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Continuity and strictly increasing functions

  1. Apr 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Let f:[0,1] →ℝ be a continuous function that does not take on any of its values twice and with f(0) < f(1), show that f is strictly increasing on [0,1].

    2. Relevant equations



    3. The attempt at a solution

    Assume that f is not strictly increasing on [0,1]. Therefore there exists a,b,c between 0 and 1. where 0 < a < b < c < 1. such that f(0) < f(a) < f(b) and f(c) < f(b) < f(1). By the intermediate value theorem there exist a x between (a,b) and a y between (b,c) for there exists a k where f(a) < k < f(b) and f(c) < k < f(b) and f(x) = k and f(y) = k. Contradiction Therefore f is strictly increasing on [0,1].

    Does this proof work or am I missing something again?
     
  2. jcsd
  3. Apr 2, 2012 #2

    micromass

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    I don't think this is necessarily true. How do you conclude the existence of those a,b and c??

    The idea of your proof is good though.
     
  4. Apr 2, 2012 #3
    If [0,1] is an interval then there exists a value between 0 and 1 that I will call a, and therefore there exists a value between a and 1 since it is a interval which I call b and then there is a value between b and 1 called c as [b,1] is also an interval.
     
  5. Apr 2, 2012 #4
    micromass' question seemed to be asking how you can select a, b and c to give the relationships you state for f(a), f(b) and f(c). Certainly there are non-monotonic functions that do not allow such a selection: ##x^2-0.5x##, for example.
     
  6. Apr 2, 2012 #5
    Am I not allowed to assume and/or define their relationship? I mean for it to be non-mono there must be a point in the middle (not literally) of the interval where f(x) is either the min/max or at least curve up and then back down (or down then back up). So why can't I pick that number to be the value of f(b)?
     
    Last edited: Apr 2, 2012
  7. Apr 2, 2012 #6
    Well, you have to define how you pick the points properly for a proof. And my example function ##x^2-0.5x## does not allow you to pick points in the way you defined them. Neither does ##1.5x-x^2##. So you have to deal with those possibilities.

    I think you are on the right track, but you're not quite nailling down the proof yet.
     
  8. Apr 2, 2012 #7

    jgens

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    You are allowed to define their relationship so long as their relationship holds for all non-monotonic functions in this case. Consider the example that Joffan posted. Making the selection you mentioned is not possible.
     
  9. Apr 2, 2012 #8

    Office_Shredder

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    Essentially b is a local maximum in your proof. If a function goes down and then up, it has no local maximum, but the function isn't strictly increasing, So you can't assume there is a local maximum.

    Instead of trying to pick three points, just pick two points that break the definition of increasing
     
  10. Apr 2, 2012 #9
    Actually, scrub that. By all means look at those functions but only to see that there are alternative shapes you might need to think about in [0,1].

    The basic problem as that you are picking three points, and you really only need to pick two, based on some interval where the function is not increasing.

    [[ edit: :smile: as Office Shredder said ]]
     
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