Holder's inequality for infinite products

Solution 1:

Yes, it is true. The result (to the best of my search-fu) first appeared in Karakostas' 2008 paper. I quote the theorem in its entirety:

Theorem Let $p_0$ be a positive real number and $p_i$ be a sequence of positive extended real numbers such that $1 \leq p_i \leq +\infty$ for all $i = 1,2,3\ldots$ and assume $$ \frac1{p_0} = \sum_i \frac{1}{p_i}$$ where $1/+\infty = 0$ by convention. Let $(X,\mu)$ be a $\sigma$-finite measure space, with $\mu(X) < +\infty$. Assume for each $i = 1,2,\ldots $ there exists a function $f_i \in L^{p_i}(X,\mu)$. If the infinite product $\prod \|f_i\|_{p_i}$ converges to some number $\in (0,+\infty]$, and if $\prod f_i$ converges a.e. on $X$ to some function $f$, then $f \in L^{p_0}(X,\mu)$ and $$ \|f\|_{p_0} \leq \prod \| f_i\|_{p_i}$$

Remark: The proof goes through the classical Hölder's inequality, and is very straightforward in the case where $(X,\mu)$ is a probability space ($\mu(X) = 1$). (In a probability space we have that $\|f\|_{L^p} \leq \|f\|_{L^q}$ if $q \geq p$. This handy fact allows one to "upgrade" the finite-term estimates and apply Fatou's lemma.) Using $\sigma$-finiteness we can exhaust $X$ be finite measure subsets, which by monotone convergence allows us to upgrade from the probability space case.