Minkowski eşitsizliğini kanıtlayınız.

Question

Minkowski Eşitsizliği: $p\geq 1$ ve $(x_n),(y_n)\in \mathcal{l}^p$ olmak üzere 

$$\left(\sum_{i=1}^{\infty}|x_i+y_i|^p\right)^{1/p}\leq \left(\sum_{i=1}^{\infty}|x_i|^p\right)^{1/p}+\left(\sum_{i=1}^{\infty}|y_i|^p\right)^{1/p}.$$

 

Not: 

$$\mathcal{l}^p:=\left\{(x_n)\in\mathbb{R}^{\mathbb{N}}\Big{|}\sum_{n=1}^{\infty}|x_n|^p<\infty\right\}$$

Comments

$L^p$ uzaylarındaki ispattan faydalanılabilir: https://en.wikipedia.org/wiki/Minkowski_inequality.

Answer 1

$p=1$ için eşitsizliğin doğru olduğu bariz.

$p>1$ durumunu inceleyelim. $z_i:=x_i+y_i$ diyelim.

$\begin{array}{rcl} |z_i|^p =|x_i+y_i|^p & = & |x_i+y_i|\cdot |x_i+y_i|^{p-1} \\ \\ & \leq &  (|x_i|+|y_i|)\cdot |z_i|^{p-1} \\ \\ & = & |x_i|\cdot |z_i|^{p-1}+|y_i|\cdot |z_i|^{p-1}\end{array}$

 

$\begin{array}{rcl} \Rightarrow \sum_{i=1}^{\infty}|z_i|^p & \leq & \sum_{i=1}^{\infty}\left(|x_i|\cdot |z_i|^{p-1}+|y_i|\cdot |z_i|^{p-1}\right)  \\ \\ & = & \sum_{i=1}^{\infty}|x_i|\cdot |z_i|^{p-1}+\sum_{i=1}^{\infty}|y_i|\cdot |z_i|^{p-1}  \\ \\ & \overset{\text{Hölder Eşitsizliği}}{\leq} & \left(\sum_{i=1}^{\infty}|x_i|^p\right)^{1/p}\cdot \left(\sum_{i=1}^{\infty}\left(|z_i|^{p-1}\right)^q\right)^{1/q} +\left(\sum_{i=1}^{\infty}|y_i|^p\right)^{1/p}\cdot \left(\sum_{i=1}^{\infty}\left(|z_i|^{p-1}\right)^q\right)^{1/q}  \\ \\ & = & \left[\left(\sum_{i=1}^{\infty}|x_i|^p\right)^{1/p} +\left(\sum_{i=1}^{\infty}|y_i|^p\right)^{1/p} \right]\cdot \left(\sum_{i=1}^{\infty}\left(|z_i|^{p-1}\right)^q\right)^{1/q}  \\ \\ & = & \left[\left(\sum_{i=1}^{\infty}|x_i|^p\right)^{1/p} +\left(\sum_{i=1}^{\infty}|y_i|^p\right)^{1/p} \right]\cdot \left(\sum_{i=1}^{\infty}|z_i|^{\underset{p}{\underbrace{pq-q}}}\right)^{1/q} \end{array}$

 

$$\Rightarrow\frac{\sum_{i=1}^{\infty}|z_i|^p}{\left(\sum_{i=1}^{\infty}|z_i|^p\right)^{1/q}}\leq \left(\sum_{i=1}^{\infty}|x_i|^p\right)^{1/p} +\left(\sum_{i=1}^{\infty}|y_i|^p\right)^{1/p}$$

$$\Rightarrow \left(\sum_{i=1}^{\infty}|z_i|^p\right)^{1-\frac1q} = \left(\sum_{i=1}^{\infty}|z_i|^p\right)^{\frac1p}\leq \left(\sum_{i=1}^{\infty}|x_i|^p\right)^{1/p} +\left(\sum_{i=1}^{\infty}|y_i|^p\right)^{1/p}.$$