Some inequalities for Cesàro means of double Vilenkin...

Tepnadze and Persson Journal of Inequalities and Applications (2018) 2018:352 https://doi.org/10.1186/s13660-018-1929-y

R E S E A R C H Open Access

Some inequalities for Cesàro means ofdouble Vilenkin–Fourier seriesT. Tepnadze1 and L.E. Persson1*

*Correspondence:[email protected] Artic University of Norway,Narvik, Norway

AbstractIn this paper, we state and prove some new inequalities related to the rate of Lp

approximation by Cesàro means of the quadratic partial sums of doubleVilenkin–Fourier series of functions from Lp.

MSC: 42C10; 42B25

Keywords: Inequalities; Approximation; Vilenkin system; Vilenkin–Fourier series;Cesàro means; Convergence in norm

1 IntroductionLet N+ denote the set of positive integers, and let N := N+ ∪ {0}. Let m := (m0, m1, . . .) be asequence of positive integers not less than 2. Denote by Zmk := {0, 1, . . . , mk –1} the additivegroup of integers modulo mk . Define the group Gm as the complete direct product of thegroups Zmj with the product of the discrete topologies of Zmj .

The direct product of the measures

μk({j}) :=

1mk

(j ∈ Zmk )

is the Haar measure on Gm with μ(Gm) = 1. If the sequence m is bounded, then Gm iscalled a bounded Vilenkin group. In this paper, we consider only bounded Vilenkin groups.The elements of Gm can be represented by sequences x := (x0, x1, . . . , xj, . . .) (xj ∈ Zmj ). Thegroup operation + in Gm is given by

x + y =((x0 + y0) mod m0, . . . , (xk + yk) mod mk , . . .

)

for x := (x0, . . . , xk , . . .) and y := (y0, . . . , yk , . . .) ∈ Gm. The inverse of + will be denoted by –.It is easy to give a base for the neighborhoods of Gm:

I0(x) := Gm,

In(x) := {y ∈ Gm|y0 = x0, . . . , yn–1 = xn–1}

for x ∈ Gm and n ∈ N . Define In := In(0) for n ∈ N+. Set en := (0, . . . , 0, 1, 0, . . .) ∈ Gm, wherethe nth coordinate of which is 1, and the rest are zeros (n ∈ N ).

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We define the so-called generalized number system based on m as follows: M0 := 1,Mk+1 := mkMk (k ∈ N ). Then every n ∈ N can be uniquely expressed as n =

∑∞j=0 njMj,

where nj ∈ Zmj (j ∈ N+), and only a finite number of nj differ from zero. We also use thefollowing notation: |n| := max{k ∈ N : nk �= 0} (i.e., M|n| ≤ n < M|n|+1, n �= 0). For x ∈ Gm, wedenote |x| :=

∑∞j=0

xjMj+1

(xj ∈ Zmj ).Next, we introduce on Gm an orthonormal system, which is called the Vilenkin sys-

tem. First, we define the complex-valued functions rk(x) : Gm → C, the generalizedRademacher functions, as follows:

rk(x) := exp2π ixk

mk

(i2 = –1, x ∈ Gm, k ∈ N

).

Now we define the Vilenkin system ψ := (ψn : n ∈ N) on Gm as

ψn(x) :=∞∏

k=0

rnkk (x) (n ∈ N).

In particular, if m = 2, then we call this system the Walsh–Paley system. Each ψn is acharacter of Gm, and all characters of Gm are of this norm. Moreover, ψn(–x) = ψ̄n(x).

The Dirichlet kernels are defined by

Dn :=n–1∑

k=0

ψk (n ∈ N+).

Recall that (see [20] or [23])

DMn (x) =

⎧⎨

⎩Mn if x ∈ In,

0 if x /∈ In.(1)

The Vilenkin system is orthonormal and complete in L1(Gm) (see [1]).Next, we introduce some notation from the theory of two-dimensional Vilenkin system.

Let m̃ be a sequence like m. The relation between the sequences (m̃n) and (M̃n) is the sameas between sequences (mn) and (Mn). The group Gm × Gm̃ is called a two-dimensionalVilenkin group. The normalized Haar measure is denoted by μ as in the one-dimensionalcase. We also suppose that m = m̃ and Gm × Gm̃ = G2

m.The norm of the space Lp(G2

m) is defined by

‖f ‖p :=(∫

G2m

∣∣f (x, y)∣∣p dμ(x, y)

)1/p

(1 ≤ p < ∞).

Denote by C(G2m) the class of continuous functions on the group G2

m endowed with thesupremum norm.

For brevity in notation, we write L∞(G2m) instead of C(G2

m).The two-dimensional Fourier coefficients, the rectangular partial sums of the Fourier

series, and the Dirichlet kernels with respect to the two-dimensional Vilenkin system are


defined as follows:

f̂ (n1, n2) :=∫

G2m

f (x, y)ψ̄n1 (x)ψ̄n2 (y) dμ(x, y),

Sn1,n2 (x, y, f ) :=n1–1∑

k1=0

n2–1∑

k2=0

f̂ (k1, k2)ψk1 (x)ψk2 (y),

Dn1,n2 (x, y) := Dn1 (x)Dn2 (y),

Denote

S(1)n (x, y, f ) :=

n–1∑

l=0

f̂ (l, y)ψ̄l(x),

S(2)m (x, y, f ) :=

m–1∑

r=0

f̂ (x, r)ψ̄r(y),

where

f̂ (l, y) =∫

Gm

f (x, y)ψl(x) dμ(x)

and

f̂ (x, r) =∫

Gm

f (x, y)ψr(y) dμ(y).

The (C, –α) means of the double Vilenkin–Fourier series are defined as follows:

σ –αn (f , x, y) =

1A–α

n–1

n∑

j=1

A–α–1n–j Sj,j(f , x, y),

where

Aα0 = 1, Aα

n =(α + 1) · · · (α + n)

n!.

It is well known that (see [28])

Aαn =

n∑

k=0

Aα–1k , (2)

Aαn – Aα

n–1 = Aα–1n , (3)

and

c1(α)nα ≤ Aαn ≤ c2(α)nα , (4)

where positive constants c1 and c2 depend on α.


The dyadic partial moduli of continuity of a function f ∈ Lp(G2m) in the Lp-norm are

defined by

ω1

(f ,

1Mn

)

p= sup

u∈In

∥∥f (· + u, ·) – f (·, ·)∥∥p

and

ω2

(f ,

1Mn

)

p= sup

v∈In

∥∥f (·, · + v) – f (·, ·)∥∥p,

whereas the dyadic mixed modulus of continuity is defined as follows:

ω1,2

(f ,

1Mn

,1

Mm

)

p

= sup(u,v)∈In×Im

∥∥f (· + u, · + v) – f (· + u, ·) – f (·, · + v) + f (·, ·)∥∥p.

It is clear that

ω1,2

(f ,

1Mn

,1

Mm

)

p≤ ω1

(f ,

1Mn

)

p+ ω2

(f ,

1Mm

)

p.

The dyadic total modulus of continuity is defined by

ω

(f ,

1Mn

)

p= sup

(u,v)∈In×In

∥∥f (· + u, · + v) – f (·, ·)∥∥p.

The problems of summability of partial sums and Cesàro means for Walsh–Fourier serieswere studied in [2, 13–19, 21, 22, 25, 26].

The convergence issue of Fejér (and Cesàro) means on the Walsh and Vilenkin groupsfor unbounded case were studied in [3–11].

In his monograph [27], Zhizhinashvili investigated the behavior of Cesàro (C,α)-meansfor double trigonometric Fourier series in detail. Goginava [18] studied the analogousquestion in the case of the Walsh system. In particular, the following theorems wereproved.

Theorem A Let f belong to Lp(G2) for some p ∈ [1,∞] and α ∈ (0, 1). Then, for any 2k ≤n < 2k+1 (k, n ∈ N ), we have the inequality

∥∥σ –α

2k (f ) – f∥∥

p ≤ c(α)

{

2kαω1(f , 1/2k–1)

p + 2kαω2(f , 1/2k–1)

p

+k–2∑

r=0

2r–kω1(f , 1/2r)

p +k–2∑

s=0

2s–kω2(f , 1/2s)

p

}

.


Theorem B Let f belong to Lp(G2) for some p ∈ [1,∞] and α ∈ (0, 1). Then, for any 2k ≤n < 2k+1 (k, n ∈ N ), we have the inequality

∥∥σ –αn (f ) – f

∥∥p ≤ c(α)

{

2kαkω1(f , 1/2k–1)

p + 2kαkω2(f , 1/2k–1)

p

+k–2∑

r=0

2r–kω1(f , 1/2r)

p +k–2∑

s=0

2s–kω2(f , 1/2s)

p

}

.

In this paper, we state and prove analogous results in the case of double Vilenkin–Fourierseries. Our main results are the following theorems.

Theorem 1 Let f belong to Lp(G2m) for some p ∈ [1,∞] and α ∈ (0, 1). Then, for any Mk ≤

n < Mk+1 (k, n ∈ N ), we have the inequality

∥∥σ –α

Mk(f ) – f

∥∥

p ≤ c(α)

(

ω1(f , 1/Mk–1)pMαk + ω2(f , 1/Ml–1)pMα

k

+k–2∑

r=0

Mr

Mkω1(f , 1/Mr)p +

k–2∑

s=0

Ms

Mkω2(f , 1/Ms)p

)

.

Theorem 2 Let f belong to Lp(G2m) for some p ∈ [1,∞] and α ∈ (0, 1). Then, for any Mk ≤

n < Mk+1 (k, n ∈ N ), we have the inequality

∥∥σ –αn (f ) – f

∥∥p

≤ c(α)

(

ω1(f , 1/Mk–1)pMαk log n + ω2(f , 1/Ml–1)pMα

k log n

+k–2∑

r=0

Mr

Mkω1(f , 1/Mr)p +

k–2∑

s=0

Ms

Mkω2(f , 1/Ms)p

)

.

To make the proofs of these theorems clearer, we formulate some auxiliary lemmas inSect. 2. Some of these lemmas are new and of independent interest. Detailed proofs canbe found in Sect. 3.

2 Auxiliary lemmasTo prove Theorems 1 and 2, we need the following three lemmas (see [1, 12], and [8],respectively)

Lemma 1 Let α1,α2, . . . ,αn be real numbers. Then

1n

∫

G

∣∣∣∣∣

n∑

k=1

αkDk(x)

∣∣∣∣∣dμ(x) ≤ c√

n

( n∑

k=1

α2k

)1/2

.

Lemma 2 Let α1,α2, . . . ,αn be real numbers. Then

1n

∫

G2m

∣∣∣∣∣

n∑

k=1

αkDk(x)Dk(y)

∣∣∣∣∣dμ(x, y) ≤ c√

n

( n∑

k=1

α2k

)1/2

.


Lemma 3 Let 0 ≤ j < nsMs and 0 ≤ ns < ms. Then

DnsMs–j = DnsMs – ψnsMs–1D̄j.

We also need the following new nemmas of independent interest.

Lemma 4 Let f belong to Lp(G2m) for some p ∈ [1,∞]. Then, for every α ∈ (0, 1), we have

the inequality

I :=1

A–αn

∥∥∥∥∥

∫

G2m

Mk–1∑

i=1

A–α–1n–i Di(u)Di(v)

[f (· – u, ·–) – f (·, ·)]dμ(u, v)

∥∥∥∥∥

p

≤k–2∑

r=0

Mr

Mkω1(f , 1/Mr)p +

k–2∑

s=0

Ms

Mkω2(f , 1/Ms)p,

where Mk ≤ n < Mk+1.

Lemma 5 Let α ∈ (0, 1) and p = Mk , Mk + 1, . . . . Then

II :=∫

G2m

∣∣∣∣∣

Mk∑

i=1

A–α–1p–i Di(u)Di(v)

∣∣∣∣∣dμ(u, v) ≤ c(α) < ∞, k = 1, 2, . . . .

Lemma 6 We have the inequality

III :=∫

G2m

∣∣∣∣∣

n∑

i=1


∣∣∣∣∣dμ(u, v) ≤ c(α) log n

3 The detailed proofs

Proof of Lemma 3 Applying Abel’s transformation, from (2) we get

I ≤ 1A–α

n

∥∥∥∥∥

∫

G2m

Mk–1–1∑

i=1

A–α–2n–i

i∑

l=1

Di(u)Di(v)[f (· – u, · – v) – f (·, ·)]dμ(u, v)

∥∥∥∥∥

p

+1

A–αn

∥∥∥∥∥

∫

G2m

A–α–1n–Mk–1

Mk–1∑

i=1

Di(u)Di(v)[f (· – u, · – v) – f (·, ·)]dμ(u, v)

∥∥∥∥∥

p

:= I1 + I2, (5)

where the first and second terms on the right side of inequality (5) are denoted by I1 andI2, respectively.


For I2, we have the estimate

I2 ≤ 1A–α

n

∥∥∥∥∥

∫

G2m


k–2∑

r=1

Mr+1–1∑

i=Mr

Di(u)Di(v)

× [f (· – u, · – v) – f (·, ·)]

∥∥∥∥∥

p

dμ(u, v)

≤ 1A–α

n

∥∥∥∥∥

∫

G2m


k–2∑

r=1

Mr+1–1∑

i=Mr

Di(u)Di(v)

× [f (· – u, · – v) – SMr ,Mr (· – u, · – v, f )

]dμ(u, v)

∥∥∥∥∥

p

+1

A–αn

∥∥∥∥∥

∫

G2m


k–2∑

r=1

Mr+1–1∑

i=Mr

Di(u)Di(v)

× [SMr ,Mr (· – u, · – v, f ) – SMr ,Mr (·, ·, f )

]dμ(u, v)

∥∥∥∥∥

p

+1

A–αn

∥∥∥∥∥

∫

G2m


k–2∑

r=1

Mr+1–1∑

i=Mr

Di(u)Di(v)

× [SMr ,Mr (·, ·, f ) – f (·, ·)]dμ(u, v)

∥∥∥∥∥

p

:= I21 + I22 + I23, (6)

where the first, second, and third terms on the right side of inequality (6) are denoted byI21, I22, and I23, respectively.

It is evident that

∫

G2m

Mr+1–1∑

i=Mr

Di(u)Di(v)[SMr ,Mr (· – u, · – v, f ) – SMr ,Mr (·, ·, f )

]dμ(u, v)

=Mr+1–1∑

i=Mr

(∫

G2m

Di(u)Di(v)SMr ,Mr (· – u, · – v, f ) dμ(u, v) – SMr ,Mr (·, ·, f ))

=Mr+1–1∑

i=Mr

(Si

(·, ·, SMr ,Mr (f ))

– SMr ,Mr (·, ·, f ))

=Mr+1–1∑

i=Mr

(SMr ,Mr (·, ·, f ) – SMr ,Mr (·, ·, f )

)= 0.

Hence

I22 = 0. (7)


Moreover, by the generalized Minkowski inequality, Lemma 2, and by (1) and (4) weobtain

I21 ≤ 1A–α

n

∣∣A–α–1n–Mk–1

∣∣k–2∑

r=1

∫

G2m

∣∣∣∣∣

Mr+1–1∑

i=Mr

Di(u)Di(v)

∣∣∣∣∣

× ∥∥f (· – u, · – v) – SMr ,Mr (· – u, · – v, f )∥∥

p dμ(u, v)

≤ c(α)Mk

k–2∑

r=1

(ω1(f , 1/Mr)p + ω2(f , 1/Mr)p

)

×∫

G2m

∣∣∣∣∣

Mr+1–1∑

i=Mr

Di(x)Di(y)

∣∣∣∣∣dμ(u, v)

≤ c(α)k–2∑

r=1

Mr

Mk

(ω1(f , 1/Mr)p + ω2(f , 1/Mr)p

). (8)

The estimation of I23 is analogous to that of I21:

I23 ≤ c(α)k–2∑

r=1

Mr

Mk

(ω1(f , 1/Mr)p + ω2(f , 1/Mr)p

). (9)

Analogously, we can estimate I1 as follows:

I1 ≤ 1A–α

n

k–2∑

r=1

∥∥∥∥∥

∫

G2m

Mr+1–1∑

i=Mr

A–α–2n–i

i∑

l=1

Dl(u)Dl(v)

× [f (· – u, · – v) – SMr ,Mr (· – u, · – v, f )

]dμ(u, v)

∥∥∥∥∥

p

+1

A–αn

k–2∑

r=1

∥∥∥∥∥

∫

G2m

Mr+1–1∑

i=Mr

A–α–2n–i

i∑

l=1

Dl(u)Dl(v)

× [SMr ,Mr (· – u, · – v, f ) – SMr ,Mr (·, ·, f )

]∥∥∥∥∥

p

dμ(u, v)

+1

A–αn

k–2∑

r=1

∥∥∥∥∥

∫

G2m

Mr+1–1∑

i=Mr

A–α–2n–i

i∑

l=1

Dl(u)Dl(v)

× [SMr ,Mr (·, ·, f ) – f (·, ·)]dμ(u, v)

∥∥∥∥∥

p

≤ 1A–α

n

k–2∑

r=1

∫

G2m

∣∣∣∣∣

Mr+1–1∑

i=Mr

A–α–2n–i

i∑

l=1

Dl(u)Dl(v)

∣∣∣∣∣

× ∥∥f (· – u, · – v) – SMr ,Mr (· – u, · – v, f )∥∥

p dμ(u, v)


+1

A–αn

k–2∑

r=1

∫

G2m

∣∣∣∣∣

Mr+1–1∑

i=Mr

A–α–2n–i

i∑

l=1

Dl(u)Dl(v)

∣∣∣∣∣

× ∥∥SMr ,Mr (·, ·, f ) – f (·, ·)∥∥p dμ(u, v)

≤ c(α)Mαk

k–2∑

r=1

Mr+1–1∑

i=Mr

(n – i)–α–2i(ω1(f , 1/Mr)p + ω2(f , 1/Mr)p

)

≤ c(α)Mαk

k–2∑

r=1

Mr+1–1∑

i=Mr

(n – Mr+1 – 1)–α–2i(ω1(f , 1/Mr)p + ω2(f , 1/Mr)p

)

≤ c(α)k–2∑

r=0

Mr

Mk

(ω1(f , 1/Mr)p + ω2(f , 1/Mr)p

). (10)

By combining (7)–(9) with (10) for I we find that

I ≤ c(α)k–2∑

r=0

Mr

Mk

(ω1(f , 1/Mr)p + ω2(f , 1/Mr)p

). (11)

The proof of Lemma 3 is complete. �

Proof of Lemma 4 It is evident that

II ≤∫

G2m

∣∣∣∣∣

Mk –1∑

i=1

A–α–1p–Mk +iDMk –i(u)DMk –i(v)

∣∣∣∣∣dμ(u, v)

+∣∣A–α–1

p–Mk

∣∣∫

G2m

DMk (u)DMk (v) dμ(u, v)

:= II1 + II2, (12)

where the first and second terms on the right side of inequality (12) are denoted by II1 andII2, respectively.

From (1) by |A–α–1p–Mk

| ≤ 1 we get that

II2 ≤ 1. (13)

Moreover, by Lemma 3 we have that

II1 ≤∫

G2m

∣∣∣∣∣

Mk –1∑

i=1

A–α–1p–Mk +iD̄i(u)D̄i(v)

∣∣∣∣∣dμ(u, v)

+∫

G2m

DMk (u)

∣∣∣∣∣

Mk –1∑

i=1

A–α–1p–Mk +iD̄i(v)

∣∣∣∣∣dμ(u, v)

+∫

G2m

DMk (v)

∣∣∣∣∣

Mk –1∑

i=1

A–α–1p–Mk +iD̄i(u)

∣∣∣∣∣dμ(u, v)


+

∣∣∣∣∣

Mk –1∑

i=1

A–α–1p–Mk +i

∣∣∣∣∣

∫

G2m

DMk (u)DMk (v) dμ(u, v)

:= II11 + II12 + II13 + II14, (14)

where the first, second, third, and fourth terms on the right side of inequality (14) aredenoted by II11, II12, II13, and II14 respectively.

From (1) and (4) it follows that

II14 ≤ c(α)∞∑

v=1

v–α–1 < ∞. (15)

By Applying Abel’s transformation, in view of Lemma 2, we have that

II11 ≤∫

G2m

∣∣∣∣∣

Mk –2∑

i=1

A–α–2p–Mk +i

i∑

l=1

D̄l(u)D̄l(v)

∣∣∣∣∣dμ(u, v)

+∫

G2m

∣∣∣∣∣A–α–1

p–1

Mk –1∑

i=1

D̄i(u)D̄i(v)

∣∣∣∣∣dμ(u, v)

≤ c(α)

{Mk –2∑

v=1

(p – Mk + i)–α–2i + (p – 1)–α–1Mk

}

≤ c(α)

{ ∞∑

i=1

i–α–1 + M–αk

}

< ∞. (16)

The estimation of II12 and II13 are analogous to the estimation of II11. Applying Abel’stransformation, in view of Lemma 1, we find that

II12 ≤∫

G2m

DMk (u)

∣∣∣∣∣

Mk –2∑

i=1

A–α–2p–Mk +i

i∑

l=1

D̄l(v)

∣∣∣∣∣dμ(u, v)

+∫

G2m

DMk (u)

∣∣∣∣∣A–α–1

p–1

Mk –1∑

i=1

D̄i(v)

∣∣∣∣∣dμ(u, v)

≤ c(α)

{Mk –2∑

v=1

(p – Mk + i)–α–2i + (p – 1)–α–1Mk

}

≤ c(α)

{ ∞∑

i=1


}

< ∞ (17)

and

III12 ≤∫

G2m

DMk (v)

∣∣∣∣∣

Mk –2∑

i=1

A–α–2p–Mk +i

i∑

l=1

D̄l(u)

∣∣∣∣∣dμ(u, v)

+∫

G2m

DMk (v)

∣∣∣∣∣A–α–1

p–1

Mk –1∑

i=1

D̄i(u)

∣∣∣∣∣dμ(u, v)


≤ c(α)

{Mk –2∑

v=1

(p – Mk + i)–α–2i + (p – 1)–α–1Mk

}

≤ c(α)

{ ∞∑

i=1


}

< ∞. (18)

The proof is complete by combining (12)–(18). �

Proof of Lemma 5 Let

n = nk1 Mk1 + · · · + nks Mks , k1 > · · · > ks ≥ 0.

Denote

n(i) = nki Mki + · · · + nks Mks , i = 1, 2, . . . , s.

Since (see [20])

Dj+nAMA = DnAMA + ψnAMA Dj, (19)

we find that

III ≤∫

G2m

∣∣∣∣∣

nk1 Mk1∑

i=1


∣∣∣∣∣dμ(u, v)

+∫

G2m

∣∣∣∣∣

n(2)∑

i=1

A–α–1n(2)–iDi(u)Di(v)

∣∣∣∣∣dμ(u, v)

+∫

G2m

Dnk1 Mk1(u)Dnk1 Mk1

(v)

∣∣∣∣∣

n(2)∑

i=1

A–α–1n(2)–i

∣∣∣∣∣dμ(u, v)

+∫

G2m

Dnk1 Mk1(u)

∣∣∣∣∣

n(2)∑

i=1

A–α–1n(2)–iDi(v)

∣∣∣∣∣dμ(u, v)

+∫

G2m

Dnk1 Mk1(v)

∣∣∣∣∣

n(2)∑

i=1

A–α–1n(2)–iDi(u)

∣∣∣∣∣dμ(u, v)

:= III1 + III2 + III3 + III4 + III5, (20)

where the first, second, third, fourth, and fifth terms on the right side of inequality (20)are denoted by III1, III2, III3, III4, and III5, respectively.

By (1) we have that

III3 ≤ c(α). (21)

Moreover, since (see [24])∣∣∣∣∣

n∑

i=1

A–α–1n–i Di(u)

∣∣∣∣∣

= O(|u|α–1), (22)


for III4, we get that

III4 ≤∫

G2m

Dnk1 Mk1(u)|v|α–1 dμ(u, v)

≤∫

Gm

|v|α–1 dμ(v) =1α

< ∞. (23)

Analogously, we find that

III5 ≤∫

G2m

Dnk1 Mk1(v)|u|α–1 dμ(u, v)

≤∫

Gm

|u|α–1 dμ(v) =1α

< ∞. (24)

For r ∈ {0, . . . mA – 1} and 0 ≤ j < MA (see [20]), this yields that

Dj+rMA =

( r–1∑

q=0

ψqMA

)

DMA + ψ rMA

Dj.

Thus we have

∫

G2m

nk1 Mk1 –1∑

i=1

A–α–1n–i Di(u)Di(v) dμ(u, v)

≤∫

G2m

nk1 –1∑

r=0

Mk1 –1∑

i=0

A–α–1n–i–rMk1

Di+rMk1(u)Di+rMk1

(v) dμ(u, v)

≤∫

G2m

nk1 –1∑

r=0

Mk1 –1∑

i=0


( r–1∑

q=0

ψqMk1

)

DMk1(u)

×( r–1∑

q=0

ψqMk1

)

DMk1(v) dμ(u, v)

+∫

G2m

nk1 –1∑

r=0

Mk1 –1∑

i=0


( r–1∑

q=0

ψqMk1

)

DMk1(u)ψ r

MADi(v) dμ(u, v)

+∫

G2m

nk1 –1∑

r=0

Mk1 –1∑

i=0


ψ rMA

Di(u)

( r–1∑

q=0

ψqMk1

)

DMk1(v) dμ(u, v)

+∫

G2m

nk1 –1∑

r=0

Mk1 –1∑

i=0


ψ rMA

Di(u)ψ rMA

Di(v) dμ(u, v).

On the other hand, by (1) and (4) we obtain that

∫

G2m

A–α–1n–nk1 Mk1

Dnk1 Mk1(u)Dnk1 Mk1

(v) dμ(u, v) ≤ c(α).


Consequently, for III1, we have the estimate

III1 ≤∫

G2m

DMk1(u)DMk1

(v)

∣∣∣∣∣

nk1 –1∑

r=0

Mk1∑

i=1


∣∣∣∣∣dμ(u, v)

+∫

G2m

DMk1(u)

∣∣∣∣∣

nk1 –1∑

r=0

Mk1∑

i=1


Di(v)

∣∣∣∣∣dμ(u, v)

+∫

G2m

DMk1(v)

∣∣∣∣∣

nk1 –1∑

r=0

Mk1∑

i=1


Di(u)

∣∣∣∣∣dμ(u, v)

+∫

G2m

∣∣∣∣∣

nk1 –1∑

r=0

Mk1∑

i=1


Di(u)Di(v)

∣∣∣∣∣dμ(u, v) + c(α)

:= III11 + III12 + III13 + III14 + c(α), (25)

where the first, second, third, and fourth terms on the right side of inequality (25) aredenoted by III11, III12, III13, and III14, respectively.

From Lemma 4 we have that

III14 ≤ c(α). (26)

The estimation of III11 is analogous to that of III3, and we find that

III11 ≤ c(α). (27)

The estimation of III12 and III13 is analogous to that of III4, and we obtain that

III12 < ∞ (28)

and

III13 < ∞. (29)

After substituting (21) and (23)–(29) into (20), we conclude that

∫

G2m

∣∣∣∣∣

n∑

i=1


∣∣∣∣∣dμ(u, v)

≤∫

G2m

∣∣∣∣∣

n(2)∑

i=1

A–α–1n(2)–iDi(u)Di(v)

∣∣∣∣∣dμ(u, v) + c(α)

≤ · · · ≤∫

G2m

∣∣∣∣∣

n(s)∑

i=1

A–α–1n(s)–iDi(u)Di(v)

∣∣∣∣∣dμ(u, v) + c(α)s

≤ c(α) + c(α)s ≤ c(α) log n.

The proof is complete. �


Now we are ready to prove the main results.

Proof of Theorem 1 It is evident that∥∥σ –α

Mk(f ) – f

∥∥p

≤ 1A–α

Mk –1

∥∥∥∥∥

∫

G2m

Mk–1∑

i=1

A–α–1Mk –iDi(u)Di(v)

[f (· – u, · – v) – f (·, ·)]dμ(u, v)

∥∥∥∥∥

p

+1

A–αMk –1

∥∥∥∥∥

∫

G2m

Mk∑

i=Mk–1+1


[f (· – u, · – v) – f (·, ·)]dμ(u, v)

∥∥∥∥∥

p

:= I + II. (30)

From Lemma 5 it follows that

I ≤ c(α)k–2∑

r=0

Mr

Mk

(ω1(f , 1/Mr)p + ω2(f , 1/Mr)p

). (31)

Moreover, for II , we have the estimate

II ≤ 1A–α

Mk –1

∥∥∥∥∥

∫

G2m

Mk∑

i=Mk–1+1


× [f (· – u, · – v) – S(1)

Mk–1(· – u, · – v, f )

]dμ(u, v)

∥∥∥∥∥

p

+1

A–αMk –1

∥∥∥∥∥

∫

G2m

Mk∑

i=Mk–1+1


× [S(1)

Mk–1(· – u, · – v, f ) – f (·, ·)]dμ(u, v)

∥∥∥∥∥

p

:= II1 + II2, (32)

where the first and second terms on the right side of inequality (32) are denoted by II1 andII2, respectively.

In view of the generalized Minkowski inequality, by (4) and Lemma 5 we get that

II1 ≤ 1A–α

Mk –1

∫

G2m

∣∣∣∣∣

Mk∑

i=Mk–1+1


∣∣∣∣∣

× ∥∥f (· – u, · – v) – S(1)

Mk–1(· – u, · – v, f )

∥∥

p dμ(u, v)

≤ c(α)Mαk ω1(f , 1/Mk–1)p. (33)

The estimation of II2 is analogous to that of II1, and we find that

II2 ≤ c(α)Mαk ω2(f , 1/Mk–1)p. (34)

Combining (30)–(34), we obtain the proof of Theorem 1. �


Proof of Theorem 2 It is evident that

∥∥σ –α

n (f ) – f∥∥

p

≤ 1A–α

n–1

∥∥∥∥∥

∫

G2m

Mk–1∑

i=1


[f (· – u, · – v) – f (·, ·)]dμ(u, v)

∥∥∥∥∥

p

+1

A–αn–1

∥∥∥∥∥

∫

G2m

Mk∑

i=Mk–1+1


[f (· – u, · – v) – f (·, ·)]dμ(u, v)

∥∥∥∥∥

p

+1

A–αn–1

∥∥∥∥∥

∫

G2m

n∑

i=Mk +1


[f (· – u, · – v) – f (·, ·)]dμ(u, v)

∥∥∥∥∥

p

:= I + II + III, (35)

where the first, second, and third terms on the right side of inequality (35) are denoted byI , II , and III , respectively.

From Lemma 4 it follows that

I ≤ c(α)k–2∑

r=0

Mr

Mk

(ω1(f , 1/Mr)p + ω2(f , 1/Mr)p

). (36)

Next, we repeat the arguments just in the same way as in the proof of Theorem 1 andfind that

II ≤ c(α)Mαk(ω1(f , 1/Mk–1)p + ω2(f , 1/Mk–1)p

). (37)

On the other hand, for III, we have

III ≤ 1A–α

n–1

∥∥∥∥∥

∫

G2m

n∑

i=Mk +1


× [f (· – u, · – v) – f (·, ·)]

∥∥∥∥∥

p

dμ(u, v)

≤ 1A–α

n

∥∥∥∥∥

∫

G2m

n∑

i=Mk +1


× [f (· – u, · – v) – SMk ,Mk (· – u, · – v, f )

]dμ(u, v)

∥∥∥∥∥

p

≤ 1A–α

n

∥∥∥∥∥

∫

G2m

n∑

i=Mk +1


× [SMk ,Mk (· – u, · – v, f ) – SMk ,Mk (·, ·, f )

]dμ(u, v)

∥∥∥∥∥

p


≤ 1A–α

n

∥∥∥∥∥

∫

G2m

n∑

i=Mk +1


× [SMk ,Mk (·, ·, f ) – f (·, ·)]dμ(u, v)

∥∥∥∥∥

p

:= III1 + III2 + III3, (38)

where the first, second, and third terms on the right side of inequality (38) are denoted byIII1, III2, and III3, respectively.

It is easy to show that

III2 = 0. (39)

By the generalized Minkowski inequality and Lemma 5, for III1, we obtain that

III1 ≤ 1A–α

n

∫

G2m

∣∣∣∣∣

n∑

i=Mk +1


∣∣∣∣∣

× ∥∥f (· – u, · – v) – SMr ,Mr (· – u, · – v, f )∥∥

p dμ(u, v)

≤ c(α)Mαk(ω1(f , 1/Mk–1)p + ω2(f , 1/Mk–1)p

)

×∫

G2m

∣∣∣∣∣

n∑

v=Mk +1

A–α–1n–v Dv(u)Dv(v)

∣∣∣∣∣dμ(u, v)

≤ c(α)Mαk log n

(ω1(f , 1/Mk–1)p + ω2(f , 1/Mk–1)p

). (40)

The estimation of III3 is analogous to that of III2, and we find that

III3 ≤ c(α)Mαk log n

(ω1(f , 1/Mk–1)p + ω2(f , 1/Mk–1)p

). (41)

After substituting (36)–(37) and (41) into (35), we obtain the proof of Theorem 2. �

AcknowledgementsThe authors would like to thank the referees for helpful suggestions.

FundingNot applicable.

Competing interestsThe authors declare that they have no competing interests.

Authors’ contributionsThe authors contributed equally to the writing of this paper. Both authors approved the final version of the manuscript.

Publisher’s NoteSpringer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Received: 22 August 2018 Accepted: 2 December 2018

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Some inequalities for Cesàro means of double Vilenkin...

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