Limit of a locally uniformly convergent sequence of continuous functions












0












$begingroup$


I have two questions:

1. I know that the uniform limit of a continuous functions is continuous. But I'm wondering whether this is true if the convergence is locally uniform. That is the uniform convergence is true within any bounded interval.

2. Also I'm confused with the uniqueness of the limit.

For example:

Define
$f_n:[0,1]rightarrow[0,1]$ such that
$$f_n(x) = begin{cases} 1-nx &, 0 leq xleq frac{1}{n} \ \0 &, frac{1}{n} leq xleq 1 end{cases}$$
Define $f:[0,1]rightarrow [0,1]$ to be zero function.

Define $h:[0,1]rightarrow [0,1]$ to be
$$h(x)=begin{cases}
1 &, x=0\ \
0 &, xneq 0
end{cases}$$

and we can prove that: $forall x,yin (0,1]$, $f_n$ converges uniformly to $f$ as well as to $h$ within $[x,y]$ (That is locally uniformly). Please point out the mistake I have done






analysis uniform-convergence sequence-of-function







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I have two questions:

    1. I know that the uniform limit of a continuous functions is continuous. But I'm wondering whether this is true if the convergence is locally uniform. That is the uniform convergence is true within any bounded interval.

    2. Also I'm confused with the uniqueness of the limit.

    For example:

    Define
    $f_n:[0,1]rightarrow[0,1]$ such that
    $$f_n(x) = begin{cases} 1-nx &, 0 leq xleq frac{1}{n} \ \0 &, frac{1}{n} leq xleq 1 end{cases}$$
    Define $f:[0,1]rightarrow [0,1]$ to be zero function.

    Define $h:[0,1]rightarrow [0,1]$ to be
    $$h(x)=begin{cases}
    1 &, x=0\ \
    0 &, xneq 0
    end{cases}$$

    and we can prove that: $forall x,yin (0,1]$, $f_n$ converges uniformly to $f$ as well as to $h$ within $[x,y]$ (That is locally uniformly). Please point out the mistake I have done






    analysis uniform-convergence sequence-of-function







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have two questions:

      1. I know that the uniform limit of a continuous functions is continuous. But I'm wondering whether this is true if the convergence is locally uniform. That is the uniform convergence is true within any bounded interval.

      2. Also I'm confused with the uniqueness of the limit.

      For example:

      Define
      $f_n:[0,1]rightarrow[0,1]$ such that
      $$f_n(x) = begin{cases} 1-nx &, 0 leq xleq frac{1}{n} \ \0 &, frac{1}{n} leq xleq 1 end{cases}$$
      Define $f:[0,1]rightarrow [0,1]$ to be zero function.

      Define $h:[0,1]rightarrow [0,1]$ to be
      $$h(x)=begin{cases}
      1 &, x=0\ \
      0 &, xneq 0
      end{cases}$$

      and we can prove that: $forall x,yin (0,1]$, $f_n$ converges uniformly to $f$ as well as to $h$ within $[x,y]$ (That is locally uniformly). Please point out the mistake I have done






      analysis uniform-convergence sequence-of-function







      share|cite|improve this question











      $endgroup$




      I have two questions:

      1. I know that the uniform limit of a continuous functions is continuous. But I'm wondering whether this is true if the convergence is locally uniform. That is the uniform convergence is true within any bounded interval.

      2. Also I'm confused with the uniqueness of the limit.

      For example:

      Define
      $f_n:[0,1]rightarrow[0,1]$ such that
      $$f_n(x) = begin{cases} 1-nx &, 0 leq xleq frac{1}{n} \ \0 &, frac{1}{n} leq xleq 1 end{cases}$$
      Define $f:[0,1]rightarrow [0,1]$ to be zero function.

      Define $h:[0,1]rightarrow [0,1]$ to be
      $$h(x)=begin{cases}
      1 &, x=0\ \
      0 &, xneq 0
      end{cases}$$

      and we can prove that: $forall x,yin (0,1]$, $f_n$ converges uniformly to $f$ as well as to $h$ within $[x,y]$ (That is locally uniformly). Please point out the mistake I have done






      analysis uniform-convergence sequence-of-function




      analysis uniform-convergence sequence-of-function






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 14:55







      Charith Jeewantha

















      asked Dec 8 '18 at 14:47









      Charith JeewanthaCharith Jeewantha

      827




      827






















          1 Answer
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          $begingroup$

          Since continuity is a local property (that is, in order to determine whether $f$ is continuous at $a$, all that matters is how $f$ behaves near $a$), yes, local uniform convergence preserves continuity.



          Concerning your other question, $(f_n)_{ninmathbb N}$ neither uniformly to no function and converges pointwise to $h$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the answer for the continuity. But for the other part, my concern is about Locally uniformly convergence.
            $endgroup$
            – Charith Jeewantha
            Dec 8 '18 at 14:57










          • $begingroup$
            But I wrote in answer that local uniform convergence preserves continuity. Wasn't that the question?
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 14:59










          • $begingroup$
            Yes I agree with you about the continuity. But how about the uniqueness of the limit
            $endgroup$
            – Charith Jeewantha
            Dec 8 '18 at 15:02










          • $begingroup$
            The limit is unique (if it exists) because, for each $x$ in the domain of the $f_n$'s, $f(x)=lim_{ntoinfty}f_n(x)$ and the limit of a sequence of numbers is unique (again, if it exists).
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 15:06













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          1 Answer
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          1 Answer
          1






          active

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          active

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          1












          $begingroup$

          Since continuity is a local property (that is, in order to determine whether $f$ is continuous at $a$, all that matters is how $f$ behaves near $a$), yes, local uniform convergence preserves continuity.



          Concerning your other question, $(f_n)_{ninmathbb N}$ neither uniformly to no function and converges pointwise to $h$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the answer for the continuity. But for the other part, my concern is about Locally uniformly convergence.
            $endgroup$
            – Charith Jeewantha
            Dec 8 '18 at 14:57










          • $begingroup$
            But I wrote in answer that local uniform convergence preserves continuity. Wasn't that the question?
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 14:59










          • $begingroup$
            Yes I agree with you about the continuity. But how about the uniqueness of the limit
            $endgroup$
            – Charith Jeewantha
            Dec 8 '18 at 15:02










          • $begingroup$
            The limit is unique (if it exists) because, for each $x$ in the domain of the $f_n$'s, $f(x)=lim_{ntoinfty}f_n(x)$ and the limit of a sequence of numbers is unique (again, if it exists).
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 15:06


















          1












          $begingroup$

          Since continuity is a local property (that is, in order to determine whether $f$ is continuous at $a$, all that matters is how $f$ behaves near $a$), yes, local uniform convergence preserves continuity.



          Concerning your other question, $(f_n)_{ninmathbb N}$ neither uniformly to no function and converges pointwise to $h$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the answer for the continuity. But for the other part, my concern is about Locally uniformly convergence.
            $endgroup$
            – Charith Jeewantha
            Dec 8 '18 at 14:57










          • $begingroup$
            But I wrote in answer that local uniform convergence preserves continuity. Wasn't that the question?
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 14:59










          • $begingroup$
            Yes I agree with you about the continuity. But how about the uniqueness of the limit
            $endgroup$
            – Charith Jeewantha
            Dec 8 '18 at 15:02










          • $begingroup$
            The limit is unique (if it exists) because, for each $x$ in the domain of the $f_n$'s, $f(x)=lim_{ntoinfty}f_n(x)$ and the limit of a sequence of numbers is unique (again, if it exists).
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 15:06
















          1












          1








          1





          $begingroup$

          Since continuity is a local property (that is, in order to determine whether $f$ is continuous at $a$, all that matters is how $f$ behaves near $a$), yes, local uniform convergence preserves continuity.



          Concerning your other question, $(f_n)_{ninmathbb N}$ neither uniformly to no function and converges pointwise to $h$.






          share|cite|improve this answer









          $endgroup$



          Since continuity is a local property (that is, in order to determine whether $f$ is continuous at $a$, all that matters is how $f$ behaves near $a$), yes, local uniform convergence preserves continuity.



          Concerning your other question, $(f_n)_{ninmathbb N}$ neither uniformly to no function and converges pointwise to $h$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 14:53









          José Carlos SantosJosé Carlos Santos

          157k22126227




          157k22126227












          • $begingroup$
            Thank you for the answer for the continuity. But for the other part, my concern is about Locally uniformly convergence.
            $endgroup$
            – Charith Jeewantha
            Dec 8 '18 at 14:57










          • $begingroup$
            But I wrote in answer that local uniform convergence preserves continuity. Wasn't that the question?
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 14:59










          • $begingroup$
            Yes I agree with you about the continuity. But how about the uniqueness of the limit
            $endgroup$
            – Charith Jeewantha
            Dec 8 '18 at 15:02










          • $begingroup$
            The limit is unique (if it exists) because, for each $x$ in the domain of the $f_n$'s, $f(x)=lim_{ntoinfty}f_n(x)$ and the limit of a sequence of numbers is unique (again, if it exists).
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 15:06




















          • $begingroup$
            Thank you for the answer for the continuity. But for the other part, my concern is about Locally uniformly convergence.
            $endgroup$
            – Charith Jeewantha
            Dec 8 '18 at 14:57










          • $begingroup$
            But I wrote in answer that local uniform convergence preserves continuity. Wasn't that the question?
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 14:59










          • $begingroup$
            Yes I agree with you about the continuity. But how about the uniqueness of the limit
            $endgroup$
            – Charith Jeewantha
            Dec 8 '18 at 15:02










          • $begingroup$
            The limit is unique (if it exists) because, for each $x$ in the domain of the $f_n$'s, $f(x)=lim_{ntoinfty}f_n(x)$ and the limit of a sequence of numbers is unique (again, if it exists).
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 15:06


















          $begingroup$
          Thank you for the answer for the continuity. But for the other part, my concern is about Locally uniformly convergence.
          $endgroup$
          – Charith Jeewantha
          Dec 8 '18 at 14:57




          $begingroup$
          Thank you for the answer for the continuity. But for the other part, my concern is about Locally uniformly convergence.
          $endgroup$
          – Charith Jeewantha
          Dec 8 '18 at 14:57












          $begingroup$
          But I wrote in answer that local uniform convergence preserves continuity. Wasn't that the question?
          $endgroup$
          – José Carlos Santos
          Dec 8 '18 at 14:59




          $begingroup$
          But I wrote in answer that local uniform convergence preserves continuity. Wasn't that the question?
          $endgroup$
          – José Carlos Santos
          Dec 8 '18 at 14:59












          $begingroup$
          Yes I agree with you about the continuity. But how about the uniqueness of the limit
          $endgroup$
          – Charith Jeewantha
          Dec 8 '18 at 15:02




          $begingroup$
          Yes I agree with you about the continuity. But how about the uniqueness of the limit
          $endgroup$
          – Charith Jeewantha
          Dec 8 '18 at 15:02












          $begingroup$
          The limit is unique (if it exists) because, for each $x$ in the domain of the $f_n$'s, $f(x)=lim_{ntoinfty}f_n(x)$ and the limit of a sequence of numbers is unique (again, if it exists).
          $endgroup$
          – José Carlos Santos
          Dec 8 '18 at 15:06






          $begingroup$
          The limit is unique (if it exists) because, for each $x$ in the domain of the $f_n$'s, $f(x)=lim_{ntoinfty}f_n(x)$ and the limit of a sequence of numbers is unique (again, if it exists).
          $endgroup$
          – José Carlos Santos
          Dec 8 '18 at 15:06




















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