Prove sequence of continuous functions is Cauchy












2












$begingroup$


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







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  • $begingroup$
    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










  • $begingroup$
    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


















2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















2












2








2





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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




















  • $begingroup$
    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












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.






share|cite|improve this answer









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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.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 22 '16 at 3:40









        Phillip HamiltonPhillip Hamilton

        892513




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