Hardy–Steklov Integral Operators: Part I

...

1 downloads 52 Views 1MB Size

ISSN 0081-5438, Proceedings of the Steklov Institute of Mathematics, 2018, Vol. 300, Suppl. 2, pp. S1–S112. © Pleiades Publishing, Ltd., 2018. Original Russian Text © D.V. Prokhorov, V.D. Stepanov, E.P. Ushakova, 2016, published in Sovremennye Problemy Matematiki, 2016, Vol. 22, pp. 3–122.

Hardy–Steklov Integral Operators: Part I D. V. Prokhorov*, V. D. Stepanov** , and E. P. Ushakova*** Steklov Mathematical Institute, Russian Academy of Sciences, Moscow, 119991 Russia Received June 15, 2016

DOI: 10.1134/S008154381803001X

INTRODUCTION This monograph is a complete self-contained study of integral transformations of the form b(x) k(x, y)f (y)v(y) dy, x > 0, Kf (x) := w(x)

(0.0.1)

a(x)

where v and w are weight functions locally integrable on (0, ∞) and the kernel k(x, y) is nonnegative on the domain {(x, y) : x > 0, a(x) < y < b(x)} and has certain generalized monotonicity properties with respect to each variable. In the model case, the boundary functions a(x) and b(x) satisfy the following conditions: (i) a(x) and b(x) are diﬀerentiable and strictly increase on (0, ∞); (ii) a(0) = b(0) = 0, a(x) < b(x) for 0 < x < ∞, and a(∞) = b(∞) = ∞. The main problems concerning transformations (0.0.1) are to ﬁnd criteria (necessary and suﬃcient conditions) for these transformations to be Lp –Lq bounded and compact on Lebesgue spaces and determine the behavior of their approximation and entropy numbers. The best studied case is that of Hardy–Steklov operators, which arise when k(x, y) ≡ 1. The study of transformations (0.0.1) with a(x) = 0 and b(x) = x, which are known as the Hardy operators (for k(x, y) ≡ 1) and Hardy-type operators (for k(x, y) ≡ 1), has been well covered in monographs (see, e.g., [1–5]). General transformations (0.0.1) were considered only in [2, Chapter 3]; in this paper, we signiﬁcantly improve results of [2, Chapter 3]. The studies mentioned above originate, in particular, from weighted Lp –Lq inequalities for Hardytype operators, whose role in real analysis and its applications is well known. Chapter 1 contains a comprehensive study of these questions in Lebesgue spaces with arbitrary measures for 0 < p, q < +∞. The main results concerning Hardy–Steklov operators are contained in Chapter 2. In this chapter, we introduce the notion of a fairway-function, which is used to ﬁnd criteria for the Lp –Lq boundedness of Hardy–Steklov operators in several alternative forms, each of which is important for applications. In Section 3.1 of Chapter 3 (Part 2), applying the boundedness criteria for Hardy–Steklov operators found in Chapter 2, we solve a similar problem for geometric mean operators of the form b(x) 1 log f (y) dy , f (y) ≥ 0. Gf (x) := exp b(x) − a(x) a(x) *

E-mail: [email protected] E-mail: [email protected] *** E-mail: [email protected] **

S1

S2

PROKHOROV et al.

Next, in Section 3.2, we characterize function spaces associated with weighted Sobolev spaces of the ﬁrst order on the real axis. We discover a number of unexpected eﬀects, which distinguish the situation under consideration from the case of Banach function spaces [6]. The concluding Sections 3.3 and 3.4 are concerned with criteria for the validity of theorems on embedding of weighted Sobolev spaces in weighted Lebesgue spaces. The content of the monograph is based on results of its authors, largely published in journals. Throughout this paper, products of the form 0 · ∞ are set to zero. The relation A B means that A ≤ cB, where c is a constant depending only on summation parameters and, possibly, some other inessential parameters. We write A ≈ B instead of A B A or A = cB. The symbol Z denotes the set of all integers, N is the set of all positive integers, and χE is the characteristic (indicator) function of E. We use the signs := and =: to deﬁne new quantities. 1. HARDY’S INEQUALITIES WITH MEASURES 1.1. Preliminary Remarks Let X ⊂ R; we assume X to be endowed with the topology induced by the Euclidean topology of R. By B := B(X) we denote the σ-algebra of Borel subsets of X, and by M := M(X), a σ-algebra of subsets of X containing B. We use {M}+ to denote the class of all M-measurable functions f : X → [0, +∞) ∪ {+∞}. When saying “μ is a measure on X,” we mean that there exists a σalgebra M such that μ is a countably additive function deﬁned on M and taking values in the extended half-line [0, +∞) ∪ {+∞}; we also assume that there exists at least one E ∈ M with μ(E) < +∞. For the σ-algebra on which a given measure λ is deﬁned we use the notation Mλ . We also set Mλ,μ := Mλ ∩ Mμ . On the extended half-line [0, +∞) ∪ {+∞}, the following axioms hold: 0 + (+∞) = a + (+∞) = a · (+∞) = +∞ for a ∈ (0, +∞) ∪ {+∞}; 0 · (+∞) = 0; (1.1.1) (+∞)α = 0−α = +∞, (+∞)−α = 0α = 0

for α ∈ (0, +∞).

For uniformity, given any a, b ∈ R ∪ {−∞, +∞}, a ≤ b, we set [a, b] := {x ∈ R | a ≤ x ≤ b},

[a, b) := [a, b] \ {b},

(a, b] := [a, b] \ {a}.

Lemma 1.1. Given a σ-ﬁnite measure λ on [a, b] and a function f ∈ {Mλ }+ , let Λf (x) := Then, for γ > 0, Λf (b)γ+1 Λf (b)γ+1 ≤ . f (x)Λf (x)γ dλ(x) ≤ max{1, γ + 1} min{1, γ + 1} [a,b]

[a,x] f

dλ.

(1.1.2)

If γ ∈ (−1, 0) and Λf (b) < +∞, then (1.1.2) holds. Proof. If γ > 0, then the second inequality in (1.1.2) follows from the monotonicity of Λf . Let us prove the ﬁrst inequality. Suppose that f (x)Λf (x)γ dλ(x) < +∞. Then

[a,b] {x} f

dλ < +∞ for any x ∈ [a, b].

Suppose that Λf (b) = +∞. We set E := {x ∈ [a, b] | Λf (x) = +∞}

and

If there exists a ξ ∈ (c, b] for which (ξ,b] f dλ = 0, then f (x)Λf (x)γ dλ(x) ≥ Λf (ξ)γ [a,b]

inf E c := b

if E = ∅, if E = ∅.

f dλ = +∞. (ξ,b]

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S3

This contradiction implies that (ξ,b] f dλ = 0 for any ξ ∈ (c, b]. Thus, by the monotone convergence theorem, we have (c,b] f dλ = 0 and, therefore, Λf (c) = +∞. The ﬁniteness of the integral {x} f dλ for any x ∈ [a, b] implies c > a and [a,c) f dλ = +∞. Hence there exists a t ∈ [a, c) for which Λf (t) > 0. Thus, γ γ f (x)Λf (x) dλ(x) ≥ Λf (t) f dλ = +∞ [a,b]

(t,c)

by the choice of c. We have obtained a contradiction. Suppose that Λf (b) < +∞. Then γ f (x)Λf (x) dλ(x) = γ f (x) [a,b]

s

0

[a,b]

Λf (b)

s

=γ

Λf (x)

γ−1

0

[a,b]

γ−1

ds dλ(x)

f (x)χ[0,Λf (x)] (s) dλ(x) ds.

Given s ≥ 0, we set Es := {x ∈ [a, b] | Λf (x) < s}. If Es = ∅, then f (x)χ[0,Λf (x)] (s) dλ(x) = Λf (b) ≥ Λf (b) − s. [a,b]

Suppose that Es = ∅, cs := sup Es , and a sequence {tn }∞ 1 ⊂ Es is such that tn ↑ cs as n → ∞. If cs ∈ Es , then f (x)χ[0,Λf (x)] (s) dλ(x) = f dλ = Λf (b) − Λf (cs ) ≥ Λf (b) − s, (s)

[a,b]

(cs ,b]

and if cs ∈ Es , then f (x)χ[0,Λf (x)] (s) dλ(x) = [a,b]

Thus,

(s)

f dλ = Λf (b) − lim Λf (t(s) n ) ≥ Λf (b) − s. n→∞

[cs ,b]

f (x)Λf (x)γ dλ(x) ≥ Λf (b)γ+1 − [a,b]

Λf (b)γ+1 γΛf (b)γ+1 = ; γ+1 γ+1

this completes the proof of the case γ > 0. Suppose that γ ∈ (−1, 0) and Λf (b) < +∞. Then the ﬁrst inequality in (1.1.2) follows from the monotonicity of Λf . Let us prove the second inequality. We have +∞ γ γ−1 f (x)Λf (x) dλ(x) = −γ f (x) s ds dλ(x) [a,b]

[a,b]

= −γ

Λf (x)

f (x)

= −γ

Λf (b)

s 0

s

0

[a,b]

Λf (b)

γ−1 [a,b]

γ−1

χ[Λf (x),Λf (b)] (s) ds +

+∞

s

γ−1

ds dλ(x)

Λf (b)

f (x)χ[Λf (x),Λf (b)] (s) dλ(x) ds + Λf (b)γ+1 .

Given s ≥ 0, we set Es := {x ∈ [a, b] | Λf (x) ≤ s}. If Es = ∅, then f (x)χ[Λf (x),Λf (b)] (s) dλ(x) = 0 ≤ s. [a,b]

Suppose that Es = ∅, cs := sup Es , and a sequence {tn }∞ 1 ⊂ Es is such that tn ↑ cs as n → ∞. If cs ∈ Es , then f (x)χ[Λf (x),Λf (b)] (s) dλ(x) = f dλ = Λf (cs ) ≤ s, (s)

[a,b]

(s)

[a,cs ]

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S4

PROKHOROV et al.

and if cs ∈ Es , then

[a,b]

Thus,

f (x)χ[Λf (x),Λf (b)] (s) dλ(x) =

[a,cs )

Λf (b)

f (x)Λf (x) dλ(x) ≤ −γ γ

f dλ = lim Λf (t(s) n ) ≤ s. n→∞

sγ ds + Λf (b)γ+1 =

0

[a,b]

Λf (b)γ+1 , γ+1

which completes the proof of the lemma. For an integral with variable lower limit, a similar assertion is valid. Lemma 1.2. Given a σ-ﬁnite measure λ on [a, b] and a function f ∈ {Mλ }+ , let Λf (x) :=

[x,b] f

dλ.

If either γ > 0 or γ ∈ (−1, 0) and Λf (a) < +∞, then Λf (a)γ+1 Λf (a)γ+1 ≤ . (1.1.3) f (x)Λf (x)γ dλ(x) ≤ max{1, γ + 1} min{1, γ + 1} [a,b] Obviously, in the case where γ ∈ (−1, 0) and [a,b] f dλ = +∞, the second inequalities in (1.1.2) and (1.1.3) remain valid, but the ﬁrst ones may fail. For example, [a,b] f (x)Λf (x)γ dλ(x) = 0 if Λf (x) = +∞ for any x ∈ [a, b]. The following lemma was proved in [7, Theorem 190] for the case of the Lebesgue measure. Its proof in the general case is essentially contained in the proof of the coincidence of the norm of an integral functional on Lp with the norm of a function in Lp (see, e.g., [8, Section IV.8]). Lemma 1.3. Suppose that X is a space with σ-ﬁnite measure p > 1, g is a nonnegative pλ, measurable function on X, and E is a λ-measurable set. If E g dλ = +∞, then there exists a nonnegative λ-measurable function f on X such that E f p dλ < +∞ and E f g dλ = +∞. Proof. Note that λ(E) > 0, because E gp dλ = +∞. Since λ is a σ-ﬁnite measure, it follows that An , where λ(An ) < +∞ for n ∈ N E= n∈N

n .

Let F := {x ∈ E | g(x) = +∞}. If λ(F ) > 0, then there exists an n ∈ N and An ∩ An = ∅ for n = such that λ(An ∩ F ) > 0. We set f = χAn ∩F . The function f is as required. Suppose that λ(F ) = 0. We set En,k := {x ∈ An | 2k ≤ g(x) < 2k+1 },

n ∈ N,

k ∈ Z.

Let us enumerate the elements of the countable family {En,k }n∈N,k∈Z by a single index as {En,k }n∈N,k∈Z = {Ej }j∈N . For each j ∈ N, we set aj := 2k , where k is the integer for which Ej = En,k ; we also let s := min{j ∈ N | λ(Ej ) = 0} − 1. For any x ∈ Ej , we have aj ≤ g(x) ≤ 2aj and p −p as+i λ(Es+i ) ≥ 2 i∈N

i∈N

The function f :=

j∈N

−1 aps+j

p

Es+i

j

−p

gp dλ = +∞.

g dλ = 2

E

−1

aps+i λ(Es+i )

χEs+j

i=1

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S5

is as required, because p

f dλ = E

j

aps+j λ(Es+j )

−p

aps+i λ(Es+i )

< +∞

i=1

j∈N

and f g dλ = E

Es+j

j∈N

f g dλ ≥

j

aps+j λ(Es+j )

−1

aps+i λ(Es+i )

= +∞

i=1

j∈N

by virtue of Abel’s theorem (see [7, Theorem 162]). Lemma 1.4. Suppose given L ∈ Mλ (R). Let Lt := L ∩ (−∞, t], and let Lt := L ∩ [t, +∞), t ∈ R. (1) If λ(Lt ) = 0 for any t ∈ L, then λ(L) = 0. (2) If λ(Lt ) = 0 for any t ∈ L, then λ(L) = 0. Proof. If L = ∅, then λ(L) = 0. Suppose that L = ∅, s := sup L, i := inf L, and sequences {tn }∞ 1 , ∞ {tn }1 ⊂ L are suchthat tn ↑ s and tn ↓ i as n → ∞. (1) If s ∈ L, then L = Ls and λ(L) = λ(Ls ) = 0. If s ∈ L, then L := ∞ 1 Ltn , and the required assertion follows from properties of measures. The case (2) is considered in a similar way by using the sequence {tn }. Lemma 1.5. Suppose that 0 < p, q < +∞ and λ and μ are σ-ﬁnite measures on [a, b]; suppose also that u, τ ∈ {Mλ }+ , v ∈ {Mμ }+ , and k is a function on [a, b]2 belonging to the class {Mμ × Mλ }+ and such that k(x1 , y) ≤ αk(x2 , y) for x1 < x2 , where α > 0 is a constant. Then the following inequalities are equivalent:

q

1/q v(x) k(x, y)u(y)f (y)τ (y) dλ(y) dμ(x) [a,b]

[a,x]

≤C v(x)

[a,b]

for all f ∈ {Mλ }+ ,

f dλ [a,b]

1/p p

q

1/q k(x, y)u(y)g(y) dλ(y) dμ(x)

[a,x] p −p

≤C

(1.1.4)

g τ

1/p for all g ∈ {Mλ }+ .

dλ

(1.1.5)

[a,b]

Proof. Suppose that inequality (1.1.5) holds. Fix any function f ∈ {Mλ }+ . Substituting g = f τ into (1.1.5), we obtain q 1/q v(x) k(x, y)u(y)f (y)τ (y) dλ(y) dμ(x) [a,b]

[a,x] p −p

≤C

(f τ ) τ

1/p dλ

1/p

≤C

[a,b]

p

f dλ

,

[a,b]

because τ p τ −p ≤ 1. Now, suppose that (1.1.4) holds. For t ∈ [a, b], we set Ft := {x ∈ [a, t] | τ (x) = +∞ ∧ k(t, x)u(x) = 0},

E := {t ∈ [a, b] | λ(Ft ) > 0}.

Since the function k is quasi-increasing in the ﬁrst variable, it follows that Ft1 ⊂ Ft2 for t1 < t2 , and hence λ(Ft ) is nondecreasing on [a, b]. Therefore, E is Borel. If E = ∅, then E v dμ = 0. Suppose that PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S6

PROKHOROV et al.

E = ∅. Fix any t ∈ E. According to Lemma 6.9 in [9], there exists an Mλ -measurable function f such that [a,b] f p dλ < +∞ and f (x) ∈ (0, 1), x ∈ [a, b]. Hence, k(t, x)u(x)f (x)τ (x) dλ(x) = +∞. Ft

Substituting f χFt into (1.1.4), we obtain 1/q v dμ [t,b]

k(t, x)u(x)f (x)τ (x) dλ(x) < +∞.

Ft

Therefore, [t,b] v dμ = 0. Since t is an arbitrary element of E, it follows by Lemma 1.4 that E v dμ = 0; thus, inequality (1.1.5) is equivalent to

q

1/q v(x) k(x, y)u(y)g(y) dλ(y) dμ(x) [a,b]\E

≤C

[a,x] p −p

g τ

1/p dλ

for all

g ∈ {Mλ }+ .

(1.1.6)

[a,b]

Note that λ(Fx ) = 0 for any x ∈ [a, b] \ E. Now, we ﬁx any function g ∈ {Mλ }+ . If [a,b] gp τ −p dλ = +∞, then (1.1.5) holds. Suppose that p −p dλ < +∞. Then the function g p τ −p is ﬁnite λ-almost everywhere; in particular, for [a,b] g τ E1 := {x ∈ [a, b] | g(x) = 0 ∧ τ (x) = 0}, we have λ(E1 ) = 0. Substituting f (y) = g(y)τ (y)−1 , y ∈ [a, b], into (1.1.4), we obtain q 1/q −1 v(x) k(x, y)u(y)g(y)τ (y) τ (y) dλ(y) dμ(x) [a,b]\E

[a,x]

1/p ) dλ(y) .

−1 p

≤C

(g(y)τ (y) [a,b]

This inequality implies (1.1.6), because

y ∈ [a, x] | k(x, y)u(y)g(y)τ (y)−1 τ (y) = k(x, y)u(y)g(y) ⊂ E1 ∪ Fx for any x ∈ [a, b] \ E. Lemma 1.6. Let 0 < p, q < +∞. Suppose that μ, λ, and ν are σ-ﬁnite measures on [a, b] such that λ k ∈ {Mμ × Mλ }+ . and ν are deﬁned on the same σ-algebra Mλ ; suppose also that w ∈ {Mλ }+ and Let (νa , νs ) be the Lebesgue decomposition of ν with respect to λ, i.e., ν = νa + νs , where νa is absolutely continuous with respect to λ (νa λ) and νs and λ are mutually singular (νs ⊥ λ). Then the following inequalities are equivalent: q 1/q 1/p p k(x, y)f (y) dλ(y) dμ(x) ≤C f w dν for all f ∈ {Mλ }+ , [a,b]

[a,x]

q 1/q k(x, y)f (y) dλ(y) dμ(x) ≤C

[a,b]

[a,b]

[a,x]

(1.1.7)

1/p f p w dνa

for all f ∈ {Mλ }+ .

[a,b]

(1.1.8) Proof. Since ν = νa + νs , inequality (1.1.8) implies (1.1.7). Conversely, suppose that (1.1.7) holds. Fix any f ∈ {Mλ }+ . Since νs and λ are mutually singular, it follows that there exists a set A ∈ Mλ such that λ(A) = 0 and νs is concentrated on A, and since νa is absolutely continuous with respect to λ PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S7

and λ(E) = 0 for any E ∈ Mλ , E ⊂ A, it follows that the measure νa is concentrated on [a, b] \ A. Let f := f χ . Substituting the function f into (1.1.7), we obtain [a,b]\A

[a,b]

q 1/q k(x, y)f (y) dλ(y) dμ(x)

[a,x]

q 1/q k(x, y)f (y) dλ(y) dμ(x) ≤C

= [a,b]

[a,x]

p

=C

f w dνa + [a,b]

f w dνs

f w dν [a,b]

1/p

p

1/p

p

1/p

p

=C

f w dνa

[a,b]

,

[a,b]

i.e., inequality (1.1.8) holds.

1.2. Hardy’s Inequality with Three Measures In this part, we prove criteria for the validity of Hardy’s inequality of the form q 1/q 1/p p v(x) f u dλ dμ(x) ≤C f w dν for all f ∈ {Mλ }+ , [a,b]

[a,x]

[a,b]

where 1 < p < +∞; 0 < q < +∞; 1/r = 1/q − 1/p; μ, λ, and ν are σ-ﬁnite measures on [a, b]; λ and ν are deﬁned on a σ-algebra Mλ ; u, w ∈ {Mλ }+ ; and v ∈ {Mμ }+ . The case where the measures λ, μ, and ν are absolutely continuous with respect to the Lebesgue measure (i.e., the case of a weighted Hardy-type inequality) was completely characterized by many authors (see monographs [5] and [2]). Theorems 1.1 and 1.2 are concerned with the case λ = ν. Theorem 1.4 gives a criterion for the validity of Hardy’s inequality in the general case on the basis of these results. Theorem 1.1. Let 1 < p ≤ q < +∞. Suppose that μ and λ are σ-ﬁnite measures on [a, b], u ∈ {Mλ }+ , and v ∈ {Mμ }+ . Then q 1/q 1/p p v(x) f u dλ dμ(x) ≤C f dλ for all f ∈ {Mλ }+ (1.2.1) [a,b]

[a,x]

[a,b]

if and only if A < +∞, where

1/q

A := sup A(t) := sup t∈[a,b]

v dμ

t∈[a,b]

[t,b]

1/p

p

u dλ

.

[a,t]

Moreover, the least constant C in (1.2.1) satisﬁes the relation C ≈ A. Proof. We set

I0 := x ∈ [a, b]

u dλ = 0 ,

I∞

[a,x]

:= x ∈ [a, b]

u dλ = +∞ , p

[a,x]

I := [a, b] \ (I0 ∪ I∞ ).

Note that I0 , I∞ , I ∈ B. Moreover, by virtue of Lemma 1.4, we have I0 u dλ = 0, because I0 ∩ (−∞, t] ⊂ [a, t] and [a,t] u dλ = 0 for any t ∈ I0 . We set p u dλ, V (x) := v dμ, x ∈ [a, b]. U (x) := [a,x]

[x,b]

Let us show that (1.2.1) holds if and only if I∞ v dμ = 0, I∞ v dμ = 0, and q 1/q 1/p p v(x) f u dλ dμ(x) ≤C f dλ for all f ∈ {Mλ }+ . I

[a,x]

[a,b]

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

(1.2.2)

S8

PROKHOROV et al.

Suppose that (1.2.1) holds. Then so does (1.2.2). Next, if I∞ = ∅, then I∞ v dμ = 0. Suppose that I∞ = ∅. Fix any t ∈ I∞ . By Lemma 1.3, there exists an f ∈ {Mλ }+ for which [a,t] f p dλ < +∞ and [a,t] f u dλ = +∞. Substituting the function f χ[a,t] into (1.2.1), we obtain

1/p

f u dλ ≤ C

1/q

V (t)

[a,t]

p

f dλ

.

[a,t]

It follows that [t,b] v dμ = 0, which implies I∞ v dμ = 0 by virtue of Lemma 1.4. Conversely, if I∞ v dμ = 0, then I0 u dλ = 0 implies that the left-hand sides of inequalities (1.2.2) and (1.2.1) are equal. Thus, (1.2.2) implies (1.2.1). Moreover, relation A < +∞ is equivalent to v dμ = 0,

A := sup A(t) < +∞. t∈I

I∞

Indeed, (1.2.3) implies A = A < +∞. Conversely, if A < +∞, then According to Lemma 1.4, this implies I∞ v dμ = 0. Moreover, A = A.

(1.2.3) [x,b] v dμ

= 0 for any x ∈ I∞ .

It also follows from these considerations that the theorem is true if I = ∅. Therefore, in what follows, we assume that I = ∅. Suﬃciency. Suppose that A < +∞ and, in particular, I∞ v dμ = 0. Fix any f ∈ {Mλ }+ . Since ¨ inequality with exponents p and p that, for any x ∈ I, we I0 u dλ = 0, it follows by virtue of Holder’s have f u dλ = f u dλ = f (y)u(y)U (y)1/(pp ) U (y)−1/(pp ) dλ(y) [a,x]

[a,x]∩I

[a,x]∩I

≤

1/p

p

f (y) U (y)

1/p dλ(y)

[a,x]∩I

−1/p

u(y) U (y)

1/p dλ(y) .

[a,x]

Therefore, by Lemma 1.1, we have f u dλ [a,x]

p

p

1/p

f (y) U (y) [a,x]

≤ A1/p

1/p 2 dλ(y) U (x)1/(p )

1/p f (y)p U (y)1/p dλ(y) V (x)−1/(qp )

[a,x]

for any x ∈ I. Applying this relation, Minkowski’s inequality, and Lemma 1.2, we obtain the following estimate of the left-hand side of (1.2.2):

q

1/q v(x) f u dλ dμ(x) I

[a,x] 1/p

v(x)

A

1/p

I

≤A

A1/p

p

f (y) U (y)

[a,x] p

1/p

1/p

−1/p

f (y) U (y)

[a,b]

[a,b]

q/p

1/q −1/p dλ(y) V (x) dμ(x) V (x)

p/q

1/p v(x) dμ(x) dλ(y)

I∩[y,b]

1/p f (y)p U (y)1/p V (y)1/q dλ(y) ≤A

1/p f p dλ

.

[a,b]

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

Necessity. Suppose that (1.2.1) holds. Then f = up −1 χ[a,t] in (1.2.2), we obtain

I∞

S9

v dμ = 0 and (1.2.2) is valid. Fix any t ∈ I. Setting

V (t)1/q U (t) ≤ CU (t)1/p . Hence A ≤ C, which proves the theorem. Theorem 1.2. Suppose that 0 < q 1, and 1/r = 1/q − 1/p. Suppose also that μ and λ are σ-ﬁnite measures on [a, b], u ∈ {Mλ }+ , and v ∈ {Mμ }+ . Then (1.2.1) holds if and only if B < +∞, where r/p r/p 1/r p u dλ v dμ v(x) dμ(x) . B := [a,b]

[a,x]

[x,b]

Moreover, the least constant C in (1.2.1) satisﬁes the relation C ≈ B. Proof. Let I0 , I∞ , and I be the intervals deﬁned in the proof of Theorem 1.1. We set U (x) := [a,x] up dλ and V (x) := [x,b] v dμ, x ∈ [a, b]. Arguing as in the proof of Theorem 1.1, we conclude that I0 u dλ = 0 and that inequality (1.2.1) is satisﬁed if and only if I∞ v dμ = 0 and (1.2.2) holds. Moreover, B < +∞ if and only if I∞ v dμ = 0 and

r/p

B :=

U (x)

r/p

V (x)

1/r v(x) dμ(x) < +∞.

I

Indeed, it follows from I0 u dλ = 0, I∞ v dμ = 0, and B < +∞ that B = B < +∞. Conversely, suppose that B < +∞. Then, by Lemma 1.2, we have 1/r r/p r/p U (x) V (x) v(x) dμ(x) U (t)1/p V (t)1/q = A(t) for any t ∈ [a, b]. B≥ [t,b]

Therefore, A B < +∞; as shown in Theorem 1.1, this implies the relation I∞ v dμ = 0, which, together with I0 u dλ = 0, gives B = B < +∞. It follows, in particular, that the assertion of theorem holds if I = ∅. Therefore, hereafter, we assume that I = ∅.

Suﬃciency. Suppose that B < +∞. Consider the function q/r r/q r/p p U (s) V (s) u(s) dλ(s) . h(x) = χI (x) [a,x]

+ q/p q/p Note that h ∈ {B} . According to Lemma 1.1, for each x ∈I, we have h(x) V (x) U (x) , and if h(x) = 0, then [x,b] v dμ = 0. Hence, Lemma 1.4 implies {x∈I|h(x)=0} v dμ = 0. Interchanging the integrals and applying Lemmas 1.2 and 1.1, we obtain

r/q r/q r/p p vh dμ ≤ v(x) U (s) V (s) u(s) dλ(s) dμ(x) [a,b]

[a,b]

[a,x]

U (s)r/q V (s)r/q u(s)p dλ(s)

= [a,b]

r/q

r/p

U (s) [a,b]

v(x)V (x) [s,b] r/q

U (s)

= [a,b]

dμ(x) u(s)p dλ(s)

u(s) dλ(s) V (x)r/p v(x) dμ(x) B r . p

[a,x]

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S10

PROKHOROV et al.

This estimate implies, in particular, that {x∈I|h(x)=+∞} v dμ = 0. Using this relation and the relation ¨ inequality with exponents r/q and p/q, obtain {x∈I|h(x)=0} v dμ = 0 proved above and applying Holder’s the following estimate of the left-hand side of the inequality (1.2.2):

q

1/q v(x) f u dλ dμ(x) I

[a,x]

v(x)h(x)h(x)−1

= I

q f u dλ

1/q dμ(x)

[a,x]

1/r

≤

r/q

vh [a,b]

dμ p

f u dλ

[a,b]

p

v(x)h(x) I

B

−p/q

f u dλ

1/p dμ(x)

[a,x]

1/p d μ(x) ,

[a,x]

μ := χI vh−p/q dμ. where μ is the measure deﬁned on Mμ by d Fix any t ∈ I. The ﬁniteness of the constant A implies [t,b] v dμ < +∞, and if [t,b] v dμ = 0, then μ ([t, b]) = 0. Suppose that [t,b] v dμ > 0. Then the monotonicity of h on I and Lemma 1.1 imply χI vh−p/q dμ μ ([t, b]) = [t,b]

≤

[t,b]

≤

v(x)

r/q

U (s)

r/p

V (s)

[a,x]

v(x)

[t,b]

r/q

r/p

U (s)

V (s)

[a,t]

r/p ≤ V (t) V (t)

r/q

U (s)

−p/r u(s) dλ(s) dμ(x) p

−p/r u(s) dλ(s) dμ(x) p

−p/r u(s) dλ(s) ≈ U (t)−p/p < +∞. p

[a,t]

It follows, in particular, that

([t, b])1/p U (t)1/p 1. sup μ

(1.2.4)

t∈I

Next, let c = inf I. If c ∈ I, then I ⊂ [c, b] and −p/q dμ = μ ([a, b]) = vh I

χI vh−p/q dμ = μ ([c, b]) < +∞.

[c,b]

If c ∈ I, then I ⊂ (c, b] and μ ([a, c]) = 0. In this case, we ﬁxany sequence {tn }∞ 1 ⊂ I such that tn ↓ c E and μ (E ) < +∞ for any n ∈ N. as n → ∞ and set En = [a, c] ∪ [tn , b]. We have [a, b]= ∞ n n 1 (I∞ ) = 0. Taking into Therefore, μ is a σ-ﬁnite measure on [a, b]. Moreover, I∞ v dμ = 0 implies μ account (1.2.4) and applying Theorem 1.1, we obtain p 1/p 1/p p f u dλ d μ(x) f dλ , [a,b]

[a,x]

[a,b]

which proves suﬃciency. Necessity. Suppose that inequality (1.2.1) is true; then I∞ v dμ = 0 and (1.2.2) holds. The ∞ σ-ﬁniteness of μ implies the existence of a sequence {En }∞ 1 of sets such that [a, b] = 1 En and μ(En ) < +∞, n ∈ N. Without loss of generality, we can assume also that En ⊂ En+1 and En ∩ I = ∅ for any n ∈ N. Let {tn }∞ 1 ⊂ I be such that tn ↑ sup I; for each n ∈ N, we set En ∩ I if sup I ∈ I, Fn := En ∩ I ∩ [a, tn ] if sup I ∈ I. PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S11

∞

Then 1 Fn = I; moreover, for any n ∈ N, we have Fn ⊂ Fn+1 and μ(Fn ) < +∞. For each n ∈ N, we set vn := min{v, n}, dμn := vn χFn dμ, and 1/r r/p r/p U (x) μn ([x, b]) dμn (x) Bn :=

I

r/p

=

r/p

U (x)

vn χFn dμ

Fn

1/r vn (x) dμ(x) .

[x,b]

If Fn = ∅, then Bn < +∞. If Fn = ∅, then sup Fn ∈ I and 1/p r/p 1/r p u dλ vn χFn dμ vn (x) dμ(x) Bn ≤

[a,sup Fn ]

≤

up dλ

[a,sup Fn ]

[x,b]

μn (Fn )1/q

1/p

p

=

Fn

1/p

(nμ(Fn ))1/q < +∞.

u dλ [a,sup Fn ]

Let f (s) = μn ([s, b])r/(pq) U (s)r/(pq ) u(s)p −1 χI (s). Then, by Lemma 1.1, we have q 1/q f u dλ dμn (x) I

[a,x]

r/(pq)

=

μn ([s, b])

I

≥

[a,x]∩I

r/(pq )

μn ([x, b]) I

U (s) [a,x]∩I

r/(pq )

r/p

=

U (s)

r/p

r/(pq )

μn ([x, b]) I

U (s)

q

1/q u(s) dλ(s) dμn (x) p

q

1/q u(s) dλ(s) dμn (x) p

q

1/q u(s) dλ(s) dμn (x) ≈ Bnr/q . p

[a,x]

On the other hand, applying Lemmas 1.2 and 1.1, we obtain 1/p 1/p f p dλ = μn ([s, b])r/q U (s)r/q u(s)p dλ(s) [a,b]

I

≈

r/p

μn ([x, b])

I

[s,b]

1/p r/q p dμn (x) U (s) u(s) dλ(s)

r/q

r/p

=

μn ([x, b]) [a,b]

U (s)

1/p u(s) dλ(s) dμn (x) Bnr/p . p

[a,x]∩I r/q

r/p

Substituting f into (1.2.2) and using the estimates found above, we see that Bn CBn . Therefore, Bn C, because the Bn are ﬁnite. Finally, since ∞ 1 Fn = I and Fn ⊂ Fn+1 , it follows that vn χFn ↑ v as n → ∞, and the monotone convergence theorem implies B C. A criterion for the validity of the dual inequality is proved in a similar way. Theorem 1.3. Suppose that 1 < p < +∞, 0 < q < +∞, and 1/r = 1/q − 1/p. Suppose also that μ and λ are σ-ﬁnite measures on [a, b], u ∈ {Mλ }+ , and v ∈ {Mμ }+ . If p ≤ q, then q 1/q 1/p p v(x) f u dλ dμ(x) ≤C f dλ for all f ∈ {Mλ }+ (1.2.5) [a,b]

[x,b]

[a,b]

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S12

PROKHOROV et al.

if and only if A∗ < +∞, where 1/q

∗

A := sup

v dμ

t∈[a,b]

p

1/p

u dλ

[a,t]

.

[t,b]

Moreover, the least constant C in (1.2.5) satisﬁes the relation C ≈ A∗ . If q < p, then inequality (1.2.5) holds if and only if B ∗ < +∞, where r/p r/p 1/r ∗ p B := u dλ v dμ v(x) dμ(x) . [a,b]

[x,b]

[a,x]

Moreover, the least constant C in (1.2.5) satisﬁes the relation C ≈ B ∗ . The general result is as follows. Theorem 1.4. Suppose that 1 < p < +∞, 0 < q < +∞, and 1/r = 1/q − 1/p. Suppose also that μ, λ, and ν are σ-ﬁnite measures on [a, b], λ and ν are deﬁned on the σ-algebra Mλ , u, w ∈ {Mλ }+ , and v ∈ {Mμ }+ . Let (νa , νs ) be the Lebesgue decomposition of the measure ν with respect to λ, i.e., ν = νa + νs , where the measure νa is absolutely continuous with respect to λ (νa λ), the measures νs and λ are mutually singular (νs ⊥ λ), and dνa /dλ is the Radon–Nikodym derivative of the measure νa with respect to λ. If p ≤ q, then q 1/q 1/p p v(x) f u dλ dμ(x) ≤C f w dν for all f ∈ {Mλ }+ (1.2.6) [a,b]

[a,x]

[a,b]

if and only if A < +∞, where A := sup t∈[a,b]

1/p dνa 1−p u w dλ . dλ [a,t]

1/q

v dμ [t,b]

p

Moreover, the least constant C in (1.2.6) satisﬁes the relation C ≈ A. If q < p, then inequality (1.2.6) holds if and only if B < +∞, where r/p r/p 1/r dνa 1−p p u w dλ v dμ v(x) dμ(x) . B := dλ [a,b] [a,x] [x,b] Moreover, the least constant C in (1.2.6) satisﬁes the relation C ≈ B. Proof. According to Lemma 1.6, inequality (1.2.6) is equivalent to 1/p q 1/q dνa dλ v(x) f u dλ dμ(x) ≤C f pw dλ [a,b] [a,x] [a,b]

for all f ∈ {Mλ }+ .

In turn, by virtue of Lemma 1.5, this inequality is equivalent to q 1/q dνa −1/p v(x) fu w dλ dμ(x) dλ [a,b] [a,x] 1/p p ≤C f dλ for all f ∈ {Mλ }+ . [a,b]

The application of Theorems 1.1 and 1.2 to the last inequality completes the proof of Theorem 1.4. The general result for the dual inequality is as follows. Theorem 1.5. Suppose that 1 < p < +∞, 0 < q < +∞, and 1/r = 1/q − 1/p. Suppose also that μ, λ, and ν are σ-ﬁnite measures on [a, b], λ and ν are deﬁned on the σ-algebra Mλ , u, w ∈ {Mλ }+ , PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S13

and v ∈ {Mμ }+ . Let (νa , νs ) be the Lebesgue decomposition of the measure ν with respect to λ, i.e., ν = νa + νs , where the measure νa is absolutely continuous with respect to λ (νa λ), the measures νs and λ are mutually singular (νs ⊥ λ), and dνa /dλ is the Radon–Nikodym derivative of the measure νa with respect to λ. If p ≤ q, then q 1/q 1/p p v(x) f u dλ dμ(x) ≤C f w dν for all f ∈ {Mλ }+ (1.2.7) [a,b]

[x,b]

[a,b]

if and only if A∗ < +∞, where ∗

1/q

A := sup t∈[a,b]

p

v dμ

u

[a,t]

[t,b]

1−p

dνa w dλ

1/p dλ

.

Moreover, the least constant C in (1.2.7) satisﬁes the relation C ≈ A∗ . If q < p, then inequality (1.2.7) holds if and only if B ∗ < +∞, where r/p r/p 1/r dνa 1−p ∗ p u w dλ v dμ v(x) dμ(x) . B := dλ [a,b] [x,b] [a,x] Moreover, the least constant C in (1.2.7) satisﬁes the relation C ≈ B ∗ .

1.3. Hardy’s Inequality for Nuclear Operators In this part, we consider the inequality 1/q q v(Kf ) dμ ≤C [a,b]

1/p p

f w dν

for all f ∈ {Mλ }+ ,

(1.3.1)

[a,b]

where μ, λ, and ν are σ-ﬁnite measures on [a, b], λ and ν are deﬁned on the same σ-algebra Mλ , k(x, y)u(y)f (y) dλ(y), (Kf )(x) := [a,x]

u, w ∈ {Mλ }+ , and v ∈ {Mμ }+ , and the kernel k is a nonnegative (Mμ,λ × Mμ,λ )-measurable function on [a, b]2 satisfying Oinarov’s condition, i.e., there exists a constant D such that D−1 (k(x, z) + k(z, y)) ≤ k(x, y) ≤ D(k(x, z) + k(z, y))

if x ≥ z ≥ y.

(1.3.2)

Standard examples of kernels with the above properties are (i) the kernel of the Riemann–Liouville integral operator in the case α ≥ 1: k(x, y) = (max{x − y, 0})α−1 ; (ii) certain kernels with logarithmic functions, such as k(x, y) = log (1 + max{x − y, 0}), β

k(x, y) = log

β

x ,1 , max y

β ≥ 0;

and their various combinations. The case where λ, μ, and ν are absolutely continuous with respect to the Lebesgue measure (i.e., the case of a weighted Hardy-type inequality) has been completely characterized in diﬀerent ways by Oinarov [10], Bloom and Kerman [11], and Stepanov [12]. In this section, relations A B and B A mean that A ≤ cB, where the constant c depends on p, q, and D (D is the constant in condition (1.3.2)). PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S14

PROKHOROV et al.

Given s > 0, f ∈ {Mλ }+ , g ∈ {Mμ }+ , and x, y ∈ [a, b], we set k(x, y)s u(y)s f (y) dλ(y), (K0 f )(x) := (Ks f )(x) := [a,x]

(Ks∗ g)(y) :=

u(y)f (y) dλ(y),

[a,x]

(K0∗ g)(y) :=

v(x)k(x, y)s g(x) dμ(x), [y,b]

v(x)g(x) dμ(x). [y,b]

Lemma 1.7. Let 1 < p, q < +∞, and let μ and λ be σ-ﬁnite measures on [a, b]. Suppose that u ∈ {Mλ }+ and v ∈ {Mμ }+ and that k is a nonnegative (Mμ × Mλ )-measurable function on [a, b]2 such that there exist constants α1 > 0 and α2 > 0 for which k(x, y1 ) ≥ α1 k(x, y2 ) if y1 < y2 and k(x1 , y) ≤ α2 k(x2 , y) if x1 < x2 . Then, for the sets I0 := {x ∈ [a, b] | (K1)(x) = 0}, I 0 := {x ∈ [a, b] |

(K1∗ 1)(x)

I∞ := {x ∈ [a, b] | (Kp 1)(x) = +∞}, I ∞ := {x ∈ [a, b] | (Kq∗ 1)(x) = +∞},

= 0},

I := [a, b] \ (I0 ∪ I∞ ),

I := [a, b] \ (I 0 ∪ I ∞ )

the following conditions hold: (1) x < y < z for all x ∈ I0 , y ∈ I, and z ∈ I∞ ; (2) x < y < z for all x ∈ I ∞ , y ∈ I, and z ∈ I 0 . In particular, all of these sets are connected and, hence, Borel. Proof. (1) Take x ∈ I0 , y ∈ I, and z ∈ I∞ . The points x, y, and z are diﬀerent and belong to [a, b]. If t ∈ [a, x), then k(t, s)u(s) dλ(s) ≤ α2 k(x, s)u(s) dλ(s) = 0 [a,t]

[a,x]

and t ∈ I0 , whence y > x and z > x. If t ∈ (z, b], then p p p k(t, s) u(s) dλ(s) ≥ k(z, s)p u(s)p dλ(s) = +∞, α2 [a,t]

[a,z]

i.e., t ∈ I∞ , whence y < z. Thus, x < y < z. Assertion (2) is proved in a similar way. The proofs of Lemma 1.8 and Theorems 1.6 and 1.7 are based on an idea of paper [12], in which a similar result was obtained for measures absolutely continuous with respect to the Lebesgue measure. Lemma 1.8. Suppose that μ and λ are σ-ﬁnite measures on [a, b], u ∈ {Mλ }+ , v ∈ {Mμ }+ , and k is a nonnegative (Mμ,λ × Mμ,λ )-measurable function on [a,b]2 satisfying Oinarov’s condition (1.3.2). Suppose also that 1 < q < +∞, f ∈ {Mλ }+ , and J := [a,b] v(Kf )q dμ < +∞. Then uf (K0 f )q−1 (Kq∗ 1) dλ + uf (Kf )q−1 (K1∗ 1) dλ. (1.3.3) J≈ [a,b]

Proof. We have

[a,b]

q−1

v(Kf )(Kf )

J=

dμ =

[a,b]

Setting w := f u and J1 :=

[a,b]

v(x)

w(y) dλ(y) [a,b]

[y,b]

uf (K1∗ [(Kf )q−1 ]) dλ.

q−1 k(x, z)w(z) dλ(z) k(x, y) dμ(x),

[a,y]

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

v(x)

J2 :=

w(y) dλ(y) [a,b]

[y,b]

S15

q−1 k(x, z)w(z) dλ(z) k(x, y) dμ(x),

[y,x]

we see that J J1 and J J1 + J2 . Since k satisﬁes condition (1.3.2), it follows that k(x, z)w(z) dλ(z) ≈ (k(x, y) + k(y, z))w(z) dλ(z); [a,y]

[a,y]

therefore,

J1 ≈

q−1

w(K0 f )

(Kq∗ 1) dλ

+

[a,b]

[a,b]

w(Kf )q−1 (K1∗ 1) dλ.

This relation implies, in particular, the lower bound for J in (1.3.3). Note that, in the case q > 2, we also have

1/(q−1) q−1 ∗ q−1 ∗ w(K0 f ) (Kq 1) dλ + w(Kf ) (K1 1) dλ J (q−2)/(q−1) . (1.3.4) J1 [a,b]

[a,b]

Suppose that q = 2. Twice changing the order of integration and applying (1.3.2), we obtain w(z) dλ(z) w(y) dλ(y) v(x)k(x, z)k(x, y) dμ(x) J2 = [a,b]

≈

[a,z]

w(z) dλ(z)

= [a,b]

[z,b]

w(y) dλ(y)

[a,b]

[a,z]

w(K0 f )(K2∗ 1) dλ

+ [a,b]

v(x)k(x, z)[k(x, z) + k(z, y)] dμ(x) [z,b]

w(Kf )(K1∗ 1) dλ.

This proves (1.3.3) for q = 2. Consider the case q = 2. For each y ∈ [a, b], we set

q−1 v(x)k(x, y) k(x, z)w(z) dλ(z) dμ(x). h(y) := [y,b]

[y,x]

Since J < +∞, it follows that μ{x ∈ [a, b] | (Kf )(x) = +∞ ∧ v(x) = 0} = 0, which implies

1+(q−2) v(x)k(x, y) k(x, t)w(t) dλ(t) dμ(x) h(y) =

[y,b]

[y,x]

v(x)k(x, z)k(x, y) dμ(x)

w(z) dλ(z)

=

[y,b]

≈

[z,b]

w(z) dλ(z) [y,b]

2 v(x)k(x, z) dμ(x)

[z,b]

q−2 k(x, t)w(t) dλ(t)

v(x)k(x, z)k(z, y) dμ(x)

w(z) dλ(z) [y,b]

[y,x]

[y,x]

+

q−2 k(x, t)w(t) dλ(t)

[z,b]

q−2 k(x, t)w(t) dλ(t)

[y,x]

=: h1 (y) + h2 (y) any y ∈ [a, b]. ¨ Suppose that q > 2. Applying Holder’s inequality with exponents q − 1 and (q − 1)/(q − 2), we obtain w(z) dλ(z) v(x)k(x, z)2 (Kf )(x)q−2 dμ(x) h1 (y) ≤ [y,b]

[z,b]

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S16

PROKHOROV et al.

w(z)(Kq∗ 1)(z)1/(q−1) dλ(z)

≤ [y,b]

=

[y,b]

(q−2)/(q−1) v(x)k(x, z)(Kf )(x)q−1 dμ(x)

[z,b]

w(z)(Kq∗ 1)(z)1/(q−1) [(K1∗ [(Kf )q−1 ])(z)](q−2)/(q−1) dλ(z).

¨ Using this estimate and Holder’s inequality with exponents q − 1 and (q − 1)/(q − 2), we arrive at wh1 dλ J2,1 :=

[a,b]

[a,b]

≤

w(y) dλ(y)

= [a,b]

[y,b]

w(z)(Kq∗ 1)(z)1/(q−1) [(K1∗ [(Kf )q−1 ])(z)](q−2)/(q−1) dλ(z)

w(z)(K0 f )(z)(Kq∗ 1)(z)1/(q−1) [(K1∗ [(Kf )q−1 ])(z)](q−2)/(q−1) dλ(z)

≤

q−1

w(K0 f )

(Kq∗ 1) dλ

1/(q−1) J (q−2)/(q−1) .

[a,b]

Similarly,

h2 (y) ≤

[z,b]

k(z, y)w(z)(K1∗ 1)(z)1/(q−1) [(K1∗ [(Kf )q−1 ])(z)](q−2)/(q−1) dλ(z)

≤ [y,b]

and

v(x)k(x, z)(Kf )(x)q−2 dμ(x)

k(z, y)w(z) dλ(z) [y,b]

J2,2 :=

wh2 dλ [a,b]

≤

w(y) dλ(y) [a,b]

=

[a,b]

[y,b]

k(z, y)w(z)(K1∗ 1)(z)1/(q−1) [(K1∗ [(Kf )q−1 ])(z)](q−2)/(q−1) dλ(z)

w(z)(Kf )(z)(K1∗ 1)(z)1/(q−1) [(K1∗ [(Kf )q−1 ])(z)](q−2)/(q−1) dλ(z)

≤

q−1

w(Kf ) [a,b]

(K1∗ 1) dλ

1/(q−1) J (q−2)/(q−1) .

However, J2 J2,1 + J2,2 . Therefore, w(K0 f )q−1 (Kq∗ 1) dλ + J2 [a,b]

[a,b]

w(Kf )q−1 (K1∗ 1) dλ

1/(q−1) J (q−2)/(q−1) .

This relation, together with (1.3.4), implies (1.3.3) for q > 2. Suppose that q ∈ (1, 2). We have −1 k(x, t)w(t) dλ(t) ≥ k(x, t)w(t) dλ(t) ≥ D k(x, z) [y,x]

[y,z]

for x ≥ z ≥ y; therefore,

[y,b]

w dλ

dλ(z).

[y,z]

Thus, by Lemma 1.2, we have w(y) dλ(y) J2,1 [a,b]

q−2

w(z)(Kq∗ 1)(z)

h1 (y)

w dλ [y,z]

q−2

w(z)(Kq∗ 1)(z) [y,b]

w dλ

dλ(z)

[y,z]

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

∗ w(z)(Kq 1)(z)

= [a,b]

w(y) [a,z]

q−2

S17

dλ(y) dλ(z)

w dλ

[y,z]

w(K0 f )q−1 (Kq∗ 1) dλ.

[a,b]

In a similar way, using the fact that k(x, t)w(t) dλ(t) ≥ [y,x]

k(x, t)w(t) dλ(t) ≥ D

−1

k(z, t)w(t) dλ(t)

[y,z]

for x ≥ z ≥ y, we obtain

h2 (y) [y,b]

k(z, y)w(z)(K1∗ 1)(z)

[a,b]

= [a,b]

[a,b]

q−2 k(z, t)w(t) dλ(t) dλ(z)

[y,z]

and, by Lemma 1.2, J2,2 w(y) dλ(y)

[y,z]

[y,b]

k(z, y)w(z)(K1∗ 1)(z)

w(z)(K1∗ 1)(z) dλ(z)

q−2 k(z, t)w(t) dλ(t) dλ(z)

[y,z]

k(z, y)w(y) dλ(y)

[a,z]

q−2 k(z, t)w(t) dλ(t)

[y,z]

w(Kf )q−1 (K1∗ 1) dλ.

Thus,

J2 J2,1 + J2,2

q−1

w(K0 f )

(Kq∗ 1) dλ

+

[a,b]

[a,b]

w(Kf )q−1 (K1∗ 1) dλ,

which proves (1.3.3) for q ∈ (1, 2). Theorem 1.6. Let 1 < p ≤ q < +∞. Suppose that μ and λ are σ-ﬁnite measures on [a, b], ν = λ, u ∈ {Mλ }+ , v ∈ {Mμ }+ , w ≡ 1, and k is a nonnegative (Mμ,λ × Mμ,λ )-measurable function on [a, b]2 satisfying Oinarov’s condition (1.3.2). Then there exists a constant C ≥ 0 for which inequality (1.3.1) holds if and only if A < +∞, where A := max{A1 , A2 } for 1/q 1/p q p v(x)k(x, t) dμ(x) u dλ , A1 := sup A1 (t) := sup t∈[a,b]

t∈[a,b]

[t,b]

A2 := sup A2 (t) := sup t∈[a,b]

t∈[a,b]

[a,t]

1/q

1/p k(t, y) u(y) dλ(y) . p

v dμ [t,b]

p

[a,t]

Moreover, the least possible constant C in (1.3.1) satisﬁes the relation A ≈ C. Proof. Let I0 , I∞ , I, I 0 , I ∞ , and I be the intervals deﬁned in Lemma 1.7. Necessity. Fix any t ∈ [a, b]. If t ∈ I0 , then A2 (t) = 0 ≤ C. Suppose that t ∈ I∞ . By Lemma 1.3, there exists a function ft ∈ {Mλ }+ such that p ft dλ < +∞ and k(t, y)u(y)ft (y) dλ(y) = +∞. [a,t]

[a,t]

Substituting this function into inequality (1.3.1), we obtain 1/q −1 v dμ D k(t, y)u(y)ft (y) dλ(y) ≤ C which implies

[t,b] [t,b] v dμ

[a,t]

[a,t]

1/p ftp dλ

,

= 0 and A2 (t) = 0 ≤ C.

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S18

PROKHOROV et al.

Suppose that t ∈ I. Substituting f (y) = χ[a,t] (y)k(t, y)p −1 u(y)p −1 into (1.3.1), we obtain q 1/q 1/p p −1 p p p v(x) k(x, y)k(t, y) u(y) dλ(y) dμ(x) ≤C k(t, y) u(y) dλ(y) . [t,b]

[a,t]

[a,t]

This inequality, together with the properties of the kernel k, imply

1/q

1/p p p v dμ [k(t, y)u(y)] dλ(y) C [k(t, y)u(y)] dλ(y) . [t,b]

[a,t]

[a,t]

Thus, A2 (t) C. Similar considerations of the inequality 1/p p ∗ p u (K1 g) dλ ≤C [a,b]

1/q

q

for all g ∈ {Mμ }+ ,

g dμ

(1.3.5)

[a,b]

which is dual to (1.3.1), yield A1 (t) C.

Suﬃciency. Suppose that A < +∞. Then [t,b] v dμ = 0 for any t ∈ I∞ ; according to Lemma 1.4, this implies I∞ v dμ = 0. Therefore, (1.3.1) is equivalent to 1/q 1/p q p v(Kf ) dμ ≤C f dλ for all f ∈ {Mλ }+ . (1.3.6) I

[a,b]

If I = ∅, then inequality (1.3.6) does hold. Suppose that I = ∅. Since μ is a σ-ﬁnite measure, we can ∞ ﬁnd a sequence {Fn }∞ 1 ⊂ Mμ of sets such that [a, b] = 1 Fn and μ(Fn ) < +∞,

Fn ⊂ Fn+1 ,

Fn ∩ I = ∅

for each

n ∈ N.

∞ Take two number sequences {tn }∞ 1 ⊂ I and {sn }1 ⊂ I such that tn ↑ sup I and sn ↓ inf I. Setting ⎧ Fn ∩ I if sup I ∈ I ∧ inf I ∈ I, ⎪ ⎪ ⎪ ⎨ Fn ∩ I ∩ [a, tn ] if sup I ∈ I ∧ inf I ∈ I, Fn := ⎪ Fn ∩ I ∩ [sn , b] if sup I ∈ I ∧ inf I ∈ I, ⎪ ⎪ ⎩ Fn ∩ I ∩ [sn , tn ] if sup I ∈ I ∧ inf I ∈ I ∞ for each n ∈ N, we see that 1 Fn = I and, for any n ∈ N, Fn ⊂ Fn+1 and μ(Fn ) < +∞. Let dμn := χFn dμ. Fix any function f ∈ {Mλ }+ for which [a,b] f p dλ < +∞ and take any n ∈ N. If Fn = ∅, then

1/q

q

v(Kf ) dμn

1/p

≤ c(p, q, D)A

[a,b]

p

f dλ

(1.3.7)

.

[a,b]

Suppose that Fn = ∅. We set αn = inf Fn and βn = sup Fn . We have αn , βn ∈ I, and the application ¨ of Holder’s inequality with exponents p and p yields q q q v(Kf ) dμn ≤ D v(x) k(βn , y)u(y)f (y) dλ(y) dμ(x) [a,b]

Fn

[a,βn ]

≤D

q

v dμ [αn ,b]

q/p k(βn , y) u(y) dλ(y) p

[a,βn ]

[a,b]

p

f dλ

< +∞;

[a,b]

by virtue of Lemma 1.8, this implies v(Kf )q dμn ≈ uf (K0 f )q−1 (Kq∗ χFn ) dλ + [a,b]

q/p

p

[a,b]

uf (Kf )q−1 (K1∗ χFn ) dλ.

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

(1.3.8) 2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S19

¨ Applying Holder’s inequality to the ﬁrst term in (1.3.8), we obtain 1/p 1/p q−1 ∗ p uf (K0 f ) (Kq χFn ) dλ ≤ f dλ H1 , [a,b]

[a,b]

where

up (K0 f )p (q−1) (Kq∗ χFn )p dλ

H1 := [a,b]

p

p (q−1)

u(t)

=

uf dλ

[a,b]

(Kq∗ χFn )(t)p dλ(t).

[a,t]

By virtue of Theorem 1.4, we have 1/(p (q−1)) H1

where

s∈[a,b]

H1

p

f dλ

,

[a,b]

up (Kq∗ χFn )p dλ

H1 = sup

1/p

1/(p (q−1))

[s,b]

.

[a,s]

It follows from the assumptions of the theorem that p q p ∗ p p ∗ p u (Kq χFn ) dλ ≤ u (Kq 1) dλ ≤ A1 [s,b]

1/p

up dλ

[s,b]

u (t) p

[s,b]

−q

p

u dλ

dλ(t)

[a,t]

for any s ∈ [a, b]. We set U (t) := [a,t] up dλ, +∞ and estimate the last relation: −q u(t)p up dλ dλ(t) = u(t)p qx−q−1 dx dλ(t) [s,b]

[a,t]

[s,b]

u(t)p

=

[s,b]

U (t)

qx−q−1 χU (t) (x) dx dλ(t) U (s)

x−q−1 dx

=q

U (s)

u(t)p χU (t) (x) dλ(t) [s,b]

x−q−1 L(x) dx.

=: q U (s)

Given x ∈ U (s), let Ex := {t ∈ [s, b] | [a,t] up dλ ≤ x}. If Ex = ∅, then L(x) = 0 ≤ x. Suppose that Ex = ∅, cx := sup Ex , and {tm }∞ 1 ⊂ Ex is a sequence such that tm ↑ cx as m → ∞. If cx ∈ Ex , then up dλ ≤ up dλ ≤ x. L(x) = [s,cx ]

If cx ∈ Ex , then

[a,cx ]

p

u dλ ≤

L(x) = [s,cx )

p

u dλ = lim

m→∞ [a,t ] m

[a,cx )

up dλ ≤ x.

Thus,

p

p

u(t) [s,b]

u dλ [a,t]

−q

−q

dλ(t) ≤ q

x

dx = q

U (s)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

p

1−q

u dλ

,

[a,s]

Vol. 300 Suppl. 2

2018

(1.3.9)

S20

PROKHOROV et al.

which implies

p

u

(Kq∗ χFn )p

1/(p (q−1))

Aq1

dλ

[s,b]

−1/p

p

u dλ [a,s]

for any s ∈ [a, b]. Therefore, H1 Aq1 and q q−1 ∗ uf (K0 f ) (Kq χFn ) dλ ≤ A1 [a,b]

¨ Applying Holder’s inequality, we obtain q−1 ∗ uf (Kf ) (K1 χFn ) dλ ≤ [a,b]

q/p p

f dλ

1/p p

f dλ

[a,b]

where

H2 := [a,b]

(1.3.10)

.

[a,b]

1/p

H2

,

(1.3.11)

up (Kf )p (q−1) (K1∗ χFn )p dλ.

Given y ≤ x, let k(x, y) := supy≤s≤x supy≤t≤s k(s, t). Then the function k does not decrease in the ﬁrst variable and does not increase in the second, so that condition (1.3.2) implies k(x, y) ≤ k(x, y) ≤ D2 k(x, y). For t ∈ [a, b], we set p (q−1) k(t, y)u(y)f (y) dλ(y) ;

g(t) := [a,t]

note that the function g does not decrease on [a, b]. In particular, g is Borel, and the kernel χ[0,g(t)] (s), where t ∈ [a, b] and s ∈ [0, g(b)], is a (B × B)-measurable function on the Cartesian product [a, b] × [0, g(b)]. Next, we have up g(K1∗ χFn )p dλ H2 ≤ [a,b]

p

u(t)

= [a,b]

[0,g(b)]

u(t)p χ[0,g(t)] (s)

=

[0,g(b)]

χ[0,g(t)] (s) ds (K1∗ χFn )(t)p dλ(t)

[a,b]

p

v(x)k(x, t) dμn (x) dλ(t) ds.

[t,b]

Applying Minkowski’s inequality, recalling that the function g does not decrease, and using the assumptions of the theorem, we obtain

1/p

p p v(x) [k(x, t)u(t)] χ[0,g(t)] (s) dλ(t) dμn (x) ds H2 ≤ [0,g(b)]

[a,b]

≤ [0,g(b)]

[a,x]

v(x)χ[0,g(x)] (s)

[a,b]

[a,x]

≤

1/p

p k(x, t) u(t) dλ(t) dμn (x) ds p

v(x)χ[0,g(x)] (s)A2 [0,g(b)] p

=: A2

[a,b]

p

−1/q v dμn

p dμn (x) ds

[x,b] p

M (s) ds. [0,g(b)]

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S21

For s ≥ 0, we set Es := {x ∈ [a, b] | g(x) ≥ s}. If Es = ∅, then M (s) = 0. Suppose that Es = ∅, cs := inf Es , and {tm }∞ 1 ⊂ Es is a sequence such that tm ↓ cs as m → ∞. In the case cs ∈ Es , Lemma 1.2 implies −1/q v(x) v dμn dμn (x) M (s) = [cs ,b]

[x,b]

1/q

v dμn

1/q v(x)χ[0,g(x)] (s) dμn (x) .

=

[cs ,b]

[a,b]

Es , the same lemma implies In the case cs ∈ −1/q M (s) = v(x) v dμn dμn (x) = lim (cs ,b]

[x,b]

lim

m→∞

1/q

v dμn

m→∞ [t ,b] m

1/q

=

v dμn

dμn (x)

1/q v(x)χ[0,g(x)] (s) dμn (x) .

=

(cs ,b]

−1/q

[x,b]

v dμn

[tm ,b]

v(x)

[a,b]

In any case, by virtue of Minkowski’s inequality, we have p /q p M (s) ds v(x)χ[0,g(x)] (s) dμn (x) ds [0,g(b)]

[0,g(b)]

≤

[a,b]

v(x)

[a,b]

q /p χ[0,g(x)] (s) ds [0,g(b)] q /p

=

v(x)g(x)

p /q dμn (x)

p /q 2p (q−1) dμn (x) ≤D

[a,b]

H2 q−1

uf (Kf ) [a,b]

Ap2

(K1∗ χFn ) dλ

.

p /q q

v(Kf ) dμn

,

[a,b]

1/p

A2

q

f dλ

v(Kf ) dμn

[a,b]

[a,b]

1/q

p

.

[a,b]

This estimate and relation (1.3.10) give

q/p q q p v(Kf ) dμn A f dλ +A [a,b]

v(Kf ) dμn

[a,b]

Therefore,

whence

p /q q

1/p p

f dλ

[a,b]

1/q q

v(Kf ) dμn

,

[a,b]

i.e., (1.3.7) holds. By virtue of the monotone convergence theorem, relation (1.3.7) with n → ∞ implies (1.3.6) and the estimate C A of the constant. Theorem 1.7. Suppose that 1 < q < p < +∞, 1/r = 1/q − 1/p, μ and λ are σ-ﬁnite measures on [a, b], ν = λ, u ∈ {Mλ }+ , v ∈ {Mμ }+ , w ≡ 1; suppose also that k is a nonnegative (Mμ,λ × Mμ,λ )measurable function on [a, b]2 satisfying Oinarov’s condition (1.3.2). Then the existence of a constant C ≥ 0 for which inequality (1.3.1) holds is equivalent to B < +∞, where B := max{B1 , B2 } for r/q r/q 1/r q p p v(x)k(x, t) dμ(x) u dλ u(t) dλ(t) , B1 := [a,b]

[t,b]

B2 :=

[a,t]

r/p

r/p 1/r k(t, y) u(y) dλ(y) v(t) dμ(t) . p

v dμ [a,b]

[t,b]

p

[a,t]

Moreover, the least possible constant C in (1.3.1) satisﬁes the relation B ≈ C. PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S22

PROKHOROV et al.

Proof. Let I0 , I∞ , I, I 0 , I ∞ , and I be the intervals deﬁned in Lemma 1.7. Suﬃciency. Suppose that B < +∞. Note that, by virtue of Lemma 1.1, we have r/q r/q r q p v(x)k(x, t) dμ(x) u dλ u(t)p dλ(t) B1 ≥ [a,s]

[t,b]

[a,t]

r/q q v(x)k(x, s) dμ(x)

[s,b]

r/q

p

[a,s]

u(t)p dλ(t) ≈ A1 (s)r

u dλ [a,t]

for any s ∈ [a, b]; similarly, Lemma 1.2 implies B2 A2 (s). Therefore, A < +∞, and considerations similar to those in the proof of Theorem 1.6 imply that inequality (1.3.1) is equivalent to (1.3.6). Moreover, constructing the same sequence {Fn } as in Theorem 1.6, we obtain (1.3.8). We begin with estimating the ﬁrst term on the right-hand side of (1.3.8). Since A1 < +∞, it follows that, for each s ∈ [a, b], [a,s] up dλ = +∞ implies (Kq∗ χFn )(s) = 0; moreover, obviously, p [a,s] u dλ = 0 implies uf (K0 f )q−1 (Kq∗ χFn ) dλ = 0. [a,s]

p

Let E := {s ∈ [a, b] | [a,s] u dλ > 0}. Then uf (K0 f )q−1 (Kq∗ χFn ) dλ [a,b]

uf (K0 f )q−1 (Kq∗ χFn ) dλ

= [a,b]∩E

u(t)f (t)(K0 f )(t)

= [a,b]∩E

(1−q)+(q−1)

p

q−1

u dλ

(Kq∗ χFn )(t) dλ(t).

[a,t]

¨ Holder’s inequality with three exponents p, p/(q − 1), and r/q yields uf (K0 f )q−1 (Kq∗ χFn ) dλ [a,b]

1/p

≤

p

f dλ

(K0 f )(t)

[a,b]

p

p

(K0 f )(t) [a,b]

[a,t]

−p

u dλ

(q−1)/p u(t) dλ(t) Bq1 . p

1/p u(t) dλ(t)

p

−p

[a,t]

1/p

p

[a,t]

u dλ [s,b]

−p

p

u dλ

[a,b]

Theorem 1.4 implies

because

p

p

f dλ

,

[a,b]

1/p u(t) dλ(t) p

1/p

p

u dλ

≤ (p )1/p

[a,s]

for any s ∈ [a, b] (see the proof of (1.3.9)). Therefore, q q−1 ∗ uf (K0 f ) (Kq χFn ) dλ B1 [a,b]

q/p p

f dλ

.

(1.3.12)

[a,b]

¨ Applying Holder’s inequality to the second term on the right-hand side of (1.3.8), we obtain (1.3.11). Let k be the same kernel as in Theorem 1.6, and let g denote the function k(t, y)u(y)f (y) dλ(y), t ∈ [a, b]; g(t) := [a,t]

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S23

note that g is nondecreasing on [a, b]. Then up g(q−1)p (K1∗ χFn )p dλ H2 ≤ [a,b]

p

≈

u(t)

χ[0,g(t)] (s)s

[a,b]

(q−1)p −1

[0,g(b)]

=

s

(q−1)p −1

[0,g(b)]

p

u(t) [a,b]

ds (K1∗ χFn )(t)p dλ(t)

χ[0,g(t)] (s)(K1∗ χFn )(t)p

dλ(t) ds.

Applying Minkowski’s inequality and taking into account the monotonicity of g, we obtain p (q−1)p −1 1/p s v(x)χ[0,g(x)] (s)(Kp 1)(x) dμn (x) ds. H2 [0,g(b)]

[a,b]

Since A2 < +∞, it follows that, for any t ∈ [a, b], the relation [t,b] v dμn = +∞ implies (Kp 1)(t) = 0; moreover, [t,b] v dμn = 0 implies v(x)χ[0,g(x)] (s)(Kp 1)(x)1/p dμn (x) = 0. [t,b]

Let E := {t ∈ [a, b] | [t,b] v dμn > 0}. Then v(x)χ[0,g(x)] (s)(Kp 1)(x)1/p dμn (x) [a,b]

v(x)χ[0,g(x)] (s)(Kp 1)(x)1/p

=

1/p−1/p v dμn

[a,b]∩E

dμn (x).

[x,b]

¨ inequality with parameters γ and γ , we obtain Take a number γ ∈ (p , r). Applying Holder’s v(x)χ[0,g(x)] (s)(Kp 1)(x)1/p dμn (x) [a,b]

v(x)χ[0,g(x)] (s)

≤ Ω(s)

1/γ [a,b]

v dμn

1/γ dμn (x) ,

γ/p

−γ /p [x,b]

where

γ/p

v(x)χ[0,g(x)] (s)(Kp 1)(x)

Ω(s) := [a,b]

v dμn

However (see the proof of Theorem 1.6),

−γ /p v(x)χ[0,g(x)] (s) v dμn dμn (x) [a,b]

dμn (x).

[x,b]

[x,b]

1−γ /p v(x)χ[0,g(x)] (s) dμn (x) . [a,b]

Therefore, H2

s

(q−1)p −1

[0,g(b)]

Since

p /γ

Ω(s)

1−p /γ v(x)χ[0,g(x)] (s) dμn (x) ds.

[a,b]

1 1 1 1 + − 1 + p (q − 1) − , (q − 1)p − 1 = p (q − 1) p γ p γ

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S24

PROKHOROV et al.

¨ it follows by Holder’s inequality with parameters γ/p and γ/(γ − p ) that p /γ γ(q−1)(1/p+1/γ)−γ/p s Ω(s) ds H2 [0,g(b)]

×

s

ds

[0,g(b)]

[a,b]

γ/p

≈

1−p /γ v(x)χ[0,g(x)] (s) dμn (x)

q−1

v(x)(Kp 1)(x) [a,b]

γ/p v dμn

q(1−γ/r)

(Kf )(x)

p /γ dμn (x)

[x,b]

1−p /γ

×

q

v(Kf ) dμn

.

[a,b]

¨ Applying Holder’s inequality with parameters r/γ and r/(r − γ) to the ﬁrst factor, we obtain 1−p /r v(Kf )q dμn . H2 Bp2 [a,b]

This estimate, together with (1.3.11), (1.3.8), and (1.3.12), implies

q/p

1/p q q p p v(Kf ) dμn B f dλ +B f dλ [a,b]

[a,b]

Thus, we have

[a,b]

1/q

vχFn (Kf ) dμ [a,b]

v(Kf ) dμn

.

[a,b]

1/p

≤ c(p, q, D)B

q

1/q q

p

f dλ

.

[a,b]

According to the monotone convergence theorem, this inequality with n → ∞ implies (1.3.6) and the estimate C B of the constant. Necessity . Suppose that (1.3.1) holds. Arguing as in the proof of Theorem1.6, we see that A < +∞. Hence, [a,t] up dλ = 0 for any t ∈ I ∞ ; according to Lemma 1.4, this implies I ∞ up dλ = 0. Therefore, r/q 1/r ∗ r/q p p (Kq 1)(t) u dλ u(t) dλ(t) . B1 =

I

[a,t]

p

dλ < +∞ for any t ∈ I. If I = ∅, then B1 = 0 ≤ C. Suppose that I = ∅. n n }∞ such that [a, b] = ∞ G Since λ is a σ-ﬁnite measure, it follows that there exists a sequence {G 1 1 n ) < +∞, G n ⊂ G n+1 , and G n ∩ I = ∅. Take sequences {tn }∞ ⊂ I and and, for each n ∈ N, λ(G

Moreover, we have

[a,t] u

{sn }∞ 1 ⊂ I such that tn ↑ sup I and sn ↓ inf I. For each n ∈ N, we set ⎧ n ∩ I G if sup I ∈ I ∧ inf I ∈ I, ⎪ ⎪ ⎪ ⎨ Gn ∩ I ∩ [a, tn ] if sup I ∈ I ∧ inf I ∈ I, Gn := n ∩ I ∩ [sn , b] if sup I ∈ I ∧ inf I ∈ I, ⎪ G ⎪ ⎪ ⎩ Gn ∩ I ∩ [sn , tn ] if sup I ∈ I ∧ inf I ∈ I. ∞ We have 1 Gn = I and, for each n ∈ N, Gn ⊂ Gn+1 and λ(Gn ) < +∞. Let r/q 1/r ∗ r/q p p (Kq 1)(t) u χGn dλ u(t) χGn (t) dλ(t) . B1 (n) := I

1

[a,t]

If Gn = ∅, then B1 (n) = 0. Suppose that Gn = ∅. Setting αn := inf Gn and βn := sup Gn , we obtain r/q 1/r ∗ r/q p p (Kq 1)(t) u dλ u(t) χGn (t) dλ(t) B1 (n) ≤ [αn ,βn ]

[a,t]

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

≤ (Kq∗ 1)(αn )1/q

(Kq∗ 1)(αn )1/q

r/q

up dλ

[a,βn ]

[a,t] p

S25

1/r u(t)p dλ(t)

1/p

u dλ

< +∞.

[a,βn ]

Next, the function (Kq∗ 1)(t)r/(pq)

fn (t) :=

r/(pq )

p

u χGn dλ

u(t)p −1 χGn (t)

[a,t] p [a,b] fn dλ

= B1 Moreover,

q−1 ∗ q−1 v(x) k(x, z)u(z)fn (z) dλ(z) k(x, y) dμ(x) (K1 [(Kfn ) ])(y) ≥

satisﬁes the condition

(n)r .

[y,b]

[a,y]

∗ (Kq 1)(y)

q−1 ufn dλ

,

[a,y]

and Lemma 1.1 implies ufn dλ = [a,y]

(Kq∗ 1)(t)r/(pq) [a,y]

(Kq∗ 1)(y)r/(pq)

[a,t]

r/(pq )

p

[a,t]

u(t)p χGn (t) dλ(t)

u χGn dλ

u(t)p χGn (t) dλ(t)

u χGn dλ

[a,y]

(Kq∗ 1)(y)r/(pq)

r/(pq )

p

r/(p q)

p

u χGn dλ

.

[a,y]

Therefore, q v(Kfn ) dμ = [a,b]

[a,b]

ufn (K1∗ [(Kfn )q−1 ]) dλ

u(y)fn (y)(Kq∗ 1)(y)

[a,b]

u(y)fn (y)(Kq∗ 1)(y)1+r/(pq )

[a,b]

r/(p q )

p

u χGn dλ

q−1 ufn dλ

dλ(y)

[a,y]

dλ(y) = B1 (n)r .

[a,y]

Substituting fn into (1.3.1), we obtain B1 (n) ≤ c(p, q, D) C. By the monotone convergence theorem, this inequality with n → ∞ implies B1 C. Similar considerations of inequality (1.3.5), which is dual to (1.3.1), give B2 C. The general case is characterized by the following theorem. Theorem 1.8. Suppose that 1 < p, q < +∞ and 1/r = 1/q − 1/p. Suppose also that μ, λ, and ν are σ-ﬁnite measures on [a, b], λ and ν are deﬁned on the same σ-algebra Mλ , u, w ∈ {Mλ }+ , and v ∈ {Mμ }+ . Let (νa , νs ) be the Lebesgue decomposition of the measure ν with respect to λ, and let dνa /dλ be the Radon–Nikodym derivative of νa with respect to λ. Suppose that k is a nonnegative (Mμ,λ × Mμ,λ )-measurable function on [a, b]2 satisfying Oinarov’s condition (1.3.2). Let ω := up (w dνa /dλ)1−p . If p ≤ q, then the existence of a constant C ≥ 0 for which inequality (1.3.1) holds is equivalent to A := max{A1 , A2 } < +∞, where 1/q 1/p v(x)k(x, t)q dμ(x) ω dλ , A1 := sup t∈[a,b]

[t,b]

A2 := sup t∈[a,b]

[a,t]

1/q

1/p k(t, y) ω(y) dλ(y) . p

v dμ [t,b]

[a,t]

Moreover, the least possible constant C in (1.3.1) satisﬁes the relation A ≈ C. PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S26

PROKHOROV et al.

If q < p, then the existence of a constant C ≥ 0 for which (1.3.1) holds is equivalent to B := max{B1 , B2 } < +∞, where r/q r/q 1/r q v(x)k(x, t) dμ(x) ω dλ ω(t) dλ(t) , B1 :=

[a,b]

[t,b]

[a,t]

r/p

B2 :=

r/p 1/r k(t, y) ω(y) dλ(y) v(t) dμ(t) . p

v dμ [a,b]

[t,b]

[a,t]

Moreover, the least possible constant C in (1.3.1) satisﬁes the relation B ≈ C. Proof. According to Lemma 1.6, inequality (1.3.1) is equivalent to q 1/q v(x) k(x, y)u(y)f (y) dλ(y) dμ(x) [a,b]

≤C

[a,x]

dνa f w dλ [a,b] p

1/p for all f ∈ {Mλ }+ .

dλ

In turn, by Lemma 1.5, this inequality is equivalent to −1/p q 1/q dνa (y) v(x) k(x, y)u(y)f (y) w(y) dλ(y) dμ(x) dλ [a,b] [a,x] 1/p p ≤C f dλ for all f ∈ {Mλ }+ . [a,b]

The application of Theorems 1.6 and 1.7 yields the required result. This result makes it possible to characterize the following inequality in the case 1 < p, q < +∞: q 1/q 1/p p v(x) f u dλ dμ(x) ≤C f w dν for all f ∈ {Mλ }+ . (1.3.13) [a,b]

[a,x)

[a,b]

Corollary 1.1. Suppose that 1 < p, q < +∞ and 1/r = 1/q − 1/p. Suppose also that μ, λ, and ν are σ-ﬁnite measures on [a, b], λ and ν are deﬁned on the same σ-algebra Mλ , u, w ∈ {Mλ }+ , and v ∈ {Mμ }+ . Let (νa , νs ) be the Lebesgue decomposition of the measure ν with respect to λ, and let dνa /dλ be the Radon–Nikodym derivative of νa with respect to λ. Suppose that k is a nonnegative (Mμ,λ × Mμ,λ )-measurable function on [a, b]2 satisfying Oinarov’s condition (1.3.2). Let ω := up (w dνa /dλ)1−p . If p ≤ q, then there exists a constant C ≥ 0 for which inequality (1.3.13) holds if and only if A < +∞, where 1/q 1/p v dμ ω dλ . A := sup t∈[a,b]

[t,b]

[a,t)

Moreover, the least possible constant C in (1.3.13) satisﬁes the relation A ≈ C. If q < p, then there exists a constant C ≥ 0 for which inequality (1.3.13) holds if and only if B < +∞, where r/p r/p 1/r v dμ ω dλ v(t) dμ(t) . B := [a,b]

[t,b]

[a,t)

Moreover, the least possible constant C in (1.3.13) satisﬁes the relation B ≈ C. PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

Proof. The function

S27

0, x = y, k(x, y) = 1, x = y

satisﬁes condition (1.3.2) and belongs to the class {B × B}+ . Therefore, by Theorem 1.8, if p ≤ q, then inequality (1.3.13) is equivalent to max{A1 , A2 } < +∞, where 1/q 1/p v dμ ω dλ , A1 := sup t∈[a,b]

A2

(t,b]

1/q

:= sup

[a,t]

v dμ

ω dλ

[t,b]

t∈[a,b]

1/p .

[a,t)

If q < p, then (1.3.13) is equivalent to the ﬁniteness of the constant max{B1 , B2 }, where r/q r/q 1/r B1 := v dμ ω dλ ω(t) dλ(t) , B2

[a,b]

(t,b]

:=

r/p

[a,t]

v dμ [a,b]

r/p ω dλ

[t,b]

1/r v(t) dμ(t) .

[a,t)

Let us show the equivalence of A1 and A2 and of B1 and B2 . Suppose that, e.g., the constant A1 is ﬁnite. For t = a, we have [a,t) ω dλ = 0. Fix any t ∈ (a, b]. Let {tn }∞ 1 be an increasing sequence of points in the interval [a, t) converging to t as n → ∞. Then ω dλ = lim ω dλ. [a,t)

n→∞ [a,t ] n

If (tn ,b] v dμ = +∞ for all n ∈ N, then [a,tn ] ω dλ = 0 for all n ∈ N and, therefore, [a,t) ω dλ = 0. Otherwise, we have 1/q 1/p 1/q 1/p v dμ ω dλ = lim v dμ ω dλ .

[t,b]

A2

n→∞

[a,t)

A1 .

A1

≤ The proof of the relation ≤ Thus, According to Lemma 1.1, we have r/p ω dλ ≈ [a,t)

A2

(tn ,b]

[a,tn ]

is similar. r/q

ω dλ

[a,t)

ω(s) dλ(s)

[a,s]

for any t ∈ [a, b]. Interchanging the integrals and applying Lemma 1.2, we obtain

r/p

r/q v dμ ω dλ ω(s) dλ(s) v(t) dμ(t) [B2 ]r ≈

[a,b]

[t,b]

[a,t)

r/q

[a,s]

ω dλ

= [a,b]

[a,s]

r/p

v dμ (s,b]

[t,b]

v(t) dμ(t) ω(s) dλ(s) ≈ [B1 ]r .

To complete the proof of the corollary, it remains to recall that A = A2 and B ≈ B2 . Example 1.1. Suppose that a = 0, each Borel set E ⊂ [0, 2], we put ⎧ ⎪ ⎨2, μ(E) = 0, ⎪ ⎩ 1

b = 2, u ≡ 1, v(1) = +∞, and v(x) = 1 for x ∈ [0, 2] \ {1}. For {1, 2} ⊂ E, {1, 2} ∩ E = ∅, otherwise,

1, λ(E) = 0,

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

1 ∈ E, 1 ∈ E.

Vol. 300 Suppl. 2

2018

S28

PROKHOROV et al.

According to Corollary 1.1, we have

2

1/2 v(x) f u dλ dμ(x) ≤C [0,2]

[0,x)

1/2 for all f ∈ {B}+ ,

f 2 dλ [0,2]

A

equals 1. However, the inequality because the corresponding constant

2

1/2

1/2 2 v(x) f u dλ dμ(x) ≤C f dλ for all [0,2]

[0,x]

f ∈ {B}+

[0,2]

does not hold, because for the constant A in Theorem 1.1, which characterizes this inequality, we have 1/2 1/2 2 v dμ u dλ = +∞. A≥ [1,2]

[0,1]

1.4. The Case 0 < p < 1 For measures absolutely continuous with respect to the Lebesgue measure, the case 0 < p < 1 is not of great interest, because, in this case, inequality (1.3.1) holds only if the left-hand side of (1.3.1) vanishes. Hardy’s inequality for sequences was characterized by Bennett [13], Grosse-Erdmann [14], and Gol’dman [15]. In particular, in [14, Theorem 9.2], the following result was obtained. Theorem 1.9. Suppose that 0 < q < p < 1, 1/r = 1/q − 1/p, an ≥ 0, wn ≥ 0, and un ≥ 0. Then q 1/q 1/p ∞ n ∞ p wn um am ≤C an for any {an }∞ (1.4.1) 1 n=1

m=1

n=1

if and only if F < +∞, where 1/r ∞ r r r Un (Wn − Wn+1 ) , F :=

Un := sup um ,

Moreover, F ≈ C. Proof. If 0 < q < ∞ and

Wn :=

1≤m≤n

n=1

∞

1/q wm

.

m=n

∞

n=1 wn = 1, then ∞ n

wn

n=1

q

uk ak

q ∞ 1 ≈ uk ak , 2ν ν=0

k=1

where

(1.4.2)

k∈Iν

1/q ∞ 1 mν := min n : wk ≤ ν . 2

Iν := [mν , mν+1 − 1],

k=n

We have Jν :=

wn

n

n∈Iν

q uk ak

k=1

≤

n∈Iν

wn

mν+1 −1

q uk ak

k=1

1 ≤ qν 2

ν

q uk ak

.

μ=0 k∈Iμ

¨ If q > 1, then, by virtue of Holder’s inequality, q q ν ν 1 1 2ν(q−1) 2μ 2ν · uk ak qν uk ak . Jν ≤ qν μ(q−1) 2 2ν 2μ 2 2 μ=0 μ=0 k∈I k∈I μ

This implies ∞ n=1

wn

n k=1

q uk ak

μ

q q ∞ ∞ ν 1 2ν(q−1) 1 = Jν uk ak uk ak . 2qν μ=0 2μ(q−1) 2μ ν=0 ν=0 μ=0 k∈I k∈I ∞

μ

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

μ

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

For 0 < q ≤ 1, we have

ν

q

ν

≤

uk ak

μ=0 k∈Iμ

therefore, ∞

wn

n

n=1

q uk ak

k=1

μ=0

S29

q uk ak

;

k∈Iμ

q q ∞ ∞ ν 1 1 ≤ uk ak ≈ uk ak . 2qν μ=0 2μ ν=0 μ=0 k∈Iμ

k∈Iμ

Conversely, it follows from the deﬁnition of mν that 1 1 < wn ≤ νq . 2 2(ν+1)q n≥m ν

Hence,

1 1 1 ≈ (ν+2)q − (ν+3)q < wn − wn = νq 2 2 2 n≥m n≥m ν+1

Therefore, q 1 uk ak 2ν k∈Iν

wn

ν=0

n∈[mν+1 ,mν+3 −1]

Applying (1.4.3), we obtain q ∞ ∞ 1 uk ak 2ν

ν+3

q uk ak

wn

ν=0 n∈[mν+1 ,mν+3 −1]

k∈Iν

wn

n

n∈[mν+1 ,mν+3 −1]

n

q uk ak

∞

wn

n=1

k=1

(1.4.3)

wn .

n∈[mν+1 ,mν+3 −1]

≤

k∈Iν

q uk ak

.

k=1

n

q uk ak

,

k=1

which proves relation (1.4.2). Now, let us prove that the condition F < ∞ is suﬃcient for the validity of (1.4.1). Using Jensen’s and ¨ Holder’s inequalities and applying (1.4.2), we obtain q q q/p ∞ n ∞ ∞ 1 1 q wn uk ak ≈ uk ak ≤ sup u apk 2ν 2qν k∈Iν k n=1 ν=0 ν=0 k=1 k∈Iν k∈Iν q/r q/p q/p ∞ ∞ ∞ 1 p p r q/r ≤ sup u ak =: J ak . 2rν k∈Iν k ν=0

ν=0 k∈Iν

k=0

It follows from (1.4.3) that mν+3 mν+3 −1 ∞ mν+3 ∞ −1 r/q −1 r/p r r wn sup uk ≈ sup uk wn wk J≤ ν=0

≤

k∈Iν

mν+1

∞ mν+3 −1

wn

∞

ν=0 mν+1

ν=0 k∈Iν

mν+1

k=n

r/p wk

sup urk ≈ F r .

1≤k≤n

k=n

To prove the lower bound, note that, by virtue of (1.4.2), inequality (1.4.1) is equivalent to q 1/q 1/p ∞ ∞ 1 p uk ak ≤C an . 2ν ν=0

(1.4.4)

n=1

k∈Iν

Let kν ∈ Iν be such that ukν = supk∈Iν uk . Applying (1.4.4) to the sequence ak deﬁned by ak = 0 for k = kν and akν = βν ≥ 0, where βν is any sequence, we obtain 1/p ∞ ∞ 1 q q 1/q p u β ≤C βν . 2qν kν ν ν=0

ν=1

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S30

PROKHOROV et al.

We have r

F =

∞

(Wnr

−

r Wn+1 )

ν=0 n∈Iν

Gν

1 2rν

sup 1≤k
urk =

sup 1≤k≤n

urk

=:

∞

Gν ,

ν=0

ν 1 1 1 r r r sup sup u = sup u ≤ u . 2rν 0≤μ≤ν k∈Iμ k 2rν 0≤μ≤ν kμ 2rν μ=0 kμ

Therefore, Fr

∞ ∞ ν 1 r 1 r u ≈ u ≤ C r. kμ rν rμ kμ 2 2 ν=0 μ=0 μ=0

Deﬁnition 1.1. A measure λ on [a, b] is said to be (a) continuous if λ({x}) = 0 for any x ∈ [a, b]; (b) discrete if there exists an at most countable set Eλ such that λ([a, b] \ Eλ ) = 0 and λ({x}) > 0 for each x ∈ Eλ . Lemma 1.9. Let λ be a σ-ﬁnite measure on [a, b]. Then (a) the set G := {x ∈ [a, b] | λ({x}) > 0} is at most countable; (b) there exist measures λc and λd deﬁned on the same σ-algebra Mλ and such that λc is continuous, λd is discrete, and λ = λc + λd . Proof. (a) The σ-ﬁniteness of the measure λ implies the existence of a sequence {En } ⊂ Mλ such that λ(En ) < +∞ and [a, b] = ∪n En . We deﬁne a measure λn by setting λn (E) := λ(E ∩ En ) for each E ∈ Mλ and let Fn denote the distribution function of the measure λn : Fn (x) := λn ([a, x)), x ∈ [a, b]. Since Fn is a bounded monotone function, it follows that the set of its discontinuity points is at most countable. Fix any point x0 ∈ G. There exists an n0 ∈ N for which x0 ∈ En0 . Moreover, λn0 ({x0 }) > 0. Hence, x0 is a point of discontinuity of the function Fn0 . Thus, the set G is contained in the union of the sets of discontinuity points of the functions Fn , which is at most countable. (b) For each E ∈ Mλ , we put λd (E) := λ(G ∩ E), where G is the same set as in (a), and deﬁne λc by λc := λ − λd . The support of the measure λd is the at most countable set G, and λd ({x}) > 0 for any x ∈ G, which means that the measure λd is discrete. Moreover, the deﬁnition of G implies the continuity of the measure λc . Theorem 1.10. Suppose that 0 < p < 1, 0 < q < +∞, λ and μ are σ-ﬁnite measures on [a, b], and λ continuous; suppose also that k is a nonnegative (Mμ × Mλ )-measurable function on [a, b]2 . Let k(x, y) dλ(y) = 0 . E := x ∈ [a, b] [a,b]

Then [a,b]

q 1/q k(x, y)f (y) dλ(y) dμ(x) ≤C

[a,b]

1/p p

f dλ

for all

f ∈ {Mλ }+ (1.4.5)

[a,b]

if and only if μ(E) = 0. PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S31

Proof. Suﬃciency is obvious. Necessity. If λ([a, b]) = 0, then E = ∅ and μ(E) = 0. Suppose that λ([a, b]) the > 0. First, consider ) < +∞ case where the measure μ is ﬁnite. Since λ is σ-ﬁnite, it follows that [a, b] = ∞ E and λ(E j j 1 for any j ∈ N. We set j Ei ∩ [−j, j], j ∈ N, Ej := Es+j , j ∈ N. s := min{i ∈ N | λ(Ei ) > 0} − 1, Ej := i=1

We have [a, b] = ∞ 1 Ej ; moreover, for each j ∈ N, 0 < λ(Ej ) < +∞ and Ej is bounded. Fix any number n ∈ N. Let Ij be an interval containing Ej . There exists an mj ∈ N such that, mj partitioning the interval Ij into 2mj equal intervals Ij1 , . . . , Ij2 , we obtain λ(Ej ∩ Ijs ) < np/(p−1) for each s ∈ {1, . . . , 2mj }. Otherwise, we could construct a nested sequence {Jk } of intervals such that their lengths tend to 0 and λ(Ej ∩ Jk ) ≥ np/(p−1) , k ∈ N. The intersection ∞ k=1 (Ej ∩ Jk ) is either ∅ or a one-point set. But since λ(Ej ) < +∞, it follows that ∞ (Ej ∩ Jk ) = lim λ(Ej ∩ Jk ) ≥ np/(p−1) > 0, λ k→∞

k=1

which contradicts the continuity of the measure λ. We set N 0 (n) :=

Ej ∩ Ijs

j∈N, s∈{1,...,2mj } : λ(Ej ∩Ijs )=0

and enumerate all Ej ∩ Ijs such that λ(Ej ∩ Ijs ) = 0 by a single index as Fi , i ∈ N. Thus, [a, b] \ N 0 (n) = ∞ p/(p−1) , i ∈ N. 1 Fi , and 0 < λ(Fi ) < n Fix any i ∈ N and any Mpλ -measurable F ⊂ Fi with λ(F ) > 0. Let γ := min{q, 1}, and let f := −1/p . We have [a,b] f dλ = 1. For q ≤ 1, by virtue of Minkowski’s inequality, the left-hand side χF λ(F ) of (1.4.5) is estimated from below as q 1/q q 1/q k(x, y)f (y) dλ(y) dμ(x) = λ(F )−1/p k(x, y) dλ(y) dμ(x) [a,b]

[a,b]

[a,b]

−1/p

F

1/q k(x, y) dμ(x) dλ(y).

≥ λ(F )

q

F

[a,b]

¨ For q > 1, applying Holder’s inequality with exponents 1/q and 1/(1 − q), we obtain q 1/q k(x, y)f (y) dλ(y) dμ(x) [a,b]

[a,b]

= λ(F )−1/p

[a,b]

q 1/q k(x, y) dλ(y) dμ(x)

F

≥ λ(F )−1/p μ([a, b])1/q−1

[a,b]

−1/p

= λ(F )

k(x, y) dλ(y) dμ(x)

F

1/q−1

μ([a, b])

F

k(x, y) dμ(x) dλ(y).

[a,b]

In both cases, we have q 1/q 1/γ nC1 γ k(x, y)f (y) dλ(y) dμ(x) ≥ k(x, y) dμ(x) dλ(y), λ(F ) F [a,b] [a,b] [a,b] where C1 := min{1, μ([a, b])1/q−1 }. PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S32

PROKHOROV et al.

Therefore, substituting f into (1.4.5), we obtain 1/γ C2 1 γ , k(x, y) dμ(x) dλ(y) ≤ λ(F ) F [a,b] n where C2 = C · C1−1 . Since F is any Mλ -measurable subset of Fi with λ(F ) > 0, it follows (by Theorem 1.40 in [9]) that there exists a set Ni ⊂ Fi such that λ(Ni ) = 0 and, for any y ∈ Fi \ Ni , 1/γ C2 γ k(x, y) dμ(x) ≤ . (1.4.6) n [a,b] n 0 n Let N n := ∞ i=1 Ni . Then λ(N ) = 0 and (1.4.6) holds for any y ∈ [a, b] \ (N (n) ∪ N ). Let N := ∞ 0 n n=1 (N (n) ∪ N ). Then λ(N ) = 0 and (1.4.6) holds for any y ∈ [a, b] \ N and any n ∈ N. Therefore, for any y ∈ [a, b] \ N , we have k(x, y) dμ(x) = 0, [a,b]

whence

k(x, y) dλ(y) dμ(x) =

[a,b]

[a,b]

k(x, y) dμ(x) dλ(y) = 0,

[a,b]

[a,b]

i.e., μ(E) = 0. Now, consider the general case. Since the measure μ is σ-ﬁnite on [a, b], it follows that there exists a sequence {Gn } of Mμ -measurable sets such that n∈N Gn = [a, b] and, for each n ∈ N, Gn ⊂ Gn+1 and μ(Gn ) < +∞. Given any Mμ -measurable set G, we put μn (G) := μ(G ∩ Gn ) for n ∈ N. Inequality (1.4.5) implies a similar inequality with μn instead of μ. It follows from the above considerations that μ(E ∩ Gn ) = μn (E) = 0 for any n ∈ N. Therefore, μ(E) = 0. Theorem 1.11. Suppose that 0 < p ≤ 1 and p ≤ q < +∞. Let λ and μ be σ-ﬁnite measures on [a, b] such that λ is a discrete measure with countable support {cn }∞ 1 ⊂ [a, b]. Suppose also that u ∈ {Mλ }+ , v ∈ {Mμ }+ , and k is a nonnegative (Mμ × Mλ )-measurable function on [a, b]2 for which there exists a constant α > 0 such that k(x, y1 ) ≥ αk(x, y2 ) for y1 < y2 . Then the inequality q 1/q v(x) k(x, y)f (y)u(y) dλ(y) dμ(x) [a,b]

[a,x]

≤C

1/p p

f dλ

for all

f ∈ {Mλ }+

(1.4.7)

[a,b]

holds if and only if F < +∞, where

1/q v(x)k(x, t) dμ(x) H(t),

F := sup

q

t∈[a,b]

H(x) :=

[t,b]

sup n: cn ∈[a,x]

u(cn )λ({cn })1/p ,

x ∈ [a, b].

Moreover, the least constant C in (1.4.7) satisﬁes the condition α F ≤ C ≤ F. Proof. Suﬃciency. Take any function f ∈ {Mλ }+ . Jensen’s inequality implies k(x, y)f (y)u(y) dλ(y) = k(x, cn )f (cn )u(cn )λ({cn }) [a,x]

n : cn ∈[a,x]

≤

k(x, cn )f (cn )λ({cn })1/p H(cn )

n : cn ∈[a,x]

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

1/p

≤

S33

k(x, cn )p f (cn )p λ({cn })H(cn )p

n : cn ∈[a,x]

1/p k(x, y)p f (y)p H(y)p dλ(y) .

= [a,x]

Applying Minkowski’s inequality, we obtain

q 1/q v(x) k(x, y)f (y)u(y) dλ(y) dμ(x) [a,b]

[a,x]

v(x)

≤

[a,b]

≤

q/p 1/q k(x, y) f (y) H(y) dλ(y) dμ(x) p

[a,x]

f (y)p H(y)p [a,b]

p

p

p/q 1/p v(x)k(x, y)q dμ(x) dλ(y) ≤F

[y,b]

1/p f p dλ

.

[a,b]

Necessity. Suppose that (1.4.7) holds. Fix any t ∈ [a, b] and any positive integer n for which cn ∈ [a, t]. Setting f := λ({cn })−1/p χ{cn } in (1.4.7), we obtain 1/q 1/p q v(x)k(x, t) dμ(x) ≤ C, α u(cn )λ({cn }) [t,b]

which implies the required estimate. Theorem 1.12. Suppose that 0 < q < p ≤ 1 and 1/r = 1/q − 1/p. Let λ and μ be σ-ﬁnite measures on [a, b] such that λ is discrete and has countable support {cn }∞ 1 ⊂ [a, b]. Suppose also that + + u ∈ {Mλ } , v ∈ {Mμ } , and k ≡ 1. Then inequality (1.4.7) holds if and only if G < +∞, where r/p 1/r r v(x) v dμ H(x) dμ(x) . G := [a,b]

[x,b]

Moreover, the least constant C in (1.4.7) satisﬁes the relation C ≈ G. Proof. Suﬃciency. Take any function f ∈ {Mλ }+ for which the right-hand side of (1.4.7) is ﬁnite. For each t ∈ [a, b], we set

r/p r/q r−1 s χ[0,H(t)] (s) v(x)χ[0,H(x)] (s) dμ(x) ds. h(t) := [0,H(b)]

[a,b]

Note that the function H does not decrease on [a, b]. In particular, H is Borel, and the kernel χ[0,H(t)] (s), where t ∈ [a, b] and s ∈ [0, H(b)], is a (B × B)-measurable function on the Cartesian product [a, b] × [0, H(b)]. First, for each t ∈ [a, b], we have

1/p

1/p −p/q vh dμ ≤ H(t) v dμ h(t)−1/q H(t) [t,b]

≤ H(t) [t,b]

[t,b]

1/p

−1/p

v dμ

v dμ [t,b]

s

r−1

−1/r χ[0,H(t)] (s) ds

≈ 1,

[0,H(b)]

because χ[t,b] (x) = 1 and s ∈ [0, H(t)] imply s ∈ [0, H(x)] and, therefore, χ[0,H(x)] (s) = 1. By virtue of Theorem 1.11, this relation implies the estimate p 1/q 1/p −p/q p v(x)h(x) f u dλ dμ(x) f dλ . [a,b]

[a,x]

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

[a,b]

Vol. 300 Suppl. 2

2018

S34

PROKHOROV et al.

In turn, this estimate implies, in particular,

v(x)

{x∈[a,b]|h(x)=0}

Secondly, we have r/q v(x)h(x) dμ(x) = [a,b]

≈

s

=

[a,b]

v(x) [a,b]

≤

[a,b]

s

[a,b]

r/p v(x)χ[0,H(t)] (s) dμ(t) sr−1 ds dμ(x) [x,b]

r/p

v dμ

s

[x,b]

r/p v(x)χ[0,H(t)] (s) dμ(t) dμ(x) ds

[x,b]

χ[0,H(x)] (s)

[0,H(b)]

v(x)

dμ(x) = 0.

r/q v(x)χ[0,H(x)] (s) dμ(x) ds

r−1

v(x)χ[0,H(x)] (s)

r−1

[0,H(b)]

f u dλ

[a,x]

[0,H(b)]

q

r−1

χ[0,H(x)] (s) ds dμ(x) ≈ G r .

[0,H(b)]

This estimate implies, in particular,

v dμ = 0. {x∈[a,b]|h(x)=+∞}

¨ Applying Holder’s inequality with exponents r/q and p/q and combining the estimates obtained above, we arrive at q 1/q

q 1/q −1 v(x) f u dλ dμ(x) = v(x)h(x)h(x) f u dλ dμ(x) [a,b]

[a,x]

≤

r/q

vh

−p/q

dμ

[a,b]

p

v(x)h(x) [a,b]

1/p

G

[a,b]

1/r

p

f dλ

f u dλ

[a,x]

1/p dμ(x)

[a,x]

.

[a,b]

Necessity. The σ-ﬁniteness of the measure μ on [a, b] implies the existence of a sequence {Gn } of Mμ -measurable sets such that Gn ⊂ Gn+1 and μ(Gn ) < +∞ for n ∈ N and n∈N Gn = [a, b]. For each n ∈ N, we set vn := min{v, n}χGn and un := min{u, n}, and for any k ∈ N and any Mλ -measurable set E, we set λk (E) := λ(E ∩ {c1 , . . . , ck }). Inequality (1.4.7) implies 1/q q vn (x) gun dλk dμ(x) ≤C [a,b]

[a,x]

1/p for all g ∈ {Mλ }+ .

p

g dλk

(1.4.8)

[a,b]

Let {ci }ki=1 be the increasing rearrangement of the sequence {ci }ki=1 ; we set Δ0 := [a, c1 ),

Δj := [cj , cj+1 ), 1 ≤ j ≤ k − 1. Take any sequence {dj }kj=1 of nonnegative real numbers and let g := kj=1 dj λ({cj })−1/p χΔj . We have

q

q k vn (x) gun dλk dμ(x) = vn (x) gun dλk dμ(x) [a,b]

Δk := [ck , b],

[a,x]

=

k i=1

vn dμ Δi

i j=1

i=0

Δi

[a,x]

q q k i g(cj )un (cj )λ({cj }) = wi dj ϕj , i=1

j=1

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

where

wi :=

S35

ϕj := un (cj )λ({cj })1/p .

vn dμ, Δi

Moreover, gp dλk =

k

[a,b]

Thus, inequality (1.4.8) implies k

wi

Δi

i=0

i

i=1

gp dλk =

k

dpi .

i=1

q 1/q ≤C

dj ϕj

k

j=1

1/p dpi

i=1

for all sets {dj }kj=1 of nonnegative numbers. Applying Theorem 1.9, we obtain r/q r/q

k k k r wj − wj sup ϕm Cr 1≤m≤i

i=1

=

k

sup 1≤m≤i

i=1

j=i+1

r/q

sup ϕm

≥

[ci+1 ,b]

un (cm )λ({cm })1/p

i=1

=

v(x)

r/p

vn dμ

vn dμ

Δi

v(x) Δi

r

1≤m≤i

k

r/q

vn dμ

[ci ,b]

The monotone convergence theorem implies k r sup u(cm )λ({cm })1/p Cr i=1

−

vn dμ

1≤m≤i

i=1

j=i

r

k

.

[ci ,b]

r/p

v dμ Δi

v dμ [ci ,b]

r/p Hk (x)r dμ(x)

v dμ [x,b]

[a,b]

[x,b]

Hk (x) :=

sup

r/p

v dμ

Hk (x)r dμ(x),

where m : 1≤m≤k, cm ∈[a,x]

u(cm )λ({cm })1/p ,

x ∈ [a, b].

For each ﬁxed x ∈ [a, b], the sequence {Hk (x)}k increases and converges to H(x) as k → ∞; therefore, the monotone convergence theorem implies C G. Theorem 1.13. Suppose that 0 < p < 1, 0 < q < +∞, and 1/r = 1/q − 1/p. Let μ, λ, and ν be σ-ﬁnite measures on [a, b] such that λ and ν are deﬁned on the same σ-algebra Mλ and λ = λc + λd , where λc and λd are, respectively, the continuous and discrete components of λ. Suppose also that u, w ∈ {Mλ }+ , v ∈ {Mμ }+ , (νa , νs ) is the Lebesgue decomposition of the measure ν with respect to λ, i.e., ν = νa + νs ,

νa λ,

νs ⊥ λ,

and dνa /dλ is the Radon–Nikodym derivative of νa with respect to λ. Finally, suppose that k is a nonnegative (Mμ × Mλ )-measurable function on [a, b]2 for which there exist constants α1 > 0 and α2 > 0 such that k(x, y1 ) ≥ α1 k(x, y2 ) if y1 < y2 and k(x1 , y) ≤ α2 k(x2 , y) if x1 < x2 , and −1/p dνa (x) , x ∈ [a, b], ω(x) := u(x) w(x) dλ PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S36

PROKHOROV et al.

E = x ∈ [a, b] v(x) H(x) :=

k(x, y)ω(y) dλc (y) = 0 ,

[a,x]

ω(t)λd ({t})1/p ,

sup

x ∈ [a, b].

t∈[a, x],λd ({t}) =0

If p ≤ q, then

v(x)

[a,b]

q 1/q k(x, y)f (y)u(y) dλ(y) dμ(x) [a,x]

≤C

1/p p

f wdν

for all

f ∈ {Mλ }+

(1.4.9)

[a,b]

if and only if μ(E) = 0 and F < +∞, where 1/q q v(t)k(t, x) dμ(t) H(x). F := sup x∈[a,b]

[x,b]

If q < p and k ≡ 1, then inequality (1.4.9) holds if and only if μ(E) = 0 and F < +∞, where r/p 1/r r v(x) v dμ H(x) dμ(x) . F := [a,b]

[x,b]

Moreover, in the case p ≤ q, the least constant C in (1.4.9) satisﬁes the relations α1 F ≤ C ≤ F, and in the case q < p, it satisﬁes the relation C ≈ F. Proof. By virtue of Lemmas 1.5 and 1.6, inequality (1.4.9) is equivalent to q 1/q v(x) k(x, y)f (y)ω(y) dλ(y) dμ(x) [a,b]

[a,x]

≤C

1/p for all f ∈ {Mλ }+ .

p

f dλ [a,b]

This inequality implies the pair of inequalities (with the same constant C)

q 1/q v(x) k(x, y)f (y)ω(y) dλc (y) dμ(x) [a,b]

[a,x]

≤C [a,b]

[a,b]

for all

f dλc

v(x)

1/p p

q 1/q k(x, y)f (y)ω(y) dλd (y) dμ(x)

[a,x]

≤C

f ∈ {Mλ }+ .

1/p p

f dλd

for all

f ∈ {Mλ }+ .

[a,b]

The application of Theorem 1.10 to the ﬁrst inequality and Theorems 1.11 and 1.12 to the second proves necessity. Now, suppose that μ(E) = 0 and F < +∞. Fix any function f ∈ {Mλ }+ . Since μ(E) = 0, it follows that q 1/q v(x) k(x, y)f (y)ω(y) dλ(y) dμ(x) [a,b]

[a,x]

v(x)

= [a,b]\E

q 1/q k(x, y)f (y)ω(y) dλ(y) dμ(x) [a,x]

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

v(x)

= [a,b]

q

1/q k(x, y)f (y)ω(y) dλd (y) dμ(x) [a,x]

1/p

F

S37

p

f dλd

1/p

≤F

[a,b]

p

f dλ

,

[a,b]

which proves suﬃciency.

1.5. Further Results This section contains additional results, some of which follows from those of Sections 1.2 and 1.3. Part of these results are used in what follows to study properties of integration operators with two variable limits. Hereafter, M denotes the σ-algebra of Lebesgue measurable subsets of [a, b]. 1.5.1. The weighted inequality. In this part, we consider inequality (1.3.1) in which a = 0, λ is the Lebesgue measure, λ = ν = μ, w = 1, and 0 < q < 1 ≤ p < ∞, i.e., a weighted Hardy-type inequality with Oinarov kernel of the form 1/q 1/p b b q p [(Kf )(x)] v(x) dx ≤ CK f (x) dx , f ∈ {M}+ , (1.5.1) 0

0

where the integral operator K is deﬁned by x k(x, y)u(y)f (y) dy, (Kf )(x) :=

x ∈ [0, b].

0

b Let α := D q + 1 and suppose that t v < ∞ for all t ∈ (0, b). Consider the function ζ : [0, b) → [0, b) deﬁned by b b v≥α v , x ∈ [0, b) ζ(x) := sup y ∈ (0, b) : y

x

(we assume that sup ∅ = 0). For each m ∈ N, by ζm we denote the composition of m functions ζ. Then ζm is a nondecreasing function on [0, b). b Theorem 1.14. Suppose that 1 ≤ p < ∞, 0 < q < ∞, 1/r = 1/q − 1/p, u, v ∈ {M}+ , and t u < ∞ for all t ∈ (0, b); suppose also that the kernel k is a Borel function on [0, b]2 satisfying Oinarov’s condition (1.3.2). Then inequality (1.5.1) holds if and only if ζ2 (x) q 1/q b 1/p b q v(x)k(x, ζ2 (x)) f u dx ≤ Cζ fp , f ∈ {M}+ , (1.5.2) 0

0

and B < ∞, where

B :=

0

⎧ b 1/q ⎪ ⎪ ⎪ v χ(a,t) ( · )k(t, · )u( · )p , sup ⎪ ⎨t∈(0,b) t

p ≤ q,

b r/p 1/r b ⎪ ⎪ ⎪ r ⎪ v(x) v χ(ζ3 (x),x) ( · )k(x, · )u( · )p dx , q < p. ⎩ 0

x

Moreover, CK ≈ B + Cζ . Proof. For each n ∈ Z, we set

b −n v≥α an := sup y ∈ (0, b) : y

(it is assumed that sup ∅ = 0). Note that ζ(an+1 ) = an . PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

(1.5.3)

S38

PROKHOROV et al.

Suﬃciency. We have (Kf )v 1/q qq

=

an+1

n∈Z an

α−n

I1 :=

I2 :=

k(x, y)f (y)u(y) dy

dx

q

an+1

I1 + I2 ,

k(an+1 , y)f (y)u(y) dy 0

α−n

q

an+1

k(an+1 , y)f (y)u(y) dy an−1

n∈Z

q

x 0

n∈Z

where

v(x)

−n

q

an−1

α

,

k(an+1 , y)f (y)u(y) dy

.

0

n∈Z

¨ Applying Holder’s integral inequality, we obtain I1 ≤

−n

α

χ(an−1 ,an+1 ) ( · )k(an+1 , · )u( · )qp

f

an+1

n∈Z

p

.

an−1

n∈Z

If p ≤ q, then we use Jensen’s inequality, which gives

b q v χ(0,an+1 ) ( · )k(an+1 , · )u( · )p I1

q/p

an+1

q/p

an+1

f

p

B q f qp .

an−1

¨ If q < p, then Holder’s inequality for sums implies

q/r −n r/q r [α ] χ(an−1 ,an+1 ) ( · )k(an+1 , · )u( · )p f qp ; I1 n∈Z

moreover, the sum on the right-hand side is estimated from above as [α−n ]r/q χ(an−1 ,an+1 ) ( · )k(an+1 , · )u( · )rp n∈Z

an+2

v

n∈Z

an+1

n∈Z

an+1

r/p

b

v an+2

χ(ζ3 (an+2 ),an+1 ) ( · )k(an+1 , · )u( · )rp

b r/p an+2 r v(x) v χ(ζ3 (an+2 ),x) ( · )k(x, · )u( · )p dx ≤ B r . x

Let us estimate I2 . Since k(ai−2 , y) = 0 for y ∈ (ai−2 , ai−1 ), it follows that an−1 ai−1 k(an+1 , y)f (y)u(y) dy = k(an+1 , y)f (y)u(y) dy 0

=

n ai−1 i≤n

ai−2

i≤n

D

ai−2

(n+1)−(j+1)

k(aj+1 , y) − D

(n+1)−j

k(aj , y) f (y)u(y) dy.

j=i−2

By condition (1.3.2), we have n (D(n+1)−(j+1) k(aj+1 , y) − D(n+1)−j k(aj , y)) j=i−2

≤ D (n+1)−(i−1) k(ai−1 , y) + D(n+1)−(i−1) k(ai , ai−1 ) +

n

D(n+1)−j k(aj+1 , aj );

j=i

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

therefore,

an−1

S39

k(an+1 , y)f (y)u(y) dy ≤ S1 + S2 ,

0

where S1 :=

D

(n+1)−(i−1)

i≤n

S2 :=

D(n+1)−(i−1)

ai−2

fu

ai−2

ai

k(ai , y)f (y)u(y) dy =: J1,n ; ai−2

i≤n

ai−1

k(ai−1 , y)f (y)u(y) dy + k(ai , ai−1 )

i≤n

ai−1

n ai−1

ai−2

D

(n+1)−j

k(aj+1 , aj ) f (y)u(y) dy.

j=i

Permuting the sums, we estimate S2 as (n+1)−j k(aj+1 , aj ) D S2 = j≤n

ai−1 ai−2

i≤j

Using the well-known relation

(n+1)−j fu = D k(aj+1 , aj )

λn

s μi

≈

i≥n

n∈Z

f u =: J2,n . 0

j≤n

aj−1

λn μsn ,

(1.5.4)

n∈Z

which holds for any λ ∈ (0, 1), μi ≥ 0, and s > 0, we obtain q ai Dq n −n q 2q −i α J1,n = D D k(ai , y)f (y)u(y) dy α ai−2 i≤n n∈Z n∈Z q an Dq n −n ≈ k(an , y)f (y)u(y) dy D α an−2 n∈Z q an −n = α k(an , y)f (y)u(y) dy B q f qp ; an−2

n∈Z

the proof of the last inequality is similar to that of the upper estimate of I1 . In the same way, we obtain aj−1 q Dq n q α−n J2,n = Dq D−j k(aj+1 , aj ) fu α 0 j≤n n∈Z n∈Z an−1 q an−1 q Dq n −n −n q ≈ fu = α k(an+1 , an ) fu , D k(an+1 , an ) α 0 0 n∈Z

whence

n∈Z

q α−n J2,n

≈

n∈Z

n∈Z

n∈Z

≤

q

q fu

0

an+2

ζ2 (x)

q

v(x)k(x, ζ2 (x))

an+1

b

q

f u dx

0

ζ2 (x)

v(x)k(x, ζ2 (x))q 0

an−1

v k(an+1 , an )

an+1

an+2

0

q f u dx ≤ Cζq f qp .

Necessity. Suppose that p ≤ q. Fix any θ ∈ (0, 1) and t ∈ (0, b). There exists a nonnegative function f ∈ Lp ([0, t]) such that f p = 1 and t k(t, y)f (y)u(y) dy ≥ θχ(0,t) ( · )k(t, · )u( · )p . 0

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S40

PROKHOROV et al.

Substituting f into (1.5.1), we obtain CK

θ sup ≥ D t∈(0,b)

Therefore, CK B if p ≤ q. Now, suppose that q < p. We have an+1 r v B n∈Z

1/q

b

r/p

b

χ(ζ3 (an ),an+1 ) ( · )k(an+1 , · )u( · )rp

v an

an

χ(0,t) k(t, · )u( · )p .

v t

[α−n ]r/q χ(an−3 ,an+1 ) ( · )k(an+1 , · )u( · )rp =: B r .

n∈Z

Fix any θ ∈ (0, 1). For any n ∈ Z, there exists an fn ∈ {M}+ such that supp fn ⊂ [an−3 , an+1 ], fn p = 1, and an+1 k(an+1 , y)u(y)fn (y) dy ≥ θχ(an−3 ,an+1 ) ( · )k(an+1 , · )u( · )p . an−3

For the functions gn := (α−n )r/(pq) χ(an−3 ,an+1 ) ( · )k(an+1 , · )u( · )p fn r/p

and

g :=

gn χ(a,b) ,

n∈Z

we have gpp =

m∈Z

am

p gn (x) dx =

n∈Z

am+1

gm (x)p dx =

am−3

m∈Z

and

am+1

b

m∈Z am

p gn (x) dx

n=m

(α−m )r/q χ(am−3 ,am+1 ) ( · )k(am+1 , · )u( · )rp = B r

m∈Z

(Kg) v q

0

am+1 m+3

an+1

n∈Z

≥θ

an+2

q

an+1

v

k(an+1 , y)u(y)gn (y) dy an−3

[α−n ]r/q χ(an−3 ,an+1 ) ( · )k(an+1 , · )u( · )rp = B r .

n∈Z

B r/p

θ 1/q B r/q , and hence CK B. It follows from (1.5.1) that CK Moreover, x q b b v(x) k(x, y)f (y)u(y) dy dx v(x)k(x, ζ2 (x))q 0

0

0

ζ2 (x)

q f u dx,

0

which implies inequality (1.5.2). Hardy’s discrete inequality readily implies a criterion for the validity of Hardy’s inequality (1.5.2) with variable upper limit in the discrete form in terms of the sequence {an }. Imposing an additional condition on the weight v, namely, requiring that v > 0 almost everywhere on (0, b), we obtain the following criterion for the validity of (1.5.1) in the integral form. Corollary 1.2. Suppose that 1 ≤ p < ∞, 0 < q 0 b almost everywhere on (0, b), and t v < ∞ for all t ∈ (0, b). Suppose also that the kernel k is a Borel function on [0, b]2 satisfying Oinarov’s condition (1.3.2). Then (1.5.1) holds if and only if B + B0 < ∞, where b r/p 1/r b q r q v(x)k(x, ζ2 (x)) χ(0,ζ2 (x)) up v(s)k(s, ζ2 (s)) ds dx B0 := 0

x

and B is deﬁned by (1.5.3). Moreover, CK ≈ B + B0 . PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S41

Proof. Let a := sup{t ∈ [0, b) : ζ2 (t) = 0}. Since v > 0 almost everywhere on (0, b), it follows that ζ2 is a continuous strictly increasing function on [a , b). On each [a , c] ⊂ (a , b), ζ2 can be pointwise approximated from below by a sequence {ϕn } of strictly increasing absolutely continuous functions on [a , c], whose inverses are absolutely continuous as well. For example, we can take x ζ2 (t)χ[a ,b) (t) dt, x ∈ [a , c] x → ϕn (x) := n x−1/n

(see [16]). Making the change of variable x = ϕ−1 n (s), applying the criterion for the validity of Hardy’s inequality (see [17, Section 1.3], [18, 19]), and making the inverse change, we obtain a two-sided estimate of the exact constant Cn in the inequality ϕn (x) q 1/q b 1/p c q v(x)k(x, ζ2 (x)) f u dx ≤ Cn fp , f ∈ {M}+ . (1.5.5) a

0

0

Namely, Cn ≈ Bn , where c c r/p 1/r q r q v(x)k(x, ζ2 (x)) χ(0,ϕn (x)) up v(s)k(s, ζ2 (s)) ds dx . Bn := a

x

Suppose that (1.5.1) holds. Then, according to Theorem 1.14, we have B CK , and inequality (1.5.2) holds with Cζ CK . Since ϕn (x) ≤ ζ2 (x), it follows that (1.5.5) holds for any c ∈ [a , b) and n ∈ N and that Cn ≤ Cζ . Hence, Bn CK , and passing to the limit as c → b and n → ∞, we obtain B0 CK by the monotone convergence theorem. Conversely, suppose that B0 < ∞. Since Bn ≤ B0 , it follows that (1.5.5) holds for any c ∈ [a , b) and n ∈ N, and that Cn B0 . Passing to the limit as c → b and n → ∞, we obtain (1.5.2) with Cζ B0 by the monotone convergence theorem. Similar considerations yield results for the operator b )(x) = k(x, y)f (x)u(x) dx, (Kf Suppose that

t 0

y ∈ (0, b).

y

v < ∞ for all t ∈ (0, b) and α = Dq + 1. We set x t v≥α v , σ(x) := inf t ∈ (0, b) : 0

x ∈ [0, b)

0

(it is assumed that inf ∅ = b). For each m ∈ N, by σm we denote the composition of m functions σ. t Theorem 1.15. Suppose that 1 ≤ p < ∞, 0 < q < ∞, 1/r = 1/q − 1/p, u, v ∈ {M}+ , and 0 v < ∞ for all t ∈ (0, b). Suppose also that the kernel k is a Borel function on [0, b]2 satisfying Oinarov’s condition (1.3.2). Then 1/q b 1/p b )q v (Kf ≤ CK fp , f ∈ {M}+ , (1.5.6) 0

if and only if b

0

q

q 1/q b 1/p f u dy ≤ Cσ fp ,

b

v(y)k(σ2 (y), y) 0

σ2 (y)

< ∞, where and B := B

f ∈ {M}+ ,

⎧ t 1/q ⎪ ⎪ ⎪ v χ(t,b) ( · )k( · , t)u( · )p , ⎨ sup t∈(0,b)

0

b ⎪ ⎪ ⎪ ⎩ v(y) 0

v 0

p ≤ q,

r/p

y

(1.5.7)

0

(1.5.8)

1/r χ(y,σ3 (y)) ( · )k( · , y)u( · )rp

dy

, q < p.

+ Cσ . Moreover, CK ≈ B PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S42

PROKHOROV et al.

Corollary 1.3. Suppose that 1 ≤ p < ∞, 0 < q 0 t almost everywhere on (0, b), and 0 v < ∞ for all t ∈ (0, b). Suppose also that the kernel k is a Borel function on [0, b]2 satisfying Oinarov’s condition (1.3.2). Then inequality (1.5.6) holds if +B 0 < ∞, where and only if B b r/p 1/r y q r q v(y)k(σ2 (y), y) χ(σ2 (y),b) up v(s)k(σ2 (s), s) ds dy B0 := 0

0

is deﬁned by (1.5.8). Moreover, C ≈ B +B 0 . and B K 1.5.2. Hardy’s operators with one variable integration limit. Suppose that p > 1 and q > 0. Let v, w ∈ {M}+ be weight functions on (0, ∞), and let kD be a (M × M)-measurable function (kernel) nonnegative on (0, ∞)2 . We set ∞ kD (x, y)f (y)v(y) dy, 0 ≤ c ≤ x ≤ d ≤ ∞. (1.5.9) Df (x) = w(x) 0

We say that f ∈

Lρ (0, ∞)

for 0 < ρ ≤ ∞ if f ρ < ∞, where ⎧ ∞ 1/ρ ⎪ ρ ⎨ |f (t)| dt , ρ < ∞, f ρ := 0 ⎪ ⎩ ess supt∈(0,∞) |f (t)|, ρ = ∞.

Suppose that the operator D acts from Lp (0, ∞) to Lq (0, ∞), where the parameters p and q are as speciﬁed above. The operator D is bounded for 0 < p < 1 and 0 < q ≤ ∞ only in the trivial case (see Theorem 1.10 or [20, Theorem 2]). For the extreme values of the summation parameter, i.e., p = 1, ∞ or q = 1, ∞, the exact value of the norm DLp (0,∞)→Lq (0,∞) is found by using the general theorem given in [21, Chapter 11, Section 1.5, Theorem 4]. Theorem 1.16. Suppose that v, w ∈ {M}+ are weight functions on (0, ∞) and the kernel kD (x, y) is a (M × M)-measurable function on (0, ∞)2 . Then the norm DLp (0,∞)→Lq (0,∞) satisﬁes the relations 1 ≤ q ≤ ∞, DL1 (0,∞)→Lq (0,∞) = ess sup v(t)kD ( · , t)w( · )q , t>0

DLp (0,∞)→L∞ (0,∞) = ess sup w(t)kD (t, · )v( · )p ,

1 0

Moreover, if kD (x, y) ≥ 0 on (0, ∞)2 , then ∞ q w (x) DL∞ (0,∞)→Lq (0,∞) =

0 ∞

DLp (0,∞)→L1 (0,∞) =

v (y)

kD (x, y)v(y) dy 0

p

0

q

∞

1/q dx

p

∞

kD (x, y)w(x) dx

1 ≤ q < ∞;

, 1/p

dy

,

1 < p < ∞.

0

We begin with criteria for the boundedness of the operator D in the cases kD (x, y) = χ[c,x](y) and kD (x, y) = χ[x,d] (y) implied by Theorems 1.1–1.3. Corollary 1.4. Suppose that 1 < p < ∞ and 0 < q < ∞. Consider the operators x f (y)v(y) dy, 0 ≤ c ≤ x ≤ d ≤ ∞, Hf (x) = w(x) ∗

(1.5.10)

c d

H f (x) = w(x)

f (y)v(y) dy,

0 ≤ c ≤ x ≤ d ≤ ∞.

(1.5.11)

x

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S43

(a) If p ≤ q, then HLp (c,d)→Lq (c,d) ≈ AM and H ∗ Lp (c,d)→Lq (c,d) ≈ A∗M , where 1/q t 1/p d wq (x) dx v p (y) dy , AM := sup c≤t≤d

A∗M

t

1/q

t

c d

q

:= sup

w (x) dx

c≤t≤d

1/p

p

v (y) dy

c

(1.5.12) (1.5.13)

.

t

∗ , where (b) For q < p, HLp (c,d)→Lq (c,d) ≈ BM R and H ∗ Lp (c,d)→Lq (c,d) ≈ BM R d d

r/p t

r/p 1/r q p q w (x) dx v (y) dy w (t) dt , BM R := c

∗ BM R

d

t

r/p

t

c d

q

:=

w (x) dx c

r/p

p

1/r q

v (y) dy

c

(1.5.14)

w (t) dt

(1.5.15)

.

t

The boundedness characteristics AM and A∗M deﬁned by (1.5.12) and (1.5.13) are customarily ∗ called the Muckenhoupt functionals, and the characteristics BM R and BM R deﬁned by (1.5.14) and (1.5.15) are called the Maz’ya–Rosin functionals. The alternative quantities AT and A∗T (see (1.5.16) and (1.5.17)) and BP S and BP∗ S (see (1.5.18) and (1.5.19)), which are also equivalent to the norms HLp (c,d)→Lq (c,d) and H ∗ Lp (c,d)→Lq (c,d) , are referred to as the Tomaselli functionals [5] and the Persson–Stepanov functionals [22], respectively. Statement 1. Let H and H ∗ be the operators deﬁned by (1.5.10) and (1.5.11), respectively. (a) If 1 < p ≤ q < ∞, then HLp (c,d)→Lq (c,d) ≈ AT and H ∗ Lp (c,d)→Lq (c,d) ≈ A∗T , where

q 1/q t −1/p t x v p (y) dy wq (x ) dx v p (y) dy , (1.5.16) AT := sup c≤t≤d

A∗T

c

c

d

d

:= sup

c

q

p

1/q

d

q

v (y) dy w (x ) dx

c≤t≤d

t

−1/p

p

v (y) dy

x

(1.5.17)

.

t

(b) If 0 < q 1, then HLp (c,d)→Lq (c,d) ≈ BP S and H ∗ Lp (c,d)→Lq (c,d) ≈ BP∗ S , where d t x q

r/p t

q−r/p 1/r p q p q v (y) dy w (x) dx v (y) dy w (t) dt , (1.5.18) BP S := c

∗ BM R

c

c

d d

d

:=

p

t

r/p

d

q

v (y) dy c

c

q w (x) dx

x

q−r/p

p

1/r q

v (y) dy

w (t) dt

.

(1.5.19)

t

Let kD (x, y) = χ[c,x](y)k(x, y), where k(x, y) is a jointly measurable function on (0, ∞)2 satisfying Oinarov’s condition (1.3.2) for c ≤ y ≤ z ≤ x ≤ d. To state results concerning the boundedness of operators (1.5.9) with Oinarov kernels of the form kD (x, y) = χ[c,x](y)k(x, y), where k ∈ (1.3.2), we introduce the quantities x x x vp , V1 (x) := k(x, y)v p (y) dy, Vp (x) := kp (x, y)v p (y) dy, V (x) := c

d

wq ,

W (y) :=

c

k(x, y)wq (x) dx,

W1 (y) :=

y

d

c d

kq (x, y)wq (x) dx

Wq (y) :=

y

y

and the functionals A := max(A0 , A1 ),

A0 := sup A0 (t) = sup [Wq (t)]1/q [V (t)]1/p , t∈(c,d)

t∈(c,d)

A1 := sup A1 (t) = sup [W (t)]1/q [Vp (t)]1/p ; t∈(c,d)

t∈(c,d)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S44

PROKHOROV et al.

A := max(A0 , A1 ), 1/q t q −1/p q V1 (x)w (x) dx , A0 := sup A0 (t) = sup [V (t)] t∈(c,d)

c

t∈(c,d)

−1/p

A1 := sup A1 (t) = sup [Vp (t)] t∈(c,d)

1/q

t

Vpq (x)wq (x) dx

,

c

t∈(c,d)

A := max(A0 , A1 ), 1/p d −1/q p p Wq (y)v (y) dy , A0 := sup A0 (t) = sup [Wq (t)] t∈(c,d)

t∈(c,d) −1/q

t d

A1 := sup A1 (t) = sup [W (t)] t∈(c,d)

t

t∈(c,d)

W1p (y)v p (y) dy

1/p ;

we also introduce B := max(B0 , B1 ), d 1/r 1/r d r/q r/p r/p r/q [Wq (t)] d[V (t)] , B1 := [Vp (t)] d(−[W (t)] ) ,

B0 := c

c

B0 :=

c

B := max(B0 , B1 ), t r/q 1/r d q −r/p q [V (t)] d V1 (x)w (x) dx , c

d

B1 :=

t r/q 1/r −r/p q q [Vp (t)] d Vp (x)w (x) dx ;

c

B0 :=

B := max(B0 , B1 ), d r/p 1/r d −r/q p p [Wq (t)] d − Wq (y)v (y) dy ,

c

B1 :=

c

d

−r/q [W (t)] d −

c

t d t

W1p (y)v p (y) dy

r/p 1/r .

Using these functionals, we give three alternative boundedness criteria for (1.5.9) with kernels kD (x, y) = χ[c,x](y)k(x, y), where k ∈ (1.3.2), in each of the two cases 1 < p ≤ q < ∞ and 1 < q < p < ∞ of relations between the summation parameters. Theorem 1.17. For the operator

x

k(x, y)f (y)v(y) dy,

Kf (x) = w(x)

0 ≤ c ≤ x ≤ d ≤ ∞,

(1.5.20)

c

with kernel k(x, y) satisfying Oinarov’s condition (1.3.2), the following relations hold. If 1 < p ≤ q < ∞, then KLp (c,d)→Lq (c,d) ≈ A ≈ A ≈ A.

(1.5.21)

KLp (c,d)→Lq (c,d) ≈ B ≈ B ≈ B.

(1.5.22)

If 1 < q < p < ∞, then The equivalence coeﬃcients in (1.5.21) and (1.5.22) depend only on p, q, and the constant D in condition (1.3.2). PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S45

Remark 1.18. (1) The characterization of the validity of (1.5.21) and (1.5.22) by the conditions A < ∞ and B < ∞, respectively, follows directly from Theorems 1.6 and 1.7. (2) For 1 < p ≤ q < ∞, the equivalence of (1.5.20) to the condition A < ∞ was proved in [11], to the condition A < ∞, in [10], and to the conditions A < ∞ and A < ∞, in [12]. In [10]–[12], the kernel k(x, y) was required to satisfy certain additional constraints of monotonicity or continuity type, which were subsequently removed [16]. Indeed, without loss of generality, the kernel k(x, y) can be assumed to be nonincreasing in x and nondecreasing in y; otherwise, it can be replaced by the equivalent kernel k0 (x, y) := sup

sup k(t, z)

y≤z≤x z≤t≤x

with these properties (see [16]), for which k(x, y) ≤ k0 (x, y) ≤ D2 k(x, y). (3) In the opposite case 1 < q < p < ∞, equivalence (1.5.22) with constant B was proved in [10] and [12]. The other criteria, condition (1.5.21) with constant A and (1.5.22) with constants B and B, were found in [23]. (4) Criteria for the compactness of operator (1.5.20) with kernel k(x, y) satisfying condition (1.3.2) were found in [12]. (5) The components of each of the constants of the types A and B in Theorem 1.17 are generally independent (see, e.g., [24], [12, Section 4], and [23]) and related by (i) (iii) (v)

A0 < ∞ ⇐⇒ A0 < ∞, A0 < ∞ =⇒ A0 < ∞, A0 < ∞ =⇒ A0 < ∞,

(ii) (iv) (vi)

A1 < ∞ A1 < ∞ A1 < ∞

⇐⇒ =⇒ =⇒

A1 < ∞, A1 < ∞, A1 < ∞;

(ii) (iv) (vi)

B1 < ∞ B1 < ∞ B1 < ∞

⇐⇒ B1 < ∞, =⇒ B1 < ∞, =⇒ B1 < ∞

similarly, for the other constants, we have (i) (iii) (v)

B0 < ∞ ⇐⇒ B0 < ∞, B0 < ∞ =⇒ B0 < ∞, B0 < ∞ =⇒ B0 < ∞,

in each group, the implications reverse to (iii)–(vi) are not generally true. Let φ : [c, d] → [0, ∞) be a strictly increasing diﬀerentiable function. Below, we give some criteria for a generalization of operator (1.5.20). Let kD (x, y) = χ[φ(c),φ(x)] (y)k(x, y). Assume that, for c ≤ φ−1 (y) ≤ z ≤ x ≤ d, the jointly measurable function k(x, y) on (0, ∞)2 satisﬁes the condition D −1 k(x, y) ≤ k(x, φ(z)) + k(z, y) ≤ Dk(x, y)

(1.5.23)

with a constant D ≥ 1 not depending on the variables x, y, and z. Corollary 1.5. Let Kφ : Lp (φ(c), φ(d)) → Lq (c, d), where φ(x) k(x, y)f (y)v(y) dy, Kφ f (x) = w(x)

0 ≤ c ≤ x ≤ d ≤ ∞,

(1.5.24)

φ(c)

and the kernel k(x, y) satisﬁes the Oinarov-type condition (1.5.23) for c ≤ φ−1 (y) ≤ z ≤ x ≤ d. (a) If 1 < p ≤ q < ∞, then Kφ Lp (φ(c),φ(d))→Lq (c,d) ≈ Aφ,0 + Aφ,1 , where

d

Aφ,0 := sup c≤t≤d

t

1/q

d

Aφ,1 := sup c≤t≤d

1/q kq (x, φ(t))wq (x) dx w

t

φ(t)

φ(c) φ(t)

q

p

p

k (t, y)v (y) dy

(1.5.25)

vp

1/p ,

1/p .

φ(c)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S46

PROKHOROV et al.

(b) For 1 < q < p < ∞, Kφ Lp (φ(c),φ(d))→Lq (c,d) ≈ Bφ,0 + Bφ,1 , where

r/q

φ(d) d q

Bφ,0 :=

φ−1 (t)

φ(c)

d Bφ,1 :=

w c

k (x, t)w (x) dx

r/p

d

r/q

1/r

p

v (t) dt

φ(t)

p

r/p

p

1/r q

k (t, y)v (y) dy

,

w (t) dt

.

φ(c)

p

φ(d) φ(d) d

Bφ,0 ≈ Bφ,0 :=

q

φ(c)

φ−1 (y)

t

× Bφ,1 ≈ Bφ,1 :=

v

p

φ(c)

q

t

Moreover,

t

q

(1.5.26)

q

k (x, y)w (x) dx

q

φ−1 (t)

d t

q

k (x, t)w (x) dx

φ(x)

p

c

q

p

×

φ(c) φ(t)

p

1/r

p

v (t) dt

k (x, y)v (y) dy c

v (y) dy

p −r/q

d

(1.5.27)

r/p w (x) dx 1/r q

k (t, y)v (y) dy

,

q

q−r/p

p

r/q

p

w (t) dt

.

(1.5.28)

φ(c)

Proof. Substituting τ = φ−1 (y) into 1/q d [Kφ f (x)]q dx ≤C c

1/p

φ(d)

f p (y) dy

we obtain the equivalent inequality 1/q

q d x q ≤C k(x, τ )f (τ ) v (τ ) dτ w (x) dx c

,

φ(c)

c

d

fp (τ ) dτ

1/p

c

with f(τ ) = f (φ(τ ))[φ (τ )]1/p ,

v(τ ) = v(φ(τ ))[φ (τ )]1/p ,

and Oinarov kernel k(x, τ ) = k(x, φ(τ )) (see (1.3.2)). Relations (1.5.25) and (1.5.26) follow from Theorem 1.17 (see also Theorems 1.6 and 1.7). For (1.5.27) and (1.5.28), see Corollary 1.18 (5) or [23, Theorem 2.2]. Similar results are valid for the operator with variable lower limit. Statement 2. Let Kφ∗ : Lp (φ(c), φ(d)) → Lq (c, d) be the operator given by Kφ∗ f (x)

φ(d)

= w(x)

k(x, y)f (y)v(y) dy,

0 ≤ c ≤ x ≤ d ≤ ∞,

(1.5.29)

φ(x)

with kernel k(x, y) ≥ 0 satisfying condition (1.5.23) with D ≥ 1 not depending on x, y, and z for all c ≤ x ≤ z ≤ φ−1 (y) ≤ d. (a) If 1 < p ≤ q < ∞, then Kφ∗ Lp (φ(c),φ(d))→Lq (c,d) ≈ A∗φ,0 + A∗φ,1 , PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

where A∗φ,0 A∗φ,1

1/q

t q

:= sup

φ(d)

q

k (x, φ(t))w (x) dx

c≤t≤d

c

1/q

t

:= sup

w

c≤t≤d

v φ(t)

φ(d)

q

p

p

1/p ,

1/p

p

k (t, y)v (y) dy

c

S47

.

φ(t)

(b) For 1 < q < p < ∞, Kφ∗ Lp (φ(c),φ(d))→Lq (c,d) ≈ B∗φ,0 + B∗φ,1 , where B∗φ,0

:=

c

d

w c

≈

φ(d)

q

∗ Bφ,0

r/p

t

:=

v t

φ(d)

p

q

p

≈

v (t) dt

,

1/r q

w (t) dt

φ(d) t

φ−1 (y)

:=

p q

q

k (x, y)w (x) dx φ(c)

φ(c)

c

φ−1 (t)

q

q

;

p

p

k (x, t)w (x) dx

d d

φ(d)

:= t

×

q

φ(x) φ(d)

1/r

p

v (t) dt

k (x, y)v (y) dy c

r/q

p

v (y) dy

p −r/q

c

B∗φ,1

1/r

p

φ(t)

× B∗φ,1

r/q

p

r/p

k (t, y)v (y) dy

c

moreover, B∗φ,0

q

k (x, t)w (x) dx φ(c)

B∗φ,1

r/q

φ(d) φ−1 (t)

r/p q

w (x) dx

q−r/p

kp (t, y)v p (y) dy

(1.5.30)

,

1/r wq (t) dt

(1.5.31)

.

φ(t)

Proof. The proof of this theorem is similar to that of Corollary 1.5. Lemmas 1.10–1.13 give estimates of the norms of some other Hardy-type operators generalizing transformations (1.5.10) and (1.5.11). We begin with the case 1 < p ≤ q < ∞. Lemma 1.10. Suppose that 1 < p ≤ q < ∞, 0 ≤ c < d ≤ ∞, and 0 ≤ a < ∞. Let b(x) be a diﬀerentiable strictly increasing function on [c, d] such that a ≤ b(x) < ∞ for x ∈ [c, d), and let b(x) f (y)v(y) dy; (1.5.32) Sf (x) := w(x) a

the function b(x) is shown in Fig. 1.5.1. Then, for the norm S := SLp (a,b(d))→Lq (c,d) of the operator S, the following two-sided estimates with equivalence constants depending only on p and q hold: d 1/q b(t) 1/p wq vp , (1.5.33) S ≈ Ab := sup c≤t≤d

S ≈ Ab := sup

t

t

c≤t≤d

a

b(x)

v c

p

1/q

q q

w (x)dx

a

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

b(t)

v

p

−1/p (1.5.34)

.

a

Vol. 300 Suppl. 2

2018

S48

PROKHOROV et al. 6 b(d) b(x) b(c)

a c

-

d Fig. 1.5.1.

Proof. Necessity (or, equivalently, the lower estimates of S) in (1.5.33) and (1.5.34) is proved by substituting the test function ft (y) = [v(y)]p −1 χ[a,b(t)] (y) into the inequality

q 1/q ≤C Sf (x) dx

d c

1/p

b(d) p

f (y) dy

(1.5.35)

.

a

To prove suﬃciency (or the upper estimate of S) in (1.5.33), note that the best constant C ≥ 0 in (1.5.35) coincides with the norm S of the operator and C ≈ C1 + C2 , where C1 and C2 are the best constants in the inequalities d 1/q 1/p b(d) b(c) q p f (y)v(y) dy w ≤ C1 f (y) dy , d

a

c

q

b(x)

1/q

f (y)v(y) dy w (x) dx

f (y) dy

,

a

a

w

1/q

p

f (y) dy

≤ C4

f (y)v(y) dy w (x) dx b(c)

1/p

b(c)

≤ C3

q

c

q

b(x)

1/q

d

q

c

1/p

b(d) p

b(c)

which, in turn, are equivalent to the inequalities b(c) f (y)v(y) dy d

a

≤ C2

q

c

a

,

(1.5.36)

1/p

b(d) p

f (y) dy

.

(1.5.37)

b(c)

¨ inequality implies Thus, C ≈ C3 + C4 . Inequality (1.5.36) combined with reverse Holder’s d 1/q b(c) 1/p wq vp ≤ Ab ; C3 = c

a

for (1.5.37), by virtue of Corollary 1.5 (a) with k(x, y) = 1 (see also Corollary 1.4 (a)), we have d 1/q b(t) 1/p wq vp ≤ Ab . C4 ≈ sup c≤t≤d

t

b(c)

This completes the proof of suﬃciency in (1.5.33). To obtain the upper estimates of S in (1.5.34), we write the inequality dual to (1.5.35): b(d) d

p 1/p d 1/q 1/p := gw dV (y) ≤C gq , I a

b−1 0 (y)

c

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

y

S49

−1 ¨ where V (y) := a v p and b−1 in0 (y) = max{c, b (y)}. Integrating by parts and applying Holder’s equality with exponents p and p , we arrive at p −1 b(d) d gw V (y)g(b−1 (y))w(b−1 (y)) db−1 (y) I≤p b−1 (y)

b(c)

≤p

b(d)

−1

[g(b b(c)

=: p

d

g

q

q

(y))] db

−1

1/q (y)

b−1 (y)

b(c)

1/q

q/(p−1)

b(d) d

q

gw

q

−1

V (y)w (b

(y)) db

−1

1/q (y)

1/q

I1 .

c

Next,

b(d) b(d)

I1 = b(c)

b(d)

d −

y

b−1 (t)

t q

q

−1

V (y)w (b

= b(c)

b(c)

b(d) b−1 (t)

= b(c)

s

= Aqb

(y)) db

−1

(y) d −

[V (b(x))] w (x) dx d − q

c

d c

b(s)

q

q/p v d −

b−1 (t)

d b−1 (t)

q/(p−1)

gw

q/(p−1)

d

gw

gw

.

b(d) d

d −

a

gw

s q/(p−1)

d

p

q/(p−1)

d

q

[V (b(x))] w (x) dx d − q

c

q/(p−1) gw V q (y)wq (b−1 (y)) db−1 (y)

c

d

≤

d

s

By virtue of Minkowski’s inequality, we have d b(s) q/p d q/(p−1)

vp d − gw c

a

s

q/(p−1)

d

b(c)

gw c

=

v q/(p−1)

d

a b(c)

gw c

p

v

p

q/p +

q/p

b−1 (y)

b(c)

b(d) d

+

a

b−1 (y)

b(c)

p

gw

q/(p−1) p/q

d

gw

q/p dV (y)

s

q/p dV (y) ≈ I q/p .

Therefore, I Ab

d

g

q

1/q I 1/p ,

c

which completes the proof of the upper estimate C Ab . Similar considerations give estimates of the norm of the integral operator with variable lower limit. Lemma 1.11. Suppose that 1 < p ≤ q < ∞, 0 ≤ c < d ≤ ∞, and 0 < b ≤ ∞. Let a(x) be a diﬀerentiable strictly increasing function such that 0 < a(x) ≤ b, x ∈ (c, d], and let b f (y)v(y) dy; (1.5.38) T f (x) := w(x) a(x)

the function a(x) is shown in Fig. 1.5.2. PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S50

PROKHOROV et al.

Then

T := T Lp (a(c),b)→Lq (c,d) ≈ Aa := sup T ≈ Aa := sup

d

b

v

c≤t≤d

t

p

w

c≤t≤d

q

1/q

t

b

q

v

c

1/p (1.5.39)

,

a(t)

1/q

b

q

w (x) dx

v

a(x)

p

p

−1/p (1.5.40)

.

a(t)

6 b

a(d) a(x) a(c) c

-

d Fig. 1.5.2.

The following two lemmas consider the case where 0 < q 1. Lemma 1.12. Suppose that 0 < q 1, 1/r = 1/q − 1/p, 0 ≤ c < d ≤ ∞, and 0 ≤ a < ∞. Let b(x) be a strictly increasing diﬀerentiable function on [c, d] such that a ≤ b(x) < ∞ for x ∈ [c, d), and let S be the operator from Lp (a, b(d)) to Lq (c, d) deﬁned by (1.5.32). Then d d r/p b(t) r/p r r wq vp wq (t) dt (1.5.41) S ≈ Bb := ≈

c

t

a

r/q

d

w

b(c)

q

v

c

p

r/p

d

S ≈ Brb := ≈

c

d t

b(x)

vp c

c

a

d

b(x)

v c

p

r/q w (x) dx v

c

vp

q

+

b(t)

wq (x) dx

b(x)

p

q

b(t)

v

p

r/p

a

r/p

q

a

c

t

q

d t

r/q w d q

+

a

and

d

a b(d)

v

p

(1.5.42)

q−r/p wq (t) dt

(1.5.43)

−r/p

a

r/q w (x) dx d − q

a

r =: B b

b(t)

v

p

−r/p

.

(1.5.44)

a

¨ Proof. As in the proof of Lemma 1.10, we ﬁrst apply reverse Holder’s inequality, obtaining d 1/q b(c) 1/p wq vp = C3 , E0 : = c

a

and then Corollary 1.4 (b) with the change φ−1 (y) = τ , which gives d d r/p b(t) r/p 1/r q p q w v w (t) dt ≈ C4 ; E1 : = c

t

b(c)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

thereby, we obtain

d

C ≈ (E0 + E1 ) ≈ r

r/p

d

r

w c

b(c)

q

v

t

r/p

p

b(t)

+

a

b(c)

S51

r/p v wq (t) dt ≈ Bbr . p

Let us prove the equivalence of functional (1.5.41) to the sum in (1.5.42). First, we assume that Bbr < ∞. Then d wq < ∞ for any t ∈ (c, d] (1.5.45) t

and, therefore, d 0 = lim s→d s

≥ lim

s→d

b(t) a

b(s)

v

r/p v d −

a

w

q

t

r/p

p

r/q

d

p

d −

d s

r/q

d

w

q

b(s)

v

= lim

s→d

t

p

r/p

r/q

d

w

a

q

.

(1.5.46)

s

r . To prove the converse, suppose that B r =: I1 + I2 < ∞. Integrating by parts, we obtain ∞ > Bbr ≈ B b b Then d r/q b(t) r/p d d b(t) r/p d r/q

+ ∞ > I1 + I2 = I1 + wq vp vp d − wq ≥

c

Therefore,

t

d

a

b(t) a

r/q

d

BrP S : =

a

t

≈ Bbr .

wq t

r B b

< ∞. Hence, Bbr ≈ b(t) set Va (t) : = a v p

Bbr

Next, we

c

c

r/p p v d −

by virtue of (1.5.45)–(1.5.46).

and d

r/q

t

[Va (x)]q wq (x) dx c

d(−[Va (t)]−r/p ).

c

According to [22, Theorem 3], we have Bb ≈ BP S and

Brb

r/q

d

≈

Vaq wq

if Va (d) = ∞

[Va (d)]−r/p + BrP S

(1.5.47)

if 0 < Va (d) < ∞.

(1.5.48)

c

Suppose that Bb < ∞. Let us prove the upper estimate C Bb . First, we assume that Va (d) = ∞. ¨ By virtue of Holder’s inequality with exponents r/q and p/q, we have d b(x) q f v wq (x) dx J := c

a

d

q

b(x)

fv

=q c

=q ≤

−q−1

[Va (s)] c d b(s)

q

d

a

p

b(s)

fv c

a

b(x) a

−q−1

[Va (s)]

dVa (s) dx =: J0

x q

fv c

d

(x)Vaq (x)

s

fv c

w

a d

q

w

Va−q (s)

q

s

w c

(x)Vaq (x) dx

q

Vaq

dVa (s)

Va−1 (s)

dVa (s)

q/p

Va−p (s) dVa (s)

BqP S .

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S52

PROKHOROV et al.

Using estimate (1.5.33) of Lemma 1.10, we obtain q J BP S

q/p

b(d)

f

p

.

a

This, together with (1.5.47), implies C Bb . If Va (d) < ∞, then d −q −q [Va (s)]−q−1 dVa (s), Va (x) = Va (d) + q x

and therefore J = J0 + Va−q (d)

d

q

b(x)

wq (x)Vaq (x) dx = : J0 + J1 .

fv c

a

Relation (1.5.48) and the estimates of J0 obtained above imply b(d) q/p q fp . J0 Bb a

b(xk )

To estimate J1 , we choose {xk }k∈Z so that b(c) f v = 2k , k ≤ N . We have d b(c) b(x) q d fv + f v wq (x)[Va (x)]q dx [Va (d)] J1 = c

a

≈

b(c)

q

b(c)

d q

fv

q

w (x)[Va (x)] dx +

a

c

wq (x)[Va (x)]q dx

fv b(c)

xk

k≤N

q

xk+1 b(x)

=: J1,1 + J1,2 . ¨ Applying Holder’s inequality with p and p and taking into account (1.5.48), we arrive at b(c) q/p b(c) q/p d p q/p q q q q f [Va (c)] w Va ≤ [Va (d)] Bb fp . J1,1 ≤ a

c

a

¨ inequality with exponents r/q and p/q implies For J1,2 , Holder’s xk+1 2q(k+1) wq (x)[Va (x)]q dx J1,2 ≤ xk

k≤N

≤4

q

≤4

b(d)

fp

q/p

b(xk )

f

v [Va (xk )]r/p

f

w

[Va (d)]

Bqb

c

wq Vaq

xk

xk+1

r/q q/r wq Vaq

q

Vaq

q+r/p

q

w (t)[Va (t)]

q/r dt

c

q/p

b(d)

f

xk+1

r/p

t

p

q

q/p

xk

k≤N

b(c)

p

b(xk−1 )

q/p d

b(d)

b(xk )

p

q/p

b(c)

wq (x)[Va (x)]q dx

xk

b(xk−1 )

k≤N

xk+1

fv

b(xk−1 )

k≤N q

q

b(xk )

p

.

b(c)

Combining all estimates obtained above, we see that C Bb . To prove the reverse inequality, suppose that C < ∞ and note that C Bb by virtue of (1.5.41). First, we show that Bb Bb , provided that Va (d) = ∞. For this purpose, we write t t t t t t q q q q q q w (x)[Va (x)] dx = Va (x) d − w = Va (c) w +q wq [Va (x)]q−1 dVa (x). c

c

x

c

c

x

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

We have Brb

≈

BrP S

≈ [Va (c)]

d

t

r

S53

r/q −r/p w d −[Va (t)] q

c

c

d t

t

w

+ c

c

q

q−1

[Va (x)]

r/q dVa (x) d(−[Va (t)]−r/p )

x

=: I1 + I2 . It is easy to see that r/p

r/q

d

I1 ≤ [Va (c)]

w

Bbr .

q

c

To estimate I2 , we write t t wq [Va (x)]q−1 dVa (x) c x

t t q q−1+q/(2p) w [Va (x)] [Va (x)]−q/(2p) dVa (x) = c

≤

x

t

w c

q/r t q/p −1/2 dVa (x) [Va (x)] dVa (x)

r/q

t

x

q

(q−1+q/(2p))(r/q)

[Va (x)]

t

[Va (t)]

q/(2p)

w c

c

r/q

t q

(q−1+q/(2p))(r/q)

[Va (x)]

q/r dVa (x) .

x

These relations and the equivalence between (1.5.41) and (1.5.42) imply d t t r/q q (q−1+q/(2p))(r/q) w [Va (x)] dVa (x) [Va (t)]r/(2p) d(−[Va (t)]−r/p ) I2 c

d

x

r/q

d

w c

c

d

x

(q−1+q/(2p))(r/q)

d r/(2p)−r/p−1

[Va (x)]

[Va (t)]

dVa (t) dVa (x)

x

r/q

d

w c

q

[Va (x)]r/q dVa (x) Bbr .

q

x

Since Va (d) = ∞, it follows that Brb ≈ I1 + I2 Bbr C r . In the case 0 < Va (d) < ∞, we substitute f (x) = v p −1 (x) into inequality (1.5.35) and obtain 1/q d −1/p q q w (x)Va (x) dx . C ≥ [Va (d)] c

Combining this with (1.5.48) and with the estimate of BP S , we arrive at the inequality Bb C, which completes the proof of (1.5.43). The equivalence of (1.5.43) and (1.5.44) follows from (1.5.48). Lemma 1.13. Suppose that 0 < q 1, 1/r = 1/q − 1/p, 0 ≤ c < d ≤ ∞, and 0 < b ≤ ∞. Let a(x) be a diﬀerentiable strictly increasing function such that 0 < a(x) ≤ b for x ∈ (c, d], and let T be an operator of the form (1.5.38) acting from Lp (a(c), b) to Lq (c, d). Then r/p d t r/p b r q p w v wq (t) dt (1.5.49) T ≈ Ba := c

c

r/q

d

≈

w

q

v

c

and T ≈ Ba :=

a(t) b

p

r/p

d d

c

b

v c

t

a(x)

p

t

r/q w d −

b

q

+

a(d)

r

d c

r/p

(1.5.50)

a(t)

r/p

q

v

p

b

q

w (x) dx

v

p

q−r/p wq (t) dt,

(1.5.51)

a(t)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S54

PROKHOROV et al.

≈

d

b

vp c

r/q

q

a(x)

d d

b

+

v c

t

b

wq (x) dx

q

p

vp

−r/p

a(c)

r/q w (x) dx d

b

q

a(x)

v

p

−r/p .

(1.5.52)

a(t)

2. INTEGRAL OPERATORS WITH TWO VARIABLE LIMITS

2.1. The Block-Diagonal Method Lemma sets, and let 2.1. Let U = k Uk and V = p k Vk be qtwo unions of disjoint measurable T = k Tk be a linear operator from L (U ) in L (V ) with linear summands Tk : Lp (Uk ) → Lq (Vk ). Then

T Lp (U )→Lq (V ) = sup Tk Lp (Uk )→Lq (Vk )

(2.1.1)

k

in the case 0 < p q < ∞. For 0 < q < p < ∞, 1/r Tk rLp (Uk )→Lq (Vk ) , T Lp (U )→Lq (V ) ≈

(2.1.2)

k

where 1/r = 1/q − 1/p. Proof. Suppose that 0 < p q < ∞. The lower estimate in (2.1.1) follows from the fact that T f Lq (V ) Tk f Lq (Vk ) for all k. The upper estimate is obtained by applying Jensen’s inequality: q Tk f qLq (Vk ) ≤ sup Tk Lp (Uk )→Lq (Vk ) f χk qLp (Uk ) T f qLq (V ) = k

k

≤ sup Tk Lp (Uk )→Lq (Vk ) k

q

k

f qLp (U ) .

(2.1.3)

Let 0 < q < p < ∞. The upper estimate in (2.1.2) is obtained by analogy with (2.1.3) with the use ¨ of Holder inequality. Let us prove the lower estimate. Note that, for any λ ∈ (0, 1), there exist functions p fk ∈ L (Uk ) such that λTk Lp (Uk )→Lq (Vk ) fk Lp (Uk ) ≤ Tk fk Lp (Uk ) . By virtue of homogeneity, we can choose the fk so that r/p

If f =

fk Lp (Uk ) = Tk Lp (Uk )→Lq (Vk ) .

χUk fk , then, since r/q = r/p + 1, it follows that Tk rLp (Uk )→Lq (Vk ) = λq (Tk Lp (Uk )→Lq (Vk ) fk Lp (Uk ) )q ≤ Tk fk qLq (Vk ) λq k

k

k

k

=

T f qLq (V )

=

T qLp (U )→Lq (V )

≤

T qLp (U )→Lq (V ) f qLp (U )

Tk rLp (Uk )→Lq (Vk )

q/p .

k

Passing to the limit as λ → 1, we obtain the lower estimate in (2.1.2). This chapter studies integral operators with two variable limits of integration of the form b(x) k(x, y)f (y)v(y) dy, x > 0, Kf (x) := w(x)

(2.1.4)

a(x)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S55

where v and w are locally integrable weight functions on (0, ∞), k(x, y) is an Oinarov-type (see Deﬁnitions 2.1 and 2.2 below) nonnegative kernel on R := {(x, y) : x > 0, a(x) < y < b(x)}, and the boundary functions a(x) and b(x) satisfy the following conditions: (i) a(x) and b(x) diﬀerentiable and strictly increase on (0, ∞); (ii) a(0) = b(0) = 0, a(x) < b(x) for 0 < x < ∞ and a(∞) = b(∞) = ∞.

(2.1.5)

We construct a block-diagonal representation of the operators speciﬁed above by using special sequences. We set ξ0 = η0 = ζ0 = 1 and deﬁne points ξk , ηk , and ζk for each k ∈ Z as follows (see Fig. 2.1.1): ξk+1 = a−1 (b(ξk )),

ξk−1 = b−1 (a(ξk )),

i.e., a(ξk+1 ) = b(ξk );

(2.1.6)

b(x) a(x)

ξk−1

ξk

x

ξk+1

Fig. 2.1.1.

if 0 < a(x) < x for all x > 0 (see Fig. 2.1.2), then we set ηk+1 = a−1 (ηk ),

(2.1.7)

ηk−1 = a(ηk );

x

a(x)

ηk−1

ηk

ηk+1

x

Fig. 2.1.2.

similarly, if x < b(x) for all x > 0 (see Fig. 2.1.3), then we set ζk+1 = b(ζk ),

ζk−1 = b−1 (ζk ).

(2.1.8)

Deﬁnition 2.1. We say that k(x, y) ≥ 0 satisﬁes the Oinarov-type condition Ob if there exists a constant D ≥ 1 not depending on the variables x, y, and z and such that D −1 k(x, y) ≤ k(x, b(z)) + k(z, y) ≤ Dk(x, y) for all

b−1 (y)

≤z≤x≤

(2.1.9)

a−1 (y).

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S56

PROKHOROV et al. b(x) x

ζk−1

ζk

x

ζk+1

Fig. 2.1.3.

Given boundary functions a and b satisfying conditions (2.1.5), we set ξ0 = 1 and deﬁne ξk for all k ∈ Z according to (2.1.6) (see Fig. 2.1.4). Breaking the half-axis (0, ∞) by the points {ξk }k∈Z into the intervals [ξk , ξk+1 ], we represent the operator K in the form K =T +S

(2.1.10)

of the sum of two block-diagonal operators T and S such that Tk and S= Sk , T = k∈Z

where

k∈Z

a(ξk+1 )

Tk f (x) = w(x)

k(x, y)f (y)v(y) dy,

(2.1.11)

Tk : Lp (a(ξk ), a(ξk+1 )) → Lq (ξk , ξk+1 ),

a(x) b(x)

Sk f (x) = w(x)

Sk : Lp (b(ξk ), b(ξk+1 )) → Lq (ξk , ξk+1 ).

k(x, y)f (y)v(y) dy, b(ξk )

The operators K, T , and S have nonnegative kernels; therefore, KLp (0,∞)→Lq (0,∞) ≈ T Lp (0,∞)→Lq (0,∞) + SLp (0,∞)→Lq (0,∞) .

6

b(x) a(x)

Sk

Tk

ξk

ξk+1

-

Fig. 2.1.4. PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

Since

S57

[a(ξk ), a(ξk+1 )) = [b(ξk ), b(ξk+1 )) = [ξk , ξk+1 ) = (0, ∞), k

k

k

we can estimate T Lp (0,∞)→Lq (0,∞) and SLp (0,∞)→Lq (0,∞) in terms of the norms Tk and Sk (see Lemma 2.1). Moreover, the kernels k(x, y) of the operators Tk and Sk satisfy condition (2.1.9) for z ≤ x and x ∈ [ξk , ξk+1 ] and for a(x) ≤ y ≤ a(ξk+1 ) and b(ξk ) ≤ y ≤ b(z), respectively. This also makes it possible to use some results of Section 1.5 to estimate the norms of the operators Tk and Sk , each of which has only one variable limit of integration. Theorem 2.1. If the kernel k(x, y) of an operator K of the form (2.1.4) satisﬁes the condition Ob and 1 0 A0 (t), A1 := supt>0 A1 (t), 1/q t q q sup k (x, b(s))w (x) dx A0 (t) := b−1 (a(t))≤s≤t

A1 (t) :=

s

sup

w

b−1 (a(t))≤s≤t

v a(t)

1/q

t

b(s)

b(s)

q

p

p

p

1/p ,

1/p

k (s, y)v (y) dy

s

.

a(t)

The operator K is compact if and only if A < ∞ and lim Ai (t) = lim Ai (t) = 0,

i = 0, 1.

t→∞

t→0

In the case 1 < q < p < ∞, KLp →Lq ≈ B :=

1/r r [Bk,1

+

r Bk,2

+

r Bk,3

+

r Bk,4 ]

,

k

where

r Bk,1

:=

r := Bk,2

a(ξk )

ξk

ξk+1 t

ξk

r/p wq

ξk

q

b(ξk )

ξk+1

ξk+1

:=

w ξk

r/q

v p (t) dt,

r/p

q

wq (t) dt,

t

k (x, t)w (x) dx r/p

p

kp (ξk , y)v p (y) dy r/q

q

b−1 (t)

a(ξk+1 )

v t

a(ξk+1 ) a(t)

b(ξk+1 ) ξk+1

:=

r Bk,4

q

k (x, b(ξk ))w (x) dx

r Bk,3

r/q

b(ξk ) a−1 (t)

v

p

r/q

v p (t) dt,

a(ξk+1 ) b(t)

q

p

p

k (t, y)v (y) dy

r/p wq (t) dt;

b(ξk )

t

here, ξ0 = 1 and the remaining ξk , k ∈ Z, are chosen by rule (2.1.6). Moreover, the operator K is compact if and only if B < ∞. Proof. We set K := KLp (0,∞)→Lq (0,∞) and Tk := Tk Lp (a(ξk ),a(ξk+1 ))→Lq (ξk ,ξk+1 ) ,

Sk := Sk Lp (b(ξk ),b(ξk+1 ))→Lq (ξk ,ξk+1 ) .

Suppose that 1 < p ≤ q < ∞. Relations (2.1.10)–(2.1.11) and Lemma 2.1 imply K ≈ sup Tk + sup Sk . k

(2.1.13)

k

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S58

PROKHOROV et al.

The norm of the operator Sk is characterized in Corollary 1.5 (a): 1/q ξk+1 q q k (x, b(s))w (x) dx Sk ≈ sup ξk ≤s≤ξk+1

+

s

sup

w

ξk ≤s≤ξk+1

v

p

1/p

b(ξk )

1/q

ξk+1

b(s)

b(s)

q

p

1/p

p

k (s, y)v (y) dy

.

b(ξk )

s

Using (2.1.9) and taking into account the relations z = ξk , x ∈ [ξk , ξk+1 ], and a(x) ≤ y ≤ a(ξk+1 ) = b(ξk ), we can decompose the kernel of the operator Tk as k(x, y) ≈ k(x, b(ξk )) + k(ξk , y). Therefore, Tk f (x) ≈ Tk,1 f (x) + Tk,2 f (x), where a(ξk+1 ) f (y)v(y) dy, Tk,1 f (x) = w(x)k(x, b(ξk ))

x ∈ [ξk , ξk+1 ],

a(x)

a(ξk+1 )

Tk,2 f (x) = w(x)

x ∈ [ξk , ξk+1 ].

k(ξk , y)f (y)v(y) dy, a(x)

Lemma 1.11 implies the following estimates of the norms of Tk,1 and Tk,2 : 1/q b(ξk ) 1/p t q q k (x, b(ξk ))w (x) dx vp Tk ≈ sup ξk ≤t≤ξk+1

+

ξk

t

sup

w

ξk ≤t≤ξk+1

Consider the functions

a(t)

1/q

b(ξk )

q

1/q

b(s)

q

k (x, b(s))w (x) dx

1/q

t

A1 (s, t) =

w

.

a(t)

q

s

1/p

p

k (ξk , y)v (y) dy

ξk

t

A0 (s, t) =

p

v a(t)

b(s)

q

p

p

p

1/p

1/p

k (s, y)v (y) dy

s

(2.1.14)

,

(2.1.15)

,

a(t)

where b−1 (a(t)) < s < t. We have Sk ≈ Tk ≈

sup

A0 (s, ξk+1 ) +

sup

A0 (ξk , t) +

ξk ≤s≤ξk+1 ξk ≤t≤ξk+1

sup

ξk ≤s≤ξk+1

sup

ξk ≤t≤ξk+1

A1 (s, ξk+1 ),

(2.1.16)

A1 (ξk , t),

(2.1.17)

and A0 = sup

sup

t>0 b−1 (a(t))
A0 (s, t),

A1 = sup

sup

t>0 b−1 (a(t))
A1 (s, t).

Moreover, sup

ξk ≤s≤ξk+1

sup

Ai (s, ξk+1 ) ≤

ξk ≤t≤ξk+1

Ai (ξk , t) ≤

sup

b−1 (a(ξk+1 ))
sup

b−1 (a(t))≤ξk ≤t

Ai (s, ξk+1 ) ≤ Ai ,

i = 0, 1,

Ai (ξk , t) ≤ Ai ,

i = 0, 1.

Using (2.1.13)–(2.1.17), we obtain K A0 + A1 . Let us prove the lower estimate K A0 + A1 . Suppose that K < ∞. If

f (y) = χ[a(t),b(s)] (y)v p −1 (y), PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S59

where b−1 (a(t) < s < x < t, then Kf (x) ≥ w(x)

b(s)

k(x, y)v p (y) dy. a(t)

Condition (2.1.9) with z = s implies k(x, y) k(x, b(s)). Therefore, b(s) v p (y) dy. Kf (x) w(x)k(x, b(s)) a(t)

This implies K

Kf q ≥ f p

1/q

t

b(s)

kq (x, b(s))wq (x) dx

1/p

v p (y) dy

s

,

a(t)

and hence K A0 (s, t) for all s < t < a−1 (b(s), which implies the required estimate K A0 . The inequality K A1 is proved in a similar way by substituting g(x) = χ[s,t] (x)wq−1 (x) into the inequality dual to Kf q ≤ Cf p . Let us prove the required estimates of the norm K in the case 1 < q < p < ∞. Note that, by Lemma 2.1, we have 1/r 1/r r r Tk + Sk . (2.1.18) K ≈ k

k

Lemma 1.13 and assertion (b) of Corollary 1.5 imply Tk ≈ Bk,1 + Bk,2 ,

Sk ≈ Bk,3 + Bk,4 ,

(2.1.19)

which proves the required estimates. The compactness criterion for the operator K : Lp (0, ∞) → Lq (0, ∞) in the case 1 < p ≤ q < ∞ is proved by using a representation of the given operator as the sum of a compact operator and operators ` theorem (see [16] with small norm. For 1 < q < p < ∞, the required assertion follows from Ando’s and [12]). An operator K of the form (2.1.4) can be regarded as a successor of, on the one hand, Volterratype operator (1.5.24) with variable upper limit and, on the other hand, operator (1.5.29) with variable lower limit. Theorem 2.1 is an analogue of Corollary 1.5 for a Volterra-type operator of the form (1.5.24) with Oinarov-type kernel k(x, y) ∈ (1.5.23). In conclusion of this section, we state a theorem for operator (2.1.4) with kernel k(x, y) satisfying condition (2.1.20), which is formally dual to (2.1.9). The counterpart of an operator with such a kernel and only one (lower) limit of integration is precisely transformation (1.5.29), which is characterized by Corollary 2. The proof of this theorem is similar to that of Theorem 2.1; it is based on Corollary 2 and Lemmas 1.10 and 1.12. Deﬁnition 2.2. We say that k(x, y) ≥ 0 satisﬁes an Oinarov-type condition Oa if there exists a D ≥ 1 not depending on x, y, and z and satisfying the condition D −1 k(x, y) ≤ k(x, a(z)) + k(z, y) ≤ Dk(x, y)

(2.1.20)

for all b−1 (y) ≤ x ≤ z ≤ a−1 (y). Theorem 2.2. Suppose that the kernel k(x, y) of an operator K of the form (2.1.4) satisﬁes condition Oa . If 1 0 A∗0 (s), A∗1 := sups>0 A∗1 (s) with t 1/q ∗ q q sup k (x, a(t))w (x) dx A0 (s) = s≤t≤a−1 (b(s))

A∗1 (s)

=

s

1/q

t

sup

w

s≤t≤a−1 (b(s))

b(s)

v

a(t)

b(s)

p

q

p

1/p

1/p

p

k (t, y)v (y) dy

s

, ;

a(t)

moreover, K is compact if and only if A∗ < ∞ and lim A∗i (s) = lim A∗i (s) = 0,

i = 0, 1.

s→∞

s→0

If 1 < q < p < ∞, then ∗

K Lp (0,∞)→Lq (0,∞)

1/r ∗ r ∗ r ∗ r ∗ r ≈ B := [(Bk,1 ) + (Bk,2 ) + (Bk,3 ) + (Bk,4 ) ] , ∗

k

where ∗ r ) (Bk,1 ∗ r ) (Bk,2 ∗ r ) (Bk,3 ∗ r ) (Bk,4

r/q

b(ξk ) a−1 (t)

=

q

q

b(ξk )

k (x, t)w (x) dx

a(ξk )

ξk

ξk+1 t

=

r/p w

ξk

ξk+1

=

w ξk

t

r/q

wq (t) dt, t

q

k (x, b(ξk ))w (x) dx r/p

v p (t) dt,

r/p

p

r/q

q

b−1 (t)

ξk+1

p

a(t)

b(ξk+1 ) ξk+1

a(ξk+1 )

t b(ξk )

k (t, y)v (y) dy

ξk

=

q

v

p

v

p

r/q

v p (t) dt,

a(ξk+1 ) b(t)

q

p

p

r/p

k (ξk+1 , y)v (y) dy

wq (t) dt

b(ξk )

for ξ0 = 1 and the ξk deﬁned by (2.1.6) for all k ∈ Z. Moreover, the operator K is compact if and only if B ∗ < ∞. Remark 2.1. Theorems 2.1 and 2.2 are dual to each other. Moreover, conditions (2.1.9) and (2.1.20) are generally independent, i.e., it may happen that one of them holds while the other does not. Accordingly, the criteria for the validity of (2.1.4) are determined by which of the conditions (2.1.9) and (2.1.20) on the kernel k(x, y) holds. As mentioned above, this is because (2.1.9) is an extension of (1.5.23) for the Volterra-type operator (1.5.24) with variable upper limit and (2.1.20) is the same condition for operator (1.5.29) with variable lower limit. Having lost the Volterra form, operator (2.1.4) may “forget” its origin, but its kernel “remembers” it.

2.2. Hardy–Steklov Operators Consider an integral transformation of the form b(x) f (y)v(y) dy Hf (x) := w(x)

(2.2.1)

(x > 0)

a(x)

with locally integrable weight functions v and w on (0, ∞) and limits a, b ∈ (2.1.5). This part of the paper is devoted to the study of the boundedness and compactness properties of Hardy–Steklov operators of the form (2.2.1) acting from Lp (0, ∞) to Lq (0, ∞). The corresponding criteria use so-called fairway – a curve balancing between graphs a(x) and b(x) and satisfying a certain balance condition, which makes it possible to join together the discrete portions obtained by applying the block-diagonal partition of operators of type (2.2.1). PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S61

2.2.1. Fairway-functions. Deﬁnition 2.3. Given boundary functions a(x) and b(x) satisfying conditions (2.1.5), a number p ∈ (1, ∞), and a weight function v(x) such that 0 < v(x) < ∞ almost everywhere on x ∈ (0, ∞) and v p (x) is locally integrable on (0, ∞), a fairway-function σ(x) is a function satisfying the conditions a(x) < σ(x) < b(x) and b(x) σ(x) v p (y) dy = v p (y) dy for all x ∈ (0, ∞). (2.2.2) a(x)

σ(x)

The fairway-function is strictly increasing, continuous, and diﬀerentiable on (0, ∞). Indeed, assume that σ(x1 ) ≥ σ(x2 ) for x1 < x2 . Then σ(x1 ) b(x1 ) b(x2 ) σ(x2 ) p p p v = v < v = vp a(x1 )

σ(x1 )

σ(x2 )

a(x2 )

by virtue of (2.2.2) and the conditions that b(x1 ) < b(x2 ) and 0 < v(x) < ∞ almost everywhere on (0, ∞), which contradicts our assumption, because a(x1 ) < a(x2 ). To prove the continuity of σ(x) on (0, ∞), note that, for σ(x) ∈ [a(z), b(z)], we have a(z) b(z) σ(z) p p v = v + vp , x = z, (2.2.3) 2 σ(x)

a(x)

b(x)

by virtue of (2.2.2) (see (2.3.3)–(2.3.9) for more details). This implies the required property, because a and b are continuous, v p is locally integrable, and 0 < v < ∞ almost everywhere on (0, ∞). The balance condition (2.2.2) implies also the diﬀerentiability of σ on (0, ∞); moreover,

v p (a(x))a (x) + v p (b(x))b (x) dx, dσ(x) = 2v p (σ(x))

x > 0.

For what follows, we need several technical lemmas. Lemma 2.2. Suppose that functions a(x) and b(x) satisfy conditions (2.1.5) and σ(x) is the corresponding fairway (see Fig. 2.2.1). Given c ∈ (0, ∞), let c− = σ −1 (a(c)), and let [N − ] denote the integer part of the number b(c) vp a(c) ; N − := log2 − b(c )

vp

a(c)

then

b(c)

vp 2[N

−]

≤

a(c) b(c− )

< 2[N v

− ]+1

.

p

a(c) b , where Consider the sequence {xj }jj=0

jb =

[N − ] [N − ] + 1

if N − = [N − ], if N − > [N − ]

deﬁned as follows: (1) x0 = c− and xjb = c; PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S62

PROKHOROV et al.

(2) if [N − ] = 0 or N − = [N − ] = 1, then jb = 1; (3) if N − > 1, then the points xj with 1 ≤ j ≤ jb − 1 = [N − ] are chosen so that b(xj−1 ) b(xj ) p v =2 vp . a(c)

(2.2.4)

a(c)

6

b(x) σ(x)

b(c)

a(x)

b(c− ) a(c) c− x1

x2

-

c Fig. 2.2.1.

Then

b(t)

p

v ≈

a(t)

b(xj+1 )

vp

(2.2.5)

a(xj )

for all t ∈ [xj , xj+1 ] and 0 ≤ j ≤ jb − 1, and b(t) p v < a(t)

b(c) a(c− )

p

v ≈

b(c)

vp

(2.2.6)

a(c)

for all t ∈ [c− , c]. Moreover, if jb ≥ 2 and 0 ≤ j ≤ jb − 2, then b(xj ) b(xj+l ) v p = 2l vp for all l ∈ {1, . . . , jb − j − 1}. a(c)

(2.2.7)

a(c)

Proof. For any t ∈ [xj , xj+1 ], where 0 ≤ j ≤ jb − 1, we have b(t) b(xj+1 ) b(xj ) p p v < v < vp . a(xj+1 )

a(t)

(2.2.8)

a(xj )

First, we prove (2.2.5) for t = xj . By the deﬁnition of xj+1 (see (2.2.4)), we have a(c) b(xj ) b(xj ) b(xj+1 ) p p p v ≤ v +2 v ≤3 vp . a(xj )

a(xj )

a(c)

Estimates (2.2.8) and (2.2.9) imply (2.2.5) for t = xj . If t = xj+1 , then b(xj+1 ) b(xj+1 ) b(xj ) b(xj+1 ) p p p p v ≥ v = v ≥ v + a(xj+1 )

a(c)

=

1 2

b(xj )

a(xj )

vp +

σ(c− )

b(xj+1 )

b(xj )

(2.2.9)

a(xj )

σ(xj )

v p (y) dy ≥

1 2

b(xj+1 )

vp

b(xj ) b(xj+1 )

vp .

(2.2.10)

a(xj )

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S63

These relations and (2.2.8) imply (2.2.5) for t = xj+1 . For t ∈ (xj , xj+1 ), we have b(xj ) b(t) 1 b(xj+1 ) p 1 b(xj+1 ) p p p v ≥ v ≥ v = v 2 a(c) 2 σ(c− ) a(t) a(c) 1 b(xj+1 ) p 1 b(xj+1 ) p (2.2.10) 1 b(xj+1 ) p ≥ v = v ≥ v . 2 σ(xj+1 ) 4 a(xj+1 ) 8 a(xj ) These relations and (2.2.8) give (2.2.5) for all t ∈ [xj , xj+1 ].

b(c− ) a(c) The proof of the inequality in (2.2.6) is elementary. Since a(c) v p = a(c− ) v p , it follows that b(c) b(c− ) b(c) b(c) p p p v = v + v <2 vp , (2.2.11) a(c− )

a(c)

a(c)

a(c)

which proves (2.2.6). Relation (2.2.7) follows from (2.2.4). The following lemma is proved in a similar way. Lemma 2.3. Suppose that functions a(x) and b(x) satisfy conditions (2.1.5) and σ(x) is the corresponding fairway (see Fig. 2.2.2). Given c ∈ (0, ∞), let c+ = σ −1 (b(c)), and let [N + ] denote the integer part of the number b(c) vp a(c) ; N + := log2 b(c) vp

a(c+ )

then

b(c)

vp 2[N

+]

≤

a(c) b(c)

< 2[N v

+ ]+1

.

p

a(c+ ) a , where Consider the sequence {xj }jj=0 [N + ] if N + = [N + ], ja = [N + ] + 1 if N + > [N + ],

deﬁned as follows: (1) x0 = c and xja = c+ ; (2) if [N + ] = 0 or N + = [N + ] = 1, then ja = 1; (3) if N + > 1, then the points xj with 1 ≤ j ≤ ja − 1 = [N + ] are chosen so that b(c) 1 b(c) p v = vp . 2 a(xj−1 ) a(xj ) Then (2.2.5) holds for all t ∈ [xj , xj+1 ] with 0 ≤ j ≤ ja − 1 and b(c+ ) b(c) b(t) p p v ≤ v ≈ vp a(t)

a(c)

(2.2.13)

a(c)

for all t ∈ [c, c+ ]. Moreover, if ja ≥ 2 and 0 ≤ j ≤ ja − 2, then b(c) b(c) l p v = vp for all l ∈ {1, . . . , ja − j − 1}. 2 a(xj+l )

(2.2.12)

(2.2.14)

a(xj )

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S64

PROKHOROV et al. 6

b(x) σ(x) a(x)

b(c) a(c+ )

a(c) c

x1

x2 c+

-

Fig. 2.2.2.

Similar assertions are valid for partitions of a type somewhat diﬀerent from that used in Lemmas 2.2 and 2.3. Lemma 2.4. Suppose that functions a(x) and b(x) satisfy conditions (2.1.5) and σ(x) is the corresponding fairway-function. Given d ∈ (0, ∞), let d− = σ −1 (a(d)) and d+ = σ −1 (b(d)). Consider the sequence {xj }0j=−ja deﬁned as follows: (1) x−ja = d and x0 = d+ ; (2) if (d+ )− ≤ d, then ja = 1; (3) if (d+ )− > d, then ja > 1 and xj−1 = (xj )− , where (xj )− > d and −ja + 2 ≤ j ≤ 0. Then (2.2.5) holds for any t ∈ [xj , xj+1 ] with −ja ≤ j ≤ −1. Moreover, if d ≤ x− ≤ t ≤ x ≤ d+ , then b(x) b(x) b(t) p p v ≈ v ≈ vp . (2.2.15) a(t)

a(x− )

a(x)

b(x− ) a(x) Proof. First, we prove (2.2.15). Since a(x− ) v p = a(x) v p and d ≤ x− ≤ t ≤ x ≤ d+ , it follows that b(x) b(x− ) b(x) b(x) b(t) p p p p v ≤ v = v + v <2 vp , (2.2.16) a(t)

a(x− )

a(x)

a(x)

a(x)

which implies the second equivalence in (2.2.15). Next, by virtue of the assumptions of the lemma and the relations d ≤ x− ≤ t ≤ x ≤ d+ , we have b(d) σ(d+ ) σ(x) b(t) 1 b(x) p (2.2.16) 1 b(x) p vp ≥ vp = vp ≥ vp = v < v ; (2.2.17) 2 a(x) 4 a(x− ) a(t) a(x) a(x) a(x) this, together with (2.2.16), gives (2.2.15). Since d ≤ xj = (xj+1 )− ≤ t ≤ xj+1 ≤ d+ , equivalences (2.2.15) imply (2.2.5). Lemma 2.5. Suppose that functions a(x) and b(x) satisfy conditions (2.1.5) and σ(x) is the corresponding fairway. Given d ∈ (0, ∞), let d− = σ −1 (a(d)) and d+ = σ −1 (b(d)). Consider the b of points deﬁned as follows: sequence {xj }jj=0 PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

6

S65

b(x) σ(x) a(x)

b(d)

a(d) d

x−3

-

x−2 x−1 d+

Fig. 2.2.3.

(1) x0 = d− and xjb = d; (2) if (d− )+ ≥ d, then jb = 1; (3) if (d− )+ < d, then jb > 1 and xj+1 = (xj )+ , where (xj )+ < d and 0 ≤ j ≤ jb − 2. 6

b(x) σ(x) a(x)

b(d)

a(d) d− x1

-

x2 d Fig. 2.2.4.

Then (2.2.5) holds for any t ∈ [xj , xj+1 ], where 0 ≤ j ≤ jb − 1. Moreover, if d− ≤ x ≤ t ≤ x+ ≤ d, then b(x+ ) b(x) b(t) vp ≈ vp ≈ vp . (2.2.18) a(t)

a(x)

a(x)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S66

PROKHOROV et al.

2.2.2. Criteria of the Muckenhoupt and Maz’ya–Rosin type. In this subsection, we present complete analogues of criteria (1.5.12)–(1.5.15) for operator (2.2.1). The obtained results are based on the notion of a fairway-function σ (see Deﬁnition 2.3). b(x) σ(x)

6

a(x) 6 Δ(x) ?

δ(x) -

-

x Fig. 2.2.5.

We set δ(x) = [b−1 (σ(x)), a−1 (σ(x))],

Δ(x) = [a(x), b(x)],

where a−1 (y) and b−1 (y) denote the functions inverse to y = a(x) and y = b(x), respectively. Theorem 2.3. Suppose that the operator H has the form (2.2.1). If 1 0

t>0

(2.2.19)

1/q w

q

v

δ(t)

p

1/p ;

(2.2.20)

Δ(t)

moreover, H : Lp (0, ∞) → Lq (0, ∞) is compact if and only if AM < ∞ and lim AM (t) = lim AM (t) = 0. t→∞

t→0

If 0 < q 1, then HLp (0,∞)→Lq (0,∞) ≈ BM R , where

BM R :=

∞

r/p w

0

q

v

δ(t)

p

(2.2.21) 1/r

r/p q

(2.2.22)

w (t)dt

Δ(t)

and 1/r = 1/q − 1/p; moreover, H is compact if and only if BM R < ∞. Proof. Consider the case 1 0 b−1 (a(t))≤s≤t

where

1/q

t

A(s, t) =

w s

b(s)

q

v

p

A(s, t),

(2.2.23)

1/p .

a(t)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

Using (2.2.2), we obtain AqM (t)

q/p

=2

t

w

b−1 (σ(t))

σ(t)

q

v

p

q/p

+

a(t)

a−1 (σ(t))

w

S67

b(t)

q

v

t

p

q/p

σ(t)

= 2q/p [Aq (b−1 (σ(t)), t) + Aq (t, a−1 (σ(t)))] ≤ 21+q/p Aq for all t > 0. This implies AM A. To prove the reverse inequality, we set τ = σ −1 (a(t)). We have A ≤ sup

sup

t>0 b−1 (a(t))≤s≤τ

A(s, t) + sup sup A(s, t) AM . t>0 τ ≤s≤t

Indeed, if b−1 (a(t)) ≤ s ≤ τ ≤ t, then (s, t) ⊂ (b−1 (a(t)), t) = (b−1 (σ(τ )), a−1 (σ(τ ))),

(a(t), b(s)) ⊂ (σ(τ ), b(τ )).

Therefore, sup

sup

t>0 b−1 (a(t))≤s≤τ

A(s, t) ≤ AM .

If τ ≤ s ≤ t, then (s, t) ⊂ (b−1 (σ(s)), a−1 (σ(s))) and (a(t), b(s)) ⊂ (a(s), b(s)), and we obtain sup sup A(s, t) ≤ AM . t>0 τ ≤s≤t

This implies AM ≈ A, and the required estimate (2.2.19) follows from (2.2.23). The compactness criterion for the operator H follows from Theorem 2.1. Consider the case 0 < q 1. Let us prove (2.2.21). First, we show that H BM R . We set ξ0 = 1 and deﬁne a sequence {ξk }k∈Z by (2.1.6). We also set ξk− = σ −1 (a(ξk )), + Δk = [ξk− , ξk+ ] = Δ− k ∪ Δk ,

ξk+ = σ −1 (b(ξk )), − Δ− k = [ξk , ξk ],

+ Δ+ k = [ξk , ξk ].

For x ∈ Δk , the operator H decomposes into the sum of four operators as x ∈ Δk ,

Hf (x) = Tk,1 f (x) + Tk,2 f (x) + Sk,1 f (x) + Sk,2 f (x), where

a(ξk )

Tk,1 f (x) = w(x)

f v, x ∈

a(x)

x ∈ Δ+ k,

f v, b(ξk )

b(ξk )

Tk,2 f (x) = w(x)

b(x)

Sk,1 f (x) = w(x)

Δ− k,

a(x)

f v, x ∈ Δ+ k,

b(x)

Sk,2 f (x) = w(x)

x ∈ Δ− k.

f v, a(ξk )

Applying estimate (1.5.49), we obtain r/p a(ξk ) r/p ξk t r q w vp wq (t) dt Tk,1 ≈ ξk−

q ≤ r

ξk−

a(t)

σ(ξk− ) a(ξk− )

v

p

r/p

r/q

ξk ξk−

w

q

(2.2.2)

=

b(ξk− ) σ(ξk− )

v

p

r/p

ξk ξk

ξk−

r/p w

q

q

w (t) dt .

t

For t ∈ [ξk− , ξk ], we have a(t) ≤ σ(ξk− ), b(ξk− ) ≤ b(t), and ξk ≤ a−1 (σ(t)). This implies the inclusions [σ(ξk− ), b(ξk− )] ⊆ Δ(t) and [t, ξk ] ⊂ [t, a−1 (σ(t))] ⊂ δ(t). Therefore, r/p r/p ξk r q p w v wq (t) dt =: Jk,1 . (2.2.24) Tk,1 ξk−

δ(t)

Again applying (1.5.49), we see that Tk,2 r ≈

Δ(t)

ξk+ t

ξk

r/p w

ξk

q

b(ξk )

v

p

r/p wq (t) dt.

a(t)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S68

PROKHOROV et al.

b(x) σ(x) a(x)

6 Sk,1 b(ξk )

Tk,2

Sk,2

a(ξk ) Tk,1 ξk−

- ξk

Δ− k

- ξ+ k

Δ+ k

-

Fig. 2.2.6.

Since ξk ≥ b−1 (σ(t)) and b(ξk ) ≤ b(t) for t ∈ [ξk , ξk+ ], it follows that [ξk , t] ⊂ (b−1 (σ(t)), t) ⊂ δ(t), Therefore, Tk,2 r

ξk+

r/p w

ξk

[a(t), b(ξk )] ⊂ Δ(t).

q

v

δ(t)

p

r/p wq (t) dt =: Jk,2 .

(2.2.25)

Δ(t)

Similarly, using (1.5.41), we arrive at Sk,1 r Jk,2 ,

Sk,2 r Jk,1 . (2.2.26) The block-diagonal structure of the operators Ti := k∈Z Tk,i , i = 1, 2, and Si := k∈Z Sk,i , i = 1, 2, together with Lemma 2.1 and estimates (2.2.24)–(2.2.26), implies the required inequality H BM R . Let us prove the inequality H BM R . First, we suppose that (2.2.27)

σ(x) = x, i.e., the fairway is the bisector of the ﬁrst quadrant. In this case, we set x− = a(x),

x+ = b(x),

Δ(x) = [x− , x+ ] = Δ− (x) ∪ Δ+ (x),

Δ− (x) = [x− , x],

Δ+ (x) = [x, x+ ]

and refer to the intervals of the form [x− , x+ ] as the base intervals. We need the following deﬁnition. Deﬁnition 2.4. Suppose that p ∈ (1, ∞); functions a(x), b(x), and v(x) satisfy the conditions in Deﬁnition 2.3; and (2.2.27) holds. By L we denote the set of all absolutely continuous functions F on (0, ∞) with F /vp < ∞ for which there exist pairwise disjoint intervals Ik = (αk , βk ) ⊂ (0, ∞) and + base intervals Jk = [c− k Ik , and F (αk ) = F (βk ) = 0 for all k ∈ Z. k , ck ] such that Ik ⊂ Jk , suppF ⊂ Consider the inequality F wq ≤ C

F v

,

F ∈ L,

(2.2.28)

p

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S69

with a constant C ≥ 0 not depending on F ∈ L. Let us show that the least possible constant C satisﬁes the condition C ≤ H.

(2.2.29)

For this purpose, we take any function F ∈ L and write ∞ |F (x)|q wq (x) dx = |F (x)|q wq (x) dx. 0

k

Ik

+ Since Ik = (αk , βk ) ⊂ [c− k , ck ] = Jk by the deﬁnition of L, it follows that there are only three possible cases:

(i) βk ≤ ck ; (ii) ck ≤ αk ; (iii) ck ∈ Ik . In case (i), we have |F (x)|q wq (x) dx ≤ Ik

x

q |F | wq (x) dx ≤

αk

Ik

Ik

q |F | wq (x) dx Δ(x)

by virtue of the inclusions (αk , x) ⊂ (a(x), x) ⊂ Δ(x), which are implied by αk ≥ c− k = a(ck ) ≥ a(βk ) ≥ a(x). Similarly, in case (ii), we have q q |F (x)| w (x) dx ≤ Ik

Ik

βk

q q |F | w (x) dx ≤

q |F | wq (x) dx

x

Ik

Δ(x)

by virtue of the inclusions (x, βk ) ⊂ (x, b(x)) ⊂ Δ(x), which are implied by the chain of inequalities βk ≤ c+ k = b(ck ) ≤ b(αk ) ≤ b(x). Finally, in case (iii), it follows from the arguments used in cases (i) and (ii) that q q c k x β k β k |F (x)|q wq (x) dx ≤ |F | wq (x) dx + |F | wq (x) dx Ik

α

αk

Ik

Δ(x)

k

q |F | wq (x) dx.

ck

x

Therefore, F ∈ L, we have ∞ F q F q q |F (x)| w (x) dx H (x) dx ≤ H v v 0 Ik k

q

≤ H

q

q

F v

q

, p

which proves (2.2.29). Now, to prove the theorem under condition (2.2.27), it suﬃces to show that BM R C.

(2.2.30)

± In turn, to show this, it suﬃces to prove the relation BM R C, where ∞ 1/r

r/p

r/p ± q p q w v w (t) dt , BM R = 0

δ± (t)

Δ(t)

+ − δ− (t) = [b−1 (t), t], and δ+ (t) = [t, a−1 (t)]. We give the proof of BM R C; the proof of BM R C is similar.

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S70

PROKHOROV et al.

b(x) σ(x) = x a(x)

6

ηk−1

ηk

ηk+1

-

∗ ηk+2 ηk+2

Fig. 2.2.7.

Let η0 = 1, and let {ηk }k∈Z ⊂ (0, ∞) be the sequence deﬁned by (2.1.7). We set ηk∗ := min{ηk+ , ηk+1 }. ∗ and let For each k ∈ Z, consider the ﬁve neighboring points ηk−1 , ηk , ηk+1 , ηk+2 , and ηk+2 r/(pq ) t ηk+2 r/(pq) s q p w v v p (s) ds. gk (t) = χ[ηk−1 ,ηk+2 ] (t) ηk−1

ηk−1

s

∗ ], we deﬁne the function On the interval [ηk−1 , ηk+2 gk (t), hk (t) = gk (ηk+2 )Ωk+2 (t),

where

ηl∗

Ωl (t) =

v

p

∗ ], t∈ / [ηk+2 , ηk+2 ∗ ], t ∈ [ηk+2 , ηk+2

−1

ηl

We set

ηk+2 ηk+2

λk := ηk−1

v p (s) ds.

t

r/q w

ηl∗

s

q

v

p

r/q

v p (s) ds.

(2.2.31)

ηk−1

s

Let us construct a representation hk (t) =

3

(2.2.32)

hk,i (t),

i=1

where hk,i ∈ L, i = 1, 2, 3. To this end, we deﬁne hk,1 by ⎧ ⎪ t∈ / [ηk−1 , ηk∗ ], ⎨0, hk,1 (t) = hk (t), t ∈ [ηk−1 , ηk ], ⎪ ⎩ gk (ηk )Ωk (t), t ∈ [ηk , ηk∗ ]. Since ηk− = ηk−1 , it follows that supp hk,1 ⊆ [ηk− , ηk∗ ] ⊆ [ηk− , ηk+ ]; therefore, hk,1 ∈ L. Next, we set (1)

hk (t) := hk (t) − hk,1 (t) PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

and

S71

⎧ ∗ ], ⎪ t∈ / [ηk , ηk+1 ⎨0, hk,2 (t) = h(1) t ∈ [ηk , ηk+1 ], k (t), ⎪ ⎩ (1) ∗ ]. hk (ηk+1 )Ωk+1 (t), t ∈ [ηk+1 , ηk+1

Obviously, hk,2 ∈ L. Finally, for the function (1)

hk,3 (t) = hk (t) − hk,2 (t) we have hk,3 ∈ L, and decomposition (2.2.32) holds. We also need the estimates hk,i

p

λk ,

v Let us write κ1 :=

hk,1 v

p

ηk

|hk (s)|p v −p (s) ds

= p

ηk−1

(2.2.33)

i = 1, 2, 3.

p

+

gkp (ηk )

ηk∗

|Ωk (s)|p v −p (s) ds =: κ1,1 + κ1,2 .

ηk

Obviously,

ηk

κ1,1 = ¨ Holder’s inequality implies p gk (ηk ) ≤

w ηk−1

s

ηk+2

ηk

v

r/q

ηk∗ ηk

s

q

p

r/q

v p (s) ds ≤ λk .

v

p

r/q

p

|Ωk (s)|p v −p (s) ds

it follows that

κ1,2 ≤ λk

ηk ηk−

v

p

v

ηk−1+j

p

p−1 .

ηk−1

ηk∗

=

v

p

1−p ,

ηk

p−1

ηk∗

v

p

1−p .

ηk

If ηk∗ = ηk+ , then κ1,2 ≤ λk by virtue of (2.2.2) and (2.2.27). Note that η+ η+ ηk−1+j k−1+j k+j vp = vp ≤ vp = 2 − ηk−1+j

ηk

v (s) ds

ηk−1

s

s

q ηk−1

w

ηk−1

Since

r/q

ηk+2

− ηk+j

ηk+j

vp

(2.2.34)

ηk−1+j

for j ≥ 0. For j = 1, this implies κ1,2 ≤ 2p−1 λk in the case ηk∗ = ηk+1 . Therefore, (2.2.33) holds for (1)

i = 1. Since hk (ηk+1 ) = hk (ηk+1 ), it follows that, for i = 2 in (2.2.33), we can write ηk+1 η∗ hk,2 p k+1 = |hk (s) − hk,1 (s)|p v −p (s) ds + gkp (ηk+1 ) |Ωk+1 (s)|p v −p (s) ds κ2 := v p ηk ηk+1 η∗ ηk+1 k |hk (s)|p v −p (s) ds + gkp (ηk ) |Ωk (s)|p v −p (s) ds ≤ ηk

+ gkp (ηk+1 )

ηk

∗ ηk+1

|Ωk+1 (s)|p v −p (s) ds

ηk+1

=: κ2,1 + κ2,2 + κ2,3 . PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S72

PROKHOROV et al.

We have κ2,2 = κ1,2 λk ; it can be shown by analogy with the case of κ1,i , i = 1, 2, that κ2,i λk for i = 1, 3. Next, the relations ηk+2 r/(pq) t s r/(pq ) wq vp v p (s) ds gk (t) ≥ ηk−1

t

≈

r/(pq)

ηk+2

w

ηk−1

t

q

v

p

r/(p q) t ∈ [ηk−1 , ηk+2 ],

,

ηk−1

t

imply the lower estimate ηk+2 q q q |gk (t)| w (t) dt hk wq ≥ ηk−1

ηk+2 ηk+2

r/p w

ηk−1

t

q

v

p

r/p wq (t) dt;

ηk−1

t

integrating by parts, we obtain hk wqq λk . Consider the functions hi =

hk,i =

|k|≤N

h2k,i +

|k|≤N

(2.2.35)

h2k+1,i = : F1,i + F2,i ,

|k|≤N

where N ∈ N. The supports of h2k,i , k ∈ Z, with ﬁxed i = 1, 2, 3 are pairwise disjoint. Therefore, F1,i ∈ L; for the same reason, F2,i ∈ L. Note that χsupphk (x) ≤ 4, x∈ supp hk ; 1≤ |k|≤N

|k|≤N

hence, setting

ΛN :=

λk

|k|≤N

and using (2.2.32)–(2.2.35) and (2.2.28), we obtain 1/q ΛN

≤

hk w q

|k|≤N

3 2

Fj,i wq ≤ C

i=1 j=1

3 2 Fj,i i=1 j=1

1/r

Therefore, C ΛN . Letting N → ∞, we arrive at C ηk+1 ηk+2 r/p wq λk ηk

≥

ηk+1

ηk

v

s

r/p wq

δ+ (s)

C

p

v

p

p 1/p

1/p

CΛN . p

where

r/p wq (s) ds

ηk−1

Δ− (s)

v

|k|≤N,i

k∈Z λk ,

s

hk,i

vp

r/p wq (s) ds.

+ Now, the required estimate C BM R follows from (2.2.2) and (2.2.27). − The inequality C BM R is proved by performing a similar construction for the intervals determined by the sequence (2.1.8). Thus, estimate (2.2.30) does hold; taking into account (2.2.29), we see that so does the inequality H BM R in the case σ(x) = x. Let us get rid of constraint (2.2.27). Consider the functions

f(t) = f (σ(t))[σ (t)]1/p ,

a(x) = σ −1 (a(x)),

b(x) = σ −1 (b(x)).

(2.2.36)

Making a change of variables on both sides of the inequality Hf q ≤ Hf p , PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

(2.2.37) Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

we obtain an inequality of the form ∞ q w (x)

b(x)

a(x)

0

q 1/q ≤ H f v dx

∞

|f|p

S73

1/p (2.2.38)

0

with v(y) = v(σ(y))[σ (y)]1/p . It is easy to see that σ(x) x p p v = v = a(x)

a(x)

b(x)

p

b(x)

v =

vp ;

(2.2.39)

x

σ(x)

therefore, the fairway for a, b, and v is σ (x) = x. It follows from the above considerations that BM R H, where r/p r/p ∞ r q p w v wq (t) dt. BM R = δ(t)

0

Δ(t)

The relations

and

Δ(t)

vp =

: = [b−1 ( δ(t) σ (t)), a−1 ( σ (t))] = [b−1 (σ(t)), a−1 (σ(t))] = δ(t) Δ(t)

v p imply BM R = BM R .

For q < p, the compactness theorem is an immediate corollary of the boundedness criterion obtained ` theorem. above and Ando’s 2.2.3. Criteria of the Tomaselli and Persson–Stepanov type. In this subsection, we give characteristics for the Hardy–Steklov operator (2.2.1) which are alternative to (1.5.16)–(1.5.19). b(x)

6

σ(x) a(x)

6 Δ(x) ?

θ(x) -

-

x Fig. 2.2.8.

By σ −1 (y) we denote the function inverse to the fairway σ(x). We set θ(t) = (σ −1 (a(t)), σ −1 (b(t))).

Δ(t) = (a(t), b(t)),

Theorem 2.4. Suppose that the assumptions of Theorem 2.3 hold. If 1 0

v θ(t)

p

(2.2.40)

1/q

q q

w (x) dx

Δ(x)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

v

p

−1/p .

Δ(t)

Vol. 300 Suppl. 2

2018

S74

PROKHOROV et al.

If 0 < q 1, then HLp (0,∞)→Lq (0,∞) ≈ BP S , where 1/r = 1/q − 1/p and ∞ BP S := 0

v θ(t)

p

(2.2.41)

r/p

q q

w (x) dx

v

Δ(x)

p

1/r

q−r/p q

w (t) dt

.

Δ(t)

Proof. An upper bound for HLp (0,∞)→Lq (0,∞) . Given boundary functions a(x) and b(x) satisfying conditions (2.1.5), let {ξk }k∈Z ⊂ (0, ∞) be the sequence of points deﬁned by (2.1.6) (with ξ0 = 1). We set ξk+ = σ −1 (b(ξk )), ξk− = σ −1 (a(ξk )), + Δk = [ξk− , ξk+ ] = Δ− k ∪ Δk ,

− Δ− k = [ξk , ξk ],

By virtue of (2.2.2), we have b(t) 1 p p v ≤ v ≤ vp , 2 Δ(t) a(ξk ) Δ(t) b(ξk ) 1 vp ≤ vp ≤ vp , 2 Δ(t) a(t) Δ(t)

+ Δ+ k = [ξk , ξk ].

if t ∈ Δ− k,

(2.2.42)

if t ∈ Δ+ k.

(2.2.43)

As in the proof of Theorem 2.3, we represent the initial operator H as the sum of four sequences of block-diagonal operators: Tk,1 f (x) + Tk,2 f (x) + Sk,1 f (x) + Sk,2 f (x). (2.2.44) Hf (x) = k∈Z

We have Hf qq =

k∈Z

≈

k∈Z

Hf qLq (Δk )

k

k∈Z

Tk,1 f qLq (Δ− ) + k

k∈Z

Sk,2 f qLq (Δ− ) + k

k∈Z

Tk,2 f qLq (Δ+ ) + k

k∈Z

Sk,1 f qLq (Δ+ ) .

k∈Z

k

(2.2.45) First, consider the trapezoidal operators Sk,2 and Tk,2 . In the case 1 < p ≤ q < ∞, using estimate (1.5.34), we derive the following relation from Lemma 1.10: b(t) −q/p t b(x) q q p q p v w (x) dx v . Sk,2 ≈ sup ξk−

t∈Δ− k

a(ξk )

a(ξk )

We have (ξk− , t) ⊂ [σ −1 (a(t)), t] ⊂ θ(t) for each t ∈ [ξk− , ξk ] and (a(ξk ), b(x)) ⊂ Δ(x) for each x ∈ (ξk− , t) ⊂ (ξk− , ξk ). Therefore, (2.2.42) implies

q −q/p q p q p v w (x) dx v ≤ AqT . (2.2.46) Sk,2 sup t∈Δ− k

θ(t)

Δ(x)

Δ(t)

Similarly, using estimate (1.5.40) of Lemma 1.11 and relation (2.2.43), we obtain

q −q/p v p wq (x) dx vp AqT . Tk,2 q sup t∈Δ+ k

θ(t)

Δ(x)

(2.2.47)

Δ(t)

If 0 < q 1, then it follows from Lemma 1.12 with (1.5.43) taken into account that r/p b(t) q−r/p ξk t b(x) q r p q p v w (x) dx v wq (t) dt. Sk,2 ≈ ξk−

ξk−

a(ξk )

a(ξk )

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S75

We have (ξk− , t) ⊂ θ(t) and (a(ξk ), b(t)) ⊂ Δ(t) for any t ∈ [ξk− , ξk ] and (a(ξk ), b(x)) ⊂ Δ(x) for any x ∈ (ξk− , t) ⊂ (ξk− , ξk ). Moreover, a(ξk ) ≤ σ(t) for t ∈ [ξk− , ξk ]. Therefore, taking into account (2.2.42), we obtain r/p

q q b(t) −r/p ξk r p q p v w (x)dx v vp wq (t) dt Sk,2 ≤ ξk−

θ(t)

Δ(x)

σ(t)

Δ(t)

! r BΔ− ,

(2.2.2)

(2.2.48)

k

where !

BΔ−

r

:=

ξk

v

ξk−

k

θ(t)

p

r/p

q q

w (x) dx

v

Δ(x)

p

q−r/p wq (t) dt.

Δ(t)

Similarly, applying (1.5.51) in Lemma 1.13 and estimate (2.2.43), we ﬁnd r/p

q q−r/p ξ + k ! r r p q p Tk,2 BΔ+ := v w (x) dx v wq (t) dt. k

ξk

θ(t)

Δ(x)

(2.2.49)

Δ(t)

To estimate the norms Tk,1 and Sk,1 of the corresponding triangular operators Tk,1 and Sk,1 , we construct sequences {ti }, 0 ≤ i ≤ ib (k) − 1, and {sj }, 0 ≤ j ≤ ja (k) − 1, as in Lemmas 2.2 and 2.3 for + c = ξk . The operators Tk,1 f (x), x ∈ Δ− k , and Sk,1 f (x), x ∈ Δk , decompose into sums as

ib (k)−1

Tk,1 f (x) =

ib (k)−1 (i)

Tk,1 f (x) :=

i=0

i=0

ja (k)−1

ja (k)−1

Sk,1 f (x) =

[Tk,1 f (x)χ(ti ,ti+1 ) (x)],

(j)

Sk,1 f (x) :=

j=0

[Sk,1 f (x)χ(sj ,sj+1 ) (x)].

j=0

In the case 1 < p ≤ q < ∞, applying equivalence (1.5.39) of Lemma 1.11, we derive the following (i) estimate for Tk,1 , 0 ≤ i ≤ ib − 1: t a(ξk ) q/p t σ(ξ − ) q/p k (i) q q p w v ≤ sup wq vp Tk,1 ≈ sup ti ≤t≤ti+1

(2.2.2)

ti+1

=

ti

wq

a(t)

b(ξk− )

vp

q/p

ti ≤t≤ti+1

(3.2.16)

=

2−i q/p

a(ξk )

ti

a(ξk− )

ti

ti+1

b(ti )

wq

vp

q/p .

a(ξk )

ti

Since q/p = q − q/p and (ti , t) ⊆ (ξk− , t) ⊂ θ(t) for t ∈ [ξk− , ξk ], it follows from (2.2.4) that ti+1 b(ti ) q b(ti+1 ) −q/p (2.2.4) (i) wq vp vp 2i q/p Tk,1 q 2q/p ti

t

sup

ti ≤t≤ti+1

(2.2.42)

a(ξk )

b(x)

v ti

p

sup

v θ(t)

b(t)

q

w (x) dx

a(ξk )

ti ≤t≤ti+1

a(ξk )

q

p

v

p

−q/p

a(ξk )

q q

w (x) dx

Δ(x)

v

p

−q/p

Δ(t)

≤ AqT .

(2.2.50)

Therefore, by virtue of p ≤ q, we have Tk,1 f qLq (Δ− ) k

=

i b −1

(i) Tk,1 f qLq (ti ,ti+1 )

≤

i=0

(2.2.50)

AqT f qLp (a(ξ − ),a(ξ k

i b −1

i=0 −1 i b k ))

(i)

Tk,1 qLp (a(ti ),a(ξk ))→Lq (ti ,ti+1 ) f qLp (a(ti ),a(ξk ))

2−i q/p AqT f qLp (a(ξ − ),a(ξ

i=0

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

k

k ))

(2.2.51)

.

Vol. 300 Suppl. 2

2018

S76

PROKHOROV et al.

Similarly, using equivalence (1.5.33) of Lemma 1.10 and relations (2.2.14), (2.2.12), and (2.2.43), we obtain Sk,1 f qLq (Δ+ ) AqT f qLp (b(ξ k

+ k ),b(ξk ))

(2.2.52)

.

In the case 0 < q 1, applying equivalence (1.5.49) of Lemma 1.13, we obtain the following (i) estimate for Tk,1 , 0 ≤ i ≤ ib − 1: ti+1 t r/p a(ξk ) r/p (i) r Tk,1 ≈ wq vp wq (t) dt ti

ti

≤

σ(ξk− )

v

(2.2.2)

p

a(ξk− ) b(ξk− )

v −i r/p

p

q

ti

r/p

ti+1

r/p

t

w

b(ti )

= 2

v

p

wq (t) dt

q

wq (t) dt

ti

ti

r/p w

a(ξk ) (2.2.7)

ti+1 t

ti

=

a(t)

r/p

r/p

ti+1 t

r/p w

a(ξk )

q

wq (t) dt.

ti

ti

The relation r/p = q · r/p + q − r/p allows us to write b(ti ) q·r/p+q−r/p ti+1 t r/p (i) r i r/p Tk,1 vp wq wq (t) dt 2 a(ξk )

(2.2.4)

b(ti )

r/p

= 2

≤ 2r/p (2.2.42)

≈

v

p

q·r/p+q

a(ξk )

ti

ti

a(ξk )

ti+1 t

v ti

(ξk− , t)

p

vp

q

p

a(ξk )

−r/p

⊂ θ(t) for t ∈ ti+1 ! (i) r (i) r i r/p Tk,1 BΔ− = 2 ti

b(t)

vp

r/p

wq (t) dt

ti

ti

r/p

wq (x) dx

q

r/p q

q−r/p wq (t) dt

a(ξk )

q

w (x) dx

v

p

q−r/p wq (t) dt.

Δ(t)

θ(t)

ti+1 t

w

Δ(x)

[ξk− , ξk ],

k

b(ti+1 )

v

ti+1 t b(x)

ti

Since (ti , t) ⊆

ti

ti

it follows that r/p

q p q v w (x) dx

Δ(x)

v

p

q−r/p wq (t) dt. (2.2.53)

Δ(t)

¨ Therefore, Holder’s inequality with exponents r/q and p/q yields Tk,1 f qLq (Δ− ) k

=

i b −1

(i) Tk,1 f qLq (ti ,ti+1 )

≤

i=0

(2.2.53)

i b −1

(i)

Tk,1 qLp (a(ti ),a(ξk ))→Lq (ti ,ti+1 ) f qLp (a(ti ),a(ξk ))

i=0

f qLp (a(ξ − ),a(ξ k

≤ f qLp (a(ξ − ),a(ξ k

i b −1 k ))

i b −1 ! k ))

k

i=0

(i)

BΔ− k

i=0

! q BΔ− f qLp (a(ξ − ),a(ξ k

! (i) q 2−i q/p BΔ−

k

k ))

r

q/r i b −1

2−i p/p

q/p

i=0

(2.2.54)

.

In a similar way, using equivalence (1.5.41) of Lemma 1.12 and relations (2.2.43), (2.2.14), and (2.2.12), we arrive at ! q (2.2.55) Sk,1 f qLq (Δ+ ) BΔ+ f qLp (b(ξ ),b(ξ + )) . k

k

k

k

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S77

Since each of the sequences {Tk,1 }, {Tk,2 }, {Sk,1 }, and {Sk,2 } consists of block-diagonal operators, Lemma 2.1 gives the following estimates for each term in sum (2.2.45): if 1 1, then, by virtue of Holder’s inequality, r 1/r (2.2.54) q Tk,1 f Lq (Δ− ) f p ≤ BP S f p . BΔ− k

k

k∈Z

k∈Z

Similar estimates are derived for {Tk,2 }, {Sk,1 }, and {Sk,2 } from (2.2.47), (2.2.52), and (2.2.46) in the case 1 1. Taking into account (2.2.45), we obtain the upper estimates in (2.2.40) and (2.2.41).

A lower bound for HLp (0,∞)→Lq (0,∞) . Suppose that 1 0 b−1 (σ(t))≤s≤t

and AT,1(s, t) :=

AT,1 (s, t),

t v s

AT,2(t, s) :=

p

vp t

1/q

AT,1

≈ sup s>0

= sup s>0

b(s)

v

p

w (x) dx

v 1/q

q

vp

−1/p .

b−1 (σ(t)) ≤ s ≤ t;

t sup

v

p

(2.2.56)

t ≤ s ≤ a−1 (σ(t)).

s≤t≤σ−1 (b(s))

σ−1 (b(s))

,

σ(t)

Δ(x)

a(s)

b(s)

wq (x) dx

AT,2(t, s)

−1/p

p

a(s)

−1/p

s

σ(t)

q

Indeed, by virtue of (2.2.2), we have σ(t) b(s) 1 b(s) p p v ≤ v ≤ vp , 2 a(s) a(s) a(s) b(s) b(s) b(s) 1 vp ≤ vp ≤ vp , 2 a(s) σ(t) a(s) Therefore,

sup

t>0 t≤s≤a−1 (σ(t))

q

Δ(x)

s

AT,2 := sup

q

v s

p

1/q

q q

w (x) dx

Δ(x)

1/q

q

w (x) dx

v

Δ(x)

(2.2.57)

p

−1/p

Δ(s)

and, similarly, AT,2

≈ sup

s

σ−1 (a(s))

s>0

v

p

1/q

q q

w (x) dx

v

Δ(x)

p

−1/p .

Δ(s)

Suppose that H < ∞. Let us substitute the function

fs,t(y) = v p −1 (y)χ[a(s),σ(t)] (y),

b−1 (σ(t)) ≤ s ≤ t,

into the inequality

∞

1/q q

(Hf ) (x) dx 0

≤ H

∞

1/p p

f (y) dy

.

0

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S78

PROKHOROV et al.

Since a(s) ≤ a(x) and σ(x) ≤ σ(t) for s ≤ x ≤ t, it follows by (2.2.2) that 1/q σ(t) −1/p t b(x) q 1 p q v w (x) dx vp ≈ AT,1(s, t). H ≥ 2 s a(x) a(s) Therefore, H AT,1 (s, t) for all s ≤ t, whence H AT,1. Substituting the function

ft,s (y) = v p −1 (y)χ[σ(t),b(s)] (y),

t ≤ s ≤ a−1 (σ(t)),

we similarly obtain H AT,2 . Now suppose that 0 < q 1. As in the proof of Theorem 2.3, we ﬁrst prove the estimate H BP S under condition (2.2.27). Recall that, in this special case, we set x− = a(x),

Δ(x) = [x− , x+ ]

x+ = b(x),

and prove the inequality C BP S ,

(2.2.58)

where C is the best constant in the diﬀerential inequality (2.2.28) in the class L of functions speciﬁed in Deﬁnition 2.4. To derive (2.2.58), we successively prove the estimates C BP±S , where ∞

r/p 1/r q

q−r/p ± p q p q v w (x) dx v w (t) dt BP S := Δ± (t)

0

Δ− (t)

(t− , t)

Δ+ (t)

Δ(x)

Δ(t)

(t, t+ ).

with = and = We use part of notations of the proof of Theorem 2.3, retaining the meaning of some of them and changing that of others. To prove C BP−S , we deﬁne sequence (2.1.7) with η0 = 1 and set ηk∗ = min{ηk+ , ηk+1 }. ∗ . Let Let us ﬁx k ∈ Z and consider the ﬁve neighboring points ηk−2 , ηk−1 , ηk , ηk+1 , and ηk+1 fk (t) = χ[ηk−2 ,ηk+1 ] (t)[gk (t) + hk (t)], where

t

r/(pq)

ηk+1

gk (t) =

w ηk−2

s

ηk+1

hk (t) =

v

p

s

q

v

−r/(pq)

p

r/(pq )

ηk−2

ηk+1 s

v

ηk−2

v p (s) ds, p

r/(pq)

q w (s) ds

v

ηk−2

ηk−2

t

q

p

.

ηk−2

∗ ], we deﬁne the function On each interval [ηk−2 , ηk+1 fk (t), t ∈ [ηk−2 , ηk+1 ], φk (t) = ∗ ], fk (ηk+1 )Ωk+1 (t), t ∈ [ηk+1 , ηk+1

where

ηl∗

Ωl (t) =

v

p

−1

ηl

ηl∗

vp .

t

We set νk := λk + μk , where

ηk+1 ηk+1

λk := ηk−2

s

ηk+1

μk :=

v ηk−2

p

r/q wq

−r/p

s

vp

r/q

ηk−2

ηk+1 s

v ηk−2

v p (s) ds, p

r/q

q q

w (s) ds

.

ηk−2

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S79

To ﬁnd a representation of the form φk (t) =

3

(2.2.59)

φk,i (t),

i=1

where φk,i ∈ L, i = 1, 2, 3, we consider the function φk,1 (t) deﬁned by ⎧ ∗ ], ⎪ t∈ / [ηk−2 , ηk−1 ⎨0, φk,1 (t) = φk (t), t ∈ [ηk−2 , ηk−1 ], ⎪ ⎩ ∗ ]. φk (ηk−1 )Ωk−1 (t), t ∈ [ηk−1 , ηk−1 − − ∗ ] ⊆ [η − , η + ]; therefore, φ = ηk−2 , it follows that supp φk,1 ⊆ [ηk−1 , ηk−1 Since ηk−1 k,1 ∈ L. We set k−1 k−1 (1)

φk (t) := φk (t) − φk,1 (t) and deﬁne

⎧ ⎪ ⎨0, φk,2 (t) = φ(1) k (t), ⎪ ⎩ (1) φk (ηk )Ωk (t),

t∈ / [ηk−1 , ηk∗ ], t ∈ [ηk−1 , ηk ], t ∈ [ηk , ηk∗ ].

(1)

The function φk,2 , as well as φk,3 (t) := φk (t) − φk,2 (t), belongs to the class L. Therefore, (2.2.59) holds. Next, let us show that φk,i

p

νk ,

v

(2.2.60)

i = 1, 2, 3.

p

For this purpose, we set ηk−1 η∗ φk,1 p k−1 = |φk (s)|p v −p (s) ds + fkp (ηk−1 ) |Ωk−1 (s)|p v −p (s) ds =: κ1,1 + κ1,2 . κ1 := v p ηk−2 ηk−1 We have

ηk−1 ηk+1

κ1,1 =

r/q w

ηk−2

s

ηk+1

v

+

p

s

q

v

−r/q

p

r/q

v p (s) ds

ηk−2

ηk+1 s

v

ηk−2

p

r/q

q w (s) ds

v

ηk−2

ηk−2

ηk

q

p

ηk−2

≤ λk + μ k = ν k . ¨ Holder’s inequality with p and p gives ηk−1 ηk+1 r/(pq) p wq fk (ηk−1 ) ≈ ηk−2

s

ηk+1

+ ≤

v

p

−r/q r/q

w

ηk−1

vp

∗ ηk−1

ηk−1

p

r/q

q

s

p

ηk−2 p−1

r/q

w (s) ds

p

v

ηk−1

vp

p

p

ηk−2

ηk−1

v (s) ds

= νk

ηk−1

q

ηk−2

v

s

p

p

v (s) ds

ηk+1 s

q

ηk−2

Since

r/(pq )

ηk−2

ηk−2

ηk−1 ηk+1

≤ (λk + μk )

v

p

v

ηk−2

ηk−2

s

v

p

p−1

+ μk

ηk−2

p−1

ηk−1

v ηk−2

.

ηk−2

|Ωk−1 (s)|p v −p (s) ds

∗ ηk−1

=

v

p

1−p ,

ηk−1

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

p

p−1

S80

PROKHOROV et al.

it follows that

ηk−1

κ1,2 νk

v

p−1

p

∗ ηk−1

v

ηk−2

p

1−p ,

ηk−1

and (2.2.2), (2.2.27), and (2.2.34) with j = 0 imply κ1,2 νk . This proves (2.2.60) the case i = 1. In (1)

view of the relations φk (ηk ) = φk (ηk ) and φk,3 (ηk+1 ) = φk (ηk+1 ), the remaining cases in (2.2.60) are proved by analogy with i = 1. The estimate ηk+1 r/(pq) t s r/(pq ) q p w v v p (s) ds gk (t) ≥ ηk−2

t

r/(pq)

ηk+1

≈

w

ηk−2

t

q

v

p

r/(p q) t ∈ [ηk−2 , ηk+1 ],

,

ηk−2

t

implies the inequality φk wqq νk .

(2.2.61)

Next, we deﬁne the functions φk,i = φ2k,i + φ2k+1,i = : F1,i + F2,i , Fi = |k|≤N

|k|≤N

|k|≤N

where N ∈ N. For each i = 1, 2, 3, the supports of the φ2k,i , k ∈ Z, are pairwise disjoint. Therefore, F1,i ∈ L. For the same reason, F2,i ∈ L. Note also that χsuppφk (x) ≤ 4, x∈ supp φk . 1≤ |k|≤N

|k|≤N

Relations (2.2.59), (2.2.60), and (2.2.61) give 1/q νk φk w |k|≤N

≤ q

|k|≤N

C

2 3 i=1 j=1 p 1/p

φk,i

C

v

|k|≤N, i

p

λ∗k Note that t

s

v ηk−2

p

ηk−2

ηk+1

t

ηk−2 s

:=

v ηk−2

ηk−2

q

p

s

q

w (s) ds =

ηk−2

ηk−2

≤q

ηk−2

w (s) ds

(2.2.62)

.

v

p

−r/q

v p (t) dt,

q−r/p wq (t) dt.

ηk−2

q v d −

ηk+1

s

vp

ηk+1

w

s

wq

ηk−2 s

q

v ηk−1

ηk+1

p

s

s

t

q

w ηk−2

v

p

ηk−2

ηk−2

t

t

1/p νk

t

r/p

q

ηk−2

t

p

|k|≤N

. We set Letting N → ∞, we obtain C k∈Z νk r/q

q ηk+1 t s p q v w (s) ds λk := ηk−2

i=1 j=1

1/r

!

2 3 Fj,i v

Fj,i wq ≤ C

p

q−1

q

v p (s) ds q−1+q/2p

s

v

p

−q/(2p)

v p (s) ds;

ηk−1

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

¨ Holder’s inequality with exponents r/q and p/q yields t ηk+1 r/q wq ≤ ηk−2

s

t

× This implies ηk+1 λk

t

w

ηk−2

ηk−2

r/q

ηk+1

ηk+1

ηk+1

= ηk−2

s

s

q

v

r/q

p

s

s

vp

p

q/r

p

v (s) ds

−1/2

q/p

p

v (s) ds

r/q +r/(2p)

.

p

t

v (s) ds

r/q +r/(2p)

ηk−2

s

r/q +r/(2p)

ηk−2

ηk−2

wq

v

p

ηk−2

v ηk−2

s

S81

v

p

r/(2p)−r/q

v p (t) dt

ηk−2

ηk+1

t

vp

r/(2p)−r/q

v p (t) dt v p (s) ds

ηk−2

s

λk . Since k = p λ r

ηk+1 t

s

v ηk−2

ηk−2

p

q

r/q w (s) ds d −

t

q

ηk−2

v

p

−r/p ,

ηk−2

k = −μk p/r + λ∗ p/q. Therefore, it follows that λ k q q q q ∗ , (λk + μk ) λk + μk = νk . λk = λk + μk ≤ max p r p r We obtain

1/r

C

νk

k∈Z

λ∗k

1/r .

k∈Z

Applying (2.2.34) with j = 0 and taking into account (2.2.2), we see that ηk−1 ηk t+ t p p p p v ≤ v + v + v ≤4 ηk−2

ηk−2

ηk−1

ηk

t+

vp

(2.2.63)

ηk

for t ∈ [ηk , ηk+1 ]. This, together with (2.2.2) and (2.2.27), implies r/p t

q q−r/p ηk+1 t s ∗ p q p v w (s) ds v wq (t) dt λk ≥ ηk

(2.2.63)

≥

ηk−2

ηk+1

t

ηk+1 t

ηk−1

ηk−1

Δ− (t)

p

ηk−2

ηk+1

ηk

s

v ηk

ηk

ηk−2

Δ− (s)

v

p

vp

ηk−2

r/p

q

t

q

w (s) ds

v

r/p

q

r/p

q

Δ− (t)

wq (s) ds

Δ(s)

q

ηk−2

q

w (s) ds

p

t+

v

p

−r/p

ηk

v

vp

p

q

q−r/p

t+

t−

v

p

wq (t) dt

−r/p

wq (t) dt

wq (t) dt,

Δ(t)

BP−S .

The inequality C BP+S can be proved by a similar method which proves the required estimate C by using intervals determined by sequence (2.1.8). Thus, estimate (2.2.58) does hold, and (2.2.29) implies the inequality H BP S in the case σ(x) = x. In the general case, this inequality is proved by making the change of variables (2.2.36) on both sides of (2.2.37). Passing to (2.2.38) with σ (x) = x for a, b, and v (see (2.2.39)), we derive BP S H from the preceding estimates for (2.2.38), where r/p

q q−r/p ∞ r p q p v w (s) ds v wq (t) dt. BP S = 0

Δ(t)

Δ(s)

Δ(t)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S82

PROKHOROV et al.

The relations Δ(t) = [ a(t), b(t)] = θ(t) and (2.2.39) imply BP S = BP S .

2.3. Generalized Hardy–Steklov Operators In this section, we present new boundedness and compactness characteristics for operators (2.1.4) with kernels k(x, y) ∈ Ob and k(x, y) ∈ Oa . Deﬁnition 2.5. Given boundary functions a(x) and b(x) satisfying conditions (2.1.5), numbers p, q ∈ (1, ∞), a function 0 < k(x, y) < ∞ jointly continuous almost everywhere on R, and weight functions 0 < v, w < ∞ deﬁned almost everywhere on (0, ∞) and such that the function k p (x, y)v p (y) is locally q q integrable on (0, ∞) with respect to y for any ﬁxed x > 0 and the product k (x, y)w (x) is locally integrable on (0, ∞) with respect to x for any y > 0, we deﬁne two k-fairways σk (x) and ρk (y) as functions satisfying the conditions a(x) < σk (x) < b(x), b−1 (y) < ρk (y) < a−1 (y), and b(x) σk (x) p p k (x, y)v (y) dy = kp (x, y)v p (y) dy, x > 0, (2.3.1) a(x)

ρk (y)

b−1 (y)

q

σk (x) a−1 (y)

q

k (x, y)w (x) dx =

kq (x, y)wq (x) dx,

y > 0.

(2.3.2)

ρk (y)

We set vk (x, y) := k(x, y)v(y) and wk (x, y) := k(x, y)w(x). By virtue of (2.3.1), for σk (x) ∈ [a(z), b(z)], we have a(z) σk (x) σk (x) p p vk (x, y) dy = vk (x, y) dy + vkp (x, y) dy, (2.3.3) a(x)

σk (z)

a(z)

Therefore, σk (x) a(x)

a(z) a(x)

b(x) σk (x)

vkp (z, y) dy =

vkp (x, y) dy −

= Similarly,

a(z)

vkp (x, y) dy

vkp (x, y) dy

σk (z)

σk (z)

= σk (x)

−

σk (x) a(z)

vkp (z, y) dy +

σk (z) σk (x)

vkp (z, y) dy.

(2.3.4)

σk (z) σk (x)

σk (z)

a(z)

vkp (z, y) dy

−

b(z)

a(x)

vkp (z, y) dy

σk (x)

+ a(z)

[vkp (x, y) − vkp (z, y)] dy.

vkp (z, y) dy

vkp (x, y) dy −

b(z) b(x)

vkp (z, y) dy +

b(x)

σk (z)

[vkp (x, y) − vkp (z, y)] dy.

By virtue of (2.3.1), this implies σk (z) σk (z) vkp (x, y) dy + vkp (z, y) dy σk (x)

σk (x)

a(z)

= a(x)

vkp (x, y) dy

(2.3.5)

b(z)

+ b(x)

vkp (z, y) dy

(2.3.6)

(2.3.7)

σk (x)

+

−

a(z)

b(x) σk (z)

[vkp (x, y) − vkp (z, y)] dy; (2.3.8)

again applying (2.3.1), we obtain σk (z) σk (z) p vk (x, y) dy + vkp (z, y) dy σk (x)

σk (x)

a(z)

= a(x)

vkp (x, y) dy

b(z)

+ b(x)

vkp (z, y) dy

σk (z)

+ a(z)

−

b(x)

σk (x)

[vkp (x, y) − vkp (z, y)] dy. (2.3.9)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S83

Note that, for k ≡ const, the last two relations take the form (2.2.3). By virtue of the assumptions in Deﬁnition 2.3.1, the function σk (x) is continuous on (0, ∞). Similarly, the k-fairway ρk (y) is continuous on (0, ∞) as well. We set Θ(t) := Θ− (t) ∪ Θ+ (t),

ϑ(t) := ϑ− (t) ∪ ϑ+ (t),

δ(t) := δ− (t) ∪ δ+ (t),

Δ(t) := Δ− (t) ∪ Δ+ (t),

Θ− (t) := [b−1 (t), ρk (t)),

Θ+ (t) := [ρk (t), a−1 (t)),

ϑ− (t) := [a(ρk (t)), t),

ϑ+ (t) := [t, b(ρk (t))),

δ− (t) := [b−1 (σk (t)), t),

δ+ (t) := [t, a−1 (σk (t))),

where

Δ− (t) := [a(t), σk (t)),

Δ+ (t) := [σk (t), b(t)).

We also set r := p q/(p − q),

1/q

± := sup A (t) := sup A± ρk ρk t>0

A± σk (Bρ±k )r (Bσ±k )r

sup A± σk (t) t>0

:=

∞

:=

0 ∞

:= 0

Aρk

t>0

Bρ±k (t) dt Bσ±k (t) dt

Θ(t)

:= sup ∞

:= 0

Θ(t)

Bσr k

∞

:=

t>0

:=

0

∞

Bσk (t) dt :=

0

1/q

wkq (x, t) dx

Θ(t)

v

1/q

r/p

δ(t)

q Δ(t)

1/p

v

vkp (t, y) dy

wq (t) dt, 1/p ,

1/p

p

ϑ(t)

, v p (t) dt,

vkp (t, y) dy

r/q

,

r/q

p

Δ(t)

1/p

r/p

ϑ(t)

wkq (x, t) dx

∞

v

p

vkp (t, y) dy

wq

w 0

q

δ(t)

∞

ϑ± (t)

Δ(t)

Θ(t)

t>0

Bρk (t) dt :=

0

w

:= sup Aρk (t) := sup

t>0

r/q r/p

vkp (t, y) dy

wkq (x, t) dx

δ± (t)

0

q Δ(t)

∞

:=

t>0

w

vp

ϑ± (t)

1/q

δ± (t)

t>0

Aσk := sup Aσk (t) := sup Bρrk

wkq (x, t) dx

r/q

,

v p (t) dt,

r/p wq (t) dt.

Theorem 2.5. Suppose that the kernel k(x, y) of an operator K of the form (2.1.4) is nonnegative on R, satisﬁes all conditions in Deﬁnition 2.5, and belongs to the Oinarov-type class Ob . Suppose also that ρk (y) and σk (x) are strictly increasing k-fairways. (a) If 1 < p ≤ q < ∞, then + A− ρk + Aσk KLp (0,∞)→Lq (0,∞) Aρk + Aσk .

(2.3.10)

If K is compact, then + A− ρk , Aσk < ∞,

+ lim A− ρk (t) = lim Aσk (t) = 0;

t→0 t→∞

t→0 t→∞

conversely, K is compact if Aρk , Aσk < ∞,

lim Aρk (t) = lim Aσk (t) = 0.

t→0 t→∞

t→0 t→∞

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S84

PROKHOROV et al.

(b) If 1 < q < p < ∞, then Bρ−k + Bσ+k KLp (0,∞)→Lq (0,∞) Bρk + Bσk .

(2.3.11)

Moreover, the operator K is compact if Bρk , Bσk < ∞, and if K is a compact operator acting from Lp (0, ∞) to Lq (0, ∞), then Bρ−k , Bσ+k < ∞. Proof. (a) A lower bound. According to equivalence (2.1.12) of Theorem 2.1, we have KLp →Lq ≈ sup

sup

t>0 b−1 (a(t))≤s≤t

[A0 (s, t) + A1 (s, t)],

(2.3.12)

where A0 (s, t) and A1 (s, t) are deﬁned by (2.1.14) and (2.1.15), i.e., t 1/q b(s) 1/p wkq (x, b(s)) dx vp , A0 (s, t) :=

s

A1 (s, t) :=

w

A− ρk (t)

a(t)

Θ− (t)

vkp (s, y) dy

1/q

1/q

=2

b(s)

q

s

Using (2.3.2), we obtain

a(t)

1/q

t

wkq (x, t) dx

s

b−1 (ρ−1 k (s))

ϑ− (t)

wkq (x, ρ−1 k (s)) dx

v

.

p

1/q

= A0 (b−1 (ρ−1 k (s)), s) ≤ sup

1/p

1/p

ρ−1 k (s)

v

p

1/p

a(s)

sup

t>0 b−1 (a(t))≤s≤t

A0 (s, t).

This, together with (2.3.12), implies A− ρk KLp (0,∞)→Lq (0,∞) . Similarly, the estimate 1/p 1/q (2.3.1) 1/p p + q w vk (t, y) dy Aσk (t) = 2 δ+ (t)

1/p

=2

−1

A1 (t, a

Δ+ (t)

(σk (t))) sup

sup

t>0 b−1 (a(t))≤s≤t

A1 (s, t)

gives A+ σk KLp (0,∞)→Lq (0,∞) , which proves the inequality on the left-hand side of (2.3.10).

An upper bound. Setting τ0 := ρk (a(t)), we can write sup

b−1 (a(t))≤s≤t

A0 (s, t) ≤ ≤

sup

b−1 (a(t))≤s≤τ0
A0 (s, t) + sup A0 (s, t) τ0 ≤s≤t

a−1 (ρ−1 k (τ0 ))

sup

b−1 (ρ−1 k (τ0 ))≤s≤τ0

s

a−1 (ρ−1 k (s))

+ sup

τ0 ≤s≤t

s

1/q

wkq (x, b(s)) dx

ρ−1 k (τ0 )

1/q wkq (x, b(s)) dx

b(s)

b(s)

v

p

v

p

1/p

1/p

a(s)

=: H1 (τ0 ) + H2 (t). Indeed, if t = a−1 (ρ−1 k (τ0 )), then (s, t) = (s, a−1 (ρ−1 k (τ0 )))

and

(a(t), b(s)) = (ρ−1 k (τ0 ), b(s)).

If τ0 ≤ s ≤ t, then (s, t) ⊂ (s, a−1 (ρ−1 k (s)))

and

(a(t), b(s)) ⊂ (a(s), b(s)).

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S85

To estimate H1 (τ0 ), we apply (2.1.9) with y = ρ−1 k (τ0 ) and z = s. Since b−1 (ρ−1 (τ0 )) ≤ s ≤ x ≤ a−1 (ρ−1 (τ0 )), i.e., b−1 (a(t)) ≤ s ≤ x ≤ t, it follows that k(x, b(s)) k(x, ρ−1 k (τ0 )). Therefore, a−1 (ρ−1 (τ0 )) 1/q b(τ0 ) 1/p k q −1 p wk (x, ρk (τ0 )) dx v H1 (τ0 ) =

b−1 (ρ−1 k (τ0 )) a−1 (z)

b−1 (z)

ρ−1 k (τ0 )

1/q

wkq (x, z) dx

b(ρk (z))

v

p

1/p

+ = A+ ρk (z) ≤ Aρk .

z

−1 −1 −1 −1 Since s ≤ x and a(x) ≤ ρ−1 k (s) ≤ b(s), i.e., b (ρk (s)) ≤ s ≤ x ≤ a (ρk (s)), we obtain the following relation for H2 (t), taking into account (2.1.9) with z = s and y = ρ−1 k (s): 1/q b(s) 1/p a−1 (ρ−1 (s)) k sup wkq (x, ρ−1 vp H2 (t) k (s)) dx s

ρk (a(t))≤s≤t

=

a−1 (z)

sup

a(t)≤z≤ρ−1 k (t)

ρk (z)

1/q wkq (x, z) dx

a(s)

b(ρk (z))

v

p

1/p

a(ρk (z))

≤ Aρk .

Therefore, sup

sup

t>0 b−1 (a(t))≤s≤t

A0 (s, t) Aρk .

(2.3.13)

Similarly, setting τ1 := σk−1 (a(t)), we write sup

b−1 (a(t))≤s≤t

A1 (s, t) ≤ ≤

A1 (s, t) + sup A1 (s, t)

sup

b−1 (a(t))≤s≤τ1
τ1 ≤s≤t

w

sup

b−1 (σk (τ1 ))≤s≤τ1

s

a−1 (σk (s))

+ sup

τ1 ≤s≤t

1/q

a−1 (σk (τ1 ))

b(s)

q

1/q

σk (τ1 ) b(s)

wq

s

a(s)

vkp (s, y) dy

1/p

1/p

vkp (s, y) dy

=: H3 (τ1 ) + H4 (t). + It is easy to see that H4 (t) ≤ supσ−1 (a(t))≤s≤t A+ σk (s) ≤ Aσk . To estimate H3 (τ1 ), we apply (2.1.9) with k z = s ≤ τ1 = x and a(τ1 ) < σk (τ1 ) ≤ y ≤ b(s), which yields 1/p a−1 (σk (τ1 )) 1/q b(s) p q sup w vk (τ1 , y) dy ≤ Aσk (τ1 ) ≤ Aσk . H3 (τ1 ) b−1 (σk (τ1 ))≤s≤τ1

s

σk (τ1 )

Therefore, sup

sup

t>0 b−1 (a(t))≤s≤t

A1 (s, t) Aσk .

(2.3.14)

Combining (2.3.12) and (2.3.13) with (2.3.14), we derive the required upper estimate of K in (2.3.10). The compactness conditions in the case 1 < p ≤ q < ∞ follow from Theorem 2.1. (b) To prove an upper bound for KLp (0,∞)→Lq (0,∞) , we set ξ0 = 1 in (2.3.11) and deﬁne points ξi by (2.1.6) for all i ∈ Z. We use the notation Δi = [ξi , ξi+1 ),

δi = [a(ξi ), b(ξi )],

i ∈ Z.

Breaking the half-axis (0, ∞) by the points {ξi }i∈Z , we decompose the operator K as K =T +S PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

(2.3.15) Vol. 300 Suppl. 2

2018

S86

PROKHOROV et al.

into the sum of the two block-diagonal operators Ti and T =

S=

i∈Z

where

k(x, y)f (y)v(y) dy,

(2.3.16)

Si ,

i∈Z

a(ξi+1 )

Ti f (x) = w(x)

Ti : Lp (δi ) → Lq (Δi ),

a(x) b(x)

Si f (x) = w(x)

k(x, y)f (y)v(y) dy,

Si : Lp (δi+1 ) → Lq (Δi ).

b(ξi )

The kernels k(x, y) of the operators Ti and Si satisfy condition (2.1.9) for z ≤ x, x ∈ [ξi , ξi+1 ], and Since

a(x) ≤ y ≤ b(ξi ),

b(ξi ) ≤ y ≤ b(z).

(2.3.17)

= (0, ∞), it follows from (2.3.15) and (2.3.16) that Kf qLq (Δi ) ≈ Ti f qLq (Δi ) + Si f qLq (Δi ) . Kf qq =

i∈Z [ξi , ξi+1 )

i∈Z

i∈Z

(2.3.18)

i∈Z

Given i, to estimate the norm of the operator Si , we ﬁx the two points sρ := b−1 (ρ−1 k (ξi+1 ))

and

sσ := σk−1 (b(ξi )) = σk−1 (a(ξi+1 ))

and consider their possible arrangements: (i) sρ < sσ ; (ii) sρ = sσ ; (iii) sρ > sσ . In case (i), we have 3

Si f (x) =

(2.3.19)

Si,j f (x) + Hi,0 f (x),

j=1

where Si,1 f (x) = χ[ξi ,sσ ] (x)Si (f χ[b(ξi ),b(sρ )] )(x), Si,2 f (x) = χ[sρ ,sσ ] (x)Si (f χ[b(sρ ),b(sσ )] )(x), Si,3 f (x) = χ[sσ ,ξi+1 ] (x)Si (f χ[b(sρ ),b(ξi+1 )] )(x), Hi,0 f (x) = χ[sσ ,ξi+1 ] (x)Si (f χ[b(ξi ),b(sρ )] )(x). Applying Corollary 1.5 to K φ = Si,1 with φ = b and taking into account (1.5.28), we obtain Si,1 r := Si,1 rLp (b(ξi ),b(sρ ))→Lq (ξi ,sσ ) (B b,0 )ri,1 + (Bb,1 )ri,1 r/q t r/q b(sρ ) sσ q p = wk (x, t) dx v v p (t) dt b(ξi )

b−1 (t)

sσ t

b(ξi )

min{b(x),b(sρ )}

+ ξi

×

ξi

b(ξi )

min{b(t),b(sρ )}

b(ξi )

q r/p vkp (x, y) dy wq (x) dx

vkp (t, y) dy

q−r/p wq (t) dt.

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

(2.3.20)

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S87

6

b(x) σk (x) −1 ρk (x)

Si,3

Si,2

a(x) Si,1

Hi,0

b(ξi )

w Ti,2

w Ti,1

a(ξi ) sρ

ξi

ξi+1

s σ tρ Fig. 2.3.1.

b,0 Since a(sσ ) < b(ξi ) = a(ξi+1 ) ≤ t ≤ b(sρ ) = ρ−1 k (ξi+1 ) in (B )i,1 , it follows that

[b−1 (t), sσ ] ⊂ [b−1 (t), a−1 (t)] Therefore,

and

(B b,0 )ri,1 ≤

b(sρ )

b(ξi )

[b(ξi ), t] ⊆ [a(ρk (t)), t].

Bρ−k (t) dt.

(2.3.21)

Let us estimate (Bb,1 )i,1 . Note that, by virtue of (2.1.9), we have k(x, y) k(t, y) (see also (2.3.17)) for b−1 (y) ≤ x ≤ t ≤ sσ < ξi+1 and, hence, r/p sσ t r/p b(t) p b,1 r q w vk (t, y) dy wq (t) dt. (B )i,1 ξi

ξi

b(ξi )

Here, [b(ξi ), b(t)] = [a(ξi+1 ), b(t)] ⊂ [a(t), b(t)], because t ≤ sσ < ξi+1 . Moreover, we have [ξi , t] ⊂ [b−1 (σk (t)), t], because t ≤ sσ = σk−1 (b(ξi )) < ξi+1 , i.e., b−1 (σk (t)) < ξi . This implies sσ b,1 r Bσ−k (t) dt. (2.3.22) (B )i,1 ξi

To estimate the norm of Si,2 , we use Corollary 1.5 with φ = b, taking into account (1.5.27) and (1.5.28): Si,2 r := Si,2 rLp (b(sρ ),b(sσ ))→Lq (sρ ,sσ ) (B b,0 )ri,2 + (Bb,1 )ri,2

p r/q b(sσ ) b(sσ ) sσ q p = wk (x, y) dx v (y) dy b(sρ )

b−1 (y)

t

sσ t b(x)

+ sρ

sρ

b(sρ )

sσ

b−1 (t)

q

vkp (x, y) dy wq (x) dx

r/p

b(t)

b(sρ )

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

wkq (x, t) dx

p −r/q

(2.3.23)

q−r/p

vkp (t, y) dy Vol. 300 Suppl. 2

v p (t) dt

wq (t) dt. (2.3.24) 2018

S88

PROKHOROV et al.

Since b−1 (t) ≤ b−1 (y) =: z ≤ x ≤ sσ < ξi+1 in (2.3.23), it follows that k(x, y) = k(x, b(z)) k(x, t) in view of (2.1.9) or (2.3.17). Therefore, r/q b(sσ ) b(sσ ) r/q sσ q b,0 r p v wk (x, t)dx v p (t) dt. (B )i,2 b(sρ )

b−1 (t)

t

−1 −1 The inequalities a(sσ ) < ρ−1 k (sσ ) < ρk (ξi+1 ) = b(sρ ) ≤ t imply sσ < a (t) and sσ < ρk (t), whence b(sσ ) < b(ρk (t)). Therefore, b(sσ ) b,0 r Bρ+k (t) dt. (2.3.25) (B )i,2 b(sρ )

b−1 (y)

≤ x ≤ t ≤ sσ < ξi+1 ; this and (2.1.9) imply k(x, y) k(t, y) (see also In (2.3.24), we have (2.3.17)). Hence, r/p sσ t r/p b(t) sσ p b,1 r q q w vk (t, y) dy w (t) dt ≤ Bσ−k (t) dt, (2.3.26) (B )i,2 sρ

sρ

b(sρ )

sρ

because ξi < sρ ≤ t ≤ sσ = σk−1 (b(ξi )) < ξi+1 , i.e., b−1 (σk (t)) ≤ ξi < sρ , and a(t) < a(ξi+1 ) = b(ξi ) < b(sρ ). Let Hi,0 := Hi,0 Lp (b(ξi ),b(sρ ))→Lq (sσ ,ξi+1 ) . By virtue of Corollary 1.5, we have Hi,0 ≈ (B b,0 )ri,0 + (B b,1 )ri,0 , where

(B b,0 )ri,0

wkq (x, t) dx

= b(ξi )

(B b,1 )ri,0

r/q

b(sρ ) ξi+1 ξi+1

sσ

r/p

ξi+1

=

w sσ

q

t

v

p

r/q

b(ξi ) b(sρ )

b(ξi )

t

(2.3.27)

vkp (t, y) dy

v p (t) dt r/p wq (t) dt.

b,0 Since a(ξi+1 ) = b(ξi ) ≤ t ≤ b(sρ ) = ρ−1 k (ξi+1 ) ≤ b(sσ ) in (B )i,0 , it follows that

[sσ , ξi+1 ] ⊂ [b−1 (t), a−1 (t)] i.e.,

(B b,0 )ri,0

≤

b(sρ )

b(ξi )

Moreover,

[b(ξi ), t] ⊂ [a(ρk (t)), t],

and

(B b,1 )ri,0 ≤

ξi+1 sσ

Bρ−k (t) dt.

(2.3.28)

Bσ+k (t) dt,

(2.3.29)

because sρ < σk−1 (a(ξi+1 )) = sσ ≤ t ≤ ξi+1 in (B b,1 )i,0 . For Si,3 := Si,3 Lp (b(sρ ),b(ξi+1 ))→Lq (sσ ,ξi+1 ) , Corollary 1.5, together with (1.5.27), implies Si,3 r (Bb,0 )ri,3 + (Bb,1 )ri,3 b(ξi+1 ) b(ξi+1 ) = b(sρ )

b−1 (y)

t

×

ξi+1 b−1 (t)

r/p w

+ t

p wkq (x, y)χ[sσ ,ξi+1 ] (x) dx

wkq (x, t)χ[sσ ,ξi+1 ] (x) dx

ξi+1 ξi+1 sσ

ξi+1

b(t)

q b(sρ )

p −r/q

vkp (t, y) dy

p

r/q

v (y) dy

v p (t) dt

r/p wq (t) dt.

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

(2.3.30)

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S89

As above, we obtain k(x, y) k(x, t) for b−1 (t) ≤ b−1 (y) = z ≤ x ≤ ξi+1 in (B b,0 )i,3 . Moreover, we −1 have b(ξi+1 ) < b(ρk (t)), because ρ−1 k (ξi+1 ) = b(sρ ) ≤ t, and ξi+1 ≤ a (t), because a(ξi+1 ) = b(ξi ) ≤ b(sρ ) ≤ t. Therefore, r/q b(ξi+1 ) b(ξi+1 ) r/q ξi+1 q r p v wk (x, t) dx v p (t) dt (Bb,0 )i,3

b(sρ )

b(ξi+1 )

≤

b(sρ )

b−1 (t)

t

Bρ+k (t) dt.

(2.3.31)

The estimate

(Bb,1 )ri,3

≤

ξi+1

sσ

Bσ+k (t) dt

(2.3.32)

follows from the inequalities a(t) ≤ a(ξi+1 ) = b(ξi ) < b(sσ ) and σk−1 (a(ξi+1 )) = sσ ≤ t. case (i), on the basis of estimates (2.3.19)–(2.3.32) we derive r Bρk (t) dt + Bσk (t) dt. Si Lp (δi+1 )→Lq (Δi ) δi+1

Thus, in (2.3.33)

Δi

6

b(x) ρ−1 k (x)

Si,3

σk (x) a(x)

Si,2 Si,1 b(ξi )

v Ti,1

v Ti,2

a(ξi ) ξi

s σ = tσ

ξi+1

sρ

Fig. 2.3.2.

In case (iii), we have Si f (x) =

3

(2.3.34)

Si,j f (x),

j=1

where Si,1 f (x) = χ[ξi ,sσ ] (x)Si f (x), Si,2 f (x) = χ[sσ ,ξi+1 ] (x)Si (f χ[b(ξi ),b(sρ )] )(x), PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S90

PROKHOROV et al.

Si,3 f (x) = χ[sρ ,ξi+1 ] (x)Si (f χ[b(sρ ),b(ξi+1 )] )(x). The estimate

Si,1 := r

Si,1 rLp (b(ξi ),b(sσ ))→Lq (ξi ,sσ )

b(sσ )

b(ξi )

Bρ−k (t) dt

sσ

+ ξi

Bσ−k (t) dt

(2.3.35)

is obtained in the same way as in case (i). To estimate Si,2 := Si,2 Lp (b(ξi ),b(sρ ))→Lq (sσ ,ξi+1 ) , we apply Corollary 1.5: Si,2 r ≈ (B b,0 )ri,2 + (B b,1 )ri,2 r/q b(sρ ) ξi+1 q = wk (x, t)χ[sσ ,ξi+1 ] dx b(ξi )

b−1 (t)

ξi+1 ξi+1

r/p w

+ sσ

q

v

p

r/q

v p (t) dt

b(ξi )

min{b(t),b(sρ )}

b(ξi )

t

t

vkp (t, y) dy

r/p wq (t) dt.

(2.3.36)

b,0 −1 Since a(ξi+1 ) = b(ξi ) ≤ t ≤ b(sρ ) = ρ−1 k (ξi+1 ) in (B )i,2 , it follows that ξi+1 ≤ a (t) and ρk (t) ≤ ξi+1 , i.e., a(ρk (t)) ≤ a(ξi+1 ) = b(ξi ). Therefore, b(sρ ) b,0 r Bρ−k (t) dt. (2.3.37) (B )i,2 ≤ b(ξi )

For sσ = σk−1 (a(ξi+1 ) ≤ t ≤ ξi+1 , we have a(t) ≤ a(ξi+1 ) = b(ξi ) and ξi+1 ≤ a−1 (σk (t)) in the expression for (B b,1 )i,2 . Hence, ξi+1 b,1 r Bσ+k (t) dt. (2.3.38) (B )i,2 ≤ sσ

By analogy with (i), we also obtain Si,3 := r

Si,3 rLp (b(sρ ),b(ξi+1 ))→Lq (sρ ,ξi+1 )

b(ξi+1 ) b(sρ )

Bρ+k (t) dt

ξi+1

+ sρ

Bσ+k (t) dt.

(2.3.39)

Summing (2.3.34)–(2.3.39), we arrive at estimate (2.3.33) in case (iii), too. Case (ii) follows from (i) or (iii). Finally, by virtue of Lemma 1.10, (2.3.16) implies 1/r r Si Lp (δi+1 )→Lq (Δi ) Bρk + Bσk . (2.3.40) S ≈ i

To estimate the norm of the operator Ti , we decompose it into a sum by using (2.1.9) (or (2.3.17)): b(ξi ) b(ξi ) f (y)v(y) dy + w(x) k(ξi , y)f (y)v(y) dy Ti f (x) ≈ w(x)k(x, b(ξi )) a(x)

a(x)

=: Tiw f (x) + Tiv f (x).

(2.3.41)

−1 Fixing the points tρ := a−1 (ρ−1 k (ξi )) and tσ := σk (b(ξi )), we can write the representations w w f (x) + Ti,2 f (x), Tiw f (x) = Tiw (f χ[a(ξi ),ρ−1 (ξi )] )(x) + Tiw (f χ[ρ−1 (ξi ),b(ξi )] )(x) =: Ti,1

(2.3.42)

v v f (x) + Ti,2 f (x) Tiv f (x) = Tiv f (x)χ[ξi ,tσ ] (x) + Tiv f (x)χ[tσ ,ξi+1 ] (x) =: Ti,1

(2.3.43)

k

k

(see ﬁgures above). By virtue of duality, taking into account (1.5.43) and applying Lemma 1.12 −1 with c = a(ξi ), d = ρ−1 k (ξi ), a = ξi , b(x) → a (y), v(y) → w(x)k(x, b(ξi )), w(x) → v(y), q = p , and p = q , we obtain w r w r := Ti,1 Lp (a(ξ Ti,1

−1 q i ),ρk (ξi ))→L (ξi ,tρ )

≈ (Ba−1 )ri,1

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

t ρ−1 k (ξi )

a−1 (y)

= a(ξi )

a(ξi )

ξi

a−1 (t)

×

ξi

p wkq (x, b(ξi )) dx

wkq (x, b(ξi )) dx

S91

r/q

v p (y) dy

p −r/q

v p (t) dt.

ρ−1 k (ξi )

Since ξi ≤ x and a(x) ≤ y ≤ t ≤ < b(ξi ) in (Ba−1 )i,1 , it follows that k(x, b(ξi )) k(x, t) in view −1 of (2.1.9). Moreover, the inequalities t ≤ ρ−1 k (ξi ) < b(ξi ) imply b (t) < ξi and a(ρk (t)) ≤ a(ξi ), i.e., r/q r/q a−1 (t) ρ−1 (ξi ) t ρ−1 (ξi ) k k q w r p p v wk (x, t) dx v (t) dt ≤ Bρ−k (t) dt. (2.3.44) Ti,1 a(ξi )

a(ξi )

ξi

a(ξi )

Similarly, by virtue of duality, using a−1 (ρ−1 k (ξi )) = tρ , taking into account (1.5.41), and applying (ξ ), d = b(ξ ), a = ξi , b(x) → a−1 (y), v(y) → w(x)k(x, b(ξi )), w(x) → v(y), Lemma 1.12 with c = ρ−1 i i k q = p , and p = q , we obtain w r w r := Ti,2 Lp (ρ−1 (ξ ),b(ξ ))→Lq (ξ ,ξ ) ≈ (Ba−1 )ri,2 Ti,2 i i i i+1 k r/q b(ξi ) r/q b(ξi ) a−1 (t) = wkq (x, b(ξi )) dx vp v p (t) dt. ρ−1 k (ξi )

Since

b−1 (t)

w r Ti,2

ξi

t

a−1 (t),

≤ ξi ≤ x ≤ b(ξi ) a−1 (t) ρ−1 k (ξi )

ξi

it follows from (2.1.9) that k(x, b(ξi )) k(x, t); moreover, r/q b(ξi ) r/q b(ξi ) wkq (x, t) dx vp v p (t) dt ≤ Bρ+k (t) dt, ρ−1 k (ξi )

t

(2.3.45)

−1 because ρ−1 k (ξi ) ≤ t ≤ b(ξi ) in (Ba−1 )i,2 , i.e., b (t) ≤ ξi and b(ξi ) ≤ b(ρk (t)). Next, applying equivalence (1.5.49) of Lemma 1.13 and inequalities (2.1.9) with ξi ≤ t and a(t) ≤ y ≤ b(ξk ), we obtain v r v r := Ti,1 Lp (a(ξi ),b(ξi ))→Lq (ξi ,tσ ) ≈ (Ba )ri,1 Ti,1 r/p tσ t r/p b(ξi ) = wq vkp (ξi , y) dy wq (t) dt

ξi

ξi

tσ t

r/p w

b(ξi )

q

ξi

ξi

a(t)

a(t)

vkp (t, y) dy

r/p

tσ

wq (t) dt ≤

ξi

Bσ−k (t) dt,

(2.3.46)

because ξi ≤ t ≤ tσ in (Ba )i,1 , i.e., b−1 (σk (t)) ≤ ξi and b(ξi ) ≤ b(t). Finally, using equivalences (1.5.51) of Lemma 1.13 and inequalities (2.1.9) with ξi < tσ ≤ t and a(t) ≤ a(x) ≤ y ≤ b(ξi ), we arrive at v r v r : = Ti,2 Lp (a(tσ ),b(ξi ))→Lq (tσ ,ξi+1 ) ≈ (Ba )ri,2 Ti,2

q r/p b(ξi ) q−r/p ξi+1 ξi+1 b(ξi ) = vkp (ξi , y) dy wq (x) dx vkp (ξi , y) dy wq (t) dt

tσ

t

ξi+1 ξi+1

tσ

a(x)

r/p

wq

b(ξi ) a(t)

t

a(t)

r/p

vkp (t, y) dy

wq (t) dt ≤

ξi+1 tσ

Bσ+k (t) dt,

(2.3.47)

because ξi < tσ = σk−1 (a(ξi+1 )) ≤ t ≤ ξi+1 in (Ba )i,2 , i.e., ξi+1 ≤ a−1 (σk (t)) and b(ξi ) < b(t). From (2.3.41)–(2.3.47) we obtain the estimate Bρk (t) dt + Bσk (t) dt. (2.3.48) Ti rLp (δi )→Lq (Δi ) δi

Δi

Combining this with (2.3.16) on the basis of Lemma 1.10, we arrive at 1/r r Ti Lp (δi )→Lq (Δi ) Bρk + Bσk . T ≈ i

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S92

PROKHOROV et al.

By virtue of (2.3.15), this, together with (2.3.40), implies the upper estimate in (2.3.11).

A lower bound for KLp (0,∞)→Lq (0,∞) . Suppose that Kf q ≤ K · f p .

(2.3.49)

KLp (0,∞)→Lq (0,∞) Bρ− ,

(2.3.50)

To show that we set ξ0 = 1 and deﬁne ξi for all other i ∈ Z by (2.1.6). Let us introduce the notation a−1 (t) t q − wk (x, t) dx, Vρ (t) = v p (y) dy; Wρ (t) = b−1 (t)

r/(pq )

note that [Wρ (t)]r/(pq) [Vρ− (t)]

a(ρk (t))

p −1

= Bρ−k (t)1/p . Moreover, for

[v(t)]

fρ (t) := [Wρ (t)]r/(pq) [Vρ− (t)]r/(pq ) [v(t)]p −1 , we have fρ p = (Bρ−k )r/p . 6

b(x) ρ−1 k (x)

ρ−1 k (ξi+1 )

a(x) t

ρ−1 k (ξi ) a(ρk (t))

a(ξi ) −1 ρk (ξi−1 ) ξi−1

b−1 (ρ−1 k (ξi ))

ξi

ξi+1

-

Fig. 2.3.3.

Since

= (0, ∞), it follows from (2.3.49) that ξi+1 1/q − r/p q (Kfρ ) (x) dx K(Bρk ) ≥ Kfρ q =

i [ξi , ξi+1 )

i −1/q

≥2

ξi

i

ξi+1

b−1 (ρ−1 k (ξi ))

1/q q

(Kfρ ) (x) dx

.

(2.3.51)

Substituting the exact expression for the operator K, we obtain ξi+1 (Kfρ )q b−1 (ρ−1 k (ξi ))

=

ξi+1

b−1 (ρ−1 k (ξi ))

w (x)

q

b(x)

q

vk (x, y)fρ (y) dy

dx

a(x)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

=

ξi+1

b−1 (ρ−1 k (ξi ))

≥

≥

b(x)

wq (x) a(x)

fρ (t)v(t)

ρ−1 k (ξi+1 )

ρ−1 k (ξi )

fρ (t)v(t)

ρ−1 k (ξi )

≥

fρ (t)v(t)

ρ−1 k (ξi )

=

ρ−1 k (ξi+1 )

ρ−1 k (ξi+1 )

ρ−1 k (ξi )

min{a−1 (t),ξi+1 } b−1 (t) ρk (t)

b−1 (t) ρk (t) b−1 (t)

[Wρ (t)]

×

dx

k(x, t)w (x)

q−1

b(x)

q

vk (x, y)fρ (y) dy a(x)

k(x, t)w (x)

q−1

t

q

dx dt

vk (x, y)fρ (y) dy

dx dt

a(x)

ρk (t)

r/(pq)

vk (x, y)fρ (y) dy a(x)

b−1 (t)

(by (2.1.9) with z = b−1 (t)) ρ−1 (ξi+1 ) k fρ (t)v(t)

S93

q−1

b(x)

vk (x, t)fρ (t) dt

ρ−1 k (ξi+1 )

ρ−1 k (ξi )

wkq (x, t)

wkq (x, t) dx

t r/(pq)

fρ (y)v(y) dy

dx dt

t

q−1

a(x)

fρ (y)v(y) dy

dt

a(ρk (t))

[Vρ− (t)]r/(pq )

[Wρ (y)]

q−1

t

ρk (t)

b−1 (t)

wkq (x, t) dx

[Vρ− (y)]r/(pq ) v p (y) dy

q−1

v p (t) dt

a(ρk (t))

(by property (2.3.1)) ρ−1 (ξi+1 ) k [Vρ− (t)]r/(pq ) [Wρ (t)]r/(pq)+1 ρ−1 k (ξi )

×

t a(ρk (t))

ρk (t)

b−1 (t)

r/(pq) wkq (z, y) dz

[Vρ− (y)]r/(pq ) v p (y) dy

q−1

v p (t) dt.

From (2.1.9) we obtain k(z, y) k(z, t) = k(z, b(τ )) for τ = b−1 (t), because b−1 (y) ≤ b−1 (t) = τ ≤ z ≤ ρk (t) ≤ a−1 (y). Therefore, in view of (2.3.1), we have ξi+1 (Kfρ )q (x) dx b−1 (ρ−1 k (ξi ))

≈

ρ−1 k (ξi+1 ) ρ−1 k (ξi )

[Vρ− (t)]r/(pq ) [Wρ (t)]r/q

ρ−1 k (ξi )

t

y

v a(ρk (t))

ρ−1 k (ξi+1 )

It follows from

p

r/(pq )

p

v (y) dy

q−1

v p (t) dt

a(ρk (t))

Bρ−k (t) dt.

−1 −1 i [ρk (ξi ), ρk (ξi+1 ))

i

= (0, ∞) that

ξi+1

1/q

q

b−1 (ρ−1 k (ξi ))

(Kfρ ) (x) dx

(Bρ−k )r/q .

The relation r/q − r/p = 1, together with (2.3.51), gives (2.3.50). To prove KLp →Lq Bσ+k ,

(2.3.52)

we pass to the following inequality dual to (2.3.49):

p 1/p ∞ a−1 (y) p wk (x, y)g(x) dx v (y) dy ≤ K · gq , 0

(2.3.53)

b−1 (y)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S94

PROKHOROV et al.

where k(x, y) satisﬁes the condition D−1 k(x, y) ≤ k(x, z) + k(b−1 (z), y) ≤ Dk(x, y),

(2.3.54)

which follows from (2.1.9), for all a(x) ≤ y ≤ z ≤ b(x). As previously, we break (0, ∞) by the points ξi of the sequence (2.1.6) with ξ0 = 1. Setting a−1 (σk (t)) b(t) + q Wσ (t) = w (x) dx and Vσ (t) = vkp (t, y) dy, t

a(t)

we arrive at [Wσ+ (t)]r/(pq ) [Vσ (t)]r/(p q ) [w(t)]q−1 = Bσ+k (t)1/q . Let

gσ (t) = [Wσ+ (t)]r/(pq ) [Vσ (t)]r/(p q ) [w(t)]q−1 ; then

gσ q = (Bσ+k )r/q . 6

b(x) σk (x) a(x)

b(ξi+1 )

σk (ξi ) ξi

a−1 (σk (t)) a−1 (σk (ξi+1 ))

ξi+1

t

ξi+2

-

Fig. 2.3.4.

Applying this relation and

= (0, ∞) to (2.3.53), we obtain σk (ξi+1 ) 1/p p ≥ Kgσ p = (Kgσ ) (y) dy

i [σk (ξi ), σk (ξi+1 ))

K(Bσ+k )r/q

−1/p

≥2

i

where Kg(y) := v(y) b(ξi+1 ) (Kgσ )p σk (ξi )

b(ξi+1 )

= σk (ξi )

a−1 (y) b−1 (y)

b(ξi+1 )

p

(Kgσ ) (y) dy

1/p (2.3.55)

,

σk (ξi )

wk (x, y)g(x) dx. Therefore,

v (y) p

σk (ξi )

i

p

a−1 (y) b−1 (y)

wk (x, y)gσ (x) dx

dy

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

b(ξi+1 )

=

v p (y)

σk (ξi )

≥

wk (t, y)gσ (t) dt

b−1 (y)

ξi+1

a−1 (y)

ξi+1

≥

b−1 (y)

k(t, y)v (y)

b(t)

p

σk (t)

dy p −1

a−1 (y)

b−1 (y)

max{a(t),σk (ξi )}

gσ (t)w(t) ξi

wk (x, y)gσ (x) dx

p

gσ (t)w(t) ξi

p −1

a−1 (y)

k(t, y)v (y)

b(t)

S95

wk (x, y)gσ (x) dx

dy dt

p −1

a−1 (y)

wk (x, y)gσ (x) dx

dy dt

t

(by (2.3.54) with z = b(t))

ξi+1

b(t)

gσ (t)w(t)

≥

ξi

σk (t)

ξi+1

b(t)

gσ (t)w(t) σk (t)

ξi

ξi+1

=

vkp (t, y)

gσ (x)w(x) dx

gσ (x)w(x) dx

a−1 (σk (t))

×

dt

t b(t)

σk (t)

p −1

a−1 (σk (t))

[Wσ+ (t)]r/(pq ) [Vσ (t)]r/(p q )

ξi

dy dt

t

vkp (t, y) dy

p −1

a−1 (y)

vkp (t, y) dy

[Wσ+ (x)]r/(pq ) [Vσ (x)]r/(p q ) wq (x) dx

p −1

wq (t) dt

t

(by property (2.3.2)) ξi+1 [Wσ+ (t)]r/(pq ) [Vσ (t)]r/(p q )+1 ξi

a−1 (σk (t)) b(t)

× t

σk (t)

vkp (x, z) dz

r/(p q )

[Wσ+ (x)]r/(pq ) wq (x) dx

p −1

wq (t) dt.

Inequalities (2.3.54) imply k(x, z) k(t, z), because a(x) ≤ σ(t) ≤ z ≤ z = b(t) ≤ b(x). Thus, in view of (2.3.2), we have b(ξi+1 ) (Kgk )p σ(ξi )

ξi+1

ξi ξi+1

[Wσ+ (t)]r/(pq ) [Vσ (t)]r/p

≈

ξi

t

a−1 (σk (t)) a−1 (σk (t))

r/(pq ) wq

wq (x) dx

p −1

wq (t) dt

x

Bσ+k (t) dt.

This and (2.3.55) imply KLp (0,∞)→Lq (0,∞) = KLq (0,∞)→Lp (0,∞) Bσ+k .

(2.3.56)

Combining (2.3.50) and (2.3.56), we arrive at the required inequality on the left-hand side of (2.3.11). The assertions concerning compactness in the case q < p follows directly from the necessary and ` theorem. suﬃcient boundedness conditions proved above and Ando’s Theorem 2.6 stated below is proved in a similar way by using Theorem 2.2, Statement 2, and Lemmas 1.12 and 1.13. Theorem 2.6. Suppose that the kernel k(x, y) > 0 of an operator K of the form (2.1.4) satisﬁes the conditions in Deﬁnition 2.5 and belongs to the Oinarov-type class Oa . Let ρk (y) and σk (x) be strictly increasing k-fairways (see Deﬁnition 2.5). PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S96

PROKHOROV et al.

(a) If 1 < p ≤ q < ∞, then − A+ ρk + Aσk KLp (0,∞)→Lq (0,∞) Aρk + Aσk .

Moreover, if K is compact, then − A+ ρk , Aσk < ∞,

− lim A+ ρk (t) = lim Aσk (t) = 0,

t→0 t→∞

t→0 t→∞

and, conversely, the operator K is compact if Aρk , Aσk < ∞,

lim Aρk (t) = lim Aσk (t) = 0.

t→0 t→∞

t→0 t→∞

(b) If 1 < q < p < ∞, then Bρ+k + Bσ−k KLp (0,∞)→Lq (0,∞) Bρk + Bσk . Moreover, K is a compact operator from Lp (0, ∞) to Lq (0, ∞) if Bρk , Bσk < ∞, and, conversely, if K : Lp (0, ∞) → Lq (0, ∞) is compact, then Bρ+k , Bσ−k < ∞. Combining Theorems 2.5 and 2.6, we obtain several boundedness criteria for (2.1.4) as an operator from Lp (0, ∞) to Lq (0, ∞). Theorem 2.7. Let K be the operator deﬁned by (2.1.4) with kernel k(x, y) > 0 satisfying the conditions in Deﬁnition 2.5 on the domain R, and let ρk (x) and σk (x) be strictly increasing k-fairways (see Deﬁnition 2.5). (a) If k(x, y) ∈ Ob and p v ≈ vp , t > 0, (2.3.57) ϑ− (t)

ϑ(t)

δ+ (t)

then KLp →Lq

wq ≈

wq ,

(2.3.58)

t > 0,

δ(t)

Aρk + Aσk , 1 < p ≤ q < ∞, ≈ Bρk + Bσk , 1 < q < p < ∞.

(2.3.59)

Moreover, for 1 < p ≤ q < ∞, the operator K : Lp (0, ∞) → Lq (0, ∞) is compact if and only if Aρk , Aσk < ∞,

lim Aρk (t) = lim Aσk (t) = 0,

t→0 t→∞

t→0 t→∞

and for 1 < q 0, ϑ+ (t)

w ≈ q

δ− (t)

(2.3.60)

ϑ(t)

wq ,

(2.3.61)

t > 0,

δ(t)

then estimate (2.3.59) holds. Moreover, for 1 < p ≤ q < ∞, K is compact as an operator from Lp (0, ∞) to Lq (0, ∞) if and only if Aρk , Aσk < ∞,

lim Aρk (t) = lim Aσk (t) = 0,

t→0 t→∞

t→0 t→∞

and for 1 < q < p < ∞, this operator is compact if and only if Bρk , Bσk < ∞. PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S97

Theorem 2.8. Suppose that the kernel 0 < k(x, y) ∈ Oa ∩ Ob of an operator K of the form (2.1.4) satisﬁes the conditions in Deﬁnition 2.5. Let ρk (x) and σk (x) be strictly increasing k-fairways (see Deﬁnition 2.5). (a) If 1 < p ≤ q < ∞, then KLp (0,∞)→Lq (0,∞) ≈ Aρk + Aσk ,

(2.3.62)

and K : Lp (0, ∞) → Lq (0, ∞) is compact if and only if Aρk , Aσk < ∞,

lim Aρk (t) = lim Aσk (t) = 0.

t→0 t→∞

t→0 t→∞

(b) If 1 < q < p < ∞, then KLp (0,∞)→Lq (0,∞) ≈ Bρk + Bσk ,

(2.3.63)

and K is compact as an operator from Lp (0, ∞) to Lq (0, ∞) if and only if Bρk , Bσk < ∞. In conclusion of this section, we give several examples. The ﬁrst is related to Theorem 2.7. Example 2.1. Let p = q = 2 and suppose that v(y) = y −3/2 , w(x) = 1, k(x, y) = (y − a(x))1/2 , a(x) = x/2, and b(x) = 2x, i.e., 2x x 1/2 f (y)y −3/2 dy. y− K1 f (x) = 2 x/2 By virtue of (2.3.1), we have Lv (x) :=

σ(x) x/2

2x x −3 x −3 y dy = y dy = Rv (x). y− y− 2 2 σ(x)

Integrating by parts, we obtain 1 σ(x) x dy −2 y− Lv (x) = − 2 x/2 2 x −2 1 1 σ(x) −1 1 x σ (x) − = − σ(x) − dy = − σ −1 (x) + σ −2 (x), 2 2 2 x/2 x 4 2x 1 5 x x dy −2 = − + σ −1 (x) − σ −2 (x). y− Rv (x) = − 2 σ(x) 2 8x 4 Since x/2 ≤ σ(x) ≤ 2x, it follows that

√ 2(4 + 3 ) x. σ(x) = 13

Similarly, by virtue of (2.3.2), we have 2y ρ(y) x x dx = dx =: Rw (y). y− y− Lw (y) := 2 2 y/2 ρ(y) We also have

3 2 ρ(y) 2 y − y− dz = , Lw (y) = 4 2 y−ρ(y)/2 y−ρ(y)/2 ρ(y) 2 2 dz = y − . Rw (y) = 2 0

y−y/4

2

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S98

PROKHOROV et al.

Therefore,

√ 4 2−3 √ y. ρ(y) = 2 2

The operator K1 with kernel k(x, y) = (y − a(x))1/2 ∈ Oa is bounded as an operator from L2 (0, ∞) to L2 (0, ∞). Indeed, it follows from (2.3.10) that 1/2 4√√2−3 t 1/2 2t 2 x −3 dx y dy t− Aρk := sup √ 4 √ 2−3 2 t>0 t/2 t 4 2 √ 3 t 1/2 4√√2−3 t 1/2 4 4 2 3 15 1 2 −2 √ < ∞, = sup √ dz dy = √ 4 √2−3 2 0 4(4 2 − 3) t>0 t 2 Aσk

√ 4(4+ 3 ) 13

1/2 t −3 y dy dx y− √ 4+ 3 2 t>0 t/2 t 13 " √ √ 3(4 + 3 ) 1/2 15 1/2 3 5(4 + 3 ) t < ∞; = sup = 13 16t 4 13 t>0

:= sup

1/2

t

2t

therefore, K1 L2 (0,∞)→L2 (0,∞) < ∞. Moreover, the relations

p

v = ϑ+ (t)

t

√ 4 √2−3 2

t

√ 4 √2−3 2

t

y

−3

√ 39 − 24 2 −2 √ t , dy = 2(41 − 24 2 )

15 √ t−2 , 41 − 24 2 ϑ(t) t √ t 9− 3 t, wq = dx = √ 4+ 3 13 δ− (t) t 13 √ √ 4(4+ 3 ) t 13 3(4 + 3 ) q t w = dx = √ 4+ 3 13 δ(t) t p

v =

√ 4 √ 2−3 4 2

y −3 dy =

13

imply (2.3.60) and (2.3.61). Thus, according to Theorem 2.7 (b), the conditions Aρk < ∞ and Aσk < ∞ are necessary and suﬃcient for the boundedness of the operator K1 : L2 (0, ∞) → L2 (0, ∞). Note that the proof of the upper estimates of KLp (0,∞)→Lq (0,∞) in Theorems 2.5 and 2.6 does not use the integral property (2.3.1) and (2.3.2) of k-fairways. This means that, to verify only the suﬃcient boundedness conditions for K, as σk (x) and ρ−1 k (x) we can take any, not necessarily satisfying (2.3.1) or (2.3.2), strictly increasing function φ(x) such that a(x) < φ(x) < b(x) for x > 0. We demonstrate this by the following example. γ x with γ > 0, v(y) = [h(y)]2/3 , w(x) = Example 2.2. Suppose that p = 3, q = 2, k(x, y) = y h √ [h(x)]1/2 with 0 ≤ h ∈ Lloc (0, ∞), a(x) = ln x + 1, and b(x) = exp 2x − 1, i.e., exp 2x−1 x γ h f (y)[h(y)]2/3 dy K2 f (x) = h(x)1/2 √

ln x+1 y x x

1/2

= h(x)

ln

√

γ 2/3 1/2 h f (y)[h(y)] dy + h(x)

x+1

y

exp 2x−1 y

x

γ h f (y)h(y)2/3 dy

x

=: K2,1 f (x) + K2,2 f (x), PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S99

! x γ p = 3/2, q = 2, and r = 6. Note that the kernel k2,1 (x, y) := y h in the deﬁnition of the opera! y γ tor K2,1 belongs to the class Ob , while k2,2 (x, y) := x h in the deﬁnition of K2,2 is of type Oa .

To prove the boundedness of the operator K2,1 in the case h(z) = exp(−z), we use the function √ φ1 (x) = x/2, where a(x) = ln x + 1 < φ1 (x) < x =: b1 (x), instead of fairways ρ−1 k (x) and σk (x). −1 (t) = exp 2t − 1, and φ−1 (t) = 2t, it follows that (t) = t, a Since b−1 1 3 2t 3 ∞ exp 2t−1 x 2γ 6 h h(x) dx h h(t) dt Bρk =φ−1 := √

0

t

t

∞ exp 2t−1 x

= 0

Bσ6k =φ

t

∞

2γ h d

t

t

=

0

=

t/2

∞

exp t−1 2 h

t/2

2 3γ + 2

4 0

ln

∞

t ln

√

t+1

t ln

√

t+1

exp t−1 t/2

√ 2t+1

2t ln

√

3 h h(t) dt 3 h h(t) dt,

2t+1

t

3γ/2 4 h h(y) dy h(t) dt

t

3γ/2 t 4 h d − h h(t) dt

y

2t+1

2t

exp 2t−1 3(2γ+1) h

1 = (2γ + 1)3 0 t ∞ exp t−1 2 := h 0

ln

3 h

x

y

2 h

y

t

ln

√ t+1

6γ+4 h h(t) dt.

Let h(z) = exp(−z). Then Bρ6k =φ−1

Bσ6k =φ

∞ 1 exp(−2(3γ + 2)t) ≤ dt (2γ + 1)3 0 (2t + 1)3/2 ∞ 1 1 exp(−2(3γ + 2)t) dt = < ∞, ≤ 3 (2γ + 1) 0 2(3γ + 2)(2γ + 1)3 4 ∞ 8 2 exp(−2t) ≤ dt ≤ < ∞. 3γ+2 3γ + 2 (t + 1) (3γ + 2)4 0

By Theorem 2.5, it follows that the operator K2,1 is bounded and compact as an operator from L3 to L2 . Similarly, Theorem 2.6 with φ2 (x) = 2x instead of σk and ρ−1 k implies the boundedness and compactness of the transformation K2,2 . Therefore, if h(z) = exp(−z), then K2 is bounded and compact as an operator from L3 to L2 .

2.4. Further Results A function σk introduced in Deﬁnition 2.5 extends the notion of a fairway σ in Deﬁnition 2.3. Together with ρk , the extended k-fairway σk is a tool for obtaining new forms of boundedness criteria for the generalized Hardy–Steklov operator (2.1.4). Unfortunately, the conditions imposed on the structure of σk in Section 2.3 are stronger and, at that, σk is less eﬀective than the original fairway σ. The main advantage of results of Section 2.3 is that they provide much more concise boundedness conditions for the operator K than those given by Theorems 2.1 and 2.2. However, to become both necessary and suﬃcient, these conditions must be supplemented by additional requirements (see Theorem 2.7 and 2.8). Obviously, this does not best aﬀect the possibility of applying results obtained for K by using σk and ρk . At the same time, the fairway method has brought out many interesting facts useful for the study of not only the properties of the Hardy–Steklov operator H : Lp → Lq of the form (2.2.1) but also its numerous applications (see, e.g., [25–31]). In particular, two diﬀerent types of boundedness criteria in terms of σ for (2.2.1) as an operator from Lp (0, ∞) to Lq (0, ∞) for all p > 1 and q > 0 have been obtained (see Theorem 2.3 and 2.4 or [30, Theorems 4.1 and 4.2] and [25, Theorem 2.1]). Subsequently, these results have been applied to solve some related problems (see Chapter 3), such as those of characterizing PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S100

PROKHOROV et al.

the geometric mean operator with two variable limits of integration (see Section 3.1), Sobolev-type embedding inequalities (Sections 3.3 and 3.4.2), and so on. Note that the operator a−1 (y) ∗ g(x)w(x) dx (2.4.1) H g(y) = v(y) b−1 (y)

from Lq to Lp , which is dual to the transformation H : Lp → Lq , is also classiﬁed with Hardy–Steklov operators. The reformulation of Theorem 2.3 for H∗ leads us to the following deﬁnition and results. Deﬁnition 2.6. Given boundary functions a(x) and b(x) satisfying conditions (2.1.5), a number q > 0, and a weight function w(x) locally qth-power integrable on (0, ∞) and such that 0 < w(x) < ∞ for almost all x ∈ (0, ∞), a dual fairway-function ρ(y) is deﬁned so that b−1 (y) < ρ(y) < a−1 (y) on (0, ∞) and a−1 (y) ρ(y) q w (x) dx = wq (x) dx for all y > 0. (2.4.2) b−1 (y)

ρ(y)

We can assume that the functions a−1 and b−1 inverse to the given boundary functions a, b ∈ (2.1.5) satisfy conditions (2.1.5) as well. Therefore, by analogy with σ (see also the comments on the Deﬁnition 2.5 of k-fairways σk and ρk ), the dual fairway ρ is diﬀerentiable and strictly increasing on (0, ∞) as well. The following theorem follows from Theorems 2.3 and 2.4 by duality. However, presently, we can assert its validity only for p > 1 and q > 1. Theorem 2.9. Let H be an operator of the form (2.2.1) from Lp (0, ∞) in Lq (0, ∞), where 1 0

∗ AT := sup t>0

b−1 (t)

a(ρ(t))

ρ−1 (a−1 (t))

ρ−1 (b−1 (t))

a−1 (y) b−1 (y)

w

p

q

p

1/p

v (y) dy

a−1 (t)

b−1 (t)

w

q

− 1 q

,

and A∗M ≈ A∗T ≈ HLp (0,∞)→Lq (0,∞) . ∗ For 1 < q < p < ∞, the operator H : Lp (0, ∞) → Lq (0, ∞) is bounded if and only if BM R < ∞, ∗ ∗ or if and only if BP S < ∞, where BM R ≈ BP S ∗ ≈ HLp (0,∞)→Lq (0,∞) and ∞ b(ρ(t)) r/q a−1 (t) r/q 1/r ∗ p q p v w v (t) dt , (2.4.4) BM R := 0

BP∗ S

:=

0

∞

a(ρ(t)) ρ−1 (a−1 (t)) ρ−1 (b−1 (t))

b−1 (t)

a−1 (y)

b−1 (y)

w

p q

p

v (y) dy

r/q

a−1 (t) b−1 (t)

w

q

p −r/q

p

1/r

v (t) dt

.

In this section, we prove that criteria dual to the Muckenhoupt-type criterion (2.4.3) and the Maz’ya– Rosin-type criterion (2.4.4) for the boundedness of (2.2.1) as an operator from Lp (0, ∞) to Lq (0, ∞) remain valid and in the case where p > 1 and 0 < q < 1 (see Theorem 2.10 (c, e)) and can be obtained in terms of a fairway σ for all p > 1 and q > 0. In addition, we ﬁnd boundedness criteria for (2.2.1) in their initial (nondual) form (see (2.2.20) and (2.2.22)) but in terms of the dual fairway-function ρ instead of σ. Note that, unlike Theorem 2.3, the suﬃcient boundedness conditions for H : Lp (0, ∞) → Lq (0, ∞) presented in this section can be proved without requiring that σ satisﬁes (2.2.2) or ρ satisﬁes (3.2.28) (see Theorem 2.10 (a, b)) for p, q > 1. The presence of several forms of boundedness characteristics for an operator may facilitate solving problems related to this operator. In particular, this is the case for problems concerning the embedding inequality (3.3.1), which is studied in Section 3.3. Inequality (3.3.1) is related to the Hardy–Steklov PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S101

operator (see Section 3.3) and, depending on the forms of the weight functions v and w, deriving an explicit expression for the fairway ρ from condition (2.4.2) may be easier than that for the fairway σ determined by Eq. (2.2.2), and vice versa. Given boundary functions a(x) and b(x) satisfying (2.1.5), we assume that a function y = ς(x) is continuous, strictly increases on (0, ∞), and satisﬁes the condition a(x) < ς(x) < b(x) for x > 0. Then the inverse function x = ς −1 (y) =: ς ∗ (y) is continuous and strictly increasing on (0, ∞) as well, and b−1 (y) < ς ∗ (y) < a−1 (y) for y > 0. For ς, we set Δ(t) := Δ− (t) ∪ Δ+ (t),

Δ− (t) := [a(t), ς(t)),

Δ+ (t) := [ς(t), b(t)),

δ(t) := δ− (t) ∪ δ+ (t),

δ− (t) := [b−1 (ς(t)), t),

δ+ (t) := [t, a−1 (ς(t))),

Θ(t) := Θ− (t) ∪ Θ+ (t),

Θ− (t) := [b−1 (t), ς ∗ (t)),

Θ+ (t) := [ς ∗ (t), a−1 (t)),

ϑ− (t) := [a(ς ∗ (t)), t), ϑ+ (t) := [t, b(ς ∗ (t))). ϑ(t) := ϑ− (t) ∪ ϑ+ (t), We deﬁne WI (t) := I(t) wq and VI (t) := I(t) v p for I(t) ⊂ (0, ∞). We also use the following notation:

(Aς ∗ )∗ := sup[WΘ (t)]1/q [Vϑ (t)]1/p ,

Aς := sup[Wδ (t)]1/q [VΔ (t)]1/p , t>0

A± ς

1/q

:= sup[Wδ± (t)]

1/p

[VΔ (t)]

t>0

∞

Bς :=

r/p

[Wδ ]

r/p

[VΔ ]

∞

,

1/r w

q

0

Bς± :=

t>0 ∗ (A± ς∗ )

:= sup[WΘ (t)]1/q [Vϑ± (t)]1/p , t>0

∗

,

∞

(Bς ∗ ) :=

r/q

[WΘ ]

r/q p

[Vϑ ]

1/r

v

,

0

1/r

[Wδ± ]r/p [VΔ ]r/p wq

(Bς±∗ )∗ :=

,

0

∞

[WΘ ]r/q [Vϑ± ]r/q v p

1/r .

0

± ∗ ∗ Here, ( · )∗ emphasizes the dual character of the functionals (Aς ∗ )∗ , (Bς ∗ )∗ , (A± ς ∗ ) , and (Bς ∗ ) . If ς = σ, then Aσ = AM , Bσ = BM R (see (2.2.20) and (2.2.22)), and a−1 (t) 1/q b(σ−1 (t)) 1/p ∗ wq vp , (Aσ−1 ) = sup t>0

∗

(Bσ−1 ) = 0

b−1 (t)

∞

a(σ−1 (t))

r/q

a−1 (t)

b−1 (t)

w

q

b(σ−1 (t)) a(σ−1 (t))

v

p

r/q

p

1/r

v (t) dt

.

∗ For ς ∗ = ρ, we have (Aρ )∗ = A∗M , (Bρ )∗ = BM R (see Theorem 2.9), and a−1 (ρ−1 (t)) 1/q b(t) 1/p wq vp , Aρ−1 = sup t>0

Bρ−1 =

0

b−1 (ρ−1 (t))

a(t)

∞ a−1 (ρ−1 (t)) b−1 (ρ−1 (t))

r/p w

q

b(t)

v

p

(2.4.5) 1/r

r/p q

w (t) dt

(2.4.6)

.

a(t)

± ± ± ± ± ∗ ∗ ± ∗ ± ∗ In a similar way, we form functionals A± σ , Bσ , (Aσ−1 ) , (Bσ−1 ) , Aρ−1 , Bρ−1 , (Aρ ) , and (Bρ ) . Note that + ∗ + ∗ ∗ (A− A− ς + Aς ≈ Aς , ς ∗ ) + (Aς ∗ ) ≈ (Aς ∗ ) ,

Bς− + Bς+ ≈ Bς ,

(Bσ−−1 )∗ + (Bσ+−1 )∗ ≈ (Bσ−1 )∗

for all p > 1 and q > 0, while (Bρ− )∗ + (Bρ+ )∗ ≈ (Bρ )∗ only if p > 1 and q > 1. Theorem 2.10. Suppose that p > 1, q > 0, q = 1, and H is an operator of the form (2.2.1). Let σ(x) be a continuous strictly increasing function on (0, ∞) such that a(x) < σ(x) < b(x) for x > 0, and let ρ(y) be a continuous strictly increasing function on (0, ∞) satisfying the condition b−1 (y) < ρ(y) < a−1 (y) for all y > 0. Finally, let ς(x) denote either σ(x) or ρ−1 (x) on (0, ∞). (a) If 1 1 and 0 < q < p < ∞, then HLp (0,∞)→Lq (0,∞) Bς . Moreover, if 1 < q < p < ∞, then HLp (0,∞)→Lq (0,∞) (Bς ∗ )∗ . (c) If 0 < q < 1 < p < ∞ and ρ is a dual fairway-function satisfying condition (2.4.2), then HLp (0,∞)→Lq (0,∞) [(Bρ− )∗ + (Bρ+ )∗ ]. If σ is a fairway-function satisfying condition (2.2.2), then HLp (0,∞)→Lq (0,∞) (Bσ−1 )∗ . (d) If 1 < p ≤ q < ∞ and σ is a fairway-function satisfying (2.2.2), then HLp (0,∞)→Lq (0,∞) (Aσ−1 )∗ .

HLp (0,∞)→Lq (0,∞) Aσ , If ρ is a dual fairway satisfying (2.4.2), then

HLp (0,∞)→Lq (0,∞) (Aρ )∗ .

HLp (0,∞)→Lq (0,∞) Aρ−1 ,

(e) If 0 < q 1, and σ is a fairway-function satisfying condition (2.2.2), then HLp (0,∞)→Lq (0,∞) [(Bσ−−1 )∗ + (Bσ+−1 )∗ ] ≈ (Bσ−1 )∗ .

HLp (0,∞)→Lq (0,∞) Bσ ,

If ρ is a dual fairway-function satisfying (2.4.2), then HLp (0,∞)→Lq (0,∞) [(Bρ− )∗ + (Bρ+ )∗ ] ≥ (Bρ )∗ .

HLp (0,∞)→Lq (0,∞) Bρ−1 , Proof. (a) To prove the inequalities

HLp (0,∞)→Lq (0,∞) A∗ς ,

HLp (0,∞)→Lq (0,∞) Aς , we apply (2.2.23). Setting τ := ς −1 (a(t)), we write sup

b−1 (a(t))≤s≤t

where H1 (t) =

A(s, t) ≤

a−1 (ς(τ ))

sup

b−1 (ς(τ ))≤s≤τ

sup

b−1 (a(t))≤s≤τ
1/q

τ ≤s≤t

b(s)

wq

s

A(s, t) + sup A(s, t) =: H1 (t) + H2 (t),

vp

1/p ,

H2 (t) ≤ sup [Wδ+ (s)]1/q [VΔ (s)]1/p ,

ς(τ )

τ ≤s≤t

because t ≤ a−1 (ς(s))) and a(s) ≤ a(t) if τ ≤ s ≤ t. Therefore,

H1 (t) + H2 (t) ≤ [Wδ (τ )]1/q [VΔ+ (τ )]1/p + sup[Wδ+ (s)]1/q [VΔ (s)]1/p Aς , s>0

which proves the inequality HLp (0,∞)→Lq (0,∞) Aς . By duality, we have HLp (0,∞)→Lq (0,∞) = H∗ Lq (0,∞)→Lp (0,∞) (Aς ∗ )∗ . (b) To prove the estimate HLp (0,∞)→Lq (0,∞) Bς for 0 < q 1, we use the sequence (2.1.6) with ξ0 = 1. We set Δk := [ξk , ξk+1 ),

δk := [a(ξk ), b(ξk )).

Breaking the half-axis (0, ∞) by the points {ξk }k∈Z , we represent the operator H in the form H=T +S of the sum of the two block-diagonal operators T = k∈Z Tk and S = k∈Z Sk , where a(ξk+1 ) f (y)v(y) dy, Tk : Lp (δk ) → Lq (Δk ), Tk f (x) = w(x)

(2.4.7)

a(x)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

b(x)

Sk f (x) = w(x)

k∈Z Δk

= (0, ∞), and

S103

Sk : Lp (δk+1 ) → Lq (Δk ),

f (y)v(y) dy, b(ξk )

k∈Z δk

= (0, ∞). We deﬁne the points τk := ς −1 (b(ξk )) = ς −1 (a(ξk+1 )) and

write Sk f (x) = Sk− f (x) + Sk+ f (x),

(2.4.8)

where Sk− f (x) = χ(ξk ,τk ) (x)Sk f (x),

Sk− : Lp (b(ξk ), b(τk )) → Lq (ξk , τk ),

Sk+ f (x) = χ(τk ,ξk+1 ) (x)Sk f (x),

Sk+ : Lp (δk+1 ) → Lq (τk , ξk+1 ).

Applying estimate (1.5.43) in Lemma 1.12 to the norm of the operator Sk− and using the relation q − r/p + qr/p = r/p , we obtain r/p b(t) q−r/p τk t b(x) q − r p q v w (x) dx vp wq (t) dt Sk ≈ ≤

ξk

ξk

τk t

b(ξk )

r/p

wq

vp

ξk

ξk

b(t)

b(ξk )

r/p

wq (t) dt.

b(ξk )

Since ξk ≤ t ≤ ς −1 (b(ξk )), it follows that b−1 (ς(t)) ≤ ξk and ς(t) ≤ b(ξk ). Therefore, τk − r [Wδ− (t)]r/p [VΔ+ (t)]r/p wq (t) dt. Sk

(2.4.9)

ξk

Similarly, by virtue of (1.5.41), it follows from Lemma 1.12 that ξk+1 ξk+1 r/p b(t) r/p wq vp wq (t) dt Sk+ r ≈ τk

≤

b(ξk )

t

ξk+1

[Wδ+ (t)]r/p [VΔ (t)]r/p wq (t) dt,

(2.4.10)

τk

because ς −1 (a(ξk+1 )) ≤ t ≤ ξk+1 , and, therefore, ξk+1 ≤ a−1 (ς(t)) and a(t) ≤ a(ξk+1 ) = b(ξk ). In view of (2.4.8)–(2.4.10), this implies τk ξk+1 r r/p r/p q [Wδ− ] [VΔ+ ] w + [Wδ+ ]r/p [VΔ ]r/p wq . (2.4.11) Sk ξk

τk

Let us estimate Tk . Consider the decomposition Tk f (x) = χ(ξk ,τk ) (x)Tk f (x) + χ(τk ,ξk+1 ) (x)Tk f (x) =: Tk− f (x) + Tk+ f (x). Using estimate (1.5.49) in Lemma 1.13, we obtain τk t r/p − r wq Tk ≈ ξk

ξk

Since ξk ≤ t ≤

ς −1 (b(ξ

k )),

b(ξk )

v

p

(2.4.12)

r/p wq (t) dt.

a(t)

b−1 (ς(t))

it follows that ≤ ξk and b(ξk ) ≤ b(t), whence τk [Wδ− (t)]r/p [VΔ (t)]r/p wq (t) dt. Tk− r

(2.4.13)

ξk

Next, using (1.5.51) in ‘Lemma 1.13 and taking into account the relation q − r/p + qr/p = r/p , we arrive at the estimate r/p b(ξk ) q−r/p ξk+1 ξk+1 b(ξk ) q + r p q v w (x) dx vp wq (t) dt Tk ≈ τk

t

a(x)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

a(t)

Vol. 300 Suppl. 2

2018

S104

PROKHOROV et al.

≤ ≤

ξk+1 ξk+1 τk ξk+1

r/p

b(ξk )

wq

vp

r/p wq (t) dt

a(t)

t

[Wδ+ (t)]r/p [VΔ− (t)]r/p wq (t) dt,

(2.4.14)

τk

because ς −1 (a(ξk+1 )) ≤ t ≤ ξk+1 ; as a consequence, ξk+1 ≤ a−1 (ς(t)), b(ξk ) = a(ξk+1 ) ≤ ς(t). Relations (2.4.12)–(2.4.14) imply τk ξk+1 #r/p q Tk r [Wδ− ]r/p [VΔ ]r/p wq + [Wδ+ ]r/p VΔ− w . ξk

τk

Using this relation, relations (2.4.11) and (2.4.7), and Lemma 2.1, we obtain [Sk r + Tk r ] Hr ≈ T r + Sr ≈ k

k

≤

τk

r/p

[Wδ− ]

r/p

[VΔ ]

q

ξk+1

w +

ξk

r/p

[Wδ+ ]

r/p

[VΔ ]

w

q

τk

ξk+1

[Wδ ]r/p [VΔ ]r/p wq = Bςr ,

ξk

k

which proves the required estimate HLp (0,∞)→Lq (0,∞) Bς . Thus, for q > 1, we have HLp (0,∞)→Lq (0,∞) = H∗ Lq (0,∞)→Lp (0,∞) (Bς ∗ )∗ by duality. (c) Suppose that 0 < q < 1 < p < ∞. Let {ξk }k∈Z , ξk = 1, be a sequence of the form (2.1.6), and let ς denote either a dual fairway-function ρ−1 satisfying condition (2.4.2) or a fairway-function σ satisfying (2.2.2). We set ξk− := ς −1 (a(ξk )),

− Δ− k := [ξk , ξk ),

ξk+ := ς −1 (b(ξk )),

+ Δ+ k := [ξk , ξk )

and represent our operator in the form of the sum of operators H ± =

H = H− + H+

± k∈Z Hk ,

Hk± f (x)

(2.4.15)

where

= χΔ± (x)w(x) k

b(x)

f (y)v(y) dy a(x)

+ + + p q and Hk− : Lp [a(ξk− ), b(ξk )) → Lq (Δ− k ) and Hk : L [a(ξk ), b(ξk )) → L (Δk ).

Note that if ς = σ, then (Bσ−−1 )∗ + (Bσ+−1 )∗ ≈ (Bσ−1 )∗ by virtue of (2.2.2). Therefore, to prove the estimates of HLp (0,∞)→Lq (0,∞) required in this part of the theorem, it suﬃces to show that HLp (0,∞)→Lq (0,∞) [(Bς−∗ )∗ + (Bς+∗ )∗ ]. Suppose that (Bς±∗ )∗ < ∞. Since q < 0, we can assume for simplicity that WΘ (t) < ∞,

0 < Vϑ± (t) < ∞

for any

t > 0.

Note that (Hk− |f |)(x)

≤ w(x)

b(x) a(ξk− )

|f |v,

(Hk+ |f |)(x)

b(ξk+ )

≤ w(x)

|f |v,

a(x)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

and Lemmas 1.12 and 1.13 give Hk− r

Hk+ r Since

r/p

r/q

Δ− k

r/p

ξk

w

z

w

Δ+ k

a(ξk− ) b(ξk+ )

r/p

z

b(z)

q

q

v

ξk

p

r/p wq (z) dz, r/p wq (z) dz.

a(z)

q

+ 1 and < 0, it follows that ξk r/p b(ξk ) ξk − r wq Hk =

v

p

a(ξk− )

≤

b(ξk )

a(ξk− )

= =:

max{ξk− ,b−1 (t)}

max{ξk− ,b−1 (t)}

a(ξk− )

I− k

+

r/q

w

q

+

v

p

q

r/q

w

w (z) dz v p (t) dt v

q

r/q

a(ξk− ) max{b(ξk− ),t}

max{ξk− ,b−1 (t)}

a(ξk )

b(z)

a(ξk− )

b(ξk ) ξk

a(ξk )

z

ξk

S105

p

r/q

v p (t) dt

max{b(ξk− ),t}

v

a(ξk− )

p

r/q

v p (t) dt

II− k.

Note that I− k

=

and II− k

(2.4.16)

a(ξk− )

≤

a(ξk )

p

v (t) dt

ξk−

b(ξk ) ξk b−1 (t)

a(ξk )

r/q

ξk

w

v

a(ξk− )

r/q w

b(ξk− )

q

t

q a(ξk− )

v

p

p

r/q

r/q

v p (t) dt

in view of q < 0. Since a(ξk− ) ≤ a(ς −1 (t)) and ξk ≤ a−1 (t) if a(ξk ) ≤ t ≤ b(ξk ), it follows that b(ξk ) − [WΘ (t)]r/q [Vϑ− (t)]r/q v p (t) dt. (2.4.17) IIk ≤ a(ξk )

If ς = σ, then condition (2.2.2) and the relation a(ξk ) = σ(ξk− ) imply ξk r/q b(ξ − ) r/q b(ξ − ) k k − p v (t) dt wq vp . Ik = σ(ξk− )

ξk−

a(ξk− )

Since a(ξk− ) ≤ a(σ −1 (t)) for σ(ξk− ) ≤ t ≤ b(ξk− ) and [ξk− , ξk ) ⊆ Θ(t), it follows that b(ξ − ) k #r/q #r/q p − v (t) dt. WΘ (t) Vϑ− (t) Ik ≤

(2.4.18)

a(ξk )

If ς = ρ−1 , then, by virtue of (2.4.2), we have a(ξk ) − p v (t) dt Ik = a(ξk− )

r/q

ξk−

w

ξk−1

q

b(ξk− )

a(ξk− )

v

p

r/q ,

and since b(ρ(t)) ≤ b(ξk− ) for a(ξk− ) ≤ t ≤ a(ξk ) and [ξk−1 , ξk− ) ⊆ Θ(t), it follows that a(ξk ) − [WΘ (t)]r/q [Vϑ+ (t)]r/q v p (t) dt. Ik ≤

(2.4.19)

a(ξk− )

Using (2.4.16)–(2.4.19) and Lemma 2.1, we arrive at − − r [H2k−1 r + H2k ] H − r ≈ k

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S106

PROKHOROV et al.

b(ξk )

r/q

[WΘ ]

r/q p

[Vϑ− ]

v +

a(ξk )

k

a(ξk ) a(ξk− )

k

[WΘ ]r/q [Vϑ+ ]r/q v p

≤ [(Bς−∗ )∗ ]r + [(Bς+∗ )∗ ]r .

(2.4.20)

By analogy with (2.4.16)–(2.4.19), we obtain b(ξk ) + r r/q r/q p Hk [WΘ ] [Vϑ+ ] v + a(ξk )

therefore, H + r

k

b(ξk+ )

[WΘ ]r/q [Vϑ− ]r/q v p ;

b(ξk )

b(ξk )

r/q

[WΘ ]

r/q p

[Vϑ+ ]

v +

a(ξk )

b(ξk+ )

[WΘ ]r/q [Vϑ− ]r/q v p

b(ξk )

k

≤ [(Bς−∗ )∗ ]r + [(Bς+∗ )∗ ]r .

(2.4.21)

Thus, HLp (0,∞)→Lq (0,∞) [(Bς−∗ )∗ + (Bς+∗ )∗ ] by virtue of (2.4.15), (2.4.20), and (2.4.21). (d) Let ς denote either a fairway σ with property (2.2.2) or a dual fairway ρ−1 satisfying (2.4.2). To prove the lower estimates required in this part theorem in the case 1 0, we have

[Wδ− (t)]1/q [VΔ− (t)]1/p + [Wδ+ (t)]1/q [VΔ+ (t)]1/p

= A(b−1 (ς(t)), t) + A(t, a−1 (ς(t))) sup

sup

t>0 b−1 (a(t))≤s≤t

A(s, t).

This relation and (2.2.23) imply HLp (0,∞)→Lq (0,∞) Aσ for ς = σ ∈ (2.2.2). By duality, for ς ∗ = ρ ∈ (2.4.2), we have HLp (0,∞)→Lq (0,∞) = H∗ Lq (0,∞)→Lp (0,∞) (Aρ )∗ . Substituting σ(t) = τ and ρ(t) = τ , we obtain HLp (0,∞)→Lq (0,∞) (Aσ−1 )∗ from HLp (0,∞)→Lq (0,∞) Aσ for ς ∗ = σ −1 and HLp (0,∞)→Lq (0,∞) Aρ−1 from HLp (0,∞)→Lq (0,∞) (Aρ )∗ for ς = ρ−1 . (e) We know from Theorem 2.3 that HLp (0,∞)→Lq (0,∞) Bσ in the case where 0 < q 1, provided that the fairway-function σ satisﬁes condition (2.2.2). The argument used to prove this estimate for HLp (0,∞)→Lq (0,∞) in Theorem 2.3 implies also the inequality HLp (0,∞)→Lq (0,∞) (Bσ−1 )∗ . Indeed, suppose that σ(x) = x. Then 1/r λk (2.4.22) HLp →Lq lim N →∞

(see pp. S68–S72), where

ηk+2 ηk+2

λk =

r/q w

ηk−1

t

|k|≤N

t

q

v

p

r/q

v p (t) dt

ηk−1

and the sequence {ηk }k∈Z is deﬁned by (2.1.7) for η0 = 1. For q > 0, using the relation (2.2.2) σ(x) = x and arguing as in the proof of the estimate HLp (0,∞)→Lq (0,∞) Bσ in Theorem 2.3, we obtain r/q r/q ηk+1 q p w v v p (t) dt, (2.4.23) λk ≥ ηk

Θ+ (t)

ϑ(t)

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S107

because Θ+ (t) = [t, a−1 (t)) ⊆ [t, ηk+2 ) and ϑ− (t) = [a(t), t) ⊆ [ηk−1 , t) for t ∈ [ηk , ηk+1 ). The same estimate holds for λk in the case q < 0, because ηk t b(ηk ) t b(t) b(t) t p p p p p p v = v + v = v + v ≤2 v ≤2 vp , t ∈ [ηk , ηk+1 ). ηk−1

ηk−1

ηk

ηk

ηk

ηk

a(t)

Combining (2.4.22) and (2.4.23), we see that because

HLp (0,∞)→Lq (0,∞) (Bσ+−1 )∗ ,

k∈Z [ηk , ηk+1 )

= (0, ∞). The inequality HLp (0,∞)→Lq (0,∞) (Bσ−−1 )∗

can be proved in a similar way but by using the intervals [ζk , ζk+1 ) determined by sequence (2.1.8) with ζ0 = 1. Thus, HLp (0,∞)→Lq (0,∞) (Bσ−−1 )∗ + (Bσ+−1 )∗ ≈ (Bσ−1 )∗

for σ(x) = x.

The same relation for any σ ∈ (2.2.2) is derived from the case σ(x) = x considered above by using the substitutions b(x) = σ −1 b((x)), a(x) = σ −1 (a(x)), f(y) = f (σ(t))[σ (t)]1/p , v(x) = v(σ(x))[σ (x)]1/p . Now, let ρ be a dual fairway satisfying condition (2.4.2). If 1 < q < p < ∞, then the estimates HLp (0,∞)→Lq (0,∞) (Bρ )∗

and

HLp (0,∞)→Lq (0,∞) Bρ−1

follow by duality from HLp (0,∞)→Lq (0,∞) Bσ

and

HLp (0,∞)→Lq (0,∞) (Bσ−1 )∗ ,

respectively (see Theorem 2.9). Moreover, since r/q > 0, we have (Bρ )∗ ≈ (Bρ− )∗ + (Bρ+ )∗ . It remains to prove the estimates HLp (0,∞)→Lq (0,∞) [(Bρ− )∗ + (Bρ+ )∗ ],

HLp (0,∞)→Lq (0,∞) Bρ−1

in the case where 0 < q < 1 < p < ∞ and ρ ∈ (2.4.2). Suppose that HLp (0,∞)→Lq (0,∞) < ∞. To obtain the required inequalities HLp (0,∞)→Lq (0,∞) [(Bρ− )∗ + (Bρ+ )∗ ] it suﬃces to show that HLp (0,∞)→Lq (0,∞) (Bρ± )∗

and

HLp (0,∞)→Lq (0,∞) Bρ−1 ,

and

HLp (0,∞)→Lq (0,∞) Bρ±−1 ,

and

HLp (0,∞)→Lq (0,∞) Bρ−−1 ,

where Bρ−1 ≈ Bρ−−1 + Bρ+−1 . First, consider ρ(y) = y. To prove HLp (0,∞)→Lq (0,∞) (Bρ+ )∗

we take sequence (2.1.7) with η0 = 1 and construct intervals κjk , j = 1, . . . , jb , jb = jb (k), on each [ηk , ηk+1 ) as follows: (1) if ηk+1 ≤ b(ηk ), then we set jb = 1 and [ηk , ηk+1 ) = κ1k ; (2) if b(ηk ) < ηk+1 ≤ b(b(ηk )), then we set jb = 2 and [ηk , ηk+1 ) = and κ2k = [b−1 (ηk+1 ), ηk+1 );

2

k j=1 κj ,

(3) if b(b(ηk )) < ηk+1 ≤ b(b(b(ηk ))), then we set jb = 3 and [ηk , ηk+1 ) = [ηk , b(ηk )), κ2k = [b(ηk ), b(b(ηk ))), and κ3k = [b−1 (ηk+1 ), ηk+1 );

where κ1k = [ηk , b(ηk )) 3

k j=1 κj ,

where κ1k =

... ................................................................................................... . PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S108

PROKHOROV et al.

j b

Finally, we obtain [ηk , ηk+1 ) = j=1 κjk , where κjk = [b(j−1) (ηk ), b(j) (ηk )) for j = 1, . . . , jb − 1 and κjkb = [b−1 (ηk+1 ), ηk+1 ). For each k ∈ Z and any j ∈ {1, . . . , jb }, we have wq , y ∈ κjk (2.4.24) WΘ (y) ≈ κjk

(more details can be extracted from Lemma 2.4 with b−1 , a−1 , and ρ instead of σ). Let r/(pq) mk r/(pq ) j q p w v , lk,j := κjk

a(mkj )

where mkj is the rightmost point of the interval κjk . Consider the function N j b (k)

fa (t) :=

χ[a(mk ),mk ] (t)lk,j v p −1 (t), j

N ∈ N.

j

k=−N j=1

By construction, we have [a(x), b(x)] ⊃ [a(mkj ), mkj ] if x ∈ κjk and each x ∈ (0, ∞) occurs in at most two intervals κjk . Since r/(pq ) + 1 = r/(p q) and r/p + 1 = r/q, it follows that, for any N ∈ N, 2Hfa qq

≥

N j b (k) k=−N j=1

≥

N j b (k)

κjk

k=−N j=1

=

N j b (k) k=−N j=1

κjk

b(x)

a(x)

q lk,j

fa (y)v(y) dy

wq (x) dx

a(x)

N j b (k) k=−N j=1

=

κjk

q

b(x)

j

mkj

v

p

r/q w

wq (x) dx

j

q

a(mkj ) q

q

lk,j v p (y)χ[a(mk ),mk ] (y) dy wq (x) dx κjk mkj

v

p

r/p =:

a(mkj )

N

jb (k)−1

k=−N

j=0

(2.4.25)

λk,j .

The intervals [mkj−1 , mkj ) formed by the rightmost points mkj of the intervals κjk with mk0 = ηk are disjoint; therefore, jb (k) mk N j |fa (y)|p dy fa pp = k=−N −1 j=1 N

=

jb (k) mk j

k=−N −1 j=1

=:

mkj−1

N

mkj−1

j b (k) i=j

jb (k+1) p lk,i χ[a(mk ),mk ] (y) + i i

i=1

p lk+1,i χ[a(mk+1 ),mk+1 ] (y) i i

v p (y) dy

jb (k)

(2.4.26)

Ik,j + IIk,j .

k=−N −1 j=1

We introduce the sets Ξ(k) := {i = j, . . . , jb (k) : (mkj−1 , mkj ) ∩ (a(mki ), mki ) = ∅}, ), mk+1 ) = ∅} Ξ(k + 1) := {i = 1, . . . , jb (k + 1) : (mkj−1 , mkj ) ∩ (a(mk+1 i i and ﬁrst consider k with |k| ≤ N . Note that Ξ(k) = {j, . . . , jb } for all such k, because, by construction, k a(mkjb (k) ) = a(mk+1 0 ) = a(ηk+1 ) = ηk = m0 .

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S109

Moreover, Ξ(k + 1) is empty if jb (k + 1) = 1 and Ξ(k + 1) = {1, . . . , jb (k + 1) − 1} if jb (k + 1) > 1. Therefore, for any ﬁxed k with |k| ≤ N and any j ∈ {j, . . . , jb (k)}, we have

jb (k)

Ik,j + IIk,j =

p lk,i

mkj

v + mkj−1

i=j

Since q < 0, it follows that r/(pq) q w lk+1,i ≤ κik+1

p

p lk+1,i

max{mkj−1 ,a(mk+1 )} i

Moreover, by virtue of (2.4.2) (see also (2.4.24)), ηk+2 ηk+1 q w =

vp .

r/(pq ) i ∈ Ξ(k + 1).

,

w ≈ q

b−1 (ηk+1 )

ηk+1

v

p

max{mkj−1 ,a(mk+1 )} i

i∈Ξ(k+1)

mkj

mkj

wq . κjk b

This relation and the inequalities r/q + 1 = r/p > 0 and r/q > 1 imply r/q mk r/p j q p w v IIk,j ≤ κik+1

i∈Ξ(k+1)

≤

mkj

b

v

p

max{mkj−1 ,a(mk+1 )} i

r/p

a(mkj )

≤2

b

vp

r/p

a(mkj ) b

r/q w

κik+1

i∈Ξ(k+1)

b

mkj

≤

q

mkj

b

r/p

r/q

k+1 i∈Ξ(k+1) κi

a(mkj ) b

r/q

ηk+2

v

p

w

q

≈ λk,jb .

wq

ηk+1

(2.4.27)

Since q < 0 and a(mkjb (k) ) = mk0 , it follows that lk,i ≤

r/(pq) w

mkj

q

v

κik

p

r/(pq ) i ∈ Ξ(k).

,

mkj−1

Therefore, using (2.4.2), we obtain the following relations by analogy with (2.4.27): mk mk r/p j r/q r/p mk r/q b (k) j j jb p q p q v w ≤2 v w Ik,j ≤ ≤

mkj−1

i=j

mkj

v

p

r/p

κik

r/q

a−1 (mkj )

w

a(mkj )

a(mkj )

q

r/q

mkj−1

mkj

=2

v

mkj−1

p

r/p

r/q

mkj

w

a(mkj )

q

mkj−1

λk,j .

(2.4.28)

If k = −N − 1, then Ξ(k) = ∅ by the deﬁnition of fa . If Ξ(k + 1) = Ξ(−N ) = ∅, then, for the same reason, we have r/p m−N r/q m−N p m−N 1 1 i p p l−N,i v ≤ v wq II−N −1,j ≤ max{m−N−1 ,a(m−N )} j−1 i

i∈Ξ(−N )

≤

m−N 1

a(m−N ) 1

v

p

r/p

i∈Ξ(−N )

i∈Ξ(−N )

m−N i−1

r/q

m−N i

m−N i−1

a(m−N ) i

w

q

λ−N,1 .

(2.4.29)

Combining (2.4.26)–(2.4.28) and (2.4.29), we see that fa pp

N j b (k)

λk,j .

k=−N j=1

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S110

PROKHOROV et al.

In view of (2.4.25), letting N → ∞, we obtain HLp (0,∞)→Lq (0,∞)

j b (k)

1/r λk,j

(2.4.30)

.

k∈Z j=1

Let us derive the estimate HLp (0,∞)→Lq (0,∞) (Bρ+ )∗ from (2.4.30). Note that r/p mk r/q b(t) r/q j p p p v ≈ v v (t) dt ≥ vp v p (t) dt κjk

κjk

κjk

t

t

by virtue of the construction of κjk and the inequality q < 0. Taking into account (2.4.24), we obtain r/q r/q q p λk,j w v v p (t) dt ≈

κjk

κjk

r/q

κjk

[WΘ (t)]

ϑ+ (t)

r/q p

v (t) dt ≥

[Vϑ+ (t)]

mkj

mkj−1

[WΘ (t)]r/q [Vϑ+ (t)]r/q v p (t) dt.

Now, the required estimate HLp (0,∞)→Lq (0,∞) (Bρ+ )∗ in the case ρ(y) = y follows from (2.4.30), because k,j [mkj , mkj+1 ) = (0, ∞). jk . By To show that HLp (0,∞)→Lq (0,∞) Bρ−−1 for ρ(y) = y, consider any κjk and its successor κ virtue of (2.4.24), the integrals κk wq and κ k wq are equivalent. Therefore, the sum γk,j of the j

j

k,j is estimated from below as corresponding quantities λk,j and λ r/q mk r/p j+1 q p w v =: μk,j γk,j k κj+1

a(mkj )

k (here, mkjb +1 = mk+1 and κj+1 = κ1k+1 if j = jb ). Hence, taking into account (2.4.24), we see that 1 mk r/p j+1 p v [Wδ (t)]r/p wq (t) dt μk,j ≈

≥

k κj+1

a(mkj )

r/p

k κj+1

[Wδ (t)]

r/p

[VΔ− (t)]

w (t) dt ≥ q

mkj+1

mkj

[Wδ (t)]r/p [VΔ− (t)]r/p wq (t) dt.

Finally, applying (2.4.30), we obtain j 1/r j 1/r b (k) b (k) 2 λk,j μk,j HLp (0,∞)→Lq (0,∞) k∈Z j=1

k∈Z j=1

j b (k) k∈Z j=1

mkj+1 mkj

r/p

[Wδ ]

r/p

[VΔ− ]

1/r w

q

= Bρ−−1 ,

(2.4.31)

which means that HLp (0,∞)→Lq (0,∞) Bρ−−1 in the case ρ(y) = y. Similarly, it can be shown by using sequence (2.1.8) with ζ0 = 1 instead of (2.1.7) that HLp (0,∞)→Lq (0,∞) (Bρ− )∗ ,

HLp (0,∞)→Lq (0,∞) Bρ+−1

for ρ(y) = y.

The case of a general function ρ is derived from that of ρ(y) = y by using the substitutions a(x) = 1/q a(ρ(x)), b(x) = b(ρ(x)), and w(x) = w(ρ(x))[ρ (x)] . PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

HARDY–STEKLOV INTEGRAL OPERATORS: PART I

S111

Corollary 2.1. Suppose that p > 1, q > 0, q = 1, and H is of the form (2.2.1). Let σ(x) be a fairwayfunction satisfying condition (2.2.2) on (0, ∞), and let ρ(y) be a dual fairway satisfying condition (2.4.2) on (0, ∞). Then, in the case 1 1, HLp (0,∞)→Lq (0,∞) ≈ Bσ ≈ (Bσ−1 )∗ ≈ Bρ−1 ≈ (Bρ− )∗ + (Bρ+ )∗ ,

(2.4.33)

where (Bρ− )∗ + (Bρ+ )∗ ≈ (Bρ )∗ if 1 < q < p < ∞. Boundedness criteria for a Hardy–Steklov operator H from Lp (0, ∞) to Lq (0, ∞) in the Tomaselli and Persson–Stepanov forms alternative to those given by Corollary 2.1 can also be obtained in terms of a fairway σ and its dual ρ both in the direct and dual forms [32, Theorem 2.1]. ACKNOWLEDGMENTS This work was supported by the Russian Science Foundation, project no. 14-11-00443. REFERENCES 1. D. E. Edmunds and W. D. Evans, Hardy Operators, Function Spaces and Embeddings (Springer-Verlag, Berlin, 2004). 2. A. Kufner and L.-E. Persson, Weighted Inequalities of Hardy Type (World Sci. Publ., River Edge, NJ, 2003). 3. A. Kufner, L. Maligranda, and L.-E. Persson, The Hardy Inequality. About Its History and Some Related Results (Vydavatelsky´ Servis, Pilsen, 2007). 4. M. A. Lifshits and W. Linde, Approximation and Entropy Numbers of Volterra Operators with Application to Brownian Motion (Amer. Math. Soc., Providence, RI, 2002). 5. B. Opic and A. Kufner, Hardy-Type Inequalities (Longman Sci. & Tech., Harlow, 1990). 6. C. Bennett and R. Sharpley, Interpolation of Operators (Academic, Boston, MA, 1988). ´ 7. G. H. Hardy, J. E. Littlewood, and G. Polya, Inequalities (Cambridge Univ. Press, Cambridge, 1934; Inostrannaya Literatura, Moscow, 1948). 8. N. Dunford and J. T. Schwartz, Linear Operators. Vol. 1: General Theory (with the assistance of William G. Bade and Robert G. Bartle) (Intersci., New York, 1958, Inostrannaya Literatura, Moscow, 1962). 9. W. Rudin, Real and Complex Analysis (McGraw-Hill, New York, 1987). 10. R. Oinarov, “Two-sided norm estimates for certain classes of integral operators,” Proc. Steklov Inst. Math. 204, 205–214 (1994). 11. S. Bloom and R. Kerman, “Weighted norm inequalities for operators of Hardy type,” Proc. Am. Math. Soc. 113, 135–141 (1991). 12. V. D. Stepanov, “Weighted norm inequalities of Hardy type for a class of integral operators,” J. London Math. Soc. (2) 50 (1), 105–120 (1994). 13. G. Bennett, “Some elementary inequalities. III,” Quart. J. Math. Oxford Ser. (2) 42 (166), 149–174 (1991). 14. K.-G. Grosse-Erdmann, The Blocking Technique, Weighted Mean Operators and Hardy’s Inequality (Springer-Verlag, Berlin, 1998). 15. M. L. Goldman, “Sharp estimates for the norms of Hardy-type operators on the cones of quasimonotone functions,” Proc. Steklov Inst. Math. 232, 109–137 (2001). 16. E. Lomakina and V. Stepanov, “On the Hardy-type integral operators in Banach function spaces,” Publ. Mat. 42 (1), 165–194 (1998). 17. V. G. Maz’ya, Sobolev Spaces (Izd. Leningrad. Univ., Leningrad, 1985) [in Russian]. 18. G. J. Sinnamon, “Weighted Hardy and Opial-type inequalities,” J. Math. Anal. Appl. 160 (2), 434–445 (1991). 19. G. Sinnamon and V. D. Stepanov, “The weighted Hardy inequality: new proofs and the case p = 1,” J. London Math. Soc. (2) 54 (1), 89–101 (1996). 20. D. V. Prokhorov and V. D. Stepanov, “Weighted estimates for the Riemann-Liouville operators and applications,” Proc. Steklov Inst. Math. 243, 278–301 (2003). 21. L. V. Kantorovich and G. P. Akilov, Functional Analysis (Nauka, Moscow, 1984) [in Russian]. 22. L.-E. Persson and V. D. Stepanov, “Weighted integral inequalities with the geometric mean operator,” J. Inequal. Appl. 7 (5), 727–746 (2002). PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

S112

PROKHOROV et al.

23. V. D. Stepanov and E. P. Ushakova, “ Alternative criteria for the boundedness of Volterra integral in Lebesgue spaces,” Math. Inequal. Appl. 12 (4), 873–889 (2009). 24. P. J. Martin-Reyes and E. T. Sawyer, “Weighted inequalities for Riemann–Liouville fractional integrals of order one and greater,” Proc. Am. Math. Soc. 106 (3), 727–733 (1989). 25. V. D. Stepanov and E. P. Ushakova, “On integral operators with variable limits of integration,” Proc. Steklov Inst. Math. 232, 290–309 (2001). 26. E. N. Lomakina, “Estimates for the approximation numbers of one class of integral operators. I,” Sib. Math. J. 44 (1), 147–159 (2003). 27. E. N. Lomakina, “Estimates for the approximation numbers of one class of integral operators. II,” Sib. Math. J. 44 (2), 298–310 (2003). 28. V. D. Stepanov and E. P. Ushakova, “ Hardy operator with variable limits on monotone functions,” J. Funct. Spaces Appl. 1 (1), 1–15 (2003). 29. V. D. Stepanov and E. P. Ushakova, “On the geometric mean operator with variable limits of integration,” Proc. Steklov Inst. Math. 260, 254–278 (2008). 30. V. D. Stepanov and E. P. Ushakova, “Kernel operators with variable intervals of integration in Lebesgue spaces and applications,” Math. Inequal. Appl. 13 (3), 449–510 (2010). 31. E. P. Ushakova, “Estimates for Schatten–von Neumann norms of Hardy–Steklov operators,” J. Approx. Theory 173, 158–175 (2013). 32. E. P. Ushakova, “Alternative boundedness characteristics for the Hardy–Steklov operator,” Eurasian Math. J. 8 (2), 74–96 (2017).

Translated by O. Sipacheva

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS

Vol. 300 Suppl. 2

2018

Hardy–Steklov Integral Operators: Part I

Recommend Documents