イェンセンの不等式の証明

イェンセンの不等式

${\varphi }_1,{\varphi }_2,\cdots {\varphi }_n$が$0<{\varphi }_i$かつ$\sum _{i=1}^n{\varphi }_i=1$を満たしまた，$x_1,x_2,\cdots x_n$を実数の列とするとき，凸凾数$f(x)$に対して以下のことが成り立つ. $$\begin{array}{r}\sum _{i=1}^n{\varphi }_{i}f(x_i)\ge f\left(\sum _{i=1}^n{\varphi }_i{x}_i\right)\cdots f(x)\mathrm{が}\mathrm{凸}\mathrm{凾}\mathrm{数}(\mathrm{下}\mathrm{に}\mathrm{凸}\mathrm{の}\mathrm{凾}\mathrm{数})\mathrm{の}\mathrm{と}\mathrm{き}.\\ (\sum _{i=1}^n{\varphi }_{i}f(x_i)\le f\left(\sum _{i=1}^n{\varphi }_i{x}_i\right)\cdots f(x)\mathrm{が}\mathrm{凹}\mathrm{凾}\mathrm{数}(\mathrm{上}\mathrm{に}\mathrm{凸}\mathrm{の}\mathrm{凾}\mathrm{数})\mathrm{の}\mathrm{と}\mathrm{き}.)\end{array}$$

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凸凾数(下に凸の凾数)

凸凾数(下に凸の凾数)の性質

任意の$x_1,x_2$に対して凸凾数(下に凸の凾数)$f(x)$では，$(x_1,f(x_1)),(x_2,f(x_2))$を結ぶ線分$g(x)$は凾数$f(x)$の上側にある.
逆に，$(x_1,f(x_1)),(x_2,f(x_2))$を結ぶ線分$g(x)$が凾数$f(x)$の上側にあるので，$f(x)$は凸凾数(下に凸の凾数)である． 

凸凾数(下に凸の凾数)にかかる線分$g(x)$の式

$$\begin{array}{rcl}{x}_t& =& (1-t)x_1+t{x}_2(\mathrm{た}\mathrm{だ}\mathrm{し}0\le t\le 1)\cdots x_1{x}_2\mathrm{間}\mathrm{を}\mathrm{単}\mathrm{位}\mathrm{と}\mathrm{し}{x}_1\mathrm{を}\mathrm{基}\mathrm{準}\mathrm{点}\mathrm{と}\mathrm{す}\mathrm{る}t\mathrm{を}\mathrm{用}\mathrm{い}\mathrm{た}x\mathrm{の}\mathrm{表}\mathrm{現}\mathrm{と}\mathrm{な}\mathrm{る}.\\ & =& \sum _{i=1}^2{\varphi }_i{x}_i\cdots {\varphi }_i=\{1-t,t\}\mathrm{で}，0\le {\varphi }_i\mathrm{か}\mathrm{つ}\sum _{i=1}^2{\varphi }_i=1\\ \\ \\ g(x)-f(x_1)& =& \frac{f(x_2)-f(x_1)}{x_2-x_1}(x-x_1)\cdots \mathrm{傾}\mathrm{き}\mathrm{が}\frac{f(x_2)-f(x_1)}{x_2-x_1}\mathrm{で}\mathrm{点}(x_1,f(x_1))\mathrm{を}\mathrm{通}\mathrm{る}\mathrm{直}\mathrm{線}\mathrm{の}\mathrm{方}\mathrm{程}\mathrm{式}\\ g(x)& =& \frac{f(x_2)-f(x_1)}{x_2-x_1}(x-x_1)+f(x_1)\\ & =& \frac{f(x_2)-f(x_1)}{x_2-x_1}(x_t-x_1)+f(x_1)\cdots x\mathrm{を}t\mathrm{で}\mathrm{表}\mathrm{現}\mathrm{す}\mathrm{る}{x}_t\mathrm{と}\mathrm{す}\mathrm{る}\\ & =& \frac{f(x_2)-f(x_1)}{x_2-x_1}\{(1-t)x_1+t{x}_2-x_1\}+f(x_1)\cdots x_t=(1-t)x_1+t{x}_2\\ & =& \frac{f(x_2)-f(x_1)}{x_2-x_1}(x_1-t{x}_1+t{x}_2-x_1)+f(x_1)\\ & =& \frac{f(x_2)-f(x_1)}{x_2-x_1}(-t{x}_1+t{x}_2)+f(x_1)\\ & =& \frac{f(x_2)-f(x_1)}{x_2-x_1}t(x_2-x_1)+f(x_1)\\ & =& \{f(x_2)-f(x_1)\}t+f(x_1)\\ & =& tf(x_2)-tf(x_1)+f(x_1)\\ & =& tf(x_2)+(1-t)f(x_1)\\ & =& (1-t)f(x_1)+tf(x_2)\\ & =& \sum _{i=1}^2{\varphi }_{i}f(x_i)\cdots {\varphi }_i=\{1-t,t\}\mathrm{で}，0\le {\varphi }_i\mathrm{か}\mathrm{つ}\sum _{i=1}^2{\varphi }_i=1\end{array}$$

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数学的帰納法によるイェンセンの不等式の証明

$n=2$の証明

$$\begin{array}{rcl}g(x_t)& \ge & f(x_t)\cdots \mathrm{凸}\mathrm{凾}\mathrm{数}(\mathrm{下}\mathrm{に}\mathrm{凸}\mathrm{の}\mathrm{凾}\mathrm{数})\mathrm{の}\mathrm{性}\mathrm{質}\mathrm{よ}\mathrm{り}g(x)\mathrm{が}f(x)\mathrm{よ}\mathrm{り}\mathrm{上}\mathrm{側}\mathrm{に}\mathrm{あ}\mathrm{る}.\\ (1-t)f(x_1)+tf(x_2)& \ge & f((1-t)x_1+t{x}_2)\cdots x_t=(1-t)x_1+t{x}_2\\ \sum _{i=1}^2{\varphi }_{i}f(x_i)& \ge & f\left(\sum _{i=1}^2{\varphi }_i{x}_i\right)\cdots n=2\mathrm{の}\mathrm{証}\mathrm{明}.\end{array}$$

$n=k$を認める状況で$n=k+1$を証明

$n=k+1$の式を変形していく． $$\begin{array}{rcl}\sum _{i=1}^{k+1}{\theta }_{i}f(x_i)& =& \sum _{i=1}^k{\theta }_{i}f(x_i)+{\theta }_{k+1}f(x_{k+1})\cdots 0\le {\theta }_i\mathrm{か}\mathrm{つ}\sum _{i=1}^{k+1}{\theta }_i=1\\ & =& {\mathrm{\Theta }}_k\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}f(x_i)+{\theta }_{k+1}f(x_{k+1})\cdots {\mathrm{\Theta }}_k=\sum _{i=1}^k{\theta }_i=1-{\theta }_{k+1}<1,\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}=\frac{1}{{\mathrm{\Theta }}_k}\sum _{i=1}^k{\theta }_i=1\\ & \ge & {\mathrm{\Theta }}_{k}f\left(\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}{x}_i\right)+{\theta }_{k+1}f(x_{k+1})\cdots n=k\mathrm{は}\mathrm{認}\mathrm{め}\mathrm{る}\mathrm{の}\mathrm{で}(\sum _{i=1}^k{\varphi }_{i}f(x_i)\ge f\left(\sum _{i=1}^k{\varphi }_i{x}_i\right)),\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}f(x_i)\ge f\left(\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}{x}_i\right).\end{array}$$

$n=2$の式としてみる． $$\begin{array}{rcl}{\mathrm{\Theta }}_{k}f\left(\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}{x}_i\right)+{\theta }_{k+1}f(x_{k+1})& =& {\lambda }_{1}f\left(a_1\right)+{\lambda }_{2}f(a_2)\cdots {\lambda }_j=\{{\mathrm{\Theta }}_k,{\theta }_{k+1}\},a_j=\{\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}{x}_i,x_{k+1}\}\\ & =& \sum _{j=1}^2{\lambda }_{j}f(a_j)\\ & \ge & f\left(\sum _{j=1}^2{\lambda }_j{a}_j\right)\cdots n=2\mathrm{は}\mathrm{上}\mathrm{記}\mathrm{で}\mathrm{証}\mathrm{明}\mathrm{済}\mathrm{み}.\end{array}$$

$\sum _{j=1}^2{\lambda }_j{a}_j$を変形していく． $$\begin{array}{rcl}\sum _{j=1}^2{\lambda }_j{a}_j& =& {\mathrm{\Theta }}_k\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}{x}_i+{\theta }_{k+1}{x}_{k+1}\\ & =& {\mathrm{\Theta }}_k\frac{1}{{\mathrm{\Theta }}_k}\sum _{i=1}^k{\theta }_i{x}_i+{\theta }_{k+1}{x}_{k+1}\cdots \sum _{i=1}^{n}C{x}_i=C\sum _{i=1}^n{x}_i\\ & =& \sum _{i=1}^k{\theta }_i{x}_i+{\theta }_{k+1}{x}_{k+1}\\ & =& \sum _{i=1}^{k+1}{\theta }_i{x}_i\end{array}$$

変形結果を戻す．

$$\begin{array}{rcl}f\left(\sum _{j=1}^2{\lambda }_j{a}_j\right)& =& f\left(\sum _{i=1}^{k+1}{\theta }_i{x}_i\right)\end{array}$$

以上より$n=k$を認めれば$n=k+1$でもイェンセンの不等式が成り立つ. $$\begin{array}{rcl}\sum _{i=1}^{k+1}{\theta }_{i}f(x_i)& \ge & f\left(\sum _{i=1}^{k+1}{\theta }_i{x}_i\right)\end{array}$$

$n=2$及び$n=k+1$の証明からの結論

数学的帰納法によりイェンセンの不等式が成り立つ.