


So , I know that $c_0^*$ and $ell_1$ are isomorphic and I guess I need to use it(or the ideas in the proof) in order to solve the following question :



Let $x= (x_n)$ be a sequence s.t for all $y in c_0$ , $sum x_ky_k$ is finite.



I need to prove that $xin ell_1$ .



So $x$ defines a functional -$phi_x$ - on $c_0$, taking $y$ to $sum x_ky_k$.



In order to imitate the proof that $c_0^*$ isomorphic to $ell_1$ I need to use a bounded functional(in order to bound the sum of absolute values of the $x_n$'s) , and im not sure if $phi_x$
is bounded (if so ,how to show it? )



Any ideas?



Thanks for helping!






functional-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This is actually easy to do without $c_0*=l^1.$
    $endgroup$
    – zhw.
    Dec 9 '18 at 21:49










  • $begingroup$
    Can you explain? @zhw.
    $endgroup$
    – user123
    Dec 9 '18 at 21:50
















-1












$begingroup$


So , I know that $c_0^*$ and $ell_1$ are isomorphic and I guess I need to use it(or the ideas in the proof) in order to solve the following question :



Let $x= (x_n)$ be a sequence s.t for all $y in c_0$ , $sum x_ky_k$ is finite.



I need to prove that $xin ell_1$ .



So $x$ defines a functional -$phi_x$ - on $c_0$, taking $y$ to $sum x_ky_k$.



In order to imitate the proof that $c_0^*$ isomorphic to $ell_1$ I need to use a bounded functional(in order to bound the sum of absolute values of the $x_n$'s) , and im not sure if $phi_x$
is bounded (if so ,how to show it? )



Any ideas?



Thanks for helping!






functional-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This is actually easy to do without $c_0*=l^1.$
    $endgroup$
    – zhw.
    Dec 9 '18 at 21:49










  • $begingroup$
    Can you explain? @zhw.
    $endgroup$
    – user123
    Dec 9 '18 at 21:50














-1












-1








-1





$begingroup$


So , I know that $c_0^*$ and $ell_1$ are isomorphic and I guess I need to use it(or the ideas in the proof) in order to solve the following question :



Let $x= (x_n)$ be a sequence s.t for all $y in c_0$ , $sum x_ky_k$ is finite.



I need to prove that $xin ell_1$ .



So $x$ defines a functional -$phi_x$ - on $c_0$, taking $y$ to $sum x_ky_k$.



In order to imitate the proof that $c_0^*$ isomorphic to $ell_1$ I need to use a bounded functional(in order to bound the sum of absolute values of the $x_n$'s) , and im not sure if $phi_x$
is bounded (if so ,how to show it? )



Any ideas?



Thanks for helping!






functional-analysis







share|cite|improve this question









$endgroup$




So , I know that $c_0^*$ and $ell_1$ are isomorphic and I guess I need to use it(or the ideas in the proof) in order to solve the following question :



Let $x= (x_n)$ be a sequence s.t for all $y in c_0$ , $sum x_ky_k$ is finite.



I need to prove that $xin ell_1$ .



So $x$ defines a functional -$phi_x$ - on $c_0$, taking $y$ to $sum x_ky_k$.



In order to imitate the proof that $c_0^*$ isomorphic to $ell_1$ I need to use a bounded functional(in order to bound the sum of absolute values of the $x_n$'s) , and im not sure if $phi_x$
is bounded (if so ,how to show it? )



Any ideas?



Thanks for helping!






functional-analysis




functional-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 '18 at 21:37









user123user123

1,306316




1,306316








  • 1




    $begingroup$
    This is actually easy to do without $c_0*=l^1.$
    $endgroup$
    – zhw.
    Dec 9 '18 at 21:49










  • $begingroup$
    Can you explain? @zhw.
    $endgroup$
    – user123
    Dec 9 '18 at 21:50














  • 1




    $begingroup$
    This is actually easy to do without $c_0*=l^1.$
    $endgroup$
    – zhw.
    Dec 9 '18 at 21:49










  • $begingroup$
    Can you explain? @zhw.
    $endgroup$
    – user123
    Dec 9 '18 at 21:50








1




1




$begingroup$
This is actually easy to do without $c_0*=l^1.$
$endgroup$
– zhw.
Dec 9 '18 at 21:49




$begingroup$
This is actually easy to do without $c_0*=l^1.$
$endgroup$
– zhw.
Dec 9 '18 at 21:49












$begingroup$
Can you explain? @zhw.
$endgroup$
– user123
Dec 9 '18 at 21:50




$begingroup$
Can you explain? @zhw.
$endgroup$
– user123
Dec 9 '18 at 21:50










2 Answers
2






active

oldest

votes


















0












$begingroup$

If ${T_n}$ is a sequence of bounded linear functionals from one Banach space $X$ to another Banach space $Y$ such that $Tx=lim_n T_n x$ exists for every $x$ then $T$ is necessarily a bounded linear functional. This is a consequence of Banach - Steinhaus Theorem (which is also known as Uniform Boundedness Principle) : just note that there is a finite constant $M$ such that $|T_n| leq M$ for all $n$ and so $|Tx|leq M|x|$.



It follows that $phi_x$ is a continuous linear functional from which it is easy to see (using $c_0^{*}=ell^{1})$ that $x in ell^{1}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Nice. Just to make sure - your $T_n$'s are $T_n(y) = sum_{k=1}^n y_kx_k$ ?
    $endgroup$
    – user123
    Dec 10 '18 at 10:14










  • $begingroup$
    @Liad You are right; we apply the theorem to this sequence.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 at 10:17










  • $begingroup$
    Very nice. Thank you!
    $endgroup$
    – user123
    Dec 10 '18 at 10:18



















0












$begingroup$

Sketch: Suppose $sum |x_n|=infty.$ Then ${1,2,dots}$ can be partitioned into consecutive blocks of integers $B_1 < B_2 <cdots$ such that



$$sum_{nin B_k}|x_n| > 1,,,k=1,2,dots$$



For $k=1,2,dots $ and $nin B_k,$ set $y_n=(text {sgn } x_n)/k.$ Then $(y_n)in c_0,$  and



$$sum_{n=1}^{infty}x_ny_n = infty.$$






share|cite|improve this answer









$endgroup$













    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    If ${T_n}$ is a sequence of bounded linear functionals from one Banach space $X$ to another Banach space $Y$ such that $Tx=lim_n T_n x$ exists for every $x$ then $T$ is necessarily a bounded linear functional. This is a consequence of Banach - Steinhaus Theorem (which is also known as Uniform Boundedness Principle) : just note that there is a finite constant $M$ such that $|T_n| leq M$ for all $n$ and so $|Tx|leq M|x|$.



    It follows that $phi_x$ is a continuous linear functional from which it is easy to see (using $c_0^{*}=ell^{1})$ that $x in ell^{1}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Nice. Just to make sure - your $T_n$'s are $T_n(y) = sum_{k=1}^n y_kx_k$ ?
      $endgroup$
      – user123
      Dec 10 '18 at 10:14










    • $begingroup$
      @Liad You are right; we apply the theorem to this sequence.
      $endgroup$
      – Kavi Rama Murthy
      Dec 10 '18 at 10:17










    • $begingroup$
      Very nice. Thank you!
      $endgroup$
      – user123
      Dec 10 '18 at 10:18
















    0












    $begingroup$

    If ${T_n}$ is a sequence of bounded linear functionals from one Banach space $X$ to another Banach space $Y$ such that $Tx=lim_n T_n x$ exists for every $x$ then $T$ is necessarily a bounded linear functional. This is a consequence of Banach - Steinhaus Theorem (which is also known as Uniform Boundedness Principle) : just note that there is a finite constant $M$ such that $|T_n| leq M$ for all $n$ and so $|Tx|leq M|x|$.



    It follows that $phi_x$ is a continuous linear functional from which it is easy to see (using $c_0^{*}=ell^{1})$ that $x in ell^{1}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Nice. Just to make sure - your $T_n$'s are $T_n(y) = sum_{k=1}^n y_kx_k$ ?
      $endgroup$
      – user123
      Dec 10 '18 at 10:14










    • $begingroup$
      @Liad You are right; we apply the theorem to this sequence.
      $endgroup$
      – Kavi Rama Murthy
      Dec 10 '18 at 10:17










    • $begingroup$
      Very nice. Thank you!
      $endgroup$
      – user123
      Dec 10 '18 at 10:18














    0












    0








    0





    $begingroup$

    If ${T_n}$ is a sequence of bounded linear functionals from one Banach space $X$ to another Banach space $Y$ such that $Tx=lim_n T_n x$ exists for every $x$ then $T$ is necessarily a bounded linear functional. This is a consequence of Banach - Steinhaus Theorem (which is also known as Uniform Boundedness Principle) : just note that there is a finite constant $M$ such that $|T_n| leq M$ for all $n$ and so $|Tx|leq M|x|$.



    It follows that $phi_x$ is a continuous linear functional from which it is easy to see (using $c_0^{*}=ell^{1})$ that $x in ell^{1}$.






    share|cite|improve this answer











    $endgroup$



    If ${T_n}$ is a sequence of bounded linear functionals from one Banach space $X$ to another Banach space $Y$ such that $Tx=lim_n T_n x$ exists for every $x$ then $T$ is necessarily a bounded linear functional. This is a consequence of Banach - Steinhaus Theorem (which is also known as Uniform Boundedness Principle) : just note that there is a finite constant $M$ such that $|T_n| leq M$ for all $n$ and so $|Tx|leq M|x|$.



    It follows that $phi_x$ is a continuous linear functional from which it is easy to see (using $c_0^{*}=ell^{1})$ that $x in ell^{1}$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 9 '18 at 23:47

























    answered Dec 9 '18 at 23:41









    Kavi Rama MurthyKavi Rama Murthy

    57.9k42160




    57.9k42160












    • $begingroup$
      Nice. Just to make sure - your $T_n$'s are $T_n(y) = sum_{k=1}^n y_kx_k$ ?
      $endgroup$
      – user123
      Dec 10 '18 at 10:14










    • $begingroup$
      @Liad You are right; we apply the theorem to this sequence.
      $endgroup$
      – Kavi Rama Murthy
      Dec 10 '18 at 10:17










    • $begingroup$
      Very nice. Thank you!
      $endgroup$
      – user123
      Dec 10 '18 at 10:18


















    • $begingroup$
      Nice. Just to make sure - your $T_n$'s are $T_n(y) = sum_{k=1}^n y_kx_k$ ?
      $endgroup$
      – user123
      Dec 10 '18 at 10:14










    • $begingroup$
      @Liad You are right; we apply the theorem to this sequence.
      $endgroup$
      – Kavi Rama Murthy
      Dec 10 '18 at 10:17










    • $begingroup$
      Very nice. Thank you!
      $endgroup$
      – user123
      Dec 10 '18 at 10:18
















    $begingroup$
    Nice. Just to make sure - your $T_n$'s are $T_n(y) = sum_{k=1}^n y_kx_k$ ?
    $endgroup$
    – user123
    Dec 10 '18 at 10:14




    $begingroup$
    Nice. Just to make sure - your $T_n$'s are $T_n(y) = sum_{k=1}^n y_kx_k$ ?
    $endgroup$
    – user123
    Dec 10 '18 at 10:14












    $begingroup$
    @Liad You are right; we apply the theorem to this sequence.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 at 10:17




    $begingroup$
    @Liad You are right; we apply the theorem to this sequence.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 at 10:17












    $begingroup$
    Very nice. Thank you!
    $endgroup$
    – user123
    Dec 10 '18 at 10:18




    $begingroup$
    Very nice. Thank you!
    $endgroup$
    – user123
    Dec 10 '18 at 10:18











    0












    $begingroup$

    Sketch: Suppose $sum |x_n|=infty.$ Then ${1,2,dots}$ can be partitioned into consecutive blocks of integers $B_1 < B_2 <cdots$ such that



    $$sum_{nin B_k}|x_n| > 1,,,k=1,2,dots$$



    For $k=1,2,dots $ and $nin B_k,$ set $y_n=(text {sgn } x_n)/k.$ Then $(y_n)in c_0,$  and



    $$sum_{n=1}^{infty}x_ny_n = infty.$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Sketch: Suppose $sum |x_n|=infty.$ Then ${1,2,dots}$ can be partitioned into consecutive blocks of integers $B_1 < B_2 <cdots$ such that



      $$sum_{nin B_k}|x_n| > 1,,,k=1,2,dots$$



      For $k=1,2,dots $ and $nin B_k,$ set $y_n=(text {sgn } x_n)/k.$ Then $(y_n)in c_0,$  and



      $$sum_{n=1}^{infty}x_ny_n = infty.$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Sketch: Suppose $sum |x_n|=infty.$ Then ${1,2,dots}$ can be partitioned into consecutive blocks of integers $B_1 < B_2 <cdots$ such that



        $$sum_{nin B_k}|x_n| > 1,,,k=1,2,dots$$



        For $k=1,2,dots $ and $nin B_k,$ set $y_n=(text {sgn } x_n)/k.$ Then $(y_n)in c_0,$  and



        $$sum_{n=1}^{infty}x_ny_n = infty.$$






        share|cite|improve this answer









        $endgroup$



        Sketch: Suppose $sum |x_n|=infty.$ Then ${1,2,dots}$ can be partitioned into consecutive blocks of integers $B_1 < B_2 <cdots$ such that



        $$sum_{nin B_k}|x_n| > 1,,,k=1,2,dots$$



        For $k=1,2,dots $ and $nin B_k,$ set $y_n=(text {sgn } x_n)/k.$ Then $(y_n)in c_0,$  and



        $$sum_{n=1}^{infty}x_ny_n = infty.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 23:35









        zhw.zhw.

        72.6k43175




        72.6k43175






























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