EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 13, No. 3, 2020, 390-402
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

E-J Summability of Orthogonal Series

F. Aydin Akgun1,∗, B. E. Rhoades2

1 Department of Mathematical Engineering, Yildiz Technical University, 34210 Esenler,
Istanbul, Turkey
2 Department of Mathematics, Indiana University, Bloomington, IN 47405-7106, U.S.A.

Abstract. In this paper we obtain a sufficient condition for the E-J summability of certain orthog-
onal series. Our results generalize the corresponding theorems for ordinary Hausdorff summability
obtained by Kalaivana and Youvaraj.

2020 Mathematics Subject Classifications: 40G05

Key Words and Phrases: E-J matrices, orthogonal series

1. Preliminaries

The set of all real or complex sequences {xn} for which An(x) :=
∑

k ankxk converges is
called the convergence domain of A, written cA, where A is an infinite matrix. A matrix A
is said to be conservative if it maps each convergent sequence into a convergent sequence,
not necessarily with the same limit. If the limit is also preserved, then the matrix is
called regular. Silverman and Toeplitz established necessary and sufficient conditions for
a matrix to be conservative[4]. They are

(i) ‖A‖∞ := supn
∑

k |ank| <∞,
(ii) t := limn

∑
k ank exists,

(iii) ak := limn ank exists for each k.

A Hausdorff matrix H = (hnk) is a lower triangular matrix with nonzero entries

hnk =

(
n

k

)
∆n−kµk,

where {µn} is any real sequence and ∆ is the forward difference operator defied by ∆µk =
µk − µk+1 and ∆n+1µk = ∆(∆nµk). For every Hausdorff matrix each row sum is equal to
µ0[3].

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v13i3.3687

Email addresses: fakgun@yildiz.edu.tr (F. Aydin Akgun), rhoades@indiana.edu (B. E. Rhoades)

https://www.ejpam.com 390 c© 2020 EJPAM All rights reserved.



F. Aydin Akgun, B. E. Rhoades / Eur. J. Pure Appl. Math, 13 (3) (2020), 390-402 391

F. Hausdorff [2] proved that a Hausdorff matrix is conservative if and only if

µn =

∫ 1

0
xndχ(x), (1)

where the mass function χ ∈ BV [0, 1].

The E-J generalized Hausdorff matrices, denoted by Hα
µ = (h

(α)
nk ), were defined inde-

pendently by Endl [1] and Jakimovski [5], with nonzero entries

h
(α)
nk =

(
n+ α

n− k

)
∆n−kµ

(α)
k , 0 ≤ k ≤ n,

for any α ≥ 0. For α = 0, the E-J matrices reduce to the ordinary Hausdorff matrices.

If the µ
(α)
n satisfy the condition

µ(α)n =

∫ 1

0
xn+αdχ(x),

where χ ∈ BV [0, 1], then the corresponding E-J matrix is conservative.

Definition 1. Let γ : [1,∞)→ [0,∞) be a nondecreasing function, A = (ank) an infinite
matrix. Then a series

∑
n bn is said to be |A, γ|k summable, if

∞∑
n=1

γ(n)knk−1|σn − σn−1|k

converges, where σn :
∑

n ankbk.

Definition 2. Let γ := {γn} be a positive sequence, β a real positive number. Then γ
is called quasi- β-power monotone decreasing if there exists a number M = M(β, γ) ≥ 1
such that

nβγ(n) ≤Mmβγ(m)

for each m ≤ n.

For any real number β,Γβ denotes the set of all increasing functions Γβ : [1,∞) →
[0,∞) such that each {γn} is a quasi β-power monotone decreasing sequence.

2. Main Results

Theorem 1. Let {ϕn}∞n=0 ⊂  L2[0, 1] be an orthonormal system, Hα
µ an E-J Hausdorff

matrix with χ monotone decreasing, γ ∈ Γβ for β > 1 − 1/k, 1 ≤ k ≤ 2. Then every
orthogonal series

∑∞
n=0 bnϕn is |Hα, γ| summable.

The following lemmas will be needed in the proof of Theorem 1.



F. Aydin Akgun, B. E. Rhoades / Eur. J. Pure Appl. Math, 13 (3) (2020), 390-402 392

Lemma 1. Let Hα be an E-J matrix with entries (hnk), where χ is a monotonically
increasing mass function on [0, 1] associated with the µn. Then

(i) amn = K

(
n− 1 + α

m− 1 + α

)
ξm+α(1− ξ)n−m for some ξ ∈ (0, 1),

(ii)

n∑
m=0

|amn|2|bm|2 ≤ K2
n∑

m=0

|bm|2 for all bn ∈ C and n ∈ N, where K = χ(1)− χ(0)

and amn =
∑n

k=m |h
(α)
nk − h

(α)
n−1,k|.

Here C = complex numbers and N = natural numbers.
Proof. (i) We consider

h
(α)
nk − h

(α)
n−1,k (2)

=
[ ∫ 1

0
µk+α(1− µ)n−k

(
n+ α

k + α

)
dχ(µ)−

∫ 1

0
µk+α(1− µ)n−1−k

(
n− 1 + α

k + α

)
dχ(µ)

]
=

∫ 1

0
µk+α(1− µ)n−k

[(n+ α

k + α

)
−
(
n− 1 + α

k + α

)
1

1− µ

]
dχ(µ),

where 0 ≤ k ≤ n. Since (
n+ α

k + α

)
−
(
n− 1 + α

k + α

)
=

(
n− 1 + α

k − 1 + α

)
,

from (2),∫ 1

0
µk+α(1− µ)n−1−k

[(
n+ α

k + α

)
(1− µ)−

(
n− 1 + α

k + α

)]
dχ(µ)

=

∫ 1

0
µk+α(1− µ)n−1−k

[(
n+ α

k + α

)
−
(
n− 1 + α

k + α

)
− µ

(
n+ α

k + α

)]
dχ(µ)

=

∫ 1

0
µk+α(1− µ)n−1−k

[(
n− 1 + α

k − 1 + α

)
− µ

(
n+ α

k + α

)]
dχ(µ).

Thus

amn =
n∑

k=m

(h
(α)
nk − h

(α)
n−1,k) =

n∑
k=m

∫ 1

0
µk+α(1− µ)n−1−k

[(
n− 1 + α

k − 1 + α

)
− µ

(
n+ α

k + α

)]
dχ(µ)

=

∫ 1

0

n∑
k=m

µk+α(1− µ)n−1−k
[(
n− 1 + α

k − 1 + α

)
− µ

(
n+ α

k + α

)]
dχ(µ).

From the above inequality,

n∑
k=m

µk+α(1− µ)n−1−k
[(
n− 1 + α

k − 1 + α

)
− µ

(
n+ α

k + α

)]



F. Aydin Akgun, B. E. Rhoades / Eur. J. Pure Appl. Math, 13 (3) (2020), 390-402 393

=

n∑
k=m

µk+α(1− µ)n−1−k
(
n− 1 + α

k − 1 + α

)
−

n∑
k=m

µk+α+1(1− µ)n−1−k
(
n+ α

k + α

)

=
n∑

k=m

µk+α(1− µ)n−1−k
(
n− 1 + α

k − 1 + α

)
−

n∑
k=m

µk+α+1(1− µ)n−1−k
[(
n− 1 + α

k − 1 + α

)
+

(
n− 1 + α

k + α

)]

=
n∑

k=m

µk+α(1− µ)n−k
(
n− 1 + α

k − 1 + α

)
−

n∑
k=m

µk+α+1(1− µ)n−1−k
(
n− 1 + α

k + α

)

=

n∑
k=m

µk+α(1− µ)n−k
(
n− 1 + α

k − 1 + α

)
−

n+1∑
g=m+1

µg+α(1− µ)n−g
(
n− 1 + α

g − 1 + α

)
= µm+α(1− µ)n−m

(
n− 1 + α

m− 1 + α

)
,

and

0 ≤ amn ≤
∫ 1

0
µm+α(1− µ)n−m

(
n− 1 + α

m− 1 + α

)
dχ(µ). (3)

Using the first mean value theorem for integrals, for some 0 < ξ < 1,∫ 1

0
µm+α(1− µ)n−m

(
n− 1 + α

m− 1 + α

)
dχ(µ) = ξm+α(1− ξ)n−m

(
n− 1 + α

m− 1 + α

)∫ 1

0
dχ(µ)

= Kξm+α(1− ξ)n−m
(
n− 1 + α

m− 1 + α

)
,

where 0 < K ≤ 1, and (i) is satisfied.

To prove (ii) we need the following lemma.

Lemma 2. For 0 < K < 1, amn ≤ 1.

Proof. From (3)

amn = K

∫ 1

0
µm+α(1− µ)n−m

(
n− 1 + α

m− 1 + α

)
dµ

= K

(
n− 1 + α

m− 1 + α

)∫ 1

0
µm+α(1− µ)n−mdµ

= K

(
n− 1 + α

m− 1 + α

)
Γ(m+ α+ 1)Γ(n−m+ 1)

Γ(n+ α+ 2)

= K
Γ(n+ α)

Γ(m+ α)Γ(n−m+ 1)

Γ(m+ α+ 1)Γ(n−m+ 1)

Γ(n+ α+ 2)

= K
(m+ α)

(n+ α+ 1)(n+ α)
.

Since n ≥ m,
amn ≤ K < 1. (4)



F. Aydin Akgun, B. E. Rhoades / Eur. J. Pure Appl. Math, 13 (3) (2020), 390-402 394

Using equation (4) we can write

n∑
m=0

|amn|2|bm|2 ≤ K2
n∑

m=0

|bm|2 ≤
n∑

m=0

|bm|2, (5)

which is a proof of (ii).

Lemma 3. Let {ϕn}∞n=0 ⊂ L2[0, 1] be an orthonormal system, Hα
µ be an E-J Hausdorff

matrix with monotonically increasing function χ on [0, 1]. Then, for n ∈ N and

K =

∫ 1

0
dχ(µ),

(i) there exists ξ ∈ (0, 1) such that∫ 1

0
|σn(x)− σn−1(x)|2 = K2

n∑
m=0

ξ2m+2α(1− ξ)2n−2m
(
n− 1 + α

m− 1 + α

)2

|bm|2

and

(ii)

∫ 1

0
|σn(x)− σn−1(x)|2dx = K2

n∑
m=0

|bm|2,

for all bm ∈ C where, for n ∈ N, σn(x) =
∑n

k=0 hnkSk(x), where Sk denotes the kth partial
sum of the orthogonal series

∑∞
m=0 bmϕm.

Proof.

σn(x)− σn−1(x) =
n∑
k=0

(h
(α)
nk − h

(α)
n−1,k)Sk(x)

=
n∑
k=0

(h
(α)
nk − h

(α)
n−1,k)

k∑
m=0

bmϕm

=
n∑

m=0

n∑
k=m

(h
(α)
nk − h

(α)
n−1,k)bmϕm

=
n∑

m=0

amnbmϕm.

Since {ϕn}∞n=0 is an orthonormal system, using Parseval’s identity,∫ 1

0
|σn(x)− σn−1(x)|2dx =

n∑
m=0

|amn|2|bm|2

= K2
n∑

m=0

ξ2m+2α(1− ξ)2n−2m
(
n− 1 + α

m− 1 + α

)2

|bm|2.



F. Aydin Akgun, B. E. Rhoades / Eur. J. Pure Appl. Math, 13 (3) (2020), 390-402 395

From Lemma 2,∫ 1

0
|σn(x)− σn−1(x)|2dx =

n∑
m=0

|amn|2|bm|2 ≤ K2
n∑

m=0

|bm|2.

Proof. To prove Theorem 1, from Definition 1 we need to show that

∞∑
n=1

γ(n)knk−1|σn − σn−1|k

converges for 1 ≤ k < 2, where, for n ∈ N, σn(x)
∑n

k=0 h
(α)
nk Sk(x).

Using Lemma 3, and Hölder’s inequality with p = 2/k, for any 1 ≤ k ≤ 2, and for all
b ∈ `2(Z+), we have

∞∑
n=1

γ(n)knk
∫ n

0
|σn(x)− σn−1(x)|kdx ≤

∞∑
n=1

γ(n)knk−1{K2‖b‖22}k/2

≤ {K‖b‖2}k
∞∑
n=1

γ(n)knk−1.

Here {γ(n)} is a quasi β-power monotone decreasing sequence with β > 1−1/k, and, since
for ε = β − 1 + 1/k, the sequence {nk−1γ(n)k} is quasi kε-power monotone decreasing.
Using Lemma 1 of [6], we have

≤ {K‖b‖2}k
∞∑
n=1

γ(2n)k(2n)k−1

≤ {K‖b‖2}kBγ(2)k(2)k−1,

where B ≥ 1.

Theorem 2. Let {ϕ}∞n=0 ⊂ L2[0, 1] be an orthogonal system and Hα
µ the corresponding

E-J Hausdorff matrix. For 1 ≤ k ≤ 2 and γ ∈ Γ(β) with β > 1 − 1/k, every orthogonal
series

∑∞
n=0 bnϕn is |Hα, γ|k summable.

Proof. Let χ ∈ BV [0, 1] be the mass function corresponding to the E-J matrix Hα. By
the Jordan decomposition theorem, χ = χ1−χ2, where χ1 and χ2 are monotone increasing
functions. To prove the theorem we apply Theorem 1 to χ1 and χ2. Theorems 1 and 2
are generalizations of Theorems 1 and 2, respectively, in [6].

Theorem 3. Let {ϕ}∞n=0 ⊂ L2[0, 1] be an orthogonal system and Hα
µ an E-J Hausdorff

matrix with χ ∈ [0, 1] and monotone increasing. For 1 ≤ k ≤ 2 and γ ∈ Γ(β) with
β > 1 − 1/k, a sufficient condition for the orthogonal series

∑∞
n=0 bnϕn to be |Hα, γ|k

summable is
∞∑
s=0

γ(2s)k


2s+1∑

m=2s+1

√
m+ α|bm|2


k/2

<∞. (6)



F. Aydin Akgun, B. E. Rhoades / Eur. J. Pure Appl. Math, 13 (3) (2020), 390-402 396

Proof. Let χ ∈ BV [0, 1] and monotonically increasing on [0, 1]. By Lemma 1(i) there
exists a ξ ∈ (0, 1) such that∫ 1

0
|σn(x)− σn−1(x)|dx ≤ K2

n∑
m=0

(
n+ α− 1

m+ α− 1

)2

ξ2m+2α(1− ξ)2n−2m|bm|2, (7)

where

σn(x) =
n∑
k=0

h
(α)
nk Sk.

For 1 ≤ k ≤ 2, by using Hölder’s inequality and equation (1),

∞∑
n=2

γ(n)knk−1
{∫ 1

0
|σn(x)− σn−1(x)|dx

}k

≤
∞∑
n=2

γ(n)knk−1

{
K2

n∑
m=0

(
n+ α− 1

m+ α− 1

)2

ξ2m+2α(1− ξ)2n−2m|bm|2
}k/2

. (8)

Replacing ξ by 1/(1 + q) in (8), we obtain

= Kk
∞∑
r=0

2r+1∑
n=2r+1

γ(n)knk−1

{
n∑
n=0

(
n+ α

m+ α

)2(m+ α

n+ α

)2

q2n−2m(1 + q)−2n−2α|bm|2
}k/2

.

(9)

O.A. Ziza [7] proved that, for q > 0, there exists a constant Cq > 0 such that

max
0≤k≤n

(
n

k

)
qk ≤ Cq

(1 + q)n√
n

, n = 1, 2, ... .

We shall generalize this Lemma for E-J matrices.

Lemma 4. For q > 0 there exists a Cq > 0 such that

max
0≤k≤n

(
n+ α

k + α

)
qk+α ≤ Cq

(1 + q)n+α√
n+ α

, n = 1, 2, ...

Proof. (
n+α+1
k+α

)(
n+α
k+α−1

) =
n+ α+ 1

k + α
.

Let

dk =
(n+ α+ 1

k + α
− 1
)
q.

The dn are decreasing in k. Let kn denote the largest value of k + α for which dkn ≥ 1.
Then dkn+1 < 1, and

max
0≤k≤n

(
n+ α

k + α

)
qk+α =

(
n+ α

kn

)
qkn .



F. Aydin Akgun, B. E. Rhoades / Eur. J. Pure Appl. Math, 13 (3) (2020), 390-402 397

It then follows that one can write

kn =
q

1 + q
(n+ α) + νn,

where 0 < νn < 1.
Then(

n+ α

kn

)
≤ C1

(n+ α)!

kn!(n+ α− kn)!
=

(n+ α)n+αe−(n+α)
√
n+ α

(kn)kne−(kn)
√
kn(n+ α− kn)(n+α−kn)e−(n+α−kn)

√
n+ α− kn

.

(10)
With p = q/(1 + q), the right hand side of (10) equals

(n+ α)(n+α)e−(n+α)
√
n+ α

(p(n+ α) + νn)(p(n+α)+νn)e−(p(n+α)+νn)
√

(p(n+ α) + νn)

× 1

(n+ α− p(n+ α)− νn)(n+α−p(n+α)−νn)e−(n+α−p(n+α)−νn)
√

(n+ α− p(n+ α)− νn)

=
(n+ α)p(n+α)+νn

(p(n+ α) + νn)(p(n+α)+νn)
× (n+ α)n+α−p(n+α)−νn

(n+ α− p(n+ α)− νn)n+α−p(n+α)−νn
(11)

×
√
n+ α√

(p(n+ α) + νn)(n+ α− p(n+ α)− νn)
.

Note that

(n+ α)

(p(n+ α) + νn)(n+ α− p(n+ α)− νn)
=

(n+ α)

(n+ α)2(p+ νn
n+α)(1− (p+ νn

n+α))
.

Set a = p + νn/(n + α) and define a function f by f(a) = a(1 − a). Then f(a) has a
minimum value of 1/4 at a = 1/2. Therefore 1/

√
f(a) ≤ 2. From (11)(

n+ α

kn

)
≤ 2

1(
p+ νn

n+α

)p(n+α)+νn × 1(
1− p− νn

n+α

)(1−p)(n+α)−νn × 1√
n+ α

= 2
1

pp(n+α)+νn
(

1 + νn
p(n+α)

)p(n+α)+νn × 1

(1− p)(1−p)(n+α)−νn
(

1− νn
(1−p)(n+α)

)(1−p)(n+α)−νn
× 1√

n+ α
.

Since (1−p)/p = (1/p)−1 and p is a fixed positive constant between 0 and 1, (p/(1−p))−νn
is clearly bounded. So also is (1 + νn/(p(n+ α)))−p(n+α)−νn .

Let g(p) = 1− νn/(1− p)(n+ α). Then

g′(p) =
νn

(1− p)2(n+ α)
,



F. Aydin Akgun, B. E. Rhoades / Eur. J. Pure Appl. Math, 13 (3) (2020), 390-402 398

and g is decreasing in p. Since 0 < p < 1 and fixed, g(p) is bounded, Using the above
facts,

(1− p
p

)−νn
and

( p

1− p

)−νn (
1 +

νn
p(n+ α)

)−(p(n+α))+νn (
1− νn

(1− p)n+ α

)−((1−p)(n+α)+νn)
are bounded. Also,(

n+ α

kn

)
≤ C3

1

pp(n+α)
× 1

(1− p)(1−p)(n+α)
× 1√

n+ α
. (12)

We can write (12) as(
n+ α

kn

)
≤ C3

1
q

1+q
( q
1+q

)(n+α)
× 1

( 1
1+q )

( 1
1+q

)(n+α)
× 1√

n+ α
. (13)

From (13)(
n+ α

kn

)qkn
≤ C3

1

( q
1+q )

( q
1+q

)(n+α)
× 1

( 1
1+q )

( 1
1+q

)(n+α)
× 1√

n+ α
qkn

= C3
1

( q
1+q )

( q
1+q

)(n+α)
× 1

( 1
1+q )

( 1
1+q

)(n+α)
× 1√

n+ α
q
( 1
1+q

)(n+α)+νn

= C3
(1 + q)n+α√

n+ α
qνn ≤ Cq

(1 + q)n+α√
n+ α

.

Thus

max
0≤k≤n

(
n+ α

k + α

)
qk+α ≤ Cq

(1 + q)n+α√
n+ α

.

From (9)

Kk
∞∑
r=0

2r+1∑
n=2r+1

γ(n)knk−1

{
n∑

m=0

(
n+ α

m+ α

)2(m+ α

n+ α

)2

q2n−2m(1 + q)−2n−2α|bm|2
}k/2

= Kk
∞∑
r=0

2r+1∑
n=2r+1

γ(n)knk−1

{
n∑

m=0

(
n+ α

m+ α

)(
m+ α

n+ α

)2(n+ α

n−m

)
qn−mqn−m(1 + q)−2n−2α|bm|2

}k/2
.

(14)
Using Lemma 4, equation (14) can be written as

≤ KkCk/2q

∞∑
r=0

2r+1∑
n=2r+1

γ(n)knk−1

{
n∑

m=0

(
n+ α

m+ α

)(
m+ α

n+ α

)2 1√
n+ α

qn−m(1 + q)−2n−2α|bm|2
}k/2



F. Aydin Akgun, B. E. Rhoades / Eur. J. Pure Appl. Math, 13 (3) (2020), 390-402 399

= KkCk/2q

∞∑
r=0

2r+1∑
n=2r+1

γ(n)knk−1

{
n∑

m=0

(
n+ α

m+ α

)
(m+ α)2

1

(n+ α)5/2
qn−m(1 + q)−2n−2α|bm|2

}k/2
.

(15)
For k = 2, the above inequality becomes

Ω ≤ K2Cq

∞∑
r=0

2r+1∑
n=2r+1

γ(n)2n(n+α)−5/2

{
n∑

m=0

(
n+ α

m+ α

)
(m+ α)2qn−m(1 + q)−2n−2α|bm|2

}5/2

.

(16)

Lemma 5. There exists a Dq > 0 such that

∞∑
n=m

(
n+ α

m+ α

)
qn−m(1 + q)−n−α ≤ Dq

for all m ∈ Z+ and 1 ≤ k ≤ 2.

Proof. The proof of the lemma is easy to verify and it is a generalization of Theorem
B of [6] to E-J matrices. Using Theorem B of [6], we can write (16) as

Ω ≤ K2CqDq

∞∑
r=0

γ(2r+1)22r(2r + α)−5/2
2r+1∑
m=0

(m+ α)2|bm|2

≤ K2CqDq

∞∑
r=0

γ(2r+1)2(2r + α)−3/2
2r+1∑
m=0

(m+ α)2|bm|2, α > 0.

Let p = 2/k. By Hölder’s inequality, for α > 0 and 1 ≤ k ≤ 2,

Ω = KkCk/2q

∞∑
r=0

( 2r+1∑
n=2r+1

γ(n)kqnq(
−k
4
−1)
)1/q{ 2r+1∑

n=2r+1

n∑
m=0

(
n+ α

m+ α

)
(m+α)2qn−m(1+q)−n−α|bm|2

}
.

(17)
Since γ ∈ Γ(β) and, for β ∈ R, by Theorem A in [6], we can write the expression in the
first bracket in (17) as 2r+1∑

n=2r+1

γ(n)kqnq(
−k
4
−1)

1/q

≤ K1/qγ(2r + 1)k(2r + 1)(−
k
4
−1)

≤ K1/qγ(2r + 1)k(2r)(−
k
4
−1).

Thus, from (17),

Ω ≤ Kk+ 1
qCk/2q

∞∑
r=0

γ(2r+1)k(2r)(
−k
4
−1)


2r+1∑

n=2r+1

n∑
m=0

(
n+ α

m+ α

)
(m+ α)2qn−m(1 + q)−n−α|bm|2


k/2

.

(18)



F. Aydin Akgun, B. E. Rhoades / Eur. J. Pure Appl. Math, 13 (3) (2020), 390-402 400

Changing the order of summation inside the brackets in the above inequality, (18) is equal
to

= K
k+ 1

qCk/2q

∞∑
r=0

{
γ(2r+1)kq(2r)q(−

k
4
−1)
} 1

q


2r+1∑
m=0

2r+1∑
n=2r+1

(
n+ α

m+ α

)
(m+ α)2qn−m(1 + q)−n−α|bm|2


k/2

≤ Kk+ 1
qCk/2q

∞∑
r=0

γ(2r+1)k(2r)(−
k
4
−1)


2r+1∑
m=0

2r+1∑
n=2r+1

(
n+ α

m+ α

)
(m+ α)2qn−m(1 + q)−n−α|bm|2


k/2

+K
k+ 1

qCk/2q

∞∑
r=0

γ(2r+1)k(2r)(−
k
4
−1)


2r+1∑

m=2r+1

2r+1∑
n=2r+1

(
n+ α

m+ α

)
(m+ α)2qn−m(1 + q)−n−α|bm|2


k/2

.

Using Lemma 5,

Ω ≤ Kk+ 1
qCk/2q

∞∑
r=0

γ(2r+1)k(2r)(−
k
4
−1)

{
2r+1∑
m=0

Dq(m+ α)2|bm|2
}k/2

+K
k+ 1

qCk/2q

∞∑
r=0

γ(2r+1)k(2r)(−
k
4
−1)


2r+1∑

m=2r+1

Dq(m+ α)2|bm|2

k/2

≤ Kk+ 1
qCk/2q

∞∑
r=0

γ(2r+1)k(2r)(−
k
4
−1)


2r+1∑
m=0

Dq(m+ α)2|bm|2

k/2

≤ Kk+ 1
qCk/2q

∞∑
r=0

γ(2r+1)k(2r)(−
3k
4
)


2r+1∑
m=0

Dq(m+ α)2|bm|2

k/2

≤ Kk+ 1
qCk/2q Dk/2

q

∞∑
r=0

γ(2r+1)k(2r)(−
3k
4
)


2r+1∑
m=0

(m+ α)2|bm|2

k/2

.

For 1 ≤ k ≤ 2,

Ω ≤ L
∞∑
r=0

γ(2r+1)k(2r)(−
3k
4
)
{ 2r+1∑
m=0

(m+ α)2|bm|2
}k/2

, (19)

where L = K
k+ 1

qC
k/2
q D

k/2
q . From (19),

Ω ≤ L
∞∑
r=0

γ(2r+1)k(2r)(−
3k
4
)

α2|b0|2 +
r∑
s=0

2s+1∑
m=2s+1

(m+ α)2|bm|2

k/2



F. Aydin Akgun, B. E. Rhoades / Eur. J. Pure Appl. Math, 13 (3) (2020), 390-402 401

≤ L
∞∑
s=0

∞∑
r=s

γ(2r+1)k(2r)(−
3k
4
)


2s+1∑

m=2s+1

(m+ α)2|bm|2

k/2

+ L
∞∑
r=0

γ(2r+1)k(2r)(−
3k
4
)αk|b0|k.

Here {γ(n)} is quasi β-power monotone decreasing and {n−3/4γ(n)} is quasi ε-power
monotone decreasing, where β > 3/4 and ε = β + 3/4. Thus, by using Lemma 1 of [6],

∞∑
n=m

γ(2n)k(2n)−3k/4 ≤Mγ(2m)k(2m)−3k/4,M ∈ Z+.

Therefore

Ω ≤ L
∞∑
s=0

γ(2s+1)k(2s)(−3k/4)


2s+1∑

m=2s+1

(m+ α)2|bm|2

k/2

+ Lγ(2)kαk|b0|k

≤ 2kL
∞∑
s=0

γ(2s)k(2s)(−3k/4)


2s+1∑

m=2s+1

(m+ α)2|bm|2

k/2

+ Lγ(2)kαk|b0|k

≤ 2kL
∞∑
s=0

γ(2s)k


2s+1∑

m=2s+1

(m+ α)2|bm|2

k/2

+ Lγ(2)kαk|b0|k

and thus
∞∑
n=2

γ(n)knk−1
∫ 1

0
|σn(x)− σn−1(x)|kdx

≤ 2(−
3k
4
)L
∞∑
s=0

γ(2s)k


2s+1∑

m=2s+1

(m+ α)2|bm|2

k/2

+ Lγ(2)kαk|b0|k.

The following Corollaries can be verified by taking α = 0 in the above theorems.

Corollary 1. Every orthogonal series
∑∞

n=0 cnψn, cn ∈ `2(Z+) is |H,ψ|k summable for
1 ≤ k ≤ 2 and γ ∈ Γβ with β > 1 − l/k, where {ψn}∞n=0 ⊂ L2[0, 1] and H is a Hausdorff
matrix with entries (hnk)n,k ∈ Z+.

This is Theorem 2 of [6].

Corollary 2. Let 1 ≤ k ≤ 2 and γ ∈ Γβ with β > −3/4, where {φn}∞n=0 ⊂ L2[0, 1] and H
is a Hausdorff matrix. Then, for any cn ∈ `2(Z+), a sufficient condition for the orthogonal
series

∑∞
n=0 cnψn to be |H, γ|k summable is

∞∑
m=0

γ(2m)k
{ 2m+1∑
n=2m+1

√
n|cn|2

}k/2
<∞. (20)

This includes the results of Theorem 3 of [6].



REFERENCES 402

References

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[3] F Hausdorff. Summationmethoden und momentfolgen ii.. Math. Z., 9:280–299, 1921.

[4] W A Hurwitz and L L Silverman. On the consistency and equivalence of certain
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[5] A Jakimovski. The product of summability methods; new classes of transformations
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[6] K Kalaivana and G P Youvaraj. Generalized absolute Hausdorff summability of or-
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[7] O A Ziza. On the summation of orthogonal series by Euler’s method. Matem. Sb.,
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