generalized Hölder inequality

Theorem Let and , where ${\sum }_{j=1}^n\frac{1}{p_j}=\frac{1}{r}$. If $f_j\in L^{p_j}$ for $1\le j\le n$, then

${\prod }_{j=1}^n{f}_j\in L^r$ and

$${||\prod _{j=1}^n{f}_j||}_r\le \prod _{j=1}^n{||f_j||}_{p_j}.$$

The usual Hölder inequality has $n=2$ and $r=1$.

Let $X$ be a finite set, say $X=\{x_1,\mathrm{\dots },x_m\}$ and $\mu$ is the counting measure on $X$, so that $\mu (\{x_i\})=1$ for all $i$. Let $f_j(x_i)=a_{ij}\ge 0$ for $j=1,\mathrm{\dots },n$ and take $r=1$. Then the inequality becomes:

$$\sum _{i=1}^m\prod _{j=1}^n{a}_{ij}\le \prod _{j=1}^n{(\sum _{i=1}^{m}a_{ij}{}^{p_j})}^{\frac{1}{p_j}}\mathit{\hspace{1em}}.$$

.

Now let ${\alpha }_j=\frac{1}{p_j}$, and $b_{ij}=a_{ij}{}^{p_j}$. Then the inequality becomes:

$$\sum _{i=1}^m\prod _{j=1}^{n}b_{ij}{}^{{\alpha }_j}\le \prod _{j=1}^n{(\sum _{i=1}^m{b}_{ij})}^{{\alpha }_j}.$$

Title	generalized Hölder inequality
Canonical name	GeneralizedHolderInequality
Date of creation	2013-03-22 16:54:35
Last modified on	2013-03-22 16:54:35
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	8
Author	Mathprof (13753)
Entry type	Theorem
Classification	msc 46E30