** The Vector Maximal Function **

DefinitionGiven with component functions , define (the scalar-valued function)

Note that each function gets to choose its own maximizing radius at each point.

Theorem

- , where .
- for .

Remark

- is not bounded on . Indeed, consider defined by so that at each . Now if and , then . Thus on .
- For this operator it is the bound that is easy:

Before going on to the proof of the theorem, we first introduce some definitions,

DefinitionThe family ofdyadic cubesisNote that the cubes of each generation (i.e., of fixed side-length ) tile . The nice nesting properties of the dyadic cubes make them very convenient for covering and/or stopping time arguments (cf. if , then or vice versa). Let

be the -algebra generated by cubes of side-length . It is equal to . Note that , which makes this afiltration. Note that where and [ is measurable with respect to and whenever ]. As we say form aMartingale.

Remark

- Note that by Lebesgue differentiation Lebesgue a.e.
- Honest martingales are rare in analysis, but approximate martingales abound. For example, say is harmonic on , and on and consider .

Lemma (A Calderón–Zygmund-type decomposition)Suppose is . Then there is a family of disjoint dyadic cubes with so that

*Proof:* If is large enough then

e.g. . Beginning with the tessellation of cubes of that size, we subdivide: if any cubes has

then we put into our collection and do not subdivide it any further. If a cube ended up in our collection, means that its parent did not:

Summing in , we get . By Lebesgue differentiation, a.e. Off the cubes those averages are all less than and so a.e. there.

*Morals*

- is good: it lies in , so we can use some kind of “trivial bound.”
- The bad part has small support. Indeed a neighborhood of this set has small measure.
- Far from , looks good.

*Proof of assertion (1)*: Do a Calderón–Zygmund decomposition at height

and

which is acceptable. As so we do not need to worry about on (which is not a dyadic cube). We need to estimate

What would happen if we replaced by defined via

- For , then
Therefore

- If for some , then is ? Well, , there exists such that
(Misc notes: , , ) Therefore, .

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