The following *spherical maximal theorem* was proved by E.M. Stein in the 1970s in dimensions 3 and higher, and by Bourgain in the 1980s in dimension 2.

**Theorem 1.*** Define the spherical maximal operator in by*

where is the normalized Lebesgue measure on the unit sphere . Then

for all

The purpose of this post is to explain some of the main ideas behind Bourgain’s proof. It’s a beautiful geometric argument that deserves to be well known; I will also have to refer to it when I get around to describing my recent joint work with Malabika Pramanik on density theorems. Among other things, I will try to explain why the case of Theorem 1 is in fact the hardest.

Note that Theorem 1 is trivial for ; the challenge is to prove it for some finite . It is known that the range of in the theorem is the best possible, but we will not worry about optimizing it in this exposition. (Not much, anyway.)

Let’s first try to get a general idea of what kind of geometric considerations might be relevant. Fix and . For the sake of the argument, let's pretend that we are looking for a counterexample to Theorem 1, i.e. a function with small but large. Let's also restrict our attention for the moment to characteristic functions of sets, so that for some set . Then . On the other hand, let be the set of all for which there exists a circle centered at such that a fixed proportion (say, 1/10-th) of is contained in . Then

for all ,

and in particular . If we could construct examples of such sets with fixed, but arbitrarily small, this would contradict Theorem 1. In particular, if we could construct a set of measure 0 such that for every (or some other set of positive measure) there is a circle centered at and contained in , Theorem 1 would fail spectacularly. Thus one of the consequences of Bourgain’s circular maximal theorem is that such sets can't exist. (This was also proved independently by Marstrand.)

Let’s now see if we can use this type of arguments to prove the theorem.

First, a couple of technical reductions. It’s hard to get quantitative results while working directly with circles, hence we will be working instead with thin annuli of thickness instead, then take the limit as . Furthermore, we will restrict the radii of the annuli to the interval . Thus instead of the maximal operator

we will be considering operators of the form

where each is some annulus of thickness and radius between 1 and 2, centered at . (We may as well assume that $f$ is non-negative.) We want to prove that is bounded on for some finite . Equivalently, we want to prove that

for all -functions . That is in turn equivalent to proving that the “dual operator”

is bounded on . We now let be the characteristic function of a set , then

It’s very easy to see that this is bounded on :

This is in fact the dual equivalent of the trivial bound on that I mentioned earlier. And now we get to the hard part: we need to prove that there is a similar bound on . By interpolation, we get that is bounded on all with (for all , not just characteristic functions). That implies Bourgain's theorem by duality.

We want to prove that . We expand the left side as

then rewrite is as

We would like this to be bounded by . In fact will also do, since we may assume that is bounded (that’s another technical reduction). Thus if we could prove that

we’d be home. Unfortunately, that’s just not true. Each annulus has thickness about and area of the same order, so that the right side of (2) is of the order . However, if the two annuli are “internally tangent” – in a clamshell configuration, one inside the other with

– the area of the intersection on the left side of (2) can easily be much larger than that. (“External” tangencies aren’t much good, either, but it’s easier to eliminate them from consideration in other ways.)

But there’s also good news. A *generic* intersection of two annuli – “transverse” intersection, as opposed to a tangency – does have area of the order . If we could just remove those pesky internal tangencies from the estimate, we’d be done.

It turns out that we can! The key observation is that, even though the internal tangencies are bad for us, there can’t be many of them. Namely, (3) is a major restriction on the set of for which internal tangencies are possible, and it turns out to be possible to translate this into an improvement in the estimate. More precisely, Bourgain divides the region of integration into two parts, one of which involves only transversal intersections (hence a good bound), the other covers all internal tangencies and is small in . Through the magic of interpolation, the estimate follows.

That’s about all I can say about the proof in a blog post, except for a couple of follow-up comments.

First, why is the case easier? Think about the size of the intersection of two spherical shells of thickness . A transverse intersection has size about . An internal tangency with has size about . When , internal tangencies are much larger than transverse intersections. That’s no longer true when , hence there is no need to consider the two cases separately.

Second, what if we try to extend this argument to ? . If is integer, then we could replace (1) by a similar expression involving -fold intersections. Unfortunately, a transverse intersection of 3 annuli – assuming that all three do intersect – has just about the same size as a transverse intersection of two annuli, and that’s not good for us. However, in higher dimensions one does obtain an additional gain by considering multiple intersections, and that’s the geometric reason why the range of exponents improves in higher dimensions.