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Math Help - Young's Inequality

  1. #1
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    Young's Inequality

    Hello,

    Could someone please give me help with the following problem?

    Suppose f is a continuous increasing function with f(0) = 0. Prove that for a,b > 0, we have a form of Young's Inequality,
    ab \leq \int_0^af(x)dx + \int_0^b f^{-1}(x)dx
    and that equality holds if and only if b = f(a)

    Thanks!
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  2. #2
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    Write the second integral as \int_0^b\!\!f^{-1}(y)\,dy, and make the substitution y = f(x). Then it becomes \int_0^{f^{-1}(b)}\!\!xf'(x)\,dx. Integrate that by parts, to get \int_0^b\!\!f^{-1}(y)\,dy = \Bigl[xf(x)\Bigr]_0^{f^{-1}(b)} - \int_0^{f^{-1}(b)}\!\!f(x)\,dx = b{f^{-1}(b)} - \int_0^{f^{-1}(b)}\!\!f(x)\,dx.

    Now suppose that b = f(a). Then the previous equation says that ab = \int_0^a\!\!f(x)\,dx + \int_0^b\!\!f^{-1}(x)\,dx. Thus equality holds for the desired inequality in that case.

    Next, suppose that b > f(a). By the result in the "equality" case, we know that af(a) = \int_0^a\!\!f(x)\,dx + \int_0^{f(a)}\!\!f^{-1}(x)\,dx. But f^{-1} is an increasing function, taking its minimum value a at the left end of the interval [f(a),b]. So \int_{f(a)}^b\!\!f^{-1}(x)\,dx > a(b-f(a)). Hence \begin{aligned}ab = af(a) + a(b-f(a)) &< \int_0^a\!\!f(x)\,dx + \int_0^{f(a)}\!\!f^{-1}(x)\,dx + \int_{f(a)}^b\!\!f^{-1}(x)\,dx \\ &= \int_0^a\!\!f(x)\,dx + \int_0^b\!\!f^{-1}(x)\,dx.\end{aligned}

    A similar argument shows that the strict inequality also holds when b < f(a).

    [But I don't see how this has anything to do with Young's inequality (except that they both say that ab is less than something).]
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Where do  f(x) is increasing and  f(0)=0 come into play?

    Is it to show  f(x) has a continuous inverse?
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  4. #4
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    Quote Originally Posted by chiph588@ View Post
    Where do  f(x) is increasing and  f(0)=0 come into play?

    Is it to show  f(x) has a continuous inverse?
    The condition that f is increasing is used in two ways. First, to ensure that f has a continuous, increasing inverse; and second, in the last part of the proof, to get the estimates for the integrals \int_{f(a)}^b\!\!f^{-1}(x)\,dx > a(b-f(a)) and (in the case when b < f(a), where I did not give the proof) \int_{f^{-1}(b)}^a\!\!f(x)\,dx > b(a - f^{-1}(b))).
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