Holder Inequality

Definition

$\dfrac{1}{p} + \dfrac{1}{q} = 1$ satisfies and for two constants $p, q$ and $\mathbf{u}, \mathbf{v} \in \mathbb{C}^n$ larger than 1, the following inequality holds:

$| \left\langle \mathbf{u}, \mathbf{v} \right\rangle | = |\mathbf{u} ^{\ast} \mathbf{v}| \le ||\mathbf{u}||_{p} ||\mathbf{v}||_{q}$

This is called the Hölder’s inequality.

Explanation

Although it should be written as Hölder’s inequality, it was transliterated due to the umlaut. It’s a marvelous inequality where the $p$ -norm and the $q$ -norm are mixed together. It’s not particularly significant in terms of usage or proof methods, but when $p=q=2$ , it becomes the Cauchy-Schwarz inequality.

Proof

Consider the case where neither is $\mathbb{0}$ as it is trivial if $\mathbf{u} = \mathbb{0}$ or $\mathbf{v} = \mathbb{0}$ .

Young’s inequality
Satisfies $\dfrac{1}{p} + \dfrac{1}{q} = 1$ and for two constants greater than 1, $p,q$ and two positive numbers $a,b$ :
$ab \le { {a^{p}} \over {p} } + {{b^{q}} \over {q}}$

By substituting $a = \dfrac{ | u_{i} | }{|| \mathbf{u}||_{p} }, b = \dfrac{ | v_{i} | }{|| \mathbf{v} || _{q} }$ into Young’s inequality, we get the following inequality:

${{| u_{i} v_{i} |} \over {||\mathbf{u}||_{p} ||\mathbf{v}||_{q} }} \le {{1} \over {p}} {{ |u_{i}|^{p} } \over {|| \mathbf{u}||_{p}^{p} }} + {{1} \over {q}} {{ |v_{i}|^q } \over {|| \mathbf{v}||_{q}^{q} }}$

Taking $\displaystyle \sum_{i = 1}^{n}$ of both sides of the equation gives:

$\begin{align*} & \sum_{i=1}^{n} \dfrac{| u_{i} v_{i} |}{||\mathbf{u}||_{p} ||\mathbf{v}||_{q} } \le & \sum_{i = 1}^{n} \left( \dfrac{1}{p} \dfrac{ |u_{i}|^{p} }{|| \mathbf{u}||_{p}^{p} } + \dfrac{1}{q} \dfrac{ |v_{i}|^q }{|| \mathbf{v}||_{q}^{q} } \right) \\ \implies && \dfrac{ \sum_{i=1}^{n} | u_{i} v_{i} |}{||\mathbf{u}||_{p} ||\mathbf{v}||_{q} } \le & \dfrac{1}{p} \dfrac{ \sum_{i=1}^{n}|u_{i}|^{p} }{|| \mathbf{u}||_{p}^{p} } + \dfrac{1}{q} \dfrac{ \sum_{i=1}^{n}|v_{i}|^q }{|| \mathbf{v}||_{q}^{q} } \\ \implies && \dfrac{ |\left\langle \mathbf{u}, \mathbf{v} \right\rangle | }{||\mathbf{u}||_{p} ||\mathbf{v}||_{q} } \le & \dfrac{1}{p} \dfrac{ || \mathbf{u}||_{p}^{p} }{|| \mathbf{u}||_{p}^{p} } + \dfrac{1}{q} \dfrac{ || \mathbf{v}||_{q}^{q} }{|| \mathbf{v}||_{q}^{q} } = \dfrac{1}{p} + \dfrac{1}{q} = 1 \end{align*}$

The third line follows from the definition of the $p$ -norm $\left( \sum_{i=1}^{n}|u_{i}|^{p} \right)^{1/p} = \left\| \mathbf{u} \right\|_{p}$ . Multiplying both sides by $||\mathbf{u}||_{p} ||\mathbf{v}||_{q}$ gives:

$| \left\langle \mathbf{u}, \mathbf{v} \right\rangle | \le ||\mathbf{u}||_{p} ||\mathbf{v}||_{q}$

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