Does strong convergence in $H$ (or $L^2$) imply convergence in $V$ (or $W_0^{1,2}$)?












0












$begingroup$


Suppose:





  • $mathbf{V}$ and $mathbf{H}$ are Hilbert spaces.


  • $mathbf{V} hookrightarrow mathbf{H}$ is compact embedding.


  • $mathbf{V}$ is dense in $mathbf{H}$.


For example $mathbf{V} = W_0^{1,2}(Omega)$ and $mathbf{H} = L^2(Omega)$ for bounded $Omega in mathbb{R}^n$.



Let $mathbf{u}_n,mathbf{u} in mathbf{V}$. It's implied from the Rellich-Kondrachov embeddidng theorem that bounded $mathbf{u}_n$ in $mathbf{V}$ has convergent subsequence in $mathbf{H}$.



I am asking the opposite question:



Consider a bounded sequence $mathbf{u}_n in mathbf{V}$. If $mathbf{u}_n to mathbf{u}$ strongly in $mathbf{H}$, does it weakly convergent in $mathbf{V}$? Does it converge to the same $mathbf{u}$?



A more general question:



What requirements the sequence $(mathbf{u}_n)_{n in mathbb{N}}$ should have in $mathbf{H}$, so that $mathbf{u}_n$ is




  • weakly convergenent in $mathbf{V}$?

  • Strongly convergent in $mathbf{V}$?


Here is a similar question, but I do not know how this is shown.






functional-analysis hilbert-spaces sobolev-spaces weak-convergence







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Suppose:





    • $mathbf{V}$ and $mathbf{H}$ are Hilbert spaces.


    • $mathbf{V} hookrightarrow mathbf{H}$ is compact embedding.


    • $mathbf{V}$ is dense in $mathbf{H}$.


    For example $mathbf{V} = W_0^{1,2}(Omega)$ and $mathbf{H} = L^2(Omega)$ for bounded $Omega in mathbb{R}^n$.



    Let $mathbf{u}_n,mathbf{u} in mathbf{V}$. It's implied from the Rellich-Kondrachov embeddidng theorem that bounded $mathbf{u}_n$ in $mathbf{V}$ has convergent subsequence in $mathbf{H}$.



    I am asking the opposite question:



    Consider a bounded sequence $mathbf{u}_n in mathbf{V}$. If $mathbf{u}_n to mathbf{u}$ strongly in $mathbf{H}$, does it weakly convergent in $mathbf{V}$? Does it converge to the same $mathbf{u}$?



    A more general question:



    What requirements the sequence $(mathbf{u}_n)_{n in mathbb{N}}$ should have in $mathbf{H}$, so that $mathbf{u}_n$ is




    • weakly convergenent in $mathbf{V}$?

    • Strongly convergent in $mathbf{V}$?


    Here is a similar question, but I do not know how this is shown.






    functional-analysis hilbert-spaces sobolev-spaces weak-convergence







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose:





      • $mathbf{V}$ and $mathbf{H}$ are Hilbert spaces.


      • $mathbf{V} hookrightarrow mathbf{H}$ is compact embedding.


      • $mathbf{V}$ is dense in $mathbf{H}$.


      For example $mathbf{V} = W_0^{1,2}(Omega)$ and $mathbf{H} = L^2(Omega)$ for bounded $Omega in mathbb{R}^n$.



      Let $mathbf{u}_n,mathbf{u} in mathbf{V}$. It's implied from the Rellich-Kondrachov embeddidng theorem that bounded $mathbf{u}_n$ in $mathbf{V}$ has convergent subsequence in $mathbf{H}$.



      I am asking the opposite question:



      Consider a bounded sequence $mathbf{u}_n in mathbf{V}$. If $mathbf{u}_n to mathbf{u}$ strongly in $mathbf{H}$, does it weakly convergent in $mathbf{V}$? Does it converge to the same $mathbf{u}$?



      A more general question:



      What requirements the sequence $(mathbf{u}_n)_{n in mathbb{N}}$ should have in $mathbf{H}$, so that $mathbf{u}_n$ is




      • weakly convergenent in $mathbf{V}$?

      • Strongly convergent in $mathbf{V}$?


      Here is a similar question, but I do not know how this is shown.






      functional-analysis hilbert-spaces sobolev-spaces weak-convergence







      share|cite|improve this question











      $endgroup$




      Suppose:





      • $mathbf{V}$ and $mathbf{H}$ are Hilbert spaces.


      • $mathbf{V} hookrightarrow mathbf{H}$ is compact embedding.


      • $mathbf{V}$ is dense in $mathbf{H}$.


      For example $mathbf{V} = W_0^{1,2}(Omega)$ and $mathbf{H} = L^2(Omega)$ for bounded $Omega in mathbb{R}^n$.



      Let $mathbf{u}_n,mathbf{u} in mathbf{V}$. It's implied from the Rellich-Kondrachov embeddidng theorem that bounded $mathbf{u}_n$ in $mathbf{V}$ has convergent subsequence in $mathbf{H}$.



      I am asking the opposite question:



      Consider a bounded sequence $mathbf{u}_n in mathbf{V}$. If $mathbf{u}_n to mathbf{u}$ strongly in $mathbf{H}$, does it weakly convergent in $mathbf{V}$? Does it converge to the same $mathbf{u}$?



      A more general question:



      What requirements the sequence $(mathbf{u}_n)_{n in mathbb{N}}$ should have in $mathbf{H}$, so that $mathbf{u}_n$ is




      • weakly convergenent in $mathbf{V}$?

      • Strongly convergent in $mathbf{V}$?


      Here is a similar question, but I do not know how this is shown.






      functional-analysis hilbert-spaces sobolev-spaces weak-convergence




      functional-analysis hilbert-spaces sobolev-spaces weak-convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 12 '18 at 6:28







      Sia

















      asked Dec 12 '18 at 1:29









      SiaSia

      1064




      1064






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          If $(u_n)$ is bounded in $V$ and $u_nto u$ in $H$, then $u_nrightharpoonup u$ in $V$:



          As $(u_n)$ is bounded in $V$, it has a weakly converging subsequence $u_{n_k}rightharpoonup v$ in $V$. The continuity of the embedding into $H$ implies $v=u$ and $uin V$.



          Using this argument, one can prove that every subsequence of $(u_n)$ has a subsequence that converges weakly to $u$ in $V$. Hence $u_nrightharpoonup u$ in $V$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you @daw. As I understood from your answer, we do not need $H$ then. As long as $u_n in V$ is bounded, it has a weakly convergent subsequence $u_n rightharpoonup u in V$. Furthermore, if $u_n to v$ in $H$, then $u = v$.
            $endgroup$
            – Sia
            Dec 13 '18 at 1:33












          • $begingroup$
            A furthure question @daw: is it a consequence of Banach-Alaoglu theorem?
            $endgroup$
            – Sia
            Dec 13 '18 at 1:42












          • $begingroup$
            yes to both......
            $endgroup$
            – daw
            Dec 13 '18 at 7:14



















          0












          $begingroup$

          From ${u_n},u in H$ we can't imply ${u_n},u in V$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I assumed $u_n in V$, and $u_n$ strongly converge to $u$ in $H$. I edited the question.
            $endgroup$
            – Sia
            Dec 12 '18 at 5:10














          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%2f3036120%2fdoes-strong-convergence-in-h-or-l2-imply-convergence-in-v-or-w-01-2%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          If $(u_n)$ is bounded in $V$ and $u_nto u$ in $H$, then $u_nrightharpoonup u$ in $V$:



          As $(u_n)$ is bounded in $V$, it has a weakly converging subsequence $u_{n_k}rightharpoonup v$ in $V$. The continuity of the embedding into $H$ implies $v=u$ and $uin V$.



          Using this argument, one can prove that every subsequence of $(u_n)$ has a subsequence that converges weakly to $u$ in $V$. Hence $u_nrightharpoonup u$ in $V$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you @daw. As I understood from your answer, we do not need $H$ then. As long as $u_n in V$ is bounded, it has a weakly convergent subsequence $u_n rightharpoonup u in V$. Furthermore, if $u_n to v$ in $H$, then $u = v$.
            $endgroup$
            – Sia
            Dec 13 '18 at 1:33












          • $begingroup$
            A furthure question @daw: is it a consequence of Banach-Alaoglu theorem?
            $endgroup$
            – Sia
            Dec 13 '18 at 1:42












          • $begingroup$
            yes to both......
            $endgroup$
            – daw
            Dec 13 '18 at 7:14
















          1












          $begingroup$

          If $(u_n)$ is bounded in $V$ and $u_nto u$ in $H$, then $u_nrightharpoonup u$ in $V$:



          As $(u_n)$ is bounded in $V$, it has a weakly converging subsequence $u_{n_k}rightharpoonup v$ in $V$. The continuity of the embedding into $H$ implies $v=u$ and $uin V$.



          Using this argument, one can prove that every subsequence of $(u_n)$ has a subsequence that converges weakly to $u$ in $V$. Hence $u_nrightharpoonup u$ in $V$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you @daw. As I understood from your answer, we do not need $H$ then. As long as $u_n in V$ is bounded, it has a weakly convergent subsequence $u_n rightharpoonup u in V$. Furthermore, if $u_n to v$ in $H$, then $u = v$.
            $endgroup$
            – Sia
            Dec 13 '18 at 1:33












          • $begingroup$
            A furthure question @daw: is it a consequence of Banach-Alaoglu theorem?
            $endgroup$
            – Sia
            Dec 13 '18 at 1:42












          • $begingroup$
            yes to both......
            $endgroup$
            – daw
            Dec 13 '18 at 7:14














          1












          1








          1





          $begingroup$

          If $(u_n)$ is bounded in $V$ and $u_nto u$ in $H$, then $u_nrightharpoonup u$ in $V$:



          As $(u_n)$ is bounded in $V$, it has a weakly converging subsequence $u_{n_k}rightharpoonup v$ in $V$. The continuity of the embedding into $H$ implies $v=u$ and $uin V$.



          Using this argument, one can prove that every subsequence of $(u_n)$ has a subsequence that converges weakly to $u$ in $V$. Hence $u_nrightharpoonup u$ in $V$.






          share|cite|improve this answer









          $endgroup$



          If $(u_n)$ is bounded in $V$ and $u_nto u$ in $H$, then $u_nrightharpoonup u$ in $V$:



          As $(u_n)$ is bounded in $V$, it has a weakly converging subsequence $u_{n_k}rightharpoonup v$ in $V$. The continuity of the embedding into $H$ implies $v=u$ and $uin V$.



          Using this argument, one can prove that every subsequence of $(u_n)$ has a subsequence that converges weakly to $u$ in $V$. Hence $u_nrightharpoonup u$ in $V$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 15:18









          dawdaw

          25k1745




          25k1745












          • $begingroup$
            Thank you @daw. As I understood from your answer, we do not need $H$ then. As long as $u_n in V$ is bounded, it has a weakly convergent subsequence $u_n rightharpoonup u in V$. Furthermore, if $u_n to v$ in $H$, then $u = v$.
            $endgroup$
            – Sia
            Dec 13 '18 at 1:33












          • $begingroup$
            A furthure question @daw: is it a consequence of Banach-Alaoglu theorem?
            $endgroup$
            – Sia
            Dec 13 '18 at 1:42












          • $begingroup$
            yes to both......
            $endgroup$
            – daw
            Dec 13 '18 at 7:14


















          • $begingroup$
            Thank you @daw. As I understood from your answer, we do not need $H$ then. As long as $u_n in V$ is bounded, it has a weakly convergent subsequence $u_n rightharpoonup u in V$. Furthermore, if $u_n to v$ in $H$, then $u = v$.
            $endgroup$
            – Sia
            Dec 13 '18 at 1:33












          • $begingroup$
            A furthure question @daw: is it a consequence of Banach-Alaoglu theorem?
            $endgroup$
            – Sia
            Dec 13 '18 at 1:42












          • $begingroup$
            yes to both......
            $endgroup$
            – daw
            Dec 13 '18 at 7:14
















          $begingroup$
          Thank you @daw. As I understood from your answer, we do not need $H$ then. As long as $u_n in V$ is bounded, it has a weakly convergent subsequence $u_n rightharpoonup u in V$. Furthermore, if $u_n to v$ in $H$, then $u = v$.
          $endgroup$
          – Sia
          Dec 13 '18 at 1:33






          $begingroup$
          Thank you @daw. As I understood from your answer, we do not need $H$ then. As long as $u_n in V$ is bounded, it has a weakly convergent subsequence $u_n rightharpoonup u in V$. Furthermore, if $u_n to v$ in $H$, then $u = v$.
          $endgroup$
          – Sia
          Dec 13 '18 at 1:33














          $begingroup$
          A furthure question @daw: is it a consequence of Banach-Alaoglu theorem?
          $endgroup$
          – Sia
          Dec 13 '18 at 1:42






          $begingroup$
          A furthure question @daw: is it a consequence of Banach-Alaoglu theorem?
          $endgroup$
          – Sia
          Dec 13 '18 at 1:42














          $begingroup$
          yes to both......
          $endgroup$
          – daw
          Dec 13 '18 at 7:14




          $begingroup$
          yes to both......
          $endgroup$
          – daw
          Dec 13 '18 at 7:14











          0












          $begingroup$

          From ${u_n},u in H$ we can't imply ${u_n},u in V$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I assumed $u_n in V$, and $u_n$ strongly converge to $u$ in $H$. I edited the question.
            $endgroup$
            – Sia
            Dec 12 '18 at 5:10


















          0












          $begingroup$

          From ${u_n},u in H$ we can't imply ${u_n},u in V$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I assumed $u_n in V$, and $u_n$ strongly converge to $u$ in $H$. I edited the question.
            $endgroup$
            – Sia
            Dec 12 '18 at 5:10
















          0












          0








          0





          $begingroup$

          From ${u_n},u in H$ we can't imply ${u_n},u in V$.






          share|cite|improve this answer









          $endgroup$



          From ${u_n},u in H$ we can't imply ${u_n},u in V$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 5:08









          Trần Quang MinhTrần Quang Minh

          3347




          3347












          • $begingroup$
            I assumed $u_n in V$, and $u_n$ strongly converge to $u$ in $H$. I edited the question.
            $endgroup$
            – Sia
            Dec 12 '18 at 5:10




















          • $begingroup$
            I assumed $u_n in V$, and $u_n$ strongly converge to $u$ in $H$. I edited the question.
            $endgroup$
            – Sia
            Dec 12 '18 at 5:10


















          $begingroup$
          I assumed $u_n in V$, and $u_n$ strongly converge to $u$ in $H$. I edited the question.
          $endgroup$
          – Sia
          Dec 12 '18 at 5:10






          $begingroup$
          I assumed $u_n in V$, and $u_n$ strongly converge to $u$ in $H$. I edited the question.
          $endgroup$
          – Sia
          Dec 12 '18 at 5:10




















          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%2f3036120%2fdoes-strong-convergence-in-h-or-l2-imply-convergence-in-v-or-w-01-2%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

          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

          Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

          Image
          up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
          Read more