# Stability of a Bi-Additive Functional Equation in Banach Modules Over aC⋆-Algebra

Authors: Won-Gil Park; Jae-Hyeong Bae
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

We solve the bi-additive functional equationf(x+y,z−w)+f(x−y,z+w)=2f(x,z)−2f(y,w) and prove that every bi-additive Borel function is bilinear. And we investigate the stability of a bi-additive functional equation in Banach modules over a unital C⋆-algebra.


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## Body

## 1. Introduction

In 1940, Ulam proposed the stability problem (see [1]).LetG1 be a group, and let G2 be a metric group with the metric d(·,·). Given ε>0, does there exist a δ>0 such that if a mapping h:G1→G2 satisfies the inequality d(h(xy),h(x)h(y))<δ for all x,y∈G1 then there is a homomorphism H:G1→G2 with d(h(x),H(x))<ε for all x∈G1?In 1941, this problem was solved by Hyers [2] in the case of Banach space. Thereafter, many authors investigated solutions or stability of various functional equations (see [3–21]).LetX and Y be real or complex vector spaces. In 1989, Aczél and Dhombres [22] proved that a mapping g:X→Y satisfies the quadratic functional equation
(1.1)g(x+y)+g(x-y)=2g(x)+2g(y)
if and only if there exists a symmetric bi-additive mapping S:X×X→Y such that g(x)=S(x,x), where
(1.2)S(x,y):=14[g(x+y)-g(x-y)]
for all x,y∈X. For a mapping f:X×X→Y, consider the bi-additive functional equation:
(1.3)f(x+y,z-w)+f(x-y,z+w)=2f(x,z)-2f(y,w).
For a mapping g:X→Y satisfying (1.1), the Aczél's bi-additive mapping S:X×X→Y given by (1.2) is a solution of (1.3).In this paper, we find out the general solution of the bi-additive functional equation (1.3) and investigate the linearity of bi-additive Borel functions. And we investigate the stability of (1.3) in Banach modules over a unital C⋆-algebra.


## 2. Solution of the Bi-Additive Functional Equation (1.3)

The general solution of the bi-additive functional equation (1.3) is as follows.Theorem 2.1.
A mappingf:X×X→Y satisfies (1.3) if and only if the mapping f is bi-additive.Proof.
Assume that the mappingf satisfies (1.3). Letting x=y=z=w=0 in (1.3), we gain f(0,0)=0. Putting w=z in (1.3), we get
(2.1)f(x+y,0)+f(x-y,2z)=2f(x,z)-2f(y,z)
for all x,y,z∈X. Setting y=x in (2.1), we have
(2.2)f(x,0)=-f(0,z)
for all x,z∈X. Taking z=0 (resp., x=0) in the above equation, we obtain
(2.3)f(x,0)=0(resp.,f(0,z)=0)
for all x∈X (resp., for all z∈X). Letting x=w=0 in (1.3) and using (2.3), we gain
(2.4)f(-y,z)=-f(y,z)
for all y,z∈X. Putting y=0 in (2.1) and using (2.3), we get
(2.5)f(x,2z)=2f(x,z)
for all x,z∈X. Replacing y by -y in (2.1) and using (2.3), (2.4), and (2.5) and the above equation, we see that f(x+y,z)=f(x,z)+f(y,z) for all x,y,z∈X.
On the other hand, lettingy=x in (1.3) and using (2.3), we gain
(2.6)f(2x,z-w)=2f(x,z)-2f(x,w)
for all x,z,w∈X. Putting y=z=0 in (1.3) and using (2.3), we get
(2.7)f(x,-w)=-f(x,w)
for all x,w∈X. Setting w=0 in (2.6) and using (2.3), we have
(2.8)f(2x,z)=2f(x,z)
for all x,z∈X. Replacing w by -w in (2.6) and using (2.7) and (2.8), we obtain that f(x,z+w)=f(x,z)+f(x,w) for all x,z,w∈X.
The converse is trivial.The bi-additive functional equation (1.3) is related to the quadratic functional equation (1.1).Iff:X×X→Y is a mapping satisfying (1.3) and g:X→Y is the mapping given by g(x):=f(x,x) for all x∈X, then one can easily obtain that g satisfies (1.1).Leta∈ℝ and g:X→Y be a mapping satisfying (1.1). If f:X×X→Y is the mapping given by f(x,y):=(a/4)[g(x+y)-g(x-y)] for all x,y∈X, then one can easily prove that f satisfies (1.3). Furthermore, g(x)=f(x,x) holds for all x∈X if a=1.The following is a result on bi-additive Borel functions.Theorem 2.2.
Letψ:ℝ×ℝ→ℝ be a bi-additive Borel function; then it is bilinear, that is, it satisfies ψ(s,t)=stψ(1,1) for all s,t∈ℝ.Proof.
Since the functionψ is bi-additive, we gain
(2.9)ψ(pu,qv)=pqψ(u,v)
for all p,q∈ℚ and all u,v∈ℝ. Letting p=v=1 in equality (2.9), we get
(2.10)ψ(u,q)=qψ(u,1)
for all q∈ℚ and all u∈ℝ. Putting u=v=1 in equality (2.9) again, we have
(2.11)ψ(p,q)=pqψ(1,1)
for all p,q∈ℚ. Note that the function v→ψ(u,v) is measurable for each fixed u∈ℝ (see [23, Proposition 2.34]). Since the function v→ψ(u,v) is additive for each fixed u∈ℝ, by [24], it is continuous for each fixed u∈ℝ. By the same reasoning, the function u→ψ(u,v) is also continuous for each fixed v∈ℝ. Let s,t∈ℝ be fixed. Since ψ is measurable, by [25, Theorem 7.14.26], for every m∈ℕ there is a closed set Fm⊂[s,s+1] such that μ([s,s+1]∖Fm)<1/m and ψ|Fm×ℝ is continuous. Since μ(Fm)→1, one can choose um∈Fm satisfying um→s. Take a sequence {qn} in ℚ converging to t. For each fixed m∈ℕ, take a sequence {pn} in ℚ converging to um. By equalities (2.10) and (2.11), we see that
(2.12)ψ(um,t)=ψ(um,lim⁡n→∞qn)=lim⁡n→∞ψ(um,qn)=lim⁡n→∞qnψ(um,1)=tψ(um,1)=tψ(lim⁡n→∞pn,1)=tlim⁡n→∞ψ(pn,1)=tlim⁡n→∞pnψ(1,1)=tumψ(1,1)
for all m∈ℕ. Hence we obtain that
(2.13)ψ(s,t)=ψ(lim⁡m→∞um,t)=lim⁡m→∞ψ(um,t)=lim⁡m→∞tumψ(1,1)=stψ(1,1),
as desired.


## 3. Stability of the Bi-Additive Functional Equation (1.3)

From now on, letX be a normed space, Y a complete normed space, and r≠2 a nonnegative real number. In this section, we investigate the stability of the bi-additive functional equation (1.3).Lemma 3.1.
Letf:X×X→Y be a mapping such that
(3.1)‖f(x+y,z-w)+f(x-y,z+w)-2f(x,z)+2f(y,w)‖≤{4ε,(r=0),ε(‖x‖r+‖y‖r+‖z‖r+‖w‖r),(0<r≠2)
for all x,y,z,w∈X. Then there exists a unique bi-additive mapping F:X×X→Y satisfying (1.3) such that
(3.2)‖f(x,y)-F(x,y)‖≤{2ε+‖f(0,0)‖,(r=0),3ε4-2r(‖x‖r+‖y‖r),(0<r<2),3ε2r-4(‖x‖r+‖y‖r),(r>2)
for all x,y∈X. The mapping F is given by
(3.3)F(x,y):={lim⁡j→∞14jf(2jx,2jy),(0≤r<2),lim⁡j→∞4jf(x2j,y2j),(r>2)
for all x,y∈X.Proof.
Consider the caser∈(0,2). Letting y=x and w=-z in (3.1), we gain
(3.4)‖f(2x,2z)+f(0,0)-2f(x,z)+2f(x,-z)‖≤2ε(‖x‖r+‖z‖r)
for all x,z∈X. Putting x=z=0 in (3.4), we get f(0,0)=0. Putting x=z=0 in (3.1), we get
(3.5)‖f(y,-w)+f(-y,w)+2f(y,w)‖≤ε(‖y‖r+‖w‖r)
for all y,w∈X. Replacing y by x and w by z in the above inequality, we have
(3.6)∥f(x,-z)+f(-x,z)+2f(x,z)∥≤ε(∥x∥r+∥z∥r)
for all x,z∈X. Setting y=-x and w=z in (3.1), we obtain
(3.7)‖f(2x,2z)-2f(x,z)+2f(-x,z)‖≤2ε(‖x‖r+‖z‖r)
for all x,z∈X. By (3.4) and (3.6), we gain
(3.8)‖f(2x,2z)-4f(x,z)+f(x,-z)-f(-x,z)‖≤3ε(‖x‖r+‖z‖r)
for all x,z∈X. By (3.4) and (3.7), we get
(3.9)‖f(x,-z)-f(-x,z)‖≤2ε(‖x‖r+‖z‖r)
for all x,z∈X. By (3.4), (3.6), and (3.7), we have
(3.10)‖f(2x,2z)-4f(x,z)‖≤3ε(‖x‖r+‖z‖r)
for all x,z∈X. Replacing x by 2jx and z by 2jz and dividing 4j+1, we obtain that
(3.11)‖14jf(2jx,2jz)-14j+1f(2j+1x,2j+1z)‖≤3ε⋅2rj4j+1(‖x‖r+‖z‖r)
for all x,z∈X and all j=0,1,2,…. For given integers l,m(0≤l<m), we obtain that
(3.12)‖14lf(2lx,2lz)-14mf(2mx,2mz)‖≤∑j=lm-1‍3ε⋅2rj4j+1(‖x‖r+‖z‖r)
for all x,z∈X. By (3.12), the sequence {(1/4j)f(2jx,2jy)} is a Cauchy sequence for all x,y∈X. Since Y is complete, the sequence {(1/4j)f(2jx,2jy)} converges for all x,y∈X. Define F:X×X→Y by F(x,y):=limj→∞(1/4j)f(2jx,2jy) for all x,y∈X. By (3.1), we have
(3.13)‖14jf(2j(x+y),2j(z-w))+14jf(2j(x-y),2j(z+w))-24jf(2jx,2jz)+24jf(2jy,2jw)‖≤ε2rj4j(‖x‖r+‖y‖r+‖z‖r+‖w‖r)
for all x,y,z,w∈X and all j=0,1,2,…. Letting j→∞ in the above inequality, we see that F satisfies (1.3). Setting l=0 and taking m→∞ in (3.12), one can obtain inequality (3.2). If G:X×X→Y is another mapping satisfying (1.3) and (3.2), by Theorem 2.1, we obtain that
(3.14)‖F(x,y)-G(x,y)‖=14n‖F(2nx,2ny)-G(2nx,2ny)‖≤14n‖F(2nx,2ny)-f(2nx,2ny)‖+14n‖f(2nx,2ny)-G(2nx,2ny)‖≤6ε⋅2n(r-2)4-2r(‖x‖r+‖y‖r)⟶0asn⟶∞
for all x,y∈X. Hence the mapping F is the unique bi-additive mapping satisfying (1.3), as desired.
The proof of the caser∈{0}∪(2,∞) is similar to that of the case r∈(0,2).From now on, letA be a unital C⋆-algebra with a norm |·|, and let ℳA and 𝒩A be left Banach A-modules with norms ||·|| and ∥·∥, respectively. Put A1:={a∈A∣|a|=1}.A bi-additive mappingF:ℳA×ℳA→𝒩A satisfying (1.3) is called A-quadratic if F(ax,ay)=a2F(x,y) for all a∈A and all x,y∈ℳA.Theorem 3.2.
Letf:ℳA×ℳA→𝒩A be a mapping such that
(3.15)‖f(ax+ay,az-aw)+f(ax-ay,az+aw)-2a2f(x,z)+2a2f(y,w)‖≤{4ε,(r=0),ε(∥x∥r+∥y∥r+∥z∥r+∥w∥r),(0<r≠2)
for all a∈A1 and all x,y,z,w∈ℳA. If f(tx,ty) is continuous in t∈ℝ for each fixed x,y∈ℳA, then there exists a unique bi-additive A-quadratic mapping F:ℳA×ℳA→𝒩A satisfying (1.3) and inequality (3.2).Proof.
Consider the caser∈(0,2). By Lemma 3.1, it follows from the inequality of the statement for a=1 that there exists a unique bi-additive mapping F:ℳA×ℳA→𝒩A satisfying (1.3) and inequality (3.2). Let x0,y0∈ℳA be fixed. And let L:𝒩A→ℝ be any real continuous linear functional, that is, L is an arbitrary real functional element of the dual space of 𝒩A restricted to the scalar field ℝ. For n∈ℕ, consider the functions ψn:ℝ→ℝ defined by ψn(t):=(1/4n)L[f(2ntx0,2nty0)] for all t∈ℝ. By the assumption that f(tx,ty) is continuous in t∈ℝ for each fixed x,y∈ℳA, the function ψn is continuous for all n∈ℕ. Note that ψn(t)=(1/4n)L[f(2ntx0,2nty0)]=L[(1/4n)f(2ntx0,2nty0)] for all n∈ℕ and all t∈ℝ. By the proof of Lemma 3.1, the sequence ψn(t) is a Cauchy sequence for all t∈ℝ. Define a function ψ:ℝ→ℝ by ψ(t):=limn→∞ψn(t) for all t∈ℝ. Note that ψ(t)=L[F(tx0,ty0)] for all t∈ℝ. Since F is bi-additive, we get
(3.16)ψ(s+t)+ψ(s-t)=L(F[(s+t)x0,(s+t)y0])+L(F[(s-t)x0,(s-t)y0])=L(F[(s+t)x0,(s+t)y0]+F[(s-t)x0,(s-t)y0])=L[F(sx0+tx0,sy0+ty0)+F(sx0-tx0,sy0-ty0)]=L[2F(sx0,sy0)+2F(tx0,ty0)]=2L[F(sx0,sy0)]+2L[F(tx0,ty0)]=2ψ(s)+2ψ(t)
for all s,t∈ℝ. Since ψ is the pointwise limit of continuous functions, it is a Borel function. Thus the function ψ as a measurable quadratic function is continuous (see [26]) so has the form ψ(t)=t2ψ(1) for all t∈ℝ. Hence we have
(3.17)L[F(tx0,ty0)]=ψ(t)=t2ψ(1)=t2L[F(x0,y0)]=L[t2F(x0,y0)]
for all t∈ℝ. Since L is any continuous linear functional, the bi-additive mapping F:ℳA×ℳA→𝒩A satisfies F(tx0,ty0)=t2F(x0,y0) for all t∈ℝ. Therefore we obtain
(3.18)F(tx,ty)=t2F(x,y)
for all t∈ℝ and all x,y∈ℳA. Let j be an arbitrary positive integer. Replacing x and z by 2jx and 2jz, respectively, and letting y=w=0 in inequality (3.15), we gain
(3.19)‖f(2jax,2jaz)-a2f(2jx,2jz)+a2f(0,0)‖≤2rj-1ε(‖x‖r+‖z‖r)
for all a∈A1 and all x,z∈ℳA. Note that there is a constant K>0 such that the condition
(3.20)‖av‖≤K|a|‖v‖
for each a∈A and each v∈𝒩A (see [27, Definition 12]). For all a∈A1 and all x,y∈ℳA, we get
(3.21)14j‖f(2jax,2jay)-a2f(2jx,2jy)‖≤2(r-2)j-1ε(‖x‖r+‖z‖r)+K|a|24j‖f(0,0)‖⟶0
as j→∞. Hence we have
(3.22)F(ax,ay)=lim⁡j→∞14jf(2jax,2jay)=a2lim⁡j→∞14jf(2jx,2jy)=a2F(x,y)
for all a∈A1 and all x,y∈ℳA. Since F(ax,ay)=a2F(x,y) for each a∈A1, by (3.18), we obtain
(3.23)F(ax,ay)=F(|a|a|a|x,|a|a|a|y)=|a|2F(a|a|x,a|a|y)=a2F(x,y)
for all nonzero a∈A and all x,y∈ℳA. By (3.18), we get F(0x,0y)=02F(x,y) for all x,y∈ℳA. Therefore the bi-additive mapping F is the unique A-quadratic mapping satisfying the inequality (3.2).
The proof of the caser∈{0}∪(2,∞) is similar to that of the case r∈(0,2).We obtain the Hyers-Ulam stability of (1.3) as a corollary of Theorem 3.2.Corollary 3.3.
LetE be a complex normed space and f:E×E→ℂ a function such that
(3.24)‖f(λx+λy,λz-λw)+f(λx-λy,λz+λw)-2λ2f(x,z)+2λ2f(y,w)‖≤ε
for all λ∈𝕋:={λ∈ℂ:|λ|=1} and all x,y,z,w∈E. If f(tx,ty) is continuous in t∈ℝ for each fixed x,y∈E, then there exists a unique bi-additive ℂ-quadratic mapping F:E×E→ℂ satisfying (1.3) such that ∥f(x,y)-F(x,y)∥≤ε/2+∥f(0,0)∥ for all x,y∈E.PutAin:={a∈A∣aisinvertibleinA}, Asa:={a∈A∣a⋆=a}, A+:={a∈Asa∣Sp(a)⊂[0,∞)}, and A1+:=A1∩A+.A unitalC⋆-algebra A is said to have  real rank0 (see [28]) if the invertible self-adjoint elements are dense in Asa.For any elementa∈A, a=a1+ia2, where a1:=(a+a⋆)/2 and a2:=(a-a⋆)/2i are self-adjoint elements, furthermore, a=a1+-a1-+ia2+-ia2-, where a1+,a1-,a2+, and a2- are positive elements (see [27, Lemma 38.8]).Theorem 3.4.
LetA be of real rank 0, and let f:ℳA×ℳA→𝒩A be a mapping such that
(3.25)‖f(ax+ay,bz-bw)+f(ax-ay,bz+bw)-2abf(x,z)+2ab(y,w)‖≤{4ε,(r=0),ε(‖x‖r+‖y‖r+‖z‖r+‖w‖r),(0<r≠2)
for all a,b∈(A1+∩Ain)∪{i} and all x,y,z,w∈ℳA. For each fixed x,y∈ℳA, let the sequence {(1/4j)f(2jax,2jby)} converge uniformly on A1×A1. If f(ax,by) is continuous in (a,b)∈(A1∪ℝ)2 for each fixed x,y∈ℳA, then there exists a unique bi-additive A-quadratic mapping F:ℳA×ℳA→𝒩A satisfying (1.3) and inequality (3.2) such that F(ax,by)=abF(x,y) for all a,b∈A1+∪{i} and all x,y∈ℳA.Proof.
Consider the caser∈(0,2). By Lemma 3.1, there exists a unique bi-additive mapping F:ℳA×ℳA→𝒩A satisfying (1.3) and inequality (3.2) on ℳA×ℳA. Let x0,y0∈ℳA be fixed. And let L be an arbitrary real functional element of the dual space of A𝒩 restricted to the scalar field ℝ. For n∈ℕ, consider the functions ψn:ℝ×ℝ→ℝ defined by ψn(s,t):=(1/4n)L[f(2nsx0,2nty0)] for all s,t∈ℝ. By the assumption that f(ax,by) is continuous in (a,b)∈(A1∪ℝ)2 for each fixed x,y∈ℳA, the function ψn is continuous for all n∈ℕ. Note that ψn(s,t)=(1/4n)L[f(2nsx0,2nty0)]=L[(1/4n)f(2nsx0,2nty0)] for all n∈ℕ and all s,t∈ℝ. By the proof of Lemma 3.1, the sequence ψn(s,t) is a Cauchy sequence for all s,t∈ℝ. Define a function ψ:ℝ×ℝ→ℝ by ψ(s,t):=limn→∞ψn(s,t) for all s,t∈ℝ. Note that ψ(s,t)=L[F(sx0,ty0)] for all s,t∈ℝ. Since the mapping F is bi-additive, we have
(3.26)ψ(s1+s2,t1-t2)+ψ(s1-s2,t1+t2)=L(F[(s1+s2)x0,(t1-t2)y0])+L(F[(s1-s2)x0,(t1+t2)y0])=L(F[(s1+s2)x0,(t1-t2)y0]+F[(s1-s2)x0,(t1+t2)y0])=L[F(s1x0+s2x0,t1y0-t2y0)+F(s1x0-s2x0,t1y0+t2y0)]=L[2F(s1x0,t1y0)-2F(s2x0,t2y0)]=2L[F(s1x0,t1y0)]-2L[F(s2x0,t2y0)]=2ψ(s1,t1)-2ψ(s2,t2)
for all s1,s2,t1,t2∈ℝ. Since ψ is the pointwise limit of continuous functions, it is a Borel function. By Theorem 2.2, we gain ψ(s,t)=stψ(1,1) for all s,t∈ℝ. Hence we get
(3.27)L[F(sx0,ty0)]=ψ(s,t)=stψ(1,1)=stL[F(x0,y0)]=L[stF(x0,y0)]
for all s,t∈ℝ. Since L is any continuous linear functional, the bi-additive mapping F:ℳA×ℳA→𝒩A satisfies F(sx0,ty0)=stF(x0,y0) for all s,t∈ℝ. Therefore we obtain
(3.28)F(sx,ty)=stF(x,y)
for all s,t∈ℝ and all x,y∈ℳA. Let j be an arbitrary positive integer. Replacing x and z by 2jx and 2jz, respectively, and letting y=w=0 in inequality (3.25), we get
(3.29)‖f(2jax,2jbz)-abf(2jx,2jz)+abf(0,0)‖≤2rj-1ε(‖x‖r+‖z‖r)
for all a,b∈(A1+∩Ain)∪{i} and all x,z∈ℳA. By inequality (3.20) and the above inequality, for all a,b∈(A1+∩Ain)∪{i} and all x,z∈ℳA, we have
(3.30)14j‖f(2jax,2jbz)-abf(2jx,2jz)‖≤2(r-2)j-1ε(‖x‖r+‖z‖r)+K|a||b|4j‖f(0,0)‖⟶0asj⟶∞.
Hence we obtain that
(3.31)F(ax,by)=lim⁡j→∞14jf(2jax,2jby)=ablim⁡j→∞14jf(2jx,2jy)=abF(x,y)
for all a,b∈(A1+∩Ain)∪{i} and all x,y∈ℳA. Let c,d∈A1+∖Ain. Since Ain∩Asa is dense in Asa, there exists two sequences {cj} and {dj} in Ain∩Asa such that cj→c and dj→d as j→∞. Put pj:=(1/|cj|)cj and qj:=(1/|dj|)dj for all j∈ℕ. Then pj→c and qj→d as j→∞. Set aj:=pj⋆pj and bj:=qj⋆qj for all j∈ℕ. Then aj→c and bj→d as j→∞ and aj,bj∈A1+∩Ain. Since {(1/4j)f(2jax,2jby)} is uniformly converges on A1×A1 for each x,y∈ℳA and f(ax,by) is continuous in a,b∈A1 for each x,y∈ℳA, we see that F(ax,by) is also continuous in a,b∈A1 for each x,y∈ℳA. In fact, we gain
(3.32)lim⁡(a,b)→(c,d)F(ax,by)=lim⁡(a,b)→(c,d)lim⁡j→∞14jf(2jax,2jby)=lim⁡j→∞lim⁡(a,b)→(c,d)14jf(2jax,2jby)=lim⁡j→∞14jf(2jcx,2jdy)=F(cx,dy)
for all x,y∈ℳA. Thus we get
(3.33)lim⁡j→∞F(ajx,bjy)=F(lim⁡j→∞ajx,lim⁡j→∞bjy)=F(cx,dy)
for all x,y∈ℳA. By equality (3.31), we have
(3.34)‖F(ajx,bjy)-cdF(x,y)‖=‖ajbjF(x,y)-cdF(x,y)‖⟶‖cdF(x,y)-cdF(x,y)‖=0
as j→∞ for all x,y∈ℳA. By equality (3.33) and the above convergence, we see that
(3.35)‖F(cx,dy)-cdF(x,y)‖≤‖F(cx,dy)-F(ajx,bjy)‖+‖F(ajx,bjy)-cdF(x,y)‖⟶0
as j→∞ for all x,y∈ℳA. By equality (3.31) and the above convergence, we obtain
(3.36)F(ax,by)=abF(x,y)
for all a,b∈A1+∪{i} and all x,y∈ℳA. Since the mapping F is bi-additive, we see that
(3.37)F(ax,ay)=F(a1+x-a1-x+ia2+x-ia2-x,a1+y-a1-y+ia2+y-ia2-y)=F(a1+x,a1+y)-F(a1+x,a1-y)+F(a1+x,ia2+y)-F(a1+x,ia2-y)-F(a1-x,a1+y)+F(a1-x,a1-y)-F(a1-x,ia2+y)+F(a1-x,ia2-y)+F(ia2+x,a1+y)-F(ia2+x,a1-y)+F(ia2+x,ia2+y)-F(ia2+x,ia2-y)-F(ia2-x,a1+y)+F(ia2-x,a1-y)-F(ia2-x,ia2+y)+F(ia2-x,ia2-y)
for all a∈A and all x,y∈ℳA. By (3.28) and equality (3.36), we have
(3.38)F(px,qy)=F(|p|p|p|x,|q|q|q|y)=|p||q|F(p|p|x,q|q|y)=pqF(x,y)
for all p,q∈{a1+,a1-,a2+,a2-} and all x,y∈ℳA. Note that a1+a1-=a1-a1+=a2+a2-=a2-a2+=0. Hence we obtain that
(3.39)F(ax,ay)=(a2+)2F(x,y)+ia1+a2+F(x,y)-ia1+a2-F(x,y)+(a1-)2F(x,y)-ia1-a2+F(x,y)+ia1-a2-F(x,y)+ia2+a1+F(x,y)-ia2+a1-F(x,y)-(a2+)2F(x,y)-ia2-a1+F(x,y)+ia2-a1-F(x,y)-(a2-)2F(x,y)=[(a1+)2+ia1+a2+-ia1+a2-+(a1-)2-ia1-a2++ia1-a2-+ia2+a1+-ia2+a1--(a2+)2-ia2-a1++ia2-a1--(a2-)2]F(x,y)=(a1+-a1-+ia2+-ia2-)2F(x,y)=a2F(x,y)
for all a∈A and all x,y∈ℳA.
The proof of the caser∈{0}∪(2,∞) is similar to that of the case r∈(0,2).



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*Source: 835893-2012-07-05.xml*