GEORGIAN MATHEMATICAL JOURNAL: Vol. 2, No. 3, 1995, 277-290

WEIGHTED REVERSE WEAK TYPE INEQUALITY FOR GENERAL MAXIMAL FUNCTIONS

J. GENEBASHVILI

Abstract. Necessary and sufficient conditions are found to be imposed on a pair of weights, for which a weak type two-weighted reverse inequality holds, in the case of general maximal functions defined in homogeneous type spaces.

§ 1. Definition and Formulation of the Basic Results

By a homogeneous type space (X, ρ, μ) we mean a topological space X with measure μ and a quasimetric, i.e., a function ρ : X×X → R_{+}^{1} satisfying the conditions

(1) ρ(x, y) = 0 ⟺ x = y;

(2) ρ(x, y) = ρ(y, x) for all x, y ∈ X;

(3) ρ(x, y)≤η(ρ(x, z)+ρ(z, y)), where η >0 does not depend on x, y, z ∈ X. Furthermore, it is assumed that

(4) all balls B(x, r) = {y ∈ X : ρ(x, y) < r} are μ-measurable and the measure μ satisfies the doubling condition

0 < μB(x, 2r) ≤ d_{2}μB(x, r) < ∞, x ∈ X, 0 < r < ∞;

(5) for any open set U ⊂ X and point x ∈ U there exists a ball B(x, r) with the condition B(x, r) ⊂ U;

(6) continuous functions with compact support are dense in L^{1}(X, dμ). In addition to this, it is required that the space X have no atoms, i.e.,

μ{x} = 0 for any point x from X.

Let f be a locally summable function, x ∈ X and t ≥ 0. We introduce the following maximal function:

Mf(x, t) = sup \frac{1}{μB} ∫_{B} |f| dμ,

277 where the lowest upper bound is taken over all balls B containing the point x and having a radius greater than t/2.

1991 Mathematics Subject Classification. 42B25,46E30.

Key words and phrases. Maximal functions, homogeneous spaces, weights, reverse weak type inequality, covering lemma.

1072-947X/95/0500-0277$07.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1995 Plenum Publishing Corporation

If X = ℝ^{n}, μ is the Lebesgue measure, ρ is the Euclidean metric, and t = 0, then Mf(x, 0) transforms to the classical Hardy–Littlewood maximal function and for n = 1 and t ≥ 0 it transforms to the maximal function considered by Carleson when estimating the Poisson integrals.

By a weight function w we shall mean a locally summable nonnegative function w : X → R_{+}^{1} and by a measure β a measure in X × [0, ∞) defined in the product of σ-algebras generated by balls in X and by intervals in [0, ∞).

The merit of this paper is in finding the criterion for the existence of a weak type reverse two-weighted inequality for the maximal functions Mf(x, t). We thereby generalize the results obtained by K. Anderson and W.-S. Young [1] and B. Muckenhoupt [2] for the classical Hardy–Littlewood maximal function and improve the result obtained in [3].

It should also be noted that the criterion for straight two-weighted inequalities of the weak type was obtained by F. Ruiz and J. Torrea [4].

In what follows B̂ will denote a cylinder B × [0, 2 rad B), N the absolute constant N = η(1 + 2η), NB the ball NB = NB(x, r) = B(x, Nr), and d_{N} a minimal constant for which μ(NB) ≤ d_{N} μB; c, c_{1}, c_{2}, … are positive constants.

This paper gives the proofs of the following theorems.

Theorem 1. Let B_{0} be some ball in X. The following conditions are equivalent:

(1) for any function f ∈ L^{1}(X, w dμ), supp f ⊂ B_{0}, and any λ, λ ≥ λ_{0} = \frac{d_{N}}{μB_{0}} ∫_{B_{0}} |f| dμ,

β\{(x, t) ∈ B̂_{0} : Mf(x, t) > λ\} ≥ \frac{c_{1}}{λ}  ∫_{\{x∈B_{0}:|f(x)|>λ\}} |f|w dμ;  (1)

(2) for any ball B such that B ∩ B_{0} ≠ ∅ and B ⊂ NB_{0}

\frac{β(N̂B ∩ B̂_{0})}{μB} ≥ c_{2} ess sup_{x∈B∩B_{0}} w(x).  (2)

Theorem 2. Let μX = ∞. The following conditions are equivalent:

(1) for any function f ∈ L^{1}(X, w dμ) and any λ > 0

β\{(x, t) ∈ X × [0, ∞) : Mf(x, t) > λ\} ≥ \frac{c_{3}}{λ}  ∫_{\{x∈X:|f(x)|>λ\}} |f|w dμ; (3)

WEIGHTED REVERSE WEAK TYPE INEQUALITY 279

(2) for any ball B

\frac{β(N̂B)}{μB} ≥ c_{4} ess sup_{x∈B} w(x).  (4)

Theorem 3. Let B_{0}, w, and β satisfy condition (2). Then if

∫ ; _{B_{0}} Mf(x, t)dβ < ∞

for the function f, we have

∫_{B_{0}} |f|(1 + log^{+} |f|)w dμ < ∞.

Theorem 4. Let w and β satisfy condition (4). Then if

∫_{{(x,t):Mf(x,t)≥1}} Mf(x, t)dβ < ∞

for the function f ∈ L^{1}(X, w dμ), we have

∫_{X} |f| log^{+} |f|w dμ < ∞.

.

Corollary. For nontrivial w and β the pair of inequalities

\frac{c_{5}}{λ} ∫_{\{x∈X:|f(x)|>λ\}} |f|w dμ ≤ β\{(x, t) ∈ X × [0, ∞) : Mf(x, t) > λ\} ≤

≤ \frac{c_{6}}{λ} ∫_{\{x∈X:|f(x)|> \frac{λ}{2} \}} |f|w dμ

hold for all f ∈ L^{1}(x, w dμ) if and only if

βB̂ ∼ μB, 0 < c_{7} ≤ w(x) ≤ c_{8} < ∞

for any ball B and any point x ∈ X.

§ 2. The Covering Lemma

In the first place note that the following statement holds in quasimetric spaces: from any covering of a set E ⊂ X we can find at most a countable subcovering. Further we have (see [5])

Lemma 1. Let E be a bounded set from X and a ball B_{x} = B(x, r_{x}) (with center at x) be given for any point x ∈ E. Then from the covering {B_{x}}_{x∈E} we can find at most a countable subfamily of nonintersecting balls (B_{k})_{k≥1} such that

∪_{k≥1}NB_{k} ⊃ E.

The essence of the requirement that μ{x} = 0, x ∈ X, mentioned in §1 becomes clear after formulating

Lemma 2. A homogeneous type space has no atoms if and only if for any δ > 0 an arbitrary set E with positive measure has a subset E_{δ} ⊂ E with the condition 0 < μE_{δ} < δ.

Proof. Let μ{x_{0}} > 0. Then the set E = {x_{0}} does not contain a subset of a positive measure smaller than μE. One aspect of the proof of the lemma becomes thereby obvious.

Let, conversely, μ{x} = 0 for all x ∈ X and E be an arbitrary set of positive measure. The continuity of measure implies that for each x ∈ E there exists a ball B_{x} with center at x such that μB_{x} < δ. According to the remark made at the beginning of this section, from the system of balls {B_{x}}_{x∈E} we can find a countable subfamily (B_{k})_{k≥1} covering B_{0}. Hence we have

μE = μ( ∪_{k≥1}(B_{k} ∩ E)) ≤ ∑_{k≥1} μ(B_{k} ∩ E).

Therefore there exists k_{0} ≥ 1 such that μ(B_{k_{0}} ∩ E) > 0. So, assuming E_{δ} = B_{k_{0}} ∩ E, we obtain E_{δ} ⊂ E and

0 < μE_{δ} ≤ μB_{k_{0}} < δ.

Lemma 3. Let Ω ⊂ X × [0, ∞) be a set such that if (x, t) ∈ Ω, then (x, τ) ∈ Ω for all τ, 0 ≤ τ < t. Let the projection Ω_{X} of the set Ω on X be a bounded set and Ω_{0} ⊂ Ω_{X} be a set of all x from Ω_{X} for which B̂(x, r) ⊂ Ω with some radius r > 0. Then there exists a sequence of balls (B_{i})_{i≥1} such that

(1) \frac{1}{N} B_{i} ∩ \frac{1}{N} B_{j} = ∅, i ≠ j;

(2) Ω_{0} = ∪_{i}B_{i} = ∪_{i}NB_{i};

(3) ∪_{i}N̂B_{i} ⊂ Ω;

(4) { \atop 3ηNB_{i}} ∩ (X × [0, ∞)\Ω) ≠ ∅, i = 1, 2, …;

(5) ∑_{i} χ_{NB_{i}}(x, t) ≤ θχ_{Ω}(x, t),

where θ ≥ 1 does not depend on x ∈ X and t ≥ 0.

Proof. Let F = X × [0, ∞)\Ω. We introduce the value

dist(x, F) =^{def} sup{r : B̂(x, r) ⊂ Ω}, x ∈ Ω_{X}.

It is clear that

0 < dist(x, F) < ∞

for any point x ∈ Ω_{0}.

Let us take

r_{x} = \frac{dist(x, F)}{2ηN^{2}}

for any x ∈ Ω_{0}. The system of balls {B(x, r_{x})}_{x∈Ω_{0}} covers Ω_{0}. By Lemma 1 there exists a sequence (B(x_{i}, r_{x_{i}}))_{i≥1} of nonintersecting balls such that

Ω_{0} ⊂ ∪_{i≥1}B(x_{i}, Nr_{x_{i}}).

Setting r_{i} = Nr_{x_{i}}, B_{i} = B(x_{i}, r_{i}), we shall have

Ω_{0} ⊂ ∪_{i≥1}B_{i} and \frac{1}{N} B_{i} ∩ \frac{1}{N} B_{j} = ∅ for i ≠ j.

Statement (1) is thereby proved.

To prove statement (3) note that

Nr_{i} = N^{2}r_{x_{i}} = \frac{dist(x_{i}, F)}{2η} < dist(x_{i}, F).

Therefore, by definition of the value “dist,” we shall have

{ \atop NB_{i}} ⊂ Ω

for each i ≥ 1.

Further, for the cylinder { \atop 3ηNB_{i}} we obtain

rad(3ηNB_{i}) = 3ηN^{2}r_{x_{i}} = \frac{3}{2} dist(x_{i}, F) > dist(x_{i}, F).

Therefore statement (4) is true.

Now we shall prove statement (2). Since Ω_{0} ⊂ ∪_{i≥1}B_{i}, it is sufficient for

us to prove that NB_{i} ⊂ Ω_{0} for all i = 1, 2, . . ..

Let us fix NB_{i} and show that dist(x, F) > 0 for any point x ∈ NB_{i}. Assume the opposite: dist(x, F) = 0. Then B̂(x, α) ∩ F ≠ ∅ for any

α > 0. Therefore there is (y, t) ∈ B̂(x, α) ∩ F. We shall consider two cases:

(a) t ≥ 2 dist(x_{i}, F); then

Nr_{i} = \frac{dist(x_{i}, F)}{2η} ≤ \frac{t}{4η} < \frac{α}{2η} < α.

(b) t < 2 dist(x_{i}, F); then y ∉ B(x_{i}, dist(x_{i}, F)), since otherwise (y, t) ∈ B̂(x_{i}, dist(x_{i}, F)) ⊂ Ω.

Thus we have

2ηNr_{i} = dist(x_{i}, F) ≤ ρ(x_{i}, y) ≤ η(ρ(x_{i}, x) + ρ(x, y)) < η(Nr_{i} + α).

Therefore Nr_{i} < α. So in both cases we find that if x ∈ NB_{i}, then rad NB_{i} < α for any

α > 0, i.e., rad NB_{i} = 0, which leads to the contradiction.

We have thereby proved that dist(x, F) > 0 for any x ∈ NB_{i} and therefore x ∈ Ω_{0}.

Finally, we shall prove the validity of statement (5). Let x ∈ NB_{i}. As shown above, dist(x, F) > 0. Consider the cylinder

B̂(x, 2 dist(x, F)). By the definition of the value “ dist ” we have B̂(x, 2 dist(x, F)) ∩ F ≠ ∅ and therefore there exists

(y, t) ∈ B̂(x, 2 dist(x, F)) ∩ F.

We shall consider two cases:

(a) t ≥ 2 dist(x_{i}, F); then

Nr_{i} = \frac{dist(x_{i}, F)}{2η} ≤ \frac{t}{4η} < \frac{dist(x, F)}{η} < 2 dist(x, F);

(b) t < 2 dist(x_{i}, F); then y ∉ B(x_{i}, dist(x_{i}, F)), since otherwise (y, t) ∈ B̂(x_{i}, dist(x_{i}, F)) ⊂ Ω.

Thus we have

2ηNr_{i} = dist(x_{i}, F) ≤ ρ(x_{i}, y) ≤ η(ρ(x_{i}, x) + ρ(x, y)) <
< η(Nr_{i} + 2 dist(x, F)).

Therefore Nr_{i} < 2 dist(x, F). So in both cases we find that if x ∈ NB_{i}, then

Nr_{i} < 2 dist(x, F).

Fix an arbitrary point x. Let NB_{i} ∋ x and y ∈ NB_{i}. Then

ρ(x, y) ≤ η(ρ(x, x_{i}) + ρ(x_{i}, y)) ≤ 2ηNr_{i} < 4η dist(x, F)

from which we conclude that

NB_{i} ⊂ B(x, 4η dist(x, F))  (5)

for any ball NB_{i} such that NB_{i} ∋ x.

Taking now y ∈ B(x_{i}, 2 dist(x_{i}, F)), we obtain

ρ(x, y) ≤ η(ρ(x, x_{i}) + ρ(x_{i}, y)) ≤ η(Nr_{i} + 2 dist(x_{i}, F)) =

= η(\frac{dist(x_{i}, F)}{2η} + 2 dist(x_{i}, F)) = (2η + \frac{1}{2}) dist(x_{i}, F).

Therefore

B(x, (2η + \frac{1}{2}) dist(x_{i}, F)) ⊃ B(x_{i}, 2 dist(x_{i}, F)).

Hence

B(x, (2η + \frac{1}{2}) dist(x_{i}, F)) ∩ F ≠ ∅.

Thus

dist(x, F) < (2η + \frac{1}{2}) dist(x_{i}, F) = (4η^{2} + η)Nr_{i}.

Therefore

rad NB_{i} > \frac{1}{4η^{2} + η} dist(x, F).  (6)

From (5) and (6) we conclude that balls NB_{i} containing the fixed point x are included in the fixed ball B(x, 4η dist(x, F)) and their radii are bounded from below by the fixed positive value \frac{1}{4η^{2}+η} dist(x, F). Therefore, since \frac{1}{N} B_{i} do not intersect pairwise, the number of such balls NB_{i} is bounded from above by some absolute constant θ. As a result,

∑_{i} χ{ \atop NB_{i}}(x, t) ≤ θχ_{Ω}(x, t).

§ 3. Proof of the Main Results

Proof of Theorem 1. Let us show that (1) ⇒ (2).

Take an arbitrary ball B ⊂ NB_{0}, B ∩ B_{0} ≠ ∅. Let y ∉ NB = B(x, Nr) and some ball B′ = B(x′, r′) contain the point y and intersect with B. We shall prove that then r′ > r.

Assume the opposite: r′ ≤ r. Let z ∈ B ∩ B′. Then

ρ(x, y) ≤ η(ρ(x, z) + ρ(z, y)) < η(r + η(ρ(z, x′) + ρ(x′, y)) <
< η(r + 2ηr′) ≤ η(1 + 2η)r = Nr,

which leads to the contradiction. Therefore r′ > r.

If now y ∈ B and z ∈ B ∩ B′, then

ρ(x′, y) ≤ η(ρ(x′, z) + ρ(z, y)) < η(r′ + η(ρ(z, x) + ρ(x, y)) <
< η(r′ + 2ηr) < η(1 + 2η)r′ = Nr′.

Therefore B ⊂ NB′.

Fix an arbitrary ε > 0. There is a set E_{ε} ⊂ B ∩ B_{0} such that w(x) > ess sup_{t∈B∩B_{0}} w(t)−ε for any point x ∈ E_{ε}. By Lemma 2 it can be assumed that

0 < μE_{ε} < \frac{μB}{d_{N}^{2}} .

Let f(x) = χ_{Eε} (x) and λ = \frac{d_{N}^{2} μE_{ε}}{μB}. Then λ < 1 and

λ_{0} = \frac{d_{N}}{μB_{0}} ∫_{B_{0}} |f| dμ = d_{N} \frac{μE_{ε}}{μB_{0}} ≤ d_{N}^{2} \frac{μE_{ε}}{μNB_{0}} ≤ d_{N}^{2} \frac{μE_{ε}}{μB} = λ.

Let further (y, t) ∉ N̂B. Consider two cases:

(a) y ∉ NB; then

Mf(y, t) = sup_{\begin{matrix}B′∋y \\ rad B′> \frac{t}{2} \end{matrix}} \frac{1}{μB′} ∫_{B′} |f| dμ ≤ sup_{\begin{matrix}B′∋y \\ B′∩B≠∅ \end{matrix}} \frac{μE_{ε}}{μB′} ≤

≤ sup_{\begin{matrix}B′∩B≠∅ \\ r′>r \end{matrix}} d_{N} \frac{μE_{ε}}{μNB′} ≤ d_{N} \frac{μE_{ε}}{μB} < d_{N}^{2} \frac{μE_{ε}}{μB} = λ.

(b) y ∈ NB, t ≥ 2Nr; then

Mf(y, t) = sup_{\begin{matrix}B′∋y \\ B′∩B≠∅ \\ rad B′> \frac{t}{2} \end{matrix}} \frac{1}{μB′} ∫_{B′} |f| dμ ≤ sup_{\begin{matrix}B′∋y \\ r′>Nr \\ B′∩B≠∅ \end{matrix}} \frac{μE_{ε}}{μB′} ≤

≤ sup_{\begin{matrix}r′>r \\ B′∩B≠∅ \end{matrix}} d_{N} \frac{μE_{ε}}{μNB′} ≤ λ.

Thus

N̂B ⊃ {(y, t) : Mf(y, t) > λ}.

Now in view of the above reasoning condition (1) leads to

β(B̂_{0} ∩ N̂B) ≥ β{(x, t) ∈ B̂_{0} : Mf(x, t) > λ} ≥

≥ \frac{c_{1}}{d_{N}^{2}} \frac{μB}{μE_{ε}} ∫_{\{x∈B∩B_{0}:χ_{Eε} (x)>λ\}} χ_{E_{ε}}(x)w(x) dμ =

= c_{2} \frac{μB}{μE_{ε}} ∫_{E_{ε}} w dμ ≥ c_{2}μB(ess sup_{x∈B∩B_{0}} w(x) − ε).

By making ε → 0 we get (2).

Now we shall prove that (2) ⇒ (1).

Fix f and assume that supp f ⊂ B_{0} and λ ≥ λ_{0} = \frac{d_{N}}{μB_{0}} ∫_{B_{0}} |f| dμ. Consider the sets

Ω = {(x, t) ∈ X × [0, ∞) : Mf(x, t) > λ},
Ω_{c} = {x ∈ X : M_{c}f(x) > λ},

where

M_{c}f(x) = sup_{r>0} \frac{1}{μB(x, r)} ∫_{B(x,r)} |f| dμ.

The set Ω satisfies the conditions of Lemma 3. Indeed, if (x, t) ∈ Ω, then it is obvious that (x, τ) ∈ Ω, 0 ≤ τ < t. Moreover, by familiar arguments Ω ⊂ N̂B_{0}. Therefore Ω_{X} is the bounded set.

Let x ∈ Ω_{c}. Then there exists r > 0 such that

\frac{1}{μB(x, r)} ∫_{B(x,r)} |f| dμ > λ.

Obviously, Mf(y, t) > λ for any (y, t) ∈ B̂(x, r) and therefore B̂(x, r) ⊂ Ω. Thus Ω_{c} ⊂ Ω_{0}, where Ω_{0} is the set mentioned in Lemma 3. By the latter lemma there exists a sequence of balls (B_{k})_{k≥1} satisfying the statements of the lemma. Since B_{k} ⊂ Ω_{0} ⊂ NB_{0} for each k ≥ 1, from the condition (2) we get

β(Ω ∩ B̂_{0}) = ∫ ; _{B_{0}} χ_{Ω}(x, t)dβ ≥ \frac{1}{θ} ∑_{k≥1} ∫ ; _{B_{0}} χ{ \atop NB_{k}}(x, t)dβ =

= \frac{1}{θ} ∑_{k≥1} β(N̂B_{k} ∩ B̂_{0}) ≥ c ∑_{k≥1} μB_{k} ess sup_{x∈B_{k}∩B_{0}} w(x).  (7)

Since { \atop 3ηNB_{k}} ∩ (X × [0, ∞)\Ω) ≠ ∅, there exists (x, t) ∈ { \atop 3ηNB_{k}} such that Mf(x, t) ≤ λ. Therefore

\frac{1}{μB_{k}} ∫_{B_{k}} |f| dμ ≤ \frac{d_{3ηN}}{(3ηNB_{k})} ∫_{3ηNB_{k}} |f| dμ < d_{3ηN}λ.

Now (7) takes the form

β(Ω ∩ B̂_{0}) ≥ \frac{c_{1}}{λ} ∑_{k≥1} ess sup_{x∈B_{k}∩B_{0}} w(x) ∫_{B_{k}} |f| dμ =

= \frac{c_{1}}{λ} ∑_{k≥1} ess sup_{x∈B_{k}∩B_{0}} w(x) ∫_{B_{k}∩B_{0}} |f| dμ ≥ \frac{c_{1}}{λ} ∑_{k≥1} ∫_{B_{k}∩B_{0}} |f|w dμ ≥

≥ \frac{c_{1}}{λ} ∫_{∪_{k≥1}(B_{k}∩B_{0})} |f|w dμ = \frac{c_{1}}{λ} ∫_{Ω_{0}∩B_{0}} |f|w dμ ≥ \frac{c_{1}}{λ} ∫_{Ω_{c}∩B_{0}} |f|w dμ =

= \frac{c_{1}}{λ}  ∫_{\{x∈B_{0}:M_{c}f(x)>λ\}} |f|w dμ ≥ \frac{c_{1}}{λ}  ∫_{\{x∈B_{0}:|f(x)|>λ\}} |f|w dμ.

Proof of Theorem 2. First of all note that the implication (3) ⇒ (4) can be proved in the same manner as the implication (1) ⇒ (2) in the preceding theorem. So we shall prove that (4) ⇒ (3).

Fix an arbitrary ball B′ and assume that f ∈ L^{1}(X, w dμ). For l > 0 we introduce the function

f_{l}(x) = \begin{cases}f(x) · χ_{lB′}(x), if |f(x)| < l, \\ l · sign f(x) · χ_{lB′}(x), if |f(x)| ≥ l, \\ 0 · χ_{X\lB′} (x). \end{cases}

Let λ > 0. Then there exists a number R > Nl such that

\frac{d_{N}^{2}}{μ(RB′)} ∫_{X} |f_{l}| dμ ≤ λ.

Let B_{0} = NRB′, β_{R}E = βE for E ⊂ R̂B′, and β_{R}{(x, t)} = ∞ for any point (x, t) ≠ R̂B′.

We shall show that if β and w satisfy (4), then B_{0}, β_{R}, and w satisfy condition (2) of Theorem 1.

Indeed, consider an arbitrary ball B ⊂ NB_{0}, B ∩B_{0} ≠ ∅. If N̂B ⊂ R̂B′, then

\frac{β_{R}(N̂B ∩ B̂_{0})}{μB} = \frac{β_{R}(N̂B)}{μB} = \frac{βN̂B}{μB} ≥ c_{4} ess sup_{x∈B} w(x) = c_{2} ess sup_{x∈B∩B_{0}} w(x).

Let N̂B ⊄ R̂B′. If N̂B ⊂ B̂_{0}, then

\frac{β_{R}(N̂B ∩ B̂_{0})}{μB} = \frac{β_{R}(N̂B)}{μB} = ∞ ≥ c_{2} ess sup_{x∈B∩B_{0}} w(x).

Thus it remains for us to consider the case with N̂B ⊄ B̂_{0}. We shall show that β_{R}(N̂B ∩ B̂_{0}) = ∞ in that case, too. To this end we have to prove that

(N̂B ∩ B̂_{0})\R̂B′ ≠ ∅.  (8)

If there exists a point z ∈ (NB ∩ B_{0})\RB′, then (8) holds. If such a point does not exists, i.e., NB ∩ (B_{0}\RB′) = ∅, then, since NB ∩ B_{0} ≠ ∅, there is a point y ∈ NB ∩ RB′.

On the other hand, since N̂B ⊄ B̂_{0}, we have either NB ⊂ B_{0} and then

rad(NB) > rad B_{0} > rad(NB′)

or NB ⊄ B_{0}, which together with the condition NB ∩ \frac{1}{N} B_{0} = NB ∩ RB′ ≠ ∅, by familiar arguments, gives

rad(NB) > rad (\frac{1}{N} B_{0}) = rad(NB′).

Therefore, if N̂B ⊄ B̂_{0}, there exists a point y ∈NB∩RB′ and rad(NB) > rad(NB′). Then

(y, 2R rad B′) ∈ N̂B\R̂B′.

Since (y, 2R rad B′) ∈ B̂_{0}, we have (8).

We have thereby shown that B_{0}, β_{R}, and w satisfy the condition (2) of Theorem 1.

As to λ, we have

λ_{0} = \frac{d_{N}}{μB_{0}} ∫_{B_{0}} |f_{l}| dμ < \frac{d_{N}^{2}}{μ(RB′)} ∫_{B_{0}} |f_{l}| dμ ≤ λ.

Now according to Theorem 1 we have

β_{R}\{(x, t) ∈ B̂_{0} : Mf_{l}(x, t) > λ\} ≥ \frac{c_{3}}{λ}  ∫_{\{x∈B_{0}:|f_{l}(x)|>λ\}} |f_{l}|w dμ. (9)

But since supp f_{l} ⊂ \frac{R}{N} B′, for (x, t) ∉ R̂B′ we shall have

Mf_{l}(x, t) ≤ \frac{d_{N}}{μ(\frac{R}{N} B′)} ∫_{\frac{R}{N} B′} |f_{l}| dμ ≤ \frac{d_{N}^{2}}{μ(RB′)} ∫_{X} |f_{l}| dμ ≤ λ.

Hence (9) takes the form

β\{(x, t) ∈ X × [0, ∞) : Mf_{l}(x, t) > λ\} ≥ \frac{c_{3}}{λ} ∫_{\{x∈X:|f_{l}(x)|>λ\}} |f_{l}|w dμ.

The more so

β\{(x, t) ∈ X × [0, ∞) : Mf(x, t) > λ\} ≥ \frac{c_{3}}{λ} ∫_{\{x∈X:|f_{l}(x)|>λ\}} |f_{l}|w dμ.

By making l tend to infinity we obtain the required inequality (3).

Proof of Theorem 3. Let w(x) > 0 on some subset B_{0} of positive measure (otherwise there is nothing to prove). Then from (2) we conclude that βB̂_{0} > 0. If f ≠ 0 almost everywhere on B_{0}, then

Mf(x, t) ≥ \frac{1}{μB_{0}} ∫_{B_{0}} |f| dμ > 0

for each (x, t) ∈ B̂_{0}. Hence from the condition

∫ ; _{B_{0}} Mf(x, t)dβ < ∞

we obtain f ∈ L(B_{0}, dμ) and βB̂_{0} < ∞. Therefore again from (2) we conclude that w is bounded on B_{0} and f ∈ L(B_{0}, w dμ).

Now we have

∫_{B_{0}} |f| log^{+} |f|w dμ = ∫_{{|f|>1}} |f| log |f|w dμ =

= ∫_{\{|f|>λ_{0}\}}|f| log \frac{|f|}{λ_{0}}w dμ + ∫_{\{1<|f|≤λ_{0}\}}|f| log |f|w dμ+log λ_{0} ∫_{\{|f|>λ_{0}\}}|f|w dμ,

where λ_{0} is taken from condition (1) of Theorem 1. (If λ_{0} < 1, then the latter expansion is unnecessary.)

By virtue of the above reasoning we see that the last two integrals are finite. Applying Theorem 1, we shall show the finiteness of the first integral:

∫_{\{|f|>λ_{0}\}} |f| log \frac{|f|}{λ_{0}}w dμ = ∫_{\{|f|>λ_{0}\}} |f| ∫_{λ_{0}}^{|f|} \frac{dλ}{λ}w dμ = ∫_{λ_{0}}^{∞} \frac{1}{λ} ∫_{\{|f|>λ\}}|f|w dμ dλ ≤

≤ c ∫_{λ_{0}}^{∞}β{(x, t) ∈ B̂_{0} : Mf(x, t) > λ}dλ ≤

≤ c ∫_{0}^{∞}β{(x, t) ∈ B̂_{0} : Mf(x, t) > λ}dλ = c ∫ ; _{B_{0}} Mf(x, t) dμ < ∞.

Proof of Theorem 4. The proof follows from Theorem 2 and the estimate

∫_{X} |f| log^{+} |f|w dμ = ∫_{\{|f|>1\}} |f| log |f|w dμ = ∫_{\{|f|>1\}} |f| ∫_{1}^{|f|} \frac{dλ}{λ}w dμ =

= ∫_{1}^{∞} \frac{1}{λ} ∫_{\{|f|>λ\}} |f|w dμdλ ≤ c ∫_{1}^{∞}β\{(x, t) ∈ X × [0, ∞) : Mf(x, t) > λ\}dλ =

= c ∫_{{(x,t):Mf(x,t)>1}} Mf(x, t)dβ < ∞.

Proof of the Corollary. Following the result of F. Ruiz and J. Torrea [4] and Theorem 2 of this paper, for the inequalities

\frac{c_{1}}{λ} ∫_{\{|f|>λ\}} |f|w dμ ≤ β\{(x, t) : Mf(x, t) > λ\} ≤ \frac{c_{2}}{λ} ∫_{\{f|> \frac{λ}{2} \}} |f|w dμ

to hold, it is necessary and sufficient that the inequalities

\frac{βB̂}{μB} ≤ c_{3} ess inf_{x∈B} w(x) and \frac{β(N̂B)}{μB} ≥ c_{4} ess sup_{x∈B} w(x)

be fulfilled simultaneously. Hence for any ball B we have

c_{5} ess sup_{x∈ \frac{1}{N} B} w(x) ≤ \frac{βB̂}{μB} ≤ c_{3} ess inf_{x∈B} w(x).

From here on the proof of the corollary is clear.

References

1. K. F. Anderson and W.-S. Young, On the reverse weak type inequality for the Hardy maximal function and the weighted classes L(log L)^{k}. Pacific J. Math. 112(1984), No. 2, 257-264.

2. B. Muckenhoupt, Weighted reverse weak type inequalities for the Hardy-Littlewood maximal function. Pacific J. Math. 117(1985), No. 2, 371-378.

3. I. Z. Genebashvili, Weighted reverse inequalities for maximal functions. Reports Extended Session Sem. Vekua Inst. Appl. Math. Tbilisi St. Univ. 7(1992), No. 2, 10-13.

4. F. J. Ruiz and J. L. Torrea, Vector-valued Calderon-Zygmund theory and Carleson measures on spaces of homogeneous nature. Studia Math. 88(1988), 221-243.

5. R. R. Coifman and G. Weiss, Analyse harmonique non-commutative sur certans espaces homogenes. Lect. Notes Math., v. 242, Springer-Verlag, Berlin etc., 1971, 1-158.

(Received 06.12.93)

Author’s address: A. Razmadze Mathematical Institute Georgian Academy of Sciences 1, Z. Rukhadze St., Tbilisi 380093 Republic of Georgia