GEORGIAN MATHEMATICAL JOURNAL: Vol. 3, No. 1, 1996, 81-96

ON STRONG MAXIMAL OPERATORS CORRESPONDING TO DIFFERENT FRAMES

G. ONIANI

Abstract. The problem is posed and solved whether the conditions f ∈ L(1 + ln^{+} L)^{2}(ℝ^{2}) and sup_{θ∈[0,π/2)} ∫_{{M_{2,θ}(f)>1}} M_{2,θ}(f) < ∞ are equivalent for functions f ∈ L(ℝ^{2}) (where M_{2,θ} denotes the strong maximal operator corresponding to the frame {OX_{θ}, OY_{θ}}).

The results obtained represent a general solution of M. de Guzmán’s problem that was previously studied by various authors.

1. Notation

Let B_{1} be a family of all cubic intervals in ℝ^{n}. We denote by M_{1} a maximal Hardy-Littlewood operator which is defined as follows:

M_{1}(f)(z) := sup \{\frac{1}{|I|} ∫_{I} |f| : z ∈ I, I ∈ B_{1}\}, z ∈ ℝ^{n},

for f ∈ L_{loc}(ℝ^{n}).

Let B_{2} be a family of all open rectangles in ℝ^{2} whose sides are parallel to the coordinate axes; OX_{θ} be the straight line obtained by rotating the OX-coordinate axis through the angle θ about the point O in the positive direction (OY_{θ} is defined analogously); B_{2,θ} be a family of all open rectangles with the sides parallel to the straight lines OX_{θ} and OY_{θ}.

For a rectangle I ⊂ ℝ^{2} we shall denote by n(I) a number θ ∈ [0, π/2) for which one of the sides of I is parallel to OX_{θ}. The regularity factor of the rectangle I will be defined as the ratio of the length of the larger side of I to the length of the smaller side of I and will be denoted by r(I).

For θ ∈ ℝ we shall denote by θ (mod π/2) a number such that 0 ≤ θ (mod π/2) < π/2 and θ − θ (mod π/2) = πk/2 for some k ∈ ℕ. One can easily verify that B_{2,θ} = B_{2,θ (mod π/2)}, θ ∈ ℝ.

The sets {OX_{θ}, OY_{θ}}, θ ∈ [0, π/2) will be called frames.

81 For f ∈ L_{loc}(ℝ^{2}) let

1991 Mathematics Subject Classification. 28A15.

Key words and phrases. Strong maximal operators, different frames, regularity factor of a rectangle.

1072-947X/96/0100-0081$09.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1996 Plenum Publishing Corporation

M_{2}(f)(z) := sup \{\frac{1}{|I|} ∫_{I} |f| : z ∈ I, I ∈ B_{2}\}, z ∈ ℝ^{2},

M_{2,θ}(f)(z) := sup \{\frac{1}{|I|} ∫_{I} |f| : z ∈ I, I ∈ B_{2,θ}\}, z ∈ ℝ^{2}.

M_{2} is called the strong maximal operator, while M_{2,θ} is called the strong maximal operator corresponding to the frame {OX_{θ (mod π/2)}, OY_{θ (mod π/2)}}. This definition is correct because by virtue of the equality B_{2,θ} =B_{2,θ (mod π/2)} we have M_{2,θ} = M_{2,θ (mod π/2)}. The latter equality implies that the family {M_{2,θ}}_{θ∈[0,π/2)} exhausts the family of all operators M_{2,θ} (θ ∈ ℝ).

2. Formulation of the Question

As is known, the space L(1 + ln^{+} L)(ℝ^{n}) can be characterized by the maximal operator M_{1} as follows (see [1], [2]):

Theorem 1^{0}. Let f ∈ L(ℝ^{n}). Then the following two conditions are equivalent:

1. f ∈ L(1 + ln^{+} L)(ℝ^{n});

2. ∫_{{M_{1}(f)>1}} M_{1}(f) < ∞.

The implication 1 ⇒ 2 was proved by Hardy and Littlewood [3] for n = 1 and by Wiener [4] for n ≥ 2. The results of the reverse nature were obtained for the first time by Stein [5] and Tsereteli [6, 7]. Guzman and Welland [1, 2] improved the above results by formulating Theorem 1^{0}.

It is known that if f ∈ L(1 + ln^{+} L)^{2}(ℝ^{2}) then (see [1])

∫_{{M_{2}(f)>1}} M_{2}(f) < ∞.  (2.1)

Guzman (see [1]) posed the question whether it was possible to characterize the space L(1 + ln^{+} L)^{2}(ℝ^{2}) by the operator M_{2} as was done for the space L(1 + ln^{+} L) using the operator M_{1}. Gogoladze [8, 9] and Bagby [10] answered this question in the negative. Their results give rise to

Theorem 2^{0}. For any functions f ∉ L(1 + ln^{+} L)^{2}(ℝ^{2}) and f ∈ L(1 + ln^{+} L)(ℝ^{2}) there exists a Lebesgue measure preserving an invertible mapping ω : ℝ^{2} → ℝ^{2} such that

∫_{{M_{2}(f∘ω)>1}} M_{2}(f ∘ ω) < ∞.

Now we proceed directly to formulating our problem. It is easy to verify that if f ∈ L(1 + ln^{+} L)^{2}(ℝ^{2}) then

∫_{{M_{2,θ}(f)>1}} M_{2,θ}(f) < ∞ for any θ ∈ [0, π/2),  (2.2)

and, moreover,

sup_{θ∈[0,π/2)}  ∫_{{M_{2,θ}(f)>1}} M_{2,θ}(f) < ∞.  (2.3)

Clearly, conclusion (2.3) is stronger than (2.1) and hence there is a better chance for us to improve the integral properties of the function f when (2.3) is fulfilled than in the case of fulfillment of (2.1). Having given this information, we formulate the problem:

Let f ∈ L(ℝ^{2}) and sup_{θ∈[0,π/2)} ∫_{{M_{2,θ}(f)>1}} M_{2,θ}(f) < ∞. Is the inclusion f ∈ L(1 + ln^{+} L)^{2}(ℝ^{2}) then valid?

We would like to note here that the functions constructed in [8–10] do not satisfy condition (2.2) and thus [8–10] do not provide a solution of the above problem. So we shall prove the following theorem which as a particular case contains the answer to the problem.

Theorem 1. For any functions f ∉ L(1 + ln^{+} L)^{2}(ℝ^{2}) and f ∈ L(1 + ln^{+} L)(ℝ^{2}) there exists a Lebesgue measure preserving an invertible mapping ω : ℝ^{2} → ℝ^{2} such that

1. the set {|f ∘ ω| > 1} is a square interval;

2. sup_{θ∈[0,π/2)} ∫_{{M_{2,θ}(f∘ω)>1}} M_{2,θ}(f ∘ ω) < ∞.

3. Auxiliary Statements

To prove Theorem 1 we shall need several lemmas. If Δ ⊂ ℝ is some interval and (length Δ) < π/2, then by M_{2,Δ}^{∗} we denote the following operator: for each f ∈ L_{loc}(ℝ^{2})

M_{2,Δ}^{∗}(f)(z) :=

=sup {M_{2,θ}(f)(z):θ ∈[cen Δ−π/4, inf Δ] ∪ [sup Δ, cen Δ+π/4)}, z ∈ℝ^{2}; where cen Δ = (inf Δ + sup Δ)/2.

Lemma 1. Let 0 < θ < π/4, H > 0, 0 < λ < H. It is assumed that the regularity factor of rectangle I satisfies the inequality r(I) ≥ \frac{H}{λ sin^{2} θ}. Then there is a rectangle I_{I,H,θ,λ} such that

{M_{2,Δ}^{∗}(Hχ_{I}) > λ} ⊂ I_{I,H,θ,λ},

|I_{I,H,θ,λ}| ≤ c_{1} \frac{H}{λ} |I|,

∫_{I_{I,H,θ,λ}} M_{2,Δ}^{∗}(Hχ_{I}) ≤ c_{2}H(1 + ln^{+} \frac{H}{λ} )|I|,

where Δ = (n(I) − θ, n(I) + θ), while c_{1} and c_{2} are the positive constants not depending on I, H, θ, and λ.

Proof. We begin by considering the case λ = 1, H > 1. Without loss of generality it will be assumed (see Fig. 1) that n(I) = 0, I is the rectangle ABCD the sides AB and BC of which have the lengths I_{1} and I_{2} respectively and I_{2}/I_{1} ≥ H/ sin^{2} θ.

Fig. 1

The strip bounded by the straight lines containing the segments AD and BC respectively will be denoted by Ĩ. Using the convexity property of the rectangle, it is easy to prove the inequality

M_{2,Δ}^{∗}(Hχ_{I})(z) ≤ 3H \frac{I_{1}}{dist(z, Ĩ)} for dist(z, Ĩ) ≥ I_{1}.  (3.1)

A minimal number α ≥ 0 for which the straight line l is parallel to the OX_{α}-axis will be denoted by n(l). Let l_{1} (n(l_{1}) = θ) and l_{2} (n(l_{2}) = π − θ) be the straight lines passing through the points C and D, respectively. The straight lines l_{1} and l_{2} divide the plane into parts. We denote the right-hand part by E_{l_{1},l_{2}}. Due to the definition of the operator M_{2,θ}^{∗} it is easy to show that

M_{2,Δ}^{∗}(Hχ_{I})(z) ≤ \frac{H|I|}{dist(z, l_{1}) dist(z, l_{2})} for z ∈ E_{l_{1},l_{2}}.  (3.2)

Above and below the rectangle I let us draw the straight lines l′_{1} and l′_{2} which are parallel to the segment AD and situated at a distance (3H + 1)I_{1} from I. Now the rectangle A′B′C′D′ can be chosen so that A′D′ ⊂ l′_{1}, B′C′ ⊂ l′_{2}, and dist(A′B′, I) = dist(C′D′, I) = 10I_{2}. We shall prove that the rectangle A′B′C′D′ can be taken as I_{I,H,θ,1}.

The strip bounded by the straight lines l′_{1} and l′_{2} will be denoted by E; the part of E lying on the right of C′D′ will be denoted by E^{+}, while the part of E lying on the left of A′B′ will be denoted by E^{−}.

By (3.1) we conclude that

{M_{2,Δ}^{∗}(Hχ_{I}) > 1} ⊂ E.  (3.3)

By virtue of the inequality I_{2}/I_{1} ≥ H/ sin^{2} θ and the definition of the rectangle A′B′C′D′ it is easy to show that

E^{+} ⊂ E_{l_{1},l_{2}};  (3.4)
dist(D′, l_{1}) ≥ \sqrt{H|I|};  (3.5)
dist(C′, l_{2}) ≥ \sqrt{H|I|}.  (3.6)

Obviously, for any z ∈ E^{+}

dist(z, l_{1}) ≥ dist(D′, l_{1});  (3.7)
dist(z, l_{2}) ≥ dist(C′, l_{2}).  (3.8)

Using (3.2) and (3.4)–(3.8), we find that

M_{2,Δ}^{∗}(Hχ_{I})(z) ≤ \frac{H|I|}{dist(z, l_{1}) dist(z, l_{2})} ≤

≤ \frac{H|I|}{dist(D′, l_{1}) dist(C′, l_{2})} ≤ \frac{H|I|}{\sqrt{H|I|}\sqrt{H|I|}} = 1  (3.9)

for any z ∈ E^{+}.

From (3.9) we readily obtain

M_{2,Δ}^{∗}(Hχ_{I})(z) ≤ 1 for all z ∈ E^{−}.  (3.10)

For this it is enough to ascertain that the following equality holds:

M_{2,Δ}^{∗}(Hχ_{I})(z) = M_{2,Δ}^{∗}(Hχ_{I})(z′), z ∈ ℝ^{2},

where z′ denotes a point symmetrical to z with respect to the center of the rectangle I.

By (3.3), (3.9), and (3.10) we have

{M_{2,Δ}^{∗}(Hχ_{I}) > 1} ⊂ A′B′C′D′.  (3.11)

Clearly,

|A′B′C′D′| ≤ (6H + 3)I_{1}21I_{2} < 200H|I|.  (3.12)

It is easy to prove the inequality

∫_{E_{x}} M_{2,Δ}^{∗}(Hχ_{I})(x, y)dy < 15H(1 + ln H)I_{1}, x ∈ ℝ,  (3.13)

where E_{x} := {(x, y) : y ∈ ℝ} ∩ E, x ∈ ℝ.

By (3.13) we obtain

∫_{A′B′C′D′} M_{2,Δ}^{∗}(Hχ_{I})(x, y)dxdy ≤

≤ 15H(1 + ln H)I_{1}( length A′D′) ≤ 315H(1 + ln H)|I|. (3.14)

By virtue of (3.11), (3.12), and (3.14) the lemma is proved for the case λ = 1, H > 1. Now it is easy to obtain the proof for the general case if we take into consideration the following obvious equalities:

M_{2,Δ}^{∗}(Hχ_{I})(z) = λM_{2,Δ}^{∗}(\frac{H}{λ} χ_{I} )(z), z ∈ ℝ^{2};  (3.15)
\{M_{2,Δ}^{∗}(Hχ_{I}) > λ\} = \{M_{2,Δ}^{∗}(\frac{H}{λ} χ_{I} ) > 1\}.  (3.16)

As is well known (see [1]), if f ∈ L(1 + ln^{+} L)(ℝ^{2}), then for each λ > 0 we have the inequality

|\{M_{2}(f) > λ\}| ≤ c_{3} ∫_{ℝ^{2}} \frac{|f|}{λ}(1 + ln^{+} \frac{|f|}{λ}),  (3.17)

where c_{3} is a positive constant not depending on f and λ.

Let Γ_{θ} be rotation of the plane through the angle θ about the point O in the positive direction. It is easy to verify that M_{2,θ}(f)(z) = M_{2}(f ∘ Γ_{θ})(Γ_{θ}^{−1}(z)). Hence we obtain |{M_{2,θ}(f) > λ}| = |{M_{2}(f ∘ Γ_{θ}) > λ}| which by virtue of (3.17) implies that Lemma 2 is valid.

Lemma 2. Let f ∈ L(1 + ln^{+} L)(ℝ^{2}); then the inequality

|\{M_{2,θ}(f) > λ\}| ≤ c_{3} ∫_{ℝ^{2}} \frac{|f|}{λ}(1 + ln^{+} \frac{|f|}{λ})

holds for each θ ∈ [0, π/2) and λ > 0.

By using this lemma we can easily prove

Lemma 3. Let f ∈ L(1 + ln^{+} L)^{2}(ℝ^{2}); then the inequality

∫_{{M_{2,θ}(f)>1}} M_{2,θ}(f) ≤ c_{4} ∫_{ℝ^{2}} |f|(1 + ln^{+} |f|)^{2},

where c_{4} is the positive constant not depending on f and θ, holds for each θ ∈ [0, π/2).

Let Ω ⊂ [0, π/2), B = ∪_{θ∈Ω}B_{2,θ} and for each f ∈ L_{loc}(ℝ^{2})

M_{B}(f)(z) := sup \{\frac{1}{|I|} ∫_{I} |f| : z ∈ I, I ∈ B\}, z ∈ ℝ^{2}.

We have

Lemma 4. Let I_{k} be a rectangle, H_{k} > 1, k ∈ ℕ, and ∑_{k=1}^{∞} H_{k}|I_{k}| < ∞. It is assumed that there exists a sequence {I_{k}} such that I_{k}∩I_{m} = ∅, k ≠ m, {M_{B}(H_{k}χ_{I_{k}}) > λ} ⊂ I_{k}, k ∈ ℕ, where λ ∈ (0, 1] is a fixed number. Then

{M_{B}( ∑_{k=1}^{∞} H_{k}χ_{I_{k}} ) > λ} ⊂ ⋃_{k=1}^{∞} I_{k}.  (3.18)

Proof. Let f_{k} := H_{k}χ_{I_{k}} (k ∈ ℕ) and f := ∑_{k=1}^{∞} H_{k}χ_{I_{k}} = sup_{k∈ℕ} H_{k}χ_{I_{k}}. Clearly, (3.18) is equivalent to the inequality

M_{B}(f)(z) ≤ λ, z ∉ ∪_{k=1}^{∞}I_{k}.  (3.19)

Let us assume that z ∉ ∪_{k=1}^{∞}I_{k}, R ∋ z, R ∈ B and prove the inequality

∫_{R∩I_{k}} f_{k} ≤ λ|R ∩ I_{k}|, k ∈ ℕ.  (3.20)

Let n(R) = θ and R_{x_{θ}} := {(ξ, η) : ξ_{θ} = x_{θ}, η_{θ} ∈ ℝ} ∩ R, for each x_{θ} ∈ ℝ (where (ξ_{θ}, η_{θ}) denotes the coordinates of the point (ξ, η) in the X_{θ}OY_{θ}- system).

We consider an arbitrarily fixed number k ∈ ℕ. The following notation is introduced:

E_{k}^{1} := {x_{θ} : R_{x_{θ}} ∩ I_{k} ≠ ∅, R_{x_{θ}}\I_{k} ≠ ∅};

E_{k}^{2} := {x_{θ} : R_{x_{θ}} ⊂ I_{k}};

R_{k}^{1} := ⋃_{x_{θ}∈E_{k}^{1}} (R_{x_{θ}} ∩ I_{k});

R_{k}^{2} := ⋃_{x_{θ}∈E_{k}^{2}} (R_{x_{θ}} ∩ I_{k}).

Clearly,

R_{k}^{1} ∩ R_{k}^{2} = ∅ and R ∩ I_{k} = R_{k}^{1} ∪ R_{k}^{2}.  (3.21)

For our further discussion we need the inequality

M_{B}(f)(z) ≤ λ, z ∈ ∂I_{k},  (3.22)

which immediately follows from the condition of the lemma (∂ denotes the boundary of sets).

If x_{θ} ∈ E_{k}^{1}, then one can easily find that one end of the segment R_{x_{θ}} ∩ I_{k} belongs to ∂I_{k} and hence by virtue of (3.22) one readily obtains

∫_{R_{x_{θ}} ∩I_{k}} f_{k}(x_{θ}, y_{θ})dy_{θ} ≤ λ|R_{x_{θ}} ∩ I_{k}|_{1},

where | · |_{1} denotes the Lebesgue measure on the straight line.

The above inequality implies

∫_{R_{k}^{1}} f_{k} = ∫_{E_{k}^{1}} dx_{θ} ∫_{R_{x_{θ}} ∩E_{k}} f_{k}(x_{θ}, y_{θ})dy_{θ} ≤

≤ ∫_{E_{k}^{1}} λ|R_{x_{θ}} ∩ I_{k}|_{1} = λ|R_{k}^{1}|.  (3.23)

It is easy to prove the following facts:

R_{k}^{2} is a rectancle included in B;  (3.24)
R_{k}^{2} has a vertex belonging to ∂I_{k}.  (3.25)

From (3.22), (3.24), (3.25) we obtain the inequality

∫_{R_{k}^{2}} f_{k} ≤ λ|R_{k}^{2}|.  (3.26)

ON STRONG MAXIMAL OPERATORS 89

Taking into account (3.12), (3.23), and (3.26), we have

∫_{R∩I_{k}} f_{k} = ∫_{R_{k}^{1}} f_{k} + ∫_{R_{k}^{2}} f_{k} ≤ λ|R_{k}^{1}| + λ|R_{k}^{2}| = λ|R ∩ I_{k}|.  (3.27)

Since k ∈ ℕ has been chosen arbitrarily, (3.27) implies the estimate

∫_{R} f = ∑_{k=1}^{∞} ∫_{R∩I_{k}} f_{k} = ∑_{k=1}^{∞} ∫_{R∩I_{k}} f_{k} ≤ ∑_{k=1}^{∞} λ|R ∩ I_{k}| ≤ λ|R|;

Because of an arbitrarily chosen R (z ∈ R, R ∈ B) we hence conclude that

M_{B}(f)(z) ≤ λ, z ∉ ∪_{k=1}^{∞}I_{k}.

For each θ ∈ [0, π/2) let A(θ) denote a set of all f ∈ L(ℝ^{2}) for which ∫_{{M_{2,θ}(f)>1}} M_{2,θ}(f) < ∞. We have

Lemma 5. Let f and g belong to the set A(θ). Then f + g ∈ A(θ) and

∫_{{M_{2,θ}(f+g)>1}} M_{2,θ}(f + g) ≤

≤ c_{5}(  ∫_{{M_{2,θ}(f)>1}} M_{2,θ}(f) +  ∫_{{M_{2,θ}(g)>1}} M_{2,θ}(g)) +
+c_{6}(∥f(1 + ln^{+} |f|)∥_{1} + ∥g(1 + ln^{+} |g|)∥_{1}),  (3.28)

where c_{5} and c_{6} are positive constants not depending on f, g, and θ. Proof. One can easily verify the following inequalities:

∫_{{M_{2,θ}(f+g)>1}} M_{2,θ}(f + g) ≤

≤  ∫_{{M_{2,θ}(f)>1/2}∪{M_{2,θ}(g)>1/2}} (M_{2,θ}(f) + M_{2,θ}(g));  (3.29)

∫_{{M_{2,θ}(f)>1/2}∪{M_{2,θ}(g)>1/2}} M_{2,θ}(f) ≤

≤  ∫_{\{M_{2,θ}(f)>1/2\}} M_{2,θ}(f) + \frac{1}{2}|\{M_{2,θ}(g) > 1/2\}|;  (3.30)

∫_{{M_{2,θ}(g)>1/2}∪{M_{2,θ}(g)>1/2}} M_{2,θ}(f) ≤

≤  ∫_{\{M_{2,θ}(g)>1/2\}} M_{2,θ}(g) + \frac{1}{2}|\{M_{2,θ}(f) > 1/2\}|.  (3.31)

By virtue of (3.29), (3.30), (3.31) and Lemma 2 we conclude that (3.28) is valid.

Let A denote a set of all functions f ∈ L(ℝ^{2}) satisfying condition (2.3). Lemma 5 immediately gives rise to

Lemma 6. Let f and g belong to the set A. Then f + g ∈ A.

4. Proof of Theorem 1

It is assumed without loss of generality that f is positive, |{f > 1}| = 1. Let E_{k} := {k − 1 ≤ f < k}, k ∈ ℕ. Clearly, there exists k_{0} ≥ 3 for which

∑_{k=k_{0}}^{∞} c_{1}k|E_{k}| < 1.  (4.1)

Choose a sequence {m_{k}} ⊂ ℕ such that

\frac{k ln^{2} k|E_{k}|}{m_{k}} ≤ 1, k ∈ ℕ.  (4.2)

Let N_{0} := \{k ∈ ℕ : k ≥ k_{0}, |E_{k}| > 0\} and the sequence \{Δ_{k,m}\}_{k∈N_{0},m=\overline{1,m_{k}}} consist of pairwise nonintersecting intervals lying on the segment [0, π/2).

For each k ∈ N_{0} and m ∈ [1, m_{k}] choose a rectangle I_{k,m} such that

|I_{k,m}| = \frac{|E_{k}|}{m_{k}};  (4.3)
n(I_{k,m}) = cen Δ_{k,m};  (4.4)
r(I_{k,m}) ≥ \frac{k}{sin^{2} |Δ_{k,m}|_{1}/2} .  (4.5)

By virtue of (4.5) and Lemma 1 there exists a rectangle I_{k,m} such that

{M_{2,Δ_{k,m}}^{∗}(kχ_{I_{k,m}} ) > 1} ⊂ I_{k,m};  (4.6)
|I_{k,m}| ≤ c_{1}k|I_{k,m}|;  (4.7)
∫_{I_{k,m}} M_{2,Δ_{k,m}}^{∗}(kχ_{I_{k,m}} ) ≤ 2c_{2}k ln k|I_{k,m}|.  (4.8)

Let \{Q_{k,m}\}_{k∈N_{0},m=\overline{1,m_{k}}} be a sequence of pairwise rectangular intervals lying on [0, 1]^{2} and each having height equal to 1, and

|Q_{k,m}| = |I_{k,m}|, k ∈ N_{0}, m = \overline{1, m_{k}}  (4.9)

(such a sequence exists by virtue of (4.1), (4.3) and (4.7)).

For each k ∈ N_{0} and m ∈ [1, m_{k}] we complete the rectangle Q_{k,m} with pairwise nonintersecting rectangles {I_{k,m,q}} which are homothetic to the rectangle I_{k,m}, i.e., we have

I_{k,m,q} = P_{k,m,q}(I_{k,m})
where P_{k,m,q} is the homothety (q ∈ ℕ);  (4.10)
I_{k,m,q} ⊂ Q_{k,m}, q ∈ ℕ;  (4.11)
I_{k,m,q} ∩ I_{k,m,q′} = ∅, q ≠ q′;  (4.12)
|Q_{k,m}\ ⋃_{q∈ℕ} I_{k,m,q}| = 0.  (4.13)

Let I_{k,m,q} := P_{k,m,q}(I_{k,m}), k ∈ N_{0}, m = \overline{1, m_{k}}, q ∈ ℕ. Clearly,

∑_{q∈ℕ} |I_{k,m,q}| = |I_{k,m}| = \frac{|E_{k}|}{m_{k}} , k ∈ N_{0}, m ∈ [1, m_{k}].  (4.14)

Since P_{k,m,q} is a homotopy, by (4.6)–(4.8) we conclude that for each k ∈ N_{0}, m ∈ [1, m_{k}], and q ∈ ℕ

{M_{2,Δ_{k,m}}^{∗}(kχ_{I_{k,m,q}} ) > 1} ⊂ I_{k,m,q};  (4.15)
|I_{k,m,q}| ≤ c_{1}k|I_{k,m,q}|;  (4.16)
∫_{I_{k,m,q}} M_{2,Δ_{k,m}}^{∗}(kχ_{I_{k,m,q}} ) ≤ 2c_{2}k ln k|I_{k,m,q}|.  (4.17)

We introduce the notation g_{k,m} := sup_{q∈ℕ}(kχ_{I_{k,m,q}}), k ∈ N_{0}, m_{0} = \overline{1, m_{k}}, and g := sup{g_{k,m} : k ∈ N_{0}, m ∈ [1, m_{k}]}. Let us prove that g ∈ A.

It is assumed that θ ∈ [0, π/2) is an arbitrary fixed number. Two cases are possible:

(a) θ ∈ ∪_{k,m}Δ_{k,m};

(b) θ ∉ ∪_{k,m}Δ_{k,m};

(a) Let θ ∈ Δ_{k(θ),m(θ)}. We introduce the notation T := {(k, m, q) : k ∈ N_{0}, m ∈ [1, m_{k}], q ∈ ℕ, (k, m) ≠ (k(θ), m(θ))}.

Since θ ∉ Δ_{k,m} for (k, m) ≠ (k(θ), m(θ)), by the definition of M_{2,Δ}^{∗} and (4.15) we obtain

{M_{2,θ}(kχ_{I_{k,m,q}} ) > 1} ⊂ {M_{2,Δ_{k,m}}^{∗}(kχ_{I_{k,m,q}} ) > 1} ⊂ I_{k,m,q},

(k, m, q) ∈ T.

92 G. ONIANI

Hence by (4.12) and Lemma 4 we conclude that

{M_{2,θ}(g − g_{k(θ),m(θ)}) > 1} ⊂ ⋃_{(k,m,q)∈T} I_{k,m,q},  (4.18)

M_{2,θ}(g − g_{k(θ),m(θ)})(z) ≤ M_{2,θ}(kχ_{I_{k,m,q}} )(z) + 1

for z ∈ I_{k,m,q}, (k, m, q) ∈ T.  (4.19)

On account of (4.14) and (4.16)–(4.19) we write

∫_{{M_{2,θ}(g−g_{k(θ),m(θ)})>1}} M_{2,θ}(g − g_{k(θ),m(θ)}) ≤

≤ ∑_{(k,m,q)∈T} ∫_{I_{k,m,q}} M_{2,θ}(g − g_{k(θ),m(θ)}) ≤

≤ ∑_{(k,m,q)∈T} ∫_{I_{k,m,q}} [M_{2,θ}(kχ_{I_{k,m,q}} ) + 1] ≤

≤ ∑_{(k,m,q)∈T} [2c_{2}k ln k|I_{k,m,q}| + |I_{k,m,q}|] ≤

≤ 1 + ∑_{k∈N_{0}} 2c_{2}k ln k|E_{k}| ≤ 1 + 8c_{2}∥f ln^{+} f∥_{1}.  (4.20)

By (4.2), (4.14) and Lemma 3 we have

∫_{{M_{2,θ}(g_{k(θ),m(θ)})>1}} M_{2,θ}(g_{k(θ),m(θ)}) ≤ c_{4} ∫_{ℝ^{2}} g_{k(θ),m(θ)}(1 + ln^{+} g_{k(θ),m(θ)})^{2} =

= c_{4} ∑_{q∈ℕ} k(θ)(1 + ln k(θ))^{2}|I_{k(θ),m(θ),q}| =

= c_{4}k(θ)(1 + ln k(θ))^{2} \frac{|E_{k(θ)}|}{m_{k(θ)}} ≤ 4c_{4}.  (4.21)

From the construction we easily find that

∥(g − g_{k(θ),m(θ)})(1 + ln^{+} (g − g_{k(θ),m(θ)})∥_{1} ≤

≤ ∥g(1 + ln^{+} g)∥_{1} ≤ 4∥f ln^{+} f∥_{1},  (4.22)

Analogously,

∥g_{k(θ),m(θ)}(1 + ln^{+} g_{k(θ),m(θ)})∥_{1} ≤ 4∥f ln^{+} f∥_{1}  (4.23)

Using the representation g = (g − g_{k(θ),m(θ)}) + g_{k(θ),m(θ)}, by virtue of (4.20)–(4.23) and Lemma 5 we obtain

∫_{{M_{2,θ}(g)>1}} M_{2,θ}(g) ≤

≤ c_{5}(1 + 8c_{2}∥f ln^{+} f∥_{1} + 4c_{4}) + 8∥f ln^{+} f∥_{1}c_{6}.  (4.24)

In proving the case (b) we have no “dangerous” term g_{k(θ),m(θ)} and therefore, applying the same reasoning as for (4.20), we can write

∫_{{M_{2,θ}(g)>1}} M_{2,θ}(g) ≤

≤ ∑_{(k,m,q)∈T_{0}} ∫_{I_{k,m,q}} M_{2,θ}(g) ≤ ∑_{(k,m,q)∈T_{0}} ∫_{I_{k,m,q}} [M_{2,θ}(kχ_{I_{k,m,q}}) + 1] ≤

≤ ∑_{(k,m,q)∈T_{0}} [2c_{2}k ln k|I_{k,m,q}| + |I_{k,m,q}|] ≤

≤ 1 + ∑_{k∈N_{0}} 2c_{2}k ln k|E_{k}| ≤ 1 + 8c_{2}∥f ln^{+} f∥,

where T_{0} := \{(k, m, q) : k ∈ N_{0}, m = \overline{1, m_{k}}, q ∈ ℕ\}.  (4.25)

Since θ ∈ [0, π/2) was chosen arbitrarily, by virtue of (4.24) and (4.25) we conclude that g ∈ A.

We shall now find the desired mapping of ω. From the construction it easily follows that

E_{k} ∩ E_{k′} = ∅, k ≠ k′,  (4.26)
( ⋃_{(m,q)∈T_{k}} I_{k,m,q}) ⋂ ( ⋃_{(m,q)∈T_{k′}} I_{k′,m,q}) = ∅, k = k′,  (4.27)

where T_{k} := {(m, q) : m ∈ [1, m_{k}], q ∈ ℕ} for k ∈ N_{0};

|E_{k}| = | ⋃_{(m,q)∈T_{k}} I_{k,m,q}| > 0, k ∈ N_{0},  (4.28)

|{f > 1}\ ⋃_{k∈N_{0}} E_{k}| = |(0, 1)^{2}\ ⋃_{(k,m,q)∈T_{0}} I_{k,m,q}| > 0.  (4.29)

By conditions (4.26)–(4.29) and one familiar result on measure-preserving transformations (see, e.g., [11, Chapter: Uniform Approximation]) we conclude that there exists a Lebesgue measure preserving an invertible mapping ω : ℝ^{2} → ℝ^{2} such that

ω( ⋃_{(m,q)∈T_{k}} I_{k,m,q}) = E_{k}, k ∈ N_{0};  (4.30)

ω((0, 1)^{2}\ ⋃_{(k,m,q)∈T_{0}} I_{k,m,q}) = {f > 1}\ ⋃_{k∈N_{0}} E_{k};  (4.31)
ω(ℝ^{2}\(0, 1)^{2}) = {f ≤ 1}.  (4.32)

(4.30) and (4.31) clearly imply

{f ∘ ω > 1} = ω^{−1}({f > 1}) = (0, 1)^{2}.  (4.33)

From the construction and conditions (4.30)–(4.32) we obtain

(f ∘ ω)χ_{{∪ \atop k∈N_{0}}E_{k}} ≤ g,
(f ∘ ω)χ_{ℝ^{2}\{∪ \atop k∈N_{0}}E_{k}} ∈ L(1 + ln^{+} L)^{2}(ℝ^{2}).

Hence, taking into account f ∘ ω = (f ∘ ω)χ_{{∪ \atop k∈N_{0}}E_{k}} + (f ∘ ω)χ_{ℝ^{2}\{∪ \atop k∈N_{0}}E_{k}},

inclusions g ∈ A, and Lemmas 3 and 6, we conclude that f ∘ ω ∈ A, i.e.,

sup_{θ∈[0,π/2)}  ∫_{{M_{2,θ}(f∘ω)>1}} M_{2,θ}(f ∘ ω) < ∞.  (4.34)

By (4.33) and (4.34) ω is the desired mapping.

5. Remarks

Remark 1. On overcoming certain technical difficulties, we can prove by a technique similar to that used to prove Theorem 1 the following generalization.

Theorem 2. Let a function f ∉ L(1+ln^{+} L)^{2}(ℝ^{2}), f ∈ L(1+ln^{+} L)(ℝ^{2}). It is assumed that a set G_{1}, |G_{1}| > 0, is such that fχ_{ℝ^{2}\G_{1}} ∈ L(1 + ln^{+} L)^{2}(ℝ^{2}). Then for any set G_{2}, |G_{2}| = |G_{1}|, there exists a Lebesgue measure preserving an invertible mapping ω : ℝ^{2} → ℝ^{2} such that

1. ω(G_{1}) = G_{2},

2. {z : ω(z) ≠ z} ⊂ G_{1} ∪ G_{2},

3. sup_{θ∈[0,π/2)} ∫_{{M_{2,θ}(f∘ω)>1}} M_{2,θ}(f ∘ ω) < ∞.

Theorem 2 yields as a corollary

Theorem 3. Let a function f ∉ L(1+ln^{+} L)^{2}(ℝ^{2}), f ∈ L(1+ln^{+} L)(ℝ^{2}). It is assumed that a set G is such that fχ_{ℝ^{2}\G} ∈ L(1 + ln^{+} L)^{2}(ℝ^{2}). Then there exists a Lebesgue measure preserving an invertible mapping ω : ℝ^{2} → ℝ^{2} such that

1. {z : ω(z) ≠ z} ⊂ G,

2. sup_{θ∈[0,π/2)} ∫_{{M_{2,θ}(f∘ω)>1}} M_{2,θ}(f ∘ ω) < ∞.

Remark 2. Let X be a set and R an equivalence relation on X. A subset Y ⊂ X will be called an R-set if the fact that y ∈ Y implies that [y]_{R} ⊂ Y, where [y]_{R} denotes a set of all elements from X R-equaivalent to y. The following problem is posed in [6]: Given an equivalence relation R on the set X, characterize the set E ⊂ X from the standpoint of R, i.e., give in explicit terms the kernel \underline{E}(R) (the greatest R-set contained in E) and the hull \overline{E}(R) (the least R-set containing E).

Consider an arbitrary set G ⊂ ℝ^{2} and choose |G| > 0, X_{G}, R_{G} and E_{G} in the following manner: X_{G} := {f ∈ L(ℝ^{2}), supp f ⊂ G}, E_{G} := {f ∈ X_{G}, f ∈ A}, f and g ∈ R_{G} will be called R_{G}-equivalent if there exists a Lebesgue measure preserving an invertible mapping ω : ℝ^{2} → ℝ^{2}, {z : ω(z) ≠ z} ⊂ G, such that g = f ∘ ω.

Let us agree that φ(L)(G) denotes a class of functions f : ℝ^{2} → ℝ with the following properties: supp f ⊂ G, ∫_{G} φ(|f|) < ∞. One can easily show that \underline{E}_{G}(R_{G}) = L(1 + ln^{+} L)^{2}(G), while by virtue of (2.3) and Theorem 3 we have the equality \overline{E}_{G}(R_{G}) = L(1 + ln^{+} L)(G). Thus the next theorem is valid.

Theorem 4. For each G ⊂ ℝ^{2}, |G| > 0, we have

\underline{E}_{G}(R_{G}) = L(1 + ln^{+} L)^{2}(G) and \overline{E}_{G}(R_{G}) = L(1 + ln^{+} L)(G).

Aknowledgment

The author wishes to express his thanks to T. Zerekidze for useful hints and discussions.

References

1. M. de Guzmán, Differentiation of integrals in ℝ^{n}. Lecture Notes in Mathematics 481, Springer, 1975.

2. M. de Guzmán and G. V. Welland, On the differentiation of integrals. Rev. Un. Mat. Argentina 25(1971), 253–276.

3. G. H. Hardy and J. E. Littlewood, A maximal theorem with functiontheoretic applications. Acta Math. 54(1930), 81–116.

4. N. Wiener, The ergodic theorem. Duke Math. J. 5(1939), 1–18.

5. E. M. Stein, Note on class L log L. Studia Math. 31(1969), 305–310.

6. O. D. Tsereteli, On the inversion of some Hardy and Littlewood theorems. (Russian) Bull. Acad. Sci. Georgian SSR 56(1969), No. 2, 269–271.

7. O. D. Tsereteli, A metric characteristic of the set of functions whose maximal functions are summable. (Russian) Trudy Tbilissk. Mat. Inst. 42(1972), 104–118.

8. L. D. Gogoladze, The Hardy-Littlewood maximal functions. (Russian) Bull. Acad. Sci. Georgian SSR 111(1983), 257–259.

9. L. D. Gogoladze, On the summation of multiple trigonometric series and conjugate functions. (Russian) Dissertation for the Doctor of Science Degree, Tbilisi, 1984.

10. R. J. Bagby, A note on the strong maximal function. Proc. Amer. Math. Soc. 88(1983), 648–650.

11. P. R. Halmos, Lectures on the ergodic theory. Math. Soc. Japan, Tokyo, 1956.

(Received 23.05.1994)

Author’s address: Faculty of Mechanics and Mathematics I. Javakhishvili Tbilisi State University 2, University St., Tbilisi 380043 Republic of Georgia