On central identities equipped with skew Lie product involving generalized derivations

Lemma 1

Lemma 1 [2020, Abbasi et al Lemma 4]

Let $\mathfrak{R}$ be a prime ring with involution of the second kind such that $\mathit{char}(\mathfrak{R})\ne 2$ . If $▿[x,x^{\ast }]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ , then $\mathfrak{R}$ is commutative.

Lemma 2

Lemma 2 [2017, Nejjar et al Lemma 2.2]

Let $\mathfrak{R}$ be a prime ring with involution of the second kind. Then $x\circ x^{\ast }\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ if and only if $\mathfrak{R}$ is commutative.

Lemma 3

Lemma 3 [2017, Nejjar et al Lemma 2.1]

Let $\mathfrak{R}$ be a prime ring with involution of the second kind. Then $[x,x^{\ast }]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ if and only if $\mathfrak{R}$ is commutative.

Lemma 4

[Posner, 1957, Lemma 3] Let $\mathfrak{R}$ be a prime ring and I a nonzero left ideal. If $\mathfrak{R}$ admits a nonzero derivation d such that $[d(x),x]\in Z(\mathfrak{R})$ for all $x\in I$ , then $\mathfrak{R}$ is commutative.

Theorem 1

Let $\mathfrak{R}$ be a 2-torsion free prime ring with involution $*$ of the second kind. If $\mathfrak{R}$ admits a generalized derivation $(\mathcal{F},d)$ such that $\mathcal{F}(▿[x,x^{\ast }])\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}$ , then $\mathfrak{R}$ is commutative or $\mathcal{F}=0$

Proof

By the assumption, we have

$$\mathcal{F}(▿[x,x^{\ast }])\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

By linearizing (2.1), we get

$$\mathcal{F}(▿[x,y^{\ast }])+\mathcal{F}(▿[y,x^{\ast }])\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$

Replacing x by xh in (2.2), where $0\ne h\in H(\mathfrak{R})\cap Z(\mathfrak{R})$ , we obtain

$$(\mathcal{F}(▿[x,y^{\ast }])+\mathcal{F}(▿[y,x^{\ast }]))h+(▿[x,y^{\ast }]+▿[y,x^{\ast }])d(h)\in Z(\mathfrak{R})\text{.}$$

Application of (2.2) gives

$$(▿[x,y^{\ast }]+▿[y,x^{\ast }])d(h)\in Z(\mathfrak{R})\text{.}$$

This implies that $▿[x,y^{\ast }]+▿[y,x^{\ast }]\in Z(\mathfrak{R})$ for all $x,y\in \mathfrak{R}$ or $d(h)=0$ for all $h\in S(\mathfrak{R})\cap Z(\mathfrak{R})$ . Suppose $▿[x,y^{\ast }]+▿[y,x^{\ast }]\in Z(\mathfrak{R})$ for all $x,y\in \mathfrak{R}$ . In particular, for $y=x$ , we have $2▿[x,x^{\ast }]\in Z(\mathfrak{R})$ . Since $\mathfrak{R}$ is 2-torsion free, so $▿[x,x^{\ast }]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ . Therefore, $\mathfrak{R}$ is commutative by Lemma 1.

Now, if $d(h)=0$ for all $h\in H(\mathfrak{R})\cap Z(\mathfrak{R})$ , then $d(k)=0$ for all $k\in S(\mathfrak{R})\cap Z(\mathfrak{R})$ . Thus from (2.2), we have

$$\mathcal{F}({\mathit{xy}}^{\ast }-y^{\ast }{x}^{\ast })+\mathcal{F}({\mathit{yx}}^{\ast }-x^{\ast }{y}^{\ast })\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$

Replacing x by xk in (2.5), we get $$(\mathcal{F}({\mathit{xy}}^{\ast }+y^{\ast }{x}^{\ast })-\mathcal{F}({\mathit{yx}}^{\ast }-x^{\ast }{y}^{\ast }))k\in Z(\mathfrak{R})\text{.}$$

The primeness of $\mathfrak{R}$ yields that

$$\mathcal{F}({\mathit{xy}}^{\ast }+y^{\ast }{x}^{\ast })-\mathcal{F}({\mathit{yx}}^{\ast }-x^{\ast }{y}^{\ast })\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$

Combining (2.5), (2.6) and using the fact that $\mathfrak{R}$ is 2-torsion free, we obtain

$$\mathcal{F}({\mathit{xy}}^{\ast })\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$ Replacing y by h in (2.7), where $0\ne h\in H(\mathfrak{R})\cap Z(\mathfrak{R})$ , we get $\mathcal{F}(x)\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ . Hence in view of Hvala (1998, Lemma 3), we conclude that $\mathfrak{R}$ is commutative or $\mathcal{F}=0$ . This completes the proof of the theorem.

Corollary 1

Corollary 1 [Abbasi et al., 2020, Theorem 2]

Let $\mathfrak{R}$ be a 2-torsion free prime ring with involution $*$ of the second kind. If $\mathfrak{R}$ admits a nonzero left centralizer $\mathcal{T}$ such that $\mathcal{T}(▿[x,x^{\ast }])\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}$ , then $\mathfrak{R}$ is commutative.

Corollary 2

Let $\mathfrak{R}$ be a 2-torsion free prime ring with involution $*$ of the second kind. If $\mathfrak{R}$ admits generalized derivations $(\mathcal{F},d)$ and $(\mathcal{G},g)$ such that $\mathcal{F}(▿[x,x^{\ast }])\pm \mathcal{G}(▿[x,x^{\ast }])\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}$ , then $\mathfrak{R}$ is commutative or $\mathcal{F}=\mp \mathcal{G}$ .

Proof

By the hypothesis, we have $$\mathcal{F}(▿[x,x^{\ast }])\pm \mathcal{G}(▿[x,x^{\ast }])\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

This implies that $$(\mathcal{F}\pm \mathcal{G})(▿[x,x^{\ast }])\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

Set $\mathcal{H}=\mathcal{F}\pm \mathcal{G}$ . Then, we have $$\mathcal{H}(▿[x,x^{\ast }])\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$ Since $\mathcal{H}=\mathcal{F}\pm \mathcal{G}$ is generalized derivation of $\mathfrak{R}$ associated with the derivation $d\pm g$ respectively. Therefore, in view of Theorem 1, we conclude that $\mathfrak{R}$ is commutative or $\mathcal{F}=\mp \mathcal{G}$ .

Corollary 3

Corollary 3 [Ali and Abbasi, 2020, Theorem 2]

Let $\mathfrak{R}$ be a 2-torsion free prime ring with involution $*$ of the second kind. If $\mathfrak{R}$ admits a generalized derivation $(\mathcal{F},d)$ such that $\mathcal{F}(▿[x,x^{\ast }])\pm ▿[x,x^{\ast }]\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}$ , then $\mathfrak{R}$ is commutative or $\mathcal{F}=\mp I_{\mathfrak{R}}$ where $I_{\mathfrak{R}}$ is the identity mapping of $\mathfrak{R}$ .

Proof

If $\mathcal{F}=0$ , then $▿[x,x^{\ast }]\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}$ . Thus $\mathfrak{R}$ is commutative by Lemma 1. Now, we may assume that $\mathcal{F}\ne 0$ . By the assumption, we have $$\mathcal{F}(▿[x,x^{\ast }])\pm ▿[x,x^{\ast }]\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

This can be written as $$(\mathcal{F}\pm I_{\mathfrak{R}})(▿[x,x^{\ast }])\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R},$$ where $I_{\mathfrak{R}}$ is the identity mapping of $\mathfrak{R}$ . Since $\mathcal{F}\pm I_{\mathfrak{R}}$ is generalized derivation of $\mathfrak{R}$ associated with the derivation d, so from Theorem 1, we get the required result.

Corollary 4

Corollary 4 [Mozumder et al., 2021, Theorem 1.11]

Let $\mathfrak{R}$ be a prime ring with involution of the second kind such that $\mathit{char}(\mathfrak{R})\ne 2$ . If $\mathfrak{R}$ admits a derivation d such that $d(▿[x,x^{\ast }])\pm ▿[x,x^{\ast }]\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}$ , then $\mathfrak{R}$ is commutative.

Theorem 2

Let $\mathfrak{R}$ be a 2-torsion free prime ring with involution $*$ of the second kind. If $\mathfrak{R}$ admits a generalized derivation $(\mathcal{F},d)$ such that $▿[x,\mathcal{F}(x)]\in Z(\mathfrak{R})$ , then $\mathfrak{R}$ is commutative or $\mathcal{F}=0$ .

Proof

By the assumption, we have

$$▿[x,\mathcal{F}(x)]\in Z(\mathfrak{R})$$ for all $x\in \mathfrak{R}$ . By linearizing (2.8), we get $$▿[x,\mathcal{F}(y)]+▿[y,\mathcal{F}(y)]\in Z(\mathfrak{R})$$ for all $x,y\in \mathfrak{R}$ . This implies that

(2.9)

$$x\mathcal{F}(y)-\mathcal{F}(y)x^{\ast }+y\mathcal{F}(x)-\mathcal{F}(x)y^{\ast }\in Z(\mathfrak{R})\text{.}$$

Replacing x by xh in (2.9), where $0\ne h\in H(\mathfrak{R})\cap Z(\mathfrak{R})$ and using it, we obtain

$$(\mathit{yx}-{\mathit{xy}}^{\ast })d(h)\in Z(\mathfrak{R})\text{.}$$

The primeness of $\mathfrak{R}$ yields that $\mathit{yx}-{\mathit{xy}}^{\ast }\in Z(\mathfrak{R})$ or $d(h)=0$ . In the first case, we have $▿[x,x^{\ast }]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ . Therefore, $\mathfrak{R}$ is commutative in view of Lemma 1. Now, if $d(h)=0$ for all $h\in H(\mathfrak{R})\cap Z(\mathfrak{R})$ , then $d(k)=0$ for all $k\in S(\mathfrak{R})\cap Z(\mathfrak{R})$ . Replacing x by xk in (2.9) and using the fact that $d(k)=0$ , where $k\in S(\mathfrak{R})\cap Z(\mathfrak{R})$ , we get $$(x\mathcal{F}(y)+\mathcal{F}(y)x^{\ast }+y\mathcal{F}(x)-\mathcal{F}(x)y^{\ast })k\in Z(\mathfrak{R})\text{.}$$

By the primeness of $\mathfrak{R}$ , we obtain

$$x\mathcal{F}(y)+\mathcal{F}(y)x^{\ast }+y\mathcal{F}(x)-\mathcal{F}(x)y^{\ast }\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$

Subtracting (2.11) from (2.9), we have

$$2\mathcal{F}(y)x^{\ast }\in Z(\mathfrak{R})$$ for all $x,y\in \mathfrak{R}$ . Since $\mathfrak{R}$ is 2-torsion free, we deduce that

(2.13)

$$\mathcal{F}(y)x^{\ast }\in Z(\mathfrak{R})\text{.}$$

Putting h in place of x, in (2.13) and using the primeness of $\mathfrak{R}$ , we get

$$\mathcal{F}(y)\in Z(\mathfrak{R})\text{for all}y\in \mathfrak{R}\text{.}$$ Therefore, in view of Lemma 3 in Hvala (1998), we conclude that either $\mathfrak{R}$ is commutative or $\mathcal{F}=0$ . Thereby the proof is completed.

Corollary 5

Let $\mathfrak{R}$ be a 2-torsion free prime ring with involution $*$ of the second kind. If $\mathfrak{R}$ admits nonzero generalized derivations $(\mathcal{F},d)$ and $(\mathcal{G},g)$ such that $▿[x,\mathcal{F}(x)]\pm ▿[x,\mathcal{G}(x)]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ , then either $\mathfrak{R}$ is commutative or $\mathcal{F}=\mp \mathcal{G}$ .

Proof

By the hypothesis, we have $$▿[x,\mathcal{F}(x)]\pm ▿[x,\mathcal{G}(x)]\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

This implies that $$▿[x,(\mathcal{F}\pm \mathcal{G})(x)]\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

Set $\mathcal{H}=\mathcal{F}\pm \mathcal{G}$ , then we obtain $$▿[x,\mathcal{H}(x)]\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$ Since $\mathcal{H}=\mathcal{F}\pm \mathcal{G}$ is generalized derivation of $\mathfrak{R}$ associated with the derivation $d\pm g$ respectively. Hence application of Theorem 2 yields that either $\mathfrak{R}$ is commutative or $\mathcal{F}=\mp \mathcal{G}$ .

Corollary 6

Let $\mathfrak{R}$ be a 2-torsion free prime ring with involution $*$ of the second kind. If $\mathfrak{R}$ admits a generalized derivation $(\mathcal{F},d)$ such that $▿[x,\mathcal{F}(x)]\pm ▿[x,x]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ , then either $\mathfrak{R}$ is commutative or $\mathcal{F}=\mp I_{\mathfrak{R}}$ , where $I_{\mathfrak{R}}$ is the identity mapping of $\mathfrak{R}$ .

Proof

By the assumption, we have $$▿[x,\mathcal{F}(x)]\pm ▿[x,x]\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

This can be written as $$▿[x,(\mathcal{F}\pm I_{\mathfrak{R}})(x)]\text{for all}x\in \mathfrak{R},$$ where $I_{\mathfrak{R}}$ is the identity mapping of $\mathfrak{R}$ . Since $\mathcal{F}\pm I_{\mathfrak{R}}$ are generalized derivations of $\mathfrak{R}$ associated with the derivation d, so from Theorem 2, we get either $\mathfrak{R}$ is commutative or $\mathcal{F}=\mp I_{\mathfrak{R}}$ , where $I_{\mathfrak{R}}$ is the identity mapping of $\mathfrak{R}$ .

Theorem 3

Let $\mathfrak{R}$ be a 2-torsion free prime ring with involution $*$ of the second kind. If $\mathfrak{R}$ admits generalized derivations $(\mathcal{F},d)$ and $(\mathcal{G},g)$ such that $\mathcal{F}(▿[x,x^{\ast }])+▿[x,\mathcal{G}(x)]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ , then $\mathfrak{R}$ is commutative or $\mathcal{F}=\mathcal{G}=0$ .

Proof

By the assumption, we have

$$\mathcal{F}(▿[x,x^{\ast }]+▿[x,\mathcal{G}(x)]\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

Expansion of (2.15) gives that

$$\mathcal{F}(x)x^{\ast }+\mathit{xd}(x^{\ast })-\mathcal{F}(x^{\ast })x^{\ast }-(x^{\ast })d(x^{\ast })+x\mathcal{G}(x)-\mathcal{G}(x)x^{\ast }\in Z(\mathfrak{R})\text{.}$$

Replacing x by xh in (2.16), where $0\ne h\in H(\mathfrak{R})\cap Z(\mathfrak{R})$ , we get

$$(2{\mathit{xx}}^{\ast }-{(x^{\ast })}^2-x^2)d(h)+(x^2-{\mathit{xx}}^{\ast })g(h)\in Z(\mathfrak{R})\text{.}$$

Substituting xk in place of x in (2.17), where $0\ne k\in S(\mathfrak{R})\cap Z(\mathfrak{R})$ , we obtain $$((-2{\mathit{xx}}^{\ast }-{(x^{\ast })}^2-x^2)d(h)+(x^2+{\mathit{xx}}^{\ast })g(h))k^2\in Z(\mathfrak{R})\text{.}$$

This implies that

$$(-2{\mathit{xx}}^{\ast }-{(x^{\ast })}^2-x^2)d(h)+(x^2+{\mathit{xx}}^{\ast })g(h)\in Z(\mathfrak{R})\text{.}$$

Subtracting (2.18) from (2.17), we arrive $${\mathit{xx}}^{\ast }(2d(h)-g(h))\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

This implies that either ${\mathit{xx}}^{\ast }\in Z(\mathfrak{R})$ or $2d(h)-g(h)=0$ . If ${\mathit{xx}}^{\ast }\in Z(\mathfrak{R})$ . Also $x^{\ast }x\in Z(\mathfrak{R})$ , hence $x\circ x^{\ast }\in Z(\mathfrak{R}),\mathfrak{R}$ is commutative by Lemma 2.

Now, if $(2d-g)(h)=0$ , then $(2d-g)(z)=0$ for all $z\in Z(\mathfrak{R})$ and so

$$2d(z)=g(z)\text{for all}z\in Z(\mathfrak{R})\text{.}$$

Application of (2.19) in (2.17) gives that $$(x^2-{(x^{\ast })}^2)d(h)\in Z(\mathfrak{R})\text{.}$$

This implies that $x^2-{(x^{\ast })}^2\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ or $d(h)=0$ for all $h\in H(\mathfrak{R})\cap Z(\mathfrak{R})$ . Consider the first case $x^2-{(x^{\ast })}^2\in Z(\mathfrak{R})$ . Linearizing the last relation, we get

$$\mathit{xy}+\mathit{yx}-x^{\ast }{y}^{\ast }-y^{\ast }{x}^{\ast }\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$

Replacing x by xk in (2.20) and $0\ne k\in S(\mathfrak{R})\cap Z(\mathfrak{R})$ , we obtain $$(\mathit{xy}+\mathit{yx}+x^{\ast }{y}^{\ast }+y^{\ast }{x}^{\ast })k\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$

This gives that

$$\mathit{xy}+\mathit{yx}+x^{\ast }{y}^{\ast }+y^{\ast }{x}^{\ast }\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$

From (2.20) and (2.21), we conclude that $$\mathit{xy}+\mathit{yx}\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$

In particular, $x\circ x^{\ast }\in Z(\mathfrak{R})$ . Therefore, $\mathfrak{R}$ is commutative by Lemma 2. Now, if $d(h)=0$ , then $g(h)=0$ and so $d(k)=g(k)=0$ for all $k\in S(\mathfrak{R})\cap Z(\mathfrak{R})$ and hence for all $z\in Z(\mathfrak{R})$ . Now, replacing x by xk in (2.16), we have $$(-\mathcal{F}(x)x^{\ast }-\mathit{xd}(x^{\ast })-\mathcal{F}(x^{\ast })x^{\ast }-x^{\ast }d(x^{\ast })+x\mathcal{G}(x)+\mathcal{G}(x)x^{\ast })k^2\in Z(\mathfrak{R})\text{.}$$

Primeness of $\mathfrak{R}$ forces that

$$-\mathcal{F}(x)x^{\ast }-\mathit{xd}(x^{\ast })-\mathcal{F}(x^{\ast })x^{\ast }-x^{\ast }d(x^{\ast })+x\mathcal{G}(x)+\mathcal{G}(x)x^{\ast }\in Z(\mathfrak{R})$$ for all $x\in \mathfrak{R}$ . Combining (2.16) and (2.22), we obtain

(2.23)

$$\mathcal{F}(x^{\ast })x^{\ast }+x^{\ast }d(x^{\ast })-x\mathcal{G}(x)\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

Linearizing (2.23) and using it, we get

$$\mathcal{F}(x^{\ast })y^{\ast }+\mathcal{F}(y^{\ast })x^{\ast }+x^{\ast }d(y^{\ast })+y^{\ast }d(x^{\ast })-x\mathcal{G}(y)-y\mathcal{G}(x)\in Z(\mathfrak{R})\text{.}$$

Replacing x by xk in (2.24), we get $$(-\mathcal{F}(x^{\ast })y^{\ast }-\mathcal{F}(y^{\ast })x^{\ast }-x^{\ast }d(y^{\ast })-y^{\ast }d(x^{\ast })-x\mathcal{G}(y)-y\mathcal{G}(x))k\in Z(\mathfrak{R})\text{.}$$

That is,

$$-\mathcal{F}(x^{\ast })y^{\ast }-\mathcal{F}(y^{\ast })x^{\ast }-x^{\ast }d(y^{\ast })-y^{\ast }d(x^{\ast })-x\mathcal{G}(y)-y\mathcal{G}(x)\in Z(\mathfrak{R})$$ for all $x,y\in \mathfrak{R}$ . Adding (2.24) and (2.25), we have

(2.26)

$$x\mathcal{G}(y)+y\mathcal{G}(x)\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$

Put $x=y=h$ in (2.26), where $0\ne h\in H(\mathfrak{R})\cap Z(\mathfrak{R})$ we have $$2h\mathcal{G}(h)\in Z(\mathfrak{R})\text{.}$$

This implies that

$$\mathcal{G}(h)\in Z(\mathfrak{R})\text{for all}h\in H(\mathfrak{R})\cap Z(\mathfrak{R})\text{.}$$

Replacing y by h in (2.26) where $0\ne h\in H(\mathfrak{R})\cap Z(\mathfrak{R})$ , we get $$x\mathcal{G}(h)+h\mathcal{G}(x)\in Z(\mathfrak{R})\text{.}$$

Taking the Lie product of above with x, we arrive at $h[\mathcal{G}(x),x]=0$ and hence

$$[\mathcal{G}(x),x]=0\text{for all}x\in \mathfrak{R}\text{.}$$

A linearization of (2.28) gives that

$$[\mathcal{G}(x),y]+[\mathcal{G}(y),x]=0\text{for all}x,y\in \mathfrak{R}\text{.}$$

Replacing x by hy in (2.29) and using (2.28), we get $$h[g(y),y]=0,$$ which implies $$[g(y),y]=0\text{for all}y\in \mathfrak{R}\text{.}$$

Therefore, in view of Lemma 4 either $g=0$ or $\mathfrak{R}$ is commutative. We proceed the proof for the case $g=0$ , that is, $\mathfrak{R}$ is noncommutative. Substitute xw for x in (2.29), we get

$$\mathcal{G}(x)[w,y]+x[\mathcal{G}(y),w]=0\text{for all}x,y,w\in \mathfrak{R}\text{.}$$

Replacing x by tx in (2.30), we obtain

$$\mathcal{G}(t)x[w,y]+\mathit{tx}[\mathcal{G}(y),w]=0\text{for all}x,y,w,t\in \mathfrak{R}\text{.}$$

Left multiplying by t in (2.30), we have

$$t\mathcal{G}(x)[w,y]+\mathit{tx}[\mathcal{G}(y),w]=0\text{for all}x,y,w,t\in \mathfrak{R}\text{.}$$

Subtracting (2.32) from (2.31), we get $$(\mathcal{G}(t)x-t\mathcal{G}(x))[w,y]=0\text{for all}x,y,w,t\in \mathfrak{R}\text{.}$$

Replacing w by rw in last relation, we get $$(\mathcal{G}(t)x-t\mathcal{G}(x))r[w,y]=0\text{for all}x,y,w,t,r\in \mathfrak{R}\text{.}$$

Invoking the primeness of $\mathfrak{R}$ , we deduce that

$$\mathcal{G}(t)x=t\mathcal{G}(x)\text{for all}x,t\in \mathfrak{R}\text{.}$$

By using (2.33) in (2.26), we have $$2y\mathcal{G}(x)\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$

In particular for $y=h$ , we have

$$\mathcal{G}(x)\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$ Thus $\mathcal{G}=0$ by Hvala (1998, Lemma 3). Now, from (2.15), we have $▿[x,\mathcal{F}(x)]\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}$ . Therefore, in view of Theorem 1, $\mathcal{F}=0$ . This completes the proof of the theorem.

Corollary 7

Let $\mathfrak{R}$ be a 2-torsion free prime ring with involution $*$ of the second kind. If $\mathfrak{R}$ admits a generalized derivation $\mathcal{F}$ such that $\mathcal{F}(▿[x,x^{\ast }])+▿[x,\mathcal{F}(x)]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ , then $\mathfrak{R}$ is commutative or $\mathcal{F}=0$ .

Theorem 4

Let $\mathfrak{R}$ be a 2-torsion free prime ring with involution $*$ of the second kind. If $\mathfrak{R}$ admits a generalized derivation $(\mathcal{F},d)$ such that $▿[x,\mathcal{F}(x^{\ast })]\pm ▿[x,x^{\ast }]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ , then $\mathfrak{R}$ is commutative or $\mathcal{F}=\mp I_{\mathfrak{R}}$ , where $I_{\mathfrak{R}}$ is the identity mapping of $\mathfrak{R}$ .

Proof

Firstly, we suppose that

$$▿[x,\mathcal{F}(x^{\ast })]+▿[x,x^{\ast }]\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

If $\mathcal{F}=0$ , then $▿[x,x^{\ast }]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ and hence $\mathfrak{R}$ is commutative by Lemma 1. Next, we assume that $\mathcal{F}\ne 0$ . By linearizing (2.35), we get $$▿[x,\mathcal{F}(y^{\ast })]+▿[y,\mathcal{F}(x^{\ast })]+▿[x,y^{\ast }]+▿[y,x^{\ast }]\in Z(\mathfrak{R})\text{.}$$

This can be written as

$$x\mathcal{F}(y^{\ast })-\mathcal{F}(y^{\ast })x^{\ast }+y\mathcal{F}(x^{\ast })-\mathcal{F}(x^{\ast })y^{\ast }+{\mathit{xy}}^{\ast }-y^{\ast }{x}^{\ast }+{\mathit{yx}}^{\ast }-x^{\ast }{y}^{\ast }\in Z(\mathfrak{R})\text{.}$$

Replacing x by xh in (2.36), where $0\ne h\in H(\mathfrak{R})\cap Z(\mathfrak{R})$ , we get $$(x\mathcal{F}(y^{\ast })-\mathcal{F}(y^{\ast })x^{\ast }+y\mathcal{F}(x^{\ast })-\mathcal{F}(x^{\ast })y^{\ast }+{\mathit{xy}}^{\ast }-y^{\ast }{x}^{\ast }+{\mathit{yx}}^{\ast }-x^{\ast }{y}^{\ast })h+({\mathit{yx}}^{\ast }-x^{\ast }{y}^{\ast })d(h)\in Z(\mathfrak{R})\text{.}$$

By using (2.36) in last relation, we have $$({\mathit{yx}}^{\ast }-x^{\ast }{y}^{\ast })d(h)\in Z(\mathfrak{R})\text{.}$$

This implies that $({\mathit{yx}}^{\ast }-x^{\ast }{y}^{\ast })\in Z(\mathfrak{R})$ or $d(h)=0$ . Now, if $({\mathit{yx}}^{\ast }-x^{\ast }{y}^{\ast })\in Z(\mathfrak{R})$ for all $x,y\in \mathfrak{R}$ , then $\mathfrak{R}$ is commutative by Lemma 1. Now, if $d(h)=0$ then we have $d(k)=0$ for all $k\in S(\mathfrak{R})\cap Z(\mathfrak{R})$ . Replacing x by xk in (2.36), where $0\ne k\in S(\mathfrak{R})\cap Z(\mathfrak{R})$ and using the fact that $d(k)=0$ , we get $$(x\mathcal{F}(y^{\ast })+\mathcal{F}(y^{\ast })x^{\ast }-y\mathcal{F}(x^{\ast })+\mathcal{F}(x^{\ast })y^{\ast }+{\mathit{xy}}^{\ast }+y^{\ast }{x}^{\ast }-{\mathit{yx}}^{\ast }+x^{\ast }{y}^{\ast })k\in Z(\mathfrak{R})\text{.}$$

This gives $$x\mathcal{F}(y^{\ast })+\mathcal{F}(y^{\ast })x^{\ast }-y\mathcal{F}(x^{\ast })+\mathcal{F}(x^{\ast })y^{\ast }+{\mathit{xy}}^{\ast }+y^{\ast }{x}^{\ast }-{\mathit{yx}}^{\ast }+x^{\ast }{y}^{\ast }\in Z(\mathfrak{R})\text{.}$$

Combining the last expression with (2.36), we arrive at $$2x\mathcal{F}(y^{\ast })+2{\mathit{xy}}^{\ast }\in Z(\mathfrak{R})$$ for all $x,y\in \mathfrak{R}$ . Since $\mathfrak{R}$ is 2-torsion free prime ring, we deduce that

$$x\mathcal{F}(y^{\ast })+{\mathit{xy}}^{\ast }\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$

Replacing x by h and y by $y^{\ast }$ in (2.37), where $0\ne h\in H(\mathfrak{R})\cap Z(\mathfrak{R})$ , we get $$\mathcal{F}(y)+y\in Z(\mathfrak{R})\text{for all}y\in \mathfrak{R}\text{.}$$

That is, $$(\mathcal{F}+I_{\mathfrak{R}})(y)\in Z(\mathfrak{R})\text{for all}y\in \mathfrak{R},$$ where $I_{\mathfrak{R}}$ is the identity mapping of $\mathfrak{R}$ . Since $\mathcal{F}+I_{\mathfrak{R}}$ is a generalized derivation of $\mathfrak{R}$ , in view of Hvala (1998, Lemma 3) either $\mathfrak{R}$ is commutative or $\mathcal{F}=-I_{\mathfrak{R}}$ . Now consider the case $▿[x,\mathcal{F}(x^{\ast })]-▿[x,x^{\ast }]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ . Proceeding the same lines with necessary variations, we get the required result.

Corollary 8

Corollary 8 [Abbasi et al., 2020, Theorem 3]

Let $\mathfrak{R}$ be a 2-torsion free prime ring with involution $*$ of the second kind. If $\mathfrak{R}$ admits a left centralizer $\mathcal{T}$ such that $▿[x,\mathcal{T}(x^{\ast })]\pm ▿[x,x^{\ast }]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ , then either $\mathcal{T}$ is a centralizer or $\mathfrak{R}$ is commutative.

Corollary 9

Corollary 9 [Mozumder et al., 2021, Theorem 1.5]

Let $\mathfrak{R}$ be a prime ring with involution of the second kind such that $\mathit{char}(\mathfrak{R})\ne 2$ . If $\mathfrak{R}$ admits a derivation d such that $▿[x,d(x^{\ast })]\pm ▿[x,x^{\ast }]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ , then $\mathfrak{R}$ is commutative.

Theorem 5

Let $\mathfrak{R}$ be a 2-torsion free prime ring with involution $*$ of the second kind. If $\mathfrak{R}$ admits a generalized derivation $(\mathcal{F},d)$ such that $▿[x,\mathcal{F}(x)]+▿[x,x^{\ast }]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ , then $\mathfrak{R}$ is commutative.

Proof

We have

$$▿[x,\mathcal{F}(x)]+▿[x,x^{\ast }]\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

Expansion of (2.38) gives that

$$x\mathcal{F}(x)-\mathcal{F}(x)x^{\ast }+{\mathit{xx}}^{\ast }-{(x^{\ast })}^2\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

Replacing x by xh in (2.39), where $0\ne h\in H(\mathfrak{R})\cap Z(\mathfrak{R})$ , we get $$(x\mathcal{F}(x)-\mathcal{F}(x)x^{\ast }+{\mathit{xx}}^{\ast }-{(x^{\ast })}^2)h^2+(x^2-{\mathit{xx}}^{\ast })d(h)\in Z(\mathfrak{R})\text{.}$$

By using (2.39) in last relation, we have $$(x^2-{\mathit{xx}}^{\ast })d(h)\in Z(\mathfrak{R})\text{.}$$

The primeness of $\mathfrak{R}$ yields that $x^2-{\mathit{xx}}^{\ast }\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ or $d(h)=0$ for all $h\in H(\mathfrak{R})\cap Z(\mathfrak{R})$ . First we consider the case

$$x^2-{\mathit{xx}}^{\ast }\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

Substituting xk in place of x in (2.40), where $0\ne k\in S(\mathfrak{R})\cap Z(\mathfrak{R})$ , we find that $$(x^2+{\mathit{xx}}^{\ast })k^2\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R},$$ which gives that

$$x^2+{\mathit{xx}}^{\ast }\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

Subtracting (2.40) from (2.41) and using the fact that $\mathfrak{R}$ is 2-torsion free, we obtain ${\mathit{xx}}^{\ast }\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}$ . This further implies $x\circ x^{\ast }\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}$ . Therefore $\mathfrak{R}$ is commutative by Lemma 2.

On the other hand, if $d(h)=0$ , then $d(k)=0$ for all $k\in S(\mathfrak{R})\cap Z(\mathfrak{R})$ and so $d(z)=0$ for all $z\in Z(\mathfrak{R})$ . Replacing x by xk in (2.39), we get $$(x\mathcal{F}(x)+\mathcal{F}(x)x^{\ast }-{\mathit{xx}}^{\ast }-{(x^{\ast })}^2)k^2\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

This implies that

$$x\mathcal{F}(x)+\mathcal{F}(x)x^{\ast }-{\mathit{xx}}^{\ast }-{(x^{\ast })}^2\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

Combining (2.39) and (2.42), we get

$$x\mathcal{F}(x)-{(x^{\ast })}^2\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

Linearization of (2.43) gives that

$$x\mathcal{F}(y)+y\mathcal{F}(x)-x^{\ast }{y}^{\ast }-y^{\ast }{x}^{\ast }\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$

Replacing x by xk in last relation, where $0\ne k\in S(\mathfrak{R})\cap Z(\mathfrak{R})$ , we get $$(x\mathcal{F}(y)+y\mathcal{F}(x)+x^{\ast }{y}^{\ast }+y^{\ast }{x}^{\ast })k\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$

By the primeness of $\mathfrak{R}$ , we have

$$x\mathcal{F}(y)+y\mathcal{F}(x)+x^{\ast }{y}^{\ast }+y^{\ast }{x}^{\ast }\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$ Combining (2.44), (2.45) and using the fact that $\mathfrak{R}$ is 2-torsion free, we obtain $x^{\ast }{y}^{\ast }+y^{\ast }{x}^{\ast }\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}$ . In particular, we have $x\circ x^{\ast }\in Z(\mathfrak{R})$ . Therefore, $\mathfrak{R}$ is commutative by Lemma 2.

Theorem 6

Let $\mathfrak{R}$ be a 2-torsion free prime ring with involution $*$ of the second kind. If $\mathfrak{R}$ admits a generalized derivation $(\mathcal{F},d)$ such that $▿[x,\mathcal{F}(x)]+[x,x^{\ast }]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ , then $\mathfrak{R}$ is commutative.

Proof

Assume that

$$▿[x,\mathcal{F}(x)]+[x,x^{\ast }]\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}\text{.}$$

By linearizing (2.46), we get $$▿[x,\mathcal{F}(y)]+▿[y,\mathcal{F}(y)]+[x,y^{\ast }]+[y,x^{\ast }]\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\text{.}$$

Expansion of the above expression gives that

$$x\mathcal{F}(y)-\mathcal{F}(y)x^{\ast }+y\mathcal{F}(x)-\mathcal{F}(x)y^{\ast }+{\mathit{xy}}^{\ast }+{\mathit{yx}}^{\ast }-x^{\ast }y-y^{\ast }x\in Z(\mathfrak{R})$$ for all $x,y\in \mathfrak{R}$ . Replacing x by xh in (2.47), where $0\ne h\in H(\mathfrak{R})\cap Z(\mathfrak{R})$ , we get

(2.48)

$$(\mathit{yx}-{\mathit{xy}}^{\ast })d(h)\in Z(\mathfrak{R})\text{for all}x,y\in \mathfrak{R}\mathrm{and}h\in H(\mathfrak{R})\cap Z(\mathfrak{R})\text{.}$$

This implies $\mathit{yx}-{\mathit{xy}}^{\ast }\in Z(\mathfrak{R})$ or $d(h)=0$ . In first case, we have $▿[x,x^{\ast }]\in Z(\mathfrak{R})$ for all $x\in \mathfrak{R}$ and hence $\mathfrak{R}$ is commutative in view of Lemma 1.

Now, if $d(h)=0$ , then $d(k)=0$ for all $k\in S(\mathfrak{R})\cap Z(\mathfrak{R})$ . Replacing x by xk, where $0\ne k\in S(\mathfrak{R})\cap Z(\mathfrak{R})$ in (2.47), we obtain $$(x\mathcal{F}(y)+\mathcal{F}(y)x^{\ast }+y\mathcal{F}(x)-\mathcal{F}(x)y^{\ast }+{\mathit{xy}}^{\ast }-{\mathit{yx}}^{\ast }+x^{\ast }y-y^{\ast }x)k\in Z(\mathfrak{R})\text{.}$$

Invoking the primeness of $\mathfrak{R}$ , we have

$$x\mathcal{F}(y)+\mathcal{F}(y)x^{\ast }+y\mathcal{F}(x)-\mathcal{F}(x)y^{\ast }+{\mathit{xy}}^{\ast }-{\mathit{yx}}^{\ast }+x^{\ast }y-y^{\ast }x\in Z(\mathfrak{R})$$ for all $x,y\in \mathfrak{R}$ . Subtracting (2.49) from (2.47), we arrive

(2.50)

$$2\mathcal{F}(y)x^{\ast }-2{\mathit{yx}}^{\ast }+2x^{\ast }y\in Z(\mathfrak{R})$$ for all $x,y\in \mathfrak{R}$ . Since $\mathfrak{R}$ is 2-torsion free prime ring, we deduce that

(2.51)

$$\mathcal{F}(y)x^{\ast }-{\mathit{yx}}^{\ast }+x^{\ast }y\in Z(\mathfrak{R})\text{.}$$ for all $x,y\in \mathfrak{R}$ . Putting h in place of $x^{\ast }$ in (2.51), we get $$\mathcal{F}(y)h\in Z(\mathfrak{R})\mathrm{for}\mathrm{all}y\in \mathfrak{R},$$ which implies $$\mathcal{F}(y)\in Z(\mathfrak{R})\mathrm{for}\mathrm{all}y\in \mathfrak{R}\text{.}$$ In view of Hvala (1998, Lemma 3), we conclude that either $\mathfrak{R}$ is commutative or $\mathcal{F}=0$ . In the latter case from (2.46), we have $[x,x^{\ast }]\in Z(\mathfrak{R})\text{for all}x\in \mathfrak{R}$ , hence $\mathfrak{R}$ is commutative by Lemma 3. This completes the proof.