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Function has a fixed point?

  1. Nov 5, 2004 #1
    I'm trying to work out how an existance of a fixed point is linked to the constraint on the differential of that function.

    For example, i need to prove f has a fixed point if f'(x)=>2.

    I understand that what I have is a monotone increasing function so it is 1-1. All the fixed points are on the line f(x) = x. So conceptually it must be true that these two lines should intersect somewhere, but I can't prove this rigorously.

    I have a feeling I should be using the Mean Value Theorem ( f(b) - f(a)) = f'(c)(b-a) but can't get much further than that.
     
  2. jcsd
  3. Nov 5, 2004 #2

    AKG

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    You want to prove that there is some c such that f(c) = c. f is continuous on R (I'm guessing). Now you should be able to prove that there are contradictions if f(x) > x for all x, or f(x) < x. Then, by intermediate value theorem, you should be able to finish the proof.
     
  4. Nov 6, 2004 #3

    mathwonk

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    let g(x) = f(x) - x. then you want to prove that g(x) = 0 has a solution for some x.

    But you, know that g'(x) >= 1, so for every pair of successive integers a = n,b = n+1, it follows from MVT that g(b)-g(a) >= 1. Do you see why?

    Hence if g(0) = c>0 say, and n < a < n+1, then g(-n-1) < 0. Do you see why?

    then since g is differentiable it is also contrinuous, and has both a negative value and a positive value, hence is somewhere zero.

    similar arguments work for g(0) = c<0.
     
  5. Nov 9, 2004 #4
    I was reading through that proof, and I did not understand how

    "Hence if g(0) = c>0 say, and n < a < n+1, then g(-n-1) < 0" this step came to be?
    thanx
     
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