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  • 闵可夫斯基(Minkowski)不等式

    对于任意的 $n$ 维向量 $a = left { x_{1},x_{2},...,x_{n} ight }$,$b = left { y_{1},y_{2},...,y_{n} ight }$,$p geq 1$,有

    $$left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{p} ight )^{frac{1}{p}} leq left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}} + left ( sum_{i=1}^{n}|y_{i}|^{p} ight )^{frac{1}{p}}$$

    证明:

    $$sum_{i=1}^{n}|x_{i} + y_{i}|^{p} = sum_{i=1}^{n}|x_{i} + y_{i}|cdot |x_{i} + y_{i}|^{p - 1} leq sum_{i=1}^{n}(|x_{i}| + |y_{i}|)cdot |x_{i} + y_{i}|^{p - 1} \
    = sum_{i=1}^{n}|x_{i}| cdot |x_{i} + y_{i}|^{p - 1} + sum_{i=1}^{n}|y_{i}| cdot |x_{i} + y_{i}|^{p - 1} $$

       必然存在一个 $q geq 1$,使得 $frac{1}{p} + frac{1}{q} = 1$,则根据 $Holder$ 不等式有

    $$sum_{i=1}^{n}|x_{i}| cdot |x_{i} + y_{i}|^{p - 1} + sum_{i=1}^{n}|y_{i}| cdot |x_{i} + y_{i}|^{p - 1}  leq left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}} cdot left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{q(p-1)} ight )^{frac{1}{q}} + left ( sum_{i=1}^{n}|y_{i}|^{p} ight )^{frac{1}{p}} cdot left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{q(p-1)} ight )^{frac{1}{q}} \
    = left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}} cdot left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{p} ight )^{frac{1}{q}} + left ( sum_{i=1}^{n}|y_{i}|^{p} ight )^{frac{1}{p}} cdot left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{p} ight )^{frac{1}{q}} \
    = left { left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}} + left ( sum_{i=1}^{n}|y_{i}|^{p} ight )^{frac{1}{p}} ight } cdot left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{p} ight )^{frac{1}{q}}$$

       于是有

    $$sum_{i=1}^{n}|x_{i} + y_{i}|^{p}
     leq  left { left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}} + left ( sum_{i=1}^{n}|y_{i}|^{p} ight )^{frac{1}{p}} ight } cdot left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{p} ight )^{frac{1}{q}}$$

       两边同除以最后一项,便可得

    $$left ( sum_{i=1}^{n}|x_{i} + y_{i}|^{p} ight )^{frac{1}{p}} leq left ( sum_{i=1}^{n}|x_{i}|^{p} ight )^{frac{1}{p}} + left ( sum_{i=1}^{n}|y_{i}|^{p} ight )^{frac{1}{p}}$$

    证毕

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  • 原文地址:https://www.cnblogs.com/yanghh/p/13343764.html
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