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

            Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

            Image
            2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
            Read more

            ComboBox Display Member on multiple fields

            Image
            2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
            Read more

            Is it possible to collect Nectar points via Trainline?

            Image
            2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
            Read more