Research article

Action of projections on Banach algebras

  • Received: 26 December 2022 Revised: 04 May 2023 Accepted: 10 May 2023 Published: 19 May 2023
  • MSC : 47L10, 47B48, 46J10, 17C65

  • Let A be a Banach algebra and n>1, a fixed integer. The main objective of this paper is to talk about the commutativity of Banach algebras via its projections. Precisely, we prove that if A is a prime Banach algebra admitting a continuous projection P with image in Z(A) such that P(an)=anfor allaG, the nonvoid open subset of A, then A is commutative and P is the identity mapping on A. Apart from proving some other results, as an application we prove that, a normed algebra is commutative iff the interior of its center is non-empty. Furthermore, we provide some examples to show that the assumed restrictions cannot be relaxed. Finally, we conclude our paper with a direction for further research.

    Citation: Shakir Ali, Amal S. Alali, Naira Noor Rafiquee, Vaishali Varshney. Action of projections on Banach algebras[J]. AIMS Mathematics, 2023, 8(8): 17503-17513. doi: 10.3934/math.2023894

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  • Let A be a Banach algebra and n>1, a fixed integer. The main objective of this paper is to talk about the commutativity of Banach algebras via its projections. Precisely, we prove that if A is a prime Banach algebra admitting a continuous projection P with image in Z(A) such that P(an)=anfor allaG, the nonvoid open subset of A, then A is commutative and P is the identity mapping on A. Apart from proving some other results, as an application we prove that, a normed algebra is commutative iff the interior of its center is non-empty. Furthermore, we provide some examples to show that the assumed restrictions cannot be relaxed. Finally, we conclude our paper with a direction for further research.



    This research has been motivated by the work's of Ali-Khan[2] and Khan[9]. All over this paper unless otherwise stated, A denotes a Banach algebra with the center Z(A), M be a closed linear subspace of A and Int(Z(A)), the interior of Z(A). For any a,bA, the symbol [a, b] will denote the commutator, abba and ab represents the anticommutator, ab+ba. An algebra A is said to be prime if for any a,bA, aAb=(0) implies a=0orb=0 and A is semiprime if for any aA, aAa=(0) implies a=0. A linear mapping D:AA is called a derivation if D(ab)=D(a)b+aD(b) holds for all a,bA. In particular, D defined by D(a)=[λ,a] for all aA is a derivation, called an inner derivation induced by an element λA. Let R be an associative ring and an additive subgroup U of R is known to be a Lie ideal of R if [u,a]U, for all uU and aR. Let M be a subspace of A, the linear operator P:AA is said to be a projection of A on M if P(a)M for all aA and P(a)=a for all aM. Let M and N be two subspaces of the Banach algebra A such that MalN is an algebraic direct sum of A, we provide the two subspaces M and N with the induced topology of A and M×N by the product topology. We say that MtN is a topological direct sum of A, if the mapping ψ:M×NA, defined by ψ(a,b)=a+b is a homeomorphism, and we say that M is complemented in A and N is its topological complement. In this case, there is a unique continuous projection P from A to M. Moreover, Sobezyk [12], established that if A is separable and M is a subspace of A isomorphic to C0 (the subspace of sequences in C which converge to 0), then M is complemented in A. In view of [11], every infinite-dimensional Banach space that is not isomorphic to a Hilbert space contains a closed uncomplemented subspace (see also [6] for details).

    Many results in literature concerning commutativity of a prime and semiprime Banach algebra are proved, for example, in the year 1990, Yood [15] showed that if G is a nonvoid open subset of a Banach algebra A, for each a,bG, if positive integers m=m(a,b) and n=n(a,b) exist such that [am,bn]=0. Then there is a positive integer r so that arZ(A) for all aA. In addition if A has no nonzero nilpotent ideal, it is sufficient to have [am,bn]Z(A) and a,bG, with m,n as above. Then A is commutative. Yood [13] also proved that if A is a Banach algebra and G1, G2 are two non-empty open subsets of A such that for each aG1 and bG2, there is an integer n=n(a,b)>1 where either (ab)n=anbn or (ab)n=bnan, then A is commutative (see [4,7,8] for recent works). Motivated by Yood's work, Ali and Khan [2] established the commutativity of Banach algebra via derivations. Moreover, they proved that if A is a unital prime Banach algebra and has a nonzero continuous linear derivation d:AA such that either d((ab)n)anbn or d((ab)n)bnan is in the center of A for an integer n=n(a,b)>1, then A is commutative (see, also [1,3,9,10] for recent works).

    In this paper, we will continue the study of the problems on Banach algebras involving projections instead of derivations. The key aim of this work is to discuss the commutativity of prime Banach algebras via its projections. Precisely, we prove that if a prime Banach algebra A admits a nonzero continuous projection P from A to Z(A) such that P(an)=an for all aG, the non-empty open subset of A and nN, then A is commutative and P is the identity map on A. Furthermore, apart from proving some other interesting results, we discuss some applications of our study.

    We recall some well known results which will be helpful in order to prove our results. We begin with the following:

    Lemma 2.1. [5] Let A be a real or complex Banach algebra and p(t)=nk=0bktk a polynomial in the real variable t with coefficients in A. If for an infinite set of real values of t, p(t)M, where M is a closed linear subspace of A, then every bk lies in M.

    Lemma 2.2. [14,Theorem 2] Suppose that there are non-empty open subsets G1,G2 of A (where A denotes a Banach algebra over the complex field with center Z) such that for each xG1 and yG2 there are positive integers n=n(x,y), m=m(x,y) depending on x and y, n>1,m>1, such that either [xn,ym]Z or xnymZ. Then A is commutative if A is semiprime.

    These lemmas will come handy while proving our results. We shall be proving the following results:

    Theorem 2.3. Let n>1 be a fixed integer. Next, let A be a real or complex semiprime Banach algebra and G be a nonvoid open subset of A. If A admits a continuous projection P whose image lies in Z(A) such that

    P(an)=anforallaG,

    then P is the identity mapping on A and A is commutative.

    Theorem 2.4. Let n>1 be a fixed integer. Next, let A be a real or complex prime Banach algebra and G1, G2 be two nonvoid open subsets of A admitting a continuous projection P whose image lies in Z(A). If for all aG1, there exists bG2 such that

    P(an)=bn,

    then P is the identity mapping on A and A is commutative.

    In [2,14], it was observed that the authors used simple multiplication and Lie product. Motivated by this, in our next result we will be using the two symbols "T" and "" representing either the Lie product "[., .]", or the Jordan product "", or the simple multiplication "" of the algebras.

    Theorem 2.5. Let n,m>1 be fixed integers. Next, let A be a real or complex prime Banach algebra and G1, G2 be two nonvoid open subsets of A. If A admits a continuous projection P whose image lies in Z(A) such that

    P(anTbm)=anbmforall(a,b)G1×G2,

    then P is the identity mapping on A and A is commutative.

    The next theorem provides the necessary and sufficient condition for the commutativity of a real or complex normed algebras.

    Theorem 2.6. The normed algebra A over C or R is commutative if and only if the interior of its center is non-empty.

    In order to prove the above mentioned theorems, we need the following auxiliary result. This result will help to bridge the gap between a complemented subspace and that of a projection map.

    Proposition 2.7. Let A be a Banach space and M be a closed subspace of A. M is complemented if there is a continuous projection P of A on M, and its complement is (IP)(A), where I is the identity mapping on A.

    Proof. For any aA, we can write a=P(a)+(aP(a)), since P(a)M and aP(a)(IP)(A), we observe that A=M+(IP)(A). Now, we will show that M(IP)(A)={0}. Let aM(IP)(A). Then a(IP)(A), and hence there is a xA such that a=(IP)(x)=xP(x). Since aM and P(x)M, we conclude that xM and P(x)=x, so we obtain a=0. Therefore, Mal(IP)(A) is an algebraic direct sum of the Banach space A. It remains for us to show that (IP)(A) is closed in A. For this, consider a sequence (bn)nN(IP)(A) that converges to bA as n. Since Mal(IP)(A) is an algebraic direct sum of A, then there is c1M and c2(IP)(A) such that b=c1+c2. Thus, we obtain

    P(b)=c1. (2.1)

    On the other hand, we have P(bn)=0 for all nN, as P is a continuous linear operator then P(limbn)=0. This implies

    c1=P(b)=0. (2.2)

    In view of (2.1) and (2.2), we obtain b=c2(IP)(A), and hence we conclude that (IP)(A) is a closed subspace. Consequently, Mt(IP)(A) is a topological direct sum of the Banach space A. This proves the proposition.

    Proof of Theorem 2.3. Let A be a semiprime Banach algebra and P be a continuous projection satisfying

    P(an)=anforallaG, (3.1)

    for a fixed nN. Then, clearly P is not zero. Thus, Z(A) forms a closed subspace of A and P is a continuous projection onto Z(A), and hence in view of Proposition 1, we conclude that Z(A) is complemented in A and a topological direct sum of A, that is,

    A=Z(A)t(IP)(A). (3.2)

    Now, let a0G and aA, then a0+kaG for any sufficiently small real k, so P((a0+ka)n)(a0+ka)n=0. We can write

    (a0+ka)n=A0+A1k+A2k2++Ankn.

    We take p(k)=P((a0+ka)n)(a0+ka)n, since P is a linear operator, we can write

    p(k)=(P(A0)A0)+(P(A1)A1)k+(P(A2)A2)k2++(P(An)An)kn.

    The coefficient of kn in the above polynomial is P(An)An. Since An=an, then this coefficient becomes P(an)an. Using the Lemma 2.1, we obtain P(an)an=0 and thus P(an)=an for all aA. Since P(a)Z(A) for all aA, we conclude that anZ(A) for all aA. This implies anbnZ(A) for all (a,b)A×A. Hence the required result follows from the Lemma 2.2, i.e., A is commutative. From Eq (3.2), we obtain (IP)(A)=0 and hence P=I, which completes the proof of theorem.

    As an immediate consequence of Theorem 2.3, we have the following results.

    Corollary 3.1. Let n>1 be a fixed integer. Next, let A be a real or complex prime Banach algebra and G be a nonvoid open subset of A. If A admits a continuous projection P whose image lies in Z(A) such that

    P(an)=anforallaG,

    then P is the identity mapping on A and A is commutative.

    Corollary 3.2. Let J be a part dense in a prime Banach algebra A. If A admits a continuous projection P whose image lies in Z(A) and there is an nN such that

    P(an)=anforallaJ,

    then A is commutative and P=I, the identity mapping on A.

    Proof. Let aA, there exists a sequence (ak)kN in J converging to a. Since (ak)kNJ, so for each kN

    P((ak)n)(ak)n=0.

    By the continuity of P, we conclude that P(an)an=0. Consequently, there exists nN such that

    P(an)an=0forallaA.

    Application of Theorem 2.3, yields the required result.

    Proof of Theorem 2.4. Let us consider two sets,

    Gn={(a,b)A×A|P(an)bn}andHn={(a,b)A×A|P(an)=bn},

    for some nN. Observe that (Gn)(G1×G2)=. Indeed, if there exists aG1 and bG2 such that (a,b)Gn for all nN, then P(an)bn for all nN, which is absurd by the hypothesis of the theorem.

    Now we claim that each Gn is open in A×A. We show that Hn, the complement of Gn, is closed. Consider a sequence (aj,bj)jNHn converging to (a,b)A×A. Since (aj,bj)jNHn, so

    P((aj)n)=(bj)nforalljN.

    As P is continuous, we deduce that P(an)=bn. Therefore, (a,b)Hn, making Hn closed and Gn open. By Baire category theorem, we imply that the intersection of all Gn's is dense if each Gn is dense, contradicting the fact that (Gn)(G1×G2)=. Hence there exists pN such that Gp is not dense in A and a nonvoid open subset G×G in Hp, such that

    P(ap)=bpforall(a,b)G×G.

    Now, let (a0,b0)G×G and (a,b)A×A. Then a0+kaG and b0+kbG for any sufficiently small real k, making P((a0+ka)p)(b0+kb)p=0. We have,

    (a0+ka)p=Ap,0(a0,a)+Ap1,1(a0,a)k+Ap2,2(a0,a)k2++A0,p(a0,a)kp,

    and

    (b0+kb)p=Bp,0(b0,b)+Bp1,1(b0,b)k+Bp2,2(b0,b)k2++B0,p(b0,b)kp.

    We put p(k)=P((a0+ka)p)(b0+kb)p, since P is a projection, we can write p(k)=P(Ap,0(a0,a))Bp,0(b0,b)+P(Ap1,1(a0,a))Bp1,1(b0,b)k +P(Ap2,2(a0,a))Bp2,2(b0,b)k2++P(A0,p(a0,a))B0,p(b0,b)kp. The coefficient of kp in the above polynomial is just P(ap)bp, according to the Lemma 2.1, we obtain P(ap)bp=0 and hence, P(ap)=bp for all (a,b)A×A. In particular, for a=b we have P(ap)=ap for all aA. In view of Theorem 2.3, we conclude A is commutative and P is the identity mapping of A.

    Proof of Theorem 2.5. We know that Z(A) forms a closed subspace of A and P is a given continuous projection onto Z(A), by Proposition 2.7, Z(A) is complemented in A. That is,

    A=Z(A)t(IP)(A), (3.3)

    a topological direct sum of Z(A) and its complement. For any pair n,mN, we define two sets:

    Gn,m={(a,b)A×A|P(anTbm)anbm}

    and

    Hn,m={(a,b)A×A|P(anTbm)=anbm}.

    We observe that (Gn,m)(G1×G2)=. If not, there exists some (a,b)G1×G2 such that (a,b)Gn,m for all n,mN, then P(anTbm)anbm for all n,mN which contradicts the hypothesis of the theorem.

    Now we claim that each Gn,m is open in A×A. We show that Hn,m, the complement of Gn,m, is closed. For this, consider a sequence ((aj,bj))jNHn,m converging to (a,b)A×A. Since ((aj,bj))jNHn,m, we have

    P((aj)nT(bj)m)=(aj)n(bj)mforalljN.

    We conclude that P(anTbm)=anbm, as P is continuous. Therefore, (a,b)Hn,m and Hn,m is closed (i.e., Gn,m is open). If every Gn,m is dense, we know that their intersection is also dense (by Baire category theorem), which contradicts with (Gn,m)(G1×G2)=. Hence there exists p,qN such that Gp,q is not dense and a nonvoid open subset G×G in Hp,q, such that

    P(apTbq)=apbqforall(a,b)G×G.

    Since P(a)Z(A) for all aA, we conclude that apbqZ(A) for all (a,b)G×G, hence the result follows from the Lemma 2.2.

    The following are the immediate consequences of Theorem 2.5.

    Remark 3.3. Let n,m>1 be fixed integers. Next, let A be a real or complex prime Banach algebra and G1, G2 be two nonvoid open subsets of A. If A admits a continuous projection P whose image lies in Z(A) such that

    P([an,bm])=[an,bm]forall(a,b)G1×G2,

    then P is the identity mapping on A and A is commutative.

    Remark 3.4. Let n,m>1 be fixed integers. Next, let A be a real or complex prime Banach algebra and G1, G2 be two nonvoid open subsets of A. If A admits a continuous projection P whose image lies in Z(A) such that

    P(anbm)=[an,bm]forall(a,b)G1×G2,

    then P is the identity mapping on A and A is commutative.

    Remark 3.5. Let n,m>1 be fixed integers. Next, let A be a real or complex prime Banach algebra and G1, G2 be two nonvoid open subsets of A. If A admits a continuous projection P whose image lies in Z(A) such that

    P([an,bm])=anbmforall(a,b)G1×G2,

    then P is the identity mapping on A and A is commutative.

    Proof of Theorem 2.6. Suppose A is commutative, then A=Z(A) and hence the interior of Z(A) is the interior of A itself, A being open implies Int(Z((A))=Int(A)=A0.

    Now, we prove the other way round. If Int(Z(A)), then there exists 0aInt(Z(A)). Let cA, we have a+kcInt(Z(A)) for any sufficiently small nonzero real k, therefore, we have

    [a+kc,b]=0forallbA,

    that is,

    [a,b]+k[c,b]=0forallbA.

    Since,

    [a,b]=0forallbA,

    we obtain

    k[c,b]=0forallbA.

    This implies that

    [c,b]=0forallbA.

    Hence, A is commutative. This completes the proof.

    Corollary 3.6. If Z(A) contains an isolated point of A, then A is commutative.

    Proof. Let x be an isolated point of A contained in Z(A). We have {x}Z(A). This gives Int({x})Int(Z(A)). Since x is an isolated point of A, the singleton set {x} is open in A, that is, Int({x})={x}. That is, Int(Z(A)). Hence, A is commutative by Theorem 2.6.

    In particular, we get the following result.

    Corollary 3.7. Let A be a normed algebra over R or C. If 0 is an isolated point in A, then A is commutative.

    Example 3.8. Let R be the field of real numbers. Next, let us consider

    A={(a11a12a21a22)|aijR,1i,j2}.

    Clearly, A is a real prime Banach algebra under the norm defined by ||A||1=maxj(2i=1|aij|) for all A=(aij)1i,j2A.

    Consider the sets, S1={(s00s)|s>0} and S2={(s00s)|sR}. Clealy, S1 and S1 are not open in A. Take E1=(1001), E2=(0010), E3=(0100) and E4=(0001), we observe that the family O={E1,E2,E3,E4} is a basis of A and Z(A)=span(E1), so we can write A=Z(A)tspan(E2,E3,E4). The mapping P defined from A to Z(A) by P(M)=a1E1 for all M=4i=0aiEiA is a continuous projection of A on Z(A). For all A=(a00a)S1, B=(b00b)S2 and for all n,mN, it is easy to see that

    An=(an00an)andBm=(bm00bm).

    Moreover, we compute

    AnBm=(anbm00anbm),AnBm=(2anbm002anbm)and[An,Bm]=(0000).

    This implies that

    AnBm=anbmE1;AnBm=2anbmE1;[An,Bm]=0E1.

    Thus, we have

    (1) P(AnBm)=AnBm

    (2) P(AnBm)=AnBm

    (3) P([An,Bm])=[An,Bm]

    (4) P(An)=An.

    One might think that the projection map P=I, but this is not true as P(E3)=0AE3. So the conditions of G1 and G2 to be open in Theorem 2.5 is indispensable.

    Our next example shows that we cannot replace R or C in Theorem 2.5 by the finite field F3=Z/3ZZ3.

    Example 3.9. Let A={(a11a12a21a22)|aijZ3}. It is easy to check that A forms a prime Banach algebra under the norm, ||A||=maxi(2j=1|aij|) for all A=(aij)1i,j2A, where |.| on Z3 is defined as,

    |ˉ0|=0,|ˉ1|=1and|ˉ2|=2.

    Next, let G={(s00s)|sZ3} be an open set in A. Consider AG, the open ball B(A,1)={YAsuchthat||AY||<1}={A}G, so G is a nonvoid open subset of A. Take, E1=(1001), E2=(0010), E3=(0100) and E4=(0001), observe that the family O={E1,E2,E3,E4} forms a basis of A and Z(A)=span(E1), so we can write A=Z(A)tspan(E2,E3,E4). The mapping P defined from A to Z(A) by P(M)=a1E1 for all M=4i=0aiEiA is a nonzero continuous projection of A on Z(A). For all A=(a00a), B=(b00b)G and for all n,mN, we have

    An=(an00an)andBm=(bm00bm).

    Thus, we obtain

    AnBm=(anbm00anbm),AnBm=(2anbm002anbm)and[An,Bm]=(0000).

    So we can write,

    AnBm=anbmE1;AnBm=2anbmE1;[An,Bm]=0E1.

    Thus, it is easy to see that

    (1) P(AnBm)=AnBm

    (2) P(AnBm)=AnBm

    (3) P([An,Bm])=[An,Bm]

    (4) P(An)=An.

    Observe that PI as P(E2)=0AE2. Hence we conclude that R or C cannot be replaced by the field F3 in case of Theorem 2.5.

    In this section, we will discuss an application of Theorem 2.5.

    Let C be the field of complex numbers. Next, let us consider

    A={(a11a12a21a22)|aijC,1i,j2}.

    Clearly, A is prime algebra over C under the norm defined by ||A||1=maxj(2i=1|aij|) for all A=(aij)1i,j2A. Consider the sets, G1={(eit00eit)|tR} and G2={(eit00eit)|tR}. Clearly, G1 and G2 are open in A. Take E1=(1001), E2=(0010), E3=(0100) and E4=(0001), we observe that the family O={E1,E2,E3,E4} is a basis of A and Z(A)=span(E1), so we can write A=Z(A)tspan(E2,E3,E4). The mapping P defined from A to Z(A) by P(M)=a1E1 for all M=4i=0aiEiA is a continuous projection of A on Z(A). For all A=(eia00eia)G1, B=(eib00eib)G2 and for all n,mN, it is easy to see that

    An=(eina00eina)andBm=(eimb00eimb).

    Moreover, we compute

    [An,Bm]=(0000).

    This implies that

    [An,Bm]=0E1.

    Thus, we have P([An,Bm])=[An,Bm]. Observe that P(E3)=0AE3. Hence, it follows from Theorem 2.5, that A is not a Banach algebra under the defined norm.

    Several papers in the literature evidence how the behaviour of some linear mappings is closely connected to the structure of the Banach algebras (cf. [1,2,3] and [9,10]). In our main results (Theorems 2.3 and 2.5), we investigate the structure of prime Banach algebras equipped with a continuous projections. Nevertheless, there are various interesting open problems related to our work. In this final section, we will propose a direction for future further research. In view of [2] and Theorems 2.3 and 2.4, the following problems remain unanswered.

    Problem 5.1. Let A be a real or complex prime Banach algebra and P:AZ(A) be a nonzero continuous projection. Suppose that there is an open subset G of A such that P(an)anZ(A) for each aG and an integer n>1. Then, what we can say about the structure of A and P?

    Problem 5.2. Let A be a real or complex prime Banach algebra and P:AZ(A) be a nonzero continuous projection. Suppose that there are open subsets G1,G2 of A such that P(an)bnZ(A) for each (a,b)G and an integer n=n(a,b)>1. Then, what we can say about the structure of A and P?

    Problem 5.3. Let A be a real or complex prime Banach algebra and P:AZ(A) be a nonzero continuous projection. Suppose that there are open subsets G1, G2 of A such that P((ab)n)anbnZ(A) for each aG1 and bG2 and an integer n=n(x,y)>1. Then, what we can say about the structure of A and P?

    Problem 5.4. Let A be a commutative Banach algebra such that it admits a continuous projection P from A to Z(A) satisfying P(a)n=an for all aG, where G is a non-empty open subset of A. Then, what we can say about the structure of A and P?

    In this paper, we discussed new criteria to study the commutativity of Banach algebras. Particularly, we described the commutativity of prime Banach algebras over the field of real and complex via its projections. In this direction, Yood [13,14] established the commutativity of Banach algebras using the polynomial identities. Similarly, taking this idea forward [2,3,10] were able to accomplished the commutativity of Banach algebras via derivations. It would be interesting to discuss the commutativity of Banach algebras involving more general functional identities via projections (see Open Problems 1–4).

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors are very thankful to the anonymous referees for their valuable comments and suggestions which have improved the manuscript immensely.

    Moreover, the authors extend their appreciation to Princess Nourah Bint Abdulrahman University(PNU), Riyadh, Saudi Arabia for funding this research under Researchers Supporting Project Number(PNURSP2023R231).

    The authors declare no conflict of interest.



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