Ann. Funct. Anal. 2 (2011), no. 1, 114–127
A nnals of F unctional A nalysis
ISSN: 2008-8752 (electronic)
URL: www.emis.de/journals/AFA/
CRITERIA ON BOUNDEDNESS OF MATRIX OPERATORS IN
WEIGHTED SPACES OF SEQUENCES AND THEIR
APPLICATIONS
ZHANAR TASPAGANBETOVA1 AND AINUR TEMIRKHANOVA2∗
Communicated by C. P. Niculescu
Abstract. In this paper we prove a new discrete Hardy type inequality involving a kernel which has a more general form than those known in the literature.
We obtain necessary and sufficient conditions for the boundedness of a matrix
operator from the weighted lp,v space into the weighted lq,u space defined by
∞
P
(Af )j :=
ai,j fi , for all f = {fi }∞
i=1 ∈ lp,v in case 1 < q < p < ∞ and
i=j
ai,j ≥ 0. Then we deduce a corresponding dual statement.
1. Introduction and preliminaries
∞
Let 1 < p, q < ∞, p1 + p10 = 1 and u = {ui }∞
i=1 , v = {vi }i=1 be positive sequences
of real numbers, which will be referred to as weight sequences. Let 1 < p < ∞.
We denote by lp,v the space of sequences f = {fi }∞
i=1 of real numbers such that
! p1
∞
X
kf kp,v :=
|vi fi |p
< ∞.
i=1
Moreover, let (ai,j ) be a non-negative triangular matrix with entries ai,j ≥ 0, if
i ≥ j ≥ 1 and ai,j = 0, if i < j.
We consider an estimate of the following form
kAf kq,u ≤ Ckf kp,v , ∀f ∈ lp,v ,
(1.1)
Date: Received: 19 February 2011; Accepted: 7 May 2011.
∗
Corresponding author.
2010 Mathematics Subject Classification. Primary 26D15; Secondary 47B37.
Key words and phrases. Inequalities, discrete Hardy-type inequalities, weights, matrix
operators.
114
CRITERIA ON BOUNDEDNESS OF MATRIX OPERATORS
115
where the matrix operator A is defined by
(Af )i :=
i
X
ai,j fj , i ≥ 1,
(1.2)
ai,j fi , j ≥ 1,
(1.3)
j=1
or
(Af )j :=
∞
X
i=j
and C is a positive finite constant independent of f .
When one of parameters p or q is equal to 1 or ∞, necessary and sufficient
conditions of the validity of (1.1) with the exact value of the best constant C > 0
have been obtained in [8]. In case 1 < p, q < ∞ inequalities as (1.1) have not
been established yet for arbitrary matrices (ai,j ). Instead inequality (1.1) has been
established with certain restrictions on the matrix (ai,j ).
When ai,j = 1, i ≥ j ≥ 1, the operators (1.2), (1.3) coincide with the discrete
i
∞
P
P
Hardy operators of the forms (A0 f )i :=
fj , (A0 f )j :=
fi , respectively. Refj=1
i=j
erences about generalizations of the original forms of the discrete and continuous
Hardy inequalities can be found in different books, see e.g. [1].
In [4], [5] necessary and sufficient conditions for the validity of (1.1) have been
obtained for 1 < p, q < ∞ under the assumption that there exists d ≥ 1 such
that the inequalities
1
(ai,k + ak,j ) ≤ ai,j ≤ d(ai,k + ak,j ), i ≥ k ≥ j ≥ 1.
(1.4)
d
hold.
A sequence {ai }∞
i=1 is called almost non-decreasing (non-increasing), if there
exists c > 0 such that cai ≥ ak (ak ≤ caj ) for all i ≥ k ≥ j ≥ 1.
In [6] estimate (1.1) has been studied under the assumption that there exist
d ≥ 1 and a sequence of positive numbers {ωk }∞
k=1 , and a non-negative matrix
(bi,j ), where bi,j is almost non-decreasing in i and almost non-increasing in j, such
that the inequalities
1
(bi,k ωj + ak,j ) ≤ ai,j ≤ d(bi,k ωj + ak,j ),
(1.5)
d
hold for all i ≥ k ≥ j ≥ 1.
In this paper we consider inequality (1.1) under the following assumption.
Assumption A: There exist d ≥ 1, a sequence of positive numbers {ωk }∞
k=1 ,
and a non-negative matrix (bi,j ), whose entries bi,j are almost non-decreasing in
i and almost non-increasing in j such that the inequalities
1
(ai,k + bk,j ωi ) ≤ ai,j ≤ d(ai,k + bk,j ωi ),
(1.6)
d
hold for all i ≥ k ≥ j ≥ 1.
116
ZH. TASPAGANBETOVA, A. TEMIRKHANOVA
Let α > 0. Let ai,j = (bi − dj )α , if i ≥ j ≥ 1, where the sequences {bi }∞
i=1 and
∞
{di }∞
are
such
that
b
is
a
non-decreasing
sequence
≥
d
,
i
≥
j
≥
1.
If
{b
}
i
j
i i=1
i=1
and {di }∞
is
an
arbitrary
sequence,
then
the
entries of the matrix (ai,j ) satisfy
i=1
condition (1.5), i.e. ai,j ≈ (bi − bk )α + ak,j , i ≥ k ≥ j ≥ 1. In general, the entries
ai,j do not satisfy condition (1.6). If {di }∞
i=1 is a non-decreasing sequence and
{bi }∞
is
an
arbitrary
sequence,
then
the
entries
ai,j satisfy condition (1.6), but
i=1
in general, condition (1.5) does not hold for the entries of the matrix (ai,j ).
Thus, conditions (1.5), (1.6) include condition (1.4) and complement each
other.
We also note that from (1.6) it easily follows that
dai,j ≥ ai,k ,
(1.7)
dai,j ≥ bk,j ωi ,
(1.8)
for i ≥ k ≥ j ≥ 1.
A continuous analogue of (1.5)-(1.6) has been considered by R. Oinarov in [3].
Notation: If M and K are real valued functionals of sequences, then we understand that the symbol M K means that there exists c > 0 such that M ≤ cK,
where c is a constant which may depend only on parameters such as p, q and d.
If M K M , then we write M ≈ K.
For the proof of our main theorem we need the following well-known result for
the discrete weighted Hardy inequality (see [1], [7]). For the sake of completeness,
we include a statement of such result.
Theorem A. Let 1 < q < p < ∞. Then the inequality
q ! 1
! p1
∞ X
∞
∞
q
X
X
µj fj uqi
≤C
|vi fi |p
∀f ∈ lp,v ,
i=1
j=i
i=1
holds if and only if

p
! p−q
∞
k
X
X
H=
uqi
k=1
(1.9)
i=1
∞
X
 p−q
pq
! p(q−1)
p−q
0
0
µpj vj−p
0
0
µpk vk−p 
< ∞.
j=k
Moreover, H ≈ C, where C is the best constant in (1.9).
We also need the following well-known result (see [4]).
Lemma A. Let γ > 0. Then there exists c > 0 such that
!γ
!γ−1
!γ
j
j
j
k
X
X
X
1 X
βk
≤
βk
βi
≤c
βk
∀j ∈ N,
c k=1
i=1
k=1
k=1
(1.10)
CRITERIA ON BOUNDEDNESS OF MATRIX OPERATORS
for all sequences {βk }∞
k=1 of positive real numbers.
Moreover, there exists c1 ≥ 1 such that
!γ
!γ−1
N
N
N
X
X
1 X
βk
≤
βk
βi
≤ c1
c1 k=j
k=j
i=k
N
X
117
!γ
βk
(1.11)
k=j
for all j, k ∈ {1, 2, ..., N }, N ∈ N ∪ {∞} and for all sequences {βk }∞
k=1 of positive
∞
P
real numbers such that
βk < ∞.
k=1
2. Main results
Theorem 2.1. Let 1 < q < p < ∞. Let the entries of the matrix (ai,j ) satisfy
Assumption A. Then estimate (1.1) for the operator defined by (1.3) holds if and
only if F = max{F1 , F2 } < ∞, where

 p−q
p
! p−q
! p(q−1)
pq
∞
i
∞
p−q
X X q q
X p0 −p0
0
0
F1 = 
bi,j uj
ωk vk
ωip vi−p 
i=1
j=1
k=i
and

F2 = 
∞
X
i
X
i=1
j=1
q
! p−q
uqj
∞
X
 p−q
pq
! q(p−1)
p−q
0
0
apk,i vk−p
uqi 
.
k=i
Moreover F ≈ C, where C is the best constant in (1.1).
Proof. Necessity. Let us assume that (1.1) holds for a finite constant C. Let
e ∞
m ≥ 1. Then we take a test sequence ff
m = {fm,k }k=1 such that
1
! p−q
m
q−1
k
P
P p0 −p0 p−q p0 −1 −p0
q
q
e
fm,k =
b u
ω v
ω
v
if 1 ≤ k ≤ m,
k,j j
j=1
i
i
k
k
i=k
fem,k = 0 if k > m.
Then
kfem kp,v =
∞
X
! p1
p
fem,k
vkp

=
k=1
m
X
k
X
k=1
j=1
p
! p−q
m
X
bqk,j uqj
 p1
! p(q−1)
p−q
0
0
0
0
ωkp vk−p  (. 2.1)
ωip vi−p
i=k
Substituting fem in the left hand side of inequality (1.1) and using (1.11) and
(1.8), we deduce that
!q−1
!q−1
j
m X
m
m
m
m
X
X
X
X
X
kAfem kqq,u aj,k fem,j
ai,k fem,i
uqk =
fem,j
uqk aj,k
ai,k fem,i
i=j
k=1 j=k
m
X
j=1
fem,j ωj
j
X
k=1
uqk bqj,k
m
X
i=j
j=1
!q−1
ωi fem,i
=
m
X
j=1
fem,j ωj
j
X
k=1
i=j
k=1

uqk bqj,k 
m
X
i=j
ωi
i
X
s=1
1
! p−q
bqi,s uqs
118
ZH. TASPAGANBETOVA, A. TEMIRKHANOVA
×
m
X
q−1
q−1
! p−q
p0
−p0
p0 −1
ωk vk
ωi
−p0
vi 
m
X
fem,j ωj
j=1
k=i

×
m
X
0
m
X
0
ωip vi−p
i=j
m
X
=
j
m
X
X
bqj,s uqs
s=1
q−1 q−1
! p−q
0
0
ωkp vk−p
! p−1
p−q
m
X
bqj,s uqs

! (p−1)(q−1)
p−q
p0
−p0
ωi vi
i=j
p
! p−q
m
X
bqj,s uqs
s=1
j=1
uqk bqj,k
k=1
s=1
j=1
q−1
! p−q
j
X
k=i
j
X
fem,j ωj
j
X
! p(q−1)
p−q
0
0
0
0
ωjp vj−p ,
ωip vi−p
i=j
i.e.
kAfem kq,u

p
! p−q
j
m
X
X

bqj,s uqs
m
X
s=1
i=j
j=1
 1q
! p(q−1)
p−q
0
0
0
0
ωip vi−p
ωjp vj−p  .
! p(q−1)
p−q
 p−q
pq
(2.2)
From (1.1), (2.1) and (2.2) it follows that

p
! p−q
j
m
X
X
j=1
s=1

bqj,s uqs
m
X
0
0
0
ωip vi−p
0
ωjp vj−p 
C
i=j
for all m ≥ 1. Since m ≥ 1 is arbitrary we have that
F1 C.
(2.3)
Inequality (1.1) holds if and only if the following dual inequality
kA∗ gkp0 ,v−1 ≤ Ckgkq0 ,u−1 , g ∈ lq0 ,u−1
(2.4)
holds for the conjugate operator A∗ , which is defined by (1.2).
Moreover, the best constants in (1.1) and (2.4) coincide.
Now let m ≥ 1. By taking a test sequence gem = {e
gm,k }∞
k=1 such that
q−1
! p−q (q−1)(p−1)
m
k
p−q
P
P
0
0
api,k vi−p
uqk for 1 ≤ k ≤ m,
gem,k =
uqj
j=1
i=k
gem,k = 0 for k > m,
we have that

ke
gm kq0 ,u−1 = 
m
k
X
X
k=1
j=1
q
! p−q
uqj
m
X
i=k
! q(p−1)
p−q
0
0
api,k vi−p
 q10
uqk  .
(2.5)
CRITERIA ON BOUNDEDNESS OF MATRIX OPERATORS
By using (1.10) and (1.7) we have that
!p0
m
i
m X
i
X
X
X
0
0
vi−p kA∗ gem kpp0 ,v−1 ≥
ai,j gem,j
ai,j gem,j
i=1
≥
m
X
gem,j
j=1
m
X
j=1
j
X
ai,j
i=j
i=1 j=1
!p0 −1
−p0
ai,k gem,k
m
X
vi
m
X
j=1
i=j
gem,j
m
X
0
0
api,j vi−p
i=j
m
X
gem,j
j=1
m
X
q−1 
! p−q
0
0
−p0
ai,j vi
j
X
!p0 −1
gem,k
k=1
0
p0 −1
(q−1)(p−1)
p−q
0
api,k vi−p
j
uqk 

k
X
uqk
!
0
p−1
p−q
j
X
0
api,j vi−p
0
q−1 p −1
! p−q
uqs

s=1
k=1
1
! p−q
uqk
k=1
q
! p−q
uqk
p0
!
X
api,j vi−p
m
X
0
vi−p
ai,k gem,k
i=k
i=j
j
m
X
X
j=1
m
X
i=j
i=j
=
gem,j
m
X
s=1
k=1
!p0 −1
k=1
j=1
k=1

q−1
! p−q
j
m
m
k
X
X
X
X
0
0
=
gem,j
api,j vi−p 
uqs
j=1
j
X
119
m
X
! q(p−1)
p−q
p0
−p0
uqj ,
ai,j vi
i=j
k=1
i.e.,

kA∗ gem kp0 ,v−1 
q
! p−q
j
m
X
X
j=1
k=1
uqk
m
X
! q(p−1)
p−q
0
0
api,j vi−p
 p10
uqj  .
(2.6)
i=j
Since m ≥ 1 is arbitrary, then (2.4), (2.5), (2.6) imply that F2 C. Hence,
(2.3) implies that
F C.
(2.7)
The proof of the necessity is thus complete.
Sufficiency. Let F < ∞ and 0 ≤ f ∈ lp,v .
For all j ≥ 1 we define the following set:
Tj = {k ∈ Z : (d + 1)−k ≤ (Af )j },
where d is the constant from (1.6) and Z is the set of integers. We assume that
inf Tj = ∞, if Tj = ∅ and kj = inf Tj , if Tj 6= ∅. We can clearly assume that
(Af )1 6= 0. Without loss of generality, we may assume that ai,j is non-increasing
in j, otherwise we take ai,j ≈ e
ai,j = sup ai,k . Therefore kj < kj+1 . If kj < ∞,
j≤k≤i
then
(d + 1)−kj ≤ (Af )j < (d + 1)−(kj −1) , j ≥ 1.
(2.8)
Let m1 = 0, k1 = km1 +1 and M1 = {j ∈ N : kj = k1 = km1 +1 }, where N is the
set of natural numbers. Suppose that m2 is such that sup M1 = m2 . Obviously
120
ZH. TASPAGANBETOVA, A. TEMIRKHANOVA
m2 > m1 and if the set M1 is upper bounded, then m2 < ∞ and m2 = max M1 .
We now define inductively the numbers 0 = m1 < m2 < . . . < ms < ∞, s ≥ 1.
We set ms+1 = sup Ms , where Ms = {j ∈ N : kj = kms +1 }.
Let N0 = {s ∈ N : ms < ∞}. Further, we assume that kms+1 = ns+1 , s ∈ N0 .
From the definition of ms and from (2.8) it follows that
(d + 1)−ns+1 ≤ (Af )j < (d + 1)−ns+1 +1 , ms + 1 ≤ j ≤ ms+1
(2.9)
for all s ∈ N0 . Then
[
N=
[ms + 1, ms+1 ), where [ms + 1, ms+1 ) ∩ [ml + 1, ml+1 ) = ∅, s 6= l.
s∈N0
Therefore
kAf kqq,u
=
s+1
X mX
(Af )qj uqj .
(2.10)
s∈N0 j=ms +1
mP
s+1
We assume that
= 0, if ms = ∞.
j=ms +1
There are two possible cases: N0 = N and N0 6= N.
1. If N0 = N, then we estimate the left hand side of (1.1) in the following way.
Clearly inequalities ns+1 < ns+2 < ns+3 imply that −ns+3 + 1 ≤ −ns+1 − 1 for
all s ∈ N. Hence, (2.9), (1.6) imply that
(d + 1)−ns+1 −1 = (d + 1)−ns+1 − d(d + 1)−ns+1 −1
≤ (d + 1)−ns+1 − d(d + 1)−ns+3 +1 < (Af )ms+1 − d (Af )ms+3
∞
∞
X
X
=
ai,ms+1 fi − d
ai,ms+3 fi
i=ms+1
i=ms+3
ms+3
X
≤
ai,ms+1 fi +
i=ms+1
X
∞
X
ai,ms+1 fi +
i=ms+1
∞
X
[ai,ms+1 − dai,ms+3 ]fi
i=ms+3
ms+3
≤
(2.11)
[d(ai,ms+3 + bms+3 ,ms+1 ωi ) − dai,ms+3 ]fi
i=ms+3
ms+3
=
X
∞
X
ai,ms+1 fi + dbms+3 ,ms+1
i=ms+1
ωi fi .
i=ms+3
Now, by using (2.9) and (2.11), we can estimate the summand on the left hand
side in (1.1) in the following way:
s+1
X mX
(Af )qj uqj <
s∈N j=ms +1
s+1
X mX
(d + 1)(−ns+1 +1)q uqj
s∈N j=ms +1
ms+1
= (d + 1)
2q
X
s∈N
(d + 1)
(−ns+1 −1)q
X
j=ms +1
uqj
CRITERIA ON BOUNDEDNESS OF MATRIX OPERATORS

ms+3
X
X
i=ms+1
s∈N
×
uqj
j=ms +1
X
X

S1 =
q
ms+1
X
ωi fi 
q
ms+3
X
i=ms+1

S2 =
q
∞
X
ms+1
X
ωi fi 

i=ms+3
s∈N
uqj ,
j=ms +1
and
bqms+3 ,ms+1
(2.12)
ms+1
ai,ms+1 fi 

X
uqj := S1 + S2 ,
j=ms +1
X
s∈N
uqj
j=ms +1
where
X
X
i=ms+1
i=ms+3
s∈N
ms+1
ai,ms+1 fi 

∞
X
bqms+3 ,ms+1 
q
ms+3
X
s∈N

+
ωi fi 
i=ms+3

ms+1
X
q
∞
X
ai,ms+1 fi + dbms+3 ,ms+1

121
uqj .
j=ms +1
To estimate S1 , we apply the Hölder Inequality in the inner summand with the
p
, pq , and we obtain
powers p, p0 and in the outer summand with the powers p−q
that

S1 ≤
 q0 
p
ms+3
0
0
api,ms+1 vi−p 
X
X

X

|vi fi |
i=ms+1
 q(p−1)
p−q

ms+3
0
X  X p0
≤
ai,ms+1 vi−p 
i=ms+1
s∈N
p

i=ms+1
s∈N
 pq
ms+3

ms+3
×
X X
ms+1
X
uqj
j=ms +1
p
! p−q
ms+1
X
uqj
 p−q
p


j=ms +1
 pq
f2 )
|vi fi |p  (F
p−q
p
kf kqp,v .
(2.13)
s∈N i=ms+1
f2 as follows:
By (1.11) and (1.7) we can estimate F

 q(p−1)
p
! p−q
p−q
ms+3
ms+1
X
X p0
X
0
f2 =

F
ai,ms+1 vi−p 
uqj
s∈N

X
i=ms+1
 q(p−1)
p−q
ms+3
X

s∈N
j=ms +1
i=ms+1
p0
−p0
ai,ms+1 vi 
ms+1
ms+1
X
X
j=ms +1
k=j
q
! p−q
uqk
uqj
122
ZH. TASPAGANBETOVA, A. TEMIRKHANOVA
X
s∈N j=ms +1
≤
j
∞
X
X
j=1
q 
! p−q
ms+1
s+1
X mX
uqk
uqk
0
0
api,j vi−p 

uqj
i=ms+1
k=1
q
! p−q
 q(p−1)
p−q
∞
X
∞
X
! q(p−1)
p−q
0
0
api,j vi−p
pq
uqj = F2p−q .
(2.14)
i=j
k=1
By (2.13) and (2.14) we deduce that
S1 F2q kf kqp,v .
(2.15)
q
Next we introduce the sequence {∆j }∞
j=1 such that ∆j = bms+3 ,ms+1
mP
s+1
uqi ,
i=ms +1
j = ms+3 and ∆j = 0, j 6= ms+3 , s ∈ N. Hence, we can rewrite S2 in the following
form:

q
!q
ms+1
∞
∞
∞
X
X
X q X
X

S2 =
ωi fi  bqms+3 ,ms+1
ui =
ωi fi ∆j .
s∈N
i=ms+3
i=ms +1
j=1
i=j
Thus, by Theorem A, we have that
fq kf kq ,
S2 H
p,v
(2.16)
where

e =
H
∞
X
k
X
k=1
i=1
p
! p−q
∞
X
∆i
 p−q
pq
! p(q−1)
p−q
0
0
0
ωjp vj−p
0
ωkp vk−p 
.
(2.17)
j=k
By Assumption A, bi,j is almost non-decreasing in i and almost non-increasing in
j, and accordingly,
k
X
ms+1
∆i =
X
ms+3 ≤k
i=1
X
ms+3 ≤k j=ms +1
uqj j=ms +1
ms+1
X
X
bqms+3 ,ms+1
bqk,j uqj ≤
k
X
(2.18)
bqk,j uqj .
j=1
By combining (2.16), (2.17) and (2.18), we obtain
S2 F1q kf kqp,v .
(2.19)
Thus, from (2.10), (2.12), (2.15) and (2.19) it follows that
kAf kq,u F kf kp,v , f ≥ 0,
(2.20)
i.e inequality (1.1) is valid and we see that the best constant in (1.1) C F .
2. If N0 6= N, i.e. max N0 < ∞ and N0 = {1, 2, ..., s0 }, s0 ≥ 1. Therefore
n
n
n
P
P
P
ms0 < ∞ and ms0 +1 = ∞. We assume that
= 0, if k > n and
=
, if
s=k
s=k
s=1
CRITERIA ON BOUNDEDNESS OF MATRIX OPERATORS
123
k ≤ 0. We have two possible cases: ns0 +1 < ∞ and ns0 +1 = ∞. We consider
such cases separately:
1) If ns0 +1 < ∞, then from (2.10) it follows that
kAf kqq,u
s+1
X mX
=
(Af )qj
uqj
ms+1
s0
X
X
=
=
ms+1
X
(Af )qj
uqj
(2.21)
s=1 j=ms +1
s∈N0 j=ms +1
sX
0 −3
(Af )qj uqj
ms+1
s0
X
+
X
(Af )qj uqj = I1 + I2 .
s=s0 −2 j=ms +1
s=1 j=ms +1
If I1 6= 0 then we estimate I1 using (2.11) and the previous proof for the case
N0 = N . Hence, we obtain
I1 F q kf kqp,v .
(2.22)
By using (2.9) and applying the Hölder Inequality with the powers p, p0 and
p
with the powers p−q
, pq , we obtain the following inequality
ms+1
s0
X
I2 =
X
(Af )qj
uqj
<
s=s0 −2 j=ms +1
= (d + 1)
(d + 1)

s0
X
∞
X

s=s0 −2

X
−ns+1 q
uqj
0
0
api,ms+1 vi−p 
s=s0 −2
i=ms+1


X
uqj
j=ms +1
p

j=ms +1
ms+1
ai,ms+1 fi 
i=ms+1
 pq
∞
X
i=ms+1
p
! p−q
i=ms+1
j=ms +1
×
∞
X
uqj
j=ms +1
ms+1
s0
X
X
|vi fi |

 q(p−1)
p−q
ms+1
p
s0
∞
0
 X  X p0
≤
ai,ms+1 vi−p 

X
(Af )qms+1
s=s0 −2
q
 q0 
∞
X
s=s0 −2
X
uqj
 p−q
p


 pq
c2 )
|vi fi |p  (F
p−q
p
kf kqp,v .
s=s0 −2 i=ms+1
c2 as follows:
Using (1.11) and (1.7) we can estimate F

 q(p−1)
p
! p−q
p−q
ms+1
s
∞
0
X
X p0
X
0
c2 =

uqj
F
ai,ms+1 vi−p 
s=s0 −2
s0
X

i=ms+1
∞
X

s=s0 −2
(2.23)
ms+1
s0
X
j=ms +1
=
≤
(d + 1)(−ns+1 +1)q uqj
s=s0 −2 j=ms +1
s=s0 −2
s0
X
X
ms+1
s0
X
q
ms+1
s0
X
i=ms+1
j=ms +1
 q(p−1)
p−q
p0
−p0
ai,ms+1 vi 
ms+1
ms+1
X
X
j=ms +1
k=j
q
! p−q
uqk
uqj
uqj
124
ZH. TASPAGANBETOVA, A. TEMIRKHANOVA
s0
X
ms+1
ms+1
X
X
s=s0 −2 j=ms +1
≤
j
∞
X
X
j=1
q 
! p−q
uqk
∞
X

 q(p−1)
p−q
0
0
api,j vi−p 
uqj
i=ms+1
k=1
q
! p−q
∞
X
uqk
! q(p−1)
p−q
p0
−p0
pq
uqj = F2p−q .
ai,j vi
(2.24)
i=j
k=1
From (2.23) and (2.24) we obtain
I2 F2q kf kqp,v .
(2.25)
From (2.21), (2.22) and (2.25) we have (2.20).
2) If ns0 +1 = ∞, which means that kms0 +1 = ∞, then we have kj = ∞ and
ms0
P
Tj = ∅, if j ≥ ms0 + 1, i.e. (Af )j = 0, if j ≥ ms0 +1 and (Af )j =
ai,j fi ,
i=j
1 ≤ j ≤ ms0 . Therefore m2 < ∞ and s0 ≥ 2. Then from (2.10) we have
kAf kqq,u
=
s+1
X mX
(Af )qj
uqj
=
sX
0 −1
ms+1
X
(Af )qj uqj
(2.26)
s=1 j=ms +1
s∈N0 j=ms +1
Similarly, we can exploit (2.26) to prove (2.20). Then (2.20) together with (2.7)
implies that C ≈ F and thus the proof is complete.
It is known that inequality (1.1) holds if and only if the dual inequality defined by (2.4) holds for the conjugate operator A∗ , which coincides with operator
defined by (1.2). Moreover, the best constants in (1.1) and (2.4) coincide.
Therefore by using Theorem 2.1 with p0 , q 0 , v −1 and u−1 replaced by q, p, u and
v, respectively, we obtain the following dual version of Theorem 2.1:
Theorem 2.2. Let 1 < q < p < ∞. Let the entries of the matrix (ai,j ) satisfy
Assumption A. Then estimate (1.1) for the operator defined by (1.2) holds if and
only if F ∗ = max{F1∗ , F2∗ } < ∞, where

F1∗ = 
∞
X
i
X
i=1
j=1

F2∗ = 
! q(p−1)
p−q
0
0
bpi,j vj−p
∞
X
i
X
i=1
j=1
∞
X
 p−q
pq
q
! p−q
ωkq uqk
ωiq uqi 
,
k=i
! p(q−1)
p−q
0
vj−p
∞
X
 p−q
pq
p
! p−q
aqk,i uqk
k=i
Moreover F ∗ ≈ C, where C is the best constant in (1.1).
0
vi−p 
.
CRITERIA ON BOUNDEDNESS OF MATRIX OPERATORS
125
3. Application of the main results
Our main results can be used to derive other inequalities. We consider an
additive estimate of the form
kAf kq,u ≤ C (kf kp,v + kA0 f kp,ρ ) , ∀f ≥ 0,
(3.1)
where the matrix operator A is defined by (1.2) and the operator A0 is defined
i
P
by (A0 f )i :=
fj , i ≥ 1.
j=1
We assume that the weighted sequences v and ρ satisfy the following conditions
vk > 0, k ≥ 1,
∞
X
ρk < ∞.
k=1
We denote by ∆ϕi the difference ϕi − ϕi−1 and we set


! p1 −1
!− 10
∞
n
p


X
X
p
−p0


+
ρi
ϕn = min
vi
, ϕ0 = 0,

1≤k≤n
i=k
i=k
for all n ≥ 1.
Next we introduce the following result of R. Oinarov [2] on the equivalence of
inequalities (2.17) and (1.1) which we exploit below.
Theorem C Let 1 < p, q < ∞. Let the entries of the matrix (ak,i ) of the operator
A be non-negative and non-increasing in i, i.e. ak,i+1 ≤ ak,i , if k ≥ 1, i ≥ 1.
Then inequality (3.1) holds if and only if the inequality
!q ! 1q
!1
∞
∞
k
0
1−p p
X
X
X
0
e
ωq
ak,i fi
≤C
f p ϕp − ϕp
, f ≥0
(3.2)
k
k
k=1
i=1
k
k−1
k=1
e where C and C
e are the best constants in (3.1) and
holds. Moreover, C ≈ C,
(3.2), respectively.
By exploiting Theorem C, we obtain the following statement:
Theorem 3.1. Let 1 < q < p < ∞. Let the entries of the matrix (ai,j ) satisfy
Assumption A. Then inequality (3.1) holds if and only if E = max{E1 , E2 } < ∞,
where
 p−q

q
pq
! p−q
! q(p−1)
p−q
∞
∞
i
X
X X p0
0
ωiq uqi 
,
E1 = 
bi,j ∆ϕpj
ωkq uqk
i=1
j=1

E2 = 
k=i
∞
X
i=1
pq
p−q
∆ϕi
∞
X
p−q
p 
! p−q
pq
aqk,i uqk

k=i
Moreover E ≈ C, where C is the best constant in (3.1).
.
126
ZH. TASPAGANBETOVA, A. TEMIRKHANOVA
Proof. We set sup ai,k = e
ai,j . Obviously,
j≤k≤i
ai,j ≤ e
ai,j .
(3.3)
dai,j ≥ sup ai,k = e
ai,j .
(3.4)
According to (1.7) we have
j≤k≤i
From (3.3) and (3.4) it follows that ai,j ≈ e
ai,j . The matrix operator
=
i
i
P
e
e
ai,j fj , i ≥ 1, is equivalent to the operator A, i.e. (Af )i ≤ Af
≤ d (Af )i or
i
j=1
e
(Af )i ≈ Af
for all f ≥ 0, i ≥ 1. Then inequality (3.1) is equivalent to
e
Af
i
e kq,u ≤ C1 (kf kp,v + kA0 f kp,ρ ) ∀f ≥ 0,
kAf
(3.5)
Moreover, C ≈ C1 , where C and C1 are the best constants in (3.1) and (3.5),
respectively. It is easy to see that the entries of the matrix (e
ai,j ) satisfy the
following condition e
ai,j ≥ e
ai,k , i ≥ k ≥ j ≥ 1. Then according to Theorem C
inequality (3.5) holds if and only if the inequality
!q ! 1q
! p1
∞
∞
k
X
X
X
1−p
0
uqk
e
ak,i fi
≤ C2
fkp ∆ϕpk
∀f ≥ 0,
(3.6)
k=1
i=1
k=1
holds. Moreover, C1 ≈ C2 , where C2 is the best constant in (3.6).
Since (3.5) is equivalent to inequality (3.1), inequality (3.6) is equivalent to
inequality (3.1). By Theorem 2.2 inequality (3.6) (and, thus, (3.5) and (3.1))
holds if and only if E = max{E1 , E2 } < ∞.
Hence, the proof is complete.
Acknowledgement. The authors wish to thank Professor Ryskul Oinarov
for several discussions and for several suggestions, which have improved the final
version of this paper.
References
1. A. Kufner, L. Maligranda and L-E. Persson, The Hardy Inequality. About its History and
Some Related Results, Vydavatelsky Servis Publishing House, Pilsen, 2007.
2. R. Oinarov, A dual inequality of the additive estimate of matrix operator, Proc. Intern.Conf.
”Achievement and development of mathematics in framework of program-Kazakhstan in
third millenium”, (Almaty, October 26-28, 2000), 111–115. (in Russian)
3. R. Oinarov, Boundedness and compactness of integral operators of Volterra type, Sibirskii
Matematicheskii Zhurnal, T.48 (2007), No 5, 1100–1115. (in Russian)
4. R. Oinarov, C.A. Okpoti and L-E. Persson, Weighted inequalities of Hardy type for matrix
operators: the case q < p, Math. Inequal. Appl. 10 (2007), no. 4, 843–861.
5. R. Oinarov and S.Kh. Shalgynbaeva, Weighted additive estimate of a class of matrix operators, Izvestiya NAN RK, serial Phys.-Mat. (1) (2004), 39–49. (in Russian)
CRITERIA ON BOUNDEDNESS OF MATRIX OPERATORS
127
6. R. Oinarov and A. Temirkhanova, Boundedness and compactness of a class of matrix operators: the case p ≤ q, Research report (2008-05), Department of Mathematics, Luleȧ
University of Technology, 1–17.
7. C.A. Okpoti, L-E. Persson and A. Wedestig, Scales of weight characterizations for the discrete Hardy and Carleman type inequalities, In: Proc. Conf. ”Function spaces, Differential
operators and Nonlinear Analysis”, FSDONA 2004 (Milovy, May 28-Jun 2, 2004), Math.
Inst. Acad. Sci. Chech Republic, Prague 2005, 236–258.
8. M. Stieglitz and H. Tietz, Matrixtransformationen von Folgenräumen Eine
Ergebnisübersicht, Math. Z. 154 (1977), 1–16.
1
Eurasian National University, Munaitpasov st., 010008 Astana, Kazakhstan.
E-mail address: zhanara.t.a@mail.ru
2
Eurasian National University, Munaitpasov st., 010008 Astana, Kazakhstan;
Eurasian National University (ENU), Astana, Kazakhstan.
E-mail address: ainura-t@yandex.ru