Ann. Funct. Anal. 4 (2013), no. 2, 118–130
A nnals of F unctional A nalysis
ISSN: 2008-8752 (electronic)
URL:www.emis.de/journals/AFA/
APPLICATION OF MULTIDIMENSIONAL HARDY OPERATOR
AND ITS CONNECTION WITH A CERTAIN NONLINEAR
DIFFERENTIAL EQUATION IN WEIGHTED VARIABLE
LEBESGUE SPACES
ROVSHAN A. BANDALIEV
Communicated by C. Cuevas
Abstract. In this paper a two weight criterion for multidimensional geometric mean operator in variable exponent Lebesgue space is proved. Also, we
found a criterion on weight functions expressing one-dimensional Hardy inequality via a certain nonlinear differential equation. In particular, considered
nonlinear differential equation is nonlinear integro-differential equation.
1. Introduction and preliminaries
It is well known that the variable exponent Lebesgue space appeared in the literature for the first time already in [18]. Further development of this theory was
connected with the theory of modular function spaces. The first systematic study
of modular spaces is presented in [17]. In the appendix, Nakano mentions explicitly variable exponent Lebesgue spaces as an example of more general spaces he
considers. Somewhat later, a more explicit version of these spaces, namely modular function spaces, were investigated by many mathematicians (see [16]). The
next step in the investigation of variable exponent spaces was given in [19] and
in [14]. The study of these spaces has been stimulated by problems of elasticity,
fluid dynamics, calculus of variations and differential equations with non-standard
growth conditions (see [7]). Recently in [1] was investigated converse theorems
of trigonometric approximation in variable exponent Lebesgue spaces with some
Muckenhoupt weights.
Date: Received: 16 December 2012; Accepted: 24 January 2013 .
2010 Mathematics Subject Classification. Primary 46E30; Secondary 47B38, 26D15.
Key words and phrases. Variable Lebesgue space, geometric mean operator, Hardy operator,
nonlinear differential equation.
118
HARDY OPERATOR AND CERTAIN NONLINEAR DIFFERENTIAL EQUATION
119
Inequalities are one of the most important instruments in many branches of
mathematics such as functional analysis, theory of differential and integral equations, interpolation theory, harmonic analysis, probability theory etc. They are
also useful in mechanics, physics and other sciences. It is well known that the classical two weight inequality for the geometric mean operator is closely connected
to the one-dimensional Hardy inequality (see [9]). Analogously, the Pólya-Knopp
type inequalities with multidimensional geometric mean operator is connected
with multidimensional Hardy type operator. In the paper [8] the connection
of the Hardy inequality with a nonlinear differential equation having a solution
with certain special properties was considered. Therefore, the consideration of
this problems in variable exponent Lebesgue space is actual.
Let Rn be the n-dimensional Euclidean space of points x = (x1 , · · · , xn ) and
!1/2
n
X
Ω be a Lebesgue measurable subset in Rn and |x| =
x2i
. Suppose that
i=1
p is a Lebesgue measurable function on Ω such that 0 < p ≤ p(x) < ∞, p =
ess inf x∈Ω p(x) and ω is a weight function on Ω, i.e. ω is a non-negative, almost
everywhere (a.e.) positive function on Ω. The Lebesgue measure of a set Ω will
n
π2
, where B(0, 1) =
be denoted by |Ω|. It is well known that |B(0, 1)| =
Γ n2 + 1
{x ∈ Rn ; |x| < 1} . Further, in this paper all sets and functions are supposed to be
Lebesgue measurable. By AC(0, ∞) we denote the set of absolutely continuous
functions on (0, ∞). For the sake of simplicity, the letter C always denotes a
positive constant which may change from one step to the next.
Definition 1.1. By Lp(x), ω (Ω) we denote the set of measurable functions f on Ω
such that for some λ0 > 0
p(x)
Z |f (x)| ω(x)
dx < ∞.
λ0
Ω
Note that the expression
kωf kLp(x) (Ω) = kf kLp(·), ω (Ω)



p(x)
Z 
f (x) ω(x) = inf λ > 0 :
dx ≤ 1 .


λ
Ω
defines a quasi-Banach spaces. In particular, for 1 ≤ p(x) < ∞ the space
Lp(x), ω (Ω) is a Banach function space (see [7]) with respect to the expression
kf kLp(x), ω (Ω) .
For ω = 1 the space Lp(x),ω (Ω) coincides with the variable Lebesgue space
Lp(x) (Ω).
We reduce two examples which characterize the norm of this space.
Example 1.2. Let p(x) =
2
3
f or
f or
T
x∈Ω
and f ∈ L2 (Rn ) L3 (Rn ) .
n
x∈R \Ω
120
R.A. BANDALIEV
We calculate kf kLp(x) (Rn ) . By the definition we have


Z
Z 
f (x) 2
kf kLp(x) (Rn ) = inf λ > 0 :
λ dx +


n
Ω
R \Ω


3

f (x) dx ≤ 1 =
λ 

n
o
a1
a2
= inf λ > 0 :
+
≤
1
= inf λ > 0 : λ3 − a1 λ − a2 ≥ 0 ,
2
3
λ
λ
Z
Z
2
|f (x)|3 dx. Now we solve the inequality
where a1 = f (x) dx and a2 =
Ω
Rn \Ω
3
λ − a1 λ − a2 ≥ 0. We consider three different cases.
a2 a3
Case 1. Let 2 − 1 > 0. Then the cubic equation λ3 − a1 λ − a2 = 0 has
4
27
s
r
3 a2
a22 a31
one real root and two complex conjugate roots. Namely, λ1 =
+
−
2
4
27
r
r
q
q
s
3 a
3
a22
a3
a22
a3
r
2
+
− 271 − a22 −
− 271
√
2
4
4
3 a2
λ1
a22 a31
−
− , λ2 = − + i 3
and
+
2
4
27
2
2
λ3 = λ2 . It is obvious that λ3 − a1 λ − a2 = (λ − λ1 ) λ2 + λ1 λ + |λ2 |2 and λ2 +
λ1 λ+|λ2 |2 > 0 for all λ ∈ (−∞, +∞). Therefore the inequality λ3 −a1 λ−a2 ≥ 0
holds if and only if λ ≥ λ1 and
s
s
r
r
3
2
3 a2
3 a2
a2 a1
a22 a31
kf kLp(x) (Rn ) =
+
−
+
−
− .
2
4
27
2
4
27
r 2
a2
λ−2
λ+ 3
r 2
a2
λ3 − a1 λ − a2 ≥ 0 holds if and only if λ ≥ 2 3
and
2
a2 a3
Case 2. Let 2 − 1 = 0. Then λ3 − a1 λ − a2 =
4
27
r
3
a2
2
and the inequality
r
a2
kf kLp(x) (Rn ) = 2 3
.
2
a2 a3
Case 3. Let 2 − 1 < 0. Then the equation λ3 − a1 λ − a2 = 0 has three dis4
27
tinct real roots. We denote by α1 , α2 and α3 the roots of this equation. By
Viète’s formulas one root of this equation is positive and two roots are negative. Let α1 > 0. Then λ3 − a1 λ − a2 = (λ − α1 ) λ2 + α1 λ + α12 − a1 = 0,
p
p
√
−α1 + 4 a1 − 3 α12
−α1 − 4 a1 − 3 α12
2 √
a1 .
α2 =
, α3 =
and a1 < α1 < √
2
2
3
It is obvious that α3 < α2 <Sα1 . Therefore the inequality λ3 − a1 λ − a2 ≥ 0 holds
if and only if λ ∈ [α3 , α2 ] [α1 , ∞) and by the definition of the norm we have
λ ≥ α1 . Thus, kf kLp(x) (Rn ) = α1 .
Example 1.3. Let n = 1, x ∈ [1, ∞), p(x) = x and f (x) = 1.
HARDY OPERATOR AND CERTAIN NONLINEAR DIFFERENTIAL EQUATION
121
We calculate k1kLp(x) ([1,∞)) . We have


Z∞


1
k1kLp(x) ([1,∞)) = inf λ > 0 :
dx
≤
1
.


λx
1
Z∞
It is obvious that
1
1
if λ > 1 and
dx =
x
λ
λ ln λ
1
inf



Z∞
λ>0:
1


1
1
λ
≤
1
=
inf
λ
>
1
:
λ
≥
e
.
dx
≤
1
=
inf
λ
>
1
:

λx
λ ln λ
Thus, k1kLp(x) ([1,∞)) = 1, 7712 · · ·
In [2] the following theorem is proved.
Theorem 1.4. Let 1 ≤ p ≤ p(x) ≤ q(y) ≤ q < ∞ for all x ∈ Ω1 ⊂ Rn and
y ∈ Ω2 ⊂ Rm . If p ∈ C (Ω1 ) , then the inequality
2
p q−p p +
≤
kf kLq(·) (Ω2 ) kf kLp(·) (Ω1 ) q
q
Lp(·) (Ω1 )
Lq(·) (Ω2 )
is valid, where q = ess inf q(x), q = ess sup q(x) and C (Ω1 ) is the space of
Ω2
Ω2
continuous functions in Ω1 and f : Ω1 × Ω2 → R is any measurable function such
that


p(x)
Z 

kf (x, ·)kq(·),Ω2
dx ≤ 1 < ∞.
kkf kq,Ω2 kp,Ω1 = inf µ > 0 :


µ
Ω1
Z
Let Hf (x) =
f (y) dy, where f ≥ 0 and B(0, |x|) = {y ∈ Rn ; |y| < |x|} .
|y|<|x|
Now we formulate the criteria on boundedness of multidimensional Hardy type
operator in weighted variable Lebesgue spaces. In [3] the following theorem is
proved.
Theorem 1.5. Let q(·) be a measurable function on Rn , 1 < p ≤ q(x) ≤ q < ∞
p
and p0 =
. Suppose that v and w are weights on Rn . Then the inequality
p−1
kHf kLq(·), w (Rn ) ≤ C kf kLp, v (Rn )
(1.1)
holds, for every f ≥ 0 if and only if there exists α ∈ (0, 1) such that

 α0  1−α

0 p
p
Z
Z
0

 

−p0
−p
A(α, p, q) = sup 
v (y) dy  v
(y)
dy
< ∞. (1.2)


t>0
|y|<t
|y|<|·|
Lq(·), w (|x|>t)
122
R.A. BANDALIEV
Moreover, if C > 0 is the best possible constant in (1.1) then
p 0 A(α, p, q)
i1/p ≤ C ≤
h 0 p
0<α<1
p
1
(1 − α) 1−α + α(p−1)
sup
≤
p q−p
+
q
q
p2
A(α, p, q)
0.
0<α<1 (1 − α)1/p
inf
Remark 1.6. Note that Theorem 1.5 in the case n = 1, q(x) = q = const for x ∈
s−1
(1 < s < p) was proved in [20] and in multidimensional Hardy
(0, ∞) and α = p−1
type operators it was proved in [3]. Two-weighted criterion for one-dimensional
Hardy operator in weighted variable Lp(x), w ([0, 1]) spaces was proved in [13]. Note
that Theorem 1.5 in the case n = 1, p(x) = p = const q(x) = q = const for
x ∈ (0, ∞) was proved in [4], [15] and etc.
2. Main results
We consider the multidimensional geometric mean operator defined as


Z
1


Gf (x) = exp 
ln f (y) dy  ,
|B(0, |x|)|
B(0, |x|)
where f > 0 and |B(0, |x|)| = |B(0, 1)| |x|n . It is obvious that G (f1 · f2 ) (x) =
Gf1 (x) · Gf2 (x).
Now we formulate a two-weight criterion on boundedness of multidimensional
geometric mean operator in variable Lebesgue spaces.
Theorem 2.1. Let q(·) be a measurable function on Rn and 0 < p ≤ q(x) ≤ q <
∞. Suppose that v and w are weights on Rn . Then the inequality
kGf kLq(·), w (Rn ) ≤ C kf kLp, v (Rn )
(2.1)
holds, for every f > 0 if and only if there exists s ∈ (1, p) such that
= sup |B(0, t)|
t>0
s−1
p
D(s, p, q)
!
R
w(·)
1
1
exp
ln
dy
s
|B(0,|·|)|
v(y)
|B(0,|·|)| p
B(0,|·|)
< ∞.(2.2)
Lq(·) (|x|>t)
Moreover, if C > 0 is the best possible constant in (2.1), then
s
2
s−1
ep
p q−p p
p D(s, p, q).
D(s,
p,
q)
≤
C
≤
sup
+
inf
e
1/p
s>1
q
q
s>1 es + 1
s−1
s−1
, where 1 < s < p. We replace f with f β , v with v β , w with
p−1
wβ (x)
p
q(x)
, 0 < β < p, and p with and q(x) with
in (1.1), (1.2). We find
|B(0, |x|)|
β
β
Proof. Let α =
HARDY OPERATOR AND CERTAIN NONLINEAR DIFFERENTIAL EQUATION
123
p
β
that for 1 < s <
β
w
β
H(f
)
|B(0, | · |)|
Lq(·)/β (Rn )

1/β β
Z

1

β
= f
(y)
dy


|B(0, | · |)|
B(0, |·|)
Lq(·), w (Rn )
β/p

Z
≤ Cβ 
[f (y)v(y)]p dy 
.
Rn
Then the inequality

1/β Z

1

β

f (y) dy  |B(0, | · |)|
B(0, |·|)
1/p

Z
1/β
≤ Cβ

[f (y)v(y)]p dy 
(2.3)
Rn
Lq(·), w (Rn )
holds if and only if
s−1 p q
A
, ,
p−1 β β


Z

βp


= sup 
[v(y)]− p−β
 t>0
|y|<t
β

 s−1
 p−βs
p
βp Z

βp
1


− p−β
[v(y)]
dy 
dy 
p

|B(0, | · |)| p−βs
|y|<|·|




Lq(·), w (|x|>t)
= B β (s, p, q, β) < ∞
and


sup  p
1<s< β
p
p−sβ
p
p−sβ
βp
≤
β/p
βp
+
1
s−1
p q−p
+
q
q
where q is replaced by
B β (s, p, q, β) ≤ Cβ


2pβ
inf
1<s< βp
β→+0
p−β
p
B β (s, p, q, β) ,
(2.4)
q
q
and q is replaced by . By L’Hospital rule, we get
β
β
 p−βs
βp


lim 
p−β
p − sβ
Z
1
|B(0, |x|)|
p
p−βs
|y|<|x|
βp

[v(y)]− p−β dy 
124
R.A. BANDALIEV

 p ln

= lim exp 

β→+0


1
|B(0,|x|)|
+ (p − βs) ln
pβ
(p − βs)

βp
− p−β
[v(y)]

dy  +
p
p−β
2
βp
− p−β
R
[v(y)]
ln
|y|<|x|
βp
− p−β
R
p
|y|<|x|
[v(y)]
1
v(y)
dy





dy
|y|<|x|
R

dy 




[v(y)]
|y|<|x|

Z
 s


= lim exp − ln 
β→+0
 p
!
βp
− p−β
R
1
dy
ln v(y)
1
|y|<|x|
s
= exp  ln
+
p
|B(0, |x|)|
|B(0, |x|)|




=
Z
1
1

s exp 
|B(0, |x|)|
|B(0, |x|)| p
ln
1

dy  .
v(y)
B(0,|x|)
Therefore
lim B (s, p, q, β)
β→+0
= sup |B(0, t)|


Z
w(·)
1
1


ln
dy 
s exp 
|B(0, | · |)| p
|B(0, | · |)|
v(y)
B(0,|·|)
s−1
p
t>0
Lq(·) (|x|>t)
= D(s, p, q) < ∞
and
s
ep
sup
es +
s>1
1
s−1
lim
1/p D(s, p, q) ≤ β→+0
Further, we have


lim 
β→+0
1/β
Cβ
≤
1/β
Z
1
|B(0, |x|)|

f β (y) dy 
p q−p
+
q
q
p2
inf e
s−1
p
s>1
D(s, p, q). (2.5)


= exp 

Z
1
|B(0, |x|)|
B(0, |x|)

ln f (y) dy  .
B(0, |x|)
1/β
Formula (2.4) implies lim Cβ = 1, and according to (2.2) and (2.5) lim Cβ
β→+0
β→+0
= C < ∞. Therefore the inequality (2.3) is valid. Moreover, from (2.3) for
β → +0 we obtain
kGf kLq(·), w (Rn ) ≤ C kf kLp, v (Rn )
and by (2.5)
s
sup
s>1
ep
es +
1
s−1
1/p D(s, p, q) ≤ C ≤
p q−p
+
q
q
This completes the proof of Theorem 2.1.
p2
inf e
s>1
s−1
p
D(s, p, q).
HARDY OPERATOR AND CERTAIN NONLINEAR DIFFERENTIAL EQUATION
125
Remark 2.2. Let q(x) = q = const and n = 1. Note that the simplest case of
(2.1) with v = w = 1 and p = q = 1 was considered in [9] and in [12]. Later on,
this inequality was generalized in various ways by many authors in [5], [6], [10],
[11], [20] and etc.
Corollary 2.3. Let q(x) = q = const, 0 < p ≤ q < ∞ and let f be a positive
function on Rn . Then
1/q
1/p


Z
Z
 [Gf (x)]q |x|γ q dx ≤ C  f p (x) |x|β p dx
(2.6)
Rn
Rn
holds with a finite constant C if and only if
γ+
n
β n
= +
q
n p
and the best constant C has the following condition:
β
1
s
1
r
+ 1q
− 1q
− p1
p (s − 1) p
q
n2
β
1
1
|B(0,
1)|
p
e
e
−
q
√
e n2 |B(0, 1)| q p sup
≤C≤
.
q
1/p
nq
n
s>1 [(s − 1)es + 1]
Remark 2.4. Note that if p = q, then the inequality (2.6) is sharp with the
β
1
e n2 + p
constant C = √
.
p
n
1 f or |x| < 1
n
Corollary 2.5. Let x ∈ R , 0 < p ≤ q(x) < ∞, q(x) =
2 f or |x| ≥ 1,
n
and let f be a positive function on R . Suppose that v(x) = 1 and w(x) = |x|β .
Then

1/p
Z
 f p (x) dx
kGf kL
n ≤ C
β (R )
q(·), |·|
Rn
holds with a finite constant C if and only if
s
1 1
pi
n
−1 ≤β ≤n
−
, s ∈ 1, 1 +
p
p 2
2
and the best constant C has the following condition:
s
2
ep
p q−p p
0
sup
+
1/p D (s, p, q) ≤ C ≤
q
q
1<s≤1+ p2 es + 1
s−1
n(s−1)
β− nsp − p1
0
p
where D (s, p, q) = |B(0, 1)| sup t
| · |
t>0
inf
1<s≤1+ p2
e
s−1
p
D0 (s, p, q),
< ∞.
Lq(·) (|·|>t)
Now we consider an application
in the theory of nonlinear ordinary differential
0
equation. Let L(t, ω, y) = ω y 1/p L (x>t) , where t ∈ (0, ∞) and ω is a weight
q(x)
function defined on (0, ∞).
126
R.A. BANDALIEV
Lemma 2.6. Let 1 < p ≤ q(x) ≤ q < ∞. Suppose that ω1 and ω2 are weight
functions defined on (0, ∞). Let the equation
1/p0
L (t, ω2 , y) − λω1 (t) (y 0 (t))
= 0, (λ > 0)
(2.7)
have a solution y such that
y(t) > 0, y 0 (t) > 0, y ∈ AC(0, ∞).
(2.8)
Then the weighted norm inequality
2
p q−p p
ku0 kLp,ω1 (0,∞)
kukLq(·),ω2 (0,∞) ≤ λ
+
q
q
holds, where u ∈ AC(0, ∞) and u(0) = lim u(t) = 0.
t→+0
Proof. Applying the Hölder inequality we have
Zx
Zx
1
−1
u(x) = u0 (t) dt = u0 (t)(y 0 (t)) p0 (y 0 (t)) p0 dt
0
0
 x
 10
p
Z
−1
0
≤  y (t) dt u0 (y 0 ) p0 Lp (0, x)
1 −1
≤ (y(x)) p0 u0 (y 0 ) p0 .
Lp (0, x)
0
Thus
kukLq(·), ω2 (0,∞)
1
1
0
0 − p0
0
p
≤ ω2 (·) (y(·)) ku (y )
kLp (0, ·) Lq(·) (0,∞)
1
1
0
0 − p0
0
p
= ω2 (·)(y(·)) u (y )
χ(0, ·) Lp (0,∞)
.
Lq(·) (0,∞)
By using Theorem 1.4 we have
1
1
−
0
0
0
0
ω2 (·)(y(·)) p u (y ) p χ(0, ·) Lp (0,∞)
Lq(·) (0,∞)
≤
p q−p
+
q
q
=
p2 1
1
ω2 (·)(y(·)) p0 u0 (y 0 )− p0 χ(0, ·) Lq(·) (0,∞)
p q−p
+
q
q
=λ
=λ
p q−p
+
q
q
p2 1
ω2 y p0 0
Lp (0,∞)
p2 1
1
0 p0
0
0 − p0 ω1 (y ) u (y ) kω1 u kLp (0,∞) = λ
This proves the Lemma 2.6.
1 0 − p0 u (y )
Lq(·) (t,∞)
p q−p
+
q
q
p2
0
Lp (0,∞)
Lp (0,∞)
p q−p
+
q
q
p2
ku0 kLp, ω
1 (0,∞)
.
HARDY OPERATOR AND CERTAIN NONLINEAR DIFFERENTIAL EQUATION
127
Put
K = p inf sup
x>0
Zx
1
0
f (x) − x − p0
Rt
0
ω10 (s) f (s)
ω1 (s)
ds
0
(f (t))1/p +1 P (t, ω2 , y, f )
dt, (2.9)
ω1 (t) (y(t))1/p0
0
where P (t, ω2 , y, f ) ≥ 0 for all t > 0, y(t) is a fixed positive solution of equation
(2.7) and the infimum is taken over the class of measurable functions such that
f (x) > x + p
0
Zt
ω10 (s) f (s)
ds for all x > 0.
ω1 (s)
0
The following lemma gives the relation between the number K and the problem
(2.7), (2.8).
Lemma 2.7. Let λ > 0 be the number from Lemma 2.6 and let K be given by
(2.9). Suppose that ω1 and ω2 are weight functions defined on (0, ∞) and the
derivative ω10 (t) exists for all t ∈ (0, ∞) and ω1 (t) ≥ ω1 (0) > 0.
Then the following statements are equivalent:
(i) if the problem (2.7), (2.8) has a solution with a locally absolutely continuous
first derivative, then λ ≥ K;
(ii) if K < +∞, then the problem (2.7), (2.8) has a solution for every λ > K.
Proof. Assume that (i) holds. Let y0 (x) be a solution of (2.7), (2.8). Let us take
y0
d
w = 0 . Assume that P (t, ω2 , y0 , w) = − L (t, ω2 , y0 ) = P (t). It is obvious that
y0
dt
P (t) ≥ 0 for all t > 0. Then by virtue of (2.7) w is a positive solution of the
equation
0
p0 ω10 (t) w(t) p0 (w(t))1/p +1 P (t)
+
w (t) =
0 + 1.
ω1 (t)
λ ω1 (t) (y0 (t))1/p
0
(2.10)
Hence (2.10) implies
Zt
w(t) ≥
0
w (s) ds = p
0
0
Zt
ω10 (s) w(s)
p0
ds +
ω1 (s)
λ
0
Zt
0
0
(w(s))1/p +1 P (s)
ω1 (s) (y0 (s))1/p
0
ds + t. (2.11)
From (2.11) implies that
p0
λ≥
Rt
w(t) − t − p0
0
Zt
ω10 (s) w(s)
ω1 (s)
ds
0
0
(w(s))1/p +1 P (s)
ω1 (s) (y0 (s))1/p
0
ds.
(2.12)
From (2.11), (2.12) and (2.9) it follows that λ ≥ K and the proof of (i) ⇒ (ii) is
complete.
128
R.A. BANDALIEV
Assume now (ii) holds. Let us fix λ > K. By the definition of K there exists
a measurable function f (x) such that
Zx 0
Zx
0
ω1 (t) f (t)
(f (t))1/p +1 P (t)
p0
0
f (x) ≥ x + p
dt +
(2.13)
0 dt.
ω1 (t)
λ
ω1 (t) (y(t))1/p
0
0
Let us define a sequence wn (x) by setting
w0 (x) = f (x),
wn (x) = x + p
0
Zt
ω10 (s) wn−1 (s)
p0
ds +
ω1 (s)
λ
Zx
0
0
0
(wn−1 (t))1/p +1 Pn−1 (t)
ω1 (t) (y(t))1/p
0
dt, (2.14)
where P0 (t) = P (t) and Pn (t) ≥ 0 for all n ∈ N. From (2.13) it follows that
w0 (x) ≥ w1 (x). We put wn−1 (x) ≥ wn (x) and let Pn (t) be decreasing sequences
with respect to n on (0, ∞), where n ∈ N. Then
0
0
Zx
Zx
(wn−1 (t))1/p +1 Pn−1 (t)
(wn (t))1/p +1 Pn (t)
dt ≥
dt
0
0
ω1 (t) (y(t))1/p
ω1 (t) (y(t))1/p
0
0
and
wn (x) − wn+1 (x) ≥ p0
Zx
ω10 (s) (wn−1 (s) − wn (s))
ds
ω1 (s)
0
Zx
≥
inf
s∈(0, ∞)
(wn−1 (s) − wn (s))
ω1 (x)
ω10 (s)
ds = inf (wn−1 (t) − wn (t)) ln
≥ 0.
t∈(0, ∞)
ω1 (s)
ω1 (0)
0
Since wn (x) ≥ 0, the sequence (2.14) converges. We denote its limit by w(x).
By the Levi monotone convergence theorem it follows that w is a nonnegative
solution of the equation
Zx 0
Zx
0
p0
ω1 (t) w(t)
(w(t))1/p +1 P (t)
0
w(x) = x + p
dt +
0 dt,
ω1 (t)
λ
ω1 (t) (y(t))1/p
0
0
where P (t) = lim Pn (t). Hence, w is absolutely continuous and satisfies the equan→∞
tion
0
p0 ω10 (x) w(x) p0 (w(x))1/p +1 P (x)
0
w (x) = 1 +
+
.
ω1 (x)
λ ω1 (x) (y(x))1/p0
Therefore the function
Rx
y0 (x) = e
a
dt
w(t)
(a be fixed in (0, ∞))
satisfies the problem (2.7), (2.8).
This completes the proof of Lemma 2.7.
Thus, we have the following
HARDY OPERATOR AND CERTAIN NONLINEAR DIFFERENTIAL EQUATION
129
Theorem 2.8. Let 1 < p ≤ q(x) ≤ q < ∞ and K < +∞. Suppose that ω1 and
ω2 are weight functions defined on (0, ∞) and the derivative ω10 (t) exists for all
t ∈ (0, ∞) and ω1 (t) ≥ ω1 (0) > 0. Then the following statements are equivalent:
a) there is a positive solution of the equation
1/p0
L (t, ω2 , y) − λω1 (t) (y 0 (t))
= 0,
y(t) > 0, y 0 (t) > 0, y ∈ AC(0, ∞),
where λ > 0;
b) the weighted norm inequality
kukLq(·),ω2 (0,∞) ≤ C0 ku0 kLp,ω1 (0,∞)
holds, where u ∈ AC(0, ∞), u(0) = lim u(t) = 0 and C0 > 0 is independent of
t→+0
u.
Remark 2.9. Note that for q(x) = q = const and 1 < p ≤ q < ∞ Theorem 2.8
was proved in [8] and the proof of Theorem 2.8 is based on the paper [8].
Acknowledgement. This work was supported by the Science Development
Foundation under the President of the Republic of Azerbaijan EIF-2010-1(1)40/06-1.
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1
Department of Mathematical Analysis,Institute of Mathematics and Mechanics of National Academy of Sciences of Azerbaijan, AZ 1141, Azerbaijan.
E-mail address: bandaliyev.rovshan@math.ab.az