Hölder's Inequality for Integrals

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

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

Let $f \in \map {\LL^p} \mu, f: X \to \R$, and $g \in \map {\LL^q} \mu, g: X \to \R$, where $\LL$ denotes Lebesgue space.

Then their pointwise product $f g$ is $\mu$-integrable, that is:
 * $f g \in \map {\LL^1} \mu$

and:
 * $\size {f g}_1 = \displaystyle \int \size {f g} \rd \mu \le \norm f_p \cdot \norm g_q$

where the $\norm {\, \cdot \,}_p$ signify $p$-seminorms.

Equality
Equality, that is:

Proof
Let $x \in X$.

Let:
 * $a_x := \dfrac {\size {\map f x} } {\norm f_p}$

and:
 * $b_x := \dfrac {\size {\map g x} } {\norm g_q}$

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


 * $\dfrac {\size {\map f x \map g x} } {\norm f_p \cdot \norm g_q} \le \dfrac {\size {\map f x}^p} {p \norm f_p^p} + \dfrac {\size {\map g x}^q} { q \norm g_q^q}$

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


 * $\ds \dfrac {\int \size {\map f x \map g x} \map \mu {\d x} } {\norm f_p \cdot \norm g_q} \le \frac {\norm f_p^p} {p \norm f_p^p} + \frac {\norm g_p^q} {q \norm g_q^q} = \frac 1 p + \frac 1 q = 1$

so:


 * $\ds \int \size {\map f x \map g x} \map \mu {\d x} \le \norm f_p \cdot \norm g_q$

If we have equality, then:
 * $\ds \int \paren {\frac {\size {\map f x}^p} {p \norm f_p^p} + \frac {\size {\map g x}^q} {q \norm g_q^q} - \frac {\size {\map f x \map g x} } {\norm f_p \cdot \norm g_q} } \map \mu {\d x} = 0$

As:
 * $\dfrac {\size {\map f x}^p} {p \norm f_p^p} + \dfrac {\size {\map g x}^q} {q \norm g_q^q} - \dfrac {\size {\map f x \map g x} } {\norm f_p \cdot \norm g_q} \ge 0$

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


 * $\dfrac {\size {\map g x}^p} {p \norm f_p^p} + \dfrac {\size {\map g x}^q} {q \norm g_q^q} = \dfrac {\size {\map f x \map g x} } {\norm f_p \cdot \norm g_q}$ a.e.

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

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


 * $\dfrac {\size {\map g x}^p} {\norm f_p^p} = \dfrac {\size {\map g x}^q} {\norm g_q^q}$

as $\paren {p - 1} q = p$.