Minkowski Inequality

Theorem

For two vectors $\mathbf{x}= (x_{1} , x_{2} , \dots , x_{n} )$ , $\mathbf{y} = (y_{1} , y_{2} , \dots , y_{n} )$ and a real number $1$ larger than $p$ , the following equation holds.

$\left( \sum_{k=1}^{n} | x_{k} + y_{k} |^{p} \right)^{{1} \over {p}} \le \left( \sum_{k=1}^{n} |x_{k}|^{p} \right)^{{1} \over {p}} + \left( \sum_{k=1}^{n} |y_{k}|^{p} \right)^{{1} \over {p}}$

This is called the Minkowski’s inequality.

Description

Minkowski’s inequality corresponds to the triangle inequality regarding the definition of $p$ -norm. Although I’m not sure if there are other ways to prove it, using Hölder’s inequality as commonly done leads to circular reasoning. This is because the definition of $p$ -norm is assumed from the beginning when describing Hölder’s inequality. Essentially, the proof of Hölder’s inequality does not need such properties of the norm, so the expression must be appropriately modified.

Proof

By the triangle inequality, the following equation is established.

$\begin{align*} \sum_{k=1}^{n} | x_{k} + y_{k} |^p \le & \sum_{k=1}^{n} |x_{k}| | x_{k} + y_{k} |^{p-1} + \sum_{k=1}^{n} |y_{k}| | x_{k} + y_{k} |^{p-1} \\ =& \left| \sum_{k=1}^{n} |x_{k}| | x_{k} + y_{k} |^{p-1} \right| + \left| \sum_{k=1}^{n} |y_{k}| | x_{k} + y_{k} |^{p-1} \right| \end{align*}$

Hölder’s Inequality
For two constants $p, q>1$ and $\mathbf{u}, \mathbf{v} \in \mathbb{C}^n$ satisfying $\displaystyle {{1 } \over {p}} + {{1} \over {q}} = 1$ ,
$\left| \sum_{k=1}^{n} u_{k} v_{k} \right| \le \left( \sum_{k=1}^{n} |u_{k}|^p \right)^{{1} \over {p}} \left( \sum_{k=1}^{n} |v_{k}|^q \right)^{{1} \over {q}}$

By Hölder’s inequality, it is as follows.

$\left| \sum_{k=1}^{n} |x_{k}| | x_{k} + y_{k} |^{p-1} \right| + \left| \sum_{k=1}^{n} |y_{k}| | x_{k} + y_{k} |^{p-1} \right| \\ \le \left( \sum_{k=1}^{n} |x_{k}|^{p} \right)^{{1}\over{p}} \left( \sum_{k=1}^{n} | x_{k} + y_{k} |^{(p-1)q} \right)^{{1}\over{q}} + \left( \sum_{k=1}^{n} |y_{k}|^{p} \right)^{{1}\over{p}} \left( \sum_{k=1}^{n} | x_{k} + y_{k} |^{(p-1)q} \right)^{{1}\over{q}}$

Since $\dfrac{1}{p} + \dfrac{1}{q} = 1$ , then $(p-1)q = p$ , and by rearranging, it is as follows.

$\sum_{k=1}^{n} | x_{k} + y_{k} |^p \le \left[ \left( \sum_{k=1}^{n} |x_{k}|^{p} \right)^{{1}\over{p}} + \left( \sum_{k=1}^{n} |y_{k}|^{p} \right)^{{1}\over{p}} \right] \left( \sum_{k=1}^{n} | x_{k} + y_{k} |^{p} \right)^{{1}\over{q}}$

Dividing both sides by $\left( \sum \limits_{k=1}^{n} | x_{k} + y_{k} |^{p} \right)^{{1}\over{q}}$ results in the following.

$\left( \sum_{k=1}^{n} | x_{k} + y_{k} |^p \right) ^{1 - {{1} \over {q}}} \le \left( \sum_{k=1}^{n} |x_{k}|^{p} \right)^{{1}\over{p}} + \left( \sum_{k=1}^{n} |y_{k}|^{p} \right)^{{1}\over{p}}$

Since $1 - \dfrac{1}{q} = \dfrac{1}{p}$ , we obtain the following result.