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Solve ivt for non-continuous functions

  1. Apr 22, 2013 #1
    1. The problem statement, all variables and given/known data

    We take f:[0,1]→[0,1] a non-increasing function, such that f(x)≥f(y) whenever x≤y; and we want to prove that there exists c∈[0,1] such that f(c)+c=1. We letA={x∈[0;1]:f(x)+x≥1} and we define c=infA.

    a) explain why c exists

    b)let xn be a sequence of elements of A that converges to c, prove that for all n,f(c)+xn≥1
    c)deduce that f(c)+c≥1
    d)if c=0 prove that f(c)+c≤1
    e)if c>0, consider xn=c−1/n prove that f(c)+c≤1

    2. Relevant equations



    3. The attempt at a solution
    Hi, I have been having trouble solving this question since the only information i could find online was using the IVT properties. Apparently this question should be solved without using the properties.
    for
    a) i have said that since the set A is bounded, and there is a lower bound, then there must be an inf for A and hence c must exsist
    b) for b i am not quite sure, as i know that xn can be written in terms of c

    So far i have solved something by the lines of:

    since f(x) + x [0,1] and we are given to find f(x) + x [itex]\geq[/itex] 1,
    f(0) + 0 [itex]\geq[/itex] 1;
    f(1) + 1 [itex]\geq[/itex] 1;
    hence f(1) [itex]\leq[/itex] c [itex]\leq[/itex] f(0)

    I just need help and some direction on where to go next. I am really stuck. Thank you in advance!
     
    Last edited: Apr 22, 2013
  2. jcsd
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