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Fixed Point For A Contraction

  1. Sep 18, 2015 #1
    Let (X, d) be a complete metric space, and suppose T : X → X is a function such that T^2 is a contraction. [By T^2, we mean the function T^2 : X → X given by T^2(x) = T(T(x))]. Show that T has a unique fixed point in X.

    So I have an answer, but I am not sure whether it is correct. It goes as follows :

    Any fixed point of T is also a fixed point of T^2, and there is only one of these. Let x∈X be the unique fixed point of T^2, and consider T^2=T(T(x))=T(x). But now T(x) is a fixed point of T^2, so T(x)=x and x is also a fixed point of T. Since x∈X, T has a unique fixed point in X

    Would that be sufficient or am I missing something?
     
  2. jcsd
  3. Sep 18, 2015 #2

    mathwonk

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    If you meant: T^2(T(x)) = T^3(x) = T(T^2(x)) = T(x), then I agree.
     
  4. Sep 19, 2015 #3

    WWGD

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    I would interpret the fixed point property for T^2 to mean that ##T^2(x)=(T \circ T)(x)=T(T(x))=x ## . The relation ## T(T(x))=T(x) ## is often called an involution, but not every contraction is an involution, e.g., ##f(x)=x/2 ## is a contraction but not an involution.
     
  5. Sep 19, 2015 #4

    HallsofIvy

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    At some point you should say "because [itex]T^2[/itex] is a contraction, [itex]T^2[/itex] has a unique fixed point". Also "[itex]T^2= T(T(x))= T(x)[/itex]" does not make sense for two reasons. The first part, [itex]T^2[/itex] is an operator while the other two [itex]T(T(x))[/itex] and [itex]T(x)[/itex] are points. I presume you meant to say [itex]T^2(x)= T(T(x))= T(x)[/itex]. But even then the second equation, [itex]T(T(x))= T(x)[/itex] assumes that x is a fixed point of T, not [itex]T^2[/itex] and that assumes what you are trying to prove. What you have proved is that "if x is a fixed point of T then it is a fixed point of [itex]T^2[/itex]". That is not what you want to prove.
     
  6. Sep 19, 2015 #5

    mathwonk

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    let me elaborate my reply; As you said, every fixed point of T is a fixed point of T^2, hence T has at most one fixed point since T^2 has only one.

    So it remains to show that T has also a fixed point at x, i.e. to show that T(x) = x. For this, since T^2 fixes x and has only one fixed point, it suffices to show that T^2 also fixes T(x).

    Then since T^2(x) = x, we get T^2(T(x)) = T^3(x) = T(T^2(x)) = T(x), and that does it.

    I assume you had some such idea but mistyped it somehow.
     
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