Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.