Weak convergence of bounded sequence $(x_n)$ in Hilbert space where $langle{x_n,yrangle}rightarrow...












2












$begingroup$


Let $H$ be a Hilbert Space endowed with the inner product $langle{.,.rangle}$ and $D$ a subset of $H$ such that span$(D)$ is dense in $H$. Show that, given a bounded sequence $(x_n)$ in $H$, such that $langle{x_n,yrangle}rightarrow langle{x_n,yrangle}$ for all $yin D$, then $x_n$ converges to $x$ weakly.



My Attempt
Let $fin H^*$ and $yin H$ (exist by Riesz): $f(x)=langle{x,yrangle}$ for all $xin H$. W.T.S $f(x_n)rightarrow f(x)$. Let $(y_m)$ sequence in span$(D)$: $y_mrightarrow y$ - (I'm not sure if I can do this as D may not countable).



begin{equation} f(x_n)= langle{x_n,lim y_mrangle}=lim langle{x_n,y_mrangle}rightarrow langle{x,yrangle}=f(x).end{equation}



Is this correct please?






functional-analysis hilbert-spaces weak-convergence







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$endgroup$

















    2












    $begingroup$


    Let $H$ be a Hilbert Space endowed with the inner product $langle{.,.rangle}$ and $D$ a subset of $H$ such that span$(D)$ is dense in $H$. Show that, given a bounded sequence $(x_n)$ in $H$, such that $langle{x_n,yrangle}rightarrow langle{x_n,yrangle}$ for all $yin D$, then $x_n$ converges to $x$ weakly.



    My Attempt
    Let $fin H^*$ and $yin H$ (exist by Riesz): $f(x)=langle{x,yrangle}$ for all $xin H$. W.T.S $f(x_n)rightarrow f(x)$. Let $(y_m)$ sequence in span$(D)$: $y_mrightarrow y$ - (I'm not sure if I can do this as D may not countable).



    begin{equation} f(x_n)= langle{x_n,lim y_mrangle}=lim langle{x_n,y_mrangle}rightarrow langle{x,yrangle}=f(x).end{equation}



    Is this correct please?






    functional-analysis hilbert-spaces weak-convergence







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      3



      $begingroup$


      Let $H$ be a Hilbert Space endowed with the inner product $langle{.,.rangle}$ and $D$ a subset of $H$ such that span$(D)$ is dense in $H$. Show that, given a bounded sequence $(x_n)$ in $H$, such that $langle{x_n,yrangle}rightarrow langle{x_n,yrangle}$ for all $yin D$, then $x_n$ converges to $x$ weakly.



      My Attempt
      Let $fin H^*$ and $yin H$ (exist by Riesz): $f(x)=langle{x,yrangle}$ for all $xin H$. W.T.S $f(x_n)rightarrow f(x)$. Let $(y_m)$ sequence in span$(D)$: $y_mrightarrow y$ - (I'm not sure if I can do this as D may not countable).



      begin{equation} f(x_n)= langle{x_n,lim y_mrangle}=lim langle{x_n,y_mrangle}rightarrow langle{x,yrangle}=f(x).end{equation}



      Is this correct please?






      functional-analysis hilbert-spaces weak-convergence







      share|cite|improve this question











      $endgroup$




      Let $H$ be a Hilbert Space endowed with the inner product $langle{.,.rangle}$ and $D$ a subset of $H$ such that span$(D)$ is dense in $H$. Show that, given a bounded sequence $(x_n)$ in $H$, such that $langle{x_n,yrangle}rightarrow langle{x_n,yrangle}$ for all $yin D$, then $x_n$ converges to $x$ weakly.



      My Attempt
      Let $fin H^*$ and $yin H$ (exist by Riesz): $f(x)=langle{x,yrangle}$ for all $xin H$. W.T.S $f(x_n)rightarrow f(x)$. Let $(y_m)$ sequence in span$(D)$: $y_mrightarrow y$ - (I'm not sure if I can do this as D may not countable).



      begin{equation} f(x_n)= langle{x_n,lim y_mrangle}=lim langle{x_n,y_mrangle}rightarrow langle{x,yrangle}=f(x).end{equation}



      Is this correct please?






      functional-analysis hilbert-spaces weak-convergence




      functional-analysis hilbert-spaces weak-convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 9:09







      Muhammad Mubarak

















      asked Jan 5 at 13:46









      Muhammad MubarakMuhammad Mubarak

      648




      648






















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          $begingroup$

          Your proof does not justify changing the order of the limits, though in fact two limits are interchangeable in this case by equicontinuity of $ymapsto langle x_n, yrangle$ on every bounded set. We will show
          $$
          K={yin H;|;lim_n langle x_n,yrangle = langle x,yrangle}
          $$           is a closed linear subspace of $H$. Linearity is obvious. To prove closedness, assume $(y_j)subset K$ converges to $yin H$. Then for all $j$, we have
          $$
          |langle x-x_n,yrangle|leq |langle x-x_n,y_jrangle|+|langle x-x_n,y-y_jrangle|leq |langle x-x_n,y_jrangle|+M|y-y_j|
          $$           where $Mleq |x|+sup_n|x_n|$. Take $nto infty$ to get
          $$
          limsup_n |langle x-x_n,yrangle|leq M|y-y_j|.
          $$           Let $jtoinfty$ to conclude
          $$
          limsup_n |langle x-x_n,yrangle|=0,
          $$           that is, $lim_n langle x_n,yrangle=langle x,yrangle$ and $yin K$.



          Now, since $Dsubset K$, we have $H = overline{text{span}} D subset K$ and hence $H=K$. This gives the desired result.






          share|cite|improve this answer











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            $begingroup$

            Your proof does not justify changing the order of the limits, though in fact two limits are interchangeable in this case by equicontinuity of $ymapsto langle x_n, yrangle$ on every bounded set. We will show
            $$
            K={yin H;|;lim_n langle x_n,yrangle = langle x,yrangle}
            $$             is a closed linear subspace of $H$. Linearity is obvious. To prove closedness, assume $(y_j)subset K$ converges to $yin H$. Then for all $j$, we have
            $$
            |langle x-x_n,yrangle|leq |langle x-x_n,y_jrangle|+|langle x-x_n,y-y_jrangle|leq |langle x-x_n,y_jrangle|+M|y-y_j|
            $$             where $Mleq |x|+sup_n|x_n|$. Take $nto infty$ to get
            $$
            limsup_n |langle x-x_n,yrangle|leq M|y-y_j|.
            $$             Let $jtoinfty$ to conclude
            $$
            limsup_n |langle x-x_n,yrangle|=0,
            $$             that is, $lim_n langle x_n,yrangle=langle x,yrangle$ and $yin K$.



            Now, since $Dsubset K$, we have $H = overline{text{span}} D subset K$ and hence $H=K$. This gives the desired result.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Your proof does not justify changing the order of the limits, though in fact two limits are interchangeable in this case by equicontinuity of $ymapsto langle x_n, yrangle$ on every bounded set. We will show
              $$
              K={yin H;|;lim_n langle x_n,yrangle = langle x,yrangle}
              $$               is a closed linear subspace of $H$. Linearity is obvious. To prove closedness, assume $(y_j)subset K$ converges to $yin H$. Then for all $j$, we have
              $$
              |langle x-x_n,yrangle|leq |langle x-x_n,y_jrangle|+|langle x-x_n,y-y_jrangle|leq |langle x-x_n,y_jrangle|+M|y-y_j|
              $$               where $Mleq |x|+sup_n|x_n|$. Take $nto infty$ to get
              $$
              limsup_n |langle x-x_n,yrangle|leq M|y-y_j|.
              $$               Let $jtoinfty$ to conclude
              $$
              limsup_n |langle x-x_n,yrangle|=0,
              $$               that is, $lim_n langle x_n,yrangle=langle x,yrangle$ and $yin K$.



              Now, since $Dsubset K$, we have $H = overline{text{span}} D subset K$ and hence $H=K$. This gives the desired result.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Your proof does not justify changing the order of the limits, though in fact two limits are interchangeable in this case by equicontinuity of $ymapsto langle x_n, yrangle$ on every bounded set. We will show
                $$
                K={yin H;|;lim_n langle x_n,yrangle = langle x,yrangle}
                $$                 is a closed linear subspace of $H$. Linearity is obvious. To prove closedness, assume $(y_j)subset K$ converges to $yin H$. Then for all $j$, we have
                $$
                |langle x-x_n,yrangle|leq |langle x-x_n,y_jrangle|+|langle x-x_n,y-y_jrangle|leq |langle x-x_n,y_jrangle|+M|y-y_j|
                $$                 where $Mleq |x|+sup_n|x_n|$. Take $nto infty$ to get
                $$
                limsup_n |langle x-x_n,yrangle|leq M|y-y_j|.
                $$                 Let $jtoinfty$ to conclude
                $$
                limsup_n |langle x-x_n,yrangle|=0,
                $$                 that is, $lim_n langle x_n,yrangle=langle x,yrangle$ and $yin K$.



                Now, since $Dsubset K$, we have $H = overline{text{span}} D subset K$ and hence $H=K$. This gives the desired result.






                share|cite|improve this answer











                $endgroup$



                Your proof does not justify changing the order of the limits, though in fact two limits are interchangeable in this case by equicontinuity of $ymapsto langle x_n, yrangle$ on every bounded set. We will show
                $$
                K={yin H;|;lim_n langle x_n,yrangle = langle x,yrangle}
                $$                 is a closed linear subspace of $H$. Linearity is obvious. To prove closedness, assume $(y_j)subset K$ converges to $yin H$. Then for all $j$, we have
                $$
                |langle x-x_n,yrangle|leq |langle x-x_n,y_jrangle|+|langle x-x_n,y-y_jrangle|leq |langle x-x_n,y_jrangle|+M|y-y_j|
                $$                 where $Mleq |x|+sup_n|x_n|$. Take $nto infty$ to get
                $$
                limsup_n |langle x-x_n,yrangle|leq M|y-y_j|.
                $$                 Let $jtoinfty$ to conclude
                $$
                limsup_n |langle x-x_n,yrangle|=0,
                $$                 that is, $lim_n langle x_n,yrangle=langle x,yrangle$ and $yin K$.



                Now, since $Dsubset K$, we have $H = overline{text{span}} D subset K$ and hence $H=K$. This gives the desired result.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 5 at 14:15

























                answered Jan 5 at 14:07









                SongSong

                7,318422




                7,318422






























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