If $f^n$ is a contraction, is $f^m(x)$ a Cauchy sequence?












1












$begingroup$


I wonder if $f: Xrightarrow X$ from the metric space $X$ to itself is such that there exists $n>1$ such that $f^n$ is a contraction, then it is true that for any point $xin X$, $(f^m(x))_m$ is a Cauchy sequence? I think of this question when I was doing an exercise that asks to show that if $X$ is complete, then $f$ has a unique fixed point. I was able to solve this exercise but still curious about the answer of my question.



Thanks in advance!






general-topology







share|cite|improve this question









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  • $begingroup$
    Do you assume that $X$ is bounded?
    $endgroup$
    – Yanko
    Dec 16 '18 at 15:04










  • $begingroup$
    @Yanko no, but I am interested in the situation in both cases.
    $endgroup$
    – Jiu
    Dec 16 '18 at 15:05










  • $begingroup$
    Just to make sure, by contraction you mean $d(f(x),f(y))<lambda d(x,y)$ where $0<lambda <1$?
    $endgroup$
    – Yanko
    Dec 16 '18 at 15:06










  • $begingroup$
    @Yanko yes, with the same $lambda$ for all $x,y$.
    $endgroup$
    – Jiu
    Dec 16 '18 at 15:07










  • $begingroup$
    @Yanko Thanks for your answer but the hypothesis is only that $f$ composed with itself $n$ times is contraction. We don’t know if $f$ is a contraction itself do we?
    $endgroup$
    – Jiu
    Dec 16 '18 at 15:29
















1












$begingroup$


I wonder if $f: Xrightarrow X$ from the metric space $X$ to itself is such that there exists $n>1$ such that $f^n$ is a contraction, then it is true that for any point $xin X$, $(f^m(x))_m$ is a Cauchy sequence? I think of this question when I was doing an exercise that asks to show that if $X$ is complete, then $f$ has a unique fixed point. I was able to solve this exercise but still curious about the answer of my question.



Thanks in advance!






general-topology







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do you assume that $X$ is bounded?
    $endgroup$
    – Yanko
    Dec 16 '18 at 15:04










  • $begingroup$
    @Yanko no, but I am interested in the situation in both cases.
    $endgroup$
    – Jiu
    Dec 16 '18 at 15:05










  • $begingroup$
    Just to make sure, by contraction you mean $d(f(x),f(y))<lambda d(x,y)$ where $0<lambda <1$?
    $endgroup$
    – Yanko
    Dec 16 '18 at 15:06










  • $begingroup$
    @Yanko yes, with the same $lambda$ for all $x,y$.
    $endgroup$
    – Jiu
    Dec 16 '18 at 15:07










  • $begingroup$
    @Yanko Thanks for your answer but the hypothesis is only that $f$ composed with itself $n$ times is contraction. We don’t know if $f$ is a contraction itself do we?
    $endgroup$
    – Jiu
    Dec 16 '18 at 15:29














1












1








1





$begingroup$


I wonder if $f: Xrightarrow X$ from the metric space $X$ to itself is such that there exists $n>1$ such that $f^n$ is a contraction, then it is true that for any point $xin X$, $(f^m(x))_m$ is a Cauchy sequence? I think of this question when I was doing an exercise that asks to show that if $X$ is complete, then $f$ has a unique fixed point. I was able to solve this exercise but still curious about the answer of my question.



Thanks in advance!






general-topology







share|cite|improve this question









$endgroup$




I wonder if $f: Xrightarrow X$ from the metric space $X$ to itself is such that there exists $n>1$ such that $f^n$ is a contraction, then it is true that for any point $xin X$, $(f^m(x))_m$ is a Cauchy sequence? I think of this question when I was doing an exercise that asks to show that if $X$ is complete, then $f$ has a unique fixed point. I was able to solve this exercise but still curious about the answer of my question.



Thanks in advance!






general-topology




general-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 16 '18 at 15:01









JiuJiu

515113




515113












  • $begingroup$
    Do you assume that $X$ is bounded?
    $endgroup$
    – Yanko
    Dec 16 '18 at 15:04










  • $begingroup$
    @Yanko no, but I am interested in the situation in both cases.
    $endgroup$
    – Jiu
    Dec 16 '18 at 15:05










  • $begingroup$
    Just to make sure, by contraction you mean $d(f(x),f(y))<lambda d(x,y)$ where $0<lambda <1$?
    $endgroup$
    – Yanko
    Dec 16 '18 at 15:06










  • $begingroup$
    @Yanko yes, with the same $lambda$ for all $x,y$.
    $endgroup$
    – Jiu
    Dec 16 '18 at 15:07










  • $begingroup$
    @Yanko Thanks for your answer but the hypothesis is only that $f$ composed with itself $n$ times is contraction. We don’t know if $f$ is a contraction itself do we?
    $endgroup$
    – Jiu
    Dec 16 '18 at 15:29


















  • $begingroup$
    Do you assume that $X$ is bounded?
    $endgroup$
    – Yanko
    Dec 16 '18 at 15:04










  • $begingroup$
    @Yanko no, but I am interested in the situation in both cases.
    $endgroup$
    – Jiu
    Dec 16 '18 at 15:05










  • $begingroup$
    Just to make sure, by contraction you mean $d(f(x),f(y))<lambda d(x,y)$ where $0<lambda <1$?
    $endgroup$
    – Yanko
    Dec 16 '18 at 15:06










  • $begingroup$
    @Yanko yes, with the same $lambda$ for all $x,y$.
    $endgroup$
    – Jiu
    Dec 16 '18 at 15:07










  • $begingroup$
    @Yanko Thanks for your answer but the hypothesis is only that $f$ composed with itself $n$ times is contraction. We don’t know if $f$ is a contraction itself do we?
    $endgroup$
    – Jiu
    Dec 16 '18 at 15:29
















$begingroup$
Do you assume that $X$ is bounded?
$endgroup$
– Yanko
Dec 16 '18 at 15:04




$begingroup$
Do you assume that $X$ is bounded?
$endgroup$
– Yanko
Dec 16 '18 at 15:04












$begingroup$
@Yanko no, but I am interested in the situation in both cases.
$endgroup$
– Jiu
Dec 16 '18 at 15:05




$begingroup$
@Yanko no, but I am interested in the situation in both cases.
$endgroup$
– Jiu
Dec 16 '18 at 15:05












$begingroup$
Just to make sure, by contraction you mean $d(f(x),f(y))<lambda d(x,y)$ where $0<lambda <1$?
$endgroup$
– Yanko
Dec 16 '18 at 15:06




$begingroup$
Just to make sure, by contraction you mean $d(f(x),f(y))<lambda d(x,y)$ where $0<lambda <1$?
$endgroup$
– Yanko
Dec 16 '18 at 15:06












$begingroup$
@Yanko yes, with the same $lambda$ for all $x,y$.
$endgroup$
– Jiu
Dec 16 '18 at 15:07




$begingroup$
@Yanko yes, with the same $lambda$ for all $x,y$.
$endgroup$
– Jiu
Dec 16 '18 at 15:07












$begingroup$
@Yanko Thanks for your answer but the hypothesis is only that $f$ composed with itself $n$ times is contraction. We don’t know if $f$ is a contraction itself do we?
$endgroup$
– Jiu
Dec 16 '18 at 15:29




$begingroup$
@Yanko Thanks for your answer but the hypothesis is only that $f$ composed with itself $n$ times is contraction. We don’t know if $f$ is a contraction itself do we?
$endgroup$
– Jiu
Dec 16 '18 at 15:29










1 Answer
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$begingroup$

Since $f^n$ is a contraction, we can find $lambdain(0,1)$ such that $d(f^nx,f^ny)leqlambda d(x,y)$ for all $x,yin X$.



Fix $xin X$, and we can partition $(f^mx)_m$ into $n$ subsequences $(x_k(r)=f^{nk+r}x)_k$ for $r=0,1,dots,n-1$. We know each $(x_k(r))_k$ is Cauchy because $f^n$ is a contraction. Let $operatorname{diam}{f^mxmid mgeq 0}=Delta$ (this is finite because we have finitely many Cauchy sequences). Then
begin{align}
operatorname{diam}{f^mxmid mgeq m_0=nk+r}
&leqoperatorname{diam}{f^mxmid mgeq nk}\
&leqlambda^kDeltato 0
end{align}
as $ktoinfty$ (equivalently $m_0toinfty$), so $(f^mx)_m$ is Cauchy.






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    1 Answer
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    1 Answer
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    3












    $begingroup$

    Since $f^n$ is a contraction, we can find $lambdain(0,1)$ such that $d(f^nx,f^ny)leqlambda d(x,y)$ for all $x,yin X$.



    Fix $xin X$, and we can partition $(f^mx)_m$ into $n$ subsequences $(x_k(r)=f^{nk+r}x)_k$ for $r=0,1,dots,n-1$. We know each $(x_k(r))_k$ is Cauchy because $f^n$ is a contraction. Let $operatorname{diam}{f^mxmid mgeq 0}=Delta$ (this is finite because we have finitely many Cauchy sequences). Then
    begin{align}
    operatorname{diam}{f^mxmid mgeq m_0=nk+r}
    &leqoperatorname{diam}{f^mxmid mgeq nk}\
    &leqlambda^kDeltato 0
    end{align}
    as $ktoinfty$ (equivalently $m_0toinfty$), so $(f^mx)_m$ is Cauchy.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      Since $f^n$ is a contraction, we can find $lambdain(0,1)$ such that $d(f^nx,f^ny)leqlambda d(x,y)$ for all $x,yin X$.



      Fix $xin X$, and we can partition $(f^mx)_m$ into $n$ subsequences $(x_k(r)=f^{nk+r}x)_k$ for $r=0,1,dots,n-1$. We know each $(x_k(r))_k$ is Cauchy because $f^n$ is a contraction. Let $operatorname{diam}{f^mxmid mgeq 0}=Delta$ (this is finite because we have finitely many Cauchy sequences). Then
      begin{align}
      operatorname{diam}{f^mxmid mgeq m_0=nk+r}
      &leqoperatorname{diam}{f^mxmid mgeq nk}\
      &leqlambda^kDeltato 0
      end{align}
      as $ktoinfty$ (equivalently $m_0toinfty$), so $(f^mx)_m$ is Cauchy.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        Since $f^n$ is a contraction, we can find $lambdain(0,1)$ such that $d(f^nx,f^ny)leqlambda d(x,y)$ for all $x,yin X$.



        Fix $xin X$, and we can partition $(f^mx)_m$ into $n$ subsequences $(x_k(r)=f^{nk+r}x)_k$ for $r=0,1,dots,n-1$. We know each $(x_k(r))_k$ is Cauchy because $f^n$ is a contraction. Let $operatorname{diam}{f^mxmid mgeq 0}=Delta$ (this is finite because we have finitely many Cauchy sequences). Then
        begin{align}
        operatorname{diam}{f^mxmid mgeq m_0=nk+r}
        &leqoperatorname{diam}{f^mxmid mgeq nk}\
        &leqlambda^kDeltato 0
        end{align}
        as $ktoinfty$ (equivalently $m_0toinfty$), so $(f^mx)_m$ is Cauchy.






        share|cite|improve this answer











        $endgroup$



        Since $f^n$ is a contraction, we can find $lambdain(0,1)$ such that $d(f^nx,f^ny)leqlambda d(x,y)$ for all $x,yin X$.



        Fix $xin X$, and we can partition $(f^mx)_m$ into $n$ subsequences $(x_k(r)=f^{nk+r}x)_k$ for $r=0,1,dots,n-1$. We know each $(x_k(r))_k$ is Cauchy because $f^n$ is a contraction. Let $operatorname{diam}{f^mxmid mgeq 0}=Delta$ (this is finite because we have finitely many Cauchy sequences). Then
        begin{align}
        operatorname{diam}{f^mxmid mgeq m_0=nk+r}
        &leqoperatorname{diam}{f^mxmid mgeq nk}\
        &leqlambda^kDeltato 0
        end{align}
        as $ktoinfty$ (equivalently $m_0toinfty$), so $(f^mx)_m$ is Cauchy.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 16 '18 at 16:41

























        answered Dec 16 '18 at 16:26









        user10354138user10354138

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