Siberian Mathematical Journal, Vol. 57, No. 5, pp. 884–904, 2016 c 2016 Stepanov V.D. and Shambilova G.E. Original Russian Text Copyright
BOUNDEDNESS OF QUASILINEAR INTEGRAL OPERATORS ON THE CONE OF MONOTONE FUNCTIONS c V. D. Stepanov and G. E. Shambilova
UDC 517.51
Abstract: We study the problem of characterizing weighted inequalities on Lebesgue cones of monotone functions on the half-axis for one class of quasilinear integral operators. DOI: 10.1134/S0037446616050190 Keywords: Hardy inequality, weighted Lebesgue space, quasilinear integral operator
1. Introduction Denote the set of all nonnegative Lebesgue measurable functions on R+ := [0, ∞) by M+ and designate the subset of all nonincreasing functions as M↓ ⊂ M+ . Given v ∈ M+ and 0 < p ≤ ∞, put ∞ 1 p p p p Lv := f ∈ M : f Lv := |f (x)| v(x) dx <∞ , L∞ v Let u, v, w, ρ ∈
M+ ,
0
:= {f ∈ M : f
L∞ v
:= ess sup v(x)|f (x)| < ∞}. x≥0
0 < p, r, q ≤ ∞. In this article we study the problem of characterizing the inequality Rf Lrρ ≤ Cf Lpv ,
f ∈ M↓ ,
(1)
where the constant C does not depend on f and is assumed the least possible. As R we consider the quasilinear integral operators of the form ∞ y q 1 q T f (x) := k1 (y, x)w(y) k2 (y, z)f (z)u(z) dz dy , f ∈ M↓ , (2) x
x T f (x) := 0
∞ Sf (x) := x
x S f (x) := 0
0
∞ q 1 q k1 (x, y)w(y) k2 (z, y)f (z)u(z) dz dy ,
f ∈ M↓ ,
(3)
f ∈ M↓ ,
(4)
f ∈ M↓ ,
(5)
y
∞ q 1 q k1 (y, x)w(y) k2 (z, y)f (z)u(z) dz dy , y
y q 1 q k1 (x, y)w(y) k2 (y, z)f (z)u(z) dz dy , 0
where the kernels ki (x, y) ≥ 0, i = 1, 2, satisfy Oinarov’s condition [1]: ki (x, y) ≈ ki (x, z) + ki (z, y),
0 ≤ y ≤ z ≤ x, i = 1, 2.
(6)
This research was carried out at the Peoples’ Friendship University of Russia and financially supported by the Russian Science Foundation (Grant 16–41–02004). Moscow. Translated from Sibirski˘ı Matematicheski˘ı Zhurnal, Vol. 57, No. 5, pp. 1131–1155, September–October, 2016; DOI: 10.17377/smzh.2016.57.519. Original article submitted February 16, 2016. Revision submitted June 16, 2016. 884
0037-4466/16/5705–0884
This problem appeared in connection with the necessity to study weighted integral inequalities on the cones of monotone and quasiconcave functions which arise in the study of the classical operators in weighted Lebesgue and Lorentz spaces (see, for example, [2–12]). It was recently been solved for k1 (x, y) = k2 (x, y) = 1, 0 ≤ y ≤ x (see [13]), by the reduction method (see, for example, [14–17] on this method) to weighted inequalities of an analogous form on cones of nonnegative functions. These are actively studied [18–24] and found various applications [25–27]. In particular, results from [20] were substantially used in [13]. However, characterizing (1) with two-kernel operators (2)–(5) required a more complicated technique of [23, 24]. As in [13], the main results have reduction form; i.e., for each of the operators (2)– (5), inequality (1) is reduced to the validity of analogous inequalities in the cone of nonnegative functions, criteria for which are known. In contrast to [13, Propositions 1–3], here we do not give an example of the exact fulfillment of these criteria in order not to burden the article with technicalities but point to the sources from which such examples can be reconstructed. Nevertheless, we give a scheme and some details of the proof of the limit cases of the integration parameters which were not accentuated in [13] since, for example, the case of q = ∞ of the so-called supremal operators has important applications (see, for example, [28] and the bibliography therein). In § 2, we obtain a criterion for the fulfillment of (1) with the operator T ; in § 3, with the operator S, and in § 4, we formulate results for the operators (3) and (5). Throughout the article, the products of the form 0 · ∞ are assumed equal to 0. The record A B means A ≤ cB with c a constant depending only on p, q, and r; A ≈ B is equivalent to A B A. Also, Z stands for the set of all integers, and χE is the characteristic function (indicator) of a set E ⊂ (0, ∞). p for 1 < p < ∞, We use the symbols := and =: for defining new quantities. If 1 ≤ p ≤ ∞ then p := p−1 p := ∞ for p = 1, and p := 1 for p = ∞. § 2. The Operator T ∞ x v, U (t) := u, W (t) := w, 0 < t < ∞. Assume for simplicity that 0 < Put V (t) := 0 0 t 0 ρ < ∞, ∞ ∞ ∞ 0 < x w < ∞ for every x > 0 and 0 ρ = ∞, 0 w = ∞. Define the sequence {an } ⊂ (0; ∞) from the equations an ρ = 2n , n ∈ Z. (7) t
t
0
σ −1
: [0; ∞) → [0; ∞) be defined by the formulas (here inf ∅ = ∞) Let σ : [0; ∞) → [0; ∞) and y x y x 1 −1 ρ , x ≥ 0. σ(x) := inf y > 0 : ρ ≥ 2 ρ , σ (x) := inf y > 0 : ρ ≥ 2 0
0
σ(x)
0
x
σ−1 (x)
0
Then σ and σ −1 are increasing functions, while 0 ρ = 2 0 ρ and 0 ρ= σ(an ) = an+1 and an = σ −1 (an+1 ). Given 0 < c < d ≤ ∞, 0 < t < ∞, h ∈ M+ , put x ∞ 1 p p k2 (x, s)u(s) h ds , Tt h(x) := χ[t,∞) (x) 0
T[c,d] h(x) := χ[c,d] (x)
1 2
x 0
(8)
ρ. In particular,
(9)
s
x
d 1 p p k2 (x, s)u(s) h ds ,
σ −1 (c)
(10)
s
∞ 1 [Tt h]q w q Tt Lpv →Lqw :=
sup
0=h∈M+
0
1 ∞ [h]p v p
(11)
0
885
and proceed similarly for T[c,d] Lpv →Lqw . Theorem 1. Let 0 < q < ∞, 0 < p < ∞, 0 < r < ∞. Then the best constant CT in ∞ 1 1 ∞ r p r p [T f (x)] ρ(x) dx ≤ CT [f (x)] v(x) dx , 0
f ∈ M↓ ,
(12)
0
satisfies CT ≈ A1 + A2 + A3 + A4 + B, where A1 , A2 , A3 , and A4 are the best constants in the inequalities: ∞
∞ r x ∞ 1 r p ∞ q p r p ρ(x) k1 (y, x)k2 (y, x)q w(y) dy h u(s) ds dx ≤ A1 hV,
0
0
x
∞
0
x
∞
0
s
∞ r x ∞ 1 r p ∞ q p r p ρ(x) k1 (y, x)w(y) dy k2 (x, s) h u(s) ds dx ≤ A2 hV,
0
0
x
∞
2
ρ(x)k1 (σ (x), x) 0
r q
∞
y ∞ 1 q r p ∞ p q r p w(y) k2 (y, s)u(s) h dy dx ≤ A4 hV 0
0
where
1 s
:=
1 r
p ≤ r, s
q L1V →L p k1 (·,σ −1 (x))w(·)
0
(16)
0
s
for h ∈ M+ and the constant B has the form ⎧ t 1 1 r ⎪ ρ T t p , sup q ⎪ ⎪ ⎨ t>0 0 L1V →Lkp (·,t)w(·) 1 B := x ∞ ⎪ ⎪ ⎪ ρ(x) ρ T[σ−1 (x),σ2 (x)] ⎩
(15)
0
y
σ 2 (x)
(14)
0
s
∞ y q ∞ q r p ∞ p q r p ρ(x) k1 (y, x)w(y) k2 (y, s)u(s) ds h dy dx ≤ A3 hV,
0
(13)
p
1 s dx , r < p,
− p1 .
Proof. The change f p → f in (12) leads to the inequality ∞
∞ y q r p ∞ q r 1 p ρ(x) k1 (y, x)w(y) k2 (y, s)f p (s)u(s) ds dy dx ≤ CT f v.
0
0
x
0
Using Proposition 2.1 in [17] and the Monotone Convergence Theorem, we obtain the equivalent inequality ∞ 0
∞ y ∞ 1 q r p ∞ p q r p ρ(x) k1 (y, x)w(y) k2 (y, s)u(s) h ds dy dx ≤ CT hV x
0
s
(17)
0
for h ∈ M+ . The upper bound. Below we will use the well-known relation (see, for example, [8, Proposition 2.1]) s 2n ai ≈ 2n asn , (18) n∈Z
886
i≥n
n∈Z
valid for all sequences of nonnegative numbers and every s > 0. Put y Tp h(y) :=
∞ 1 p p k2 (y, s)u(s) h ds .
0
s
We infer an+1 ∞ ∞ r r q q q q n p p J := ρ(x) k1 (y, x)w(y)(Tp h(y)) dy dx 2 k1 (y, an )w(y)(Tp h(y)) dy n
≈
an an+2
n
2
n
x
r
q p
q
k1 (y, an )w(y)(Tp h(y)) dy
n
J1 ≈
2
k1 (y, an )w(y)(Tp h(y)) dy
=: J1 + J2 ,
an+2
0
an
k1 (y, an )w(y)
2
n
q
n
an+2
n
r
an+2 an−1 ∞ 1 q r p q 2 k1 (y, an )w(y) k2 (y, s)u(s) h ds dy
n
q p
n
n
an
+
+
an
∞
an
s
∞
y
k2 (y, s)u(s)
an−1
1 q r p q h ds dy =: J1,1 + J1,2 .
s
Next, involving the fact that k2 (y, s) ≈ k2 (y, x) + k2 (x, s) for s ≤ x ≤ y, we have J1,1 ≈
an
an+2 an−1 ∞ 1 q r p q ρ(x) dx k1 (y, an )w(y) k2 (y, s)u(s) h ds dy
n a n−1
an
≈
ρ(x) dx
n a n−1
+
an n a n−1 ∞
∞ +
ρ(x) dx
0
an an+2
s an−1 ∞
k1 (y, an )k2q (y, x)w(y) 0
an an+2
k1 (y, an )w(y)
q r q h u(s) ds dy
s
∞ 1 q r p q k2 (x, s)u(s) h ds dy
an−1 0
an
1 p
s
∞ r x ∞ 1 r q p q ρ(x) k1 (y, x)k2 (y, x)w(y) dy h u(s) ds dx
0
0
x
s
∞ r x ∞ 1 r r q p ρ(x) k1 (y, x)w(y) dy k2 (x, s)u(s) h ds dx Ar1 + Ar2 hLp 1 .
0
V
0
x
(19)
s
Write J1,2 ≈
n
an+2 y an+2 1 q r p q 2n k1 (y, an )w(y) k2 (y, s)u(s) h ds dy an
an−1
s
an+2 y ∞ 1 q r p q n + 2 k1 (y, an )w(y) k2 (y, s)u(s) h ds dy =: J1,2,1 + J1,2,2 . n
an
an−1
an+2
887
We obtain J1,2,2 ≈
an
an+2 y q ∞ q r p q ρ(x) k1 (y, an )w(y) k2 (y, s)u(s) ds h dy
n a n−1
∞ ≤
an
an−1
an+2
∞ y q ∞ q r r p q ρ(x) k1 (y, x)w(y) k2 (y, s)u(s) ds h dy dx ≤ Ar3 hLp 1 .
0
0
x
(20)
V
y
Now, using (9) and (10), we have J1,2,1 =
an+2 p r q p q n p 2 k1 (y, an )w(y)(T[an ,an+2 ] h(y)) dy
n
≤
an
2n T[an ,an+2 ]
r p
an+2 r p hV .
q
L1V [an−1 ,an+2 ]→Lkp
n
1 (·,an )w(·)
[an ,an+2 ]
an−1
For p ≤ r, applying Jensen’s inequality, we infer J1,2,1
an−1 r ≤ sup ρ T[an ,an+2 ] p n
∞
q L1V →Lkp (·,a )w(·) n 1
0
r hV
0
J1,2,1 ≤ B
J1,2,1 ≤
2
ns r
∞
q L1V →Lkp (·,t)w(·) 1
0
r
p
hV
.
0
∞
p r
t r ≤ sup ρ Tt p t>0
Hence,
For r < p, reckoning with H¨ older’s inequality
p
1 s
p
hV. 0
=
1 r
T[an ,an+2 ]
−
1 p
, we find that r
r ∞
s p
q L1V →Lkp (·,a )w(·) n 1
n
(21)
s
hV
p
0
because
2
ns r
n
T[an ,an+2 ]
s p
an+1 an s s p ≈ ρ ρ T[σ−1 (an+1 ),σ2 (an )] p
q
L1V →Lkp
1 (·,an )w(·)
n
k1 (·,σ −1 (an+1 ))w(·)
0
an
s
an+1
x ≤ ρ(x) ρ T[σ−1 (x),σ2 (x)] n
∞ = 0
p
q L1V →L p k1 (·,σ −1 (x))w(·)
0
p r
dx
s
x ρ(x) ρ T[σ−1 (x),σ2 (x)]
Therefore,
dx = B s .
∞
J1,2,1 ≤ B p
hV. 0
888
p
q L1V →L p k1 (·,σ −1 (x))w(·)
0
an
q
L1V →L p
(22)
Consider J2 =
∞ 2n
n
r
q p
q
k1 (y, an )w(y)(Tp h(y)) dy
=
n
an+2
≈
n
2
n
i≥na i+2
n
i≥na i+2
ai+3 r q q 2 k1 (y, ai+1 )w(y)(Tp h(y)) p dy
n
+
ai+3 r q q n p 2 k1 (y, an )w(y)(Tp h(y)) dy
ai+3
q p
r
k1 (ai+1 , an )w(y)(Tp h(y)) dy
q
=: J2,1 + J2,2 .
i≥na i+2
Applying (18), we get an+3 r r q q n p 2 k1 (y, an+1 )w(y)(Tp h(y)) dy ≈ J1 ≤ Ar1 + Ar2 + Ar3 + B r hLp 1 . J2,1 ≤ V
n
(23)
an+1
To estimate J2,2 , we use the inequality (see [17, Lemma 3.1]) 1 i α α k1 (aj+1 , aj ) , k1 (ai+1 , an )
α ∈ (0, 1),
(24)
j=n
and Minkowski’s inequality to derive that ai+3 r q q n p 2 k1 (ai+1 , an ) w(y)(Tp h(y)) dy J2,2 = n
2n
i
n
≤
n
2
n
ai+2
1 ai+3 r α q q α p k1 (aj+1 , aj ) w(y)(Tp h(y)) dy
j=n
i≥n
=
i≥n
α
k1 (aj+1 , aj )
j≥n
n
2
n
i≥j a
ai+2 ai+3
q p
w(y)(Tp h(y)) dy i+2
∞ α
k1 (aj+1 , aj )
j≥n
α
q p
α
w(y)(Tp h(y)) dy
r αq
r αq
aj+2
∞ r q q (18) n p ≈ 2 k1 (an+1 , an ) w(y)(Tp h(y)) dy n
≈
an ρ(x)k1 (an+1 , an )
n a n−1 an
≤
2
ρ(x)k1 (σ (x), x)
n a n−1
∞ 0
r q
an+2 ∞
r q w(y)(Tp h(y) dy) dx q p
r q w(y)(Tp h(y) dy) dx q p
σ 2 (x) 2
ρ(x)k1 (σ (x), x)
=
r q
an+2 ∞
r q
∞
q p
w(y)(Tp h(y)) dy σ 2 (x)
r q
r
dx ≤ Ar4 hLp 1 . V
(25)
It follows from (19)–(25) that the upper bound CT A1 + A2 + A3 + A4 + B is proved. 889
The lower bound. Diminish the domains of integration in (17): (1) [0, y] → [0, x] and obtain CT ≥ A1 + A2 (since k2 (y, s) ≈ k2 (x, s) + k2 (y, x) for 0 < s ≤ x ≤ y); (2) [s, ∞) → [y, ∞) and obtain CT ≥ A3 ; (3) [x, ∞) → [σ 2 (x), ∞) and obtain CT ≥ A4 (since k1 (y, x) k1 (σ 2 (x), x) for x < σ 2 (x) ≤ y). Inequality (17) for h ∈ M+ implies that t p ∞ y ∞ 1 q p ∞ r p q p ρ k1 (y, t)w(y) k2 (y, s)u(s) h ds dy ≤ CT hV. 0
0
t
σ −1 (t)
s
Therefore, CT B, and the theorem is proved for p ≤ r. It remains to find the lower bounds for B when r < p. We have ∞ s
B =
0 an+1
0
n
≤
n
p
q L1V →L p k1 (·,σ −1 (x))w(·)
p
q L1V →L p k1 (·,σ −1 (x))w(·)
0
an an+1
an+1 s s p ρ(x) dx ρ T[σ−1 (an ),σ2 (an+1 )] p
an
≈
ns r
Bs
q
k1 (·,σ −1 (an ))w(·)
T[an−1 ,an+3 ]
s p
.
q
L1V →Lkp
n
Hence,
dx
L1V →L p
0
2
dx
s
x ρ(x) ρ T[σ−1 (x),σ2 (x)]
=
s
x ρ(x) ρ T[σ−1 (x),σ2 (x)]
1 (·,an−1 )w(·)
2
ns r
s
T[an−1 ,an+3 ] p
L1V
n
q →Lkp (·,a 1 n−1 )w(·)
=: B s .
Let θ ∈ (0, 1) be some fixed number. Then for each n ∈ Z there exists hn ∈ L1V [an−2 , an+3 ] such that q ≥ θT[an−1 ,an+3 ] . Put hn L1 [an−2 ,an+3 ] = 1 and T[an−1 ,an+3 ] hn pq p V
gn := 2
L1V →Lk
Lk
1 (·,an−1 )w(·)
ns r
s
T[an−1 ,an+3 ] p
q L1V →Lkp (·,a 1 n−1 )w(·)
Then gL1
V
hn ,
g :=
1 (·,an−1 )w(·)
gn ,
Tn := T[an−1 ,an+3 ]
1 (·,an−1 )w(·)
an+3 ns ps gn V = 2 r Tn = B s . n a n−2
n
On the other hand, ∞ D :=
n a n−2
890
0
r ∞ q q ρ(x) k1 (y, x)w(y)(Tp g(y)) p dx x
p r
an+3 q p q ρ(x) dx k1 (y, an−1 )w(y)(Tp g(y)) p
an−1
q
L1V →Lkp
an−1
.
an+3
y an+3 1 p q p r p p q p 2n k1 (y, an−1 )w(y) k2 (y, s)u(s) gn ds dy
n
an−1
an−2
s
an+3 p r q p q n p = 2 k1 (y, an−1 )w(y)(T[an−1 ,an+3 ] gn ) dy n
=
2n 2
ns p
an−1 sr p2
r
r
Tn T[an−1 ,an+3 ] hn p q
≥ θp
Lkp
n
p
2
ns r
s
r
Tnp = θ p B s .
n
1 (·,an−1 )w(·)
(17)
1
sp
1
Inequality (17) yields B s CTp CTp gL1 ≥ D r θB r . Consequently, CT θ p B ≥ θ p B. Since V θ ∈ (0, 1) is arbitrary, CT B. As regards the limit values of the parameters, we make the following Remark 1. (1) If q = ∞ then CT ≈ A1 + A2 + A3 + A4 + B, where A1 , A2 , A3 , and A4 are the best constants in x ∞ 1 ∞ r p ∞ p r p r ρ(x)[ess sup k1 (y, x)k2 (y, x)w(y)] h u(s) ds dx ≤ A1 hV, (26) 0 ∞
y≥x
x ρ(x)[ess sup k1 (y, x)w(y)]r
0
∞ 0
∞
for h ∈
M+
y≥x
0
0
s
∞ 1 r p ∞ p r p k2 (x, s) h u(s) ds dx ≤ A2 hV,
0
0
s
y ∞ 1 r p ∞ p r p ρ(x) ess sup k1 (y, x)w(y) k2 (y, s)u(s) ds h dx ≤ A3 hV, y≥x
0
ρ(x)k1 (σ 2 (x), x)r
0
0
(2) If p = ∞ or r = ∞, then
1 CT = T , v Lrρ
q
p L1V →Lw
t>0
s
p
1 s dx , r < p.
p = ∞,
1
CT ≈ sup R(t)Tt p
p ≤ r,
k1 (·,σ −1 (x))w(·)
V
0
(29)
0
s
and the ⎧ constant B has the form t 1 1 r ⎪ ρ T t Lp 1 →L∞ , ⎪ ⎨ sup V k1 (·,t)w(·) t>0 0 B := ∞ x ⎪ ⎪ ⎩ ρ(x) ρ T[σ−1 (x),σ2 (x)] L1 →L∞ 0
(28)
0
y
y ∞ 1 r p ∞ p r p ess sup w(y) k2 (y, s)u(s) h ds dx ≤ A4 hV
y≥σ 2 (x)
(27)
,
r = ∞,
where R(t) := ess sup0
+
n
y≥x
an
n
1
2n [ ess sup k1 (y, an )w(y)(Tp h(y)) p ]r y∈[an ,an+2 )
1
2n [ ess sup k1 (y, an )w(y)(Tp h(y)) p ]r =: J1 + J2 . y∈[an+2 ,∞)
891
By analogy to Theorem 1, we prove the inequality r J1 Ar1 + Ar2 + Ar3 + B r hLp 1 .
(30)
V
To estimate J2 , use the relation
2n (sup ai )s ≈ i≥n
n∈Z
2n asn
(31)
n∈Z
instead of (18). We have
J2 =
= ≈
2n [sup
+
2n [sup
k1 (y, an )w(y)(Tp h(y)) p ]r 1
ess sup
i≥n y∈[ai+2 ,ai+3 )
2n [sup
k1 (y, ai+1 )w(y)(Tp h(y)) p ]r 1
ess sup
i≥n y∈[ai+2 ,ai+3 )
n
1
ess sup
i≥n y∈[ai+2 ,ai+3 )
n
y∈[an+2 ,∞)
n
n
1
2n [ ess sup k1 (y, an )w(y)(Tp h(y)) p ]r
k1 (ai+1 , an )w(y)(Tp h(y)) p ]r =: J2,1 + J2,2 .
Applying (31) as in Theorem 1, we obtain r J2,1 ≤ Ar1 + Ar2 + Ar3 + B r hLp 1 .
(32)
V
To estimate J2,2 , we use Minkowski’s inequality and (24) to see that
J2,2 =
2
sup
i
i≥n
n
≤
2
n
k1 (aj+1 , aj )
=
n (18)
≈
≈
ess sup y∈[ai+2 ,ai+3 )
k1 (aj+1 , aj ) (sup
2n
= 0
ess sup
1 p
α
1
1 p
r
r
α
w(y)(Tp h(y)) )
k1 (aj+1 , aj ) ess sup w(y)(Tp h(y)) p
r
y≥aj+2
j≥n
1
2n k1 (an+1 , an )r [ess sup w(y)(Tp h(y)) p ]r y≥an+2
1
ρ(x) dxk1 (an+1 , an )r [ess sup w(y)(Tp h(y)) p ]r y≥an+2
1
ρ(x)k1 (σ 2 (x), x)r [ess sup w(y)(Tp h(y)) p ]r dx y≥σ 2 (x)
n a n−1
∞
k1 (ai+1 , an )w(y)(Tp h(y))
i≥j y∈[ai+2 ,ai+3 )
n an
n a n−1 an
≤
α
α
j≥n
k1 (ai+1 , an )w(y)(Tp h(y)) p ]r
1
α
j=n
n
1
ess sup
i≥n y∈[ai+2 ,ai+3 )
n
n
2n [sup
1
r
ρ(x)k1 (σ 2 (x), x)r [ess sup w(y)(Tp h(y)) p ]r dx ≤ Ar4 hLp 1 . y≥σ 2 (x)
V
(33)
Relations (30)–(33) imply the upper bound CT A1 + A2 + A3 + A4 + B. The lower bound is proved by analogy to Theorem 1. 892
Remark 2. Sharp two-sided estimates of the best constant in (13)–(15), (26)–(28) and the constants B by explicit integral functionals are found by reduction theorems in [20–22] and criteria for the boundedness of Hardy-type integral operators [1, 29–31]. But the left-hand sides of (16) and (29) contain an additional iteration; therefore, we give the reduction theorem for this case (see Lemma 1 below). Suppose that λ, μ, ν, η ∈ M+ and the kernel k(x, y) satisfies Oinarov’s condition (6), while the sequence an and the function σ(x) are defined by (7) and (8) with λ in place of ρ. Given 0 < c < d ≤ ∞, 0 < t, q < ∞, h ∈ M+ , put x ∞ q 1 q k(x, s)ν(s) h ds , Tt h(x) := χ[t,∞) (x) 0
(34)
s
x d q 1 q k(x, s)ν(s) h ds . T[c,d] h(x) := χ[c,d] (x) σ −1 (c)
(35)
s
We have Lemma 1. Let 0 < p, q, r < ∞. Then the best constant C in ∞
∞ y ∞ q r p 1 ∞ q r p λ(x) μ(y) k(y, s)ν(s) h ds dy dx ≤ C hη,
0
0
x
h ∈ M+ ,
(36)
0
s
satisfies C ≈ G1 + G2 + G3 + G, where G1 , G2 , and G3 are the best constants in ∞ 0
∞ p x ∞ q p 1 ∞ r q p r λ(x) μ(y)[k(y, x)] q dy ν(s) h ds dx ≤ G1 hη,
∞
∞
0
x
0
s
∞ p x ∞ q p 1 ∞ r q p λ(x) μ k(x, s)ν(s) h ds dx ≤ G2 hη,
0
0
x
0
s
∞ y r ∞ r p 1 ∞ q r p λ(x) μ(y) k(y, s)ν(s) ds h dy dx ≤ G3 hη
0
0
x
0
y
for h ∈ M+ and the constant G has the form ⎧ t 1 p ⎪ λ T t L1η →Lrμ , p ≥ 1, sup ⎪ ⎨ t>0 0 G := ∞ x 1−p p ⎪ 1−p p ⎪ ⎩ λ(x) λ T[σ−1 (x),σ(x)] L1η →Lrμ dx , 0 < p < 1. 0
0
Proof. The upper bound. We infer ∞ p
J :=
∞ y ∞ q r p q r λ(x) μ(y) k(y, s)ν(s) h ds dy dx
0
≈
n
x an+1
n
2
0 y
μ(y) an
s ∞
k(y, s)ν(s) 0
q r p q r h ds dy
s
893
≈
an+1 y ∞ q r p q r 2n μ(y) k(y, s)ν(s) h ds dy
n
an
an−1
s
an+1 an−1 ∞ q r p q r n + 2 μ(y) k(y, s)ν(s) h ds dy =: J1p + J2p . n
Estimate J2 : J2p
≈
0
an
s
an+1 an−1 ∞ q r p q r λ(x) dx μ(y) k(y, s)ν(s) h ds dy .
an n a n−1
0
an
s
Since k(y, s) ≈ k(y, x) + k(x, s) for s ≤ an−1 ≤ x ≤ an ≤ y, we get J2p
≈
an
an+1 p an−1 ∞ q p r q r λ(x) μ(y)[k(y, x)] q dy dx ν(s) h ds
n a n−1
+
0
an
an
λ(x)
n a n−1 an
n a n−1
+
k(x, s)ν(s) 0
=
+
s
s an+1
p r μ
an
0
x
an
0
x r
s
p x
μ(y)[k(y, x)] q dy
λ(x) 0
s
∞ p x ∞ q p r q λ(x) μ k(x, s)ν(s) h ds dx
n a n−1 ∞ ∞
∞
q h ds dx p q
∞ p x ∞ q p r q r λ(x) μ(y)[k(y, x)] q dy ν(s) h ds dx
∞
an−1
r
∞ q p q ν(s) h ds dx
0
x
s
∞ p x ∞ q p ∞ p r q p p λ(x) μ k(x, s)ν(s) h ds dx ≤ G1 + G2 hη .
0
0
x
0
s
Estimate J1 :
J1p ≈
y r p ∞ p an+1 q r n 2 μ(y) k(y, s)ν(s) ds dy h
n
an
an−1
an+1
an+1 y an+1 q r p q r p p n + 2 μ(y) k(y, s)ν(s) h ds dy =: J1,1 + J1,2 . n
an
an−1
s
Estimate J1,1 : p J1,1
≈
an
an+1 y r p ∞ p q r λ(x) dx μ(y) k(y, s)ν(s) ds dy h
n a n−1
∞
λ(x) 0
894
∞
an
y μ(y)
x
0
an−1
k(y, s)ν(s) ds
an+1 r q
∞ y
∞ p r p h dy ≤ G3 hη . p r
0
Now estimate J1,2 . Using notations (34) and (35), write p = J1,2
an+1 an+1 p p r 2n μ(y)[T[an ,an+1 ] h(y)]r dy ≤ 2n T[an ,an+1 ] pL1 →Lr hη .
n
η
n
an
μ
an−1
For p ≥ 1, applying Jensen’s inequality, we have an ∞ p ∞ p p p p λ T[an ,an+1 ] L1 →Lr hη G hη . J1,2 sup n
η
μ
0
0
0
For 0 < p < 1, reckoning with H¨ older’s inequality, we infer p J1,2
1−p ∞ p ∞ p an+1 1 p 1−p 1−p λ(x)dx T[an ,an+1 ] L1 →Lr hη Gp hη . η
n
μ
0
an
0
Therefore, the upper bound C G1 + G2 + G3 + G is proved. For the lower bound, it suffices to repeat the corresponding argument in the proof of Theorem 1. Inequality (29) is reduced similarly. § 3. The Main Results for the Operator S + (t) := ∞ u Let u ˜ := u2 , U t ˜ , and 0 < t < ∞. Given 0 < c < d ≤ ∞, 0 < t < ∞, and h ∈ M , put V
p
∞ s 1 p p Tt h(x) := χ[t,∞) (x) k2 (s, x) hV u ˜ (s) ds ,
(37)
0
x
σ(d) s 1 p p k2 (s, x) hV u ˜ (s) ds , T[c,d] h(x) := χ[c,d] (x) x
Tt Lpv →Lqw :=
(38)
c
1 ∞ [Tt h]q w q sup
0
0=h∈M+
(39)
1 ∞ [h]p v p 0
and similarly for T[c,d] Lpv →Lqw . Theorem 2. Suppose that 0 < q < ∞, 0 < p < ∞, 0 < r < ∞. Then the best constant CS in ∞ 1 1 ∞ r p r p [Sf (x)] ρ(x) dx ≤ CS [f (x)] v(x) dx , f ∈ M↓ , (40) 0
0
satisfies CS ≈ A1 + A2 + A3 + A4 + B, where A1 , A2 , A3 , and A4 are the best constants in ∞ 0
∞ 0
3
σ (x) r ∞ s 1 r p ∞ q p r p 3 q ρ(x) k1 (y, x)k2 (σ (x), x) w(y) dy hV u ˜ (s) ds dx ≤ A1 h, σ 3 (x)
x
0
0
3
σ (x) r ∞ s 1 r p ∞ q p r p 3 ρ(x) k1 (y, x)w(y) dy k2 (s, σ (x)) hV u ˜ (s) ds dx ≤ A2 h, x
σ 3 (x)
0
0
895
∞
∞ ∞ q y q r p ∞ p q r p ρ(x) k1 (y, x)w(y) k2 (s, y)˜ u(s) ds hV dy dx ≤ A3 h,
0
∞
x
0
y r
ρ(x)k1 (σ 2 (x), x) q
∞
0
0
∞ s 1 q r p ∞ p q r p w(y) k2 (s, y)˜ u(s) hV dy dx ≤ A4 h
σ 2 (x)
0
y
0
for h ∈ M+ and the constant B has the form ⎧ t 1 1 r ⎪ ρ Tt p , sup q ⎪ ⎪ ⎨ t>0 0 L1 →Lkp (·,t)w(·) 1 B := x ∞ ⎪ ⎪ ⎪ ρ(x) ρ T[σ−1 (x),σ2 (x)] ⎩ 0
p ≤ r, s
q L1 →L p k1 (·,σ −1 (x))w(·)
0
p
1 s dx , r < p.
Proof. In contrast to Theorem 1, here we apply Theorem 3.2 of [17]; then, for h ∈ M+ , inequality (40) is equivalent to the inequality ∞ ∞ 0
∞ s 1 q r p ∞ p q r p k1 (y, x)w(y) k2 (s, y) hV u ˜ (s) ds w(y) dy ρ(x) dx ≤ CS h.
x
0
y
(41)
0
The upper bound. Put ∞
s 1 p p k2 (s, y)˜ u(s) hV ds .
y
0
Tp h(y) := We infer
an+1 ∞ ∞ r r q q q q n p p ρ(x) k1 (y, x)w(y)(Tp h(y)) dy dx 2 k1 (y, an )w(y)(Tp h(y)) dy I := n
≈
an an+2
n
2
n
x q p
r
k1 (y, an )w(y)(Tp h(y)) dy
n
+
n
an
I1 ≈
2
q p
k1 (y, an )w(y)(Tp h(y)) dy
r q
=: I1 + I2 ,
an+2
an+3 s 1 q r an+2 p q 2 k1 (y, an )w(y) k2 (s, y)˜ u(s) hV ds dy an
0
y
∞ s 1 q r p q k1 (y, an )w(y) k2 (s, y)˜ u(s) hV ds dy =: I1,1 + I1,2 .
an+2
2n
n
n
n
n
+
q
an
∞
an
0
an+3
Since k2 (s, y) ≈ k2 (σ 3 (x), y) + k2 (s, σ 3 (x)) for x ≤ y ≤ σ 3 (x) ≤ s, we obtain I1,2 ≈
an
an+2 ∞ s 1 q r p q ρ(x) dx k1 (y, an )w(y) k2 (s, y)˜ u(s) hV ds dy
n a n−1
≈
an
n a n−1
896
an
0
an+3
an+2 ∞ s 1 q r p q q 3 ρ(x) dx k1 (y, an )k2 (σ (x), y)w(y) hV u ˜ (s) ds dy an
an+3
0
an+2 ∞ s 1 q r p q 3 ρ(x) dx k1 (y, an )w(y) k2 (s, σ (x))˜ u(s) hV ds dy
an
+
n a n−1
∞
an
0
an+3
σ3 (x)
ρ(x)
k1 (y, x)k2q (σ 3 (x), y)w(y) dy
0
r ∞ s q σ 3 (x)
x
∞ +
σ 3 (x)
ρ(x)
0
k1 (y, x)w(y) dy
r ∞ q
hV
p
r u ˜ (s) ds dx
0
s 1 r p k2 (s, σ (x))˜ u(s) hV ds dx 3
0
σ 3 (x)
x
1
r Ar1 + Ar2 hLp 1 .
(42)
Write
I1,1 ≈
an+2 an+3 an 1 q r p q 2 k1 (y, an )w(y) k2 (s, y)˜ u(s) hV ds dy n
n
+
0
y
an+3 s 1 q r p q k1 (y, an )w(y) k2 (s, y)˜ u(s) hV ds dy =: I1,1,1 + I1,1,2 .
2n
n
an
an+2
an
y
an
We have I1,1,1 ≈
an
an+2 an+3 q an q r p q ρ(x) k1 (y, an )w(y) k2 (s, y)u(s) ds hV dy
n a n−1
∞ ≤
an
0
y
∞ ∞ q y q r r p q ρ(x) k1 (y, x)w(y) k2 (s, y)˜ u(s) ds hV dy dx ≤ Ar3 hLp 1 .
0
x
(43)
0
y
Now, recalling (37) and (38), we have I1,1,2 =
an+2 p r q p q p 2 k1 (y, an )w(y)(T[an ,an+2 ] h(y)) dy n
n
≤
an
2n T[an ,an+2 ]
r p
an+3 r p h .
q
L1 [an ,an+3 ]→Lkp
n
1 (·,an )w(·)
[an ,an+2 ]
an
Given p ≤ r and applying Jensen’s inequality, we get I1,1,2 ≤ sup 2 T[an ,an+2 ] n
n
r p
∞ r t r p h ≤ sup ρ Tt p
q
L1 →Lkp
1 (·,an )w(·)
0
Consequently, p r
t>0
0
q
L1 →Lkp
1 (·,t)w(·)
∞ r p h . 0
∞
I1,1,2 ≤ Bp
h.
(44)
0
897
Given r < p and involving H¨ older’s inequality
I1,1,2 ≤
2
ns r
1
=
s
T[an ,an+2 ]
1 p
−
s p
, we have r ∞ pr s h .
q
L1 →Lkp
n
Since
1 r
1 (·,an )w(·)
2
ns r
s
T[an ,an+2 ] p
q
L1 →Lkp
n
1 (·,an )w(·)
an+1 an s s p ≈ ρ ρ T[σ−1 (an+1 ),σ2 (an )] p n
n
∞ =
k1 (·,σ −1 (an+1 ))w(·)
x ρ(x) ρ T[σ−1 (x),σ2 (x)]
≤
q
L1 →L p
0
an an+1
0
p
q L1 →L p k1 (·,σ −1 (x))w(·)
0
an
s
s
x ρ(x) ρ T[σ−1 (x),σ2 (x)]
0
p
q L1 →L p k1 (·,σ −1 (x))w(·)
0
dx
we infer
dx = Bs ,
∞
p r
I1,1,2 ≤ B
p
h.
(45)
0
Consider
I2 =
∞ n
2
n
≈
n
n
2
r
q p
q
k1 (y, an )w(y)(Tp h(y)) dy
=
ai+3 r q q p 2 k1 (y, an )w(y)(Tp h(y)) dy n
n
an+2 ai+3
r
q p
q
k1 (y, ai+1 )w(y)(Tp h(y)) dy
+
i≥na i+2 ai+2
n
2
n
i≥na i+2
q p
k1 (ai+1 , an )w(y)(Tp h(y)) dy
r q
i≥na i+1
=: I2,1 + I2,2 . Applying (18), we obtain an+3 r r q q n 2 k1 (y, an+1 )w(y)(Tp h(y)) p dy ≈ I1 ≤ Ar1 + Ar2 + Ar3 + Br hLp 1 . I2,1 ≤ n
an+1
To estimate I2,2 , we use (24) and Minkowski’s inequality to find that ai+3 r q q n p 2 k1 (ai+1 , an ) w(y)(Tp h(y)) dy I2,2 = n
n
2
i
n
≤
n
2
n
1 ai+3 r α q q k1 (aj+1 , aj ) w(y)(Tp h(y)) p dy α
α
k1 (aj+1 , aj )
j≥n
n
898
ai+2
j=n
i≥n
=
i≥n
n
2
j≥n
i≥j a
ai+2 ai+3
q p
α
w(y)(Tp h(y)) dy i+2
∞ α
k1 (aj+1 , aj )
q p
w(y)(Tp h(y)) dy aj+2
α
r αq
r αq
(46)
(18)
≈
∞ r q q 2 k1 (an+1 , an ) w(y)(Tp h(y)) p dy n
n
≈
an+2
an ρ(x)k1 (an+1 , an )
n a n−1 an
≤
2
n a n−1
=
∞
r
q p
r q
q
w(y)(Tp h(y)) dy an+2
ρ(x)k1 (σ (x), x)
∞
r q
∞
r
q p
w(y)(Tp h(y)) dy
q
σ 2 (x) r q
2
ρ(x)k1 (σ (x), x)
∞
q p
r
w(y)(Tp h(y)) dy
0
r
q
≤ Ar4 hLp 1 .
(47)
σ 2 (x)
It follows from (42)–(47) that the upper bound CS A1 + A2 + A3 + A4 + B is proved. The lower bound. Diminish the domains of integration in (41): (1) [x, ∞) → [x, σ 3 (x)], [y, ∞) → [σ 3 (x), ∞) and obtain CS ≥ A1 + A2 (since k2 (s, y) ≈ k2 (σ 3 (x), y) + k2 (s, σ 3 (x)) for x ≤ y ≤ σ 3 (x) ≤ s); (2) [0, s] → [0, y] and obtain CS ≥ A3 ; (3) [x, ∞) → [σ 2 (x), ∞) and obtain CS ≥ A4 (since k1 (y, x) k1 (σ 2 (x), x) for x < σ 2 (x) ≤ y). It is proved by analogy to Theorem 1 that CS ≥ B. For the limit values of the parameters, we have Remark 3. (1) CS ≈ A1 + A2 + A3 + A4 + B for q = ∞, where A1 , A2 , A3 , and A4 are the best constants in the inequalities ∞ s 1 r p ∞ ∞ p r p ρ(x)[ ess sup k1 (y, x)k2 (σ 3 (x), x)w(y)]r hV u ˜ (s) ds dx ≤ A1 h, x
0
∞ x
∞
k2 (s, σ 3 (x))
0
∞
ρ(x)k1 (σ 2 (x), x)r
0
for h ∈
M+
r p ∞ r p u ˜ (s) ds dx ≤ A2 h, 0
1 r
y k2 (s, y)˜ u(s) ds
p
hV
∞ h, 0
∞ s 1 r p ∞ p r p ess sup w(y) k2 (s, y)˜ u(s) hV ds dx ≤ A4 h
y≥σ 2 (x)
0
y
(2) For p = ∞ and r = ∞, we have 1 CS = S , p = ∞, v Lrρ
p ≤ r, s
p
k1 (·,σ −1 (x))w(·)
0
1 s dx , r < p.
1
CS ≈ sup R(t)Tt p t>0
(48)
0
and the constant B has the form ⎧ t 1 1 r p ⎪ ρ T , sup ⎪ t ⎨ L1 →L∞ k1 (·,t)w(·) t>0 0 B := ∞ x ⎪ ⎪ ⎩ ρ(x) ρ T[σ−1 (x),σ2 (x)] L1 →L∞ 0
p r dx ≤ Ap3
0
y
p
0
∞
y≥x
1 hV
σ 3 (x)
ρ(x) ess sup k1 (y, x)w(y)
0
s
∞
ρ(x)[ ess sup k1 (y, x)w(y)]r 0
0
σ 3 (x)
q
L1V →Lkp
, r = ∞,
1 (·,t)w(·)
where R(t) := ess sup0
Suppose that λ, μ, ν, η ∈ M+ , the kernel k(x, y) satisfies Oinarov’s condition (6), and the sequence an and the function σ(x) are defined by (7) and (8) with λ instead of ρ. For 0 < c < d ≤ ∞, 0 < t, q < ∞, h ∈ M+ , put ∞ s q 1 q ∗ Tt h(x) := χ[t,∞) (x) k(x, s)ν(s) h ds , 0
x ∗ h(x) T[c,d]
:= χ[c,d] (x)
s q 1 q k(x, s)ν(s) h ds .
σ(d)
x
c
In much the same way as in Remark 2, we will give a lemma that enables us to reduce the inequalities with the constants A4 . Lemma 2. Suppose that 0 < p, q, r < ∞. Then the best constant C ∗ in ∞ ∞ ∞ s q r p 1 ∞ q r p ∗ λ(x) μ(y) k(y, s)ν(s) h ds dy dx ≤C hη, h ∈ M+ , 0
satisfies
C∗
≈
x
G∗1
∞
+
G∗2
+
0
y
G∗3
+
G∗ ,
where
G∗1 ,
G∗2 ,
0
and
G∗3
are the best constants in the inequalities
2
σ (x) p ∞ ∞ q p 1 ∞ r q p r 2 ∗ q λ(x) μ(y)[k(σ (x), y)] dy ν(s) h ds dx ≤ G1 hη,
0
σ 2 (x)
x
∞
0
s
2
σ (x) p ∞ s q p 1 ∞ r q p 2 ∗ λ(x) μ k(s, σ (x))ν(s) h ds dx ≤ G2 hη,
0
∞ 0
0
σ 2 (x)
x
0
∞ ∞ r y r p 1 ∞ q r p ∗ λ(x) μ(y) k(y, s)ν(s) ds h dy dx ≤ G3 hη x
0
y
for h ∈ M+ and the constant G∗ has the form ⎧ t 1 p ⎪ λ T ∗ t L1η →Lrμ , sup ⎪ ⎨ t>0 0 G∗ := ∞ x ⎪ ⎪ ⎩ λ(x) λ T ∗−1 0
[σ
0
(x),σ(x)] L1η →Lrμ
0
p ≥ 1,
p 1−p
1−p p dx , 0 < p < 1.
Inequality (48) is reduced similarly. § 4. The Main Results for the Operators T and S The characterization of (1) proved analogously. Let us state the results. ∞ ∞for the operators T and S is We will assume that 0 < x ρ < ∞ for every x > 0 and 0 ρ = ∞. Define the sequence {bn } ⊂ (0; ∞) from the equations ∞ ρ = 2−n , n ∈ Z. bn
Let ζ : [0; ∞) → [0; ∞) and ζ −1 : [0; ∞) → [0; ∞) be defined by the formulas (here sup ∅ = 0) ∞ ∞ ∞ ∞ 1 −1 ζ(x) := sup y > 0 : ρ ≥ ρ , ζ (x) := sup y > 0 : ρ ≥ 2 ρ . 2 y
900
x
y
x
(49)
For 0 < c < d ≤ ∞, 0 < t, p < ∞, h ∈ M+ , put ∞ s 1 p p k2 (s, x)˜ u(s) hV ds , Tt h(x) := χ(0,t] (x)
(50)
0
x
ζ(d) s 1 p p k2 (s, x)˜ u(s) hV ds . T[c,d] h(x) := χ[c,d] (x) x
(51)
c
Theorem 3. Suppose that 0 < q < ∞, 0 < p < ∞, and 0 < r < ∞. Then the best constant CT in 1 1 ∞ ∞ r p r p [T f (x)] ρ(x) dx ≤ CT [f (x)] v(x) dx , f ∈ M↓ , (52) 0
0
satisfies CT ≈ A1 + A2 + A3 + A4 + B, where A1 , A2 , A3 , and A4 are the best constants in ∞ x r ∞ s 1 r p ∞ q p r p q ρ(x) k1 (x, y)k2 (x, y) w(y) dy hV u ˜ (s) ds dx ≤ A1 h, 0
∞ 0
0
∞
0
0 −2
(x))
r q
:=
1 r
0
ζ −2 (x)
∞ s 1 q r p ∞ p q r p w(y) k2 (s, y)˜ u(s) hV dy dx ≤ A4 h
0
0
y
0
the constant B has the form ⎧ ∞ 1 1 r t p ⎪ ρ T , sup q ⎪ ⎪ ⎨ t>0 t L1 →Lkp (t,·)w(·) 1 B := ∞ ∞ ⎪ ⎪ ⎪ ρ(x) ρ T[ζ −1 (x),ζ 2 (x)] ⎩ 0
1 s
0
0
y
ρ(x)k1 (x, ζ
where
0
x
∞
for h ∈
0
x ∞ q y q r p ∞ p q r p ρ(x) k1 (x, y)w(y) k2 (s, y)˜ u(s) ds hV dy dx ≤ A3 h,
0
0 + M and
0
x
x r ∞ s 1 r p ∞ q p r p ρ(x) k1 (x, y)w(y) dy k2 (s, x) hV u ˜ (s) ds dx ≤ A2 h,
p ≤ r, s
q L1 →L p 2 k1 (ζ (x),·)w(·)
x
− p1 .
p
1 s dx ,
r < p,
Remark 4. (1) CT ≈ A1 + A2 + A3 + A4 + B for q = ∞, where A1 , A2 , A3 , and A4 are the best constants in 1 r p ∞ s ∞ ∞ p r p r ρ(x)[ess sup k1 (x, y)k2 (x, y)w(y) dy] hV u ˜ (s) ds dx ≤ A1 h, 0 ∞
y∈(0,x)
∞ ρ(x)[ess sup k1 (x, y)w(y) dy]r
0 ∞ 0 ∞
y∈(0,x)
0
0
0
0
s 1 r p ∞ p r p k2 (s, x) hV u ˜ (s) ds dx ≤ A2 h,
x
∞ y 1 r p ∞ p r p ρ(x) ess sup k1 (x, y)w(y) k2 (s, y)˜ u(s) ds hV dy dx ≤ A3 h, y∈(0,x)
ρ(x)k1 (x, ζ 0
x
−2
r
(x))
y
∞
ess sup w(y)
y∈(0,ζ −2 (x))
0
k2 (s, y)˜ u(s) y
r
s hV 0
1 p
0
∞ p dx ≤ A4 h p r
0
901
for h ∈ M+ and the constant B has the form ⎧ ∞ 1 1 r ⎪ t p 1 ∞ ρ T , sup ⎪ ⎨ L →Lk (t,·)w(·) t>0 1 t B := ∞ ∞ ⎪ ⎪ ⎩ ρ(x) ρ T[ζ −1 (x),ζ 2 (x)] L1 →L∞ 0
p ≤ r, s
p
k1 (ζ 2 (x),·)w(·)
x
(2) For p = ∞ and r = ∞, we have 1 CT = T , p = ∞, v Lrρ
1 s dx ,
1
CT ≈ sup R(t)Tt p
, r = ∞,
q
L1V →Lkp
t>0
r < p.
1 (t,·)w(·)
where R(t) := ess supz≥t ρ(z).
For 0 < c < d ≤ ∞, 0 < t < ∞, h ∈ M+ , put x ∞ 1 p p t h(x) := χ(0,t] (x) k2 (x, s)u(s) h ds , T 0
T[c,d] h(x) := χ[c,d] (x)
(53)
s
x
d 1 p p k2 (x, s)u(s) h ds .
ζ −1 (c)
(54)
s
Theorem 4. Suppose that 0 < q < ∞, 0 < p < ∞, and 0 < r < ∞. Then the best constant CS in ∞ ∞ 1 1 r p r p [S f (x)] ρ(x) dx ≤ CS [f (x)] v(x) dx , f ∈ M↓ , (55) 0
0
satisfies CS ≈ A1 + A2 + A3 + A4 + B, where A1 , A2 , A3 , and A4 are the best constants in ∞
−3
x r ζ (x) ∞ 1 r p ∞ q p r q p −3 ρ(x) k1 (x, y)k2 (y, ζ (x))w(y) dy h u(s) ds dx ≤ A1 hV,
0
∞
0
ζ −3 (x) −3
x r ζ (x) ∞ 1 r p ∞ q p r p ρ(x) k1 (x, y)w(y) dy k2 (ζ −3 (x), s)u(s) h ds dx ≤ A2 hV,
0
0
ζ −3 (x)
∞
∞
0
0
ρ(x)k1 (x, ζ −2 (x))
902
1 s
:=
r q
0
y ∞ 1 q r p ∞ p q r p w(y) k2 (y, s)u(s) h dy dx ≤ A4 hV 0
1 r
− p1 .
x
0
s
and the constant B has the form ⎧ ∞ 1 1 r ⎪ t p ρ T , sup ⎪ q ⎨ t>0 t L1V →Lkp (t,·)w(·) 1 B := ∞ ∞ ⎪ ⎪ ρ(x) ρ T ⎩ [ζ −1 (x),ζ 2 (x)] 0
where
0
y
ζ −2 (x)
0
M+
0
s
x y q ∞ q r p ∞ p q r p ρ(x) k1 (x, y)w(y) k2 (y, s)u(s) ds h dy dx ≤ A3 hV,
0
for h ∈
0
s
q L1V →L p 2 k1 (ζ (x),·)w(·)
p ≤ r, s
p
1 s dx ,
r < p,
Remark 5. (1) CS ≈ A1 + A2 + A3 + A4 + B for q = ∞, where A1 , A2 , A3 , and A4 are the best constants in ∞
−3
ζ (x) ∞ 1 r p ∞ p r p −3 r ρ(x)[ ess sup k1 (x, y)k2 (y, ζ (x))w(y)] h u(s) ds dx ≤ A1 hV, y∈(ζ −3 (x),x)
0
∞
0 −3
ζ (x) ∞ 1 r p ∞ p r p r −3 ρ(x)[ ess sup k1 (x, y)w(y)] k2 (ζ (x), s)u(s) h ds dx ≤ A2 hV, y∈(ζ −3 (x),x)
0
∞
0
∞ ρ(x)k1 (x, ζ
y∈(0,x)
−2
0
r
(x))
0
M+
0
s
y ∞ 1 r p ∞ p r p ρ(x) ess sup k1 (x, y)w(y) k2 (y, s)u(s) ds h dx ≤ A3 hV,
0
for h ∈
0
s
y ∞ 1 r p ∞ p r p ess sup w(y) k2 (y, s)u(s) h ds dx ≤ A4 hV
y∈(0,ζ −2 (x))
0
x
(2) For p = ∞ and r = ∞, we have 1 CS = S v r , p = ∞, Lρ
0
s
and the constant B has the form ⎧ ∞ 1 1 r ⎪ t p 1 ⎪ ρ T , ⎨ sup LV →L∞ k1 (t,·)w(·) t B := t>0 ∞ ∞ ⎪ [ζ −1 (x),ζ 2 (x)] ⎪ ρ(x) ρ T ⎩ 0
0
y
p ≤ r, 1 p s dx ,
s
L1V →L∞
k1 (ζ 2 (x),·)w(·)
1
t p CS ≈ sup R(t)T t>0
q
L1V →Lkp
r < p.
, r = ∞,
1 (t,·)w(·)
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[email protected] G. E. Shambilova Financial University Under the Government of the Russian Federation, Moscow, Russia E-mail address:
[email protected]
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