Reverse fixed point conclusion











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If I have a function such as:



$$f:M rightarrow mathbb{R} $$



where $M$ is any metric space denoted by :



$$(M,d)$$



$$f(x) =d(x,y) $$ where $y in M$ is a fixed point.



I am trying to show that this function satisfies the Lipschitz condition.
https://en.wikipedia.org/wiki/Lipschitz_continuity
$$frac{d(f(x),f(z))}{d(x,z)}leq K$$ for $K geq 0$



Currently I am stuck at two things.



First, the distance representation is not quite clear for me , I am not sure if I can use here $d(x,y) = |x-y|$?



I only learnt two fixed points theorems, the Banach fixed point and the Brouwer fixed point.



Secondly, If I have a fixed point $y$ :
it could be a Banach case(unique fixed point), if $M$ is a complete metric space and If $f$ is a contraction i.e. $0 leq K <1$



Otherwise, it is not unique fixed point.



I am not sure what approach should I use here? contradiction or attempt some sort of reverse fixed point iteration direct proof?






real-analysis metric-spaces fixed-point-theorems







share|cite|improve this question




























    up vote
    0
    down vote

    favorite












    If I have a function such as:



    $$f:M rightarrow mathbb{R} $$



    where $M$ is any metric space denoted by :



    $$(M,d)$$



    $$f(x) =d(x,y) $$ where $y in M$ is a fixed point.



    I am trying to show that this function satisfies the Lipschitz condition.
    https://en.wikipedia.org/wiki/Lipschitz_continuity
    $$frac{d(f(x),f(z))}{d(x,z)}leq K$$ for $K geq 0$



    Currently I am stuck at two things.



    First, the distance representation is not quite clear for me , I am not sure if I can use here $d(x,y) = |x-y|$?



    I only learnt two fixed points theorems, the Banach fixed point and the Brouwer fixed point.



    Secondly, If I have a fixed point $y$ :
    it could be a Banach case(unique fixed point), if $M$ is a complete metric space and If $f$ is a contraction i.e. $0 leq K <1$



    Otherwise, it is not unique fixed point.



    I am not sure what approach should I use here? contradiction or attempt some sort of reverse fixed point iteration direct proof?






    real-analysis metric-spaces fixed-point-theorems







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      If I have a function such as:



      $$f:M rightarrow mathbb{R} $$



      where $M$ is any metric space denoted by :



      $$(M,d)$$



      $$f(x) =d(x,y) $$ where $y in M$ is a fixed point.



      I am trying to show that this function satisfies the Lipschitz condition.
      https://en.wikipedia.org/wiki/Lipschitz_continuity
      $$frac{d(f(x),f(z))}{d(x,z)}leq K$$ for $K geq 0$



      Currently I am stuck at two things.



      First, the distance representation is not quite clear for me , I am not sure if I can use here $d(x,y) = |x-y|$?



      I only learnt two fixed points theorems, the Banach fixed point and the Brouwer fixed point.



      Secondly, If I have a fixed point $y$ :
      it could be a Banach case(unique fixed point), if $M$ is a complete metric space and If $f$ is a contraction i.e. $0 leq K <1$



      Otherwise, it is not unique fixed point.



      I am not sure what approach should I use here? contradiction or attempt some sort of reverse fixed point iteration direct proof?






      real-analysis metric-spaces fixed-point-theorems







      share|cite|improve this question















      If I have a function such as:



      $$f:M rightarrow mathbb{R} $$



      where $M$ is any metric space denoted by :



      $$(M,d)$$



      $$f(x) =d(x,y) $$ where $y in M$ is a fixed point.



      I am trying to show that this function satisfies the Lipschitz condition.
      https://en.wikipedia.org/wiki/Lipschitz_continuity
      $$frac{d(f(x),f(z))}{d(x,z)}leq K$$ for $K geq 0$



      Currently I am stuck at two things.



      First, the distance representation is not quite clear for me , I am not sure if I can use here $d(x,y) = |x-y|$?



      I only learnt two fixed points theorems, the Banach fixed point and the Brouwer fixed point.



      Secondly, If I have a fixed point $y$ :
      it could be a Banach case(unique fixed point), if $M$ is a complete metric space and If $f$ is a contraction i.e. $0 leq K <1$



      Otherwise, it is not unique fixed point.



      I am not sure what approach should I use here? contradiction or attempt some sort of reverse fixed point iteration direct proof?






      real-analysis metric-spaces fixed-point-theorems




      real-analysis metric-spaces fixed-point-theorems






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited 2 days ago

























      asked 2 days ago









      JKM

      4714




      4714






















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          The triangle inequality implies that, for all $x_1,x_2in M$,
          $$
          d(x_1,x_2)+d(x_2,y)ge d(x_1,y) quadtext{and}quad
          d(x_2,x_1)+d(x_1,y)ge d(x_2,y)
          $$
          hence
          $$
          d(x_1,x_2)ge d(x_1,y)-d(x_2,y) quadtext{and}quad
          d(x_1,x_2)ge d(x_2,y)-d(x_1,y)
          $$
          and therefore
          $$
          |,f(x_1)-f(x_2)|=big|,d(x_1,y)-d(x_2,y)big|le d(x_1,x_2).
          $$
          So, $f$ is Lipschitz, with $K=1$.






          share|cite|improve this answer





















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            The triangle inequality implies that, for all $x_1,x_2in M$,
            $$
            d(x_1,x_2)+d(x_2,y)ge d(x_1,y) quadtext{and}quad
            d(x_2,x_1)+d(x_1,y)ge d(x_2,y)
            $$
            hence
            $$
            d(x_1,x_2)ge d(x_1,y)-d(x_2,y) quadtext{and}quad
            d(x_1,x_2)ge d(x_2,y)-d(x_1,y)
            $$
            and therefore
            $$
            |,f(x_1)-f(x_2)|=big|,d(x_1,y)-d(x_2,y)big|le d(x_1,x_2).
            $$
            So, $f$ is Lipschitz, with $K=1$.






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              The triangle inequality implies that, for all $x_1,x_2in M$,
              $$
              d(x_1,x_2)+d(x_2,y)ge d(x_1,y) quadtext{and}quad
              d(x_2,x_1)+d(x_1,y)ge d(x_2,y)
              $$
              hence
              $$
              d(x_1,x_2)ge d(x_1,y)-d(x_2,y) quadtext{and}quad
              d(x_1,x_2)ge d(x_2,y)-d(x_1,y)
              $$
              and therefore
              $$
              |,f(x_1)-f(x_2)|=big|,d(x_1,y)-d(x_2,y)big|le d(x_1,x_2).
              $$
              So, $f$ is Lipschitz, with $K=1$.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                The triangle inequality implies that, for all $x_1,x_2in M$,
                $$
                d(x_1,x_2)+d(x_2,y)ge d(x_1,y) quadtext{and}quad
                d(x_2,x_1)+d(x_1,y)ge d(x_2,y)
                $$
                hence
                $$
                d(x_1,x_2)ge d(x_1,y)-d(x_2,y) quadtext{and}quad
                d(x_1,x_2)ge d(x_2,y)-d(x_1,y)
                $$
                and therefore
                $$
                |,f(x_1)-f(x_2)|=big|,d(x_1,y)-d(x_2,y)big|le d(x_1,x_2).
                $$
                So, $f$ is Lipschitz, with $K=1$.






                share|cite|improve this answer












                The triangle inequality implies that, for all $x_1,x_2in M$,
                $$
                d(x_1,x_2)+d(x_2,y)ge d(x_1,y) quadtext{and}quad
                d(x_2,x_1)+d(x_1,y)ge d(x_2,y)
                $$
                hence
                $$
                d(x_1,x_2)ge d(x_1,y)-d(x_2,y) quadtext{and}quad
                d(x_1,x_2)ge d(x_2,y)-d(x_1,y)
                $$
                and therefore
                $$
                |,f(x_1)-f(x_2)|=big|,d(x_1,y)-d(x_2,y)big|le d(x_1,x_2).
                $$
                So, $f$ is Lipschitz, with $K=1$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Yiorgos S. Smyrlis

                61.5k1383161




                61.5k1383161






























                     

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