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Continous function in interval problem

  1. Nov 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Let [itex]f: [0,2]\to\mathbb{R}[/itex] be continous and [itex]f(0) = f(2)[/itex]. Show that there exist [itex]x,y\in [0,2][/itex] with the following property:
    [itex](*)\ y-x = 1[/itex] and [itex]f(x) = f(y)\ (*)[/itex]
    2. Relevant equations
    Bolzano-Cauchy theorem: If a function [itex]f[/itex] is continous in some interval [itex][a,b][/itex] and [itex]f(a) <0, f(b) > 0[/itex] (or vice versa) then there exists [itex]c\in (a,b)[/itex] such that [itex]f(c) = 0[/itex]

    3. The attempt at a solution
    If [itex]f[/itex] was constant, then it's trivial. Fix [itex]x\in [0,1][/itex] and the condition [itex]f(x+1)-f(x) = 0[/itex] is satisfied.
    Hence, assume [itex]f[/itex] is not constant.

    Let us observe function [itex]g(x) := f(x+1)-f(x)[/itex], [itex]0\leq x\leq 1[/itex]. The objective is to show that [itex]g(x)=0[/itex] is possible with which we will have proven the existence of the required [itex]x,y[/itex] (is this correct to say? )

    Let us note that:
    [itex]g(0) = f(1) - f(0)[/itex] and [itex]g(1) = f(2) - f(1) = f(0) - f(1)[/itex]. If [itex]g(0) > 0[/itex], then [itex]g(1) <0[/itex] (or vice versa), we can therefore conclude that:
    Per Bolzano-Cauchy theorem there exists [itex]c\in (0,1)[/itex] such that [itex]g(c) = f(c+1) - f(c) = 0[/itex] from which we can establish [itex]x = c[/itex] and [itex]y = c+1[/itex] and the condition [itex](*)[/itex] is satisfied

    If [itex]g(0) = 0[/itex] then also [itex]g(1) = 0[/itex] and again the condition [itex](*)[/itex] is satisfied. [itex]Q.E.D[/itex]
     
  2. jcsd
  3. Nov 7, 2015 #2

    fresh_42

    Staff: Mentor

    Looks nice. And what exactly do you expect as replies here? Does your proof still hold for arbitrary c ∈[0,d] with y - x = c and arbitrary continuous functions f : [a,b] → ℝ with f(a) = f(b) and how big can d be at most?
     
    Last edited: Nov 7, 2015
  4. Nov 7, 2015 #3
    Most of all expect criticism on presenting proof, generalizing the problem is welcome. Also would like the proof to be challenged if there is something I might have missed.
     
  5. Nov 7, 2015 #4

    fresh_42

    Staff: Mentor

    I've seen nothing wrong. And your presentation reveals that you work carefully and think about the special cases. I'ld let it go through. (Bluster me if I'm wrong!) There is only one little, tiny, small remark from my side: Don't shout QUOD ERAT DEMONSTRANDUM. A simpe qed will do and a ◊ or box is even more pleasant
     
  6. Nov 7, 2015 #5

    fresh_42

    Staff: Mentor

    I've seen nothing wrong. And your presentation reveals that you work carefully and think about the special cases. I'ld let it go through. (Bluster me if I'm wrong!) There is only one little, tiny, small remark from my side: Don't shout QUOD ERAT DEMONSTRANDUM. A simpe qed will do and a ◊ or box is even more pleasant. :wink:

    - sorry, probs with the connection and a failed search for a remove button
     
    Last edited: Nov 7, 2015
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