Jensen 不等式证明（数学归纳法）
2023-12-12 11:22 CST
Jensen 不等式定义
若 $f(x)$ 为区间 $I$ 上的下凸函数，则对于任意 $x_{i} \in I$ 和满足 $\displaystyle\sum_{i=1}^{n} \lambda_{i} = 1$ 的 $\lambda_{i} \gt 0 \left( i = 1, 2, \cdots, n \right)$，成立
\[f \left( \sum_{i=1}^{n} \lambda_{i} x_{i} \right) \leqslant \sum_{i=1}^{n} \lambda_{i}f(x_{i})\]特别地，取 $\displaystyle\lambda_{i} = \frac{1}{n} \left( i = 1, 2, \cdots, n \right)$，就有
\[f \left( \frac{1}{n} \sum_{i=1}^{n} x_{i} \right) \leqslant \frac{1}{n} \sum_{i=1}^{n} f(x_{i})\]Jensen 不等式证明
使用下凸函数的定义和数学归纳法证明。
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当 $n = 1$，有 $\lambda_{1} = 1$，则 $f(\lambda_{1}x_{1}) \leqslant \lambda_{1}f(x_{1})$，Jensen 不等式成立。
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当 $n = 2$，$f(x)$ 为下凸函数，根据下凸函数定义，有 $\forall \lambda \in \left(0,1 \right): f(\lambda x_{1} + \left(1-\lambda\right) x_{2}) \leqslant \lambda f(x_{1}) + \left(1-\lambda\right) f(x_{2})$。令 $\lambda_{1} = \lambda$，则 $\lambda_{2} = 1 - \lambda$，得 $f(\lambda_{1}x_{1} + \lambda_{2}x_{2}) \leqslant \lambda_{1}f(x_{1}) + \lambda_{2}f(x_{2})$，Jensen 不等式成立。
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假设当 $n = k$，不等式成立，即
- 当 $n = k + 1$，由命题条件 $\displaystyle\sum_{i=1}^{k+1} \lambda_{i} = 1$ 可得 $\displaystyle 1-\lambda_{k+1} = \sum_{i=1}^{k}\lambda_{i}$。$\forall \lambda_{i} \gt 0$，所以 $1- \lambda_{k+1} \neq 0$
考察 $\displaystyle\frac{\displaystyle\sum_{i=1}^{k} \lambda_{i} x_{i}}{1 - \lambda_{k+1}}$，只要其属于 $I$，就可以直接使用下凸函数定义。$x_{i}$ 是任意给定的，不妨设 $x_{1} \lt x_{2} \lt \cdots x_{k} \lt x_{k+1}$。所以有
\[\begin{equation} \begin{aligned} &\sum_{i=1}^{k} \lambda_{i} x_{1} \leqslant \sum_{i=1}^{k} \lambda_{i} x_{i} \leqslant \sum_{i=1}^{k} \lambda_{i} x_{k} \\ \implies & x_{1} \sum_{i=1}^{k} \lambda_{i} \leqslant \sum_{i=1}^{k} \lambda_{i} x_{i} \leqslant x_{k} \sum_{i=1}^{k} \lambda_{i} \\ \implies & x_{1} \frac{\displaystyle\sum_{i=1}^{k} \lambda_{i}}{1 - \lambda_{k+1}} \leqslant \frac{\displaystyle\sum_{i=1}^{k} \lambda_{i} x_{i}}{1 - \lambda_{k+1}} \leqslant x_{k} \frac{\displaystyle\sum_{i=1}^{k} \lambda_{i}}{1 - \lambda_{k+1}} \\ \implies & x_{1} \leqslant \frac{\displaystyle\sum_{i=1}^{k} \lambda_{i} x_{i}}{1 - \lambda_{k+1}} \leqslant x_{k} \end{aligned} \end{equation}\]由于 $x_{1}$ 和 $x_{k}$ 都属于 $I$，则 $\displaystyle \frac{\displaystyle\sum_{i=1}^{k} \lambda_{i} x_{i}}{1 - \lambda_{k+1}}$ 也属于 $I$。所以可以对 $(2)$ 式使用下凸函数的定义
\[\begin{equation} \begin{aligned} f \left( \sum_{i=1}^{k+1} \lambda_{i} x_{i} \right) &= f \left( \begin{split} \left( 1 - \lambda_{k+1} \right) \frac{\displaystyle\sum_{i=1}^{k} \lambda_{i} x_{i}}{1 - \lambda_{k+1}} + \lambda_{k+1}x_{k+1} \end{split} \right) \\ &\leqslant \left( 1 - \lambda_{k+1} \right) f \left( \begin{split} \frac{\displaystyle\sum_{i=1}^{k} \lambda_{i} x_{i}}{1 - \lambda_{k+1}} \end{split} \right) + \lambda_{k+1} f \left(x_{k+1}\right) \\ &= \left( 1 - \lambda_{k+1} \right) f \left( \displaystyle\sum_{i=1}^{k} \frac{\lambda_{i} x_{i}}{1 - \lambda_{k+1}} \right) + \lambda_{k+1} f \left(x_{k+1}\right) \\ \end{aligned} \end{equation}\]由于 $\displaystyle\sum_{i=1}^{k} \frac{\lambda_{i}}{1 - \lambda_{k+1}} = 1$，符合 $n=k$ 时 Jensen 不等式成立条件，所以有 $\displaystyle f \left( \displaystyle\sum_{i=1}^{k} \frac{\lambda_{i} x_{i}}{1 - \lambda_{k+1}} \right) \leqslant \sum_{i=1}^{k} \frac{\lambda_{i}}{1-\lambda_{k+1}} f \left( x_{i} \right)$，代入 $(4)$ 式可以得到 Jensen 不等式成立
\[\begin{equation} \begin{aligned} f \left( \sum_{i=1}^{k+1} \lambda_{i} x_{i} \right) &\leqslant \left( 1 - \lambda_{k+1} \right) f \left( \displaystyle\sum_{i=1}^{k} \frac{\lambda_{i} x_{i}}{1 - \lambda_{k+1}} \right) + \lambda_{k+1} f \left(x_{k+1}\right) \\ &\leqslant \left( 1 - \lambda_{k+1} \right) \sum_{i=1}^{k} \frac{\lambda_{i}}{1-\lambda_{k+1}} f \left( x_{i} \right) + \lambda_{k+1} f \left(x_{k+1}\right) \\ &= \sum_{i=1}^{k} \lambda_{i} f \left( x_{i} \right) + \lambda_{k+1} f \left(x_{k+1}\right) \\ &= \sum_{i=1}^{k+1} \lambda_{i} f \left( x_{i} \right) \end{aligned} \end{equation}\]- 综上所述，由数学归纳法得 $\forall n \left( n = 1, 2, \cdots, k, k+1, \cdots \right)$ 有
即 Jensen 不等式成立。
- 直接将 $\displaystyle\lambda_{i} = \frac{1}{n}$ 代入 $(6)$ 式，可得
Jensen 不等式证明（数学归纳法） by mkckr0 is licensed under CC BY-NC-ND 4.0
