$f(x_0)=0 forall f in X^*$ ($X^*$ is topological dual) then $x_0=0$.












1












$begingroup$


Let X be a normed vector space of infinite dimension (possibly) and $x_0 in X$.



I would like to show that if $f(x_0)=0 forall f in X^*$ (the topological dual) then $x_0=0$.
I asked this question here but I forgot to mention that the functions were in the topological dual and not the algebraic dual, so I had an explanation for why this is true in the case of algebraic dual which was still very interesting.



The suggestion for when $f$ are in the topological dual was to use Hahn-Banach theorem. I have the version that says that a functional of $Y^*$ (where $Y$ is a subspace of $X$) can be extended to a functional of $X^*$ (again here we consider topological duals). But here I am not sure how this helps me. What should I consider for the subspace $Y$?










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
    $endgroup$
    – José Carlos Santos
    Jan 12 at 17:52






  • 1




    $begingroup$
    This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
    $endgroup$
    – David C. Ullrich
    Jan 12 at 17:56












  • $begingroup$
    Okay, thanks! I am not familiar with those kind of reasoning yet...
    $endgroup$
    – roi_saumon
    Jan 12 at 18:14
















1












$begingroup$


Let X be a normed vector space of infinite dimension (possibly) and $x_0 in X$.



I would like to show that if $f(x_0)=0 forall f in X^*$ (the topological dual) then $x_0=0$.
I asked this question here but I forgot to mention that the functions were in the topological dual and not the algebraic dual, so I had an explanation for why this is true in the case of algebraic dual which was still very interesting.



The suggestion for when $f$ are in the topological dual was to use Hahn-Banach theorem. I have the version that says that a functional of $Y^*$ (where $Y$ is a subspace of $X$) can be extended to a functional of $X^*$ (again here we consider topological duals). But here I am not sure how this helps me. What should I consider for the subspace $Y$?










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
    $endgroup$
    – José Carlos Santos
    Jan 12 at 17:52






  • 1




    $begingroup$
    This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
    $endgroup$
    – David C. Ullrich
    Jan 12 at 17:56












  • $begingroup$
    Okay, thanks! I am not familiar with those kind of reasoning yet...
    $endgroup$
    – roi_saumon
    Jan 12 at 18:14














1












1








1





$begingroup$


Let X be a normed vector space of infinite dimension (possibly) and $x_0 in X$.



I would like to show that if $f(x_0)=0 forall f in X^*$ (the topological dual) then $x_0=0$.
I asked this question here but I forgot to mention that the functions were in the topological dual and not the algebraic dual, so I had an explanation for why this is true in the case of algebraic dual which was still very interesting.



The suggestion for when $f$ are in the topological dual was to use Hahn-Banach theorem. I have the version that says that a functional of $Y^*$ (where $Y$ is a subspace of $X$) can be extended to a functional of $X^*$ (again here we consider topological duals). But here I am not sure how this helps me. What should I consider for the subspace $Y$?










share|cite|improve this question











$endgroup$




Let X be a normed vector space of infinite dimension (possibly) and $x_0 in X$.



I would like to show that if $f(x_0)=0 forall f in X^*$ (the topological dual) then $x_0=0$.
I asked this question here but I forgot to mention that the functions were in the topological dual and not the algebraic dual, so I had an explanation for why this is true in the case of algebraic dual which was still very interesting.



The suggestion for when $f$ are in the topological dual was to use Hahn-Banach theorem. I have the version that says that a functional of $Y^*$ (where $Y$ is a subspace of $X$) can be extended to a functional of $X^*$ (again here we consider topological duals). But here I am not sure how this helps me. What should I consider for the subspace $Y$?







functional-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 17:58









David C. Ullrich

60.1k43994




60.1k43994










asked Jan 12 at 17:48









roi_saumonroi_saumon

54538




54538








  • 3




    $begingroup$
    Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
    $endgroup$
    – José Carlos Santos
    Jan 12 at 17:52






  • 1




    $begingroup$
    This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
    $endgroup$
    – David C. Ullrich
    Jan 12 at 17:56












  • $begingroup$
    Okay, thanks! I am not familiar with those kind of reasoning yet...
    $endgroup$
    – roi_saumon
    Jan 12 at 18:14














  • 3




    $begingroup$
    Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
    $endgroup$
    – José Carlos Santos
    Jan 12 at 17:52






  • 1




    $begingroup$
    This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
    $endgroup$
    – David C. Ullrich
    Jan 12 at 17:56












  • $begingroup$
    Okay, thanks! I am not familiar with those kind of reasoning yet...
    $endgroup$
    – roi_saumon
    Jan 12 at 18:14








3




3




$begingroup$
Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
$endgroup$
– José Carlos Santos
Jan 12 at 17:52




$begingroup$
Possible duplicate of Prove that if $f(x)=f(y)$ for all $fin X^{*},$ then $x=y$
$endgroup$
– José Carlos Santos
Jan 12 at 17:52




1




1




$begingroup$
This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
$endgroup$
– David C. Ullrich
Jan 12 at 17:56






$begingroup$
This is in whatever book you learned the H-B theorem from. Assume $x_0ne0$. Let $Y$ be the span of $x_0$. Take $Lambdain Y^*$ with $Lambda(x_0)ne0$, for example define $Lambda(cx_0)=c$.
$endgroup$
– David C. Ullrich
Jan 12 at 17:56














$begingroup$
Okay, thanks! I am not familiar with those kind of reasoning yet...
$endgroup$
– roi_saumon
Jan 12 at 18:14




$begingroup$
Okay, thanks! I am not familiar with those kind of reasoning yet...
$endgroup$
– roi_saumon
Jan 12 at 18:14










1 Answer
1






active

oldest

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2












$begingroup$

If $x_0$ is not the zero vector then consider the functional $f$
which is defined on the space that $x_0$ spans as follows:
$f(a x_0)=a|x_0|$.
It follows from the Hahn-Banach theorem that there is a functional $Fin X^*$ with $F(x)=f(x) forall xin text{span}(x_0)$ and of course $F(x_0)$ is not equal to $0$.






share|cite|improve this answer











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    1 Answer
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    2












    $begingroup$

    If $x_0$ is not the zero vector then consider the functional $f$
    which is defined on the space that $x_0$ spans as follows:
    $f(a x_0)=a|x_0|$.
    It follows from the Hahn-Banach theorem that there is a functional $Fin X^*$ with $F(x)=f(x) forall xin text{span}(x_0)$ and of course $F(x_0)$ is not equal to $0$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      If $x_0$ is not the zero vector then consider the functional $f$
      which is defined on the space that $x_0$ spans as follows:
      $f(a x_0)=a|x_0|$.
      It follows from the Hahn-Banach theorem that there is a functional $Fin X^*$ with $F(x)=f(x) forall xin text{span}(x_0)$ and of course $F(x_0)$ is not equal to $0$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        If $x_0$ is not the zero vector then consider the functional $f$
        which is defined on the space that $x_0$ spans as follows:
        $f(a x_0)=a|x_0|$.
        It follows from the Hahn-Banach theorem that there is a functional $Fin X^*$ with $F(x)=f(x) forall xin text{span}(x_0)$ and of course $F(x_0)$ is not equal to $0$.






        share|cite|improve this answer











        $endgroup$



        If $x_0$ is not the zero vector then consider the functional $f$
        which is defined on the space that $x_0$ spans as follows:
        $f(a x_0)=a|x_0|$.
        It follows from the Hahn-Banach theorem that there is a functional $Fin X^*$ with $F(x)=f(x) forall xin text{span}(x_0)$ and of course $F(x_0)$ is not equal to $0$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 12 at 19:07

























        answered Jan 12 at 19:00









        NikiforosNikiforos

        362




        362






























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