Open Access

Generalized higher (U,M)‐derivations in prime Γ‐Rings

SpringerPlus20143:100

https://doi.org/10.1186/2193-1801-3-100

Received: 5 January 2014

Accepted: 10 February 2014

Published: 19 February 2014

Abstract

Let M be a 2‐torsion free prime Γ‐ring satisfying the condition a α b β c=a β b α c,a,b,cM and α,βΓ, U be an admissible Lie ideal of M and F=(f i )iN be a generalized higher (U,M)‐derivation of M with an associated higher (U,M)‐derivation D=(d i )iN of M. Then for all nN we prove that f n ( uαm ) = i + j = n f i ( u ) α d j ( m ) , u U , m M , α Γ .

Keywords

Lie ideal Admissible Lie ideal (U, M)-derivation Generalized (U, M)-derivation Generalized higher (U, M)-derivation Prime Γ-ring

Introduction

The notion of a Γ‐ring has been developed by Nobusawa (1964), as a generalization of a ring. Following Barnes (1966) generalized the concept of Nobusawa’s Γ‐ring as a more general nature. Nowadays Γ‐ring theory is a showpiece of mathematical unification, bringing together several branches of the subject. It is the best research area for the Mathematicians and during 40 years, many classical ring theories have been generalized in Γ‐rings by many authors. The notions of derivation and Jordan derivation in Γ‐rings have been introduced by Sapanci and Nakajima (1997). Afterwards, in the light of some significant results due to Jordan left derivation of a classical ring obtained by Jun and Kim (1996), some extensive results of left derivation and Jordan left derivation of a Γ‐ring were determined by Ceven (2002). In (Halder and Paul 2012), Halder and Paul extended the results of (Ceven 2002) in Lie ideals. Let M and Γ be additive abelian groups. If there is a mapping M×Γ×MM (sending (x,α,y) into x α y) such that (i) (x+y)α z=x α z+y α z,x(α+β)y=x α y+x β y,x α(y+z)=x α y+x α z, (ii) (x α y)β z=x α(y β z), for all x,y,zM and α,βΓ, then M is called a Γ‐ring. This concept is more general than a ring and was introduced by Barnes (1966). A Γ‐ring M is called a prime Γ‐ring if a,bM,a ΓM Γb=0 implies a=0 or b=0. A Γ‐ring M is 2‐torsion free if 2a=0 implies a=0,aM. For any x,yM and αΓ, we induce a new product, the Lie product by [ x,y] α =x α yy α x. An additive subgroup UM is said to be a Lie ideal of M if whenever uU,mM and αΓ, then [ u,m] α U. In the main results of this article we assume that the Lie ideal U verifies u α uU,uU. A Lie ideal of this type is called a square closed Lie ideal. Furthermore, if the Lie ideal U is square closed and U is not contained in Z(M), where Z(M) denotes the center of M, then U is called an admissible Lie ideal of M. In (Herstein 1957), Herstein proved a well‐known result in prime rings that every Jordan derivation is a derivation. Afterwards many Mathematicians studied extensively the derivations in prime rings. In (Awter 1984), Awtar extended this result in Lie ideals. (U,R)‐derivations in rings have been introduced by Faraj et al. (2010), as a generalization of Jordan derivations on a Lie ideals of a ring. The notion of a (U,R)‐derivation extends the concept given in (Awter 1984). In this paper (Faraj et al. 2010), they proved that if R is a prime ring, char (R)≠2, U a square closed Lie ideal of R and d a (U,R)‐ derivation of R, then d(u r)=d(u)r+u d(r),,uU,rR. This result is a generalization of a result in (Awter (1984), Theorem in section 3). In this article, we introduce the concept of a (U,M)‐derivation, generalized (U,M)‐derivation and generalized higher (U,M)‐derivation, where U is a Lie ideal of a Γ‐ring M. Examples of a Lie ideal of a Γ‐ring, (U,M)‐derivation, generalized (U,M)‐derivation, higher (U,M)‐derivation and generalized higher (U,M)‐derivation are given here. A result in (Halder and Paul (2012), Theorem 2.8) is generalized in Γ‐rings by the new concept of a (U,M)‐derivation. Throughout the article, we use the condition a α b β c=a β b α c,a,b,cM and α,βΓ and this is represented by (*). We make the basic commutator identities [ x α y,z] β = [ x,z] β α y+x[ α,β] z y+x α[y,z] β , [ x,y α z] β = [ x,y] β α z+y[ α,β] x z+y α[ x,z] β , x,y,zM,α,βΓ. According to the condition (*), the above two identities reduces to [ x α y,z] β = [ x,z] β α y+x α[y,z] β ,[ x,y α z] β = [ x,y] β α z+y α[ x,z] β ,x,y,zM,α,βΓ.

Generalized (U,M)‐derivation

In view of the concept of (U,R)‐derivation of an ordinary ring developed by Faraj et al. (2010), we have been determined some important results in Rahman and Paul (2013) due to these concepts in case of certain Γ‐rings after introducing the notions of (U,M)‐derivation of Γ‐rings as defined below.

Definition 1

(Rahman and Paul (2013), Definition 2.1) Let M be a Γ‐ring and U be a Lie ideal of M. An additive mapping d :MM is said to be a (U,M)‐derivation of M if d(u α m + s α u) = d(u)α m + u α d(m) + d(s)α u + s α d(u),uU,m,sM and αΓ.

Definition 2.

(Rahman and Paul (2013), Definition 2.2) Let M be a Γ‐ring and U be a Lie ideal of M. An additive mapping f :MM is said to be a generalized (U, M)‐ derivation of M if there exists a (U,M)‐derivation d of M such that f(u α m+s α u)=f(u)α m+u α d(m)+f(s)α u+s α d(u),uU,m,sM and αΓ.

The existence of a Lie ideal of a Γ‐ring, (U,M)‐derivation and a generalized (U,M)‐derivation are confirmed by the following examples.

Example 1.

Let R be an associative ring with 1 and U a Lie ideal of R. Let M=M1,2(R) and Γ = n. 1 0 : n Z , then M is a Γ‐ring.

If N={(x,x):xR}M and U1={(u,u):uU} then N is a sub Γ‐ring of M and U1 is a Lie ideal of N. Let f:RR be a generalized (U,R)‐derivation. Then there exists a (U,R)‐derivation d:RR such that f(u α x+s α u)=f(u)α x+u α d(x)+f(s)α u+s α d(u).

If we define a mapping D :NN by D((x,x))=(d(x),d(x)), then we have D ( u , u ) n 0 ( x , x ) + ( y , y ) n 0 ( u , u ) = D ( ( unx , unx ) + ( ynu , ynu ) ) = D ( ( unx + ynu , unx + ynu ) ) = ( d ( unx + ynu ) , d ( unx + ynu ) ) .

After calculation we have D(u1α x1+y1α u1)=D(u1)α x1+u1α D(x1)+D(y1)α u1+y1α D(u1), where u 1 = ( u , u ) , α = n 0 , x 1 = ( x , x ) , y 1 = ( y , y ) .

Hence D is a (U1,N)− derivation on N.

Let F :NN be the additive mapping defined by F((x,x))=(f(x),f(x)), then considering u 1 = ( u , u ) U 1 , α = n 0 Γ , x 1 = ( x , x ) , y 1 = ( y , y ) N , we have F ( u 1 α x 1 + y 1 α u 1 ) = F ( unx + ynu , unx + ynu ) = ( f ( unx + ynu ) , f ( unx + ynu ) ) = ( f ( u ) nx + und ( x ) + f ( y ) nu + ynd ( u ) , f ( u ) nx + und ( x ) + f ( y ) nu + ynd ( u ) ) = ( f ( u ) , f ( u ) ) n 0 ( x , x ) + ( f ( y ) , f ( y ) ) n 0 ( u , u ) + ( u , u ) n 0 ( d ( x ) , d ( x ) ) + ( y , y ) n 0 ( d ( u ) , d ( u ) ) = F ( ( u , u ) ) n 0 ( x , x ) + ( u , u ) n 0 D ( ( x , x ) ) + F ( ( y , y ) ) n 0 ( u , u ) + ( y , y ) n 0 D ( ( u , u ) ) = F ( u 1 ) α x 1 + u 1 αD ( x 1 ) + F ( y 1 ) α u 1 + y 1 αD ( u 1 ) .

Hence F is a generalized (U1,N)−derivation on N.

Lemma 1

( Rahman and Paul (2013 ), Lemma 2.4) Let M be a 2‐torsion free Γ‐ring satisfying the condition (*). U be a Lie ideal of M and f be a generalized (U,M)‐derivation of M. Then
  1. (i)

    f(uαmβu)=f(u)αmβu+uαd(m)βu+uαmβd(u),uU,mM and α,βΓ.

     
  2. (ii)

    f(uαmβv+vαmβu)=f(u)αmβv+uαd(m)βv+uαmβd(v)+f(v)αmβu+vαd(m)βu+vαmβd(u),u,vU,mM and α,βΓ.

     

Definition 3.

(Rahman and Paul (2013), Definition 2.5) Let d be a (U,M)‐derivation of M, then we define Φ α (u,m)=d(u α m)−d(u)α mu α d(m),uU,mM and αΓ.

Now, we state some useful results that have already been discussed in Rahman and Paul (2013).

Lemma 2.

Let d be a (U,M)‐derivation of M, then
  1. (i)

    Φ α (u,m)=−Φ α (m,u), uU,mM and αΓ.

     
  2. (ii)

    Φ α (u+v,m)=Φ α (u,m)+Φ α (v,m),u,vU,mM and αΓ.

     
  3. (iii)

    Φ α (u,m+n)=Φ α (u,m)+Φ α (u,n),uU,m,nM and αΓ.

     
  4. (iv)

    Φα+β(u,m)=Φ α (u,m)+Φ β (u,m),uU,mM and α,βΓ.

     

The proofs are obvious by using the Definition 3.

Definition 4.

(Rahman and Paul (2013), Definition 2.7) If f is a generalized (U,M)‐derivation of M and d is a (U,M)‐derivation of M, then we define Ψ α (u,m)=f(u α m)−f(u)α mu α d(m),uU,mM and αΓ.

Also, we need the following important results that have already been discussed in Rahman and Paul (2013).

Lemma 3.

Let f be a generalized (U,M)‐derivation of M, then
  1. (i)

    Ψ α (u,m)=−Ψ α (m,u),uU,mM and αΓ.

     
  2. (ii)

    Ψ α (u+v,m)=Ψ α (u,m)+Ψ α (v,m),u,vU,mM and αΓ.

     
  3. (iii)

    Ψ α (u,m+n)=Ψ α (u,m)+Ψ α (u,n),uU,m,nM and αΓ.

     
  4. (iv)

    Ψα+β(u,m)=Ψ α (u,m)+Ψ β (u,m),uU,mM and α,βΓ.

     

The proofs are obvious by using the Definition 4.

Lemma 4.

(Rahman and Paul (2013), Lemma 2.11) Let U be a Lie ideal of a 2‐torsion free prime Γ‐ring M satisfying the condition (*) and U is not contained in Z(M). If a,bM (resp. bU and aM) such that a αUβb=0,α,βΓ, then a=0 or b=0.

Theorem 1.

(Rahman and Paul (2013), Theorem 2.13) Let M be a 2‐torsion free prime Γ‐ring satisfying the condition (*), U be an admissible Lie ideal of M and f be a generalized (U,M)‐derivation of M, then Ψ α (u,v)=0,u,vU and αΓ.

Remark 1

If we replace U by a square closed Lie ideal in the Theorem 1, then the theorem is also true.

Theorem 2.

(Rahman and Paul (2013), Theorem 2.14) Let M be a 2‐torsion free prime Γ‐ring satisfying the condition (*), U a square closed Lie ideal of M and f be a generalized (U,M)‐derivation of M, then f(u α m)=f(u)α m+u α d(m),uU mM and αΓ.

Generalized higher (U,M)‐derivation

In this section, we introduce generalized higher (U,M)‐derivations in Γ‐rings.

Definition 5.

Let M be a Γ‐ring and U be a Lie ideal of M and F = ( f i ) i N 0 be a family of additive mappings of M into itself such that f0=id M , where id M is an identity mapping on M. Then F is said to be a generalized higher (U,M)‐derivation of M if there exists an higher (U,M)‐derivation D=(d i )iN of M such that for each n N , f n ( uαm + sαu ) = i + j = n f i ( u ) α d j ( m ) + f i ( s ) α d j ( u ) , u U , m , s M and α,βΓ.

Example 2.

Let N and U1 are as in Example 1. If f n :RR be a generalized higher (U,R)‐derivation. Then there exists a higher (U1,R) derivation d n :RR such that f n ( uαx + yαu ) = i + j = n f i ( u ) α d j ( x ) + f i ( y ) α d j ( u ) .

If we define a mapping D n :NN by D n ((x,x))=(d n (x),d n (x)). Then D n is a higher (U1,N)‐derivation on N.

Let F n :NN be the additive mapping defined by F n ((x,x))=(f n (x),f n (x)). Then by the similar calculation as in Example 1, we can show that, F n is a generalized higher (U1,N)‐derivation on N.

Lemma 5.

Let M be a 2‐torsion free Γ‐ring satisfying the condition (*), U be a Lie ideal of M and F=(f i )iNbe a generalized higher (U,M)‐derivation of M. Then f n ( uαmβu ) = i + j + k = n f i ( u ) α d j ( m ) β d k ( u ) , u U , m M and α,βΓ.

Proof

Let x=u α((2u)β m+m β(2u))+((2u)β m+m β(2u))α u.

Replacing m and s by (2u)β m+m β(2u) and (2u)α m+m α(2u) respectively in f n ( uαm + sαu ) = i + j = n f i ( u ) α d j ( m ) + f i ( s ) α d j ( u ) and using the condition (*), we have f n ( x ) = i + j = n f i ( u ) α d j ( ( 2 u ) βm + ( 2 u ) ) + f i ( ( 2 u ) βm + ( 2 u ) ) α d j ( u ) = 2 i + j = n f i ( u ) α l + t = j ( d l ( u ) β d t ( m ) + d l ( m ) β d t ( u ) ) + 2 i + j = n p + q = i ( f p ( u ) β d q ( m ) + f p ( m ) β d q ( u ) ) α d j ( u ) = 2 i + l + t = n f i ( u ) α ( d l ( u ) β d t ( m ) + f i ( u ) α d l ( m ) β d t ( u ) ) + 2 p + q + j = n ( f p ( u ) β d q ( m ) α d j ( u ) + f p ( m ) β d q ( u ) α d j ( u ) ) .

Thus we have
f n ( x ) = 2 i + l + t = n f i ( u ) α ( d l ( u ) β d t ( m ) + f i ( u ) α d l ( m ) β d t ( u ) ) + 2 p + q + j = n ( f p ( u ) α d q ( m ) β d j ( u ) + f p ( m ) α d q ( u ) β d j ( u ) ) .
(1)

On the other hand by the definition of higher (U,M)‐ derivation and using the condition (*) f n ( x ) = 2 f n ( ( uαu ) βm + ( uαu ) ) + 2 f n ( uαmβu ) + 2 f n ( uβmαu ) = 2 f n ( ( uαu ) βm + ( uαu ) ) + 2 f n ( uαmβu ) + 2 f n ( uαmβu ) = 2 i + j = n ( f i ( uαu ) β d j ( m ) + f i ( m ) β d j ( uαu ) ) + 4 f n ( uαmβu ) = 2 i + j = n r + s = i f r ( u ) α d s u ) β d j ( m ) + 2 i + j = n e + k = j f i ( m ) α d e u ) β d k ( u ) + 4 f n ( uαmβu ) .

Thus we have
f n ( x ) = 2 r + s + j = n f r ( u ) α d s d j ( m ) + 2 i + e + k = n f i ( m ) β d e u ) α d k ( u ) + 4 f n ( uαmβu ) .
(2)

Now comparing (1) and (2) we get 4 f n ( uαmβu ) = 4 i + j + k = n f i ( u ) α d j ( m ) β d k ( u ) , u U , m M and α,βΓ. Using 2‐torsion freeness of M, we get the desired result.

Lemma 6.

Let M be a 2‐torsion free Γ‐ring satisfying the condition (*), U be a Lie ideal of M and F=(f i )iNbe a generalized higher (U,M)‐derivation of M. Then f n ( uαmβv + vαmβu ) = i + j + k = n f i ( u ) α d j ( m ) d k ( v ) + f i ( v ) α d j ( m ) β d k ( u ) , u , v U , m M and α,βΓ.

Proof.

Linearizing of f n ( uαmβu ) = i + j + k = n f i ( u ) α d j ( m ) β d k ( u ) with respect to u gives us f n ( ( u + v ) αmβ ( u + v ) ) = i + j + k = n f i ( u + v ) α d j ( m ) β d k ( u + v ) = i + j + k = n ( f i ( u ) α d j ( m ) β d k ( u ) + f i ( u ) α d j ( m ) β d k ( v ) + f i ( v ) α d j ( m ) β d k ( u ) + f i ( v ) α d j ( m ) β d k ( v ) ) .

On the other hand f n ( ( u + v ) αmβ ( u + v ) ) = f n ( uαmβu ) + f n ( uαmβv + vαmβu ) + f n ( vαmβv ) = i + j + k = n ( f i ( u ) α d j ( m ) β d k ( u ) + f n ( uαmβv + vαmβu ) + i + j + k = n ( f i ( v ) α d j ( m ) β d k ( v ) .

Now comparing above two expressions, we get f n ( uαmβv + vαmβu ) = i + j + k = n f i ( u ) α d j ( m ) β d k ( v ) + f i ( v ) α d j ( m ) β d k ( u ) , u , v U , m M and α,βΓ.

Definition 6.

Let M be a 2‐torsion free Γ‐ring satisfying the condition (*) and U be a Lie ideal of M. Let F=(f i )iN be a generalized higher (U,M)‐derivation of M. For every fixed nN, we define ψ n α ( u , m ) = f n ( uαm ) i + j = n f i ( u ) α d j ( m ) , u U , m M , α Γ . Also let D=(d i )iN be a higher (U,M)‐derivation of M. For every fixed nN, we define ϕ n α ( u , m ) = d n ( uαm ) i + j = n d i ( u ) α d j ( m ) , u U , m M , α Γ .

Remark 2.

ψ n α ( u , m ) = 0 , u U , m M , α Γ and nN if and only if f n ( uαm ) = i + j = n f i ( u ) α d j ( m ) , u U , m M , α Γ and nN. Also ϕ n α ( u , m ) = 0 , u U , m M , α Γ and nN if and only if d n ( uαm ) = i + j = n d i ( u ) α d j ( m ) , u U , m M , α Γ and nN.

Lemma 7.

Let M be a 2‐torsion free Γ‐ring satisfying the condition (*) and U be a Lie ideal of M. For every uU,mM,αΓ and nN, then ψ n α ( u , m ) + ψ n α ( m , u ) = 0 and ϕ n α ( u , m ) + ϕ n α ( m , u ) = 0 .

The proofs are obvious by the Definition 6, higher (U,M)‐derivation of M and generalized higher (U,M)‐derivation of M.

Lemma 8.

Let M be a 2‐torsion free prime Γ‐ring satisfying the condition (*), U be an admissible Lie ideal of M and F=(f i )iNbe a generalized higher (U,M)‐derivation of M. Then ψ n α ( u , v ) = 0 , u , v U , α Γ and nN.

Proof.

We have ψ 0 α ( u , v ) = 0 , u , v U , α Γ and by Theorem 1, ψ 1 α ( u , v ) = 0 , u , v U , α Γ .

Now we assume, by induction on nN, that ψ m α ( u , v ) = 0 , u , v U , α Γ , mN and m<n.

Let x=4(u α v β w γ v α u+v α u β w γ u α v).

Then by using Lemma 6, we have f n ( x ) = 4 f n ( uαv ) βwγvαu + 4 uαvβwγ d n ( vαu ) + i + j + k = n i , k < n f i ( uαv ) β d j ( w ) γ d k ( vαu ) + 4 f n ( vαu ) βwγuαv + 4 vαuβwγ d n ( uαv ) + i + j + k = n i , k < n f i ( vαu ) β d j ( w ) γ d k ( uαv ) .

On the other hand, by Lemma 5 and D=(d i )iN is a higher (U,M)‐derivation of M. f n ( x ) = 4 uαvβwγ s + k = n d s ( v ) α d k ( u ) + 4 i + p = n f i ( u ) α d p ( v ) βwγvαu + i + p + q + s + k = n s + k , i + p < n f i ( u ) α d p ( v ) β d q ( w ) γ d s ( v ) α d k ( u ) + 4 vαuβwγ r + k = n d r ( u ) α d k ( v ) + 4 i + l = n f i ( v ) α d l ( u ) βwγuαv + i + l + t + r + k = n i + l , r + k < n f i ( v ) α d l ( u ) β d t ( w ) γ d r ( u ) α d k ( v ) .

Now comparing the two expressions of f n (x) and using ψ m α ( u , v ) = 0 , u , v U , α Γ , m < n , we get 4 ψ n α ( u , v ) βwγvαu + 4 ψ n α ( v , u ) βwγuαv + 4 uαvβwγ ϕ n α ( v , u ) + 4 vαuβwγ ϕ n α ( u , v ) = 0 .

Using Lemma 7 and 2‐torsion freeness of M we get ψ n α ( u , v ) βwγ [ u , v ] α + [ u , v ] α βwγ ϕ n α ( u , v ) = 0 .

Since D=(d i )iN is a higher (U,M)‐derivation of M, thus we have ϕ n α ( u , v ) = 0 . Now by Lemma 4 and since U is noncentral, thus we get ψ n α ( u , v ) = 0 , u , v U , α Γ and nN.

Now we prove the main result.

Theorem 3.

Let M be a 2‐torsion free prime Γ‐ring satisfying the condition (*), U be an admissible Lie ideal of M and F=(f i )iNbe a generalized higher (U,M)‐derivation of M. Then f n ( uαm ) = i + j = n f i ( u ) α d j ( m ) , u U , m M , α Γ and nN.

Proof.

We have ψ 0 α ( u , m ) = 0 , u U , m M , α Γ and by Theorem 1, ψ 1 α ( u , m ) = 0 , u U , m M , α Γ .

Now we assume, by induction on nN, that ψ m α ( u , m ) = 0 , u U , m M , α Γ , m N and m<n.

Now since F=(f i )iN is a generalized higher (U,M)‐derivation of M, we have 0 = ψ n α ( u , uβm mβu ) = f n ( uαuβm ) f n ( uαmβu ) i + j = n f i ( u ) α d j ( uβm mβu ) .

Since D=(d i )iN is a higher (U,M)‐derivation of M, thus we have
f n ( uαuβm ) = i + l + t = n ( f i ( u ) α d l ( u ) β d t ( m ) .
(3)
Since F=(f i )iN is a generalized higher (U,M)‐derivation of M, thus we have f n ( ( uβm ) + ( uβm ) αu ) = i + j = n f i ( u ) α d j ( uβm ) + f i ( uβm ) α d j ( u ) = f n ( u ) α ( uβm ) + d n ( uβm ) + i + j = n i , j < n f i ( u ) α d j ( uβm ) .
+ f n ( uβm ) α ( u ) + ( uβm ) α d n ( u ) + i + j = n i , j < n f i ( uβm ) α d j ( u ) .
(4)
Since ψ m α ( u , m ) = 0 , u U , m M , α Γ , m < n .
f n ( ( uβm ) + ( uβm ) αu ) = f n ( u ) α ( uβm ) + d n ( uβm ) + i + j = n i , l + t < n f i ( u ) α d l ( u ) β d t ( m ) + f n ( uβm ) α ( u ) + ( uβm ) α d n ( u ) + i + j = n p + q , j < n f p ( u ) β d q ( m ) α d j ( u ) .
(5)
On the other hand, by using Equation (3) and Lemma 5, we get f n ( ( uβm ) + ( uβm ) αu ) = f n ( uαuβm ) + f n ( uβmαu ) = i + l + t = n f i ( u ) α d l ( u ) β d t ( m ) + i + j + k = n f i ( u ) β d j ( m ) α d k ( u ) = f n ( u ) α ( uβm ) + l + t = n d l ( u ) β d t ( m ) + i + l + t = n i , l + t < n f i ( u ) α d l ( u ) β d t ( m ) .
+ ( uαm ) β d n ( u ) + i + j = n f i ( u ) α d j ( m ) βu + i + j + k = n i + j , k < n f i ( u ) α d j ( m ) β d k ( u ) .
(6)
By comparing (5) and (6) and using the condition (*), we get
ψ n α ( u , m ) βu = 0 , u U , m M , α , β Γ , n N .
(7)
Linearizing of (7) with respect to u, gives us
ψ n α ( u , m ) βv + ψ n α ( v , m ) βu ) = 0 , u , v U , m M , α , β Γ , n N .
(8)

Replacing v by v α v in (8) and since ψ n α ( uαu , m ) = 0 , thus ψ n α ( u , m ) βvαv = 0 . This implies that 0 = ψ n α ( u , m ) β ( u + v ) α ( u + v ) = ψ n α ( u , m ) βvαu .

Hence by Lemma 4 and since U 0 , ψ n α ( u , m ) = 0 , u U , m M , α Γ and nN.

Thus by the Remark 2, we have f n ( uαm ) = i + j = n f i ( u ) α d j ( m ) , u U , m M , α Γ and nN.

Mathematics Subject Classification (2010)

13N15; 16W10; 17C50

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Jagannath University
(2)
Department of Mathematics, University of Rajshahi

References

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© Rahman and Paul; licensee Springer. 2014

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