Example of a sequence of functions where the limit cannot be interchanged

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Give an example of a sequence of continuous functions $f_n$ on $[0,1]$ with $f_n$ converges pointwise to a continuous function $f$ such that the following relation does't hold:



$$lim_n rightarrow infty lim_x rightarrow 0 f_n(x)=lim_x rightarrow 0lim_n rightarrow infty f_n(x)$$




I know such a convergence is not uniform. I already tried with this one: $f_n(x)= 2nx e^-nx^2$. Actually this one satisfies the given limit condition even though the convergence is not uniform! Any hint?




real-analysis uniform-convergence




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    up vote
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    down vote

    favorite













    Give an example of a sequence of continuous functions $f_n$ on $[0,1]$ with $f_n$ converges pointwise to a continuous function $f$ such that the following relation does't hold:



    $$lim_n rightarrow infty lim_x rightarrow 0 f_n(x)=lim_x rightarrow 0lim_n rightarrow infty f_n(x)$$




    I know such a convergence is not uniform. I already tried with this one: $f_n(x)= 2nx e^-nx^2$. Actually this one satisfies the given limit condition even though the convergence is not uniform! Any hint?




    real-analysis uniform-convergence




    share|cite|improve this question





















      up vote
      6
      down vote

      favorite









      up vote
      6
      down vote

      favorite












      Give an example of a sequence of continuous functions $f_n$ on $[0,1]$ with $f_n$ converges pointwise to a continuous function $f$ such that the following relation does't hold:



      $$lim_n rightarrow infty lim_x rightarrow 0 f_n(x)=lim_x rightarrow 0lim_n rightarrow infty f_n(x)$$




      I know such a convergence is not uniform. I already tried with this one: $f_n(x)= 2nx e^-nx^2$. Actually this one satisfies the given limit condition even though the convergence is not uniform! Any hint?




      real-analysis uniform-convergence




      share|cite|improve this question












      Give an example of a sequence of continuous functions $f_n$ on $[0,1]$ with $f_n$ converges pointwise to a continuous function $f$ such that the following relation does't hold:



      $$lim_n rightarrow infty lim_x rightarrow 0 f_n(x)=lim_x rightarrow 0lim_n rightarrow infty f_n(x)$$




      I know such a convergence is not uniform. I already tried with this one: $f_n(x)= 2nx e^-nx^2$. Actually this one satisfies the given limit condition even though the convergence is not uniform! Any hint?





      real-analysis uniform-convergence





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      asked Jul 28 at 8:15









      Learning Mathematics

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          Since all $f_n$ are continuous we have $lim_xto0 f_n(x) = f_n(0)$ and since $f = lim_ntoinfty f_n$ is also continuous we have $lim_xto 0 f(x) = f(0)$.



          So your relation boils down to $lim_ntoinfty f_n(0) = f(0)$, which is true because $f_n to f$ pointwise.






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          • so your conclusion is: the limit can be interchanged even the convergence is pointwise.
            – Learning Mathematics
            Jul 28 at 8:38







          • 1




            @LearningMathematics Yes, because $f$ is also assumed to be continuous. This doesn't follow from $f_n to f$ pointwise.
            – mechanodroid
            Jul 28 at 8:39











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          up vote
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          accepted










          Since all $f_n$ are continuous we have $lim_xto0 f_n(x) = f_n(0)$ and since $f = lim_ntoinfty f_n$ is also continuous we have $lim_xto 0 f(x) = f(0)$.



          So your relation boils down to $lim_ntoinfty f_n(0) = f(0)$, which is true because $f_n to f$ pointwise.






          share|cite|improve this answer





















          • so your conclusion is: the limit can be interchanged even the convergence is pointwise.
            – Learning Mathematics
            Jul 28 at 8:38







          • 1




            @LearningMathematics Yes, because $f$ is also assumed to be continuous. This doesn't follow from $f_n to f$ pointwise.
            – mechanodroid
            Jul 28 at 8:39















          up vote
          7
          down vote



          accepted










          Since all $f_n$ are continuous we have $lim_xto0 f_n(x) = f_n(0)$ and since $f = lim_ntoinfty f_n$ is also continuous we have $lim_xto 0 f(x) = f(0)$.



          So your relation boils down to $lim_ntoinfty f_n(0) = f(0)$, which is true because $f_n to f$ pointwise.






          share|cite|improve this answer





















          • so your conclusion is: the limit can be interchanged even the convergence is pointwise.
            – Learning Mathematics
            Jul 28 at 8:38







          • 1




            @LearningMathematics Yes, because $f$ is also assumed to be continuous. This doesn't follow from $f_n to f$ pointwise.
            – mechanodroid
            Jul 28 at 8:39













          up vote
          7
          down vote



          accepted







          up vote
          7
          down vote



          accepted






          Since all $f_n$ are continuous we have $lim_xto0 f_n(x) = f_n(0)$ and since $f = lim_ntoinfty f_n$ is also continuous we have $lim_xto 0 f(x) = f(0)$.



          So your relation boils down to $lim_ntoinfty f_n(0) = f(0)$, which is true because $f_n to f$ pointwise.






          share|cite|improve this answer













          Since all $f_n$ are continuous we have $lim_xto0 f_n(x) = f_n(0)$ and since $f = lim_ntoinfty f_n$ is also continuous we have $lim_xto 0 f(x) = f(0)$.



          So your relation boils down to $lim_ntoinfty f_n(0) = f(0)$, which is true because $f_n to f$ pointwise.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 28 at 8:33









          mechanodroid

          22.2k52041




          22.2k52041











          • so your conclusion is: the limit can be interchanged even the convergence is pointwise.
            – Learning Mathematics
            Jul 28 at 8:38







          • 1




            @LearningMathematics Yes, because $f$ is also assumed to be continuous. This doesn't follow from $f_n to f$ pointwise.
            – mechanodroid
            Jul 28 at 8:39

















          • so your conclusion is: the limit can be interchanged even the convergence is pointwise.
            – Learning Mathematics
            Jul 28 at 8:38







          • 1




            @LearningMathematics Yes, because $f$ is also assumed to be continuous. This doesn't follow from $f_n to f$ pointwise.
            – mechanodroid
            Jul 28 at 8:39
















          so your conclusion is: the limit can be interchanged even the convergence is pointwise.
          – Learning Mathematics
          Jul 28 at 8:38





          so your conclusion is: the limit can be interchanged even the convergence is pointwise.
          – Learning Mathematics
          Jul 28 at 8:38





          1




          1




          @LearningMathematics Yes, because $f$ is also assumed to be continuous. This doesn't follow from $f_n to f$ pointwise.
          – mechanodroid
          Jul 28 at 8:39





          @LearningMathematics Yes, because $f$ is also assumed to be continuous. This doesn't follow from $f_n to f$ pointwise.
          – mechanodroid
          Jul 28 at 8:39













           

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