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Let $M$ be the Hardy-Littlewood maximal operator. In the book "Weighted norm inequalities and Related Topics" by Rubio de Francia and J. Cuerva, page 150, theorem 2.1.2 states as the following:

*For every $1<p<\infty$, there is a constant $c_p>0$ such that, for any measurable functions on $\mathbb R^n$, $\phi\geq0$ and $f$, we have the inequality:

(2.13) $\int\limits_{\mathbb R^n}\left(Mf(x)\right)^p\phi(x)dx\leq C_p\int\limits_{\mathbb R^n}|f(x)|^p(M\phi)(x)dx$*

In the proof of this theorem the authors said that: when $M\phi(x)=\infty$ a.e., then (2.13) holds trivially.

If $|f(x)|>0$ on a subset with positive measure, everything seems right. But if I chose namely $f(x)=0$ a.e., I do not see why (2.13) is trivial, because in this sense it may be $0\cdot \infty$ is undefined.

So how should I understand (2.13)? Is it really trivial in that case and why?

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If $f=0$ a.e., then the left side is $0$. –  robjohn Nov 16 '11 at 2:02
    
This is the main part of my question, why the left side is $0$ when $f=0$ a.e. –  Hai Minh Nov 16 '11 at 2:28
1  
$Mf(x)=0$ everywhere, does it not? –  robjohn Nov 16 '11 at 2:41
    
Yes, but since may be $\phi(x)=\infty$, so we get $0\cdot\infty=???$. –  Hai Minh Nov 16 '11 at 6:22
    
Is there any reason in the text to expect that $\phi(x)$ might take non-real values? –  robjohn Nov 16 '11 at 7:30

1 Answer 1

up vote 2 down vote accepted

When dealing with functions $f, g:{\mathbb R}^n \rightarrow [0,\infty]$, in order to make Lebesgue integration theory work, you specify that if $f(x) = 0$ and $ g(x) = \infty$, then $f(x)g(x) = 0$. This is the only way you can make commutativity, associativity, and distributivity hold (i.e. $(f(x) + g(x))h(x) = f(x)h(x) + g(x)h(x)$ and so on.) If you have Rudin's "Real and Complex Analysis" there's a short discussion at the bottom of p.18 about this issue. By the way, measurable functions in this setting are defined to be functions for which $f^{-1}(a,\infty]$ are measurable for all $a$. Equivalently, $f^{-1}(\infty)$ and each $f^{-1}(a,\infty)$ is measurable.

So in the case you're worried about, your inequality reduces to $0 \leq 0$ which is obviously ok.

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Why is all that necessary? There is something positive on the RHS and $Mf = 0$ so you have $0 \leqslant \text{ something positive}$ which always holds. –  Jonas Teuwen Nov 19 '11 at 18:16
    
I was explaining why the left-hand side is zero even if $\phi(x)$ can take $\infty$ as a value. –  Zarrax Nov 19 '11 at 20:18
    
Oh right. Fine :-). +1. –  Jonas Teuwen Nov 19 '11 at 20:25
    
Excellent answer! –  Hai Minh Nov 21 '11 at 14:52

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