Jensen 不等式证明

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。

1. 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。

2. 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。

3. 假设当 \(n = k\)，不等式成立，即

\[\begin{equation}
f \left( \sum_{i=1}^{k} \lambda_{i} x_{i} \right) \leqslant \sum_{i=1}^{k} \lambda_{i}f(x_{i})
\end{equation}
\]

1. 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 \(1- \lambda_{k+1} \neq 0\)

\[\begin{equation} \label{eqn:one}
\begin{aligned}
f \left( \sum_{i=1}^{k+1} \lambda_{i} x_{i} \right) &= f \left( \sum_{i=1}^{k} \lambda_{i} x_{i} + \lambda_{k+1}x_{k+1} \right) \\
&= f \left( \begin{split} \left( 1 - \lambda_{k+1} \right) \dfrac{\displaystyle\sum_{i=1}^{k} \lambda_{i} x_{i}}{1 - \lambda_{k+1}} + \lambda_{k+1}x_{k+1} \end{split} \right) \\
\end{aligned}
\end{equation}
\]

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 \(\eqref{eqn:one}\) 式使用下凸函数的定义

\[\begin{equation} \label{eqn:two}
\begin{aligned}
f \left( \sum_{i=1}^{k+1} \lambda_{i} x_{i} \right)
&= f \left( \begin{split} \left( 1 - \lambda_{k+1} \right) \frac{\displaystyle\sum_{i=1}^{k} \lambda_{i} x_{i}}{1 - \lambda_{k+1}} + \lambda_{k+1}x_{k+1} \end{split} \right) \\
&\leqslant \left( 1 - \lambda_{k+1} \right) f \left( \begin{split} \frac{\displaystyle\sum_{i=1}^{k} \lambda_{i} x_{i}}{1 - \lambda_{k+1}} \end{split} \right) + \lambda_{k+1} f \left(x_{k+1}\right) \\
&= \left( 1 - \lambda_{k+1} \right) f \left( \displaystyle\sum_{i=1}^{k} \frac{\lambda_{i} x_{i}}{1 - \lambda_{k+1}} \right) + \lambda_{k+1} f \left(x_{k+1}\right) \\
\end{aligned}
\end{equation}
\]

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 \(\eqref{eqn:two}\) 式可以得到 Jensen 不等式成立

\[\begin{equation}
\begin{aligned}
f \left( \sum_{i=1}^{k+1} \lambda_{i} x_{i} \right)
&\leqslant \left( 1 - \lambda_{k+1} \right) f \left( \displaystyle\sum_{i=1}^{k} \frac{\lambda_{i} x_{i}}{1 - \lambda_{k+1}} \right) + \lambda_{k+1} f \left(x_{k+1}\right) \\
&\leqslant \left( 1 - \lambda_{k+1} \right) \sum_{i=1}^{k} \frac{\lambda_{i}}{1-\lambda_{k+1}} f \left( x_{i} \right) + \lambda_{k+1} f \left(x_{k+1}\right) \\
&= \sum_{i=1}^{k} \lambda_{i} f \left( x_{i} \right) + \lambda_{k+1} f \left(x_{k+1}\right) \\
&= \sum_{i=1}^{k+1} \lambda_{i} f \left( x_{i} \right)
\end{aligned}
\end{equation}
\]

1. 综上所述，由数学归纳法得 \(\forall n \left( n = 1, 2, \cdots, k, k+1, \cdots \right)\) 有

\[\begin{equation} \label{eqn:final}
f \left( \sum_{i=1}^{n} \lambda_{i} x_{i} \right) \leqslant \sum_{i=1}^{n} \lambda_{i}f(x_{i})
\end{equation}
\]

即 Jensen 不等式成立。

1. 直接将 \(\displaystyle\lambda_{i} = \frac{1}{n}\) 代入 \(\eqref{eqn:final}\) 式，可得

\[f \left( \frac{1}{n} \sum_{i=1}^{n} x_{i} \right) \leqslant \frac{1}{n} \sum_{i=1}^{n} f(x_{i})
\]