Convergence to a fixed point












3












$begingroup$


Let $f : [a,b] rightarrow [a,b]$ be a continuous function s.t. $f'(x)$ is defined on $(a,b)$ and $leftlvert f'(x)rightrvert leqq t$ where $0<t<1$. Prove that for any point $x_0$ in $[a,b]$ the sequence defined by $$ x_n=f(x_{n-1}), n>0$$
converges to one unique fixed point.



Attempt:
Frankly, I have struggled to make a real attempt due to the fact that I can't find notes relating to this.



Obviously, I am assuming that there exists $x$ in$ [a,b]$ s.t. $f(x)=x$ but how do I relate the sequence to this $x$?



I'm strictly not allowed to assume Banach's theorem in this question, nor the Cauchy sequence because they come up on the second part of the course. I rather have to PROVE this.






analysis convergence numerical-methods fixed-point-theorems fixedpoints







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  • $begingroup$
    Another related question.
    $endgroup$
    – rtybase
    5 hours ago






  • 1




    $begingroup$
    Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
    $endgroup$
    – rtybase
    4 hours ago










  • $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
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3












$begingroup$


Let $f : [a,b] rightarrow [a,b]$ be a continuous function s.t. $f'(x)$ is defined on $(a,b)$ and $leftlvert f'(x)rightrvert leqq t$ where $0<t<1$. Prove that for any point $x_0$ in $[a,b]$ the sequence defined by $$ x_n=f(x_{n-1}), n>0$$
converges to one unique fixed point.



Attempt:
Frankly, I have struggled to make a real attempt due to the fact that I can't find notes relating to this.



Obviously, I am assuming that there exists $x$ in$ [a,b]$ s.t. $f(x)=x$ but how do I relate the sequence to this $x$?



I'm strictly not allowed to assume Banach's theorem in this question, nor the Cauchy sequence because they come up on the second part of the course. I rather have to PROVE this.






analysis convergence numerical-methods fixed-point-theorems fixedpoints







share|cite|improve this question









New contributor




Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Another related question.
    $endgroup$
    – rtybase
    5 hours ago






  • 1




    $begingroup$
    Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
    $endgroup$
    – rtybase
    4 hours ago










  • $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
    $endgroup$
    – dantopa
    2 hours ago
















3












3








3





$begingroup$


Let $f : [a,b] rightarrow [a,b]$ be a continuous function s.t. $f'(x)$ is defined on $(a,b)$ and $leftlvert f'(x)rightrvert leqq t$ where $0<t<1$. Prove that for any point $x_0$ in $[a,b]$ the sequence defined by $$ x_n=f(x_{n-1}), n>0$$
converges to one unique fixed point.



Attempt:
Frankly, I have struggled to make a real attempt due to the fact that I can't find notes relating to this.



Obviously, I am assuming that there exists $x$ in$ [a,b]$ s.t. $f(x)=x$ but how do I relate the sequence to this $x$?



I'm strictly not allowed to assume Banach's theorem in this question, nor the Cauchy sequence because they come up on the second part of the course. I rather have to PROVE this.






analysis convergence numerical-methods fixed-point-theorems fixedpoints







share|cite|improve this question









New contributor




Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Let $f : [a,b] rightarrow [a,b]$ be a continuous function s.t. $f'(x)$ is defined on $(a,b)$ and $leftlvert f'(x)rightrvert leqq t$ where $0<t<1$. Prove that for any point $x_0$ in $[a,b]$ the sequence defined by $$ x_n=f(x_{n-1}), n>0$$
converges to one unique fixed point.



Attempt:
Frankly, I have struggled to make a real attempt due to the fact that I can't find notes relating to this.



Obviously, I am assuming that there exists $x$ in$ [a,b]$ s.t. $f(x)=x$ but how do I relate the sequence to this $x$?



I'm strictly not allowed to assume Banach's theorem in this question, nor the Cauchy sequence because they come up on the second part of the course. I rather have to PROVE this.






analysis convergence numerical-methods fixed-point-theorems fixedpoints




analysis convergence numerical-methods fixed-point-theorems fixedpoints






share|cite|improve this question









New contributor




Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




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share|cite|improve this question




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edited 4 hours ago







Grace













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asked 5 hours ago









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  • $begingroup$
    Another related question.
    $endgroup$
    – rtybase
    5 hours ago






  • 1




    $begingroup$
    Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
    $endgroup$
    – rtybase
    4 hours ago










  • $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
    $endgroup$
    – dantopa
    2 hours ago




















  • $begingroup$
    Another related question.
    $endgroup$
    – rtybase
    5 hours ago






  • 1




    $begingroup$
    Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
    $endgroup$
    – rtybase
    4 hours ago










  • $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
    $endgroup$
    – dantopa
    2 hours ago


















$begingroup$
Another related question.
$endgroup$
– rtybase
5 hours ago




$begingroup$
Another related question.
$endgroup$
– rtybase
5 hours ago




1




1




$begingroup$
Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
$endgroup$
– rtybase
4 hours ago




$begingroup$
Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
$endgroup$
– rtybase
4 hours ago












$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
$endgroup$
– dantopa
2 hours ago






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Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
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2 hours ago












3 Answers
3






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2












$begingroup$

The Mean Value Theorem tells you that the sequence ${x_n }$ is Cauchy because $|f(y)-f(x)| leq t|y-x|$. The space $[0, 1]$ is complete, so since the sequence is Cauchy, it must converge.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    This can be proved using the Banach Fixed Point Theorem.



    Intuitively, the BFPT tells us that if there is some function $F$ such that the distance between any two points $x$ and $y$ (when scaled by a constant $q$) is always larger than the distance between the corresponding images ($F(x)$ and $F(y)$), then the sequence
    $$x_n = F(x_{n-1}) $$
    will converge to a unique fixed point. For a more rigorous treatment of the BFPT statement: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem.



    Since in this case you know that
    $$|f'(x)| leq t $$
    This implies that



    $$Big|frac{f(x) - f(y)}{x - y}Big| leq t$$



    for all possible points $x,y in [a,b]$. Think about why this is the case (Hint: Use the mean value theorem).



    This implies that



    $$| f(x) - f(y)| < t |x-y| $$



    which is what the BFPT requires. From here, we can just apply the BFPT to state that there is a fixed point in $[a,b]$ for the sequence



    $$ x_n = f(x_{n-1})$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Should probably expressly mention that the space $[0, 1]$ is complete, since that is a necessary condition for the Banach fixed point theorem to hold. For an easy counterexample demonstrating that you need completeness, consider $f(x) = x/2$ on $(0, 1]$.
      $endgroup$
      – Robert Shore
      4 hours ago










    • $begingroup$
      @RobertShore Yes, you are right, thanks!
      $endgroup$
      – Sean Lee
      4 hours ago



















    1












    $begingroup$

    To prove this from scratch note that by MVT $$|x_n-x_m| leq |x_{m+1}-x_m|+|x_{m+2}-x_{m+1}|$$ $$+cdots+|x_{n}-x_{n-1}|leq |x_{m+1}-x_m| (1+t+t^{2}+cdots+t^{n+m-1})$$ for $n >m$. Also $|x_{m+1}-x_m| leq t^{m-1} |x_2-x_1|$. Using the convergence of the geometric series $sum t^{n}$ conclude that ${x_n}$ is Cauchy. If $x =lim x_n$ then the definition of $x_n$'s and continuity of $f$ tells you that $f(x)=x$. Uniqueness follows easily by MVT.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

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      active

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      2












      $begingroup$

      The Mean Value Theorem tells you that the sequence ${x_n }$ is Cauchy because $|f(y)-f(x)| leq t|y-x|$. The space $[0, 1]$ is complete, so since the sequence is Cauchy, it must converge.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        The Mean Value Theorem tells you that the sequence ${x_n }$ is Cauchy because $|f(y)-f(x)| leq t|y-x|$. The space $[0, 1]$ is complete, so since the sequence is Cauchy, it must converge.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          The Mean Value Theorem tells you that the sequence ${x_n }$ is Cauchy because $|f(y)-f(x)| leq t|y-x|$. The space $[0, 1]$ is complete, so since the sequence is Cauchy, it must converge.






          share|cite|improve this answer











          $endgroup$



          The Mean Value Theorem tells you that the sequence ${x_n }$ is Cauchy because $|f(y)-f(x)| leq t|y-x|$. The space $[0, 1]$ is complete, so since the sequence is Cauchy, it must converge.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 4 hours ago

























          answered 5 hours ago









          Robert ShoreRobert Shore

          2,044116




          2,044116























              1












              $begingroup$

              This can be proved using the Banach Fixed Point Theorem.



              Intuitively, the BFPT tells us that if there is some function $F$ such that the distance between any two points $x$ and $y$ (when scaled by a constant $q$) is always larger than the distance between the corresponding images ($F(x)$ and $F(y)$), then the sequence
              $$x_n = F(x_{n-1}) $$
              will converge to a unique fixed point. For a more rigorous treatment of the BFPT statement: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem.



              Since in this case you know that
              $$|f'(x)| leq t $$
              This implies that



              $$Big|frac{f(x) - f(y)}{x - y}Big| leq t$$



              for all possible points $x,y in [a,b]$. Think about why this is the case (Hint: Use the mean value theorem).



              This implies that



              $$| f(x) - f(y)| < t |x-y| $$



              which is what the BFPT requires. From here, we can just apply the BFPT to state that there is a fixed point in $[a,b]$ for the sequence



              $$ x_n = f(x_{n-1})$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Should probably expressly mention that the space $[0, 1]$ is complete, since that is a necessary condition for the Banach fixed point theorem to hold. For an easy counterexample demonstrating that you need completeness, consider $f(x) = x/2$ on $(0, 1]$.
                $endgroup$
                – Robert Shore
                4 hours ago










              • $begingroup$
                @RobertShore Yes, you are right, thanks!
                $endgroup$
                – Sean Lee
                4 hours ago
















              1












              $begingroup$

              This can be proved using the Banach Fixed Point Theorem.



              Intuitively, the BFPT tells us that if there is some function $F$ such that the distance between any two points $x$ and $y$ (when scaled by a constant $q$) is always larger than the distance between the corresponding images ($F(x)$ and $F(y)$), then the sequence
              $$x_n = F(x_{n-1}) $$
              will converge to a unique fixed point. For a more rigorous treatment of the BFPT statement: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem.



              Since in this case you know that
              $$|f'(x)| leq t $$
              This implies that



              $$Big|frac{f(x) - f(y)}{x - y}Big| leq t$$



              for all possible points $x,y in [a,b]$. Think about why this is the case (Hint: Use the mean value theorem).



              This implies that



              $$| f(x) - f(y)| < t |x-y| $$



              which is what the BFPT requires. From here, we can just apply the BFPT to state that there is a fixed point in $[a,b]$ for the sequence



              $$ x_n = f(x_{n-1})$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Should probably expressly mention that the space $[0, 1]$ is complete, since that is a necessary condition for the Banach fixed point theorem to hold. For an easy counterexample demonstrating that you need completeness, consider $f(x) = x/2$ on $(0, 1]$.
                $endgroup$
                – Robert Shore
                4 hours ago










              • $begingroup$
                @RobertShore Yes, you are right, thanks!
                $endgroup$
                – Sean Lee
                4 hours ago














              1












              1








              1





              $begingroup$

              This can be proved using the Banach Fixed Point Theorem.



              Intuitively, the BFPT tells us that if there is some function $F$ such that the distance between any two points $x$ and $y$ (when scaled by a constant $q$) is always larger than the distance between the corresponding images ($F(x)$ and $F(y)$), then the sequence
              $$x_n = F(x_{n-1}) $$
              will converge to a unique fixed point. For a more rigorous treatment of the BFPT statement: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem.



              Since in this case you know that
              $$|f'(x)| leq t $$
              This implies that



              $$Big|frac{f(x) - f(y)}{x - y}Big| leq t$$



              for all possible points $x,y in [a,b]$. Think about why this is the case (Hint: Use the mean value theorem).



              This implies that



              $$| f(x) - f(y)| < t |x-y| $$



              which is what the BFPT requires. From here, we can just apply the BFPT to state that there is a fixed point in $[a,b]$ for the sequence



              $$ x_n = f(x_{n-1})$$






              share|cite|improve this answer









              $endgroup$



              This can be proved using the Banach Fixed Point Theorem.



              Intuitively, the BFPT tells us that if there is some function $F$ such that the distance between any two points $x$ and $y$ (when scaled by a constant $q$) is always larger than the distance between the corresponding images ($F(x)$ and $F(y)$), then the sequence
              $$x_n = F(x_{n-1}) $$
              will converge to a unique fixed point. For a more rigorous treatment of the BFPT statement: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem.



              Since in this case you know that
              $$|f'(x)| leq t $$
              This implies that



              $$Big|frac{f(x) - f(y)}{x - y}Big| leq t$$



              for all possible points $x,y in [a,b]$. Think about why this is the case (Hint: Use the mean value theorem).



              This implies that



              $$| f(x) - f(y)| < t |x-y| $$



              which is what the BFPT requires. From here, we can just apply the BFPT to state that there is a fixed point in $[a,b]$ for the sequence



              $$ x_n = f(x_{n-1})$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 4 hours ago









              Sean LeeSean Lee

              483211




              483211












              • $begingroup$
                Should probably expressly mention that the space $[0, 1]$ is complete, since that is a necessary condition for the Banach fixed point theorem to hold. For an easy counterexample demonstrating that you need completeness, consider $f(x) = x/2$ on $(0, 1]$.
                $endgroup$
                – Robert Shore
                4 hours ago










              • $begingroup$
                @RobertShore Yes, you are right, thanks!
                $endgroup$
                – Sean Lee
                4 hours ago


















              • $begingroup$
                Should probably expressly mention that the space $[0, 1]$ is complete, since that is a necessary condition for the Banach fixed point theorem to hold. For an easy counterexample demonstrating that you need completeness, consider $f(x) = x/2$ on $(0, 1]$.
                $endgroup$
                – Robert Shore
                4 hours ago










              • $begingroup$
                @RobertShore Yes, you are right, thanks!
                $endgroup$
                – Sean Lee
                4 hours ago
















              $begingroup$
              Should probably expressly mention that the space $[0, 1]$ is complete, since that is a necessary condition for the Banach fixed point theorem to hold. For an easy counterexample demonstrating that you need completeness, consider $f(x) = x/2$ on $(0, 1]$.
              $endgroup$
              – Robert Shore
              4 hours ago




              $begingroup$
              Should probably expressly mention that the space $[0, 1]$ is complete, since that is a necessary condition for the Banach fixed point theorem to hold. For an easy counterexample demonstrating that you need completeness, consider $f(x) = x/2$ on $(0, 1]$.
              $endgroup$
              – Robert Shore
              4 hours ago












              $begingroup$
              @RobertShore Yes, you are right, thanks!
              $endgroup$
              – Sean Lee
              4 hours ago




              $begingroup$
              @RobertShore Yes, you are right, thanks!
              $endgroup$
              – Sean Lee
              4 hours ago











              1












              $begingroup$

              To prove this from scratch note that by MVT $$|x_n-x_m| leq |x_{m+1}-x_m|+|x_{m+2}-x_{m+1}|$$ $$+cdots+|x_{n}-x_{n-1}|leq |x_{m+1}-x_m| (1+t+t^{2}+cdots+t^{n+m-1})$$ for $n >m$. Also $|x_{m+1}-x_m| leq t^{m-1} |x_2-x_1|$. Using the convergence of the geometric series $sum t^{n}$ conclude that ${x_n}$ is Cauchy. If $x =lim x_n$ then the definition of $x_n$'s and continuity of $f$ tells you that $f(x)=x$. Uniqueness follows easily by MVT.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                To prove this from scratch note that by MVT $$|x_n-x_m| leq |x_{m+1}-x_m|+|x_{m+2}-x_{m+1}|$$ $$+cdots+|x_{n}-x_{n-1}|leq |x_{m+1}-x_m| (1+t+t^{2}+cdots+t^{n+m-1})$$ for $n >m$. Also $|x_{m+1}-x_m| leq t^{m-1} |x_2-x_1|$. Using the convergence of the geometric series $sum t^{n}$ conclude that ${x_n}$ is Cauchy. If $x =lim x_n$ then the definition of $x_n$'s and continuity of $f$ tells you that $f(x)=x$. Uniqueness follows easily by MVT.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  To prove this from scratch note that by MVT $$|x_n-x_m| leq |x_{m+1}-x_m|+|x_{m+2}-x_{m+1}|$$ $$+cdots+|x_{n}-x_{n-1}|leq |x_{m+1}-x_m| (1+t+t^{2}+cdots+t^{n+m-1})$$ for $n >m$. Also $|x_{m+1}-x_m| leq t^{m-1} |x_2-x_1|$. Using the convergence of the geometric series $sum t^{n}$ conclude that ${x_n}$ is Cauchy. If $x =lim x_n$ then the definition of $x_n$'s and continuity of $f$ tells you that $f(x)=x$. Uniqueness follows easily by MVT.






                  share|cite|improve this answer









                  $endgroup$



                  To prove this from scratch note that by MVT $$|x_n-x_m| leq |x_{m+1}-x_m|+|x_{m+2}-x_{m+1}|$$ $$+cdots+|x_{n}-x_{n-1}|leq |x_{m+1}-x_m| (1+t+t^{2}+cdots+t^{n+m-1})$$ for $n >m$. Also $|x_{m+1}-x_m| leq t^{m-1} |x_2-x_1|$. Using the convergence of the geometric series $sum t^{n}$ conclude that ${x_n}$ is Cauchy. If $x =lim x_n$ then the definition of $x_n$'s and continuity of $f$ tells you that $f(x)=x$. Uniqueness follows easily by MVT.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  Kavi Rama MurthyKavi Rama Murthy

                  65k42766




                  65k42766






















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                      CARDNET

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                      CARDNET は日本カードネットワークが運営するクレジット決済ネットワーク。日本カードネットワークの略称としてもCARDNETが使用されている。 事件・事故 2017年4月15日午前11時8分からクレジットカードの決済がしづらくなる障害が発生した。障害は同日午後5時18分に復旧した。原因は、複数あるCARDNETセンターの拠点のうち1つで、2重化してあるL3スイッチの片方が故障し、1系統にトラフィックが集中したことによる輻輳が発生したため。対処としては、故障したL3スイッチ交換とシステム再起動が行われた [1] 。 出典 ^ 金子寛人 ( 2017年4月17日 ). “CARDNETのクレジット決済に6時間強障害、原因はL3スイッチ故障”. 日経コンピュータ. 2018年10月18日 閲覧。 この項目は、コンピュータに関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（PJ:コンピュータ/P:コンピュータ）。 この項目は、経済に関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（ポータル 経済学、プロジェクト 経済）。 This page is only for reference, If you need detailed information, please check here
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                      Boot-repair Failure: Unable to locate package grub-common:i386

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                      0 I am on a system dual-booting Windows and Debian (usually using the latter), and a recent Windows update appears to have wreaked havoc on the Debian boot system. I managed to boot Windows again, but not Linux. Following online suggestions, I created a Live USB of Ubuntu, and booted from that. Then I have installed and run the boot-repair utility. The utility instructs me to type several lines including sudo chroot "/mnt/boot-sav/sda6" apt-get purge -y grub*-common grub-common:i386 To which I receive the error E: Unable to locate package grub-common:i386 I am given to understand the issue may have to do with having UEFI vs BIOS (I believe I have UEFI), but do not understand this well. In any case I have subsequently run the boot-repair BootInfo Summary utility which created the following file h
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                      濃尾地震

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                      濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸
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