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Math Help - Prove that f(x) is constant...?

  1. #1
    Super Member fardeen_gen's Avatar
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    Prove that f(x) is constant...?

    Suppose that function f : R R satisfies the inequality:
    n
    | 3^r{f(x + ry) - f(x - ry)}| ≤ 1 for every positive integer r, x, y
    r = 1
    Prove that f(x) = constant



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  2. #2
    Senior Member
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    Clarity?

    If abs(\sum_{r=1}^n 3^r[f(x+ry)-f(x-ry)])\leq 1 for all n\in\mathbb{N}, x,y\in\mathbb{R} then f is constant.

    Suppose f is not constant. Then there exists an a,b such that f(a)\neq f(b), or  f(a)=f(b)+\epsilon for some \epsilon\neq 0. Choose a positive integer n such that 3^n|\epsilon|>2. Define x,y such that a=x+ny and b=x-ny

    Now, by construction, |a_n|>2, for a_n the last term of the summation. Now we have two cases:

    Case 1: abs(\sum_{r=1}^n a_r)>1 so f does not meet the hypothesis for the choice n,x,y as prescribed.

    Case 2: abs(\sum_{r=1}^n a_r)\leq1 or, the terms a_r for r<n counterbalanced a_n to put the whole sum back under 1.

    More rigorously, \sum_{r=1}^n a_r\in[-1,1]. WLOG, suppose |a_n|=a_n>2. Then \sum_{r=1}^{n-1} a_r = \sum_{r=1}^n a_r - a_n<-1. So f does not meet the hypothesis for the choice n-1,x,y as prescribed.

    Ergo, by contrapositive, the theorem is true.
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