Convergence in a Hilbert Space












3












$begingroup$


I have a homework question I am attempting to no avail.



Let $mathcal{H}$ be a Hilbert space. Let $Xsubseteqmathcal{H}$ be a convex set. Suppose that $(x_{n})_{ngeq 1}$ is a sequence in $X$ such that,
begin{align}
lim_{nrightarrowinfty}|x_{n}|=inf_{xin X}|x|.
end{align}
Show that $x_{n}$ converges in $mathcal{H}$. Is the limit of $x_{n}$ necessarily in $X$?



I have tried to show that
begin{align}
(x_{n}-x|x_{n}-x)rightarrow 0,
end{align}
however I run into issues due to the limit of $|x_{n}|$ being defined as an infimum. In every case when the problem is reduced to $|x_{n}-x|$ I cannot make this go to zero because of the infimum. Also I have not identified the importance of $X$ being convex, this could also be the point I am missing. If this is the case could you please leave a small hint and allow me to return with more work.



EDIT: Is it true that $inf_{xin X}|x|implies P_{X}(0)$?






functional-analysis convergence hilbert-spaces







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I have a homework question I am attempting to no avail.



    Let $mathcal{H}$ be a Hilbert space. Let $Xsubseteqmathcal{H}$ be a convex set. Suppose that $(x_{n})_{ngeq 1}$ is a sequence in $X$ such that,
    begin{align}
    lim_{nrightarrowinfty}|x_{n}|=inf_{xin X}|x|.
    end{align}
    Show that $x_{n}$ converges in $mathcal{H}$. Is the limit of $x_{n}$ necessarily in $X$?



    I have tried to show that
    begin{align}
    (x_{n}-x|x_{n}-x)rightarrow 0,
    end{align}
    however I run into issues due to the limit of $|x_{n}|$ being defined as an infimum. In every case when the problem is reduced to $|x_{n}-x|$ I cannot make this go to zero because of the infimum. Also I have not identified the importance of $X$ being convex, this could also be the point I am missing. If this is the case could you please leave a small hint and allow me to return with more work.



    EDIT: Is it true that $inf_{xin X}|x|implies P_{X}(0)$?






    functional-analysis convergence hilbert-spaces







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I have a homework question I am attempting to no avail.



      Let $mathcal{H}$ be a Hilbert space. Let $Xsubseteqmathcal{H}$ be a convex set. Suppose that $(x_{n})_{ngeq 1}$ is a sequence in $X$ such that,
      begin{align}
      lim_{nrightarrowinfty}|x_{n}|=inf_{xin X}|x|.
      end{align}
      Show that $x_{n}$ converges in $mathcal{H}$. Is the limit of $x_{n}$ necessarily in $X$?



      I have tried to show that
      begin{align}
      (x_{n}-x|x_{n}-x)rightarrow 0,
      end{align}
      however I run into issues due to the limit of $|x_{n}|$ being defined as an infimum. In every case when the problem is reduced to $|x_{n}-x|$ I cannot make this go to zero because of the infimum. Also I have not identified the importance of $X$ being convex, this could also be the point I am missing. If this is the case could you please leave a small hint and allow me to return with more work.



      EDIT: Is it true that $inf_{xin X}|x|implies P_{X}(0)$?






      functional-analysis convergence hilbert-spaces







      share|cite|improve this question











      $endgroup$




      I have a homework question I am attempting to no avail.



      Let $mathcal{H}$ be a Hilbert space. Let $Xsubseteqmathcal{H}$ be a convex set. Suppose that $(x_{n})_{ngeq 1}$ is a sequence in $X$ such that,
      begin{align}
      lim_{nrightarrowinfty}|x_{n}|=inf_{xin X}|x|.
      end{align}
      Show that $x_{n}$ converges in $mathcal{H}$. Is the limit of $x_{n}$ necessarily in $X$?



      I have tried to show that
      begin{align}
      (x_{n}-x|x_{n}-x)rightarrow 0,
      end{align}
      however I run into issues due to the limit of $|x_{n}|$ being defined as an infimum. In every case when the problem is reduced to $|x_{n}-x|$ I cannot make this go to zero because of the infimum. Also I have not identified the importance of $X$ being convex, this could also be the point I am missing. If this is the case could you please leave a small hint and allow me to return with more work.



      EDIT: Is it true that $inf_{xin X}|x|implies P_{X}(0)$?






      functional-analysis convergence hilbert-spaces




      functional-analysis convergence hilbert-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 23 at 6:07







      Jack

















      asked Jan 23 at 5:59









      JackJack

      717




      717






















          1 Answer
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          5












          $begingroup$

          Note: this is very similar to the fact that in every Hilbert Space, on every closed convex set we have a projection operator.
          Let's show that $x_n$ is a Cauchy sequence, since you are in a Hilbert Space it must converge.
          Note that by the parallelogram equality we must have:



          $$ Vert x_m+x_n Vert ^2 + Vert x_m - x_n Vert^2 = 2(Vert x_n Vert^2 + Vert x_m Vert^2)$$
          So we have:
          $$ Vert x_m - x_nVert^2 = 2left(Vert x_n Vert^2 + Vert x_m Vert^2right) - Vert x_m +x_n Vert ^2 = 2left(Vert x_n Vert^2 + Vert x_m Vert^2 - 2Vert frac{x_m+x_n}{2}Vert ^2right)$$
          Since $X$ is convex, we know $frac{x_m+x_n}{2} in X$, now if $M = inf{Vert x Vert vert x in X}$ we must therfore have:



          $$ Vert x_m - x_n Vert ^2 leq 2(Vert x_n Vert^2 + Vert x_m Vert^2 - 2M^2) underset{n,mrightarrow infty}{rightarrow} 2(2M^2 - 2M^2) = 0$$
          Thus $x_n$ is Cauchy.



          The Limit doesn't have to be in $X$, unless it is closed, for example take an open ball in a Hilbert Space that doesn't contain the origin, and find an appropriate sequence.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you, I guess my whole approach was incorrect.
            $endgroup$
            – Jack
            Jan 23 at 6:28











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          1 Answer
          1






          active

          oldest

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          active

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          votes






          active

          oldest

          votes









          5












          $begingroup$

          Note: this is very similar to the fact that in every Hilbert Space, on every closed convex set we have a projection operator.
          Let's show that $x_n$ is a Cauchy sequence, since you are in a Hilbert Space it must converge.
          Note that by the parallelogram equality we must have:



          $$ Vert x_m+x_n Vert ^2 + Vert x_m - x_n Vert^2 = 2(Vert x_n Vert^2 + Vert x_m Vert^2)$$
          So we have:
          $$ Vert x_m - x_nVert^2 = 2left(Vert x_n Vert^2 + Vert x_m Vert^2right) - Vert x_m +x_n Vert ^2 = 2left(Vert x_n Vert^2 + Vert x_m Vert^2 - 2Vert frac{x_m+x_n}{2}Vert ^2right)$$
          Since $X$ is convex, we know $frac{x_m+x_n}{2} in X$, now if $M = inf{Vert x Vert vert x in X}$ we must therfore have:



          $$ Vert x_m - x_n Vert ^2 leq 2(Vert x_n Vert^2 + Vert x_m Vert^2 - 2M^2) underset{n,mrightarrow infty}{rightarrow} 2(2M^2 - 2M^2) = 0$$
          Thus $x_n$ is Cauchy.



          The Limit doesn't have to be in $X$, unless it is closed, for example take an open ball in a Hilbert Space that doesn't contain the origin, and find an appropriate sequence.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you, I guess my whole approach was incorrect.
            $endgroup$
            – Jack
            Jan 23 at 6:28
















          5












          $begingroup$

          Note: this is very similar to the fact that in every Hilbert Space, on every closed convex set we have a projection operator.
          Let's show that $x_n$ is a Cauchy sequence, since you are in a Hilbert Space it must converge.
          Note that by the parallelogram equality we must have:



          $$ Vert x_m+x_n Vert ^2 + Vert x_m - x_n Vert^2 = 2(Vert x_n Vert^2 + Vert x_m Vert^2)$$
          So we have:
          $$ Vert x_m - x_nVert^2 = 2left(Vert x_n Vert^2 + Vert x_m Vert^2right) - Vert x_m +x_n Vert ^2 = 2left(Vert x_n Vert^2 + Vert x_m Vert^2 - 2Vert frac{x_m+x_n}{2}Vert ^2right)$$
          Since $X$ is convex, we know $frac{x_m+x_n}{2} in X$, now if $M = inf{Vert x Vert vert x in X}$ we must therfore have:



          $$ Vert x_m - x_n Vert ^2 leq 2(Vert x_n Vert^2 + Vert x_m Vert^2 - 2M^2) underset{n,mrightarrow infty}{rightarrow} 2(2M^2 - 2M^2) = 0$$
          Thus $x_n$ is Cauchy.



          The Limit doesn't have to be in $X$, unless it is closed, for example take an open ball in a Hilbert Space that doesn't contain the origin, and find an appropriate sequence.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you, I guess my whole approach was incorrect.
            $endgroup$
            – Jack
            Jan 23 at 6:28














          5












          5








          5





          $begingroup$

          Note: this is very similar to the fact that in every Hilbert Space, on every closed convex set we have a projection operator.
          Let's show that $x_n$ is a Cauchy sequence, since you are in a Hilbert Space it must converge.
          Note that by the parallelogram equality we must have:



          $$ Vert x_m+x_n Vert ^2 + Vert x_m - x_n Vert^2 = 2(Vert x_n Vert^2 + Vert x_m Vert^2)$$
          So we have:
          $$ Vert x_m - x_nVert^2 = 2left(Vert x_n Vert^2 + Vert x_m Vert^2right) - Vert x_m +x_n Vert ^2 = 2left(Vert x_n Vert^2 + Vert x_m Vert^2 - 2Vert frac{x_m+x_n}{2}Vert ^2right)$$
          Since $X$ is convex, we know $frac{x_m+x_n}{2} in X$, now if $M = inf{Vert x Vert vert x in X}$ we must therfore have:



          $$ Vert x_m - x_n Vert ^2 leq 2(Vert x_n Vert^2 + Vert x_m Vert^2 - 2M^2) underset{n,mrightarrow infty}{rightarrow} 2(2M^2 - 2M^2) = 0$$
          Thus $x_n$ is Cauchy.



          The Limit doesn't have to be in $X$, unless it is closed, for example take an open ball in a Hilbert Space that doesn't contain the origin, and find an appropriate sequence.






          share|cite|improve this answer











          $endgroup$



          Note: this is very similar to the fact that in every Hilbert Space, on every closed convex set we have a projection operator.
          Let's show that $x_n$ is a Cauchy sequence, since you are in a Hilbert Space it must converge.
          Note that by the parallelogram equality we must have:



          $$ Vert x_m+x_n Vert ^2 + Vert x_m - x_n Vert^2 = 2(Vert x_n Vert^2 + Vert x_m Vert^2)$$
          So we have:
          $$ Vert x_m - x_nVert^2 = 2left(Vert x_n Vert^2 + Vert x_m Vert^2right) - Vert x_m +x_n Vert ^2 = 2left(Vert x_n Vert^2 + Vert x_m Vert^2 - 2Vert frac{x_m+x_n}{2}Vert ^2right)$$
          Since $X$ is convex, we know $frac{x_m+x_n}{2} in X$, now if $M = inf{Vert x Vert vert x in X}$ we must therfore have:



          $$ Vert x_m - x_n Vert ^2 leq 2(Vert x_n Vert^2 + Vert x_m Vert^2 - 2M^2) underset{n,mrightarrow infty}{rightarrow} 2(2M^2 - 2M^2) = 0$$
          Thus $x_n$ is Cauchy.



          The Limit doesn't have to be in $X$, unless it is closed, for example take an open ball in a Hilbert Space that doesn't contain the origin, and find an appropriate sequence.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 26 at 10:13









          Davide Giraudo

          127k17154268




          127k17154268










          answered Jan 23 at 6:18









          pitariverpitariver

          425113




          425113












          • $begingroup$
            Thank you, I guess my whole approach was incorrect.
            $endgroup$
            – Jack
            Jan 23 at 6:28


















          • $begingroup$
            Thank you, I guess my whole approach was incorrect.
            $endgroup$
            – Jack
            Jan 23 at 6:28
















          $begingroup$
          Thank you, I guess my whole approach was incorrect.
          $endgroup$
          – Jack
          Jan 23 at 6:28




          $begingroup$
          Thank you, I guess my whole approach was incorrect.
          $endgroup$
          – Jack
          Jan 23 at 6:28


















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