In this section, we write a unit element of algebra \(\mathcal {A}\) by *e*.

### Theorem 3.1

*Let*
\(\mathcal {A}\)
*be a semiprime unital Banach algebra*. *Suppose that*
\(\delta: \mathcal {A}\to \mathcal {A}\)
*is a mapping subject to the inequality* (2.8) *and for some*
\(\varepsilon\ge0\),

$$\begin{aligned} \bigl\Vert \delta\bigl(x^{2}\bigr)-2x\delta(x)\bigr\Vert \leq\varepsilon \end{aligned}$$

(3.1)

*for all*
\(x \in \mathcal {A}\). *Then*
*δ*
*is a linear derivation which maps*
\(\mathcal {A}\)
*into the intersection of its center*
\(Z(\mathcal {A})\)
*and its radical*
\(\operatorname{rad}(\mathcal {A})\).

### Proof

Employing the same way in the proof Theorem 2.4, we find that *δ* is linear. By linearization of (3.1) and additivity of *δ*, we get

$$\begin{aligned} \bigl\Vert \delta\bigl(x^{2}\bigr)+\delta(xy)+ \delta(yx)+\delta\bigl(y^{2}\bigr)-2x\delta (x)-2x\delta (y)-2y \delta(x)-2y\delta(y)\bigr\Vert \leq\varepsilon \end{aligned}$$

(3.2)

for all \(x,y \in \mathcal {A}\). Substituting −*x* for *x* in (3.2), we have

$$\begin{aligned} \bigl\Vert \delta\bigl(x^{2}\bigr)-\delta(xy)- \delta(yx)+\delta\bigl(y^{2}\bigr)-2x\delta (x)+2x\delta (y)+2y \delta(x)-2y\delta(y)\bigr\Vert \leq\varepsilon \end{aligned}$$

(3.3)

for all \(x,y \in \mathcal {A}\). Equations (3.2) and (3.3) yield

$$\begin{aligned} &\bigl\Vert 2\delta(xy)+2\delta(yx)-4x\delta(y)-4y\delta(x)\bigr\Vert \\ &\quad\leq\bigl\Vert \delta\bigl(x^{2}\bigr)+\delta(xy)+\delta(yx)+ \delta \bigl(y^{2}\bigr)-2x\delta (x)-2x\delta(y)-2y\delta(x)-2y \delta(y)\bigr\Vert \\ &\qquad{} +\bigl\Vert \delta\bigl(x^{2}\bigr)-\delta(xy)-\delta(yx)+ \delta \bigl(y^{2}\bigr)-2x\delta (x)+2x\delta(y)+2y\delta(x)-2y \delta(y)\bigr\Vert \leq2\varepsilon \end{aligned}$$

for all \(x,y \in \mathcal {A}\). We have therefore

$$\begin{aligned} \bigl\Vert \delta(xy+yx)-2x\delta(y)-2y\delta(x)\bigr\Vert \leq \varepsilon \quad\mbox{for all } x,y \in \mathcal {A}. \end{aligned}$$

(3.4)

Putting \(xy+yx\) for *y* in (3.4), we obtain

$$\begin{aligned} \bigl\Vert \delta\bigl(x(xy+yx)+(xy+yx)x\bigr)-2x \delta(xy+yx)-2(xy+yx)\delta (x)\bigr\Vert \leq \varepsilon \end{aligned}$$

(3.5)

for all \(x,y \in \mathcal {A}\). On the other hand, we have from (3.4) and the equation

$$x(xy+yx)+(xy+yx)x=x^{2}y+yx^{2}+2xyx $$

the result

$$\begin{aligned} &\bigl\Vert \delta\bigl(x(xy+yx)+(xy+yx)x\bigr)-2x^{2} \delta(y)-2y\delta \bigl(x^{2}\bigr)-2\delta (xyx)\bigr\Vert \\ &\quad= \bigl\Vert \delta\bigl(x^{2}y+yx^{2} \bigr)-2x^{2}\delta(y)-2y\delta \bigl(x^{2}\bigr)\bigr\Vert \leq \varepsilon \end{aligned}$$

(3.6)

for all \(x,y \in \mathcal {A}\). By comparing (3.5) and (3.6), we arrive at

$$\begin{aligned} &\bigl\Vert 2x\delta(xy+yx)+2(xy+yx)\delta(x)-2x^{2} \delta(y)-2y\delta \bigl(x^{2}\bigr)-2\delta(xyx)\bigr\Vert \\ &\quad\leq\bigl\Vert \delta\bigl(x(xy+yx)+(xy+yx)x\bigr)-2x^{2} \delta (y)-2y\delta \bigl(x^{2}\bigr)-2\delta(xyx)\bigr\Vert \\ &\qquad{} +\bigl\Vert \delta\bigl(x(xy+yx)+(xy+yx)x\bigr)-2x\delta (xy+yx)-2(xy+yx)\delta (x)\bigr\Vert \leq2\varepsilon \end{aligned}$$

(3.7)

for all \(x,y \in \mathcal {A}\). Applying equation (3.7) with (3.1) and (3.4), we have

$$\begin{aligned} &\bigl\Vert 2x^{2}\delta(y)+6xy\delta(x)-2yx \delta(x)-2\delta (xyx)\bigr\Vert \\ &\quad\leq2\|x\|\bigl\Vert \delta(xy+yx)-2x\delta(y)-2y\delta (x)\bigr\Vert +2\|y\|\bigl\Vert \delta \bigl(x^{2}\bigr)-2x\delta(x)\bigr\Vert \\ &\qquad{} +\bigl\Vert 2x\delta(xy+yx) +2(xy+yx)\delta(x)-2x^{2} \delta(y)-2y\delta\bigl(x^{2}\bigr)-2\delta (xyz)\bigr\Vert \\ &\quad\leq2\bigl(\Vert x\Vert +\|y\|+1\bigr)\varepsilon \end{aligned}$$

(3.8)

for all \(x,y \in \mathcal {A}\). Letting \(x=nx, y=ny\) in (3.8) and then dividing the resulting inequality by \(n^{3}\), we get

$$\begin{aligned} \bigl\Vert 2x^{2}\delta(y)+6xy\delta(x)-2yx\delta(x)-2 \delta (xyx)\bigr\Vert \leq 2 \biggl(\frac{\|x\|}{n^{2}}+\frac{\|y\|}{n^{2}}+ \frac{1}{n^{3}} \biggr)\varepsilon \end{aligned}$$

(3.9)

for all \(x,y \in \mathcal {A}\) and all positive integers *n*. Taking the limit \(n \to\infty\) of (3.9), it is reduced to the equation

$$\begin{aligned} \delta(xyx)=x^{2}\delta(y)+3xy\delta(x)-yx\delta(x) \quad\mbox{for all } x,y \in \mathcal {A}. \end{aligned}$$

(3.10)

Putting \(x=y=z=e\) in (3.10), we get \(\delta(e)=0\). Again, considering \(y=e\) in (3.10), we easily prove that

$$\delta\bigl(x^{2}\bigr)=2x\delta(x) \quad\mbox{for all } x \in \mathcal {A}. $$

This means that *δ* is a linear left Jordan derivation.

On the other hand, from Vukman’s result [16], we see that *δ* is a linear derivation with \(\delta(\mathcal {A}) \subseteq Z(\mathcal {A})\). Since \(Z(\mathcal {A})\) is a commutative Banach algebra, the Singer-Wermer theorem tells us that \(\delta|_{Z(\mathcal {A})}\) maps \(Z(\mathcal {A})\) into \(\operatorname{rad}(Z(\mathcal {A}))=Z(\mathcal {A}) \cap \operatorname{rad}(\mathcal {A})\) and thus \(\delta^{2}(\mathcal {A}) \subseteq \operatorname{rad}(\mathcal {A})\). Using the semiprimeness of \(\operatorname{rad}(\mathcal {A})\) as well as the identity

$$2\delta(x)y\delta(x)= \delta^{2}(xyx)-x\delta^{2}(yx)-\delta ^{2}(xy)x+x\delta^{2}(y)x \quad(x,y \in \mathcal {A}), $$

we have \(\delta(\mathcal {A}) \subseteq \operatorname{rad}(\mathcal {A})\). Therefore \(\delta(\mathcal {A}) \subseteq Z(\mathcal {A}) \cap \operatorname{rad}(\mathcal {A})\), which concludes the proof. □

As consequences of Theorem 3.1, we get the following.

### Corollary 3.2

*Let*
\(\mathcal {A}\)
*be a unital semisimple Banach algebra*. *Assume that a mapping*
\(\delta : \mathcal {A}\to \mathcal {A}\)
*satisfies the assumptions of Theorem*
3.1. *Then*
*δ*
*is identically zero*.

Now we consider the result which is needed in the following theorems.

### Lemma 3.3

*Let*
\(\mathcal {A}\)
*be a Banach algebra*. *Suppose that*
\(\mathcal{L}:\mathcal {A}\times \mathcal {A}\to \mathcal {A}\)
*is a bilinear mapping and that*
*ξ*
*and*
*η*
*are mappings satisfying*
\(\mathcal{L}(x,y)= x\xi(y)+ y\eta(x)\)
*for all*
\(x,y \in \mathcal {A}\). *If*
\(\mathcal {A}\)
*is semiprime or unital*, *then*
*ξ*
*and*
*η*
*are linear mappings*.

### Proof

Note that, for all \(x,y \in \mathcal {A}\) and all \(\lambda \in \mathbb{C}\),

$$\begin{aligned} x\xi(\lambda y)+ \lambda y\eta(x)=\mathcal{L}(x, \lambda y)=\lambda \mathcal{L}(x,y)=\lambda \bigl( x \xi(y)+ y\eta(x)\bigr). \end{aligned}$$

Hence we see that, for all \(x,y \in \mathcal {A}\),

$$\begin{aligned} x\bigl(\xi(\lambda y)-\lambda \xi(y)\bigr)=0. \end{aligned}$$

(3.11)

If \(\mathcal {A}\) is unital, then we see that \(\xi(\lambda y)=\lambda \eta(y)\) by letting \(x=e\) in (3.11).

If \(\mathcal {A}\) is nonunital, then \(\xi(\lambda y)-\lambda \xi(y)\) lies in the right annihilator \(\operatorname{ran}(\mathcal {A})\) of \(\mathcal {A}\). If \(\mathcal {A}\) is semiprime, then \(\operatorname{ran}(\mathcal {A})=0\), so that \(\xi(\lambda y)=\lambda \xi(y)\) for all \(y \in \mathcal {A}\) and all \(\lambda \in\mathbb{C}\).

Observe that, for all \(x,y,z \in \mathcal {A}\),

$$\begin{aligned} x\xi(y+z)+(y+z)\eta(x)&=\mathcal{L}(x, y+z)=\mathcal {L}(x,y)+\mathcal {L}(x,z) \\ &= x\xi(y)+ y\eta(x)+x\xi(z)+ z\eta(x). \end{aligned}$$

Hence \(x(\xi(y+z)-\xi(y)-\xi(z))=0\) for all \(y,z \in \mathcal {A}\). As above, we get \(\xi(x+z)=\xi(x)+\xi(z)\) for all \(x,z \in \mathcal {A}\), so that *ξ* is linear.

Similarly, one can prove that *η* is linear. □

### Theorem 3.4

*Let*
\(\mathcal {A}\)
*be a semiprime Banach algebra*. *Assume that*
\(l \ge3\)
*is a fixed integer and*
\(s_{1},s_{2}, \ldots, s_{l}\)
*are fixed positive real numbers*, *where*
\(s_{1}=\lambda s\) (\(s>1\)), \(s_{2}> 1\)
*and*
\(s_{3}=1\). *Suppose that*
\(\delta: \mathcal {A}\to \mathcal {A}\)
*is a mapping with*
\(\delta(0)=0\)
*such that*, *for some*
\(\varepsilon\ge0\),

$$\begin{aligned} \Biggl\Vert \sum_{j=1}^{l}s_{j} \delta(x_{j})\Biggr\Vert \leq\Biggl\Vert \delta \Biggl(\sum _{j=1}^{l}s_{j}x_{j} \Biggr)\Biggr\Vert +\varepsilon \end{aligned}$$

(3.12)

*for all*
\(x_{1}, x_{2}, \ldots, x_{l} \in \mathcal {A}\)
*and all*
\(\lambda\in\mathbb{T}_{\varepsilon}\), *where*
\(x_{3}= \lambda z\) (\(z \in \mathcal {A}\)) *and*

$$\begin{aligned} \bigl\Vert \delta(xy+yx)-2x\delta(y)-2y\delta(x)\bigr\Vert \leq\theta \end{aligned}$$

(3.13)

*for some*
\(\theta\ge0\)
*and all*
\(x,y \in \mathcal {A}\). *Then*
*δ*
*is a linear derivation which maps*
\(\mathcal {A}\)
*into the intersection of its center*
\(Z(\mathcal {A})\)
*and its radical*
\(\operatorname{rad}(\mathcal {A})\).

### Proof

We let \(x_{4}=x_{5}=\cdots=x_{l}=0\) in (3.12) and then put \(x_{1}=x, x_{2}=y, s_{2}=t\) to have

$$\begin{aligned} \bigl\Vert \lambda s\delta(x)+t\delta(y)+\delta(\lambda z)\bigr\Vert \le \bigl\Vert \delta (\lambda sx+ty+\lambda z)\bigr\Vert +\varepsilon \end{aligned}$$

(3.14)

for all \(x,y,z \in \mathcal {A}\) and all \(\lambda\in\mathbb{T}_{\varepsilon}\). Now we consider \(\lambda=1\) in (3.14). It follows from the result in [14] that there exists a unique additive mapping \(\mathcal{D}: \mathcal {A}\to \mathcal {A}\) defined by

$$\begin{aligned} \mathcal{D}(x):=\lim_{n\to\infty}\frac{\delta(s^{n}x)}{s^{n}} \quad\mbox{for all } x \in \mathcal {A}. \end{aligned}$$

(3.15)

Moreover, \(s\mathcal{D}(\frac{x}{s})=\mathcal{D}(x)\) holds for all \(x \in \mathcal {A}\).

Letting \(x=\frac{x}{s}, y=0\), and \(z=-x\) in (3.14), we find that

$$\begin{aligned} \biggl\Vert \lambda s\delta \biggl(\frac{x}{s} \biggr)-\delta(\lambda x) \biggr\Vert \le \varepsilon \end{aligned}$$

for all \(x\in \mathcal {A}\) and all \(\lambda\in\mathbb{T}_{\varepsilon}\). This implies that

$$\begin{aligned} \lim_{n \to\infty}\frac{1}{s^{n}}\biggl\Vert \lambda s\delta \biggl(\frac {s^{n}x}{s} \biggr)-\delta\bigl(\lambda s^{n}x\bigr)\biggr\Vert \le\lim_{n \to \infty} \frac{\varepsilon}{s^{n}}=0. \end{aligned}$$

Thus \(\lambda s\mathcal{D}(\frac{x}{s})=\mathcal{D}(\lambda x)\), so that \(\mathcal{D}(\lambda x)=\lambda\mathcal{D}(x)\) for all \(x\in \mathcal {A}\) and all \(\lambda\in\mathbb{T}_{\varepsilon}\). Thus we see that \(\mathcal{D}\) is linear [17].

By (3.13), we see that

$$\begin{aligned} \lim_{n \to\infty}\biggl\Vert \frac{\delta(s^{n}(xy+yx))}{s^{n}}- 2x \delta (y)-2 y \frac{\delta(s^{n}x)}{s^{n}}\biggr\Vert \le\lim_{n \to\infty }\frac {\theta}{s^{n}} =0. \end{aligned}$$

Hence we arrive at

$$\mathcal{D}(xy+yx)=2x\delta(y)+2y\mathcal{D}(x) \quad\mbox{for all } x,y \in \mathcal {A}. $$

It follows from Lemma 3.3 that *δ* is linear. Then we have by (3.15) that \(\mathcal {D}=\delta\). Therefore

$$\delta\bigl(x^{2}\bigr)=2x\delta(x) \quad\mbox{for all } x \in \mathcal {A}. $$

That is, *δ* is a linear left Jordan derivation.

The remainder of the proof can be carried out similarly to the corresponding part of Theorem 3.1. □

### Theorem 3.5

*Let*
\(\mathcal {A}\)
*be a unital Banach algebra*. *Assume that*
\(l \ge3\)
*is a fixed integer and*
\(s_{1},s_{2}, \ldots, s_{l}\)
*are fixed positive real numbers*, *where*
\(s_{1}=\lambda s\) (\(s>1\)), \(s_{2}> 1\)
*and*
\(s_{3}=1\). *Suppose that*
\(\delta: \mathcal {A}\to \mathcal {A}\)
*is a mapping with*
\(\delta(0)=0\)
*such that*, *for some*
\(\varepsilon\ge0\),

$$\begin{aligned} \Biggl\Vert \sum_{j=1}^{l}s_{j} \delta(x_{j})\Biggr\Vert \leq\Biggl\Vert \delta \Biggl(\sum _{j=1}^{l}s_{j}x_{j} \Biggr)\Biggr\Vert +\varepsilon \end{aligned}$$

(3.16)

*for all*
\(x_{1}, x_{2}, \ldots, x_{3} \in \mathcal {A}\)
*and all*
\(\lambda\in S:=\{1,i\}\), *where*
\(x_{3}= \lambda z\) (\(z \in \mathcal {A}\)) *and* (3.13). *If*
\(\delta(p e)=0\)
*for all irrational numbers*
*p*, *then*
*δ*
*is a linear left Jordan derivation*. *In this case*
\(\mathcal {A}\)
*is a semiprime unital Banach algebra*, *δ*
*is a linear derivation which maps*
\(\mathcal {A}\)
*into the intersection of its center*
\(Z(\mathcal {A})\)
*and its radical*
\(\operatorname{rad}(\mathcal {A})\).

### Proof

We first consider \(\lambda=1\) in (3.16). We see by the result in [14] that there is a unique additive mapping \(\mathcal{D}: \mathcal {A}\to \mathcal {A}\) defined by (3.15). In addition, \(s\mathcal{D}(\frac{x}{s})=\mathcal{D}(x)\) for all \(x \in \mathcal {A}\).

Also we set \(\lambda=i\) in (3.16). And we take \(x_{4}=x_{5}=\cdots=x_{l}=0\) in (3.16) and then let \(x_{1}=x, x_{2}=y, s_{2}=t\) to have

$$\begin{aligned} \bigl\Vert i s\delta(x)+t\delta(y)+\delta(i z)\bigr\Vert \le \bigl\Vert \delta(i sx+ty+i z)\bigr\Vert +\varepsilon \end{aligned}$$

(3.17)

for all \(x,y,z \in \mathcal {A}\). Putting \(x=\frac{x}{s}, y=0\) and \(z=-x\) in (3.17), we obtain

$$\begin{aligned} \biggl\Vert i s\delta \biggl(\frac{x}{s} \biggr)-\delta(i x)\biggr\Vert \le \varepsilon \end{aligned}$$

for all \(x\in \mathcal {A}\), which shows that

$$\begin{aligned} \lim_{n \to\infty}\frac{1}{s^{n}}\biggl\Vert i s \delta \biggl( \frac {s^{n}x}{s} \biggr)-\delta\bigl(i s^{n}x\bigr)\biggr\Vert \le \lim_{n \to\infty} \frac {\varepsilon}{s^{n}}=0. \end{aligned}$$

Hence \(i s\mathcal{D}(\frac{x}{s})=\mathcal{D}(i x)\). So we have \(\mathcal{D}(i x)=i \mathcal{D}(x)\) for all \(x\in \mathcal {A}\).

We have by (3.13)

$$\begin{aligned} \lim_{n \to\infty}\biggl\Vert \frac{\delta(s^{2n}(xy+yx))}{s^{2n}}- 2x \frac {\delta(s^{n}y)}{s^{n}}-2 y \frac{\delta(s^{n}x)}{s^{n}}\biggr\Vert \le \lim _{n \to\infty}\frac{\theta}{s^{2n}} =0. \end{aligned}$$

This implies that

$$\begin{aligned} \mathcal{D}(xy+yx)=2x\mathcal{D}(y)+2y\mathcal{D}(x) \quad\mbox{for all } x,y \in \mathcal {A}. \end{aligned}$$

(3.18)

Again, by virtue of (3.13), we see that

$$\begin{aligned} \lim_{n \to\infty}\biggl\Vert \frac{\delta(s^{n}(xy+yx))}{s^{n}}- 2x \delta (y)-2 y \frac{\delta(s^{n}x)}{s^{n}}\biggr\Vert \le\lim_{n \to\infty }\frac {\theta}{s^{n}} =0. \end{aligned}$$

This implies that

$$\begin{aligned} \mathcal{D}(xy+yx)=2x\delta(y)+2y\mathcal{D}(x) \quad\mbox{for all } x,y \in \mathcal {A}. \end{aligned}$$

(3.19)

Comparing (3.18) and (3.19), we arrive at \(x\delta (y)=x\mathcal{D}(y)\) for all \(x,y \in \mathcal {A}\). Since \(\mathcal {A}\) contains the unit element, we find that \(\mathcal {D}=\delta\). Equation (3.19) can be written

$$\begin{aligned} \delta(xy+yx)=2x\delta(y)+2y\delta(x) \quad\mbox{for all } x,y \in \mathcal {A}. \end{aligned}$$

(3.20)

Letting \(x=y=e\) in (3.20), we have \(\delta(e)=0\). Now we obtain by additivity of *δ*
\(\delta(q x)=q \delta(x)\) for all \(q \in\mathbb{Q}\) and all \(x \in \mathcal {A}\). So \(\delta(q e)=q \delta (e)=0\) for all \(q \in\mathbb{Q}\). This fact and the assumption of *δ* imply that \(\delta(t e)=0\) for all \(t \in\mathbb{R}\). Considering \(y=t e\) in (3.20), we have \(\delta(t x)=t \delta (x)\) for all \(t \in\mathbb{R}\) and all \(x \in \mathcal {A}\). Thus *δ* is \(\mathbb{R}\)-linear. Hence we see that

$$\begin{aligned} \delta(\mu x)=\delta\bigl((t_{1}+t_{2} i ) x\bigr)= \delta(t_{1} x)+t_{2} \delta( i x)=t_{1} \delta( x)+ t_{2} i\delta( x)=(t_{1} + t_{2} i)\delta( x)=\mu \delta( x) \end{aligned}$$

for all \(\mu\in\mathbb{C}\) and all \(x \in \mathcal {A}\). So we see that *δ* is \(\mathbb{C}\)-linear. In view of (3.20), we get

$$\delta\bigl(x^{2}\bigr)=2x\delta(x) \quad\mbox{for all } x \in \mathcal {A}. $$

Thereby *δ* is a linear left Jordan derivation.

On the other hand, if \(\mathcal {A}\) is semiprime unital Banach algebra, then the rest of the proof is similar to the corresponding part of Theorem 3.1. □

### Theorem 3.6

*Let*
\(\mathcal {A}\)
*be a semisimple Banach algebra*. *Assume that*
\(l \ge3\)
*is a fixed integer and*
\(s_{1},s_{2}, \ldots, s_{l}\)
*are fixed positive real numbers*, *where*
\(s_{1}=\lambda s\) (\(s>1\)), \(s_{2}> 1\)
*and*
\(s_{3}=1\). *Suppose that*, *for each*
\(k=0,1\), \(\delta_{k} : \mathcal {A}\to \mathcal {A}\)
*is a mapping with*
\(\delta_{k}(0)=0\)
*such that*, *for some*
\(\varepsilon\ge0\),

$$\begin{aligned} \Biggl\Vert \sum_{j=1}^{l}s_{j} \delta_{k}(x_{j})\Biggr\Vert \leq\Biggl\Vert \delta _{k} \Biggl(\sum_{j=1}^{l}s_{j}x_{j} \Biggr)\Biggr\Vert +\varepsilon \end{aligned}$$

(3.21)

*for all*
\(x_{1}, x_{2}, \ldots, x_{l} \in \mathcal {A}\)
*and all*
\(\lambda\in\mathbb{T}_{\varepsilon}\), *where*
\(x_{3}= \lambda z\) (\(z \in \mathcal {A}\)) *and*

$$\begin{aligned} &\bigl\Vert \delta_{0}(xy+yx)-2x\delta_{0}(y)-2y \delta_{0}(x)\bigr\Vert \leq\theta_{0}, \end{aligned}$$

(3.22)

$$\begin{aligned} &\bigl\Vert \delta_{1}(xy+yx)-x\delta_{1}(y)-y \delta_{1}(x)-x\delta _{0}(y)-y\delta_{0}(x)\bigr\Vert \leq\theta_{1} , \end{aligned}$$

(3.23)

*for some*
\(\theta_{0}, \theta_{1} \ge0\)
*and all*
\(x,y \in \mathcal {A}\). *Then*
\(\delta_{1}\)
*is a linear generalized left Jordan derivation associated with a linear left Jordan derivation*
\(\delta_{0}\). *In this case*, \(\delta_{1}\)
*is continuous*.

### Proof

It is well known that semisimple algebras are semiprime [18]. As we saw in the proof of Theorem 3.4, \(\delta_{0}\) is a linear left Jordan derivation. In addition, we see that there exists a unique linear mapping \(\mathcal {D}_{1}: \mathcal {A}\to \mathcal {A}\) defined by

$$\begin{aligned} \mathcal{D}_{1}(x):=\lim_{n\to\infty} \frac{\delta_{1}(s^{n}x)}{s^{n}} \quad \mbox{for all } x \in \mathcal {A}. \end{aligned}$$

(3.24)

According to (3.23) and (3.24), we see that

$$\begin{aligned} \lim_{n \to\infty}\biggl\Vert \frac{\delta_{1}(s^{n}(xy+yx))}{s^{n}}- x \delta_{1}(y)- y \frac{\delta_{1}(s^{n}x)}{s^{n}}-x\delta _{0}(y)-y\delta _{0}(x)\biggr\Vert \le\lim_{n \to\infty}\frac{\theta_{1}}{s^{n}} =0, \end{aligned}$$

which implies that

$$\begin{aligned} \mathcal{D}_{1}(xy+yx)=x\delta_{1}(y)+y \mathcal{D}_{1}(x)+x\delta _{0}(y)+y\delta_{0}(x) \quad\mbox{for all } x,y \in \mathcal {A}. \end{aligned}$$

(3.25)

So we obtain from (3.25)

$$\begin{aligned} \mathcal{D}_{1}(xy+yx)-x\delta_{0}(y)-y \delta_{0}(x)=x\delta _{1}(y)+y\mathcal{D}_{1}(x) \quad\mbox{for all } x,y \in \mathcal {A}. \end{aligned}$$

(3.26)

In particular, the left-side of equation (3.26) is a bilinear mapping. Lemma 3.3 guarantees that \(\delta_{1}\) is linear. By (3.24), we have \(\mathcal {D}_{1}=\delta_{1}\). Equation (3.25) gives

$$\delta_{1}\bigl(x^{2}\bigr)=x\delta_{1}(x)+x \delta_{0}(x) \quad\mbox{for all } x \in \mathcal {A}. $$

Thus \(\delta_{1}\) is a linear generalized left Jordan derivation.

Therefore, since \(\mathcal {A}\) is semisimple, we conclude that \(\delta_{1}\) is continuous; see [19]. This completes the proof. □