Characterization of derivations on strongly double triangle subspace lattice algebras by local actions

Zijie Qin

doi:10.1515/math-2025-0213

Article Open Access

Characterization of derivations on strongly double triangle subspace lattice algebras by local actions

Zijie Qin

Published/Copyright: November 27, 2025

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$Open Mathematics$

From the journal Open Mathematics Volume 23 Issue 1

Abstract

Let D be a strongly double triangle subspace lattice on a nonzero complex reflexive Banach space and Alg D the associated subspace lattice algebra. Assume that F , G ∈ Alg D and set R ( F , G ) = { ( A , B ) ∈ Alg D × Alg D : A F B = G } . A linear mapping δ : Alg D → Alg D is called to be derivable on R ( F , G ) if δ(AB) = δ(A)B+ Aδ(B) whenever ( A , B ) ∈ R ( F , G ) . In this paper, it is shown that if G is an injective operator or an operator with dense range, then every linear mapping derivable on R ( I , G ) is a derivation. It is also shown that if F satisfies certain conditions, then every linear mapping derivable on R ( F , 0 ) is a generalized derivation.

Keywords: derivations; generalized derivations; strongly double triangle subspace lattice; derivable mappings

MSC 2020: 47B47; 47L35

1 Introduction

Let A be an algebra with unit I. Recall that a linear mapping δ : A → A is a derivation if δ(AB) = δ(A)B + Aδ(B) for all A , B ∈ A , and a generalized derivation if δ(AB) = δ(A)B + Aδ(B) − Aδ(I)B for all A , B ∈ A . The characterizations of derivations by their local behaviors have attracted many authors (see [1], [2], [3], [4], [5] and references therein). One direction is to study conditions under which derivations can be completely determined by the action on some sets of elements. Recall that δ is derivable at some fixed element G ∈ A if δ(AB) = δ(A)B + Aδ(B) whenever AB = G. Clearly, a derivation from A into itself is derivable at any element in A . However, the converse is, in general, not true.

Linear (or additive) mappings derivable at some fixed point have been extensively studied in the last two decades. In [6], [7], [8], [9], [10], the authors studied the structure of linear mappings derivable at zero point for some algebras. Among others, Pang and Yang [10] proved that every linear mapping derivable at zero point on strongly double triangle subspace lattice algebras is a generalized derivation. For nonzero element case, Ghahramani [11] investigated additive mappings derivable at nontrivial idempotents on Banach algebras which contain a nontrivial idempotent. In [12], Zhu, Xiong and Li showed that every linear mapping derivable at any nonzero element on B(H) is a derivation, where B(H) denotes the algebra of all bounded linear operators on Hilbert space H. Recently, the authors in [13] extended the result in [12] to the Banach space case. They showed that if X is a Banach space, then every derivable mapping at a nonzero element in B(X) is a derivation. As a corollary, they proved that every generalized derivable mapping on B(X) is a generalized derivation. For non-prime algebras case, it was shown in [14] that every linear mapping derivable at any injective operator or operator with dense range is a derivation on nest algebras on Banach spaces. Qi and Hou [15] considered the same problem on JSL algebras.

In [16], Pan studied this problem in a new and natural way. Let F , G ∈ A . Set R ( F , G ) = { ( A , B ) ∈ A × A : A F B = G } . Then R ( F , G ) is a relation on A . A linear mapping δ from A into itself is said to be derivable on R ( F , G ) if δ(AB) = δ(A)B + Aδ(B) whenever ( A , B ) ∈ R ( F , G ) (see [16]). Obviously, for a linear mapping δ : A → A , δ is derivable at G ∈ A if and only if δ is derivable on R ( I , G ) ; every generalized derivation δ is derivable on R ( δ ( I ) , 0 ) . It is natural to ask that for what algebra A does it hold that δ : A → A is a generalized derivation if and only if δ is derivable on R ( F , 0 ) for some F ∈ A . Pan [16] showed that every linear mapping on complex matrix algebras which is derivable on R ( F , 0 ) is a generalized derivation. Later, in [17], Pan showed if N is a nest on a Hilbert space H and P is an idempotent in a nest algebra Alg N which is either left-faithful to N or right-faithful to N ⊥ , then every linear mapping δ : Alg N → B ( H ) derivable on R ( P , 0 ) is a generalized derivation. Chen and Lu [18] extended the result in [16], they proved the same result for JSL algebras.

The class of strongly double triangle subspace lattice algebras is another important kind of reflexive algebras. Note that the proofs in [14],15],18] depend heavily on the existence of rank-one operators. Since strongly double triangle subspace lattice algebras contain no rank-one operators [19], Proposition 3.1], the problem is more complicated than the previous one. In this paper, we study linear mappings on strongly double triangle subspace lattice algebras which are derivable on certain relations.

Let us introduce some notations used in this paper. Throughout, X will be a nonzero reflexive complex Banach space with topological dual X*. By B(X) we denote the algebra of all bounded linear operators on X. For A ∈ B(X), A* stands for the adjoint of A. For nonzero vectors f* ∈ X* and x ∈ X, we define the rank-one operator f* ⊗ x by y↦f*(y)x for y ∈ X. For any subset L ⊆ X, by L ^⊥ we denote the annihilator of L, that is, L ^⊥ = {f* ∈ X* : f*(x) = 0 for all x ∈ L}. Dually, for any nonempty subset M ⊆ X*, ^⊥ M denotes its pre-annihilator, that is, ^⊥ M = {x ∈ X : f*(x) = 0 for all f* ∈ M}.

A family L of closed subspaces of X is called a subspace lattice on X if it contains (0) and X, and is closed under the operations closed linear span ∨ and intersection ∩ in the sense that ∨ γ ∈ Γ L γ ∈ L and ∩ γ ∈ Γ L γ ∈ L for every family {L _γ : γ ∈ Γ} of elements in L . Given a subspace lattice L on X, the associated subspace lattice algebra Alg L is the set of operators on X leaving every subspace in L invariant, that is,

Alg L = { A ∈ B ( X ) : A L ⊆ L for every L ∈ L } .

A double triangle subspace lattice on X is a set D = { ( 0 ) , K , L , M , X } of subspaces of X satisfying K ∩ L = L ∩ M = M ∩ K = (0) and K ∨ L = L ∨ M = M ∨ K = X. If one of the three sums K + L, L + M and M + K is closed, we say that D is a strongly double triangle subspace lattice. Observe that D ⊥ = { ( 0 ) , K ⊥ , L ⊥ , M ⊥ , X * } is a double triangle subspace lattice on the reflexive Banach space X*. Put

K 0 = K ∩ ( L + M ) , L 0 = L ∩ ( M + K ) , M 0 = M ∩ ( K + L )

and

K p = K ⊥ ∩ ( L ⊥ + M ⊥ ) , L p = L ⊥ ∩ ( M ⊥ + K ⊥ ) , M p = M ⊥ ∩ ( K ⊥ + L ⊥ ) .

Notice that K _p, L _p and M _p (resp. K ^⊥, L ^⊥ and M ^⊥) play the same role for D ⊥ as K ₀, L ₀ and M ₀ (resp. K, L and M) do for D . Moreover, it follows from [20], Lemma 2.2] that dimK ₀ = dimL ₀ = dimM ₀ and dimK _p = dimL _p = dimM _p.

This paper is organized as follows. Let D = { ( 0 ) , K , L , M , X } be a strongly double triangle subspace lattice on a nonzero complex reflexive Banach space X and K + L is closed. Assume that δ is a linear mapping from Alg D into itself. In Section 2, we proved that if G ∈ Alg D is an injective operator or an operator with dense range and δ is derivable on R ( I , G ) , then δ is a derivation. As a corollary, we show that every generalized derivable mapping at G (that is, δ(AB) = δ(A)B + Aδ(B) − Aδ(I)B whenever AB = G) on Alg D is a generalized derivation. In Section 3, let F be a nonzero operator in Alg D satisfying one of the following conditions: (1) F | K 0 ∉ C I and F | L 0 ∉ C I ; (2) F | K 0 ∈ C I and F | L 0 ∈ C I . Here F | K 0 denote the restriction of F to K ₀, and I K 0 denotes the identity on K ₀. We proved that if δ is derivable on R ( F , 0 ) , then δ is a generalized derivation.

We close this section by summarizing some lemmas from [20],21] on double triangle subspace lattice algebras.

Lemma 1.1.

([20], Lemma 2.1]) Let D = { ( 0 ) , K , L , M , X } be a double triangle subspace lattice on a complex reflexive Banach space X. Then the following statements hold.

K ₀ ⊆ K ⊆^⊥ K _p, L ₀ ⊆ L ⊆^⊥ L _p and M ₀ ⊆ M ⊆^⊥ M _p.
K ₀ ∩ L ₀ = L ₀ ∩ M ₀ = M ₀ ∩ K ₀ = (0).
K _p ∩ L _p = L _p ∩ M _p = M _p ∩ K _p = (0).
K ₀ + L ₀ = L ₀ + M ₀ = M ₀ + K ₀ = K ₀ + L ₀ + M ₀.
K _p + L _p = L _p + M _p = M _p + K _p = K _p + L _p + M _p.

Lemma 1.2.

([20], Theorem 2.1]) Let D = { ( 0 ) , K , L , M , X } be a double triangle subspace lattice on a complex reflexive Banach space X. Then the following statements hold.

Every finite-rank operator of Alg D has even rank.
If R is of the form R = f* ⊗ y − g* ⊗ x, where x ∈ K ₀, y ∈ L ₀, x + y ∈ M ₀ and f* ∈ K _p, g* ∈ L _p, f* + g* ∈ M _p are nonzero, then R ∈ Alg D and every rank-two operator in Alg D is of this form.
Alg D contains a nonzero finite-rank operator if and only if dimK ₀ ≠ 0 and dimK _p ≠ 0.

Lemma 1.3.

([20], Lemma 3.2]) Let x, y ∈ X and f*, g* ∈ X* are nonzero vectors such that x ∈ K ₀, y ∈ L ₀, x + y ∈ M ₀ and f* ∈ K _p, g* ∈ L _p, f* + g* ∈ M _p. Set R = f* ⊗ y − g* ⊗ x. Then f*(y) = −g*(x) and R ² = f*(y)R.

Lemma 1.4.

([20], Theorem 2.3]) Let D = { ( 0 ) , K , L , M , X } be a strongly double triangle subspace lattice on a nonzero complex reflexive Banach space X. If K + L is closed, then

K ₀ is dense in K and L ₀ is dense in L.
K ^⊥ + L ^⊥ = X*, M ₀ = M and M _p = M ^⊥.
K ₀ + L ₀ + M ₀ is dense in X.
K _p + L _p + M _p is dense in X*.

From the above lemmas, we see that Alg D is rich in rank-two operators. Moreover, as shown in [21], rank-two operators in Alg D have the following two useful properties.

Lemma 1.5.

([21], Lemma 1.5]) Let D be a strongly double triangle subspace lattice on a nonzero complex reflexive Banach space. Then every rank-two operator in Alg D is a linear combination of rank-two idempotents in Alg D .

The following result identifies the behavior of derivable mappings on the set of rank-two operators, which is crucial for proving our main results.

Proposition 1.6.

([21], Lemma 2.4]) Suppose that D = { ( 0 ) , K , L , M , X } is a strongly double triangle subspace lattice on X with K + L = X. Let Φ : K 0 × K p → Alg D be a bilinear map. If Φ ( x , f * ) ( k e r f * ∩ k e r f 1 * ) ⊆ span { x , x 1 } for every x = x ₁ + x ₂ ∈ K ₀ and every f * = f 1 * + f 2 * ∈ K p , where x ₁ ∈ L ₀, x ₂ ∈ M ₀, f 1 * ∈ L p , f 2 * ∈ M p , then there exist linear mappings S : L ₀ → L ₀, T : K ₀ → K ₀, V : K _p → X* and W : K _p → X* such that

(1.1) Φ ( x , f * ) = f * ⊗ S x 1 + f 1 * ⊗ T x + V f * ⊗ x + W f * ⊗ x 1

for every x = x ₁ + x ₂ ∈ K ₀ with x ₁ ∈ L ₀, x ₂ ∈ M ₀ and every f * = f 1 * + f 2 * ∈ K p with f 1 * ∈ L p , f 2 * ∈ M p . Here kerf* denotes the kernel of f* and span{x, x ₁} denotes the linear span of x and x ₁.

We refer readers [19],20],22] to more properties of strongly double triangle subspace lattice algebras.

2 Derivable mappings on R ( I , G )

Let D = { ( 0 ) , K , L , M , X } be a strongly double triangle subspace lattice on a nonzero complex reflexive Banach space X. Without loss of generality, we may assume that K + L is closed. Then K + L = X. It is easy to verify that dimK ₀ ≠ 0 ≠ dimK _p. By Lemma 1.2, Alg D contains nonzero finite-rank operators. In this section, we study linear mappings derivable (resp. generalized derivable) at an injective operator or an operator with dense range on Alg D .

One of the main results reads as follows.

Theorem 2.1.

Let D be a strongly double triangle subspace lattice on a nonzero complex reflexive Banach space. Assume that G ∈ Alg D is an injective operator or an operator with dense range. If δ : Alg D → Alg D is a linear mapping derivable on R ( I , G ) , then δ is a derivation.

To prove Theorem 2.1, we need the following lemmas. In what follows, D will be a strongly double triangle subspace lattice on a nonzero complex reflexive Banach space X and G ∈ Alg D .

Lemma 2.2.

Let δ : Alg D → Alg D be a linear mapping derivable on R ( I , G ) . If G is an injective operator or an operator with dense range, then δ(I) = 0.

Proof.

Since δ is derivable on R ( I , G ) , it follows from I ⋅ G = G ⋅ I = G that δ(I)G + δ(G) = δ(G) + Gδ(I) = δ(G). This leads to δ(I)G = Gδ(I) = 0. As G is an injective operator or an operator with dense range, we have δ(I) = 0. □

Lemma 2.3.

Suppose δ : Alg D → Alg D is a linear mapping derivable on R ( I , G ) .

If G is an operator with dense range, then
1. δ(PG) = δ(P)G + Pδ(G) and δ(P) = δ(P)P + Pδ(P) for every idempotent operator P ∈ Alg D .
2. δ(NG) = δ(N)G + Nδ(G) and δ(N)N + Nδ(N) = 0 for every square-zero operator N ∈ Alg D .
If G is an injective operator, then
1. δ(GP) = δ(G)P + Gδ(P) and δ(P) = δ(P)P + Pδ(P) for every idempotent operator P ∈ Alg D .
2. δ(GN) = δ(G)N + Gδ(N) and δ(N)N + Nδ(N) = 0 for every square-zero operator N ∈ Alg D .

Proof.

We only prove (1), since the proof of the other is almost the same. Let P be any idempotent operator in Alg D . Using the fact that δ(I) = 0 by Lemma 2.2, it follows from ( I − 3 P ) I − 3 2 P G = G that

(2.1) δ ( G ) = δ ( I − 3 P ) I − 3 2 P G + ( I − 3 P ) δ I − 3 2 P G = − 3 δ ( P ) G + 9 2 δ ( P ) P G + δ ( G ) − 3 2 δ ( P G ) − 3 P δ ( G ) + 9 2 P δ ( P G ) .

On the other hand, since I − 3 2 P ( I − 3 P ) G = G , we obtain

(2.2) δ ( G ) = δ I − 3 2 P ( I − 3 P ) G + I − 3 2 P δ ( ( I − 3 P ) G ) = − 3 2 δ ( P ) G + 9 2 δ ( P ) P G + δ ( G ) − 3 δ ( P G ) − 3 2 P δ ( G ) + 9 2 P δ ( P G ) .

Comparing Eq. (2.1) and Eq. (2.2), we get δ(PG) = δ(P)G + Pδ(G). This together with Eq. (2.1) gives us (δ(P)P + Pδ(P) − δ(P))G = 0. As G is an operator with dense range, we arrive at δ(P) = δ(P)P + Pδ(P).

Let N ∈ Alg D be such that N ² = 0. Since δ is derivable on R ( I , G ) , it follows from (I + N)(I − N)G = G = (I − N)(I + N)G that

(2.3) δ ( G ) = δ ( I + N ) ( I − N ) G + ( I + N ) δ ( ( I − N ) G ) = δ ( N ) G − δ ( N ) N G + δ ( G ) − δ ( N G ) + N δ ( G ) − N δ ( N G )

and

δ ( G ) = δ ( I − N ) ( I + N ) G + ( I − N ) δ ( ( I + N ) G ) = − δ ( N ) G − δ ( N ) N G + δ ( G ) + δ ( N G ) − N δ ( G ) − N δ ( N G ) .

Comparing these two equations, we have δ(NG) = δ(N)G + Nδ(G). This together with Eq. (2.3) gives (δ(N)N + Nδ(N))G = 0. As the range of G is dense, one can conclude that δ(N)N + Nδ(N) = 0. □

Lemma 2.4.

Let A be invertible in Alg D . Suppose δ : Alg D → Alg D is a linear mapping derivable on R ( I , G ) . Then the following statements hold.

δ(G) = δ(A)A ⁻¹ G + Aδ(A ⁻¹ G).
δ(G) = δ(GA ⁻¹)A + GA ⁻¹ δ(A).

Proof.

Since δ is derivable on R ( I , G ) , it follows directly from the equation AA ⁻¹ G = G = GA ⁻¹ A. □

The proof of Theorem 2.1. For clarity of exposition, we will organize the proof into a series of claims.

Claim 1.

For nonzero vectors x ₁ ∈ L ₀, x ₂ ∈ M ₀ with x = x ₁ + x ₂ ∈ K ₀ and nonzero functionals f 1 * ∈ L p , f 2 * ∈ M p with f * = f 1 * + f 2 * ∈ K p , we have δ f * ⊗ x 1 − f 1 * ⊗ x k e r f * ∩ k e r f 1 * ⊆ span { x , x 1 } .

Set R = f * ⊗ x 1 − f 1 * ⊗ x . By Lemma 1.3, f * ( x 1 ) = − f 1 * ( x ) . Now we consider two cases.

Case 1. f * ( x 1 ) = − f 1 * ( x ) ≠ 0 .

By Lemma 1.3, we have 1 f * ( x 1 ) R 2 = 1 f * ( x 1 ) R . It follows from the linearity of δ and Lemma 2.3 that

δ ( R ) = 1 f * ( x 1 ) δ ( R ) R + 1 f * ( x 1 ) R δ ( R ) .

Then, for any z ∈ k e r f * ∩ k e r f 1 * , we get

δ ( R ) z = 1 f * ( x 1 ) R δ ( R ) z = 1 f * ( x 1 ) ( δ ( R ) * f * ) ( z ) x 1 − 1 f * ( x 1 ) ( δ ( R ) * f 1 * ) ( z ) x .

From this, δ ( R ) ( k e r f * ∩ k e r f 1 * ) ⊆ span { x , x 1 } .

Case 2. f * ( x 1 ) = f 1 * ( x ) = 0 .

Note that K ₀ ⊆ K ⊆^⊥ K _p by Lemma 1.1. Since K + L = X and L ₀ is dense in L by Lemma 1.4, there exists a vector y ₁ ∈ L ₀ such that f*(y ₁) = 1. By Lemma 1.1, there exist unique vectors y ∈ K ₀ and y ₂ ∈ M ₀ such that y ₁ = y − y ₂. Since f*(y ₁) = f*(x ₁ + y ₁) = 1, by Case 1, for every z ∈ k e r f * ∩ k e r f 1 * , there exist scalars λ 1 , λ 2 , λ 3 , λ 4 ∈ C such that

δ f * ⊗ y 1 − f 1 * ⊗ y z = λ 1 y 1 + λ 2 y

and

δ ( f * ⊗ ( x 1 + y 1 ) − f 1 * ⊗ ( x + y ) ) z = λ 3 ( x 1 + y 1 ) + λ 4 ( x + y ) .

It follows that

(2.4) δ ( R ) z = δ ( ( f * ⊗ ( x 1 + y 1 ) − f 1 * ⊗ ( x + y ) ) − f * ⊗ y 1 − f 1 * ⊗ y ) ( z ) = λ 3 x 1 + λ 4 x + ( λ 3 − λ 1 ) y 1 + ( λ 4 − λ 2 ) y .

Notice that f * ( y 1 ) = − f 1 * ( y ) = 1 by Lemma 1.3. As R ² = 0, by Lemma 2.3 and Eq. (2.4), we obtain

0 = ( R δ ( R ) + δ ( R ) R ) z = R ( λ 3 x 1 + λ 4 x + ( λ 3 − λ 1 ) y 1 + ( λ 4 − λ 2 ) y ) = ( λ 3 − λ 1 ) x 1 + ( λ 4 − λ 2 ) x .

As x ₁ and x are linearly independent, we arrive at λ ₁ = λ ₃ and λ ₂ = λ ₄. Applying Eq. (2.4) again, we see that δ(R)z = λ ₃ x ₁ + λ ₄ x ∈ span{x, x ₁}.

By Proposition 1.6 and Claim 1, there exist linear mappings S : L ₀ → L ₀, T : K ₀ → K ₀, V : K _p → X* and W : K _p → X* such that Eq. (1.1) holds for every x = x ₁ + x ₂ ∈ K ₀ and every f * = f 1 * + f 2 * ∈ K p with x ₁ ∈ L ₀, x ₂ ∈ M ₀, f 1 * ∈ L p , f 2 * ∈ M p .

Claim 2.

Let x ₁ ∈ L ₀, x ₂ ∈ M ₀, x = x ₁ + x ₂ ∈ K ₀ and f 1 * ∈ L p , f 2 * ∈ M p , f * = f 1 * + f 2 * ∈ K p be nonzero. Then the following statements hold.

(Vf*)(y ₁) = 0 for every y ₁ ∈ L and (Wf*)(y) = 0 for every y ∈ K.
(Wf*)(x ₁) + f*(Sx ₁) = 0 and ( V f * ) ( x ) + f 1 * ( T x ) = 0 .

For y ₁ ∈ L, by Eq. (1.1), we have

δ f * ⊗ x 1 − f 1 * ⊗ x y 1 = f * ⊗ S x 1 + f 1 * ⊗ T x + V f * ⊗ x + W f * ⊗ x 1 y 1 = f * ( y 1 ) S x 1 + ( V f * ) ( y 1 ) x + ( W f * ) ( y 1 ) x 1 .

Since δ is a linear mapping from Alg D into itself, δ f * ⊗ x 1 − f 1 * ⊗ x y 1 , Sx ₁ and x ₁ are vectors in L, which further implies that (Vf*)(y ₁)x ∈ L. Note that K ∩ L = (0). So, (Vf*)(y ₁) = 0 for every y ₁ ∈ L. By a similar way, we prove that (Wf*)(y) = 0 for every y ∈ K.

By Lemma 1.3, we have f * ( x 1 ) = − f 1 * ( x ) . Set R = f * ⊗ x 1 − f 1 * ⊗ x . Then R ∈ Alg D by Lemma 1.2. Now we distinguish two cases.

Case 1. f * ( x 1 ) = − f 1 * ( x ) ≠ 0 .

By Lemma 1.3, we have 1 f * ( x 1 ) R 2 = 1 f * ( x 1 ) R . This together with Lemma 2.3 gives us Rδ(R)R = 0. Applying (1) and Eq. (1.1), we see that

0 = f * ( x 1 ) ( f * ( S x 1 ) + ( W f * ) ( x 1 ) ) f * ⊗ x 1 + f 1 * ( x ) ( f 1 * ( T x ) + ( V f * ) ( x ) ) f 1 * ⊗ x .

As both {x, x ₁} and f * , f 1 * are linearly independent sets, we have ( W f * ) ( x 1 ) + f * ( S x 1 ) = ( V f * ) ( x ) + f 1 * ( T x ) = 0 .

Case 2. f * ( x 1 ) = f 1 * ( x ) = 0 .

In this case, R ² = 0. By Lemma 2.3, δ(R)R + Rδ(R) = 0. It follows from (1) and Eq. (1.1) that

0 = f * ( S x 1 ) f * ⊗ x 1 − f 1 * ( T x ) f 1 * ⊗ x − ( V f * ) ( x ) f 1 * ⊗ x + ( W f * ) ( x 1 ) f * ⊗ x 1 = ( f * ( S x 1 ) + ( W f * ) ( x 1 ) ) f * ⊗ x 1 − ( f 1 * ( T x ) + ( V f * ) ( x ) ) f 1 * ⊗ x .

From this, ( W f * ) ( x 1 ) + f * ( S x 1 ) = ( V f * ) ( x ) + f 1 * ( T x ) = 0 .

Claim 3.

For any rank-two operators R 1 , R 2 ∈ Alg D , we have δ(R ₁ R ₂) = δ(R ₁)R ₂ + R ₁ δ(R ₂).

By Lemma 1.2, there exist nonzero vectors x = x ₁ + x ₂, y = y ₁ + y ₂ ∈ K ₀ with x ₁, y ₁ ∈ L ₀, x ₂, y ₂ ∈ M ₀ and nonzero functionals f * = f 1 * + f 2 * , g * = g 1 * + g 2 * ∈ K p with f 1 * , g 1 * ∈ L p , f 2 * , g 2 * ∈ M p such that R 1 = f * ⊗ x 1 − f 1 * ⊗ x and R 2 = g * ⊗ y 1 − g 1 * ⊗ y . Then R 1 R 2 = g * ⊗ ( f * ( y 1 ) ) x 1 − g 1 * ⊗ ( − f 1 * ( y ) ) x . According to Eq. (1.1), it follows that

δ ( R 1 R 2 ) = f * ( y 1 ) g * ⊗ S x 1 − f 1 * ( y ) g 1 * ⊗ T x + f * ( y 1 ) W g * ⊗ x 1 − f 1 * ( y ) V g * ⊗ x .

Now, by Claim 2 and Eq. (1.1), one can conclude that

δ ( R 1 ) R 2 + R 1 δ ( R 2 ) = f * ( y 1 ) g * ⊗ S x 1 − f 1 * ( y ) g 1 * ⊗ T x − ( V f * ) ( y ) g 1 * ⊗ x + ( W f * ) ( y 1 ) g * ⊗ x 1 + f * ( S y 1 ) g * ⊗ x 1 + f * ( y 1 ) W g * ⊗ x 1 − f 1 * ( T y ) g 1 * ⊗ x − f 1 * ( y ) V g * ⊗ x = f * ( y 1 ) g * ⊗ S x 1 − f 1 * ( y ) g 1 * ⊗ T x − ( ( V f * ) ( y ) + f 1 * ( T y ) ) g 1 * ⊗ x + ( f * ( S y 1 ) + ( W f * ) ( y 1 ) ) g * ⊗ x 1 + f * ( y 1 ) W g * ⊗ x 1 − f 1 * ( y ) V g * ⊗ x = δ ( R 1 R 2 ) .

Claim 4.

δ is a derivation.

Let A ∈ Alg D . Since δ is linear and δ(I) = 0 by Lemma 2.2, we may assume that A is invertible in Alg D . Let P ∈ Alg D be any rank-two idempotent. For any nonzero scalar α ∈ C with α ≠ − 1, set β = −α(1 + α)⁻¹.

Case 1. G is an operator with dense range.

Since δ is derivable on R ( I , G ) , it follows from (A + αAP) ⋅ (A ⁻¹ + βPA ⁻¹)G = G that

δ ( G ) = δ ( A + α A P ) ( A − 1 + β P A − 1 ) G + ( A + α A P ) δ ( ( A − 1 + β P A − 1 ) G ) = δ ( A ) A − 1 G + β δ ( A ) P A − 1 G + α δ ( A P ) A − 1 G + α β δ ( A P ) P A − 1 G + A δ ( A − 1 G ) + β A δ ( P A − 1 G ) + α A P δ ( A − 1 G ) + α β A P δ ( P A − 1 G ) .

This together with Lemma 2.4(1) gives us

0 = β ( δ ( A ) P A − 1 G + A δ ( P A − 1 G ) ) + α ( δ ( A P ) A − 1 G + A P δ ( A − 1 G ) ) + α β ( δ ( A P ) P A − 1 G + A P δ ( P A − 1 G ) ) .

Denote

T 1 = δ ( A ) P A − 1 G + A δ ( P A − 1 G ) ,

T 2 = δ ( A P ) A − 1 G + A P δ ( A − 1 G ) ,

T 3 = δ ( A P ) P A − 1 G + A P δ ( P A − 1 G ) .

Then

− 1 1 + α T 1 + T 2 − α 1 + α T 3 = 0 .

Sending α to infinity, one can get T ₂ = T ₃. So, − 1 1 + α T 1 + T 3 − α 1 + α T 3 = 0 and hence, T ₁ = T ₃. Note that AP, as well as PA ⁻¹ Z is a rank-two operator in Alg D . Thus, by Claim 3, we have

δ ( A ) P A − 1 G + A δ ( P A − 1 G ) = δ ( A P ) P A − 1 G + A P δ ( P A − 1 G ) = δ ( A P A − 1 G ) .

As δ is linear, by Lemma 1.5, we get

(2.5) δ ( A ) R A − 1 G + A δ ( R A − 1 G ) = δ ( A R A − 1 G )

for all rank-two operators R ∈ Alg D .

Let x be arbitrary nonzero vector in K ₀. By Lemma 1.1, there exist nonzero vectors x ₁ ∈ L ₀ and x ₂ ∈ M ₀ such that x = x ₁ + x ₂. Take any nonzero functionals f 1 * ∈ L p and f 2 * ∈ M p such that f * = f 1 * + f 2 * ∈ K p , and set R = f * ⊗ x 1 − f 1 * ⊗ x . Then R ∈ Alg D by Lemma 1.2. Using Eq. (1.1) and Eq. (2.5), we have

( A − 1 G ) * f * ⊗ S A x 1 + ( A − 1 G ) * f 1 * ⊗ T A x + V ( A − 1 G ) * f * ⊗ A x + W ( A − 1 G ) * f * ⊗ A x 1 = ( A − 1 G ) * f * ⊗ δ ( A ) x 1 − ( A − 1 G ) * f 1 * ⊗ δ ( A ) x + ( A − 1 G ) * f * ⊗ A S x 1 + ( A − 1 G ) * f 1 * ⊗ A T x + V ( A − 1 G ) * f * ⊗ A x + W ( A − 1 G ) * f * ⊗ A x 1 .

This reduces to

( A − 1 G ) * f 1 * ⊗ ( δ ( A ) + T A − A T ) x = ( A − 1 G ) * f * ⊗ ( δ ( A ) + A S − S A ) x 1 .

Since the range of G is dense, it follows that

f 1 * ⊗ ( δ ( A ) + T A − A T ) x = f * ⊗ ( δ ( A ) + A S − S A ) x 1 .

Note that both {x, x ₁} and f * , f 1 * are linearly independent sets by Lemma 1.1. So, δ(A)x = (AT − TA)x for all A ∈ Alg D and all x ∈ K ₀. Hence, δ(AB)x = (δ(A)B + Aδ(B))x for all A , B ∈ Alg D and all x ∈ K ₀. Since K ₀ is dense in K by Lemma 1.4, δ(AB)x = (δ(A)B + Aδ(B))x for all A , B ∈ Alg D and all x ∈ K. Applying a similar argument as above, we can get δ(AB)x = (δ(A)B + Aδ(B))x for all x ∈ L. Because K + L = X, we have δ(AB) = δ(A)B + Aδ(B) for all A , B ∈ Alg D , and consequently, δ is a derivation.

Case 2. G is an injective operator.

Since G(A ⁻¹ + βPA ⁻¹) ⋅ (A + αAP) = G, we have

δ ( G ) = δ ( G ( A − 1 + β P A − 1 ) ) ( A + α A P ) + G ( A − 1 + β P A − 1 ) δ ( A + α A P ) = δ ( G A − 1 ) A + α δ ( G A − 1 ) A P + β δ ( G P A − 1 ) A + α β δ ( G P A − 1 ) A P + G A − 1 δ ( A ) + α G A − 1 δ ( A P ) + β G P A − 1 δ ( A ) + α β G P A − 1 δ ( A P ) .

According to Lemma 2.4(2), it follows that

0 = α ( δ ( G A − 1 ) A P + G A − 1 δ ( A P ) ) + β ( δ ( G P A − 1 ) A + G P A − 1 δ ( A ) ) + α β ( δ ( G P A − 1 ) A P + G P A − 1 δ ( A P ) ) .

Denote

T 1 = δ ( G A − 1 ) A P + G A − 1 δ ( A P ) ,

T 2 = δ ( G P A − 1 ) A + G P A − 1 δ ( A ) ,

T 3 = δ ( G P A − 1 ) A P + G P A − 1 δ ( A P ) .

Then

T 1 − 1 1 + α T 2 − α 1 + α T 3 = 0 .

Sending α to infinity, we arrive at T ₁ = T ₃. So, T 3 − 1 1 + α T 2 − α 1 + α T 3 = 0 and hence, T ₂ = T ₃. This together with Claim 3 gives us

δ ( G P A − 1 ) A + G P A − 1 δ ( A ) = δ ( G P A − 1 ) A P + G P A − 1 δ ( A P ) = δ ( G P ) .

By the linearity of δ and Lemma 1.5, we obtain

(2.6) δ ( G R A − 1 ) A + G R A − 1 δ ( A ) = δ ( G R )

for every rank-two operator R ∈ Alg D . Fix nonzero x ∈ K ₀ and take any f 1 * ∈ L p . By Lemma 1.1, there exist x ₁ ∈ L ₀, x ₂ ∈ M ₀ and f 2 * ∈ M p , f* ∈ K _p such that x = x ₁ + x ₂ and f 1 * = f * − f 2 * , respectively. Set R = f * ⊗ x 1 − f 1 * ⊗ x . Then R ∈ Alg D by Lemma 1.2. Applying Eq. (1.1) and Eq. (2.6), we see that

f * ⊗ S G x 1 + f 1 * ⊗ T G x + V f * ⊗ G x + W f * ⊗ G x 1 = f * ⊗ S G x 1 + f 1 * ⊗ T G x + A * V ( A − 1 ) * f * ⊗ G x + A * W ( A − 1 ) * f * ⊗ G x 1 + δ ( A ) * ( A − 1 ) * f * ⊗ G x 1 − δ ( A ) * ( A − 1 ) * f 1 * ⊗ G x .

It follows that

( δ ( A ) * ( A − 1 ) * f * + A * W ( A − 1 ) * f * − W f * ) ⊗ G x 1 = ( δ ( A ) * ( A − 1 ) * f 1 * − A * V ( A − 1 ) * f * + V f * ) ⊗ G x .

Since G is injective, we have

(2.7) ( δ ( A ) * ( A − 1 ) * f * + A * W ( A − 1 ) * f * − W f * ) ⊗ x 1 = ( δ ( A ) * ( A − 1 ) * f 1 * − A * V ( A − 1 ) * f * + V f * ) ⊗ x .

Let y ∈ K ₀ be arbitrary. Note that δ ( A ) * ( A − 1 ) * f * ∈ K p ⊆ K ⊥ and Ay ∈ K ₀ ⊆ K. Applying Eq. (2.7) to y, by Claim 2, we have

0 = ( ( δ ( A ) * ( A − 1 ) * f * ) ( y ) + ( W ( A − 1 ) * f * ) ( A y ) − ( W f * ) ( y ) ) x 1 = ( δ ( A ) * ( A − 1 ) * f * + A * W ( A − 1 ) * f * − W f * ) ⊗ x 1 ( y ) = ( δ ( A ) * ( A − 1 ) * f 1 * − A * V ( A − 1 ) * f * + V f * ) ⊗ x ( y ) = ( f 1 * ( A − 1 δ ( A ) y ) − ( V ( A − 1 ) * f * ) ( A y ) + ( V f * ) ( y ) ) x = f 1 * ( ( A − 1 δ ( A ) + A − 1 T A − T ) y ) x .

It follows that f 1 * ( ( A − 1 δ ( A ) + A − 1 T A − T ) y ) = 0 for all f 1 * ∈ L p . Since ^⊥ L _p = ^⊥(L ^⊥) = L, we arrive at (A ⁻¹ δ(A) + A ⁻¹ TA − T)y ∈ L. Note that K ∩ L = (0). So (A ⁻¹ δ(A) + A ⁻¹ TA − T)y = 0 for all y ∈ K ₀. Hence δ(A)y = (AT − TA)y for all A ∈ Alg D and all y ∈ K ₀. Now using the same argument as in the proof of Case 1, we see that δ is a derivation. □

Recall that a linear mapping δ : Alg D → Alg D is generalized derivable at G ∈ Alg D if δ(AB) = δ(A)B + Aδ(B) − Aδ(I)B whenever AB = G. As an immediate consequence of Theorem 2.1, we give a characterization of linear mappings generalized derivable at injective operators or operators with dense range on Alg D .

Corollary 2.5.

Let D be a strongly double triangle subspace lattice on a nonzero complex reflexive Banach space. Assume that G ∈ Alg D is an injective operator or an operator with dense range. If δ : Alg D → Alg D is a linear mapping generalized derivable at G, then δ is a generalized derivation.

Proof.

Define a linear mapping ϕ : Alg D → Alg D by

(2.8) ϕ ( A ) = δ ( A ) − δ ( I ) A

for A ∈ Alg D . Let A , B ∈ Alg D . A simple computation shows that

ϕ ( A ) B + A ϕ ( B ) = ( δ ( A ) − δ ( I ) A ) B + A ( δ ( B ) − δ ( I ) B ) = δ ( A ) B + A δ ( B ) − δ ( I ) A B − A δ ( I ) B .

This together with the fact that ϕ(AB) = δ(AB) − δ(I)AB implies that

ϕ ( A ) B + A ϕ ( B ) − ϕ ( A B ) = δ ( A ) B + A δ ( B ) − A δ ( I ) B − δ ( A B ) .

Since δ is generalized derivable at G, ϕ is derivable at G. By Theorem 2.1, ϕ is a derivation. This together with Eq. (2.8) gives

δ ( A B ) = ϕ ( A B ) + δ ( I ) A B = ( ϕ ( A ) + δ ( I ) A ) B + A ( ϕ ( B ) + δ ( I ) B ) − A δ ( I ) B = δ ( A ) B + A δ ( B ) − A δ ( I ) B .

Thus, δ is a generalized derivation. □

3 Derivable mappings on R ( F , 0 )

Throughout this section, D = { ( 0 ) , K , L , M , X } is a strongly double triangle subspace lattice on a nonzero complex reflexive Banach space X with K + L = X. In this section, we study linear mappings derivable on R ( F , 0 ) for some F ∈ Alg D .

Our main result reads as follows.

Theorem 3.1.

Let D be a strongly double triangle subspace lattice on a nonzero complex reflexive Banach space X. Suppose F is a nonzero fixed operator in Alg D satisfying one of the following conditions:

F | K 0 ∉ C I K 0 and F | L 0 ∉ C I L 0 .
F | K 0 ∈ C I K 0 and F | L 0 ∈ C I L 0 .

If δ : Alg D → Alg D is a linear mapping derivable on R ( F , 0 ) , then δ is a generalized derivation.

Proof.

We will complete the proof by checking two claims.

Claim 1. If F | K 0 ∉ C I K 0 and F | L 0 ∉ C I L 0 , then δ is a generalized derivation.

As F | K 0 ∉ C I , there exists y ∈ K ₀ such that y and Fy are linearly independent. Note that K ∨ M = X and K ∩ M = (0). According to Lemma 1.4(2), we can take g 2 * ∈ M p such that g 2 * ( y ) = 1 and g 2 * ( F y ) = 0 . By Lemma 1.1, there exist y ₁ ∈ L ₀ and y ₂ ∈ M ₀ such that y = y ₁ + y ₂ and there exist g 1 * ∈ L p and g* ∈ K _p such that g 2 * = g * − g 1 * , respectively. Since K ₀ ⊆ K ⊆^⊥ K _p, in view of Lemma 1.3, we conclude that

1 = g 2 * ( y ) = g * − g 1 * ( y ) = − g 1 * ( y ) = g * ( y 1 )

and

0 = g 2 * ( F y ) = g * − g 1 * ( F y ) = − g 1 * ( F y ) = g * ( F y 1 ) .

For every x = x ₁ + x ₂ ∈ K ₀ with x ₁ ∈ L ₀, x ₂ ∈ M ₀ and every f * = f 1 * + f 2 * ∈ K p with f 1 * ∈ L p , f 2 * ∈ M p , we have g * ⊗ x 1 − g 1 * ⊗ x ∈ Alg D and f * ⊗ y 1 − f 1 * ⊗ y ∈ Alg D by Lemma 1.2. Compute

g * ⊗ x 1 − g 1 * ⊗ x F f * ⊗ y 1 − f 1 * ⊗ y = g * ( F y 1 ) f * ⊗ x 1 + g 1 * ( F y ) f 1 * ⊗ x = 0 .

Since δ is derivable on R ( F , 0 ) , it follows that

δ f * ⊗ x 1 − f 1 * ⊗ x = δ ( g * ⊗ x 1 − g 1 * ⊗ x f * ⊗ y 1 − f 1 * ⊗ y ) = δ g * ⊗ x 1 − g 1 * ⊗ x f * ⊗ y 1 − f 1 * ⊗ y + g * ⊗ x 1 − g 1 * ⊗ x δ f * ⊗ y 1 − f 1 * ⊗ y .

This leads to

δ f * ⊗ x 1 − f 1 * ⊗ x k e r f * ∩ k e r f 1 * ⊆ span { x , x 1 } .

By Proposition 1.6, there exist linear mappings S : L ₀ → L ₀, T : K ₀ → K ₀, V : K _p → X* and W : K _p → X* such that Eq. (1.1) holds for every x = x ₁ + x ₂ ∈ K ₀ and every f * = f 1 * + f 2 * ∈ K p with x ₁ ∈ L ₀, x ₂ ∈ M ₀, f 1 * ∈ L p , f 2 * ∈ M p . Since δ is a linear mapping from Alg D into itself, by using the same manner as the proof in Claim 2 of Theorem 2.1, one can get (Vf*)(y ₁) = 0 for every y ₁ ∈ L and (Wf*)(y) = 0 for every y ∈ K.

Let x ∈ K ₀ be arbitrary. By Lemma 1.1, there exists x ₁ ∈ L ₀ and x ₂ ∈ M ₀ such that x = x ₁ + x ₂. We first show that there exist nonzero functionals f 1 * ∈ L p and f* ∈ K _p such that (I − FR)FR = 0, where R = f * ⊗ x 1 − f 1 * ⊗ x . Now distinguish two possibilities.

Case 1. Fx = 0.

In this case, since L ₀ ∩ M ₀ = (0), we have Fx ₁ = Fx ₂ = 0. Take any nonzero f * = f 1 * + f 2 * ∈ K p with f 1 * ∈ L p and f 2 * ∈ M p . The direct computation shows that (I − FR)FR = 0.

Case 2. Fx ≠ 0.

We can take f 1 * ∈ L p such that f 1 * ( F x ) = − 1 . Indeed, assume on the contrary that g 1 * ( F x ) = 0 for every g 1 * ∈ L p . Using the fact that K ₀ ⊆ K ⊆^⊥ K _p, we see that g 1 * + g * ( F x ) = 0 for every g 1 * ∈ L p and every g* ∈ K _p. Note that, by Lemma 1.1(5) and Lemma 1.4(4), K _p + L _p is dense in X*. Since X is reflexive, we arrive at Fx = 0, a contradiction. Thus, by Lemma 1.1, there exist f* ∈ K _p and f 2 * ∈ M p such that f 1 * = f * − f 2 * , and by Lemma 1.3, f * ( F x 1 ) = − f 1 * ( F x ) = 1 . It follows that

( I − F R ) F R = f * ⊗ F x 1 − f 1 * ⊗ F x − ( f * ( F x 1 ) f * ⊗ F x 1 + f 1 * ( F x ) f 1 * ⊗ F x ) = 0 ,

as desired.

Let A ∈ Alg D . Then A(I − FR)FR = 0. Since δ is derivable on R ( F , 0 ) , we have

(3.1) δ ( A ( I − F R ) R ) = δ ( A ( I − F R ) ) R + A ( I − F R ) δ ( R ) .

Notice that R ² = f*(x ₁)R by Lemma 1.3. This together with Eq. (1.1) leads to

(3.2) δ ( A ( I − F R ) R ) = δ ( A R ) − f * ( x 1 ) δ ( A F R ) = f * ⊗ S A x 1 + f 1 * ⊗ T A x + V f * ⊗ A x + W f * ⊗ A x 1 − f * ( x 1 ) ( f * ⊗ S A F x 1 + f 1 * ⊗ T A F x + V f * ⊗ A F x + W f * ⊗ A F x 1 ) .

On the other hand, by Eq. (1.1), we have

(3.3) δ ( A ( I − F R ) ) R + A ( I − F R ) δ ( R ) = δ ( A ) R − δ ( A F R ) R + A δ ( R ) − A F R δ ( R ) = f * ⊗ δ ( A ) x 1 − f 1 * ⊗ δ ( A ) x − f * ( x 1 ) f * ⊗ S A F x 1 + f 1 * ( x ) f 1 * ⊗ T A F x + V f * ( x ) f 1 ⊗ A F x − W f * ( x 1 ) f * ⊗ A F x 1 + f * ⊗ A S x 1 + f 1 * ⊗ A T x + V f * ⊗ A x + W f * ⊗ A x 1 − f * ( S x 1 ) f * ⊗ A F x 1 − f * ( x 1 ) W f * ⊗ A F x 1 + f 1 * ( T x ) f 1 * ⊗ A F x + f 1 * ( x ) V f * ⊗ A F x .

Using the fact that f * ( x 1 ) = − f 1 * ( x ) , and combining Eq. (3.1), Eq. (3.2) and Eq. (3.3), we obtain that

f 1 * ⊗ ( δ ( A ) + T A − A T − V f * ( x ) A F − f 1 * ( T x ) A F ) x = f * ⊗ ( δ ( A ) + A S − S A − W f * ( x 1 ) A F − f * ( S x 1 ) A F ) x 1 .

Since K ₀ ∩ L ₀ = (0), nonzero vectors x and x ₁ are linearly independent. So, there exists some scalar μ x ∈ C such that (δ(A) + TA − AT)x = μ _x AFx for every x ∈ K ₀. Since δ(A) + TA − AT and AF are linear, there exists a constant μ ∈ C such that (δ(A) + TA − AT)x = μAFx for all x ∈ K ₀. Taking A = I in the above equation, we have δ(I)x = μFx, which further implies that

(3.4) ( δ ( A ) + T A − A T ) x = A δ ( I ) x

for all x ∈ K ₀. Define a linear mapping Δ : Alg D → Alg D by

Δ ( A ) = δ ( A ) − A δ ( I )

for every A ∈ Alg D . Then by Eq. (3.4), Δ(AB)x = (Δ(A)B + AΔ(B))x for all x ∈ K ₀. Since K ₀ is dense in K by Lemma 1.4, Δ(AB)x = (Δ(A)B + AΔ(B))x for all x ∈ K. Applying a similar argument, we get Δ(AB)x = (Δ(A)B + AΔ(B))x for all x ∈ L. Because K + L = X, Δ is a derivation. Now, by the definition of Δ, we have

δ ( A B ) − A B δ ( I ) = Δ ( A ) B + A Δ ( B ) = ( δ ( A ) − A δ ( I ) ) B + A ( δ ( A ) − A δ ( I ) ) = δ ( A ) B − A δ ( I ) B + A δ ( B ) − A B δ ( I )

for every A , B ∈ Alg D . Consequently, δ(AB) = δ(A)B + Aδ(B) − Aδ(I)B for every A , B ∈ Alg D , that is, δ is a generalized derivation.

Claim 2. If F | K 0 ∈ C I K 0 and F | L 0 ∈ C I L 0 , then δ is a generalized derivation.

Assume that there exists a constant λ ∈ C such that Fx = λx for all x ∈ K ₀. Let P be any rank-two idempotent in Alg D . By Lemma 1.4, we have APF(I − P)|_K = λAP(I − P)|_K = 0 and A(I − P)FP|_K = λA(I − P)P|_K = 0. Applying a similar argument as above, we obtain APF(I − P)|_L = A(I − P)FP|_L = 0. Since K + L = X, we get APF(I − P) = A(I − P)FP = 0. So,

0 = δ ( A P ( I − P ) ) = δ ( A P ) ( I − P ) + A P δ ( I − P )

and

0 = δ ( A ( I − P ) P ) = δ ( A ( I − P ) ) P + A ( I − P ) δ ( P ) .

Comparing the above two equations, one can get

(3.5) δ ( A P ) = δ ( A ) P + A δ ( P ) − A P δ ( I ) .

Taking A = I in Eq. (3.5), we get Pδ(I) = δ(I)P. This together with Eq. (3.5) gives us

δ ( A P ) = δ ( A ) P + A δ ( P ) − A δ ( I ) P .

By Lemma 1.5 and the linearity of δ, we have

(3.6) δ ( A R ) = δ ( A ) R + A δ ( R ) − A δ ( I ) R

for every rank-two operator R ∈ Alg D . Let A , B ∈ Alg D . According to Eq. (3.6), we have

δ ( A B R ) = δ ( A B ) R + A B δ ( R ) − A B δ ( I ) R .

On the other hand, assoicated with Eq. (3.6), we obtain

δ ( A B R ) = δ ( A ) B R + A δ ( B R ) − A δ ( I ) B R = δ ( A ) B R + A ( δ ( B ) R + B δ ( R ) − B δ ( I ) R ) − A δ ( I ) B R = δ ( A ) B R + A δ ( B ) R + A B δ ( R ) − A B δ ( I ) R − A δ ( I ) B R .

Comparing these equations, we get

( δ ( A B ) − δ ( A ) B − A δ ( B ) + A δ ( I ) B ) f * ⊗ x 1 − f 1 * ⊗ x = 0

for every x = x ₁ + x ₂ ∈ K ₀ with x ₁ ∈ L ₀ and x ₂ ∈ M ₀ and every f * = f 1 * + f 2 * ∈ K p with f 1 * ∈ L p and f 2 * ∈ M p . Note that K ₀ ∩ L ₀ = (0) by Lemma 1.1. So, x and x ₁ are linearly independent and hence,

0 = ( δ ( A B ) − δ ( A ) B − A δ ( B ) + A δ ( I ) B ) x = ( δ ( A B ) − δ ( A ) B − A δ ( B ) + A δ ( I ) B ) x 1

for every x ∈ K ₀ and x ₁ ∈ L ₀. Since K ₀ is dense in K and L ₀ is dense in L, using the fact that K + L = X, we obtain that δ(AB) = δ(A)B + Aδ(B) − Aδ(I)B for every A , B ∈ Alg D , that is, δ is a generalized derivation. □

Applying Claim 2 of Theorem 3.1, we immediately get one of the main results in [10].

Corollary 3.2.

([10]) Let D be a strongly double triangle subspace lattice on a nonzero complex reflexive Banach space. Suppose δ : Alg D → Alg D is a linear mapping derivable at zero point, then δ is a generalized derivation.

Corresponding author: Zijie Qin, School of Mathematics and Statistics, Changshu Institute of Technology, Changshu, 215500, P.R. China, E-mail: zijieqin92@163.com

Funding source: National Natural Science Foundation of China

Award Identifier / Grant number: 12501163

Funding source: Natural Science Foundation of the Jiangsu Higher Education Institutions of China

Award Identifier / Grant number: 24KJB110001

Acknowledgments

The author would like to thank the referees for their thorough reading of the paper and many helpful comments.

Research ethics: Not applicable.
Informed consent: Not applicable.
Author contributions: The author confirms the sole responsibility for the conception of the study, presented results, and manuscript preparation.
Use of Large Language Models, AI and Machine Learning Tools: None declared.
Conflict of interest: The author states no conflict of interest.
Research funding: This work was supported by the National Natural Science Foundation of China (Grant No. 12501163) and the Natural Science Foundation of the Jiangsu Higher Education Institutions of China (Grant No. 24KJB110001).
Data availability: Not applicable.

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Received: 2024-11-20

Accepted: 2025-10-10

Published Online: 2025-11-27

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/math-2025-0213

Keywords for this article

derivations; generalized derivations; strongly double triangle subspace lattice; derivable mappings

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