About weak convergence in $L^{infty}$












2














doing my homework I'm dealing with this:



Let, for all $nin mathbb{N} quad f_n(t) := e^{-nt^2}, quad t in [-1,1]$ Show that



1)$f_n overset{ast}{rightharpoonup} 0$ in $L^infty(-1,1)$



2)$f_n$ does not converge weakly to $0$ in $L^infty (-1,1)$



So I did 1) simply considering
begin{equation}
|f_n(x)g(x)| le|g(x)| quad forall g in L^1(-1,1)
end{equation}



So the result follows easily from the dominated convergence theorem.



But for 2) I know that given $f_n, f in X^*$
begin{equation}
f_n rightharpoonup f quad Leftrightarrow quad phi(f_n)rightarrow phi(f) quad forall phi in X^{**}
end{equation}
But how can I identify the dual of $L^infty$ to solve this?






functional-analysis weak-convergence dual-spaces







share|cite|improve this question


















  • 2




    You do not need to identify the dual of $L^infty$ (which is very complicated). You only need to construct one element $phi$ of $L^infty$ such that $phi(f_n)rightarrow phi(f)$ fails. A hint for constructing continuous linear functionals: choose a Banach subspace $E subset L^infty$ that contains all the $f_n$, use a continuous functional on $E$, then cite the Hahn-Banach theorem to extend it to all of $L^infty$.
    – GEdgar
    Dec 26 at 15:18










  • but following your reasonment, shouldn't $phi$ be in $(L^{infty})^{*}$? Why you say that $phi$ is in $L^infty$? Assuming the definition I've written is right..
    – James Arten
    Dec 26 at 15:21












  • Correct, it should say $phi$ is in $(L^infty)^*$.
    – GEdgar
    Dec 26 at 15:25












  • you mean the definition I've written is wrong? Just to understand
    – James Arten
    Dec 26 at 15:26






  • 1




    Your definition is correct, where $X^*$ is $L^infty$. My correction is now corrected.
    – GEdgar
    Dec 26 at 15:28


















2














doing my homework I'm dealing with this:



Let, for all $nin mathbb{N} quad f_n(t) := e^{-nt^2}, quad t in [-1,1]$ Show that



1)$f_n overset{ast}{rightharpoonup} 0$ in $L^infty(-1,1)$



2)$f_n$ does not converge weakly to $0$ in $L^infty (-1,1)$



So I did 1) simply considering
begin{equation}
|f_n(x)g(x)| le|g(x)| quad forall g in L^1(-1,1)
end{equation}



So the result follows easily from the dominated convergence theorem.



But for 2) I know that given $f_n, f in X^*$
begin{equation}
f_n rightharpoonup f quad Leftrightarrow quad phi(f_n)rightarrow phi(f) quad forall phi in X^{**}
end{equation}
But how can I identify the dual of $L^infty$ to solve this?






functional-analysis weak-convergence dual-spaces







share|cite|improve this question


















  • 2




    You do not need to identify the dual of $L^infty$ (which is very complicated). You only need to construct one element $phi$ of $L^infty$ such that $phi(f_n)rightarrow phi(f)$ fails. A hint for constructing continuous linear functionals: choose a Banach subspace $E subset L^infty$ that contains all the $f_n$, use a continuous functional on $E$, then cite the Hahn-Banach theorem to extend it to all of $L^infty$.
    – GEdgar
    Dec 26 at 15:18










  • but following your reasonment, shouldn't $phi$ be in $(L^{infty})^{*}$? Why you say that $phi$ is in $L^infty$? Assuming the definition I've written is right..
    – James Arten
    Dec 26 at 15:21












  • Correct, it should say $phi$ is in $(L^infty)^*$.
    – GEdgar
    Dec 26 at 15:25












  • you mean the definition I've written is wrong? Just to understand
    – James Arten
    Dec 26 at 15:26






  • 1




    Your definition is correct, where $X^*$ is $L^infty$. My correction is now corrected.
    – GEdgar
    Dec 26 at 15:28
















2












2








2







doing my homework I'm dealing with this:



Let, for all $nin mathbb{N} quad f_n(t) := e^{-nt^2}, quad t in [-1,1]$ Show that



1)$f_n overset{ast}{rightharpoonup} 0$ in $L^infty(-1,1)$



2)$f_n$ does not converge weakly to $0$ in $L^infty (-1,1)$



So I did 1) simply considering
begin{equation}
|f_n(x)g(x)| le|g(x)| quad forall g in L^1(-1,1)
end{equation}



So the result follows easily from the dominated convergence theorem.



But for 2) I know that given $f_n, f in X^*$
begin{equation}
f_n rightharpoonup f quad Leftrightarrow quad phi(f_n)rightarrow phi(f) quad forall phi in X^{**}
end{equation}
But how can I identify the dual of $L^infty$ to solve this?






functional-analysis weak-convergence dual-spaces







share|cite|improve this question













doing my homework I'm dealing with this:



Let, for all $nin mathbb{N} quad f_n(t) := e^{-nt^2}, quad t in [-1,1]$ Show that



1)$f_n overset{ast}{rightharpoonup} 0$ in $L^infty(-1,1)$



2)$f_n$ does not converge weakly to $0$ in $L^infty (-1,1)$



So I did 1) simply considering
begin{equation}
|f_n(x)g(x)| le|g(x)| quad forall g in L^1(-1,1)
end{equation}



So the result follows easily from the dominated convergence theorem.



But for 2) I know that given $f_n, f in X^*$
begin{equation}
f_n rightharpoonup f quad Leftrightarrow quad phi(f_n)rightarrow phi(f) quad forall phi in X^{**}
end{equation}
But how can I identify the dual of $L^infty$ to solve this?






functional-analysis weak-convergence dual-spaces




functional-analysis weak-convergence dual-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 26 at 15:03









James Arten

9910




9910








  • 2




    You do not need to identify the dual of $L^infty$ (which is very complicated). You only need to construct one element $phi$ of $L^infty$ such that $phi(f_n)rightarrow phi(f)$ fails. A hint for constructing continuous linear functionals: choose a Banach subspace $E subset L^infty$ that contains all the $f_n$, use a continuous functional on $E$, then cite the Hahn-Banach theorem to extend it to all of $L^infty$.
    – GEdgar
    Dec 26 at 15:18










  • but following your reasonment, shouldn't $phi$ be in $(L^{infty})^{*}$? Why you say that $phi$ is in $L^infty$? Assuming the definition I've written is right..
    – James Arten
    Dec 26 at 15:21












  • Correct, it should say $phi$ is in $(L^infty)^*$.
    – GEdgar
    Dec 26 at 15:25












  • you mean the definition I've written is wrong? Just to understand
    – James Arten
    Dec 26 at 15:26






  • 1




    Your definition is correct, where $X^*$ is $L^infty$. My correction is now corrected.
    – GEdgar
    Dec 26 at 15:28
















  • 2




    You do not need to identify the dual of $L^infty$ (which is very complicated). You only need to construct one element $phi$ of $L^infty$ such that $phi(f_n)rightarrow phi(f)$ fails. A hint for constructing continuous linear functionals: choose a Banach subspace $E subset L^infty$ that contains all the $f_n$, use a continuous functional on $E$, then cite the Hahn-Banach theorem to extend it to all of $L^infty$.
    – GEdgar
    Dec 26 at 15:18










  • but following your reasonment, shouldn't $phi$ be in $(L^{infty})^{*}$? Why you say that $phi$ is in $L^infty$? Assuming the definition I've written is right..
    – James Arten
    Dec 26 at 15:21












  • Correct, it should say $phi$ is in $(L^infty)^*$.
    – GEdgar
    Dec 26 at 15:25












  • you mean the definition I've written is wrong? Just to understand
    – James Arten
    Dec 26 at 15:26






  • 1




    Your definition is correct, where $X^*$ is $L^infty$. My correction is now corrected.
    – GEdgar
    Dec 26 at 15:28










2




2




You do not need to identify the dual of $L^infty$ (which is very complicated). You only need to construct one element $phi$ of $L^infty$ such that $phi(f_n)rightarrow phi(f)$ fails. A hint for constructing continuous linear functionals: choose a Banach subspace $E subset L^infty$ that contains all the $f_n$, use a continuous functional on $E$, then cite the Hahn-Banach theorem to extend it to all of $L^infty$.
– GEdgar
Dec 26 at 15:18




You do not need to identify the dual of $L^infty$ (which is very complicated). You only need to construct one element $phi$ of $L^infty$ such that $phi(f_n)rightarrow phi(f)$ fails. A hint for constructing continuous linear functionals: choose a Banach subspace $E subset L^infty$ that contains all the $f_n$, use a continuous functional on $E$, then cite the Hahn-Banach theorem to extend it to all of $L^infty$.
– GEdgar
Dec 26 at 15:18












but following your reasonment, shouldn't $phi$ be in $(L^{infty})^{*}$? Why you say that $phi$ is in $L^infty$? Assuming the definition I've written is right..
– James Arten
Dec 26 at 15:21






but following your reasonment, shouldn't $phi$ be in $(L^{infty})^{*}$? Why you say that $phi$ is in $L^infty$? Assuming the definition I've written is right..
– James Arten
Dec 26 at 15:21














Correct, it should say $phi$ is in $(L^infty)^*$.
– GEdgar
Dec 26 at 15:25






Correct, it should say $phi$ is in $(L^infty)^*$.
– GEdgar
Dec 26 at 15:25














you mean the definition I've written is wrong? Just to understand
– James Arten
Dec 26 at 15:26




you mean the definition I've written is wrong? Just to understand
– James Arten
Dec 26 at 15:26




1




1




Your definition is correct, where $X^*$ is $L^infty$. My correction is now corrected.
– GEdgar
Dec 26 at 15:28






Your definition is correct, where $X^*$ is $L^infty$. My correction is now corrected.
– GEdgar
Dec 26 at 15:28












1 Answer
1






active

oldest

votes


















2














Ok I think I solved this.



In this case, we don't have weak convergence of $f_n$ to $0$ because we can consider the extension of Dirac $delta_0$ as linear functional of $(L^infty (-1,1))^*$, which is a linear bounded functional on $C[-1,1]$:
begin{equation}
delta_0(f) = f(0) quad forall f in C[-1,1]
end{equation}



Then if we call $T$ this extension it follows $T(f_n) = f_n(0)=1$.



(we have the extension because of Hahn Banach Theorem, thanks for your suggestions).






share|cite|improve this answer























    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%2f3053011%2fabout-weak-convergence-in-l-infty%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














    Ok I think I solved this.



    In this case, we don't have weak convergence of $f_n$ to $0$ because we can consider the extension of Dirac $delta_0$ as linear functional of $(L^infty (-1,1))^*$, which is a linear bounded functional on $C[-1,1]$:
    begin{equation}
    delta_0(f) = f(0) quad forall f in C[-1,1]
    end{equation}



    Then if we call $T$ this extension it follows $T(f_n) = f_n(0)=1$.



    (we have the extension because of Hahn Banach Theorem, thanks for your suggestions).






    share|cite|improve this answer




























      2














      Ok I think I solved this.



      In this case, we don't have weak convergence of $f_n$ to $0$ because we can consider the extension of Dirac $delta_0$ as linear functional of $(L^infty (-1,1))^*$, which is a linear bounded functional on $C[-1,1]$:
      begin{equation}
      delta_0(f) = f(0) quad forall f in C[-1,1]
      end{equation}



      Then if we call $T$ this extension it follows $T(f_n) = f_n(0)=1$.



      (we have the extension because of Hahn Banach Theorem, thanks for your suggestions).






      share|cite|improve this answer


























        2












        2








        2






        Ok I think I solved this.



        In this case, we don't have weak convergence of $f_n$ to $0$ because we can consider the extension of Dirac $delta_0$ as linear functional of $(L^infty (-1,1))^*$, which is a linear bounded functional on $C[-1,1]$:
        begin{equation}
        delta_0(f) = f(0) quad forall f in C[-1,1]
        end{equation}



        Then if we call $T$ this extension it follows $T(f_n) = f_n(0)=1$.



        (we have the extension because of Hahn Banach Theorem, thanks for your suggestions).






        share|cite|improve this answer














        Ok I think I solved this.



        In this case, we don't have weak convergence of $f_n$ to $0$ because we can consider the extension of Dirac $delta_0$ as linear functional of $(L^infty (-1,1))^*$, which is a linear bounded functional on $C[-1,1]$:
        begin{equation}
        delta_0(f) = f(0) quad forall f in C[-1,1]
        end{equation}



        Then if we call $T$ this extension it follows $T(f_n) = f_n(0)=1$.



        (we have the extension because of Hahn Banach Theorem, thanks for your suggestions).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 27 at 8:53

























        answered Dec 27 at 8:39









        James Arten

        9910




        9910






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3053011%2fabout-weak-convergence-in-l-infty%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

            Human spaceflight

            Image
            Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
            Read more

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            Image
            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
            Read more

            File:DeusFollowingSea.jpg

            Image
            File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
            Read more