Convergence in $(L_P(a,b),|.|_p)$ doesn't imply convergence in $(C(a,b),sup)$












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Why is the continuous function $f_n(x)=(1+nx)^{-1}$ pointwise convergent to zero in $(L_P(a,b),|.|_p)$?



It seems to me for the considered function that $|f_n(x)-0|_p leq frac{1}{(1-p)} (frac{(1+n)^{1-p}}{n} - frac{1}{n}) < epsilon$
then for all $n>N$, $f_n(x) to 0$ uniformly in $(L_P(a,b),|.|_p)$ as $N$ depends on $epsilon$ and $p$.



And If I'm wrong and it is pointwise convergent in $(L_P(a,b),|.|_p)$ how did he use this information to deduce that it is not convergent in $(C(a,b),sup)$?










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    $begingroup$


    enter image description here



    Why is the continuous function $f_n(x)=(1+nx)^{-1}$ pointwise convergent to zero in $(L_P(a,b),|.|_p)$?



    It seems to me for the considered function that $|f_n(x)-0|_p leq frac{1}{(1-p)} (frac{(1+n)^{1-p}}{n} - frac{1}{n}) < epsilon$
    then for all $n>N$, $f_n(x) to 0$ uniformly in $(L_P(a,b),|.|_p)$ as $N$ depends on $epsilon$ and $p$.



    And If I'm wrong and it is pointwise convergent in $(L_P(a,b),|.|_p)$ how did he use this information to deduce that it is not convergent in $(C(a,b),sup)$?










    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      enter image description here



      Why is the continuous function $f_n(x)=(1+nx)^{-1}$ pointwise convergent to zero in $(L_P(a,b),|.|_p)$?



      It seems to me for the considered function that $|f_n(x)-0|_p leq frac{1}{(1-p)} (frac{(1+n)^{1-p}}{n} - frac{1}{n}) < epsilon$
      then for all $n>N$, $f_n(x) to 0$ uniformly in $(L_P(a,b),|.|_p)$ as $N$ depends on $epsilon$ and $p$.



      And If I'm wrong and it is pointwise convergent in $(L_P(a,b),|.|_p)$ how did he use this information to deduce that it is not convergent in $(C(a,b),sup)$?










      share|cite|improve this question











      $endgroup$




      enter image description here



      Why is the continuous function $f_n(x)=(1+nx)^{-1}$ pointwise convergent to zero in $(L_P(a,b),|.|_p)$?



      It seems to me for the considered function that $|f_n(x)-0|_p leq frac{1}{(1-p)} (frac{(1+n)^{1-p}}{n} - frac{1}{n}) < epsilon$
      then for all $n>N$, $f_n(x) to 0$ uniformly in $(L_P(a,b),|.|_p)$ as $N$ depends on $epsilon$ and $p$.



      And If I'm wrong and it is pointwise convergent in $(L_P(a,b),|.|_p)$ how did he use this information to deduce that it is not convergent in $(C(a,b),sup)$?







      functional-analysis convergence proof-explanation normed-spaces lp-spaces






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      edited Dec 26 '18 at 16:55









      Davide Giraudo

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      asked Dec 26 '18 at 9:32









      Dreamer123Dreamer123

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          $begingroup$

          Since $|f_n|_pto0$, we say that $f_n$ converges to $0$ in $L^p$ (the expression uniform convergence in $L^p$ is not used.)



          It is clear the $lim_{ntoinfty}f_n(x)$ is $0$ if $xin(0,1]$ and $1$ if $x=0$. Thus, $f_n$ converges pointwise to the discontinuous function $f(x)=0$ if $xin(0,1]$, $f(0)=1$. Since each $f_n$ is continuous, the convergence cannot be uniform (in $(C([0,1]),sup)$.) You can also see this by noting that $f_n(1/n)=1/2$ and $sup|f_n|ge1/2>0$.






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            $begingroup$

            First, note that the supremum norm of $C[a,b]$ is the norm of uniform convergence. This means a sequence $f_n to f$ wrt. to $||, ., ||_infty$ if and only if $f_n to f$ uniformly. Make sure that you understand why this is true and that you know the difference between pointwise and uniform convergence.



            $f_n to f$ with respect to $||, ., ||_p$ means only that
            begin{align}
            int_a^b |f_n(t) - f(t)|, dlambda(t) to 0 qquad text{as} qquad n to infty
            end{align}

            It doesn't say anything about pointwise or uniform convergence from first sight. In fact there are examples, where $f_n to f$ with respect to the $L^p$-norm, but $f_n(x) to f(x)$ at no point $x in (a,b)$. See eg. the answer to this question: Does convergence in $L^{p}$ implies convergence almost everywhere?.



            Your counterexample is a little bit difference. You have a sequence of functions that converges in $L^p$ and that converges pointwise (i.e. $forall x in [a,b]:f_n(x) to f(x)$ for some function $f$). But the sequence does not converge uniformly. To see this, just compute the maximum distance between two elements $sup_{tin [a,b]} |f_n(t) - f_m(t)| = ||f_n - f_m||_infty$. So it is an example that convertgence in $L^p[a,b]$ does not imply convergence in $C[a,b]$.



            There are other ways to see, that this can not be true. Suppose I have $f_n to f$ in $L^p$. If I change $f$ in point, then it won't do anything to the intgrals, so still $f_n to f'$. But it can definitly change pointwise (and hence also uniform) convergence.



            I hope this helps a little bit. "Why is the continuous function $f_n(x)=(1+nx)^{−1}$ pointwise convergent to zero in $(L^p(a,b),∥.∥_p)$?" does not make a lot of sense as a question, since the $L^p$ does not care about any properties on sets of measure zero (which includes single points).






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              2 Answers
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              0












              $begingroup$

              Since $|f_n|_pto0$, we say that $f_n$ converges to $0$ in $L^p$ (the expression uniform convergence in $L^p$ is not used.)



              It is clear the $lim_{ntoinfty}f_n(x)$ is $0$ if $xin(0,1]$ and $1$ if $x=0$. Thus, $f_n$ converges pointwise to the discontinuous function $f(x)=0$ if $xin(0,1]$, $f(0)=1$. Since each $f_n$ is continuous, the convergence cannot be uniform (in $(C([0,1]),sup)$.) You can also see this by noting that $f_n(1/n)=1/2$ and $sup|f_n|ge1/2>0$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Since $|f_n|_pto0$, we say that $f_n$ converges to $0$ in $L^p$ (the expression uniform convergence in $L^p$ is not used.)



                It is clear the $lim_{ntoinfty}f_n(x)$ is $0$ if $xin(0,1]$ and $1$ if $x=0$. Thus, $f_n$ converges pointwise to the discontinuous function $f(x)=0$ if $xin(0,1]$, $f(0)=1$. Since each $f_n$ is continuous, the convergence cannot be uniform (in $(C([0,1]),sup)$.) You can also see this by noting that $f_n(1/n)=1/2$ and $sup|f_n|ge1/2>0$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Since $|f_n|_pto0$, we say that $f_n$ converges to $0$ in $L^p$ (the expression uniform convergence in $L^p$ is not used.)



                  It is clear the $lim_{ntoinfty}f_n(x)$ is $0$ if $xin(0,1]$ and $1$ if $x=0$. Thus, $f_n$ converges pointwise to the discontinuous function $f(x)=0$ if $xin(0,1]$, $f(0)=1$. Since each $f_n$ is continuous, the convergence cannot be uniform (in $(C([0,1]),sup)$.) You can also see this by noting that $f_n(1/n)=1/2$ and $sup|f_n|ge1/2>0$.






                  share|cite|improve this answer









                  $endgroup$



                  Since $|f_n|_pto0$, we say that $f_n$ converges to $0$ in $L^p$ (the expression uniform convergence in $L^p$ is not used.)



                  It is clear the $lim_{ntoinfty}f_n(x)$ is $0$ if $xin(0,1]$ and $1$ if $x=0$. Thus, $f_n$ converges pointwise to the discontinuous function $f(x)=0$ if $xin(0,1]$, $f(0)=1$. Since each $f_n$ is continuous, the convergence cannot be uniform (in $(C([0,1]),sup)$.) You can also see this by noting that $f_n(1/n)=1/2$ and $sup|f_n|ge1/2>0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 26 '18 at 11:24









                  Julián AguirreJulián Aguirre

                  69.5k24297




                  69.5k24297























                      0












                      $begingroup$

                      First, note that the supremum norm of $C[a,b]$ is the norm of uniform convergence. This means a sequence $f_n to f$ wrt. to $||, ., ||_infty$ if and only if $f_n to f$ uniformly. Make sure that you understand why this is true and that you know the difference between pointwise and uniform convergence.



                      $f_n to f$ with respect to $||, ., ||_p$ means only that
                      begin{align}
                      int_a^b |f_n(t) - f(t)|, dlambda(t) to 0 qquad text{as} qquad n to infty
                      end{align}

                      It doesn't say anything about pointwise or uniform convergence from first sight. In fact there are examples, where $f_n to f$ with respect to the $L^p$-norm, but $f_n(x) to f(x)$ at no point $x in (a,b)$. See eg. the answer to this question: Does convergence in $L^{p}$ implies convergence almost everywhere?.



                      Your counterexample is a little bit difference. You have a sequence of functions that converges in $L^p$ and that converges pointwise (i.e. $forall x in [a,b]:f_n(x) to f(x)$ for some function $f$). But the sequence does not converge uniformly. To see this, just compute the maximum distance between two elements $sup_{tin [a,b]} |f_n(t) - f_m(t)| = ||f_n - f_m||_infty$. So it is an example that convertgence in $L^p[a,b]$ does not imply convergence in $C[a,b]$.



                      There are other ways to see, that this can not be true. Suppose I have $f_n to f$ in $L^p$. If I change $f$ in point, then it won't do anything to the intgrals, so still $f_n to f'$. But it can definitly change pointwise (and hence also uniform) convergence.



                      I hope this helps a little bit. "Why is the continuous function $f_n(x)=(1+nx)^{−1}$ pointwise convergent to zero in $(L^p(a,b),∥.∥_p)$?" does not make a lot of sense as a question, since the $L^p$ does not care about any properties on sets of measure zero (which includes single points).






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        First, note that the supremum norm of $C[a,b]$ is the norm of uniform convergence. This means a sequence $f_n to f$ wrt. to $||, ., ||_infty$ if and only if $f_n to f$ uniformly. Make sure that you understand why this is true and that you know the difference between pointwise and uniform convergence.



                        $f_n to f$ with respect to $||, ., ||_p$ means only that
                        begin{align}
                        int_a^b |f_n(t) - f(t)|, dlambda(t) to 0 qquad text{as} qquad n to infty
                        end{align}

                        It doesn't say anything about pointwise or uniform convergence from first sight. In fact there are examples, where $f_n to f$ with respect to the $L^p$-norm, but $f_n(x) to f(x)$ at no point $x in (a,b)$. See eg. the answer to this question: Does convergence in $L^{p}$ implies convergence almost everywhere?.



                        Your counterexample is a little bit difference. You have a sequence of functions that converges in $L^p$ and that converges pointwise (i.e. $forall x in [a,b]:f_n(x) to f(x)$ for some function $f$). But the sequence does not converge uniformly. To see this, just compute the maximum distance between two elements $sup_{tin [a,b]} |f_n(t) - f_m(t)| = ||f_n - f_m||_infty$. So it is an example that convertgence in $L^p[a,b]$ does not imply convergence in $C[a,b]$.



                        There are other ways to see, that this can not be true. Suppose I have $f_n to f$ in $L^p$. If I change $f$ in point, then it won't do anything to the intgrals, so still $f_n to f'$. But it can definitly change pointwise (and hence also uniform) convergence.



                        I hope this helps a little bit. "Why is the continuous function $f_n(x)=(1+nx)^{−1}$ pointwise convergent to zero in $(L^p(a,b),∥.∥_p)$?" does not make a lot of sense as a question, since the $L^p$ does not care about any properties on sets of measure zero (which includes single points).






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          First, note that the supremum norm of $C[a,b]$ is the norm of uniform convergence. This means a sequence $f_n to f$ wrt. to $||, ., ||_infty$ if and only if $f_n to f$ uniformly. Make sure that you understand why this is true and that you know the difference between pointwise and uniform convergence.



                          $f_n to f$ with respect to $||, ., ||_p$ means only that
                          begin{align}
                          int_a^b |f_n(t) - f(t)|, dlambda(t) to 0 qquad text{as} qquad n to infty
                          end{align}

                          It doesn't say anything about pointwise or uniform convergence from first sight. In fact there are examples, where $f_n to f$ with respect to the $L^p$-norm, but $f_n(x) to f(x)$ at no point $x in (a,b)$. See eg. the answer to this question: Does convergence in $L^{p}$ implies convergence almost everywhere?.



                          Your counterexample is a little bit difference. You have a sequence of functions that converges in $L^p$ and that converges pointwise (i.e. $forall x in [a,b]:f_n(x) to f(x)$ for some function $f$). But the sequence does not converge uniformly. To see this, just compute the maximum distance between two elements $sup_{tin [a,b]} |f_n(t) - f_m(t)| = ||f_n - f_m||_infty$. So it is an example that convertgence in $L^p[a,b]$ does not imply convergence in $C[a,b]$.



                          There are other ways to see, that this can not be true. Suppose I have $f_n to f$ in $L^p$. If I change $f$ in point, then it won't do anything to the intgrals, so still $f_n to f'$. But it can definitly change pointwise (and hence also uniform) convergence.



                          I hope this helps a little bit. "Why is the continuous function $f_n(x)=(1+nx)^{−1}$ pointwise convergent to zero in $(L^p(a,b),∥.∥_p)$?" does not make a lot of sense as a question, since the $L^p$ does not care about any properties on sets of measure zero (which includes single points).






                          share|cite|improve this answer











                          $endgroup$



                          First, note that the supremum norm of $C[a,b]$ is the norm of uniform convergence. This means a sequence $f_n to f$ wrt. to $||, ., ||_infty$ if and only if $f_n to f$ uniformly. Make sure that you understand why this is true and that you know the difference between pointwise and uniform convergence.



                          $f_n to f$ with respect to $||, ., ||_p$ means only that
                          begin{align}
                          int_a^b |f_n(t) - f(t)|, dlambda(t) to 0 qquad text{as} qquad n to infty
                          end{align}

                          It doesn't say anything about pointwise or uniform convergence from first sight. In fact there are examples, where $f_n to f$ with respect to the $L^p$-norm, but $f_n(x) to f(x)$ at no point $x in (a,b)$. See eg. the answer to this question: Does convergence in $L^{p}$ implies convergence almost everywhere?.



                          Your counterexample is a little bit difference. You have a sequence of functions that converges in $L^p$ and that converges pointwise (i.e. $forall x in [a,b]:f_n(x) to f(x)$ for some function $f$). But the sequence does not converge uniformly. To see this, just compute the maximum distance between two elements $sup_{tin [a,b]} |f_n(t) - f_m(t)| = ||f_n - f_m||_infty$. So it is an example that convertgence in $L^p[a,b]$ does not imply convergence in $C[a,b]$.



                          There are other ways to see, that this can not be true. Suppose I have $f_n to f$ in $L^p$. If I change $f$ in point, then it won't do anything to the intgrals, so still $f_n to f'$. But it can definitly change pointwise (and hence also uniform) convergence.



                          I hope this helps a little bit. "Why is the continuous function $f_n(x)=(1+nx)^{−1}$ pointwise convergent to zero in $(L^p(a,b),∥.∥_p)$?" does not make a lot of sense as a question, since the $L^p$ does not care about any properties on sets of measure zero (which includes single points).







                          share|cite|improve this answer














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                          edited Dec 26 '18 at 11:32

























                          answered Dec 26 '18 at 11:26









                          N.BeckN.Beck

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