Linear Generalized N-derivations on C-algebras*

2. Introduction

Throughout the discussion, unless otherwise mentioned, $\mathcal{A}$ will denote $C^{*}$-algebra with $\mathcal{Z}\left(\mathcal{A}\right)$ as its center. However, $\mathcal{A}$ may or may not have unity. The symbols $[x,y]$ and $x\circ y$ denote the commutator $xy-yx$ and the anti-commutator $xy+yx$, respectively, for any $x,y\in \mathcal{A}$. An algebra $\mathcal{A}$ is said to be prime if $x\mathcal{A}y=\left\{0\right\}$ implies that either $x=0$ or $y=0$, and semiprime if $x\mathcal{A}x=\left\{0\right\}$ implies that $x=0$, where $x,y\in \mathcal{A}$. An additive subgroup U of R is said to be a Lie ideal of R if $[u,r]\in U$, for all $u\in U$, $r\in R$. U is called a square-closed Lie ideal of R if U is a Lie ideal and $u^2\in U$ for all $u\in U$.

A Banach algebra is a linear associate algebra which, as a vector space, is a Banach space with norm $\left|\right|.\left|\right|$ satisfying the multiplicative inequality; $\left|\right|xy\left|\right|\le \left|\right|x\left|\right|\left|\right|y\left|\right|$ for all x and y in $\mathcal{A}$. An involution on an algebra $\mathcal{A}$ is a linear map $x\mapsto x^{*}$ of $\mathcal{A}$ into itself such that the following conditions are hold: $\left(i\right)$${\left(xy\right)}^{*}=y^{*}{x}^{*}$, $\left(ii\right)$${\left(x^{*}\right)}^{*}=x$, and $\left(iii\right)$${(x+\lambda y)}^{*}=x^{*}+\overline{\lambda }{y}^{*}$ for all $x,y\in \mathcal{A}$ and $\lambda \in \mathbb{C}$, the field of complex numbers, where $\overline{\lambda }$ is the conjugate of $\lambda$. An algebra equipped with an involution is called a *-algebra or algebra with involution.

A Banach *-algebra is a Banach algebra $\mathcal{A}$ together with an isometric involution $\left|\right|x^{*}\left|\right|=\left|\right|x\left|\right|$ for all $x\in \mathcal{A}$ . A $C^{*}$-algebra $\mathcal{A}$ is a Banach *-algebra with the additional norm condition $\left|\right|x^{*}x\left|\right|={\left|\right|x\left|\right|}^2$ for all $x\in \mathcal{A}$.

A linear operator $\mathcal{D}$ on a $C^{*}$-algebra $\mathcal{A}$ is called a derivation if $\mathcal{D}\left(xy\right)=\mathcal{D}\left(x\right)y+x\mathcal{D}\left(y\right)$ holds for all $x,y\in \mathcal{A}$. Consider the inner derivation ${\delta }_a$ implemented by an element a in $\mathcal{A}$, which is defined as ${\delta }_a\left(x\right)=xa-ax$ for every x in $\mathcal{A}$, as a typical example of a nonzero derivation in a noncommutative algebra.

In order to broaden the scope of derivation, Maksa [13] introduced the concept of symmetric bi-derivations. A bi-linear map $\mathfrak{D}:\mathcal{A}\times \mathcal{A}\to \mathcal{A}$ is said to be a bi-derivation if

$$\mathfrak{D}(x{x}^{\prime },y)=\mathfrak{D}(x,y)x^{\prime }+x\mathfrak{D}(x^{\prime },y)$$

$$\mathfrak{D}(x,y{y}^{\prime })=\mathfrak{D}(x,y)y^{\prime }+y\mathfrak{D}(x,y^{\prime })$$

holds for any $x,x^{\prime },y,y^{\prime }\in \mathcal{A}$. The foregoing conditions are identical if $\mathfrak{D}$ is also a symmetric map, that is, if $\mathfrak{D}(x,y)=\mathfrak{D}(y,x)$ for every $x,y\in \mathcal{A}$. In this case, $\mathfrak{D}$ is referred to as a symmetric bi-derivation of $\mathcal{A}$. Vukman [20] investigated symmetric bi-derivations in prime and semiprime rings. Argac and Yenigul [3] and Muthana [15] obtained the similar type of results on Lie ideals of R.

In this paper we briefly discuss the various extensions of the notion of derivations on $C^{*}$-algebras. The most general and important one among them is the notion of a generalized symmetric linear n-derivations on $C^{*}$-algebras. The concept of derivation and symmetric bi-derivation was generalized by Park [16] as follows: a n-linear map $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ is said to be a symmetric(permuting) linear n-derivation if $\mathfrak{D}$ is permuting and $\mathfrak{D}(x_1,x_2,\dots ,x_i{x}_i^{\prime },\dots ,x_n)=\mathfrak{D}(x_1,x_2,\dots ,x_n)x_i^{\prime }+x_i\mathfrak{D}(x_1^{\prime },x_2,\cdots ,x_n)$ hold for all $x_i,x_i^{\prime }\in \mathcal{A},i=1,2,\dots ,n$. A map $\mathfrak{d}:\mathcal{A}\to \mathcal{A}$ defined by $\mathfrak{d}\left(x\right)=\mathfrak{D}(x,x,\dots ,x)$ is called the trace of $\mathfrak{D}$. If $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ is permuting and n-linear, then the trace $\mathfrak{d}$ of $\mathfrak{D}$ satisfies the relation

$$\mathfrak{d}(x+y)=\mathfrak{d}\left(x\right)+\mathfrak{d}\left(y\right)+\underset{k=1}{\overset{n-1}{{\sum }^n}}{C}_k{h}_k(x;y)$$

for all $x,y\in \mathcal{A}$, where $n{C}_k=\left(\genfrac{}{}{0pt}{}{n}{k}\right)$ and

$$h_k(x;y)=\mathfrak{D}(\underset{(n-k)-\mathrm{times}}{\underbrace{x,\dots ,x}},\underset{k-\mathrm{times}}{\underbrace{y,\dots ,y}}).$$

Ashraf et al. [4] introduced the notion of symmetric generalized n-derivations in a ring, building upon the concept of generalized derivation. Let $n\ge 1$ be a fixed positive integer. A symmetric n-linear map $\mathcal{G}:{\mathcal{A}}^n\to \mathcal{A}$ is known to be symmetric linear generalized n-derivation if there exists a symmetric linear n-derivation $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ such that $\mathcal{G}\left(x_1,x_2,\dots ,x_i{x}_i^{\prime },\dots ,x_n\right)=\mathcal{G}\left(x_1,x_2,\dots ,x_i,\dots ,x_n\right)x_i^{\prime }+x_i\mathfrak{D}(x_1,x_2,\dots ,x_i^{\prime },\dots ,x_n)$ holds for all $x_i,x_i^{\prime }\in \mathcal{A}$.

There has been considerable interest in the structure of linear derivation and linear bi-derivation on $C^{*}$-algebras. Derivations on $C^{*}$-algebras were described in various ways by different authors. For example, in 1966, Kadison [11] proved that each linear derivation of a $C^{*}$-algebra annihilates its center. In 1989, Mathieu [14] extended the Posner’s first result [17] on $C^{*}$-algebras. Basically, he proved that if the product of two linear derivations d and $d^{\prime }$ on a $C^{*}$-algebra is a linear derivation then $d{d}^{\prime }=0$. Very recently, Ekrami and Mirzavaziri [7] showed that “if $\mathcal{A}$ is a $C^{*}$-algebra admitting two linear derivations d and $d^{\prime }$ on $\mathcal{A}$, then there exists a linear derivation D on $\mathcal{A}$ such that $d{d}^{\prime }+d^{\prime }d=D^2$ if and only if d and $d^{\prime }$ are linearly dependent".

In [2], Ali and Khan proved that if $\mathcal{A}$ is a $C^{*}$-algebra admitting a symmetric bilinear generalized *-biderivation $\mathcal{H}:\mathcal{A}\times \mathcal{A}\to \mathcal{A}$ with an associated symmetric bilinear *-biderivation $\mathcal{B}:\mathcal{A}\times \mathcal{A}\to \mathcal{A}$, then $\mathcal{H}$ maps $\mathcal{A}\times \mathcal{A}$ into $Z\left(\mathcal{A}\right)$. In [19], Rehman and Ansari characterized the trace of symmetric bi-derivation and obtain more general results by considering various conditions on a subset of the ring R, viz. Lie ideal of R. Basically, they proved that: let R be a prime ring with $charR\ne 2$ and U be a square closed Lie ideal of R. Suppose that $B:R\times R\to R$ is a symmetric bi-derivation and f, the trace of B. If $\left[f\right(x),x]=0$ for all $x\in U$, then either $U\subseteq Z\left(R\right)$ or $f=0$ (see also [1,8,12,18] for recent results).

In the prospect of above motivation, we try to prove some results based on linear generalized n-derivations of $C^{*}$-algebras. We aim to achieve broader outcomes by examining various conditions within a specific subset of the $C^{*}$-algebra $\mathcal{A}$, viz. Lie ideal of $\mathcal{A}$. Precisely, we prove that if $\mathcal{A}$ is a $C^{*}$-algebra, U is a square closed Lie ideal of $\mathcal{A}$ admitting a nonzero symmetric linear generalized n-derivation $\mathcal{G}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{g}:\mathcal{A}\to \mathcal{A}$ associated with symmetric linear n-derivation $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{d}:\mathcal{A}\to \mathcal{A}$ satisfying the condition of $\left(g\right(xy)-g(yx\left)\right)\pm [x,y]\in Z\left(R\right)$ for all $x,y\in U$, then $U\subseteq Z\left(R\right)$.

3. The Results

In order to establish the proofs of our main theorems, we first state a result which we use frequently in the proof of our main results.

Lemma 1. 

[10, Corrolary 2.1] Let R be a 2-torsion free semiprime ring, U a Lie ideal of R such that $U⊈Z\left(R\right)$ and $a,b\in U$.

• If $aUa=\left\{0\right\}$, then $a=0$.
• If $aU=\left\{0\right\}$ ($Ua=\left\{0\right\}$), then $a=0$.
• If U is a square closed Lie ideal and $aUb=\left\{0\right\}$, then $ab=0$ and $ba=0$.

Lemma 2. 

[9, Lemma 1] Let R be a semiprime, 2 torsion-free ring and let U be a Lie ideal of R. Suppose that $[U,U]\subseteq Z\left(R\right)$, then $U\subseteq Z\left(R\right)$.

Lemma 3. 

[5] Let n be a fixed positive integer and R a $n!$-torsion free ring. Suppose that $y_1,y_2,\dots ,y_n\in R$ satisfy $\lambda y_1+{\lambda }^2{y}_2+\cdots +{\lambda }^n{y}_n=0$ for $\lambda =1,2,\dots ,n$. Then $y_i=0$ for $i=1,2,\dots ,n.$

Daif and Bell [6] proved that if a semiprime ring admits a derivation d such that either $xy-d\left(xy\right)=yx-d\left(yx\right)$ or $xy+d\left(xy\right)=yx+d\left(yx\right)$ holds for all $x,y\in R$, then R is commutative. In this section, apart from proving other results, we expand the previous result by demonstrating the following theorem for the traces of generalized linear n-derivation on certain subsets of $\mathcal{A}$.

Theorem 1. 

For any fixed integer $n\ge 2$, let $\mathcal{A}$ be a $C^{*}$-algebra, U be a square closed Lie ideal of $\mathcal{A}$. If $\mathcal{A}$ admits a nonzero symmetric linear generalized n-derivation $\mathcal{G}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{g}:\mathcal{A}\to \mathcal{A}$ associated with symmetric linear n-derivation $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{d}:\mathcal{A}\to \mathcal{A}$ satisfying the condition of $\left(g\right(xy)-g(yx\left)\right)\pm [x,y]\in Z\left(\mathcal{A}\right)$ for all $x,y\in U$, then $U\subseteq Z\left(\mathcal{A}\right)$.

Proof. 

It is given that

$$\left(g\right(xy)-g(yx\left)\right)\pm [x,y]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$$

Replacing y by $x+my$, where $1\le m\le n-1$ in the given condition, we get

$$\mathcal{g}\left(x\right(x+my\left)\right)-\mathcal{g}\left(\right(x+my\left)x\right)\pm [x,x+my]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U$$

which on solving, we have

$$\begin{array}{c}\mathcal{g}\left(xmy\right)-\mathcal{g}\left(myx\right)+\underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{x^2,\dots ,x^2}},\underset{t-\mathrm{times}}{\underbrace{xmy,\dots ,xmy}})-\underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{x^2,\dots ,x^2}},\underset{t-\mathrm{times}}{\underbrace{myx,\dots ,myx}})\hfill \\ \hfill \pm [x,my]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.\end{array}$$

(1)

By using hypothesis, we get

$$\underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{x^2,\dots ,x^2}},\underset{t-\mathrm{times}}{\underbrace{xmy,\dots ,xmy}})-\underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{x^2,\dots ,x^2}},\underset{t-\mathrm{times}}{\underbrace{myx,\dots ,myx}})\pm [x,my]\in Z\left(\mathcal{A}\right)$$

for all $x,y\in U$. Making use of Lemma 3, we see that

$$\mathcal{G}(x^2,\dots ,x^2,xy)-\mathcal{G}(x^2,\dots ,x^2,yx)\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$$

(2)

For $1\le m\le n$, (1) can also be written as

$$\begin{array}{c}{m}^n\mathcal{g}\left(xy\right)-m^n\mathcal{g}\left(yx\right)+\underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{x^2,\dots ,x^2}},\underset{t-\mathrm{times}}{\underbrace{xmy,\dots ,xmy}})-\underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{x^2,\dots ,x^2}},\underset{t-\mathrm{times}}{\underbrace{myx,\dots ,myx}})\hfill \\ \hfill \pm [x,my]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.\end{array}$$

Agai making use of Lemma 3, we have

$$n\{\mathcal{G}(x^2,\dots ,x^2,xy)-\mathcal{G}(x^2,\dots ,x^2,yx)\}\pm [x,y]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$$

(3)

From (2) and (3), we get $[x,y]\in Z\left(\mathcal{A}\right)$ for all $x,y\in U$. As every $C^{*}$-algebra is a semiprime ring, using Lemma 2, we get $U\subseteq Z\left(\mathcal{A}\right)$. □

Theorem 2. 

For any fixed integer $n\ge 2$, let $\mathcal{A}$ be a $C^{*}$-algebra, U be a square closed Lie ideal of $\mathcal{A}$. If $\mathcal{A}$ admits a nonzero symmetric linear generalized n-derivation $\mathcal{G}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{g}:\mathcal{A}\to \mathcal{A}$ associated with symmetric linear n-derivation $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{d}:\mathcal{A}\to \mathcal{A}$ satisfying the condition of $\left(g\right(x)\pm g(y\left)\right)\pm x\circ y\in Z\left(\mathcal{A}\right)$ for all $x,y\in U$, then $U\subseteq Z\left(\mathcal{A}\right)$.

Proof. 

Suppose on the contrary that $U⊈Z\left(\mathcal{A}\right)$. We have given that

$$\left(g\right(x)-g(y\left)\right)\pm x\circ y\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$$

Replacing y by $x+my$, where $z\in U$ and $1\le m\le n-1$ in the given condition, we get

$$\mathcal{g}\left(x\right)\pm \mathcal{g}(x+my)\pm (x\circ x+my)\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U$$

which on solving, we have

$$\begin{array}{c}\hfill \mathcal{g}\left(x\right)\pm \mathcal{g}\left(x\right)\mathcal{g}\left(my\right)\pm \underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{x,\dots ,x}},\underset{t-\mathrm{times}}{\underbrace{my,\dots ,my}})\pm x\circ x\pm x\circ my\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U.\end{array}$$

(4)

Using the given condition, we get

$$\mathcal{g}\left(x\right)\pm x^2\pm \underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{x,\dots ,x}},\underset{t-\mathrm{times}}{\underbrace{my,\dots ,my}})\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U.$$

Multiply the above equation by m which implies that

$$m{A}_1(x,y)+m^2{A}_2(x,y)+\cdots +m^{n-1}{A}_{n-1}(x,y)\in Z\left(\mathcal{A}\right)$$

for all $x,y,z\in U$ where $A_t(x,y)$ represents the term in which z appears t-times.

Making use of Lemma 3, we see that

$$\mathcal{G}(x,\dots ,x,y)\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$$

Replace y by x, we get

$$\mathcal{g}\left(x\right)\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,\in U.$$

From hypothesis, we have $x\circ y\in Z\left(\mathcal{A}\right)$ for all $x,y\in U$. Again replace x by $yx$, we have $y(x\circ y)\in Z\left(\mathcal{A}\right)$ which imply $\left[y\right(x\circ y),z]\in Z\left(\mathcal{A}\right)$. On solving, we get $[y,z](x\circ y)=0$ for all $x,y,x\in U$. Again replace x by $xz$, we have $[y,z]x[z,y]=0$ for all $x,y,z\in U$. By Lemma 1, we have $[z,y]=0$ for all $y,z\in U$. Again using Lemma 2, we get $U\subseteq Z\left(\mathcal{A}\right)$, which is a contradiction. □

Theorem 3. 

Let $\mathcal{A}$ be a $C^{*}$-algebra, U be a square closed Lie ideal of $\mathcal{A}$. Suppose that $\mathcal{A}$ admits a nonzero symmetric linear generalized n-derivation $\mathcal{G}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{g}:\mathcal{A}\to \mathcal{A}$ associated with symmetric linear n-derivation $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{d}:\mathcal{A}\to \mathcal{A}$ satisfying $\mathcal{g}\left(x^2\right)\pm x^2=0$ for all $x\in U$, then $U\subseteq Z\left(\mathcal{A}\right)$.

Proof. 

Suppose on the contrary that $U⊈Z\left(\mathcal{A}\right)$. We have given that $\mathcal{G}:{\mathcal{A}}^n\to \mathcal{A}$ be symmetric linear generalized n-derivations associated with $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ of a $C^{*}$-algebra $\mathcal{A}$ such that $\mathcal{g}\left(x^2\right)\pm x^2=0$ for all $x\in U$. Therefore, $\mathcal{A}$ is semiprime as $\mathcal{A}$ is a $C^{*}$-algebra. Now replacing x by $x+my$, $y\in U$ for $1\le m\le n-1$ in the given condition, we get

$$\mathcal{g}{(x+my)}^2\pm {(x+my)}^2=0\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$$

Further solving, we have

$$\begin{array}{c}\mathcal{g}(x^2)+\mathcal{g}\left(m(xy+yx)\right)+\underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{x^2,\dots ,x^2}},\underset{t-\mathrm{times}}{\underbrace{m(xy+yx),\dots ,m(xy+yx)}})+\hfill \\ \mathcal{g}\left({\left(my\right)}^2\right)+\underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{x^2+m(xy+yx),\dots ,x^2+m(xy+yx)}},\underset{t-\mathrm{times}}{\underbrace{{\left(my\right)}^2,\dots ,{\left(my\right)}^2}})\\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\pm x^2\pm {\left(my\right)}^2\pm m(xy+yx)=0\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.\end{array}$$

In accordance of the given condition and Lemma 3, we get

$$n\mathcal{G}(x^2,\dots ,x^2,xy+yx)\pm (xy+yx)=0\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$$

Replacing y by x, we find that

$$2n\mathcal{g}\left(x^2\right)\pm 2x^2=0,$$

or

$$x^2=0.$$

This implies that $xy+yx=0$ for all $x,y\in U$. Replacing y by $yz$, where $z\in U$, we get $[x,y]z=0$. Again replacing z by $z[x,y]$, we get $[x,y]z[x,y]=0$ for all $x,y,z\in U$. Using the Lemma 1, we get $[x,y]=0$ for all $x,y\in U$. By Lemma 2, we get $U\subseteq Z\left(\mathcal{A}\right)$, a contradiction. □

Corollary 1. 

For any fixed integer $n\ge 2$, let $\mathcal{A}$ be a $C^{*}$-algebra, U be a square closed Lie ideal of $\mathcal{A}$. If $\mathcal{A}$ admits a nonzero symmetric linear generalized n-derivation $\mathcal{G}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{g}:\mathcal{A}\to \mathcal{A}$ associated with symmetric linear n-derivation $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{d}:\mathcal{A}\to \mathcal{A}$ satisfying $\mathcal{g}(x\circ y)\pm x\circ y=0$ for all $x,y\in U$ then $U\subseteq Z\left(\mathcal{A}\right)$.

Theorem 4. 

For any fixed integer $n\ge 2$, let $\mathcal{A}$ be a $C^{*}$-algebra, U be a square closed Lie ideal of $\mathcal{A}$. If $\mathcal{A}$ admits a nonzero symmetric linear generalized n-derivation $\mathcal{G}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{g}:\mathcal{A}\to \mathcal{A}$ associated with symmetric linear n-derivation $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{d}:\mathcal{A}\to \mathcal{A}$ satisfying one of the following conditions:

(i) 

$\left[\mathcal{g}\right(x),\mathcal{g}(y\left)\right]-[x,y]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U$

(ii) 

$\left[\mathcal{g}\right(x),\mathcal{g}(y\left)\right]-[y,x]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$

Then $U\subseteq Z\left(\mathcal{A}\right)$.

Proof. (i) Given that

$$\left[\mathcal{g}\right(x),\mathcal{g}(y\left)\right]-[x,y]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$$

(5)

Consider a positive integer m; $1\le m\le n-1$. Replacing y by $y+mz$, where $z\in U$ in (5), we get

$$\left[\mathcal{g}\right(x),\mathcal{g}(y+mz\left)\right]-[x,y+mz]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U.$$

On further solving, we get

$$\begin{array}{c}[\mathcal{g}\left(x\right),\mathcal{g}\left(y\right)]+[\mathcal{g}\left(x\right),\mathcal{g}\left(mz\right)]+[\mathcal{g}\left(x\right),\underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{y,\dots ,y}},\underset{t-\mathrm{times}}{\underbrace{mz,\dots ,mz}})]-\hfill \\ \hfill -[x,mz]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U.\end{array}$$

On taking account of hypothesis, we see that

$$m{A}_1(x,y,z)+m^2{A}_2(x,y,z)+\cdots +m^{n-1}{A}_{n-1}(x,y,z)\in Z\left(\mathcal{A}\right)$$

where $A_t(x,y,z)$ represents the term in which z appears t-times.

Using Lemma 3, we have

$$\left[\mathcal{g}\right(x),\mathcal{G}(y,\dots ,y,z\left)\right]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U.$$

In particular, for $z=y$, we get

$$\left[\mathcal{g}\right(x),\mathcal{g}(y\left)\right]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$$

Now using the given condition, we find that

$$[x,y]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$$

From Lemma 2, $U\subseteq Z\left(\mathcal{A}\right)$.

$\left(ii\right)$ Follows from the first implication with a slight modification. □

Corollary 2. 

For any fixed integer $n\ge 2$, let $\mathcal{A}$ be a $C^{*}$-algebra, U be a square closed Lie ideal of $\mathcal{A}$. If $\mathcal{A}$ admits a nonzero symmetric linear generalized n-derivation $\mathcal{G}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{g}:\mathcal{A}\to \mathcal{A}$ associated with symmetric linear n-derivation $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{d}:\mathcal{A}\to \mathcal{A}$ satisfying one of the following conditions:

(i) 

$\mathcal{g}\left(x\right)\mathcal{g}\left(y\right)\pm xy\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U$

(ii) 

$\mathcal{g}\left(x\right)\mathcal{g}\left(y\right)\pm yx\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$

Then $U\subseteq Z\left(\mathcal{A}\right)$.

Corollary 3. 

For any fixed integer $n\ge 2$, let $\mathcal{A}$ be a $C^{*}$-algebra, U be a square closed Lie ideal of $\mathcal{A}$. If $\mathcal{A}$ admits a nonzero symmetric linear generalized n-derivation $\mathcal{G}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{g}:\mathcal{A}\to \mathcal{A}$ associated with symmetric linear n-derivation $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{d}:\mathcal{A}\to \mathcal{A}$ satisfying one of the following conditions:

(i) 

$\left[\mathcal{g}\right(x),\mathcal{g}(y\left)\right]=[x,y]\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U$

(ii) 

$\left[\mathcal{g}\right(x),\mathcal{g}(y\left)\right]=[y,x]\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$

Then $U\subseteq Z\left(\mathcal{A}\right)$.

Theorem 5. 

For any fixed integer $n\ge 2$, let $\mathcal{A}$ be a $C^{*}$-algebra, U be a square closed Lie ideal of $\mathcal{A}$. If $\mathcal{A}$ admits a nonzero symmetric linear generalized n-derivation $\mathcal{G}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{g}:\mathcal{A}\to \mathcal{A}$ associated with symmetric linear n-derivation $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{d}:\mathcal{A}\to \mathcal{A}$ satisfying the condition $\mathcal{g}(x\circ y)\pm [x,y]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U$, then $U\subseteq Z\left(\mathcal{A}\right)$.

Proof. 

Replacing y by $y+mz$ for $1\le m\le n-1$, $z\in U$ in the given condition, we get

$$\mathcal{g}(x\circ (y+mz\left)\right)\pm [x,y+mz]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U.$$

On further solving and using the specified condition, we get

$$\underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{x\circ y,\dots ,x\circ y}},\underset{t-\mathrm{times}}{\underbrace{x\circ mz,\dots ,x\circ mz}})\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U$$

which implies that

$$m{A}_1(x,y,z)+m^2{A}_2(x,y,z)+\cdots +m^{n-1}{A}_{n-1}(x,y,z)\in Z\left(\mathcal{A}\right)$$

for all $x,y,z\in U$ where $A_t(x,y,z)$ represents the term in which z appears t-times. Using Lemma 3, we get

$$\mathcal{G}(x\circ y,\dots ,x\circ y,x\circ z)\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U.$$

(6)

For $z=y$, we get $\mathcal{g}(x\circ y)\in Z\left(\mathcal{A}\right)$ then our hypothesis reduces to $[x,y]\in Z\left(\mathcal{A}\right)$. Using the Lemma 2, we get $U\subseteq Z\left(\mathcal{A}\right)$. □

Corollary 4. 

For any fixed integer $n\ge 2$, let $\mathcal{A}$ be a $C^{*}$-algebra, U be a square closed Lie ideal of $\mathcal{A}$. If $\mathcal{A}$ admits a nonzero symmetric linear generalized n-derivation $\mathcal{G}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{g}:\mathcal{A}\to \mathcal{A}$ associated with symmetric linear n-derivation $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{d}:\mathcal{A}\to \mathcal{A}$ satisfying the condition $\mathcal{d}(x\circ y)\pm [x,y]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U$, then $U\subseteq Z\left(\mathcal{A}\right)$.

Theorem 6. 

For any fixed integer $n\ge 2$, let $\mathcal{A}$ be a $C^{*}$-algebra, U be a square closed Lie ideal of $\mathcal{A}$. If $\mathcal{A}$ admits a nonzero symmetric linear generalized n-derivation $\mathcal{G}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{g}:\mathcal{A}\to \mathcal{A}$ associated with symmetric linear n-derivation $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{d}:\mathcal{A}\to \mathcal{A}$ satisfying one of the following conditions:

(i) 

$\mathcal{g}\left(\right[x,y\left]\right)\pm \mathcal{g}\left(x\right)\pm [x,y]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U$

(ii) 

$\mathcal{g}\left(\right[x,y\left]\right)\pm \mathcal{g}\left(y\right)\pm [x,y]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$

Then $U\subseteq Z\left(\mathcal{A}\right)$.

Proof.$\left(i\right)$ Given that

$$\mathcal{g}\left(\right[x,y\left]\right)\pm \mathcal{g}\left(x\right)\pm [x,y]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$$

Replacing x by $x+mz$, where $z\in U$ and $1\le m\le n-1$ in the given condition, we get

$$\mathcal{g}\left(\right[x+mz,y\left]\right)\pm \mathcal{g}(x+mz)\pm [x+mz,y]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U$$

which on solving and using hypothesis, we obtain

$$\begin{array}{c}\underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{[x,y],\dots ,[x,y]}},\underset{t-\mathrm{times}}{\underbrace{[mz,y],\dots ,[mz,y]}})\hfill \\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\pm \underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{x,\dots ,x}},\underset{t-\mathrm{times}}{\underbrace{mz,\dots ,mz}})\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U\end{array}$$

which implies that

$$m{A}_1(x,y,z)+m^2{A}_2(x,y,z)+\cdots +m^{n-1}{A}_{n-1}(x,y,z)\in Z\left(\mathcal{A}\right)$$

for all $x,y,z\in U$ where $A_t(x,y,z)$ represents the term in which z appears t-times.

Making use of Lemma 3 and torsion restriction, we see that

$$\mathcal{G}\left(\right[x,y],\dots ,[x,y],[z,y\left]\right)\pm \mathcal{G}(x,\dots ,x,z)\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U.$$

Replace z by x to get

$$\mathcal{g}\left(\right[x,y\left]\right)\pm \mathcal{g}\left(x\right)\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U.$$

Hence, by using the given condition, we find that $[x,y]\in Z\left(\mathcal{A}\right)$. On taking account of Lemma 2, we get $U\subseteq Z\left(\mathcal{A}\right)$.

$\left(ii\right)$ Given that

$$\mathcal{g}\left(\right[x,y\left]\right)\pm \mathcal{g}\left(x\right)\pm [x,y]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$$

Replacing y by $y+mz$, where $z\in U$ and $1\le m\le n-1$ in the given condition, we get

$$\mathcal{g}\left(\right[x,y+mz\left]\right)\pm \mathcal{g}(y+mz)\pm [x,y+mz]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U$$

which on solving and using hypothesis, we obtain

$$\begin{array}{c}\underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{[x,y],\dots ,[x,y]}},\underset{t-\mathrm{times}}{\underbrace{[x,mz],\dots ,[x,mz]}})\hfill \\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\pm \underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{y,\dots ,y}},\underset{t-\mathrm{times}}{\underbrace{mz,\dots ,mz}})\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U\end{array}$$

which implies that

$$m{A}_1(x,y,z)+m^2{A}_2(x,y,z)+\cdots +m^{n-1}{A}_{n-1}(x,y,z)\in Z\left(\mathcal{A}\right)$$

for all $x,y,z\in U$ where $A_t(x,y,z)$ represents the term in which z appears t-times.

Making use of Lemma 3 and torsion restriction, we see that

$$\mathcal{G}\left(\right[x,y],\dots ,[x,y],[x,z\left]\right)\pm \mathcal{G}(y,\dots ,y,z)\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U.$$

Replace z by y to get

$$\mathcal{g}\left(\right[x,y\left]\right)\pm \mathcal{g}\left(y\right)\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U.$$

Hence, by using the given condition, we find that $[x,y]\in Z\left(\mathcal{A}\right)$. On taking account of Lemma 2, we get $U\subseteq Z\left(\mathcal{A}\right)$.

$\left(ii\right)$ Follows from the first implication with a slight modification. □

Corollary 5. 

For any fixed integer $n\ge 2$, let $\mathcal{A}$ be a $C^{*}$-algebra, U be a square closed Lie ideal of $\mathcal{A}$. If $\mathcal{A}$ admits a nonzero symmetric linear n-derivation $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{d}:\mathcal{A}\to \mathcal{A}$ satisfying one of the following conditions:

(i) 

$\mathcal{d}\left(\right[x,y\left]\right)\pm \mathcal{d}\left(x\right)\pm [x,y]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U$

(ii) 

$\mathcal{d}\left(\right[x,y\left]\right)\pm \mathcal{d}\left(y\right)\pm [x,y]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U$

Then $U\subseteq Z\left(\mathcal{A}\right)$.

Theorem 7. 

For any fixed integer $n\ge 2$, let $\mathcal{A}$ be a $C^{*}$-algebra, U be a square closed Lie ideal of $\mathcal{A}$. If $\mathcal{A}$ admits a nonzero symmetric linear generalized n-derivation $\mathcal{G}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{g}:\mathcal{A}\to \mathcal{A}$ associated with symmetric linear n-derivation $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{d}:\mathcal{A}\to \mathcal{A}$ satisfying one of the following conditions:

(i) 

$\mathcal{g}\left(x\right)\circ \mathcal{g}\left(y\right)\pm x\circ y\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U$

(ii) 

$\mathcal{g}\left(x\right)\circ \mathcal{g}\left(y\right)\pm [x,y]\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U$

Then $U\subseteq Z\left(\mathcal{A}\right)$.

Proof. (i) Suppose on the contrary that $U⊈Z\left(\mathcal{A}\right)$. It is given that

$$\mathcal{g}\left(x\right)\circ \mathcal{g}\left(y\right)\pm x\circ y\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$$

Replacing y by $y+mz$, where $z\in U$ and $1\le m\le n-1$ in the given condition, we get

$$\mathcal{g}\left(x\right)\circ \mathcal{g}(y+mz)\pm x\circ (y+mz)\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U$$

which on solving, we have

$$\begin{array}{c}\mathcal{g}\left(x\right)\circ \mathcal{g}\left(y\right)+\mathcal{g}\left(x\right)\circ \mathcal{g}\left(mz\right)+\mathcal{g}\left(x\right)\circ \underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathcal{G}(\underset{(n-t)-\mathrm{times}}{\underbrace{y,\dots ,y}},\underset{t-\mathrm{times}}{\underbrace{mz,\dots ,mz}})\hfill \\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\pm x\circ y\pm x\circ mz\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U.\end{array}$$

By using hypothesis, we get

$$\mathcal{g}\left(x\right)\circ \underset{t=1}{\overset{n-1}{{\sum }^n}}{C}_t\mathfrak{D}(\underset{(n-t)-\mathrm{times}}{\underbrace{y,\dots ,y}},\underset{t-\mathrm{times}}{\underbrace{mz,\dots ,mz}})\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U$$

which implies that

$$m{A}_1(x,y,z)+m^2{A}_2(x,y,z)+\cdots +m^{n-1}{A}_{n-1}(x,y,z)\in Z\left(\mathcal{A}\right)$$

for all $x,y,z\in U$ where $A_t(x,y,z)$ represents the term in which z appears t-times.

Making use of Lemma 3, we see that

$$\mathcal{g}\left(x\right)\circ \mathcal{G}(y,\dots ,y,z)\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U.$$

In particular, $z=y$, we get

$$\mathcal{g}\left(x\right)\circ \mathcal{g}\left(y\right)\in Z\left(\mathcal{A}\right)\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U.$$

Hence, by using the given condition, we find that $x\circ y\in Z\left(\mathcal{A}\right)$ for all $x,y\in U$. Replacing x by $yx$, we get $y(x\circ y)\in Z\left(\mathcal{A}\right)$ for all $x,y\in U$. We can also write it as

$$\left[y\right(x\circ y),z]\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y,z\in U$$

which on solving, we get $[y,z]x\circ y=0$ for all $x,y,z\in U$. Again replace x by $xz$ and using the same equation, we get $[y,z]x[z,y]=0$ for all $x,y,z\in U$. Using Lemma 1, we have $[z,y]=0$ for all $z,y\in U$. By Lemma 2, we have $U\subseteq Z\left(\mathcal{A}\right)$ which is a contradiction.

$\left(ii\right)$ Proceeding in the same way as in $\left(i\right)$, we conclude. □

Corollary 6. 

For any fixed integer $n\ge 2$, let $\mathcal{A}$ be a $C^{*}$-algebra, U be a square closed Lie ideal of $\mathcal{A}$. If $\mathcal{A}$ admits a nonzero symmetric linear n-derivation $\mathfrak{D}:{\mathcal{A}}^n\to \mathcal{A}$ with trace $\mathcal{d}:\mathcal{A}\to \mathcal{A}$ satisfying one of the following conditions:

(i) 

$\mathcal{d}\left(x\right)\circ \mathcal{d}\left(y\right)\pm x\circ y=0\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U$

(ii) 

$\mathcal{d}\left(x\right)\circ \mathcal{d}\left(y\right)\pm [x,y]=0\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x,y\in U$

then $U\subseteq Z\left(\mathcal{A}\right)$.