1. ## An Inequality

Suppose that $p_{j}\in[1, \infty]$ for $j = 1,...,m$ and let $1/r = \sum_{j=1}^{m}1/p_{j}$.
For $f_{j}\in{L_{p_{j}}\left(X, \mu, \mathbb{K}\right)$, show that $\prod_{j=1}^{m}\in{L_{r}\left(X, \mu, \mathbb{K}\right)$ and that:

$\displaystyle \biggl\|\prod_{j=1}^{m} f_{j} \biggl\|_{r} \le \prod_{j=1}^{m} \|{f_j}\|_{p_j}$

2. This is just a generalization of Hölder's inequality, and you can prove it as follows (obviously the argument is inductive):

Assume $m=2$, and consider $s_i=\frac{p_i}{r}$ then $1=\frac{1}{s_1} + \frac{1}{s_2}$ and we get:

$\int_X |f_1f_2|^r=\| |f_1|^r|f_2|^r\| _1 \leq \| |f_1|^r \| _{s_1} \| |f_2|^r \| _{s_2} = \left( \int_X |f_1|^{p_1} \right) ^{\frac{r}{p_1}}\left( \int_X |f_2|^{p_2} \right) ^{\frac{r}{p_2}}$

This yields $\| f_1f_2\| _r^r \leq \| f_1\| _{p_1}^r\| f_2\| _{p_2}^r$ which is the result for $m=2$.

For the induction step, call $f=f_1...f_m$ then $\| f\| _r \leq \| f_1 \| _{p_1} \| f_2...f_m\| _s \leq \| f_1\| _{p_1}...\| f_m \| _{p_m}$ where the last inequality is by the induction hypothesis with $\frac{1}{s} = \sum_{j=2}^{m} \frac{1}{p_j}$