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Thread: An inequality

  1. #1
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    An inequality

    Hello!

    I have the following problem:

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

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

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

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

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

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

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

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

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