$f(x_0)=0 forall f in X^*$ ($X^*$ 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

votes


















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











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071186%2ffx-0-0-forall-f-in-x-x-is-topological-dual-then-x-0-0%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071186%2ffx-0-0-forall-f-in-x-x-is-topological-dual-then-x-0-0%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Mario Kart Wii

            What does “Dominus providebit” mean?

            The Binding of Isaac: Rebirth/Afterbirth