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Thread: proof an inequality; measure theory

  1. #1
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    proof an inequality; measure theory

    Hi there


    Let C be compakt
    $\displaystyle \mu$ a radon measure and $\displaystyle \mu(C) < \infty$.

    Let $\displaystyle v \in L^1(\mathbb{R}) , \ 0 < a < b < \infty$

    Do you know how to proof the following inequality:

    $\displaystyle \int_{a}^b |v| d\mu \le \mu([a,b]) \cdot \|v\|_\infty$

    I don't get it.
    By the way I'm not sure about the correctness of this inequaility if $\displaystyle \mu $ is an arbitrary measure. If it's correct where $\displaystyle \mu $ equals the Lebesgue measure, you can use this instead of the Radon measure.

    Any comments are welcome

    Yours Rapha
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  2. #2
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    Presumably $\displaystyle \mu $ is supposed to be a positive Radon measure.

    From the definition of the $\displaystyle L^\infty$-norm, $\displaystyle |v(t)|\leqslant \|v\|_\infty$ for almost all t in [a,b]. Then $\displaystyle \int_a^b\!\!\!|v(t)|\,d\mu(t) \leqslant \int_a^b\!\!\!\|v\|_\infty\,d\mu(t) = \mu([a,b]) \|v\|_\infty$ (integral of a constant is the constant times the measure of the interval).
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  3. #3
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    Hello Opalg

    thank you very much!

    Kind regards
    Rapha
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