AI.doll

このブログは僕のためのメモです。

$$ \newcommand{\inner}[2]{\langle #1, #2 \rangle} \newcommand{\matr}[1]{\boldsymbol{#1}} \newcommand{\pdif}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\pdifn}[3]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\dif}[2]{\frac{d #1}{d #2}} \newcommand{\bracket}[1]{\left\langle #1 \right\rangle} $$

Minkowski inequality

Minkowski 不等式

 p\geq 1のとき,  \left\langle|X+Y|^p\right\rangle^{1/p}\leq\left\langle|X|^p\right\rangle^{1/p}+\left\langle|Y|^p\right\rangle^{1/p}.

証明

 p=1のときは三角不等式になり明らかに成り立つので p>1を考える.

 
\begin{aligned}
  \left\langle|X+Y|^p\right\rangle &\leq \left\langle(|X|+|Y|)|X+Y|^{p-1}\right\rangle\\
                    &= \left\langle|X||X+Y|^{p-1}\right\rangle+\left\langle|Y||X+Y|^{p-1}\right\rangle.
\end{aligned}

 q=\frac{p}{p-1}とすると, Hölder不等式から

 
\begin{aligned}
\left\langle|X||X+Y|^{p-1}\right\rangle+\left\langle|Y||X+Y|^{p-1}\right\rangle
                    &\leq \left\langle|X|^p\right\rangle^{1/p} \left\langle|X+Y|^{(p-1)q}\right\rangle^{1/q}+\left\langle|Y|^p\right\rangle^{1/p} \left\langle|X+Y|^{(p-1)q}\right\rangle^{1/q}\\
                    &= \left(\left\langle|X|^p\right\rangle^{1/p}+\left\langle|Y|^p\right\rangle^{1/p}\right)\left\langle|X+Y|^p\right\rangle^\frac{p-1}{p}.
\end{aligned}

よって,

 
\begin{aligned}
  \left\langle|X+Y|^p\right\rangle^{1-\frac{p-1}{p}} = \left\langle|X+Y|^p\right\rangle^{1/p}
  \leq \left\langle|X|^p\right\rangle^{1/p}+\left\langle|Y|^p\right\rangle^{1/p}.
\end{aligned}