How to show non existence of an operator $A: V to V^*$











up vote
1
down vote

favorite












Let $V$ be a Banach space , $p in (1,2)$ and $mu >0.$ How can I prove that there does not exist an operator $A:V to V^*$ such that
$$langle Au-Av, u-vrangle geq mu Vert u-v Vert ^p,qquad u, v in V. $$






functional-analysis operator-theory banach-spaces







share|cite|improve this question
























  • any additional assumptions on $A$, such as linearity or continuity?
    – supinf
    7 hours ago










  • No, but it is $p in (1,2)$, edited it
    – Tesla
    7 hours ago










  • $Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
    – LeB
    7 hours ago










  • @LeB its not an inner product, its a duality product
    – supinf
    7 hours ago






  • 1




    @supinf I guess "operator" usually refers to linear operator?
    – Cave Johnson
    7 hours ago

















up vote
1
down vote

favorite












Let $V$ be a Banach space , $p in (1,2)$ and $mu >0.$ How can I prove that there does not exist an operator $A:V to V^*$ such that
$$langle Au-Av, u-vrangle geq mu Vert u-v Vert ^p,qquad u, v in V. $$






functional-analysis operator-theory banach-spaces







share|cite|improve this question
























  • any additional assumptions on $A$, such as linearity or continuity?
    – supinf
    7 hours ago










  • No, but it is $p in (1,2)$, edited it
    – Tesla
    7 hours ago










  • $Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
    – LeB
    7 hours ago










  • @LeB its not an inner product, its a duality product
    – supinf
    7 hours ago






  • 1




    @supinf I guess "operator" usually refers to linear operator?
    – Cave Johnson
    7 hours ago















up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $V$ be a Banach space , $p in (1,2)$ and $mu >0.$ How can I prove that there does not exist an operator $A:V to V^*$ such that
$$langle Au-Av, u-vrangle geq mu Vert u-v Vert ^p,qquad u, v in V. $$






functional-analysis operator-theory banach-spaces







share|cite|improve this question















Let $V$ be a Banach space , $p in (1,2)$ and $mu >0.$ How can I prove that there does not exist an operator $A:V to V^*$ such that
$$langle Au-Av, u-vrangle geq mu Vert u-v Vert ^p,qquad u, v in V. $$






functional-analysis operator-theory banach-spaces




functional-analysis operator-theory banach-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









user587192

97810




97810










asked 7 hours ago









Tesla

900426




900426












  • any additional assumptions on $A$, such as linearity or continuity?
    – supinf
    7 hours ago










  • No, but it is $p in (1,2)$, edited it
    – Tesla
    7 hours ago










  • $Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
    – LeB
    7 hours ago










  • @LeB its not an inner product, its a duality product
    – supinf
    7 hours ago






  • 1




    @supinf I guess "operator" usually refers to linear operator?
    – Cave Johnson
    7 hours ago




















  • any additional assumptions on $A$, such as linearity or continuity?
    – supinf
    7 hours ago










  • No, but it is $p in (1,2)$, edited it
    – Tesla
    7 hours ago










  • $Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
    – LeB
    7 hours ago










  • @LeB its not an inner product, its a duality product
    – supinf
    7 hours ago






  • 1




    @supinf I guess "operator" usually refers to linear operator?
    – Cave Johnson
    7 hours ago


















any additional assumptions on $A$, such as linearity or continuity?
– supinf
7 hours ago




any additional assumptions on $A$, such as linearity or continuity?
– supinf
7 hours ago












No, but it is $p in (1,2)$, edited it
– Tesla
7 hours ago




No, but it is $p in (1,2)$, edited it
– Tesla
7 hours ago












$Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
– LeB
7 hours ago




$Au$ and $u$ are in different space which are a Banach space and its Dual. How can you do the inner product of them?
– LeB
7 hours ago












@LeB its not an inner product, its a duality product
– supinf
7 hours ago




@LeB its not an inner product, its a duality product
– supinf
7 hours ago




1




1




@supinf I guess "operator" usually refers to linear operator?
– Cave Johnson
7 hours ago






@supinf I guess "operator" usually refers to linear operator?
– Cave Johnson
7 hours ago












1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










Pick an arbitrary point $vin V$ such that $|v|=1$.



Let $n$ be an arbitrary integer.
(Our goal is a proof by contradiction as $ntoinfty$).



We consider the points $v_k:= tfrac{k}{n} v$ for $k=0,ldots, n$.
Then, by assumption we have
$$
langle A v_{k+1} - Av_k,v_{k+1}-v_k rangle
geq
mu | v_{k+1}-v_k|^p
$$
for every $k=0,ldots,n-1$.



Summing up those terms will lead to the inequality
$$
tfrac 1n langle Av - A0 , v rangle geq mu n^{1-p}.
$$



For sufficiently large $n$, this cannot be true.






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',
    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%2f2995388%2fhow-to-show-non-existence-of-an-operator-a-v-to-v%23new-answer', 'question_page');
    }
    );

    Post as a guest


















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Pick an arbitrary point $vin V$ such that $|v|=1$.



    Let $n$ be an arbitrary integer.
    (Our goal is a proof by contradiction as $ntoinfty$).



    We consider the points $v_k:= tfrac{k}{n} v$ for $k=0,ldots, n$.
    Then, by assumption we have
    $$
    langle A v_{k+1} - Av_k,v_{k+1}-v_k rangle
    geq
    mu | v_{k+1}-v_k|^p
    $$
    for every $k=0,ldots,n-1$.



    Summing up those terms will lead to the inequality
    $$
    tfrac 1n langle Av - A0 , v rangle geq mu n^{1-p}.
    $$



    For sufficiently large $n$, this cannot be true.






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      Pick an arbitrary point $vin V$ such that $|v|=1$.



      Let $n$ be an arbitrary integer.
      (Our goal is a proof by contradiction as $ntoinfty$).



      We consider the points $v_k:= tfrac{k}{n} v$ for $k=0,ldots, n$.
      Then, by assumption we have
      $$
      langle A v_{k+1} - Av_k,v_{k+1}-v_k rangle
      geq
      mu | v_{k+1}-v_k|^p
      $$
      for every $k=0,ldots,n-1$.



      Summing up those terms will lead to the inequality
      $$
      tfrac 1n langle Av - A0 , v rangle geq mu n^{1-p}.
      $$



      For sufficiently large $n$, this cannot be true.






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Pick an arbitrary point $vin V$ such that $|v|=1$.



        Let $n$ be an arbitrary integer.
        (Our goal is a proof by contradiction as $ntoinfty$).



        We consider the points $v_k:= tfrac{k}{n} v$ for $k=0,ldots, n$.
        Then, by assumption we have
        $$
        langle A v_{k+1} - Av_k,v_{k+1}-v_k rangle
        geq
        mu | v_{k+1}-v_k|^p
        $$
        for every $k=0,ldots,n-1$.



        Summing up those terms will lead to the inequality
        $$
        tfrac 1n langle Av - A0 , v rangle geq mu n^{1-p}.
        $$



        For sufficiently large $n$, this cannot be true.






        share|cite|improve this answer












        Pick an arbitrary point $vin V$ such that $|v|=1$.



        Let $n$ be an arbitrary integer.
        (Our goal is a proof by contradiction as $ntoinfty$).



        We consider the points $v_k:= tfrac{k}{n} v$ for $k=0,ldots, n$.
        Then, by assumption we have
        $$
        langle A v_{k+1} - Av_k,v_{k+1}-v_k rangle
        geq
        mu | v_{k+1}-v_k|^p
        $$
        for every $k=0,ldots,n-1$.



        Summing up those terms will lead to the inequality
        $$
        tfrac 1n langle Av - A0 , v rangle geq mu n^{1-p}.
        $$



        For sufficiently large $n$, this cannot be true.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 7 hours ago









        supinf

        5,490926




        5,490926






























             

            draft saved


            draft discarded



















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2995388%2fhow-to-show-non-existence-of-an-operator-a-v-to-v%23new-answer', 'question_page');
            }
            );

            Post as a guest






































































            -

            Popular posts from this blog

            mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

            Image
            0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
            Read more

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more