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Thread: An integral inequality

  1. #1
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    An integral inequality

    I spent a day trying to prove this inequality, but I didn't finish it. Hopefully someone will prove it.
    f(x) and g(x) are monotonically increasing or decreasing simultaneously on [a,b].
    An integral inequality-222.png
    Last edited by LLWZ; Jul 2nd 2019 at 09:05 PM.
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  2. #2
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    Re: An integral inequality

    Two quick notes before starting:

    1. f and g must be either both increasing or both decreasing (presumably that's what you mean by increasing or decreasing simultaneously). A counterexample where f is decreasing and g is increasing is $\displaystyle f(x) = 1-x$ and $\displaystyle g(x)=x$.

    2. If f and g are both decreasing, then $\displaystyle -f$ and $\displaystyle -g$ are both increasing, so it suffices to do the proof with the condition that f and g are both increasing.

    So we assume that f and g are both increasing on [a,b], and we want to prove $\displaystyle \int_a^b f(x) g(x)\,dx \ge \frac{1}{b-a}\int_a^b f(x)\,dx\int_a^b g(x)\,dx$.

    Let $\displaystyle f_{ave}$ be the average value of f: $\displaystyle f_{ave}=\frac{1}{b-a}\int_a^b f(x)\,dx$. Since f is increasing, there is a c, $\displaystyle a\le c\le b$, such that $\displaystyle f(x)\le f_{ave}$ if $\displaystyle x<c$ and $\displaystyle f(x)\ge f_{ave}$ if $\displaystyle x>c$.

    We have $\displaystyle \int_a^c f_{ave}-f(x)\,dx = \int_c^b f(x)-f_{ave}\,dx$ and the integrands are always nonnegative. Since g is increasing, $\displaystyle g(x_1) \le g(x_2)$ for $\displaystyle x_1$ in [a,c] and $\displaystyle x_2$ in [c,b]. So $\displaystyle \int_a^c (f_{ave}-f(x))g(x)\,dx \le \int_c^b (f(x)-f_{ave})g(x)\,dx$. Rearranging gives $\displaystyle \int_a^b f_{ave}\,g(x)\,dx \le \int_a^b f(x)g(x)\,dx$, and the conclusion follows.

    - Hollywood
    Thanks from topsquark, LLWZ and Idea
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  3. #3
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    Re: An integral inequality

    Nice. I can't believe I have never seen that inequality nor (obviously) a proof of it before.
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    Re: An integral inequality

    $\displaystyle \displaystyle \int_a^c (f_{ave}-f(x))g(x)\,dx \le \int_c^b (f(x)-f_{ave})g(x)\,dx$

    could you please explain how you got this.

    $g$ is not necessarily a positive function so what properties of integrals are being used here?
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    Re: An integral inequality

    Quote Originally Posted by Idea View Post
    $\displaystyle \displaystyle \int_a^c (f_{ave}-f(x))g(x)\,dx \le \int_c^b (f(x)-f_{ave})g(x)\,dx$

    could you please explain how you got this.

    $g$ is not necessarily a positive function so what properties of integrals are being used here?
    Let $p = sup_{x<c}g(x) \text{ and } q = inf_{x>c} g(x)$. So $p \le q$, and if $g$ is continuous then $p=q=g(c)$. Start with:

    $\int_a^c \bar f - f(x)~dx = \int_c^b f(x) -\bar f ~dx$

    Put $p$ and $q$ in the left and right side of that to get the middle inequality below to get his inequality:

    $\int_a^c (\bar f - f(x))g(x)~dx \le \int_a^c (\bar f - f(x))p~dx \le \int_c^b (f(x) -\bar f)q~dx \le \int_c^b (f(x) -\bar f)g(x)~dx $
    Last edited by Walagaster; Jul 7th 2019 at 02:20 PM.
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  6. #6
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    Re: An integral inequality

    @LLWZ

    @hollywood

    @Walagaster

    thank you very much
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