Generalized Holder Inequality: Proving the Inequality for Arbitrary Exponents

[Mathkk]

Let $a_i \in \mathbb R^n$ with $a_i = (a_{i}^j)_{j = 1 ... n} = (a_{i}^1, ... ,a_{i}^n)$ for $i = 1, ... , k$ and let $p_1,...,p_k \in \mathbb R_{>1}$ with $\frac1{p_1}+ ... + \frac1{p_k} = 1$

Then show the following inequality by assuming that there are for every $i = 1, ... ,k$ one $N \in \mathbb N_{>1}$ and one $n_i \in \{1,...,2^N\}$ with $p_i = 2^N / n_i$:

$$\sum_{j=1}^n|\prod_{i=1}^ka_{i}^j| \leq \prod_{i=1}^k(\sum_{j=1}^n|a_{i}^j|^{p_i})^{1 \over {p_i}}$$

 

[Opalg]

Let $a_i \in \mathbb R^n$ with $a_i = (a_{i}^j)_{j = 1 ... n} = (a_{i}^1, ... ,a_{i}^n)$ for $i = 1, ... , k$ and let $p_1,...,p_k \in \mathbb R_{>1}$ with $\frac1{p_1}+ ... + \frac1{p_k} = 1$

Then show the following inequality by assuming that there are for every $i = 1, ... ,k$ one $N \in \mathbb N_{>1}$ and one $n_i \in \{1,...,2^N\}$ with $p_i = 2^N / n_i$:

$$\sum_{j=1}^n\Bigl|\prod_{i=1}^ka_{i}^j\Bigr| \leqslant \prod_{i=1}^k\Bigl(\sum_{j=1}^n\bigl|a_{i}^j\bigr|^{p_i}\Bigr)^{1 / {p_i}}$$

The usual way to prove this is as a special case ($r=1$) of the more general inequality $$\Bigl(\sum_{j=1}^n\Bigl|\prod_{i=1}^ka_{i}^j\Bigr|^r\Bigr)^{1/r} \leqslant \prod_{i=1}^k\Bigl(\sum_{j=1}^n\bigl|a_{i}^j \bigr|^{p_i}\Bigr)^{1 / {p_i}},$$ where $\frac1{p_1}+ \ldots + \frac1{p_k} = \frac1r.$ That inequality is most easily proved by induction on $k$ (see the proof here).

I do not see how it helps to approximate the numbers $p_i$ by $2^N/n_i$.