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Thread: Problem with periodic function

  1. #1
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    Problem with periodic function

    Mój problem jest następujący:My problem is as follows

    Let $\displaystyle f(x)=xh(x)$, where $\displaystyle h(x)$ is periodic function with period 1. Prove or give a counterexample to the following statement: Function $\displaystyle f$ is increasing if and only if $\displaystyle h$ is constant (and the constant is positive). We assume that $\displaystyle h$ is defined on the whole R.

    Thank you for help
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  2. #2
    Senior Member Shanks's Avatar
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    the "If" part is trivial.
    for the " only if" part: since h is periodic function with period 1, we only need to focus on the closed interval [0,1].
    To prove h is positive constant, it is surffice to prove that h is nondecreasing in [0,1] which implies h is constant in [0,1].
    this can be Proved by contradiction!
    suppose there is two points $\displaystyle x_1 < x_2$ in [0,1] such that $\displaystyle h(x_1)>h(x_2)$.
    then $\displaystyle k+x_2>k+x_1$,where $\displaystyle k$ is arbitrary positive integer. since f is increasing, we have
    $\displaystyle f(k+x_2)-f(k+x_1)=k(h(x_2)-h(x_1))+x_2h(x_2)-x_1h(x_1)\geq 0$ (*)
    hold for any arbitrary positive integer $\displaystyle k$.
    since $\displaystyle h(x_1)>h(x_2)$, let k approches infinity, then
    $\displaystyle k(h(x_2)-h(x_1))+x_2h(x_2)-x_1h(x_1)$ tend to negative infinity, which is contradict to (*).
    thus h is nondecreasing in [0,1]. combined with h(0)=h(1) gives that h is constant in [0,1], therefore constant in all real line.
    since f is increasing, the constant is definitely positive, of course.
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