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

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          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





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

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          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




















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