Problem about weak convergence












1












$begingroup$


I am reading a lecture note about Radon Riesz theorem, the resources is here: https://faculty.etsu.edu/gardnerr/5210/notes/Radon-Riesz.pdf



At page 4, fourth line from the bottom, it says convergence in measure. But if this result is implied by weak convergence of $f_n$. I can’t proof it.



I have tried some methods, but this lead to a question about the definition of weak convergence. In the definition, which of the following one is exactly the real definition of weak convergence?




  • for every $Tin X^*$, for all $varepsilon>0$ May related to $T$, there is an $N$, such that for all $n>N$, $d(Tf_n, Tf)<varepsilon$.

  • for all $varepsilon>0$, there is an $N$, such that for all $n>N$ and all $T$, $d(Tf_n, Tf)<varepsilon$.


I checked some book but they generally says like
$$Tf_nto Tf,qquad text{for all } Tin X^*$$



This makes curious for me.



Finally I almost forgot, if the definition is the second one I have proved: weak convergence implies convergence in measure, but if the definition is the first one, the prove failed, I also need a prove or a counterexample that weak convergence can’t imply convergence in measure, I also want to know how the lecture note get the result about convergence in measure if there is really an counterexample. THX.






real-analysis functional-analysis weak-convergence strong-convergence







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I am reading a lecture note about Radon Riesz theorem, the resources is here: https://faculty.etsu.edu/gardnerr/5210/notes/Radon-Riesz.pdf



    At page 4, fourth line from the bottom, it says convergence in measure. But if this result is implied by weak convergence of $f_n$. I can’t proof it.



    I have tried some methods, but this lead to a question about the definition of weak convergence. In the definition, which of the following one is exactly the real definition of weak convergence?




    • for every $Tin X^*$, for all $varepsilon>0$ May related to $T$, there is an $N$, such that for all $n>N$, $d(Tf_n, Tf)<varepsilon$.

    • for all $varepsilon>0$, there is an $N$, such that for all $n>N$ and all $T$, $d(Tf_n, Tf)<varepsilon$.


    I checked some book but they generally says like
    $$Tf_nto Tf,qquad text{for all } Tin X^*$$



    This makes curious for me.



    Finally I almost forgot, if the definition is the second one I have proved: weak convergence implies convergence in measure, but if the definition is the first one, the prove failed, I also need a prove or a counterexample that weak convergence can’t imply convergence in measure, I also want to know how the lecture note get the result about convergence in measure if there is really an counterexample. THX.






    real-analysis functional-analysis weak-convergence strong-convergence







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I am reading a lecture note about Radon Riesz theorem, the resources is here: https://faculty.etsu.edu/gardnerr/5210/notes/Radon-Riesz.pdf



      At page 4, fourth line from the bottom, it says convergence in measure. But if this result is implied by weak convergence of $f_n$. I can’t proof it.



      I have tried some methods, but this lead to a question about the definition of weak convergence. In the definition, which of the following one is exactly the real definition of weak convergence?




      • for every $Tin X^*$, for all $varepsilon>0$ May related to $T$, there is an $N$, such that for all $n>N$, $d(Tf_n, Tf)<varepsilon$.

      • for all $varepsilon>0$, there is an $N$, such that for all $n>N$ and all $T$, $d(Tf_n, Tf)<varepsilon$.


      I checked some book but they generally says like
      $$Tf_nto Tf,qquad text{for all } Tin X^*$$



      This makes curious for me.



      Finally I almost forgot, if the definition is the second one I have proved: weak convergence implies convergence in measure, but if the definition is the first one, the prove failed, I also need a prove or a counterexample that weak convergence can’t imply convergence in measure, I also want to know how the lecture note get the result about convergence in measure if there is really an counterexample. THX.






      real-analysis functional-analysis weak-convergence strong-convergence







      share|cite|improve this question











      $endgroup$




      I am reading a lecture note about Radon Riesz theorem, the resources is here: https://faculty.etsu.edu/gardnerr/5210/notes/Radon-Riesz.pdf



      At page 4, fourth line from the bottom, it says convergence in measure. But if this result is implied by weak convergence of $f_n$. I can’t proof it.



      I have tried some methods, but this lead to a question about the definition of weak convergence. In the definition, which of the following one is exactly the real definition of weak convergence?




      • for every $Tin X^*$, for all $varepsilon>0$ May related to $T$, there is an $N$, such that for all $n>N$, $d(Tf_n, Tf)<varepsilon$.

      • for all $varepsilon>0$, there is an $N$, such that for all $n>N$ and all $T$, $d(Tf_n, Tf)<varepsilon$.


      I checked some book but they generally says like
      $$Tf_nto Tf,qquad text{for all } Tin X^*$$



      This makes curious for me.



      Finally I almost forgot, if the definition is the second one I have proved: weak convergence implies convergence in measure, but if the definition is the first one, the prove failed, I also need a prove or a counterexample that weak convergence can’t imply convergence in measure, I also want to know how the lecture note get the result about convergence in measure if there is really an counterexample. THX.






      real-analysis functional-analysis weak-convergence strong-convergence




      real-analysis functional-analysis weak-convergence strong-convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 15:18







      Larry Eppes

















      asked Dec 26 '18 at 15:08









      Larry EppesLarry Eppes

      461311




      461311






















          1 Answer
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          1












          $begingroup$

          The second one is false. To see this, if $T$ is a (fixed) continuous linear functional, then any constant multiplied by $T$ is also a continuous linear functional. Then the second statement is essentially asking for
          $$|cT(f_n) - cT(f)| = c|T(f_n) - T(f)| leq epsilon$$
          for any constant $c$.



          The first one is (almost) correct, the $epsilon$ is still arbitrary, just now $N$ is allowed to depend on $T$.



          In general, weak convergence does not imply convergence in measure, take $f_n(x) = chi_{[n, n+1)}(x)$ which weakly converges to zero in $L^p(mathbb{R})$ for $1<p<infty$, but fails to converge in measure .






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I've got it, yes, for $X^*cong L^q$, so for every $gin L^q$, we can define $Tf:=int fg$ which is finite linearity(linear is obvious, finite is ensured by Holder inequality), then by $int_{[n,n+1)}gto 0$ implies weakly converges to zero.
            $endgroup$
            – Larry Eppes
            Dec 27 '18 at 2:25










          • $begingroup$
            I re-checked the notes in the link, and retried to understand the document in the fourth page, so ${f_n}to f$ in measure is false? although by $|f|^{p-2}fin L^q$, so that $Tg:=int |f|^{p-2}fg$, $forall gin L^p$ is a bounded linear functional, and weakly convergence imply $Tf_nto Tf$ directly without converge in measure.
            $endgroup$
            – Larry Eppes
            Dec 27 '18 at 2:39












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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          The second one is false. To see this, if $T$ is a (fixed) continuous linear functional, then any constant multiplied by $T$ is also a continuous linear functional. Then the second statement is essentially asking for
          $$|cT(f_n) - cT(f)| = c|T(f_n) - T(f)| leq epsilon$$
          for any constant $c$.



          The first one is (almost) correct, the $epsilon$ is still arbitrary, just now $N$ is allowed to depend on $T$.



          In general, weak convergence does not imply convergence in measure, take $f_n(x) = chi_{[n, n+1)}(x)$ which weakly converges to zero in $L^p(mathbb{R})$ for $1<p<infty$, but fails to converge in measure .






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I've got it, yes, for $X^*cong L^q$, so for every $gin L^q$, we can define $Tf:=int fg$ which is finite linearity(linear is obvious, finite is ensured by Holder inequality), then by $int_{[n,n+1)}gto 0$ implies weakly converges to zero.
            $endgroup$
            – Larry Eppes
            Dec 27 '18 at 2:25










          • $begingroup$
            I re-checked the notes in the link, and retried to understand the document in the fourth page, so ${f_n}to f$ in measure is false? although by $|f|^{p-2}fin L^q$, so that $Tg:=int |f|^{p-2}fg$, $forall gin L^p$ is a bounded linear functional, and weakly convergence imply $Tf_nto Tf$ directly without converge in measure.
            $endgroup$
            – Larry Eppes
            Dec 27 '18 at 2:39
















          1












          $begingroup$

          The second one is false. To see this, if $T$ is a (fixed) continuous linear functional, then any constant multiplied by $T$ is also a continuous linear functional. Then the second statement is essentially asking for
          $$|cT(f_n) - cT(f)| = c|T(f_n) - T(f)| leq epsilon$$
          for any constant $c$.



          The first one is (almost) correct, the $epsilon$ is still arbitrary, just now $N$ is allowed to depend on $T$.



          In general, weak convergence does not imply convergence in measure, take $f_n(x) = chi_{[n, n+1)}(x)$ which weakly converges to zero in $L^p(mathbb{R})$ for $1<p<infty$, but fails to converge in measure .






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I've got it, yes, for $X^*cong L^q$, so for every $gin L^q$, we can define $Tf:=int fg$ which is finite linearity(linear is obvious, finite is ensured by Holder inequality), then by $int_{[n,n+1)}gto 0$ implies weakly converges to zero.
            $endgroup$
            – Larry Eppes
            Dec 27 '18 at 2:25










          • $begingroup$
            I re-checked the notes in the link, and retried to understand the document in the fourth page, so ${f_n}to f$ in measure is false? although by $|f|^{p-2}fin L^q$, so that $Tg:=int |f|^{p-2}fg$, $forall gin L^p$ is a bounded linear functional, and weakly convergence imply $Tf_nto Tf$ directly without converge in measure.
            $endgroup$
            – Larry Eppes
            Dec 27 '18 at 2:39














          1












          1








          1





          $begingroup$

          The second one is false. To see this, if $T$ is a (fixed) continuous linear functional, then any constant multiplied by $T$ is also a continuous linear functional. Then the second statement is essentially asking for
          $$|cT(f_n) - cT(f)| = c|T(f_n) - T(f)| leq epsilon$$
          for any constant $c$.



          The first one is (almost) correct, the $epsilon$ is still arbitrary, just now $N$ is allowed to depend on $T$.



          In general, weak convergence does not imply convergence in measure, take $f_n(x) = chi_{[n, n+1)}(x)$ which weakly converges to zero in $L^p(mathbb{R})$ for $1<p<infty$, but fails to converge in measure .






          share|cite|improve this answer









          $endgroup$



          The second one is false. To see this, if $T$ is a (fixed) continuous linear functional, then any constant multiplied by $T$ is also a continuous linear functional. Then the second statement is essentially asking for
          $$|cT(f_n) - cT(f)| = c|T(f_n) - T(f)| leq epsilon$$
          for any constant $c$.



          The first one is (almost) correct, the $epsilon$ is still arbitrary, just now $N$ is allowed to depend on $T$.



          In general, weak convergence does not imply convergence in measure, take $f_n(x) = chi_{[n, n+1)}(x)$ which weakly converges to zero in $L^p(mathbb{R})$ for $1<p<infty$, but fails to converge in measure .







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 26 '18 at 18:47









          XiaoXiao

          4,87811636




          4,87811636












          • $begingroup$
            I've got it, yes, for $X^*cong L^q$, so for every $gin L^q$, we can define $Tf:=int fg$ which is finite linearity(linear is obvious, finite is ensured by Holder inequality), then by $int_{[n,n+1)}gto 0$ implies weakly converges to zero.
            $endgroup$
            – Larry Eppes
            Dec 27 '18 at 2:25










          • $begingroup$
            I re-checked the notes in the link, and retried to understand the document in the fourth page, so ${f_n}to f$ in measure is false? although by $|f|^{p-2}fin L^q$, so that $Tg:=int |f|^{p-2}fg$, $forall gin L^p$ is a bounded linear functional, and weakly convergence imply $Tf_nto Tf$ directly without converge in measure.
            $endgroup$
            – Larry Eppes
            Dec 27 '18 at 2:39


















          • $begingroup$
            I've got it, yes, for $X^*cong L^q$, so for every $gin L^q$, we can define $Tf:=int fg$ which is finite linearity(linear is obvious, finite is ensured by Holder inequality), then by $int_{[n,n+1)}gto 0$ implies weakly converges to zero.
            $endgroup$
            – Larry Eppes
            Dec 27 '18 at 2:25










          • $begingroup$
            I re-checked the notes in the link, and retried to understand the document in the fourth page, so ${f_n}to f$ in measure is false? although by $|f|^{p-2}fin L^q$, so that $Tg:=int |f|^{p-2}fg$, $forall gin L^p$ is a bounded linear functional, and weakly convergence imply $Tf_nto Tf$ directly without converge in measure.
            $endgroup$
            – Larry Eppes
            Dec 27 '18 at 2:39
















          $begingroup$
          I've got it, yes, for $X^*cong L^q$, so for every $gin L^q$, we can define $Tf:=int fg$ which is finite linearity(linear is obvious, finite is ensured by Holder inequality), then by $int_{[n,n+1)}gto 0$ implies weakly converges to zero.
          $endgroup$
          – Larry Eppes
          Dec 27 '18 at 2:25




          $begingroup$
          I've got it, yes, for $X^*cong L^q$, so for every $gin L^q$, we can define $Tf:=int fg$ which is finite linearity(linear is obvious, finite is ensured by Holder inequality), then by $int_{[n,n+1)}gto 0$ implies weakly converges to zero.
          $endgroup$
          – Larry Eppes
          Dec 27 '18 at 2:25












          $begingroup$
          I re-checked the notes in the link, and retried to understand the document in the fourth page, so ${f_n}to f$ in measure is false? although by $|f|^{p-2}fin L^q$, so that $Tg:=int |f|^{p-2}fg$, $forall gin L^p$ is a bounded linear functional, and weakly convergence imply $Tf_nto Tf$ directly without converge in measure.
          $endgroup$
          – Larry Eppes
          Dec 27 '18 at 2:39




          $begingroup$
          I re-checked the notes in the link, and retried to understand the document in the fourth page, so ${f_n}to f$ in measure is false? although by $|f|^{p-2}fin L^q$, so that $Tg:=int |f|^{p-2}fg$, $forall gin L^p$ is a bounded linear functional, and weakly convergence imply $Tf_nto Tf$ directly without converge in measure.
          $endgroup$
          – Larry Eppes
          Dec 27 '18 at 2:39


















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