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Convergent Sequences and Functions

  1. Feb 7, 2012 #1


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    Hello all, I am having trouble with a convergent series problem.

    The problem statement:

    Let f:ℝ→ℝ be a function such that there exists a constant 0<c<1 for which:
    |f(x)-f(y)| ≤c|x-y|
    for every x,y in ℝ. Prove that there exists a unique a in ℝ such that f(a) = a.

    There is a provided hint:
    Consider a sequence {x(n)} defined recurrently by x(n+1) = f(x(n)). Prove that it converges and its limit a satisfies f(a) = a.

    My confusion is how to show that the recursive function converges and satisfies the required limit. Should I use a specific function f?

    I am fairly certain that we need to show that this sequence (defined in the hint section) is Cauchy. This is what I have so far:

    Consider the sequence:
    x1 = f(0)
    x2 = f(x1)
    x3 = f(x2)
    x(n+1) = f(xn)

    Now let us take |xn-xm|:
    |xn-xm| = (f(x(n-1)) - f(x(m-1))) < c|x(n-1) - x(m-1)|

    However, here I don't know how to pick N sch that for every e>0:
    c|x(n-1) - x(m-1)|<e for every n,m>= N (i.e. the definition of a cauchy sequence)

    Thanks in advance for all help!
    Last edited: Feb 7, 2012
  2. jcsd
  3. Feb 7, 2012 #2


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    [itex] | x_2 - x_1 | < c | x_1 - x_0 | [/itex]. [itex] | x_3 - x_2 | < c^2 | x_1 - x_0| [/itex]. [itex] | x_4 - x_3 | < c^3 | x_1 - x_0 | [/itex]. Etc etc. Take that for a hint and see what you can do with it. Once you've show it's Cauchy and defines a fixed point prove there aren't two fixed points.
    Last edited: Feb 7, 2012
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