THE WEAK-TYPE (1, 1) OF THE PARABOLIC MAXIMAL FUNCTION

TERENCE TAO

Abstract. We reprove Mike Christ’s result on the parabolic maximal function.

1. Introduction

Let dμ_{k} be normalized smoothed out measure on the parabolic arc

{(t, t^{2}) : t ∼ 2^{k}}.

We wish to establish the bound

∥(∑_{k} |f ∗ dμ_{k}|^{2})^{1/2}∥_{L^{1,∞}} ≲ ∥f∥_{H^{1}}

where H^{1} is of course parabolic. The square function of course pointwise dominates the maximal function.

We may write f = ∑_{q} λ_{q}a_{q} where the {q} are a finite collection of dyadic parabolic cubes, λ_{q} > 0, and a_{q} are atoms. We now need to show that

|\{(∑_{k} | ∑_{q} λ_{q}a_{q} ∗ dμ_{k}|^{2})^{1/2} ≳ α\} ≲ \frac{1}{α} ∑_{q} λ_{q}.  (1)

Let E denote the set

E = \{M(∑_{q} λ_{q} \frac{χ_{q}}{|q|}) ≳ α\},

where M is the parabolic Hardy-Littlewood maximal function. One easily verifies that

∥ ∑_{q:q⊄E} χ_{q}∥_{∞} ≲ α,

1991 Mathematics Subject Classification. 42B25.

so the contribution of these cubes is acceptable. Thus we may restrict our attention to those cubes q in E. In particular, we may freely assume from the Hardy-Littlewood inequality that

| ⋃_{q} Cq| ≲ \frac{1}{α} ∑_{q} λ_{q}.  (2)

We now use an abstract form of the Calderón-Zygmund decomposition, in which no nesting or doubling properties are assumed.

Lemma 1.1. Let {Q} be a partially ordered collection of sets, and to each Q we associate a set T (Q) of finite measure. Let {q} be a finite subcollection of {Q}, and for each q ∈ {q} let λ_{q} be a non-negative number.

Assume the finiteness condition that for every q ∈ {q} and A > 0, there are at most a finite number of Q ∈ {Q} such that Q ⊃ q and |T (Q)| < A.

Let α > 0. Then to each q ∈ {q} we may associate a Q(q) ∈ {Q} such that

Q(q) ⊇ q and Q(q) ≥ q  (3)

for all q ∈ {q},

∑_{q⊆Q:Q(q)<Q} λ_{q} < α|T (Q)|  (4)

for all Q ∈ {Q}, and

| ⋃_{q} T (Q(q))| ≤ \frac{1}{α} ∑_{q} λ_{q} + | ⋃_{q} T (q)|  (5)

Proof It suffices to verify the lemma assuming that < is a total ordering, since every partial ordering can be completed to a total ordering.

Let Q denote the set of all Q such that Q ⊇ q for at least one q ∈ {q}, and such that

|T (Q)| ≤ \frac{1}{α} ∑_{q} λ_{q} + | ⋃_{q} T (q)|.

This set clearly contains {q}. By the finiteness assumption, the set of Q is finite.

We now perform the following algorithm. Initialize q to {q}, and Q to be the the largest element in Q, and let Q(q) be initially undefined for all q ∈ {q}.

If

∑_{q∈q:q⊆Q} λ_{q} ≥ α|T (Q)|,

then we set Q(q) = Q for all q ∈ q such that q ⊆ Q, and then remove all these q from q. Otherwise, if Q ∈ q then we set Q(Q) = Q, and remove Q from Q.

After this, we decrement Q to be the next largest element of Q, and repeat the process until all the elements of Q are exhausted.

For each q ∈ Q, the algorithm would have assigned a Q(q) to this q by the time Q has reached q; it is clear that (3) will hold. If the estimate (4) failed for some Q ∈ Q, then Q (or some greater set) would have been selected as Q(q) for all q ⊆ Q, a contradiction. Hence (4) holds for this Q. It is also clear that (4) holds for all Q ∉ Q by construction of Q.

To see (5), it clearly suffices to show that

| ⋃_{q:Q(q)≠q} T (Q(q))| ≤ \frac{1}{α} ∑_{q} λ_{q}.

We can estimate the left-hand side as

∑_{Q:Q=Q(q) for some q≠Q} |T (Q)|.

But by construction, if Q appears in the above sum then

∑_{q:Q(q)=Q} λ_{q} ≥ α|T (Q)|.

Combining the two statements yields the result.

Let {Q} be the space of all dyadic rectangles. We define the width σ(Q) and eccentricity τ(Q) to be the integers such that Q is a 2^{σ(Q)} × 2^{σ(Q)+τ(Q)}. Note that the parabolic cubes are those rectangles q for which σ(q) = τ(q). We partially order the set {Q} of all rectangles by defining Q′ > Q if τ(Q′) > τ(Q).

We define the tendril T (Q) to be the set

T (Q) = {x + (t, t^{2}) : x ∈ Q, 0 ≤ t ≲ 2^{τ(Q)}};

note that

|T (Q)| ∼ 2^{σ(Q)+2τ(Q)} + 2^{2σ(Q)+τ(Q)}.

The finiteness condition is easily verified, so we may obtain a map q ↦ Q(q) satisfying (3), (4), and (5). Since T (q) ⊂ Cq for all q ∈ {q}, we can combine (2) and (5) to obtain

| ⋃_{q} T (Q(q))| ≤ \frac{1}{α} ∑_{q} λ_{q}.  (6)

For any q, we see that a_{q} ∗ dμ_{k} is supported in T (Q(q)) if k < τ(Q(q)). From (6),

(1) then follows from

|\{(∑_{k} | ∑_{q:k≥τ(Q(q))} λ_{q}a_{q} ∗ dμ_{k}|^{2})^{1/2} ≳ α\} ≲ \frac{1}{α} ∑_{q} λ_{q}.

It suffices by Chebyshev to prove the L^{2} estimate

∥ ∑_{k} | ∑_{q:k≥τ(Q(q))} λ_{q}a_{q} ∗ dμ_{k}|^{2})^{1/2}∥_{2}^{2} ≲ α ∑_{q} λ_{q}.

By the triangle inequality it suffices to show

∥ ∑_{k} | ∑_{q:τ(Q(q))=k−s} λ_{q}a_{q} ∗ dμ_{k}|^{2})^{1/2}∥_{2}^{2} ≲ 2^{−s}α ∑_{q} λ_{q}

for all s ≥ 0.

Fix s. It suffices to show that

∥ ∑_{q:τ(Q(q))=k−s} λ_{q}a_{q} ∗ dμ_{k}∥_{2}^{2} ≲ 2^{−s}α ∑_{q:τ(Q(q))=k−s} λ_{q}

for each k, since the claim follows by summing in k. By scaling we may take k = 0. Expanding the left-hand side, we reduce to

∑_{q,q′:τ(Q(q))=τ(Q(q′))=−s} λ_{q}λ_{q′}|⟨a_{q} ∗ dμ_{0}, a_{q′} ∗ dμ_{0}⟩| ≲ 2^{−s}α ∑_{q:τ(Q(q))=−s} λ_{q}.

By symmetry we may assume that |q′| ≤ |q|. It then suffices to show that

∑_{q′:|q′|≤|q|,τ(Q(q′))=−s} λ_{q′}|⟨a_{q} ∗ dμ_{0}, a_{q′} ∗ dμ_{0}⟩| ≲ 2^{−s}α  (7)

for all q.

Fix q; we may center q at the origin. q has dimension 2^{τ(q)} × 2^{2τ(q)}; from (3) we note that

τ(q) ≤ τ(Q(q)) = −s.  (8)

Rewrite the left-hand side of (7) as

∑_{q′:|q′|≤|q|,τ(Q(q′))=−s} λ_{q′}|⟨a_{q} ∗ F, a_{q′}⟩|

where F = dμ_{0} ∗ d̃μ_{0}. Observe that F is supported on the sector

{(x_{1}, x_{2}) : |x_{2}| ≲ |x_{1}|}

and obeys the estimates

|∇^{α}F (x)| ≲ |x|^{−1−|α|}

for all |α| ≤ 2. From the size and moment conditions on a_{q}, this implies that the function a_{q} ∗ F satisfies

|∇^{α}(a_{q} ∗ F)(x)| ≲ 2^{τ(q)}(2^{τ(q)} + |x|)^{−2−|α|}

for all |α| ≤ 1. This in turn implies from the size and moment conditions on a_{q′} and the assumption |q′| ≤ |q| that

|⟨a_{q} ∗ F, a_{q′}⟩| ≲ 2^{2τ(q)}diam(q ∪ q′)^{−3},

where the diameter is respect to the Euclidean metric. Thus it suffices to show that

∑_{q′:τ(Q(q′))=−s} λ_{q′}diam(q ∪ q′)^{−3} ≲ 2^{−2τ(q)}2^{−s}α.

For each n ≥ 0, let Q_{n} denote the rectangle

{(x_{1}, x_{2}) : |x_{1}| ≤ 2^{τ(q)+n+s−1}, |x_{2}| ≤ 2^{τ(q)+n}}.

THE WEAK-TYPE (1, 1) OF THE PARABOLIC MAXIMAL FUNCTION 5

Observe that

diam(q ∪ q′)^{−3} ≲ 2^{−3τ(q)} ∑_{n≥0,q′⊂Q_{n}} 2^{−3n},

and so it suffices to show that

∑_{n≥0} 2^{−3n} ∑_{q′:τ(Q(q′))=−s,q′⊂Q_{n}} λ_{q′} ≲ 2^{τ(q)}α. (9)

Since τ(Q_{n}) = −s + 1, we have from (4) that

∑_{q′:τ(Q(q′))=−s,q′⊂Q_{n}} λ_{q′} ≤ ∑_{q′⊂Q_{n}:Q(q′)<Q_{n}} λ_{q′} ≤ α|T (Q_{n})| ∼ α(2^{τ(q)+n−s}+2^{2τ(q)+2n}).

This we may estimate the left-hand side of (9) by

α ∑_{n≥0} 2^{τ(q)−2n−s} + 2^{2τ(q)−n},

and (9) follows from (8).

Department of Mathematics, UCLA, Los Angeles CA 90095-1555

E-mail address: tao@math.ucla.edu