Hölder's Inequality for Integrals

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, i.e.:

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