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Prove sequence of continuous functions is Cauchy
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I have a sequence of continuous functions that, $f_{J}: [0,1] rightarrow mathbb{R}, J=1,2,3, ldots$ converges to a function $f$. $f_{J}$ is continuous function and is piecewise linear on each interval in $mathcal{P}^{J}$ (i.e. the powerset) and therefore completely defined by its values in $X^{J}$. First of all, it is defined that $f_{J+1}(t)=f_{J}(t)$ $forall$ $t in X^{J}$, where set $X^{J}={t_{0},t_{1},ldots,t_{N}}={0,2^{-J},2times 2^{-J},3times 2^{-J} ldots,1}$. For $ t=left(2k-1 right) 2^{-left(J+1right)} in X^{J+1}setminus X^{J}, k=1,ldots,2^{J}.$ It is also defined that, $$f_{J+1}(t)=f_{J}(t)+left( -1right)^{k+J+1}2^{-J/(2-1)}$$
This definition of $f_{J+1}$ is extended to the whole of [0,1].
I need to prove that the sequence $f_{1},f_{2},ldots$ converges uniformly to a continuous function, $f: [0,1] rightarrow mathbb{R}$, by showing that is it is a Cauchy sequence in the sup-norm.
$textbf{My approach:}$
If $X$ is a complete metric space, then every Cauchy sequence ${x_{n}} subset X$ converges to a limit $x$ since the function provided is complete with respect to the metric induced by the sup-norm. So, all I want to do is show that this is a Cauchy and by completeness it has to converge. I am not sure how can I go about proving it is a Cauchy, and that is what I need help with.
real-analysis continuity cauchy-sequences supremum-and-infimum
asked Feb 22 '16 at 0:44
jxjxjxjxjxjx
112
$endgroup$
add a comment |
$begingroup$
I have a sequence of continuous functions that, $f_{J}: [0,1] rightarrow mathbb{R}, J=1,2,3, ldots$ converges to a function $f$. $f_{J}$ is continuous function and is piecewise linear on each interval in $mathcal{P}^{J}$ (i.e. the powerset) and therefore completely defined by its values in $X^{J}$. First of all, it is defined that $f_{J+1}(t)=f_{J}(t)$ $forall$ $t in X^{J}$, where set $X^{J}={t_{0},t_{1},ldots,t_{N}}={0,2^{-J},2times 2^{-J},3times 2^{-J} ldots,1}$. For $ t=left(2k-1 right) 2^{-left(J+1right)} in X^{J+1}setminus X^{J}, k=1,ldots,2^{J}.$ It is also defined that, $$f_{J+1}(t)=f_{J}(t)+left( -1right)^{k+J+1}2^{-J/(2-1)}$$
This definition of $f_{J+1}$ is extended to the whole of [0,1].
I need to prove that the sequence $f_{1},f_{2},ldots$ converges uniformly to a continuous function, $f: [0,1] rightarrow mathbb{R}$, by showing that is it is a Cauchy sequence in the sup-norm.
$textbf{My approach:}$
If $X$ is a complete metric space, then every Cauchy sequence ${x_{n}} subset X$ converges to a limit $x$ since the function provided is complete with respect to the metric induced by the sup-norm. So, all I want to do is show that this is a Cauchy and by completeness it has to converge. I am not sure how can I go about proving it is a Cauchy, and that is what I need help with.
real-analysis continuity cauchy-sequences supremum-and-infimum
asked Feb 22 '16 at 0:44
jxjxjxjxjxjx
112
$endgroup$
$begingroup$
Welcome Math.SE! Take the tour to get familiar with this site.. If you receive useful answers, consider accepting one.
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– Shailesh
Feb 22 '16 at 1:04
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Your sequence of functions are defined on a compact space. Moreover you know they converge to f. So pick any 2 arbitrarily $f_j, f_k $ and then use the triangle inequality.
$endgroup$
– Phillip Hamilton
Feb 22 '16 at 1:15
$begingroup$
@ Philip, so are you suggesting some kind of induction based proof?
$endgroup$
– jxjxjx
Feb 22 '16 at 1:23
$begingroup$
I believe you have a typo in your def'n of $f_{J+1}(t)$ .The term $2^{-J/(2-1)}$ doesn't look right.Please edit. If your def'n is what I think it is (when edited) then $sup |f_{J+1}(x)-f_J(x)|$ is about $2^{-J}$ which implies it is a Cauchy sequence.
$endgroup$
– DanielWainfleet
Feb 22 '16 at 18:06
add a comment |
$begingroup$
I have a sequence of continuous functions that, $f_{J}: [0,1] rightarrow mathbb{R}, J=1,2,3, ldots$ converges to a function $f$. $f_{J}$ is continuous function and is piecewise linear on each interval in $mathcal{P}^{J}$ (i.e. the powerset) and therefore completely defined by its values in $X^{J}$. First of all, it is defined that $f_{J+1}(t)=f_{J}(t)$ $forall$ $t in X^{J}$, where set $X^{J}={t_{0},t_{1},ldots,t_{N}}={0,2^{-J},2times 2^{-J},3times 2^{-J} ldots,1}$. For $ t=left(2k-1 right) 2^{-left(J+1right)} in X^{J+1}setminus X^{J}, k=1,ldots,2^{J}.$ It is also defined that, $$f_{J+1}(t)=f_{J}(t)+left( -1right)^{k+J+1}2^{-J/(2-1)}$$
This definition of $f_{J+1}$ is extended to the whole of [0,1].
I need to prove that the sequence $f_{1},f_{2},ldots$ converges uniformly to a continuous function, $f: [0,1] rightarrow mathbb{R}$, by showing that is it is a Cauchy sequence in the sup-norm.
$textbf{My approach:}$
If $X$ is a complete metric space, then every Cauchy sequence ${x_{n}} subset X$ converges to a limit $x$ since the function provided is complete with respect to the metric induced by the sup-norm. So, all I want to do is show that this is a Cauchy and by completeness it has to converge. I am not sure how can I go about proving it is a Cauchy, and that is what I need help with.
real-analysis continuity cauchy-sequences supremum-and-infimum
asked Feb 22 '16 at 0:44
jxjxjxjxjxjx
112
$endgroup$
I have a sequence of continuous functions that, $f_{J}: [0,1] rightarrow mathbb{R}, J=1,2,3, ldots$ converges to a function $f$. $f_{J}$ is continuous function and is piecewise linear on each interval in $mathcal{P}^{J}$ (i.e. the powerset) and therefore completely defined by its values in $X^{J}$. First of all, it is defined that $f_{J+1}(t)=f_{J}(t)$ $forall$ $t in X^{J}$, where set $X^{J}={t_{0},t_{1},ldots,t_{N}}={0,2^{-J},2times 2^{-J},3times 2^{-J} ldots,1}$. For $ t=left(2k-1 right) 2^{-left(J+1right)} in X^{J+1}setminus X^{J}, k=1,ldots,2^{J}.$ It is also defined that, $$f_{J+1}(t)=f_{J}(t)+left( -1right)^{k+J+1}2^{-J/(2-1)}$$
This definition of $f_{J+1}$ is extended to the whole of [0,1].
I need to prove that the sequence $f_{1},f_{2},ldots$ converges uniformly to a continuous function, $f: [0,1] rightarrow mathbb{R}$, by showing that is it is a Cauchy sequence in the sup-norm.
$textbf{My approach:}$
If $X$ is a complete metric space, then every Cauchy sequence ${x_{n}} subset X$ converges to a limit $x$ since the function provided is complete with respect to the metric induced by the sup-norm. So, all I want to do is show that this is a Cauchy and by completeness it has to converge. I am not sure how can I go about proving it is a Cauchy, and that is what I need help with.
real-analysis continuity cauchy-sequences supremum-and-infimum
real-analysis continuity cauchy-sequences supremum-and-infimum
asked Feb 22 '16 at 0:44
jxjxjxjxjxjx
112
asked Feb 22 '16 at 0:44
jxjxjxjxjxjx
112
asked Feb 22 '16 at 0:44
jxjxjxjxjxjx
112
asked Feb 22 '16 at 0:44
jxjxjxjxjxjx
112
asked Feb 22 '16 at 0:44
jxjxjxjxjxjx
112
112
$begingroup$
Welcome Math.SE! Take the tour to get familiar with this site.. If you receive useful answers, consider accepting one.
$endgroup$
– Shailesh
Feb 22 '16 at 1:04
$begingroup$
Your sequence of functions are defined on a compact space. Moreover you know they converge to f. So pick any 2 arbitrarily $f_j, f_k $ and then use the triangle inequality.
$endgroup$
– Phillip Hamilton
Feb 22 '16 at 1:15
$begingroup$
@ Philip, so are you suggesting some kind of induction based proof?
$endgroup$
– jxjxjx
Feb 22 '16 at 1:23
$begingroup$
I believe you have a typo in your def'n of $f_{J+1}(t)$ .The term $2^{-J/(2-1)}$ doesn't look right.Please edit. If your def'n is what I think it is (when edited) then $sup |f_{J+1}(x)-f_J(x)|$ is about $2^{-J}$ which implies it is a Cauchy sequence.
$endgroup$
– DanielWainfleet
Feb 22 '16 at 18:06
add a comment |
$begingroup$
Welcome Math.SE! Take the tour to get familiar with this site.. If you receive useful answers, consider accepting one.
$endgroup$
– Shailesh
Feb 22 '16 at 1:04
$begingroup$
Your sequence of functions are defined on a compact space. Moreover you know they converge to f. So pick any 2 arbitrarily $f_j, f_k $ and then use the triangle inequality.
$endgroup$
– Phillip Hamilton
Feb 22 '16 at 1:15
$begingroup$
@ Philip, so are you suggesting some kind of induction based proof?
$endgroup$
– jxjxjx
Feb 22 '16 at 1:23
$begingroup$
I believe you have a typo in your def'n of $f_{J+1}(t)$ .The term $2^{-J/(2-1)}$ doesn't look right.Please edit. If your def'n is what I think it is (when edited) then $sup |f_{J+1}(x)-f_J(x)|$ is about $2^{-J}$ which implies it is a Cauchy sequence.
$endgroup$
– DanielWainfleet
Feb 22 '16 at 18:06
$begingroup$
Welcome Math.SE! Take the tour to get familiar with this site.. If you receive useful answers, consider accepting one.
$endgroup$
– Shailesh
Feb 22 '16 at 1:04
$begingroup$
Welcome Math.SE! Take the tour to get familiar with this site.. If you receive useful answers, consider accepting one.
$endgroup$
– Shailesh
Feb 22 '16 at 1:04
$begingroup$
Your sequence of functions are defined on a compact space. Moreover you know they converge to f. So pick any 2 arbitrarily $f_j, f_k $ and then use the triangle inequality.
$endgroup$
– Phillip Hamilton
Feb 22 '16 at 1:15
$begingroup$
Your sequence of functions are defined on a compact space. Moreover you know they converge to f. So pick any 2 arbitrarily $f_j, f_k $ and then use the triangle inequality.
$endgroup$
– Phillip Hamilton
Feb 22 '16 at 1:15
$begingroup$
@ Philip, so are you suggesting some kind of induction based proof?
$endgroup$
– jxjxjx
Feb 22 '16 at 1:23
$begingroup$
@ Philip, so are you suggesting some kind of induction based proof?
$endgroup$
– jxjxjx
Feb 22 '16 at 1:23
$begingroup$
I believe you have a typo in your def'n of $f_{J+1}(t)$ .The term $2^{-J/(2-1)}$ doesn't look right.Please edit. If your def'n is what I think it is (when edited) then $sup |f_{J+1}(x)-f_J(x)|$ is about $2^{-J}$ which implies it is a Cauchy sequence.
$endgroup$
– DanielWainfleet
Feb 22 '16 at 18:06
$begingroup$
I believe you have a typo in your def'n of $f_{J+1}(t)$ .The term $2^{-J/(2-1)}$ doesn't look right.Please edit. If your def'n is what I think it is (when edited) then $sup |f_{J+1}(x)-f_J(x)|$ is about $2^{-J}$ which implies it is a Cauchy sequence.
$endgroup$
– DanielWainfleet
Feb 22 '16 at 18:06
add a comment |
1 Answer
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A sequence {$f_n$} is said to be Cauchy if for every $epsilon >0$ there is an integer $N$ such that for $n,m in Bbb N$ then $|f_n - f_m| < epsilon$ when $n>m>N$
So we know that $f_J to f$
Then by the triangle inequality $|f_n - f_m| le |f_n - f| + |f_m - f|$
Since we know $f_J$ converges to $f$, then by definition $|f_J - f| < frac{epsilon }{2}$ where $frac{epsilon }{2}$ is arbitrary
Finally, $|f_n - f_m| le |f_n - f| + |f_m - f| < frac{epsilon }{2} + frac{epsilon }{2} to |f_n - f_m| < epsilon$
This shows that {$f_n$} is Cauchy.
You need to show that for your specific function, $|f_J - f| < frac{epsilon }{2}$ can be interpreted to be bounded above by this arbitrary $frac{epsilon }{2}$ which will show that it too converges.
answered Feb 22 '16 at 3:40
Phillip HamiltonPhillip Hamilton
892513
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$begingroup$
A sequence {$f_n$} is said to be Cauchy if for every $epsilon >0$ there is an integer $N$ such that for $n,m in Bbb N$ then $|f_n - f_m| < epsilon$ when $n>m>N$
So we know that $f_J to f$
Then by the triangle inequality $|f_n - f_m| le |f_n - f| + |f_m - f|$
Since we know $f_J$ converges to $f$, then by definition $|f_J - f| < frac{epsilon }{2}$ where $frac{epsilon }{2}$ is arbitrary
Finally, $|f_n - f_m| le |f_n - f| + |f_m - f| < frac{epsilon }{2} + frac{epsilon }{2} to |f_n - f_m| < epsilon$
This shows that {$f_n$} is Cauchy.
You need to show that for your specific function, $|f_J - f| < frac{epsilon }{2}$ can be interpreted to be bounded above by this arbitrary $frac{epsilon }{2}$ which will show that it too converges.
answered Feb 22 '16 at 3:40
Phillip HamiltonPhillip Hamilton
892513
$endgroup$
add a comment |
$begingroup$
A sequence {$f_n$} is said to be Cauchy if for every $epsilon >0$ there is an integer $N$ such that for $n,m in Bbb N$ then $|f_n - f_m| < epsilon$ when $n>m>N$
So we know that $f_J to f$
Then by the triangle inequality $|f_n - f_m| le |f_n - f| + |f_m - f|$
Since we know $f_J$ converges to $f$, then by definition $|f_J - f| < frac{epsilon }{2}$ where $frac{epsilon }{2}$ is arbitrary
Finally, $|f_n - f_m| le |f_n - f| + |f_m - f| < frac{epsilon }{2} + frac{epsilon }{2} to |f_n - f_m| < epsilon$
This shows that {$f_n$} is Cauchy.
You need to show that for your specific function, $|f_J - f| < frac{epsilon }{2}$ can be interpreted to be bounded above by this arbitrary $frac{epsilon }{2}$ which will show that it too converges.
answered Feb 22 '16 at 3:40
Phillip HamiltonPhillip Hamilton
892513
$endgroup$
add a comment |
$begingroup$
A sequence {$f_n$} is said to be Cauchy if for every $epsilon >0$ there is an integer $N$ such that for $n,m in Bbb N$ then $|f_n - f_m| < epsilon$ when $n>m>N$
So we know that $f_J to f$
Then by the triangle inequality $|f_n - f_m| le |f_n - f| + |f_m - f|$
Since we know $f_J$ converges to $f$, then by definition $|f_J - f| < frac{epsilon }{2}$ where $frac{epsilon }{2}$ is arbitrary
Finally, $|f_n - f_m| le |f_n - f| + |f_m - f| < frac{epsilon }{2} + frac{epsilon }{2} to |f_n - f_m| < epsilon$
This shows that {$f_n$} is Cauchy.
You need to show that for your specific function, $|f_J - f| < frac{epsilon }{2}$ can be interpreted to be bounded above by this arbitrary $frac{epsilon }{2}$ which will show that it too converges.
answered Feb 22 '16 at 3:40
Phillip HamiltonPhillip Hamilton
892513
$endgroup$
A sequence {$f_n$} is said to be Cauchy if for every $epsilon >0$ there is an integer $N$ such that for $n,m in Bbb N$ then $|f_n - f_m| < epsilon$ when $n>m>N$
So we know that $f_J to f$
Then by the triangle inequality $|f_n - f_m| le |f_n - f| + |f_m - f|$
Since we know $f_J$ converges to $f$, then by definition $|f_J - f| < frac{epsilon }{2}$ where $frac{epsilon }{2}$ is arbitrary
Finally, $|f_n - f_m| le |f_n - f| + |f_m - f| < frac{epsilon }{2} + frac{epsilon }{2} to |f_n - f_m| < epsilon$
This shows that {$f_n$} is Cauchy.
You need to show that for your specific function, $|f_J - f| < frac{epsilon }{2}$ can be interpreted to be bounded above by this arbitrary $frac{epsilon }{2}$ which will show that it too converges.
answered Feb 22 '16 at 3:40
Phillip HamiltonPhillip Hamilton
892513
answered Feb 22 '16 at 3:40
Phillip HamiltonPhillip Hamilton
892513
answered Feb 22 '16 at 3:40
Phillip HamiltonPhillip Hamilton
892513
answered Feb 22 '16 at 3:40
Phillip HamiltonPhillip Hamilton
892513
892513
add a comment |
add a comment |
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Welcome Math.SE! Take the tour to get familiar with this site.. If you receive useful answers, consider accepting one.
$endgroup$
– Shailesh
Feb 22 '16 at 1:04
$begingroup$
Your sequence of functions are defined on a compact space. Moreover you know they converge to f. So pick any 2 arbitrarily $f_j, f_k $ and then use the triangle inequality.
$endgroup$
– Phillip Hamilton
Feb 22 '16 at 1:15
$begingroup$
@ Philip, so are you suggesting some kind of induction based proof?
$endgroup$
– jxjxjx
Feb 22 '16 at 1:23
$begingroup$
I believe you have a typo in your def'n of $f_{J+1}(t)$ .The term $2^{-J/(2-1)}$ doesn't look right.Please edit. If your def'n is what I think it is (when edited) then $sup |f_{J+1}(x)-f_J(x)|$ is about $2^{-J}$ which implies it is a Cauchy sequence.
$endgroup$
– DanielWainfleet
Feb 22 '16 at 18:06