Let $f_{1},...,f_{n}:X \to \bar{\mathbb R}$ measurable, while $f:=f_{1}\times...\times f_{n}$, and $p_{1},...,p_{n} \in [1,\infty]$ where $f_{i}\in L^{p_{i}}(\mu), \forall i\in \{1,...,n\}$.
Note $\frac{1}{p}=\sum_{i=1}^{n}\frac{1}{p_{i}}$
Show:
i) $f_{1}\times...\times f_{n}=f \in L^{p}$
ii) $||f||_{p} \leq \prod_{i}^{n}||f_{i}||_{p_{i}}$
My ideas:
ii) In class we have proven the Hölder inequality for two functions, i.e. $||fg||_{r}\leq||f||_{p}||g||_{q}$, whereby $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$. I think that the extension of this would be to merely use induction to show that Hölder is valid for any $n \in \mathbb N$, but this seems too easy.
i) I am stuck on $f_{1}\times...\times f_{n}\in L^p$.
$(\int_{\mathbb R}|f|^p)^\frac{1}{p}=(\int_{\mathbb R}|\prod_{i=1}^{n}f_{i}|^p)^{\sum_{i=1}^{n}\frac{1}{p_{i}}}=\prod_{j=1}^{n}(\int_{\mathbb R}\prod_{i=1}^{n}|f_{i}|^p)^{\frac{1}{p_{j}}}$
and from $\frac{1}{p}=\sum_{i=1}^{n}\frac{1}{p_{i}}$, we glean that $p \leq p_{i}$ , $i=1,...,n$
Therefore, can I say $\prod_{j=1}^{n}(\int_{\mathbb R}\prod_{i=1}^{n}|f_{i}|^p)^{\frac{1}{p_{j}}}\leq \prod_{j=1}^{n}(\int_{\mathbb R}\prod_{i=1}^{n}|f_{i}|^{p_{i}})^{\frac{1}{p_{j}}}$ and that is $< \infty$
Therefore $f \in L^{p}$. I am not sure on this proof.
To show how trivial this is, let me switch the notation, something that people should have already done probably.
I use $q=p^{-1}\in[0,1]$ and $q_i=p_i^{-1}\in[0,1]$ and introduce $$ [\![f]\!]_q := \|f\|_{q^{-1}} = \|f\|_p. $$
Then Holder inequality says that $[\![f_1f_2]\!]_{q_1+q_2}\leq[\![f_1]\!]_{q_1}[\![f_2]\!]_{q_2}$ if $q_i\in[0,1]$ and $q_1+q_2\leq 1$.
Now, in our case, $q=q_1+\dots+q_n = (q_1+\dots+q_{n-1})+q_n$, so $$ [\![f_1\dotsm f_n]\!]_q \leq [\![f_1\dotsm f_{n-1}]\!]_{q_1+\dots+q_{n-1}}[\![f_n]\!]_{q_n} \leq \dots \leq [\![f_1]\!]_{q_1} \dotsm [\![f_n]\!]_{q_n} $$ by bringing outside one function at a time with Holder.
No more ugly $\frac1{\left(\frac1{p_1}+\dots+\frac1{p_{n-1}}\right)^{-1}}+\frac1{p_n}=1$ expressions for the World.