# On Double Summability of Double Conjugate Fourier Series

Authors: H. K. Nigam; Kusum Sharma
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

For the first time, a theorem on double matrix summability of double
conjugate Fourier series is established.


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## Body

## 1. Introduction

The Fourier series off(x) is given by
(1.1)f(x)~a02+∑n=1∞(ancos⁡nx+bnsin⁡nx).
Conjugate to the series (1.1) is given by
(1.2)∑n=1∞(ancos⁡nx-bnsin⁡nx)
and is known as conjugate Fourier series. It is well known that the corresponding conjugate function of (1.2) is defined as
(1.3)f̅(x)=1π∫0πf(x+t)-f(x-t)2tan(1/2)tdt.Letf(x,y) is integrable (L) over the square Q(-π,-π;π,π) and is periodic with period 2π in each variable.The double Fourier series of a functionf(x,y) which is analogue for two variables of the series (1.1), is given by
(1.4)f(x,y)=∑m=0∞∑n=0∞λm,n[αm,ncos⁡mxcos⁡ny+βm,nsin⁡mxcos⁡ny+γm,ncos⁡mxsin⁡ny+δm,nsin⁡mxsin⁡ny]=∑m=0∞∑n=0∞λm,nAm,n(x,y),(1.5)αm,n=1π2∬Qf(x,y)cos⁡mxcos⁡nydxdy
with three similar expressions form,n=0,1,2,… and for βm,n,γm,nand δm,n where Q represents the fundamental square (-π,-π;π,π).One can associate three conjugate series to the double Fourier series (1.4) in the following way:(1.6)∑m=1∞∑n=0∞λm,n[-βm,ncos⁡mxcos⁡ny+αm,nsin⁡mxcos⁡ny-δm,ncos⁡mxsin⁡ny+γm,nsin⁡mxsin⁡ny],(1.7)∑m=0∞∑n=1∞λm,n[-γm,ncos⁡mxcos⁡ny-δm,nsin⁡mxcos⁡ny+αm,ncos⁡mxsin⁡ny+βm,nsin⁡mxsin⁡ny],(1.8)∑m=1∞∑n=1∞[δm,ncos⁡mxcos⁡ny-γm,nsin⁡mxcos⁡ny-βm,ncos⁡mxsin⁡ny+αm,nsin⁡mxsin⁡ny],
where λm,n=1,λm,0=λ0,n=1/2,m,n≥1,λ0,0=1/4.The conjugate functionsf̅(1)(x,y),f̅(2)(x,y) and f̅(3)(x,y) corresponding to (1.7), and (1.8) are defined as
(1.9)f̅(1)(x,y)=-1π∫0πf(x+s,y)-f(x-s,y)2tan(1/2)sds,f̅(2)(x,y)=-1π∫0πf(x,y+t)-f(x,y-t)2tan(1/2)tdt,f̅(x,y)=1π2∫0π∫0π[f(x+s,y+t)-f(x-s,y+t)-f(x+s,y-t)-f(x-s,y-t)2tan(1/2)s2tan(1/2)t]dsdt.
We will consider the symmetric square partial sum of series (1.8).LetT=(am,j) and S=(bn,k) be two infinite triangular matrices. Let ∑m=0∞∑n=0∞pm,n be a double series with sm,n=∑j=0m∑k=0npj,k as its (m,n)th partial sums. The double matrix mean tm,n is given by
(1.10)tm,n=∑j=0m∑k=0nam,jbn,ksj,k,The double series∑m=0∞∑n=0∞pm,n with the sequence of (m,n)th partial sums (sm,n) is said to summable by double matrix summability method or summable (T,S) if tm,n tends to a limit s as m→∞ and n→∞.The regularity conditions of double matrix summability means are given by(1.11)∑j=0m∑k=0nam,jbn,k⟶1asm⟶∞,n⟶∞,lim⁡m,n∑k=0n|am,jbn,k|=0foreachj=1,2,3,…,lim⁡m,n∑j=0m|am,jbn,k|=0foreachk=1,2,3,…
We write
(1.12)ψ(x,y)=ψ(x,y;s,t)=f(x+s,y+t)-f(x-s,y+t)-f(x+s,y-t)+f(x-s,y-t),Ψ(x,y)=∫0x∫0y|ψ(s,t)|dsdt,Ψ1(x,t)=∫0x|ψ(s,t)|ds,Ψ2(s,y)=∫0y|ψ(s,t)|dt,τ=(1t)=integralpartof1t,σ=(1s)=integralpartof1s,K̅m(s)=12π∑j=0mam,jcos⁡(j+(1/2))ssin⁡(s/2),K̅n(t)=12π∑k=0nbn,kcos⁡(k+(1/2))tsin⁡(t/2),
Important particular cases of the double matrix summability method are(i)
(C,1,1) summability mean ([1]) if am,j=1/(m+1)forallm and bn,k=(1/(n+1))foralln;(ii)
(H,1,1) summability mean ([5]) if am,j=1/(m-j+1)logm and bn,k=1/(n-k+1)logn;(iii)
(N,pm,qn) summability mean ([2]) if am,j=pm-j/Pm and bn,k=qn-k/Qn, provided Pm=∑j=0mpj≠0 and Qn=∑k=0nqk≠0.Double matrix summability method(T,S) is assumed to be regular throughout this paper.


## 2. Main Theorem

Rajagopal [1] previously proved a theorem on the Nörlund summability of Fourier series. Result of Rajagopal [1] contained various results due to Hardy [2], Hirokowa [3], Hirokowa and Kayashima [4], Pati [5], Siddiqui [6], and Singh [7]. Thereafter Sharma [8] proved a theorem dealing with the harmonic summability of double Fourier series. The result of Sharma [8] is a generalization of the theorem due to Hille and Tamarkin [9] for double Fourier series and also is analogous to the theorem of Chow [10] for summability (C,1,1) of the double Fourier series. The theorem of Sharma [8] was generalized by Mishra [11] for double Nörlund summability. The result Mishra [11] was generalized by Okuyama and Miyamoto [12]. But nothing seems to have been done so far to study double matrix summability of conjugate Fourier series. Therefore, the purpose of this paper is to establish the following theorem.Theorem 2.1.
Let(am,j)j=0m and (bn,k)k=0n be two real nonnegative and nondecreasing sequences with j≤m and k≤n, respectively.
LetT=(am,j) and S=(bn,k) be two infinite triangular matrices with(2.1)am,j≥0,am,j=0,j>m;bn,k≥0,bn,k=0,k>n;Am,σ=∑j=0σam,j;Bn,τ=∑k=0τbn,k;Am,m=1∀m≥0;Bn,n=1∀n≥0.
If the conditions
(2.2)Ψ(x,y)=∫0x∫0y|ψ(s,t)|dsdt=O{xξ(1/x)⋅yχ(1/y)},(2.3)∫0πψ1(x,t)dt=O{xξ(1/x)},∫0πψ2(s,y)ds=O{yχ(1/y)}
hold, then the double conjugate Fourier series (1.8) is double matrix (T,S) summable to, f̅(x,y), where
(2.4)f̅(x,y)=-14π2∬0π1tan(1/2)stan(1/2)tdsdt
at every point where these integrals exist provided ξ(x) and χ(y) are two positive monotonic increasing functions of x and y such that ξ(m)→∞,asm→∞ andχ(n)→∞,asn→∞,
(2.5)∫1mAm,ssξ(s)ds=O(1),m⟶∞,∫1nBn,ttχ(t)dt=O(1),n⟶∞.


## 3. Lemmas

Lemma 3.1.
One has(3.1)K̅m(s)=O(1s)for0≤s≤1m.Proof.
For0≤s≤1/m, sin(t/2)≥t/π and |cosmt|≤1,
(3.2)|K̅m(s)|≤12|∑j=0mam,jcos⁡(j+(1/2))ssin⁡(s/2)|≤12{∑j=0mam,j|cos⁡(j+(1/2))s||sin⁡(s/2)|}≤πs∑j=0mam,j=O(1s).Lemma 3.2.
One has(3.3)K̅n(t)=O(1t)for0≤t≤1n.Proof.
This can be proved similar to Lemma3.1.Lemma 3.3 (see [2]).
If(am,u) is non-negative and non-decreasing with μ, then, for 0≤a≤b≤∞,0≤s≤π and any m,
(3.4)|∑μ=abam,m-μei(m-μ)s|=O(Am,σ).Lemma 3.4.
If(bn,v) is non-negative and non-decreasing with ν, then, for 0≤a≤b≤∞,0≤t≤π and any n,
(3.5)|∑ν=abbn,n-νei(n-ν)t|=O(Bn,τ).Proof.
This is similar to Lemma3.3.Lemma 3.5.
One has(3.6)K̅m(s)=O(Am,σs)for0<1m<s≤π.Proof.
Since1/m<s≤π,sin(s/2)≥s/π,
(3.7)|K̅m(s)|≤12π|∑j=1mam,jcos⁡(j+(1/2))ssin⁡(s/2)|≤12s|∑j=1mam,m-jRe{ei(m-j+(1/2))s}|≤12s|∑j=1mam,m-jReei(m-j)s||ei(s/2)|=12s|Re∑j=1mam,m-jei(m-j)s|=O(1s)⋅O(Am,σ)byLemma3.3=O(Am,σs).Lemma 3.6.
One has(3.8)K̅n(t)=O(Bn,τt),for0<1n<π.Proof.
It can be proved similar to Lemma3.5 but using Lemma 3.4.


## 4. Proof of The Theorem

The(j,k)th partial sums s̅j,k(x,y) of the series (1.8) is given by
(4.1)s̅j,k(x,y)-f̅(x,y)=14π2∫0π∫0πψ(s,t)cos⁡(j+(1/2))ssin⁡(s/2)cos⁡(k+(1/2))tsin⁡(t/2)dsdt.
Then,
(4.2)∑j=0m∑k=0nam,jbn,k{s̅j,k(x,y)-f̅(x,y)}=14π2∫0π∫0πϕ(s,t)∑j=0m∑k=0nam,jbn,kcos⁡(j+(1/2))ssin⁡(s/2)⋅cos⁡(k+(1/2))tsin⁡(t/2)dsdt
or(4.3)t̅m,n(x,y)-f̅(x,y)=∬0πψ(s,t)K̅m(s)K̅n(t)dsdt=(∫0ν∫0η+∫0ν∫ηπ+∫νπ∫0η+∫νπ∫ηπ)⋅ψ(s,t)K̅m(s)K̅n(t)dsdt=I1+I2+I3+I4(say).
We consider
(4.4)I1=(∫01/m∫01/n+∫1/mν∫01/n+∫01/m∫1/nη+∫1/mν∫1/nη)⋅ψ(s,t)K̅m(s)K̅n(t)dsdt=I1.1+I1.2+I1.3+I1.4(say),
where s≤ν, t≤η and (2.2) holds.Now consider(4.5)I1.1=∫01/m∫01/nψ(s,t)K̅m(s)K̅n(t)dsdt=O⌊∫01/m∫01/n(1s)(1t)ψ(s,t)dsdt]byLemma3.1and3.2=O(mn)⌊∫01/m∫01/nψ(s,t)dsdt]=O(mn)ψ(1m,1n)=O(mn)o{1mnξ(m)χ(n)}by(2.2)=o(1),asm⟶∞,n⟶∞.Now,(4.6)|I1.2|≤∫1/mν∫01/nO|ψ(s,t)||K̅m(s)||K̅n(t)|dsdt≤∫01/n|Kn(t)|dt∫1/mν|ψ(s,t)|Am,[1/s]sdsbyLemma3.5={∫01/n(1t)dt∫1/mν|ψ(s,t)|Am,[1/s]sds}byLemma3.2=O(n)[∫01/nAm,[1/ν]νψ1(ν,t)dt]+O(mn)(Am,m)[∫01/nψ1(1m,t)dt]+O(n)[∫01/ndt∫1/mνdds(Am,[1/s]s)|ψ(s,t)|ds]=I1.2.1+I1.2.2+I1.2.3(say).
Then,
(4.7)I1.2.1+I1.2.2=O(n)[∫01/nψ1(ν,t)dt]+O(mn)[∫01/nψ1(1m,t)dt]=O(n)Ψ(ν,1n)+O(mn)Ψ(1m,1n)=O(n)o(νξ(1/ν)⋅1/nχ(n))+O(mn)o(1/mnξ(m)χ(n))=o(1),asm⟶∞,n⟶∞.Next,(4.8)I1.2.3=O(n)∫01/ndt∫1/mνΨ1(s,t)Am,[1/s]s2ds+O(n)∫01/ndt∫1/mνΨ1(s,t)sdds(Am,[1/s])ds=O(n)[∫1/mν(∫01/nΨ1(s,t)dt)Am,[1/s]s2ds]+O(n)[∫1/mν(∫01/nΨ1(s,t))1sdds(Am,[1/s])ds]=O(n)∫1/mνΨ1(s,1n)Am,[1/s]s2ds+O(n)∫1/mνΨ1(s,1n)1sdds(Am,[1/s])ds=O(n)∫1/mνo(sξ(1/s)⋅1nχ(n))Am,[1/s]s2ds+O(n)[∫1/mνo(sξ(1/s)⋅1nχ(n))1sdds(Am,[1/s])ds]=o[1χ(n)][∫1/νmAm,ssξ(s)ds]+o[1χ(n)][∫1/νm1ξ(s)dds(Am,s)ds]=o[1χ(n)]o(1)+o[1ξ(m)χ(n)][Am,n]=o(1),asm⟶∞,n⟶∞.
Thus,
(4.9)I1.2=o(1),asm⟶∞,n⟶∞.
Similarly
(4.10)I1.3=o(1),asm⟶∞,n⟶∞.
Now,
(4.11)|I1.4|=O[∫1/mν∫1/nη|ψ(s,t)|Am,[1/s]s⋅Bn,[1/t]tdtds]=O[Bn,[1/η]Am,[1/ν]Ψ(ν,η)νη-mBn,[1/η]Am,mΨ(1/m,η)η]-Bn,[1/η]η∫1/mνψ(s,η)dds(Am,[1/s]s)ds-nBn,nAm,[1/ν]Ψ(ν,1/n)ν+mnBn,nAm,mΨ(1m,1n)+nBn,n∫1/mνΨ(s,1n)dds(Am,[1/s]s)ds+Am,[1/ν]ν∫1/mνΨ1(s,t)ddt(Bn,[1/t]t)dt-mAm,m∫1/nξΨ2(1m,t)ddt(Bn,[1/t]t)dt-{∫1/mν∫1/nηΨ(s,t)dds(Am,[1/s]s)ddt(Bn,[1/t]t)dtds}=I1.4.1+I1.4.2+I1.4.3+I1.4.4+I1.4.5+I1.4.6+I1.4.7+I1.4.8+I1.4.9(say).
Then,
(4.12)I1.4.1+I1.4.2=O[Bn,[1/η]Am,[1/ν]Ψ(ν,η)νη+mBn,[1/η]Am,mΨ(1/m,η)η]=o[Bn,[1/η]Am,[1/ν]α(1/ν)β(1/η)]+o[mBn,[1/η]{1mξ(m)⋅ηχ(1/η)}]=o[Bn,[1/η]Am,[1/ν]ξ(1/ν)χ(1/η)]+o[Bn,[1/η]1ξ(m)⋅ηχ(1/η)]=o(1)+o(1)[1ξ(m)]=o(1),asm⟶∞,n⟶∞bytheregularityof(T,S),I1.4.3=-Bn,[1/η]η∫1/mνΨ(s,η)dds(Am,[1/s]s)ds=o(Bn,[1/η]η)∫1/mν(sξ(1/s)⋅ηχ(1/η))(Am,[1/s]s2)ds+o(Bn,[1/η]η)∫1/mνo(sξ(1/s)⋅ηχ(1/η))1sdds(Am,[1/s])ds=o(Bn,[1/η]χ(1/η))∫1/mνAm,[1/s]sξ(1/s)ds+o(Bn,[1/η]χ(1/η))∫1/mν1ξ(1/s)dds(Am,[1/s])ds=o(Bn,[1/η]χ(1/η))∫1/νmAm,ssξ(s)ds+o(Bn,s/ηχ(1/η))∫1/νm1ξ(s)d(Am,s)=o(Bn,[1/η]χ(1/η))O(1)+o(Bn,[1/η]χ(1/η))O(Am,m)=o(1),asm⟶∞,n⟶∞.
Thus, we get
(4.13)I1.4.3=o(1),asm⟶∞,n⟶∞.
Similarly(4.14)I1.4.4=o(1),asm⟶∞,n⟶∞.Now,(4.15)I1.4.5=mnBn,nAm,mΨ(1m,1n)=o(mnBn,nAm,m1mnξ(m)χ(n))=o(1),asm⟶∞,n⟶∞.
Consider
(4.16)I1.4.6=nBn,n∫1/mνΨ(s,1n)dds(Am,[1/s]s)ds=nBn,n∫1/mνΨ(s,1n)Am,[1/s]s2ds+nBn,n∫1/mνΨ(s,1n)1sdds(Am,[1/s])ds=nBn,n∫1/mνo(sξ(1/s)⋅1nχ(η))Am,[1/s]s2ds+nBn,n∫1/mν(sξ(1/s)nχ(η))1sdds(Am,[1/s])ds=o(nBn,nnχ(n))∫1/mνAm,[1/s]sξ(1/s)ds+o(1χ(n))∫1/mν1ξ(1/s)dds(Am,[1/s])ds=o(1),asm⟶∞,n⟶∞.
As similar to I1.4.3,
(4.17)I1.4.7=o(1),asm⟶∞,n⟶∞.
As similar to I1.4.6,
(4.18)I1.4.8=o(1),asm⟶∞,n⟶∞.
Now,
(4.19)I1.4.9=O[{∫1/mν∫1/nηΨ(s,t)(Am,[1/s]s2)+1sdds(Am,[1/s])}⋅{Bn,[1/t]t2+1tddt(Bn,[1/t])}dtds]=I1.4.9.1+I1.4.9.2+I1.4.9.3+I1.4.9.4(say).
Then,
(4.20)I1.4.9.1+I1.4.9.2=O[∫1/mν∫1/nηΨ(s,t){Am,[1/s]s2⋅Bn,[1/t]t2}dtds]+O[∫1/mν∫1/nηΨ(s,t){Am,[1/s]s21t⋅ddt(Bn,[1/t])}dtds]=O[∫1/mν∫1/nηo{sξ(1/s)⋅tχ(1/t)}Am,[1/s]s2⋅Bn,[1/t]t2dtds]+O[∫1/mν∫1/nηo{sξ(1/s)⋅tχ(1/t)}Am,[1/s]s21t⋅ddt(Bn,[1/t])dtds]=o[∫1/mν∫1/nηAm,[1/s]Bn,[1/t]sξ(1/s)tχ(1/t)dtds]+o[∫1/mν∫1/nηAm,[1/s]sξ(1/s)χ(1/t)ddt(Bn,[1/t])dtds]=o[∫1/νmAm,ssξ(s)ds∫1/ηnBn,ttχ(t)dt]+o[∫1/νmAm,ssξ(s)ds∫1/ηn1χ(t)ddt(Bn,[1/t])dt]=o(1)+o(1)∫1/ηnO(1)ddt(Bn,[1/t])dt=o(1),asm⟶∞,n⟶∞.
Similarly,
(4.21)I1.4.9.3=o(1),asm⟶∞,n⟶∞.
Now,(4.22)I1.4.9.4=O[∫1/mν∫1/nηΨ(s,t)1sdds(Am,[1/s])1tddt(Bn,[1/t])dtds]=o[∫1/mν∫1/nη(sξ(1/s)⋅tχ(1/t))1sdds(Am,[1/s])1t⋅ddt(Bn,[1/t])dtds]=o[∫1/mν∫1/nη(1ξ(1/s)⋅χ(1/t))dds(Am,[1/s])ddt(Bn,[1/t])dtds]=o[∫1/mν1ξ(1/s)dds(Am,[1/s])ds∫1/nη1χ(1/t)ddt(Bn,[1/t])dt]=o[∫1/mνO(1)dds(Am,[1/s])ds∫1/nηO(1)ddt(Bn,[1/t])d]=o(1),asm⟶∞,n⟶∞.
Hence,
(4.23)I1.4=o(1),asm⟶∞,n⟶∞.
Therefore,(4.24)I1=o(1),asm⟶∞,n⟶∞.Now,1/m<ν<π,1/n<η<π, thus we obtain(4.25)|I3|≤∫νπ|Km(s)|ds∫01/n|ψ(s,t)||Kn(t)|dt+∫νπ|Km(s)|ds∫1/nν|ψ(s,t)||Kn(t)|dt=I3.1+I3.2(say).
Using Lemma 3.2 and Lemma 3.5, we have
(4.26)I3.1=∫νπAm,σsds∫01/n|ψ(s,t)|1tdt=O(n)∫νπAm,σsds∫01/n|ψ(s,t)|dt=O(n)∫0πΨ2(s,1n)ds=O(1χ(n))=o(1),asm⟶∞,n⟶∞.
Using Lemma 3.5 and Lemma 3.6,(4.27)I3.2=O[∫νπAm,σsds∫1/nη|ψ(s,t)|Bn,[τ]tdt]=O[∫νπds{(Ψ2(s,t)Bn,[τ]t)1/nη-∫1/nηΨ(s,t)ddt(Bn,[τ]t)dt}]=O[∫νπds{Ψ2(s,η)Bn,[1/n]η-Ψ2(s,1n)nBn,n}]+[∫νπds∫1/nηΨ2(s,t)ddt(Bn,[τ]t)dt]=O[∫νπΨ2(s,η)Bn,[1/n]ηds]+O(n)[∫0πΨ2(s,1n)ds]+O[∫νπds∫1/nηΨ2(s,t)ddt(Bn,[τ]t)dt]=o(1)+o(1)+o(1)(similartoI1.4.9)=o(1),asm→∞,n→∞.
Hence,
(4.28)I3=o(1),asm⟶∞,n⟶∞.
Similarly,(4.29)I2=o(1),asm⟶∞,n⟶∞.By the regularity conditions of matrix method and Riemann-Lebesgue theorem,(4.30)I4=o(1),asm⟶∞,n⟶∞.
Therefore,
(4.31)t̅m,n(x,y)-f̅(x,y)=o(1)
This completes the proof of the theorem.



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*Source: 104592-2012-06-20.xml*