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  • Jensen 不等式

    若f(x)为区间I上的下凸(上凸)函数,则对于任意xi∈I和满足∑λi=1的λi>0(i=1,2,...,n),成立:

    [f(sum ^{n} _{i=1} lambda _{i}x_{i})leq sum ^{n} _{i=1} lambda _{i} f(x_{i}) qquad (f(sum ^{n}_{i=1}lambda _{i}x_{i})geq sum ^{n}_{i=1}lambda _{i}f(x_{i}))]

    特别地,取λi=1/n  (i=1,2,...,n),就有

    [f(frac{1}{n}sum ^{n}_{i=1}x_{i})leq frac{1}{n}sum ^{n}_{i=1} qquad (f(frac{1}{n}sum ^{n}_{n=1})geq frac{1}{n}sum ^{n}_{i=1}f(x_{i}))]

     

    为了方便说明,以下函数均以下凸函数为例

    证明:

    在i=1,2时 Jensen不等式 显然成立:

     

    [f(lambda _{1}x_{1}+lambda _{2}x_{2})leq lambda _{1}f(x_{1})+lambda _{2}f(x_{2})]

    [f(sum ^{n} _{i=1} lambda _{i}x_{i})leq sum ^{n} _{i=1} lambda _{i} f(x_{i})]

    利用数学归纳法证明 i≥3 的情况

     

    [f(sum ^{n+1}_{i=1}lambda _{i}x_{i})=f(lambda _{n+1}x_{n+1}+sum ^{n}_{i=1}lambda _{i}x_{i})]

    由题意[sum ^{n+1}_{i=1}lambda _{i}=1],

    设[eta _{i}=frac{lambda {i}}{1-lambda _{n+1}}]

    得:

    [f(sum ^{n+1}_{i=1}lambda _{i}x_{i})=f[lambda _{n+1}x_{n+1}+(1-lambda _{n+1})sum ^{n}_{i=1}eta _{i}x_{i}]]

    由i=2时 Jensen不等式 成立,可得

    [f(sum ^{n+1}_{i=1}lambda _{i}x_{i})leq lambda _{n+1}f(x_{n+1})+(1-lambda _{n+1})f(sum ^{n}_{i=1}eta _{i}x_{i})]

    [f(sum ^{n+1}_{i=1}lambda _{i}x_{i})leq lambda _{n+1}f(x_{n+1})+(1-lambda _{n+1})sum ^{n}_{i=1}eta _{i}f(x_{i})=sum ^{n+1}_{i=1}lambda _{i}f(x_{i})]

    于是证得Jensen不等式在i≥3时也成立

    [f(sum ^{n} _{i=1} lambda _{i}x_{i})leq sum ^{n} _{i=1} lambda _{i} f(x_{i})]

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  • 原文地址:https://www.cnblogs.com/InWILL/p/10486485.html
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