1 Introduction

The symbol A shall denote a Banach algebra over the complex field \(\mathbf {C}\) with unity e. The symbols Z(A) and M will denote the center and a closed linear subspace of A. An algebra A is said to be prime if for any \(x, y\in A\) such that \(xAy=0\), either \(x=0\) or \(y=0\). An additive mapping \(d:A\rightarrow A\) is said to be a derivation if \(d(xy)=d(x)y+xd(y)\) and \(d(cx)=cd(x)\), for all \(x,y\in A\) and for all \(c\in \mathbf {C}\). An additive mapping \(g:A\rightarrow A\) is said to be a generalized derivation associated with a derivation d if \(g(xy)=g(x)y+xd(y)\) and \(g(cx)=cg(x)\), for all \(x,y\in A\) and for all \(c\in \mathbf {C}\). An additive mapping \(T:A\rightarrow A\) is said to be a left multiplier if \(T(xy)=T(x)y\) and \(T(cx)=cT(x)\), for all \(x,y\in A\) and for all \(c\in \mathbf {C}\).

Yood (1991), has shown that if A is a unital Banach algebra and \(n=n(x,y)>1\) is a positive integer such that either \((xy)^n-x^ny^n\in M\) or \((xy)^n-y^nx^n\in M\) for sufficiently many x and y, then \([x,y]\in M\). Ali and Khan (2015) have shown that if A is a unital prime Banach algebra with nonzero continuous linear derivation \(d:A\rightarrow A\) such that either \(d((xy)^m) - x^my^m\) or \(d((xy)^m) - y^mx^m\) is in the center Z(A) of A for an integer \(m = m(x, y)>1\) and sufficiently many x, y,  then A is commutative. In this article we extend these results for a generalized derivation. Let \(p(t)=\sum \nolimits _{i=0}^{n}a_it^i\) be a polynomial in real variable t with coefficients in A. As mentioned in Yood (1991), if \(p(t)\in M\) for infinitely many real t then each \(a_i\in M\).

For a derivation d on a prime ring R, Posner proved the following result:

Lemma 1

[Posner (1957), Theorem 2] Let R be a prime ring and d be a nonzero derivation of R such that [d(x), x] is in the center of R, for all \(x\in R\). Then R is commutative.

Many other authors have generalized Posner’s result in several ways for rings and algebras (see Bell 1999; Brešar 2004; Herstein 1961; Vukman 1992; Yood 1984, 1990, where further references can be found). One of the generalizations of Posner’s result by Lee and Lee is of our interest:

Lemma 2

[Lee and Lee (1983), Theorem 2] Let d be a nonzero derivation on a prime ring R and U be a lie ideal of R such that \([x, d(U)]\subseteq Z(R)\). Then either \(x\in Z(R)\) or \(U\subseteq Z(R)\).

Ashraf and Ali have given a relationship between the commutativity of a ring and its left multipliers (see Ashraf and Ali 2008). In this article we also find a connection between the commutativity of a prime Banach algebra and its left multipliers.

2 Main results

Theorem 1

Let A be a unital prime Banach algebra and \(g:A\rightarrow A\) be a nonzero continuous generalized derivation associated with nonzero continuous derivation d on A such that \(g(e)\in Z(A)\) and \(d(g(e))\ne 0\). Suppose that there are open subsets \(G_1\) and \(G_2\) of A such that either \(g((xy)^n)-d(x^n)d(y^n)\in Z(A)\) or \(g((xy)^n)-d(y^n)d(x^n)\in Z(A)\) for each \(x\in G_1\) and for each \(y\in G_2\) and an integer \(n=n(x,y)>1\). Then A is commutative.

Proof

Set

$$\begin{aligned} f(x,y,n):=g((xy)^n)-d(x^n)d(y^n) \end{aligned}$$

and

$$\begin{aligned} h(x,y,n):=g((xy)^n)-d(y^n)d(x^n). \end{aligned}$$

Let \(x\in G_1\) be an arbitrarily fixed element. For each positive integer n consider the set \(U_n=\{y\in A\mid f(x,y,n)\notin Z(A), h(x,y,n)\notin Z(A)\}\). We show that \(U_n\) is open by showing that its complement \(U_n^{c}\) is closed. Let \((s_k)\) be a sequence in \(U_n^c\) such that \(\lim s_k=s\). Since \(s_k\in U_n^c\) we have either

$$\begin{aligned} h(x,s_k,n)=g((xs_k)^n)-d(x^n)d(s_k^n)\in Z(A) \end{aligned}$$
(1)

or

$$\begin{aligned} f(x,s_k,n)=g((xs_k)^n)-d(s_k^n)d(x^n)\in Z(A). \end{aligned}$$
(2)

Since g and d are continuous it follows that

$$\begin{aligned} \displaystyle g((xs)^n)-d(x^n)d(s^n)\in Z(A) ~\text {or}~ g((xs)^n)-d(s^n)d(x^n)\in Z(A). \end{aligned}$$

This shows that \(\displaystyle s\in U_n^c\) and so \(U_n\) is open.

By the Baire category theorem, if each \(U_n\) is dense then their intersection is also dense, which contradicts the existence of \(G_2\). Therefore there exists an integer \(i=i(x)>1\) such that \(U_i\) is not dense and a nonempty open set \(G_3\) in the complement of \(U_i\) such that either \(g((xy)^i)-d(x^i)d(y^i)\in Z(A)\) or \(g((xy)^i)-d(y^i)d(x^i)\in Z(A)\), for all \(y\in G_3\). Take \(z\in G_3\) and \(w\in A\). So for sufficiently small real t, \((z+tw)\in G_3\) and either

$$\begin{aligned} g((x(z+tw))^i)-d(x^i)d((z+tw)^i)\in Z(A) \end{aligned}$$
(3)

or

$$\begin{aligned} g((x(z+tw))^i)-d((z+tw)^i)d(x^i)\in Z(A). \end{aligned}$$
(4)

Thus for infinitely many t either (3) or (4), say (3) must hold. Since \(g((x(z+tw))^i)-d(x^i)d((z+tw)^i)\) is polynomial in t which is in Z(A), each of its coefficients must be in Z(A). The coefficient of \(t^i\) is \(g((xw)^i)-d(x^i)d(w^i)\) which is in Z(A). We have therefore shown that, given \(x\in G_1\), there exists a positive integer i depending on x such that for each \(w\in A\) either

$$\begin{aligned} g((xw)^i)-d(x^i)d(w^i)\in Z(A) \end{aligned}$$

or

$$\begin{aligned} g((xw)^i)-d(w^i)d(x^i)\in Z(A). \end{aligned}$$

Next we show that for each \(y\in A\) there is an integer \(j=j(y)>1\) such that for each \(v\in A\) either

$$\begin{aligned} g((vy)^j)-d(v^j)d(y^j)\in Z(A) \end{aligned}$$

or

$$\begin{aligned} g((vy)^j)-d(y^j)d(v^j)\in Z(A). \end{aligned}$$

Fix an arbitrary \(y\in A\) and for each \(k>1\), define

$$\begin{aligned} W_k=\{z\in A\mid f(z,y,k)\notin Z(A), h(z,y,k)\notin Z(A)\}. \end{aligned}$$

As shown above each \(W_k\) is open, so by the Baire category theorem, if each \(W_k\) is dense then its intersection is also dense which contradicts the existence of \(G_1\). Hence there exists an integer \(j=j(y)>1\) and a nonempty open subset \(G_4\) in the complement of \(W_j\). If \(z\in G_4\) and \(v\in A\) then for sufficiently small real t, \((z+tv)\in G_4\) and either

$$\begin{aligned} g(((z+tv)y)^j)-d((z+tv)^j)d(y^j)\in Z(A) \end{aligned}$$
(5)

or

$$\begin{aligned} g(((z+tv)y)^j)-d(y^j)d((z+tv)^j)\in Z(A). \end{aligned}$$
(6)

With the same argument as earlier we see that for each \(v\in A\), either

$$\begin{aligned} g((vy)^j)-d(v^j)d(y^j)\in Z(A) \end{aligned}$$

or

$$\begin{aligned} g((vy)^j)-d(y^j)d(v^j)\in Z(A). \end{aligned}$$

Now let \(S_k, k>1\) be the set of all \(y\in A\) such that for each \(w\in A\) either

$$\begin{aligned} g((wy)^k)-d(w^k)d(y^k)\in Z(A) \end{aligned}$$

or

$$\begin{aligned} g((wy)^k)-d(y^k)d(w^k)\in Z(A). \end{aligned}$$

The union of \(S_k\) is A. It is obvious to see that each \(S_k\) is closed. Again, by the Baire category theorem, some \(S_n\) must contain a nonempty open subset \(G_5\). Take \(z\in G_5\) and \(x\in A\). For all sufficiently small real t and each \(w\in A\) either

$$\begin{aligned} g((w(z+tx))^n)-d(w^n)d((z+tx)^n)\in Z(A) \end{aligned}$$

or

$$\begin{aligned} g((w(z+tx))^n)-d((z+tx)^n)d(w^n)\in Z(A). \end{aligned}$$

By earlier arguments, for all \(x, w\in A\), we have either

$$\begin{aligned} g((wx)^n)-d(w^n)d(x^n)\in Z(A) \end{aligned}$$

or

$$\begin{aligned} g((wx)^n)-d(x^n)d(w^n)\in Z(A). \end{aligned}$$

Since A is unital, for all real t and for all \(x,y\in A\) we have either

$$\begin{aligned} g(((e+tx)y)^n)-d((e+tx)^n)d(y^n)\in Z(A) \end{aligned}$$
(7)

or

$$\begin{aligned} g(((e+tx)y)^n)-d(y^n)d((e+tx)^n)\in Z(A). \end{aligned}$$
(8)

Now, by collecting the coefficient of t in the above expressions, we have either

$$\begin{aligned} g(xy^n+Q)-nd(x)d(y^n)\in Z(A) \quad \textit{for all}\quad x,y\in A \end{aligned}$$
(9)

or

$$\begin{aligned} g(xy^n+Q)-nd(y^n)d(x)\in Z(A) \quad \textit{for all} \quad x,y\in A, \end{aligned}$$
(10)

where \(Q=\sum \nolimits _{k=1}^{n-1}y^kxy^{n-k}\).

Again if we start with \(g((y(e+tx))^n)\) in place of \(g(((e+tx)y)^n)\), we have either

$$\begin{aligned} g(y^nx+Q)-nd(y^n)d(x)\in Z(A) \quad \textit{for all} \quad x,y\in A \end{aligned}$$
(11)

or

$$\begin{aligned} g(y^nx+Q)-nd(x)d(y^n)\in Z(A) \quad \textit{for all} \quad x,y\in A. \end{aligned}$$
(12)

At least one of the pairs of equations {(9) (11)}, {(10) (12)},{(9) (12)} and {(10) (11)} must hold. Subtracting these pairs we have either

$$\begin{aligned} g[x,y^n]-n[d(x),d(y^n)]\in Z(A) \quad \textit{for all} \quad x,y\in A \end{aligned}$$
(13)

or

$$\begin{aligned} g[x,y^n]+n[d(x),d(y^n)]\in Z(A) \quad \textit{for all} \quad x,y\in A \end{aligned}$$
(14)

or

$$\begin{aligned} g[x,y^n]\in Z(A) \quad \textit{for all} \quad x,y\in A. \end{aligned}$$
(15)

Now if \(g[x,y^n]\in Z(A)\) then by replacing y by \((e+ty)\) we have \(g[x,y]\in Z(A)\) for all \(x,y\in A\). Replace x by ex we get \(g(e)[x,y]+d[x,y]\in Z(A)\), or equivalently,

$$\begin{aligned}{}[g(e)[x,y]+d[x,y],z]=0, \quad \textit{for all}\quad x,y,z\in A. \end{aligned}$$
(16)

This can be written as \([g(e)[x,y]+[x,dy], z]+[[dx,y], z]=0\). Replace y by [y, w] we get \([g(e)[x,[y,w]]+[x,d[y,w]],z]+[[dx,[y,w]],z]=0\). Now use (16) to get \([[dx,[y,w]],z]=0\); hence, \([dx,[y,w]]\in Z(A)\), for all \(x,y,w\in A\). In the light Lemma 2 we have either \([y,w]\in Z(A)\) or \(A\subseteq Z(A)\). In both the cases A is commutative.

Now if (13) holds then by replacing y by \((e+ty)\) we have \(g[x,y]-n[d(x),d(y)]\in Z(A)\) for all \(x,y\in A\). In this expression replace x by xg(e) in order to obtain \((g[x,y]-n[d(x),d(y)])g(e)+[x,y]d(g(e))-n[xd(g(e)),d(y)]\in Z(A)\) for all \(x,y\in A\). Consequently, since Z(A) is a linear subspace of A and \((g[x,y]-n[d(x),d(y)])g(e)\in Z(A)\), it follows that \([x,y]d(g(e))-n[xd(g(e)),d(y)]\in Z(A)\) for all \(x,y\in A\). Hence if we set \(y=x\) and observe that \(g(e)\in Z(A)\) implies \(d(g(e))\in Z(A)\), we obtain \([x,d(x)]d(g(e))\in Z(A)\) for all \(x\in A\). So, in particular, \([[x,d(x)]d(g(e)),z]=0\) for all \(x,z\in A\). Replacing z by zy now yields \([[x,d(x)],z]yd(g(e))=0\) for all \(x,y,z\in A\). Thus, since A is prime and \(d(g(e))\ne 0\), it follows that \([x,d(x)]\in Z(A)\) for all \(x\in A\). Hence, by Lemma 1 we may infer that A is commutative. Similarly it can be shown that if (14) holds then A is commutative.\(\square \)

Theorem 2

Let A be a unital prime Banach algebra and \(g:A\rightarrow A\) be a nonzero continuous generalized derivation associated with a nonzero continuous derivation d on A such that \(g(e)\in Z(A)\). Suppose that there are open subsets \(G_1\) and \(G_2\) of A such that either \(g((xy)^n)-x^ny^n\in Z(A)\) or \(g((xy)^n)-y^nx^n\in Z(A)\) for each \(x\in G_1\) and for each \(y\in G_2\) and an integer \(n=n(x,y)>1\). Then A is commutative.

Proof

Proceeding as in Theorem 1, we get either

$$\begin{aligned} g[x,y]\in Z(A) \quad \textit{for all} \quad x,y\in A \end{aligned}$$
(17)

or

$$\begin{aligned} g[x,y]-n[x,y]\in Z(A) \quad \textit{for all} \quad x,y\in A \end{aligned}$$
(18)

or

$$\begin{aligned} g[x,y]+n[x,y]\in Z(A) \quad \textit{for all} \quad x,y\in A. \end{aligned}$$
(19)

Suppose (17) holds. Replace x by ex we get \(g(e)[x,y]+d[x,y]\in Z(A)\), or equivalently,

$$\begin{aligned}{}[g(e)[x,y]+d[x,y],z]=0, \quad \textit{for all} \quad x,y,z\in A. \end{aligned}$$
(20)

This can be written as \([g(e)[x,y]+[x,dy],z]+[[dx,y],z]=0\). Replace y by [y, w] we get \([g(e)[x,[y,w]]+[x,d[y,w]],z]+[[dx,[y,w]],z]=0\). Use (20) to get \([[dx,[y,w]],z]=0\) or \([dx,[y,w]]\in Z(A)\), for all \(x,y,w\in Z(A)\). In the light Lemma 2 we have either \([y,w]\in Z(A)\) or \(A\subseteq Z(A)\). In both the cases A is commutative.

Now consider \(g[x,y]-n[x,y]\in Z(A)\), for all \(x,y\in A\). Replace x by ex we get \(g(e)[x,y]+d[x,y]-n[x,y]\in Z(A)\) or \([(g(e)-n)[x,y]+d[x,y],z]=0\) for all \(x,y,z\in A\), which is similar to Eq. (20). Thus in this case it can also be shown that A is commutative. Similarly, we can prove that if \(g[x,y]+n[x,y]\in Z(A)\) then A is commutative.\(\square \)

Theorem 3

Let A be a unital prime Banach algebra and \(g:A\rightarrow A\) be a nonzero continuous generalized derivation associated with a nonzero continuous derivation d on A such that \(g(e)\in Z(A)\). Suppose that there are open subsets \(G_1\) and \(G_2\) of A such that either \(g((xy)^n-x^ny^n)\in Z(A)\) or \(g((xy)^n-y^nx^n)\in Z(A)\) for each \(x\in G_1\) and for each \(y\in G_2\) and an integer \(n=n(x,y)>1\). Then A is commutative.

Proof

Proceeding as in Theorem 1, we obtain either

$$\begin{aligned} g[x,y]\in Z(A) \quad \textit{for all} \quad x,y\in A \end{aligned}$$
(21)

or

$$\begin{aligned} g([x,y]-n[x,y])\in Z(A) \quad \textit{for all} \quad x,y\in A \end{aligned}$$
(22)

or

$$\begin{aligned} g([x,y]+n[x,y])\in Z(A) \quad \textit{for all}\quad x,y\in A. \end{aligned}$$
(23)

Equations (22) and (23) can be written as \((1-n)g[x,y]\in Z(A)\) and \((1+n)g[x,y]\in Z(A)\), respectively. Thus the equations (21), (22) and (23) all reduce to \(g[x,y]\in Z(A)\). The result then follows using the argument in the proof of Theorem 2.\(\square \)

Theorem 4

Let A be a unital prime Banach algebra and \(T:A\rightarrow A\) be a continuous left multiplier on A such that \(T(x)\ne \pm nx\), for all nonzero \(x\in A\) and integers \(n\ge 0\). Suppose that there are open subsets \(G_1\) and \(G_2\) of A such that either \(T((xy)^n)-x^ny^n\in Z(A)\) or \(T((xy)^n)-y^nx^n\in Z(A)\) for each \(x\in G_1\) and for each \(y\in G_2\) and an integer \(n=n(x,y)>1\). Then A is commutative.

Proof

Proceeding as in Theorem 1, we get either

$$\begin{aligned} T[x,y]\in Z(A) \quad \textit{for all}\quad x,y\in A \end{aligned}$$
(24)

or

$$\begin{aligned} T[x,y]-n[x,y]\in Z(A) \quad \textit{for all} \quad x,y\in A \end{aligned}$$
(25)

or

$$\begin{aligned} T[x,y]+n[x,y]\in Z(A) \quad \textit{for all} \quad x,y\in A. \end{aligned}$$
(26)

If \(T[x,y]\in Z(A)\), for all \(x,y\in A\), then \([T[x,y],z]=0\), for all \(x,y,z\in A\). Replacing y by yx and then z by rz we have \(T[x,y]r[x,z]=0\), for all \(r,x,y,z\in A\). Hence, \(T[x,y]r[x,y]=0\), for all \(r,x,y\in A\), and so, by hypothesis on T and the fact that A is prime, we have \([x,y]=0\), for all \(x,y\in A\). If \(T[x,y]\pm n[x,y]\in Z(A)\), for all \(x,y\in A\), then, as before \((T[x,y]\pm n[x,y])r[x,z]=0\), for all \(r,x,y,z\in A\). Since \(n>1\) it follows that \(T[x,y]\ne \pm n[x,y]\), for all \(x,y\in A\), and so, \([x,z]=0\), for all \(x,z\in A\). This completes the proof.

Theorem 5

Let A be a unital prime Banach algebra and \(T:A\rightarrow A\) be a continuous left multiplier on A such that \(T(x)\ne \pm nx\), for all \(n\ge 0\). Suppose that there are open subsets \(G_1\) and \(G_2\) of A such that either \(T((xy)^n-x^ny^n)\in Z(A)\) or \(T((xy)^n-y^nx^n)\in Z(A)\) for each \(x\in G_1\) and for each \(y\in G_2\) and an integer \(n=n(x,y)>1\). Then A is commutative.

Proof

The proof is similar to the proof of Theorem 4.

Theorem 6

Let A be a unital prime Banach algebra and \(T:A\rightarrow A\) be a nonzero continuous left multiplier on A such that \(T(e)\in Z(A)\) and \(nT(e)\ne \pm e\), for all integers \(n\ge 1\). Suppose that there are open subsets \(G_1\) and \(G_2\) of A such that either \(T((xy)^n)-T(x^n)T(y^n)\in Z(A)\) or \(T((xy)^n)-T(y^n)T(x^n)\in Z(A)\) for each \(x\in G_1\) and for each \(y\in G_2\) and an integer \(n=n(x,y)>1\). Then A is commutative.

Proof

Proceeding as in Theorem 1, we obtain either

$$\begin{aligned} T[x,y]\in Z(A) \quad \textit{for all} \quad x,y\in A \end{aligned}$$
(27)

or

$$\begin{aligned} T[x,y]-n[T(x),T(y)]\in Z(A) \quad \textit{for all} \quad x,y\in A \end{aligned}$$
(28)

or

$$\begin{aligned} T[x,y]+n[T(x),T(y)]\in Z(A) \quad \textit{for all} \quad x,y\in A. \end{aligned}$$
(29)

Since \(T(e)\in Z(A)\), it follows from the fact that A is prime that \(T(x)=0\) implies \(x=0\). Now, if \(T[x,y]\in Z(A)\), for all \(x,y\in A\), then A is commutative as in the proof of Theorem 4. On the other hand, if \(T[x,y]\pm n[T(x),T(y)]\in Z(A)\), for all \(x,y\in A\), then as in the proof of Theorem 4, we have

$$\begin{aligned} (T[x,y]\pm n[T(x),T(y)])r[x,z]=0, \quad \textit{for all} \quad r,x,y,z\in A. \end{aligned}$$

Hence, \(T[x,y](e\pm nT(e))r[x,y]=0\), for all \(r,x,y\in A\). If \(T[x,y](e\pm nT(e))=0\) then \(T[x,y]r(e\pm nT(e))=0\), for all \(r\in A\) and so, \(T[x,y]=0\) and hence \([x,y]=0\). So, \([x,y]=0\), for all \(x,y\in A\). The result now follows.\(\square \)

3 Open questions

The authors would like to open the following questions for further studies:

Question 1

Can the hypothesis that \(g(e)\in Z(A)\) be removed from the assumptions in Theorem 1, Theorem 2 and Theorem 3?

Question 2

Can the hypothesis that \(T(x)\ne \pm nx\) be removed from the assumptions in Theorem 4 and Theorem 5?

Question 3

Can the hypotheses \(T(e)\in Z(A)\) and \(nT(e)\ne \pm e\), for \(n\ge 1\) be removed from the assumptions in Theorem 6?