Siberian Mathematical Journal, Vol. 52, No. 6, pp. 1042–1055, 2011 c 2011 O˘ınarov R. Original Russian Text Copyright 

BOUNDEDNESS AND COMPACTNESS IN WEIGHTED LEBESGUE SPACES OF INTEGRAL OPERATORS WITH VARIABLE INTEGRATION LIMITS c R. O˘  ınarov

UDC 517.51

Abstract: Considering the integral operators with nonnegative kernels and variable integration limits, we obtain criteria of boundedness and compactness in weighted Lebesgue spaces under some conditions on the kernels that are weaker than those studied before. Keywords: integral operator with variable integration limits, Lebesgue space, boundedness, compactness

§ 1. Introduction Suppose that I = (a, b), −∞ ≤ a < b ≤ +∞, 1 < p, q < ∞, 

1 1 p + p = 1, ρ and w are nonnegative,  ρ−p are locally integrable over I.

measurable functions finite a.e. on I such that ρp , wq , w−q , and Denote the set of all measurable functions f finite almost everywhere on I such that  b 1 p f p,ρ ≡ ρf p = |ρf |p <∞ a

by Lp,ρ ≡ Lp (ρ, I). Consider the integral operators β(x) 

K(x, s)f (s) ds,

x ∈ I,

(1)

β(s)  K(x, s)g(x) dx, K− g(s) =

s ∈ I,

(2)

K+ f (x) = α(x)

α(s)

from Lp,ρ into Lq,w . In the case when K(x, s) ≡ 1, the operator (1) is denoted by β(x)  Hf (x) = f (s) ds, x ∈ I,

(3)

α(x)

and called the Hardy–Steklov operator [1]. Impose the following conditions on the boundary functions α and β: (i) α(x) and β(x) are locally absolutely continuous and strictly increasing functions on I; (ii) α(x) < β(x) for every x ∈ I and limx→a α(x) = limx→a β(x) = a, limx→b α(x) = limx→b β(x) = b. In the last decade, the question was under intensive research of the boundedness and compactness of (1)–(3) in weighted Lebesgue spaces (see [1–7]) and Banach function spaces (see [8]). The operator (3) is studied quite well (see [1–5, 7]). Thanks to the introduction of the notion of a “fairway function,” Stepanov and Ushakova in [1, 7] obtained some criteria for the boundedness and compactness of (3) in integral form that are completely analogous to the case when one of the boundary functions of the operator (3) is constant. They found connection between the boundedness of (3) and inequalities of the form wf q ≤ C(ρf  s + vf p ). Astana. Translated from Sibirski˘ı Matematicheski˘ı Zhurnal, Vol. 52, No. 6, pp. 1313–1328, November–December, 2011. Original article submitted October 28, 2010. 1042

0037-4466/11/5206–1042

The boundedness of (1)–(3) is established by applying the block-diagonal method [1–8] whose essence consists in splitting the integral operator with two variable limits into a sum of integral operators with a single variable limit and a subsequent application of the so-obtained results for these operators. Such an approach was first applied to a Hardy–Steklov operator (3) in the article by Batuev and Stepanov [2] when α(x) = αx, β(x) = βx, a = 0, b = ∞, and the general case with conditions (i) and (ii) was presented in Batuev’s Kandidat Thesis [4] carried out and defended in 1991 under the supervision of Stepanov. Basing on the origins of this method, below we call it the Batuev–Stepanov block-diagonal method. The operators of the form of (1) were studied in [6–8] in the case when their kernels K(x, s) ≥ 0 satisfy the condition 1 (K(x, β(z)) + K(z, s)) ≤ K(x, s) ≤ d(K(x, β(z)) + K(z, s)) d

(4)

for a < z ≤ x < b,

α(x) ≤ s ≤ β(z),

(5)

and the operators of the form of (2) (see [1, 7]) under the condition 1 (K(x, z) + K(α(z), s)) ≤ K(x, s) ≤ d(K(x, z) + K(α(z), s)) d

(6)

for a < s ≤ z < b,

α(z) ≤ x ≤ β(s),

(7)

which allows for the use of the Batuev–Stepanov diagonal method, where d ≥ 1 is a constant independent of x, z, and s. In the case of α(x) = a and β(x) ≡ x, i.e., when (1) has the form x Kf (x) =

K(x, s)f (s) ds,

(8)

a

and K(x, s) ≥ 0 satisfies the condition (see [9]) 1 (K(x, z) + K(z, s)) ≤ K(x, s) ≤ d(K(x, z) + K(z, s)) d

(9)

for a < s ≤ z ≤ x < b, the operator (8) was the topic of many studies (see [10–13] and the references therein). Some criteria for the (Lp,ρ → Lq,w )-boundedness and compactness of (8) were given in [14].  It is easy to see that (4) and (5) are equivalent to the fulfillment of (9) for the kernels K(x, s) ≡ −1 K(x, β(s)) for s ≤ z ≤ x ≤ α (β(s)), s ∈ I, while (6) and (7) are equivalent to the fulfillment of (9) for  the kernels K(x, s) ≡ K(α(x), s) for β −1 (α(x)) ≤ s ≤ z ≤ x, x ∈ I. Here, under conditions on the kernels of (1) and (2) weaker than (4) and (6) we establish the (Lp,ρ → Lq,w )-boundedness and compactness of (1) and (2). In this article, we put 00 = 0 and ∞ · 0 = 0. The relation A B means that A ≤ cB, where the constant c > 0 can depend only on insignificant parameters. We write A ≈ B instead of A B A. Everywhere below, Z is the set of integers, χD (·) is the characteristic function of a set D ⊂ R. The paper is organized as follows: In § 2, we give necessary notions, notation, and some known assertions that are used in the proofs of the main results which are the contents of § 3. § 2. Notation, Notions, and Assertions Take [c, d] ⊂ I. Given an integer n ≥ 0, define the classes On± (c, d) (see [15]) of kernels of operators (1) and (2) and agree to write K(·, ·) ≡ Kn (·, ·) if K(·, ·) ∈ On± (c, d). 1043

First, define the classes On+ (c, d), n ≥ 0. Put Ωc,d = {(x, s) : c ≤ s ≤ x ≤ d} and let K + (·, ·) be a nonnegative measurable function on Ωc,d nondecreasing in the first argument. The function K + (·, ·) ≡ K0+ (·, ·) belongs to O0+ (c, d) if and only if K0+ (·, ·) has the form K0+ (x, s) ≡ v(s). Suppose that the classes Oi+ (c, d), i = 0, 1, . . . , n − 1, n ≥ 1, are defined. The function K + (·, ·) ≡ Kn+ (·, ·) belongs to On+ (c, d) if and only if there exist Ki+ (·, ·) ∈ Oi+ (c, d), i = 0, 1, . . . , n− 1, and a number hn > 0 such that n  + Kn,i (x, t)Ki+ (t, s) K + (x, s) ≡ Kn+ (x, s) ≤ hn i=0

for c ≤ s ≤ t ≤ x ≤ d, where Kn+ (x, s) , c
+ (x, t) = inf Kn,i

+ i = 0, 1, . . . , n − 1, Kn,n (·, ·) ≡ 1.

+ (x, t)Ki+ (t, s), i = 0, 1, . . . , n − 1, for c ≤ s ≤ t ≤ x ≤ d; therefore, By definition, Kn+ (x, s) ≥ Kn,i

Kn+ (x, s) ≈

n 

+ Kn,i (x, t)Ki+ (t, s),

c ≤ s ≤ t ≤ x ≤ d,

(10)

i=0

where the equivalence constants depend only on hn and n ≥ 0. It is easy to see that if to K + (·, ·) there are Ki+ (·, ·) ∈ Oi+ (c, d), i = 0, 1, . . . , n − 1, and ϕn,i (·, ·) ≥ 0, i = 0, 1, . . . , n − 1, measurable on Ωc,d such that +

K (x, s) ≈

n−1 

ϕn,i (x, t)Ki+ (t, s) + K + (t, s)

i=0

for c ≤ s ≤ t ≤ x ≤ d then K + (x, s) ≡ Kn+ (x, s) ∈ On+ (c, d). Therefore, in what follows we assume + (x, t), i = 0, 1, . . . , n − 1, to be arbitrary nonnegative measurable functions on Ωc,d satisfying (10). Kn,i Let K − (·, ·) be defined as above excepting the condition of nondecreasing in the first argument. We will require that the function be nonincreasing in the second argument instead. Define the classes On− (c, d), n ≥ 0. They consist of the functions of the form K − (x, s) ≡ K0− (x, s) = u(x). Suppose that Oi− (c, d), i = 0, 1, . . . , n − 1, n ≥ 1, are defined. The function K − (·, ·) ≡ Kn− (·, ·) belongs to On− (c, d) if and only if there exist Ki− (·, ·) ∈ Oi− (c, d), i = 0, 1, . . . , n − 1, and a number hn > 0 such that n  − K − (x, s) ≡ Kn− (x, s) ≤ hn Ki− (x, t)Ki,n (t, s), c ≤ s ≤ t ≤ x ≤ d, i=0

where − Kn,n (·, ·) ≡ 1,

Kn− (x, s) , t≤x≤d K − (x, t) i

− Ki,n (t, s) = inf

i = 0, 1, . . . , n − 1.

As in (10), each class On− (c, d), n ≥ 1, of kernels Kn− (·, ·) is characterized by a relation of the form Kn− (x, s) ≈

n 

− Ki− (x, t)Ki,n (t, s),

c ≤ s ≤ t ≤ x ≤ d,

(11)

i=0 − (·, ·), Ki,n

i = 0, 1, . . . , n − 1, may be assumed arbitrary nonnegative functions satisfying (11). where The kernels of the classes O1+ (c, d) and O1− (c, d) are characterized by the relations

1044

+ (x, t)v(s) + K1+ (t, s), K1+ (x, s) ≈ K1,0

(12)

− (t, s) K1− (x, s) ≈ K1− (x, t) + u(x)K0,1

(13)

for c ≤ s ≤ t ≤ x ≤ d; moreover, in (12) K1+ (x, s) is nondecreasing in the first argument and K1− (x, s) in (13) is nonincreasing in the second argument. We say that K ± (·, ·) belongs to On± (I), n ≥ 0, if it belongs to On± (c, d) for every [c, d] ⊂ I. We may assume a function K(·, ·) that satisfies (9) nondecreasing in the first argument and nonincreasing in the second argument, since K(·, ·) is equivalent to a function with such properties [16]. Therefore from (9), (12), and (13) it follows that a function satisfying (9) belongs to O1+ (I) ∩ O1− (I). Suppose that K + (·, ·) ≥ 0 is defined and measurable on Ω+ = {(x, s) : a < x < b, α(x) ≤ s ≤ β(x)} and is nondecreasing in the first argument. Given an integer n ≥ 0, define classes On+ (α, β(·), Ω+ ). The class O0+ (α, β(·), Ω+ ) consists of the functions of the form K + (x, s) ≡ K0+ (x, s) = v(s) for (x, s) ∈ Ω+ . Suppose that Oi+ (α, β(·), Ω+ ), i = 0, 1, . . . , n−1, n ≥ 1, are defined. The function K + (·, ·) belongs to On+ (α, β(·), Ω+ ) if and only if there + (x, t) ≥ 0, i = 0, 1, . . . , n−1, defined and measurable on Ωa,b = {(x, t) : a < t ≤ x < b} exist functions Kn,i + and functions Ki (·, ·) ∈ Oi+ (α, β(·), Ω+ ), i = 0, 1, . . . , n − 1, such that +

K (x, s) ≡

Kn+ (x, s)

≈

n 

+ Kn,i (x, t)Ki+ (t, s),

+ Kn,n (·, ·) ≡ 1,

(14)

i=0

for a < t ≤ x < b,

α(x) ≤ s ≤ β(t),

(15)

where the equivalence constants do not depend on x, t, and s. + The function Kn,i (x, t), i = 0, 1, . . . , n − 1, may be defined as + (x, t) = Kn,i

Kn+ (x, s) + α(x)≤s≤β(t) Ki (t, s) inf

if a < t ≤ x < b.

If we replace s by β(s) in (14) and (15) then we will infer that the containment K + (x, s) ∈ is equivalent to the condition K + (x, β(s)) ∈ On+ (− (t)) for all t ∈ I, where − (t) =

On+ (α, β(·), Ω+ ) [β −1 (α(t)), t].

Suppose now that K − (·, ·) ≥ 0 is defined, measurable on Ω− = {(x, s) : a < s < b, α(s) ≤ x ≤ β(s)} and nonincreasing in the second argument. Define On− (α(·), β, Ω− ), n ≥ 0. To the class O0− (α(·), β, Ω− ) we assign all functions of the form K − (x, s) ≡ K0− (x, s) = u(x) for all (x, s) ∈ Ω− . Suppose that Oi− (α(·), β, Ω− ), i = 0, 1, . . . , n − 1, n ≥ 1, are defined. Then K − (·, ·) belongs to On− (α(·), β, Ω− ) if − and only if there exist Ki,n (t, s), i = 0, 1, . . . , n − 1, defined and measurable on Ωa,b and Ki− (x, t) ∈ − Oi (α(·), β, Ω− ), i = 0, 1, . . . , n − 1, such that K − (x, s) ≡ Kn− (x, s) ≈

n 

− Ki− (x, t)Ki,n (t, s),

− Kn,n (·, ·) ≡ 1,

(16)

i=0

for a < s ≤ t < b,

α(t) ≤ x ≤ β(s),

(17)

where the equivalence constants are independent of x, t, and s. − Here Ki,n (·, ·), i = 0, 1, . . . , n − 1, may be defined as − (t, s) = Ki,n

Kn− (x, s) − α(t)≤x≤β(s) Ki (x, t) inf

if a < s ≤ t < b.

If we replace x by α(x) in (16) and (17) then, as above, we see that the condition K − (x, s) ∈ is equivalent to the condition K − (α(x), s) ∈ On− (+ (t)) for all t ∈ I, where + (t) =

On− (α(·), β, Ω− ) [t, α−1 (β(t))].

We list some known assertions that are necessary for the proof of the main results. Lemmas 2.1 and 2.2 of [7] imply: 1045

Lemma 1. Suppose that 1 < p ≤ q < ∞ and a ≤ c < d ≤ b. Then the (Lp,ρ (α(c), β(d)) →  β(x)  β(d) Lq,w (c, d))-norms of the operators H + f (x) = α(c) f (t) dt and H − g(s) = α(s) g(t) dt satisfy  d

+

H  ≈ sup

β(t) 1    1 q p −p w (x) dx ρ (s) ds , q

c
t

α(c)

 1  β(d)  1  q p q −p w (x) dx ρ (s) ds .

 t

H −  ≈ sup c
c

α(t)

Theorems 5 and 6 in [15] imply: +  x Lemma 2. Suppose that 1 < p ≤+ q < ∞, −a ≤ c < d ≤ b, and the kernel +of K f (x) = c K(x, s)f (s) ds, x ∈ (c, d), belongs to On (c, d) ∪ On (c, d), n ≥ 1. Then the operator K : Lp,ρ (c, d) → Lq,w (c, d) is bounded if and only if one of the following is fulfilled:

B1+ B2+

 z

 d

p

q

= sup

w (x)

c
K (x, s)ρ

z

 q (s) ds

p

1 q

dx

< ∞,

c

 z = sup

ρ

c
−p

−p

 d

 p q

(s)

q

K (x, s)w (x) dx

c

q

 1 p

ds

< ∞;

z

moreover, the (Lp,ρ (c, d) → Lq,w (c, d))-norm of K+ satisfies K+  ≈ B1+ ≈ B2+ . Lemma 3. Suppose that 1 < p ≤ q < ∞, a ≤ c < d ≤ b, and the kernel of K− g(s) = + − − s K(x, s)g(x) dx, s ∈ (c, d), belongs to On (c, d) ∪ On (c, d), n ≥ 1. Then K : Lp,ρ (c, d) → Lq,w (c, d) is bounded if and only if one of the following is fulfilled: d

B1− B2−

 d

 z q

= sup

w (s)

c
c
−p

K (x, s)ρ

c

 q (x) dx

p

1 q

ds

< ∞,

z

 d = sup

p

ρ

−p

 z q

(x)

z

 p q

K (x, s)w (s) ds

q

 1 dx

p

< ∞;

c

moreover, the (Lp,ρ (c, d) → Lq,w (c, d))-norm K−  of K− satisfies K−  ≈ B1− ≈ B2− . § 3. The Main Results Put A+ 1 (z)

A+ 2 (z)

1046

 z =

sup

y∈− (z)

y

 =

 β(y)  q  1  q p p −p w (x) K (x, s)ρ (s) ds dx , q

α(z) β(y) 

sup

y∈− (z)

ρ α(z)

−p

 z

 p q

(s)

q

K (x, s)w (x) dx y

q

 1 ds

p

,

A− 1 (z)

A− 2 (z)

 y =

sup

y∈+ (z)

z

 =

 β(z)  q  1  q p p −p w (s) K (x, s)ρ (x) dx ds , q

α(y) β(z) 

sup

y∈+ (z)

ρ

−p

 y q

(x)

q

K (x, s)w (s) ds

q

 1 dx

p

,

z

α(y)

A± i

 p

= sup a
A± i (z),

i = 1, 2.

Recall that + (z) = [z, α−1 (β(z))], − (z) = [β −1 (α(z)), z]. Theorem 1. Suppose that 1 < p ≤ q < ∞. If the kernel of (1) belongs to On+ (α, β(·), Ω+ ) ∪ n ≥ 1, then (i) the operator (1) is (Lp,ρ → Lq,w )-bounded iff A+ i < ∞ at least for one i = 1, 2; moreover, the + + (Lp,ρ → Lq,w )-norm of (1) satisfies K+  ≈ A1 ≈ A2 ; + + (ii) the operator (1) is (Lp,ρ → Lq,w )-compact iff A+ i < ∞ and limy→a Ai (y) = limy→b Ai (y) = 0 at least for one i = 1, 2.

On− (β −1 (·), α−1 , Ω+ ),

Theorem 2. Suppose that 1 < p ≤ q < ∞. If the kernel of (2) belongs to On− (α(·), β, Ω− ) ∪ On+ (β, α−1 (·), Ω− ), n ≥ 1, then (i) the operator (2) is (Lp,ρ → Lq,w )-bounded iff A− i < ∞ at least for one i = 1, 2; moreover, the − Lp,ρ → Lq,w -norm of (2) satisfies K−  ≈ A− ≈ A ; 1 2 − + (ii) the operator (2) is (Lp,ρ → Lq,w )-compact iff A− i < ∞ and limz→a Ai (z) = limz→b Ai (z) = 0 at least for one i = 1, 2. We first prove Theorems 3 and 4, which are special cases of Theorems 1 and 2 respectively, and then prove Theorems 1 and 2. Theorem 3. Suppose that 1 < p ≤ q < ∞. If the kernel of (1) belongs to On+ (α, β(·), Ω+ ), n ≥ 1, then (i) the operator (1) is (Lp,ρ → Lq,w )-bounded if and only if A+ i < ∞ at least for one i = 1, 2; + ≈ A ; moreover, K+  ≈ A+ 1 2 + + (ii) the operator (1) is (Lp,ρ → Lq,w )-compact iff A+ i < ∞ and limz→a Ai (z) = limz→b Ai (z) = 0 at least for one i = 1, 2. Theorem 4. Suppose that 1 < p ≤ q < ∞. If the kernel of (2) belongs to On− (α(·), β, Ω− ), n ≥ 1, then (i) the operator (2) is (Lp,ρ → Lq,w )-bounded if and only if A− i < ∞ at least for one i = 1, 2; − − moreover, K−  ≈ A1 ≈ A2 ; − + (ii) the operator (2) is (Lp,ρ → Lq,w )-compact iff A− i < ∞ and limz→a Ai (z) = limz→b Ai (z) = 0 at least for one i = 1, 2. Proof of Theorem 3. (i) Sufficiency. Suppose that A+ i < ∞ for at least one i = 1, 2. To prove the (Lp,ρ → Lq,w )-boundedness of (1), apply the Batuev–Stepanov block-diagonal method. Given t0 ∈ I, construct the sequence of points {tk }k∈Z ⊂ I : tk+1 = α−1 (β(tk )), k ∈ Z, and put k ≡ + (tk ), δk = [τk , τk+1 ], where τk = α(tk ) = β(tk−1 ), k ∈ Z. From conditions (i) and (ii) on α(·) and β(·) it follows that I = k k = k δk . Then for a function f ≥ 0 we have b wK+ f qq = a

 β(x) q  β(x) q    q q w (x) K(x, s)f (s) ds dx = w (x) K(x, s)f (s) ds dx. α(x)

k  k

(18)

α(x)

1047

Since α(tk ) ≤ α(x) ≤ α(tk+1 ) = β(tk ) ≤ β(x) ≤ β(tk+1 ) for x ∈ k , we infer α(t k+1 )

β(x) 

K(x, s)f (s) ds = α(x)

β(x) 

K(x, s)f (s) ds +

x ∈ k .

K(x, s)f (s) ds,

(19)

β(tk )

α(x)

Following [1], consider the operators α(t k+1 )

Tk+ f (x) =

Tk+ : Lp,ρ (δk ) → Lq,w (k ),

K(x, s)f (s) ds, α(x) β(x) 

Sk+ f (x) =

K(x, s)f (s) ds,

Sk+ : Lp,ρ (δk+1 ) → Lq,w (k ).

β(tk )

From (18) and (19) we have wK+ f qq

≈



q

w (x)



q Tk+ f (x)

dx +



k  k

q wq (x) Sk+ f (x) dx

k  k

   q    q   q

+ + p p |ρf | + |ρf |

max sup Tk+ , sup Sk+ ρf qp . Tk Sk ≤ q p

k

q p

k

δk

Hence,

k

δk+1

k



K+  max sup Tk+ , sup Sk+ , k

(20)

k

where Tk+ and Sk+ are the norms of Tk+ and Sk+ from Lp,ρ (δk ) into Lq,w (k ) and from Lp,ρ (δk+1 ) into Lq,w (k ). By hypothesis, the kernel K(·, ·) of (1) belongs to On+ (α, β(·), Ω+ ); therefore, for x ≥ tk , α(x) ≤ s ≤ β(tk ) = α(tk+1 ), by (14) and (15) we have K(x, s) ≡ Kn (x, s) ≈

n 

Kn,i (x, tk )Ki (tk , s).

i=0

Consequently, if x ∈ k then Tk+ f (x)

≈

n 

+ Tk,i f (x),

(21)

i=0

where + f (x) Tk,i

β(t  k)

= Kn,i (x, tk )

Ki (tk , s)f (s) ds, i = 0, 1, . . . , n,

Kn,n (·, ·) ≡ 1.

α(x)

 + , where Tk,i  is the norm of T + from Lp,ρ (δk ) into Lq,w (k ). From (21) we have Tk+ ≤ ni=0 Tk,i k,i 1048

+ Since Tk,i , i = 0, 1, . . . , n, are in fact Hardy operators, by Lemma 1 n 

Tk 

 z q wq (x)Kn,i (x, tk ) dx

sup

i=0 z∈k

 1  β(t  1  k) q p p −p Ki (tk , s)ρ (s) ds

tk

α(z)

 z

sup

 β(t  q  1  k) q p p −p w (x) K (x, s)ρ (s) ds dx q

z∈k

tk

α(z)

 z  p  1  β(t  k) q p −p q q ≈ sup ρ (s) K (x, s)w (x) dx ds . z∈k

tk

α(z)

Hence,  z sup Tk  sup

sup

a
k∈Z

≈ sup

sup

a
= sup

 β(y)  q  1  q p q p −p w (x) K (x, s)ρ (s) ds dx

y

α(z)

 z  p  1  β(y)  q q  ρ−p (s) K q (x, s)wq (x) dx ds y

α(z)

 z  p  1  β(y)  q q + −p q q ρ (s) K (x, s)w (x) dx ds = A+ 2 ≈ A1 .

sup

a
α(z)

(22)

y

Estimate Sk , k ∈ Z. The quantity Sk  is the best constant in the inequality  β(x) 

 q

w (x) k

q K(x, s)f (s) ds

1 dx

q

 ≤ Sk 

1 |ρ(s)f (s)| ds

β(tk )

p

p

.

δk+1

Here, performing the change s = β(y) and putting f (β(y))β  (y) = g(y) using the fact that δk+1 = [β(tk ), β(tk+1 )], we infer 

 x q

w (x) k

q K(x, β(y)g(y) dy

1 dx

q

 ≤ Sk 

1 |˜ ρ(y)g(y)| dy p

p

,

k

tk 1−p

where ρ ˜(y) = ρ(β(y))(β  (y)) p . As was said above, from (14) and (15) it follows that K(x, β(y)) ∈ On+ (k ); therefore, by Lemma 2,  z  q  1  tk+1 q p q p −p w (x) K (x, β(y))(˜ ρ(y)) dy dx Sk  sup z∈k

 z ≈ sup z∈k

tk

z

tk

 p  1  tk+1 q p −p q q (˜ ρ(y)) K (x, β(y))w (x) dx dy . z

1049

Hence, performing the change β(y) = s, we have Sk 

sup

z∈− (tk+1 )

  tk+1 q w (x)

p

−p

K (x, s)ρ

 q p

(s) ds

1 dx

q

α(tk+1 )

z

= Then

β(z) 

A+ 1 (tk+1 )

≈ A+ 2 (tk+1 ).

+ sup Sk  A+ 1 ≈ A2 .

(23)

+ K+  A+ 1 ≈ A2 .

(24)

k∈Z

From (20), (22), and (23) it follows that

Necessity. Suppose that K+  < ∞ and z ∈ − (t), t ∈ I. The (Lp,ρ → Lq,w )-boundedness of (1) implies the (Lq ,w−1 → Lp ,ρ−1 )-boundedness of the adjoint K∗+ g(s) =

−1 (s) α

K(x, s)g(x) dx.

(25)

β −1 (s) 

Put g0 (x) = χ[z,t] (x)wq (x); then, by the local integrability of wq and w−q , we have 0 < w−1 g0 q = wq,(z,t) = γ < ∞. Using (14) and (15), we infer K+  = K∗+  ≥ 1 ≥ γ

  β(z)  −p ρ (s)

p 

K(x, s)g0 (x) dx

 1 ds

p

 t p  1  β(z)  p −p ρ (s) K(x, s)g0 (x) dx ds z

α(t)

1  γ

−1 (s) α

β −1 (s)

α(t)

1  γ

ρ−1 K∗+ g0 p w−1 g0 q

 β(z)  1 t  p p −p Ki (z, s)ρ (s) ds Kn,i (x, z)wq (x) dx,

i = 0, 1, . . . , n.

z

α(t)

Consequently, β(z) 





Kip (z, s)ρ−p (s) ds < ∞,

i = 0, 1, . . . , n.

α(t)

Therefore, define the test functions in the form 



fi (s) ≡ fi,t (s) = χ[α(t),β(z)] (s)Kip −1 (z, s)ρ−p (s), Then

 β(z) 1  p p −p ρfi p = Ki (z, s)ρ (s) ds , α(t)

1050

i = 0, 1, . . . , n.

i = 0, 1, . . . , n.

(26)

Using (14) and (15), we infer  t wK+ fi q ≥

 β(x) q  1  q q w (x) K(x, s)fi (s) ds dx

z

 t ≥

 β(z) q  1  q p −1 −p w (x) K(x, s)Ki (z, s)ρ (s) ds dx q

z

α(t)

 t q Kn,i (x, z)wq (x) dx



α(x)

 1 β(z)  q   Kip (z, s)ρ−p (s) ds,

z

i = 0, 1, . . . , n.

(27)

α(t)

Then (26) and (27) imply: K+  

n  wK+ fi q

ρfi p

i=0



β(z) 

  Kip (z, s)ρ−p (s) ds

×

 1 p

≥

n   i=0

 t ≈

q Kn,i (x, z)wq (x) dx

z

 β(z)  q  1  q p p −p w (x) K (x, s)ρ (s) ds dx

z

 ≈

ρ

−p

q

q

α(t) β(z) 

1

t

α(t)

 t

 p q

(s)

q

K (x, s)w (x) dx

q

 1 ds

p

.

z

α(t)

Hence, by the arbitrariness of z ∈ − (t), t ∈ I, we have + ∞ > K+   A+ 1 ≈ A2 .

(28)

+ From (24) and (28) we obtain K+  ≈ A+ 1 ≈ A2 . Item (i) of Theorem 3 is proved. (ii) Necessity. Suppose that (1) is (Lp,ρ → Lq,w )-compact. Then it is bounded; consequently, by (i) f (s) + Ai < ∞, i = 1, 2. Put f¯i,t (s) = ρfi,ti,t p , i = 0, 1, . . . , n. Take g ∈ Lp ,ρ−1 . Then

b

 β(z)  1  p  −1 p g(s)f¯i,t (s) ds ≤ |ρ g| dt .

β(z) 

g(s)f¯i,t (s) ds = a

α(t)

α(t)

Since z ∈ − (t), the relations t → a (t → b) imply that z → a (z → b), α(t) → a (α(t) → b), and β(z) → a (β(z) → b), we have β(z) 

lim

t→a α(t)

p

|ρ−1 g| dt = lim

β(z) 

t→b α(t)



|ρ−1 g|p dt = 0.

Consequently, for each i = 0, 1, . . . , n, the t-family of functions f¯i,t converges weakly to zero as t → a and t → b. Then limt→a wK+ f¯i,t q = limt→b wK+ f¯i,t q = 0, i = 0, 1, . . . , n, by the (Lp,ρ → Lq,w )compactness of (1). 1051

As in (27), we have n 

wK+ f¯i,t q 

i=0

sup

n  

z∈− (t) i=0

1

t q Kn,i (x, z)wq (x) dx

q

z

 β(z)  1  p p + −p × Ki (z, s)ρ (s) ds ≈ A+ 1 (t) ≈ A2 (t). α(t) + Therefore, limt→a A+ i (t) = limt→b Ai (t) = 0, i = 1, 2. + + (ii) Sufficiency. Suppose that A+ i < ∞ and limt→a Ai (t) = limt→b Ai (t) = 0 at least for one i = 1, 2. By (i), A+ i < ∞ implies that (1) is (Lp,ρ → Lq,w )-bounded. Let a < c < d < b. Consider χ[c,d] K+ f (x) ≡ χ[c,d] (x)K+ f (x). The operator χ[c,d] K+ is bounded from Lp,ρ into Lq,w but its action from Lp,ρ into Lq,w is equivalent to the action from Lp,ρ (α(c), β(d)) into Lq,w (c, d). From A+ i < ∞ it follows that  d  β(d)  q  1  q p p p −p w (x)K (x, s)ρ (s) ds dx < ∞. c

α(c)

Consequently (see [17, Theorem 6.3]), χ(c,d) K+ is compact from Lp,ρ (α(c), β(d)) into Lq,w (c, d); this is equivalent to its (Lp,ρ → Lq,w )-compactness. Then K+ − χ[c,d] K+  ≤ χ(a,c) K+  + χ(d,b) K+ ,

(29)

where  ·  is the norm of the operators from Lp,ρ into Lq,w . The kernel χ(a,c) (x)K(x, s) (χ(d,b) (x)K(x, s)) of χ(a,c) K+ (χ(d,b) K+ ) satisfies (14) and (15); therefore, by (i), χ(a,c) K+  sup A+ i (z), a
χ(d,b) K+  sup A+ i (z),

i = 1, 2.

(30)

d
By (29) and (30) the operator (1) is the uniform limit of the compact operators χ[c,d] K+ as c → a and d → b. Consequently, (1) is compact. Theorem 3 is proved. Proof of Theorem 4. Prove the sufficiency of (i). The rest of the claim is proved like in Theorem 3. Suppose that A− i < ∞ at least for one i = 1, 2. Applying the Batuev–Stepanov block-diagonal method for g ≥ 0, we get   − q q q w (s)(Tk g(s)) ds + wq (s)(Sk− g(s))q ds. wK− gq ≈ k  k

Hence, as in Theorem 3,

k  k

K−  max{sup Tk− , sup Sk− }, k

where Tk− g(s)

β(t  k)

=

k

K(x, s)g(x) dx,

Tk− : Lp,ρ (δk ) → Lq,w (k ),

K(x, s)g(x) dx,

Sk− : Lp,ρ (δk+1 ) → Lq,w (k ).

α(s)

Sk− g(s) =

β(s) 

α(tk+1 )

1052

(31)

By hypothesis, the kernel K(·, ·) belongs to On− (α(·), β, Ω− ); therefore, by (16) and (17) K(x, s) ≡ Kn (x, s) ≈

n 

Ki (x, tk+1 )Ki,n (tk+1 , s)

i=0

for s ∈ k and α(tk+1 ) ≤ x ≤ β(s). Therefore, for s ∈ k , we have Sk− g(s)

≈

n 

β(s) 

Ki,n (tk+1 , s)

i=0

Ki (x, tk+1 )g(x) dx =

− Sk,i g(s),

i=0

α(tk+1 )

where

n 

β(s) 

− Sk,i g(s) = Ki,n (tk+1 , s)

Ki (x, tk+1 )g(x) dx,

i = 0, 1, . . . , n.

α(tk+1 )

n

− Hence, Sk− 

i=0 Sk,i , where Sk,i  is the norm of Sk,i , i = 0, 1, . . . , n, from Lp,ρ (δk+1 ) into Lq,w (k ). Since all Sk,i , i = 0, 1, . . . , n, are Hardy operators, by Lemma 1

Sk− 

n 

sup

− i=0 z∈ (tk+1 )



 1  β(z)  1  tk+1  q p q p q −p w (s)Ki,n (tk+1 , s) ds Ki (x, tk+1 )ρ (x) dx α(tk+1 )

z

sup

z∈− (tk+1 )

 β(z)  q  1  tk+1  q p q p −p w (s) K (x, s)ρ (x) dx ds

 ≈

β(z) 

sup

z∈− (tk+1 )

α(tk+1 )

z

ρ

−p

 tk+1  p  1 q p (x) K q (x, s)wq (s) ds dx .

α(tk+1 )

z

Consequently, sup Sk−  k

 t

sup

 t = sup

sup

a
z

 β(z)  q  1  p p q p −p w (s) K (x, s)ρ (x) dx ds α(t)

 β(z)  q  1  p p − p −p w (s) K (x, s)ρ (x) dx ds = A− 1 ≈ A2 . q

sup

a
z

(32)

α(t)

Estimate the (Lp,ρ (δk+1 ) → Lq,w (k ))-norms Tk−  of Tk− , k ∈ Z. Performing the change x = α(y) in the inequality 

 α(tk+1 ) q  1  1 q p − p w (s) K(x, s)g(x) dx ds ≤ Tk  |ρg| dx q

k

and then putting

δ(tk )

α(s)

g(α(y))α (y)

= f (y), we have

 tk+1 q  1  tk+1 1  tk+1 q p q p w (s) K(α(y), s)f (y) dy ds ≤ Tk  |˜ ρ(y)f (y)| dy , s

tk

where ρ ˜ = ρ(α(y))(α (y))

1−p p

(33)

tk

. 1053

Since K(α(y), s) ∈ On− (k ) by (16) and (17), by Lemma 3 (33) yields  tk+1  q  1 q p p −p w (s) K (α(y), s)˜ ρ (y) dy ds

 z Tk  sup

q

z∈k

 ≈ sup

ρ ˜

z∈k

 =

z∈+ (tk )

−p

 z

ρ

−p

 p q

(y)

q

q

K (α(y), s)w (s) ds

z

β(t  k)

sup

z

tk tk+1

 1 p

dy

tk

 z

 p q

(x)

q

q

K (x, s)w (s) ds

 1 p

dx

− = A− 2 (tk ) ≈ A1 (tk ).

tk

α(z)

Then

− sup Tk−  A− 1 ≈ A2 .

(34)

k

− From (31), (32), and (34) we have K−  A− 1 ≈ A2 . Theorem 4 is proved.

Proof of Theorem 1. It remains to prove Theorem 1 only for the case when the kernel of (1) belongs to On− (β −1 (·), α−1 , Ω+ ). Suppose that K(·, ·) ∈ On− (β −1 (·), α−1 , Ω+ ). The operator (1) is (Lp,ρ → Lq,w )-bounded and compact if and only if the adjoint operator (25) is (Lq ,w−1 → Lp ,ρ−1 )-bounded and compact respectively. By Theorem 4, the operator (25) is (Lq ,w−1 → Lp ,ρ−1 )-bounded if and only if − ≈ A  − , and is (Lq ,w−1 → Lp ,ρ−1 ) − < ∞ for at least one i = 1, 2, moreover, K+  = K∗+  ≈ A A 1 2 i  − < ∞ and compact if and only if A i  − (z) = lim A  − (z) = 0 lim A i i

z→a

(35)

z→b

for at least one i = 1, 2. Here  − (z) = A 1

−1 (z)

 t sup

ρ

z≤t≤β(α−1 (z))

 α

−p

q

(s)  t

 

q

sup

β −1 (t)

−p

q

w (x)

− A i

 1 p

ds

,

β −1 (t)

z

z≤t≤β(α−1 (z))

q

q

K (x, s)w (x) dx

α−1 (z)

 − (z) = A 2

 p

K (x, s)ρ

 q (s) ds

p

1 dx

q

,

z

= sup a
 − (z), A i

i = 1, 2.

 − ≈ A+ , i = 1, 2, and condition (35) is equivalent to the condition Thus, if we prove that A i i + + limz→a Ai (z) = limz→b Ai (z) = 0, Theorem 1 will be proved.  − (z), perform the change α−1 (z) → z and β −1 (t) → t. Then In the expression A i  − (α(z)) = A 1

β(t)  

sup

ρ

α(z)≤β(t)≤β(z) β(t)  

=

sup

β −1 (α(z))≤t≤z

ρ α(z)

−p

−p

 z

 p q

(s)

q

q

K (x, s)w (x) dx

ds

p

t

α(z)

 z

 p q

(s)

q

K (x, s)w (x) dx

q

 1 ds

p

= A+ 2 (z);

t

 − (z) = supa
 1

 − (α(z)) = A+ (z) and A  − = A+ . It is easy to see that (32) is fulfilled if and only if Similarly, A 2 1 2 1 + −  (α(z)) = 0, limz→b A+ (z) = limz→b A  − (α(z)) = 0. limz→a Ai (z) = limz→a A 2−i i 2−i The proof of Theorem 1 is complete. Theorem 2 is proved by analogy. References 1. Stepanov V. D. and Ushakova E. P., “On integral operators with variable integration limits,” Trudy Mat. Inst. Steklov., 232, 298–317 (2001). ` N. and Stepanov V. D., Weighted Inequalities of Hardy Type [Preprint], Vychisl. Tsentr, Dal nevostochn. 2. Batuev E. Nauch. Tsentr Ross. Akad. Nauk, Vladivostok (1987). ` N. and Stepanov V. D., “On weighted inequalities of Hardy type,” Sibirsk. Mat. Zh., 30, No. 1, 13–22 (1989). 3. Batuev E. ` N., Boundedness of Operators in Weighted Lebesgue Spaces on a Half-Axis [in Russian], Dis. Kand. Fiz.-Mat. 4. Batuev E. Nauk, Khabarovsk Politekh. Inst. (1991). 5. Heining H. P. and Sinnamon G., “Mapping properties of integral averaging operators,” Studia Math., 129, 157–177 (1998). 6. Chen T. and Sinnamon G., “Generalized Hardy operators and normalizing measures,” J. Ineq. Appl., 7, No. 6, 829–866 (2002). 7. Stepanov V. D. and Ushakova E. P., “Kernel operators with variable intervals of integration in Lebesgue spaces and applications,” Math. Ineq. Appl., 13, No. 3, 449–510 (2010). 8. Gogatishvili A. and Lang J., “The generalized Hardy operators with kernel and variable integral limits in Banach function spaces,” J. Ineq. Appl., 4, No. 1, 1–16 (1999). 9. O˘ınarov R., “Weighted inequalities for one class of integral operators,” Dokl. Akad. Nauk SSSR, 319, No. 5, 1076–1076 (1991). 10. Stepanov V. D., “Weighted norm inequalities for integral operators and related topics,” in: Nonlinear Analysis, Function Spaces and Applications, Prague, 1994, 5, pp. 139–175. 11. Edmunds D. E. and Stepanov V. D., “On the singular numbers of certain Volterra integral operators,” J. Funct. Anal., 134, 222–246 (1995). 12. Stepanov V. D., “On the lower bounds for Schatten–von Neumann of certain Volterra integral operators,” J. London. Math. Soc., 61, 905–922 (2000). 13. Kufner A. and Perrson L.-E., Weighted Inequalities of Hardy Type, Word Sci. Publ. Co. Inc., River Edge (2003). 14. O˘ınarov R., “Two-sided norm estimates for certain classes of integral operators,” Trudy Mat. Inst. Steklov., 204, 240–250 (1993). 15. O˘ınarov R., “Boundedness and compactness of Volterra type integral operators,” Siberian Math. J., 48, No. 5, 884–896 (2007). 16. Oinarov R., “Reversion of H¨ older type inequalities for sums of weighted norms and additive weighted estimates of integral operators,” Mat. Ineq. Appl., 6, No. 3, 221–236 (2003). 17. Krasnosel skij M. A., Zabrejko P. P., Pustylnik E. I., and Sobolevskij P. E., Integral Operators in Spaces of Summable Functions, Noordhoff International Publishing, The Netherlands (1976). R. O˘ınarov Gumilev Eurasian National University, Astana, Kazakhstan E-mail address: o [email protected]

1055