All-Derivable Subsets for Nest Algebras on Banach Spaces

Transcription

1 International Mathematical Forum, Vol. 9, 014, no. 1, 1-11 HIKARI Ltd, All-Derivable Subsets for Nest Algebras on Banach Spaces Yanfang Zhang, Jinchuan Hou 1 Department of Mathematics Taiyuan University of Technology Taiyuan 03004, P.R. China sxzyf101@sina.com, jinchuanhou@aliyun.com Xiaofei Qi Department of Mathematics Shanxi University Taiyuan , P.R. China xiaofeiqisxu@aliyun.com Copyright c 014 Yanfang Zhang, Jinchuan Hou Xiaofei Qi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, reproduction in any medium, provided the original work is properly cited. Abstract Let N be a nest on a complex Banach space X 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 Keywords: Banach spaces, nest algebras, derivations, all-derivable subsets 1 Supported by NNSFC , Supported by NNSFC YSFSPC

2 Yanfang Zhang, Jinchuan Hou Xiaofei Qi 1 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 applications, 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 [, 5, 6, 10] the references therein). One approach is to characterize derivations by their local behaviors. We say that a map ϕ : A Ais derivable at a point Z Aif ϕ(a)b + Aϕ(B) =ϕ(z) for any A, B Awith AB = Z, we call such Z a derivable point of ϕ. Obviously, a linear map is a derivation if only if it is derivable at every point. It is natural 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 Ais 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] 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 [1] recently. It was shown in [1] 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 δ : A Abe a linear map. If a subset S Asatisfies that every Z Sis 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 [1], to find some all-derivable subsets for all nest algebras on complex Banach spaces without any additional assumptions on the nests, generalize the main result obtained in [1]. 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 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

3 All-derivable subsets of nest algebras 3 AlgN, that is, a linear map δ : AlgN AlgN is derivable on the subset S if only if δ is a derivation. Particularly, as a consequence we obtain the main result in [1] again by taking S = {Z}, the singleton consisting of Z, where Z is injective or of dense range. Corollary 1.. Let N be a nest on a complex Banach space X with dim X 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 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 δ : AlgN AlgN be a linear map. Let P AlgN be an idempotent operator. Then δ is derivable at both P I P if only if δ is a derivation. The paper is organized as follows. We fix some notations preliminary lemmas in Section prove Theorem 1.1 in Section 3. Preliminaries lemmas In this section let us fix some notations give some lemmas. For a (real or complex) Banach space X, denote by X B(X) the dual of 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 sts for its adjoint operator which is the linear operator from X into X defined by (T f)(x) =f(tx) for any f X x X. If f X x X, x f sts 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 only if both x f are nonzero, every rank one operator can be represented as this form. Clearly, (x f) = f x x T =(Tx), 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) off at x. In addition, we use the symbols ran(t ) ker(t ) for the range 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} X, that is, any two elements in N can compare with each other under the partial order N is closed under the formation of arbitrary closed linear span (denoted by ) intersection (denoted by ). Denote AlgN = {A B(X) :N LatA} the associated nest algebra, where LatA sts for the invariant subspace lattice of A. When N {{0},X}, we say that N is nontrivial. It is clear that AlgN = B(X) ifn is trivial. For N N, let N = {M N M N}, N + = {M N N M} N =(N ). Also, let {0} = {0}

4 4 Yanfang Zhang, Jinchuan Hou Xiaofei Qi X + = X. Denote D 1 (N )= {N N N X} D (N )= {N N N N {0}}. Note that D 1 (N ) is dense in X D (N ) is dense in X. It is clear that D 1 (N )=X if only if X X, D (N )=X if only if {0} {0} +. It is well-known that a rank one operator x f belongs to AlgN if only if there is some N N such that x N f N. For more information on nest algebras, we refer to [3, 11]. The following lemmas are needed to prove the main result. Lemma.1-.3 are easily checked. Lemma.1. Let A be a unital ring. Suppose that δ : A Ais an additive map. If nonzero element Z A is a derivable point of δ, then, δ(i)z = Zδ(I) =0. Lemma.. Let A be a unital ring. Suppose that δ : A Ais an additive map. If nonzero element Z Ais a derivable point of δ, then, (1) for every element N Awith N =0, we have δ(nz)=δ(n)z + Nδ(Z) δ(i)nz [δ(n)n + Nδ(N) δ(i) Nδ(I)N]Z =0; () for every element N Awith N =0, we have δ(zn)=δ(z)n + Zδ(N) ZNδ(I) Z[δ(N)N + Nδ(N) δ(i) Nδ(I)N] =0. Lemma.3. Let A be a unital ring. Suppose that δ : A Ais an additive map. If nonzero element Z Ais a derivable point of δ, then, for any invertible element A A, δ(a 1 Z)=A 1 δ(z) A 1 δ(a)a 1 Z δ(za 1 )=δ(z)a 1 ZA 1 δ(a)a 1. Lemma.4. Let A be a unital complex algebra. Suppose that δ : A A is a linear map. If Z Ais a nonzero element such that δ is derivable at Z, then the following are true. (1) For any idempotent P AlgN, we have δ(pz)=δ(p )Z + Pδ(Z) δ(i)pz

5 All-derivable subsets of nest algebras 5 δ(p )Z =(δ(p )P + Pδ(P ))Z. () For any idempotent P AlgN, we have δ(zp)=δ(z)p + Zδ(P ) ZPδ(I) Zδ(P )=Z(δ(P )P + Pδ(P )). Proof. Let P Abe any idempotent element, that is, P satisfies P P. Since Z =(I 1 3iP )(I 1+ 3iP )Z, we have δ(z) = δ(i 1 3i P )(I 1+ 3i P )Z +(I 1 3i P )δ(z 1+ 3i PZ) = δ(i)z 1+ 3iδ(I)PZ 1 3i 1+ 3i δ(pz) 1 3i δ(p )Z + δ(p )PZ + δ(z) Pδ(Z)+Pδ(PZ). (.1) On the other h, Z =(I 1+ 3i P )(I 1 3i P )Z gives δ(z) = δ(i)z 1 3iδ(I)PZ 1+ 3iδ(P )Z + δ(p )PZ + δ(z) 1 3iδ(PZ) 1+ 3iPδ(Z)+Pδ(PZ). (.) It follows from Eq.(.1) Eq.(.) that δ(pz)=δ(p )Z + Pδ(Z) δ(i)pz. (.3) Substituting Eq.(.3) into Eq.(.1) notice that δ(i)z = 0, one obtains which completes the proof of (1). By considering the equation Z = Z(I 1 3i δ(p )Z =(δ(p )P + Pδ(P ))Z, (.4) P )(I 1+ 3i P )=Z(I 1+ 3i P )(I 1 3i P ), one can get (). The following Lemmas.5-.7 come from [1]. Lemma.5. Let N be a nest on a real or complex Banach space X with dim X. Assume that δ : AlgN AlgN is a linear map satisfying δ(p )=δ(p)p + Pδ(P ) for all idempotent operators P AlgN δ(n)n + Nδ(N) =0for all operators N AlgN with N =0.IfX X, then,

6 6 Yanfang Zhang, Jinchuan Hou Xiaofei Qi (1) for any x X f X, we have δ(x f) ker(f) span{x}; () there exist linear transformations B : X X C : X X such that δ(x f) =Bx f + x Cf Bx,f + x, Cf =0for all x X f X. Lemma.6. Let N be a nest on a real or complex Banach space X with dim X. Assume that δ : AlgN AlgN is a linear map satisfying δ(p )=δ(p)p + Pδ(P ) for all idempotent operators P AlgN δ(n)n + Nδ(N) =0for all operators N AlgN with N =0.If{0} {0} +, then (1) for any f X any x {0} +, we have δ(x f) (ker x ) span{f}; () there exist linear transformations B : X X C : κ({0} + ) X such that δ(x f) = Bf x + f Cx Bf,x + f,cx =0holds for all x {0} + f X. Lemma.7. Let X be a Banach space over the real or complex field F. Suppose that N is a nest on X δ : AlgN AlgN is a linear map satisfying δ(p )=δ(p)p + Pδ(P ) for all idempotent operators P AlgN δ(n)n + Nδ(N) =0for all operators N AlgN with N =0. If X = X {0} = {0} +, then (1) there exists a bilinear functional β :(D 1 (N ) D (N )) AlgN F such that (δ(x f) β x,f I) ker(f) span{x} holds for all x f AlgN ; () there exist linear transformations B : D 1 (N ) D 1 (N ) C : D (N ) D (N ) such that δ(x f) β x,f I = x Cf + Bx f holds for all x f AlgN. 3 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, δ : AlgN AlgN is a linear map derivable on S. Let us show that δ is a derivation. By Lemma.1, for any Z S, we have δ(i)z = 0. As the linear manifold spanned by the ranges of all Z Sis dense in X, we must have δ(i) =0. (3.1) It follows from Lemma. (1) Lemma.4 (1) that the following statements are true: For any N AlgN with N = 0, we have δ(n)n + Nδ(N) = 0 (3.)

7 All-derivable subsets of nest algebras 7 δ(nz)= δ(n)z + Nδ(Z) Z S; (3.3) for any idempotent operators P AlgN, we have δ(p )=δ(p )P + Pδ(P ) (3.4) δ(pz)=δ(p )Z + Pδ(Z) Z S. (3.5) We consider three cases. Case 1. X X. In this case, by Eqs.(3.) (3.4), δ satisfies the hypotheses of Lemmas.5. So, there exist linear transformations B : X X C : X X such that δ(x f) =Bx f + x Cf Bx,f + x, Cf = 0 for all x X f X. Thus, as x f is either a multiple of a idempotent or a square zero operator, by Eqs.(3.3) (3.5), the linearity of δ, for any Z S,we have δ(x fz)=(bx f)z +(x Cf)Z +(x f)δ(z). (3.6) Note that, for any T any x f AlgN, there exists some λ C such that λ > T (λi T ) 1 x f < 1. Then both λi T λi T x f =(λi T )(I (λi T ) 1 x f) are invertible with their inverses still in AlgN. It is obvious that (I (λi T ) 1 x f) 1 = I +(1 α) 1 (λi T ) 1 x f, where α = (λi T ) 1 x, f. Thus, for any T any x f AlgN with x X f X, take λ C such that λ > T (λi T ) 1 x f < 1. By Lemma.3, Eqs.(3.1) (3.6), we get δ(z) = δ(λi T x f)(i +(1 α) 1 (λi T ) 1 x f)(λi T ) 1 Z +(λi T x f)δ((i +(1 α) 1 (λi T ) 1 x f)(λi T ) 1 Z) = δ(z) (1 α) 1 B(x (λi T ) 1 f)z (1 α) 1 δ(t )(λi T ) 1 (x (λi T ) 1 f)z (x (λi T ) 1 Cf)Z +(x C(λI T ) 1 f)z +(1 α) 1 (λi T )B(λI T ) 1 (x (λi T ) 1 f)z (x (λi T ) 1 δ(t ) (λi T ) 1 f)z (1 α) 1 ( (λi T ) 1 x, Cf + B(λI T ) 1 x, f )(x (λi T ) 1 f)z. As (λi T ) 1 x, Cf + B(λI T ) 1 x, f = 0, the above equation becomes 0= (1 α) 1 B(x (λi T ) 1 f)z +(1 α) 1 δ(t )(λi T ) 1 (x (λi T ) 1 f)z +(x (λi T ) 1 Cf)Z (x C(λI T ) 1 f)z (1 α) 1 (λi T )B(λI T ) 1 (x (λi T ) 1 f)z +(x (λi T ) 1 δ(t ) (λi T ) 1 f)z.

8 8 Yanfang Zhang, Jinchuan Hou Xiaofei Qi Since the span{ran(z) :Z S}is dense in X, we must have so 0= (1 α) 1 Bx (λi T ) 1 f +(1 α) 1 δ(t )(λi T ) 1 x (λi T ) 1 f +x (λi T ) 1 Cf x C(λI T ) 1 f (1 α) 1 (λi T )B(λI T ) 1 x (λi T ) 1 f +x (λi T ) 1 δ(t ) (λi T ) 1 f, [δ(t )(λi T ) 1 (λi T )B(λI T ) 1 + B]x (λi T ) 1 f = x (1 α)[c(λi T ) 1 (λi T ) 1 C (λi T ) 1 δ(t ) (λi T ) 1 ]f holds for every x X. This entails that there is a scalar β λ such that on X. It follows that δ(t )(λi T ) 1 (λi T )B(λI T ) 1 + B = β λ I δ(t )=BT TB + β λ (λi T ). Taking different λ 1,λ in the above equation with T I, we see that, (λ 1 β λ1 λ β λ )I =(β λ1 β λ )T, which implies that β λ1 = β λ = 0. Consequently, δ(t )=BT TB holds for all T AlgN, that is, δ is a derivation. Case. {0} {0} +. In this case, δ satisfies the assumptions in Lemmas.6. A similar argument of Case 1 shows that δ is a derivation, too, we omit the details. Case 3. {0} = {0} + X = X. In this case X is infinite dimensional every nonzero element N in N is infinite dimensional. By Lemmas.7, there exists a bilinear functional β :(D 1 (N ) D (N )) AlgN C, linear transformations B : D 1 (N ) D 1 (N ) C : D (N ) D (N ) such that (δ(x f) β x,f I) ker(f) span{x} δ(x f) β x,f I = x Cf + Bx f (3.7) hold for all x f AlgN. Then, by Eqs.(3.3)-(3.5), Lemma.3 the linearity of δ, for any Z S, we have δ(x fz)=(bx f)z +(x Cf)+β x,f Z +(x f)δ(z). (3.8)

9 All-derivable subsets of nest algebras 9 Let β x,f = x, Cf + Bx,f. Then a similar argument as that in [1] shows that { βx,f =0, if x, f 0; β x,f = β x,f, if x, f =0. For any T x f AlgN, take λ such that λ > T (λi T ) 1 x f < 1. Note that I =(λi T x f)(i +(1 α) 1 (λi T ) 1 x f)(λi T ) 1, where α = (λi T ) 1 x, f. Since δ is derivable at Z S, by Lemma.3, Eqs.(3.3) (3.5), we have δ(z) = δ(λi T x f)[i +(1 α) 1 (λi T ) 1 x f](λi T ) 1 Z +(λi T x f)δ((i +(1 α) 1 (λi T ) 1 x f)(λi T ) 1 Z) = [ δ(t ) Bx f x Cf β x,f I][(λI T ) 1 Z +(1 α) 1 (λi T ) 1 (x f)(λi T ) 1 Z] +[λi T x f][(λi T ) 1 δ(z)+(λi T ) 1 δ(t )(λi T ) 1 Z +(1 α) 1 (B(λI T ) 1 (x f)(λi T ) 1 Z +(λi T ) 1 x ((λi T ) 1 ) fδ(z)+β (λi T ) 1 x,((λi T ) 1 ) fz)]. Note that (1 α) 1 α =(1 α) 1 1 β x,f = x, Cf + Bx,f. The above equation gives β x,f (λi T ) 1 Z (1 α) 1 β (λi T ) 1 x,(λi T ) 1 f(λi T )Z = (1 α) 1 δ(λi T )((λi T ) 1 x (λi T ) 1 f)z (1 α) 1 (Bx (λi T ) 1 f)z (x (λi T ) 1 Cf)Z (1 α) 1 β(λi T ) 1 x,f(x (λi T ) 1 f)z (1 α) 1 β x,f ((λi T ) 1 x (λi T ) 1 f)z +(x f(λi T ) 1 δ(λi T )(λi T ) 1 )Z +(x C(λI T ) 1 f)z +(1 α) 1 ((λi T )B(λI T ) 1 x (λi T ) 1 f)z (1 α) 1 β (λi T ) 1 x,(λi T ) 1 f(x f)z. Since the linear span of {ran(z) :Z S}is dense in X, the above equation implies that β x,f (λi T ) 1 (1 α) 1 β (λi T ) 1 x,(λi T ) 1 f(λi T ) = (1 α) 1 δ(λi T )(λi T ) 1 x (λi T ) 1 f (1 α) 1 Bx (λi T ) 1 f x (λi T ) 1 Cf (1 α) 1 β(λi T ) 1 x,fx (λi T ) 1 f (1 α) 1 β x,f (λi T ) 1 x (λi T ) 1 f +x f(λi T ) 1 δ(λi T )(λi T ) 1 + x C(λI T ) 1 f +(1 α) 1 (λi T )B(λI T ) 1 x (λi T ) 1 f (1 α) 1 β (λi T ) 1 x,(λi T ) 1 f(x f). (3.9)

10 10 Yanfang Zhang, Jinchuan Hou Xiaofei Qi By a corresponding argument as in the proof of the main result in [1], 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 [1] 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 Sfrom the left sides then applying Lemmas. The proof of Theorem 1.1 is completed. References [1] R-L. An, J-C. Hou, Characterizations of derivations on triangular ring: Additive maps derivable at idempotents, Lin. Alg. Appl., 431 (009), [] R. L. Crist, Local derivations on operator algebras, J. Func. Anal., 135 (1996), [3] K. R. Davision, Nest Algebras, Pitman Research Notes in Mathematics Series, vol. 191, Longman Scientific Technical, Burnt mill Harlow, Essex, UK [4] J-C. Hou, X-F. Qi, Additive maps derivable at some point on J -subspace lattice algebras, Lin. Alg. Appl., 49 (008), [5] R. V. Kadison, Local derivations, J. Algebra, 130 (1990), [6] D. R. Larson, A. R. Sourour, Local derivations local automorphisms of B(X), Proc. Sympo. in Pure Math., 51 (1990), [7] J. Li, Z. Pan, On derivable mappings, J. Math. Anal. Appl., 374 (011), [8] X-F. Qi, J-C. Hou, Characterizations of derivations of Banach space nest algebras: All-derivable points, Lin. Alg.. Appl., 43 (010), [9] X-F. Qi, J-C. Hou, Full-derivable points of J -subspace lattice algebras, Rocky Mountain J. Math., to appear. [10] P. Šemrl, Local automorphisms derivations on B(H), Proc. Amer. Math. Soc., 15 (1997),

11 All-derivable subsets of nest algebras 11 [11] M. Spivac, Derivations nest algebras on Banach spaces, Israel. J. Math., 50() (1985), [1] Y-F. Zhang, J-C. Hou, X-F. Qi, Characterizing derivations for any nest algebras on Banach spaces by their behaviors at an injective operator, arxiv: [math. FA] 013. [13] J. Zhou, Linear mappings derivable at some nontrivial elements, Lin. Alg. Appl., 435 (011), [14] J. Zhu, C. Xiong, R. Zhang, All-derivable points in the algebra of all upper triangular matrices, Lin. Alg. Appl., 49(4) (008), [15] J. Zhu, C. Xiong, Derivable mappings at unit operator on nest algebras, Lin. Alg. Appl., 4 (007), [16] J. Zhu, S. Zhao, Characterization of all-derivable points in nest algebras, Proc. Amer. Math. Soc., 141 (013), Received: November 5, 013