Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 1, Issue 1, Article 10, 2000
WEIGHTED MODULAR INEQUALITIES FOR HARDY-TYPE OPERATORS ON
MONOTONE FUNCTIONS
HANS P. HEINIG AND QINSHENG LAI
D EPARTMENT OF M ATHEMATICS AND S TATISTICS , M C M ASTER U NIVERSITY, H AMILTON , O NTARIO ,
L8S 4K1, C ANADA .
heinig@mcmail.cis.mcmaster.ca, jlai@comnetix.com
Received 3 November, 1999; accepted 31 January, 2000
Communicated by B. Opic
Rx
A BSTRACT. If (Kf )(x) = 0 k(x, y)f (y) dy, x > 0, is a Hardy-type operator defined on the
cone of monotone functions, then weight characterizations for which the modular inequality
Z ∞
Z ∞
Q−1
Q[θ(Kf )]w ≤ P −1
P [Cf ]v
0
0
holds, are given for a large class of modular functions P, Q. Specifically, these functions need
not both be N -functions, and the class includes the case where Q ◦ P −1 is concave. Our results
generalize those in [7, 24], where the case Q ◦ P −1 convex, with P, Q, N -function was studied.
Applications involving the Hardy averaging operator, its dual, the Hardy-Littlewood maximal
function, and the Hilbert transform are also given.
Key words and phrases: Hardy-type operators, modular inequalities, weights, N -functions, characterizations, Orlicz-Lorentz
spaces.
2000 Mathematics Subject Classification. 26D15, 42B25, 26A33, 46E30.
1. I NTRODUCTION
An integral operator K defined by
Z x
(Kf )(x) =
k(x, y)f (y) dy,
x > 0, f ≥ 0
0
is called a Hardy type operator, if the kernel k satisfies
(1.1)
(i)
(ii)
k(x, y) > 0, x > y > 0, k is increasing in x and decreasing in y.
k(x, y) ≤ D[k(x, z) + k(z, y)], 0 < y < z < x,
for some constant D > 0.
ISSN (electronic): 1443-5756
c 2000 Victoria University. All rights reserved.
The first author was supported in part by NSERC Grant A-4837. The second author’s research was supported by NSERC Grant PDF
181814.
010-99
2
H ANS P. H EINIG AND Q INSHENG L AI
k(x, y) = 1; k(x, y) = φ(x − y), φ increasing, φ(a + b) ≤ D[φ(a) + φ(b)] 0 < a, b < ∞;
and k(x, y) = ψ(y/x), ψ decreasing, ψ(ab) ≤ D[ψ(a) + ψ(b)] 0 < a, b < 1; are examples of
kernels satisfying (1.1) and hence define Hardy-type operators.
If k(x, y) has no monotonicity properties, satisfies (ii) and its reverse, then k is said to satisfy
the Oinarov condition ([22]) and we write k(x, y) ≈ k(x, z) + k(z, y), 0 < y < z < x.
In this paper we study Hardy-type operators (and its duals) defined on the cone of monotone functions. Specifically, weight functions θ, w, v are characterized for which the modular
inequality
Z ∞
Z ∞
−1
−1
(1.2)
Q
Q[θ(x)(Kf )(x)]w(x) dx ≤ P
P [Cf (x)]v(x) dx
0
0
is satisfied for a large class of modular functions P, Q, and f ≥ 0, monotone.
For example, if K = I, the identity operator and 0 ≤ f↓, then the weights are characterized
for which (1.2) holds with P , Q increasing and P weakly convex (cf. Theorem 3.1). For general
K, defined on 0 ≤ f ↓, weight characterizations are given for which (1.2) holds with P an N function, P , P̃ ∈ ∆2 and Q weakly convex (cf. Theorem 3.4). Specifically, Q ◦ P −1 may be
concave. These results together with the corresponding results where K is defined on the cone
of increasing functions are new. The case 0 < q < 1 < p for the general K, defined on 0 ≤ f↑,
was unknown until this paper.
If P (x) = xp , Q(x) = xq , 0 < p, q < ∞, θ(x) = 1, then our results reduce to weighted
Lebesgue space inequalities and in particular if k(x, y) = 1, to the weight characterizations
of Ariño-Muckenhoupt [1] (p = q > 1 w = v), Sawyer [21] (1 < p, q < ∞) and Stepanov
[23] (0 < q < p, p > 1). The general case where P and Q are N -functions, such that P and
its complementary function P̃ satisfy ∆2 with Q ◦ P −1 convex (more precisely P Q) was
studied by Sun [24] with k(x, y) the convolution kernel.
To explain the scope of our results we require some definitions and known facts.
A non-negative function P on R+ is called an N -function if it has the form
Z x
p(t) dt,
x > 0,
(1.3)
P (x) =
0
where p is non-decreasing, right continuous on (0, ∞), p(0+) = 0, p(∞) = ∞ and p(t) > 0 if
t > 0. Clearly
P (x)
x
lim
= lim
= 0.
x→0+
x→∞ P (x)
x
Given an N -function P , then its complementary function P̃ is defined by P̃ (y) = supx>0 {xy −
P (x)} and
(1.4)
t ≤ P −1 (t)P̃ −1 (t) ≤ 2t,
p(t/2)/2 ≤ P (t)/t ≤ p(t),
t>0
holds. It is easily seen that if P is an N -function so is P̃ , and the complement relation is
symmetric.
If (X, µ) is a σ-finite measure space, then a µ-measurable function f belongs to the Orliczspace LP (µ) if the Luxemburg norm
Z
|f (x)|
kf kP (µ) = inf λ > 0 :
P
dµ(x) ≤ 1
λ
X
is finite. The Orlicz norm in LP (µ) is defined by
Z
Z
0
P̃ (g)dµ ≤ 1 .
kf kP (µ) = sup f g dµ :
X
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
X
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W EIGHTED MODULAR INEQUALITIES FOR H ARDY- TYPE OPERATORS
3
We note that the Luxemburg and Orlicz norm are equivalent and
Z
(1.5)
kf kP (µ) ≤ 1 if and only if
P (f )dµ ≤ 1.
X
Given an N -function P , we always use the Luxemburg norm in LP (µ) and define that associate space LP̃ (µ) of LP (µ) consists of those µ-measurable g, for which the Orlicz norm
Z
kgkP̃ (µ) = sup f g dµ : kf kP (µ) ≤ 1
X
is finite.
A weight function u (u 6≡ 0, u 6≡ ∞) is a non-negative measurable and locally integrable
function on R+ , and if dµ(x) = u(x)dx, then we write P (µ) = P (u). The standard duality
principle in Orlicz spaces may be written as
R∞
g
fg
sup 0
= ,
g ≥ 0.
u P̃ (u)
0≤f kf kP (u)
For these and other facts see [13, 14, 20].
Definition 1.1.
a) An increasing function P : R+ → R+ is said to satisfy ∆2 , (P ∈ ∆2 ), if there is a
constant C > 1, such that P (2t) ≤ CP (t), t ≥ 0.
b) A strictly increasing function Q : R+ → R+ is weakly convex, (Q ∈ ∆2 ), if Q(0) = 0,
Q(∞) = ∞ and 2Q(t) ≤ Q(M t), t ≥ 0, for some constant M > 1.
c) ([16]) If P and Q are increasing, then we write P Q, if there is a constant A > 0,
such that
!
X
X
Q ◦ P −1 (aj ) ≤ Q ◦ P −1 A
aj
j
j
is satisfied for all non-negative sequences {aj }j∈Z .
A convex function Q satisfying Q(0) = 0, Q(∞) = ∞ is weakly convex (with M = 2).
However, the weakly convex function Q(t) = tα , t ≥ 0, 0 < α < 1, is not convex, and
Q(t) = ln(1 + t), t ≥ 0 is not weakly convex. Observe also that if Q ◦ P −1 is convex, then
P Q.
The main result of this paper (Theorem 3.4) characterizes the weights θ, w, v for which (1.2)
is satisfied for decreasing f ≥ 0 with P an N -function, P , P̃ ∈ ∆2 and Q weakly convex. This
characterization is expressed in terms of estimates involving covering sequences.
Definition 1.2. A strictly increasing positive sequence {xj }j∈Z is called a covering sequence if
M
the sequence is of the form {xj }∞
j=−∞ or of the form {xj }j=N , where M and/or N is finite. In
the latter case we define xN −1 = 0 and/or xM +1 = ∞.
R xj
In some
instances
covering
sequences
satisfy
v = 2k , k ∈ Z, where v is a weight function.
0
R
R
∞
∞
If 2N < 0 v < 2N +1 then in the case 2N < 0 v < 3 · 2N −1 we set xN = ∞ and the covering
−1
sequence is {xj }N
set xN +1 = ∞ and the covering sequence is
j=−∞ . In the remaining case
R xwe
j+1
N
k−1
≤ xj v ≤ 3 · 2k−1 for 0 < xj < ∞.
{xj }j=−∞ . Under these conventions 2
The manuscript is divided into four sections. The next section contains the weight characterization of a modular Hardy-type inequality for Young’s and weakly convex functions by
Qinsheng Lai [19]. As a consequence a corresponding result for the dual operator follows. In
addition, modular Hardy and conjugate Hardy inequalities (Lemma 2.3) are given. Section 3,
the main results, contain the weighted modular inequalities for the identity operator (Theorem
3.1) and Hardy-type operator (Theorem 3.4) defined on decreasing functions. Some special
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
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4
H ANS P. H EINIG AND Q INSHENG L AI
cases given there are needed in Section 4 and seem to be new even in the Lebesgue space case.
In the last section results for the Hardy operator on increasing functions are given. Moreover, the
boundedness of the Hardy-Littlewood maximal function and the Hilbert transform in weighted
Orlicz-Lorentz spaces are characterized.
The notation is standard, R+ and R denote the non-negative real and real numbers respectively, while Z denotes the set of integers. The symbol χE stands for the characteristic function
of a set E. All functions are Rassumed measurableRand u, v, w, θ denote
R ∞ weight functions. If u
x
∗
is a weight function u(E) = E u(x) dx, U (x) = 0 u and U (x) = x u, (x > 0). Instead of
non-increasing, non-decreasing we shall say decreasing and increasing respectively, otherwise
we shall prefix it by “strictly”. If f ≥ 0 is increasing (decreasing) we shall write 0 ≤ f ↑
(0 ≤ f↓) and similarly for sequences. Expressions of the form A ≈ B are interpreted to mean
that A/B are bounded above and below by positive constants. Constants are (with the exception
of those of Definition 1.1) denoted by B and C and they may have different values at different
places. Inequalities, such as (1.2), are interpreted to mean that if the right side is finite, so is the
left side and the inequality holds.
Other notations and concepts are introduced when needed.
2. P RELIMINARY R ESULTS
In order to prove weighted modular inequalities for Hardy type operators defined on the
cone of monotone functions, a number of results are required. The first result (Theorem 2.1)
by Q. Lai [19] is a weight characterization of the Hardy-type operator for which a weighted
modular inequality is satisfied. This theorem extends corresponding work of [3, 4, 18, 22, 24]
to Young’s functions P and weakly convex functions Q without the assumption that Q ◦ P −1
(or more precisely P Q) is convex.
Theorem 2.1. ([19, Thm. 1]) Suppose K is a Hardy-type operator, P a Young’s function and
Q weakly convex. Let θ, w, ρ and v be weight functions, then the modular inequality
Z ∞
Z ∞
−1
−1
Q
Q[θ(x)Kf (x)]w(x) dx ≤ P
P [Cρ(x)f (x)]v(x) dx
0
0
is satisfied for all f ≥ 0, if and only if there are constants B > 0, such that,
"
#
!
X Z xj+1
k(x
,
·)χ
θ(x) j
(xj−1 ,xj )
Q−1
Q
w(x) dx ≤ P −1
B
εj vρ
xj
P̃ (εj v)
j
!
X
1/εj
j
and
Q−1
XZ
j
xj+1
xj
"
#
!
χ
θ(x)k(x, xj ) (xj−1 ,xj ) Q
w(x) dx ≤ P −1
εj vρ B
P̃ (εj v)
!
X
1/εj
j
hold for all positive sequences {εj }j∈Z and all covering sequences {xj }j∈Z .
A corresponding result for the conjugate Hardy-type operator
Z ∞
∗
(K h)(x) =
k(y, x)h(y) dy,
x > 0, h ≥ 0,
x
¯ (y) = h(1/y)/y 2 ,
where k satisfies (1.1), also holds. In fact, writing k̃(x, y) = k( y1 , x1 ) and h̄
then a change of variables shows that
Z x
∗
¯ (y) dy ≡ (K h̄
¯ )(x)
(K h)(1/x) =
k̃(x, y)h̄
0
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W EIGHTED MODULAR INEQUALITIES FOR H ARDY- TYPE OPERATORS
5
is a Hardy-type operator since k̃(x, y) satisfies the same conditions as k(x, y). Writing ḡ(x) =
g(1/x) and ḡ¯(x) = g(1/x)/x2 it follows that
Z ∞
Z ∞
−1
∗
−1
¯
¯
Q
Q[θ(x)K h(x)]w(x) dx = Q
Q[θ̄(x)K h̄(x)]w̄(x) dx
0
and
P
−1
0
∞
Z
¯ (x)]v̄¯(x) dx
P [Cx ρ̄(x)h̄
2
=P
−1
0
Z
∞
P [Cρ(x)h(x)]v(x) dx .
0
Also
1
k(·, )χ(1/xj ,1/xj−1 ) xj
=
εj ρv
¯
P̃ (εj v¯)
k̃(x , ·)χ
j
(xj−1 ,xj ) εj ρ̄v̄
P̃ (εj v)
and
χ(xj−1 ,xj ) χ(1/xj ,1/xj−1 ) =
.
εj ρ̄v̄ ε
ρv
¯
j
P̃ (εj v¯)
P̃ (εj v)
Therefore, if 1/xj = y−k , k ∈ Z, then {yj }j∈Z is also a covering sequence, whenever {xj }j∈Z
is. Thus, the following characterization follows from Theorem 2.1.
Proposition 2.2. If K ∗ is the conjugate Hardy-type operator, P a Young’s function and Q
weakly convex, then
Z ∞
Z ∞
−1
∗
−1
Q
Q[θ(x)(K h)(x)]w(x) dx ≤ P
P [Cρ(x)h(x)]v(x) dx
0
0
is satisfied for all h ≥ 0, if and only if there is a constant B > 0, such that
"
#
!
X Z yj
k(·,
y
θ(x) j )χ(yj ,yj+1 ) −1
Q
Q
w(x) dx ≤ P −1
B
εj ρv
yj−1
P̃ (εj v)
j
!
X
1/εj
j
and
Q−1
j
"
#
!
χ
θ(x)k(yj , x) (yj ,yj+1 ) Q
w(x) dx ≤ P −1
εj ρv B
yj−1
P̃ (εj v)
XZ
yj
!
X
1/εj
j
holds for all positive sequences {εj }j∈Z and all covering sequences {yj }j∈Z .
Note that if Q is an N -function, then Q is convex and in particular, weakly convex. Hence
Theorem 2.1 and Proposition 2.2 hold in this case.
The following result is required in the next section.
Rx
R∞
Lemma 2.3. Suppose P and P̃ are N -functions, V (x) = 0 v, V ∗ (x) = x v and v is a weight
function.
(i) If V (∞) = ∞, then there exists a constant C > 0, such that
Z x Z ∞
Z ∞ 1
(2.1)
P
f v v(x) dx ≤
P [Cf (x)]v(x) dx,
V (x) 0
0
0
(2.2)
is satisfied for all f ≥ 0 if and only if P̃ ∈ ∆2 , and
Z ∞ Z ∞
Z ∞
fv
P
v(x) dx ≤
P [Cf (x)]v(x) dx,
V
0
x
0
is satisfied for all f ≥ 0 if and only if P ∈ ∆2 .
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H ANS P. H EINIG AND Q INSHENG L AI
(ii) If V ∗ (0) = ∞, then
Z ∞ Z ∞
Z ∞ 1
(2.3)
P
f v v(x) dx ≤
P [Cf (x)]v(x) dx,
V ∗ (x) x
0
0
(2.4)
is satisfied for all f ≥ 0 if and only if P̃ ∈ ∆2 , and
Z ∞
Z ∞ Z x
fv
P
v(x) dx ≤
P [Cf (x)]v(x) dx,
∗
0
0 V
0
is satisfied for all f ≥ 0 if and only if P ∈ ∆2 .
The conditions V (∞) = ∞ and V ∗ (0) = ∞ are only required in the necessity part of the
proof.
Proof. First observe that if f¯(x) = f (1/x), v̄¯(x) = v(1/x)/x2 then via obvious changes of
¯
variables, (2.3) reduces to (2.1) with f replaced by f¯ and
R t v by v̄ . A similar change of variable
∗
shows that (2.4) reduces to (2.2). Note that V (1/t) = 0 v̄¯. Therefore it suffices to prove only
part (i) of Lemma 2.3.
Next we observe that (2.1) is equivalent to
Z ∞
Z ∞ Z ∞
fv
(2.5)
P̃
v(x) dx ≤
P̃ [Cf (x)]v(x) dx.
V
0
x
0
To see this, recall that by [4, Prop. 2.5] (see also [11]) that (2.1) holds if and only if for every
ε > 0,
Z x
1
f v.
kT f kP (εv) ≤ Ckf kP (εv) where T f (x) =
V (x) 0
But by the standard duality principle in Orlicz spaces this is equivalent to
∗ Z ∞
g
T g g(t)
∗
≤C , where T g(x) = v(x)
dt
εv εv P̃ (εv)
V (t)
x
P̃ (εv)
is the conjugate operator of T . By homogeneity of the norm and again applying [4, Prop. 2.5]
it follows that this inequality is equivalent to
Z ∞ Z ∞ 1
Cg(x)
∗
P̃
(T g)(x) v(x) dx ≤
P̃
v(x) dx,
v(x)
v(x)
0
0
which is (2.5) with g = f v. Hence we only need to show that (2.1) is satisfied, if and only if
P̃ ∈ ∆2 .
Let P̃ ∈ ∆2 and define f + (x) = f (x) if x ≥ 0 and zero otherwise, v(|x|)dx = dµ(x), then
Z x
Z
1
1
+
f v ≤ (Mµ f )(x) := sup
f + dµ, I ∈ R.
V (x) 0
µ(I)
x∈I
I
Clearly Mµ is sublinear and of type (∞, ∞), and weak type (1, 1), with respect to dµ. Now the
argument of [5, p. 149-150] shows that P̃ ∈ ∆2 is sufficient for
Z ∞
Z ∞
+
P Mµ f (x) dµ(x) ≤
P (Cf (x)) dµ(x),
0
0
from which (2.1) follows.
To prove that (2.1) implies P̃ ∈ ∆2 , it suffices (see [5, Prop. 3]) to prove that there exists a
δ > 0, such that p(δx) ≤ 1/2 p(x), where p(x) = P 0 (x) with p(0) = 0.
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W EIGHTED MODULAR INEQUALITIES FOR H ARDY- TYPE OPERATORS
7
By Theorem 2.1, with Q = P , k(x, y) = 1, θ = 1/V , ρ = 1/v, f replaced by f v, xj = r > 0,
xj−1 = 0 and xj+1 = ∞, (2.1) implies
Z ∞ 1 χ(0,r) P
v(t) dt ≤ 1/ε
B V (t)
ε P̃ (εv)
r
for all ε > 0 and r > 0. But by the definition of the Luxemburg norm and (1.4) with t =
1/(ε V (r))
Z r χ
1
1
(0,r) = inf λ > 0 :
P̃
εv(t) dt ≤ 1
ε P̃ (εv) ε
λ
0
1
=
εP̃ −1 εV1(r)
1
V (r) −1
≥
P
.
2
εV (r)
Hence (2.1) implies
∞
Z
r
If x =
V (r)
2BV (t)
P −1
1
εV (r)
Z
V (r)
P
P −1
2BV (t)
1
εV (r)
v(t) dt ≤ 1/ε.
, this inequality is
P −1 ( εV1(r) )/(2B)
0
2B
P (x)
.
dx ≤
2
1
x
−1
εV (r)P
εV (r)
Writing
y=P
−1
1
εV (r)
one obtains
Z
y/(2B)
(2.6)
0
Then it follows from (1.4) that
ing by parts
P (x)
dx ≤ 2B P (y)/y,
x2
R y/(4B)
0
p(x)
x
Z
y > 0.
dx ≤ 4Bp(y). Now let 0 < η < 1, then on integrat-
y/(4B)
p(x)
dx
x
0
Z y/(4B)
y ≥
log
dp(t)
4Bt
0
Z ηy/(4B)
y ≥
log
dp(t)
4Bt
0
≥ log(1/η)p(ηy/(4B)).
4Bp(y) ≥
Choose η so that log(1/η) ≥ 8B and δ = η/(4B), then p(δy) ≤
implies by [5, Prop. 3] that P̃ ∈ ∆2 .
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
1
2
p(y). This, as was noted,
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H ANS P. H EINIG AND Q INSHENG L AI
3. M AIN R ESULTS
Our first result concerns the identity operator defined on monotone functions.
Theorem 3.1. Suppose P and Q are increasing and P is weakly convex. Then
Z ∞
Z ∞
−1
−1
(3.1)
Q
Q[θ(x)f (x)]w(x) dx ≤ P
P [Cf (x)]v(x) dx
0
0
holds for all 0 ≤ f↓, if and only if there is a constant B > 0, such that,
!
!
Z xj+1
i
X Z xj+1 h εj
X
−1
−1
Q
(3.2)
Q
θ(x) w(x) dx ≤ P
P (εj )
v(x) dx
B
xj
xj
j
j
is satisfied for all non-negative
decreasing sequences {εj }j∈Z and the covering sequence
Rx
{xj }j∈Z such that 0 j v = 2k , k ∈ Z.
Similarly, (3.1) holds for all 0 ≤ f ↑, if and only if (3.2) is satisfied forR all non-negative
∞
increasing sequences {εj }j∈Z and the covering sequence {xj }j∈Z satisfying xj = 2−k .
Proof. We only prove the first part of the theorem since the argument for the second part is
similar.
P
Let {εj }j∈Z be any decreasing sequence, then f (x) = j εj χ(xj ,xj+1 ) (x) is decreasing and
substituting f into (3.1), (3.2) follows with
R xj B = C.
Conversely if (3.2) holds then, since 0 v = 2k and 2P (x) ≤ P (M x), M > 1
!
Z ∞
X Z xj+1
Q−1
Q[θ(x)f (x)]w(x) dx ≤ Q−1
Q[θ(x)f (xj )]w(x) dx
0
xj
j
≤ P −1
X
Z
P (Bf (xj ))
X
Z
=P
v
xj−1
XZ
!
xj
P (M Bf (x))v(x) dx
xj−1
j
−1
!
xj
2P (Bf (xj ))
j
≤ P −1
v
xj
j
= P −1
!
xj+1
∞
Z
P [M Bf (x)]v(x) dx .
0
This proves Theorem 3.1.
If Q ◦ P −1 is convex, Theorem 3.1 has the following form:
Corollary 3.2. Suppose P and Q are increasing, P is weakly convex and P Q. Then (3.1)
holds for all 0 ≤ f↓, if and only if for all ε > 0 and r > 0, there is a constant B > 0, such that,
(3.3)
−1
Z
Q
0
r
θ(x) −1
P
Q
B
ε
Rr
0
v
w(x) dx
≤ P −1 (ε).
Similarly, (3.1) is satisfied for all 0 ≤ f↑, if and only if
Z ∞ ε
θ(x) −1
−1
R∞
(3.4)
Q
Q
P
w(x) dx ≤ P −1 (ε)
B
v
r
r
is satisfied.
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
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W EIGHTED MODULAR INEQUALITIES FOR H ARDY- TYPE OPERATORS
9
Proof. By Theorem 3.1 is suffices to show that (3.2) with increasing (decreasing) sequence
{εj }j∈Z is equivalent to (3.3) (respectively (3.4)).
Rr
First fix j = k0 ∈ Z and let xj0 = r > 0. Then for fixed ε > 0 define εm = P −1 (ε/ 0 v), if
m < k0 and zero otherwise. Clearly {εm }m∈Z is decreasing and by (3.2)
Z r θ(x) −1
ε
−1
Rr
Q
Q
P
w(x) dx
B
v
0
0
!
X Z xj+1 θ(x)
ε
−1
−1
Rr
=Q
Q
P
w(x) dx
B
v
x
j
0
j<k0
!
X Z xj+1 θ(x)εj = Q−1
Q
w(x) dx
B
x
j
j<k0
!
Z xj+1
X ε
≤ P −1
P P −1 R r
v(x) dx = P −1 (ε).
v
xj
0
j<k
0
To prove the converse, recall that since P is weakly convex, there is an M ≥ 1, such that
2P (x) ≤ P (M x). Hence with y = P (M x)
P −1 (y) ≤ M P −1 (y/2),
y > 0.
R xj
If {xj }j∈Z is a covering sequence satisfying 0 v = 2k and ηj > 0, to be determined later, then
by (3.5) and (3.3) with ε = ηj and r = xj+1 ,
!#
Z xj+1 "
θ(x) −1
ηj
R xj+1
Q
P
w(x) dx
BM
v
xj
xj
Z xj+1 θ(x) −1
ηj
R xj+1
≤
Q
P
w(x) dx ≤ Q ◦ P −1 (ηj ).
B
v
0
0
(3.5)
Since P Q, summing over k ∈ Z yields
!#
"
X Z xj+1
X
θ(x) −1
ηj
R xj+1
Q
P
w(x) dx ≤
Q ◦ P −1 (ηj ) ≤ Q ◦ P −1
BM
v
xj
xj
j
j
!
X
Aηj
,
j
where A is the constant arising from conditionRP Q (cf. Defn. 1.1c) ). Now choose ηj so that
x
{ηj /2k } is decreasing, hence εj = P −1 (Aηj / xjj+1 v) defines a decreasing sequence. Therefore
!
Z xj+1 !
X Z xj+1 θ(x)
X
P (εj )
−1
−1
−1
Q
Q
P
w(x) dx ≤ P
P (εj )
v ,
MB
A
xj
xj
j
j
and applying (3.5) α-times so that 2α /A ≥ 1, then
2P (εj )
1 −1 2
1
−1
P
≥
P
P (εj ) ≥ · · · ≥ α εj
2A
M
A
M
and the result follows.
If 0 ≤ f↑, fix k0 ∈ Z such that xj0 = r > 0 and define
Z ∞
−1
εm = P (ε/
v) if m ≥ k0 and zero otherwise.
r
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
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10
H ANS P. H EINIG
AND
Q INSHENG L AI
Then {εm }m∈Z is increasing and the previous argument
shows that (3.4) follows from (3.2).
R∞
Also if {xj }j∈Z is a covering sequence such that xj v = 2−k and ηj > 0, then by (3.5) and
Rx
(3.4), since 2 xjj+1 v = 2−k ,
Z
xj+1
xj
"
θ(x) −1
Q
P
MB
!#
ηj
R xj+1
xj
w(x) dx
v
Z
∞
≤
xj
Summing over k, and choosing ηj so that
"
θ(x) −1
P
Q
B
ηj
{ 2−k
}
Aη
R xj+1j
εj = P −1
xj
!#
ηj
R∞
xj
w(x) dx ≤ Q ◦ P −1 (ηj ).
v
is an increasing sequence, then with
!
k ∈ Z,
,
v
defines an increasing sequence, where A is the constant arising from the condition P Q. The
inequality (3.2) now follows as before.
Corollary 3.2 was proved by J. Q. Sun [24, Lemma 3.1] in the case when P and Q are
N -functions (hence convex). If P (x) = xp , Q(x) = xq , 0 < p ≤ q < ∞, one obtains
(with θ(x) = 1) the well known weight conditions ([21, 23]) which characterize (3.1). If
0 < q < p < ∞ we have:
Corollary 3.3. Let 0 < q < p < ∞ and 1/r = 1/q − 1/p, then the following are equivalent:
Z ∞
1/q
Z ∞
1/p
q
p
(3.6)
f w
≤C
f v
0
0
is satisfied for all 0 ≤ f↓.
Z
∞
(3.7)
[W 1/p V −1/p ]r w ≡ B0r < ∞,
0
X
r
w(Ej )1/q (Ej )−1/p ≡ B1r < ∞,
(3.8)
j
where w(Ej ) =
R xj+1
xj
w, v(Ej ) =
xj
v and the covering sequence {xj } satisfies V (xj ) = 2k .
#1/q
"
(3.9)
R xj+1
X
εqj w(Ej )
j
#1/p
"
≤B
X
εpj v(Ej )
j
holds for all decreasing sequences {εj }j∈Z and covering sequences {xj } with V (xj ) = 2k .
Rt
Rt
(Recall: W (t) = 0 w, V (t) = 0 v.)
If 0 ≤ f↑ the above statement holds with W and V replaced by W ∗ and V ∗ , respectively, the
covering sequence {xj } satisfies V ∗ (xj ) = 2−k and {εj } is taken to be increasing.
Proof. We only prove the corollary in the case 0 ≤ f ↓ since the case 0 ≤ f ↑ is proved, with
obvious modifications, in the same way.
The equivalence of (3.6) and (3.9) follows at once from Theorem 3.1 with Q(x) = xq ,
P (x) = xp , θ(x) = 1. Since the equivalence of (3.6) and (3.7) was proved in [21, 23], it
remains to prove (3.7) ⇒ (3.8) ⇒ (3.9).
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
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W EIGHTED MODULAR INEQUALITIES FOR H ARDY- TYPE OPERATORS
Since r/q = r/p + 1 and w(Ej ) =
r/q
w(Ej )
11
R xj+1
xj
r
=
q
r
≤
q
w, it follows that
!
Z
Z
xj+1
t
w w(t) dt
xj
Z
xj
xj+1
W (t)r/p w(t) dt
xj
on integrating. Since v(Ej ) = 2k = V (xj ) it follows therefore that
Z
X
r X xj+1 −rk/p
1/q
−1/p r
[w(Ej ) v(Ej )
] ≤
2
W (t)r/p v(t) dt
q
xj
j
j
Z
r2r/p X xj+1
V (t)−r/p W (t)r/p w(t) dt
≤
q
xj
j
=
r2r/p r
B0 .
q
Hence (3.7) ⇒ (3.8).
P
Since the dual of the sequence space `p/q is `r/q , it follows that j [w(Ej )v(Ej )−q/p ]r/q =
B1r < ∞ implies
!q/p
X
X
p/q
ηj
ηj w(Ej )v(Ej )−q/p ≤ B1q
j
j
1/q
for any positive sequence {ηj } in `p/q . Now choose {ηj } so that ηj = εj v(Ej )1/p = εj 2k/p
with {εj }j∈Z decreasing. Thus (3.8) ⇒ (3.9), which completes the proof.
Note that if 0 < q < p = 1, then with q =
α
,
α+1
α > 0, r = α and one shows that
B1
≤ B0 ≤ 2(1 + α)1/α B1 .
1/α
2(1 + α)
Here of course
∞
Z
W α V −α w
B0 =
0
!1/α
1/α
and
B1 =
X
w(Ej )α+1 v(Ej )−α
.
j
We now give the main result of this section.
Theorem 3.4. Suppose P is an N -function, P and P̃ satisfy the ∆2 condition and Q weakly
convex. If K is a Hardy-type operator defined on the cone of decreasing functions, then
Z ∞
Z ∞
−1
−1
(3.10)
Q
Q[θ(x)Kf (x)]w(x) dx ≤ P
P [Cf (x)]v(x) dx
0
0
is satisfied, if and only if there is a constant B > 0, such that for all positive sequences {εj }j∈Z
and all covering sequences {xj }j∈Z with i(x) = x
"
#
!
!
X Z xj+1
X 1
k(x
θ(x) j , ·)χ(xj−1 ,xj ) i −1
−1
(3.11) Q
Q
w(x) dx ≤ P
B
ε
V
εj
j
x
j
P̃ (εj v)
j
j
(3.12) Q−1
XZ
j
xj+1
xj
"
#
!
χ
i
θ(x)k(x, xj ) (xj−1 ,xj ) Q
w(x) dx ≤ P −1
B
εj V
P̃ (εj v)
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
X 1
εj
j
!
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12
(3.13)
H ANS P. H EINIG
Q−1
XZ
j
xj+1
xj
AND
Q INSHENG L AI
"
#
!
(K1)χ
θ(x) (xj−1 ,xj )
Q
w(x) dx ≤ P −1
B εj V
P̃ (εj v)
X 1
εj
j
!
are satisfied, and Rfor all positive decreasing sequences {εj }j∈Z and the covering sequence
x
{xj }j∈Z satisfying 0 j v = 2k
!
Z xj+1 !
X Z xj+1 εj θ(x)(K1)(x) X
−1
−1
(3.14)
Q
Q
w(x) dx ≤ P
P (εj )
v
B
xj
xj
j
j
is satisfied.
(Sufficiency.) The idea comes from [23]. We may assume that f has the form f (x) =
RProof.
∞
h,
h
> 0, for once the result has been proved Rfor such f , a limiting argument (see e.g. [24])
x
x
gives the general case. Clearly since (K1)(x) = 0 k(x, y) dy
Z x
Z ∞
(Kf )(x) =
k(x, y)
h(t) dtdy
0
y
Z
x
= (K1)(x)f (x) +
Z
h(t)
0
But since
Z x
1
1
−
=
V (y)−2 v(y) dy
V (t) V (x)
t
t
k(x, y) dy
dt.
0
Z
x
Z
h(s)V (s) ds ≤
and
0
x
f (t)v(t) dt,
0
it follows again on interchanging the order of integration that
Z x
Z t
h(t)
k(x, y) dydt
0
0
Z xZ t
Z x
1
−2
=
k(x, y)h(t)V (t)
+
V (s) v(s) ds dydt
V (x)
0
0
t
Z x
Z x
1
=
k(x, y)
h(t)V (t) dtdy
V (x) 0
y
Z t
Z x
Z s
−2
+
V (s) v(s)
h(t)V (t)
k(x, y) dy dtds
0
0
0
Z x
Z x
1
≤
k(x, y)
f (t)v(t) dtdy
V (x) 0
0
Z s
Z s
Z x
−2
+
V (s) v(s)
k(x, y) dy
f (t)v(t) dtds
0
0
0
Z x
1
≤ (K1)(x)
f (t)v(t) dt + I(x) (by definition of I(x)) respectively.
V (x) 0
Now since k(x, y) ≤ D[k(x, s) + k(s, y)], y < s < x,
Z x
Z s
−2
I(x) ≤ D
k(x, s)V (s) v(s)s
f (t)v(t) dtds
0
0
Z s
Z x
−2
+
V (s) v(s)
f (t)v(t) dt (K1)(s) ds
0
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
0
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W EIGHTED MODULAR INEQUALITIES FOR H ARDY- TYPE OPERATORS
13
and writing
s
Z
−2
F (s) = V (s) v(s)s
f v,
0
one obtains
Z x
1
f (t)v(t) dt
(Kf )(x) ≤ (K1)(x)f (x) + (K1)(x)
V (x) 0
Z x
Z x
(K1)(s)
+D
k(x, s)F (s) ds + D
F (s) ds
s
0
0
≡ (I1 + I2 + I3 + I4 )(x),
respectively. Now
θ(x)(Kf )(x) ≤ θ(x)
4
X
Ii (x) ≤ 4θ(x) max Is (x) = 4θ(x)Is(x) (x),
s=1,2,3,4
i=1
where s(x) ∈ {1, 2, 3, 4}, and since Q is increasing and satisfies 2Q(x) ≤ Q(M x), M > 1,
we have
Q[θ(x)(Kf )(x)] ≤ Q[4θ(x)Is(x) (x)]
≤
4
X
Q[4θ(x)Is (x)]
s=1
≤
4
X
1
i=1
4
Q[4M 2 θ(x)Is (x)].
Integration yields
∞
Z
Q[θ(x)(Kf )(x)]w(x) dx ≤
0
Z
4
X
1
s=1
4
∞
Q[4M 2 θ(x)Is (x)]w(x) dx
0
and therefore it suffices to prove that
Z
Z ∞
2
−1
(3.15)
Q[4M θ(x)Is (x)]w(x) dx ≤ Q ◦ P
0
∞
P [Cf (x)]v(x) dx
0
s = 1, 2, 3, 4 is satisfied.
Since I1 (x) = (K1)(x)f (x), then by Theorem 3.1 with θ replaced by θ(x)(K1)(x), (3.15)
holds if and only if (3.14) is satisfied.
Rx
Rx
1
1
Next, since 0 ≤ f ↓ so is V (x)
f
v,
and
since
I
(x)
=
(K1)(x)
f v, Theorem 3.1
2
V (x) 0
0
shows (with θ replaced by θ(x)(K1)(x)) that (3.14) is equivalent to
Z ∞ Z ∞
Z x 1
2
−1
Q[4M θ(x)I2 (x)]w(x) dx ≤ Q ◦ P
P C
f v v(x) dx
V (x) 0
0
0
Z ∞
−1
≤Q◦P
P [Cf (x)]v(x) dx .
0
Here the last inequality follows from (2.1) of Lemma 2.3.
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
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14
H ANS P. H EINIG
AND
Q INSHENG L AI
Next, I3 (x) = D(KF )(x), so that by Theorem 2.1 with ρ(x) = V (x)/(xv(x))
Z
∞
2
Q[4M θ(x)I3 (x)]w(x) ≤ Q ◦ P
−1
Z
∞
P C
1
V (x)
x
Z
f v v(x) dx
Z ∞
−1
≤Q◦P
P [Cf (x)]v(x) dx .
0
0
0
0
Here the first inequality holds if (3.11) and (3.12) are satisfied and the second follows from (2.1)
of Lemma 2.3.
Rx
Since I4 (x) = D 0 (K1)(s) F (s)
ds we apply Theorem 2.1 with k(x, y) = 1 and ρ(x) =
s
V (x)/(v(x)(K1)(x)), so that
Z
∞
2
Q[4M θ(x)I4 (x)]w(x) dx ≤ Q ◦ P
−1
Z
∞
P C
1
V (x)
x
Z
f v v(x) dx
Z ∞
−1
≤Q◦P
P [Cf (x)]v(x) dx
0
0
0
0
is satisfied if and only if (3.13) holds. The last inequality follows of course again from (2.1) of
Lemma 2.3.
(Necessity.) Since 0 ≤ f ↓, (Kf )(x) ≥ (K1)(x)f (x) so that (3.10) implies (3.1) with
θ replaced by θ(x)(K1)(x). Now Theorem 3.1 applies if P is an N -function and Q weakly
convex and so (3.14) follows.
To prove that (3.10) implies (3.11) observe first that for fixed k,
ik(xj , ·)χ(xj−1 ,xj ) εj V
P̃ (εj v)
(3.16)
is bounded. If this is not the case, then there is a sequence {fn } of non-negative functions
satisfying kCfn kP (εj v) ≤ 1, with C the constant of (3.10), and a sequence {αn } with αn → ∞,
n → ∞, such that (by definition of Orlicz norm) for each n
xj
Z
xk(xj , x)fn (x)v(x)
dx
V (x)
xj−1
Z x Z xj
k(xj , x)fn (x)v(x)
≤C
dy dx
V (x)
0
0
Z xj Z xj
k(xj , x)fn (x)v(x)
=C
dxdy
V (x)
0
y
Z xj
≤C
k(xj , y)Fn (y) dy,
αn < C
0
since k(xj , ·) is decreasing. Here Fn (y) =
Z
R∞
y
fn (x)v(x)
V (x)
dx. But since
∞
P [Cfn (x)]εj v(x) dx ≤ 1,
0
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
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W EIGHTED MODULAR INEQUALITIES FOR H ARDY- TYPE OPERATORS
15
(2.2) of Lemma 2.3 and (3.10) show that
P
−1
1
εj
≥P
−1
Z
∞
P [Cfn (x)]v(x) dx
Z ∞ C
−1
≥P
P
Fn (x) v(x) dx
C1
0
Z ∞ θ(x)
−1
≥Q
Q
(KFn )(x) w(x) dx
C1
0
!
Z xj+1 Z
θ(x) xj
−1
≥Q
Q
k(xj , y)Fn (y) dy w(x) dx
C1 0
xj
!
Z xj+1 θ(x)α
n
> Q−1
Q
w(x) dx ,
C C1
xj
0
where C1 is the constant of (2.2). But this is a contradiction since αn → ∞. Hence (3.16) is
bounded.
Now suppose (3.11) fails to be satisfied. Then for any B > 0 there exists a covering sequence
{xj }j∈Z and a positive sequence {εj }j∈Z such that
P −1
X 1
εj
j
!
< Q−1
XZ
xj+1
xj
j
"
θ(x)
Q
2BC1
#
!
k(xj , ·)χ(xj−1 ,xj ) i w(x) dx
εj V
P̃ (εj v)
where C1 is taken to be the constant of (2.2). Now for k ∈ Z, choose fj ≥ 0, such that
suppfj ⊂ (xj−1 , xj ) with
∞
Z
P [BC1 fj (x)]εj v(x) dx ≤ 1
(3.17)
0
and
Z xj
1 xk(xj , x)fj (x)v(x)
k(xj , ·)χ(xj−1 ,xj ) i ≤
dx
2BC1
εj V
V (x)
xj−1
P̃ (εj v)
Z xj
≤
k(xj , y)Fj (y) dy,
0
where
Z
Fj (y) =
y
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
∞
fj (x)v(x)
dx.
V (x)
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16
H ANS P. H EINIG
Let f (x) =
assumption
P
−1
P
fj (x) and F (x) =
j
∞
Z
0
R∞
x
AND
f (t)v(t)
V (t)
Q INSHENG L AI
dt, then by (2.2) of Lemma 2.3, (3.17) and our
P [BF (x)]v(x) dx
Z ∞
−1
≤P
P [BC1 f (x)]v(x) dx
0
XZ
= P −1
P [BC1 fj (x)]v(x) dx
xj−1
j
!
X 1
εj
j
X Z xj+1
≤ P −1
#
!
k(x
θ(x) j , ·)χ(xj−1 ,xj ) i w(x) dx
Q
2BC
ε
V
1
j
x
j
P̃ (εj v)
j
!
Z xj
X Z xj+1 Q θ(x)
k(xj , y)Fj (y) dy w(x) dx
< Q−1
≤ Q−1
Z
"
xj
j
−1
!
xj
0
∞
≤Q
Q [θ(x)(KF )(x)] w(x) dx
Z ∞
−1
≤P
P [CF (x)]v(x) dx ,
0
0
where the last inequality is (3.10). But this is impossible for B > C and hence (3.11) must be
satisfied.
To show that (3.12) and (3.13) are satisfied one proceeds as before. First one shows that both
χ(xj−1 ,xj ) i εj V P̃ (εj v)
(K1)χ(xj−1 ,xj ) εj V
P̃ (εj v)
and
are bounded for fixed k. Then with f (x) and F (x) defined as above one has
Z
xj
xj−1
Z
x fj (x)v(x)
dx ≤
V (x)
xj
Z
0
∞
y
fj (x)v(x)
dxdy
V (x)
and for (3.13)
Z
xj
xj−1
(K1)(x)fj (x)v(x)
dx ≤
V (x)
Z
Z
x
k(x, y) dy
0
Z
≤
0
xj
Z
k(xj , y)
0
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
xj
y
∞
fj (x)v(x)
dx
V (x)
fj (x)v(x)
dxdy.
V (x)
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W EIGHTED MODULAR INEQUALITIES FOR H ARDY- TYPE OPERATORS
17
Inequality (3.12) is then obtained as (3.11) was shown to hold. To prove (3.13) assume to the
contrary that (3.13) fails. Then for any B > 0, we have
Z ∞
−1
P
P [BF (x)]v(x) dx
0
Z ∞
−1
≤P
P [BC1 f (x)]v(x) dx
0
!
X 1
≤ P −1
εj
j
#
!
"
X Z xj+1
(K1)χ
θ(x)
(xj−1 ,xj )
< Q−1
Q
w(x) dx
2BC
ε
V
1
j
x
j
P̃
(ε
v)
j
j
!
Z
Z
Z
xj
∞
X xj+1
f
(t)v(t)
j
≤ Q−1
Q θ(x)
k(x, y)
dtdy w(x) dx
V (t)
xj
0
y
j
!
Z xj
X Z xj+1 ≤ Q−1
Q θ(x)
k(x, y)F (y) dy w(x) dx
xj
j
−1
Z
0
∞
≤Q
Q [θ(x)(KF )(x)] w(x) dx
Z ∞
−1
≤P
P [CF (x)]v(x) dx
0
0
from which the contradiction follows for B > C. This proves Theorem 3.4.
If k(x, y) = 1, θ(x) = xa , −1 ≤ a < ∞, the conditions (3.11), (3.12), (3.13) coincide and
since (K1)(x) = x we get the following corollary.
Corollary 3.5. Let P and Q be as in Theorem 3.4 and a ≥ −1. Then
Z ∞ Z x Z ∞
−1
−1
a
f w(x) dx ≤ P
P [Cf (x)]v(x) dx
Q
Q x
0
0
0
is satisfied for all 0 ≤ f↓,R if and only if for all decreasing sequences {εj }j∈Z and the covering
x
sequence {xj } satisfying 0 j v = 2k ,
!
Z xj+1 !
i
X Z xj+1 h εj
X
(3.18)
Q−1
Q
xa+1 w(x) dx ≤ P −1
P (εj )
v
B
xj
xj
j
j
holds, and
(3.19)
Q−1
XZ
j
xj+1
xj
"
#
!
iχ
xa (xj−1 ,xj ) Q
w(x) dx ≤ P −1
B
εj V
P̃ (εj v)
X 1
εj
j
!
is satisfied for all positive sequences {εj } and all covering sequences {xj }.
If Q is also an N -function, then a result corresponding to Corollary 3.5 holds also for the
dual operator.
Corollary 3.6. Let P and Q be N -functions and P , P̃ ∈ ∆2 . If a ≥ −1 then
Z ∞ Z ∞
Z ∞
−1
a
−1
(3.20)
Q
Q
t f (t) dt w(x) dx ≤ P
P [Cf (x)]v(x) dx
0
x
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
0
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18
H ANS P. H EINIG
AND
Q INSHENG L AI
is satisfied for all 0 ≤ f↓, if and only if
#
!
" X Z yj
k(·,
y
)χ
1 j
(yj ,yj+1 )
(3.21)
Q−1
w(x) dx ≤ P −1
Q
B
εj V
yj−1
P̃ (εj v)
j
X 1
εj
j
!
and
Q−1
(3.22)
#
!
χ
k(yj , x) (yj ,yj+1 ) w(x) dx ≤ P −1
Q
εj V B
yj−1
P̃ (εj v)
XZ
j
"
yj
X 1
εj
j
!
holds for all positive sequences {εj }j∈Z and all covering sequences {yj }j∈Z . Here
ln(y/x)
if a = −1,
k(y, x) =
a+1
a+1
y
−x
if a > −1.
Proof. By [7, Thm. 2.2], (3.20) is equivalent to
Z ∞ Z ∞ Z ∞
Z
a
−1
−1
Q
Q
t
h(s) dsdt w(x) dx ≤ P
0
x
t
∞
0
V (x)h(x)
P C
v(x) dx ,
v(x)
h ≥ 0. However, since
∞
Z
a
Z
t
x
∞
∞
Z
h(s) dsdt =
t
k(y, x)h(y) dy,
x
the result follows from Proposition 2.2 with θ(x) = 1, ρ(x) = V (x)/v(x).
Remark 3.1.
(i) Let P (x) = xp , Q(x) = xq , 0 < q < p < ∞, p > 1 and a = −1 then (3.18) is
!1/q
!1/p
X q
X p
εj w(Ej )
≤C
εj v(Ej )
,
j
where w(Ej ) =
R xj+1
xj
j
w and v(Ej ) =
R xj+1
v. But by Corollary 3.3 this is equivalent to
xj
∞
Z
[W 1/p V −1/p ]r w < ∞,
0
Also, if ηj =
X
Z
−q/p
εj
xj+1
ηj
then (3.19) takes the form
! Z
x−q w(x) dx
xj
j
xj
p0
−p0
t V (t)
1
1 1
= − .
r
q p
!q/p0
v(t) dt
!q/p
≤C
X
xj−1
p/q
ηj
.
j
But the dual space of `p/q is `r/q , where 1r = 1q − p1 and hence (3.19) is in this case

!r/q Z
!r/p0 1/r
X Z xj+1

xj
−q
p0
−p0
x w(x) dx
t V (t) v(t) dt
≤ C.


xj
xj−1
j
(Cf. [21, Thm. 2], where this was proved in case 1 < q < p < ∞ and [23] in the
remaining case.)
(ii) Considering Corollary 3.6 in the case P (x) = xp , Q(x) = xq , 1 < q < p < ∞, a = −1
we see that (3.21) takes the form
!q/p0
!q/p
Z yj ! Z yj+1
X
X
0
0
p/q
ηj
w
lnp (t/yj )V (t)−p v(t) dt
≤C
ηj
,
j
yj−1
yj
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
j
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W EIGHTED MODULAR INEQUALITIES FOR H ARDY- TYPE OPERATORS
19
−q/p
where again ηj = εj . But again since `r/q is the dual of `p/q it follows that in this
case (3.21) is equivalent to

!r/q Z
!r/p0 1/r
X Z yj

yj+1
0
0
w
≤ C,
lnp (t/yj )V (t)−p v(t) dt


yj−1
yj
j
where 1r = 1q − p1 . Similarly, (3.22) takes the form

!r/q Z
!r/p0 1/r
X Z yj

yj+1
0
lnq (yj /x)w(x) dx
V (t)−p v(t) dt
≤ C,


yj−1
yj
j
1
r
= 1q − p1 , for all covering sequences {yj }j∈Z .
Hence these two conditions are necessary and sufficient for the inequality
Z ∞
Z ∞
q 1/q
Z ∞
1/p
f (t)
p
w(x)
dt dx
≤C
f (x) v(x) dx
t
0
x
0
to be satisfied for all 0 ≤ f↓.
4. H ARDY- TYPE OPERATORS ON INCREASING FUNCTIONS
In order to obtain weight characterizations for which modular inequalities for the Hardy-type
operator
Z
x
(Kf )(x) =
0 ≤ f↑
k(x, y)f (y) dy,
0
are satisfied, we require also that the kernel k̄ defined by
Z x
(4.1)
k̄(x, y) =
k(x, t) dt
y
satisfies also conditions (i) and (ii) of (1.1). That is,
k̄(x, y) ≤ D[k̄(x, z) + k̄(z, y)],
(4.2)
0 < y < z < x.
Note that if k(x, t) = (x − t)α , α ≥ 0 then k̄ satisfies (4.2). On the other hand if k(x, t) =
ln(x/t) then k̄ does not satisfy (4.2) for any D ≥ 1.
The principal result for Hardy-type operators defined on the cone of increasing functions is
the following:
Theorem 4.1. Suppose K is a Hardy-type operator and k̄ defined by (4.1) satisfies (4.2). Let P
be an N -function with P , P̃ ∈ ∆2 and Q weakly convex. Then the modular inequality
Z ∞
Z ∞
−1
−1
(4.3)
Q
Q[θ(x)(Kf )(x)]w(x) dx ≤ P
P [Cf (x)]v(x) dx
0
0
is satisfied for all 0 ≤ f↑, if and only if there is a constant B > 0, such that,
"
#
!
X Z xj+1
k̄(x
,
·)χ
θ(x)
j
(xj−1 ,xj )
(4.4)
Q−1
Q
w(x) dx ≤ P −1
∗
B
εj V
xj
P̃ (εj v)
j
X 1
εj
j
!
X 1
εj
j
!
and
(4.5)
Q−1
XZ
j
xj+1
xj
"
#
!
χ
k̄(x, xj )θ(x) (xj−1 ,xj ) Q
w(x) dx ≤ P −1
∗
B
εj V
P̃ (εj v)
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
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20
H ANS P. H EINIG
AND
Q INSHENG L AI
holds for Rall positive sequences {εj }j∈Z and all covering sequences {xj }j∈Z . Here again
∞
V ∗ (x) = x v with V ∗ (0) = ∞.
Rx
Proof. Without loss of generality we may
assume
that
f
has
the
form
f
(x)
=
h, h ≥ 0 (cf.
0
R x ∗ −2
∗
−1
[24, Lemma 3.2]). Since V (x) = 0 V (t) v(t) dt, changing the order of integration we
show that
Z x
Z y
(Kf )(x) =
k(x, y)
h(s) dsdy
0
0
Z x
V ∗ (s)
=
h(s)k̄(x, s) ∗
ds
V (s)
0
Z x
Z s
∗
=
h(s)k̄(x, s)V (s)
V ∗ (t)−2 v(t) dtds
0
Z0 x
Z x
∗
−2
≤
V (t) v(t)k̄(x, t)
h(s)V ∗ (s) dsdt
Z0 x
Zt ∞
≤
V ∗ (t)−2 v(t)k̄(x, t)
v(y)f (y) dydt.
0
Hence if F (t) = V ∗ (t)−2 v(t)
with ρ(x) = V ∗ (x)/v(x)
Z
0
∞
R∞
t
t
f v, then Kf (x) ≤
Rx
0
k̄(x, t)F (t) dt and by Theorem 2.1
∞
Z x
Q[θ(x)(Kf )(x)]w(x) dx ≤
Q θ(x)
k̄(x, t)F (t) dt w(x) dx
0
0
Z ∞ Z ∞ C
−1
≤Q◦P
P
vf v(x) dx
V ∗ (x) x
0
Z
if and only if (4.4) and (4.5) are satisfied. Now (4.3) follows from (2.3) of Lemma 2.3.
To prove necessity one proves first that for fixed k
k̄(xj , ·)χ(xj−1 ,xj ) εj V ∗
P̃ (εj v)
is bounded. But this is proved (via contradiction) in the same way as the boundedness of (3.16)
in the proof of Theorem 3.4, only now k and V are replaced by k̄ and V ∗ , respectively. To prove
that (4.4) is satisfied assume to the contrary that for every B > 0 there exist {xj } and {εj } such
that
"
#
!
!
X Z xj+1
X 1
k̄(x
θ(x) j , ·)χ(xj−1 ,xj ) −1
−1
Q
Q
w(x) dx > P
.
∗
2BC
ε
V
εj
1
j
x
j
P̃ (εj v)
j
j
R∞
By duality of Orlicz spaces there exists fj ≥ 0 such that suppfj ⊂ (xj−1 , xj ), 0 P [BC1 fj ]εj v
≤ 1 and
Z xj
1 k̄(xj , x)fj (x)v(x)
k̄(xj , ·)χ(xj−1 ,xj ) <
dx.
∗
2BC1
εj V
V ∗ (x)
xj−1
P̃ (εj v)
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
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W EIGHTED MODULAR INEQUALITIES FOR H ARDY- TYPE OPERATORS
Now let f =
21
Rx
fj and F (x) = 0 Vf v∗ , so F↑. Also
Z xj
Z xj
Z
k̄(xj , x)fj (x)v(x)
fj (x)v(x) xj
dx ≤
k(xj , s) dsdx
V ∗ (x)
V ∗ (x)
xj−1
0
x
Z xj
Z s
fj (x)v(x)
=
k(xj , s)
dxds
V ∗ (x)
0
0
Z xj
≤
k(xj , s)F (s) ds
P
0
and therefore by (2.4) of Lemma 2.3
Z ∞
−1
P
P [BF (x)]v(x) dx
0
Z ∞
−1
≤P
P [BCf (x)]v(x) dx
0
!
X 1
−1
≤P
εj
j
"
#
!
X Z xj+1
k̄(x
,
·)χ
θ(x)
j
(x
,x
)
j−1 j < Q−1
Q
w(x) dx
∗
2BC
ε
V
1
j
x
j
P̃
(ε
v)
j
j
!
Z xj
X Z xj+1
−1
≤Q
Q[θ(x)
k(xj , s)F (s) ds]w(x) dx
xj
j
−1
Z
0
∞
≤Q
Q[θ(x)(KF )(x)]w(x) dx
Z ∞
−1
≤P
P [CF (x)]v(x) dx .
0
0
Here the last inequality is (4.3). But this is a contradiction for B > C. Hence (4.4) is satisfied.
The proof of (4.5) is similar, only now fj is chosen so that
Z xj
χ
1 fj (y)v(y)
(xj−1 ,xj ) <
dy
∗
2C1
εj V
V ∗ (y)
xj−1
P̃ (εj v)
and
Z
k̄(x, xj )
0
xj
fj (y)v(y)
dy ≤
V ∗ (y)
Z
xj
fj (y)v(y)
dy
V ∗ (y)
0
Z x
fj (y)v(y)
≤
k̄(x, y)
dy
V ∗ (y)
0
Z x
Z s
fj (y)v(y)
=
k(x, s)
dyds,
V ∗ (y)
0
0
k̄(x, y)
x ∈ (xj , xj+1 ). We omit the details. This proves the theorem.
Remark 4.1.
(i) If V ∗ (0) < ∞, Theorem 4.1 still holds, provided that in addition to (4.4) and (4.5) the
weight condition
Z ∞ 1 −1
1
1
−1
−1
(4.6)
Q
Q
P
θ(x)k̄(x, 0) w(x) dx ≤ P
∗
B
εV (0)
ε
0
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
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22
H ANS P. H EINIG
AND
Q INSHENG L AI
is also satisfied for all ε > 0.
(ii) If Q is an N -function and hence convex, the result may also be proved via the duality
principle given in [7, Thm. 2.2].
A consequence of Theorem 4.1 is the following:
Corollary 4.2.
(i) ([10, Thm. 2.1]) If 1 < p ≤ q < ∞, then
Z ∞ Z x q
Z ∞
1/p
1/q
1
p
(4.7)
≤C
f v
f w(x) dx
x 0
0
0
is satisfied for all 0 ≤ f↑, if and only if, for all t > 0
Z ∞
1/q
q −q
(x − t) x w(x) dx
V ∗ (t)−1/p
t
and
∞
Z
x−q w(x) dx
1/q Z
t
t
p0
−p0
(t − x) V ∗ (x)
1/p0
v(x) dx
0
are bounded.
(ii) If 0 < q < p < ∞, p > 1 then (4.7) is satisfied for all 0 ≤ f ↑ if and only if for all
covering sequences {xj }

!r/q Z
!r/p0 1/r
xj
X Z xj+1
0
0

 ≤C
(4.8)
x−q w(x) dx
(xj − x)p V ∗ (x)−p v(x) dx
xj
j
and

X Z

!r/q
(x − xj )q x−q w(x) dx
xj
j
1
r
xj+1
xj−1
Z
!r/p0 1/r
xj
0
V ∗ (x)−p v(x) dx
≤C

xj−1
= 1q − p1 , are satisfied.
(If V ∗ (0) < ∞ the condition (4.6) must also be taken into account.)
Proof. Let Q(x) = xq , P (x) = xp , 1 < p ≤ q < ∞, θ(x) = x1 , k(x, y) = 1 in Theorem 4.1.
Since P Q we may take in Theorem 4.1 xj = t > 0, xj−1 = 0, xj+1 = ∞ and the result
(i) follows. If 0 < q < p < ∞, p > 1, then Q is weakly convex and by Theorem 4.1, (4.7) is
satisfied for all 0 ≤ f↑, if and only if for all covering sequences {xj }

! Z
!q/p0 
!q/p
Z xj+1
xj
X
X p/q
0
0

 ≤ C1
ηj
ηj
x−q w(x) dx
(xj − x)p V ∗ (x)−p v(x) dx
xj
j
and

X

ηj
j
Z
xj−1
xj+1
(x − xj )q xq w(x) dx
xj
j
! Z
xj
xj−1
−q/p
!q/p0 
−p0
V ∗ (x)
v(x) dx
 ≤ C2
!q/p
X
j
where we have taken ηj = εj
in (4.4) and (4.5). But since the dual of `p/q is `r/q ,
the previous two estimates are equivalent to (4.8) and (4.9) respectively.
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
p/q
ηj
1
r
=
1
q
− p1 ,
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W EIGHTED MODULAR INEQUALITIES FOR H ARDY- TYPE OPERATORS
23
The result of Corollary 4.2 (ii) in the case 1 < q < p < ∞ was also proved in [10, Thm. 2.2],
but the case 0 < q < 1 < p seems to be new.
In the remaining portion of this section we apply the results of the previous section to
show that the Hardy-Littlewood maximal function and the Hilbert transform are bounded in
weighted Orlicz-Lorentz spaces. This, in particular, extends the Lorentz space results of AriñoMuckenhoupt [1] and Sawyer [21] to this general setting.
V
If P is an increasing function of R+ with P (0) = 0, then the Orlicz-Lorentz
spaces P (v),
R
∞
with weight v consist of all Lebesgue measurable f on Rn such that P −1 0 P (f ∗ (x))v(x) dx
< ∞. Here f ∗ (t) = inf{s > 0 : |{x : |f (x)| > s}| ≤ t} denotes the equimeasurable decreasing
rearrangement of |f |.
R
1
Recall that if (M f )(x) = supx∈Q |Q|
|f (y)| dy is the Hardy-Littlewood maximal function,
Q
Rx
∗
then it is well known (cf. [2]) that (M f ) (x) ≈ x1 0 f ∗ . It follows therefore from Corollary 3.5
with a = −1 that the following proposition holds:
Proposition
M :
R ∞ P , P̃ ∈ ∗∆2 and Q weakly
R ∞convex.∗ Then
V
V 4.3. Suppose P is an N -function,
−1
−1
(v)
→
(w)
is
bounded,
that
is
Q
Q((M
f
)
)w
≤
CP
P
(Cf
)v
, if and
P
Q
0
0
if there are constants B > 0, such that
!
Z xj+1 !
X εj Z xj+1
X
w ≤ P −1
P (εj )
v
(4.9)
Q−1
Q
B
xj
xj
j
j
Rx
is satisfied for all decreasing sequences {εj } and the covering sequence {xj } satisfying 0 j v =
2k , and
"
#
!
!
X Z xj+1
X 1
i
χ
1
(x
,x
)
j−1
j
Q
w(x) dx ≤ P −1
(4.10)
Q−1
εj V
xB
εj
x
j
P̃ (εj v)
j
j
is satisfied for all positive sequences {εj }j∈Z and all covering sequences {xj }.
Rx
Here again V (x) = 0 v and i(x) = x.
Another illustration involves the Hilbert transform defined by the principle value integral
Z
1 ∞ f (t)
(Hf )(x) = P.V.
dt.
π −∞ x − t
Then (see [21, (1.15)]) the rearrangement inequality
Z x
Z ∞ ∗
1
f (t)
∗
∗
(Hf ) (x) ≤ C1
f (t) dt +
dt ≤ C2 (Hf ∗ )∗ (x)
x 0
t
x
V
V
is satisfied. But this implies that the Hilbert transform is bounded from P (v) to Q (w) if and
only if
Z ∞
Z ∞
−1
−1
Q
Q[T f (x)]w(x) dx ≤ P
P [Cf (x)]v(x) dx
0
0
is satisfied, where
(4.11)
T f (x) = x
−1
Z
x
Z
f (t) dt +
0
x
∞
f (t)
dt,
t
0 ≤ f↓ .
However, (4.11) will be satisfied if and only if it is satisfied for the averaging operator and its
conjugate defined on decreasing functions. Hence Corollaries 3.5 and 3.6 apply with a = −1
and we have:
J. Ineq. Pure and Appl. Math., 1(1) Art. 10, 2000
http://jipam.vu.edu.au/
24
H ANS P. H EINIG
AND
Q INSHENG L AI
V
Proposition
4.4.
Suppose
P
and
Q
are
N
-functions
and
P
,
P̃
∈
∆
.
Then
H
:
2
P (v) →
R xj
V
k
Q (w) is bounded if and only if (4.9) (with {εj }↓, {xj } satisfying 0 v = 2 ), (4.10) and (see
(3.18), (3.19))
#
!
!
" X Z yj
X
ln(·/y
)χ
1 1
j
(yj ,yj+1 )
Q−1
Q
w(x) dx ≤ P −1
,
B
εj V
εj
yj−1
P̃ (εj v)
j
j
#
!
!
"
X Z yj
X
χ
1
ln(y
/x)
(y
,y
)
j
j j+1 Q−1
Q
w(x) dx ≤ P −1
,
εj V B
ε
j
y
j−1
P̃
(ε
v)
j
j
j
are satisfied for all positive sequences {εj }j∈Z and all covering sequences {xj }.
R EFERENCES
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