Tue, 10/11/2011 - 01:57 — wick

In last weeks lecture, we studied the basic behavior of the maximal function. Recall that the maximal function was defined with respect to balls or cubes.

In this lecture we will study the strong maximal function. This operator is defined with respect to rectangles whose sides are parallel to the coordinate axes. This slight change in the definition changes the presents some challenges as to the boundedness of the operator; however it still ends up being bounded. Again, we will show that this operator is bounded on all the corresponding $L^p$ spaces. We will give two proofs of this, one is an easy reduction to iterated maximal functions. The other proof is a more clever covering lemma.

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## Mistake in Lecture 2

Dear All,

Tim Ferguson pointed out to me that the original version of the covering lemma that I had provided had a mistake in the proof. There is a problem in the selection procedure for the rectangles, and the argument given didn't work to conclude a certain necessary inequality.

The idea I had given was essentially correct. To prove a covering lemma, we need a selection procedure on the rectangles. We then need a way to measure how much we have "lost" in this procedure, and this is where induction comes in, coupled with the geometry of the rectangles in question. We select the rectangles, then recognize that it is possible to slice away a given side. We then compare this to the maximal function and see that what we took away wasn't too large, and then integrate out the missing side. These were the steps we used, but I sliced away the missing side incorrectly.

I have corrected the version in the lecture, and uploaded a corrected version. The corrected version contains a comment about where the mistake was made so that the interested reader can compare the differences in the file. Aside from typos, I think that this fixes things.

Many thanks to Tim for reading through the notes carefully. As I had hoped, this is becoming a learning process for all involved.

Regards,

Brett

## Dear Phi, Thanks for the

Dear Phi,

Thanks for the question. Here are the answers that I am aware of. Maybe some of the other participants have more information.

1. I am not aware of people studying the difference between the centered and uncentered strong maximal function. This seems like an interesting question to explore though.

2. We are allowed to consider arbitrary rectangles; long and thin are perfectly acceptable.

The Lebesgue Differentiation Theorem is usually connected to the weak-type estimates that the corresponding maximal operator satisfies. For example, the maximal function satisfies weak (1,1) estimates, and because of this we can differentiate functions in L^1. The strong maximal function is different in that it can differentiate Llog L. This is because, using a slight variant of the covering lemma that appears in this lecture, one can prove that the two parameter strong maximal function satisfies a weak Llog L type inequality.

Regards,

Brett

## Center and un-center definition of Maximal function?

Dear Brett D. Wick,

In your first lecture, the maximal function in your definition is the center one that means your $x$ is the center of the cubes $Q$. However, in your second lecture, the maximal function is defined over the sub of all Rectangles with sides parallel to the coordinate exes.

My question are:

1) Is there any difference between the centered one and un-centered one? I believe that there are some differences.

2) Do we need some more properties of these rectangles since it seems that you cannot take the very long and very thin rectangles such that the measures go to zero? (for example the result of Lebesgue Differentiation Theorem)