Georgian Mathematical Journal 1(1994), No. 1, 9-29

CRITERIA OF GENERAL WEAK TYPE INEQUALITIES

FOR INTEGRAL TRANSFORMS WITH POSITIVE KERNELS

I. GENEBASHVILI, A. GOGATISHVILI, AND V. KOKILASHVILI

Abstract. Necessary and sufficient conditions are derived in order that an inequality of the form

φ(λ)θ(β{(x, t) ∈ X × [0, ∞) : K(fdν)(x, t) > λ}) ≤

≤ c ∫_{X} ψ (\frac{f(x)}{η(λ)}) σ(x)dν(x)

be fulfilled for some positive c independent of λ and a ν-measurable nonnegative function f : X → R^{1}, where

K(fdν)(x, t) = ∫_{X} f(y)k(x, y, t)dν(y), t ≥ 0,

k : X ×X ×[0, ∞) → R^{1} is a nonnegative measurable kernel, (X, d, μ) is a homogeneous type space, φη and ψ are quasiconvex functions, ψ ∈ Δ_{2}, and t^{−α}θ(t) is a decreasing function for some α, 0 < α < 1. A similar problem was solved in Lorentz spaces with weights.

1. Introduction

This paper presents a characterization of weight functions and kernels for which we have general weight weak type inequalities for integral transforms of the form

K(fdν)(x, t) = ∫_{X} f(y)k(x, y, t)dν(y),  (1.1)

where X is a homogeneous type space, and k : X × X × [0, ∞) → R^{1} a nonnegative measurable kernel.

The homogeneous type space (X, d, μ) is a space with measure μ such that the class of compactly supported continuous functions is dense in the space L^{1}(X, μ). Moreover, it is also assumed that there is a nonnegative real-valued function d : X × X → R^{1} satisfying the following conditions:

1991 Mathematics Subject Classification. 42B25,26D15,46E30.

(i) d(x, x) = 0 for all x ∈ X;

(ii) d(x, y) > 0 for all x ≠ y in X;

(iii) There is a constant a_{0} such that d(x, y) ≤ a_{0}d(y, x) for all x, y in X; (iv) There is a constant a_{1} such that d(x, y) ≤ a_{1}(d(x, z) + d(z, y)) for all

x, y, z in X;

(v) For each neighborhood V of x in X there is an r > 0 such that the ball B(x, r) = {y ∈ X : d(x, y) < r} is contained in V;

(vi) The balls B(x, r) are measurable for all x and r > 0;

(vii) There is a constant b such that μB(x, 2r) ≤ bμB(x, r) for all x ∈ X and r > 0 (see [1], p. 2).

In the sequel B̂(x, r) will denote the set B(x, r) × [0, 2r) for r > 0 and the one-point set {x} for r = 0 . The set B(x, 0) will be assumed to be empty. β will be a measure defined on the product of σ-algebras generated by balls in X and by intervals from [0, ∞).

Let φ, ψ, and η be nonnegative nondecreasing functions on [0, ∞). For our further discussion we will also need the following basic definitions of quasiconvex functions. We call ω a Young function if it is a nonnegative increasing convex function on [0, ∞) with ω(0) = 0, ω(∞) = ∞ and not identically zero or ∞ on (0, ∞); it may have a jump up to ∞ at some point t > 0, but in that case it should be left continuous at t (see [2]).

The function ψ is called quasiconvex if there exist a Young function ω and a constant c > 1 such that

ω(t) ≤ ψ(t) ≤ ω(ct), t ≥ 0.  (1.2)

Clearly, ψ(0) = 0, and for s ≤ t we have ψ(s) ≤ ψ(ct) . To the quasiconvex function ψ we can put into correspondence its complementary function ψ̃ defined by ψ̃(t) = sup_{s≥0}(st − ψ(s)).

The subadditivity of the supremum easily implies that ψ̃ is always a Young function and (ψ̃)^{}˜ ≤ ψ. The equality holds if ψ is itself a Young function. If ψ_{1} ≤ ψ_{2}, then ψ̃_{2} ≤ ψ̃_{1}, and if ψ_{1}(t) = aψ(bt), then ψ̃_{1}(t) = aψ̃ (\frac{t}{ab}). Hence from (1.2) we have

ω̃ (\frac{t}{c}) ≤ ψ̃(t) ≤ ω̃(t).  (1.3)

Now from the definition of ψ̃ we obtain Young’s inequality

st ≤ ψ(s) + ψ̃(t), s, t ≥ 0.  (1.4)

It should be noted that unlike ψ the function ψ̃ may jump to ∞ at some point t > 0. For example, if ψ(t) = t, then ψ̃(t) = ∞·χ_{(1,∞)}(t) . Throughout the paper we take 0 · ∞ to be zero.

We use the convention that c denotes the absolute constant which may change from line to line.

The function ψ satisfies the (global) Δ_{2} condition (ψ ∈ Δ_{2}) if there exists c > 0 such that ψ(2t) ≤ cψ(t), t > 0.

Some properties of quasiconvex functions and also of functions satisfying the Δ_{2} condition will be presented in Section 2.

Now we are ready to formulate the main results of this paper.

In the sequel θ will always be a positive nondecreasing function.

Theorem 1.1. Let k : X × X × [0, ∞) → R^{1} be any measurable nonnegative kernel, let ψ ∈ Δ_{2}, and let the function t^{−α}θ(t) decrease for some α ∈ (0, 1). Let, further, ν be a finite measure on X, and σ : X → R^{1} be an almost everywhere positive function which is locally summable with respect to measure ν.

Assume that there exist positive constants ε and c_{1} such that

∫_{X\B(a,r)} ψ̃ (ε\frac{φ(s)η(s)}{s} \frac{θ(βB̂(a, N_{0}(2r + t)))}{σ(y)}k(a, y, t)) σ(y)dν(y) ≤

≤ c_{1}φ(s)θ(βB̂(a, N_{0}(2r + t)))  (1.5)

for any s > 0, r ≥ 0, a ∈ X and t ≥ 0, where N_{0} = a_{1}(1 + 2a_{0}) .

Then there exists a positive constant c_{2} such that for any λ > 0 and any nonnegative ν-measurable function f : X → R^{1} the following inequality holds:

φ(λ)θ(β{(x, t) ∈ X × [0, ∞) : K(fdν)(x, t) > λ}) ≤
≤ c_{2} ∫_{X} ψ (\frac{f(x)}{η(λ)}) σ(x)dν(x).  (1.6)

Assume now that the nonnegative measurable kernel k satisfies the following additional condition: there exist numbers N ≥ N_{0}, N_{0} = a_{1}(1 + 2a_{0}) and c′ such that

k(a, y, t) ≤ c′k(x, y, τ)  (1.7)

when y ∈ X \ B(a, r), (x, τ) ∈ B̂(a, N(r + t)) for any a ∈ X, r ≥ 0, t ≥ 0.

Theorem 1.2. Let φη and ψ be quasiconvex functions, let ψ ∈ Δ_{2}, let the function t^{−α}θ(t) decrease for some α ∈ (0, 1), and let k satisfy the condition (1.7).

Then the inequality (1.6) is equivalent to any of the following conditions:

(i) there exist positive constants ε and c_{3} such that

∫_{X\B(a,r)} ψ̃ (ε\frac{φ(s)η(s)}{s} \frac{θ(βB̂(a, r + t))}{σ(y)}k(a, y, t)) σ(y)dν(y) ≤

≤ c_{3}φ(s)θ(βB̂(a, r + t))  (1.8)

for arbitrary s > 0, r ≥ 0, a ∈ X and t ≥ 0; (ii) there exist positive constants ε and c_{4} such that

∫_{X} ψ̃ (ε\frac{φ(s)η(s)}{s} \frac{θ(βB̂(a, t))}{σ(y)}k(a, y, t)) σ(y)dν(y) ≤

≤ c_{4}φ(s)θ(βB̂(a, t))  (1.9)

for any s > 0, a ∈ X and t ≥ 0;

(iii) there exists a positive constant c_{5} such that for any a ∈ X, r ≥ 0, t ≥ 0 and for any nonnegative ν-measurable functions F : X → R^{1}, supp F ⊂ X \ B(a, r) we have

φ (K(F dν)(a, t)) θ(βB̂(a, r + t)) ≤
≤ c_{5} ∫_{X} ψ (\frac{F (x)}{η(K(F dν)(a, t)}) σ(y)dν(y).  (1.10)

Let k : X × X → R^{1} be a nonnegative measurable function satisfying the following condition: there exist numbers N ≥ N_{0} and c′ > 0 such that k(a, y) ≤ c′k(x, y) for any a ∈ X, y ∈ X \ B(a, r), and x ∈ B(a, Nr) .

For any positive function ϱ: X → R^{1}, locally summable with respect to measure ν, it will be assumed below that

ϱE = ∫_{E} ϱ(x)dν(x)

for any ν-measurable set E ⊂ X .

We have

Theorem 1.3. Let the functions φ, η, and ψ satisfy the conditions of Theorem 1.2. Then the following statements are equivalent:

(i) there exists a positive constant c_{6} such that for any λ > 0 and for any measurable nonnegative function f

φ(λ)θ (ϱ{x ∈ X : ∫_{X} k(x, y)f(y)dν(y) > λ}) ≤

≤ c_{6} ∫_{X} ψ (\frac{f(y)}{η(λ)}) σ(y)dν(y)

for any λ>0 and for any measurable nonnegative function f:X → R^{1}; (ii) there exist positive numbers ε and c_{7} such that

∫_{X\B(a,r)} ψ̃ (ε\frac{φ(s)η(s)}{s} \frac{θ(ϱB(a, r))}{σ(y)}k(a, y)) σ(y)dν(y) ≤

≤ c_{7}φ(s)θ(ϱB(a, r))

for any s > 0, a ∈ X, and r ≥ 0 .

The above-formulated results contain the solutions of problems of description of a set of weights ensuring in Orlicz spaces the validity of both weak and extra-weak weighted inequalities for transform (1.1) which are natural analogies of inequalities of the weak type (p, q). Indeed, for φ = ψ, η ≡ 1 (1.6) becomes a weak type weighted inequality, while for φ ≡ 1, η(λ) = λ we obtain an extra-weak type weighted inequality. It is understood that an inequality of the weak type (φ, φ) is essentially stronger than an inequality of the extra-weak type (φ, φ).

The solutions of similar problems in Lorentz spaces are derived in Section 3. Section 4 contains a discussion of the interesting corollaries of Theorems 1.2 and 1.3 for integral operators such as potentials and their generalizations, Poisson integrals and their generalizations, the Hardy operators, and others. Here we give a very brief survey only of the results preceding this paper.

The solution of a weak type two weight problem for Riesz potentials in Lebesgue spaces was obtained in [3], [4], the criterion found in [4] being more easily verifiable. The latter result was extended to the integrals on homogeneous type spaces in [5]. A similar problem was treated in [6] (see also [7], Theorems 6.1.1 and 6.1.2) in Lorentz spaces over R^{n} for integral transforms

Kf(x) = ∫_{R^{n}} k(x, y)f(y)dy.

Subsequently in [8] generalizations were obtained for transforms of type (1.1) when X = R^{n}, dν(y) = dy. More particular cases of generalized potentials and Poisson integrals were considered in [8] and [9], respectively. The latter deals with Lorentz spaces and the former with Lebesgue spaces. In Orlicz classes the problem of description of a set of weights ensuring the validity of weak type weighted inequalities was previously studied mainly for maximal functions [10], [11], [12] and the Hardy operator [13], [14].

2. Proof of the Main Theorems

In this section use will be made of some properties of quasiconvex functions satisfying the Δ_{2} condition, also of the covering lemma in homogeneous type spaces.

Lemma 2.1 ([11], p. 4). The following statements are equaivalent:

(i) φ is a quasiconvex function;

(ii) there exists c > 1 such that

\frac{φ(s)}{s} ≤ \frac{φ(ct)}{t} for s < t.  (2.1)

Hence for quasiconvex functions φ we immediately obtain the estimates

δφ(t) ≤ φ(cδt), t ≥ 0, δ > 1,  (2.2)
φ(δt) ≤ δφ(ct), t ≥ 0, δ < 1.  (2.3)

For convex functions the inequalities to be given above are valid when c = 1 .

Lemma 2.2. If ω is a Young function, then

ω (\frac{ω̃(t)}{t}) ≤ ω̃(t), t ≥ 0.  (2.4)

Proof. By virtue of the equality (ω̃)˜ = ω we have

ω (\frac{ω̃(t)}{t}) = sup_{s≥0} s (\frac{ω̃(t)}{t} − \frac{ω̃(s)}{s}) ≤ sup_{0≤s<t} s\frac{ω̃(t)}{t} ≤ ω̃(t).

since the expression in the brackets is negative when t < s.

Lemma 2.3 ([11], p.17). Let ψ satisfy the Δ_{2}-condition. Then there exist p > 1 and c > 1 such that

s^{−p}ψ(s) ≤ ct^{−p}ψ(t)  (2.5)

for 0 ≤ t ≤ s.

Lemma 2.4. Let E be a bounded set in X, and for each point x ∈ E let a ball B_{x} = B(x, r_{x}) be given such that sup_{x∈E} rad B_{x} < ∞. Then from the family {B_{x}}_{x∈E} we can choose a (finite or infinite) sequence of pairwise disjoint balls (B_{j})_{j} for which E ⊂ ∪_{j≥1}N_{0}B_{j}, N_{0} = a_{1}(1 + 2a_{0}), and for each B_{x} ∈{B_{x}}_{x∈E} there exists a ball B_{j_{0}} such that x∈N_{0}B_{j_{0}} and rad B_{x} ≤ 2 rad B_{j_{0}}.

Proof. Set R_{1} = sup_{x∈E} rad B_{x}. There obviously exists a ball B_{1} = B_{x_{1}} from the family {B_{x}}_{x∈E} provided that rad B_{1} > 2^{−1}R_{1}. If x ∉ N_{0}B_{1} ∩ E, then B_{x} ∩ B_{1} = ∅. Indeed, making the opposite assumption that there exists a point y ∈B_{x} ∩ B_{1}, we will have

d(x_{1}, x) ≤ a_{1}(d(x_{1}, y) + d(y, x)) < a_{1}(rad B_{x_{1}} + a_{0}d(x, y)) <
< a_{1}(rad B_{1} + a_{0} rad B_{x}) < a_{1}(1 + 2a_{0}) rad B_{1} = N_{0} rad B_{1},

which leads to the contradiction.

Obviously, rad B_{x} ≤ 2 rad B_{1} for an arbitrary point x ∈ N_{0}B_{1} ∩ E. Assuming now that R_{2} = sup_{x∈E\N_{0}B_{1}} rad B_{x}, we can find a ball B_{2} = B_{x_{2}} from the family {B_{x}}_{x∈E\N_{0}B_{1}} provided that B_{2} ∩B_{1} = ∅, rad B_{2} > 2^{−1}R_{2} and rad B_{x} ≤ 2 rad B_{2} for each point x∈(N_{0}B_{2} ∩ E)\N_{0}B_{1}. Proceeding in this way, we arrive at the sequence {B_{j}}_{j≥1} of nonintersecting balls. If this sequence is finite, then it will be the one we wanted to obtain.

Let the sequence be infinite. If we show that for each point x ∈ E there exists a ball B_{j} for which x ∈ N_{0}B_{j}, then setting j_{0} equal to the minimal value among similar j’s, we obtain the desired covering.

Assume the opposite. Let in E there exists a point x_{0} ∈ E such that x_{0} ∉ N_{0}B_{j} for every j. Then we will have B_{x_{0}} ∈ {B_{x}}_{x∈E\∪_{j=1}^{n}N_{0}B_{j}} for any natural number n, and hence rad B_{x_{0}} ≤ R_{n} < 2 rad B_{n} for each n.

On the other hand, it is obvious that ∪_{x∈E}B_{x} is a bounded set, i.e. it is contained within some ball B_{0}. It therefore turns out that (B_{j})_{j≥1} is an infinite sequence of nonintersecting balls contained in B_{0}. Therefore rad B_{n} → 0 (see, for example, [17], p. 68). The latter result leads to the contradiction rad B_{x_{0}} = 0.

Proof of Theorem 1.1. Fix the function f ≥ 0 and λ > 0. Without loss of generality it can be assumed that

\frac{1}{c_{1}θ(β(X × [0, ∞)))φ(λ)} ∫_{X} ψ(\frac{4c_{1}}{ε} \frac{f(x)}{η(λ)})σ(x)dν < 1.  (2.6)

Otherwise we would have

φ(λ)θ(β{(x, t) : K(fdν)(x, t) > λ}) ≤ φ(λ)θ(β(X × [0, ∞))) ≤

≤ \frac{1}{c_{1}} ∫_{X} ψ(\frac{4c_{1}}{ε} \frac{f(x)}{η(λ)})σ(x)dν

and, since ψ ∈Δ_{2}, the proof is complete.

Assume (x, t) ∈ E_{λ}, where E_{λ} = {(x′, t′) ∈ X × [0, ∞) : K(fdν)(x′, t′) > λ}. By virtue of (2.6) for (x, t) there exists a finite r ≥ 0 such that

\frac{1}{c_{1}θ(βB̂(x, N_{0}(2r + t)))φ(λ)} ∫_{X} ψ(\frac{4c_{1}}{ε} \frac{f(x)}{η(λ)})σ(x)dν < 1.

If the greatest lower bound of r is positive, then there exists a positive number r_{0} = r_{0}(x, t) such that the inequalities

\frac{1}{c_{1}θ(βB̂(x, N_{0}(r_{0} + t)))φ(λ)} ∫_{X} ψ(\frac{4c_{1}}{ε} \frac{f(x)}{η(λ)})σ(x)dν ≥ 1,  (2.7)
\frac{1}{c_{1}θ(βB̂(x, N_{0}(2r_{0} + t)))φ(λ)} ∫_{X} ψ(\frac{4c_{1}}{ε} \frac{f(x)}{η(λ)})σ(x)dν < 1 (2.8)

are simultaneously fulfilled.

For such r_{0} we would have by virtue of inequality (1.4) and condition (1.5)

∫_{X\B(x,r_{0})} f(y)k(x, y, t)dν = \frac{λ}{4c_{1}φ(λ)θ(βB̂(x, N_{0}(2r_{0}+t)))} ×

× ∫_{X\B(x,r_{0})} \frac{4c_{1}}{ε} \frac{f(y)}{η(λ)}ε\frac{φ(λ)η(λ)}{λ} \frac{θ(βB̂(x, N_{0}(2r_{0} + t)))}{σ(y)}k(x, y, t)σ(y)dν ≤

≤ \frac{λ}{4c_{1}φ(λ)θ(βB̂(x, N_{0}(2r_{0} + t)))} ∫_{X\B(x,r_{0})} ψ(\frac{4c_{1}f(x)}{εη(λ)})σ(x)dν +

+\frac{λ}{4c_{1}φ(λ)θ(βB̂(x, N_{0}(2r_{0}+t)))} ×

× ∫_{X\B(x,r_{0})} ψ̃(ε\frac{φ(λ)η(λ)}{λ} \frac{θ(βB̂(x, N_{0}(2r_{0} + t)))}{σ(y)}k(x, y, t))σ(y)dν ≤

≤ \frac{λ}{4} + \frac{λ}{4} = \frac{λ}{2}.

But since (x, t)∈E_{λ}, the latter estimate implies

∫_{B(x,r_{0})} f(y)k(x, y, t)dν > \frac{λ}{2}.  (2.9)

When the measure ν is concentrated at the point x, the above-mentioned greatest lower bound may turn out to be equal to zero. Then instead of (2.9) we have

k(x, x, t)f(x)ν\{x\} > \frac{λ}{2}.  (2.10)

Therefore due to (1.4) and (1.5)

φ(λ)θ(βB̂(x, N_{0}t))) ≤ \frac{1}{2c_{1}} ψ(\frac{4c_{1}}{ε} \frac{f(x)}{η(λ)})σ(x)ν\{x\} +

+\frac{1}{2c_{1}} ψ̃(ε\frac{φ(λ)η(λ)}{λ} \frac{θ(βB̂(x, N_{0}t)))}{σ(x)}k(x, x, t))σ(x)ν\{x\} ≤

≤ \frac{1}{2c_{1}} ∫_{B(x,t)} ψ(\frac{4c_{1}}{ε} \frac{f(x)}{η(λ)})σ(y)dν(y) + \frac{1}{2}φ(λ)θ(βB̂(x, N_{0}t)).

Hence

φ(λ)θ(βB̂(x, N_{0}t)) ≤ \frac{1}{4c_{1}} ∫_{B(x,t)} ψ(\frac{4c_{1}}{ε} \frac{f(y)}{η(λ)})σ(y)dν(y). (2.11)

Let us now consider the case when ν is not concentrated at the point x. Let n be the greatest nonnegative integer for which

b =^{def} lim_{r→0} βB̂(r + t)) < 2^{−n}βB̂(x, N_{0}(r_{0} + t));

n may be equal to ∞ if b = 0. For each k, 0 < k ≤ n, we set r_{k} = sup{r : βB̂(x, N_{0}(r+t)) ≤ 2^{−k}βB̂(x, N_{0}(r_{0} +t))}. Then (r_{k})_{k} is a decreasing (finite or infinite) sequence and

βB̂(x, N_{0}(r_{k} + t)) ≤ 2^{−k}βB̂(x, N_{0}(r_{0} + t)) ≤
≤ βB̂(x, N_{0}(2r_{k} + t)).  (2.12)

Let B_{k} = B(x, r_{k}), 0 ≤ k ≤ n, and B_{n+1} = {x}. Since by the condition of the theorem u^{−α}θ(u) decreases, we have

a_{k}^{p} = \frac{βB̂(x, N_{0}(r_{k} + t))}{θ(βB̂(x, N_{0}(r_{k} + t)))} · \frac{θ(βB̂(x, N_{0}(r_{0} + t)))}{βB̂(x, N_{0}(r_{0} + t))} =

= \frac{βB̂(x, N_{0}(r_{k} + t))}{θ(βB̂(x, N_{0}(r_{k} + t)))} \frac{θ(βB̂(x, N_{0}(r_{0} + t)))}{(βB̂(x, N_{0}(r_{0} + t)))^{1−α}(βB̂(x, N_{0}(r_{0} + t)))^{α}} ≤

≤ c(\frac{βB̂(x, N_{0}(r_{k} + t))}{βB̂(x, N_{0}(r_{0} + t))})^{1−α} ≤ c(\frac{1}{2^{k}} )^{1−α}.

Because of this

∑_{k=0}^{n} a_{k} ≤ c^{\frac{1}{p}} ∑_{k=0}^{∞} (\frac{1}{2^{k}} )^{\frac{1−α}{p}} = a < ∞.

Inequality (2.9) can now be rewritten as

\frac{λ}{2} ∑_{k=0}^{n} \frac{a_{k}}{a} ≤ ∑_{k=0}^{n} ∫_{B_{k}\B_{k+1}} k(x, y, t)f(y)dν,

whence it follows that there exists k_{0}, 0 ≤ k_{0} ≤ n, such that

\frac{λ}{2} \frac{a_{k_{0}}}{a} < ∫_{B_{k_{0}} \B_{k_{0}+1}} k(x, y, t)f(y)dν.

Therefore

φ(λ)θ(βB̂(x, N_{0}(2r_{k_{0}+1} + t))) ≤

≤ \frac{1}{2c_{1}} ∫_{B_{k_{0}} \B_{k_{0}+1}} \frac{4ac_{1}}{εa_{k_{0}}} \frac{f(y)}{η(λ)}ε\frac{φ(λ)η(λ)}{λ} \frac{θ(βB̂(x, N_{0}(2r_{k_{0}+1} + t)))}{σ(y)} ×

×k(x, y, t)σ(y)dν(y) ≤ \frac{1}{2c_{1}} ∫_{B_{k_{0}}} ψ(\frac{4ac_{1}}{εa_{k_{0}}} \frac{f(y)}{η(λ)})σ(y)dν +

+\frac{1}{2c_{1}} ∫_{B_{k_{0}} \B_{k_{0}+1}} ψ̃(ε\frac{φ(λ)η(λ)}{λ} \frac{θ(βB̂(x, N_{0}(2r_{k_{0}+1} + t)))}{σ(y)} ×

×k(x, y, t))σ(y)dν ≤ \frac{1}{2c_{1}} ∫_{B_{k_{0}}} ψ(\frac{4ac_{1}}{εa_{k_{0}}} \frac{f(y)}{η(λ)})σ(y)dν +

+\frac{1}{2}φ(λ)θ(βB̂(x, N_{0}(2r_{k_{0}+1} + t))),

and as a result we have

φ(λ)θ(βB̂(x, N_{0}(2r_{k_{0}+1} + t))) ≤ ca_{k_{0}}^{−p} ∫_{B(x,r_{k_{0}} )} ψ(\frac{4c_{1}}{ε} \frac{f(y)}{η(λ)})σ(y)dν.

Next, taking into account (2.12), we obtain the estimate

φ(λ)θ(βB̂(x, N(r_{k_{0}} + t))≤ca_{k_{0}}^{−p} ∫_{B(x,r_{k_{0}} +t)}ψ(\frac{4c_{1}f(y)}{εη(λ)})σ(y)dν (2.13)

which by the definition of the number a_{k_{0}} takes the form

φ(λ)βB̂(x, N_{0}(r_{k_{0}} + t)) ≤
≤ c\frac{βB̂(x, N_{0}(r_{0} + t))}{θ(βB̂(x, N_{0}(r_{0} + t)))} ∫_{B(x,r_{k_{0}} +t)} ψ(\frac{4c_{1}f(y)}{εη(λ)})σ(y)dν. (2.14)

Rewrite (2.7) now as

θ(βB̂(x, N_{0}(r_{0} + t))) ≤ \frac{1}{c_{1}φ(λ)} ∫_{X} ψ(\frac{4c_{1}}{ε} \frac{f(y)}{η(λ)})σ(y)dν. (2.15)

If θ^{−1} is defined by θ^{−1}(u) = sup{τ : θ(τ) ≤ u}, then θ(θ^{−1}(u)) ≤ u and θ(2θ^{−1}(u)) ≥ u, i.e., \frac{u}{2} ≤ θ(θ^{−1}(u)) ≤ u. Moreover, θ^{−1}(θ(u)) = sup\{τ : θ(τ) ≤ θ(u)} ≥ u. Therefore (2.15) will yield the estimate

βB̂(x, N_{0}(r_{0} + t)) ≤ θ^{−1}(θ(βB̂(x, N_{0}(r_{0} + t)))) ≤

≤ θ^{−1}(\frac{1}{c_{1}φ(λ)} ∫_{X} ψ(\frac{4c_{1}}{ε} \frac{f(x)}{η(λ)})σ(x)dν).

Thus

θ(θ^{−1}(\frac{1}{c_{1}φ(λ)} ∫_{X} ψ(\frac{4c_{1}f(x)}{εη(λ)})σ(x)dν)) ≤

≤ \frac{1}{c_{1}φ(λ)} ∫_{X} ψ(\frac{4c_{1}f(x)}{εη(λ)})σ(x)dν,

which by virtue of the fact that \frac{θ(u)}{u} decreases yields the estimate

\frac{βB̂(x, N_{0}(r_{0} + t))}{θ(βB̂(x, N_{0}(r_{0} + t)))} ≤

≤ c\frac{θ^{−1}(\frac{1}{c_{1}φ(λ)} ∫_{X} ψ(\frac{4c_{1}}{ε} \frac{f(y)}{η(λ)})σ(y)dν)}{θ(θ^{−1}(\frac{1}{c_{1}φ(λ)} ∫_{X} ψ(\frac{4c_{1}}{ε} \frac{f(y)}{η(λ)})σ(y)dν))} ≤

≤ 2c\frac{θ^{−1}(\frac{1}{c_{1}φ(λ)} ∫_{X} ψ(\frac{4c_{1}}{ε} \frac{f(y)}{η(λ)})σ(y)dν)}{\frac{1}{c_{1}φ(λ)} ∫_{X} ψ(\frac{4c_{1}}{ε} \frac{f(y)}{η(λ)})σ(y)dν} =^{def} l(f, λ).

After the above manipulations condition (2.14) can be formulated as follows: for each (x, t)∈E_{λ} there exists a ball B_{x,t} such that x is its center, t ≤ rad B_{x,t}, and

φ(λ)β(N_{0}B̂_{x,t}) ≤ cl(f, λ) ∫_{B_{x,t}} ψ(\frac{4c_{1}}{ε} \frac{f(y)}{η(λ)})σ(y)dν.  (2.16)

Now fix a ball B_{0} and consider the sets B̂_{0}∩E_{λ} and B_{0}∩{x : K(fdν)(x, 0) > λ} ≡ B_{0} ∩ E_{λ}^{0}.

It is obvious that the latter set is contained in the former. For each x ∈ E_{λ}^{0} we set d(x) = sup{t : (x, t) ∈ B̂_{0} ∩ E_{λ}}. It is easy to verify that d(x) < 2 rad B_{0}. For each x ∈ B_{0} ∩ E_{λ}^{0} there exist t_{x} ≥ \frac{2d(x)}{N_{0}} such that (x, t_{x}) ∈ B̂_{0} ∩ E_{λ}(N_{0} > 2); consequently, for (x, t_{x}), (2.16) is valid.

As a result we have the following situation: for each x ∈ B_{0} ∩ E_{λ}^{0} there exists a ball B_{x} with center at the point x such that rad B_{x} > \frac{d(x)}{N_{0}} and (2.16) is fulfilled for B_{x} = B_{x,t}.

If sup_{x∈B_{0}∩E_{λ}^{0}} rad B_{x} = ∞, then, clearly, there exists a ball B_{1} ∈{B_{x}}_{x∈E_{λ}^{0}∩B_{0}}

such that E_{λ} ∩ B̂_{0} ⊂ B̂_{1}.

If sup rad B_{x} <∞, then, due to Lemma 2.4, from the family {B_{x}}_{x∈E_{λ}^{0}} ∩ B_{0} covering the bounded set E_{λ}^{0} ∩ B_{0} we can choose a sequence (B_{j}) of nonintersecting balls for which ∪_{j≥1}N_{0}B_{j} ⊃ E_{λ}^{0} ∩ B_{0} and (2.16) holds.

It will be shown that ({ \atop N_{0}B_{j}})_{j≥1} covers the set E_{λ} ∩ B_{0}. To this end we prove that each (x, d(x)) ∈ ∪_{j≥1}{ \atop N_{0}B_{j}}. Indeed, if x is the center of some ball B_{j}, then there is nothing to prove. Let x not be the center of B_{j}; then by Lemma 2.4 for B_{x} there exists a ball B_{j} such that x ∈ N_{0}B_{j} and rad B_{x} ≤ 2 rad B_{j}. Therefore d(x) < N_{0} rad B_{x} ≤ 2N_{0} rad B_{j} or, which is

the same, (x, d(x))∈{ \atop N_{0}B_{j}}.

On account of the foregoing reasoning we can derive estimates

φ(λ)β\{(x, t)∈B̂_{0} : K(fdν)(x, t) > λ\} ≤ ∑_{j=1}^{∞} φ(λ)β{ \atop N_{0}B_{j}} ≤

≤ cl(f, λ) ∑_{j=1}^{∞} ∫_{B_{j}} ψ(\frac{4c_{1}}{ε} \frac{f(y)}{η(λ)})σ(y)dν ≤

≤ cl(f, λ) ∫_{X} ψ(\frac{4c_{1}}{ε} \frac{f(y)}{η(λ)})σ(y)dν ≤

≤ cφ(λ)θ^{−1}(\frac{1}{c_{1}φ(λ)} ∫_{X} ψ(\frac{4c_{1}}{ε} \frac{f(y)}{η(λ)})σ(y)dν)

which yield

θ(β{(x, t)∈B̂_{0} : K(fdν)(x, t) > λ}) ≤

≤ θ(cθ^{−1}(\frac{1}{c_{1}φ(λ)} ∫_{X} ψ(\frac{4c_{1}}{ε} \frac{f(x)}{η(λ)})σ(x)dν)).

Taking into account ψ ∈ Δ_{2} and u^{−α}ψ(u) ↓, from the latter estimate we obtain the inequality

θ(β{(x, t)∈B̂_{0} : K(fdν)(x, t) > λ}) ≤

≤ c^{α}θ(θ^{−1}(\frac{1}{c_{1}φ(λ)} ∫_{X} ψ(\frac{4c_{1}}{ε} \frac{f(x)}{η(λ)})σ(x)dν)) ≤

≤ \frac{c}{φ(λ)} ∫_{X} ψ(\frac{f(x)}{η(λ)})σ(x)dν.

If we now assume that rad B_{0} tends to infinity, we obtain (1.6).

Consider the case dβ = ϱdν⊗δ_{0}, where δ_{0} is the Dirac measure supported at the origin and

k(x, y, t) = \{ \begin{matrix}k(x, y), & t = 0, \\ 0, & t > 0. \end{matrix}

In that case due to Theorem 1.1 we have

Corollary 2.1. Let the functions φ, η and ψ satisfy the conditions of Theorem 1.1. It is further assumed that there exist positive ε and c_{1} such that

∫_{X\B(a,r)} ψ̃(ε\frac{φ(s)η(s)}{s} \frac{θ(ϱB(a, 2N_{0}r))}{σ(y)}k(a, y))σ(y)dν ≤

c_{1}φ(s)θ(ϱB(a, 2N_{0}r))

for any s > 0, r ≥ 0 and a∈X.

In that case there exists c_{2} > 0 such that the inequality

φ(λ)θ(ϱ{x∈X : ∫_{X} k(x, y)f(y)dν(y) > λ}) ≤

≤ c_{2} ∫_{X} ψ(\frac{f(x)}{η(λ)})σ(x)dν

holds for any λ > 0 and any nonnegative measurable function f : X → R^{1}.

It is time to make some remarks. Taking a closer look at the proof of Theorem 1.1, we readily find that if βB̂(x, r) is continuous with respect to r for each x∈X, the factor 2 in condition (1.5) can be omitted.

Moreover, if the space (X, d, μ) possesses the Besicovitch property (consisting in the fact that for every bounded set E any family {B(y, r(y))}_{y∈E} of balls contains a countable (or finite) subfamily {B_{n}} = {B(y_{n},r(y_{n})}, n ∈ N, such that E ⊂ ∪B_{n} and ∑ χ_{B_{n}} ≤ c, where χ_{B_{n}} is the characteristic function of the set B_{n}), then in Corollary 2.1 we can set N_{0} = 1.

Finally we remark that for φ(λ) = ψ(λ) = λ^{p}, η ≡ 1, θ(u) = u^{\frac{p}{q}}, X = R^{n} Corollary 2.1 becomes the particular case of Theorem 6.1.1 from [7], p. 171.

The proof of Theorem 1.2 rests on a number of lemmas.

Lemma 2.5. Let θ be any increasing function, and let the kernel k satisfy condition (1.7). If condition (1.6) is fulfilled, then there exists a constant c > 0 such that for any a∈X, r ≥ 0, t ≥ 0, and any nonnegative measurable function F:X →R^{1}, supp F ⊂ X\B(a, r) we have the inequality

\begin{matrix}φ(K(F dν)(a, t))θ(βB̂(a, N(r + t))) ≤ \\ ≤ c ∫_{X} ψ(\frac{c′F (x)}{η(K(F dν)(a, t))})σ(x)dν, \end{matrix}  (2.17)

where N and c′ are the constants from condition (1.7).

Proof. Fix a ∈ X, r ≥ 0, t ≥ 0, and F : X → R^{1} assuming that supp F ⊂ X\B(a, r). Inequality (2.17) is obtained if in (1.6) we set f = c′F and λ = K(F dν)(a, t).

It is sufficient only to note that the inclusion

B̂(a, N(r + t)) ⊂ {(x, τ) : K(c′F dν)(x, τ) ≥ K(F dν)(a, t)}

holds by virtue of condition (1.7).

Lemma 2.6. Let φη and ψ be quasiconvex functions, and let k be a nonnegative kernel. Then condition (2.17) with the constants c, c′, and N implies the existence of ε and c_{1} such that the inequality

∫_{X\B(a,r)} ψ̃(ε\frac{φ(λ)η(λ)}{λ} \frac{θ(βB̂(a, N(r + t)))}{σ(y)}k(a, y, t))σ(y)dν(y)≤

≤ c_{1}φ(λ)θ(βB̂(a, N(r + t)))  (2.18)

holds for any λ > 0, r ≥ 0, t ≥ 0, and a∈X.

Proof. It is obvious that (2.17) is fulfilled for σ_{1} = σ + δ, too, if δ > 0. Let a ∈ X, r ≥0, t ≥ 0, λ > 0 be the fixed constants. Due to (1.8) it can be assumed without loss of generality that the function ψ is convex. For M > 0 we define the value

I = ∫_{D} ψ̃(ε\frac{φ(λ)η(λ)}{λ} \frac{θ(βB̂(a, N(r + t))}{σ_{1}(y)}k(a, y, t))σ_{1}(y)dν,

where D = B(a, R)\B(a, r) ∩ {y ∈ X : k(a, y, t) < M}, R > r, while the constant ε will be appropriately chosen later.

We introduce the notation

T_{δ}(y) = ε\frac{φ(λ)η(λ)}{λ} \frac{θ(βB̂(a, N(r + t)))}{σ_{1}(y)}k(a, y, t),

g(y) = \frac{ψ̃(T_{σ}(y))}{T_{σ}(y)}εη(λ)χ_{D}(y),

allowing us to write

I = φ(λ)θ(βB̂(a, N(r + t)))\frac{K(gdν)(a, t)}{λ}.  (2.19)

Our next step is to show that for sufficiently small ε’s the value I is finite. If lim_{t→∞} \frac{ψ(t)}{t} = ∞, then ψ̃ is finite everywhere and thus

I ≤ ψ̃(ε\frac{φ(λ)η(λ)}{λ} \frac{θ(βB̂(a, N(r + t)))}{δ}M) ∫_{B(a,R)} σ_{1}(x)dν < ∞

for any ε > 0.

Let now ψ(t) ≤ At, A > 0. Then from the condition (2.17) we obtain

φ(K(F dν)(a, t))η(K(F dν)(a, t))θ(βB̂(a, N(r + t))) ≤
≤ c ∫_{X} F (x)σ(x)dν.  (2.20)

Set

l = ∥χ_{X\B(a,r)} \frac{k(a, ·, t)}{σ_{1}} ∥_{L^{∞}}.

From the definition of the norm in L^{∞} it follows that there exists a measurable set E ⊂ X\B(a, r), νE > 0 such that

\frac{k(a, y, t)}{σ_{1}(y)} > \frac{l}{2}

for y ∈E.

Set in (2.20)

F (y) = \frac{λ}{νEk(a, y, t)}χ_{E}(y).

Recall that a and t are fixed. Obviously, K(F dν)(a, t) = λ and hence by virtue of (2.20) we obtain the estimate

\frac{φ(λ)η(λ)}{λ}θ(βB̂(a, N(r + t))) ≤ \frac{c}{νE} ∫_{E} \frac{σ(y)}{k(a, y, t)}dν ≤ \frac{2c}{l},

which yields

\frac{φ(λ)η(λ)}{λ}θ(βB̂(a, N(r + t)))χ_{X\B(a,r)} \frac{k(a, y, t)}{σ_{1}(y)} < c,

where the constant does not depend on λ, r, t, and a. Thus we conclude that

I ≤ ψ̃(εc) ∫_{B(a,r)} σ_{1}(y)dν.

If now ε is so small that ψ̃(cε) < ∞, then the value I will be finite for the respective ε.

Now it will be shown that

I ≤ bφ(λ)θ(βB̂(a, N(r + t))) + cb ∫_{X} ψ(\frac{c′g(y)}{η(λ)})σ_{1}dν,  (2.21)

where the constants b, c and c′ do not depend on λ, r, and t.

Let a ∈ X and t ≥ 0 be such that K(gdν)(a, t) < bλ, where the constant b is such that

\frac{φ(s)η(s)}{s} ≤ b\frac{φ(u)η(u)}{u}  (2.22)

for bs < u (see Lemma 2.1).

Then evidently (2.19) will yield

I ≤ bφ(λ)θ(βB̂(a, N(r + t))).

Let now K(gdν)(a, t) > bλ.

Using (2.22) and condition (2.17), from (2.19) we obtain the estimate

I = φ(λ)θ(βB̂(a, N(r + t)))\frac{K(gdν)(a, t)}{λ} ≤

≤ b\frac{φ(K(gdν)(a, t))η(K(gdν)(a, t))}{η(λ)}θ(βB̂(a, N(r + t))) ≤

≤ cb\frac{η(K(gdν)(a, t))}{η(λ)} ∫_{X} ψ(\frac{c′g(x)}{η(K(gdν)(a, t))})σ_{1}(x)dν.

Since the function ψ is convex, estimating the right-hand part of the latter inequality by means of (2.2) we conclude that

I ≤ cb ∫_{X} ψ(\frac{c′g(x)}{η(λ)})σ_{1}(x)dν.

Thus we have shown that inequality (2.21) is valid.

Rewrite (2.21) as

I ≤ bφ(λ)θ(βB̂(a, N(r + t))) +
+cb ∫_{X} ψ(c′ε\frac{ψ̃(T_{δ}(y))}{T_{δ}(y)})σ_{1}(y)dν.  (2.23)

Let ε be chosen so small that c′ε < 1. Then, by virtue of the assumption that the function ψ is convex and taking into account (2.3) and (2.4), from (2.23) we have

∫_{D} ψ̃(T_{δ}(y))σ(y)dν ≤ bφ(λ)θ(βB̂(a, N(r + t))) +
+cc′bε ∫_{D} ψ̃(T_{δ}(y))σ_{1}(y)dν.

If ε is so small that cc′bε < 1, then the latter inequality implies

∫_{D} ψ̃(T_{δ}(y))σ_{1}(y)dν ≤ cφ(λ)θ(βB̂(a, N(r + t))).

Passing here to the limit when R → ∞, M → ∞, and δ → 0, we obtain the desired inequality (2.18).

Lemma 2.7. Let the kernel k satisfy condition (1.7), and let inequality (1.9) be fulfilled. Then (1.5) is valid.

Proof. Replace t by N_{0}(2r+t) in condition (1.9) and take into consideration that by virtue of condition (1.7) we have

k(a, y, t) ≤ c′k(a, y, N_{0}(2r + t))

for any y ∈X\B(a, r).

Proof of Theorem 2.2. The proof is accomplished by the diagram

(1.6) ⇒ (1.10)
⇑  ⇓
(1.9) ⇐ (1.8)

By Lemma 2.5 (1.6) ⇒ (1.10). Then by Lemma 2.6 (1.10) ⇒ (1.8). When r = 0, condition (1.8) yields (1.9). Next, by Lemma 2.7 we obtain (1.5). Finally, using Theorem 1.1, we ascertain that the implication (1.5) ⇒ (1.6) is valid.

We make some remarks connected with the proof of Theorem 1.3. If k(x, y, t) = k(x, y), dβ = ϱdμ⊗δ_{0} Lemmas 2.5 and 2.6 can be reformulated in the respective manner. Further, proceeding from Corollary 2.1 and following the proof of Theorem 2.2, we ascertain that Theorem 1.3 is valid.

3. Criteria of General Weak Type Weighted Inequalities in Lorentz Spaces

Let (Y, ν) be a space with a positive σ-additive measure ν. When 1 ≤ p ≤ ∞, 1 ≤ s ≤ ∞, the Lorentz space L_{ν}^{ps} is the space of all ν-measurable functions f for which ∥f∥_{L^{ps}(Y,ν)} < ∞, where

∥f∥_{L^{ps}(Y,ν)} = (s ∫_{0}^{∞}(ν\{y ∈ Y : |f(y)| > τ\})^{\frac{s}{p}} τ^{s−1}dτ)^{\frac{1}{s}}
if 1 ≤ p < ∞, 1 ≤ s < ∞,

and

∥f∥_{L^{ps}(Y,ν)} = sup τ(\{y ∈ Y : |f(y)| > τ\})^{\frac{1}{p}} if 1 ≤ p < ∞, s = ∞.

If 1 < p < ∞ and 1 ≤ s ≤ ∞, or p = s = 1, or p = s = ∞, then L^{ps}(Y, ν) is a Banach space with norm equivalent to ∥ · ∥_{L^{ps}(Y,ν)}.

In the sequel X will denote a homogeneous type space, β a positive measure given on the product of σ-algebras generated by balls from X and by intervals from [0, ∞), and ν a finite positive measure on X.

Theorem 3.1. Let 1 ≤ s ≤ p < q < ∞, and let k : X ×X ×[0, ∞) → R^{1} be an arbitrarily chosen nonnegative kernel. In that case, if there exists a number c_{1} > 0 such that the inequality

(βB̂(a, N_{0}(2r + t))^{\frac{1}{q}} ∥χ_{X\B(a,r)} \frac{k(a, ·, t)}{σ}∥_{L^{p′s′} (X,σdν)} ≤ c_{1} (3.1)

holds for any a ∈ X, r ≥ 0, t ≥ 0, then there exists a positive constant c_{2} such that

β{(x, t) ∈ X × [0, ∞) : K(fdν)(x, t) > λ} ≤
≤ c_{2}λ^{−q}∥f∥_{L^{ps}(X,ν)}^{q}  (3.2)

for any measurable nonnegative f : X → R^{1} and λ > 0.

Theorem 3.2. Let 1 ≤ s ≤ p < q < ∞, and let the kernel k satisfy condition (1.7). Then the folowing statements are equivalent:

(i) (3.2) is fulfilled;

(ii) there exists a positive constant c such that

(βB̂(a, (2r + t))^{\frac{1}{q}} ∥χ_{X\B(a,r)} \frac{k(a, ·, t)}{σ}∥_{L^{p′s′} (X,σdν)} ≤ c

for any a ∈ X, r ≥ 0, t ≥ 0; (iii) there exists a number c_{1} > 0 such that

(βB̂(a, t)^{\frac{1}{q}} ∥\frac{k(a, ·, t)}{σ}∥_{L^{p′s′} (x,σdν)} ≤ c_{1}.

The proofs of these theorems are accomplished in the manner described in Section 2 using the technique from [7], Chapter 6, and we therefore leave them out. Note that the solution of the two-weight problem in the sense of [3] was previously derived in [19] in Lebesgue space for a fractional integral over a homogenous type space.

4. General Weak Type Inequalities for Classical Operators

In this section we discuss some specific examples for which the results of the previous sections are valid.

Consider the kernel

k(x, y, t) = (μB(x, d(x, y) + t))^{−δ}, δ > 0.

It is easy to verify that it satisfies condition (1.7). Let y ∈ X\B(a, r) and (x, τ) ∈ B̂(a, N(r + t)), where N is an arbitrary positive number. It is sufficient to show the inclusion

B(x, d(x, y) + τ) ⊂ B(a, c(d(a, y) + t)).

Indeed, assuming that z ∈B(x, d(x, y) + τ), we obtain a chain of inequalities

d(a, z) ≤ a_{1}(d(a, x) + d(x, z)) ≤ a_{1}(N(r + t) + d(x, z)) ≤
≤ a_{1}N(r + t) + a_{1}d(x, y) + a_{1}τ ≤ 3a_{1}N(r + t) +
+a_{1}^{2}(d(x, a) + d(a, y)) ≤ 3a_{1}N(r + t) + a_{1}^{2}a_{0}d(a, x) +
+a_{1}^{2}d(a, y) ≤ (3a_{1} + a_{1}^{2}a_{0})N(r + t) + a_{1}^{2}d(a, y) ≤
≤ (3a_{1} + a_{1}^{2}a_{0})Nt + ((3a_{1} + a_{1}^{2}a_{0})N + a_{1}^{2})d(a, y) ≤
≤ ((3a_{1} + a_{1}^{2}a_{0})N + a_{1}^{2})(d(a, y) + t).

Thus condition (1.7) is fulfilled. For such kernels we have Theorems 2.2 and 3.2 and hence we obtain the solution of the general weak type weight problem in Orlicz and Lorentz spaces for classical operators such as Riesz potentials, Poisson integrals, and others.

Let X = R^{n}, d a Euclidean distance, μ a Lebesgue measure, and

T_{γ}f(x, t) = ∫_{R^{n}} \frac{f(y)}{(|x − y| + t)^{n−γ}} dν, 0 < γ < n.

a generalized potential. Theorem 2.2 yields a solution of the general weaktype weight problem for T_{γ} in Orlicz spaces. It was previously solved in Lorentz space in [6] (see also [7], Theorem 6.5.1).

Now consider the Poisson integral in the upper half-space

ℙf(x, y) = ∫_{R^{n}} f(y)P(x − y, t)dy,

where P(x, t) = c_{n}t(t^{2} + |x|^{2})^{− \frac{n+1}{2}} is the Poisson kernel for R_{+}^{n+1}. The criterion of a two-weight inequality of the weak type (p, q) was established in [10]. From Theorem 2.2 we obtain

Corollary 4.1. Let φη and ψ be quasiconvex functions, ψ ∈ Δ_{2}, and let the function t^{−α}θ(t) decrease for some α ∈ (0, 1). Then the following statements are equivalent:

(i) there exists a constant c_{1} > 0 such that

φ(λ)θ(β\{(x, t)∈R^{n} × [0, ∞) : \frac{1}{t}ℙf(x, t) > λ\}) ≤

c_{1} ∫_{R^{n}} ψ(\frac{f(x)}{η(λ)})σ(x)dν

for any λ > 0 and any nonnegative measurable function f : R^{n} → R^{1}; (ii) there exist positive constants ε and c_{2} such that

∫_{R^{n}\B(a,r)} ψ̃(ε\frac{φ(λ)η(λ)}{λ} \frac{θ(βB̂(a, r + t))}{σ(y)} \frac{1}{t}P(a − y, t))σ(y)dν ≤

≤ c_{2}φ(λ)θ(βB̂(a, r + t))

for any λ > 0, a∈R^{n}, r ≥ 0, t ≥ 0; (iii) there exists positive constants ε and c_{3} such that

∫_{R^{n}} ψ̃(ε\frac{φ(λ)η(λ)}{λ} \frac{θ(βB̂(a, t))}{σ(y)} \frac{1}{t}P(a − y, t))σ(y)dν ≤

≤ c_{3}φ(λ)θ(βB̂(a, t))

for any λ > 0, a∈R^{n} and t ≥ 0.

Let now X = [0, ∞), d a Euclidean distance, μ a Lebesgue measure, and

k(x, y) = \{ \begin{matrix}1 & for x > y, \\ 0 & for x ≤ y. \end{matrix}

Then for the Hardy transform f ↦ ∫_{0}^{x} f(y)dν Theorem 2.3 yields the criterion of validity of the weak-type inequality figuring in this theorem. This criterion is written in the form

∫_{0}^{x} ψ̃(ε\frac{φ(λ)η(λ)}{λ} \frac{θ(ϱ(x, ∞))}{σ(y)})σ(y)dν ≤ cφ(λ)θ(ϱ(x, ∞)).

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(Received 21.08.1992)

A.Razmadze Mathematical Institute

Georgian Academy of Sciences

1 Z.Rukhadze St., Tbilisi 380093

Republic of Georgia