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Mathematical and Computer Modelling
journal homepage: www.elsevier.com/locate/mcm
An absolute double summability factor theorem
Ekrem Savaş
Department of Mathematics, Istanbul Ticaret University, Üsküdar, Istanbul, Turkey
a r t i c l e i n f o
Article history:
Received 27 February 2008 Accepted 26 June 2008
Keywords:
Absolute summability Double summability Double weighed mean Summability factors
a b s t r a c t
In an earlier paper Savas and Rhoades [Ekrem Savaş, B.E. Rhoades, A note on | A |
ksummability factors, Nonlinear Anal. 66 (2007) 1879–1883. [1]] obtained a summability factor theorem for absolute summability of order k ≥ 1. In this paper we extend that result to doubly infinite matrices.
© 2008 Elsevier Ltd. All rights reserved.
A doubly infinite matrix A = ( a
mnjk) is said to be doubly triangular if a
mnjk= 0 for j > m or k > n. The ( mn ) th term of the A-transform of a double sequence { s
mn} is defined by
T
mn:=
n
X
µ=
0 nX
ν=
0a
mnµν s µν .
For any double sequence u
mn, ∆
11is defined by
∆
11u
mn= u
mn− u
m+
1,
n− u
m,
n+
1+ u
m+
1,
n+
1. For any four-fold sequence v
mnij,
∆
11v
mnij:= v
mnij− v
m+
1,
n,
i,
j, − v
m,
n+
1,
i,
j+ v
m+
1,
n+
1,
i,
j∆
ijv
mnij:= v
mnij− v
m,
n,
i+
1,
j− v
m,
n,
i,
j+
1+ v
m,
n,
i+
1,
j+
1,
∆
0jv
mnij:= v
mnij− v
m,
n,
i,
j+
1, and (1)
∆
i 0v
mnij:= v
mnij− v
m,
n,
i+
1,
j. A series P P
b
mn, with partial sums s
mnis said to be summable | A |
k, k ≥ 1 if
∞
X
m
=
1∞
X
n
=
1( mn )
k−
1| ∆
11T
m−
1,
n−
1|
k< ∞. (2)
We may associate with A two doubly triangular matrices A and A as follows: ˆ
¯ a
mnij=
m
X
µ=
i nX
ν=
ja
mnµν , n , m = 0 , 1 , 2 , . . . ,
and
ˆ a
m,
n,
i,
j= ∆
11a ¯
m−
1,
n−
1,
i,
j, m , n = 0 , 1 , 2 , 3 , . . . . (3)
E-mail addresses:ekremsavas@yahoo.com,esavas@iticu.edu.tr.
0895-7177/$ – see front matter
©
2008 Elsevier Ltd. All rights reserved.doi:10.1016/j.mcm.2008.06.015
Note that a ˆ
0000= ¯ a
0000= a
0000.
Let y
mndenote the ( mn ) th term of the A-transform of P
mµ=
0P
nν=
0b µν λ µν . We can write y
mn=
m
X
µ=
0 nX
ν=
0a
mnµν µ
X
i
=
0X ν
j=
0b
ijλ
ij=
m
X
i
=
0 nX
j
=
0b
ijλ
ij mX
µ=
i nX
ν=
ja
mnµν
=
m
X
i
=
0 nX
j
=
0b
ijλ
ija ¯
mnij.
It then follows that
∆
11y
m−
1,
n−
1= y
m−
1,
n−
1− y
m,
n−
1− y
m−
1,
n+ y
mn=
m
−
1X
i
=
0 n−
1X
j
=
0b
ijλ
ija ¯
m−
1,
n−
1,
i,
j−
m
X
i
=
0 n−
1X
j
=
0b
ijλ
ija ¯
m,
n−
1,
i,
j−
m
−
1X
i
=
0 nX
j
=
0b
ijλ
ija ¯
m−
1,
n,
i,
j+
m
X
i
=
0 nX
j
=
0b
ijλ
ija ¯
mnij=
m
X
i
=
0 nX
j
=
0b
ijλ
ija ˆ
mnij, as in [2].
Since
a ¯
m−
1,
n−
1,
m,
j= ¯ a
m−
1,
n−
1,
i,
n= ¯ a
m,
n−
1,
i,
n= ¯ a
m−
1,
n,
m,
n= 0 . But b
mn= s
m−
1,
n−
1− s
m−
1,
n− s
m,
n−
1+ s
mn, so
∆
11y
m−
1,
n−
1=
m
X
i
=
0 nX
j
=
0a ˆ
mnijλ
ij( s
i−
1,
j−
1− s
i−
1,
j− s
i,
j−
1+ s
ij)
=
m
−
1X
i
=
0 n−
1X
j
=
0a ˆ
m,
n,
i+
1,
j+
1λ
i+
1.
j+
1s
ij−
m
−
1X
i
=
0 nX
j
=
0ˆ a
m,
n,
i+
1,
jλ
i+
1,
js
ij−
m
X
i
=
0 n−
1X
j
=
0a ˆ
m,
n,
i,
j+
1λ
i,
j+
1s
ij+
m
X
i
=
0 nX
j
=
0a ˆ
mnijλ
ijs
ij=
m
−
1X
i
=
0 n−
1X
j
=
0∆
ij(ˆ a
mnijλ
ij) s
ij−
m
−
1X
i
=
0a ˆ
m,
n,
i+
1,
nλ
i+
1,
ns
in−
n
−
1X
j
=
0a ˆ
m,
n,
m,
j+
1λ
m,
j+
1,
n+
1s
mj+
n
X
j
=
0a ˆ
mnmjλ
mjs
mj+
m
−
1X
i
=
0a ˆ
mninλ
ins
in=
m
−
1X
i
=
0 n−
1X
j
=
0∆
ij(ˆ a
mnijλ
ij) s
ij+
m
−
1X
i
=
0( ∆
i0a ˆ
mninλ
in) s
in+
n
−
1X
j
=
0( ∆
0ja ˆ
mnmjλ
mj) s
mj+ ˆ a
mnmnλ
mns
mn. (4)
Also we may write
∆
i 0a ˆ
mninλ
in= λ
in∆
i 0a ˆ
mnin+ ˆ a
m,
n,
i+
1,
n∆
i 0λ
in, and
∆
0ja ˆ
mnmjλ
mj= λ
mj∆
0ja ˆ
mnmj+ ˆ a
m,
n,
m,
j+
1∆
0jλ
mj, so that
m
−
1X
i
=
0( ∆
i0ˆ a
mninλ
in) s
in+
n
−
1X
j
=
0( ∆
0ja ˆ
mnmjλ
mj) s
mj=
m
−
1X
i
=
0[ λ
in∆
i0a ˆ
mnin+ ˆ a
m,
n,
i+
1,
n∆
i0λ
in] s
in+
n
−
1X
j
=
0[ λ
mj∆
0jˆ a
mnmj+ ˆ a
m,
n,
m,
j+
1∆
0jλ
mj] s
mj. (5)
We shall need the following lemma.
Lemma 1. Let { u
ij} , {v
ij} be two double sequences. Then
∆
ij( u
ijv
ij) = v
ij∆
iju
ij+ ( ∆
0ju
i+
1,
j)( ∆
i 0v
ij) + ( ∆
i0u
i,
j+
1)( ∆
0jv
ij) + u
i+
1,
j+
1∆
ijv
ij. (6) Proof. Just expand the right-hand side of (5).
Theorem 1. Let A be a doubly triangular matrix with nonnegative entries satisfying (i) ∆
11a
m−
1,
n−
1,
i,
j≥ 0,
(ii) P
nν=
0a
mniν = P
n−
1ν=
0a
m,
n−
1,
i,ν = b ( m , i ), P
mµ=
0a
mnµ
j= P
m−
1µ=
0a
m−
1,
n,µ,
j= a ( n , j ), (iii) mna
mnmn= O ( 1 ) ,
(iv) a
mnij≥ max { a
m,
n+
1,
i,
j, a
m+
1,
n,
i,
j} for m ≥ i , n ≥ j, and i , j = 0 , 1 , . . . , (v) P
mi
=
0P
nj
=
0a
mnij= O ( 1 ) .
If the sequence { s
mn} is bounded and sequence { λ
mn} is a double sequence such that (vi) P ∞
i
=
1P ∞
j
=
1a
mnij| λ
ij|
k< ∞ , (vii) P
m−
1i
=
0P
n−
1j
=
0| ∆
0jλ
ij| = O ( 1 ) , (viii) P ∞
i
=
0P ∞
j
=
0| ∆
i0λ
ij| < ∞ , (ix) P
m−
1i
=
0P
n−
1j
=
0| ∆
ijλ
ij| = O ( 1 ) , and (x) P
mm
=
1P
nn
=
1| λ
mn| = O ( 1 ) . Then the series P P
b
mnλ
mnis summable | A |
k, k ≥ 1.
Proof. In order to prove the theorem it is necessary, from (2), to show that
∞
X
m
=
1∞
X
n
=
1( mn )
k−
1| ∆
11y
mn|
k< ∞.
From (6),
∆
ij(ˆ a
mnijλ
ij) = λ
ij∆
ij(ˆ a
mnij) + ( ∆
0ja ˆ
m,
n,
i+
1,
j)( ∆
i0λ
ij) + ( ∆
i 0a ˆ
m,
n,
i,
j+
1)( ∆
0jλ
ij) + ˆ a
m,
n,
i+
1,
j+
1∆
ijλ
ij. (7) Using (7),
m
−
1X
i
=
0 n−
1X
j
=
0∆
ij(ˆ a
mnijλ
ij) s
ij=
m
−
1X
i
=
0 n−
1X
j
=
0λ
ij( ∆
ija ˆ
mnij) + ( ∆
0ja ˆ
m,
n,
i+
1,
j)( ∆
i0λ
ij)
+ ( ∆
i 0a ˆ
m,
n,
i,
j+
1)( ∆
0jλ
ij) + ˆ a
m,
n,
i+
1,
j+
1( ∆
ijλ
ij) s
ij. (8) Therefore, using (4), (5) and (8), we write
∆
11y
m−
1,
n−
1=
9
X
r
=
1T
mnr, say .
From Minkowski’s inequality, it is sufficient to show that
∞
X
m
=
1∞
X
n
=
1( mn )
k−
1| T
mnr|
k< ∞, for r = 1 , 2 , . . . , 9 . Using Hólder’s inequality,
I
1:=
M
+
1X
m
=
1 N+
1X
n
=
1( mn )
k−
1| T
mn1|
k= O ( 1 )
M
+
1X
m
=
1 N+
1X
n
=
1( mn )
k−
1m
−
1X
i
=
0 n−
1X
j
=
0| ( ∆
ija ˆ
mnij)||λ
ij|| s
ij|
!
k= O ( 1 )
M
+
1X
m
=
1 N+
1X
n
=
1( mn )
k−
1m
−
1X
i
=
0 n−
1X
j
=
0| ( ∆
ija ˆ
mnij)||λ
ij|
k| s
ij|
k!
m−
1X
i
=
0 n−
1X
j
=
0| ( ∆
ija ˆ
mnij)|
!
k−
1.
From (3),
a ˆ
mnij= ∆
11¯ a
m−
1,
n−
1,
i,
j= ¯ a
m−
1,
n−
1,
i,
j− ¯ a
m,
n−
1,
i,
j− ¯ a
m−
1,
n,
i,
j+ ¯ a
mnij=
m
−
1X
µ=
i n−
1X
ν=
ja
m−
1,
n−
1,
i,
j−
m
X
µ=
i n−
1X
ν=
ja
m,
n−
1,
i,
j=
m
−
1X
µ=
i nX
ν=
ja
m−
1,
n,
i,
j+
m
X
µ=
i nX
ν=
ja
mnij,
since a
m−
1,
n,
m,ν = a
m,
n−
1,µ,
n= 0.
Using (1) and property (ii),
a ˆ
mnij=
m
X
µ=
i nX
ν=
j( a
m−
1,
n−
1,µ,ν − a
m,
n−
1,µ,ν − a
m−
1,
n,µ,ν + a
m,
n,µ,ν )
=
m
−
1X
µ=
i"
b ( m − 1 , µ) −
j
−
1X
ν=
0a
m−
1,
n−
1,µ,ν − b ( m , µ) +
j
−
1X
ν=
0a
m,
n−
1,µ,ν
− b ( m − 1 , µ) +
j
−
1X
ν=
0a
m−
1,
n,µ,ν + b ( m , µ) −
j
−
1X
ν=
0a
m,
n,µ,ν
#
=
i
−
1X
µ=
0 j−
1X
ν=
0∆
11a
m−
1,
n−
1,µ,ν ≥ 0 , (9)
as in [2]. Using (1) and (9),
∆
ija ˆ
mnij=
i
−
1X
µ=
0 j−
1X
ν=
0−
i
X
µ=
0 j−
1X
ν=
0−
i
−
1X
µ=
0 jX
ν=
0+
i
X
µ=
0 jX
ν=
0!
∆
11a
m−
1,
n−
1,µ,ν
= −
j
−
1X
ν=
0∆
11a
m−
1,
n−
1,
i,ν +
j
X
ν=
0∆
11a
m−
1,
n−
1,
i,ν
= ∆
11a
m−
1,
n−
1,
i,
j≥ 0 , (10)
and, from property (ii),
m
−
1X
i
=
0 n−
1X
j
=
0∆
ija ˆ
mnij=
m
−
1X
i
=
0 n−
1X
n
=
0a
m−
1,
n−
1,
i,
j− a
m,
n−
1,
i,
j− a
m−
1,
n,
i,
j+ a
mnij=
m
−
1X
i
=
0b ( m − 1 , i ) − b ( m , i ) − b ( m − 1 , i ) + a
m−
1,
n,
i,
n+ b ( m , i ) − a
mnin=
m
−
1X
i
=
0( a
m−
1,
n,
i,
n− a
mnin)
= a ( n , n ) − a ( n , n ) + a
mnmn. Using condition (iii), and since { s
mn} is bounded
I
1= O ( 1 )
M
+
1X
m
=
1 N+
1X
n
=
1( mna
mnmn)
k−
1m
−
1X
i
=
0 n−
1X
j
=
0| ∆
ija ˆ
mnij|| λ
ij|
k= O ( 1 )
m
X
i
=
0 NX
j
=
0| λ
ij|
kM
+
1X
m
=
i+
1 N+
1X
n
=
j+
1| ∆
ijˆ a
mnij| .
Using (10) and property (vi),
0 ≤
M
+
1X
m
=
i+
1 N+
1X
n
=
j+
1| ∆
ija ˆ
m,
n,
i,
j|
=
M
+
1X
m
=
i+
1 N+
1X
n
=
j+
1( a
m−
1,
n−
1,
i,
j− a
m,
n−
1,
i,
j− a
m−
1,
n,
i,
j+ a
mnij)
=
M
+
1X
m
=
i+
1( a
m−
1,
j,
i,
j− a
m−
1,
N+
1,
i,
j− a
mjij+ a
m,
N+
1,
i,
j)
= a
ijij− a
M+
1,
j,
i,
j− a
i,
N+
1,
i,
j+ a
M+
1,
N+
1,
i,
j≤ a
ijij. (11)
Finally, using property (vi),
I
1= O ( 1 )
M
X
i
=
0 NX
j
=
0a
ijij| λ
ij|
k= O ( 1 ).
Again, using Hölder’s inequality,
I
2:=
M
+
1X
m
−
1 N+
1X
n
=
1( mn )
k−
1| T
mn2|
k=
M
+
1X
m
−
1 N+
1X
n
=
1( mn )
k−
1m
−
1X
i
=
0 n−
1X
j
=
0( ∆
0ja ˆ
m,
n,
i+
1,
j)( ∆
i 0λ
ij) s
ijk
= O ( 1 )
M+
1X
m
−
1 N+
1X
n
=
1( mn )
k−
1"
m−
1X
i
=
0 n−
1X
j
=
0| ∆
0jˆ a
m,
n,
i+
1,
jk ∆
i 0λ
ijk s
ij|
k# "
m−
1X
i
=
0 n−
1X
j
=
0| ∆
0ja ˆ
m,
n,
i+
1,
j|| ∆
i 0λ
ij|
#
k−
1.
Using (9) and condition (ii),
0 ≤ ˆ a
m,
n,
i+
1,
j=
i
X
µ=
0 j−
1X
ν=
0∆
11a
m−
1,
n−
1,µ,ν
≤
m
−
1X
µ=
0 n−
1X
ν=
0( a
m−
1,
n−
1,µ,ν − a
m,
n−
1,µ,ν − a
m−
1,
n,µ,ν + a
m,
n,µ,ν )
=
m
−
1X
µ=
0b ( m − 1 , µ) − b ( m , µ) − b ( m − 1 , µ) + a
m−
1,
n,µ,
n+ b ( m , µ) − a
mnµν
=
m
−
1X
µ=
0( a
m−
1,
n,µ,
n− a
mnµν )
= a ( n , n ) − a ( n , n ) + a
mnmn. Since
| ∆
0ja ˆ
m,
n,
i+
1,
j| ≤ ˆ a
m,
n,
i+
1,
j+ ˆ a
m,
n,
i+
1,
j+
1, using conditions (iii) and (viii),
I
2= O ( 1 )
M
+
1X
m
=
1 N+
1X
n
=
1( mna
mnmn)
k−
1m
−
1X
i
=
0 n−
1X
j
=
0| ∆
0ja ˆ
m,
n,
i+
1,
jk ∆
i 0λ
ijk s
ij|
k= O ( 1 )
M
X
i
=
0 NX
j
=
0| ∆
i 0λ
ij|| s
ij|
kM
+
1X
m
=
i+
1 N+
1X
n
=
j+
1| ∆
0ja ˆ
m,
n,
i+
1,
j| .
From (9),
M
+
1X
m
=
i+
1 N+
1X
n
=
j+
1| ∆
0ja ˆ
m,
n,
i+
1,
j| = O ( 1 )
M
+
1X
m
=
i+
1 N+
1X
n
=
j+
1(ˆ a
m,
n,
i+
1,
j+ ˆ a
m,
n,
i+
1,
j+
1)
= O ( 1 )
M+
1X
m
=
i+
1 N+
1X
n
=
j+
1 iX
µ=
0 j−
1X
ν=
0∆
11a
m−
1,
n−
1,µ,ν +
i
X
µ=
0 jX
ν=
0∆
11a
m−
1,
n−
1,µ,ν
! .
Using conditions (i), (ii), and (v),
M
+
1X
m
=
i+
1 N+
1X
n
=
j+
1 iX
µ=
0 j−
1X
ν=
0∆
11a
m−
1,
n−
1,µ,ν =
i
X
µ=
0 j−
1X
ν=
0 M+
1X
m
=
i+
1 N+
1X
n
=
j+
1a
m−
1,
n−
1,µ,ν − a
m,
n−
1,µ,ν − a
m−
1,
n,µ,ν + a
m,
n,µ,ν
=
i
X
µ=
0 j−
1X
ν=
0 M+
1X
m
=
i+
1a
m−
1,
j,µ,ν − a
m−
1,
N+
1,µ,ν − a
m,
j,µ,ν + a
m,
N+
1,µ,ν
=
i
X
µ=
0 j−
1X
ν=
0a
ijµν − a
M+
1,
j,µ,ν − a
i,
N+
1,µ,ν + a
M+
1,
N+
1,µ,ν
=
i
X
µ=
0"
b ( i , µ) − a
ijµ
j− b ( M + 1 , µ) + a
M+
1,
j,µ,
j− b ( i , µ) +
N
+
1X
ν=
ja
i,
N+
1,µ,ν + b ( M + 1 , µ) −
N
+
1X
ν=
ja
M+
1,
N+
1,µ,ν
#
=
i
X
µ=
0− a
ijµ
j+ a
M+
1.
j.µ,
j+
N
+
1X
ν=
ja
i,
N+
1,µ,ν − a
M+
1,
N+
1,µ,ν
!
= − a ( j , j ) + a ( j , j ) −
M
+
1X
µ=
i+
1a
M+
1,
j,µ,
j+
N
+
1X
ν=
ja ( N + 1 , ν) − a ( N + 1 , ν) +
M
+
1X
µ=
i+
1a
M+
1,
N+
1,µ,ν
!
≤
M
+
1X
µ=
i+
1 N+
1X
ν=
ja
M+
1,
N+
1,µ,ν = O ( 1 ).
Similarly,
M
+
1X
m
=
i+
1 N+
1X
n
=
j+
1 iX
µ=
0 jX
ν=
0∆
11a
m−
1,
n−
1,µ,ν = O ( 1 ), (12) and hence I
2= O ( 1 ) by condition (viii).
Using condition (viii), it can be shown that I
3= O ( 1 ) . Using Hölder’s inequality,
I
4:=
M
+
1X
m
=
1 N+
1X
n
=
1( mn )
k−
1| T
mn4|
k= O ( 1 )
M+
1X
m
=
1 N+
1X
n
=
1( mn )
k−
1 m−
1X
i
=
0 n−
1X
j
=
0|ˆ a
m,
n,
i+
1,
j+
1|| ∆
ijλ
ij| s
ij!
k= O ( 1 )
M+
1X
m
=
1 N+
1X
n
=
1( mn )
k−
1"
m−
1X
i
=
0 n−
1X
j
=
0|ˆ a
m,
n,
i+
1,
j+
1k ∆
ijλ
ijk s
ij|
k# "
m−
1X
i
=
0 n−
1X
j
=
0|ˆ a
m,
n,
i+
1,
j+
1|| ∆
ijλ
ij|
#
k−
1. Again from (9) and condition (ii),
0 ≤ ˆ a
m,
n,
i+
1,
j+
1=
i
X
µ=
0 jX
ν=
0∆
11a
m−
1,
n−
1,µ,ν
≤
m
−
1X
µ=
0 n−
1X
ν=
0( a
m−
1,
n−
1,µ,ν − a
m,
n−
1,µ,ν a
m−
1,
n,µ,ν + a
mnµν )
=
m
−
1X
µ=
0b ( m − 1 , µ) − b ( m , µ) − b ( m − 1 , µ) + a
m−
1,
n,µ,
n+ b ( m , µ) − a
mnµ
n=
m
−
1X
µ=
0( a
m−
1,
n,µ,
n− a
mnµ
n)
= a ( n , n ) − a ( n , n ) + a
mnmn. Using properties (iii), (ii), and (ix),
I
4= O ( 1 )
M
+
1X
m
=
1 N+
1X
n
=
1( mna
mnmn)
k−
1"
m−
1X
i
=
0 n−
1X
j
=
0|ˆ a
m,
n,
i+
1,
j+
1k ∆
ijλ
ijk s
ij|
k# "
m−
1X
i
=
0 n−
1X
j
=
0| ∆
ijλ
ij|
#
k−
1= O ( 1 )
M
+
1X
m
=
1 N+
1X
n
=
1"
m−
1X
i
=
0 n−
1X
j
=
0|ˆ a
m,
n,
i+
1,
j+
1|| ∆
ijλ
ij|
#
= O ( 1 )
N
X
i
=
0 NX
j
=
0| ∆
ijλ
ij| X
ijM
+
1X
m
=
i+
1 N+
1X
n
=
j+
1|ˆ a
m,
n,
i+
1,
j+
1|
= O ( 1 )
m
−
1X
i
=
0 NX
j
=
0| ∆
ijλ
ij| = O ( 1 ).
Using (5) and Hölder’s inequality,
I
5:=
M
+
1X
m
=
1 N+
1X
n
=
1( mn )
k−
1| T
mn5|
k=
M
+
1X
m
=
1 N+
1X
n
=
1( mn )
k−
1m
−
1X
i
=
0λ
in∆
i 0ˆ a
mnins
ink
= O ( 1 )
M
+
1X
m
=
1 N+
1X
n
=
1( mn )
k−
1m
−
1X
i
=
0| λ
in|| ∆
i0a ˆ
mnin| s
in!
k= O ( 1 )
M
+
1X
m
=
1 N+
1X
n
=
1( mn )
k−
1"
m−
1X
i
=
0| ∆
i0a ˆ
mnin| (|λ
ins
in| )
k# "
m−
1X
i
=
0| ∆
i0a ˆ
mnin|
#
k−
1. From (3),
∆
i0a ˆ
mnin= ∆
i 0∆
11¯ a
m−
1,
n−
1,
i,
n= ∆
i 0(¯ a
m−
1,
n−
1,
i,
n− ¯ a
m,
n−
1,
i,
n− ¯ a
m−
1,
n,
i,
n+ ¯ a
mnin)
= ∆
i 0−
m
−
1X
µ=
ia
m−
1,
n,ν,
n+
m
X
µ=
ia
mnµ
n!
= − a
m−
1,
n,
i,
n+ a
mnin≤ 0 . (13)
Using property (ii),
m
−
1X
i
=
0| ∆
i0a ˆ
mnin| =
m
−
1X
i
=
0( a
m−
1,
n−
1,
i,
n− a
mnin)
= a ( n , n ) − a ( n , n ) + a
mnmn. Using property (iii),
I
5= O ( 1 )
M
+
1X
m
=
1 N+
1X
n
=
1( mna
mnmn)
k−
1"
m−
1X
i
=
0| ∆
i0a ˆ
mnin|| λ
in|
k#
= O ( 1 )
M
+
1X
m
=
1 N+
1X
n
=
1 m−
1X
i
=
0| ∆
i0a ˆ
mnin|| λ
in|
k= O ( 1 )
N
+
1X
n
=
1 MX
i
=
0| λ
in|
kM
+
1X
m
=
i+
1| ∆
i 0a ˆ
mnin| .
From (13),
M
+
1X
m
=
i+
1| ∆
i 0a ˆ
mnin| =
M
+
1X
m
=
i+
1( a
m−
1,
n,
i,
n− a
mnin)
= a
inin− a
M+
1,
n,
i,
n≤ a
inin. Therefore, by property (vi), I
5= O ( 1 ) . Using Hölder’s inequality
I
6:=
M
+
1X
m
=
1 N+
1X
n
=
1( mn )
k−
1| T
mn6|
k=
M
+
1X
m
=
1 N+
1X
n
=
1( mn )
k−
1m
−
1X
i
=
0a ˆ
m,
n,
i+
1,
n( ∆
i 0λ
i n) s
ink
= O ( 1 )
M
+
1X
m
=
1 N+
1X
n
=
1( mn )
k−
1m
−
1X
i
=
0|ˆ a
m,
n,
i+
1,
n|| ( ∆
i 0λ
in)| s
in!
k= O ( 1 )
M+
1X
m
=
1 N+
1X
n
=
1( mn )
k−
1"
m−
1X
i
=
0|ˆ a
m,
n,
i+
1,
nk ∆
i 0λ
ink s
in|
k# "
m−
1X
i
=
0|ˆ a
m,
n,
i+
1,
n|| ∆
i 0λ
in|
#
k−
1.
Using (3) and condition (ii),
a ˆ
m,
n,
i+
1,
n= ¯ a
m−
1,
n−
1,
i+
1,
n− ¯ a
m,
n−
1,
i+
1,
n− ¯ a
m−
1,
n,
i+
1,
n+ ¯ a
m,
n,
i+
1,
n= −
m
−
1X
µ=
i+
1a
m−
1,
n,µ,
n+
m
X
µ=
i+
1a
m,
n,µ,
n= − a ( n , n ) +
i
X
µ=
0a
m−
1,
n,µ,
n+ a ( n , n ) −
i
X
µ=
0a
m,
n,µ,
n≥ 0
≤
m
−
1X
µ=
0( a
m−
1,
n,µ,
n− a
m,
n,µ,
n)
= a ( n , n ) − a ( n , n ) + a
mnmn. (14)
Thus, using conditions (vii) and (iii),
I
6= O ( 1 )
M
+
1X
m
=
1 N+
1X
n
=
1( mna
mnmn)
k−
1"
m−
1X
i
=
0|ˆ a
m,
n,
i+
1,
nk ∆
i 0λ
ink s
in|
k# "
m−
1X
i
=
0| ∆
i 0λ
in|
#
= O ( 1 )
M
+
1X
m
=
1 N+
1X
n
=
1"
m−
1X
i
=
0|ˆ a
m,
n,
i+
1,
n|| ∆
i 0λ
in|
#
= O ( 1 )
M
X
i
=
0 N+
1X
n
=
1| ∆
i 0λ
in|
M
+
1X
m
=
i+
1|ˆ a
m,
n,
i+
1,
n| .
Again using (9) and condition (ii),
M
+
1X
m
=
i+
1|ˆ a
m,
n,
i+
1,
n| =
M
+
1X
m
=
i+
1 iX
µ=
0( a
m−
1,
n,µ,
n− a
m,
n,µ,
n)
=
i
X
µ=
0 M+
1X
m
=
i+
1( a
m−
1,
n,µ,
n− a
m,
n,µ,
n)
=
i
