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Math Help - An Inequality

  1. #1
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    An Inequality

    Suppose that  p_{j}\in[1, \infty] for j = 1,...,m and let 1/r = \sum_{j=1}^{m}1/p_{j}.
    For f_{j}\in{L_{p_{j}}\left(X, \mu, \mathbb{K}\right), show that \prod_{j=1}^{m}\in{L_{r}\left(X, \mu, \mathbb{K}\right) and that:

    \displaystyle \biggl\|\prod_{j=1}^{m} f_{j} \biggl\|_{r} \le \prod_{j=1}^{m} \|{f_j}\|_{p_j}
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  2. #2
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    This is just a generalization of Hölder's inequality, and you can prove it as follows (obviously the argument is inductive):

    Assume m=2, and consider s_i=\frac{p_i}{r} then 1=\frac{1}{s_1} + \frac{1}{s_2} and we get:

    \int_X |f_1f_2|^r=\| |f_1|^r|f_2|^r\| _1 \leq \| |f_1|^r \| _{s_1} \| |f_2|^r \| _{s_2} = \left( \int_X |f_1|^{p_1} \right) ^{\frac{r}{p_1}}\left( \int_X |f_2|^{p_2} \right) ^{\frac{r}{p_2}}

    This yields \| f_1f_2\| _r^r \leq \| f_1\| _{p_1}^r\| f_2\| _{p_2}^r which is the result for m=2.

    For the induction step, call f=f_1...f_m then \| f\| _r \leq \| f_1 \| _{p_1} \| f_2...f_m\| _s \leq \| f_1\| _{p_1}...\| f_m \| _{p_m} where the last inequality is by the induction hypothesis with \frac{1}{s} = \sum_{j=2}^{m} \frac{1}{p_j}
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