# On the Convergence of Absolute Summability for Functions of Bounded Variation in Two Variables

Authors: Ying Mei; Dansheng Yu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

By adopting some new ideas, we obtain the estimates of an absolute convergence for the functions of the bounded variation in two variables. Our results generalize the related results of Humphreys and Bojanic (1999) and Wang and Yu (2003) from one dimension to two dimensions and can be applied to several summability methods.


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## Body

## 1. Introduction

Letf be 2π-periodic functions integrable on [-π,π]. Denote its Fourier series by
(1.1)S(f,x)~a02+∑k=1∞‍(akcoskx+bksinkx),
where the Fourier coefficients ak and bk are defined as follows:
(1.2)ak:=1π∫-ππ‍f(t)cosktdt,bk:=1π∫-ππ‍f(t)sinktdt.Denoted bySn(f,x) the nth partial sums of the Fourier series (1.1), that is,
(1.3)Sn(f,x):=a02+∑k=1n‍(akcoskx+bksinkx).Whenf is of bounded variation, Bojanić [1] obtained the following result on the rate of the convergence of approximation by Sn(f,x).Theorem A.
If a periodic functionf is of bounded variation on the interval [-π,π], then the following estimate holds for every x and n=1,2,…:
(1.4)|Sn(f,x)-f(x)|≤3n∑k=1n‍V0π/k(φx),
where φx(t):=f(x+t)+f(x-t)-f(x+0)-f(x-0) and V(φx,[0,t]) is the total variation of φx on [0,t].There are a lot of interesting generalizations that have been achieved by many authors (see [1–9]). Among them, Jenei [4] and Móricz [5] generalized Theorem A to the double Fourier series of functions of bounded variation; Humphreys and Bojanic [3], Wang and Yu [7] investigated the absolute convergence of Cesàro means of Fourier series.LetCα:=(cnkα) be the Cesàro matrix of order α, that is,
(1.5)cnkα:=An-kα-1Anα,k=0,1,…,n,
where
(1.6)Anα:=Γ(n+α+1)Γ(α+1)Γ(n+1),n=0,1,….Then, the so-called Cesàro means of orderα of (1.1) is defined by (denote by σnα(f,x)):
(1.7)σnα(f,x)=1Anα∑k=0n‍An-kα-1Sk(f,x).Humphreys and Bojanic [3] investigated the rate of the absolute convergence of Cesàro means of the series (1.1). Their result can be read as follows.Theorem B.
Letx∈[0,π] and f be 2π-periodic functions of bounded variation on [-π,π]. Then for α>0 and n≥2, one has
(1.8)Rnα(f,x)≤4αnπ∑k=1n‍V0π/k(φx),
where
(1.9)Rnα(f,x):=∑k=n+1∞‍|σkα(f,x)-σk-1α(f,x)|,φx(t):=f(x+t)+f(x-t)-f(x+0)-f(x-0)
and varab(f) be the total variation of f on [a,b].However, Wang and Yu [7] showed that Theorem A is not correct when 0<α<1. In fact, they proved the following.Theorem C.
Suppose0<α<1, x∈[0,π] and f are 2π-periodic functions of bounded variation on [-π,π]. Then for n≥2, one has
(1.10)Rnα(f,x)≤100α2nα∑k=1n‍kα-1V0π/k(φx),
and there exists a 2π-periodic function f* of bounded variation on [-π,π] and a point x∈[0,π] such that
(1.11)Rnα(f*,x)>120000αnα∑k=1n‍kα-1V0π/k(φx)(n≥8).Motivated by Theorems B and C, and the results of Jenei [4] and Móricz [5] on the double Fourier series, we will investigate the absolute convergence of a kind of very general summability of the double Fourier series. Our new results not only generalize Theorems B and C to the double case, but also can be applied to many other classical summability methods. We will present our main results on Section 2. Proofs will be given in Section 3. In Section 4, we will apply our results to some classical summability methods.


## 2. The Main Results

Letf(x,y) be a function periodic in each variable with period 2π and integrable on the two-dimensional torus 𝕋2=𝕋×𝕋 in Lebesgue’s sense, in symbol, f∈L(𝕋2). The double Fourier series of a complex-valued function f∈L(𝕋2) is defined by
(2.1)f(x,y)~∑k∈ℤ∑l∈ℤ‍f^(k,l)ei(kx+ly),
where the f^(k,l) are the Fourier coefficients of f:
(2.2)f^(k,l):=14π2∫𝕋2‍f(u,v)e-i(ku+lv)dudv,(k,l)∈ℤ2.LetR:=[a1,b1]×[a2,b2] be a bounded and closed rectangle on the plane. A function f(x,y) defined on R is said to be of bounded variation over R in the sense of Hardy-Krause, in symbol, f(x,y)∈BVH(R), if(i)
the total variationV(f,R) of f(x,y) over R is finite, that is,
(2.3)V(f,R):=sup∑j=1m∑k=1n‍|f(xj,yk)-f(xj-1,yk)-f(xj,yk-1)+f(xj,yk)|<∞,where the supremum is extended for all finite partitions(2.4)a1=x0<x1<⋯<xm=b1,a2=y0<y1<⋯<yn=b2of the intervals[a1,b1] and [a2,b2];(ii)
the marginal functionsf(·,a2) and f(a1,·) are of bounded variation over the intervals [a1,b1] and [a2,b2], respectively.For anyf(x,y)∈BVH(𝕋), define
(2.5)f(x,y):=14(f(x+0,y+0)+f(x-0,y+0)+f(x+0,y-0)+f(x-0,y-0)),φxy(u,v):=f(x+u,y+v)+f(x-u,y+v)+f(x+u,y-v)+f(x-u,y-v)-4f(x,y).For convenience, write(2.6)V00st(φxy):=V(φxy,[0,s]×[0,t]).For any double sequence{amn}, define
(2.7)Δ11amn:=amn-am-1,n-am,n-1+am-1,n-1,Δ10amn:=amn-am-1,n,Δ01amn:=amn-am,n-1.For any fourfold sequence{amnjk}, write
(2.8)Δ11amnjk:=amnjk-am,n,j+1,k-am,n,j,k+1+am,n,j+1,k+1,Δ01amnjk:=amnjk-am,n,j,k+1,Δ10amnjk:=am,n,j,k-am,n,j+1,k.A doubly infinite matrixT:=(tmnjk) is said to be doubly triangular if tmnjk=0 for j>m or k>n. The mnth term of the T-transform of the double Fourier series (2.1) is defined by
(2.9)Tmn(x,y):=∑μ=0m∑ν=0n‍tmnμνSμν(x,y),
where Sμν(x,y) is the μνth partial sum of (2.1), that is,
(2.10)Sμν(x,y):=∑|k|=0μ∑|l|=0ν‍f^(k,l)ei(kx+ly):=∑|k|=0μ∑|l|=0ν‍Akl(x,y).Write(2.11)t¯mnjk:=∑μ=jm∑ν=kn‍tmnjk,0≤j≤m,0≤k≤n,t¯mnjk=0,ifj>mork>n,t^mnjk(1,1):=t¯mnij-t¯m-1,n,i,j-t¯m,n-1,i,j+t¯m-1,n-1,i,j.Set(2.12)Rmn(1,1)(f;x,y):=∑j=m+1∞∑k=n+1∞‍|Δ11Tjk(x,y)|.Our main results are the following Theorems2.1 and 2.2.Theorem 2.1.
Letf(x,y) be a periodic function and f(x,y)∈BVH(𝕋2). Assume that T is a lower doubly triangular matrix satisfying
(2.13)t^mnk0(1,1)=t^mn0l(1,1)=0,k,l=0,1,…,(2.14)∑k=1m∑l=1n‍|t^mnkl(1,1)|=O(1),
and there exist constants α,β(0<α,β≤1) such that
(2.15)|∑k=1m∑l=1n‍t^mnkl(1,1)klsinkxsinly|={O(l-1-βxy-β),(x,y)∈[0,πm]×[πn,π],O(k-1-αx-αy),(x,y)∈[πm,π]×[0,πn],O(k-1-αl-1-βx-αy-β),(x,y)∈[πm,π]×[πn,π].
Then forn≥2, (x,y)∈𝕋2, one has
(2.16)Rmn(1,1)(f;x,y)≤C1mαnβ∑k=1m∑l=1n‍kα-1lβ-1V00(π/k)(π/l)(φxy).Theorem 2.2.
Letf(x,y) and T satisfy all the conditions of Theorem 2.1, except (2.15) is replaced by
(2.17)∑k=1m∑l=1n‍|Δk*l*(t^mnkl(1,1)kσ(k*)lσ(l*))|=O(mτ(k*)nτ(l*)),
where k*,l*=0 or 1, k*+l*≥1, τ(0)=σ(0)=0, τ(1)=σ(1)+1=2. Then for n≥2, (x,y)∈𝕋2, one has
(2.18)Rmn(1,1)(f;x,y)≤C1mn∑k=1m∑l=1n‍V00(π/k)(π/l)(φxy).


## 3. Proofs of Results

Lemma 3.1 (see [10]).
Ifg(x,y) is continuous on a rectangle R:=[a1,b1]×[a2,b2] and f(x,y)∈BVH(R), then(i)
g(x,y) is integrable with respect to f(x,y) over R in the sense of Riemann-Stieltjes integral;(ii)
f(x,y) is integrable with respect to g(x,y) over R in the sense of Riemann-Stieltjes integral and
(3.1)∫R‍f(x,y)dxdyg(x,y)=∫R‍g(x,y)dxdyf(x,y)+∫a1b1‍f(x,b2)dxg(x,b2)-∫a1b1‍f(x,a2)dxg(x,a2)+∫a2b2‍f(b1,y)dyg(b1,y)-∫a2b2‍f(a1,y)dyg(a1,y)-f(b1,b2)g(b1,b2)+f(a1,b2)g(a1,b2)+f(a2,b1)g(a2,b1)-f(a1,a2)g(a1,a2).Lemma 3.2 (see [10]).
Iff(x,y)∈BVH(R) and g(x,y) is absolutely continuous on R, that is,
(3.2)g(x,y)=∫a1x‍∫a2y‍h(u,v)dudv+C,(x,y)∈R,
for some function h(x,y)∈L(R) and constant C, then
(3.3)∫R‍f(x,y)dxdyg(x,y)=∫R‍f(x,y)h(x,y)dxdy.Proof of Theorem2.1.
By the definition ofTmn(x,y) (see (2.9)), we have
(3.4)Tjk(x,y)=∑i1=0j∑i2=0k‍tjki1i2∑|μ|=0i1∑|ν|=0i2‍Aμν(x,y)=∑|μ|=0j∑|ν|=0k‍Aμν(x,y)t¯jk|μ||ν|.
Therefore,(3.5)Δ11Tjk(x,y)=∑|μ|=0j∑|ν|=0k‍Aμν(x,y)t¯jk|μ||ν|-∑|μ|=0j-1∑|ν|=0k‍Aμν(x,y)t¯j-1,k,|μ|,|ν|-∑|μ|=0j∑|ν|=0k-1‍Aμν(x,y)t¯j,k-1,|μ|,|ν|+∑|μ|=0j-1∑|ν|=0k-1‍Aμν(x,y)t¯j-1,k-1,|μ|,|ν|=∑|μ|=0j∑|ν|=0k‍Aμν(x,y)t^jk|μ||ν|(1,1)(by(2.11))=∑μ=1j∑ν=1k‍(Aμν(x,y)+A-μ,ν(x,y)+Aμ,-ν(x,y)+A-μ,-ν(x,y))t^jkμν(1,1)(by(2.13)).
Noting that(3.6)Aμν(x,y)+A-μ,ν(x,y)+Aμ,-ν(x,y)+A-μ,-ν(x,y)=f^(μ,ν)ei(μx+νy)+f^(-μ,ν)ei(-μx+νy)+f^(μ,-ν)ei(μx-νy)+f^(-μ,-ν)ei(-μx-νy)=14π2∫𝕋2‍f(s,t)cosμ(s-x)cosν(t-y)dsdt,
we have
(3.7)Δ11Tjk(x,y)=14π2∑μ=1j∑ν=1k‍t^jkμν(1,1)∫𝕋2‍f(s,t)cosμ(s-x)cosν(t-y)dsdt=14π2∑μ=1j∑ν=1k‍t^jkμν(1,1)∬0πφxy(s,t)cosμscosνtdsdt=14π2∑μ=1j∑ν=1k‍t^jkμν(1,1)μν∬0πφxy(s,t)dsdtsinμssinνt,
where in the last equality, Lemma 3.2 is applied.
By (3.7) and Lemma 3.1, we have
(3.8)Rmn(1,1)(f;x,y)=14π2∑j=m+1∞∑k=n+1∞‍|∑μ=1j∑ν=1k‍t^jkμν(1,1)μν∬0πφxy(s,t)dsdtsinμssinνt|=14π2∑j=m+1∞∑k=n+1∞‍|∑μ=1j∑ν=1k‍t^jkμν(1,1)μν∬0πsinμssinνtdsdtφxy(s,t)|≤14π2∑j=m+1∞∑k=n+1∞‍∬0π|∑μ=1j∑ν=1k‍t^jkμν(1,1)μνsinμssinνt|dsdtV00st(φxy)=14π2∑j=m+1∞∑k=n+1∞‍(∫0π/j‍∫0π/k‍+∫0π/j‍∫π/kπ‍+∫π/jπ‍∫0π/k‍+∫π/jπ‍∫π/kπ‍)×|∑μ=1j∑ν=1k‍t^jkμν(1,1)μνsinμssinνt|dsdtV00st(φxy)=:I1+I2+I3+I4.
By (2.14), we have (Denote by χ(a1,b1)×(a2,b2)(x,y) and χ(a1,b1)(x) the characteristic functions of (a1,b1)×(a2,b2) and (a1,b1), respectively.)(3.9)I1=14π2∑j=m+1∞∑k=n+1∞‍∫0π/j‍∫0π/k‍|∑μ=1j∑ν=1k‍t^jkμνμνsinμssinνt|dsdtV00st(φxy)≤14π2∑j=m+1∞∑k=n+1∞‍∫0π/j‍∫0π/k‍∑μ=1j∑ν=1k‍|t^jkμν|stdsdtV00st(φxy)≤C∑j=m+1∞∑k=n+1∞‍∫0π/j‍∫0π/k‍stdsdtV00st(φxy)≤C∫0π/(m+1)‍∫0π/(n+1)‍∑j=m+1∞∑k=n+1∞‍stχ(0,π/j)×(0,π/k)(s,t)dsdtV00st(φxy)≤C∫0π/(m+1)‍∫0π/(n+1)‍dsdtV00st(φxy)≤CV00(π/(m+1))(π/(n+1))(φxy)≤Cmαnβ∑j=1m∑k=1n‍jα-1kβ-1V00(π/j)(π/k)(φxy).
ForI2, by the first inequality of (2.15),
(3.10)I2=14π2∑j=m+1∞∑k=n+1∞‍∫0π/j‍∫π/kπ‍|∑μ=1j‍∑ν=1k‍t^jkμνμνsinμssinνt|dsdtV00st(φxy)≤C∑j=m+1∞∑k=n+1∞‍k-1-β∫0π/j‍∫π/kπ‍st-βdsdtV00st(φxy)≤C∑j=m+1∞∑k=n+1∞‍k-1-β(∫0π/j‍∫π/kπ/(n+1)‍+∫0π/j‍∫π/(n+1)π‍)st-βdsdtV00st(φxy)≤C∫0π/(m+1)‍∫0π/(n+1)‍∑j=m+1∞‍sχ(0,π/j)(s)∑k>(π/t)∞‍k-1-βt-βdsdtV00st(φxy)+C∫0π/(m+1)‍∫π/(n+1)π‍∑j=m+1∞‍sχ(0,π/j)(s)∑k=n+1∞‍k-1-βt-βdsdtV00st(φxy)≤C∫0π/(m+1)‍∫0π/(n+1)‍dsdtV00st(φxy)+Cn-β∫0π/(m+1)‍∫π/(n+1)π‍t-βdsdtV00st(φxy)≤CV00(π/(m+1))(π/(n+1))(φxy)+Cn-β∑k=1n‍∫0π/(m+1)‍∫π/(k+1)π/k‍t-βdsdtV00st(φxy)≤CV00(π/(m+1))(π/(n+1))(φxy)+Cn-β∑k=1n‍kβ-1V00(π/(m+1))(π/k)(φxy)≤Cmαnβ∑j=1m‍∑k=1n‍jα-1kβ-1V00(π/j)(π/k)(φxy).
Similarly, we also have(3.11)I3≤Cmαnβ∑j=1m∑k=1n‍jα-1kβ-1V00(π/j)(π/k)(φxy).
By (2.15), we deduce that
(3.12)I4≤C∑j=m+1∞∑k=n+1∞‍∫π/jπ‍∫π/kπ‍j-1-αk-1-βs-αt-βdsdtV00st(φxy)≤C∑j=m+1∞∑k=n+1∞(∫π/jπ/(m+1)‍∫π/kπ/(n+1)‍+∫π/(m+1)π‍∫π/kπ/(n+1)‍+∫π/jπ/(m+1)‍∫π/(n+1)π‍+∫π/(m+1)π‍∫π/(n+1)π‍)×j-1-αk-1-βs-αt-βdsdtV00st(φxy)≤C∫0π/(m+1)‍∫0π/(n+1)‍(∑j>(π/s)∞‍j-1-αs-α)(∑k>(π/t)∞‍k-1-βt-β)dsdtV00st(φxy)+C∫0π/(m+1)‍∫π/(n+1)π‍(∑j>(π/s)∞‍j-1-αs-α)(∑k=n+1∞‍k-1-βt-β)dsdtV00st(φxy)+C∫π/(m+1)π‍∫0π/(n+1)‍(∑j=m+1∞‍j-1-αs-α)(∑k>(π/t)∞‍k-1-βt-β)dsdtV00st(φxy)+C∫π/(m+1)π‍∫π/(n+1)π‍(∑j=m+1∞‍j-1-αs-α)(∑k=n+1∞‍k-1-βt-β)dsdtV00st(φxy)≤CV00(π/(m+1))(π/(n+1))(φxy)+Cn-β∑k=1n‍kβ-1V00(π/(m+1))(π/k)(φxy)+Cm-α∑j=1m‍jα-1V00(π/j)(π/(n+1))(φxy)+Cmαnβ∑j=1m∑k=1n‍jα-1kβ-1V00(π/j)(π/k)(φxy)≤CCmαnβ∑j=1m∑k=1n‍jα-1kβ-1V00(π/(m+1))(π/(n+1))(φxy).
We get (2.16) by combining (3.8)–(3.12).Proof of Theorem2.2.
Set(3.13)D~i(t)=∑j=1i‍sinit=cos(1/2)t-cos(i+(1/2))t2sin(1/2)t,i=1,2,….
By Abel’s transformation, we have(3.14)∑k=1m∑l=1n‍t^mnkl(1,1)klsinkssinlt=∑k=1m‍sinksk∑l=1n‍Δ01t^mnkl(1,1)lD~l(t)=∑l=1n‍sinltl∑k=1m‍Δ10t^mnkl(1,1)kD~k(s)=∑k=1m∑l=1n‍Δ11t^mnkl(1,1)klD~k(s)D~l(t).
Therefore, by the following well-known inequalities:|D~i(t)|=O(t-1), |sint|≤|t|, and condition (2.17), we see that (2.15) holds with α=β=1, and thus we get Theorem 2.2 from Theorem 2.1.


## 4. Applications

### 4.1. Cesàro’s Means

LetT=(C,γ,δ)=Cγ×Cδ be the double Cesàro matrix of order (γ,δ) with γ,δ>-1, that is, T has entries
(4.1)tmnjk=Cmjγ×Cnkδ=Am-jγ-1An-kδ-1AmγAnδ.Then for0≤j≤m-1, 0≤k≤n-1,
(4.2)t^mnjk(1,1)=Δ11t¯m-1,n-1,j,k=(∑u=jm-1‍Am-1-uγ-1Am-1γ-∑u=jm‍Am-uγ-1Amγ)(∑v=kn-1‍An-1-vδ-1An-1δ-∑v=kn‍An-vδ-1Anδ)=(Am-jγAmγ-Am-1-jγAm-1γ)(An-kδAnδ-An-1-kδAn-1δ)=jkAm-jγ-1An-kδ-1mnAmγAnδ,t^mnmn(1,1)=tmnmn=1AmαAnβ.In particular,(4.3)t^mn0k(1,1)=t^mnj0(1,1)=0,j,k=0,1,….In other words,T satisfies condition (2.13) of Theorem 2.1.By the well-known inequality (e.g., see [11])
(4.4)Anγ=nγΓ(γ+1)(1+O(1n)),
we deduce that
(4.5)∑j=1m∑k=1n‍|t^mnjk(1,1)|=(∑j=1m‍jAmγ-1mAmγ+1Amγ)(∑k=1n‍kAn-kδ-1nAnδ+1Anδ)=(O(m-2)∑j=1m/2‍j+O(m-γ)∑j=m/2m‍Am-jγ-1+1Amγ)×(O(n-2)∑k=1n/2‍k+O(n-δ)∑k=n/2n‍An-kδ-1+1Anδ)=O(1),γ,δ>0,
which means condition (2.14) of Theorem 2.1.For0<γ<1, we have
(4.6)∑k=1n‍An-kγ-1nAnγsinkx=Im(1nAnγ∑k=1n‍An-kγ-1eikx)=Im(einxnAnγ∑k=0n-1‍Akγ-1e-ikx)=:Im(J1+J2),
where
(4.7)J1:=einxnAnγ(1-e-ix)-γ,J2:=einxnAnγ∑k=n∞‍Akγ-1e-ikx.By (4.4), we observe that
(4.8)|J1|=O(n-1-γx-γ).SinceAkγ-1 decreases monotonically to 0, J2 converges for 0<x≤π, and
(4.9)|J2|=O(Anγ-1nAnγ|1-e-ix|-1)=O(n-2x-1)=O(n-1-γx-γ)
for t∈[(π/n),π].By (4.4)–(4.9), we obtain that
(4.10)∑k=1n‍An-kγ-1nAnγsinkx=O(n-1-γx-γ),x∈[πn,π],0<γ≤1.On the other hand, we have(4.11)∑k=1n‍An-kγ-1nAnγsinkx=O(x)∑k=1n‍kAn-kγ-1nAnγ=O(x),x∈[0,πn],γ>0.Forγ≥1, we have that
(4.12)∑k=1n-1‍|Δ(An-kγ-1nAnγ)|=1nAnγ∑k=1n-1‍(An-kγ-1-An-k-1γ-1)+1nAnγ=O(n-2).Now, by (4.10) and (4.11), we deduce that
(4.13)|∑j=1m∑k=1n‍t^mnjk(1,1)jksinjxsinky|≤|∑j=1m‍Am-jγ-1mAmγsinjx||∑k=1n‍An-kδ-1nAnδsinky|={O(k-1-δxy-δ),(x,y)∈[0,πm]×[πn,π],O(j-1-γx-γy),(x,y)∈[πm,π]×[0,πn],O(j-1-γk-1-δx-γy-δ),(x,y)∈[πm,π]×[πn,π]
for 0<γ, δ≤1.Whenγ>1, 0<δ≤1, by (4.10) and (4.11) again,
(4.14)|∑j=1m∑k=1n‍t^mnjk(1,1)jksinjxsinky|≤|∑j=1m‍Am-jγ-1mAmγsinjx||∑k=1n‍An-kδ-1nAnδsinky|=O(k-1-δxy-δ),
for (x,y)∈[0,(π/m)]×[(π/n),π]. By (4.10)–(4.12), we have
(4.15)|∑j=1m∑k=1n‍t^mnjk(1,1)jksinjxsinky|≤|∑j=1m‍Δ(Am-jγ-1mAmγ)Dj(x)||∑k=1n‍An-kδ-1nAnδsinky|={O(j-2x-1y),(x,y)∈[πm,π]×[0,πn],O(j-2k-1-δx-1y-δ),(x,y)∈[πm,π]×[πn,π].Similarly, we have(4.16)|∑j=1m∑k=1n‍t^mnjk(1,1)jksinjxsinky|={O(k-2xy-1),(x,y)∈[0,πm]×[πn,π],O(j-1-γx-γy),(x,y)∈[πm,π]×[0,πn],O(j-1-γk-2x-γy-1),(x,y)∈[πm,π]×[πn,π],
for δ>1,0<γ≤1,and
(4.17)|∑j=1m∑k=1n‍t^mnjk(1,1)jksinjxsinky|={O(k-2xy-1),(x,y)∈[0,πm]×[πn,π],O(j-2x-1y),(x,y)∈[πm,π]×[0,πn],O(j-2k-2x-1y-1),(x,y)∈[πm,π]×[πn,π],
for γ,δ>1.Setting(4.18)γ~:={γ,0<γ≤1,1,γ>1,δ~:={δ,0<δ≤1,1,δ>1,
then we have
(4.19)|∑j=1m∑k=1n‍t^mnjk(1,1)jksinjxsinky|={O(k-1-δ~xy-1),(x,y)∈[0,πm]×[πn,π],O(j-1-γ~x-1y),(x,y)∈[πm,π]×[0,πn],O(j-1-γ~k-1-δ~x-1y-1),(x,y)∈[πm,π]×[πn,π],
by (4.13)–(4.17).From (4.19), we have the following corollary of Theorem 2.1, which generalizes Theorem B from one dimension to two dimensions.Theorem 4.1.
Letf(x,y) be a periodic function and f(x,y)∈BVH(𝕋2) and T=(C,γ,δ) be the double Cesàro matrix of order (γ,δ) with γ,δ>0. Then for n≥2, (x,y)∈𝕋2, one has
(4.20)Rmn(1,1)(f;x,y)≤C1mγ~nδ~∑k=1m∑l=1n‍kγ~-1lδ~-1V00(π/k)(π/l)(φxy).


### 4.2. Bernstein-Rogosinski’s Means

The so-called Bernstein-Rogosinski means of Fourier series (1.1) are defined by
(4.21)Bn(f,x):=12(Sn(f,x+12n+1π)+Sn(f,x-12n+1π)).Now, we introduce the following Bernstein-Rogosinski means of double Fourier series (2.1):
(4.22)Bmn(f,x):=14(Smn(f;x+π2m+1,y+π2n+1)+Smn(f;x+π2m+1,y-π2n+1)+Smn(f;x-π2m+1,y+π2n+1)+Smn(f;x-π2m+1,y-π2n+1)).Direct calculations yield that(4.23)Bmn(f,x)=∑j=0m∑k=0n‍cosjπ2m+1coskπ2n+1Δ11Sjk(x,y)=∑j=0m-1∑k=0n-1‍(cosj2m+1π-cosj+12m+1π)(cosk2n+1π-cosk+12n+1π)Sjk(x,y)+∑j=0m-1‍(cosj2m+1π-cosj+12m+1π)cosnπ2n+1Sjn(x,y)+∑k=0n-1‍(cosk2n+1π-cosk+12n+1π)cosmπ2m+1Smk(x,y)+cosmπ2m+1cosnπ2n+1Smn(x,y),Thus,Bmn(f,x) can be regarded as a T-transformation of Fourier series (1.1) with T of entries defined as
(4.24)tmnjk=(cosj2m+1π-cosj+12m+1π)(cosk2n+1π-cosk+12n+1π),0≤j<m,0≤k<n,tmnjn=(cosj2m+1π-cosj+12n+1π)cosnπ2n+1,0≤j<m,tmnmk=(cosk2n+1π-cosk+12n+1π)cosmπ2m+1,0≤k<n,tmnmn=cosmπ2m+1cosnπ2n+1.By direct calculations, we have(4.25)t^mnjk(1,1)=(cosj2m-1π-cosj2m+1π)(cosk2n-1π-cosk2n+1π),0≤j<m,0≤k<n,(4.26)t^mnjn(1,1)=(cosj2m-1π-cosj2m+1π)cosnπ2n+1,0≤j<m,(4.27)t^mnmk(1,1)=(cosk2n-1π-cosk2n+1π)cosmπ2m+1,0≤k<n,(4.28)t^mnmn(1,1)=cosmπ2m+1cosnπ2n+1.Therefore, (4.26) and (4.27) imply (2.13), while from (4.25)–(4.28), we get
(4.29)∑j=1m∑k=1n‍|t^mnjk(1,1)|≤∑j=1m-1∑k=1n-1‍|cosj2m-1π-cosj2m+1π||cosk2n-1π-cosk2n+1π|+∑j=1m-1‍|cosj2m-1π-cosj2m+1π|cosnπ2n+1+∑k=1n-1‍|cosk2n-1π-cosk2n+1π|cosmπ2m+1+cosmπ2m+1cosnπ2n+1≤4∑j=1m-1∑k=1n-1‍sin2mj4m2-1πsinj4m2-1πsin2nk4n2-1πsink4n2-1π+2∑j=1m-1‍sin2mj4m2-1πsinj4m2-1π+2∑k=1n-1‍sin2nk4n2-1πsink4n2-1π+1≤4π4∑j=1m-1∑k=1n-1‍2mj2(4m2-1)22nk2(4n2-1)2+2π2∑j=1m-1‍2mj2(4m2-1)2+2π2∑k=1n-1‍2nk2(4n2-1)2+1=O(1),
which implies (2.14).Finally, we verify thatT satisfies condition (2.17); thus T satisfies all the conditions of Theorem 2.2. By direct calculations, we deduce that
(4.30)∑j=1m∑k=1n‍|Δ01t^mnjk(1,1)k|=(∑j=1m‍|cosj2m-1π-cosj2m+1π|+cosmπ2m+1)×(∑k=1n-1‍|cos(k/(2n-1))π-cos(k/(2n+1))πk-cos((k+1)/(2n-1))π-cos((k+1)/(2n+1))πk+1|+1ncosnπ2n+1)=O(∑k=1n-1‍|cos(k/(2n-1))π-cos(k/(2n+1))πk-cos((k+1)/(2n-1))π-cos((k+1)/(2n+1))πk+1|+n-2)=O(1)∑k=1n-1‍1k((cosk2n-1π-cosk2n+1π)-(cosk+12n-1π-cosk+12n+1π))+O(1)∑k=1n-1‍(1k-1k+1)|cosk+12n-1π-cosk+12n+1π|=∑k=1n-1‍1k|(sin2nkπ4n2-1-sin2n(k+1)π4n2-1)sinkπ4n2-1+(sinkπ4n2-1-sin(k+1)π4n2-1)sin2n(k+1)π4n2-1|+∑k=1n-1‍1k(k+1)sin2n(k+1)π4n2-1sin(k+1)π4n2-1=O(1)∑k=1n-1‍1ksinnπ4n2-1sinkπ4n2-1+O(1)∑k=1n-1‍1k2sin2n(k+1)π4n2-1sin(k+1)π4n2-1=O(1)∑k=1n-1‍n-3=O(n-2).Analogously,(4.31)∑j=1m∑k=1n‍|Δ10t^mnjk(1,1)j|=O(m-2),∑j=1m∑k=1n‍|Δ11t^mnjk(1,1)j|=O(m-2n-2).Now, (4.30) and (4.31) imply (2.17).


### 4.3. Riesz’s Means

Let{pn} and {qn} be positive sequences such that Pn=p0+p1+⋯+pn→∞, Qn = q0+q1+⋯+qn → ∞, and let T:=(tmnjk) be a lower triangular matrix with entries tmnjk = pjqk/PmQn, j=0,1,…m, k=0,1,…,n; m,n=1,2,…. Then the T-transformation of Fourier series (2.1) is known as Riesz’s mean.Proposition 4.2.
Let{pn} and {qn} be positive nondecreasing sequences such that (one says An~Bn, if there are two positive constants C1 and C2 such that C1≤(An/Bn)≤C2.)(4.32)npn~Pn,nqn~Qn,(4.33)∑k=1n‍pkk=O(pn),∑k=1n‍qkk=O(qn).
ThenT satisfies all the conditions of Theorem 2.2.Proof.
Direct calculations yield that (setP-1=Q-1=0)
(4.34)t^mnjk(1,1)=(∑i=jm‍piPm-∑i=jn-1‍piPm-1)(∑i=kn‍qiQn-∑i=kn-1‍qiQn-1)=pmqnPj-1Qk-1PmPm-1QnQn-1,0≤j<m,0≤k<n,(4.35)t^mnmk(1,1)=pmPm(∑i=kn‍qiQn-∑i=kn-1‍qiQn-1)=pmqnQk-1PmQnQn-1,0≤k<n,t^mnjn(1,1)=(∑i=jm‍piPm-∑i=jn-1‍piPm-1)qnQn=pmqnPj-1PmPm-1Qn,0≤j<m,(4.36)t^mnmn(1,1)=pmqnPmQn.
It follows from (4.35) that T satisfies condition (2.13).
Since{pn} and {qn} are nondecreasing, by (4.32), (4.34)–(4.36), we have
(4.37)∑j=1m∑k=1n‍|t^mnjk(1,1)|=pmqnPmPm-1QnQn-1∑j=1m∑k=1n‍Pj-1Qk-1=O(mnpmqnPmQn)=O(1).
Hence,T satisfies condition (2.14).
Since{pn} and {qn} are nondecreasing, by (4.32)–(4.36), we have
(4.38)∑j=1m∑k=1n‍|Δ01(t^mnjk(1,1)k)|=(pmPmPm-1∑j=1m-1‍Pj-1+pmPm)×(qnQnQn-1∑k=1n-1‍|Qk-1k-Qkk+1|+qnnPn)=O(1)(qnQnQn-1(∑k=1n-1‍Qk-1k(k+1)+∑k=2n-1‍qkk+1)+O(n-2))=O(1)(1nQn-1∑k=1n-1‍qkk+n-2)=O(qn-1nQn-1)=O(n-2).
Similarly,(4.39)∑j=1m∑k=1n‍|Δ10t^mnjk(1,1)j|=O(m-2),∑j=1m∑k=1n‍|Δ11t^mnjk(1,1)j|=O(m-2n-2).
Now, by (4.38) and (4.39), we show that T also satisfies condition (2.17).


## 4.1. Cesàro’s Means

LetT=(C,γ,δ)=Cγ×Cδ be the double Cesàro matrix of order (γ,δ) with γ,δ>-1, that is, T has entries
(4.1)tmnjk=Cmjγ×Cnkδ=Am-jγ-1An-kδ-1AmγAnδ.Then for0≤j≤m-1, 0≤k≤n-1,
(4.2)t^mnjk(1,1)=Δ11t¯m-1,n-1,j,k=(∑u=jm-1‍Am-1-uγ-1Am-1γ-∑u=jm‍Am-uγ-1Amγ)(∑v=kn-1‍An-1-vδ-1An-1δ-∑v=kn‍An-vδ-1Anδ)=(Am-jγAmγ-Am-1-jγAm-1γ)(An-kδAnδ-An-1-kδAn-1δ)=jkAm-jγ-1An-kδ-1mnAmγAnδ,t^mnmn(1,1)=tmnmn=1AmαAnβ.In particular,(4.3)t^mn0k(1,1)=t^mnj0(1,1)=0,j,k=0,1,….In other words,T satisfies condition (2.13) of Theorem 2.1.By the well-known inequality (e.g., see [11])
(4.4)Anγ=nγΓ(γ+1)(1+O(1n)),
we deduce that
(4.5)∑j=1m∑k=1n‍|t^mnjk(1,1)|=(∑j=1m‍jAmγ-1mAmγ+1Amγ)(∑k=1n‍kAn-kδ-1nAnδ+1Anδ)=(O(m-2)∑j=1m/2‍j+O(m-γ)∑j=m/2m‍Am-jγ-1+1Amγ)×(O(n-2)∑k=1n/2‍k+O(n-δ)∑k=n/2n‍An-kδ-1+1Anδ)=O(1),γ,δ>0,
which means condition (2.14) of Theorem 2.1.For0<γ<1, we have
(4.6)∑k=1n‍An-kγ-1nAnγsinkx=Im(1nAnγ∑k=1n‍An-kγ-1eikx)=Im(einxnAnγ∑k=0n-1‍Akγ-1e-ikx)=:Im(J1+J2),
where
(4.7)J1:=einxnAnγ(1-e-ix)-γ,J2:=einxnAnγ∑k=n∞‍Akγ-1e-ikx.By (4.4), we observe that
(4.8)|J1|=O(n-1-γx-γ).SinceAkγ-1 decreases monotonically to 0, J2 converges for 0<x≤π, and
(4.9)|J2|=O(Anγ-1nAnγ|1-e-ix|-1)=O(n-2x-1)=O(n-1-γx-γ)
for t∈[(π/n),π].By (4.4)–(4.9), we obtain that
(4.10)∑k=1n‍An-kγ-1nAnγsinkx=O(n-1-γx-γ),x∈[πn,π],0<γ≤1.On the other hand, we have(4.11)∑k=1n‍An-kγ-1nAnγsinkx=O(x)∑k=1n‍kAn-kγ-1nAnγ=O(x),x∈[0,πn],γ>0.Forγ≥1, we have that
(4.12)∑k=1n-1‍|Δ(An-kγ-1nAnγ)|=1nAnγ∑k=1n-1‍(An-kγ-1-An-k-1γ-1)+1nAnγ=O(n-2).Now, by (4.10) and (4.11), we deduce that
(4.13)|∑j=1m∑k=1n‍t^mnjk(1,1)jksinjxsinky|≤|∑j=1m‍Am-jγ-1mAmγsinjx||∑k=1n‍An-kδ-1nAnδsinky|={O(k-1-δxy-δ),(x,y)∈[0,πm]×[πn,π],O(j-1-γx-γy),(x,y)∈[πm,π]×[0,πn],O(j-1-γk-1-δx-γy-δ),(x,y)∈[πm,π]×[πn,π]
for 0<γ, δ≤1.Whenγ>1, 0<δ≤1, by (4.10) and (4.11) again,
(4.14)|∑j=1m∑k=1n‍t^mnjk(1,1)jksinjxsinky|≤|∑j=1m‍Am-jγ-1mAmγsinjx||∑k=1n‍An-kδ-1nAnδsinky|=O(k-1-δxy-δ),
for (x,y)∈[0,(π/m)]×[(π/n),π]. By (4.10)–(4.12), we have
(4.15)|∑j=1m∑k=1n‍t^mnjk(1,1)jksinjxsinky|≤|∑j=1m‍Δ(Am-jγ-1mAmγ)Dj(x)||∑k=1n‍An-kδ-1nAnδsinky|={O(j-2x-1y),(x,y)∈[πm,π]×[0,πn],O(j-2k-1-δx-1y-δ),(x,y)∈[πm,π]×[πn,π].Similarly, we have(4.16)|∑j=1m∑k=1n‍t^mnjk(1,1)jksinjxsinky|={O(k-2xy-1),(x,y)∈[0,πm]×[πn,π],O(j-1-γx-γy),(x,y)∈[πm,π]×[0,πn],O(j-1-γk-2x-γy-1),(x,y)∈[πm,π]×[πn,π],
for δ>1,0<γ≤1,and
(4.17)|∑j=1m∑k=1n‍t^mnjk(1,1)jksinjxsinky|={O(k-2xy-1),(x,y)∈[0,πm]×[πn,π],O(j-2x-1y),(x,y)∈[πm,π]×[0,πn],O(j-2k-2x-1y-1),(x,y)∈[πm,π]×[πn,π],
for γ,δ>1.Setting(4.18)γ~:={γ,0<γ≤1,1,γ>1,δ~:={δ,0<δ≤1,1,δ>1,
then we have
(4.19)|∑j=1m∑k=1n‍t^mnjk(1,1)jksinjxsinky|={O(k-1-δ~xy-1),(x,y)∈[0,πm]×[πn,π],O(j-1-γ~x-1y),(x,y)∈[πm,π]×[0,πn],O(j-1-γ~k-1-δ~x-1y-1),(x,y)∈[πm,π]×[πn,π],
by (4.13)–(4.17).From (4.19), we have the following corollary of Theorem 2.1, which generalizes Theorem B from one dimension to two dimensions.Theorem 4.1.
Letf(x,y) be a periodic function and f(x,y)∈BVH(𝕋2) and T=(C,γ,δ) be the double Cesàro matrix of order (γ,δ) with γ,δ>0. Then for n≥2, (x,y)∈𝕋2, one has
(4.20)Rmn(1,1)(f;x,y)≤C1mγ~nδ~∑k=1m∑l=1n‍kγ~-1lδ~-1V00(π/k)(π/l)(φxy).


## 4.2. Bernstein-Rogosinski’s Means

The so-called Bernstein-Rogosinski means of Fourier series (1.1) are defined by
(4.21)Bn(f,x):=12(Sn(f,x+12n+1π)+Sn(f,x-12n+1π)).Now, we introduce the following Bernstein-Rogosinski means of double Fourier series (2.1):
(4.22)Bmn(f,x):=14(Smn(f;x+π2m+1,y+π2n+1)+Smn(f;x+π2m+1,y-π2n+1)+Smn(f;x-π2m+1,y+π2n+1)+Smn(f;x-π2m+1,y-π2n+1)).Direct calculations yield that(4.23)Bmn(f,x)=∑j=0m∑k=0n‍cosjπ2m+1coskπ2n+1Δ11Sjk(x,y)=∑j=0m-1∑k=0n-1‍(cosj2m+1π-cosj+12m+1π)(cosk2n+1π-cosk+12n+1π)Sjk(x,y)+∑j=0m-1‍(cosj2m+1π-cosj+12m+1π)cosnπ2n+1Sjn(x,y)+∑k=0n-1‍(cosk2n+1π-cosk+12n+1π)cosmπ2m+1Smk(x,y)+cosmπ2m+1cosnπ2n+1Smn(x,y),Thus,Bmn(f,x) can be regarded as a T-transformation of Fourier series (1.1) with T of entries defined as
(4.24)tmnjk=(cosj2m+1π-cosj+12m+1π)(cosk2n+1π-cosk+12n+1π),0≤j<m,0≤k<n,tmnjn=(cosj2m+1π-cosj+12n+1π)cosnπ2n+1,0≤j<m,tmnmk=(cosk2n+1π-cosk+12n+1π)cosmπ2m+1,0≤k<n,tmnmn=cosmπ2m+1cosnπ2n+1.By direct calculations, we have(4.25)t^mnjk(1,1)=(cosj2m-1π-cosj2m+1π)(cosk2n-1π-cosk2n+1π),0≤j<m,0≤k<n,(4.26)t^mnjn(1,1)=(cosj2m-1π-cosj2m+1π)cosnπ2n+1,0≤j<m,(4.27)t^mnmk(1,1)=(cosk2n-1π-cosk2n+1π)cosmπ2m+1,0≤k<n,(4.28)t^mnmn(1,1)=cosmπ2m+1cosnπ2n+1.Therefore, (4.26) and (4.27) imply (2.13), while from (4.25)–(4.28), we get
(4.29)∑j=1m∑k=1n‍|t^mnjk(1,1)|≤∑j=1m-1∑k=1n-1‍|cosj2m-1π-cosj2m+1π||cosk2n-1π-cosk2n+1π|+∑j=1m-1‍|cosj2m-1π-cosj2m+1π|cosnπ2n+1+∑k=1n-1‍|cosk2n-1π-cosk2n+1π|cosmπ2m+1+cosmπ2m+1cosnπ2n+1≤4∑j=1m-1∑k=1n-1‍sin2mj4m2-1πsinj4m2-1πsin2nk4n2-1πsink4n2-1π+2∑j=1m-1‍sin2mj4m2-1πsinj4m2-1π+2∑k=1n-1‍sin2nk4n2-1πsink4n2-1π+1≤4π4∑j=1m-1∑k=1n-1‍2mj2(4m2-1)22nk2(4n2-1)2+2π2∑j=1m-1‍2mj2(4m2-1)2+2π2∑k=1n-1‍2nk2(4n2-1)2+1=O(1),
which implies (2.14).Finally, we verify thatT satisfies condition (2.17); thus T satisfies all the conditions of Theorem 2.2. By direct calculations, we deduce that
(4.30)∑j=1m∑k=1n‍|Δ01t^mnjk(1,1)k|=(∑j=1m‍|cosj2m-1π-cosj2m+1π|+cosmπ2m+1)×(∑k=1n-1‍|cos(k/(2n-1))π-cos(k/(2n+1))πk-cos((k+1)/(2n-1))π-cos((k+1)/(2n+1))πk+1|+1ncosnπ2n+1)=O(∑k=1n-1‍|cos(k/(2n-1))π-cos(k/(2n+1))πk-cos((k+1)/(2n-1))π-cos((k+1)/(2n+1))πk+1|+n-2)=O(1)∑k=1n-1‍1k((cosk2n-1π-cosk2n+1π)-(cosk+12n-1π-cosk+12n+1π))+O(1)∑k=1n-1‍(1k-1k+1)|cosk+12n-1π-cosk+12n+1π|=∑k=1n-1‍1k|(sin2nkπ4n2-1-sin2n(k+1)π4n2-1)sinkπ4n2-1+(sinkπ4n2-1-sin(k+1)π4n2-1)sin2n(k+1)π4n2-1|+∑k=1n-1‍1k(k+1)sin2n(k+1)π4n2-1sin(k+1)π4n2-1=O(1)∑k=1n-1‍1ksinnπ4n2-1sinkπ4n2-1+O(1)∑k=1n-1‍1k2sin2n(k+1)π4n2-1sin(k+1)π4n2-1=O(1)∑k=1n-1‍n-3=O(n-2).Analogously,(4.31)∑j=1m∑k=1n‍|Δ10t^mnjk(1,1)j|=O(m-2),∑j=1m∑k=1n‍|Δ11t^mnjk(1,1)j|=O(m-2n-2).Now, (4.30) and (4.31) imply (2.17).


## 4.3. Riesz’s Means

Let{pn} and {qn} be positive sequences such that Pn=p0+p1+⋯+pn→∞, Qn = q0+q1+⋯+qn → ∞, and let T:=(tmnjk) be a lower triangular matrix with entries tmnjk = pjqk/PmQn, j=0,1,…m, k=0,1,…,n; m,n=1,2,…. Then the T-transformation of Fourier series (2.1) is known as Riesz’s mean.Proposition 4.2.
Let{pn} and {qn} be positive nondecreasing sequences such that (one says An~Bn, if there are two positive constants C1 and C2 such that C1≤(An/Bn)≤C2.)(4.32)npn~Pn,nqn~Qn,(4.33)∑k=1n‍pkk=O(pn),∑k=1n‍qkk=O(qn).
ThenT satisfies all the conditions of Theorem 2.2.Proof.
Direct calculations yield that (setP-1=Q-1=0)
(4.34)t^mnjk(1,1)=(∑i=jm‍piPm-∑i=jn-1‍piPm-1)(∑i=kn‍qiQn-∑i=kn-1‍qiQn-1)=pmqnPj-1Qk-1PmPm-1QnQn-1,0≤j<m,0≤k<n,(4.35)t^mnmk(1,1)=pmPm(∑i=kn‍qiQn-∑i=kn-1‍qiQn-1)=pmqnQk-1PmQnQn-1,0≤k<n,t^mnjn(1,1)=(∑i=jm‍piPm-∑i=jn-1‍piPm-1)qnQn=pmqnPj-1PmPm-1Qn,0≤j<m,(4.36)t^mnmn(1,1)=pmqnPmQn.
It follows from (4.35) that T satisfies condition (2.13).
Since{pn} and {qn} are nondecreasing, by (4.32), (4.34)–(4.36), we have
(4.37)∑j=1m∑k=1n‍|t^mnjk(1,1)|=pmqnPmPm-1QnQn-1∑j=1m∑k=1n‍Pj-1Qk-1=O(mnpmqnPmQn)=O(1).
Hence,T satisfies condition (2.14).
Since{pn} and {qn} are nondecreasing, by (4.32)–(4.36), we have
(4.38)∑j=1m∑k=1n‍|Δ01(t^mnjk(1,1)k)|=(pmPmPm-1∑j=1m-1‍Pj-1+pmPm)×(qnQnQn-1∑k=1n-1‍|Qk-1k-Qkk+1|+qnnPn)=O(1)(qnQnQn-1(∑k=1n-1‍Qk-1k(k+1)+∑k=2n-1‍qkk+1)+O(n-2))=O(1)(1nQn-1∑k=1n-1‍qkk+n-2)=O(qn-1nQn-1)=O(n-2).
Similarly,(4.39)∑j=1m∑k=1n‍|Δ10t^mnjk(1,1)j|=O(m-2),∑j=1m∑k=1n‍|Δ11t^mnjk(1,1)j|=O(m-2n-2).
Now, by (4.38) and (4.39), we show that T also satisfies condition (2.17).



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*Source: 513206-2012-12-30.xml*