In analysis, Holder's inequality says that if we have a sequence $p_1, p_2, \ldots, p_n$ of real numbers in $[1,\infty]$ such that $\sum_{i=1}^n \frac{1}{p_i} = \frac{1}{r}$, and a sequence of measurable functions $f_1, f_2, \ldots, f_n$, then letting $f = f_1 f_2 \cdots f_n$, we have the inequality \begin{equation} \lVert f \rVert_r \leq \lVert f_1 \rVert_{p_1} \lVert f_2 \rVert_{p_2} \cdots \lVert f_n \rVert_{p_n}. \end{equation}

In particular, if $f_i \in L^{p_i}(X,\mu)$ for all $i$, then $f \in L^r(X,\mu)$.

I'm looking for a generalization of this inequality to infinite products. That is, suppose we have an infinite sequence $(p_i)_{i \in \mathbb{N}}$ of real numbers in $[1,\infty]$ such that $\sum_{i=1}^\infty \frac{1}{p_i} = \frac{1}{r}$ and an infinite sequence of measurable functions $(f_i)_{i \in \mathbb{N}}$. Suppose moreover that the function $f(x) = \lim_{n \to \infty} \prod_{i=1}^n f_i(x)$ exists for almost every $x$. Under what conditions can I assert that \begin{equation} \lVert f \rVert_r \leq \liminf_{n \to \infty} \prod_{i=1}^n \lVert f_i \rVert_{p_i} ? \end{equation}

It seems to me that this might follow automatically from the first version of Holder's theorem I quoted above, but I'm a little uncomfortable with taking the limits. Is there anything I should watch out for?

Thanks in advance!

Davide Giraudo
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1 Answers1


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.

Willie Wong
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  • Thanks! The trick of solving the problem on a finite measure space first is a good one to keep in mind. – JHF Feb 19 '13 at 21:10