Hölder's Inequality for Integrals

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Theorem

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

Let $p, q \in \R_{>0}$ 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:

\(\ds \norm {f g}_1\) \(=\) \(\ds \int \size {f g} \rd \mu\)
\(\ds \) \(\le\) \(\ds \paren {\int \size f^p \rd \mu}^{1 / p} \paren {\int \size g^q \rd \mu}^{1 / q}\)
\(\ds \) \(=\) \(\ds \norm f_p \cdot \norm g_q\)

where:

$\size {f g}$ denotes the absolute value function applied to the pointwise product of $f$ and $g$
the $\norm {\, \cdot \,}_p$ signify $p$-seminorms.


Equality

Equality, that is:

$\ds \int \size {f g} \rd \mu = \norm f_p \cdot \norm g_q$

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

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





$\dfrac {\size {\map f x}^{p - 1} } {\size {\map g x} } = c$

for some $c \in \R$.


Generalized Hölder Inequality

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

For $i = 1, \ldots, n$ let $p_i \in \R_{>0}$ such that:

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

Let $f_i \in \map {\LL^{p_i} } \mu, f_i: X \to \R$, where $\LL$ denotes Lebesgue space.


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

$\ds \prod_{i \mathop = 1}^n f_i \in \map {\LL^1} \mu$

and:

$\ds \norm {\prod_{i \mathop = 1}^n f_i}_1 = \int \size {\prod_{i \mathop = 1}^n f_i} \rd \mu \le \prod_{i \mathop = 1}^n \norm {f_i}_{p_i}$

where the various instances of $\norm {\, \cdot \,}$ signify $p$-seminorms.


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 if and only if $b_x = {a_x}^{p - 1}$.

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

$\blacksquare$


Also known as

Hölder's Inequality for Integrals is also seen referred to just as Hölder's Inequality.

This allows it to be confused with Hölder's Inequality for Sums, so the full form is used on $\mathsf{Pr} \infty \mathsf{fWiki}$.


Also see


Source of Name

This entry was named for Otto Ludwig Hölder.


Sources