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Math Help - An inequality

  1. #1
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    An inequality

    Hello!

    I have the following problem:

    Let g:\mathbb{R}\rightarrow\mathbb{R} be a differentiable function with increasing derivative. And let (X,M,\mu) be a finite measure space such that \mu(X)=1.

    Prove: f bounded and measurable then the following holds
    g(\int_Xfd\mu)\le \int_X g(f)d\mu<\infty

    The hint is to use the fact that g lies above any of its tangent lines.
    ------------------------------------------------------------------

    Since g' is increasing it is easy to prove that g\ge g(y_0)+g'(y_0)(y-y_0). I tried to use it in the following way - but without result:

    \int_Xg(f(x))d\mu(x)\ge\int_X(g(f(x_0))+g'(f(x_0))  (f(x)-f(x_0)))d\mu(x)=g(f(x_0))+g'(f(x_0))\int_X(f(x)-f(x_0))d\mu(x)=g(f(x_0))-g'(f(x_0))f(x_0)+g'(f(x_0))\int_Xfd\mu

    What am I missing?
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  2. #2
    Super Member girdav's Avatar
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    Re: An inequality

    Apply the inequality g(x)\geq g(y_0+g'(y_0)(x-x_0) to x:=f(t) and y_0:=\int_Xfd\mu, then integrate.
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  3. #3
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    Re: An inequality

    And how do I show  \int_X g(f)d\mu<\infty?

    First I thought it had something to do with f being bounded but g is not the increasing one it's g'.
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  4. #4
    Super Member girdav's Avatar
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    Re: An inequality

    Since f is bounded and g continuous, g\circ f is bounded. Now use the fact that we integrate over a finite-measure space.
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