Minkowski's 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 $p$ greater than $1$, 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 Minkowski’s inequality.

Explanation

Minkowski’s inequality corresponds to the triangle inequality in the definition of the $p$-norm. I am not sure whether there is some other method of proof, but using Hölder’s inequality as one usually does turns into a circular argument. This is because stating Hölder’s inequality already presupposes that the $p$-norm is defined; essentially, the proof of Hölder’s inequality does not require such a property of the norm, so the expression must be modified appropriately.

Proof

By the triangle inequality, the following equation holds.

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

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

By Hölder’s inequality, we have the following.

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

Since $\dfrac{1}{p} + \dfrac{1}{q} = 1$, we have $(p-1)q = p$, and rearranging again gives the following.

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

Dividing both sides by $\left( \sum \limits_{k=1}^{n} | x_{k} + y_{k} |^{p} \right)^{{1}\over{q}}$ gives 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 $1 - \dfrac{1}{q} = \dfrac{1}{p}$ here, we obtain the following result.