Abstract. A certain two-sided maximal inequality is shown to be false.

A COUNTEREXAMPLE TO A TWO-SIDED HARDY-LITTLEWOOD MAXIMAL INEQUALITY

TERENCE TAO

1. Introduction

Let f be a non-negative function on R. For each integer j, let E_{j} be the projection operator

E_{j}(f)(x) = 2^{−j} ∫_{2^{j}k}^{2^{j}(k+1)} f

where k = ⌊x/2^{j}⌋. From the Hardy-Littlewood maximal inequality we have that

∥ sup_{j} E_{j}f∥_{2} ≲ ∥f∥_{2}.

By duality we have

∥ ∑_{k} E_{k}g_{k}∥_{2} ≲ ∥ ∑_{k} g_{k}∥_{2}.

From these two inequalities we can trivially deduce

∥ sup_{j} ∑_{k} E_{j}g_{k}∥_{2} ≲ ∥ ∑_{k} g_{k}∥_{2}

and

∥ sup_{j} ∑_{k} E_{k}g_{k}∥_{2} ≲ ∥ ∑_{k} g_{k}∥_{2}

respectively. One is then led to the following question: Conjecture 1.1. Is it true that

∥ sup_{j} ∑_{k} E_{j+k}g_{k}∥_{2} ≲ ∥ ∑_{k} g_{k}∥_{2}?

By linearizing and replacing the dyadic model with a continuous one, the conjecture can be seen to be equivalent to the estimate

∥\frac{1}{r(x)s(y)} ∫_{|x−y|∼r(x)s(y)} g(y) dy∥_{L^{2}(x)} ≲ ∥g∥_{2} (1)

for arbitrary non-negative functions r(x), s(y).

Nets Katz has informed the author that the above conjecture would imply the differentiability of Lipschitz vector fields in R^{2}. The estimate (1) would also be helpful for a maximal estimate involving the Schrödinger evolution operator. Unfortunately, the conjecture (and therefore (1)) fails, as this note will show.

By duality the conjecture is equivalent to the statement

∫ ∑_{j} ∑_{k} f_{j}E_{j+k}(g_{k}) ≲ ∥ ∑_{j} f_{j}∥_{2}∥ ∑_{k} g_{k}∥_{2},

which we can write symmetrically as

∫ ∑_{j} ∑_{k} E_{j+k}(f_{j})E_{j+k}(g_{k}) ≲ ∥ ∑_{j} f_{j}∥_{2}∥ ∑_{k} g_{k}∥_{2}

since E_{j+k} is a projection. By replacing j with 2j, k with −2k and restricting by symmetry to the range j ≥ k, we see that the above estimate is equivalent to

∑_{j≥k} ∫ E_{2(j−k)}(f_{j})E_{2(j−k)}(g_{k}) ≲ ∥ ∑_{j} f_{j}∥_{2}∥ ∑_{k} g_{k}∥_{2}. (2)

We will now find a counterexample to (2). Let N > 0 be a large integer to be chosen later, and let I be the interval [0, 2^{10^{N}}). Define the function t(x)

t(x) = ∑_{i=0}^{N−1} 10^{i}d_{i}(x),

where d_{i}(x) is the binary digit

d_{i}(x) = ⌊\frac{x}{2^{10^{i}}} ⌋ mod 2.

For any integer j, define f_{j} to be the characteristic function of

{x ∈ I : j = t(x)},

and set g_{j} = f_{j} for all j.

Proposition 1.2. For any j > k in the range of t, we have

∫ E_{2(j−k)}(f_{j})E_{2(j−k)}(g_{k}) = 2^{−d(j,k)−1}|I|

where d = d(j, k) is the unique integer such thats

10^{d+1} ≥ 2|j − k| > 10^{d}. (3)

Proof Since the decimal expansion of j and k consists purely of 1s and 0s, it is easy to check that

⌊\frac{j}{10^{d+1}} ⌋ = ⌊\frac{k}{10^{d+1}} ⌋.

From this and (3) we have

E_{2(j−k)}(f_{j}(x)) = E_{2(j−k)}(g_{k}(x)) = 2^{−d−1}

whenever

⌊\frac{t(x)}{10^{d+1}} ⌋ = ⌊\frac{j}{10^{d+1}} ⌋ = ⌊\frac{k}{10^{d+1}} ⌋, (4)

and

E_{2(j−k)}(f_{j}(x)) = E_{2(j−k)}(g_{k}(x)) = 0

The set of x for which (4) occurs has measure 2^{d+1−N}|I|. Putting these facts together gives the result.

From this proposition, the left-hand side of (2) is equal to

∑_{j>k} 2^{−d(j,k)−1}|I| = ∑_{j≠k} 2^{−d(j,k)−2}|I|

where j, k vary over the range of t. For each j and any 0 ≤ d < N, there are exactly 2^{d} values of k for which d(j, k) = d. Thus the left-hand side of (2) becomes

∑_{d=0}^{N−1} 2^{−2}|I| = N/4|I|.

On the other hand, we clearly have

∑_{j} f_{j} = ∑_{k} g_{k} = χ_{I},

so the right-hand side of (2) is simply |I|. Letting N → ∞ we obtain our desired contradiction.

We remark that the above counterexample in fact shows that the best constant C_{M} in the inequality

∥ sup_{|j|<M} ∑_{|k|<M} E_{k}g_{k}∥_{2} ≤ C_{M}∥ ∑_{|k|<M} g_{k}∥_{2}

grows at least as fast as log log M as M → ∞. On the other hand, it is easy to show that C_{M} ≲ log M. We do not know the exact growth rate of C_{M} but conjecture that C_{M} ∼ log log M.

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

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