A question about Riesz spaces












0












$begingroup$


A real vector space $E$ is said to be an ordered vector space whenever it is equipped with an order relation $ge$ that is compatible with the algebraic structure of $E$.



A Riesz space is an ordered vector space $E$ which for each pair of vectors $x,y in E$, the supremum and the infimum of the st ${x,y}$ both exist in $E$.
Following the classical notation, we shall write
$$x vee y := sup {x,y} quad , quad x wedge y := inf{x ,y } .$$
An example of Riesz space is function space $E$ of real valued functions on a set $Omega$ such that for each pair $f , g in E$ the functions $$[f vee g](w) := max {f(w),g(w)} quad, quad [f wedge g](w) := min{f(w) ,g(w) } $$ both belong to $E$.



A Riesz space is caled Dedekind complete whenever every nonempty bounded above subset has a supremum .



Here $mathcal{L}_b(E,F)$ is the vector space of all order bounded operators from $E$ to $F$.



By "postive operator" book of "Charalambos D.Aliprantis and Owen Burkinshow" we have the following theorem




Theorem(F.Riesz-Kantorovich)
. If $E$ and $F$ are Riesz spaces with $F$ Dedekind complete, thenthe ordered vector space $mathcal{L}_b(E,F)$ is a Dedekind complete Riesz space with the lattice operations $$|T| = sup{|Ty| : |y|le x },$$ $$ [S vee T](x)=sup{S(y)+T(z) : y,z in E^+ , y+z=x} ,$$ $$ [S wedge T](x)=sup{S(y)+T(z) : y,z in E^+ , y+z=x}$$ for all $S,T in mathcal{L}_b(E,F)$ and $x in E^+$.




Now By this theorem I want to prove the following exercise from the first section of this book:




Consider the positive operators $S,T : L_1[0,1] to L_1[0,1]$ defind by $$S(f)=f quad , quad T(f)=[int_0^1 f(x) dx].1$$ Then show that $S wedge T = 0$







functional-analysis operator-theory vector-lattices







share|cite|improve this question











$endgroup$












  • $begingroup$
    You have a typo in your theorem. It should be $(Swedge T)(x)=inf{S(y)+T(z):y,zin E^+,y+z=x}$.
    $endgroup$
    – Ben W
    Dec 29 '18 at 17:06
















0












$begingroup$


A real vector space $E$ is said to be an ordered vector space whenever it is equipped with an order relation $ge$ that is compatible with the algebraic structure of $E$.



A Riesz space is an ordered vector space $E$ which for each pair of vectors $x,y in E$, the supremum and the infimum of the st ${x,y}$ both exist in $E$.
Following the classical notation, we shall write
$$x vee y := sup {x,y} quad , quad x wedge y := inf{x ,y } .$$
An example of Riesz space is function space $E$ of real valued functions on a set $Omega$ such that for each pair $f , g in E$ the functions $$[f vee g](w) := max {f(w),g(w)} quad, quad [f wedge g](w) := min{f(w) ,g(w) } $$ both belong to $E$.



A Riesz space is caled Dedekind complete whenever every nonempty bounded above subset has a supremum .



Here $mathcal{L}_b(E,F)$ is the vector space of all order bounded operators from $E$ to $F$.



By "postive operator" book of "Charalambos D.Aliprantis and Owen Burkinshow" we have the following theorem




Theorem(F.Riesz-Kantorovich)
. If $E$ and $F$ are Riesz spaces with $F$ Dedekind complete, thenthe ordered vector space $mathcal{L}_b(E,F)$ is a Dedekind complete Riesz space with the lattice operations $$|T| = sup{|Ty| : |y|le x },$$ $$ [S vee T](x)=sup{S(y)+T(z) : y,z in E^+ , y+z=x} ,$$ $$ [S wedge T](x)=sup{S(y)+T(z) : y,z in E^+ , y+z=x}$$ for all $S,T in mathcal{L}_b(E,F)$ and $x in E^+$.




Now By this theorem I want to prove the following exercise from the first section of this book:




Consider the positive operators $S,T : L_1[0,1] to L_1[0,1]$ defind by $$S(f)=f quad , quad T(f)=[int_0^1 f(x) dx].1$$ Then show that $S wedge T = 0$







functional-analysis operator-theory vector-lattices







share|cite|improve this question











$endgroup$












  • $begingroup$
    You have a typo in your theorem. It should be $(Swedge T)(x)=inf{S(y)+T(z):y,zin E^+,y+z=x}$.
    $endgroup$
    – Ben W
    Dec 29 '18 at 17:06














0












0








0





$begingroup$


A real vector space $E$ is said to be an ordered vector space whenever it is equipped with an order relation $ge$ that is compatible with the algebraic structure of $E$.



A Riesz space is an ordered vector space $E$ which for each pair of vectors $x,y in E$, the supremum and the infimum of the st ${x,y}$ both exist in $E$.
Following the classical notation, we shall write
$$x vee y := sup {x,y} quad , quad x wedge y := inf{x ,y } .$$
An example of Riesz space is function space $E$ of real valued functions on a set $Omega$ such that for each pair $f , g in E$ the functions $$[f vee g](w) := max {f(w),g(w)} quad, quad [f wedge g](w) := min{f(w) ,g(w) } $$ both belong to $E$.



A Riesz space is caled Dedekind complete whenever every nonempty bounded above subset has a supremum .



Here $mathcal{L}_b(E,F)$ is the vector space of all order bounded operators from $E$ to $F$.



By "postive operator" book of "Charalambos D.Aliprantis and Owen Burkinshow" we have the following theorem




Theorem(F.Riesz-Kantorovich)
. If $E$ and $F$ are Riesz spaces with $F$ Dedekind complete, thenthe ordered vector space $mathcal{L}_b(E,F)$ is a Dedekind complete Riesz space with the lattice operations $$|T| = sup{|Ty| : |y|le x },$$ $$ [S vee T](x)=sup{S(y)+T(z) : y,z in E^+ , y+z=x} ,$$ $$ [S wedge T](x)=sup{S(y)+T(z) : y,z in E^+ , y+z=x}$$ for all $S,T in mathcal{L}_b(E,F)$ and $x in E^+$.




Now By this theorem I want to prove the following exercise from the first section of this book:




Consider the positive operators $S,T : L_1[0,1] to L_1[0,1]$ defind by $$S(f)=f quad , quad T(f)=[int_0^1 f(x) dx].1$$ Then show that $S wedge T = 0$







functional-analysis operator-theory vector-lattices







share|cite|improve this question











$endgroup$




A real vector space $E$ is said to be an ordered vector space whenever it is equipped with an order relation $ge$ that is compatible with the algebraic structure of $E$.



A Riesz space is an ordered vector space $E$ which for each pair of vectors $x,y in E$, the supremum and the infimum of the st ${x,y}$ both exist in $E$.
Following the classical notation, we shall write
$$x vee y := sup {x,y} quad , quad x wedge y := inf{x ,y } .$$
An example of Riesz space is function space $E$ of real valued functions on a set $Omega$ such that for each pair $f , g in E$ the functions $$[f vee g](w) := max {f(w),g(w)} quad, quad [f wedge g](w) := min{f(w) ,g(w) } $$ both belong to $E$.



A Riesz space is caled Dedekind complete whenever every nonempty bounded above subset has a supremum .



Here $mathcal{L}_b(E,F)$ is the vector space of all order bounded operators from $E$ to $F$.



By "postive operator" book of "Charalambos D.Aliprantis and Owen Burkinshow" we have the following theorem




Theorem(F.Riesz-Kantorovich)
. If $E$ and $F$ are Riesz spaces with $F$ Dedekind complete, thenthe ordered vector space $mathcal{L}_b(E,F)$ is a Dedekind complete Riesz space with the lattice operations $$|T| = sup{|Ty| : |y|le x },$$ $$ [S vee T](x)=sup{S(y)+T(z) : y,z in E^+ , y+z=x} ,$$ $$ [S wedge T](x)=sup{S(y)+T(z) : y,z in E^+ , y+z=x}$$ for all $S,T in mathcal{L}_b(E,F)$ and $x in E^+$.




Now By this theorem I want to prove the following exercise from the first section of this book:




Consider the positive operators $S,T : L_1[0,1] to L_1[0,1]$ defind by $$S(f)=f quad , quad T(f)=[int_0^1 f(x) dx].1$$ Then show that $S wedge T = 0$







functional-analysis operator-theory vector-lattices




functional-analysis operator-theory vector-lattices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 19:56







I-love-math

















asked Dec 29 '18 at 16:38









I-love-mathI-love-math

666




666












  • $begingroup$
    You have a typo in your theorem. It should be $(Swedge T)(x)=inf{S(y)+T(z):y,zin E^+,y+z=x}$.
    $endgroup$
    – Ben W
    Dec 29 '18 at 17:06


















  • $begingroup$
    You have a typo in your theorem. It should be $(Swedge T)(x)=inf{S(y)+T(z):y,zin E^+,y+z=x}$.
    $endgroup$
    – Ben W
    Dec 29 '18 at 17:06
















$begingroup$
You have a typo in your theorem. It should be $(Swedge T)(x)=inf{S(y)+T(z):y,zin E^+,y+z=x}$.
$endgroup$
– Ben W
Dec 29 '18 at 17:06




$begingroup$
You have a typo in your theorem. It should be $(Swedge T)(x)=inf{S(y)+T(z):y,zin E^+,y+z=x}$.
$endgroup$
– Ben W
Dec 29 '18 at 17:06










1 Answer
1






active

oldest

votes


















1












$begingroup$

Fix $fin L_1[0,1]$. For each $epsilon>0$ find $deltain(0,1)$ so that if $Asubset[0,1]$ has measure $<delta$ then $|fboldsymbol{1}_A|_{L_1[0,1]}<epsilon$. For each such $A$, write
$$f=fboldsymbol{1}_{[0,1]setminus A}+fboldsymbol{1}_A.$$
Observe that
$$(Swedge T)(f)leq fboldsymbol{1}_{[0,1]setminus A}+epsilonboldsymbol{1}.$$
In particular, $(Swedge T)(f)leqepsilon$ on $A$. Since $A$ is an arbitrary set of measure $<delta$, we have $(Swedge T)(f)leqepsilon$ on $[0,1]$. But $epsilon>0$ was arbitrary too, so $(Swedge T)(f)=0$.






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%2f3056020%2fa-question-about-riesz-spaces%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









    1












    $begingroup$

    Fix $fin L_1[0,1]$. For each $epsilon>0$ find $deltain(0,1)$ so that if $Asubset[0,1]$ has measure $<delta$ then $|fboldsymbol{1}_A|_{L_1[0,1]}<epsilon$. For each such $A$, write
    $$f=fboldsymbol{1}_{[0,1]setminus A}+fboldsymbol{1}_A.$$
    Observe that
    $$(Swedge T)(f)leq fboldsymbol{1}_{[0,1]setminus A}+epsilonboldsymbol{1}.$$
    In particular, $(Swedge T)(f)leqepsilon$ on $A$. Since $A$ is an arbitrary set of measure $<delta$, we have $(Swedge T)(f)leqepsilon$ on $[0,1]$. But $epsilon>0$ was arbitrary too, so $(Swedge T)(f)=0$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Fix $fin L_1[0,1]$. For each $epsilon>0$ find $deltain(0,1)$ so that if $Asubset[0,1]$ has measure $<delta$ then $|fboldsymbol{1}_A|_{L_1[0,1]}<epsilon$. For each such $A$, write
      $$f=fboldsymbol{1}_{[0,1]setminus A}+fboldsymbol{1}_A.$$
      Observe that
      $$(Swedge T)(f)leq fboldsymbol{1}_{[0,1]setminus A}+epsilonboldsymbol{1}.$$
      In particular, $(Swedge T)(f)leqepsilon$ on $A$. Since $A$ is an arbitrary set of measure $<delta$, we have $(Swedge T)(f)leqepsilon$ on $[0,1]$. But $epsilon>0$ was arbitrary too, so $(Swedge T)(f)=0$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Fix $fin L_1[0,1]$. For each $epsilon>0$ find $deltain(0,1)$ so that if $Asubset[0,1]$ has measure $<delta$ then $|fboldsymbol{1}_A|_{L_1[0,1]}<epsilon$. For each such $A$, write
        $$f=fboldsymbol{1}_{[0,1]setminus A}+fboldsymbol{1}_A.$$
        Observe that
        $$(Swedge T)(f)leq fboldsymbol{1}_{[0,1]setminus A}+epsilonboldsymbol{1}.$$
        In particular, $(Swedge T)(f)leqepsilon$ on $A$. Since $A$ is an arbitrary set of measure $<delta$, we have $(Swedge T)(f)leqepsilon$ on $[0,1]$. But $epsilon>0$ was arbitrary too, so $(Swedge T)(f)=0$.






        share|cite|improve this answer









        $endgroup$



        Fix $fin L_1[0,1]$. For each $epsilon>0$ find $deltain(0,1)$ so that if $Asubset[0,1]$ has measure $<delta$ then $|fboldsymbol{1}_A|_{L_1[0,1]}<epsilon$. For each such $A$, write
        $$f=fboldsymbol{1}_{[0,1]setminus A}+fboldsymbol{1}_A.$$
        Observe that
        $$(Swedge T)(f)leq fboldsymbol{1}_{[0,1]setminus A}+epsilonboldsymbol{1}.$$
        In particular, $(Swedge T)(f)leqepsilon$ on $A$. Since $A$ is an arbitrary set of measure $<delta$, we have $(Swedge T)(f)leqepsilon$ on $[0,1]$. But $epsilon>0$ was arbitrary too, so $(Swedge T)(f)=0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 17:05









        Ben WBen W

        2,276615




        2,276615






























            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%2f3056020%2fa-question-about-riesz-spaces%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

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Image
            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
            Read more

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
            Read more