$L(V,mathbb{R})$ with $|cdot|_1$ complete?












3












$begingroup$


Let $(V,|cdot|_V)$ and $(W,|cdot|_W)$ be normed vector spaces. By $L(V,W)$ we denote the space of bounded linear operators from $V$ to $W.$ If $W$ is complete, then $L(V,W)$ with the operator norm $|T| := sup left{|Tu|_W : |u|_V leq 1right}$ is a Banach space.



Now assume in addition that $gamma$ is probability measure on $V$ with $mathrm{supp}, gamma = V$ and with existing first moment.



My Question: Is $L(V,mathbb{R})$ closed subspace in $(mathrm{L}_1(V),|cdot|_1)$?



In this case $L(V,mathbb{R})$ with $|cdot|_1$ would also be a Banach space.






general-topology functional-analysis lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    do you define $Vert f Vert_1:=Vert f Vert + Vert f' Vert$? What is $L_1(V)$?
    $endgroup$
    – noctusraid
    Dec 30 '18 at 13:22












  • $begingroup$
    $mathrm{L}_1(V)$ denotes the usual Lebesgue space that contains all measurable $f : V to mathbb{R}$ such that $$ |f|_1 := int_V |f(v)| , gamma(mathrm{d}v) < infty$$
    $endgroup$
    – H17
    Dec 30 '18 at 13:25












  • $begingroup$
    @H17 The norm on $L(V,mathbb{R})$ is irrelevant, right?
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:33










  • $begingroup$
    @H17 Also, there is a technicality that is bothering me. $L_1(V)$ consists of equivalence classes of functions whereas $L(V,mathbb{R})$ consists of genuine functions. So, I don't really view $L(V,mathbb{R})$ as a subset of $L_1(V)$.
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:35








  • 1




    $begingroup$
    Actually the random variable (construction) is just the identity $x mapsto x.$ Random variables are very useful to work with (describing events etc ...). However, in probability theory the interesting object is actually the probability measure. Random variables are in fact rarely necessary.
    $endgroup$
    – H17
    Dec 30 '18 at 15:30
















3












$begingroup$


Let $(V,|cdot|_V)$ and $(W,|cdot|_W)$ be normed vector spaces. By $L(V,W)$ we denote the space of bounded linear operators from $V$ to $W.$ If $W$ is complete, then $L(V,W)$ with the operator norm $|T| := sup left{|Tu|_W : |u|_V leq 1right}$ is a Banach space.



Now assume in addition that $gamma$ is probability measure on $V$ with $mathrm{supp}, gamma = V$ and with existing first moment.



My Question: Is $L(V,mathbb{R})$ closed subspace in $(mathrm{L}_1(V),|cdot|_1)$?



In this case $L(V,mathbb{R})$ with $|cdot|_1$ would also be a Banach space.






general-topology functional-analysis lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    do you define $Vert f Vert_1:=Vert f Vert + Vert f' Vert$? What is $L_1(V)$?
    $endgroup$
    – noctusraid
    Dec 30 '18 at 13:22












  • $begingroup$
    $mathrm{L}_1(V)$ denotes the usual Lebesgue space that contains all measurable $f : V to mathbb{R}$ such that $$ |f|_1 := int_V |f(v)| , gamma(mathrm{d}v) < infty$$
    $endgroup$
    – H17
    Dec 30 '18 at 13:25












  • $begingroup$
    @H17 The norm on $L(V,mathbb{R})$ is irrelevant, right?
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:33










  • $begingroup$
    @H17 Also, there is a technicality that is bothering me. $L_1(V)$ consists of equivalence classes of functions whereas $L(V,mathbb{R})$ consists of genuine functions. So, I don't really view $L(V,mathbb{R})$ as a subset of $L_1(V)$.
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:35








  • 1




    $begingroup$
    Actually the random variable (construction) is just the identity $x mapsto x.$ Random variables are very useful to work with (describing events etc ...). However, in probability theory the interesting object is actually the probability measure. Random variables are in fact rarely necessary.
    $endgroup$
    – H17
    Dec 30 '18 at 15:30














3












3








3





$begingroup$


Let $(V,|cdot|_V)$ and $(W,|cdot|_W)$ be normed vector spaces. By $L(V,W)$ we denote the space of bounded linear operators from $V$ to $W.$ If $W$ is complete, then $L(V,W)$ with the operator norm $|T| := sup left{|Tu|_W : |u|_V leq 1right}$ is a Banach space.



Now assume in addition that $gamma$ is probability measure on $V$ with $mathrm{supp}, gamma = V$ and with existing first moment.



My Question: Is $L(V,mathbb{R})$ closed subspace in $(mathrm{L}_1(V),|cdot|_1)$?



In this case $L(V,mathbb{R})$ with $|cdot|_1$ would also be a Banach space.






general-topology functional-analysis lp-spaces







share|cite|improve this question











$endgroup$




Let $(V,|cdot|_V)$ and $(W,|cdot|_W)$ be normed vector spaces. By $L(V,W)$ we denote the space of bounded linear operators from $V$ to $W.$ If $W$ is complete, then $L(V,W)$ with the operator norm $|T| := sup left{|Tu|_W : |u|_V leq 1right}$ is a Banach space.



Now assume in addition that $gamma$ is probability measure on $V$ with $mathrm{supp}, gamma = V$ and with existing first moment.



My Question: Is $L(V,mathbb{R})$ closed subspace in $(mathrm{L}_1(V),|cdot|_1)$?



In this case $L(V,mathbb{R})$ with $|cdot|_1$ would also be a Banach space.






general-topology functional-analysis lp-spaces




general-topology functional-analysis lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 14:21









Davide Giraudo

127k17154268




127k17154268










asked Dec 30 '18 at 13:16









H17H17

313




313












  • $begingroup$
    do you define $Vert f Vert_1:=Vert f Vert + Vert f' Vert$? What is $L_1(V)$?
    $endgroup$
    – noctusraid
    Dec 30 '18 at 13:22












  • $begingroup$
    $mathrm{L}_1(V)$ denotes the usual Lebesgue space that contains all measurable $f : V to mathbb{R}$ such that $$ |f|_1 := int_V |f(v)| , gamma(mathrm{d}v) < infty$$
    $endgroup$
    – H17
    Dec 30 '18 at 13:25












  • $begingroup$
    @H17 The norm on $L(V,mathbb{R})$ is irrelevant, right?
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:33










  • $begingroup$
    @H17 Also, there is a technicality that is bothering me. $L_1(V)$ consists of equivalence classes of functions whereas $L(V,mathbb{R})$ consists of genuine functions. So, I don't really view $L(V,mathbb{R})$ as a subset of $L_1(V)$.
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:35








  • 1




    $begingroup$
    Actually the random variable (construction) is just the identity $x mapsto x.$ Random variables are very useful to work with (describing events etc ...). However, in probability theory the interesting object is actually the probability measure. Random variables are in fact rarely necessary.
    $endgroup$
    – H17
    Dec 30 '18 at 15:30


















  • $begingroup$
    do you define $Vert f Vert_1:=Vert f Vert + Vert f' Vert$? What is $L_1(V)$?
    $endgroup$
    – noctusraid
    Dec 30 '18 at 13:22












  • $begingroup$
    $mathrm{L}_1(V)$ denotes the usual Lebesgue space that contains all measurable $f : V to mathbb{R}$ such that $$ |f|_1 := int_V |f(v)| , gamma(mathrm{d}v) < infty$$
    $endgroup$
    – H17
    Dec 30 '18 at 13:25












  • $begingroup$
    @H17 The norm on $L(V,mathbb{R})$ is irrelevant, right?
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:33










  • $begingroup$
    @H17 Also, there is a technicality that is bothering me. $L_1(V)$ consists of equivalence classes of functions whereas $L(V,mathbb{R})$ consists of genuine functions. So, I don't really view $L(V,mathbb{R})$ as a subset of $L_1(V)$.
    $endgroup$
    – mathworker21
    Dec 30 '18 at 13:35








  • 1




    $begingroup$
    Actually the random variable (construction) is just the identity $x mapsto x.$ Random variables are very useful to work with (describing events etc ...). However, in probability theory the interesting object is actually the probability measure. Random variables are in fact rarely necessary.
    $endgroup$
    – H17
    Dec 30 '18 at 15:30
















$begingroup$
do you define $Vert f Vert_1:=Vert f Vert + Vert f' Vert$? What is $L_1(V)$?
$endgroup$
– noctusraid
Dec 30 '18 at 13:22






$begingroup$
do you define $Vert f Vert_1:=Vert f Vert + Vert f' Vert$? What is $L_1(V)$?
$endgroup$
– noctusraid
Dec 30 '18 at 13:22














$begingroup$
$mathrm{L}_1(V)$ denotes the usual Lebesgue space that contains all measurable $f : V to mathbb{R}$ such that $$ |f|_1 := int_V |f(v)| , gamma(mathrm{d}v) < infty$$
$endgroup$
– H17
Dec 30 '18 at 13:25






$begingroup$
$mathrm{L}_1(V)$ denotes the usual Lebesgue space that contains all measurable $f : V to mathbb{R}$ such that $$ |f|_1 := int_V |f(v)| , gamma(mathrm{d}v) < infty$$
$endgroup$
– H17
Dec 30 '18 at 13:25














$begingroup$
@H17 The norm on $L(V,mathbb{R})$ is irrelevant, right?
$endgroup$
– mathworker21
Dec 30 '18 at 13:33




$begingroup$
@H17 The norm on $L(V,mathbb{R})$ is irrelevant, right?
$endgroup$
– mathworker21
Dec 30 '18 at 13:33












$begingroup$
@H17 Also, there is a technicality that is bothering me. $L_1(V)$ consists of equivalence classes of functions whereas $L(V,mathbb{R})$ consists of genuine functions. So, I don't really view $L(V,mathbb{R})$ as a subset of $L_1(V)$.
$endgroup$
– mathworker21
Dec 30 '18 at 13:35






$begingroup$
@H17 Also, there is a technicality that is bothering me. $L_1(V)$ consists of equivalence classes of functions whereas $L(V,mathbb{R})$ consists of genuine functions. So, I don't really view $L(V,mathbb{R})$ as a subset of $L_1(V)$.
$endgroup$
– mathworker21
Dec 30 '18 at 13:35






1




1




$begingroup$
Actually the random variable (construction) is just the identity $x mapsto x.$ Random variables are very useful to work with (describing events etc ...). However, in probability theory the interesting object is actually the probability measure. Random variables are in fact rarely necessary.
$endgroup$
– H17
Dec 30 '18 at 15:30




$begingroup$
Actually the random variable (construction) is just the identity $x mapsto x.$ Random variables are very useful to work with (describing events etc ...). However, in probability theory the interesting object is actually the probability measure. Random variables are in fact rarely necessary.
$endgroup$
– H17
Dec 30 '18 at 15:30










1 Answer
1






active

oldest

votes


















3












$begingroup$

The answer is no. Consider $V=l^2(mathbb{N}, mathbb{R})$ and let $e_i$ be the i-th basis vector of the usual Schauder basis. Set
$$ gamma = sum_{ngeq 1} 2^{-n} delta_{e_n}$$
And consider the following sequence of bounded linear functionals
$$l_m (x) = sum_{j=1}^m j cdot x_j $$
This forms a Cauchy sequence as for $m>k$ holds
$$ Vert l_m - l_k Vert = sum_{j=k+1}^m jcdot 2^{-j}$$
I leave it to you to you to show that this sequence does not converge to some bounded operator (assume it does and then do the computation for the norm to get a contradiction).



Added: I forgot to check that my probability measure admits the first moment (I did not even know how it was defined, thanks to H17 for telling me). In fact not only does the first moment exist, but all moments do. Indeed, for $kgeq 1$ we have
$$ int Vert x Vert^k gamma(dx) = sum_{ngeq 1} Vert e_n Vert^k cdot 2^{-n}
= sum_{ngeq 1} 2^{-n} = 1. $$






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%2f3056821%2flv-mathbbr-with-cdot-1-complete%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









    3












    $begingroup$

    The answer is no. Consider $V=l^2(mathbb{N}, mathbb{R})$ and let $e_i$ be the i-th basis vector of the usual Schauder basis. Set
    $$ gamma = sum_{ngeq 1} 2^{-n} delta_{e_n}$$
    And consider the following sequence of bounded linear functionals
    $$l_m (x) = sum_{j=1}^m j cdot x_j $$
    This forms a Cauchy sequence as for $m>k$ holds
    $$ Vert l_m - l_k Vert = sum_{j=k+1}^m jcdot 2^{-j}$$
    I leave it to you to you to show that this sequence does not converge to some bounded operator (assume it does and then do the computation for the norm to get a contradiction).



    Added: I forgot to check that my probability measure admits the first moment (I did not even know how it was defined, thanks to H17 for telling me). In fact not only does the first moment exist, but all moments do. Indeed, for $kgeq 1$ we have
    $$ int Vert x Vert^k gamma(dx) = sum_{ngeq 1} Vert e_n Vert^k cdot 2^{-n}
    = sum_{ngeq 1} 2^{-n} = 1. $$






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      The answer is no. Consider $V=l^2(mathbb{N}, mathbb{R})$ and let $e_i$ be the i-th basis vector of the usual Schauder basis. Set
      $$ gamma = sum_{ngeq 1} 2^{-n} delta_{e_n}$$
      And consider the following sequence of bounded linear functionals
      $$l_m (x) = sum_{j=1}^m j cdot x_j $$
      This forms a Cauchy sequence as for $m>k$ holds
      $$ Vert l_m - l_k Vert = sum_{j=k+1}^m jcdot 2^{-j}$$
      I leave it to you to you to show that this sequence does not converge to some bounded operator (assume it does and then do the computation for the norm to get a contradiction).



      Added: I forgot to check that my probability measure admits the first moment (I did not even know how it was defined, thanks to H17 for telling me). In fact not only does the first moment exist, but all moments do. Indeed, for $kgeq 1$ we have
      $$ int Vert x Vert^k gamma(dx) = sum_{ngeq 1} Vert e_n Vert^k cdot 2^{-n}
      = sum_{ngeq 1} 2^{-n} = 1. $$






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        The answer is no. Consider $V=l^2(mathbb{N}, mathbb{R})$ and let $e_i$ be the i-th basis vector of the usual Schauder basis. Set
        $$ gamma = sum_{ngeq 1} 2^{-n} delta_{e_n}$$
        And consider the following sequence of bounded linear functionals
        $$l_m (x) = sum_{j=1}^m j cdot x_j $$
        This forms a Cauchy sequence as for $m>k$ holds
        $$ Vert l_m - l_k Vert = sum_{j=k+1}^m jcdot 2^{-j}$$
        I leave it to you to you to show that this sequence does not converge to some bounded operator (assume it does and then do the computation for the norm to get a contradiction).



        Added: I forgot to check that my probability measure admits the first moment (I did not even know how it was defined, thanks to H17 for telling me). In fact not only does the first moment exist, but all moments do. Indeed, for $kgeq 1$ we have
        $$ int Vert x Vert^k gamma(dx) = sum_{ngeq 1} Vert e_n Vert^k cdot 2^{-n}
        = sum_{ngeq 1} 2^{-n} = 1. $$






        share|cite|improve this answer











        $endgroup$



        The answer is no. Consider $V=l^2(mathbb{N}, mathbb{R})$ and let $e_i$ be the i-th basis vector of the usual Schauder basis. Set
        $$ gamma = sum_{ngeq 1} 2^{-n} delta_{e_n}$$
        And consider the following sequence of bounded linear functionals
        $$l_m (x) = sum_{j=1}^m j cdot x_j $$
        This forms a Cauchy sequence as for $m>k$ holds
        $$ Vert l_m - l_k Vert = sum_{j=k+1}^m jcdot 2^{-j}$$
        I leave it to you to you to show that this sequence does not converge to some bounded operator (assume it does and then do the computation for the norm to get a contradiction).



        Added: I forgot to check that my probability measure admits the first moment (I did not even know how it was defined, thanks to H17 for telling me). In fact not only does the first moment exist, but all moments do. Indeed, for $kgeq 1$ we have
        $$ int Vert x Vert^k gamma(dx) = sum_{ngeq 1} Vert e_n Vert^k cdot 2^{-n}
        = sum_{ngeq 1} 2^{-n} = 1. $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 30 '18 at 15:10

























        answered Dec 30 '18 at 14:23









        Severin SchravenSeverin Schraven

        6,5301935




        6,5301935






























            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%2f3056821%2flv-mathbbr-with-cdot-1-complete%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