Hölder's Inequality

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Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $p, q \in \R$ such that $\dfrac 1 p + \dfrac 1 q = 1$.

Let $f \in \mathcal{L}^p \left({\mu}\right), f: X \to \R$, and $g \in \mathcal{L}^q \left({\mu}\right), g: X \to \R$, where $\mathcal L$ denotes Lebesgue space.


Then their pointwise product $f g$ is integrable, i.e. $f g \in \mathcal{L}^1 \left({\mu}\right)$, and:

$\left\Vert{f g}\right\Vert_1 = \displaystyle \int \left\vert{f g}\right\vert \, \mathrm d \mu \le \left\Vert{f}\right\Vert_p \cdot \left\Vert{g}\right\Vert_q$

where the $\left\Vert{\cdot}\right\Vert_p$ signify $p$-seminorms.


Equality

Equality, that is:

$\displaystyle \int \left\vert{f g}\right\vert \, \mathrm d \mu = \left\Vert{f}\right\Vert_p \cdot \left\Vert{g}\right\Vert_q$

holds if and only if, for almost all $x \in X$:

$\dfrac {\left\vert{f \left({x}\right)}\right\vert^p} {\left\Vert{f}\right\Vert_p^p} = \dfrac {\left\vert{g \left({x}\right)}\right\vert^q} {\left\Vert{g}\right\Vert_q^q}$


Hölder's Inequality for Sums

Let $p, q \in \R_{>0}$ be strictly positive real numbers such that:

$\dfrac 1 p + \dfrac 1 q = 1$

Let:

$\mathbf x = \sequence {x_n} \in \ell^p$
$\mathbf y = \sequence {y_n} \in \ell^q$

where $\ell^p$ denotes the $p$-sequence space.

Let $\norm {\mathbf x}_p$ denote the $p$-norm of $\mathbf x$.


Then $\mathbf x \mathbf y = \sequence {x_n y_n} \in \ell^1$, and:

$\norm {\mathbf x \mathbf y}_1 \le \norm {\mathbf x}_p \norm {\mathbf y}_q$


Generalized Hölder Inequality

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

For $i = 1, \ldots, n$ let $p_i \in \R$ such that:

$\displaystyle \sum_{i \mathop = 1}^n \frac 1 {p_i} = 1$

Let $f_i \in \mathcal L^{p_i} \left({\mu}\right), f_i: X \to \R$, where $\mathcal L$ denotes Lebesgue space.


Then their pointwise product $\displaystyle \prod_{i \mathop = 1}^n f_i$ is integrable, that is:

$\displaystyle \prod_{i \mathop = 1}^n f_i \in \mathcal L^1 \left({\mu}\right)$

and:

$\displaystyle \left\Vert{\prod_{i \mathop = 1}^n f_i}\right\Vert_1 = \int \left\vert{\prod_{i \mathop = 1}^n f_i}\right\vert \, \mathrm d \mu \le \prod_{i \mathop = 1}^n \left\Vert{f_i}\right\Vert_{p_i}$

where the $\left\Vert{\cdot}\right\Vert$ signify $p$-seminorms.


Proof

Let $x \in X$.

Let:

$a_x := \dfrac {\left\vert{f \left({x}\right)}\right\vert} {\left\Vert{f}\right\Vert_p}$

and:

$b_x := \dfrac {\left\vert{g \left({x}\right)}\right\vert} {\left\Vert{g}\right\Vert_q}$


Applying Young's Inequality for Products to $a_x$ and $b_x$:

$\dfrac {\left\vert{f \left({x}\right) g \left({x}\right)}\right\vert} {\left\Vert{f}\right\Vert_p \cdot \left\Vert{g}\right\Vert_q} \le \dfrac{\left\vert{f \left({x}\right)}\right\vert^p} {p \left\Vert{f}\right\Vert_p^p} + \dfrac{\left\vert{g \left({x}\right)}\right\vert^q}{ q \left\Vert{g}\right\Vert_q^q}$


By Integral of Positive Measurable Function is Monotone, integrating both sides of this inequality over x yields:

$\displaystyle \dfrac {\int \left\vert{f \left({x}\right) g \left({x}\right)}\right\vert \ \mu \left({\mathrm d x}\right)} {\left\Vert{f}\right\Vert_p \cdot \left\Vert{g}\right\Vert_q} \le \frac{\left\Vert{f}\right\Vert_p^p} {p \left\Vert{f}\right\Vert_p^p} + \frac{\left\Vert{g}\right\Vert_p^q} {q \left\Vert{g}\right\Vert_q^q} = \frac 1 p + \frac 1 q = 1$

so:

$\displaystyle \int \left\vert{f \left({x}\right) g \left({x}\right)}\right\vert \ \mu \left({\mathrm d x}\right) \le \left\Vert{f}\right\Vert_p \cdot \left\Vert{g}\right\Vert_q$


If we have equality, then:

$\displaystyle \int \frac {\left\vert{f \left({x}\right)}\right\vert^p} {p \left\Vert{f}\right\Vert_p^p} + \frac{\left\vert{g \left({x}\right)}\right\vert^q} {q \left\Vert{g}\right\Vert_q^q} - \frac{\left\vert{f \left({x}\right) g \left({x}\right)}\right\vert} {\left\Vert{f}\right\Vert_p \cdot \left\Vert{g}\right\Vert_q} \ \mu \left({\mathrm d x}\right) = 0$


As :

$\displaystyle \frac{\left\vert{f \left({x}\right)}\right\vert^p} {p \left\Vert{f}\right\Vert_p^p} + \frac{\left\vert{g \left({x}\right)}\right\vert^q} {q \left\Vert{g}\right\Vert_q^q} - \frac{\left\vert{f \left({x}\right) g \left({x}\right)}\right\vert} {\left\Vert{f}\right\Vert_p \cdot \left\Vert{g}\right\Vert_q} \ge 0$

it follows from Integrable Function Zero A.E. iff Absolute Value has Zero Integral that:

$\displaystyle \frac{\left\vert{f \left({x}\right)}\right\vert^p} {p \left\Vert{f}\right\Vert_p^p} + \frac{\left\vert{g \left({x}\right)}\right\vert^q} {q \left\Vert{g}\right\Vert_q^q} = \frac{\left\vert{f \left({x}\right) g \left({x}\right)}\right\vert} {\left\Vert{f}\right\Vert_p \cdot \left\Vert{g}\right\Vert_q}$ a.e.


By Young's Inequality for Products, we have equality iff $b_x = a_x^{p-1}$.

Raising both sides to the $q$th power gives:

$\displaystyle \frac{\left\vert{f \left({x}\right)}\right\vert^p}{\left\Vert{f}\right\Vert_p^p} = \frac{\left\vert{g \left({x}\right)}\right\vert^q} {\left\Vert{g}\right\Vert_q^q}$

as $\left({p - 1}\right) q = p$.

$\blacksquare$


Source of Name

This entry was named for Otto Ludwig Hölder.


Sources