Why is $(f_n)_{ninmathbb{N}}$ with $f_n :[0,1]rightarrowmathbb{R}, f_n(x)=x^n$ not uniformly convergent?...












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  • Prove $x^n$ is not uniformly convergent

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If I choose $f:[0,1]rightarrow mathbb{R}$ with $f(1)=1$ and otherwise $f(x)=0$ then $||f_n-f||_{[0,1]}rightarrow0$ therefore it is uniformly convergent and because it is uniformly convergent $f$ would have to be continuous but it is not, where is the mistake?






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 Jan 8 at 23:19


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    Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
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    – copper.hat
    Jan 8 at 21:58
















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This question already has an answer here:




  • Prove $x^n$ is not uniformly convergent

    2 answers




If I choose $f:[0,1]rightarrow mathbb{R}$ with $f(1)=1$ and otherwise $f(x)=0$ then $||f_n-f||_{[0,1]}rightarrow0$ therefore it is uniformly convergent and because it is uniformly convergent $f$ would have to be continuous but it is not, where is the mistake?






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 Jan 8 at 23:19


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
    $endgroup$
    – copper.hat
    Jan 8 at 21:58














0












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0





$begingroup$



This question already has an answer here:




  • Prove $x^n$ is not uniformly convergent

    2 answers




If I choose $f:[0,1]rightarrow mathbb{R}$ with $f(1)=1$ and otherwise $f(x)=0$ then $||f_n-f||_{[0,1]}rightarrow0$ therefore it is uniformly convergent and because it is uniformly convergent $f$ would have to be continuous but it is not, where is the mistake?






real-analysis







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Prove $x^n$ is not uniformly convergent

    2 answers




If I choose $f:[0,1]rightarrow mathbb{R}$ with $f(1)=1$ and otherwise $f(x)=0$ then $||f_n-f||_{[0,1]}rightarrow0$ therefore it is uniformly convergent and because it is uniformly convergent $f$ would have to be continuous but it is not, where is the mistake?





This question already has an answer here:




  • Prove $x^n$ is not uniformly convergent

    2 answers







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edited Jan 8 at 21:48









Bernard

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asked Jan 8 at 21:46









RM777RM777

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 Jan 8 at 23:19


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
    $endgroup$
    – copper.hat
    Jan 8 at 21:58


















  • $begingroup$
    Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
    $endgroup$
    – copper.hat
    Jan 8 at 21:58
















$begingroup$
Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
$endgroup$
– copper.hat
Jan 8 at 21:58




$begingroup$
Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
$endgroup$
– copper.hat
Jan 8 at 21:58










1 Answer
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It is not uniformly convergent. For each $ninmathbb{N}$ you can choose $x=frac{1}{sqrt[n]{2}}$ and get $|f_n(x)-f(x)|geqfrac{1}{2}$. So it doesn't matter how far you go in the sequence, the distance will never be less than $frac{1}{2}$ for all $x$.






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    1 Answer
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    1 Answer
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    active

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    active

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    2












    $begingroup$

    It is not uniformly convergent. For each $ninmathbb{N}$ you can choose $x=frac{1}{sqrt[n]{2}}$ and get $|f_n(x)-f(x)|geqfrac{1}{2}$. So it doesn't matter how far you go in the sequence, the distance will never be less than $frac{1}{2}$ for all $x$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      It is not uniformly convergent. For each $ninmathbb{N}$ you can choose $x=frac{1}{sqrt[n]{2}}$ and get $|f_n(x)-f(x)|geqfrac{1}{2}$. So it doesn't matter how far you go in the sequence, the distance will never be less than $frac{1}{2}$ for all $x$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        It is not uniformly convergent. For each $ninmathbb{N}$ you can choose $x=frac{1}{sqrt[n]{2}}$ and get $|f_n(x)-f(x)|geqfrac{1}{2}$. So it doesn't matter how far you go in the sequence, the distance will never be less than $frac{1}{2}$ for all $x$.






        share|cite|improve this answer









        $endgroup$



        It is not uniformly convergent. For each $ninmathbb{N}$ you can choose $x=frac{1}{sqrt[n]{2}}$ and get $|f_n(x)-f(x)|geqfrac{1}{2}$. So it doesn't matter how far you go in the sequence, the distance will never be less than $frac{1}{2}$ for all $x$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 21:52









        MarkMark

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        6,848416















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