Is the completion of $C_0^infty(mathbb{R}^n)$ with respect to $int_{mathbb{R}^n}| nabla varphi|^2dx$...












1












$begingroup$


Equip $C_0^infty(mathbb{R}^n)$ with the norm
$$ |varphi|^2_1 := int_{mathbb{R}^n}| nabla varphi|^2 text{d} x.$$



Indeed, $| cdot |_1$ is a norm on $C_0^infty(mathbb{R}^n)$ because any constant function with compact support must be identically zero. Let $tilde{H}_0^1(mathbb{R}^n)$ denote the completion of $C_0^infty(mathbb{R}^n)$ with respect to the above norm. Note that $tilde{H}_0^1(mathbb{R}^n)$ is not quite the usual space $H_0^1(mathbb{R}^n)$, which is the closure of $C_0^infty(mathbb{R}^n)$ with respect to the norm $(int_{mathbb{R}^n} |varphi|^2 + | nabla varphi|^2 text{d} x)^{frac{1}{2}}$.




Are there any references that describe properties of $tilde{H}_0^1(mathbb{R}^n)$? In particular, is it true that $tilde{H}_0^1(mathbb{R}^n) subseteq L^2(mathbb{R^n})$?




I have noticed that



$$| varphi |^2_1 = int_{mathbb{R}^n} | xi|^2 |hat{varphi }(xi)|^2 text{d} x.$$



Which has led me to conjecture that functions in $tilde{H}_0^1(mathbb{R}^n)$ may be allowed to have some blow-up near zero, hence it may be that $tilde{H}_0^1(mathbb{R}^n)$ is not contained in $L^2(mathbb{R}^n)$.



Solutions or reference suggestions are greatly appreciated.






real-analysis hilbert-spaces sobolev-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    @user1952009 mathworld.wolfram.com/BanachCompletion.html
    $endgroup$
    – Thomas
    Jun 1 '16 at 18:13










  • $begingroup$
    @thomas : yes I didn't express myself correctly, the completion of any normed vector space is a Banach space, but here do we get a weird Banach space of class of equivalence of functions, or a Banach space of measurable functions with the usual almost everywhere equivalence ?
    $endgroup$
    – reuns
    Jun 1 '16 at 18:16








  • 1




    $begingroup$
    @user1952009 Strictly speaking the metric completion is always comprised of equivalence classes of Cauchy sequences rather than elements of the space being completed. In fact this is true even when the space was already complete; the completion of $mathbb{R}$ is not $mathbb{R}$, from the set theory perspective. But we tend to ignore this because it does us little good.
    $endgroup$
    – Ian
    Jun 1 '16 at 18:18












  • $begingroup$
    @Ian : it seems related to the question : is it $subset L^2$ ? to me it means that if $F= G$ in his space then $F = G$ in $L^2$ ???
    $endgroup$
    – reuns
    Jun 1 '16 at 18:21












  • $begingroup$
    @user1952009 You can be a bit pedantic about the set theory but analysts know what is meant by "$subset L^2$". For example "$C^infty_c subset L^2$" is something we say all the time but it isn't true in the set-theoretic world.
    $endgroup$
    – Ian
    Jun 1 '16 at 18:22


















1












$begingroup$


Equip $C_0^infty(mathbb{R}^n)$ with the norm
$$ |varphi|^2_1 := int_{mathbb{R}^n}| nabla varphi|^2 text{d} x.$$



Indeed, $| cdot |_1$ is a norm on $C_0^infty(mathbb{R}^n)$ because any constant function with compact support must be identically zero. Let $tilde{H}_0^1(mathbb{R}^n)$ denote the completion of $C_0^infty(mathbb{R}^n)$ with respect to the above norm. Note that $tilde{H}_0^1(mathbb{R}^n)$ is not quite the usual space $H_0^1(mathbb{R}^n)$, which is the closure of $C_0^infty(mathbb{R}^n)$ with respect to the norm $(int_{mathbb{R}^n} |varphi|^2 + | nabla varphi|^2 text{d} x)^{frac{1}{2}}$.




Are there any references that describe properties of $tilde{H}_0^1(mathbb{R}^n)$? In particular, is it true that $tilde{H}_0^1(mathbb{R}^n) subseteq L^2(mathbb{R^n})$?




I have noticed that



$$| varphi |^2_1 = int_{mathbb{R}^n} | xi|^2 |hat{varphi }(xi)|^2 text{d} x.$$



Which has led me to conjecture that functions in $tilde{H}_0^1(mathbb{R}^n)$ may be allowed to have some blow-up near zero, hence it may be that $tilde{H}_0^1(mathbb{R}^n)$ is not contained in $L^2(mathbb{R}^n)$.



Solutions or reference suggestions are greatly appreciated.






real-analysis hilbert-spaces sobolev-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    @user1952009 mathworld.wolfram.com/BanachCompletion.html
    $endgroup$
    – Thomas
    Jun 1 '16 at 18:13










  • $begingroup$
    @thomas : yes I didn't express myself correctly, the completion of any normed vector space is a Banach space, but here do we get a weird Banach space of class of equivalence of functions, or a Banach space of measurable functions with the usual almost everywhere equivalence ?
    $endgroup$
    – reuns
    Jun 1 '16 at 18:16








  • 1




    $begingroup$
    @user1952009 Strictly speaking the metric completion is always comprised of equivalence classes of Cauchy sequences rather than elements of the space being completed. In fact this is true even when the space was already complete; the completion of $mathbb{R}$ is not $mathbb{R}$, from the set theory perspective. But we tend to ignore this because it does us little good.
    $endgroup$
    – Ian
    Jun 1 '16 at 18:18












  • $begingroup$
    @Ian : it seems related to the question : is it $subset L^2$ ? to me it means that if $F= G$ in his space then $F = G$ in $L^2$ ???
    $endgroup$
    – reuns
    Jun 1 '16 at 18:21












  • $begingroup$
    @user1952009 You can be a bit pedantic about the set theory but analysts know what is meant by "$subset L^2$". For example "$C^infty_c subset L^2$" is something we say all the time but it isn't true in the set-theoretic world.
    $endgroup$
    – Ian
    Jun 1 '16 at 18:22
















1












1








1





$begingroup$


Equip $C_0^infty(mathbb{R}^n)$ with the norm
$$ |varphi|^2_1 := int_{mathbb{R}^n}| nabla varphi|^2 text{d} x.$$



Indeed, $| cdot |_1$ is a norm on $C_0^infty(mathbb{R}^n)$ because any constant function with compact support must be identically zero. Let $tilde{H}_0^1(mathbb{R}^n)$ denote the completion of $C_0^infty(mathbb{R}^n)$ with respect to the above norm. Note that $tilde{H}_0^1(mathbb{R}^n)$ is not quite the usual space $H_0^1(mathbb{R}^n)$, which is the closure of $C_0^infty(mathbb{R}^n)$ with respect to the norm $(int_{mathbb{R}^n} |varphi|^2 + | nabla varphi|^2 text{d} x)^{frac{1}{2}}$.




Are there any references that describe properties of $tilde{H}_0^1(mathbb{R}^n)$? In particular, is it true that $tilde{H}_0^1(mathbb{R}^n) subseteq L^2(mathbb{R^n})$?




I have noticed that



$$| varphi |^2_1 = int_{mathbb{R}^n} | xi|^2 |hat{varphi }(xi)|^2 text{d} x.$$



Which has led me to conjecture that functions in $tilde{H}_0^1(mathbb{R}^n)$ may be allowed to have some blow-up near zero, hence it may be that $tilde{H}_0^1(mathbb{R}^n)$ is not contained in $L^2(mathbb{R}^n)$.



Solutions or reference suggestions are greatly appreciated.






real-analysis hilbert-spaces sobolev-spaces







share|cite|improve this question











$endgroup$




Equip $C_0^infty(mathbb{R}^n)$ with the norm
$$ |varphi|^2_1 := int_{mathbb{R}^n}| nabla varphi|^2 text{d} x.$$



Indeed, $| cdot |_1$ is a norm on $C_0^infty(mathbb{R}^n)$ because any constant function with compact support must be identically zero. Let $tilde{H}_0^1(mathbb{R}^n)$ denote the completion of $C_0^infty(mathbb{R}^n)$ with respect to the above norm. Note that $tilde{H}_0^1(mathbb{R}^n)$ is not quite the usual space $H_0^1(mathbb{R}^n)$, which is the closure of $C_0^infty(mathbb{R}^n)$ with respect to the norm $(int_{mathbb{R}^n} |varphi|^2 + | nabla varphi|^2 text{d} x)^{frac{1}{2}}$.




Are there any references that describe properties of $tilde{H}_0^1(mathbb{R}^n)$? In particular, is it true that $tilde{H}_0^1(mathbb{R}^n) subseteq L^2(mathbb{R^n})$?




I have noticed that



$$| varphi |^2_1 = int_{mathbb{R}^n} | xi|^2 |hat{varphi }(xi)|^2 text{d} x.$$



Which has led me to conjecture that functions in $tilde{H}_0^1(mathbb{R}^n)$ may be allowed to have some blow-up near zero, hence it may be that $tilde{H}_0^1(mathbb{R}^n)$ is not contained in $L^2(mathbb{R}^n)$.



Solutions or reference suggestions are greatly appreciated.






real-analysis hilbert-spaces sobolev-spaces




real-analysis hilbert-spaces sobolev-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 12:15









Falrach

1,701224




1,701224










asked Jun 1 '16 at 17:48









JZShapiroJZShapiro

1,99511125




1,99511125












  • $begingroup$
    @user1952009 mathworld.wolfram.com/BanachCompletion.html
    $endgroup$
    – Thomas
    Jun 1 '16 at 18:13










  • $begingroup$
    @thomas : yes I didn't express myself correctly, the completion of any normed vector space is a Banach space, but here do we get a weird Banach space of class of equivalence of functions, or a Banach space of measurable functions with the usual almost everywhere equivalence ?
    $endgroup$
    – reuns
    Jun 1 '16 at 18:16








  • 1




    $begingroup$
    @user1952009 Strictly speaking the metric completion is always comprised of equivalence classes of Cauchy sequences rather than elements of the space being completed. In fact this is true even when the space was already complete; the completion of $mathbb{R}$ is not $mathbb{R}$, from the set theory perspective. But we tend to ignore this because it does us little good.
    $endgroup$
    – Ian
    Jun 1 '16 at 18:18












  • $begingroup$
    @Ian : it seems related to the question : is it $subset L^2$ ? to me it means that if $F= G$ in his space then $F = G$ in $L^2$ ???
    $endgroup$
    – reuns
    Jun 1 '16 at 18:21












  • $begingroup$
    @user1952009 You can be a bit pedantic about the set theory but analysts know what is meant by "$subset L^2$". For example "$C^infty_c subset L^2$" is something we say all the time but it isn't true in the set-theoretic world.
    $endgroup$
    – Ian
    Jun 1 '16 at 18:22




















  • $begingroup$
    @user1952009 mathworld.wolfram.com/BanachCompletion.html
    $endgroup$
    – Thomas
    Jun 1 '16 at 18:13










  • $begingroup$
    @thomas : yes I didn't express myself correctly, the completion of any normed vector space is a Banach space, but here do we get a weird Banach space of class of equivalence of functions, or a Banach space of measurable functions with the usual almost everywhere equivalence ?
    $endgroup$
    – reuns
    Jun 1 '16 at 18:16








  • 1




    $begingroup$
    @user1952009 Strictly speaking the metric completion is always comprised of equivalence classes of Cauchy sequences rather than elements of the space being completed. In fact this is true even when the space was already complete; the completion of $mathbb{R}$ is not $mathbb{R}$, from the set theory perspective. But we tend to ignore this because it does us little good.
    $endgroup$
    – Ian
    Jun 1 '16 at 18:18












  • $begingroup$
    @Ian : it seems related to the question : is it $subset L^2$ ? to me it means that if $F= G$ in his space then $F = G$ in $L^2$ ???
    $endgroup$
    – reuns
    Jun 1 '16 at 18:21












  • $begingroup$
    @user1952009 You can be a bit pedantic about the set theory but analysts know what is meant by "$subset L^2$". For example "$C^infty_c subset L^2$" is something we say all the time but it isn't true in the set-theoretic world.
    $endgroup$
    – Ian
    Jun 1 '16 at 18:22


















$begingroup$
@user1952009 mathworld.wolfram.com/BanachCompletion.html
$endgroup$
– Thomas
Jun 1 '16 at 18:13




$begingroup$
@user1952009 mathworld.wolfram.com/BanachCompletion.html
$endgroup$
– Thomas
Jun 1 '16 at 18:13












$begingroup$
@thomas : yes I didn't express myself correctly, the completion of any normed vector space is a Banach space, but here do we get a weird Banach space of class of equivalence of functions, or a Banach space of measurable functions with the usual almost everywhere equivalence ?
$endgroup$
– reuns
Jun 1 '16 at 18:16






$begingroup$
@thomas : yes I didn't express myself correctly, the completion of any normed vector space is a Banach space, but here do we get a weird Banach space of class of equivalence of functions, or a Banach space of measurable functions with the usual almost everywhere equivalence ?
$endgroup$
– reuns
Jun 1 '16 at 18:16






1




1




$begingroup$
@user1952009 Strictly speaking the metric completion is always comprised of equivalence classes of Cauchy sequences rather than elements of the space being completed. In fact this is true even when the space was already complete; the completion of $mathbb{R}$ is not $mathbb{R}$, from the set theory perspective. But we tend to ignore this because it does us little good.
$endgroup$
– Ian
Jun 1 '16 at 18:18






$begingroup$
@user1952009 Strictly speaking the metric completion is always comprised of equivalence classes of Cauchy sequences rather than elements of the space being completed. In fact this is true even when the space was already complete; the completion of $mathbb{R}$ is not $mathbb{R}$, from the set theory perspective. But we tend to ignore this because it does us little good.
$endgroup$
– Ian
Jun 1 '16 at 18:18














$begingroup$
@Ian : it seems related to the question : is it $subset L^2$ ? to me it means that if $F= G$ in his space then $F = G$ in $L^2$ ???
$endgroup$
– reuns
Jun 1 '16 at 18:21






$begingroup$
@Ian : it seems related to the question : is it $subset L^2$ ? to me it means that if $F= G$ in his space then $F = G$ in $L^2$ ???
$endgroup$
– reuns
Jun 1 '16 at 18:21














$begingroup$
@user1952009 You can be a bit pedantic about the set theory but analysts know what is meant by "$subset L^2$". For example "$C^infty_c subset L^2$" is something we say all the time but it isn't true in the set-theoretic world.
$endgroup$
– Ian
Jun 1 '16 at 18:22






$begingroup$
@user1952009 You can be a bit pedantic about the set theory but analysts know what is meant by "$subset L^2$". For example "$C^infty_c subset L^2$" is something we say all the time but it isn't true in the set-theoretic world.
$endgroup$
– Ian
Jun 1 '16 at 18:22












1 Answer
1






active

oldest

votes


















1












$begingroup$

The answer is no in general: Take for instance $u(x)= eta(x) |x|^{-n/p}$, where $eta$ is a smooth function that is zero in a neighborhood of the origin and is identically one outside of some big ball centered at the origin. You can check by polar coordinates that $nabla uin L^p$, but $unotin L^p$ (to check that test functions approximate $u$ you can use the usual cut-off functions $psi_R(x)=psi(x/R)$ for $psiin C_c^infty$ that is identically one in the unit ball).



What is true however is that this space is contained in $L^{2n/(n-2)}$ by the Gagliardo-Nirenberg inequality.






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%2f1808634%2fis-the-completion-of-c-0-infty-mathbbrn-with-respect-to-int-mathbbr%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












    $begingroup$

    The answer is no in general: Take for instance $u(x)= eta(x) |x|^{-n/p}$, where $eta$ is a smooth function that is zero in a neighborhood of the origin and is identically one outside of some big ball centered at the origin. You can check by polar coordinates that $nabla uin L^p$, but $unotin L^p$ (to check that test functions approximate $u$ you can use the usual cut-off functions $psi_R(x)=psi(x/R)$ for $psiin C_c^infty$ that is identically one in the unit ball).



    What is true however is that this space is contained in $L^{2n/(n-2)}$ by the Gagliardo-Nirenberg inequality.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The answer is no in general: Take for instance $u(x)= eta(x) |x|^{-n/p}$, where $eta$ is a smooth function that is zero in a neighborhood of the origin and is identically one outside of some big ball centered at the origin. You can check by polar coordinates that $nabla uin L^p$, but $unotin L^p$ (to check that test functions approximate $u$ you can use the usual cut-off functions $psi_R(x)=psi(x/R)$ for $psiin C_c^infty$ that is identically one in the unit ball).



      What is true however is that this space is contained in $L^{2n/(n-2)}$ by the Gagliardo-Nirenberg inequality.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The answer is no in general: Take for instance $u(x)= eta(x) |x|^{-n/p}$, where $eta$ is a smooth function that is zero in a neighborhood of the origin and is identically one outside of some big ball centered at the origin. You can check by polar coordinates that $nabla uin L^p$, but $unotin L^p$ (to check that test functions approximate $u$ you can use the usual cut-off functions $psi_R(x)=psi(x/R)$ for $psiin C_c^infty$ that is identically one in the unit ball).



        What is true however is that this space is contained in $L^{2n/(n-2)}$ by the Gagliardo-Nirenberg inequality.






        share|cite|improve this answer









        $endgroup$



        The answer is no in general: Take for instance $u(x)= eta(x) |x|^{-n/p}$, where $eta$ is a smooth function that is zero in a neighborhood of the origin and is identically one outside of some big ball centered at the origin. You can check by polar coordinates that $nabla uin L^p$, but $unotin L^p$ (to check that test functions approximate $u$ you can use the usual cut-off functions $psi_R(x)=psi(x/R)$ for $psiin C_c^infty$ that is identically one in the unit ball).



        What is true however is that this space is contained in $L^{2n/(n-2)}$ by the Gagliardo-Nirenberg inequality.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jun 3 '16 at 1:49









        Jose27Jose27

        6,30011012




        6,30011012






























            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%2f1808634%2fis-the-completion-of-c-0-infty-mathbbrn-with-respect-to-int-mathbbr%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

            Index of /

            This page is only for reference, If you need detailed information, please check here
            Read more

            Tribalistas

            Image
              Nota: Para o álbum homônimo, veja Tribalistas (álbum de 2002). Para o segundo álbum homônimo, veja Tribalistas (álbum de 2017). Para o tipo de organização tribal, veja Tribalismo. Tribalistas Informação geral Origem Salvador e Rio de Janeiro País   Brasil Gênero(s) MPB Pop Samba Período em atividade 2002–2004 2017–atualmente Gravadora(s) EMI Phonomotor Records Universal Music Afiliação(ões) Dadi Pretinho da Serrinha Margareth Menezes Integrantes Arnaldo Antunes Carlinhos Brown Marisa Monte Página oficial www2.uol.com.br/tribalistas/ ...
            Read more

            Listed building

            Image
            Listed building (literalmente: prédio listado ) é toda estrutura protegida por lei pelo complexo sistema de tombamento do Reino Unido. Um listed building é qualquer estrutura oficialmente designada como tendo especial interesse arquitetônico, histórico ou cultural. É um estatuto amplamente usado, aplicado a centenas de milhares de estruturas. Estruturas tombadas não podem ser alteradas sem permissão especial, e desfrutam de isenções fiscais e outros benefícios. Cada um dos países do Reino Unido trata tombamento de forma diferente: Índice 1 Inglaterra 2 País de Gales 3 Irlanda do Norte 4 Escócia 5 Referências Inglaterra | O Palácio de Buckingham, Londres. As estruturas tombadas são listadas no Statutory List of Buildings of Special Architectural or Historic Interest , e incluem campos, sinais de trânsito, piers e áreas geográficas. A principal legislação pertinente é o Ancient Monuments and Archaeological Areas Act 1979 , [ 1 ] embora legisla...
            Read more