Georgian Mathematical Journal 1(1994), No. 1, 77-97

WEIGHTED ESTIMATES FOR THE HILBERT TRANSFORM OF ODD FUNCTIONS

LUBOŠ PICK

Abstract. The aim of the present paper is to characterize the classes of weights which ensure the validity of one-weighted strong, weak or extra-weak type estimates in Orlicz classes for the integral operator

H_{0}f(x) = \frac{2}{π} ∫_{0}^{∞} \frac{yf(y)}{x^{2} − y^{2}} dy, x ∈ (0, ∞).

1. Introduction. The Hilbert transform is given for any function f satisfying

∫_{−∞}^{∞} |f(x)| (1 + |x|)^{−1} dx < ∞

by the Cauchy principal value integral

Hf(x) = \frac{1}{π} lim_{ε→0+} ∫_{R\(x−ε,x+ε)} \frac{f(y)}{x − y} dy.

If f is an odd function, then Hf is even, and Hf(x) = H_{o}f(|x|), where

H_{o}f(x) = \frac{2}{π} ∫_{0}^{∞} \frac{yf(y)}{x^{2} − y^{2}} dy, x ∈ (0, ∞).

The Hilbert transform is closely related to the Hardy–Littlewood maximal operator

Mf(x) = sup_{I∋x} \frac{1}{|I|} ∫_{I} |f(y)| dy.

1991 Mathematics Subject Classification. 42B20,42B25,46E30.

If ϱ is a weight (measurable and nonnegative function) and 1 ≤ p < ∞, strong type inequalities

∫_{Ω} |T f(x)|^{p}ϱ(x) dx ≤ C ∫_{Ω} |f(x)|^{p} ϱ(x) dx,  (1.1)

as well as weak type inequalities

ϱ({|T f| > λ}) ≤ C λ^{−p} ∫_{Ω} |f(x)|^{p} ϱ(x) dx,  (1.2)

have been widely studied by many authors. The pioneering result of Muckenhoupt [13] stated that (1.1) holds with Ω = R, T = M and p > 1 if and only if ϱ ∈ A_{p}, that is,

sup_{I} ϱ_{I} · ((ϱ^{1−p′})_{I})^{p−1} ≤ C,

and (1.2) holds with Ω = R, T = M and p ≥ 1 if and only if ϱ ∈ A_{p}, where ϱ ∈ A_{1} means ϱ_{I} ≤ Cess inf_{I}ϱ. Hunt, Muckenhoupt and Wheeden [10] proved the same result for Ω = R and T = H. The class of good weights for (1.1) or (1.2) with Ω = (0, ∞) and T = H_{o} appears to be strictly larger than A_{p}. This result is due to Andersen who showed that (1.1) with Ω = (0, ∞), p > 1, and T = H_{o} holds if and only if ϱ ∈ A_{p}^{o}, that is,

ϱ(a, b)(∫_{a}^{b} ϱ^{1−p′}(x)x^{p′} dx)^{p−1} ≤ C(b^{2} − a^{2})^{p},  (a, b) ⊂ R, (1.3)

and (1.2) with Ω = (0, ∞), p ≥ 1, and T = H_{o} holds if and only if ϱ ∈ A_{p}^{o}, where ϱ ∈ A_{1}^{o} means

\frac{ϱ(a, b)}{b^{2} − a^{2}} ≤ Cess inf_{(a,b)} \frac{ϱ(x)}{x}.

Our aim is to study analogous inequalities where the power function t^{p} is replaced by a general convex function Φ(t). More precisely, we shall study the inequalities

∫_{Ω} Φ(|T f(x)|) ϱ(x) dx ≤ C ∫_{Ω} Φ(C|f(x)|) ϱ(x) dx,  (1.4)

ϱ({|T f| > λ}) · Φ(λ) ≤ C ∫_{Ω} Φ(C|f(x)|) ϱ(x) dx,  (1.5)

and

ϱ({|T f| > λ}) ≤ C ∫_{Ω} Φ(Cλ^{−1}|f(x)|) ϱ(x) dx.  (1.6)

We call (1.4) strong type inequality, (1.5) weak type inequality, and (1.6) extra–weak type inequality. While (1.4) is an analogue of (1.1), (1.5) and (1.6) are two different analogues of (1.2). It is always true that (1.4) ⇒ (1.5) ⇒ (1.6), and none of these implications is reversible in general. The interest in these types of inequalities stems from their use in various problems of Fourier analysis. For example, extra–weak type inequalities have interesting interpolation applications (see [2]).

We assume throughout that Φ is a convex nondecreasing function on (0, ∞), Φ(0) = 0. In fact, it is not hard to prove that for all the above operators the inequalities (1.4) or (1.5) always imply at least quasiconvexity of Φ. For more detaile, see [9].

Weak and extra–weak type inequalities together were apparently firstly studied in [14] for T = M and Ω = R^{n}. In [9] the following results were obtained (for definitions see Section 2 below):

Theorem A. The inequality (1.5) holds with T = H and Ω = R if and only if Φ ∈ Δ_{2} and ϱ ∈ A_{Φ}.

Theorem B. Let Φ ∈ Δ_{2}^{0}. Then (1.6) holds with T = H and Ω = R if and only if ϱ ∈ E_{Φ}.

The main aim of the present paper is to characterize the classes of weights for which the inequalities (1.4–6) hold with Ω = (0, ∞) and T = H_{o} (Theorems 3–5 in Section 4). Moreover, we get a characterization for the strong type inequality (1.4) with Ω = R and T = H. This is given in Section 3 (Theorem 2) as well as the similar assertion for T = M (Theorem 1). However, in the case T = M we do not obtain a full characterization but are left with a small but significant gap between the necessary and the sufficient condition.

It should be mentioned that Andersen obtained in [1] L^{p}–results also for the operator

H_{e}f(x) = ∫_{0}^{∞} \frac{xf(y)}{x^{2} − y^{2}} dy,  (1.7)

the Hilbert transform for even functions. However, our methods do not provide analogous results for H_{e} with t^{p} replaced by Φ(t).

Let us finally mention that the result of Andersen was generalized to the case of multiple Hilbert transform in [17]. For other related results see also [15], [16]. Some of the results of this paper were included in the comprehensive monograph [18].

2. Preliminaries. Let Φ be a convex nondecreasing function on [0, ∞), Φ(0) = 0, which does not vanish identically on [0, ∞), but it is allowed that Φ ≡ 0 on [0, a] and/or Φ ≡ ∞ on (a, ∞) for some a > 0 provided that Φ(a−) is finite. The complementary function to Φ, Φ̃(t) = sup_{s>0}(st − Φ(s)), has the

same properties as Φ (for example, convexity of Φ̃ follows easily from the subadditivity of supremum). Moreover, the Young inequality

st ≤ Φ(s) + Φ̃(t)  (2.1)

holds for all s, t positive. Both Φ and Φ̃ are invertible on (0, ∞) and it follows immediately from (2.1) that

Φ^{−1}(t) · Φ̃^{−1}(t) ≤ 2t,  t > 0.  (2.2)

We say that Φ satisfies the Δ_{2} condition, (Φ ∈ Δ_{2}), if Φ(2t) ≤ CΦ(t). If this estimate holds merely near 0 (near ∞), we write Φ ∈ Δ_{2}^{0} (Φ ∈ Δ_{2}^{∞}). We recall that Φ∈Δ_{2} is equivalent to 2Φ^{−1}(t) ≤ Φ^{−1}(Ct).

The functions

R_{Φ}(t) = Φ(t)/t, S_{Φ}(t) = Φ̃(t)/t

will play a crucial role in the sequel. Clearly, R_{Φ} and S_{Φ} are nondecreasing on [0, ∞). It is known [14], [9] that

Φ(S_{Φ}(t)) ≤ CΦ̃(t),  t ≥ 0,  (2.3)

and, by convexity,

Φ(λS_{Φ}(t)) ≤ CλΦ̃(t),  t ≥ 0, λ ∈ (0, 1).  (2.4)

We say that Φ is of bounded type near zero (near infinity), and write Φ ∈ B_{0} (Φ ∈ B_{∞}) if R_{Φ}(t) ≥ a > 0 (or R_{Φ}(t) ≤ a < ∞) for all t > 0. This classification was introduced in [9]. It was proved in [9] that

R_{Φ}(t) ≥ a,  t > 0  ⇔  Φ̃(t) ≡ 0,  t ∈ [0, a],
R_{Φ}(t) ≤ a,  t > 0  ⇔  Φ̃(t) ≡ ∞,  t ∈ (a, ∞),
Φ(t) ≡ 0,  t ∈ [0, a]  ⇔  S_{Φ}(t) ≥ a,  t > 0,
Φ(t) ≡ ∞,  t ∈ (a, ∞)  ⇔  S_{Φ}(t) ≤ a,  t > 0.

The functions R_{Φ} and S_{Φ} need not be injective. However, thanks to convexity of Φ, they can be constant on intervals only in a few special cases (this is the main difference between R_{Φ} and Φ′), namely, if R_{Φ} is equal to a constant on an interval (a, b), then it must be a = 0 (b may be ∞). On the rest of its domain R_{Φ} is strictly increasing and thus invertible. Of course, the same holds for S_{Φ}.

It follows easily from (2.3) that

R_{Φ}(t) ≤ S_{Φ}^{−1}(Ct)  (2.5)

holds for admissible t (that is, for t such that Ct belongs to the range of the invertible part of S_{Φ}). We shall also make use of the (converse) estimate

S_{Φ}^{−1}(t) ≤ 2R_{Φ}(2t),  admissible t.  (2.6)

To prove (2.6) substitute in (2.2) t→Φ̃(t) to get Φ̃(t) ≤ Φ(2S_{Φ}(t)). The complementary version of the last inequality reads as Φ(t) ≤ Φ̃(2R_{Φ}(t)), which yields t ≤ 2S_{Φ}(2R_{Φ}(t)). Putting now t → 2t and assuming that 2t is admissible we get (2.6).

Let us introduce the notion of index of a nondecreasing function.

Putting h(λ) = sup_{t>0} Φ(λt)/Φ(t), λ ≥ 0, we define the lower index of Φ as i(Φ) = lim_{λ→0+} log h(λ)/ log λ and the upper index of Φ as

I(Φ) = lim_{λ→∞} log h(λ)/ log λ.

It follows easily from the definitions that for every ε > 0 there exists C_{ε} ≥ 1 such that

Φ(λt) ≤ C_{ε} max {λ^{i(Φ)−ε}, λ^{I(Φ)+ε}} Φ(t), t ≥ 0, λ ≥ 0, (2.7)

and

min {μ^{i(Φ)−ε}, μ^{I(Φ)+ε}} Φ(t) ≤ C_{ε}Φ(μt), t ≥ 0, μ ≥ 0. (2.8)

Let us recall that Φ ∈ Δ_{2} is equivalent to I(Φ) < ∞, and Φ̃ ∈ Δ_{2} is equivalent to i(Φ) > 1.

We define the weighted modular by m_{ϱ}(f, Φ) = ∫_{−∞}^{∞}Φ(|f(x)|)ϱ(x) dx; then

the weighted Orlicz space L_{Φ,ϱ} is the set of all functions f for which m_{ϱ}(f/λ, Φ) is finite for some λ > 0. This space can be equipped with the Orlicz norm

|||f|||_{Φ,ϱ} = sup { ∫_{−∞}^{∞}fgϱ, m_{ϱ}(g, Φ̃) ≤ 1} ,

and also with the Luxemburg norm

∥f∥_{Φ,ϱ} = inf{λ > 0, m_{ϱ}(f/λ, Φ) ≤ 1}.

The norms are equivalent, and the unit ball in L_{Φ,ϱ} with respect to the Luxemburg norm coincides with the set of all functions f such that m_{ϱ}(f, Φ) ≤ 1. The Hölder inequality

∫_{−∞}^{∞} fgϱ dx ≤ ∥f∥_{Φ,ϱ} · |||g|||_{Φ̃,ϱ}

holds, and is saturated in the sense that

∥f∥_{Φ,ϱ} = sup { ∫_{−∞}^{∞}fgϱ dx, |||g|||_{Φ̃,ϱ} ≤ 1}

and

|||f|||_{Φ,ϱ} = sup { ∫_{−∞}^{∞}fgϱ dx, ∥g∥_{Φ̃,ϱ} ≤ 1} .

The norm topology is stronger than the modular one, whence the modular inequality ∫ Φ(T f)ϱ ≤ C ∫ Φ(C|f|)ϱ implies its norm counterpart ∥T f∥_{Φ,ϱ} ≤ C∥f∥_{Φ,ϱ}, where T is any positive homogeneous operator.

As usual, given measurable functions h, g and a measurable set E, |E| means ∫_{E} dx, h(E) means ∫_{E} h, h_{E} means |E|^{−1}h(E), and h ( { g > λ }) means ∫_{{x∈R,g(x)>λ}} h(t)dt.

The letter I will always denote an interval in R, and if I = (a, b), we put I′ = (b, 2b − a), I^{∗} = (a, 2b − a), and let αI, α > 0, be the interval concentric with I and α times as long.

If ϱ(2I) ≤ Cϱ(I) for all I, we say that ϱ is a doubling weight.

We say that ϱ ∈ A_{Φ} if either Φ ∉ B_{0} ∪ B_{∞} and there exist C, ε such that

sup_{α>0} sup_{I} αϱ_{I}R_{Φ}(\frac{ε}{|I|} ∫_{I} S_{Φ}(\frac{1}{αϱ(x)}) dx) ≤ C,

or Φ ∈ B_{0} ∪ B_{∞} and ϱ ∈ A_{1}.

We say that ϱ ∈ E_{Φ} if there exist C, ε > 0 such that

sup_{I} \frac{1}{|I|} ∫_{I} S_{Φ}(ε\frac{ϱ_{I}}{ϱ(x)}) dx ≤ C.

3. Strong type inequalities for the maximal operator and the Hilbert transform. We start by considering the strong type inequality for the operator M. As known [6], the nonweighted inequality

∫ Φ(Mf) ≤ C ∫ Φ(|f|)

holds if and only if Φ̃ ∈ Δ_{2}. Kerman and Torchinsky [11] proved that under the assumption that both Φ and Φ̃ satisfy the Δ_{2} condition the weighted inequality

∫ Φ(Mf)ϱ ≤ C ∫ Φ(|f|) ϱ

is equivalent to the condition

sup_{α,I} (\frac{α}{|I|} ∫_{I} ϱ(x) dx) ϕ (\frac{1}{|I|} ∫_{I} ϕ^{−1} (\frac{1}{αϱ(x)}) dx) ≤ C,

where ϕ = Φ′.

As we shall see, the assumption Φ ∈ Δ_{2} can be removed. On the other hand, the assumption Φ̃ ∈ Δ_{2}, at least near infinity, is necessary.

Theorem 1. Assume that ϱ and Φ are such that ϱ ∈ A_{Φ} and Φ̃ ∈ Δ_{2}. Then there exists C such that for every f the inequality

∫_{−∞}^{∞} Φ(Mf(x))ϱ(x) dx ≤ C ∫_{−∞}^{∞} Φ(|f(x)|)ϱ(x) dx  (3.1)

holds.

Conversely, if (3.1) holds with C independent of f, then ϱ ∈ A_{Φ} and Φ̃ ∈ Δ_{2}^{∞}.

We shall need the following two observations:

Lemma 1. If

|||Mf|||_{Φ,ϱ} ≤ C |||f|||_{Φ,ϱ},  (3.2)

then Φ̃ ∈ Δ_{2}^{∞}.

Lemma 2. If Φ̃ ∈ Δ_{2} and ϱ ∈ A_{Φ}, then there exists a function Φ_{0} such that ϱ ∈ A_{Φ_{0}} and i(Φ_{0}) < i(Φ).

Proof of Theorem 1. Necessity. As mentioned above, the modular inequality (3.1) implies (3.2). Necessity of Φ ∈ Δ_{2}^{∞} thus follows from Lemma 1. As proved in [14], ϱ ∈ A_{Φ} is necessary even for the weak type inequality, the more so for (3.1).

Sufficiency. Let Φ_{0} be the function from Lemma 2. Put p = i(Φ_{0}) and F_{p}(t) = Φ(t^{1/p}). Then i(F_{p}) = \frac{1}{p}i(Φ) > 1, whence the weighted maximal operator M_{ϱ}f = sup ϱ(I)^{−1} ∫_{I} |f|ϱ is bounded on L_{F_{p},ϱ} [6]. Moreover, ϱ ∈ A_{Φ_{0}} implies ϱ ∈ A_{p}, and (Mf)^{p} ≤ CM_{ϱ}(f^{p}) [11]. Thus,

∫ Φ(Mf)ϱ = ∫ F_{p} ((Mf)^{p}) ϱ ≤ ∫ F_{p} (M_{ϱ} (C|f|^{p})) ϱ ≤
≤ C ∫ F_{p} (C|f|^{p}) ϱ = C ∫ Φ(C|f|)ϱ. ■

Proof of Lemma 1. Fix a K > 0 such that the set E = { K^{−1} ≤ ϱ(x)≤K} has positive measure. Let x be a density point of E; with no loss of generality let x = 0. Fix a_{0} such that |E ∩ [0, a)| ≥ \frac{3}{4}a for all a ≤ a_{0}. Then, for such a,

|E ∩ (4^{−1}a, a)| ≥ \frac{1}{2}a.  (3.3)

Indeed, we have

|E ∩ (4^{−1}a, a)| = |E ∩ (0, a)| − |E ∩ (0, 4^{−1}a)| ≥ \frac{3}{4}a − \frac{1}{4}a.

From this we obtain the following observation to be used below: Since \frac{1}{x} is a decreasing function, we have for every a ∈ (0, a_{0})

∫_{E∩(4^{−1}a,a)} \frac{dx}{x} ≥ ∫_{2^{−1}a}^{a} \frac{dx}{x} = log 2.  (3.4)

Moreover, for every a ∈ (0, a_{0}),

|E ∩ (0, a)| ≤ a = 4^{m}|(0, 4^{−m}a)| ≤ 3^{−1}4^{m+1}|E ∩ (0, 4^{−m}a)|,

and so, by the definition of E,

ϱ(E ∩ (0, a)) ≤ K|E ∩ (0, a)| ≤ K3^{−1}4^{m+1}|E ∩ (0, 4^{−m}a)| ≤
≤ K^{2}3^{−1}4^{m+1}ϱ(E ∩ (0, 4^{−m}a)).  (3.5)

For m ∈ N and a fixed b ∈ (0, a_{0}) put f_{m}(x) = χ_{E∩(0,4^{−m}b)}(x). Then, by (3.5),

|||f_{m}|||_{Φ,ϱ} = ϱ(E ∩ (0, 4^{−m}b)) · Φ̃^{−1} (\frac{1}{ϱ(E ∩ (0, 4^{−m}b))}) ≤
≤ K|E ∩ (0, 4^{−m}b)| · Φ̃^{−1}(\frac{K^{2}3^{−1}4^{m+1}}{ϱ(E ∩ (0, b))}).  (3.6)

Moreover, for x ∈ (4^{−m}b, b), Mf_{m}(x) ≥ x^{−1}|E ∩ (0, 4^{−m}b)|. Therefore, setting

g(x) = Φ̃^{−1} (\frac{1}{ϱ(E ∩ (0, b))}) · χ_{E∩(0,b)}(x),

we get ∫ Φ̃(g)ϱ = 1, and thus

|||Mf_{m}|||_{Φ,ϱ} ≥ ∫_{−∞}^{∞} Mf_{m}(x)g(x)ϱ(x) dx ≥

≥ |E ∩ (0, 4^{−m}b)|Φ̃^{−1} (\frac{1}{ϱ(E ∩ (0, b))}) ∫_{E∩(4^{−m}b,b)} \frac{ϱ(x)}{x} dx ≥

≥ |E ∩ (0, 4^{−m}b)|Φ̃^{−1} (\frac{1}{ϱ(E ∩ (0, b))}) K^{−1} ∑_{n=1}^{m} ∫_{E∩(4^{−n}b,4^{−n+1}b)} \frac{dx}{x} ≥

≥ by (3.4) ≥

≥ |E ∩ (0, 4^{−m}b)|Φ̃^{−1} (\frac{1}{ϱ(E ∩ (0, b))}) K^{−1}m log 2.  (3.7)

Combining (3.2), (3.6) and (3.7), we arrive at

Φ̃^{−1} (\frac{1}{ϱ(E ∩ (0, b))}) · m ≤ \frac{CK^{2}}{log 2} · Φ̃^{−1} (\frac{K^{2}3^{−1}4^{m+1}}{ϱ(E ∩ (0, b))}) .

Choose m = 2CK^{2}/ log 2. Since m does not depend on b, the last inequality can be rewritten as

2Φ̃^{−1}(t) ≤ Φ̃^{−1}(C_{0}t), t ≥ t_{0},

with C_{0} = 3^{−1}4^{m+1}K^{2} and t_{0} = (ϱ(E ∩ (0, a_{0})))^{−1}. In other words, Φ̃ ∈ Δ_{2}^{∞}. ■

Proof of Lemma 2. Fix α > 0 and I and define v = S_{Φ}(1/αϱ). We claim that v ∈ A_{∞}, that is, there exist α and β independent of I such that the set

E_{β} = {x ∈ I; v(x) > βv_{I}}

satisfies |E_{β}| > α|I|.

We have to distinguish several cases. First assume that i(Φ) = ∞. Since ϱ ∈ A_{Φ} always implies ϱ ∈ A_{∞} [9], and ϱ ∈ A_{∞} always implies ϱ ∈ A_{p} for certain p < ∞, in this case the assertion of the lemma is easily satisfied.

Suppose i(Φ) < ∞. Since Φ̃ ∈ Δ_{2}, Φ cannot be of any bounded type. However, this is not true for Φ̃, so it can be either

(i) S_{Φ}(0, ∞) = (0, ∞);

(ii) S_{Φ}(0, ∞) = (0, a); (iii) S_{Φ}(0, ∞) = (a, ∞); or (iv) S_{Φ}(0, ∞) = [a, ∞); with some positive a. Note that in case (ii) S_{Φ} is invertible on (0, a),

and in cases (iii) and (iv) S_{Φ} is invertible on (a, ∞). Choose γ ∈ (1, i(Φ)) arbitrarily. Then, by (2.7),

Φ(λt) ≤ C_{γ} · λ^{γ} · Φ(t),  t ≥ 0, λ ∈ (0, 1).  (3.8)

Let ε be the constant from A_{Φ}. Choose β ≤ ε/2 in order that

C_{ϱ} · 2^{γ} · C_{γ} · (\frac{β}{ε} )^{γ−1} ≤ \frac{1}{2},  (3.9)

where C_{ϱ} is an A_{Φ} constant for the weight ϱ. Given fixed I, suppose that βv_{I} is admissible for S_{Φ}^{−1}. We may then conclude from A_{Φ} that

\frac{C_{ϱ}}{R_{Φ}(εv_{I})} ≥ \frac{1}{|I|} ∫_{I} \frac{dx}{S_{Φ}^{−1}(v(x))} ≥ \frac{|I\E_{β}|}{|I|} \frac{1}{S_{Φ}^{−1}(βv_{I})}.  (3.10)

Hence, by (3.10), (2.6), (3.8) and (3.9),

\frac{|I\E_{β}|}{|I|} ≤ C_{ϱ} \frac{S_{Φ}^{−1}(βv_{I})}{R_{Φ}(εv_{I})} ≤ 2C_{ϱ} \frac{R_{Φ}(2βv_{I})}{R_{Φ}(εv_{I})} = \frac{C_{ϱ}ε}{β} \frac{Φ(\frac{2β}{ε} εv_{I})}{Φ(εv_{I})} ≤
≤ C_{ϱ}2^{γ}C_{γ}(\frac{β}{ε} )^{γ−1} ≤ \frac{1}{2},  (3.11)

or |E_{β}| > \frac{1}{2}|I|.

Now suppose that βv_{I} is not admissible for S_{Φ}^{−1}. This is possible only in case (iii) or (iv) if βv_{I} ≤ a. But then, of course, E_{β} = I, and the desired estimate is trivial. Therefore, v ∈ A_{∞}.

Now, as known [5], v satisfies the reverse Hölder inequality, that is, there are positive C and δ such that

(\frac{1}{|I|} ∫_{I} v^{1+δ}(x) dx)^{1/(1+δ)} ≤ \frac{C}{|I|} ∫_{I} v(x) dx  (3.12)

for all I.

We define the function Φ_{0} by means of its complementary function: put

S_{Φ_{0}} = S_{Φ}^{1+δ}, that is, Φ̃_{0}(t) = t · [S_{Φ}(t)]^{1+δ} .

Then, obviously,

I(Φ̃_{0}) = I(Φ̃) + δ(I(Φ̃) − 1).

The case I(Φ̃) = 1 (that is, i(Φ) = ∞) was already excluded at the beginning. On the other hand, I(Φ̃) cannot be ∞, since Φ̃ ∈ Δ_{2}. Consequently, I(Φ̃_{0})>I(Φ̃), which is of course equivalent to i(Φ_{0}) < i(Φ).

It remains to prove ϱ ∈ A_{Φ_{0}}. We start by rewriting (3.12) as

\frac{ε}{2C} (\frac{1}{|I|} ∫_{I} v^{1+δ}(x) dx)^{1/(1+δ)} ≤ \frac{ε}{2|I|} ∫_{I} v(x) dx.  (3.13)

Suppose first that everything is admissible for S_{Φ}^{−1}. Then, as S_{Φ}^{−1} is nondecreasing,

S_{Φ}^{−1} (\frac{ε}{2C} (\frac{1}{|I|} ∫_{I} v^{1+δ}(x) dx)^{1/(1+δ)})

≤ S_{Φ}^{−1} (\frac{ε}{2|I|} ∫_{I} v(x) dx) .  (3.14)

Note that S_{Φ_{0}}^{−1}(t) = S_{Φ}^{−1}(t^{1/(1+δ)}) for admissible t. So, (3.14) gives

S_{Φ_{0}}^{−1} ((\frac{ε}{2C} )^{1+δ} \frac{1}{|I|} ∫_{I} v^{1+δ}(x) dx) ≤

≤ S_{Φ}^{−1} (\frac{ε}{2|I|} ∫_{I} v(x) dx) ,  (3.15)

which by means of (2.5) and (2.6) yields

R_{Φ_{0}} ((\frac{ε}{2C} )^{1+δ} \frac{1}{|I|} ∫_{I} S_{Φ_{0}} (\frac{1}{αϱ(x)}) dx) ≤

≤ 2R_{Φ} (\frac{ε}{|I|} ∫_{I} S_{Φ} (\frac{1}{αϱ(x)}) dx) .  (3.16)

Now assume that it was not possible to apply S_{Φ}^{−1} in (3.13). This can happen only in case (iii) or (iv) and

\frac{ε}{2C} (\frac{1}{|I|} ∫_{I} v^{1+δ}(x) dx)^{1/(1+δ)} < a.  (3.17)

Note that in the cases (iii) or (iv) it is for all t S_{Φ}(t) ≥ a, that is, S_{Φ_{0}}(t) ≥ a^{1+δ}, which is equivalent to R_{Φ_{0}}(t) = 0 for t ≤ a^{1+δ}. Thus, in this case (3.16) holds trivially. It is clear that from A_{Φ} and (3.16) A_{Φ_{0}} already follows. The proof is finished. ■

The method of the proof of Lemma 2 is the same as that in [11], the only slight change being that we have replaced derivatives of Φ and Φ̃ by R_{Φ} and S_{Φ}. Actually, our proof shows that the condition Φ̃ ∈ Δ_{2} is not required as an assumption, and allows us to insert it as part of the statement of the strong maximal theorem.

Let us turn our attention to the Hilbert transform.

It follows easily from the Kerman – Torchinsky theorem that if Φ ∈ Δ_{2} and Φ̃ ∈ Δ_{2}, then ϱ ∈ A_{Φ} is necessary and sufficient for

∫ Φ(|Hf|)ϱ ≤ C ∫ Φ(C|f|)ϱ.  (3.18)

Indeed, for sufficiency we use Coifman’s inequality [4]

∫ Φ(|Hf|)ϱ ≤ C ∫ Φ(Mf)ϱ

which is valid provided that Φ ∈ Δ_{2} and ϱ ∈ A_{∞}. However, Φ ∈ Δ_{2} is an assumption, and ϱ ∈ A_{∞} follows from ϱ ∈ A_{Φ} [9].

It may be of some interest that both Φ ∈ Δ_{2} and Φ̃ ∈ Δ_{2} are also necessary for (3.18). We have the following characterization of the strong type inequality for the Hilbert transform.

Theorem 2. The inequality

∫_{−∞}^{∞} Φ(|Hf(x)|)ϱ(x) dx ≤ C ∫_{−∞}^{∞} Φ(|f(x)|)ϱ(x) dx  (3.19)

holds if and only if Φ ∈ Δ_{2}, Φ̃ ∈ Δ_{2}, and ϱ ∈ A_{Φ}.

We shall make use of the following assertion.

Lemma 3. Let us define the operator

G_{ϱ}f(x) = \frac{1}{ϱ(x)} · H(fϱ)(x).

Then the following statements are equivalent.

(i) There is C such that

∫_{−∞}^{∞} Φ(|Hf(x)|)ϱ(x) dx ≤ C ∫_{−∞}^{∞} Φ(C|f(x)|)ϱ(x) dx;

(ii)There is C such that

∫_{−∞}^{∞} Φ̃(|G_{ϱ}f(x)|)ϱ(x) dx ≤ C ∫_{−∞}^{∞} Φ̃(C|f(x)|)ϱ(x) dx.  (3.20)

Proof of Theorem 2. The “if part” has already been established. It thus remains to show Φ ∈ Δ_{2} and Φ̃ ∈ Δ_{2}. It was proved in [9] that even the weak type inequality with Hilbert transform implies Φ ∈ Δ_{2} and ϱ ∈ A_{Φ}.

It remains to prove Φ̃ ∈ Δ_{2}. By Lemma 3, (3.19) is equivalent to (3.20). Of course, (3.20) implies the weak type inequality

ϱ({|G_{ϱ}f| > λ}) · Φ(λ) ≤ C ∫_{−∞}^{∞} Φ(Cf)ϱ.  (3.21)

Take K positive such that the set E = {K^{−1} ≤ ϱ(x) ≤ K} has positive measure, and for any λ > 0 define f = \frac{λ}{2C} χ_{E_{0}}, where E_{0} is any bounded subset of E. Inserting f into (3.21) we get

Φ̃(λ) ≤ C \frac{ϱ(E_{0})}{ϱ(\{|G_{ϱ}f| > 2C\})} · Φ̃(\frac{λ}{2}),

in other words, Φ̃ ∈ Δ_{2}. The idea is due to A. Gogatishvili [8]. ■

To prove Lemma 3 we employ the following result of D. Gallardo which was communicated to the author personally [7]. We give a sketch of the proof since as far as we know the author has not published it. When this manuscript was written, we learned that the same assertion was proved in a preprint by Bloom and Kerman ([3], Proposition 2.5).

Lemma 4. Let T be a positively homogeneous operator. Then the modular estimate

∫ Φ(|T f(x)|)ϱ(x) dx ≤ C ∫ Φ(C|f(x)|)ϱ(x) dx

is equivalent to the existence of C such that for all ε and f the norm inequality ∥T f∥_{Φ,εϱ} ≤ C∥f∥_{Φ,εϱ} holds.

Proof of Lemma 4. Let the norm inequality be satisfied with C independent of ε and f. Then, by the definition of the Luxemburg norm, ∫ Φ((C∥f∥_{Φ,εϱ})^{−1}|T f(x)|)εϱ(x) dx ≤ 1. Fix f, a function with finite modular, and put ε = (∫ Φ(C|f|)ϱ)^{−1} > 0. Then ∥Cf∥_{Φ,εϱ} = 1, and so

∫ Φ(|T f|)εϱ = ∫ Φ (\frac{C|T f(x)|}{C∥Cf∥_{Φ,εϱ}} ) εϱ ≤ 1.

Inserting ε, we are done. The converse implication is obvious. ■

Proof of Lemma 3. By Lemma 4, (i) is equivalent to

∥Hf∥_{Φ,εϱ} ≤ C∥f∥_{Φ,εϱ}, all ε.

That is,

C ≥ sup_{∥f∥_{Φ,εϱ}≤1} ∥Hf∥_{Φ,εϱ} =

= sup_{∥f∥_{Φ,εϱ}≤1} sup_{∥g∥_{Φ̃,εϱ}≤1} ∫_{−∞}^{∞} |Hf(x)|g(x)εϱ(x) dx =

= sup_{∥g∥_{Φ̃,εϱ}≤1} sup_{∥f∥_{Φ,εϱ}≤1} ∫_{−∞}^{∞} |f(x)| · \frac{1}{ϱ(x)}|H(gϱ)(x)|εϱ(x) dx =

= sup_{∥g∥_{Φ̃,εϱ}≤1} ∥G_{ϱ}g∥_{Φ̃,εϱ},

which is, again by Lemma 4, equivalent to (ii). ■

4. The Hilbert transform for odd functions. In this section we shall make use of the measure ν defined on (0, ∞) by dν(x) = x dx.

We say that ϱ ∈ A_{Φ}^{o} if either Φ ∉ B_{0} ∪ B_{∞} and there exist positive C, ε such that

sup_{α,I} (\frac{α}{ν(I)} ∫_{I} \frac{ϱ(x)}{x} dν) R_{Φ} (\frac{ε}{ν(I)} ∫_{I} S_{Φ} (ε\frac{x}{αϱ(x)}) dν) ≤ C

or Φ ∈ B_{0} ∪ B_{∞} and ϱ ∈ A_{1}^{o}, that is,

\frac{ϱ(I)}{ν(I)} ≤ Cess inf_{I} \frac{ϱ(x)}{x}.

We say that ϱ ∈ E_{Φ}^{o} if there exist positive C, ε such that

sup_{I} \frac{1}{ν(I)} ∫_{I} S_{Φ} (ε\frac{x}{ϱ(x)} \frac{ϱ(I)}{ν(I)}) dν ≤ C.

Remark. Obviously, ϱ ∈ A_{1}^{o} implies ϱ ∈ A_{Φ}^{o} for any Φ. Further, putting α = \frac{ν(I)}{ϱ(I)} we obtain that ϱ ∈ A_{Φ}^{o} implies ϱ ∈ E_{Φ}^{o} for any Φ.

We shall prove the following theorems.

WEIGHTED ESTIMATES FOR THE HILBERT TRANSFORM 91

Theorem 3. The strong type inequality

∫_{0}^{∞}Φ(|H_{o}f(x)|)ϱ(x) dx ≤ C ∫_{0}^{∞}Φ(C|f(x)|)ϱ(x) dx  (4.1)

holds with C independent of f if and only if Φ ∈ Δ_{2}, Φ̃ ∈ Δ_{2}, and ϱ ∈ A_{Φ}^{o}. Theorem 4. The weak type inequality

ϱ({|H_{o}f| > λ}) · Φ(λ) ≤ C ∫_{0}^{∞}Φ(C |f(x)|) ϱ(x) dx  (4.2)

holds with C independent of f and λ if and only if Φ ∈ Δ_{2}, and ϱ ∈ A_{Φ}^{o}. Theorem 5. Let Φ ∈ Δ_{2}^{0}. Then the extra–weak type inequality

ϱ(\{|H_{o}f| > λ\}) ≤ C ∫_{0}^{∞}Φ(\frac{C |f(x)|}{λ}) ϱ(x) dx  (4.3)

holds with C independent of f and λ if and only if ϱ ∈ E_{Φ}^{o}.

The following auxiliary assertion is a modification of Lemma 1 from [1]. Lemma 5. Define

σ(x) = \frac{ϱ(\sqrt{|x|})}{2\sqrt{|x|}}, x ≠ 0.

Then ϱ ∈ A_{Φ}^{o} if and only if σ ∈ A_{Φ}, and ϱ ∈ E_{Φ}^{o} if and only if σ ∈ E_{Φ}.

Proof of Lemma 5. Let I = (a, b), a > 0, and put J = (\sqrt{a}, \sqrt{b}). Easily, |I| = 2ν(J) and σ(I) = ϱ(J). Therefore,

ασ_{I}R_{Φ} (\frac{γ}{|I|} ∫_{I} S_{Φ} (\frac{1}{ασ(x)}) dx) =

= \frac{αϱ(J)}{2ν(J)}R_{Φ} (\frac{γ}{ν(J)} ∫_{J} S_{Φ} (\frac{2y}{αϱ(y)}) dν(y)) ,

and analogously

∫_{I} Φ̃ (ε\frac{σ_{I}}{σ(x)}) \frac{σ(x)}{σ(I)}dx = ∫_{J} Φ̃ (ε\frac{yϱ(J)}{ϱ(y)ν(J)}) \frac{ϱ(y)}{ϱ(J)}dy.

Similar argument holds for b < 0, and, in the remaining case, we split the interval into two. ■

Proof of Theorem 4. Necessity. First we claim that if (4.3) holds, then ϱ is a doubling weight.

Let I = (a, b), 0 < a < b < ∞. Assume that supp f ⊂ I. Since ν(I^{∗}) = 2b(b − a) > (b^{2} − a^{2}),

H_{o}f(x) ≥ \frac{1}{πν(I^{∗})} ∫_{I} f(y) dν(y),  x ∈ I′.  (4.4)

Now, easily 4ν(I) > ν(I^{∗}), so inserting f = χ_{I} in (4.4) yields H_{o}χ_{I}(x) > (4π)^{−1} for all x ∈ I′. This together with (4.3) leads to ϱ(I′) ≤ Cϱ(I). By symmetry,

C^{−1}ϱ(I) ≤ ϱ(I′) ≤ Cϱ(I),  (4.5)

and the doubling condition follows.

Now we shall show, using again the idea from [8], that Φ ∈ Δ_{2}. Given λ, set f = (2C)^{−1}λχ_{I}, where C is from (4.2) and I is an appropriate interval. It then follows from (4.2) that

Φ(λ) ≤ C \frac{ϱ(I)}{ϱ(\{|H_{o}χ_{I}| > 2C\})}Φ(λ/2), λ > 0,

that is, Φ ∈ Δ_{2}.

It remains to show that ϱ ∈ A_{Φ}^{o}. Given α > 0, and I = (a, b), put

f = C^{−1}S_{Φ} (\frac{γ}{α} \frac{x}{ϱ(x)}) χ_{I}(x),

where C is from (4.2), and

λ = \frac{1}{2πν(I^{∗})} ∫_{I} f dν.

Then by (4.4), (2.1), (4.2) and (2.3)

ϱ(I′) · Φ(λ) ≤ C ∫_{I} ΦS_{Φ} (\frac{γ}{α} \frac{x}{ϱ(x)}) ϱ(x) dx ≤ Cγα^{−1} ∫_{I} f dν,

or, by (4.5),

\frac{αϱ(I)}{ν(I)}Φ (\frac{1}{8πCν(I)} ∫_{I} S_{Φ} (\frac{γ}{α} \frac{x}{ϱ(x)}) dν(x)) ≤

≤ \frac{Cγ}{ν(I)} ∫_{I} S_{Φ} (\frac{γ}{α} \frac{x}{ϱ(x)}) dν(x) .  (4.6)

Denote A = ∫_{I} f dν. Obviously, A > 0. Assume that A = ∞. Then ∫_{I} Φ̃(\frac{γ}{α} \frac{x}{ϱ(x)})ϱ(x)dx = ∞, and there must exist a function g ∈ L_{Φ,ϱ}(I) such

that

∞ = ∫_{I} \frac{γ}{α} \frac{x}{ϱ(x)}g(x) ϱ(x) dx = \frac{γ}{α} ∫_{I} g dν.

This and (4.4) would give H_{o}(εg)(x) = ∞ for all x ∈ I′ and ε > 0 and, by (4.2),

ϱ(I′)Φ(λ) ≤ C ∫_{I} Φ(εCg(x))ϱ(x) dx, λ, ε > 0.

Since g ∈ L_{Φ,ϱ}, there must be ε such that the last integral is finite, and so it follows that ϱ(I′) = 0. However, since ϱ is doubling and nontrivial, this is impossible. Hence 0 < A < ∞ and we can divide both sides of (4.6) by ν(I)^{−1}A to get ϱ ∈ A_{Φ}^{o}. ■

Sufficiency. By Lemma 5, ϱ ∈ A_{Φ}^{o} implies σ ∈ A_{Φ}. We thus have from Theorem A

σ({Hg > λ}) · Φ(λ) ≤ C ∫_{−∞}^{∞} Φ(C|g(x)|)σ(x) dx.

For given f on (0, ∞) put g(x) = f(\sqrt{x}) for x > 0, and 0 elsewhere. Then Hg(x) = (H_{o}f)(\sqrt{x}) [1], and therefore

ϱ({x > 0, |H_{o}f(x)| > λ}) ≤ σ({x ∈ R, |Hg(x)| > λ}) ≤

≤ \frac{C}{Φ(λ)} ∫_{0}^{∞}Φ(Cf(y))ϱ(y) dy. ■

Proof of Theorem 3. Necessity. By Theorem 4, ϱ ∈ A_{Φ} and Φ ∈ Δ_{2} are necessary even for the weak type inequality. It remains to prove Φ̃ ∈ Δ_{2}. In the same way as in Lemma 3 and Lemma 4 we can prove that (4.1) implies

ϱ(\{\frac{1}{ϱ(x)}|H_{e}(fϱ)(x)| > λ\})Φ̃(λ) ≤ C ∫_{0}^{∞}Φ̃(C|f(x)|)ϱ(x) dx.

For the definition of H_{e} see (1.7). Putting f = \frac{λ}{2C} χ_{E_{0}} similarly as in the proof of Lemma 3 we get Φ̃ ∈ Δ_{2}.

Sufficiency. By Lemma 5 and Lemma 2, ϱ ∈ A_{Φ}^{o} implies σ ∈ A_{Φ_{0}} with i(Φ_{0}) < i(Φ). By Theorem A, σ ∈ A_{Φ_{0}} and Φ_{0} ∈ Δ_{2} imply the weak type inequality

σ({x ∈ R; Hg(x) > λ})Φ_{0}(λ) ≤ C ∫_{−∞}^{∞} Φ_{0}(Cg(x))σ(x) dx

for every g. Given f on (0, ∞), we put g = f(\sqrt{x})·χ_{\{x>0\}}, x ∈ R. A change of variables then gives

ϱ({x > 0; H_{o}f > λ})Φ_{0}(λ) ≤ C ∫_{0}^{∞}Φ_{0}(Cf(x))ϱ(x) dx,

which yields the assertion by the usual interpolation argument. ■

Proof of Theorem 5. Necessity. First assume that Φ ∉ B_{∞}. Note that then S_{Φ} is finite on (0, ∞). Fix k ∈ N and an interval I, put I_{k} = {x ∈ I, x ≤ kϱ(x)}, and define

h(x) = h_{k}(x) = S_{Φ} (ε\frac{x}{ϱ(x)} \frac{ϱ(I)}{ν(I)}) χ_{I_{k}}(x)

with ε to be specified later. Put

β_{I} = \frac{1}{ν(I)} ∫_{I} h dν.

Now, assume that K is the biggest of the constants C from (2.4), (4.3), and (4.5). We then have from (4.5) that ϱ(I) ≤ Kϱ({|H_{0}h| ≥ (4π)^{−1}β_{I}}). Therefore, by (4.3) with f = h and λ < (4π)^{−1}β_{I},

∫_{I_{k}} Φ̃(ε\frac{x}{ϱ(x)} \frac{ϱ(I)}{ν(I)})ϱ(x) dx = ε\frac{ϱ(I)}{ν(I)} ∫_{I_{k}} S_{Φ}(ε\frac{x}{ϱ(x)} \frac{ϱ(I)}{ν(I)}) dν =
= εϱ(I)β_{I} ≤ 4πKεϱ(I) + δ_{I},

where δ_{I} = 0 if β_{I} ≤ 4πK, and

δ_{I} = K^{2}εβ_{I} ∫_{I_{k}} Φ(\frac{4πK}{β_{I}} h(x))ϱ(x) dx, if β_{I} > 4πK.

In any case, using (2.4) with λ = 4πK/β_{I} we get

∫_{I_{k}} Φ̃(ε\frac{x}{ϱ(x)} \frac{ϱ(I)}{ν(I)})ϱ(x) dx ≤

≤ 4πKεϱ(I) + 4πK^{3}ε ∫_{I_{k}} Φ̃(ε\frac{x}{ϱ(x)} \frac{ϱ(I)}{ν(I)})ϱ(x) dx.

Now, since S_{Φ} is finite, we have

∫_{I_{k}} Φ̃(ε\frac{x}{ϱ(x)} \frac{ϱ(I)}{ν(I)})ϱ(x) dx = ε\frac{ϱ(I)}{ν(I)} ∫_{I_{k}} S_{Φ}(ε\frac{x}{ϱ(x)} \frac{ϱ(I)}{ν(I)}) dν ≤

≤ εϱ(I)S_{Φ}(εk \frac{ϱ(I)}{ν(I)}) < ∞,

and hence we can put ε < (4πK^{3})^{−1} and subtract to get

∫_{I_{k}} Φ̃(ε\frac{x}{ϱ(x)} \frac{ϱ(I)}{ν(I)})ϱ(x) dx ≤ \frac{4πKε}{1 − 4πK^{3}ε}ϱ(I),

which yields ϱ ∈ E_{Φ}^{o} sonce the constant on the right does not depend on k.

If Φ ∈ B_{∞}, then Φ(t) ≤ Ct for all t and therefore, inserting f = χ_{E} and λ = ν(E)/(2πν(I^{∗})) into (4.3) we obtain

ϱ(I) ≤ Cϱ(E)Φ(C \frac{ν(I^{∗})}{ν(E)}) ≤ Cϱ(E)\frac{ν(I^{∗})}{ν(E)},

that is, ϱ ∈ A_{1}^{o}. Therefore, in this case ϱ ∈ A_{Φ}^{o} for any Φ (see Remark above).

Sufficiency. By Lemma 5, ϱ ∈ E_{Φ}^{o} implies σ ∈ E_{Φ} , whence, using Theorem B, we have

σ(\{Hg > λ\}) ≤ C ∫_{−∞}^{∞} Φ(\frac{C|g(x)|}{λ})σ(x) dx.

The same argument as in the proof of Theorem 4 now leads to the assertion. ■

Acknowledgment. I am grateful to Professor V. Kokilashvili who directed my attention to the problems concerning odd and even functions.

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(Received 5.10.1992)

Author’s address: Mathematical Institute of the Czechoslovak Academy of Sciences, Žitná 25, 115 67 Praha 1, Czechoslovakia.

Current address: School of Mathematics, University of Wales College of Cardiff, Senghennydd Road, Cardiff CF2 4AG, Wales, U.K.