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    Real Analysis : Young's Inequality

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    Note: * = infinite

    Suppose that the function f:[0,*)->R is continuous and strictly increasing, with f(0) = 0 and f([0,*)) = [0,*). Then define

    F(x) = the integral from 0 to x of f and
    G(x) = the integral from 0 to x of f^-1 for all x>=0

    (a) Prove Young's Inequality:
    ab <= F(a) + G(b) for all a >= 0 and b >= 0

    (b) Now use Young's Inequality with f(x) = x^(p-1) for all x>=0, and p>1 fixed, to prove that if the number q is chosen to have the property that 1/p + 1/q = 1, then

    ab <= a^p/p + b^q/q for a >= 0 and b >= 0.

    © BrainMass Inc. brainmass.com November 29, 2021, 11:55 pm ad1c9bdddf
    https://brainmass.com/math/real-analysis/real-analysis-young-s-inequality-10031

    Solution Summary

    Young's inequality is proven and a functional property is proven using Young's inequality.

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