Proof that pointwise convergence can disrupt convergence












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$begingroup$


I'm trying to get a grasp on point-wise convergence and am hoping to prove something to give a concrete example of why it's weak. The lemma goes as follows .



Suppose $f _ { n } : [ a , b ] rightarrow mathbb { R }$ is a sequence of continuous functions
that converge point wise , but NOT UNIFORMLY to a continuous function $ f : [ a , b ] rightarrow mathbb { R } $ .



Then there exists a convergence sequence $ x _ { n } rightarrow x operatorname { in } [ a , b ] $ such that $f _ { n } left( x _ { n } right)$ does not converge to $ f(x) $ .






real-analysis analysis pointwise-convergence







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I'm trying to get a grasp on point-wise convergence and am hoping to prove something to give a concrete example of why it's weak. The lemma goes as follows .



    Suppose $f _ { n } : [ a , b ] rightarrow mathbb { R }$ is a sequence of continuous functions
    that converge point wise , but NOT UNIFORMLY to a continuous function $ f : [ a , b ] rightarrow mathbb { R } $ .



    Then there exists a convergence sequence $ x _ { n } rightarrow x operatorname { in } [ a , b ] $ such that $f _ { n } left( x _ { n } right)$ does not converge to $ f(x) $ .






    real-analysis analysis pointwise-convergence







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I'm trying to get a grasp on point-wise convergence and am hoping to prove something to give a concrete example of why it's weak. The lemma goes as follows .



      Suppose $f _ { n } : [ a , b ] rightarrow mathbb { R }$ is a sequence of continuous functions
      that converge point wise , but NOT UNIFORMLY to a continuous function $ f : [ a , b ] rightarrow mathbb { R } $ .



      Then there exists a convergence sequence $ x _ { n } rightarrow x operatorname { in } [ a , b ] $ such that $f _ { n } left( x _ { n } right)$ does not converge to $ f(x) $ .






      real-analysis analysis pointwise-convergence







      share|cite|improve this question











      $endgroup$




      I'm trying to get a grasp on point-wise convergence and am hoping to prove something to give a concrete example of why it's weak. The lemma goes as follows .



      Suppose $f _ { n } : [ a , b ] rightarrow mathbb { R }$ is a sequence of continuous functions
      that converge point wise , but NOT UNIFORMLY to a continuous function $ f : [ a , b ] rightarrow mathbb { R } $ .



      Then there exists a convergence sequence $ x _ { n } rightarrow x operatorname { in } [ a , b ] $ such that $f _ { n } left( x _ { n } right)$ does not converge to $ f(x) $ .






      real-analysis analysis pointwise-convergence




      real-analysis analysis pointwise-convergence






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 2:22







      Andrew Hardy

















      asked Dec 16 '18 at 2:17









      Andrew HardyAndrew Hardy

      125




      125






















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          $begingroup$

          Since the convergence is not uniform there exists $epsilon>0$ and a subsequence $f_{n_i}(x_{n_i})$ (where we can assume the $x_{n_i}$ are convergent to some $x$ because $[a,b]$ is compact) such that $|f_{n_i}(x_{n_i})-f(x_{n_i})|ge epsilon$. For $i$ large enough, $|f(x)-f(x_{n_i})|<epsilon/2$. By the Reverse Triangle Inequality:
          $$ |f_{n_i}(x_{n_i})-f(x)|ge||f_{n_i}(x_{n_i})-f(x_{n_i})| -|f(x_{n_i})-f(x)||ge epsilon/2$$
          So $f_{n_i}(x_{n_i})$ cannot converge to $f(x)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, I think this is 99% of it! Do you mind clarifying the initial claim. From lacking uniform convergence, we're finding a small enough $ epsilon $ so that the norm is not always less than it for all the elements in the sequence? That doesn't seem trivial to me. I'm also confused why you seemed to take two norms in your final inequality. Finally a proposed an edit about redefining $ 2 epsilon $ so your final inequality is more obviously sufficient?
            $endgroup$
            – Andrew Hardy
            Dec 16 '18 at 3:50










          • $begingroup$
            $f_nto f$ uniformly iff $lim max_{xin [a,b]} |f_n(x)-f(x)|=0$. If we do not have uniform convergence, this limit does not exist so there must be some $epsilon>0$ s.t. for any $N$, there is some $nge N$ with $max_{xin [a,b]} |f_n(x)-f(x)|ge epsilon$, so you can find an $x_n$ with $|f_n(x_n)-f(x_n)|geepsilon$. You can then repeat this process to obtain the subsequence in the initial claim.
            $endgroup$
            – Guacho Perez
            Dec 16 '18 at 4:19













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          $begingroup$

          Since the convergence is not uniform there exists $epsilon>0$ and a subsequence $f_{n_i}(x_{n_i})$ (where we can assume the $x_{n_i}$ are convergent to some $x$ because $[a,b]$ is compact) such that $|f_{n_i}(x_{n_i})-f(x_{n_i})|ge epsilon$. For $i$ large enough, $|f(x)-f(x_{n_i})|<epsilon/2$. By the Reverse Triangle Inequality:
          $$ |f_{n_i}(x_{n_i})-f(x)|ge||f_{n_i}(x_{n_i})-f(x_{n_i})| -|f(x_{n_i})-f(x)||ge epsilon/2$$
          So $f_{n_i}(x_{n_i})$ cannot converge to $f(x)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, I think this is 99% of it! Do you mind clarifying the initial claim. From lacking uniform convergence, we're finding a small enough $ epsilon $ so that the norm is not always less than it for all the elements in the sequence? That doesn't seem trivial to me. I'm also confused why you seemed to take two norms in your final inequality. Finally a proposed an edit about redefining $ 2 epsilon $ so your final inequality is more obviously sufficient?
            $endgroup$
            – Andrew Hardy
            Dec 16 '18 at 3:50










          • $begingroup$
            $f_nto f$ uniformly iff $lim max_{xin [a,b]} |f_n(x)-f(x)|=0$. If we do not have uniform convergence, this limit does not exist so there must be some $epsilon>0$ s.t. for any $N$, there is some $nge N$ with $max_{xin [a,b]} |f_n(x)-f(x)|ge epsilon$, so you can find an $x_n$ with $|f_n(x_n)-f(x_n)|geepsilon$. You can then repeat this process to obtain the subsequence in the initial claim.
            $endgroup$
            – Guacho Perez
            Dec 16 '18 at 4:19


















          0












          $begingroup$

          Since the convergence is not uniform there exists $epsilon>0$ and a subsequence $f_{n_i}(x_{n_i})$ (where we can assume the $x_{n_i}$ are convergent to some $x$ because $[a,b]$ is compact) such that $|f_{n_i}(x_{n_i})-f(x_{n_i})|ge epsilon$. For $i$ large enough, $|f(x)-f(x_{n_i})|<epsilon/2$. By the Reverse Triangle Inequality:
          $$ |f_{n_i}(x_{n_i})-f(x)|ge||f_{n_i}(x_{n_i})-f(x_{n_i})| -|f(x_{n_i})-f(x)||ge epsilon/2$$
          So $f_{n_i}(x_{n_i})$ cannot converge to $f(x)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, I think this is 99% of it! Do you mind clarifying the initial claim. From lacking uniform convergence, we're finding a small enough $ epsilon $ so that the norm is not always less than it for all the elements in the sequence? That doesn't seem trivial to me. I'm also confused why you seemed to take two norms in your final inequality. Finally a proposed an edit about redefining $ 2 epsilon $ so your final inequality is more obviously sufficient?
            $endgroup$
            – Andrew Hardy
            Dec 16 '18 at 3:50










          • $begingroup$
            $f_nto f$ uniformly iff $lim max_{xin [a,b]} |f_n(x)-f(x)|=0$. If we do not have uniform convergence, this limit does not exist so there must be some $epsilon>0$ s.t. for any $N$, there is some $nge N$ with $max_{xin [a,b]} |f_n(x)-f(x)|ge epsilon$, so you can find an $x_n$ with $|f_n(x_n)-f(x_n)|geepsilon$. You can then repeat this process to obtain the subsequence in the initial claim.
            $endgroup$
            – Guacho Perez
            Dec 16 '18 at 4:19
















          0












          0








          0





          $begingroup$

          Since the convergence is not uniform there exists $epsilon>0$ and a subsequence $f_{n_i}(x_{n_i})$ (where we can assume the $x_{n_i}$ are convergent to some $x$ because $[a,b]$ is compact) such that $|f_{n_i}(x_{n_i})-f(x_{n_i})|ge epsilon$. For $i$ large enough, $|f(x)-f(x_{n_i})|<epsilon/2$. By the Reverse Triangle Inequality:
          $$ |f_{n_i}(x_{n_i})-f(x)|ge||f_{n_i}(x_{n_i})-f(x_{n_i})| -|f(x_{n_i})-f(x)||ge epsilon/2$$
          So $f_{n_i}(x_{n_i})$ cannot converge to $f(x)$.






          share|cite|improve this answer









          $endgroup$



          Since the convergence is not uniform there exists $epsilon>0$ and a subsequence $f_{n_i}(x_{n_i})$ (where we can assume the $x_{n_i}$ are convergent to some $x$ because $[a,b]$ is compact) such that $|f_{n_i}(x_{n_i})-f(x_{n_i})|ge epsilon$. For $i$ large enough, $|f(x)-f(x_{n_i})|<epsilon/2$. By the Reverse Triangle Inequality:
          $$ |f_{n_i}(x_{n_i})-f(x)|ge||f_{n_i}(x_{n_i})-f(x_{n_i})| -|f(x_{n_i})-f(x)||ge epsilon/2$$
          So $f_{n_i}(x_{n_i})$ cannot converge to $f(x)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 3:19









          Guacho PerezGuacho Perez

          3,92911132




          3,92911132












          • $begingroup$
            Thanks, I think this is 99% of it! Do you mind clarifying the initial claim. From lacking uniform convergence, we're finding a small enough $ epsilon $ so that the norm is not always less than it for all the elements in the sequence? That doesn't seem trivial to me. I'm also confused why you seemed to take two norms in your final inequality. Finally a proposed an edit about redefining $ 2 epsilon $ so your final inequality is more obviously sufficient?
            $endgroup$
            – Andrew Hardy
            Dec 16 '18 at 3:50










          • $begingroup$
            $f_nto f$ uniformly iff $lim max_{xin [a,b]} |f_n(x)-f(x)|=0$. If we do not have uniform convergence, this limit does not exist so there must be some $epsilon>0$ s.t. for any $N$, there is some $nge N$ with $max_{xin [a,b]} |f_n(x)-f(x)|ge epsilon$, so you can find an $x_n$ with $|f_n(x_n)-f(x_n)|geepsilon$. You can then repeat this process to obtain the subsequence in the initial claim.
            $endgroup$
            – Guacho Perez
            Dec 16 '18 at 4:19




















          • $begingroup$
            Thanks, I think this is 99% of it! Do you mind clarifying the initial claim. From lacking uniform convergence, we're finding a small enough $ epsilon $ so that the norm is not always less than it for all the elements in the sequence? That doesn't seem trivial to me. I'm also confused why you seemed to take two norms in your final inequality. Finally a proposed an edit about redefining $ 2 epsilon $ so your final inequality is more obviously sufficient?
            $endgroup$
            – Andrew Hardy
            Dec 16 '18 at 3:50










          • $begingroup$
            $f_nto f$ uniformly iff $lim max_{xin [a,b]} |f_n(x)-f(x)|=0$. If we do not have uniform convergence, this limit does not exist so there must be some $epsilon>0$ s.t. for any $N$, there is some $nge N$ with $max_{xin [a,b]} |f_n(x)-f(x)|ge epsilon$, so you can find an $x_n$ with $|f_n(x_n)-f(x_n)|geepsilon$. You can then repeat this process to obtain the subsequence in the initial claim.
            $endgroup$
            – Guacho Perez
            Dec 16 '18 at 4:19


















          $begingroup$
          Thanks, I think this is 99% of it! Do you mind clarifying the initial claim. From lacking uniform convergence, we're finding a small enough $ epsilon $ so that the norm is not always less than it for all the elements in the sequence? That doesn't seem trivial to me. I'm also confused why you seemed to take two norms in your final inequality. Finally a proposed an edit about redefining $ 2 epsilon $ so your final inequality is more obviously sufficient?
          $endgroup$
          – Andrew Hardy
          Dec 16 '18 at 3:50




          $begingroup$
          Thanks, I think this is 99% of it! Do you mind clarifying the initial claim. From lacking uniform convergence, we're finding a small enough $ epsilon $ so that the norm is not always less than it for all the elements in the sequence? That doesn't seem trivial to me. I'm also confused why you seemed to take two norms in your final inequality. Finally a proposed an edit about redefining $ 2 epsilon $ so your final inequality is more obviously sufficient?
          $endgroup$
          – Andrew Hardy
          Dec 16 '18 at 3:50












          $begingroup$
          $f_nto f$ uniformly iff $lim max_{xin [a,b]} |f_n(x)-f(x)|=0$. If we do not have uniform convergence, this limit does not exist so there must be some $epsilon>0$ s.t. for any $N$, there is some $nge N$ with $max_{xin [a,b]} |f_n(x)-f(x)|ge epsilon$, so you can find an $x_n$ with $|f_n(x_n)-f(x_n)|geepsilon$. You can then repeat this process to obtain the subsequence in the initial claim.
          $endgroup$
          – Guacho Perez
          Dec 16 '18 at 4:19






          $begingroup$
          $f_nto f$ uniformly iff $lim max_{xin [a,b]} |f_n(x)-f(x)|=0$. If we do not have uniform convergence, this limit does not exist so there must be some $epsilon>0$ s.t. for any $N$, there is some $nge N$ with $max_{xin [a,b]} |f_n(x)-f(x)|ge epsilon$, so you can find an $x_n$ with $|f_n(x_n)-f(x_n)|geepsilon$. You can then repeat this process to obtain the subsequence in the initial claim.
          $endgroup$
          – Guacho Perez
          Dec 16 '18 at 4:19




















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