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Math Help - Unique Fixed Point for B-Contraction

  1. #1
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    Unique Fixed Point for B-Contraction

    Having trouble with this one...

    Question: (S,rho) a metric space. The function L: S --> S is a contraction with modulus A. L(S) belongs to K, a compact space with at least one element. I need to show that there's one fixed point in S.


    Any ideas?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by galahad View Post
    Having trouble with this one...

    Question: (S,rho) a metric space. The function L: S --> S is a contraction with modulus A. L(S) belongs to K, a compact space with at least one element. I need to show that there's one fixed point in S.


    Any ideas?
    Hint:

    Spoiler:


    Consider the sequence \left\{f^n(k)\right\}_{n\in\mathbb{N}} where k\in K.

    Uniqueness also clearly follows since if one assumes that x\ne y but f(x)=x,f(y)=y then d(f(x),f(y))=d(x,y)\leqslant A d(x,y)<d(x,y)

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    Hm, let me see...

    If f(x0)=x1, f(x1)=x2 and so on...

    d(x_n,x_n-1) <= A^n d(x1,x0)


    For n high enough, we should have A^n-->0, so:

    d(x_n,x_n-1) = 0
    So x_n=x_n-1, so f^n(x0)=f^(n-1)(x0)

    This implies that f(x_n-1)=x_n-1, so lim x_n-1 = v is a fixed point? Can I do that?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by galahad View Post
    Hm, let me see...

    If f(x0)=x1, f(x1)=x2 and so on...

    d(x_n,x_n-1) <= A^n d(x1,x0)


    For n high enough, we should have A^n-->0, so:

    d(x_n,x_n-1) = 0
    So x_n=x_n-1, so f^n(x0)=f^(n-1)(x0)
    My bad!!! I was thinking of a different argument. Try thinking about the mapping which maps each point of K to it's distance between itself and it's image under f

    Spoiler:
    Note that since f:K\to K is continuous so is the mapping \text{id}\oplus f:K\to K\times K:k\mapsto (k,f(k)) (to see this look at the projections). But, it is an elementary fact that the mapping

    k_1,k_2)\mapsto d_K(k_1,k_2)" alt="d:K\times K\to \mathbb{R}k_1,k_2)\mapsto d_K(k_1,k_2)" /> is continuous. It then follows that d\circ\left(\text{id}\oplus f\right):K\to\mathbb{R}:k\overset{\text{id}\oplus f}{\longmapsto}(k,f(k))\overset{d}{\longmapsto} d(k,f(k)) is continuous. In particular, by the compactness of K and

    the extreme value theorem the mapping d\circ\left(\text{id}\oplus f\right) assumes a minimum at some point k_0\in K. Now, assume that d\circ\left(\text{id}\oplus f\right)(k_0)=d(k_0,f(k_0))\ne 0. Then,

    d(f(f(k_0)),f(k_0))\leqslant A\text{ }d(k_0,f(k_0))<d(k_0,f(k_0)) which contradicts the minimality of k_0
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