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Fixed point theorem knowing that ##T^n## is a contraction

  1. Nov 13, 2013 #1
    The problem statement, all variables and given/known data.
    Let ##X## be a complete metric space and let ##T:X \to X## such that there exists ##n \in \mathbb N##: ##T^n## is a contraction. Prove that there is a unique ##x \in X## such that ##T(x)=x##.

    The attempt at a solution.
    Sorry but I am completely lost with this exercise and I am a little bit confused about the following: if ##T^n## is a contraction, would this mean that for any ##x,y \in X## ##d(T^n(x), T^n(y))<αd(x,y)## or ##d(T^n(x), T^n(y))<αd(T^{n-1}(x),T^{n-1}(y))##, for ##0<α<1##?. How could I begin the exercise to find the ##x## such that ##T(x)=x##?.
     
  2. jcsd
  3. Nov 13, 2013 #2

    Dick

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    Let's just write ##C## instead of ##T^n## so we can skip some exponents. Here's a hint. Pick ##x_0## to be any point. Now define ##x_1=C(x_0), x_2=C(x_1)##, etc. Now you've got things like ##d(x_1,x_2)=d(C(x_0),C(x_1)) \le αd(x_0,x_1)##. The first step is going to be trying to prove that sequence is Cauchy.
     
  4. Nov 13, 2013 #3

    HallsofIvy

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    BOTH. The definition of "contraction map" is that for any x, y, [itex]d(T(x), T(y))\le \alpha d(x, y)[/itex] (with [itex]\alpha< 0[/itex]). For any positive integer, [itex]d(T^n(x), T^n(y))<\alpha d(T^{n-1}(x), T^{n-1}(y))[/itex] is just that same statement with [itex]T^{n-1}(x)[/itex] in place of x and [itex]T^{n-1}(y)[/itex] in place of y.

    Take [itex]x_0[/itex] to be any point in X. Look at the sequence [itex]\{T^n(x_0)\}[/itex]. Show that it is a Cauchy sequence so converges. See what happens when you apply T to the limit of that sequence.
     
  5. Nov 13, 2013 #4

    Dick

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    Not quite, I don't think. The problem (for some reason) didn't say T is a contraction, it says ##T^n## is a contraction.
     
  6. Nov 20, 2013 #5
    Ok, but with that sequence, what I am goint to prove is that ##T^n## has a fixed point. If I call that point ##x##, would ##x## help me to find the fixed point of ##T## that I am looking for?
     
  7. Nov 20, 2013 #6

    HallsofIvy

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    In my first response I said "(with [itex]\alpha< 0[/itex])" when, of course, I meant "(with [itex]\alpha< 1[/itex])".

    Now, in your first post you said "there exist [itex]n\in N[/itex] such that [itex]T^n[/itex] is contraction". The first thing I would do is define [itex]S= T^n[/itex] for that n so that S is a contraction. You can then show that the sequence [itex]T^m x_0[/itex] is a Cauchy sequence and so converges to some point [itex]x[/itex].

    The point is that applying T to that x, Tx, is the same as applying T to each term in the sequence. But that just shifts the sequence by one place. It is still the same sequence and still converges to [itex]x[/itex]. That is, Tx=x.
     
  8. Nov 20, 2013 #7

    Dick

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    First you want to show that the fixed point is unique. If T^n(x)=x and T^n(y)=y, then x=y. Now can you show that if x is a fixed point then T(x) is also a fixed point?
     
  9. Nov 20, 2013 #8
    Suppose there are ##x,y \in X## such that ##T^n(x)=x## and ##T^n(y)=y##, ##d(x,y)=d(T^n(x),T^n(y))\leq \alpha d(x,y)## and this inequality holds if and only if ##x=y##.

    Now let ##x## be the fixed point of ##T^n##, ##T^n(T(x))=T(T^n(x))=T(x)##. It follows that ##T(x)## is a fixed point but by unicity we have that ##T(x)=x##, which means ##x## is a fixed point of ##T##. One can prove unicity using the fact that ##T^n## is a contraction.

    Thank you very much!
     
  10. Nov 20, 2013 #9

    Dick

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    BTW it might be tempting to think that if T^n is a contraction, then T must be a contraction. That's not true. Take X=R^2 and T to be the linear mapping represented by the matrix [[0,1],[1/2,0]]. T is not a contraction. T^2 is a contraction. Show that if you are interested.
     
  11. Nov 20, 2013 #10

    Dick

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    Very welcome. Nicely done!
     
  12. Nov 20, 2013 #11
    Thanks for the remark, I appreciate it. Another example could be ##f:[\frac{1}{4},\frac{1}{2}] \to [\frac{1}{4},\frac{1}{2}]## defined as ##f(x)=x##. ##f## is clearly not a contraction but ##f^2## is a contraction.
     
  13. Nov 20, 2013 #12

    Dick

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    f^2(x) means f(f(x)). f(f(x))=f(x)=x. I'm not sure that's a good example. You really have to go to more than one dimension to get a linear example.
     
  14. Nov 20, 2013 #13
    Oops.
     
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