Weak* Convergence exercise











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I'm dealing with this exercise about weak* convergence and I'm literally getting lost with indexes. I have this:



Let $X := c_0(mathbb{N}), hspace{3mm}x_0 in X^*=ell^1(mathbb{N}), hspace{3mm}{x_n}_{n in mathbb{N}} subset X^*$ bounded. Show that



begin{equation}
x_n rightharpoonup^* x_0 Longleftrightarrow x_n(k) rightarrow x_0(k)
end{equation}

for fixed $k in mathbb{N}$.










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    up vote
    0
    down vote

    favorite












    I'm dealing with this exercise about weak* convergence and I'm literally getting lost with indexes. I have this:



    Let $X := c_0(mathbb{N}), hspace{3mm}x_0 in X^*=ell^1(mathbb{N}), hspace{3mm}{x_n}_{n in mathbb{N}} subset X^*$ bounded. Show that



    begin{equation}
    x_n rightharpoonup^* x_0 Longleftrightarrow x_n(k) rightarrow x_0(k)
    end{equation}

    for fixed $k in mathbb{N}$.










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I'm dealing with this exercise about weak* convergence and I'm literally getting lost with indexes. I have this:



      Let $X := c_0(mathbb{N}), hspace{3mm}x_0 in X^*=ell^1(mathbb{N}), hspace{3mm}{x_n}_{n in mathbb{N}} subset X^*$ bounded. Show that



      begin{equation}
      x_n rightharpoonup^* x_0 Longleftrightarrow x_n(k) rightarrow x_0(k)
      end{equation}

      for fixed $k in mathbb{N}$.










      share|cite|improve this question













      I'm dealing with this exercise about weak* convergence and I'm literally getting lost with indexes. I have this:



      Let $X := c_0(mathbb{N}), hspace{3mm}x_0 in X^*=ell^1(mathbb{N}), hspace{3mm}{x_n}_{n in mathbb{N}} subset X^*$ bounded. Show that



      begin{equation}
      x_n rightharpoonup^* x_0 Longleftrightarrow x_n(k) rightarrow x_0(k)
      end{equation}

      for fixed $k in mathbb{N}$.







      functional-analysis weak-convergence






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      asked Nov 15 at 11:19









      James Arten

      579




      579






















          2 Answers
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          What you are supposed to prove is the following: suppose $sum_j |a_j| <infty, sum_j |a_{nj}|$ is bounded and $a_{nj} to a_j$ as $n to infty$ for each $j$; then $sum_j a_{nj} c_j to sum_j a_jc_j$ for every sequence $(c_j)$ which tends to $0$. To prove this let $epsilon >0$ and choose $N$ such that $|c_j| <epsilon$ for all $j geq N$. Then $|sum_j a_{nj} c_j - sum_j a_jc_j| leq |sum_j^{N-1} a_{nj} c_j - sum_j^{N-1} a_jc_j|+epsilonsum_{j=N}^{infty} |a_{nj}|$. Can you now complete the proof?



          [I am writing $a_{nj}$ for the j-th component of $x_n$ and $a_j$ for the j-th component of $x_0$].






          share|cite|improve this answer




























            up vote
            1
            down vote













            To avoid confusion with the indices, you can also tackle this problem abstractly. Recall the following general result:



            Theorem. Let $X$ be a topological space and $fin C(X)$. If $(f_n)$ is an equicontinuous sequence of continuous functions from $X$ to $mathbb{R}$ that converges to $f$ on a dense subset, then $(f_n)$ converges to $f$ everywhere.



            The proof is straightforward if you know the definition of equicontinuity.



            Now, if $X$ is a normed space and all the functions $f_n$ and $f$ are linear, then it suffices to have convergence on a total set (one which has a dense linear hull) instead of a dense set and equicontinuity is satisfied if $sup_nlVert f_nrVert<infty$. This solves the problem in question (as well as a bunch of similar exercises you might encounter in the future).






            share|cite|improve this answer





















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              2 Answers
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              2 Answers
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              active

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              up vote
              2
              down vote













              What you are supposed to prove is the following: suppose $sum_j |a_j| <infty, sum_j |a_{nj}|$ is bounded and $a_{nj} to a_j$ as $n to infty$ for each $j$; then $sum_j a_{nj} c_j to sum_j a_jc_j$ for every sequence $(c_j)$ which tends to $0$. To prove this let $epsilon >0$ and choose $N$ such that $|c_j| <epsilon$ for all $j geq N$. Then $|sum_j a_{nj} c_j - sum_j a_jc_j| leq |sum_j^{N-1} a_{nj} c_j - sum_j^{N-1} a_jc_j|+epsilonsum_{j=N}^{infty} |a_{nj}|$. Can you now complete the proof?



              [I am writing $a_{nj}$ for the j-th component of $x_n$ and $a_j$ for the j-th component of $x_0$].






              share|cite|improve this answer

























                up vote
                2
                down vote













                What you are supposed to prove is the following: suppose $sum_j |a_j| <infty, sum_j |a_{nj}|$ is bounded and $a_{nj} to a_j$ as $n to infty$ for each $j$; then $sum_j a_{nj} c_j to sum_j a_jc_j$ for every sequence $(c_j)$ which tends to $0$. To prove this let $epsilon >0$ and choose $N$ such that $|c_j| <epsilon$ for all $j geq N$. Then $|sum_j a_{nj} c_j - sum_j a_jc_j| leq |sum_j^{N-1} a_{nj} c_j - sum_j^{N-1} a_jc_j|+epsilonsum_{j=N}^{infty} |a_{nj}|$. Can you now complete the proof?



                [I am writing $a_{nj}$ for the j-th component of $x_n$ and $a_j$ for the j-th component of $x_0$].






                share|cite|improve this answer























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  What you are supposed to prove is the following: suppose $sum_j |a_j| <infty, sum_j |a_{nj}|$ is bounded and $a_{nj} to a_j$ as $n to infty$ for each $j$; then $sum_j a_{nj} c_j to sum_j a_jc_j$ for every sequence $(c_j)$ which tends to $0$. To prove this let $epsilon >0$ and choose $N$ such that $|c_j| <epsilon$ for all $j geq N$. Then $|sum_j a_{nj} c_j - sum_j a_jc_j| leq |sum_j^{N-1} a_{nj} c_j - sum_j^{N-1} a_jc_j|+epsilonsum_{j=N}^{infty} |a_{nj}|$. Can you now complete the proof?



                  [I am writing $a_{nj}$ for the j-th component of $x_n$ and $a_j$ for the j-th component of $x_0$].






                  share|cite|improve this answer












                  What you are supposed to prove is the following: suppose $sum_j |a_j| <infty, sum_j |a_{nj}|$ is bounded and $a_{nj} to a_j$ as $n to infty$ for each $j$; then $sum_j a_{nj} c_j to sum_j a_jc_j$ for every sequence $(c_j)$ which tends to $0$. To prove this let $epsilon >0$ and choose $N$ such that $|c_j| <epsilon$ for all $j geq N$. Then $|sum_j a_{nj} c_j - sum_j a_jc_j| leq |sum_j^{N-1} a_{nj} c_j - sum_j^{N-1} a_jc_j|+epsilonsum_{j=N}^{infty} |a_{nj}|$. Can you now complete the proof?



                  [I am writing $a_{nj}$ for the j-th component of $x_n$ and $a_j$ for the j-th component of $x_0$].







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 15 at 12:04









                  Kavi Rama Murthy

                  43k31751




                  43k31751






















                      up vote
                      1
                      down vote













                      To avoid confusion with the indices, you can also tackle this problem abstractly. Recall the following general result:



                      Theorem. Let $X$ be a topological space and $fin C(X)$. If $(f_n)$ is an equicontinuous sequence of continuous functions from $X$ to $mathbb{R}$ that converges to $f$ on a dense subset, then $(f_n)$ converges to $f$ everywhere.



                      The proof is straightforward if you know the definition of equicontinuity.



                      Now, if $X$ is a normed space and all the functions $f_n$ and $f$ are linear, then it suffices to have convergence on a total set (one which has a dense linear hull) instead of a dense set and equicontinuity is satisfied if $sup_nlVert f_nrVert<infty$. This solves the problem in question (as well as a bunch of similar exercises you might encounter in the future).






                      share|cite|improve this answer

























                        up vote
                        1
                        down vote













                        To avoid confusion with the indices, you can also tackle this problem abstractly. Recall the following general result:



                        Theorem. Let $X$ be a topological space and $fin C(X)$. If $(f_n)$ is an equicontinuous sequence of continuous functions from $X$ to $mathbb{R}$ that converges to $f$ on a dense subset, then $(f_n)$ converges to $f$ everywhere.



                        The proof is straightforward if you know the definition of equicontinuity.



                        Now, if $X$ is a normed space and all the functions $f_n$ and $f$ are linear, then it suffices to have convergence on a total set (one which has a dense linear hull) instead of a dense set and equicontinuity is satisfied if $sup_nlVert f_nrVert<infty$. This solves the problem in question (as well as a bunch of similar exercises you might encounter in the future).






                        share|cite|improve this answer























                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          To avoid confusion with the indices, you can also tackle this problem abstractly. Recall the following general result:



                          Theorem. Let $X$ be a topological space and $fin C(X)$. If $(f_n)$ is an equicontinuous sequence of continuous functions from $X$ to $mathbb{R}$ that converges to $f$ on a dense subset, then $(f_n)$ converges to $f$ everywhere.



                          The proof is straightforward if you know the definition of equicontinuity.



                          Now, if $X$ is a normed space and all the functions $f_n$ and $f$ are linear, then it suffices to have convergence on a total set (one which has a dense linear hull) instead of a dense set and equicontinuity is satisfied if $sup_nlVert f_nrVert<infty$. This solves the problem in question (as well as a bunch of similar exercises you might encounter in the future).






                          share|cite|improve this answer












                          To avoid confusion with the indices, you can also tackle this problem abstractly. Recall the following general result:



                          Theorem. Let $X$ be a topological space and $fin C(X)$. If $(f_n)$ is an equicontinuous sequence of continuous functions from $X$ to $mathbb{R}$ that converges to $f$ on a dense subset, then $(f_n)$ converges to $f$ everywhere.



                          The proof is straightforward if you know the definition of equicontinuity.



                          Now, if $X$ is a normed space and all the functions $f_n$ and $f$ are linear, then it suffices to have convergence on a total set (one which has a dense linear hull) instead of a dense set and equicontinuity is satisfied if $sup_nlVert f_nrVert<infty$. This solves the problem in question (as well as a bunch of similar exercises you might encounter in the future).







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 15 at 14:08









                          MaoWao

                          2,268416




                          2,268416






























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