# Bourgain’s circular maximal theorem: an exposition

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 ${\mathbb R}^d$ by $M f(x)=\sup_{t>0}\int_{S^{d-1}}|f(x+ty)|d\sigma(y),$

where $\sigma$ is the normalized Lebesgue measure on the unit sphere $\mathbb S^{d-1}$. Then $\| M f(x) \|_{p} \leq C\| f \|_{p}$ for all $p > \frac{d}{d-1}.$

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 $d=2$ case of Theorem 1 is in fact the hardest.

Note that Theorem 1 is trivial for $p=\infty$; the challenge is to prove it for some finite $p$. It is known that the range of $p$ 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 $d=2$ and $1. For the sake of the argument, let's pretend that we are looking for a counterexample to Theorem 1, i.e. a function $f$ with $\| f \|_p$ small but $\| Mf \|_p$ large. Let's also restrict our attention for the moment to characteristic functions of sets, so that $f = {\bf 1}_\Theta$ for some set $\Theta \subset {\mathbb R}^2$. Then $\| f \|_p = | \Theta |^{1/p}$. On the other hand, let $\Omega$ be the set of all $x$ for which there exists a circle $C_x$ centered at $x$ such that a fixed proportion (say, 1/10-th) of $C_x$ is contained in $\Theta$. Then $Mf(x) \geq .1$ for all $x \in \Omega$,

and in particular $\| Mf \|_p \geq .1 | \Omega |^{1/p}$. If we could construct examples of such sets with $|\Omega |$ fixed, but $|\Theta |$ arbitrarily small, this would contradict Theorem 1. In particular, if we could construct a set $\Theta$ of measure 0 such that for every $x \in [0,1]^2$ (or some other set of positive measure) there is a circle $C_x$ centered at $x$ and contained in $\Theta$, Theorem 1 would fail spectacularly. Thus one of the consequences of Bourgain’s circular maximal theorem is that such sets $\Theta$ 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 $2^{-k}$ instead, then take the limit as $k \to \infty$. Furthermore, we will restrict the radii of the annuli to the interval $[1,2]$. Thus instead of the maximal operator $M$ we will be considering operators of the form $\Phi_k f(x) = \frac{1}{|E_{x,k}|} \int_{E_{x,k}} f(z) dz,$

where each $E_{x,k}$ is some annulus of thickness $2^{-k}$ and radius $r_x$ between 1 and 2, centered at $x$. (We may as well assume that $f$ is non-negative.) We want to prove that $\Phi_k$ is bounded on $L^p$ for some finite $p$. Equivalently, we want to prove that $|\langle \Phi_k f, g\rangle|= | \int \Phi_k f(x) g(x) dx | \leq C \|f\|_{p} \|g\|_{p'}$

for all $L^{p'}$-functions $g$. That is in turn equivalent to proving that the “dual operator” $\Phi^*_k g(z) = \int g(x) \frac{1}{|E_{x,k}|} \chi_{E_{x,k}}(z) dx$

is bounded on $L^{p'}$. We now let $g$ be the characteristic function of a set $\Omega$, then $\Phi^*_k g(z) = \int_{\Omega} \frac{1}{|E_{x,k}|} \chi_{E_{x,k}}(z) dx.$

It’s very easy to see that this is bounded on $L^1$: $\int \Phi^*_k g(z) dz = \int_{\Omega} \int \frac{1}{|E_{x,k}|} \chi_{E_{x,k}}(z) dz dx =\int_\Omega 1 dx = |\Omega|.$

This is in fact the dual equivalent of the trivial $L^\infty$ bound on $M$ that I mentioned earlier. And now we get to the hard part: we need to prove that there is a similar bound on $L^2$. By interpolation, we get that $\Phi^*_k$ is bounded on all $L^{p'}$ with $1 < p^{\prime} < 2$ (for all $g \in L^{p^{\prime}}$, not just characteristic functions). That implies Bourgain's theorem by duality.

We want to prove that $\| \Phi^*_k g\|_2^2 \leq \| g \|_2^2 = |\Omega|$. We expand the left side as $\int \Big| \int_\Omega \frac{1}{|E_{x,k}|} \chi_{E_{x,k}}(z) dx \Big| dx = \int \int_{\Omega\times\Omega} \frac{1}{|E_{x,k}|\, |E_{y,k}| } \chi_{E_{x,k}}(z) \chi_{E_{y,k}}(z) dx\,dy\,dz,$

then rewrite is as $(1) \quad \int_{\Omega\times\Omega} \frac{1}{|E_{x,k}|\, |E_{y,k}| } |E_{x,k} \cap E_{y,k} | dx\,dy\, .$

We would like this to be bounded by $| \Omega |$. In fact $|\Omega \times \Omega |$ will also do, since we may assume that $\Omega$ is bounded (that’s another technical reduction). Thus if we could prove that $(2) \quad |E_{x,k} \cap E_{y,k} | \leq C |E_{x,k}|\, |E_{y,k}|,$

we’d be home. Unfortunately, that’s just not true. Each annulus has thickness about $2^{-k}$ and area of the same order, so that the right side of (2) is of the order $2^{-2k}$. However, if the two annuli are “internally tangent” – in a clamshell configuration, one inside the other with $(3) \quad |x - y| = |r_x - r_y|,$

– 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 $2^{-2k}$. If we could just remove those pesky internal tangencies from the $L^2$ 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 $(x, y)\in\Omega^2$ for which internal tangencies are possible, and it turns out to be possible to translate this into an improvement in the $L^1$ estimate. More precisely, Bourgain divides the region of integration into two parts, one of which involves only transversal intersections (hence a good $L^2$ bound), the other covers all internal tangencies and is small in $L^1$. Through the magic of interpolation, the $L^{p^{\prime}}$ 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 $d\geq 3$ easier? Think about the size of the intersection of two spherical shells of thickness $2^{-k}$. A transverse intersection has size about $2^{-2k}$. An internal tangency with $|x-y|\sim 1$ has size about $2^{-k} \cdot 2^{-(d-1) k/2} =2^{-k (d+1) /2}$. When $d=2$, internal tangencies are much larger than transverse intersections. That’s no longer true when $d\geq 3$, hence there is no need to consider the two cases separately.

Second, what if we try to extend this argument to $p > 2$? . If $p' = n$ is integer, then we could replace (1) by a similar expression involving $n$-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.

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