# generalized Hölder inequality

Theorem Let $1\leq r<\infty$ and $1\leq p_{j}<\infty$, where $\sum_{j=1}^{n}\frac{1}{p_{j}}=\frac{1}{r}$. If $f_{j}\in L^{p_{j}}$ for $1\leq j\leq n$, then

$\prod_{j=1}^{n}f_{j}\in L^{r}$ and

 $||\prod_{j=1}^{n}f_{j}||_{r}\leq\prod_{j=1}^{n}||f_{j}||_{p_{j}}.$

Let $X$ be a finite set, say $X=\{x_{1},\ldots,x_{m}\}$ and $\mu$ is the counting measure on $X$, so that $\mu(\{x_{i}\})=1$ for all $i$. Let $f_{j}(x_{i})=a_{ij}\geq 0$ for $j=1,\ldots,n$ and take $r=1$. Then the inequality becomes:

 $\sum_{i=1}^{m}\prod_{j=1}^{n}a_{ij}\leq\prod_{j=1}^{n}(\sum_{i=1}^{m}{a_{ij}}^% {p_{j}})^{\frac{1}{p_{j}}}\quad.$

.

Now let $\alpha_{j}=\frac{1}{p_{j}}$, and $b_{ij}={a_{ij}}^{p_{j}}$. Then the inequality becomes:

 $\sum_{i=1}^{m}\prod_{j=1}^{n}{b_{ij}}^{\alpha_{j}}\leq\prod_{j=1}^{n}(\sum_{i=% 1}^{m}b_{ij})^{\alpha_{j}}.$
Title generalized Hölder inequality GeneralizedHolderInequality 2013-03-22 16:54:35 2013-03-22 16:54:35 Mathprof (13753) Mathprof (13753) 8 Mathprof (13753) Theorem msc 46E30