# A Characterization of Bi-Jordan Homomorphisms on Banach Algebras

Authors: Abbas Zivari-Kazempour
Journal: International Journal of Analysis
 (2017)

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## Abstract

For Banach algebrasA and B, we show that if U=A×B is unital and commutative, each bi-Jordan homomorphism from U into a semisimple commutative Banach algebra D is a bihomomorphism.


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## Body

## 1. Introduction

LetA and B be complex Banach algebras and let φ:A→B be a linear map. Then φ is called n-homomorphism if for all a1,a2,…,an∈A, (1)φa1a2⋯an=φa1φa2⋯φan. The concept of n-homomorphism was studied for complex algebras by Hejazian et al. in [1]. A 2-homomorphism is then just a homomorphism, in the usual sense. We refer the reader to [2], for certain properties of 3-homomorphisms.In [3], Eshaghi Gordji introduced the concept of an n-Jordan homomorphism. A linear map φ between Banach algebras A and B is called an n-Jordan homomorphism if (2)φan=φan,a∈A. A 2-Jordan homomorphism is called simply a Jordan homomorphism.It is obvious that eachn-homomorphism is an n-Jordan homomorphism, but in general the converse is false. The converse statement may be true under certain conditions. For example, Zelazko in [4] proved that every Jordan homomorphism from Banach algebra A into a semisimple commutative Banach algebra B is a homomorphism. See also [5] for another approach to the same result. The reader is referred to [6], for characterization of 3-Jordan homomorphism.Also it is shown in [3] that every n-Jordan homomorphism between two commutative Banach algebras is an n-homomorphism for n∈{2,3,4} and this result is extended to the case n=5 in [7]. Some investigation has been done about Jordan derivations and Jordan centralizers in [8, 9].Throughout the paper, letU=A×B. Then U is a Banach algebra for the multiplication (3)a,bx,y=ax,by,a,b,x,y∈U, and with norm (4)a,b=a+b. Clearly, U is commutative if and only if both A and B are commutative, and it is unital if and only if both A and B are unital. Without any confusion we denote by e, the unit element of both A and B.LetD be a complex Banach algebra. A bilinear map is a function φ:U→D such that for any a∈A the map b↦φ(a,b) is linear map from B to D, and for any b∈B the map a↦φ(a,b) is linear map from A to D.A bilinear mapφ is called bihomomorphism if for all (a,b),(x,y)∈U, (5)φax,by=φa,bφx,y, and it is called bi-Jordan homomorphism (BJH, for short) if (6)φa2,b2=φa,b2,a,b∈U.It is obvious that each bihomomorphism is BJH, but in general the converse is not true. For example, let(7)A=ab00:a,b∈R, and B=A∪{I}, where I=1001. Define φ:A×B→A by φ(x,y)=xy. Then φ is bilinear and (8)φx2,y2=x2y2=xy2=φx,y2, for all (x,y)∈U=A×B. Thus, φ is a bi-Jordan homomorphism, but it is not bihomomorphism. For instance, let (9)u=ab00,x=cd00,v=st00,y=I. Then (u,v),(x,y)∈U and (10)φux,vy=acsact00≠ascasd00=φu,vφx,y.In this paper, we show that the converse statement holds under certain conditions.


## 2. Main Results

We commence with some useful lemmas.Lemma 1.
Suppose thatφ:U→C is a BJH. Then (1)
φ(xy+yx,b2)=2φ(x,b)φ(y,b);(2)
φ(a2,xy+yx)=2φ(a,x)φ(a,y).Proof.
The proof is straightforward.Lemma 2.
Suppose thatφ:U→C is a BJH. If U is unital and commutative, then (1)
φ(ax,e)=φ(a,e)φ(x,e);(2)
φ(e,by)=φ(e,b)φ(e,y).Proof.
The proof follows from Lemma1.Lemma 3.
LetU be unital, and let φ:U→C be a nonzero BJH. Then φ(e,e)≠0.Proof.
Letφ be a BJH. Then for all (a,b)∈U, we get(11)φa2,b2=φa,b2. Replacing a by x+e and b by y+e in (11) gives(12)φx2+2x+e,y2+2y+e=φx+e,y+e2. Assume that φ(e,e)=0; then by Lemma 1,(13)φx,e=φe,y=0,x,y∈U. It follows from (12) and (13) that(14)φ2x,2y+φ2x,y2+φx2,2y=0, for all (x,y)∈U. By Lemma 1,(15)φ2x,y2=2φx,yφe,y=0,φx2,2y=2φx,eφx,y=0. Thus, by (14) and (15), we get φ(x,y)=0, for all (x,y)∈U, which is contradiction.Now we state and prove the main theorem. The main idea of the proof can be found in [6].Theorem 4.
LetU be unital and commutative, and let φ be a BJH from U into a semisimple commutative Banach algebra D. Then φ is a bihomomorphism.Proof.
We first assume thatD=C and let φ:U→C be a BJH. Then, for all (a,b)∈U, we get(16)φa2,b2=φa,b2. Replacing a by x+e and b by y+e in (16), we have(17)φx2+2x+e,y2+2y+e=φx+e,y+e2. By Lemma 3, φ(e,e)≠0, so (16) gives(18)φe,e=1. It follows from Lemma 1 that(19)φ2x,y2=2φx,yφe,y,φx2,2y=2φx,eφx,y. Thus, by (17), (18), and (19) we get(20)φx,y=φx,eφe,y, for all (x,y)∈U. Replacing x by ax and y by by in (20) gives(21)φax,by=φax,eφe,by. By (20) and Lemma 2 we deduce(22)φa,bφx,y=φa,eφx,eφe,bφe,y=φax,eφe,by. By (21) and (22), (23)φax,by=φa,bφx,y, for all (a,b),(x,y)∈U, and so φ is bihomomorphism.
Now suppose thatD is semisimple and commutative. Let M(D) be the maximal ideal space of D. We associate with each f∈M(D) a function φf:U→C defined by (24)φfa,b≔fφa,b,a,b∈U. Pick f∈M(D) arbitrary. It is easy to see that φf is a BJH, so by the above argument it is a bihomomorphism. Thus by the definition of φf we have (25)fφax,by=fφa,bfφx,y=fφa,bφx,y. Since f∈M(D) was arbitrary and D is assumed to be semisimple, we obtain (26)φax,by=φa,bφx,y, for all (a,b),(x,y)∈U.The second dual spaceA′′ of a Banach algebra A admits two Banach algebra multiplications known as the first and second Arens products, each extending the product on A. These products which we denote by □ and ♢, respectively, can be defined as follows: (27)F□G=w∗-limi⁡limj⁡aibj,F♢G=w∗-limj⁡limi⁡aibj, where (ai) and (bj) are nets in A that converge, in w∗-topologies, to F and G, respectively. The Banach algebra A is said to be Arens regular if F□G=F♢G on the whole of A′′. Some significant results related to the Arens regularity of certain bilinear maps and Banach algebra obtained in [10]. For more information on the Arens products, we refer the reader to [11, 12], for example.It is shown in [11] that every C∗-algebra A is Arens regular and semisimple. Also the second dual of each C∗-algebra is also a C∗-algebra.Theorem 5.
Ifk is the natural embedding of a Banach space X into X′′, then k(X) is w∗-dense in X′′.Proof.
See [13].Theorem 6.
Letφ:U→D be a continuous BJH. Then the second adjoint φ′′:U′′→D′′ of φ is also a BJH.Proof.
By Theorem5, there are bounded nets (ai) and (bj) in A and B that converge, in w∗-topologies, to F and G, respectively. Then(28)φ′′F2,G2=φ′′w∗-limi⁡ai2,w∗-limj⁡bj2=φw∗-limi⁡ai2,w∗-limj⁡bj2=w∗-limi⁡limj⁡φai2,bj2=w∗-limi⁡limj⁡φai,bj2=w∗-limi⁡limj⁡φ′′ai,bj2=φ′′w∗-limi⁡ai,w∗-limj⁡bj2=φ′′F,G2. Thus, φ′′ is BJH, as claimed.Since the second dual of everyC∗-algebra is unital [14], we deduce the following result from Theorems 4 and 6.Corollary 7.
Letφ:U→D be a continuous BJH between commutative C∗-algebras. Then φ′′:U′′→D′′ is a bihomomorphism.Theorem 8.
Suppose thatφ is a BJH from the unital Banach algebra U into C. Then φ is a n-BJH; that is, (29)φan,bn=φa,bn, for all n≥3.Proof.
Letφ:U→C be a BJH. Then, for all (a,b)∈U,(30)φa2,b2=φa,b2. Replacing a by a+x and b by b+y in (30), we have(31)φax+xa,by+yb=2φa,bφx,y+2φa,yφx,b. Replacing a by a2 and b by b2 in (31) gives(32)φa2x+xa2,b2y+yb2=2φa2,b2φx,y+2φa2,yφx,b2. Replacing x by a and y by b in (32), we obtain(33)φ2a3,2b3=2φa2,b2φa,b+2φa2,bφa,b2. Replacing x and y by e in (31), we get(34)φ2a,2b=2φa,bφe,e+2φa,eφe,b. Since φ(e,e)=1, so (34) gives(35)φa,b=φa,eφe,b, for all (a,b)∈U. By Lemma 1,(36)φa2,b=φa,bφa,e,φa,b2=φa,bφe,b. It follows from (35) and (36) that(37)φa2,bφa,b2=φa,b2φa,eφe,b=φa,b3. By (33) and (37), (38)φa3,b3=φa,b3, for all (a,b)∈U. Thus, the result is established for n=3. An easy induction argument now finishes the proof.As a consequence of Theorem8 we have the next result.Corollary 9.
Suppose thatφ is a BJH from the unital Banach algebra U into a semisimple commutative Banach algebra D. Then φ is an n-BJH.Baker in [15] proved that an almost multiplicative function is either bounded or multiplicative. Now we prove an analogous result of Baker’s theorem for bihomomorphism.Theorem 10.
Letδ>0 and φ:U→C satisfy(39)φax,by-φa,bφx,y≤δ, for all (a,b),(x,y)∈U. Then either φ is bihomomorphism or (40)φx,y≤121+1+4δ≔ε, for all (x,y)∈U.Proof.
Letε2-ε=δ and ε>1. Suppose that there exist (s,t)∈U such that |φ(s,t)|>ε, so that |φ(s,t)|=ε+p, where p>0. Then we have (41)φs2,t2≥φs,t2-φs,t2-φs2,t2≥φs,t2-φs,t2-φs2,t2; hence(42)φs2,t2≥φs,t2-δ=ε+p2-δ=2εp+p2+ε>ε+2p. Now make the induction assumption:(43)φs2n,t2n>ε+n+1p. Let (44)Mn=φs2n,t2n. Then(45)Mn+1=Mn2-Mn2-Mn+1≥Mn2-δ=Mn2+ε-ε2, whence(46)Mn+1≥ε+n+1p2+ε-ε2=2n+2εp+ε+n+12p2>ε+n+2p, and (43) is established for all n∈N.
Given that(a,b),(x,y),(u,v)∈U, we have (47)φaxu,byv-φa,bφxu,yv≤δ,φaxu,byv-φax,byφu,v≤δ. Thus, (48)φax,byφu,v-φa,bφxu,yv≤2δ. Hence (49)φax,byφu,v-φa,bφx,yφu,v≤φax,byφu,v-φa,bφxu,yv+φa,bφxu,yv-φa,bφx,yφu,v≤2δ+δφa,b. Therefore (50)φu,vφax,by-φa,bφx,y≤2δ+δφa,b. Now put (u,v)=(s2n,t2n) to obtain (51)φax,by-φa,bφx,y≤2δ+δφa,bMn≤2δ+δφa,bε+n+1p, for all n∈N and (a,b)∈U. Letting n→∞, we get φ(ax,by)=φ(a,b)φ(x,y) and so φ is a bihomomorphism.As Baker pointed out in his article, the above proof works for functionsφ:U→D, where D is a Banach algebra in which the norm is multiplicative; that is, xy=xy for all x,y∈D. For norm algebra D for which the norm is not multiplicative, the situation is false. For example, let δ>0 and choose ε>0 so that |ε-ε2|=δ, and define φ:R×R→M3(R) by (52)φx,y=x000y000εx,y∈R. Then with the usual matrix norm (53)φax,by-φa,bφx,y=δ, for all (a,b),(x,y)∈R×R. Clearly, φ is unbounded, but φ is not bihomomorphism.



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*Source: 4206871-2017-01-31.xml*