Annales Academiæ Scientiarum Fennicæ Series A. I. Mathematica Volumen 18, 1993, 211–248

BILIPSCHITZ MAPPINGS AND STRONG A_{∞} WEIGHTS

Stephen Semmes

Rice University, Department of Mathematics, Wiess School of Natural Sciences P.O. Box 1892, Houston, TX 77251, U.S.A.; semmes@rice.edu

Abstract. Given a doubling measure μ on R^{n}, one can define an associated quasidistance on R^{n} by

δ(x, y) = μ(B_{x,y})^{1/n},

where B_{x,y} is the smallest ball which contains the points x and y. This paper is concerned with the resulting geometry which is induced by μ. The main result provides a condition on μ under which R^{n} equipped with the quasidistance δ(·, ·) admits a bilipschitz embedding into R^{N} with the standard Euclidean structure for some N < ∞. This sufficient condition is not so far from being necessary, but of course the reader knows how it is.

This paper is closely related to an earlier one by Guy David and the author [DS1].

1. Introduction

Let μ be a doubling measure on R^{n}, so that there is a C > 0 such that

(1.1)  0 < μ(2B) ≤ Cμ(B)

for all balls B. (Here 2B denotes the ball with the same center as B but twice the diameter.) We can associate to μ the quasidistance δ(x, y) on R^{n} defined by

(1.2)  δ(x, y) = μ(B_{x,y})^{1/n},

where

(1.3)  B_{x,y} is the smallest ball that contains both x and y.

To say that δ(x, y) is a quasidistance means that it is nonnegative and symmetric, that it vanishes exactly when x = y, and that it satisfies the following weakened form of the triangle inequality:

(1.4)  δ(x, z) ≤ C(δ(x, y) + δ(y, z))

1991 Mathematics Subject Classification: Primary 30C99; Secondary 42B99, 51K99.

The author is partially supported by the National Science Foundation, the Alfred P. Sloan Foundation, and the Marian and Speros Martel Foundation.

for some C > 0 and all x, y, z ∈ R^{n}. If (1.4) were true with C = 1 then δ(x, y) would be a distance function (as opposed to a quasidistance).

This paper is concerned with the following type of issue. We restrict ourselves to the case where n > 1 from now on, because the n = 1 case is trivial for the topics that will be considered here.

Question 1.5. Under what conditions on μ is δ(·, ·) bilipschitz equivalent to the standard distance on R^{n}? In other words, when is it true that there is a C > 0 and a mapping f: R^{n} → R^{n} such that

(1.6)  C^{−1}δ(x, y) ≤ |f(x) − f(y)| ≤ Cδ(x, y)

for all x, y ∈ R^{n}?

This problem has a natural reformulation in terms of quasiconformal mappings. If a mapping f as above exists, then it must be quasiconformal, and μ will be comparable in size to the pull-back of Lebesgue measure on R^{n} by f. Gehring’s theorem [Ge] then implies that μ must be absolutely continuous, μ = ω(x) dx, with ω(x) ∈ L_{loc}^{1}(R^{n}), and ω must be comparable in size to the Jacobian of f. Conversely, if f: R^{n} → R^{n} is quasiconformal and ω(x) is comparable in size to its Jacobian, then μ = ω(x) dx is a doubling measure and f satisfies (1.6). The problem of characterizing the functions ω that arise in this fashion is notoriously difficult.

Here is a variation of Question 1.5 that is somewhat easier.

Question 1.7. Under what conditions on μ does (R^{n}, δ(·, ·)) admit a bilipschitz embedding into R^{n} for some N? In other words, for which μ do there exist constants C, N > 0 and a mapping f: R^{n} → R^{N} such that (1.6) holds?

The following is a simple necessary condition for there to exist a mapping f as in Questions 1.5 or 1.7. (1.8) There exists a C > 0 and a distance function δ′(·, ·) on R^{n} such that

C^{−1}δ(x, y) ≤ δ′(x, y) ≤ Cδ(x, y).

The point here is that δ′(·, ·) should satisfy the actual triangle inequality, i.e., the analogue of (1.4) with C = 1. The necessity of this condition is trivial; simply set

δ′(x, y) = |f(x) − f(y)|.

The main result of this paper (Theorem 5.2) is that a “slightly” stronger version of (1.8) is also a sufficient condition for Question 1.7.

Condition (1.8)—in a different formulation—is considered already in [DS1]. It was noted there that Gehring’s argument in [Ge] implies that if (1.8) holds, then μ is absolutely continuous, and that its density is an A_{∞} weight. (See Proposition 3.4 below.) Let us give a formal name for the class of these weights.

Bilipschitz mappings and strong A_{∞} weights 213

Definition. A positive locally integrable function on R^{n} is called a strong A_{∞} weight if it is the density of a doubling measure μ which satisfies (1.8).

There is another problem which is closely related to Questions 1.5 and 1.7, that will not be addressed in this paper, but which should be mentioned for conceptual reasons. Let M be a connected hypersurface in R^{d+1} , d ≥ 2. Assume a priori that M is smooth, and even that M ∪{∞} is a smooth (closed) embedded submanifold of R^{d+1} ∪ {∞} ≅ S^{d+1} . Let n(x) be a smooth choice of the unit normal to M. (A theorem in algebraic topology implies that M is orientable under these hypotheses.) Let ∥n∥_{∗} denote the BMO norm of n on M, i.e.,

∥n∥_{∗} = sup_{{x∈M \atop R>0}} \frac{1}{|B(x, r) ∩ M|} ∫_{B(x,r)∩M} |n(y) − n_{x,R}| dy,

where n_{x,R} denotes the average of n over B(x, r) ∩ M.

Question 1.9. Do there exist constants ε, k > 0, depending only on d, so that if M is as above and ∥n∥_{∗} ≤ ε, then there exists a mapping g from R^{d} onto M such that

(1.10)  (1 + k)^{−1}|x − y| ≤ |g(x) − g(y)| ≤ (1 + k)|x − y|

for all x, y ∈ R^{d}?

If the answer is yes, it would be better if we could choose k so that k → 0 as ∥n∥_{∗} → 0.

The condition that ∥n∥_{∗} be small is quite natural. In particular there are various equivalent conditions concerning the geometry of M and analysis on M given in [S1, 3]. One of these equivalences implies a converse to Question 1.9: if M is the image of a g that satisfies (1.10) with k small, then ∥n∥_{∗} is small. Notice that ∥n∥_{∗} is unaffected by translations, rotations, or dilations of M.

On the other hand, it follows from [S2, 3] that if ∥n∥_{∗} is small enough, then M is homeomorphic to R^{d}, and that one can even construct parameterizations of M with L^{p} bounds on their differentials and their inverses, with p → ∞ as ∥n∥_{∗} → 0. These bounds do not depend on our a priori smoothness assumptions on M in a quantitative way. It is much more difficult, however, to build a bilipschitz parameterization of M.

When d = 2 uniformization can be used to produce a conformal mapping h from R^{2} onto M. Using extremal length estimates it can be shown that if ∥n∥_{∗} is small enough, then there is a doubling measure μ on R^{2} such that h satisfies (1.6) with C and the doubling constant depending only on d. This is proved in [S2], along with related facts, e.g., μ comes from an A_{∞} weight ω with the BMO norm of log ω tending to zero as ∥n∥_{∗} → 0. It follows from this that if Question 1.5 can be answered affirmatively when n = 2, then Question 1.9 can also be answered affirmatively when d = 2. Conversely, if there is a counterexample for Question 1.5, that might provide insight on how to find a counterexample for Question 1.9.

The organization of this paper is as follows. Some basic facts about A_{p} weights (including the definitions) will be reviewed in the next section. In Section 3 there are some simple reformulations of (1.8) as well as other useful properties of strong A_{∞} weights, and in Section 4 some examples and counterexamples pertaining to strong A_{∞} weights are given. The main theorem is stated in Section 5, and the remainder of the paper is devoted to its proof.

Throughout this paper we use the notation

∫_{A} g  for  \frac{1}{|A|} ∫_{A} g,

where |A| denotes the Lebesgue measure of A.

The author is grateful to Guy David for many relevant conversations.

2. A review of basic facts about A_{p} weights

General references include [CF], [Ga] (especially Section 6 of Chapter 6), and [JL] (especially pp. 19–28).

Let w be a nonnegative locally integrable function on R^{n} with w > 0 a.e.

Definition 2.1: (i) w is an A_{∞} weight if there exist C, γ > 0 such that

(2.2)  \frac{∫_{A} w}{∫_{Q} w} ≤ C(\frac{|A|}{|Q|})^{γ}

for all cubes Q and all measurable subsets A of Q.

(ii) w is an A_{p} weight, 1 < p < ∞, if there exists C > 0 so that

(2.3)  (∫_{Q} w)(∫_{Q} w^{−1/(p−1)})^{p−1} ≤ C

for all cubes Q.

(iii) w is an A_{1} weight if there is a constant C > 0 so that

(2.4)  ∫_{Q} w ≤ C( essinf_{Q} w).

In each case you get an equivalent condition if you use balls instead of cubes.

Facts about A_{p} weights:

(a) A_{1} ⊆ A_{p} ⊆ A_{q} ⊆ A_{∞} when 1 < p < q < ∞.

(b) A_{∞} = ∪_{p<∞}A_{p}.

(c) If w ∈ A_{∞}, then σ(E) = ∫_{E} w(x) dx defines a doubling measure on R^{n}.

(d) If w_{1} , w_{2} ∈ A_{p}, 1 ≤ p ≤ ∞, and if λ ∈ [0, 1], then w_{1}^{λ}w_{2}^{1−λ} ∈ A_{p}. (This can easily be derived from Hölder’s inequality.)

(e) w ∈ A_{∞} if and only if there exist C > 0 and p > 1 such that the reverse Hölder inequality

(2.5)  (∫_{Q} w^{p})^{1/p} ≤ C ∫_{Q} w

is satisfied by all cubes Q.

(f) If w ∈ A_{p}, then there is an r > 1 such that w^{r} ∈ A_{p}.

(g) If w ∈ A_{∞} then there is a C > 0 so that

(2.6)  exp ( ∫_{Q} log w) ≤ ∫_{Q} w ≤ C exp ( ∫_{Q} log w)

holds for all cubes Q (and similarly for all balls). Indeed, the first inequality follows from Jensen’s theorem, while the second can be derived, for example, from (b) and Jensen’s theorem. The converse is also true: if w satisfies (2.6), then w is an A_{∞} weight. (See [H], and also Proposition 3.5 below.)

(h) Suppose that u and v are A_{1} weights, and let t be any positive real number. Then uv^{−t} is an A_{∞} weight. Conversely, if w is an A_{∞} weight, then there exist u, v ∈ A_{1} and t > 0 so that w = uv^{−t}. This is the factorization theorem of Peter Jones [J].

3. Basic facts about strong A_{∞} weights

A large proportion of the content of this section can already be found in [DS1], but the presentation here is somewhat different, and in a few instances it is also more detailed.

We begin with a couple of easy and useful reformulations of (1.8).

Lemma 3.1. (1.8) holds if and only if (3.2) there is a constant C > 0 so that for any finite sequence x_{1}, …, x_{k}

of points in R^{n} we have

δ(x_{1}, x_{k}) ≤ C ∑_{j=1}^{k−1} δ (x_{j}, x_{j+1}) .

It is immediate that (1.8) implies (3.2). To prove the converse define δ′(·, ·) by

δ′(x, y) = inf { ∑_{j=1}^{k−1} δ (x_{j}, x_{j+1}) },

where the infimum is taken over all finite sequences {x_{j}}_{j=1}^{k} such that x_{1} = x and x_{k} = y. Clearly δ′(x, y) ≤ δ(x, y) for all x, y, and (3.2) implies that C^{−1}δ(x, y) ≤ δ′(x, y). It is also easy to see that δ′(·, ·) is a distance function.

Lemma 3.3. Let μ be a doubling measure on R^{n}. Then (3.2) holds if and only if the same condition is true but with the additional constraint that x_{j} ∈ B(x_{1}, 2|x_{k} − x_{1}|) for all j.

The “only if” part is automatic, and so we need only concern ourselves with the “if” part. Suppose that the above constrained version of (3.2) is satisfied, and let us show that that implies the unconstrained version. Let x_{1}, …, x_{k} be given, and suppose that there is a j, 1 ≤ j ≤ k, such that

x_{j} ∉ B(x_{1}, 2|x_{k} − x_{1}|).

Let j_{0} be the smallest such j. Then

x_{j} ∈ B(x_{1}, 2|x_{j_{0}} − x_{1}|)

when 1 ≤ j ≤ j_{0} , and so the constrained version of (3.2) can be applied to give

δ(x_{1}, x_{j_{0}}) ≤ C ∑_{i=1}^{j_{0}−1} δ(x_{i}, x_{i+1}) ≤ C ∑_{i=1}^{k−1} δ(x_{i}, x_{i+1}).

On the other hand we have

δ(x_{1}, x_{k}) ≤ Cδ(x_{1}, x_{j_{0}})

because x_{k} ∈ B(x_{1}, |x_{j_{0}} − x_{1}|). This uses also the definition (1.2) of δ(·, ·) and the doubling condition on μ. Combining these inequalities yields

δ(x_{1}, x_{k}) ≤ C ∑_{i=1}^{k−1} δ(x_{i}, x_{i+1}),

as desired.

Next we discuss the absolute continuity of the μ’s for which (1.8) holds.

Proposition 3.4. If μ is a doubling measure that satisfies (1.8), then μ is absolutely continuous and it is given by ω(x)dx, where ω is an A_{∞} weight.

Proposition 3.4 is essentially due to Gehring [Ge]. He stated it only in a special case, but his proof works in general. A proof will now be given for the convenience of the reader. One of the key ingredients is the following criterion for membership in A_{∞}.

Proposition 3.5. Suppose that ω ∈ L_{loc}^{1}(R^{n}), ω > 0 a.e., and that there exist constants C_{1} > 0 and r ∈ (0, 1) so that

(3.6)  ∫_{Q} ω ≤ C_{1}(∫_{Q} ω^{r})^{1/r}

for all cubes Q. Then there exist C_{2} > 0 and p > 1, depending only on C_{1} , r, and n, so that

(∫_{Q} ω^{p})^{1/p} ≤ C_{2} ∫_{Q} ω

for all cubes Q. In particular, ω is an A_{∞} weight with constants that depend only on C_{1} and r (by Fact (e) of Section 2).

See [Ge] or [Ga] (Theorem 6.9, p. 260) for a proof.

Let us now use this to prove Proposition 3.4. Let θ(x) be a smooth bump function on R^{n}, with θ ≥ 0, supp θ ⊆ B(0, 1), and ∫_{R^{n}} θ = 1. Set θ_{t}(x) = t^{−n}θ (x/t), and define ω_{t} on R^{n} for each t > 0 by

(3.7)  ω_{t} = θ_{t} ∗ μ.

It is not hard to check that μ will satisfy the conclusions of Proposition 3.4 if we can show that each ω_{t} is in A_{∞} with constants that do not depend on t.

According to Proposition 3.5 it suffices to show that there is a C_{1} so that each ω_{t} satisfies (3.6) with r = 1/n. To do this we fix t and consider separately the cases where diam Q ≤ t and diam Q > t. The first case is trivial, because the doubling condition on μ ensures that ω_{t} is roughly constant on the scale of t. To treat the situation where diam Q > t we use the following estimate.

Lemma 3.8. Let μ be a doubling measure that satisfies (1.8). Then there is a C > 0 so that if L is any line segment with length > t and endpoints x and y, then

δ(x, y) ≤ C ∫_{L} ω_{t}(z)^{1/n}dz.

The “dz ” on the right side denotes arclength measure.

This result is easily derived from Lemma 3.1. Let k be the integer part of length(L)/t, and let z_{0}, …, z_{k} be points on L such that z_{0} = x, z_{k} = y, and t ≤ |z_{j+1} − z_{j}| ≤ 2t for each j. Using the doubling condition on μ and (3.2) we get

\begin{aligned}∫_{L} ω_{t}(z)^{1/n}dz & = ∑_{j=1}^{k} ∫_{[z_{j−1},z_{j}]} ω_{t}(z)^{1/n}dz ≥ ∑_{j=1}^{k} C^{−1}μ (B_{z_{j−1},z_{j}})^{1/n} \\ & = C^{−1} ∑_{j=1}^{k} δ (z_{j−1}, z_{j}) ≥ C^{−1}δ(x, y). \end{aligned}

This proves Lemma 3.8.

Once you have the lemma it is easy to check that (3.6) holds for r = 1/n and ω = ω_{t} when diam Q > t, using Fubini’s theorem, the definition of δ(x, y), and the doubling condition on μ. When diam Q ≤ t (3.6) is trivial. This finishes the proof of Proposition 3.4.

Next we give a characterization of strong A_{∞} weights in terms of lengths of curves. This is in fact how the strong A_{∞} condition was defined in [DS1].

By a path we mean a continuous map from an interval into R^{n}. Given a path γ: [a, b] → R^{n} we define its μ-length as follows. To each partition {t_{j}}_{j=0}^{l} of [a, b] (i.e., a = t_{0} < t_{1} < ⋯ < t_{l} = b) we associate the real number

(3.9)  ∑_{j=1}^{l} δ(γ(t_{j−1}), γ(t_{j})).

Given η > 0 consider the quantity obtained by taking the infimum of (3.9) over all partitions {t_{j}} of [a, b] such that |t_{j} − t_{j−1}| ≤ η for all j. This quantity gets larger as η gets smaller, and its limit as η → 0 is defined to be the μ-length of γ.

(Note that the “lim sup” in the definition of the ω -length of a path in [DS1] should have been a “lim inf ”, in order to be consistent with the usual definition of Hausdorff measures. However, it is not hard to show that, in the end, this change does not really matter.)

It is easy to check that this definition of the μ-length of a path has many of the normal properties. For example, if γ_{1}: [a, b] → R^{n} and γ_{2}: [b, c] → R^{n} are two paths with γ_{1}(b) = γ_{2}(b), and if γ: [a, c] → R^{n} is obtained by joining γ_{1} to γ_{2} in the obvious way, then

(3.10)  μ-length(γ) = μ-length(γ_{1}) + μ-length(γ_{2}).

The deduction of this fact from the definition of the μ-length is straightforward.

Having defined the μ-length of a path we can define the associated geodesic distance in the obvious way, i.e.,

(3.11)  δ_{g}(x, y) = inf {μ-length(γ) : γ is a path that joins x to y} .

(Here the subscript “g ” standard for “geodesic”.)

Proposition 3.12. Let μ be a doubling measure on R^{n}.

(a) μ satisfies (1.8) if and only if there is a constant C > 0 such that

(3.13)  C^{−1}δ(x, y) ≤ δ_{g}(x, y) ≤ Cδ(x, y)

for all x, y ∈ R^{n}.

(b) Suppose that μ is absolutely continuous and that its density is an A_{∞} weight. Then there is a C > 0 so that

(3.14)  δ_{g}(x, y) ≤ Cδ(x, y)

for all x, y ∈ R^{n}.

Let us begin with the “only if” portion of (a). It is easy to check from the definitions that the first inequality in (3.13) holds when μ satisfies (1.8) and hence (3.2). The second inequality in (3.13) follows from (b) and Proposition 3.4.

Conversely, (3.13) imples (1.8), because the geodesic distance always satisfies the triangle inequality.

It remains to prove (b). Let ω denote the density of μ, and fix x, y ∈ R^{n}. We need to find a path that joins x to y whose μ-length we can control. Set z = \frac{1}{2}(x + y) and R = |x − y|, and let H denote the hyperplane through z that is orthogonal to the line segment that joins x to y. For each u ∈ B(z, R) ∩ H let γ_{u}: [0, R] → R^{n} be the path such that γ_{u}(0) = x, γ_{u}(R/2) = u, γ_{u}(R) = y, and γ̇_{u} is constant on (0, R/2) and on (R/2, R). To prove (3.14) it suffices to show that

μ-length(γ_{u}) ≤ Cδ(x, y)

for some u ∈ B(z, R) ∩ H. To do this we shall estimate the average of the left hand side.

For each t > 0 define ω_{t} by (3.7). It is not hard to check that if {s_{j}}_{j=0}^{l} is a partition of [0, R] with t ≤ |s_{j} − s_{j−1}| ≤ 2t for all j, then

∑_{j=1}^{l} δ(γ_{u}(s_{j}), γ_{u}(s_{j−1})) ≤ C ∑_{j=1}^{l} ∫_{s_{j−1}}^{s_{j}} ω_{t}(γ_{u}(s))^{1/n}ds ≤ C ∫_{0}^{R} ω_{t}(γ_{u}(s))^{1/n}ds.

(Note that |γ̇_{u}| ≈ 1.) Using this and the definition of the μ-length of a path it is easy to see that

μ-length(γ_{u}) ≤ C lim inf_{t→0} ∫_{0}^{R} ω_{t}(γ_{u}(s))^{1/n}ds.

Hence

(3.15)

δ_{g}(x, y) ≤ C ∫_{B(z,R)∩H} ( lim inf_{t→0} ∫_{0}^{R} ω_{t}(γ_{u}(s))^{1/n}ds) du

≤ C lim inf_{t→0} ∫_{B(z,R)∩H} ∫_{0}^{R} ω_{t}(γ_{u}(s))^{1/n}ds du.

using Fatou’s lemma. Standard computations permit us to bound this last expression by

(3.16)  C lim inf_{t→0} ∫_{B(z,10R)} ω_{t}(q)^{1/n}L(x, y, q) dq,

where L(x, y, q) = max (|x − q|^{1−n}, |y − q|^{1−n}). (The iterated integral on the right side of (3.15) can be split into two pieces corresponding to s in [0, R/2] and [R/2, R], each of which is basically an integral in polar co-ordinates centered at x or y.)

To control (3.16) we use Fact (e) in Section 2, that ω satisfies a reverse Hölder inequality like (2.5). Let p > 1 be the corresponding exponent, as in (2.5), and let r be the exponent conjugate to np, so that r < n/(n − 1). Then, using Hölder’s inequality, we get that (3.16) is dominated by

CR^{n} lim inf_{t→0}( ∫_{B(z,10R)} ω_{t}^{p})^{1/(np)}( ∫_{B(z,10R)} L(x, y, q)^{r}dq)^{1/r}

≤ CR^{n}( ∫_{B(z,10R)} ω)^{1/n}R^{1−n}

≤ Cμ(B(z, 10R))^{1/n} ≤ Cδ(x, y).

Combining this with (3.15) gives (3.14). This proves Proposition 3.12.

Before leaving this section we record the following useful criterion for being strong A_{∞}.

Lemma 3.17. Assume that u ∈ A_{1} , ω is a strong A_{∞} weight, and that r is a positive real number. If u^{r}ω ∈ A_{∞}, then u^{r}ω is a strong A_{∞} weight.

Let μ, μ̃ be the doubling measures that correspond to ω, u^{r}ω, respectively, and let δ(·, ·) and δ̃(·, ·) be the associated quasidistances. Let x_{1}, …, x_{k} ∈ R^{n} be given, with x_{j} ∈ B(x_{1}, 2|x_{k} − x_{1}|) for all j. According to Lemma 3.3 it is enough to show that

(3.18)  δ̃(x_{1}, x_{k}) ≤ C ∑_{j=1}^{k−1} δ̃(x_{j}, x_{j+1}).

Set B = B_{x_{1},x_{k}} and B_{j} = B_{x_{j},x_{j+1}} , and notice that B_{j} ⊆ 100B for all j. By definitions we have

δ̃(x_{j}, x_{j+1}) = ( ∫_{B_{j}} ω̃)^{1/n} ≥ ( essinf_{B_{j}} u^{r/n})( ∫_{B_{j}} ω)^{1/n} ≥ ( essinf_{100B} u^{r/n})δ(x_{j}, x_{j+1}),

and therefore

(3.19)

∑_{j=1}^{k−1} δ̃(x_{j}, x_{j−1}) ≥ ( essinf_{100B} u^{r/n}) ∑_{j=1}^{k−1} δ(x_{j}, x_{j+1})

≥ C^{−1}( essinf_{100B} u^{r/n})δ(x_{1}, x_{k})

= C^{−1}( essinf_{100B} u^{r/n})( ∫_{B} ω)^{1/n}.

In the second inequality the assumption that ω is a strong A_{∞} weight is used. From Jensen and the assumption that u is an A_{1} weight we have

C ∫_{B} ω ≥ CR^{n} ∫_{B} ω ≥ CR^{n} exp (∫_{B} log ω)

and

C essinf_{100B} u ≥ exp (∫_{B} log u).

In combination with (3.19) this yields (with ω̃ = u^{r}ω)

∑_{j=1}^{k−1} δ̃(x_{j}, x_{j+1}) ≥ C^{−1}R exp (\frac{1}{n} ∫_{B} log ω̃) ≥ C^{−1}R(∫_{B} ω̃)^{1/n} ≥ C^{−1}δ̃(x_{1}, x_{k}).

The second inequality uses Fact (g) from Section 2 applied to ω̃. This gives (3.18), which is what we wanted.

4. Examples and counterexamples

Roughly speaking, the difference between A_{∞} and strong A_{∞} weights is that the latter must satisfy some additional (rather subtle) constraints on how they can be small. Here are some simple examples to illustrate this point.

Examples. (a) If ω is an A_{1} weight, then it is a strong A_{∞} weight. This can be derived easily from Lemma 3.3, or it can be viewed as a special case of Lemma 3.17. The point is that the A_{1} condition prevents ω from ever being small.

(b) Suppose that ω is a continuous nonnegative function on R^{n}. Then ω cannot be a strong A_{∞} weight if it vanishes on a rectifiable curve. (Otherwise the curve would have μ-length equal to zero, μ = ωdx, contrary to Proposition 3.12(a).)

(c) If ω(x) = |x_{1}|^{t}, x = (x_{1}, …, x_{n}), then ω is an A_{∞} weight when t > −1, but it is a strong A_{∞} weight only when −1 < t ≤ 0.

(d) If ω(x) = |x|^{p}, then ω is both A_{∞} and strong A_{∞} exactly when p > −n. Thus a strong A_{∞} weight can vanish to large order at a single point.

(e) Let Z be a closed set in R^{n} that has measure zero, and let p > 0 be given. A sufficient condition for

ω(x) = dist(x, Z)^{p}

to be an A_{∞} weight is that Z should be uniformly thin, which means that there is an a > 0 so that for each x ∈ Z and R > 0 there is a y ∈ B(x, R) such that B(y, aR)∩Z = ∅. This condition is not adequate to ensure that ω be a strong A_{∞} weight. (Consider Z = a line in R^{2} .) A sufficient condition for ω to be strong A_{∞} is that Z should be uniformly scattered, which means that there is a b > 0 so that if x and y lie in Z and if E is a closed connected set that contains them, then there is a ball B centered on E with radius b|x − y| such that B ∩ Z = ∅.

It is not so difficult to build Cantor sets in R^{n} that are uniformly scattered and that have Hausdorff dimension as close to n as you like. Thus strong A_{∞} weights can vanish on rather large sets, and to large order.

Although the uniformly scattered condition is convenient for producing examples of strong A_{∞} weights, it is by no means essential. Strong A_{∞} weights can vanish on nontrivial connected sets, as we shall soon see. These connected sets must be pretty crooked, though, because they must not contain a rectifiable arc.

Let us consider now some natural questions about strong A_{∞} weights which are motivated by classical results for A_{∞} weights.

Question 4.1. If ω is a strong A_{∞} weight, is the same true of ω^{s}, 0 < s < 1? Question 4.2. If ω is a strong A_{∞} weight, must there be a p > 1 such that

ω^{p} is also strong A_{∞}?

In order to state the remaining questions we need a definition.

Definition 4.3. ω = uv is called an A_{1} factorization of ω if u, v^{−t} ∈ A_{1} for some t ≥ 0.

As mentioned in Fact (h) of Section 2, every A_{∞} weight admits an A_{1} factorization, and the existence of such a factorization implies that ω ∈ A_{∞}.

Question 4.4. Does every strong A_{∞} weight ω admit an A_{1} factorization ω = uv such that v is also a strong A_{∞} weight?

Question 4.5. Does every strong A_{∞} weight ω admit an A_{1} factorization ω = uv such that there is an ε > 0 so that u^{1−ε}v^{1+ε} is also a strong A_{∞} weight?

In other words, this last question asks whether every strong A_{∞} weight can be made smaller, in a certain way, and still be strong A_{∞}.

The answers to each of these questions is no. The counterexamples presented below were obtained jointly with Guy David. They have the additional feature that they arise as the Jacobians of quasiconformal mappings in the plane.

Let us begin with the counterexample to Question 4.1. Let λ be a completely singular doubling measure on R, and define h: R → R by

h(x) = ∫_{0}^{x} dλ.

Let H be a Beurling–Ahlfors extension of h to a quasiconformal map of the complex plane onto itself. (See [BA] for an example of such a λ as well as the definition of H.) Let ω_{1} be the Jacobian of H. Thus ω_{1} is a strong A_{∞} weight (because it is the Jacobian of a q.c. mapping), but ω_{1}^{s} is not a strong A_{∞} weight for any s ∈ (0, 1), as we now show.

From the definition of H it follows that H is C^{1} away from R and that

|∇H(z)| ≈ \frac{1}{|y|} ∫_{x}^{x+|y|}dλ, z = x + iy,

for all z ∈ C with y ≠ 0. Because λ is completely singular we have that

(4.6)  lim_{y→0} |∇H(x + iy)| = 0 a.e..

Fix s ∈ (0, 1), and let μ_{1,s} be the doubling measure on C associated to ω_{1}^{s}. To prove that ω_{1}^{s} is not strong A_{∞} it suffices to show that

(4.7)  lim_{t→0} μ_{1,s}-length(γ_{t}) = 0,

where γ_{t}: [0, 1] → C is the path defined by γ_{t}(u) = u + it.

Because ω_{1} is continuous on C \ R we have

μ_{1,s}-length(γ_{t}) = ∫_{0}^{1} ω_{1}(u + it)^{s/2}du ≤ ∫_{0}^{1} |∇H(u + it)|^{s}du

when t ≠ 0. For 0 < |t| ≤ 1 we have

∫_{0}^{1} |∇H(u + it)| du ≤ C ∫_{0}^{2} dλ,

so that |∇H(u + it)|^{s} lies in L^{1/s}([0, 1]) as a function of u, with norm that is uniformly bounded in t, 0 < |t| < 1. Combining this with (4.6) and standard arguments we get that

lim_{t→0} ∫_{0}^{1} |∇H(u + it)|^{s}du = 0,

which implies (4.7).

The counterexample for Question 4.2 will be obtained through sort of a dual construction. This time the starting point will be a quasisymmetric embedding f: R → C whose image is very much not rectifiable (e.g., all nontrivial subarcs will have infinite length). Recall that f: R → C is said to be quasisymmetric if there is a C > 0 so that

|f(u) − f(v)| ≤ C|f(u) − f(w)|

whenever u, v, w ∈ R satisfy |u − v| ≤ |u − w|. Instead of the Beurling–Ahlfors theorem we shall employ the following result of Tukia [T].

Theorem 4.8. If f: R → C is quasisymmetric, then there is a quasiconformal mapping F: C → C which extends f, is C^{1} on C \ R, and satisfies

(4.9)  C^{−1}|∇F (x + iy)| ≤ \frac{|f(x + y) − f(x − y)|}{|y|} ≤ C|∇F (x + iy)|

for some C > 0 and all x + iy ∈ C \ R.

We want to apply this extension theorem to an f whose image is far from rectifiable, e.g., an f whose image is the Van Koch snowflake. The conditions described in the next lemma will be adequate for our purposes.

Given f: R → C and an interval I ⊆ R, I = [a, b], set

Δ_{I}f = \frac{|f(a) − f(b)|}{|a − b|}.

Lemma 4.10. There exists a quasisymmetric embedding f: R → C such that Δ_{I}f → ∞ uniformly as |I| → 0 and

Δ_{I}f ≥ C^{−1}Δ_{J}f

whenever I ⊆ J, for some C > 0.

It is not difficult to build such an f by standard methods. For instance you can take f to be the parameterization of a self-similar quasicircle with Hausdorff dimension larger than 1. You can take f to be “self-similar” too.

Let f be as in Lemma 4.10, and let F be the extension of f promised by Theorem 4.8. Set G = F^{−1} , and let ω_{2} be the Jacobian of G, so that ω_{2} is a strong A_{∞} weight. Fix p > 1, and consider ω_{2}^{p}. If ω_{2}^{p} is not an A_{∞} weight, then we do not have to do anything. Suppose that it is an A_{∞} weight, and let μ_{2,p} denote the corresponding doubling measure. For each t > 0 define α_{t}: [0, 1] → C by α_{t}(x) = F (x + it). If we can show that

(4.11)  lim_{t→0} μ_{2,p}-length(α_{t}) = 0,

then it follows that ω_{2}^{p} is not a strong A_{∞} weight.

By definitions ω_{2} is continuous on C \ f(R) and so

μ_{2,p}-length(α_{t}) = ∫_{0}^{1} ω_{2}(α_{t}(u))^{p/2}|α′_{t}(u)| du

when t > 0. Set

ε_{t} = sup {ω_{2}(α_{t}(u)) : 0 ≤ u ≤ 1},

so that

μ_{2,p}-length(α_{t}) ≤ ε_{t}^{(p−1)/2} ∫_{0}^{1} ω_{2}(α_{t}(u))^{1/2}|α′_{t}(u)| du

≤ ε_{t}^{(p−1)/2} ∫_{0}^{1} ω_{2}(α_{t}(u))^{1/2}|∇F (u + it)| du

≤ Cε_{t}^{(p−1)/2}.

For this last inequality we have used the simple fact that

(4.12)  ω_{2}(z) ≈ |∇F (G(z))|^{−2}.

This fact together with (4.9) and the information that Δ_{I}f → ∞ uniformly as |I| → 0 also implies that ε_{t} → 0 as t → 0. This gives (4.11).

Notice that ω_{2} is continuous on C and vanishes precisely on the connected set f(R).

The counterexample for Question 4.4 is the same weight ω_{1} used to dispatch Question 4.1. Indeed, suppose that ω_{1} admits an A_{1} factorization ω_{1} = uv with v a strong A_{∞} weight. Then this contradicts Lemma 3.17, because v is a strong A_{∞} weight, ω_{1}^{s} = (u^{s}v^{s−1})v is an A_{∞} weight, and u^{s}v^{s−1} can be written as a positive power of an A_{1} weight when s < 1, but ω_{1}^{s} is not a strong A_{∞} weight when 0 < s < 1.

Similarly, the counterexample for Question 4.2 also provides a counterexample to Question 4.5. Let ω_{2} be as above, and assume that ω_{2} admits an A_{1} factorization ω_{2} = uv such that u^{1−ε}v^{1+ε} is a strong A_{∞} weight. To get a contradiction we will use the following.

Lemma 4.13. Under these conditions u^{1−η}v^{1+η} is a strong A_{∞} weight for all η ∈ (0, ε).

Set ω_{2,η} = u^{1−η}v^{1+η}. Then
ω_{2,η} = ω_{2,0}^{1−η/ε}ω_{2,ε}^{η/ε},

and so ω_{2,η} is an A_{∞} weight when 0 < η < ε, because of Fact (d) in Section 2. On the other hand,

ω_{2,η} = (u^{ε−η}v^{η−ε})ω_{2,ε},

and so Lemma 3.17 implies that ω_{2,η} is a strong A_{∞} weight when 0 < η < ε, because u^{ε−η}v^{η−ε} is a positive power of an A_{1} weight in that case. This proves Lemma 4.13.

Choose η ∈ (0, ε) small enough so that ω_{2}^{1+η} is an A_{∞} weight. This is possible, because of Fact (f) from Section 2. Then Lemmas 3.17 and 4.13 imply that ω_{2}^{1+η} is also a strong A_{∞} weight, because

ω_{2}^{1+η} = u^{1+η}v^{1+η} = u^{2η}(u^{1−η}v^{1+η}).

However, we already know that ω_{2}^{p} is not a strong A_{∞} weight for any p > 1. Therefore ω_{2} is a counterexample for Question 4.5.

Remark 4.14. It can be shown that ω_{2}^{−t} ∈ A_{1} for some t > 0. This can be derived from (4.12) and the fact that |∇F |^{2} ∈ A_{1} . The latter can itself be proved using (4.9) and Lemma 4.10.

5. The statement of the main result

Definition 5.1. ω is a stronger A_{∞} weight if it is a strong A_{∞} weight and if it admits an A_{1} factorization (see Definition 4.3) ω = uv such that u^{1−ε}v^{1+ε} is a strong A_{∞} weight for some ε > 0.

Thus, in order for a strong A_{∞} weight to be a stronger A_{∞} weight, it has to be possible to make it substantially smaller (“more degenerate”) without destroying the strong A_{∞} property. The strong A_{∞} weights which are not stronger A_{∞} weights should be thought of as living at the boundary of the set of all strong A_{∞} weights. This is made more precise by Example (d) below. In other words, the set of stronger A_{∞} weights can be viewed as being a kind of “interior” of the space of strong A_{∞} weights.

Theorem 5.2. If ω is a stronger A_{∞} weight, then there exist N > n, C > 0, and f: R^{n} → R^{N} that satisfies (1.6), with μ = ωdx.

The proof of this theorem is given in Sections 6–10. The proof is somewhat complicated, and the reader might find it helpful to consider the important special case of A_{1} weights, which is much simpler.

Although every A_{∞} weight admits an A_{1} factorization, not every strong A_{∞} weight is a stronger A_{∞} weight, as we saw in Section 4. (Note that the conclusion of Theorem 5.2 does hold for that example—i.e., ω_{2} —from Section 4.) Still, there are plenty of stronger A_{∞} weights, as the following examples indicate.

Examples.

(a) Every A_{1} weight is a stronger A_{∞} weight.

(b) The examples of strong A_{∞} weights discussed in (d) and (e) of Section 4 are also stronger A_{∞} weights. This is not hard to check, using the observation that ω^{−t} ∈ A_{1} for some t > 0 when ω is as in (e) and also when ω is as in (d) and p ≥ 0. (If p < 0 then ω is already an A_{1} weight.)

(c) Suppose that ω^{−t} ∈ A_{1} for some t > 0. Then ω is a stronger A_{∞} weight if ω^{1+ε} is a strong A_{∞} weight. This uses the fact that if v^{−t} ∈ A_{1} for some t > 0 and v is a strong A_{∞} weight, then v^{s} is a strong A_{∞} weight for all s ∈ (0, 1). This simple observation can be derived from Lemma 3.17, for instance.

(d) Every strong A_{∞} weight can be approximated by a stronger A_{∞} weight, in the following sense. Suppose that ω is a strong A_{∞} weight, and that ω = uv is an A_{1} factorization of ω. Consider ω_{ε} = u^{1/(1−ε)}v^{1/(1+ε)} , ε > 0. This lies in A_{∞} when ε > 0 is small enough, by Facts (f) and (h) in Section 2. When ω_{ε} is an A_{∞} weight it is also strong A_{∞}, because of Lemma 3.17, and in that case it is clearly stronger A_{∞} as well because ω is strong A_{∞}.

There are of course a number of issues left unresolved by Theorem 5.2. One of the most basic questions is whether the strong A_{∞} condition alone is sufficient to imply the conclusions of Theorem 5.2. I do not have a strong opinion on this, but I am willing to go out on a limb for the analogous issue for Question 1.5.

Conjecture 5.3. There is a strong A_{∞} weight ω on R^{2} for which there does not exist a mapping f: R^{2} → R^{2} that satisfies (1.6).

Persons of a more reckless (or insightful?) nature might be willing to speculate, in print, about the situation in which the stronger A_{∞} (or even the A_{1} ) condition is assumed.

Unfortunately there do not seem to be any good methods available for proving the nonexistence of a mapping f as in Conjecture 5.3. One way to try to generate necessary conditions on ω for the existence of such a mapping f is to look for analytical results on R^{2} that would have to have a counterpart for ω if f were to exist. For example, you could take the classical Sobolev and Poincaré inequalities. These do not work, because their counterparts for ω are true for all strong A_{∞} weights (in all dimensions), by the results of [DS1].

Another problem which is connected to these issues is the following.

Question 5.4. Is Theorem 5.2 still true if we demand that N depend on n alone?

The proof given below has N depending on ω in a substantial way. This is true even when ω is an A_{1} weight; in that case the proof is substantially simpler, but N still depends on ω.

Next we discuss ω -regular mappings and their relationship with the topics in this paper.

A continuous mapping f: R^{n} → R^{N} is said to be ω -regular for some A_{∞} weight ω if there is a C > 0 so that: (5.5) f has locally integrable first (distributional) derivatives, and

|∇f(x)| ≤ Cω(x)^{1/n} a.e.; and

(5.6) ∫_{f^{−1}(B)} ω(x) dx ≤ CR^{n} for every ball B with radius R in R^{N}.

This class of mappings arose in [D] in connection with the behavior of singular integral operators on their images. As pointed out in [DS1], under the assumption that ω is an A_{∞} weight it can be shown that (5.5) is equivalent to the requirement that there be a C > 0 so that

(5.7)  |f(x) − f(y)| ≤ Cδ(x, y)  for all x, y ∈ R^{n}.

The proof of this equivalence is pretty straightforward, although perhaps a little bit technical. For passing from (5.5) to (5.7) it is useful to remember Proposition 3.12(b), for instance.

This reformulation of (5.7) makes it easy to see that f: R^{n} → R^{N} is ω -regular if it satisfies (1.6) with μ = ω dx. The converse is not true, because ω -regular mappings need not even be injective. However, if f is ω -regular, then f^{−1}(p) has only a bounded number of elements for all p ∈ R^{N}. (This is not hard to check, by estimating the ω -mass of f^{−1}(B(p, r)) as r → 0.)

It is natural to hope that it might be easier to characterize the A_{∞} weights ω on R^{n} for which there is an ω -regular mapping f: R^{n} → R^{N} for some N than to answer Question 1.7 completely, but so far nothing has occurred to reinforce this hope. A necessary condition for the existence of an ω -regular mapping is that ω be strongly A_{∞} (see [DS1]), but there is not a better sufficient condition known than the one provided by Theorem 5.2.

The case of A_{1} weights is of particular interest in the context of ω -regular mappings, because of the following result from [DS2]. Suppose that f: R^{n} → R^{N} is an ω -regular mapping for some A_{∞} weight ω. Then there is an A_{1} weight ω′ and an ω′ -regular mapping g: R^{n} → R^{N+1} such that

f(R^{n}) ⊆ g(R^{n}).

For this we identify R^{N} with a subset of R^{N+1} in the obvious way.

6. Conceptual preliminaries

Lemma 6.1. Let μ be a doubling measure on R^{n}. Then μ satisfies (1.8) if and only if there is a constant C > 0 so that for each x, y ∈ R^{n} there is a function f: R^{n} → R with the property that

(6.2)  |f(z) − f(w)| ≤ Cδ(z, w)  for all z, w ∈ R^{n}, and
(6.3)  |f(x) − f(y)| ≥ C^{−1}δ(x, y).

It is important to remember here that f is allowed to depend on x and y. The conclusion of Theorem 5.2 concerns the existence of a single R^{N} -valued function f that works for all pairs of points x, y in R^{n}.

The proof of Lemma 6.1 is easy. Suppose first that (1.8) holds. Let δ′(·, ·) be as in (1.8), and let x, y ∈ R^{n} be given. Define f: R^{n} → R by f(z) = δ′(x, z). Then f satisfies (6.2) and (6.3) because of (1.8) and the fact that δ′(·, ·) satisfies the triangle inequality.

Conversely, suppose that for each x, y ∈ R^{n} we can find f as above. Define δ′(·, ·) by

δ′(z, w) = sup {|g(z) − g(w)| : g is a real-valued function on R^{n}such that
|g(a) − g(b)| ≤ δ(a, b) for all a, b ∈ R^{n}}.

Then δ′(·, ·) is nonnegative, symmetric, and satisfies the triangle inequality and also δ′(z, w) ≤ δ(z, w) for all z and w. The existence of the good f ’s implies that

δ(x, y) ≤ Cδ′(x, y)

for all x, y ∈ R^{n} as well, so that (1.8) is true. This proves the lemma.

It is not hard to improve Lemma 6.1 by showing that if μ satisfies (1.8), then there exist C, m > 0 so that for each t > 0 there is a mapping f: R^{n} → R^{m} so that (6.2) holds, and also (6.3) for all x, y ∈ R^{n} such that t < |x − y| < 2t. In other words, instead of simply separating well a given pair of points, we can separate any pair of points at a given scale.

This is still not good enough for Theorem 5.2; for that we must deal with all scales at the same time. There does not seem to be any simple way to combine the mappings that work for the various individual scales. Similar difficulties arise when you try to get L^{∞} estimates for \overline{∂} problems, and Peter Jones’ work in that subject provided part of the inspiration for the methods employed in the next sections.

There is a variant of Lemma 6.1 that will be useful later for constructing partitions of unity that are adapted to a given strong A_{∞} weight, and which we record now.

Lemma 6.4. Suppose that μ satisfies (1.8). Then there is a C > 0 so that for each cube Q in R^{n} there is a function θ: R^{n} → R such that 0 ≤ θ ≤ 1, θ = 1 on 2Q, θ = 0 on R^{n} \ 3Q, and

|θ(x) − θ(y)| ≤ Cδ(x, y)μ(Q)^{−1/n} for all x, y ∈ R^{n}.

Let δ′(·, ·) be as in (1.8), and consider

δ′(x, R^{n} \ 3Q) = inf {δ′(x, y) : y ∈ R^{n} \ 3Q}.

Standard computations imply that this function satisfies (6.2), because δ′(x, y) satisfies (6.2) as a function of x for each y ∈ R^{n}. Define θ by

θ(x) = min (1, {δ′(2Q, R^{n} \ 3Q)^{−1}δ′(x, R^{n} \ 3Q)}),

where, as usual,

δ′ (2Q, R^{n} \ 3Q) = inf{δ′(x, R^{n} \ 3Q) : x ∈ 2Q}.

Then it is easy to verify that θ has the required properties.

7. Technical preliminaries

We shall use the following notations from now on.

Let ω be the given stronger A_{∞} weight on R^{n}, and let ω = uv be as in Definition 5.1. Set

(7.1)  ω̃ = u^{1−ε}v^{1+ε},

where ε > 0 is as in Definition 5.1, so that ω̃ is a strong A_{∞} weight. (If ω is simply an A_{1} weight, then we can take ω̃ ≡ 1.) As in Lemma 4.13 we can shrink ε in (7.1) and still have a strong A_{∞} weight, and so we can (and do) assume that ε > 0 is small enough so that

α = ω/ω̃ = u^{ε}v^{−ε}

is an A_{1} weight. Let μ and μ̃ be the measures on R^{n} corresponding to the weights ω and ω̃, and let δ(·, ·) and δ̃(·, ·) be as in (1.2).

Let L > 0 be large, to be chosen later, depending on ε, n, and the relevant constants that arise in the various conditions on ω, ω̃, u, v, and α. In what follows all of the constants “C ” that occur are permitted to depend on any of the constants just mentioned except L; the presence of any dependence on L will always be made clear.

For each j ∈ Z set

(7.2)  Ω_{j} = {x ∈ R^{n} : α^{∗}(x) > L^{jεn}},

where α^{∗} denotes the Hardy–Littlewood maximal function of α, i.e.,

α^{∗}(x) = sup_{B∋x} ∫_{B} α,

where the supremum is taken over all balls that contain x. In order to build the mapping f promised in Theorem 5.2 we shall first define some auxiliary mappings in the next section which, roughly speaking, do a good job of separating pairs of points that lie in a given Whitney cube of some Ω_{j}. In this section we derive some estimates concerning the various weights and the Ω_{j} ’s.

Define j_{0} by

j_{0} = inf{j : Ω_{j+1} ≠ R^{n}} = sup{j : Ω_{j} = R^{n}}.

Clearly, j_{0} < +∞, but it could happen that j_{0} = −∞. The j ’s for which j < j_{0} will not be playing a role in what we do.

Set

(7.3)  d_{j}(x) = dist(x, R^{n} \ Ω_{j}),  x ∈ R^{n},

for j > j_{0} . Here “dist” refers to the Euclidian distance. Let δ_{j}(x) and δ̃_{j}(x) be the counterparts of (7.3) associated to δ(·, ·) and δ̃(·, ·), i.e.,

(7.4)  δ_{j}(x) = δ(x, R^{n} \ Ω_{j}) = inf {δ(x, y) : y ∈ R^{n} \ Ω_{j}}

and

(7.5)  δ̃_{j}(x) = δ̃(x, R^{n} \ Ω_{j})

for x ∈ R^{n} and j > j_{0} . When j ≤ j_{0} it is convenient to define d_{j}, δ_{j}, and δ̃_{j} to all be identically equal to +∞.

One of the important features of the Ω_{j} ’s is that Ω_{j+1} is sparse inside of Ω_{j} when j > j_{0} and L is large. This is made precise by the following.

Lemma 7.6. There exist C, η > 0 so that

(7.7)  d_{j+1}(x) ≤ CL^{−ε}d_{j}(x),

(7.8)  δ_{j+1}(x) ≤ CL^{−η}δ_{j}(x),

and

(7.9)  δ̃_{j+1}(x) ≤ CL^{−η}δ̃_{j}(x)

for all x ∈ R^{n} and j > j_{0} .

The first inequality can be derived from the definitions as follows. Because α is an A_{1} weight we have

(7.10)  α ≥ C^{−1}α^{∗} ≥ C^{−1}L^{jεn} a.e. on Ω^{j}.

On the other hand

(7.11)  ∫_{B(x,R)} α ≤ L^{jεn}  when R ≥ d_{j}(x)

by definition of Ω_{j} and α^{∗}. Hence

C^{−1}L^{(j+1)εn}d_{j+1}(x)^{n} ≤ ∫_{B(x,d_{j+1}(x))} α ≤ ∫_{B(x,d_{j}(x))} α ≤ CL^{jεn}d_{j}(x)^{n},

(by (7.11)). This implies (7.7).

The other two estimates—(7.8) and (7.9)—follow from (7.7) and the observation that if σ is a doubling measure and B_{1} , B_{2} are two balls with B_{1} ⊆ B_{2} , then

\frac{σ(B_{1})}{σ(B_{2})} ≤ C(\frac{|B_{1}|}{|B_{2}|})^{a}

for some C, a > 0. This is easily checked. (Notice that we are only asking that this inequality hold for balls. That is why you only need a doubling condition on σ, and not an A_{∞} condition.)

Next we record a lemma on the relationship between μ and μ̃ in terms of the Ω_{j} ’s.

Lemma 7.12. Let B be a ball in R^{n}.

(a) If B ⊆ Ω_{j}, then

μ(B) ≥ C^{−1}L^{jεn}μ̃(B).

(b) If B intersects R^{n} \ Ω_{j}, then

μ(B) ≤ CL^{jεn}μ̃(B).

Part (a) follows immediately from (7.10) and the various definitions. To prove (b) we use Fact (g) in Section 2 to get that

∫_{B} ω ≤ C(∫_{B} ω̃)(∫_{B} α).

The definition of Ω_{j} implies that

(7.13)  ∫_{B} α ≤ L^{jεn}  when B ∩ (R^{n} \ Ω_{j}) ≠ ∅,

and from here Part (b) follows directly.

The last topic that we take up in this section deals, roughly speaking, with controlling ω on Ω_{j} \ Ω_{j+1} .

Lemma 7.14. Let B be a ball and Q be a cube. Assume that

(7.15)  Q ⊆ Ω_{j}  and  Q \ Ω_{j+1} ≠ ∅

and that

(7.16)  B ⊆ 100Q  and  B intersects Q \ Ω_{j+1}.

Then

(7.17)  | ∫_{B} log ω − ∫_{Q} log ω |≤ log L^{n} + C.

The point here is that B could be much smaller than Q, so that (7.17) would not follow from the doubling condition on μ. Notice in particular that (7.17) implies that

(7.18)  | log ω(x) − ∫_{Q} log ω |≤ log L^{n} + C a.e. on Q \ Ω_{j+1}.

We assume for the rest of this section that B and Q are as in Lemma 7.14. Observe first that

(7.19)  α ≥ C^{−1}L^{jεn} a.e. on Q

and

(7.20)  ∫_{Q} α ≤ CL^{(j+1)εn}.

These inequalities follow from (7.15) and the definition of Ω_{j}. With the aid of Jensen they imply that

(7.21)  log L^{jεn} − C ≤ ∫_{Q} log α ≤ log L^{(j+1)εn} + C.

Similarly, (7.16) implies

(7.22)  ∫_{B} log α ≤ log (∫_{B} α) ≤ L^{(j+1)εn} + C.

Next we consider u^{ε} and v^{−ε}. Because they are both A_{1} weights we have that

(7.23)

(a)  ∫_{Q} log u^{ε} ≤ essinf_{100Q} (log u^{ε}) + C,

(b)  ∫_{Q} log v^{−ε} ≤ essinf_{100Q} (log v^{−ε}) + C.

Hence, by (7.21)–(7.23),

∫_{B} log u^{ε} + ∫_{B} log v^{−ε} = ∫_{B} log α ≤ log L^{(j+1)εn} + C

≤ log L^{εn} + ∫_{Q} log α + C

= log L^{εn} + ∫_{Q} log u^{ε} + ∫_{Q} log v^{−ε} + C

≤ log L^{εn} + essinf_{100Q} (log u^{ε}) + essinf_{100Q} (log v^{−ε}) + C.

Because B ⊆ 100Q we get from here that

(7.24)

(a)  ∫_{B} log u^{ε} ≤ log L^{εn} + essinf_{100Q} (log u^{ε}) + C and

(b)  ∫_{B} log v^{−ε} ≤ log L^{εn} + essinf_{100Q} (log v^{−ε}) + C.

Now let us look at ω. We have, by (7.24)(a) and (7.23)(b),

∫_{B} log ω = ∫_{B} log u + ∫_{B} log v

≤ log L^{n} + essinf_{100Q} log u + esssup_{100Q} log v + C

≤ log L^{n} + ∫_{Q} log u + ∫_{Q} log v + C

≤ log L^{n} + ∫_{Q} log ω + C.

Similarly,

∫_{B} log ω ≥ − log L^{n} + ∫_{Q} log ω − C,

using (7.24)(b) and (7.23)(a). These estimates imply (7.17), as desired. This proves Lemma 7.14.

8. Building blocks for defining f

In this section we construct the building blocks that will be used to define f (in the next section). To do that we must first set up some Whitney decompositions and partitions of unity.

For each j > j_{0} let {Q_{j,k}} be an enumeration of the maximal dyadic cubes Q contained in Ω_{j} that satisfy

(8.1)  3 diam Q ≤ (10n)^{−10n} inf_{3Q} d_{j}(x).

Thus ∪_{k}\overline{Q}_{j,k} = Ω_{j}, Q_{j,k} and Q_{j,l} have disjoint interiors unless k = l, (8.2) 3 diam Q_{j,k} ≤ (10n)^{−10n} dist (3Q_{j,k}, R^{n} \ Ω_{j}), (8.3) diam Q_{j,k} ≥ C^{−1} dist (3Q_{j,k}, R^{n} \ Ω_{j}), and (8.4) for each j, {10Q_{j,k}}_{k} is a sequence of cubes with bounded overlap.

We can associate partitions of unity to these Whitney decompositions in the usual way, except that we want the bump functions to be Lipschitz relative to δ̃(·, ·). For each j, k let θ_{j,k} be the function promised to Q = Q_{j,k} by Lemma 6.4, except that μ and δ(·, ·) are replaced by μ̃ and δ̃(x, y). Define φ_{j,k}: R^{n} → R by

φ_{j,k}(x) = 0 when x ∉ Ω_{j},

and

φ_{j,k} = θ_{j,k}( ∑_{l} θ_{j,l})^{−1} on Ω_{j}.

Then each φ_{j,k} satisfies

(8.5)  supp φ_{j,k} ⊆ 3Q_{j,k} ⊆ Ω_{j},
(8.6)  0 ≤ φ_{j,k} ≤ 1,  ∑_{k} φ_{j,k} = χ_{Ω_{j}},

and

(8.7)  |φ_{j,k}(x) − φ_{j,k}(y)| ≤ Cδ̃(x, y)μ̃(Q_{j,k})^{−1/n}

for all x, y ∈ R^{n}. Note that

(8.8)  μ̃(Q_{j,k})^{−1/n} ≈ δ̃_{j}(x)  for all x ∈ Q_{j,k},

because of (8.2), (8.3), and the fact that μ̃ is a doubling measure.

It is very important that the Lipschitz condition (8.7) be in terms of δ̃(·, ·) instead of δ(·, ·). This is the sort of thing that will allow us to sum in j without losing control, as we shall have to do. It is for this purpose that we need ω to be a stronger A_{∞} weight, rather than merely strong A_{∞}.

Set

(8.9)  M_{0} = inf {(∫_{B} ω)^{1/n} : B ∩ (R^{n} \ Ω_{j_{0}+1}) ≠ ∅}

and

(8.10)  M_{j,k} = ( ∫_{Q_{jk}} ω)^{1/n}.

Notice that M_{j,k} is comparable to the ratio of the δ(·, ·)-diameter of Q to its Euclidean diameter. Also, M_{0} > 0, as one can show using Lemma 7.14. (See Section 10 for more details, especially the estimates between (10.13) and (10.14).)

The next 3 lemmas concern the previously-heralded building blocks. Let q_{j,k} denote the center of Q_{j,k}.

Lemma 8.11. Assume that j_{0} > −∞. Then there is a continuous mapping f_{0}: R^{n} → R^{n} such that

(8.12)  |f_{0}(x) − M_{0}x| ≤ M_{0}(10n)^{−10n}d_{j_{0}+1}(x),

and

(8.13)  |f_{0}(x) − f_{0}(y)| ≤ CL^{(j_{0}+1)ε}δ̃(x, y)  for all x, y ∈ R^{n}.

Lemma 8.14. For each j > j_{0} and each k there is a continuous mapping h_{j,k}: 3Q_{j,k} → R^{n} such that

(8.15)  |h_{j,k}(x) − M_{j,k}(x − q_{j,k})| ≤ M_{j,k}(10n)^{−10n}d_{j+1}(x)

and

(8.16)  |h_{j,k}(x) − h_{j,k}(y)| ≤ CL^{(j+1)ε+1}δ̃(x, y)

for all x, y ∈ R^{n}.

Lemma 8.17. Let h_{j,k} : 3Q_{j,k} → R^{n} be as in Lemma 8.14, and define f_{j,k}: R^{n} → R^{n} by

\begin{aligned}f_{j,k}(x) & = 0  when x ∈ R^{n} \ 3Q_{j,k}, \\ f_{j,k} & = θ_{j,k}h_{j,k}  on 3Q_{j,k}. \end{aligned}

Then

(8.18)  |f_{j,k}(x)| ≤ Cδ_{j}(x)

and

(8.19)  |f_{j,k}(x) − f_{j,k}(y)| ≤ CL^{(j+1)ε+1}δ̃(x, y)  for all x, y ∈ R^{n}.

Again, for the purposes of a future summation in j it is important that the Lipschitz conditions in (8.13) and (8.19) be in terms of δ̃(·, ·) rather than δ(·, ·).

The rest of this section will be devoted to the proofs of these lemmata.

Let us begin with Lemma 8.11. Suppose that j_{0} > −∞, and define f_{0}: R^{n} → R^{n} by

(8.20)

f_{0}(x) = M_{0}x  when x ∈ R^{n} \ Ω_{j_{0}+1}  and
f_{0} = M_{0} ∑_{k} φ_{j_{0}+1,k} q_{j_{0}+1,k}  on Ω_{j_{0}+1}.

It is easy to see that f_{0} is continuous. We must verify (8.12) and (8.13).

Of course (8.12) is automatic when x ∈ R^{n} \ Ω_{j_{0}+1} , and so we may as well assume that x ∈ Ω_{j_{0}+1} . Because ∑_{k} φ_{j_{0}+1,k}(x) = 1 we have

|f_{0}(x) − M_{0}x| ≤ M_{0} ∑_{k} φ_{j_{0}+1,k}(x)|x − q_{j_{0}+1,k}|.

If φ_{j_{0}+1,k}(x) ≠ 0, then x ∈ 3Q_{j_{0}+1,k}, and so

|f_{0}(x) − M_{0}x| ≤ M_{0} ∑_{k} φ_{j_{0}+1,k}(x) (3 diam Q_{j_{0}+1,k})

≤ M_{0} ∑_{k} φ_{j_{0}+1,k}(x)(10n)^{−10n}d_{j_{0}+1}(x)

≤ M_{0}(10n)^{−10n}d_{j_{0}+1}(x).

In the second inequality we are using the fact that each Q_{j,k} satisfies (8.2). This proves (8.12).

Consider now (8.13). Suppose first that

(8.21)  |x − y| ≥ \frac{1}{10}d_{j_{0}+1}(x).

Then

(8.22)  \begin{aligned}|f_{0}(x) − f_{0}(y)| & ≤ |f_{0}(x) − M_{0}x| + M_{0}|x − y| + |f_{0}(y) − M_{0}y| \\ & ≤ M_{0}(d_{j_{0}+1}(x) + |x − y| + d_{j_{0}+1}(y)) \\ & ≤ 2M_{0}(d_{j_{0}+1}(x) + |x − y|) ≤ CM_{0}|x − y|. \end{aligned}

In the third inequality we have used the fact that

(8.23)  d_{j_{0}+1}(y) ≤ d_{j_{0}+1}(x) + |x − y|.

On the other hand, (8.21) implies that 20B_{x,y} intersects R^{n} \ Ω_{j_{0}+1} , and so from (8.9) and Lemma 7.12(b) we get

(8.24)  \begin{aligned}M_{0}|x − y| & ≤ ( ∫_{20B_{x,y}} ω)^{1/n}|x − y| ≤ C( ∫_{20B_{x,y}} ω)^{1/n} \\ & ≤ CL^{(j_{0}+1)ε}μ̃ (20B_{x,y})^{1/n} ≤ CL^{(j_{0}+1)ε}δ̃(x, y). \end{aligned}

Combining this with (8.22) gives (8.13) when (8.21) holds. Suppose now that

(8.25)  |x − y| < \frac{1}{10}d_{j_{0}+1}(x).

In particular both x and y lie in Ω_{j_{0}+1} . Using (8.20) we get

(8.26)  \begin{aligned}|f_{0}(x) − f_{0}(y)| & = M_{0}| ∑_{k} (φ_{j_{0}+1,k}(x) − φ_{j_{0}+1,k}(y))q_{j_{0}+1,k}| \\ & = M_{0}| ∑_{k} (φ_{j_{0}+1,k}(x) − φ_{j_{0}+1,k}(y)) (q_{j_{0}+1,k} − x) |. \end{aligned}

In order for a term in the sum not to vanish either x or y must lie in 3Q_{j_{0}+1,k}. In either case we must have

(8.27)  3Q_{j_{0}+1,k} ∩ B(x, \frac{1}{10}d_{j_{0}+1}(x)) ≠ ∅,

by (8.25). This can occur for at most a bounded number of k ’s (for a given x). For each k for which (8.27) is true we have that

|x − q_{j_{0}+1,k}| ≤ Cd_{j_{0}+1}(x)

and

|φ_{j_{0}+1,k}(x) − φ_{j_{0}+1,k}(y)| ≤ Cδ̃(x, y)μ̃ (Q_{j_{0}+1,k})^{−1/n} ,

by (8.7). Before inserting these estimates into (8.26) it is convenient to set

B_{x} = B(x, 2d_{j_{0}+1}(x)),

so that

Cμ̃ (Q_{j_{0}+1,k}) ≥ μ̃(B_{x}).

Now combining with (8.26) we get

(8.28)  |f_{0}(x) − f_{0}(y)| ≤ CM_{0}δ̃(x, y)μ̃ (B_{x})^{−1/n} d_{j_{0}+1}(x).

Because B_{x} intersects R^{n} \ Ω_{j_{0}+1} we have from (8.9) and Lemma 7.12(b) that

(8.29)  M_{0}d_{j_{0}+1}(x) ≤ μ (B_{x})^{1/n} ≤ CL^{(j_{0}+1)ε}μ̃ (B_{x})^{1/n} .

This together with (8.28) gives (8.13) when (8.25) holds. This completes the proof of Lemma 8.11.

The proof of Lemma 8.14 is quite similar. Fix j > j_{0} . Define h_{j,k}: 3Q_{j,k} → R^{n} by

h_{j,k}(x) = M_{j,k}(x − q_{j,k}) when x ∈ 3Q_{j,k} \ Ω_{j+1}

and

(8.30)  h_{j,k} = M_{j,k} ∑_{l} φ_{j+1,l} (q_{j+1,l} − q_{j,k})  on 3Q_{j,k} ∩ Ω_{j+1}.

It is easy to check that h_{j,k} is continuous.

The proof of (8.15) is almost identical to that of (8.12), and so we omit it. The proof of (8.16) closely resembles that of (8.13), but there are a couple of changes, and so we sketch the argument.

Suppose that x, y ∈ 3Q_{j,k} satisfy

(8.31)  |x − y| ≥ \frac{1}{10}d_{j+1}(x).

Analogous to (8.22) we have

(8.32)  |h_{j,k}(x) − h_{j,k}(y)| ≤ CM_{j,k}|x − y|.

On the other hand

(8.33)  M_{j,k} ≤ CL^{(j+1)ε+1}( ∫_{20B_{x,y}} ω̃)^{1/n}.

To see this we first observe that

(8.34)  M_{j,k} = ( ∫_{Q_{j,k}} ω)^{1/n} ≤ CL( ∫_{20B_{x,y}} ω)^{1/n}.

This follows from the doubling condition on μ if 20B_{x,y} is not contained in 100Q_{j,k} (so that B_{x,y} is not too small compared to Q_{j,k}), and from Lemma 7.14 and Fact (g) in Section 2 when 20B_{x,y} ⊆ 100Q_{j,k}. Clearly (8.33) follows from (8.34), Lemma 7.12(b), and the fact that 20B_{x,y} intersects R^{n} \ Ω_{j+1} . Combining (8.32) with (8.33) gives (8.16) when (8.31) holds.

Assume now that x, y ∈ 3Q_{j,k} satisfy

|x − y| < \frac{1}{10}d_{j+1}(x),

so that x, y ∈ Ω_{j+1} in particular. Just as in (8.28) we have

|h_{j,k}(x) − h_{j,k}(y)| ≤ CM_{j,k}δ̃(x, y)μ̃(B_{x})^{−1/n}d_{j+1}(x),

where now

B_{x} = B (x, 2d_{j+1}(x)) .

We also have

M_{j,k} ≤ CL^{(j+1)ε+1}(∫_{B_{x}} ω̃)^{1/n}

for essentially the same reasons as for (8.33). These inequalities combine to give (8.16) in this case as well. This proves Lemma 8.14.

It remains to prove Lemma 8.17. Observe first that

(8.35)  |h_{j,k}(x)| ≤ Cμ (Q_{j,k})^{1/n}  when x ∈ 3Q_{j,k}.

This is easily derived from the definition of h_{j,k} and M_{j,k}. With this in hand (8.18) follows directly. To get (8.19) you use Lemma 6.4, (8.16), (8.35), and also the second inequality in the next lemma.

Lemma 8.36. There is a C > 0 so that

C^{−1}L^{jεn}μ̃(Q_{j,k}) ≤ μ(Q_{j,k}) ≤ CL^{jεn}μ̃(Q_{j,k})

for all j, k.

This is an easy consequence of Lemma 7.12, and (8.3), and the doubling condition on μ̃.

9. The definition of f

Basically f is going to be defined by combining the f_{j,k} ’s from Section 8. This is slightly tricky, because we are only allowing f to take values in a finite dimensional space, and so we have to arrange the f_{j,k} ’s into a finite set of piles. We have to be careful about how we do this, in order to make certain that a pair of f_{j,k} ’s from the same pile do not interact too much.

Set N_{2} = N_{0}N_{1} , where N_{0} is a large constant to be chosen later (independently of L), and N_{1} is a large constant that will be selected very soon (depending only on n).

Lemma 9.1. If N_{1} is large enough then we can define a mapping

(9.2)  Q_{j,k} ↦ a_{j,k},  a_{j,k} ∈ {1, 2, …, N_{2}}

in such a way that a_{j,k} = a_{j′,k′} implies:

(i) j ≡ j′ mod N_{0} ;

(ii) k = k′ if also j = j′ and

(9.3)  dist (3Q_{j,k}, 3Q_{j′,k′}) < \frac{1}{2} dist (Q_{j,k} ∪ Q_{j′,k′}, R^{n} \ Ω_{j}) .

The proof of this requires just a small coding argument. Choose N_{1} so that for each j and k the set

{k′ : (9.3) holds with j′ = j}

has less than N_{1} elements. Then the correspondence (9.2) can be defined one cube Q_{j,k} at a time (in any order), and our condition on N_{1} ensures that at each stage there is a choice for the value of a_{j,k} that is consistent with the previous choices and the requirements listed in the lemma.

We are going to take f to be a mapping from R^{n} into R^{N} with N = (N_{2} + 1)n + N_{2} and with f of the form

f = (f_{0}, f_{1}, …, f_{N_{2}}, g_{1}, …, g_{N_{2}}),

where the f_{i} ’s take values in R^{n} and the g_{i} ’s take values in R. Here f_{0} is as in Lemma 8.11 when j_{0} > −∞, and we take f_{0} ≡ 0 otherwise. The remaining f_{i} ’s and g_{i} ’s are defined as follows.

Given i, 1 ≤ i ≤ N_{2} , set

(9.4)  f_{i} = ∑_{{(j,k):a_{j,k}=i}} f_{j,k}

and

(9.5)  g_{i} = ∑_{{(j,k):a_{j,k}=i}} g_{j,k},

where the f_{j,k} ’s are as in Lemma 8.17, and where g_{j,k}: R^{n} → R is defined by

(9.6)  g_{j,k}(x) = L^{jε}δ̃′(x, R^{n} \ 2Q_{j,k}).

Here δ̃′(·, ·) is a distance function on R^{n} that is comparable to δ̃(·, ·), as in (1.8).

In the next section we shall prove that this choice of f works. Before leaving this section we record some estimates on the g_{j,k} ’s.

Lemma 9.7. Each g_{j,k} satisfies

(9.8)  g_{j,k}(x) ≤ Cδ_{j}(x)

and

(9.9)  |g_{j,k}(x) − g_{j,k}(y)| ≤ CL^{jε}δ̃(x, y)  for all x, y ∈ R^{n}.

Also,

(9.10)  g_{j,k}(x) ≥ C^{−1}δ_{j}(x)  when x ∈ \overline{Q_{j,k}}.

It is easy to derive (9.8) and (9.10) from (9.6), Lemma 8.36, and the doubling condition on μ̃. As usual, (9.9) follows from (1.8) and the fact that δ̃′(·, ·) satisfies the triangle inequality.

10. The end of the proof of Theorem 5.2

Lemma 10.1. If L is large enough, then there is a C(L) > 0 so that

(10.2)  |f(x) − f(y)| ≤ C(L)δ(x, y)

whenever x, y ∈ R^{n}.

Consider first f_{0} . We may as well assume that j_{0} > −∞. Thus Ω_{j_{0}} = R^{n}, and so Lemma 7.12(a) implies that

L^{j_{0}ε}δ̃(x, y) ≤ Cδ(x, y)

for all x, y ∈ R^{n}, which in turn implies that the contribution of f_{0} to (10.2) is fine, because of (8.13).

Next we look at the f_{i} ’s and g_{i} ’s, 1 ≤ i ≤ N_{2} . We begin by deriving estimates for

(10.3)  ∑_{j_{0}<j≤j_{1}} ∑_{k} (|f_{j,k}(x) − f_{j,k}(y)| + |g_{j,k}(x) − g_{j,k}(y)|)

and

(10.4)  ∑_{j≥j_{1}} ∑_{k} (|f_{j,k}(x)| + |g_{j,k}(x)|),

where j_{1} is any integer such that j_{1} > j_{0} .

Each term in the sum in (10.3) is bounded by CL^{(j+1)ε+1}δ̃(x, y), by (8.19) and (9.9). Also, for each j there are at most a bounded number of terms in the k sum in (10.3) which are nonzero, because f_{j,k} and g_{j,k} are both supported in 3Q_{j,k}. Hence

(10.3) ≤ ∑_{j_{0}<j≤j_{1}} CL^{(j+1)ε+1}δ̃(x, y) ≤ CL^{(j_{1}+1)ε+1} (1 − L^{−ε})^{−1} δ̃(x, y).

We may as well require L to be so large that

(10.5)  L^{−ε} ≤ \frac{1}{2},

whence

(10.6)  (10.3) ≤ CL^{(j_{1}+1)ε+1}δ̃(x, y).

For (10.4) we notice that each term in the sum is bounded by Cδ_{j}(x), by (8.18) and (9.8), and that for each j there are at most a bounded number of terms in the k sum that do not vanish, because of the conditions on the supports of f_{j,k} and g_{j,k}. Hence

(10.4) ≤ ∑_{j≥j_{1}} Cδ_{j}(x).

From (7.8) it follows that

(10.7)  (10.4) ≤ Cδ_{j_{1}}(x),

at least if L is large enough so that CL^{−η} ≤ \frac{1}{2} .

To finish the proof of Lemma 10.1 it suffices to show that

(10.8)  ∑_{j>j_{0}} ∑_{k} (|f_{j,k}(x) − f_{j,k}(y)| + |g_{j,k}(x) − g_{j,k}(y)|) ≤ C(L)δ(x, y)

for all x, y ∈ R^{n}. Fix x, y ∈ R^{n}. Define j(x, y) to be the largest integer j such that

(10.9)  x, y ∈ Ω_{j}  and  B(x, 10|x − y|) ⊆ Ω_{j}.

Notice that j(x, y) ≥ j_{0} and j(x, y) > −∞, but that j(x, y) = j_{0} is possible when j_{0} > −∞. To prove (10.8) we split the j sum into two pieces, according to whether or not j ≤ j(x, y).

Using (10.6) with j_{1} = j(x, y) we get

\begin{aligned}∑_{j_{0}<j≤j(x,y)} & ∑_{k} (|f_{j,k}(x) − f_{j,k}(y)| + |g_{j,k}(x) − g_{j,k}(y)|) \\ & ≤ CL^{(j(x,y)+1)ε+1}δ̃(x, y) ≤ CL^{1+ε}δ(x, y). \end{aligned}

The second inequality comes from Lemma 7.12(a) and the fact that B_{x,y} ⊆ Ω_{j(x,y)} . From (10.7) we have

\begin{aligned}∑_{j>j(x,y)} & ∑_{k} (|f_{j,k}(x) − f_{j,k}(y)| + |g_{j,k}(x) − g_{j,k}(y)|) \\ & ≤ C(δ_{j(x,y)+1}(x) + δ_{j(x,y)+1}(y)) ≤ Cδ(x, y). \end{aligned}

The last inequality uses the fact that (10.9) fails for j = j(x, y) + 1, as well as the doubling condition on μ.

Combining these estimates gives (10.8). This proves Lemma 10.1.

Lemma 10.10. If N_{0} and L are large enough, then there is a C(L) > 0 so that

(10.11)  |f(x) − f(y)| ≥ C(L)^{−1}δ(x, y)

for all x, y ∈ R^{n}.

Fix x, y ∈ R^{n}, and let j(x, y) be as above. In particular

(10.12)  |x − y| > \frac{1}{10}d_{j(x,y)+1}(x).

Suppose first that j(x, y) = j_{0} (and hence j_{0} > −∞). Then, by (8.12),

|f(x) − f(y)| ≥ |f_{0}(x) − f_{0}(y)|
≥ M_{0}|x − y| − |f_{0}(x) − M_{0}x| − |f_{0}(y) − M_{0}y|

≥ M_{0}\{|x − y| − (10n)^{−10n}(d_{j_{0}+1}(x) + d_{j_{0}+1}(y))\} ≥ \frac{1}{2}M_{0}|x − y|.

For the last inequality we have used (10.12) and

(10.13)  d_{j_{0}+1}(y) ≤ d_{j_{0}+1}(x) + |x − y| ≤ 11|x − y|.

We need to show that M_{0}|x − y| ≥ C(L)^{−1}δ(x, y).

Let B_{x,y} be as in (1.3), as usual. From (10.12) we get that 20B_{x,y} intersects R^{n} \ Ω_{j_{0}+1} , and so

(∫_{B} ω)^{1/n} ≥ C^{−1}L^{−2}( ∫_{20B_{x,y}} ω)^{1/n}

for any other ball B that intersects R^{n} \ Ω_{j_{0}+1} , because of Lemma 7.14 and Fact (g) from Section 2. (In applying Lemma 7.14 we may take Q to be any cube that contains B and 20B_{x,y}, because Ω_{j_{0}} = R^{n}.) Using this inequality and the definition (8.9) of M_{0} we get

M_{0} ≥ C^{−1}L^{−2}( ∫_{20B_{x,y}} ω)^{1/n} ≥ C^{−1}L^{−2}δ(x, y)|x − y|^{−1}.

Combining this with our previous estimates yields (10.11) when j(x, y) = j_{0} .

Now suppose that j(x, y) > j_{0} . Set J = j(x, y), and choose K so that \overline{Q_{J,K}} contains x. (We can do this because x ∈ Ω_{J}.) Set i = a_{J,K}. We have to distinguish between the cases where y does or does not belong to 2Q_{J,K}.

Suppose first that y ∈ 2Q_{J,K}. Then

(10.14)  \begin{aligned}|f(x) − f(y)| & ≥ |f_{i}(x) − f_{i}(y)| \\ & ≥ |f_{J,K}(x) − f_{J,K}(y)| − ∑_{(j,k)∈Z} |f_{j,k}(x) − f_{j,k}(y)|, \end{aligned}

where Z = {(j, k) : a_{j,k} = i but (j, k) ≠ (J, K)}. We are going to show that the first term on the right side of (10.14) is large, by construction, while the sum is comparitively small if N_{0} is large enough.

Because x, y ∈ 2Q_{J,K} we have θ_{J,K}(x) = θ_{J,K}(y) = 1, and so

(10.15)  \begin{aligned}|f_{J,K}(x) & − f_{J,K}(y)| = |h_{J,K}(x) − h_{J,K}(y)| \\ & ≥ M_{J,K}|x − y| − |h_{J,K}(x) − M_{J,K}(x − q_{J,K})| \\ & − |h_{J,K}(y) − M_{J,K}(y − q_{J,K})| \\ & ≥ M_{J,K}\{|x − y| − (10n)^{−10n}(d_{J+1}(x) + d_{J+1}(y))\} \\ & ≥ \frac{1}{2}M_{J,K}|x − y|, \end{aligned}

by (8.15). For the last inequality we used (10.12) and the analogue of (10.13) in this situation. To control this from below by δ(x, y) we use

(10.16)  M_{J,K} = ( ∫_{Q_{J,K}} ω)^{1/n} ≥ C^{−1}L^{−1}( ∫_{20B_{x,y}} ω)^{1/n}.

This follows from the doubling condition on μ when 20B_{x,y} is not contained in 100Q_{J,K} (so that B_{x,y} is not too small compared to Q_{J,K}), and from Lemma 7.14 and Fact (g) in Section 2 when 20B_{x,y} ⊆ 100Q_{J,K}. (To apply Lemma 7.14 we need to know that 20B_{x,y} intersects R^{n} \ Ω_{J+1} , which is true, by (10.12).) Combining (10.15) with (10.16) gives

(10.17)  |f_{J,K}(x) − f_{J,K}(y)| ≥ C^{−1}L^{−1}δ(x, y).

Let E denote the sum on the right side of (10.14), which we want to be small compared to the right side of (10.17). Observe that

(10.18)  E = ∑_{(j,k)∈W} |f_{j,k}(x) − f_{j,k}(y)|,

where W = {(j, k) : a_{j,k} = i and j ≠ J}. Indeed, W ⊆ Z, and (j, k) ∈ Z \ W only when j = J, a_{j,k} = i, but k ≠ K. Under these circumstances Lemma 9.1 implies that Q_{j,k} is far enough way from Q_{J,K} that f_{j,k}(x) − f_{j,k}(y) = 0, since x, y ∈ 2Q_{J,K}. Hence such a (j, k) does not contribute to E, and we have (10.18).

Using Lemma 9.1 again we have

E ≤ E_{+} + E_{−},

where

E_{+} = ∑_{j≥J+N_{0}} ∑_{k} |f_{j,k}(x) − f_{j,k}(y)|

and

E_{−} = ∑_{j_{0}<j≤J−N_{0}} ∑_{k} |f_{j,k}(x) − f_{j,k}(y)|.

Applying (10.7) and (10.6) we get

E_{+} ≤ C(δ_{J+N_{0}}(x) + δ_{J+N_{0}}(y))

and

E_{−} ≤ CL^{(J−N_{0}+1)ε+1}δ̃(x, y).

We can use Lemma 7.6 to obtain

E_{+} ≤ C(CL^{−η})^{N_{0}−1}(δ_{J+1}(x) + δ_{J+1}(y)) ≤ C(CL^{−η})^{N_{0}−1}δ(x, y).

The second inequality is a consequence of the failure of (10.9) when j = J + 1. On the other hand,

E_{−} ≤ CL^{1−(N_{0}−1)ε}δ(x, y),

because of Lemma 7.12(a) and the fact that B_{x,y} ⊆ Ω_{J} (since (10.9) holds with j = J). Altogether we have

(10.19)  E ≤ C[(CL^{−η})^{N_{0}−1} + L^{1−(N_{0}−1)ε}]δ(x, y).

Thus if N_{0} satisfies

(10.20)  (N_{0} − 1)η, (N_{0} − 1)ε − 1 > 1,

then we can choose L so large that

E ≤ \frac{1}{2} |f_{J,K}(x) − f_{J,K}(y)|,

because of (10.17). Combining this with (10.17) and (10.14) we get (10.11) in this case (where j(x, y) > j_{0} , y ∈ 2Q_{J,K}).

It remains to take care of (10.11) when j(x, y) > j_{0} and y ∈ R^{n} \ 2Q_{J,K}. In this case we proceed using g_{i} instead of f_{i}, but the arguments are otherwise essentially identical.

We have

(10.21)  \begin{aligned}|f(x) − f(y)| & ≥ |g_{i}(x) − g_{i}(y)| \\ & ≥ |g_{J,K}(x) − g_{J,K}(y)| − ∑_{(j,k)∈Z} |g_{j,k}(x) − g_{j,k}(y)|, \end{aligned}

where Z is the same as before. Because x ∈ \overline{Q_{J,K}} but y ∉ 2Q_{J,K} we have g_{J,K}(x) ≥ C^{−1}δ_{J}(x) and g_{J,K}(y) = 0, by (9.10) and (9.6). We also have δ(x, y) ≤ Cδ_{J}(x) because (10.9) holds with j = J. Thus

(10.22)  |g_{J,K}(x) − g_{J,K}(y)| ≥ C^{−1}δ(x, y).

248 Stephen Semmes

Let E′ denote the sum on the right side of (10.21). Essentially the same arguments as before imply that E′ is dominated by the right side of (10.19). (The proof of the analogue of (10.18) is slightly different this time, because we do not have y ∈ 2Q_{J,K} now. However, the fact that (10.9) holds with j = J provides an adequate substitute.) Hence

E′ ≤ \frac{1}{2} |g_{J,K}(x) − g_{J,K}(y)|

if N_{0} satisfies (10.20) and L is large enough. Combining this with (10.22) and (10.21) gives (10.11) in this case.

This completes the proof of Lemma 10.10, and also of Theorem 5.2.

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Received 15 April 1991