Prove this version of Banach differential fixed point












3












$begingroup$


Let $I$ a closed interval and $f:Ito mathbb{R}$ a differentiable function such that:




  1. $f(I)subseteq I$

  2. Exists $pinmathbb{N}$ and $cinmathbb{R}$ such that $g=fcirc f cdots circ f=f^p$ satisfy $|g'(x)|leq c< 1$, $forall xin I$.


Prove that:



a. Exists a unique $ain I$ such that $f(a)=a$.



b. For all $xin I$,$lim_{nto infty} f^n (x)=a$.






calculus fixed-point-theorems







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Let $I$ a closed interval and $f:Ito mathbb{R}$ a differentiable function such that:




    1. $f(I)subseteq I$

    2. Exists $pinmathbb{N}$ and $cinmathbb{R}$ such that $g=fcirc f cdots circ f=f^p$ satisfy $|g'(x)|leq c< 1$, $forall xin I$.


    Prove that:



    a. Exists a unique $ain I$ such that $f(a)=a$.



    b. For all $xin I$,$lim_{nto infty} f^n (x)=a$.






    calculus fixed-point-theorems







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Let $I$ a closed interval and $f:Ito mathbb{R}$ a differentiable function such that:




      1. $f(I)subseteq I$

      2. Exists $pinmathbb{N}$ and $cinmathbb{R}$ such that $g=fcirc f cdots circ f=f^p$ satisfy $|g'(x)|leq c< 1$, $forall xin I$.


      Prove that:



      a. Exists a unique $ain I$ such that $f(a)=a$.



      b. For all $xin I$,$lim_{nto infty} f^n (x)=a$.






      calculus fixed-point-theorems







      share|cite|improve this question









      $endgroup$




      Let $I$ a closed interval and $f:Ito mathbb{R}$ a differentiable function such that:




      1. $f(I)subseteq I$

      2. Exists $pinmathbb{N}$ and $cinmathbb{R}$ such that $g=fcirc f cdots circ f=f^p$ satisfy $|g'(x)|leq c< 1$, $forall xin I$.


      Prove that:



      a. Exists a unique $ain I$ such that $f(a)=a$.



      b. For all $xin I$,$lim_{nto infty} f^n (x)=a$.






      calculus fixed-point-theorems




      calculus fixed-point-theorems






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 7 at 7:36









      Raul_MFerRaul_MFer

      333




      333






















          1 Answer
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          0












          $begingroup$

          First off, we may apply the fixed point theorem to $g$, since $|g'(x)|leq c<1$ for all $x$ and $g(I)subset I$, and we obtain that $g^{n}(x)to a$ for all $x$, for some $ain I$ with $g(a)=a$.



          Now fix $xin I$, take the sequence $left{f^{n}(x)right}_{nin mathbb{N}}$ and write it as a union of a finite amount of subsequences in the following way
          $$left{f^{n}(x)right}_{nin mathbb{N}}=bigcup_{j=0}^{p-1}left{f^{np+j}(x)right}_{nin mathbb{N}}=bigcup_{j=0}^{p-1}left{g^{n}(f^{j}(x))right}_{nin mathbb{N}} $$
          since the sequences $left{g^{n}(f^{j}(x))right}_{nin mathbb{N}}$ all converge to the same number $a$ for $j=0,dots, p-1$, then the whole sequence $left{f^{n}(x)right}$ must converge to $a$.



          To see that $f(a)=a$, notice that $g(a)=a$ implies that $f^{np}(a)=a$ for all $nin mathbb{N}$. Therefore we have $f^{np+j}(a)=f^{j}(f^{np}(a))=f^{j}(a)$ for all $nin mathbb{N}$ and $jin left{0,dots,p-1right}$, i.e. the sequence $left{f^{n}(a)right}$ is periodic with period $p$. On the other hand we know that $lim_{nto +infty}f^n(a)=a$, and so necessarily $f^{n}(a)=a$ for all $nin mathbb{N}$. In particular, $f(a)=a$.



          Alternatively, $$g(f(a))=f^{p}(f(a))=f(f^{p}(a))=f(g(a))=f(a)$$
          i.e. $f(a)$ is a fixed point for $g$. But $a$ is the unique fixed point for $g$, thus $f(a)=a$.






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            First off, we may apply the fixed point theorem to $g$, since $|g'(x)|leq c<1$ for all $x$ and $g(I)subset I$, and we obtain that $g^{n}(x)to a$ for all $x$, for some $ain I$ with $g(a)=a$.



            Now fix $xin I$, take the sequence $left{f^{n}(x)right}_{nin mathbb{N}}$ and write it as a union of a finite amount of subsequences in the following way
            $$left{f^{n}(x)right}_{nin mathbb{N}}=bigcup_{j=0}^{p-1}left{f^{np+j}(x)right}_{nin mathbb{N}}=bigcup_{j=0}^{p-1}left{g^{n}(f^{j}(x))right}_{nin mathbb{N}} $$
            since the sequences $left{g^{n}(f^{j}(x))right}_{nin mathbb{N}}$ all converge to the same number $a$ for $j=0,dots, p-1$, then the whole sequence $left{f^{n}(x)right}$ must converge to $a$.



            To see that $f(a)=a$, notice that $g(a)=a$ implies that $f^{np}(a)=a$ for all $nin mathbb{N}$. Therefore we have $f^{np+j}(a)=f^{j}(f^{np}(a))=f^{j}(a)$ for all $nin mathbb{N}$ and $jin left{0,dots,p-1right}$, i.e. the sequence $left{f^{n}(a)right}$ is periodic with period $p$. On the other hand we know that $lim_{nto +infty}f^n(a)=a$, and so necessarily $f^{n}(a)=a$ for all $nin mathbb{N}$. In particular, $f(a)=a$.



            Alternatively, $$g(f(a))=f^{p}(f(a))=f(f^{p}(a))=f(g(a))=f(a)$$
            i.e. $f(a)$ is a fixed point for $g$. But $a$ is the unique fixed point for $g$, thus $f(a)=a$.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              First off, we may apply the fixed point theorem to $g$, since $|g'(x)|leq c<1$ for all $x$ and $g(I)subset I$, and we obtain that $g^{n}(x)to a$ for all $x$, for some $ain I$ with $g(a)=a$.



              Now fix $xin I$, take the sequence $left{f^{n}(x)right}_{nin mathbb{N}}$ and write it as a union of a finite amount of subsequences in the following way
              $$left{f^{n}(x)right}_{nin mathbb{N}}=bigcup_{j=0}^{p-1}left{f^{np+j}(x)right}_{nin mathbb{N}}=bigcup_{j=0}^{p-1}left{g^{n}(f^{j}(x))right}_{nin mathbb{N}} $$
              since the sequences $left{g^{n}(f^{j}(x))right}_{nin mathbb{N}}$ all converge to the same number $a$ for $j=0,dots, p-1$, then the whole sequence $left{f^{n}(x)right}$ must converge to $a$.



              To see that $f(a)=a$, notice that $g(a)=a$ implies that $f^{np}(a)=a$ for all $nin mathbb{N}$. Therefore we have $f^{np+j}(a)=f^{j}(f^{np}(a))=f^{j}(a)$ for all $nin mathbb{N}$ and $jin left{0,dots,p-1right}$, i.e. the sequence $left{f^{n}(a)right}$ is periodic with period $p$. On the other hand we know that $lim_{nto +infty}f^n(a)=a$, and so necessarily $f^{n}(a)=a$ for all $nin mathbb{N}$. In particular, $f(a)=a$.



              Alternatively, $$g(f(a))=f^{p}(f(a))=f(f^{p}(a))=f(g(a))=f(a)$$
              i.e. $f(a)$ is a fixed point for $g$. But $a$ is the unique fixed point for $g$, thus $f(a)=a$.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                First off, we may apply the fixed point theorem to $g$, since $|g'(x)|leq c<1$ for all $x$ and $g(I)subset I$, and we obtain that $g^{n}(x)to a$ for all $x$, for some $ain I$ with $g(a)=a$.



                Now fix $xin I$, take the sequence $left{f^{n}(x)right}_{nin mathbb{N}}$ and write it as a union of a finite amount of subsequences in the following way
                $$left{f^{n}(x)right}_{nin mathbb{N}}=bigcup_{j=0}^{p-1}left{f^{np+j}(x)right}_{nin mathbb{N}}=bigcup_{j=0}^{p-1}left{g^{n}(f^{j}(x))right}_{nin mathbb{N}} $$
                since the sequences $left{g^{n}(f^{j}(x))right}_{nin mathbb{N}}$ all converge to the same number $a$ for $j=0,dots, p-1$, then the whole sequence $left{f^{n}(x)right}$ must converge to $a$.



                To see that $f(a)=a$, notice that $g(a)=a$ implies that $f^{np}(a)=a$ for all $nin mathbb{N}$. Therefore we have $f^{np+j}(a)=f^{j}(f^{np}(a))=f^{j}(a)$ for all $nin mathbb{N}$ and $jin left{0,dots,p-1right}$, i.e. the sequence $left{f^{n}(a)right}$ is periodic with period $p$. On the other hand we know that $lim_{nto +infty}f^n(a)=a$, and so necessarily $f^{n}(a)=a$ for all $nin mathbb{N}$. In particular, $f(a)=a$.



                Alternatively, $$g(f(a))=f^{p}(f(a))=f(f^{p}(a))=f(g(a))=f(a)$$
                i.e. $f(a)$ is a fixed point for $g$. But $a$ is the unique fixed point for $g$, thus $f(a)=a$.






                share|cite|improve this answer











                $endgroup$



                First off, we may apply the fixed point theorem to $g$, since $|g'(x)|leq c<1$ for all $x$ and $g(I)subset I$, and we obtain that $g^{n}(x)to a$ for all $x$, for some $ain I$ with $g(a)=a$.



                Now fix $xin I$, take the sequence $left{f^{n}(x)right}_{nin mathbb{N}}$ and write it as a union of a finite amount of subsequences in the following way
                $$left{f^{n}(x)right}_{nin mathbb{N}}=bigcup_{j=0}^{p-1}left{f^{np+j}(x)right}_{nin mathbb{N}}=bigcup_{j=0}^{p-1}left{g^{n}(f^{j}(x))right}_{nin mathbb{N}} $$
                since the sequences $left{g^{n}(f^{j}(x))right}_{nin mathbb{N}}$ all converge to the same number $a$ for $j=0,dots, p-1$, then the whole sequence $left{f^{n}(x)right}$ must converge to $a$.



                To see that $f(a)=a$, notice that $g(a)=a$ implies that $f^{np}(a)=a$ for all $nin mathbb{N}$. Therefore we have $f^{np+j}(a)=f^{j}(f^{np}(a))=f^{j}(a)$ for all $nin mathbb{N}$ and $jin left{0,dots,p-1right}$, i.e. the sequence $left{f^{n}(a)right}$ is periodic with period $p$. On the other hand we know that $lim_{nto +infty}f^n(a)=a$, and so necessarily $f^{n}(a)=a$ for all $nin mathbb{N}$. In particular, $f(a)=a$.



                Alternatively, $$g(f(a))=f^{p}(f(a))=f(f^{p}(a))=f(g(a))=f(a)$$
                i.e. $f(a)$ is a fixed point for $g$. But $a$ is the unique fixed point for $g$, thus $f(a)=a$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 7 at 8:47

























                answered Jan 7 at 8:29









                Lorenzo QuarisaLorenzo Quarisa

                3,312518




                3,312518






























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