Sequence of continuous function converging pointwise to continuous function is equicontinuous?
$begingroup$
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 $|f_n(x) - f_n(y)| < epsilon$.
real-analysis analysis continuity sequence-of-function equicontinuity
asked Dec 15 '18 at 16:21
warpfelwarpfel
1067
$endgroup$
add a comment |
$begingroup$
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 $|f_n(x) - f_n(y)| < epsilon$.
real-analysis analysis continuity sequence-of-function equicontinuity
asked Dec 15 '18 at 16:21
warpfelwarpfel
1067
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $|f_n(x) - f_n(y)| < epsilon$.
real-analysis analysis continuity sequence-of-function equicontinuity
asked Dec 15 '18 at 16:21
warpfelwarpfel
1067
$endgroup$
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
real-analysis analysis continuity sequence-of-function equicontinuity
asked Dec 15 '18 at 16:21
warpfelwarpfel
1067
asked Dec 15 '18 at 16:21
warpfelwarpfel
1067
asked Dec 15 '18 at 16:21
warpfelwarpfel
1067
asked Dec 15 '18 at 16:21
warpfelwarpfel
1067
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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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!
answered Dec 15 '18 at 16:27
MindlackMindlack
4,885211
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add a comment |
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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!
answered Dec 15 '18 at 16:27
MindlackMindlack
4,885211
$endgroup$
add a comment |
$begingroup$
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!
answered Dec 15 '18 at 16:27
MindlackMindlack
4,885211
$endgroup$
add a comment |
$begingroup$
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!
answered Dec 15 '18 at 16:27
MindlackMindlack
4,885211
$endgroup$
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!
answered Dec 15 '18 at 16:27
MindlackMindlack
4,885211
answered Dec 15 '18 at 16:27
MindlackMindlack
4,885211
answered Dec 15 '18 at 16:27
MindlackMindlack
4,885211
answered Dec 15 '18 at 16:27
MindlackMindlack
4,885211
4,885211
add a comment |
add a comment |
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$begingroup$
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