Lie triple derivations of standard operator algebras

Abstract

Let X be a Banach space over the field $\mathbb F$ ($\mathbb R$ or $\mathbb C)$. Denote by B(X) the set of all bounded linear operators on X and by F(X) the set of all finite rank operators on X. A subalgebra $\mathcal A\subseteq B(X)$ is called a standard operator algebra if $F(X)\subseteq \mathcal A$. Suppose that $\delta$ is a mapping from $\mathcal A$ into B(X). First, we prove that if $\delta$ is a Lie triple derivation, then $\delta$ is standard. Next, we show that if $\delta$ is a local Lie triple derivation and $\mathrm {dim}(X)\ge 3$, then $\delta$ is a Lie triple derivation. Finally, we prove that if $\delta$ is a 2-local Lie triple derivation, then $\delta =d+\tau$, where d is a derivation, and $\tau$ is a homogeneous mapping from $\mathcal A$ into $\mathbb {F}I$ such that $\tau (A+B)=\tau (A)$ for each A, B in $\mathcal A$ where B is a sum of double commutators.

1 Introduction

Let $\mathcal {A}$ be an associative algebra over the field $\mathbb F$ ($\mathbb R$ or $\mathbb C)$ and $\mathcal M$ be an $\mathcal A$-bimodule. A linear mapping $\delta$ from $\mathcal A$ into $\mathcal M$ is called a derivation if $\delta (AB)=\delta (A)B+A\delta (B)$ for each A, B in $\mathcal A$, and $\delta$ is called an inner derivation if there exists an element M in $\mathcal M$ such that $\delta (A)=AM-MA$ for every A in $\mathcal A$. Clearly, every inner derivation is a derivation. In [13, 24], Kadison and Sakai independently proved that every derivation on a von Neumann algebra is inner. In [6], Chernoff proved that every derivation from a standard operator algebra $\mathcal A$ into B(X) is inner for a Banach space X. In [8], Christensen showed that every derivation on nest algebras is inner.

In 1990, Kadison [14], Larson and Sourour [15] independently introduced the concept of local derivations. A linear mapping $\delta$ from $\mathcal {A}$ into $\mathcal {M}$ is called a local derivation if for every A in $\mathcal {A}$ there exists a derivation $\delta _A$ (depending on A) from $\mathcal {A}$ into $\mathcal {M}$ such that $\delta (A)=\delta _A(A)$. In [14], Kadison proved that every continuous local derivation from a von Neumann algebra into its dual Banach module is a derivation. In [15], Larson and Sourour proved that every local derivation on B(X) is a derivation for a Banach space X. In [12], Johnson proved that every local derivation from a $C^*$-algebra into its Banach bimodule is a derivation. In [29], Zhu and Xiong proved that every local derivation from a unital standard operator algebra $\mathcal A$ into B(X) is a derivation.

In 1997, $\check{\text {S}}$emrl [25] introduced the concept of 2-local derivations. A mapping (not necessarily linear) $\delta$ from $\mathcal {A}$ into $\mathcal {M}$ is called a 2-local derivation if for each A, B in $\mathcal {A}$, there exists a derivation $\delta _{A,B}$ (depending on A, B) from $\mathcal {A}$ into $\mathcal {M}$ such that $\delta (A)=\delta _{A,B}(A)$ and $\delta (B)=\delta _{A,B}(B)$. In [25], $\check{\text {S}}$emrl proved that every 2-local derivation on $B(\mathcal H)$ is a derivation for a separable Hilbert space H. In [2], Ayupov and Kudaybergenov proved that every 2-local derivation on a von Neumann algebra is a derivation. In [10], we showed that every 2-local derivation from a standard operator algebra $\mathcal A$ into B(X) is a derivation.

A linear mapping $\delta$ from $\mathcal A$ into $\mathcal M$ is called a Lie derivation if $\delta ([A,B])=[\delta (A),B]+[A,\delta (B)]$ for each A, B in $\mathcal A$, where $[A,B]=AB-BA$ is called a commutator on $\mathcal A$. A Lie derivation $\delta$ is said to be standard if it can be decomposed as $\delta =d+\tau$, where d is a derivation from $\mathcal {A}$ into $\mathcal M$ and $\tau$ is a linear mapping from $\mathcal {A}$ into $\mathcal {Z}(\mathcal {M},\mathcal A)$ with $\tau ([A,B])=0$ for each A, B in $\mathcal {A}$, where $\mathcal {Z}(\mathcal {M},\mathcal A)=\{M\in \mathcal {M}:MA=AM~\text{ for } \text{ every }~A~\text{ in }~\mathcal {A}\}$.

An interesting problem is to identify those algebras on which every Lie derivation is standard. In [22], Mathieu and Villena proved that every Lie derivation on a $C^*$-algebra is standard. In [7], Cheung characterized Lie derivations on triangular algebras. In [20, 21], Lu studied Lie derivations on CDCSL algebras and reflexive algebras, respectively. In [3], Benkovi${\check{\text {c}}}$ proved that every Lie derivation on a matrix algebra $M_n(\mathcal A)$ is standard, where $n\ge 2$ and $\mathcal A$ is a unital algebra.

Similarly to local derivations and 2-local derivations, in [4], Chen et al. introduced the concepts of local Lie derivations and 2-local Lie derivations. A linear mapping $\delta$ from $\mathcal {A}$ into $\mathcal {M}$ is called a local Lie derivation if for every A in $\mathcal {A}$ there exists a Lie derivation $\delta _{A}$ (depending on A) from $\mathcal {A}$ into $\mathcal {M}$ such that $\delta (A)=\delta _{A}(A)$. A mapping (not necessarily linear) $\delta$ from $\mathcal {A}$ into $\mathcal {M}$ is called a 2-local Lie derivation if for every A, B in $\mathcal {A}$ there exists a Lie derivation $\delta _{A,B}$ (depending on A, B) from $\mathcal {A}$ into $\mathcal {M}$ such that $\delta (A)=\delta _{A,B}(A)$ and $\delta (B)=\delta _{A,B}(B)$.

In [4], Chen et al. study local Lie derivations and 2-local Lie derivations on B(X). In [5], Chen and Lu proved that every local Lie derivation on nest algebras is a Lie derivation. In [18, 19], Liu and Zhang proved that under certain conditions every local Lie derivation on triangular algebras is a Lie derivation, and every local Lie derivation on factor von Neumann algebras with dimension exceeding 1 is a Lie derivation. In [9], He et al. proved that every local Lie derivation on some algebras such as finite von Neumann algebras, nest algebras, Jiang–Su algebras and UHF algebras is a Lie derivation, and every 2-local Lie derivation on on some algebras such as factor von Neumann algebras, Jiang–Su algebra and UHF algebras is also a Lie derivation. In [16, 17], Liu proved that under certain conditions every local Lie derivation on generalized matrix algebras is a Lie derivation, and he showed that every 2-local Lie derivation of nest subalgebras of factors is a Lie derivation.

A linear mapping $\delta$ from $\mathcal A$ into $\mathcal M$ is a Lie triple derivation if $\delta ([[A,B],C])=[[\delta (A),B],C]+[[A,\delta (B)],C]+[[A,B],\delta (C)]$ for each A, B and C in $\mathcal A$. We call [[A, B], C] a double commutator on $\mathcal A$. It is clear that every Lie derivation is a Lie triple derivation. A Lie triple derivation $\delta$ from $\mathcal A$ into $\mathcal M$ is said to be standard if it can be decomposed as $\delta =d+\tau$, where d is a derivation from $\mathcal {A}$ into $\mathcal M$ and $\tau$ is a linear mapping from $\mathcal {A}$ into $\mathcal {Z}(\mathcal {M},\mathcal {A})$ with $\tau ([[A,B],C])=0$ for each A, B and C in $\mathcal A$.

Similarly to Lie derivations, the authors always consider the problem of identifying those algebras on which every Lie triple derivation is standard. In [23], Miers proved that if $\mathcal A$ is a von Neumann algebra with no central abelian summands, then every Lie triple derivation on $\mathcal A$ is standard. In [11], Ji and Wang proved that every continuous Lie triple derivation on TUHF algebras is standard. In [28], Zhang et al. proved that if $\mathcal N$ is a nest on a complex separable Hilbert space $\mathcal H$, then every Lie triple derivation on the nest algebra $\mathrm {Alg}\,\mathcal {N}$ is standard. In [27], Yu and Zhang studied the Lie triple derivations on commutative subspace lattice algebras. In [3], Benkovi${\check{\text {c}}}$ showed that if $\mathcal A$ is a unital algebra with a nontrivial idempotent, then under suitable assumptions every Lie triple derivation d on $\mathcal A$ is of the form $d=\Delta +\delta +\tau$, where $\Delta$ is a derivation on $\mathcal A$, $\delta$ is a Jordan derivation on $\mathcal A$ and $\tau$ is a linear mapping from $\mathcal A$ into its center $\mathcal {Z}(\mathcal {A})$ that vanishes on $[[\mathcal A,\mathcal A],\mathcal A]$. In [1], Ashraf and Akhtar proved that every Lie triple derivation on a generalized matrix algebra is standard. In [26], Wani proved that every Lie triple derivation from standard operator algebra into itself is standard.

Now we give the concepts of local Lie triple derivations and 2-local Lie triple derivations. A linear mapping $\delta$ from $\mathcal {A}$ into $\mathcal {M}$ is called a local Lie triple derivation if for every A in $\mathcal {A}$ there exists a Lie triple derivation $\delta _{A}$ (depending on A) from $\mathcal {A}$ into $\mathcal {M}$ such that $\delta (A)=\delta _{A}(A)$. A mapping (not necessarily linear) $\delta$ from $\mathcal {A}$ into $\mathcal {M}$ is called a 2-local Lie triple derivation if for every A, B in $\mathcal {A}$ there exists a Lie triple derivation $\delta _{A,B}$ (depending on A, B) from $\mathcal {A}$ into $\mathcal {M}$ such that $\delta (A)=\delta _{A,B}(A)$ and $\delta (B)=\delta _{A,B}(B)$.

In this paper, we always suppose that X is a Banach space over the field $\mathbb F$ ($\mathbb R$ or $\mathbb C)$. Denote by B(X) the set of all linear mappings on X and by F(X) the set of all finite rank operators on X. A subalgebra $\mathcal A\subseteq B(X)$ is called a standard operator algebra if $F(X)\subseteq \mathcal A$. Suppose that $\delta$ is a mapping from $\mathcal A$ into B(X). In Sect. 2, we prove that if $\delta$ is a Lie triple derivation, then $\delta$ is standard. In Sect. 3, we prove that if $\delta$ is a local Lie triple derivation and $\mathrm {dim}(X)\ge 3$, then $\delta$ is a Lie triple derivation. In Sect. 4, we prove that if $\delta$ is a 2-local Lie triple derivation, then $\delta =d+\tau$, where d is a derivation and $\tau$ is a homogeneous mapping from $\mathcal A$ into $\mathbb {F}I$ such that $\tau (A+B)=\tau (A)$ for each A, B in $\mathcal A$ where B is a sum of double commutators.

We shall review some simple properties of rank one operators and finite rank operators. Denote by $X^*$ the set of all bounded linear functionals on X. For each x in X and f in $X^{*}$, one can define an operator $x\otimes f$ by $(x\otimes f)y=f(y)x$ for every y in X. Obviously, $x\otimes f\in B(X)$. If both x and f are nonzero, then $x\otimes f$ is an operator of rank one. The following properties are evident and will be used frequently in this paper.

Proposition 1.1

Suppose that X is a Banach space and $\mathcal A\subseteq B(X)$ is a standard operator algebra. For each x, y in X, f, g in $X^{*}$ and A, B in B(X), the following statements hold:

• (1) $(x\otimes f)A=x\otimes (fA)$ and $A(x\otimes f)=(Ax)\otimes f$;

• (2) $(x\otimes f)(y\otimes g)=f(y)(x\otimes g)$;

• (3) $\mathcal {Z}(B(X),\mathcal A)=\mathbb {F}I$.

2 Lie triple derivations

In this section, we choose $x_0\in X$ and $f_0\in X^*$ such that $f_0(x_0)=1$, and denote by I the unit operator in B(X). For the convenience of expression, we give some symbols firstly. Let $P_1=x_0\otimes f_0$ and $P_2=I-P_1$. It is easy to see that $P_1$ and $P_2$ are two idempotents in B(X). Denote $P_i\mathcal A P_j$ and $P_iB(X)P_j$ by $\mathcal A_{ij}$ and $B(X)_{ij}$, respectively, denote $P_iAP_j$ by $A_{ij}$ for every A in $\mathcal A$, where $1\le i,j\le 2$.

Lemma 2.1

$P_1AP_1=f_0(Ax_0)P_1=f_0(P_1AP_1x_0)P_1$ for every A in B(X). Moreover, $B(X)_{11}$ is commutative.

Proof

For every A in B(X), by Proposition 1.1 (1) and (2), we have

$$\begin{aligned} P_1AP_1=x_0\otimes f_0Ax_0\otimes f_0=f_0(Ax_0)x_0\otimes f_0=f_0(Ax_0)P_1. \end{aligned}$$

(2.1)

Replacing A by $P_1AP_1$ in (2.1), we get

$$\begin{aligned} P_1AP_1=P_1P_1AP_1P_1=f_0(P_1AP_1x_0)P_1. \end{aligned}$$

It follows that $B(X)_{11}$ is commutative. $\square$

Lemma 2.2

• (1) If $BA_{21}=0$ for every $A_{21}$ in $\mathcal A_{21}$, then $BP_2=0$.

• (2) If $A_{12}B=0$ for every $A_{12}$ in $\mathcal A_{12}$, then $P_2B=0$.

Proof

(1) Let $A_{21}=P_2x\otimes f_0P_1$, where x is an arbitrary element in X. We obtain

$$\begin{aligned} 0=BP_2x\otimes f_0P_1x_0=f_0(P_1x_0)BP_2x=BP_2x. \end{aligned}$$

It follows that $BP_2=0$.

(2) Let $A_{12}=P_1x_0\otimes fP_2$, where f is an arbitrary element in $X^*$. We obtain

$$\begin{aligned} 0=P_1x_0\otimes fP_2Bx=f(P_2Bx)P_1x_0=f(P_2Bx)x_0 \end{aligned}$$

for every x in X. It follows that $f(P_2Bx)=0$ for each $f\in X^*$ and x in X. Thus, $P_2B=0$.

$\square$

Next we consider Lie triple derivations from a unital standard operator algebra $\mathcal A$ into B(X). The following theorem is the main result in this section.

Theorem 2.3

Let X be a Banach space and $\mathcal A\subseteq B(X)$ be a unital standard operator algebra. If $\delta$ is a Lie triple derivation $\delta$ from $\mathcal A$ into B(X), then $\delta$ is standard.

Before we prove Theorem 2.3, we present some lemmas.

Lemma 2.4

$\delta (I)\in \mathbb {F}I$.

Proof

Let P be an idempotent in $\mathcal A$. We have

$$\begin{aligned} 0=\delta ([[I,P],P])=[[\delta (I),P],P]=[\delta (I)P-P\delta (I),P]=\delta (I)P+P\delta (I)-2P\delta (I)P. \end{aligned}$$

Multiplying the above equation by P from the right, we obtain $P\delta (I)P=\delta (I)P$. It means that $(I-P)\delta (I)P=0$. Thus, $P_1\delta (I)P_2=P_2\delta (I)P_1=0$; it follows that $\delta (I)\in B(X)_{11}+B(X)_{22}$. By Lemma 2.1, we know that $\mathcal A_{11}$ is commutative, so $[\delta (I),A_{11}]=0$ for every $A_{11}$ in $\mathcal A_{11}$. In the following, we show

$$\begin{aligned}{}[\delta (I),A_{22}]=[\delta (I),A_{12}]=[\delta (I),A_{21}]=0 \end{aligned}$$

for every $A_{22}$ in $\mathcal A_{22}$, $A_{12}$ in $\mathcal A_{12}$ and $A_{21}$ in $\mathcal A_{21}$.

For each A, B in $\mathcal A$, we have

$$\begin{aligned}{}[[A,B],\delta (I)]=\delta ([[A,B],I])-[[A,\delta (B)],I]-[[\delta (A),B],I]=0. \end{aligned}$$

By $A_{12}=[P_1,A_{12}]$ and $A_{21}=[A_{21},P_1]$, we have

$$\begin{aligned}{}[\delta (I),A_{12}]=[\delta (I),A_{21}]=0. \end{aligned}$$

(2.2)

By (2.2), it follows that

$$\begin{aligned} 0=[\delta (I),A_{22}B_{21}]=[\delta (I),A_{22}]B_{21}+A_{22}[\delta (I),B_{21}]=[\delta (I),A_{22}]B_{21} \end{aligned}$$

for every $A_{22}$ in $\mathcal A_{22}$ and $B_{21}$ in $\mathcal A_{21}$. By Lemma 2.2, we have $[\delta (I),A_{22}]P_{2}=0$. By $\delta (I)\in B(X)_{11}+B(X)_{22}$, we obtain $[\delta (I),A_{22}]\in B(X)_{22}$, it follows that $[\delta (I),A_{22}]=0$. Hence by Proposition 1.1 (3), we have $\delta (I)\in \mathcal Z(B(X),\mathcal A)=\mathbb {F}I$. $\square$

Lemma 2.5

$P_1\delta (P_1)P_1+P_2\delta (P_1)P_2\in \mathbb {F}I$.

Proof

By Lemma 2.1, we know that $P_1\delta (P_1)P_1=\lambda P_1$, where $\lambda =f_0(P_1\delta (P_1)P_1x_0)\in \mathbb {F}$. Let x be in X and let $P_2x\otimes f_0P_1=A_{21}$. It follows that

$$\begin{aligned} -\delta (A_{21})&=\delta ([[P_2,A_{21}],P_2])\nonumber \\&=[[\delta (P_2),A_{21}],P_2]+[[P_2,\delta (A_{21})],P_2]+[[P_2,A_{21}],\delta (P_2)]\nonumber \\&=-A_{21}\delta (P_2)P_2-P_2\delta (P_2)A_{21}+A_{21}\delta (P_2)+2P_2\delta (A_{21})P_2\nonumber \\&\quad -\delta (A_{21})P_2-P_2\delta (A_{21})+A_{21}\delta (P_2)-\delta (P_2)A_{21}. \end{aligned}$$

(2.3)

Multiplying (2.3) by $P_2$ from the left and by $P_1$ from the right, we obtain

$$\begin{aligned} P_2\delta (P_2)A_{21}=A_{21}\delta (P_2)P_1. \end{aligned}$$

That is,

$$\begin{aligned} P_2\delta (P_2)P_2x\otimes f_0P_1=P_2x\otimes f_0P_1\delta (P_2)P_1. \end{aligned}$$

(2.4)

By letting both sides of (2.3) act on $x_0$ in X, we have

$$\begin{aligned} f_0(P_1x_0)P_2\delta (P_2)P_2x=f_0(P_1\delta (P_2)P_1x_0)P_2x. \end{aligned}$$

Since $f_0(P_1x_0)=f_0(x_0)=1$, it follows that

$$\begin{aligned} P_2\delta (P_2)P_2=f_0(P_1\delta (P_2)P_1x_0)P_2. \end{aligned}$$

(2.5)

By Lemma 2.4, we know that $\delta (I)\in \mathbb {F}I$. It follows that

$$\begin{aligned} P_2\delta (I)P_2=\delta (I)P_2=\delta (I)f_0(x_0)P_2=f_0(\delta (I)x_0)P_2=f_0(P_1\delta (I)P_1x_0)P_2. \end{aligned}$$

Now replacing $\delta (P_2)$ by $\delta (I)-\delta (P_1)$ in (2.5), we obtain

$$\begin{aligned} P_2\delta (P_1)P_2=f_0(P_1\delta (P_1)P_1x_0)P_2=\lambda P_2. \end{aligned}$$

This implies $P_1\delta (P_1)P_1+P_2\delta (P_1)P_2=\lambda (P_1+P_2)=\lambda I$. $\square$

Let $G=P_1\delta (P_1)P_2-P_2\delta (P_1)P_1$ and define a mapping $\Delta$ from $\mathcal A$ into B(X) by

$$\begin{aligned} \Delta (A)=\delta (A)-[A,G] \end{aligned}$$

for every A in $\mathcal A$. Obviously, $\Delta$ is also a Lie triple derivation from $\mathcal A$ into B(X). Moreover,

$$\begin{aligned} \Delta (P_1)=\delta (P_1)-[P_1,G]=P_1\delta (P_1)P_1+P_2\delta (P_1)P_2 \end{aligned}$$

and, by Lemma 2.5, we know that $\Delta (P_1)\in \mathbb {F}I$. In Lemmas 2.6, 2.7 and 2.8, we show some properties of $\Delta$.

Lemma 2.6

$\Delta (\mathcal A_{ij})\subseteq B(X)_{ij}$, where $1\le i,j\le 2$ and $i\ne j$.

Proof

Since $\Delta (P_1)\in \mathbb {F}I$, for each $A_{12}$ in $\mathcal A_{12}$, we have

$$\begin{aligned} \Delta (A_{12})&=\Delta ([[A_{12},P_1],P_1])\nonumber \\&=[[\Delta (A_{12}),P_{1}],P_1]+[[A_{12},\Delta (P_1)],P_1]+[[A_{12},P_1],\Delta (P_1)]\nonumber \\&=[[\Delta (A_{12}),P_{1}],P_1]\nonumber \\&=P_1\Delta (A_{12})P_2+P_2\Delta (A_{12})P_1. \end{aligned}$$

(2.6)

In the following, we show that $P_{2}\Delta (A_{12})P_1=0$.

Let $B_{12}$ be in $\mathcal A_{12}$, then $[A_{12},B_{12}]=0$. Thus,

$$\begin{aligned} 0=\Delta (0)=\Delta ([[A_{12},B_{12}],C])&=[[\Delta (A_{12}),B_{12}],C]+[[A_{12},\Delta (B_{12})],C]\nonumber \\&=[[\Delta (A_{12}),B_{12}]+[A_{12},\Delta (B_{12})],C] \end{aligned}$$

for every C in $\mathcal A$. It means that $J=[\Delta (A_{12}),B_{12}]+[A_{12},\Delta (B_{12})]\in \mathbb {F}I$. Since $A_{12}=[P_1,A_{12}]$, we have

$$\begin{aligned}{}[\Delta (A_{12}),B_{12}]&=J-[A_{12},\Delta (B_{12})]=J-[[P_1,A_{12}],\Delta (B_{12})]\nonumber \\&=J-(\Delta ([[P_1,A_{12}],B_{12}])-[[\Delta (P_1),A_{12}],B_{12}]-[[P_1,\Delta (A_{12})],B_{12}])\nonumber \\&=J+[[P_1,\Delta (A_{12})],B_{12}]. \end{aligned}$$

By (2.6), we have

$$\begin{aligned}{}[P_1\Delta (A_{12})P_2+P_2\Delta (A_{12})P_1,B_{12}]&=J+[[P_1,P_1\Delta (A_{12})P_2+P_2\Delta (A_{12})P_1],B_{12}]\nonumber \\&=J+[P_1\Delta (A_{12})P_2-P_2\Delta (A_{12})P_1,B_{12}]. \end{aligned}$$

Hence

$$\begin{aligned}{}[P_2\Delta (A_{12})P_1,B_{12}]=\frac{1}{2}J\in \mathbb {F}I. \end{aligned}$$

It is well known that $[P_2\Delta (A_{12})P_1,B_{12}]=0$. Thus, $P_2\Delta (A_{12})B_{12}=B_{12}\Delta (A_{12})P_1=0$ for every $B_{12}$ in $\mathcal A_{12}$. By Lemma 2.2, we know that $P_{2}\Delta (A_{12})P_1=0$. Similarly, we have $\Delta (\mathcal A_{21})\subseteq B(X)_{21}$. $\square$

Lemma 2.7

$\Delta (\mathcal A_{11})\subseteq \mathbb {F}I.$

Proof

For every $A_{11}$ in $\mathcal A_{11}$, by Lemma 2.1, we have

$$\begin{aligned} \Delta (A_{11})=\Delta (P_1A_{11}P_1)=\Delta (f_0(A_{11}x_0)P_1)=f_0(A_{11}x_0)\Delta (P_1). \end{aligned}$$

Since $\Delta (P_1)\in \mathbb {F}I$, it follows that $\Delta (A_{11})\in \mathbb {F}I$. $\square$

Lemma 2.8

$\Delta (A_{22})-f_0(\Delta (A_{22})x_0)I\in B(X)_{22}$ for every $A_{22}$ in $\mathcal A_{22}$. In particular, $\Delta (P_2)=f_0(\Delta (P_2)x_0)I.$

Proof

Through simple calculation, we get

$$\begin{aligned} 0=\Delta ([[P_1,A_{22}],P_1])&=[[P_1,\Delta (A_{22})],P_1]=-P_1\Delta (A_{22})P_2-P_2\Delta (A_{22})P_1. \end{aligned}$$

It follows that $\Delta (A_{22})\in B(X)_{11}+ B(X)_{22}$. By Lemma 2.1, we obtain

$$\begin{aligned} \Delta (A_{22})=P_1\Delta (A_{22})P_1+P_2\Delta (A_{22})P_2=f_0(\Delta (A_{22})x_0)P_1+P_2\Delta (A_{22})P_2, \end{aligned}$$

that is,

$$\begin{aligned} \Delta (A_{22})-f_0(\Delta (A_{22})x_0)I=-f_0(\Delta (A_{22})x_0)P_2+P_2\Delta (A_{22})P_2\in B(X)_{22}. \end{aligned}$$

Since $\Delta (P_2)=\Delta (I)-\Delta (P_1)\in \mathbb {F}I$, we have

$$\begin{aligned} \Delta (P_2)-f_0(\Delta (P_2)x_0)I\in \mathbb {F}I\cap B(X)_{22}=\{0\}. \end{aligned}$$

Thus, $\Delta (P_2)=f_0(\Delta (P_2)x_0)I.$ $\square$

In the following, we prove Theorem 2.3.

Proof

Define two mappings $\tau$ and D on from $\mathcal A$ into B(X) by

$$\begin{aligned} \tau (A)=f_0(P_1AP_1x_0)\Delta (P_1)+f_0(\Delta (P_2AP_2)x_0)I \end{aligned}$$

and

$$\begin{aligned} D(A)=\Delta (A)-\tau (A) \end{aligned}$$

for every A in $\mathcal A$. It is clear that $\tau$ is a linear mapping from $\mathcal A$ into $\mathcal Z(B(X),\mathcal A)$ and D is a linear mapping from $\mathcal A$ into B(X). Moreover, according to the previous lemmas and the definitions of $\tau$ and D, we have

• (1) $D(A_{ij})=\Delta (A_{ij})\in B(X)_{ij}$ for every $A_{ij}$ in $\mathcal A_{ij}$, where $1\le i,j\le 2$ and $i\ne j$;

• (2) $D(P_1)=D(P_2)=D(I)=0$;

• (3) $D(A_{11})=0$ for every $A_{11}$ in $\mathcal A_{11}$;

• (4) $D(A_{22})\in B(X)_{22}$ for every $A_{22}$ in $\mathcal A_{22}$.

To prove that $\Delta$ is standard, it is sufficient to show that D is a derivation and $\tau ([[A,B],C])=0$ for each A, B and C in $\mathcal A$.

In the following we show

$$\begin{aligned} D(A_{ij}B_{sk})=D(A_{ij})B_{sk}+A_{ij}D(B_{sk}) \end{aligned}$$

for every $A_{ij}$ in $\mathcal A_{ij}$ and $B_{sk}$ in $\mathcal A_{sk}$, where $1\le i,j,s,k\le 2$.

Since $D(\mathcal A_{ij})\in B(X)_{ij}$, we have

$$\begin{aligned} D(A_{ij}B_{sk})=D(A_{ij})B_{sk}+A_{ij}D(B_{sk}) \end{aligned}$$

for $j\ne s$. Thus, we only need to prove the following 8 cases:

• (1) $D(A_{11}B_{11})=D(A_{11})B_{11}+A_{11}D(B_{11})$;

• (2) $D(A_{11}B_{12})=D(A_{11})B_{12}+A_{11}D(B_{12})$;

• (3) $D(A_{12}B_{22})=D(A_{12})B_{22}+A_{12}D(B_{22})$;

• (4) $D(A_{21}B_{11})=D(A_{21})B_{11}+A_{21}D(B_{11})$;

• (5) $D(A_{22}B_{21})=D(A_{22})B_{21}+A_{22}D(B_{21})$;

• (6) $D(A_{22}B_{22})=D(A_{22})B_{22}+A_{22}D(B_{22})$;

• (7) $D(A_{12}B_{21})=D(A_{12})B_{21}+A_{12}D(B_{21})$;

• (8) $D(A_{21}B_{12})=D(A_{21})B_{12}+A_{21}D(B_{12})$.

Since $D(A_{11})=0$ for every $A_{11}$ in $\mathcal A_{11}$, case (1) is trivial.

For each A, B in $\mathcal A$, by $\Delta (A)-D(A)=\tau (A)\in \mathcal Z(B(X),\mathcal A)$, we have $[\Delta (A),B]=[D(A),B]$. Therefore

$$\begin{aligned} D(A_{11}B_{12})&=\Delta (A_{11}B_{12})=-\Delta ([[P_1,B_{12}],A_{11}])\\&=-[[P_1,\Delta (B_{12})],A_{11}]-[[P_1,B_{12}],\Delta (A_{11})]\\&=-[\Delta (B_{12}),A_{11}]-[B_{12},\Delta (A_{11})]\\&=[A_{11},D(B_{12})]+[D(A_{11}),B_{12}]\\&=A_{11}D(B_{12})+D(A_{11})B_{12} \end{aligned}$$

for each $A_{11}$ in $\mathcal A_{11}$ and $B_{12}$ in $\mathcal A_{12}$. Thus, case (2) holds. The cases (3), (4) and (5) are similar to case (2), so we omit the proofs.

For every $C_{21}$ in $\mathcal A_{21}$, according to case (5), we have the following two equations:

$$\begin{aligned} D(A_{22}B_{22}C_{21})=D(A_{22}B_{22})C_{21}+A_{22}B_{22}D(C_{21}) \end{aligned}$$

(2.7)

and

$$\begin{aligned} D(A_{22}B_{22}C_{21})&=D(A_{22})B_{22}C_{21}+A_{22}D(B_{22}C_{21})\nonumber \\&=D(A_{22})B_{22}C_{21}+A_{22}D(B_{22})C_{21}+A_{22}B_{22}D(C_{21}) \end{aligned}$$

(2.8)

for each $A_{22},B_{22}$ in $\mathcal A_{22}$. Comparing (2.7) and (2.8), we have

$$\begin{aligned} D(A_{22}B_{22})C_{21}=D(A_{22})B_{22}C_{21}+A_{22}D(B_{22})C_{21}. \end{aligned}$$

It follows that $(D(A_{22}B_{22})-D(A_{22})B_{22}-A_{22}D(B_{22}))C_{21}=0$ for every $C_{21}$ in $\mathcal A_{21}$. By Lemma 2.2 and $D(A_{22})\in \mathcal A_{22}$, we know that

$$\begin{aligned} D(A_{22}B_{22})-D(A_{22})B_{22}-A_{22}D(B_{22})=0. \end{aligned}$$

Finally, we show cases (7) and (8). Let $A_{12}$ be in $\mathcal A_{12}$ and $B_{21}$ be in $\mathcal A_{21}$. Through simple calculation, we obtain

$$\begin{aligned}&\Delta ([[A_{12},P_2],B_{21}])-D([[A_{12},P_2,],B_{21}])\\&\quad =[[\Delta (A_{12}),P_2],B_{21}]+[[A_{12},P_2],\Delta (B_{21})]-D([[A_{12},P_2,],B_{21}])\\&\quad =[\Delta (A_{12}),B_{21}]+[A_{12},\Delta (B_{21})]-D[A_{12},B_{21}]\\&\quad =[D(A_{12}),B_{21}]+[A_{12},D(B_{21})]-D(A_{12}B_{21}-B_{21}A_{12})\\&\quad =D(A_{12})B_{21}-B_{21}D(A_{12})+A_{12}D(B_{21})-D(B_{21})A_{12}-D(A_{12}B_{21})+D(B_{21}A_{12})\\&\quad =(D(A_{12})B_{21}+A_{12}D(B_{21})-D(A_{12}B_{21}))+(D(B_{21}A_{12})-B_{21}D(A_{12})-D(B_{21})A_{12}). \end{aligned}$$

Since $\Delta ([A_{12},B_{21}])-D([A_{12},B_{21}])$ belongs to $\mathbb {F}I$, we may assume that

$$\begin{aligned} \Delta ([A_{12},B_{21}])-D([A_{12},B_{21}])=\lambda I \end{aligned}$$

holds for some $\lambda$ in $\mathbb {F}$. That is,

$$\begin{aligned} \lambda I&=(D(A_{12})B_{21}+A_{12}D(B_{21})-D(A_{12}B_{21}))\nonumber \\&\quad +(D(B_{21}A_{12})-B_{21}D(A_{12})-D(B_{21})A_{12}). \end{aligned}$$

(2.9)

Since $D(\mathcal A_{ij})\in B(X)_{ij}$, we get

$$\begin{aligned} D(A_{12})B_{21}+A_{12}D(B_{21})-D(A_{12}B_{21})\in B(X)_{11} \end{aligned}$$

and

$$\begin{aligned} D(B_{21}A_{12})-B_{21}D(A_{12})-D(B_{21})A_{12}\in B(X)_{22}. \end{aligned}$$

Multiplying (2.9) by $P_1$ and $P_2$ respectively from the right, we obtain the following two equations:

$$\begin{aligned} D(A_{12}B_{21})=D(A_{12})B_{21}+A_{12}D(B_{21})-\lambda P_1 \end{aligned}$$

(2.10)

and

$$\begin{aligned} D(B_{21}A_{12})=B_{21}D(A_{12})+D(B_{21})A_{12}+\lambda P_2. \end{aligned}$$

(2.11)

By case (2) and equation (2.10), we obtain

$$\begin{aligned}&D(A_{12}B_{21}A_{12})\nonumber \\&\quad =D(A_{12}B_{21})A_{12}+A_{12}B_{21}D(A_{12})\nonumber \\&\quad =D(A_{12})B_{21}A_{12}+A_{12}D(B_{21})A_{12}-\lambda A_{12}+A_{12}B_{21}D(A_{12}). \end{aligned}$$

(2.12)

By case (3) and equation (2.9), we obtain

$$\begin{aligned}&D(A_{12}B_{21}A_{12})\nonumber \\&\quad =D(A_{12})B_{21}A_{12}+A_{12}D(B_{21}A_{12})\nonumber \\&\quad =D(A_{12})B_{21}A_{12}+A_{12}D(B_{21})A_{12}+A_{12}B_{21}D(A_{12})+\lambda A_{12}. \end{aligned}$$

(2.13)

Comparing (2.12) and (2.13), we have $\lambda A_{12}=0$. Thus, $\lambda =0$. By (2.10) and (2.9), cases (7) and (8) hold.

By cases (1)–(8), this implies immediately that D is a derivation. Now we show that $\tau ([[A,B],C])=0$ for each A, B and C in $\mathcal A$. Indeed,

$$\begin{aligned} \tau ([[A,B],C])&=\, \Delta ([[A,B],C])-D([[A,B],C])\nonumber \\&=\, [[\Delta (A),B],C]+[[A,\Delta (B)],C]+[[A,B],\Delta (C)]-D([[A,B],C])\nonumber \\&=\, [[D(A),B],C]+[[A,D(B)],C]+[[A,B],D(C)]-D([[A,B],C])\nonumber \\&=\, 0.\nonumber \end{aligned}$$

It follows that $\Delta (A)=D(A)+\tau (A)$ is a standard Lie triple derivation from $\mathcal A$ into B(X). Define a linear mapping from $\mathcal A$ into B(X) by

$$\begin{aligned} d(A)=D(A)+[A,G] \end{aligned}$$

for every A in $\mathcal A$. Thus, we have

$$\begin{aligned} \delta (A)=\Delta (A)+[A,G]=D(A)+\tau (A)+[A,G]=d(A)+\tau (A), \end{aligned}$$

where d is a derivation from $\mathcal A$ into B(X) and $\tau$ is a linear mapping from $\mathcal A$ into $\mathcal Z(B(X),\mathcal A)$ such that $\tau ([[A,B],C])=0$ for each A, B and C in $\mathcal A$. $\square$

For a non-unital standard operator algebra, the following result holds.

Corollary 2.9

Let X be a Banach space and $\mathcal A\subseteq B(X)$ be a non-unital standard operator algebra. If $\delta$ is a Lie triple derivation $\delta$ from $\mathcal A$ into B(X), then $\delta$ is standard.

Proof

Denote the unital algebra $\mathcal A\oplus \mathbb {F}I$ by $\widetilde{\mathcal A}$. Thus, $\widetilde{\mathcal A}$ is a unital standard operator algebra. Define a linear mapping $\widetilde{\delta }$ from $\widetilde{\mathcal A}$ into B(X) by

$$\begin{aligned} \widetilde{\delta }(A+\lambda I)=\delta (A) \end{aligned}$$

for every A in $\mathcal A$ and $\lambda$ in $\mathbb F$. Through a simple calculation, it is easy to show that $\widetilde{\delta }$ is also a Lie triple derivation. By Theorem 2.3, we know that $\widetilde{\delta }$ is standard, and so is $\delta$. $\square$

3 Local Lie triple derivations

In this section, we study local Lie triple derivations and the following theorem is the main result.

Theorem 3.1

Let X be a Banach space of dimension at least 3 and $\mathcal A\subseteq B(X)$ be a unital standard operator algebra. If $\delta$ is a local Lie triple derivation $\delta$ from $\mathcal A$ into B(X), then $\delta$ is a Lie triple derivation.

Proof

For every A in B(X), there is a Lie triple derivation $\delta _A$ from $\mathcal A$ into B(X) such that $\delta (A)=\delta _A(A).$ By Theorem 2.3, we know $\delta _A(A)$ is standard, then there exist a derivation $d_A$ from $\mathcal A$ into B(X) and a scalar operator $\tau _A(A)$ in $\mathbb {F}I$ such that $\delta (A)=d_A(A)+\tau _A(A).$ By [6, Corollary 3.4], we know that $d_A$ is an inner derivation, then there exists an element $T_A$ in B(X) such that $d_A(A)=[A,T_A]$. Thus, we have

$$\begin{aligned} \delta (A)=d_A(A)=[A,T_A]+\tau _A(A). \end{aligned}$$

We claim that $\tau _A(A)$ is unique. In fact, if

$$\begin{aligned} \delta (A)=[A,S_A]+\tau '_A(A) \end{aligned}$$

for some $S_A$ in B(X) and $\tau '_A(A)$ in $\mathbb {F}I$, then

$$\begin{aligned}{}[A,S_A-T_A]=\tau _A(A)-\tau '_A(A)=\lambda I \end{aligned}$$

for some $\lambda$ in $\mathbb {F}$. It is well known that $\tau _A(A)=\tau '_A(A)$. Hence we can define a mapping from $\mathcal A$ into $\mathbb {F}I$ by

$$\begin{aligned} \tau (A)=\tau _A(A) \end{aligned}$$

for every A in $\mathcal {A}$. Moreover, by the definition of $\tau$ and Theorem 2.3, we know that $\tau (A)=\tau _A(A)=0$ if A is a sum of double commutators.

For each x in X and f in $X^*$, define $\psi (x,f)=\tau (x\otimes f)$. Then we have

$$\begin{aligned} \delta (x\otimes f)=[x\otimes f,T_{x\otimes f}]+\psi (x,f) \end{aligned}$$

(3.1)

for some $T_{x\otimes f}$ in B(X). In the following we show that $\psi (x,f)$ is a bilinear mapping.

Firstly, we show the homogeneity of $\psi$. For each x in X, f in $X^*$ and $\lambda$ in $\mathbb {F}$, by (3.1), we have

$$\begin{aligned} \delta (x\otimes f)=[x\otimes f,T_{x\otimes f}]+\psi (x,f)~~ \mathrm {and} ~~\delta (\lambda x\otimes f)=[\lambda x\otimes f,T_{\lambda x\otimes f}]+\psi (\lambda x,f). \end{aligned}$$

By $\delta (\lambda x\otimes f)=\lambda \delta (x\otimes f)$, we infer

$$\begin{aligned}{}[\lambda x\otimes f,T_{x\otimes f}-T_{\lambda x\otimes f}]=\psi (\lambda x,f)-\lambda \psi (x,f)\in \mathbb {F}I. \end{aligned}$$

Thus, $\lambda \psi (x,f)=\psi (\lambda x,f)$. This proved that $\psi$ is homogenous in the first variable. In the same way, we can show that $\psi$ is homogenous in the second variable.

Secondly, we show that $\psi (x,f)$ is biadditive. We note that $\psi (x,f)=0$ for x in X and f in $X^*$ with $f(x)=0$. Indeed, we may choose an element z in X such that $f(z)=1$, then $x\otimes f=[[x\otimes f,z\otimes f],z\otimes f]$ is a double commutator and hence $\psi (x,f)=\tau (x\otimes f)=0$.

Let $x_1, x_2$ be in X and f be in $X^*$. If both $x_1$ and $x_2$ belong to $\mathrm {ker}f$, then

$$\begin{aligned} \psi (x_1,f)=\psi (x_2,f)=\psi (x_1+x_2,f)=0 \end{aligned}$$

and so

$$\begin{aligned} \psi (x_1+x_2,f)=\psi (x_1,f)+\psi (x_2,f). \end{aligned}$$

If one of $x_1$ and $x_2$ is not in $\mathrm {ker}f$, then $\mathrm {dim}(\mathrm {span}\{x_1,x_2\}\cap \mathrm {ker}f)\le 1$. Since $\mathrm {dim}(X)\ge 3$, we know that $\mathrm {dim}(\mathrm {ker}f)\ge 2$. Thus, we can take $y\in \mathrm {ker}f$ such that $y\notin \mathrm {span}\{x_1,x_2\}$. By (3.1), we have the following equations:

$$\begin{aligned} \delta (x_1\otimes f)y=\psi (x_1,f)y+\mu _1x_1~~~~\delta (x_2\otimes f)y=\psi (x_2,f)y+\mu _2x_2 \end{aligned}$$

and

$$\begin{aligned} \delta ((x_1+x_2)\otimes f)y=\psi (x_1+x_2,f)y+\mu (x_1+x_2) \end{aligned}$$

for some $\mu ,\mu _1,\mu _2\in \mathbb {F}$. Since $\delta$ is an additive mapping, we know that

$$\begin{aligned} (\psi (x_1+x_2,f)-\psi (x_1,f)-\psi (x_2,f))y=\mu _1x_1+\mu _2x_2-\mu (x_1+x_2). \end{aligned}$$

Since $y\notin$ span $\{x_1,x_2\}$, it follows that

$$\begin{aligned} \psi (x_1+x_2,f)=\psi (x_1,f)+\psi (x_2,f). \end{aligned}$$

It means that $\psi$ is additive in the first variable.

Let $f_1,f_2$ be in $X^*$ and x be in X. If $x\in \mathrm {ker}f_1\cap \mathrm {ker}f_2$, then

$$\begin{aligned} \psi (x,f_1+f_2)=\psi (x,f_1)=\psi (x,f_2)=0 \end{aligned}$$

and so

$$\begin{aligned} \psi (x,f_1+f_2)=\psi (x,f_1)+\psi (x,f_2). \end{aligned}$$

If $x\notin \mathrm {ker}f_1\cap \mathrm {ker}f_2$, then we can take $z\in \mathrm {ker}f_1\cap \mathrm {ker}f_2$ which is linearly independent of x, By (3.1), we have

$$\begin{aligned} \delta (x\otimes f_1)z=\psi (x,f_1)z+\lambda _1x~~~~\delta (x\otimes f_2)z=\psi (x,f_2)z+\lambda _2x \end{aligned}$$

and

$$\begin{aligned} \delta (x\otimes (f_1+f_2))z=\psi (x,f_1+f_2)z+\lambda x \end{aligned}$$

for some $\lambda ,\lambda _1,\lambda _2\in \mathbb {F}$. Since $\delta$ is an additive mapping, we know that

$$\begin{aligned} (\psi (x,f_1+f_2)-\psi (x,f_1)-\psi (x,f_2))z=(\lambda _1+\lambda _2-\lambda )x. \end{aligned}$$

Since z and x are linearly independent, it follows that

$$\begin{aligned} \psi (x,f_1+f_2)=\psi (x,f_1)+\psi (x,f_2). \end{aligned}$$

The next goal is to show that there is an element J in B(X) such that

$$\begin{aligned} \delta (x\otimes f)=[x\otimes f,J]+\psi (x,f) \end{aligned}$$

for every rank one operator $x\otimes f$ in B(X).

For each x in X and f in $X^*$, define

$$\begin{aligned} \phi (x,f)=[x\otimes f,T_{x\otimes f}]=\delta (x\otimes f)-\psi (x,f). \end{aligned}$$

(3.2)

It is easy to see that $\phi (x,f)$ is a bilinear mapping and $\phi (x,f)\mathrm {ker}f\subseteq \mathbb {F}x$. Hence by [21, Proposition 1.1], there are two linear mappings $T:X\rightarrow X$ and $S^*:X^*\rightarrow X^*$ such that

$$\begin{aligned} \phi (x,f)=[x\otimes f,T_{x\otimes f}]=Tx\otimes f+x\otimes S^*f \end{aligned}$$

(3.3)

for each x in X and f in $X^*$. It follows that

$$\begin{aligned} (T+T_{x\otimes f})x\otimes f=x\otimes (T^*_{x\otimes f}-S^*)f \end{aligned}$$

(3.4)

for each x in X and f in $X^*$.

We claim that $S^*=-T^*$. We only have to show that $S^*f(x)=-f(Tx)$ for each x in X and f in $X^*$. It is trivial if one of x and f is zero. Suppose that neither of x and f is zero. If both sides of (3.4) are zeros, then

$$\begin{aligned} (T+T_{x\otimes f})x=(T^*_{x\otimes f}-S^*)f=0. \end{aligned}$$

It follows that

$$\begin{aligned} S^*f(x)=T^*_{x\otimes f}f(x)=f(T_{x\otimes f}x)=-f(Tx). \end{aligned}$$

If both sides of (3.4) are not zeros, then we have

$$\begin{aligned}{}[(T+T_{x\otimes f})x\otimes f]^2=[x\otimes (T^*_{x\otimes f}-S^*)f]^2, \end{aligned}$$

that is,

$$\begin{aligned} f((T+T_{x\otimes f})x)((T+T_{x\otimes f})x\otimes f)=((T^*_{x\otimes f}-S^*)f)(x)(x\otimes (T^*_{x\otimes f}-S^*)f). \end{aligned}$$

It follows that

$$\begin{aligned} f((T+T_{x\otimes f})x)=((T^*_{x\otimes f}-S^*)f)(x) \end{aligned}$$

and then $S^*f(x)=-f(Tx)$. Consequently, we always have $S^*=-T^*$. By (3.2) and (3.3), we have

$$\begin{aligned} \delta (x\otimes f)=Tx\otimes f+x\otimes S^*f +\psi (x,f)=[x\otimes f,-T]+\psi (x,f) \end{aligned}$$

for every $x\otimes f$ in $\mathcal A$. Let $J=-T$ and by $\psi (x,f)=\tau (x\otimes f)\in \mathbb {F}I$, we obtain

$$\begin{aligned} \delta (A)=[A,J]+\lambda _{A}I \end{aligned}$$

(3.5)

for every $A=x\otimes f$ in $\mathcal A$ and some $\lambda _{A}\in \mathbb {F}$. Finally, we show that

$$\begin{aligned} \delta (A)=[A,J]+\lambda _AI \end{aligned}$$

holds for every A in $\mathcal {A}$. Suppose that P, Q are two idempotents of rank one and let $P^\perp =I-P$, $Q^\perp =I-Q$. By Proposition 1.1(1) and (3.5), it follows that

$$\begin{aligned} \delta (A)&=\delta (PA+P^\perp AQ+P^\perp AQ^\perp )\nonumber \\&=\delta (PA)+\delta (P^\perp AQ)+\delta (P^\perp AQ^\perp )\nonumber \\&=[PA,J]+\lambda _{PA}I+[P^\perp AQ,J]+\lambda _{P^\perp AQ}I+[P^\perp AQ^\perp ,T_{P^\perp AQ^\perp }]+\lambda _{P^\perp AQ^\perp }I\nonumber \\&=PAJ-JPA+P^\perp AQJ-JP^\perp AQ+[P^\perp AQ^\perp ,T_{P^\perp AQ^\perp }]+\lambda _A I\nonumber \\&=PAJ-JAQ+P^\perp AQJ-JPAQ^\perp +[P^\perp AQ^\perp ,T_{P^\perp AQ^\perp }]+\lambda _A I, \end{aligned}$$

(3.6)

where $\lambda _A=\lambda _{PA}+\lambda _{P^\perp AQ}+\lambda _{P^\perp AQ^\perp }$. Multiplying (3.6) by P on the left and by Q on the right, we have

$$\begin{aligned} P\delta (A)Q=P[A,J]Q+\lambda PQ, \end{aligned}$$

that is,

$$\begin{aligned} P(\delta (A)-[A,J]-\lambda _A I)Q=0. \end{aligned}$$

By the arbitrariness of P and Q, it follows that $\delta (A)=[A,J]+\lambda _A I$, where J is a fixed element and $\lambda _A$ is depends on A. By the uniqueness of $\tau$, we know that $\tau (A)=\lambda _A I$ and $\tau$ is a linear mapping from $\mathcal A$ into $\mathbb {F}I$ vanishing on every double commutator, which means that $\delta$ is a Lie triple derivation. $\square$

Corollary 3.2

Let X be a Banach space of dimension at least 3 and $\mathcal A\subseteq B(X)$ be a non-unital standard operator algebra. If $\delta$ is a local Lie triple derivation $\delta$ from $\mathcal A$ into B(X), then $\delta$ is a Lie triple derivation.

Proof

Denote the unital algebra $\mathcal A\oplus \mathbb {F}I$ by $\widetilde{\mathcal A}$. Thus, $\widetilde{\mathcal A}$ is a unital standard operator algebra. Define a linear mapping $\widetilde{\delta }$ from $\widetilde{\mathcal A}$ into B(X) by

$$\begin{aligned} \widetilde{\delta }(A+\lambda I)=\delta (A) \end{aligned}$$

for every A in $\mathcal A$ and $\lambda$ in $\mathbb F$.

Since $\delta$ is a local Lie triple derivation from $\mathcal A$ into B(X), for each $A\in \mathcal A$ and $\lambda \in \mathbb F$, there exists a Lie triple derivation $\delta _A$ such that $\delta (A)=\delta _A(A)$. Define a linear mapping $\widetilde{\delta _A}$ from $\widetilde{\mathcal A}$ into B(X) by

$$\begin{aligned} \widetilde{\delta _A}(B+\lambda I)=\delta _A(B) \end{aligned}$$

for every B in $\mathcal A$ and $\lambda$ in $\mathbb F$. It is easy to show that $\widetilde{\delta _A}$ is also a Lie triple derivation. Moreover, we have

$$\begin{aligned} \widetilde{\delta }(A+\lambda I)=\delta (A)=\delta _A(A)=\widetilde{\delta _A}(A+\lambda I). \end{aligned}$$

It means that $\widetilde{\delta }$ is a local Lie triple derivation from $\widetilde{\mathcal A}$ into B(X). By the result of the case that $\mathcal A$ contains the unit, $\widetilde{\delta }$ is a Lie triple derivation. Hence $\delta$ is also a Lie triple derivation. $\square$

4 2-Local Lie triple derivations

In this section, we study the 2-local Lie triple derivations and the following theorem is the main result.

Theorem 4.1

Let X be a Banach space and $\mathcal A\subseteq B(X)$ be a unital standard operator algebra. If $\delta$ is a 2-local Lie triple derivation from $\mathcal A$ into B(X), then $\delta =d+\tau$, where d is a derivation and $\tau$ is a homogeneous mapping from $\mathcal A$ into $\mathbb {F}I$ such that $\tau (A+B)=\tau (A)$ for each A, B in $\mathcal A$ where B is a sum of double commutators.

Proof

Similarly to the proof of Theorem 3.1, we can show that $\delta$ has a unique decomposition at each point A in $\mathcal {A}$, i.e.

$$\begin{aligned} \delta (A)=\delta _A(A)=d_A(A)+\tau _A(A), \end{aligned}$$

where $\delta _{A}$ is a Lie triple derivation, $d_A$ is a derivation and $\tau _A$ is a linear mapping from $\mathcal {A}$ into $\mathbb {F}I$ such that $\tau _A[[X,Y],Z]=0$ each X, Y and Z in $\mathcal A$.

Thus, we can define

$$\begin{aligned} d(A)=d_A(A)~~\mathrm {and}~~\tau (A)=\tau _A(A) \end{aligned}$$

for every A in $\mathcal {A}$.

In the following we show that d is a derivation and $\tau$ is a homogeneous mapping. Given A and B in $\mathcal {A}$, there exists a Lie triple derivation $\delta _{A,B}$ from $\mathcal A$ into B(X) such that

$$\begin{aligned} \delta (A)=\delta _{A,B}(A)=d_{A,B}(A)+\tau _{A,B}(A), \end{aligned}$$

and

$$\begin{aligned} \delta (B)=\delta _{A,B}(B)=d_{A,B}(B)+\tau _{A,B}(B), \end{aligned}$$

where $d_{A,B}$ + $\tau _{A,B}$ is the standard decomposition of $\delta _{A,B}$. By the uniqueness of the decomposition, $d(A)=d_{A,B}(A)$ and $d(B)=d_{A,B}(B)$. Hence d is a 2-local derivation and by [10, Theorem 3.1], we know d is a derivation from $\mathcal A$ into B(X).

For every A in $\mathcal A$ and $\lambda$ in $\mathbb {F}$, there exists a Lie triple derivation $\delta _{A,\lambda A}$ from $\mathcal A$ into B(X) such that

$$\begin{aligned} \delta (A)=\delta _{A,\lambda A}(A)~~\mathrm {and}~~\delta (\lambda A)=\delta _{A,\lambda A}(\lambda A). \end{aligned}$$

It follows that $\delta$ is homogeneous, and so is $\tau$.

Moreover, for each A, B in $\mathcal A$ where B is a sum of double commutators, there is a linear mapping $\tau _{A,A+B}$ from $\mathcal {A}$ into $\mathbb {F}I$ vanishing on every double commutator such that

$$\begin{aligned} \tau (A+B)=\tau _{A,A+B}(A+B)=\tau _{A,A+B}(A)=\tau (A). \end{aligned}$$

$\square$