Prove $x^n$ is not uniformly convergent











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This question pertains to the sequence of functions $f_n(x)=x^n$ on the interval $[0,1]$.



It can be shown this sequence of functions ${f_n}$ converges point-wise to the limit $f$ where $f$ is defined by $f(x)=0$ on $[0,1)$ and $f(x)=1$ at $x=1$.



However, this sequence of functions ${f_n}$ does not converge uniformly to $f$. One way to prove this (which I have seen) is via a theorem which proves that if a sequence of functions ${f_n}$ converges uniformly to $f$, then $f$ is continuous. And clearly it is not the case that $f$ is continuous in our example, so our convergence is not uniform.



However, I have seen another test for uniform convergence on an interval $s$. That is that:



$${rm lim} [{rm sup} { | f_n(x)-f(x)| : xin S }]=0$$



This conception of uniform convergence can be found in my Walter Rudin analysis book for example. My question is, how can we use this definition of uniform convergence to show $x^n$ is not uniformly convergent?



I am having trouble seeing how to apply this definition to this example, especially since the limit $f$ is defined piece-wise. But I know I need to find the sup of the difference between $f_n$ and $f$ over the interval $[0,1]$ (how do I find this sup?). And then once I find it, I need to take the limit as $ntoinfty$ and show it does not equal $0$.



Thank you for your help!










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  • You probably want something explicit. Pick $epsilon<1$ small. Check that for any $delta>0$ there is an $N$ (you can explicitly find a formula for $N$ in terms of $epsilon,delta$) such that if $n>N$, then $|f_n(x)-f(x)|<epsilon$ if $0le xle 1-delta$. Use this to explicitly verify that uniform convergence is violated.
    – Andrés E. Caicedo
    Jan 27 '14 at 2:19















up vote
4
down vote

favorite
5












This question pertains to the sequence of functions $f_n(x)=x^n$ on the interval $[0,1]$.



It can be shown this sequence of functions ${f_n}$ converges point-wise to the limit $f$ where $f$ is defined by $f(x)=0$ on $[0,1)$ and $f(x)=1$ at $x=1$.



However, this sequence of functions ${f_n}$ does not converge uniformly to $f$. One way to prove this (which I have seen) is via a theorem which proves that if a sequence of functions ${f_n}$ converges uniformly to $f$, then $f$ is continuous. And clearly it is not the case that $f$ is continuous in our example, so our convergence is not uniform.



However, I have seen another test for uniform convergence on an interval $s$. That is that:



$${rm lim} [{rm sup} { | f_n(x)-f(x)| : xin S }]=0$$



This conception of uniform convergence can be found in my Walter Rudin analysis book for example. My question is, how can we use this definition of uniform convergence to show $x^n$ is not uniformly convergent?



I am having trouble seeing how to apply this definition to this example, especially since the limit $f$ is defined piece-wise. But I know I need to find the sup of the difference between $f_n$ and $f$ over the interval $[0,1]$ (how do I find this sup?). And then once I find it, I need to take the limit as $ntoinfty$ and show it does not equal $0$.



Thank you for your help!










share|cite|improve this question
























  • You probably want something explicit. Pick $epsilon<1$ small. Check that for any $delta>0$ there is an $N$ (you can explicitly find a formula for $N$ in terms of $epsilon,delta$) such that if $n>N$, then $|f_n(x)-f(x)|<epsilon$ if $0le xle 1-delta$. Use this to explicitly verify that uniform convergence is violated.
    – Andrés E. Caicedo
    Jan 27 '14 at 2:19













up vote
4
down vote

favorite
5









up vote
4
down vote

favorite
5






5





This question pertains to the sequence of functions $f_n(x)=x^n$ on the interval $[0,1]$.



It can be shown this sequence of functions ${f_n}$ converges point-wise to the limit $f$ where $f$ is defined by $f(x)=0$ on $[0,1)$ and $f(x)=1$ at $x=1$.



However, this sequence of functions ${f_n}$ does not converge uniformly to $f$. One way to prove this (which I have seen) is via a theorem which proves that if a sequence of functions ${f_n}$ converges uniformly to $f$, then $f$ is continuous. And clearly it is not the case that $f$ is continuous in our example, so our convergence is not uniform.



However, I have seen another test for uniform convergence on an interval $s$. That is that:



$${rm lim} [{rm sup} { | f_n(x)-f(x)| : xin S }]=0$$



This conception of uniform convergence can be found in my Walter Rudin analysis book for example. My question is, how can we use this definition of uniform convergence to show $x^n$ is not uniformly convergent?



I am having trouble seeing how to apply this definition to this example, especially since the limit $f$ is defined piece-wise. But I know I need to find the sup of the difference between $f_n$ and $f$ over the interval $[0,1]$ (how do I find this sup?). And then once I find it, I need to take the limit as $ntoinfty$ and show it does not equal $0$.



Thank you for your help!










share|cite|improve this question















This question pertains to the sequence of functions $f_n(x)=x^n$ on the interval $[0,1]$.



It can be shown this sequence of functions ${f_n}$ converges point-wise to the limit $f$ where $f$ is defined by $f(x)=0$ on $[0,1)$ and $f(x)=1$ at $x=1$.



However, this sequence of functions ${f_n}$ does not converge uniformly to $f$. One way to prove this (which I have seen) is via a theorem which proves that if a sequence of functions ${f_n}$ converges uniformly to $f$, then $f$ is continuous. And clearly it is not the case that $f$ is continuous in our example, so our convergence is not uniform.



However, I have seen another test for uniform convergence on an interval $s$. That is that:



$${rm lim} [{rm sup} { | f_n(x)-f(x)| : xin S }]=0$$



This conception of uniform convergence can be found in my Walter Rudin analysis book for example. My question is, how can we use this definition of uniform convergence to show $x^n$ is not uniformly convergent?



I am having trouble seeing how to apply this definition to this example, especially since the limit $f$ is defined piece-wise. But I know I need to find the sup of the difference between $f_n$ and $f$ over the interval $[0,1]$ (how do I find this sup?). And then once I find it, I need to take the limit as $ntoinfty$ and show it does not equal $0$.



Thank you for your help!







real-analysis sequences-and-series analysis convergence uniform-convergence






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edited Jan 27 '14 at 2:54









HK Lee

13.8k51957




13.8k51957










asked Jan 27 '14 at 1:33









Mathemanic

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  • You probably want something explicit. Pick $epsilon<1$ small. Check that for any $delta>0$ there is an $N$ (you can explicitly find a formula for $N$ in terms of $epsilon,delta$) such that if $n>N$, then $|f_n(x)-f(x)|<epsilon$ if $0le xle 1-delta$. Use this to explicitly verify that uniform convergence is violated.
    – Andrés E. Caicedo
    Jan 27 '14 at 2:19


















  • You probably want something explicit. Pick $epsilon<1$ small. Check that for any $delta>0$ there is an $N$ (you can explicitly find a formula for $N$ in terms of $epsilon,delta$) such that if $n>N$, then $|f_n(x)-f(x)|<epsilon$ if $0le xle 1-delta$. Use this to explicitly verify that uniform convergence is violated.
    – Andrés E. Caicedo
    Jan 27 '14 at 2:19
















You probably want something explicit. Pick $epsilon<1$ small. Check that for any $delta>0$ there is an $N$ (you can explicitly find a formula for $N$ in terms of $epsilon,delta$) such that if $n>N$, then $|f_n(x)-f(x)|<epsilon$ if $0le xle 1-delta$. Use this to explicitly verify that uniform convergence is violated.
– Andrés E. Caicedo
Jan 27 '14 at 2:19




You probably want something explicit. Pick $epsilon<1$ small. Check that for any $delta>0$ there is an $N$ (you can explicitly find a formula for $N$ in terms of $epsilon,delta$) such that if $n>N$, then $|f_n(x)-f(x)|<epsilon$ if $0le xle 1-delta$. Use this to explicitly verify that uniform convergence is violated.
– Andrés E. Caicedo
Jan 27 '14 at 2:19










2 Answers
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It all boils down to proving that $$sup_{xin [0,1]}|x^n-chi_{{1}}|=sup_{xin [0,1)}|x^n-0|=sup_{xin[0,1)}x^n=1notto 0$$






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  • 2




    @Pedro Tamaroff: What is $chi_{{1}}$?
    – Sujaan Kunalan
    Jul 7 '16 at 14:54










  • It is the indicator function for $x=1$.
    – Brofessor
    Sep 5 '17 at 13:45










  • Can you explain, why you can go from $[0,1]$ to $[0,1)$?
    – philmcole
    Jan 8 at 17:51




















up vote
2
down vote













I think the best way to see that this function doesn't converge uniformly on $x in [0,1]$ is to note that the limiting function is discontinuous for $x in [0,a], a<1$ and $x=1$:
$$
lim_{n to infty} f_n(x)= left{
begin{array}{rl}
0 &x in [0,a]\
1 & x=1
end{array}
right.
$$






share|cite|improve this answer





















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    2 Answers
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    2 Answers
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    up vote
    5
    down vote



    accepted










    It all boils down to proving that $$sup_{xin [0,1]}|x^n-chi_{{1}}|=sup_{xin [0,1)}|x^n-0|=sup_{xin[0,1)}x^n=1notto 0$$






    share|cite|improve this answer



















    • 2




      @Pedro Tamaroff: What is $chi_{{1}}$?
      – Sujaan Kunalan
      Jul 7 '16 at 14:54










    • It is the indicator function for $x=1$.
      – Brofessor
      Sep 5 '17 at 13:45










    • Can you explain, why you can go from $[0,1]$ to $[0,1)$?
      – philmcole
      Jan 8 at 17:51

















    up vote
    5
    down vote



    accepted










    It all boils down to proving that $$sup_{xin [0,1]}|x^n-chi_{{1}}|=sup_{xin [0,1)}|x^n-0|=sup_{xin[0,1)}x^n=1notto 0$$






    share|cite|improve this answer



















    • 2




      @Pedro Tamaroff: What is $chi_{{1}}$?
      – Sujaan Kunalan
      Jul 7 '16 at 14:54










    • It is the indicator function for $x=1$.
      – Brofessor
      Sep 5 '17 at 13:45










    • Can you explain, why you can go from $[0,1]$ to $[0,1)$?
      – philmcole
      Jan 8 at 17:51















    up vote
    5
    down vote



    accepted







    up vote
    5
    down vote



    accepted






    It all boils down to proving that $$sup_{xin [0,1]}|x^n-chi_{{1}}|=sup_{xin [0,1)}|x^n-0|=sup_{xin[0,1)}x^n=1notto 0$$






    share|cite|improve this answer














    It all boils down to proving that $$sup_{xin [0,1]}|x^n-chi_{{1}}|=sup_{xin [0,1)}|x^n-0|=sup_{xin[0,1)}x^n=1notto 0$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 27 '14 at 2:41

























    answered Jan 27 '14 at 1:34









    Pedro Tamaroff

    95.5k10149295




    95.5k10149295








    • 2




      @Pedro Tamaroff: What is $chi_{{1}}$?
      – Sujaan Kunalan
      Jul 7 '16 at 14:54










    • It is the indicator function for $x=1$.
      – Brofessor
      Sep 5 '17 at 13:45










    • Can you explain, why you can go from $[0,1]$ to $[0,1)$?
      – philmcole
      Jan 8 at 17:51
















    • 2




      @Pedro Tamaroff: What is $chi_{{1}}$?
      – Sujaan Kunalan
      Jul 7 '16 at 14:54










    • It is the indicator function for $x=1$.
      – Brofessor
      Sep 5 '17 at 13:45










    • Can you explain, why you can go from $[0,1]$ to $[0,1)$?
      – philmcole
      Jan 8 at 17:51










    2




    2




    @Pedro Tamaroff: What is $chi_{{1}}$?
    – Sujaan Kunalan
    Jul 7 '16 at 14:54




    @Pedro Tamaroff: What is $chi_{{1}}$?
    – Sujaan Kunalan
    Jul 7 '16 at 14:54












    It is the indicator function for $x=1$.
    – Brofessor
    Sep 5 '17 at 13:45




    It is the indicator function for $x=1$.
    – Brofessor
    Sep 5 '17 at 13:45












    Can you explain, why you can go from $[0,1]$ to $[0,1)$?
    – philmcole
    Jan 8 at 17:51






    Can you explain, why you can go from $[0,1]$ to $[0,1)$?
    – philmcole
    Jan 8 at 17:51












    up vote
    2
    down vote













    I think the best way to see that this function doesn't converge uniformly on $x in [0,1]$ is to note that the limiting function is discontinuous for $x in [0,a], a<1$ and $x=1$:
    $$
    lim_{n to infty} f_n(x)= left{
    begin{array}{rl}
    0 &x in [0,a]\
    1 & x=1
    end{array}
    right.
    $$






    share|cite|improve this answer

























      up vote
      2
      down vote













      I think the best way to see that this function doesn't converge uniformly on $x in [0,1]$ is to note that the limiting function is discontinuous for $x in [0,a], a<1$ and $x=1$:
      $$
      lim_{n to infty} f_n(x)= left{
      begin{array}{rl}
      0 &x in [0,a]\
      1 & x=1
      end{array}
      right.
      $$






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        I think the best way to see that this function doesn't converge uniformly on $x in [0,1]$ is to note that the limiting function is discontinuous for $x in [0,a], a<1$ and $x=1$:
        $$
        lim_{n to infty} f_n(x)= left{
        begin{array}{rl}
        0 &x in [0,a]\
        1 & x=1
        end{array}
        right.
        $$






        share|cite|improve this answer












        I think the best way to see that this function doesn't converge uniformly on $x in [0,1]$ is to note that the limiting function is discontinuous for $x in [0,a], a<1$ and $x=1$:
        $$
        lim_{n to infty} f_n(x)= left{
        begin{array}{rl}
        0 &x in [0,a]\
        1 & x=1
        end{array}
        right.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 27 '14 at 1:37









        Alex

        14.2k42133




        14.2k42133






























             

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