Hölder's Inequality for Integrals/General

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

For $i = 1, \ldots, n$ let $p_i \in \R$ such that:
 * $\displaystyle \sum_{i \mathop = 1}^n \dfrac 1 {p_i} = 1$

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
We use the Principle of Mathematical Induction.

Assume that the result holds for $i = n-1$. We show that the result holds for $i=n$.

Define:
 * $q_n := \dfrac {p_n} {p_n-1}$

and for $i = 1, \ldots, n-1$, define:
 * $r_i := p_i \cdot \left({1 - \dfrac 1 {p_n} }\right)$

Then:
 * $\dfrac 1 {p_n} + \dfrac 1 {q_n} = 1$
 * $\displaystyle \sum_{i \mathop = 1}^{n-1} \dfrac 1 {r_i} = 1$

and:
 * $q_n \cdot r_i = p_i$

Applying Hölder's Inequality to $\displaystyle f := \prod_{i \mathop = 1}^{n-1} f_i$ and $g := f_n$, we find: