Sequence of continuous function converging pointwise to continuous function is equicontinuous?












0












$begingroup$


I've proven the following "theorem":



Let $I subset mathbb{R}$ be an interval, $(f_n: I rightarrow mathbb{R})_{n in mathbb{N}}$ be a family of continuous functions converging pointwise to a continuous function $f: I rightarrow mathbb{R}$ on $I$. Then: $(f_n)_{n in mathbb{N}}$ is equicontinuous on I.



Now my problem is, that here Equicontinuity of a pointwise convergent sequence of monotone functions with continuous limit additionally the $f_n$ have to be monotonic. So is my proof a generalization, or am I just missing something? Here is my proof:



Proof: Let $epsilon > 0$. Observe first:
begin{equation}
| f_n(x) - f_n(y) | leq |f_n(x) - f(x)| + |f_n(y) - f(y)| + |f(x)- f(y)|
end{equation}

Now there is by pointwise convergence of $(f_n)_{n in mathbb{N}}$ a $N in mathbb{N}$ such that for all $n geq N$ we have $|f_n(x) - f(x)|<frac{epsilon}{3}$ and $|f_n(y) - f(y)| < frac{epsilon}{3}$. Further there is a $delta > 0$ such that $|f(x) - f(y)| < frac{epsilon}{3}$ for $|x-y| < delta$ by continuity of $f$. Hence we have shown, that there is a $N in mathbb{N}$ and a $delta > 0$ such that for all $n geq N$
begin{equation}
|f_n(x) - f_n(y)| < frac{epsilon}{3} + frac{epsilon}{3} + frac{epsilon}{3} = epsilon
end{equation}

holds. Now let $n < N$. Then, by continuity of $f_n$ there is a $delta_n$ such that $(x-y) < delta_n$ implies $|f_n(x) - f_n(y)| < epsilon$. Setting
begin{equation}
tilde{delta} = min_{n < N} delta_n
end{equation}

(which exists and is greater than $0$) we obtain, that for all $n < N$ the following holds:
begin{equation}
|x - y| < tilde{delta} Rightarrow |f_n(x) - f_n(y) | < epsilon
end{equation}

Setting now $hat{delta} = min {delta, tilde{delta} }$ we have, that for all $n in mathbb{N}$ the following holds:
begin{equation}
|x-y| < hat{delta} Rightarrow |f_n(x)- f_n(y) | < epsilon
end{equation}

Hence we have shown, that for all $epsilon > 0$ there is a $hat{delta} > 0$ such that forall $n in mathbb{N}$ we have, that $|x-y| < delta$ implies $|f_n(x) - f_n(y)| < epsilon$.










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$endgroup$












  • $begingroup$
    Can your interval be open?
    $endgroup$
    – kimchi lover
    Dec 15 '18 at 16:24










  • $begingroup$
    Yes, i stated no assumptions regarding the interval.
    $endgroup$
    – warpfel
    Dec 15 '18 at 16:27










  • $begingroup$
    The functions $f_n(x)=x^n$ converge to 0 pointwise on $(0,1)$ but not uniformly.
    $endgroup$
    – kimchi lover
    Dec 15 '18 at 16:28
















0












$begingroup$


I've proven the following "theorem":



Let $I subset mathbb{R}$ be an interval, $(f_n: I rightarrow mathbb{R})_{n in mathbb{N}}$ be a family of continuous functions converging pointwise to a continuous function $f: I rightarrow mathbb{R}$ on $I$. Then: $(f_n)_{n in mathbb{N}}$ is equicontinuous on I.



Now my problem is, that here Equicontinuity of a pointwise convergent sequence of monotone functions with continuous limit additionally the $f_n$ have to be monotonic. So is my proof a generalization, or am I just missing something? Here is my proof:



Proof: Let $epsilon > 0$. Observe first:
begin{equation}
| f_n(x) - f_n(y) | leq |f_n(x) - f(x)| + |f_n(y) - f(y)| + |f(x)- f(y)|
end{equation}

Now there is by pointwise convergence of $(f_n)_{n in mathbb{N}}$ a $N in mathbb{N}$ such that for all $n geq N$ we have $|f_n(x) - f(x)|<frac{epsilon}{3}$ and $|f_n(y) - f(y)| < frac{epsilon}{3}$. Further there is a $delta > 0$ such that $|f(x) - f(y)| < frac{epsilon}{3}$ for $|x-y| < delta$ by continuity of $f$. Hence we have shown, that there is a $N in mathbb{N}$ and a $delta > 0$ such that for all $n geq N$
begin{equation}
|f_n(x) - f_n(y)| < frac{epsilon}{3} + frac{epsilon}{3} + frac{epsilon}{3} = epsilon
end{equation}

holds. Now let $n < N$. Then, by continuity of $f_n$ there is a $delta_n$ such that $(x-y) < delta_n$ implies $|f_n(x) - f_n(y)| < epsilon$. Setting
begin{equation}
tilde{delta} = min_{n < N} delta_n
end{equation}

(which exists and is greater than $0$) we obtain, that for all $n < N$ the following holds:
begin{equation}
|x - y| < tilde{delta} Rightarrow |f_n(x) - f_n(y) | < epsilon
end{equation}

Setting now $hat{delta} = min {delta, tilde{delta} }$ we have, that for all $n in mathbb{N}$ the following holds:
begin{equation}
|x-y| < hat{delta} Rightarrow |f_n(x)- f_n(y) | < epsilon
end{equation}

Hence we have shown, that for all $epsilon > 0$ there is a $hat{delta} > 0$ such that forall $n in mathbb{N}$ we have, that $|x-y| < delta$ implies $|f_n(x) - f_n(y)| < epsilon$.










share|cite|improve this question









$endgroup$












  • $begingroup$
    Can your interval be open?
    $endgroup$
    – kimchi lover
    Dec 15 '18 at 16:24










  • $begingroup$
    Yes, i stated no assumptions regarding the interval.
    $endgroup$
    – warpfel
    Dec 15 '18 at 16:27










  • $begingroup$
    The functions $f_n(x)=x^n$ converge to 0 pointwise on $(0,1)$ but not uniformly.
    $endgroup$
    – kimchi lover
    Dec 15 '18 at 16:28














0












0








0





$begingroup$


I've proven the following "theorem":



Let $I subset mathbb{R}$ be an interval, $(f_n: I rightarrow mathbb{R})_{n in mathbb{N}}$ be a family of continuous functions converging pointwise to a continuous function $f: I rightarrow mathbb{R}$ on $I$. Then: $(f_n)_{n in mathbb{N}}$ is equicontinuous on I.



Now my problem is, that here Equicontinuity of a pointwise convergent sequence of monotone functions with continuous limit additionally the $f_n$ have to be monotonic. So is my proof a generalization, or am I just missing something? Here is my proof:



Proof: Let $epsilon > 0$. Observe first:
begin{equation}
| f_n(x) - f_n(y) | leq |f_n(x) - f(x)| + |f_n(y) - f(y)| + |f(x)- f(y)|
end{equation}

Now there is by pointwise convergence of $(f_n)_{n in mathbb{N}}$ a $N in mathbb{N}$ such that for all $n geq N$ we have $|f_n(x) - f(x)|<frac{epsilon}{3}$ and $|f_n(y) - f(y)| < frac{epsilon}{3}$. Further there is a $delta > 0$ such that $|f(x) - f(y)| < frac{epsilon}{3}$ for $|x-y| < delta$ by continuity of $f$. Hence we have shown, that there is a $N in mathbb{N}$ and a $delta > 0$ such that for all $n geq N$
begin{equation}
|f_n(x) - f_n(y)| < frac{epsilon}{3} + frac{epsilon}{3} + frac{epsilon}{3} = epsilon
end{equation}

holds. Now let $n < N$. Then, by continuity of $f_n$ there is a $delta_n$ such that $(x-y) < delta_n$ implies $|f_n(x) - f_n(y)| < epsilon$. Setting
begin{equation}
tilde{delta} = min_{n < N} delta_n
end{equation}

(which exists and is greater than $0$) we obtain, that for all $n < N$ the following holds:
begin{equation}
|x - y| < tilde{delta} Rightarrow |f_n(x) - f_n(y) | < epsilon
end{equation}

Setting now $hat{delta} = min {delta, tilde{delta} }$ we have, that for all $n in mathbb{N}$ the following holds:
begin{equation}
|x-y| < hat{delta} Rightarrow |f_n(x)- f_n(y) | < epsilon
end{equation}

Hence we have shown, that for all $epsilon > 0$ there is a $hat{delta} > 0$ such that forall $n in mathbb{N}$ we have, that $|x-y| < delta$ implies $|f_n(x) - f_n(y)| < epsilon$.










share|cite|improve this question









$endgroup$




I've proven the following "theorem":



Let $I subset mathbb{R}$ be an interval, $(f_n: I rightarrow mathbb{R})_{n in mathbb{N}}$ be a family of continuous functions converging pointwise to a continuous function $f: I rightarrow mathbb{R}$ on $I$. Then: $(f_n)_{n in mathbb{N}}$ is equicontinuous on I.



Now my problem is, that here Equicontinuity of a pointwise convergent sequence of monotone functions with continuous limit additionally the $f_n$ have to be monotonic. So is my proof a generalization, or am I just missing something? Here is my proof:



Proof: Let $epsilon > 0$. Observe first:
begin{equation}
| f_n(x) - f_n(y) | leq |f_n(x) - f(x)| + |f_n(y) - f(y)| + |f(x)- f(y)|
end{equation}

Now there is by pointwise convergence of $(f_n)_{n in mathbb{N}}$ a $N in mathbb{N}$ such that for all $n geq N$ we have $|f_n(x) - f(x)|<frac{epsilon}{3}$ and $|f_n(y) - f(y)| < frac{epsilon}{3}$. Further there is a $delta > 0$ such that $|f(x) - f(y)| < frac{epsilon}{3}$ for $|x-y| < delta$ by continuity of $f$. Hence we have shown, that there is a $N in mathbb{N}$ and a $delta > 0$ such that for all $n geq N$
begin{equation}
|f_n(x) - f_n(y)| < frac{epsilon}{3} + frac{epsilon}{3} + frac{epsilon}{3} = epsilon
end{equation}

holds. Now let $n < N$. Then, by continuity of $f_n$ there is a $delta_n$ such that $(x-y) < delta_n$ implies $|f_n(x) - f_n(y)| < epsilon$. Setting
begin{equation}
tilde{delta} = min_{n < N} delta_n
end{equation}

(which exists and is greater than $0$) we obtain, that for all $n < N$ the following holds:
begin{equation}
|x - y| < tilde{delta} Rightarrow |f_n(x) - f_n(y) | < epsilon
end{equation}

Setting now $hat{delta} = min {delta, tilde{delta} }$ we have, that for all $n in mathbb{N}$ the following holds:
begin{equation}
|x-y| < hat{delta} Rightarrow |f_n(x)- f_n(y) | < epsilon
end{equation}

Hence we have shown, that for all $epsilon > 0$ there is a $hat{delta} > 0$ such that forall $n in mathbb{N}$ we have, that $|x-y| < delta$ implies $|f_n(x) - f_n(y)| < epsilon$.







real-analysis analysis continuity sequence-of-function equicontinuity






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asked Dec 15 '18 at 16:21









warpfelwarpfel

1067




1067












  • $begingroup$
    Can your interval be open?
    $endgroup$
    – kimchi lover
    Dec 15 '18 at 16:24










  • $begingroup$
    Yes, i stated no assumptions regarding the interval.
    $endgroup$
    – warpfel
    Dec 15 '18 at 16:27










  • $begingroup$
    The functions $f_n(x)=x^n$ converge to 0 pointwise on $(0,1)$ but not uniformly.
    $endgroup$
    – kimchi lover
    Dec 15 '18 at 16:28


















  • $begingroup$
    Can your interval be open?
    $endgroup$
    – kimchi lover
    Dec 15 '18 at 16:24










  • $begingroup$
    Yes, i stated no assumptions regarding the interval.
    $endgroup$
    – warpfel
    Dec 15 '18 at 16:27










  • $begingroup$
    The functions $f_n(x)=x^n$ converge to 0 pointwise on $(0,1)$ but not uniformly.
    $endgroup$
    – kimchi lover
    Dec 15 '18 at 16:28
















$begingroup$
Can your interval be open?
$endgroup$
– kimchi lover
Dec 15 '18 at 16:24




$begingroup$
Can your interval be open?
$endgroup$
– kimchi lover
Dec 15 '18 at 16:24












$begingroup$
Yes, i stated no assumptions regarding the interval.
$endgroup$
– warpfel
Dec 15 '18 at 16:27




$begingroup$
Yes, i stated no assumptions regarding the interval.
$endgroup$
– warpfel
Dec 15 '18 at 16:27












$begingroup$
The functions $f_n(x)=x^n$ converge to 0 pointwise on $(0,1)$ but not uniformly.
$endgroup$
– kimchi lover
Dec 15 '18 at 16:28




$begingroup$
The functions $f_n(x)=x^n$ converge to 0 pointwise on $(0,1)$ but not uniformly.
$endgroup$
– kimchi lover
Dec 15 '18 at 16:28










1 Answer
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$begingroup$

Your proof is wrong, because it basically impmies that every pointwise convergence of continuous functions to a continuous function is uniform. I think the flaw is that your $N$ depends both on $x$ (not important, it is fixed) but in $y$ as well!






share|cite|improve this answer









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    $begingroup$

    Your proof is wrong, because it basically impmies that every pointwise convergence of continuous functions to a continuous function is uniform. I think the flaw is that your $N$ depends both on $x$ (not important, it is fixed) but in $y$ as well!






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Your proof is wrong, because it basically impmies that every pointwise convergence of continuous functions to a continuous function is uniform. I think the flaw is that your $N$ depends both on $x$ (not important, it is fixed) but in $y$ as well!






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your proof is wrong, because it basically impmies that every pointwise convergence of continuous functions to a continuous function is uniform. I think the flaw is that your $N$ depends both on $x$ (not important, it is fixed) but in $y$ as well!






        share|cite|improve this answer









        $endgroup$



        Your proof is wrong, because it basically impmies that every pointwise convergence of continuous functions to a continuous function is uniform. I think the flaw is that your $N$ depends both on $x$ (not important, it is fixed) but in $y$ as well!







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 15 '18 at 16:27









        MindlackMindlack

        4,585210




        4,585210






























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