1 Introduction and preliminaries

Let \(\mathcal {A}\) and \(\mathcal {B}\) be two algebras, let \(\mathfrak {M}\) be a \(\mathcal {B}\)-bimodule and let \(\psi , \phi :\mathcal {A} \rightarrow \mathcal {B}\) be two mappings. A linear mapping \(d:\mathcal {A} \rightarrow \mathfrak {M}\) is called a \((\psi , \phi\))-derivation if \(d(ab) = d(a)\psi (b) + \phi (a)d(b)\) for all \(a, b \in \mathcal {A}\). If \(\mathcal {A} \subseteq \mathcal {B}\) and \(\phi = I = \psi\), the identity mapping on \(\mathcal {A}\), then we reach an ordinary derivation. The main objective of this study is to investigate the automatic continuity of \((\psi , \phi\))-derivations on some topological algebras. Generally, the automatic continuity of a certain class of mappings, e.g. \((\psi , \phi\))-derivations, is the study of (algebraic) conditions on a category, e.g. Banach algebras, which guarantee that every \((\psi , \phi\))-derivation is continuous. Let us give a brief background in this regard. The theory of automatic continuity of derivations has a long history. Results on automatic continuity of linear mappings defined on Banach algebras comprise a fruitful area of research developed during the last sixty years. The reader is referred to [2, 3, 14] for a deep and extensive study on this subject. In 1958, Kaplansky [11] conjectured that every derivation on a \(C^{*}\)-algebra is continuous. Two years later, Sakai [15] answered to this conjecture. Indeed, he proved that every derivation on a \(C^{*}\)-algebra is automatically continuous and later in 1972, Ringrose [13], by using the pioneering work of Bade and Curtis [1] concerning the automatic continuity of a module homomorphism between bimodules over C(K)-spaces, showed that every derivation from a \(C^{*}\)-algebra \(\mathcal {A}\) into a Banach \(\mathcal {A}\)-bimodule is automatically continuous. Also, Johnson and Sinclair [10] investigated the continuity of derivations on semisimple Banach algebras. In [12], it is shown that if \(\psi , \phi\) are continuous \(*\)-linear mappings, then every \((\psi , \phi )\)-derivation from a \(C^{*}\)-algebra into \(B(\mathcal {H})\) is automatically continuous, and in [8] the assumption of linearity of \(\psi , \phi\) were deleted. Moreover, Hou et al. [9] proved that if \(\mathcal {X}\) is simple and \(\psi , \phi\) are surjective and continuous mappings on \(B(\mathcal {X})\), then every \((\psi , \phi )\)-derivation on \(B(\mathcal {X})\) is continuous. For more material concerning the automatic continuity of mappings, see, e.g. [4,5,6, 16].

This paper consists of two sections. The main results of the paper are presented in the second section. In this section, e denotes the identity element of any unital algebra. First, we obtain a characterization of \((\psi , \phi )\)-derivations as follows: Let \(\mathcal {A}\) and \(\mathcal {B}\) be two unital algebras and let \(d:\mathcal {A} \rightarrow \mathcal {B}\) be a \((\psi , \phi\))-derivation such that \(d(e) \in Inv(\mathcal {B})\), where \(Inv(\mathcal {B})\) denotes the set of all invertible elements of \(\mathcal {B}\). If either \([\psi (e), d(e)] = 0\) or \([\phi (e), d(e)] = 0\), where \([a, b] = ab - ba\) \((a, b \in \mathcal {A})\), then \(d(ab) = d(a) (d(e))^{-1}d(b)\) for all \(a, b \in \mathcal {A}\). In particular, if \(d(e) = e\), then d is a homomorphism. Using this result, we prove that if \(\mathcal {A}\) and \(\mathcal {B}\) are two topological unital algebras and \(d:\mathcal {A} \rightarrow \mathcal {B}\) is a \((\psi , \phi\))-derivation such that \(d(e) = e\) and also if we have all the conditions under which every homomorphism from \(\mathcal {A}\) into \(\mathcal {B}\) is continuous, then d, \(\psi\) and \(\phi\) are continuous mappings. In addition, we obtain some results concerning the continuity of \((\psi , \phi\))-derivations on unital involutive topological algebras. Suppose that \((\mathcal {A}, *)\) and \((\mathcal {B}, \star )\) are two unital, involutive topological algebras and \(d:\mathcal {A} \rightarrow \mathcal {B}\) is a \((\psi , \phi\))-derivation such that d, \(\psi\) and \(\phi\) are \((*, \star )\)-mappings and \(d(e) \in Inv(\mathcal {B})\). Assume that either \([\psi (e), d(e)] = 0\) or \([\phi (e), d(e)] = 0\). If we have all the conditions under which every homomorphism from \(\mathcal {A}\) into \(\mathcal {B}\) is continuous, then d, \(\psi\) and \(\phi\) are continuous mappings. Another result in this regard reads as follows: Let \((\mathcal {A}, *)\) and \((\mathcal {B}, \star )\) be two unital involutive algebras and let \(d_1, d_2:\mathcal {A} \rightarrow \mathcal {B}\) be two \((*, \star )-(\psi , \phi )\)-derivations such that \(d_1(e_\mathcal {A})d_2(a_0) = e_\mathcal {B}\) or \(d_1(a_0)d_2(e_\mathcal {A}) = e_\mathcal {B}\) for some \(a_0 \in \mathcal {A}\). Suppose that \([\psi (a), d_2(b)] = 0 = [\phi (a), d_1(b)]\) for all \(a, b \in \mathcal {A}\). Then, the mappings \(\psi\) and \(\phi\) are linear. Moreover, suppose we have the conditions under which every homomorphism from \(\mathcal {A}\) into \(\mathcal {B}\) is continuous. Then, \(d_1\), \(d_2\), \(\psi\) and \(\phi\) are continuous linear mappings.

2 Main results

In this section, without further mention, e denotes the identity element of any unital algebra. If \(\mathcal {A}\) is a unital algebra, \(Inv(\mathcal {A})\) denotes the set of all invertible elements of \(\mathcal {A}\). Let \(\mathcal {A}\) and \(\mathcal {B}\) be two algebras, let \(\mathfrak {M}\) be a \(\mathcal {B}\)-bimodule and let \(\psi , \phi :\mathcal {A} \rightarrow \mathcal {B}\) be two mappings. Recall that a linear mapping \(d:\mathcal {A} \rightarrow \mathfrak {M}\) is called a \((\psi , \phi\))-derivation if \(d(ab) = d(a)\psi (b) + \phi (a)d(b)\) for all \(a, b \in \mathcal {A}\). We now provide an example of this notion.

Example 2.1

Let \(\mathcal {A}\) and \(\mathcal {B}\) be two algebras (finite dimensional or not). It is easy to see that \(\mathfrak {A} = \mathcal {A} \times \mathcal {B}\) is an algebra by the following operations:

$$\begin{aligned}&(a_1, b_1) \bullet (a_2, b_2) = (a_1 a_2, b_1 b_2); \\&(a_1, b_1) + (a_2, b_2) = (a_1 + a_2, b_1 + b_2); \\&\lambda (a_1, b_1) = (\lambda a_1, \lambda b_1) \end{aligned}$$

for all \(a_1, a_2 \in \mathcal {A}\), \(b_1, b_2 \in \mathcal {B}\) and \(\lambda \in \mathbb {C}\). Let \(F, G: \mathcal {B} \rightarrow \mathcal {B}\) be two mappings. Define the mappings \(d, \psi , \phi :\mathfrak {A} \rightarrow \mathfrak {A}\) by

$$\begin{aligned}&d((a,b)) = (a, 0), \\&\psi ((a,b)) = (\frac{a}{2},F(b)), \\&\phi ((a,b)) = (\frac{a}{2}, G(b)), \end{aligned}$$

A routine calculation shows that d is a linear \((\psi , \phi )\)-derivation on \(\mathfrak {A}\).

We begin with the following theorem, which provides a characterization for \((\psi , \phi\))-derivations.

Theorem 2.2

Let \(\mathcal {A}\) and \(\mathcal {B}\) be two unital algebras and let \(d:\mathcal {A} \rightarrow \mathcal {B}\) be a \((\psi , \phi\))-derivation such that \(d(e) \in Inv(\mathcal {B})\). If either \([\psi (e), d(e)] = 0\) or \([\phi (e), d(e)] = 0\), then both \(\phi\) and \(\psi\) are linear mappings and \(d(ab) = d(a) (d(e))^{-1}d(b)\) for all \(a, b \in \mathcal {A}\). In particular, if \(d(e) = e\), then d is a homomorphism.

Proof

Suppose that \([\phi (e), d(e)] = 0\). Then, it is easy to see that \([\phi (e), (d(e))^{-1}] =0\). So, \(d(e) = d(e) \psi (e) + \phi (e) d(e) = d(e) (\psi (e) + \phi (e))\). This equality with the assumption that d(e) is an invertible element of \(\mathcal {B}\) implies that \(\psi (e) + \phi (e) = e\). We have \(d(a) = d(a) \psi (e) + \phi (a) d(e)\) for any \(a \in \mathcal {A}\). So, \(d(a)(e - \psi (e)) = \phi (a) d(e)\) and consequently,

$$\begin{aligned} \phi (a) = d(a) \phi (e)(d(e))^{-1}, \ \ \ \ \ \ \ \ \ (a \in \mathcal {A}). \end{aligned}$$
(2.1)

Similarly, we can get that

$$\begin{aligned} \psi (a) = (d(e))^{-1} \psi (e) d(a), \ \ \ \ \ \ \ \ \ (a \in \mathcal {A}). \end{aligned}$$
(2.2)

It follows from (2.1) and (2.2) that both \(\phi\) and \(\psi\) are linear mappings and also we have

$$\begin{aligned} d(ab)&= d(a) \psi (b) + \phi (a) d(b) \\&= d(a) (d(e))^{-1} \psi (e) d(b) + d(a) \phi (e)(d(e))^{-1} d(b) \\&= d(a) (d(e))^{-1} \psi (e) d(b) + d(a) (d(e))^{-1} \phi (e) d(b) \\&= d(a) (d(e))^{-1}(\psi (e) + \phi (e))d(b) \\&= d(a)(d(e))^{-1}d(b), \end{aligned}$$

which means that

$$\begin{aligned} d(ab) = d(a)(d(e))^{-1}d(b), \ \ \ \ \ \ \ \ \ \ (a, b \in \mathcal {A}). \end{aligned}$$
(2.3)

Clearly, if \(d(e) = e\), then d is a homomorphism. Besides, we can prove that \([d(e), \psi (e)] = 0\). In view of (2.2) and (2.3), we have

$$\begin{aligned} \psi (ab)&= (d(e))^{-1} \psi (e) d(ab) \\&= (d(e))^{-1} \psi (e)d(a)(d(e))^{-1}d(b) \\&= \psi (a) (d(e))^{-1}d(b). \end{aligned}$$

Letting \(a = e\) in the above equations, we get that

$$\begin{aligned} \psi (b) = \psi (e) (d(e))^{-1}d(b), \ \ \ \ \ \ \ \ \ (b \in \mathcal {A}). \end{aligned}$$
(2.4)

Comparing (2.2) and (2.4), we obtain that \(\psi (e) (d(e))^{-1}d(a) = (d(e))^{-1} \psi (e) d(a)\) for all \(a \in \mathcal {A}\). Putting \(a = e\) in the previous equation, we get that \([d(e), \psi (e)] = 0\). So, each of the equations \([d(e), \psi (e)] = 0\) or \([\phi (e), d(e)] = 0\) implies the other. \(\square\)

In the following, there are some consequences of the previous theorem.

Corollary 2.3

Let \(\mathcal {A}\) and \(\mathcal {B}\) be two unital normed algebras and let \(d:\mathcal {A} \rightarrow \mathcal {B}\) be a \((\psi , \phi\))-derivation such that \(d(e) \in Inv(\mathcal {B})\). Suppose that either \([\psi (e), d(e)] = 0\) or \([\phi (e), d(e)] = 0\). Then, the continuity of d implies the continuity of both \(\psi\) and \(\phi\).

Proof

Using (2.1) and (2.2), we obtain the required result. \(\square\)

In the following, we present some conditions that provide the continuity of \((\psi , \phi\))-derivations.

Corollary 2.4

Let \(\mathcal {A}\) and \(\mathcal {B}\) be two topological unital algebras and let \(d:\mathcal {A} \rightarrow \mathcal {B}\) be a \((\psi , \phi\))-derivation such that \(d(e) = e\). If we have all the conditions under which every homomorphism from \(\mathcal {A}\) into \(\mathcal {B}\) is continuous, then d, \(\psi\) and \(\phi\) are continuous mappings.

Proof

It follows from Theorem 2.2 that d is a homomorphism from \(\mathcal {A}\) into \(\mathcal {B}\). Since we are assuming all the conditions under which every homomorphism from \(\mathcal {A}\) into \(\mathcal {B}\) is continuous, we deduce that d is continuous. This fact along with (2.1) and (2.2) implies the continuity of \(\psi\) and \(\phi\). \(\square\)

Remark 2.5

There are many different conditions under which a homomorphism is continuous. For instance, if \(\mathcal {A}\) is a Banach \(*\)-algebra and \(\mathcal {B}\) is a \(C^{*}\)-algebra, then it follows from [3, Corollary 3.2.4] that every \(*\)-homomorphism \(\theta :\mathcal {A} \rightarrow \mathcal {B}\) is automatically continuous. For more material about the continuity of homomorphisms and other results, see, e.g. [2, Proposition 5.1.1, Theorem 5.1.8, Theorem 5.2.4, Coroolary 5.2.5].

Let \(\mathcal {A}\) be a complex algebra. Recall that an involution over \(\mathcal {A}\) is a map \(*: \mathcal {A} \rightarrow \mathcal {A}\) satisfying the following conditions for all \(a, b \in \mathcal {A}\) and all \(\lambda \in \mathbb {C}\):

  1. 1.

    \((a^{*})^{*} = a\),

  2. 2.

    \((ab)^{*} = b^{*}a^{*}\),

  3. 3.

    \((a +b)^{*} = a^{*} + b^{*}\),

  4. 4.

    \((\lambda a)^{*} = \overline{\lambda } a^{*}\).

An algebra \(\mathcal {A}\) equipped with an involution \(*\) is called an involutive algebra or \(*\)-algebra and is denoted, as an ordered pair, by \((\mathcal {A}, *)\). Let \((\mathcal {A}, *)\) and \((\mathcal {B}, \star )\) be two involutive algebras. A mapping \(T:\mathcal {A }\rightarrow \mathcal {B}\) is called a \((*, \star )\)-map if \(T(a) = (T(a^{*}))^{\star }\) for all \(a \in \mathcal {A}\).

Theorem 2.6

Let \((\mathcal {A}, *)\) and \((\mathcal {B}, \star )\) be two unital, involutive algebras and let \(d:\mathcal {A} \rightarrow \mathcal {B}\) be a \((\psi , \phi\))-derivation such that d, \(\psi\) and \(\phi\) are \((*, \star )\)-mappings and \(d(e) \in Inv(\mathcal {B})\). If either \([\psi (e), d(e)] = 0\) or \([\phi (e), d(e)] = 0\), then \(\theta = \phi + \psi :\mathcal {A} \rightarrow \mathcal {B}\) is a homomorphism and further, \(d(a) = d(e) \theta (a) = \theta (a) d(e)\) for all \(a \in \mathcal {A}\).

Proof

It follows from Theorem 2.2 that \(\phi (e) + \psi (e) = e\) and also it follows from (2.1) that \(\phi (a) = d(a) \phi (e)(d(e))^{-1}\) for all \(a \in \mathcal {A}\). Since \(\phi\) and d are \((*, \star )\)-mappings, we have

$$\begin{aligned} \phi (a) = (\phi (a^{*}))^{\star } = (d(a^{*}) \phi (e^{*})(d(e^{*}))^{-1})^{\star } = (d(e))^{-1}\phi (e)d(a) \end{aligned}$$
(2.5)

Considering \(\theta = \phi + \psi\) and using (2.2) and (2.5), we have

$$\begin{aligned} \theta (a) = \psi (a) + \phi (a) = (d(e))^{-1} \psi (e) d(a) + (d(e))^{-1}\phi (e)d(a) = (d(e))^{-1} d(a). \end{aligned}$$
(2.6)

It is observed that

$$\begin{aligned} d(a) = d(e) \theta (a), \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a \in \mathcal {A}). \end{aligned}$$
(2.7)

Similarly, we get that

$$\begin{aligned} \theta (a) = d(a) (d(e))^{-1}, \ \ \ \ \ \ \ \ \ \ (a \in \mathcal {A}) \end{aligned}$$

and so, we have

$$\begin{aligned} d(a) = \theta (a)d(e), \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a \in \mathcal {A}). \end{aligned}$$

Our next task is to show that \(\theta\) is a homomorphism. Using (2.3) and (2.7), we get that

$$\begin{aligned} d(a)(d(e))^{-1} d(b) = d(ab) = d(e)\theta (ab), \ \ \ \ \ \ \ \ (a, b \in \mathcal {A}). \end{aligned}$$
(2.8)

Left multiplication of (2.8) by \((d(e))^{-1}\) and using (2.6) give

$$\begin{aligned} \theta (ab) = (d(e))^{-1} d(a)(d(e))^{-1} d(b) = \theta (a)\theta (b), \ \ \ \ \ \ \ \ (a, b \in \mathcal {A}). \end{aligned}$$

which means that \(\theta\) is a homomorphism. This proves the theorem, completely. \(\square\)

An immediate corollary reads as follows:

Corollary 2.7

Suppose that \((\mathcal {A}, *)\) and \((\mathcal {B}, \star )\) are two unital, involutive topological algebras and \(d:\mathcal {A} \rightarrow \mathcal {B}\) is a \((\psi , \phi\))-derivation such that d, \(\psi\) and \(\phi\) are \((*, \star )\)-mappings and \(d(e) \in Inv(\mathcal {B})\). Assume that either \([\psi (e), d(e)] = 0\) or \([\phi (e), d(e)] = 0\). If we have all the conditions under which every homomorphism from \(\mathcal {A}\) into \(\mathcal {B}\) is continuous, then d, \(\psi\) and \(\phi\) are continuous mappings.

Proof

It follows from Theorem 2.6 that there exists a homomorphism \(\theta :\mathcal {A} \rightarrow \mathcal {B}\) such that \(d(a) = d(e) \theta (a) = \theta (a) d(e)\) for all \(a \in \mathcal {A}\). Since we are assuming all the conditions under which every homomorphism from \(\mathcal {A}\) into \(\mathcal {B}\) is continuous, we obtain the continuity of d. Now, Eqs. (2.1) and (2.2) imply the continuity of \(\phi\) and \(\psi\), respectively. \(\square\)

In the following, we provide an example that shows that the conditions of Theorem 2.6 are not superfluous.

Example 2.8

Let \((\mathcal {A}, *)\) be an involutive algebra. Set \(\mathcal {U} = \mathbb {C} \bigoplus \mathcal {A}\). Consider \(\mathcal {U}\) as an algebra with pointwise addition, scalar multiplication and the product defined by

$$\begin{aligned} (\alpha , a) \bullet (\beta , b) = (\alpha \beta , \alpha b + \beta a), \ \ \ \ \ \ \ \ \ \ \ (\alpha , \beta \in \mathbb {C}, a, b \in \mathcal {A}). \end{aligned}$$

\(\mathcal {U}\) is also an involutive algebra when we define \(\star : \mathcal {U} \rightarrow \mathcal {U}\) as follows:

$$\begin{aligned} (\alpha , a)^{\star }= (\overline{\alpha }, a^{*}), \ \ \ \ \ \ \ \ \ \ \ (\alpha \in \mathbb {C}, a\in \mathcal {A}). \end{aligned}$$

Furthermore, \(e = (1, 0)\) is the identity of \(\mathcal {U}\). Let \(R, S, T : \mathcal {A} \rightarrow \mathcal {A}\) be \(*\)-linear mappings. We define the mappings \(d, \psi , \phi :\mathcal {U} \rightarrow \mathcal {U}\) by \(d((\alpha , a)) = (0, T(a))\), \(\psi ((\alpha , a)) = (\alpha , S(a))\) and \(\phi ((\alpha , a)) = (\alpha , R(a))\) for all \((\alpha , a) \in \mathcal {U}\). It is clear that \(d, \psi , \phi\) are \(\star\)-mappings and also d is a \((\psi , \phi )\)-derivation. A straightforward verification shows that \((\alpha , a)^{-1} = (\alpha ^{-1}, \frac{-a}{\alpha ^2})\) for all \((\alpha , a) \in \mathbb {C}\backslash \{0\}\bigoplus \mathcal {A}\). So, \(Inv(\mathcal {U}) = \mathbb {C}\backslash \{0\}\bigoplus \mathcal {A}\) and obviously, \(d(e) \notin Inv(\mathcal {U})\). As can be seen, \(\theta = \psi + \phi\) is not a homomorphism and further \(d \ne d(e) \theta\). Note that if \(\mathcal {A}\) is a normed algebra, then so is \(\mathcal {U}\) with the following norm:

$$\begin{aligned} \Vert (\alpha , a)\Vert = |\alpha | + \Vert a\Vert , \ \ \ \ \ \ \ \ \ \ \ (\alpha \in \mathbb {C}, a\in \mathcal {A}). \end{aligned}$$

Theorem 2.9

Suppose that \(\mathcal {A}\) and \(\mathcal {B}\) are two unital algebras such that \(\mathcal {B}\) is commutative. Let \(d:\mathcal {A} \rightarrow \mathcal {B}\) be a \((\psi , \phi\))-derivation such that \(d(e) \in Inv(\mathcal {B})\). Then, \(\theta = \phi + \psi :\mathcal {A} \rightarrow \mathcal {B}\) is a homomorphism and also \(d(a) = d(e) \theta (a)\) for all \(a \in \mathcal {A}\).

Proof

By using an argument similar to the proof of Theorem 2.6, we get the desired result. \(\square\)

Corollary 2.10

Suppose that \(\mathcal {A}\) and \(\mathcal {B}\) are two unital topological algebras such that \(\mathcal {B}\) is commutative. Let \(d:\mathcal {A} \rightarrow \mathcal {B}\) be a \((\psi , \phi\))-derivation such that \(d(e) \in Inv(\mathcal {B})\). If we have all the conditions under which every homomorphism from \(\mathcal {A}\) into \(\mathcal {B}\) is continuous, then d, \(\psi\) and \(\phi\) are continuous mappings.

Note that if \(\mathcal {A}\) is an \(*\)-algebra, then a straightforward verification shows that \(\mathcal {A} \times \mathcal {A}\) is also an \(*\)-algebra by regarding the following structure:

  1. 1.

    \((a, b) + (c, d) = (a + c, b + d)\);

  2. 2.

    \(\lambda (a, b) = (\lambda a, \lambda b)\):

  3. 3.

    \((a, b). (c, d) = (ac, bd)\);

  4. 4.

    \((a, b)^{*} = (a^{*}, b^{*})\);

for \(a, b \in \mathcal {A}\) and \(\lambda \in \mathbb {C}\).

Similar to the \((*, \star )\)-mappings, a bi-mapping \(\Omega :\mathcal {A} \times \mathcal {A} \rightarrow \mathcal {B}\) is a \((*, \star )\)-mapping if \(\Omega (a, b) = (\Omega (a^{*}, b^{*}))^{\star }\) for all \(a, b \in \mathcal {A}\). Let \(\psi , \phi :\mathcal {A} \rightarrow \mathcal {B}\) be two mappings. A bi-linear mapping (i.e., linear in both arguments) \(\Omega :\mathcal {A} \times \mathcal {A} \rightarrow \mathcal {B}\) is called a left two variable \((\psi , \phi )\)-derivation if \(\Omega (ab, c) = \Omega (a,c) \psi (b) + \phi (a) \Omega (b,c)\) for all \(a, b, c \in \mathcal {A}\). A right two variable \((\psi , \phi )\)-derivation is defined, similarly. A bi-linear mapping \(\Omega :\mathcal {A} \times \mathcal {A} \rightarrow \mathcal {B}\) is said to be a two variable \((\psi , \phi )\)-derivation if it is both a left-and a right two variable \((\psi , \phi )\)-derivation. A \((*, \star )\)-left two variable \((\psi , \phi )\)-derivation means a left two variable \((\psi , \phi )\)-derivation \(\Omega :\mathcal {A} \times \mathcal {A} \rightarrow \mathcal {B}\), whenever \(\Omega\), \(\psi\) and \(\phi\) are \((*, \star )\)-mappings.

Theorem 2.11

Let \((\mathcal {A}, *)\) and \((\mathcal {B}, \star )\) be two unital involutive algebras and let \(d_1, d_2:\mathcal {A} \rightarrow \mathcal {B}\) be two \((*, \star )-(\psi , \phi )\)-derivations such that \(d_1(e)d_2(a_0) = e\) or \(d_1(a_0)d_2(e) = e\) for some \(a_0 \in \mathcal {A}\). Suppose that \([\psi (a), d_2(b)] = 0 = [\phi (a), d_1(b)]\) for all \(a, b \in \mathcal {A}\). Then, the mappings \(\psi\) and \(\phi\) are linear. Moreover, suppose we have the conditions under which every homomorphism from \(\mathcal {A}\) into \(\mathcal {B}\) is continuous. Then, \(d_1\), \(d_2\), \(\psi\) and \(\phi\) are continuous linear mappings.

Proof

Define \(\Omega :\mathcal {A} \times \mathcal {A} \rightarrow \mathcal {B}\) by \(\Omega (a,b) = d_1(a)d_2(b)\). It is easy to see that \(\Omega\) is a \((*, \star )-\)two variable \((\psi , \phi )\)-derivation. So, we have

$$\begin{aligned} \Omega (ab, c)&= \Big (\Omega (b^{*}a^{*}, c^{*})\Big )^{\star } \\&= \Big (\Omega (b^{*}, c^{*})\psi (a^{*}) + \phi (b^{*})\Omega (a^{*}, c^{*})\Big )^{\star } \\&= \psi (a) \Omega (b, c) + \Omega (a, c) \phi (b) \end{aligned}$$

Moreover, we know that \(\Omega (ab, c) = \Omega (a, c) \psi (b) + \phi (a) \Omega (b, c)\) for all \(a, b, c \in \mathcal {A}\). Hence, we have the following expressions:

$$\begin{aligned} \Omega (ab, c)&= \frac{1}{2}\Omega (ab, c) + \frac{1}{2}\Omega (ab, c) \\&= \frac{\Omega (a, c) \psi (b) + \phi (a) \Omega (b, c)}{2} + \frac{\psi (a) \Omega (b, c) + \Omega (a, c) \phi (b)}{2} \end{aligned}$$

So,

$$\begin{aligned} \Omega (ab, c) = \Omega (a, c) \left( \frac{\psi (b) + \phi (b)}{2}\right) + \left( \frac{\psi (a) + \phi (a)}{2}\right) \Omega (b, c), \end{aligned}$$

for all \(a, b, c \in \mathcal {A}\). Considering \(\frac{\psi + \phi }{2} = \Sigma\), we see that

$$\begin{aligned} \Omega (ab, c) = \Omega (a, c) \Sigma (b) + \Sigma (a) \Omega (b, c), \end{aligned}$$

for all \(a, b, c \in \mathcal {A}\). Let \(a_0\) be an element of \(\mathcal {A}\) such that \(d_1(e)d_2(a_0) = e\). So, it is observed that \(\Omega (e, a_0) = e\). It follows from [7, Theorem 2.16] that there exists a unital homomorphism \(\Theta :\mathcal {A} \rightarrow \mathcal {B}\) defined by \(\Theta (a) = \Omega (a, a_0)\) such that \(\Omega (a, b) = \Theta (ab)(\Theta (a_0))^{-1}\) for all \(a, b \in \mathcal {A}\) and also \(\frac{\psi (a) + \phi (a)}{2} = \Sigma (a) = \frac{\Omega (a, a_0)}{2} = \frac{\Theta (a)}{2}\) for all \(a \in \mathcal {A}\). Consequently, \(\Theta = \phi + \psi\). Since \(d_1(e)d_2(a_0) = e\) and \([\psi (a), d_2(b)] = 0\) for all \(a, b \in \mathcal {A}\), we have

$$\begin{aligned} e = d_1(e)d_2(a_0)&= d_1(e)\psi (e)d_2(a_0) + \phi (e) d_1(e) d_2(a_0) \\&= d_1(e)d_2(a_0)\psi (e) + \phi (e) d_1(e) d_2(a_0)\\&= \psi (e) + \phi (e). \end{aligned}$$

So, we have the following statements:

$$\begin{aligned} \Theta (a)&= \Omega (a, a_0) = d_1(a)d_2(a_0) \\&= d_1(a)\psi (e)d_2(a_0) + \phi (a) d_1(e) d_2(a_0) \\&= d_1(a)d_2(a_0)\psi (e) + \phi (a) \\&= \Theta (a) \psi (e) + \phi (a), \end{aligned}$$

which means that

$$\begin{aligned} \phi (a)&= \Theta (a) \phi (e), \ \ \ \ \ \ \ \ \ (a \in \mathcal {A}). \end{aligned}$$
(2.9)

Reasoning like above, one can easily get that

$$\begin{aligned} \psi (a)&= \psi (e)\Theta (a), \ \ \ \ \ \ \ \ \ (a \in \mathcal {A}). \end{aligned}$$
(2.10)

Putting \(\Theta (a) = \psi (a) + \phi (a)\) (\(a \in \mathcal {A}\)) in Eq. (2.9) and using \(\psi (e) + \phi (e) = e\), we obtain that

$$\begin{aligned} \psi (a)&= \Theta (a) \psi (e), \ \ \ \ \ \ \ \ \ (a \in \mathcal {A}). \end{aligned}$$
(2.11)

Similarly, we get that

$$\begin{aligned} \phi (a)&= \phi (e) \Theta (a), \ \ \ \ \ \ \ \ \ (a \in \mathcal {A}). \end{aligned}$$
(2.12)

Since \(\Theta\) is a linear mapping, the above discussion implies that both \(\psi\) and \(\phi\) are linear mappings and further, if \(\psi (e) = \phi (e)\), then \(\phi = \psi\). Now, we are going to prove that \(d_1\), \(d_2\), \(\psi\) and \(\phi\) are continuous linear mappings. Note that \(\Theta\) is continuous, since we are assuming the conditions under which every homomorphism from \(\mathcal {A}\) into \(\mathcal {B}\) is continuous. This fact with Eqs. (2.9) and (2.10) (or (2.11) and (2.12)) imply that both \(\psi\) and \(\phi\) are continuous linear mappings. Now, our task is to prove that \(d_1\) and \(d_2\) are continuous. We know that \(\Sigma (a) = \frac{\Omega (a, a_0)}{2} = \frac{\Theta (a)}{2}\) for all \(a \in \mathcal {A}\). So, \(\Sigma\) is continuous and also \(\Sigma (e) = \frac{e}{2}\). Using an argument similar to the one given above, it can be shown that

$$\begin{aligned}&d_1(ab) = d_1(a) \Sigma (b) + \Sigma (a) d_1(b) \\&d_2(ab) = d_2(a) \Sigma (b) + \Sigma (a) d_2(b) \end{aligned}$$

for all \(a, b \in \mathcal {A}\) Thus, we have \(d_1(a) = d_1(a)\Sigma (e) + \Sigma (a) d_1(e) = \frac{d_1(a)}{2} + \Sigma (a) d_1(e)\) for all \(a \in \mathcal {A}\). Hence,

$$\begin{aligned} d_1(a) = 2 \Sigma (a) d_1(e), \ \ \ \ \ \ \ (a \in \mathcal {A}) \end{aligned}$$
(2.13)

and similarly, we get that

$$\begin{aligned} d_2(a) = 2 \Sigma (a) d_2(e), \ \ \ \ \ \ \ (a \in \mathcal {A}) \end{aligned}$$
(2.14)

Equations (2.13) and (2.14) with the continuity of \(\Sigma\) give that both \(d_1\) and \(d_2\) are continuous. Thereby, our proof is complete. \(\square\)

An immediate consequence of Theorem 2.11 reads as follows:

Corollary 2.12

Let \((\mathcal {A}, *)\) and \((\mathcal {B}, \star )\) be two unital involutive algebras and let \(d:\mathcal {A} \rightarrow \mathcal {B}\) be a \((*, \star )-(\psi , \phi )\)-derivation such that \(d(e)d(a_0) = e\) or \(d(a_0)d(e) = e\) for some \(a_0 \in \mathcal {A}\). Suppose that \([\psi (a), d(b)] = 0 = [\phi (a), d(b)]\) for all \(a, b \in \mathcal {A}\). Then, the mappings \(\psi\) and \(\phi\) are linear. Moreover, suppose we have the conditions under which every homomorphism from \(\mathcal {A}\) into \(\mathcal {B}\) is continuous. Then, d, \(\psi\) and \(\phi\) are continuous linear mappings.