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Some further extensions of absolute Cesàro summability for double series

Abstract

In a recent paper [Savaş and Rhoades in Appl. Math. Lett. 22:1462-1466, 2009], the authors extended the result of Flett [Proc. Lond. Math. Soc. 7:113-141, 1957] to double summability. In this paper, we consider some further extensions of absolute Cesàro summability for double series.

MSC:40F05, 40G05.

Let m = 0 n = 0 a m n be an infinite double series with real or complex numbers, with partial sums

s m n = i = 0 m j = 0 n a i j .

For any double sequence ( x m n ) we shall define

Δ 11 x m n = x m n x m + 1 , n x m , n + 1 + x m + 1 , n + 1 .

Denote by A k 2 the sequence space defined by

A k 2 = { ( s m n ) m , n = 0 : m = 1 n = 1 ( m n ) k 1 | a m n | k < ; a m n = Δ 11 s m 1 , n 1 }

for k1.

A four-dimensional matrix T=( t m n i j :m,n,i,j=0,1,) is said to be absolutely k th power conservative for k1, if TB( A k 2 ), i.e., if

m = 1 n = 1 ( m n ) k 1 | Δ 11 s m 1 , n 1 | k <,

then

m = 1 n = 1 ( m n ) k 1 | Δ 11 t m 1 , n 1 | k <,

where

t m n = i = 0 j = 0 t m n i j s i j (m,n=0,1,).

A double infinite Cesáro matrix C(α,β) is a double infinite Hausdorff matrix with entries

h m n i j = E m i α 1 E n j β 1 E m α E n β ,

where

E m α = ( m + α m ) = Γ ( α + m + 1 ) Γ ( m + 1 ) Γ ( α + 1 ) m α Γ ( α + 1 ) .

The series a m n is said to be summable |C(α,β) | k , k1, α,β>1, if (see [1])

m = 1 n = 1 ( m n ) k 1 | Δ 11 σ m 1 , n 1 α β | k <,
(1)

where σ m n α β denotes the mn-term of the C(α,β) transform of a sequence ( s m n ); i.e.,

σ m n α β = 1 E m α E n β i = 0 m j = 0 n E m i α 1 E n j β 1 s i j .
(2)

Quite recently, Savaş and Rhoades [2] extended the result of Flett [3] to double summability. Their theorem is as follows.

Theorem 1 Let αγ>1, βδ>1, and m n a m n be a double series with partial sums s m n . If m n a m n is |C(γ,δ) | k -summable, then it is also |C(α,β) | k -summable, k1.

It then follows that if one sets γ=δ=0, then C(α,β)B( A k 2 ) for each α,β0. In this paper, we consider some further extensions of absolute Cesàro summability for double series.

We shall use the following lemmas.

Lemma 1 If θ>1, ϕ>1, θφ>0 and ϕψ>0, then

m = i n = j E m i φ E n j ψ m n E m θ E n ϕ = 1 i j E i θ φ 1 E j ϕ ψ 1 .
(3)

Proof For α>1, n1 since

1 E n α = 0 1 ( 1 x ) α x n 1 dx

and

( 1 x ) α = n = 0 E n α 1 x n ,

we obtain

m = i n = j E m i φ E n j ψ m n E m θ E n ϕ = m = i E m i φ m E m θ n = 0 E n ψ ( n + j ) E n + j ϕ = m = i E m i φ m E m θ n = 0 E n ψ 0 1 ( 1 x ) ϕ x n + j 1 d x = m = i E m i φ m E m θ 0 1 ( 1 x ) ϕ x j 1 ( n = 0 E n ψ x n ) d x = 0 1 ( 1 x ) ϕ ψ 1 x j 1 d x m = 0 E m φ ( m + i ) E m + i θ = 1 j E j ϕ ψ 1 m = 0 E m φ 0 1 ( 1 x ) θ x m + i 1 d x = 1 j E j ϕ ψ 1 0 1 ( 1 x ) θ φ 1 x i 1 d x = 1 i j E i θ φ 1 E j ϕ ψ 1 .

 □

For single series, Lemma 1 due to Chow [4].

Lemma 2 Let 1kr< and α,β>1. For i,j1, let

A i j = A i j (α,β)= m = i n = j | E m i α 1 | r / k | E n j β 1 | r / k m n ( E m α ) r / k ( E n β ) r / k .
(4)

Then, if k=r,

A i j = { O ( i α 1 j β 1 ) , if α , β < 0 , O ( i α 1 j 1 ) , if α < 0 , β 0 , O ( i 1 j β 1 ) , if α 0 , β < 0 , O ( i 1 j 1 ) , if α , β 0 .
(5)

If k<r<, then

A i j = { O ( i α r k 1 j β r k 1 ) , if α < 1 k / r , β < 1 k / r , O ( i α r k 1 j r k log j ) , if α < 1 k / r , β = 1 k / r , O ( i α r k 1 j r k ) , if α < 1 k / r , β > 1 k / r , O ( i α r k 1 j β r k 1 log i ) , if α = 1 k / r , β < 1 k / r , O ( i r k j r k log i log j ) , if α = 1 k / r , β = 1 k / r , O ( i α r k 1 j r k log i ) , if α = 1 k / r , β > 1 k / r , O ( i r k j β r k 1 ) , if α > 1 k / r , β < 1 k / r , O ( i r k j β r k 1 log j ) , if α > 1 k / r , β = 1 k / r , O ( i r k j r k ) , if α > 1 k / r , β > 1 k / r .
(6)

Proof Since | E n α |K(α) n α for all α, and E n α M(α) n α for α>1, where K(α) and M(α) are positive constants depending only on α, if k<r<, then

A i j = O ( 1 ) m = i n = j m α r k 1 n β r k 1 ( m i + 1 ) ( α 1 ) r k ( n j + 1 ) ( β 1 ) r k = O ( 1 ) ( m = i 2 i 1 m α r k 1 ( m i + 1 ) ( α 1 ) r k + m = 2 i m α r k 1 ( m i + 1 ) ( α 1 ) r k ) × ( n = j 2 j 1 n β r k 1 ( n j + 1 ) ( β 1 ) r k + n = 2 j n β r k 1 ( n j + 1 ) ( β 1 ) r k ) = O ( 1 ) ( i α r k 1 m = i 2 i 1 ( m i + 1 ) ( α 1 ) r k + m = 2 i m r k 1 ) × ( j β r k 1 n = j 2 j 1 ( n j + 1 ) ( β 1 ) r k + n = 2 j n r k 1 ) = O ( 1 ) i α r k 1 j β r k 1 m = 1 i n = 1 j m ( α 1 ) r k n ( β 1 ) r k + O ( 1 ) i α r k 1 j r / k m = 1 i m ( α 1 ) r k + O ( 1 ) i r / k j β r k 1 n = 1 j n ( β 1 ) r k + O ( 1 ) ( i j ) r / k .

According as ( α 1 ) r k and ( β 1 ) r k =1, <1 or >1, we have (6). The case k=r is proved similarly. □

For single series, Lemma 2 due to Mehdi [5].

We now prove the following theorem.

Theorem 2 Let rk1.

(i) It holds that C(α,β)( A k 2 , A r 2 ) for each α,β>1k/r.

(ii) If α,β=1k/r and the condition m = 1 n = 1 ( m n ) k 1 logmlogn| a m n | k =O(1) is satisfied then C(α,β)( A k 2 , A r 2 ).

(iii) If the condition m = 1 n = 1 m k + ( 1 α ) r k 2 n k + ( 1 β ) r k 2 | a m n | k =O(1) is satisfied then C(α,β)( A k 2 , A r 2 ) for each k/r<α,β<1k/r.

(iv) If β=1k/r and the condition m = 1 n = 1 m k + ( 1 α ) r k 2 n k 1 logn| a m n | k =O(1) is satisfied then C(α,β)( A k 2 , A r 2 ) for each k/r<α<1k/r.

(v) If the condition m = 1 n = 1 m k + ( 1 α ) r k 2 n k 1 | a m n | k =O(1) is satisfied then C(α,β)( A k 2 , A r 2 ) for each k/r<α<1k/r and β>1k/r.

(vi) If α=1k/r and the condition m = 1 n = 1 m k 1 n k + ( 1 β ) r k 2 logm| a m n | k =O(1) is satisfied then C(α,β)( A k 2 , A r 2 ) for each k/r<β<1k/r.

(vii) If the condition m = 1 n = 1 m k 1 n k 1 logm| a m n | k =O(1) is satisfied then C(α,β)( A k 2 , A r 2 ) for each α>1k/r, β<1k/r.

(viii) If the condition m = 1 n = 1 m k 1 n k + ( 1 β ) r k 2 | a m n | k =O(1) is satisfied then C(α,β)( A k 2 , A r 2 ) for each α>1k/r and k/r<β<1k/r.

(ix) If β=1k/r and the condition m = 1 n = 1 m k 1 n k 1 logn| a m n | k =O(1) is satisfied then C(α,β)( A k 2 , A r 2 ) for each α>1k/r.

Proof We shall show that ( σ m n α β ) A r 2 , i.e.,

m = 1 n = 1 ( m n ) r 1 | Δ 11 σ m 1 , n 1 α β | r <.

Let τ m n α β denote the mn-term of the C(α,β)-transform in terms of (mn a m n ), i.e.,

τ m n α β = 1 E m α E n β i = 1 m j = 1 n E m i α 1 E n j β 1 ij a i j .

For α,β>1, since

τ m n α β =mn ( σ m n α β σ m , n 1 α β σ m 1 , n α β + σ m 1 , n 1 α β ) ,

to prove the theorem, it will be sufficient to show that

m = 1 n = 1 ( m n ) 1 | τ m n α β | r <.
(7)

Using Hölder’s inequality, we have

m = 1 n = 1 1 m n | τ m n α β | r = m = 1 n = 1 1 m n | 1 E m α E n β i = 1 m j = 1 n E m i α 1 E n j β 1 i j a i j | r m = 1 n = 1 1 m n ( E m α ) r ( E n β ) r { i = 1 m j = 1 n | E m i α 1 | | E n j β 1 | i k j k | a i j | k } r / k × { i = 1 m j = 1 n | E m i α 1 | | E n j β 1 | } ( k 1 ) r / k .

Since

i = 1 m j = 1 n | E m i α 1 | | E n j β 1 | = ( | E 0 α 1 | + i = 1 m 1 | E m i α 1 | ) ( | E 0 β 1 | + j = 1 n 1 | E n j β 1 | ) = ( | E 0 α 1 | + | i = 1 m 1 E m i α 1 | ) ( | E 0 β 1 | + | j = 1 n 1 E n j β 1 | ) = ( | E 0 α 1 | + | i = 0 m E m i α 1 E m α 1 E 0 α 1 | ) × ( | E 0 β 1 | + | j = 0 n E n j β 1 E n β 1 E 0 β 1 | ) = ( | E 0 α 1 | + | E m 1 α E 0 α 1 | ) ( | E 0 β 1 | + | E n 1 β E 0 β 1 | ) ,

and using the fact that

| E m 1 α E m α |=O(1)and| E n 1 β E n β |=O(1),

we obtain

m = 1 n = 1 1 m n | τ m n α β | r m = 1 n = 1 ( E m α ) ( k 1 ) r / k ( E n β ) ( k 1 ) r / k m n ( E m α ) r ( E n β ) r × { i = 1 m j = 1 n | E m i α 1 | | E n j β 1 | i k j k | a i j | k } r / k m = 1 n = 1 ( E m α ) r / k ( E n β ) r / k m n × { i = 1 m j = 1 n | E m i α 1 | | E n j β 1 | ( i j ) 1 k / r + k 2 / r | a i j | k 2 / r ( i j ) ( r k ) + k ( r k ) / r | a i j | k ( r k ) / r } r / k .

Applying Hölder’s inequality with indices r/k, r/(rk), we deduce that

m = 1 n = 1 1 m n | τ m n α β | r m = 1 n = 1 ( E m α ) r / k ( E n β ) r / k m n i = 1 m j = 1 n | E m i α 1 | r / k | E n j β 1 | r / k ( i j ) k 1 + r / k | a i j | k × { i = 1 m j = 1 n ( i j ) k 1 | a i j | k } ( r k ) / k .

Since ( s m n ) A k 2 , we have

m = 1 n = 1 1 m n | τ m n α β | r =O(1) i = 1 j = 1 ( i j ) k 1 | a i j | k ( i j ) r / k m = i n = j | E m i α 1 | r / k | E n j β 1 | r / k m n ( E m α ) r / k ( E n β ) r / k .

(i) From Lemma 1, if α,β>1k/r, then

m = i n = j ( E m i α 1 ) r / k ( E n j β 1 ) r / k m n ( E m α ) r / k ( E n β ) r / k = ( i j ) 1 1 E i r k 1 E j r k 1 = O ( ( i j ) r / k ) .

Therefore, for the case α,β>1k/r, we have

m = 1 n = 1 1 m n | τ m n α β | r = O ( 1 ) i = 1 j = 1 ( i j ) k 1 | a i j | k ( i j ) r / k ( i j ) r / k = O ( 1 ) i = 1 j = 1 ( i j ) k 1 | a i j | k = O ( 1 ) ,

since ( s m n ) A k 2 .

(ii) If α,β=1k/r, from Lemma 2, then

m = i n = j | E m i α 1 | r / k | E n j β 1 | r / k m n ( E m α ) r / k ( E n β ) r / k =O ( i r k j r k log i log j ) .

Hence,

m = 1 n = 1 1 m n | τ m n α β | r =O(1) i = 1 j = 1 ( i j ) k 1 logilogj| a i j | k =O(1).

(iii) If k/r<α, β<1k/r, from Lemma 2, then

m = i n = j | E m i α 1 | r / k | E n j β 1 | r / k m n ( E m α ) r / k ( E n β ) r / k =O ( i α r k 1 j β r k 1 ) ,

and then

m = 1 n = 1 1 m n | τ m n α β | r = O ( 1 ) i = 1 j = 1 ( i j ) k + r / k 1 i α r k 1 j β r k 1 | a i j | k = O ( 1 ) i = 1 j = 1 i k + ( 1 α ) r k 2 j k + ( 1 β ) r k 2 | a i j | k = O ( 1 ) .

(iv) If β=1k/r and k/r<α<1k/r, from Lemma 2, then

m = i n = j | E m i α 1 | r / k | E n j β 1 | r / k m n ( E m α ) r / k ( E n β ) r / k =O ( i α r k 1 j r k log j ) ,

therefore, we have

m = 1 n = 1 1 m n | τ m n α β | r = O ( 1 ) i = 1 j = 1 ( i j ) k 1 | a i j | k ( i j ) r / k i α r k 1 j r k log j = O ( 1 ) i = 1 j = 1 i k + ( 1 α ) r k 2 j k 1 log j | a i j | k = O ( 1 ) .

(v) If k/r<α<1k/r and β>1k/r, then

m = 1 n = 1 1 m n | τ m n α β | r = O ( 1 ) i = 1 j = 1 ( i j ) k 1 | a i j | k ( i j ) r / k i α r k 1 j r k = O ( 1 ) i = 1 j = 1 i k + ( 1 α ) r k 2 j k 1 | a i j | k = O ( 1 ) ,

by using Lemma 2.

(vi) If α=1k/r and k/r<β<1k/r, then

m = 1 n = 1 1 m n | τ m n α β | r = O ( 1 ) i = 1 j = 1 ( i j ) k 1 | a i j | k ( i j ) r / k i r k j β r k 1 log i = O ( 1 ) i = 1 j = 1 i k 1 j k + ( 1 β ) r k 2 log i | a i j | k = O ( 1 ) ,

by using Lemma 2.

(vii) If α=1k/r and β>1k/r, then

m = 1 n = 1 1 m n | τ m n α β | r = O ( 1 ) i = 1 j = 1 ( i j ) k 1 | a i j | k ( i j ) r / k i r k j r k log i = O ( 1 ) i = 1 j = 1 i k 1 j k 1 log i | a i j | k = O ( 1 ) ,

by using Lemma 2.

(viii) If α>1k/r and k/r<β<1k/r, then

m = 1 n = 1 1 m n | τ m n α β | r = O ( 1 ) i = 1 j = 1 ( i j ) k 1 | a i j | k ( i j ) r / k i r k j β r k 1 = O ( 1 ) i = 1 j = 1 i k 1 j k + ( 1 β ) r k 2 | a i j | k = O ( 1 ) ,

by using Lemma 2.

(ix) If α>1k/r and β=1k/r, then

m = 1 n = 1 1 m n | τ m n α β | r = O ( 1 ) i = 1 j = 1 ( i j ) k 1 | a i j | k ( i j ) r / k i r k j β r k 1 log j = O ( 1 ) i = 1 j = 1 i k 1 j k 1 log j | a i j | k = O ( 1 ) ,

by using Lemma 2. □

The one-dimensional version of Theorem 2 appears in [6]. By (5), Theorem 2 includes the following theorem with the special case r=k.

Theorem 3 Let k1.

(i) It holds that C(α,β)B( A k 2 ) for each α,β0.

(ii) If the condition m = 1 n = 1 m k α 1 n k β 1 | a m n | k =O(1) is satisfied then C(α,β)B( A k 2 ) for each 1<α<0 and 1<β<0.

(iii) If the condition m = 1 n = 1 m k α 1 n k 1 | a m n | k =O(1) is satisfied then C(α,β)B( A k 2 ) for each 1<α<0 and β0.

(iv) If the condition m = 1 n = 1 m k 1 n k β 1 | a m n | k =O(1) is satisfied then C(α,β)B( A k 2 ) for each α0 and 1<β<0.

Remark 1 Theorem 3 moderates Theorem 1 of [7]. Since Holder’s inequality is valid if each of the terms is nonnegative, it should be added the absolute values of the binomial coefficients in the proof of Theorem 1 of [7], when 1<α<0 and/or 1<β<0. Therefore, if we replace the binomial coefficients with their absolute values, then the inequality (15) of [7] will be true. So, we should add the conditions, given above in (ii), (iii) and (iv) of Theorem 3 in the statement of Theorem 1 of [7], for the cases 1<α<0 and/or 1<β<0.

Corollary 1 Let θ m n α = 1 E m α i = 0 m E m i α 1 s i n =C(α,0)( s m n ).

(i) It holds that C(α,0)B( A k 2 ) for each α0.

(ii) If the condition m = 1 n = 1 m k α 1 n k 1 | a m n | k =O(1) is satisfied then C(α,0)B( A k 2 ) for each 1<α<0.

Corollary 2 Let θ m n β = 1 E n β j = 0 n E n j β 1 s m j =(C,0,β)( s m n ).

(i) It holds that C(0,β)B( A k 2 ) for each β0.

(ii) If the condition m = 1 n = 1 m k 1 n k β 1 | a m n | k =O(1) is satisfied then C(0,β)B( A k 2 ) for each 1<β<0.

Corollary 3 Let σ m n = 1 ( m + 1 ) ( n + 1 ) i = 0 m j = 0 n s i j =(C,1,1)( s m n ). Then C(1,1)B( A k 2 ).

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Acknowledgements

Dedicated to Professor Hari M. Srivastava.

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Şevli, H., Savaş, E. Some further extensions of absolute Cesàro summability for double series. J Inequal Appl 2013, 144 (2013). https://doi.org/10.1186/1029-242X-2013-144

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Keywords

  • absolute summability
  • bounded operator
  • Cesáro matrix
  • double series
  • Hausdorff matrix