All-Derivable Subsets for Nest Algebras on Banach Spaces

Let N be a nest on a complex Banach space X and let AlgN be the associated nest algebra. We say that a subset S ⊂ AlgN is an all-derivable subset of AlgN if every linear map δ from AlgN into itself derivable on S (i.e. δ satisfies that, for each Z ∈ S, δ(A)B + Aδ(B) = δ(Z) for any A,B ∈ AlgN with AB = Z) is a derivation. In this paper, we show that S is an all-derivable subset of AlgN if span{ran(Z) : Z ∈ S} is dense in X or ∩{ker Z : Z ∈ S} = {0}. Mathematics Subject Classification: 47B47, 47L35


Introduction
A linear map δ : A → A, where A is an algebra, is a derivation if δ(AB) = δ(A)B + Aδ(B) for all A, B ∈ A. The derivations are very important linear maps both in theory and applications, and were studied intensively.Particularly, the question of under what conditions that a linear (even additive) map becomes a derivation attracted much attention of authors (for instance, see [2,5,6,10] and the references therein).One approach is to characterize derivations by their local behaviors.We say that a map ϕ : A → A is derivable at a point Z ∈ A if ϕ(A)B + Aϕ(B) = ϕ(Z) for any A, B ∈ A with AB = Z, and we call such Z a derivable point of ϕ.Obviously, a linear map is a derivation if and only if it is derivable at every point.It is natural and interesting to ask the question whether or not a linear map is a derivation if it is derivable only at one given point.As usual, we say that an element Z ∈ A is an all-derivable point of A if every linear map on A derivable at Z is in fact a derivation.So far, we have known that there exist many all-derivable points (or full-drivable points) for certain (operator) algebras (see [1,4,7,8,13,14,16] and the references therein).However, zero point 0 is not an all-derivable point for any algebra because the generalized derivations are derivable at 0 [8].
The invertible elements are all-derivable points for many algebras.For instance, invertible elements are all-derivable points of J -subspace lattice algebras ( [4,9]).As nest algebras are of an important class of operator algebras, there are many papers on finding all-derivable points of nest algebras.But so far all known results were obtained under some additional assumptions on nests or the spaces (Ref.[1,8,15]).The problem of finding all-derivable points of any nest algebras on any Banach spaces was attacked by [12] recently.It was shown in [12] that every injective operator as well as every operator of dense range is an all-derivable point of any nest algebra on any Banach space.
More generally, we may introduce a notion of derivable subsets.Let A be an algebra and δ : A →→ A be a linear map.If a subset S ⊂ A satisfies that every Z ∈ S is a derivable point of δ, we say that S is a derivable subset of δ or δ is derivable on S; if, for any linear map ϕ : A → A, ϕ is derivable on S will imply that ϕ is a derivation, we say that S is an all-derivable subset of the algebra A. Clearly, a derivation is derivable on any subset.
The purpose of the present paper is, based on the study in [12], to find some all-derivable subsets for all nest algebras on complex Banach spaces without any additional assumptions on the nests, and generalize the main result obtained in [12].
The following is the main result of this paper.Theorem 1.1.Let N be a nest on a complex Banach space X with dim X ≥ 2 and S ⊂ AlgN .If ∩{ker Z : Z ∈ S} = {0} or if span{ran(Z) : Z ∈ S} is dense in X, then S is an all-derivable subset of the nest algebra AlgN , that is, a linear map δ : AlgN → AlgN is derivable on the subset S if and only if δ is a derivation.
Particularly, as a consequence we obtain the main result in [12] again by taking S = {Z}, the singleton consisting of Z, where Z is injective or of dense range.
Corollary 1.2.Let N be a nest on a complex Banach space X with dim X ≥ 2 and δ : AlgN → AlgN be a linear map.Let Z ∈ AlgN be an injective operator or an operator with dense range.Then δ is derivable at the operator Z if and only if δ is a derivation.
The following corollary is also immediate.Corollary 1.3.Let N be a nest on a complex Banach space X with dim X ≥ 2 and δ : AlgN → AlgN be a linear map.Let P ∈ AlgN be an idempotent operator.Then δ is derivable at both P and I − P if and only if δ is a derivation.
The paper is organized as follows.We fix some notations and preliminary lemmas in Section 2 and prove Theorem 1.1 in Section 3.

Preliminaries and lemmas
In this section let us fix some notations and give some lemmas.
For a (real or complex) Banach space X, denote by X * and B(X) the dual of and the algebra of all bounded linear operators on X, respectively.The canonical embedding from X into X * * is denoted by κ.For any T ∈ B(X), T * stands for its adjoint operator which is the linear operator from X * into X * defined by (T * f )(x) = f (T x) for any f ∈ X * and x ∈ X.If f ∈ X * and x ∈ X, x ⊗ f stands for the operator defined by (x ⊗ f )(y) = f (y)x for any y ∈ X.It is easily seen that rank(x ⊗ f ) ≤ 1; x ⊗ f is of rank one if and only if both x and f are nonzero, and every rank one operator can be represented as this form.Clearly, (x ⊗ f ) * = f ⊗ x * * and x * * T * = (T x) * * , where x * * = κ(x).For any non-empty subset N ⊆ X, denote by N ⊥ the annihilator of N, that is, N ⊥ = {f ∈ X * : f (x) = 0 for every x ∈ X}.Some times we use x, f to present the value f (x) of f at x.In addition, we use the symbols ran(T ) and ker(T ) for the range and the kernel of operator T , respectively.
A nest N in X is a complete chain of closed linear subspaces of X containing the trivial subspaces {0} and X, that is, any two elements in N can compare with each other under the partial order "⊆" and N is closed under the formation of arbitrary closed linear span (denoted by ) and intersection (denoted by ).Denote AlgN = {A ∈ B(X) : N ⊆ LatA} the associated nest algebra, where LatA stands for the invariant subspace lattice of A.
It is well-known that a rank one operator x ⊗ f belongs to AlgN if and only if there is some N ∈ N such that x ∈ N and f ∈ N ⊥ − .For more information on nest algebras, we refer to [3,11].
The following lemmas are needed to prove the main result.Lemma 2.1-2.3 are easily checked.Lemma 2.1.Let A be a unital ring.Suppose that δ : Lemma 2.2.Let A be a unital ring.Suppose that δ : (1) for every element N ∈ A with N 2 = 0, we have (2) for every element N ∈ A with N 2 = 0, we have

Lemma 2.3. Let A be a unital ring. Suppose that δ : A → A is an additive map. If nonzero element Z ∈ A is a derivable point of δ, then, for any invertible element
Lemma 2.4.Let A be a unital complex algebra.Suppose that δ : A → A is a linear map.If Z ∈ A is a nonzero element such that δ is derivable at Z, then the following are true.

Proof of the main result
In this section we complete the proof of Theorem 1.1 by using the lemmas in the previous section.
Proof of Theorem 1.1.The "if " part is obvious.We give the proof of the "only if" part here.
Assume that S ⊂ AlgN is a subset with span{ran(Z) : Z ∈ S} is dense in X, and δ : AlgN → AlgN is a linear map derivable on S. Let us show that δ is a derivation.
By Lemma 2.1, for any Z ∈ S, we have δ(I)Z = 0.As the linear manifold spanned by the ranges of all Z ∈ S is dense in X, we must have We consider three cases.Case 1. X − = X.In this case, by Eqs.(3.2) and (3.4), δ satisfies the hypotheses of Lemmas 2.5.So, there exist linear transformations B : X → X and C : − .Thus, as x ⊗ f is either a multiple of a idempotent or a square zero operator, by Eqs.(3.3) and (3.5), and the linearity of δ, for any Z ∈ S, we have Note that, for any T and any x ⊗ f ∈ AlgN , there exists some λ ∈ C such that |λ| > T and (λI Thus, for any T and any x⊗f ∈ AlgN with x ∈ X and f ∈ X ⊥ − , take λ ∈ C such that |λ| > T and (λI − T ) −1 x f < 1.By Lemma 2.3, Eqs.(3.1) and (3.6), we get Since the span{ran(Z) : Z ∈ S} is dense in X, we must have and so holds for every x ∈ X.This entails that there is a scalar β λ such that Taking different λ 1 , λ 2 in the above equation with T = I, we see that, (λ T , which implies that β λ 1 = β λ 2 = 0. Consequently, δ(T ) = BT − T B holds for all T ∈ AlgN , that is, δ is a derivation.Case 2. {0} = {0} + .In this case, δ satisfies the assumptions in Lemmas 2.6.A similar argument of Case 1 shows that δ is a derivation, too, and we omit the details.
In this case X is infinite dimensional and every nonzero element N in N is infinite dimensional.
By Lemmas 2.7, there exists a bilinear functional β : hold for all x ⊗ f ∈ AlgN .Then, by Eqs.(3.3)-(3.5),Lemma 2.3 and the linearity of δ, for any Z ∈ S, we have Let βx,f = x, Cf + Bx, f .Then a similar argument as that in [12] shows that βx,f = 0 For any T and x ⊗ f ∈ AlgN , take λ such that |λ| > T and (λI − T ) −1 x f < 1.Note that Since δ is derivable at Z ∈ S, by Lemma 2.3, Eqs.(3.3) and (3.5), we have Since the linear span of {ran(Z) : Z ∈ S} is dense in X, the above equation implies that (3.9)By a corresponding argument as in the proof of the main result in [12], we see that β x,f = 0 for any x ⊗ f ∈ AlgN .Then by Eq.(3.7), we get that δ(x ⊗ f ) = x ⊗ Cf + Bx ⊗ f holds for all x ⊗ f ∈ AlgN .Now, a similar argument as in [12] ensures that δ is a derivation.
Therefore, every subset S ⊂ AlgN satisfying span{ran(Z) : Z ∈ S} is dense in X is an all-derivable subset of AlgN .
The fact that every subset S ⊂ AlgN satisfying ∩{ker Z : Z ∈ S} = {0} is also an all-derivable subset of AlgN can be proved similarly by multiplying Z ∈ S from the left sides and then applying Lemmas.
The proof of Theorem 1.1 is completed.2