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Math Help - Contraction mapping theorem

  1. #1
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    Contraction mapping theorem

    0<c<1 and |f(x)-f(y)<c|c-y|
    a) Show f is continuous on all of R
    b)Pick some point y1 in R and construct the sequence (y1,f(y1),f(f(y1)),....)
    In general if y_(n+1)=f(yn) show that the resulting sequence yn is a Cauchy sequence. hence we may let y=limyn
    c)Prove that y is a fixed point of f and that is unique in this regard.
    d) Finally prove that if x is any arbitrary point in R then the sequence (x,f(x),f(f(x)),...) converges to y defined in (b).



    a) want to show if |x-c|<delta then |f(x)-f(c)|<epsilon
    b) A sequence is Cauchy if |an-am|<epsilon
    c)want to show f(y)=y
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kathrynmath View Post
    0<c<1 and |f(x)-f(y)<c|c-y|
    a) Show f is continuous on all of R
    It's much, much stronger than continuous. If you're allowed to use the limit definition of continuity just note that \displaystyle 0\leqslant \lim_{x\to y}|f(x)-f(y)|\leqslant \lim_{x\to y}c|x-y|=0
    b)Pick some point y1 in R and construct the sequence (y1,f(y1),f(f(y1)),....)
    In general if y_(n+1)=f(yn) show that the resulting sequence yn is a Cauchy sequence. hence we may let y=limyn
    Note that by induction d\left(f^{m+1}(y_1),f^{m}(y_1)\right)<c^m d\left(f(y_1),y_1\right). Thus, we see that

    \begin{aligned}d\left(f^{n+k}(y_1),f^n(y_1)\right) &\leqslant \sum_{j=0}^{k-1}d\left(f^{n+j+1}(y_1),f^{n+j}(y_1)\right)\\ &< \sum_{j=0}^{k-1}c^{n+j}d\left(f(y_1),y_1\right)\\ &< \frac{c^n}{1-c}d\left(f(y_1),y_1\right)\end{aligned}

    From where it evidently follows, letting n\to\infty that \left\{f^{n}(y_1)\right\}_{n\in\mathbb{N}} is Cauchy, and thus by the completeness of \mathbb{R} it follows that y=\lim f^{n}(y_1) exists.
    c)Prove that y is a fixed point of f and that is unique in this regard.
    Uniqueness is clear in general since if x\ne y but f(x)=x,f(y)=y then we see our conditions state that d(x,y)=d\left(f(x),f(y)\right)<cd(x,y)<d(x,y)

    Now, to see that y is a fixed point we merely note that since \lim f^n(y_1)=y that \lim f^{n+1}(y_1)=y and so by the continuity of f we have that f(y)=f\left( \lim f^n(y_1)\right)=\lim f^{n+1}(y_1)=y

    d) Finally prove that if x is any arbitrary point in R then the sequence (x,f(x),f(f(x)),...) converges to y defined in (b).
    Merely note that y_1 was arbitrary to begin with.


    Remark: There is a much more elegant way of proving that a contractive map from a complete metric space to itself has a fixed point.
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  3. #3
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    What are all the d's? If they are derivatives, we haven't gotten to derivatives yet.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kathrynmath View Post
    What are all the d's? If they are derivatives, we haven't gotten to derivatives yet.
    Distance function...absolute value.
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  5. #5
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    ok what about the lim f^n and lim f^(n+1). I guess I'm not seeing what f^n is
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kathrynmath View Post
    ok what about the lim f^n and lim f^(n+1). I guess I'm not seeing what f^n is
    Look here.
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  7. #7
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    I'm not sure how d follows from y1 being arbitrary in b. I do notice we have similar sequences.
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