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Cauchy Sequence

  1. Dec 14, 2011 #1
    Let f : [a,b] → [a,b] satisfy

    |f(x)-f(y)| ≤ λ|x-y|

    where 0<λ<1. Prove f is continuous. Choose any Xo ε [a,b] and for n ≥ 1 define X_n+1 = f(Xn). Prove that the sequence (Xn) is convergent and that its limit L is a 'fixed point' of f, namely f(L)=L
  2. jcsd
  3. Dec 14, 2011 #2


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    The continuity part seems pretty straightforward, unless I'm mistaken. Let c be an arbitraty point in [a, b]. Let ε > 0 be given. Let δ = ε. Them for all x such that |x - c| < δ, we have: |f(x) - f(c)| <= ...
  4. Dec 14, 2011 #3
    reckon you can help me further please i am really struggling
  5. Dec 14, 2011 #4


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    No. You must make your own efforts on homework problems. We are not here to solve the questions for you. We will help you as soon as you make an attempt.
  6. Dec 14, 2011 #5
    Well, you initially begin with
    [tex]\left|x_2 - x_1\right|=\left|f(x_1) - f(x_0)\right|\le \lambda\left|x_1 - x_0\right|[/tex]
    if you take one more step down the sequence, you can see that
    [tex]\left|x_3 - x_2\right|=\left|f(x_2) - f(x_1)\right|\le \lambda\left|x_2 - x_1\right|\le \lambda^2\left|x_1 - x_0\right|[/tex]
    I don't think it's difficult to see how this generalizes.

    Can you see how this implies the sequence is Cauchy? What do we know about Cauchy sequences? Finally, what do we know about the limit of a sequence in a continuous function?
  7. Dec 14, 2011 #6


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    And for the first part, can you continue where I wrote "..."? What does your function satisfy, by defintion?
  8. Dec 14, 2011 #7
    sorry the answers no :/
    satisfy's |f(x)-f(y)| ≤ λ|x-y) where 0<λ<1
    Last edited: Dec 14, 2011
  9. Dec 14, 2011 #8


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    OK. Now, we're looking at all x such that |x - c| < δ, right? So, apply your function property to these x's...
  10. Dec 14, 2011 #9
    ok i get that, so what do i plug in to the x's?
  11. Dec 14, 2011 #10


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    I already gave you the initial push in post #2 after "we have:". You only need to adjust your function inequality a bit now.
  12. Dec 14, 2011 #11
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