On Transitive and Localizing Operator Algebras

Transcription

1 International Mathematical Forum, Vol. 6, 2011, no. 60, On Transitive and Localizing Operator Algebras Ömer Gök and Elif Demir Yıldız Technical University Faculty of Arts and Sciences Mathematics Department, Davutpaşa Campus İstanbul, Turkey Abstract. In this paper we are interested in several conditions of operator algebras on Banach spaces to be transitive or localizing and to have an invariant subspaces. Mathematics Subject Classification: 47A15, 46B25, 47L10 Keywords: Operator algebra, localizing algebra, transitive algebra, invariant subspace 1. Introduction Throughout this paper X will be a real Banach space and L(X) be the space of all continuous linear operators on X. If T L(X) we say that T has an invariant subspace if there exists a closed non-zero proper subspace Y of X such that T (Y ) Y. We say that a subspace Y is hyperinvariant for T if Y is invariant under every operator in {T } c. Here S c is the commutant of a set S L(X), that is S c = {A L(X) B S, AB = BA}. A subset S L(X) is said to be transitive if Sx is dense in X for every non-zero x X, where Sx = {Ax A S}. In this work, we deal with the conditions of a subalgebra A of L(X) and its dual algebra A of L(X ) being localizing or transitive. Also according to these conditions we are interested in a quasinilpotent operator T and its adjoint operator T about having hyperinvariant subspace.

2 2956 Ö. Gök and E. Demir Definition 1. Let X be a Banach space, X be the topological dual of X and T L(X). Then its adjoint operator T L(X ) is defined via the duality formula: <x,t x >=< T x, x > for all x X and x X. Definition 2. Let X be a Banach space and T L(X).T is said to be a quasinilpotent operator if lim n T n x 1 n = 0 for all x X. Definition 3. Let X be a Banach space.we know that L(X) is a vector space with the usual algebraic operations. Also L(X) is an operator algebra and multiplication is defined as for T,S L(X), T.S = T S L(X). That is a vector subspace A of L(X) is called an algebra of operators ( or a subalgebra of L(X) ) whenever T,S A imply T S A. We will shortly write TS instead of T S. Lemma 4.([5]) Let X be a vector space, R, S, T L(X) and scalar. Then: (a)r(st)=(rs)t (b)r(s + T )=RS + RT (c)(s + T )R = SR + TR (d)i X T = TI X = T (e)( S)T = (ST)=S( T ). These properties are exactly the extra axioms which a vector space must satisfy in order to be an algebra. Let A L(X) be an algebra of operators.then its dual algebra is the subalgebra A of L(X ) defined by A = {T L(X ) T A}. A is an operator algebra. To see this, firstly all S,T A must imply S T A, [1]. Since A is an algebra, for all T,S A we have TS A and S T =(TS) so that S T A. In addition to this, A satisfies the properties listed in Lemma4. Definition 5. Let A L(X) be an algebra of operators. A is transitive if it has no common invariant subspaces. Also we can say that for an operator T L(X), {T } c is transitive if T has no hyperinvariant subspaces, [4]. Definition 6. Let A L(X) be an algebra of operators. A is said to be localizing if there exists a closed ball B in X such that 0 / B and for every

3 On transitive and localizing operator algebras 2957 sequence {x n } in B there is a subsequence {x ni } and a sequence {S i } in A such that S i 1 and {S i x ni } converges in norm to a nonzero vector. Definition 7. Let A L(X) be an algebra of operators. A is said to be sesquitransitive if for every non-zero z X there is a positive real C = C(z) such that for every x linearly independent of z, for every y X, and every ε>0 there exists A S such that Ax y <εand Az C z. Theorem 8. ([4]) If X is a real Banach space then every sesquitransitive subalgebra A of L(X) containing a compact operator is WOT-dense in L(X). (WOT means weak operator topology) Definition 9.Let X be a Banach space and (X,., ) be ordered vector space. If x y implies x y then X is called a Banach lattice. If X is an ordered vector space, then the set X + = {x X : x 0} is referred to as the positive cone of X. Definition 10.A Banach lattice X is said to be an AL-space if x + y = x + y for all x, y X + with x y =0. Definition 11.A Banach lattice X is said to be an AM-space if x y = max{ x, y } for all x, y X + with x y =0. Corollary 12. If X is a AL or AM-space then every sesquitransitive subalgebra A of L(X) containing a weak compact operator is WOT-dense in L(X). If T is an injective compact operator, then {T } c is a localizing algebra, [4]. Theorem 13. If T is a one-to-one,onto and compact operator on a Banach space X, then {T } c is localizing. Proof. If T is one-to-one, onto and compact operator then its adjoint is also one-to-one, onto and compact operator. Hence {T } c is localizing. We use the method of minimal vectors and minimal functionals based on [6] in the proof of the following theorem. Let T be a bounded operator on Banach space X and T be the adjoint of T. Since we will be interested in the hyperinvariant subspaces of T, without loss of generality, we can assume T is one-to-one and has dense range, that is Range(T )=X. Otherwise kert or Range(T ) would be hyperinvariant for T.

4 2958 Ö. Gök and E. Demir Fix a vector (functional) f 0 0inX and a positive real ε< f 0. Let K =(T ) 1 B[f 0,ε]. K is a convex closed set. To see this, write K =(T ) 1 B[f 0,ε]={(T ) 1 f : f B[f 0,ε]} X. For λ [0, 1] and all (T ) 1 f 1, (T ) 1 f 2 K, λ(t ) 1 f 1 +(1 λ)(t ) 1 f 2 K must be satisfied. To achieve this, it is enough to show λf 1 +(1 λ)f 2 B[f 0,ε] from the definition of K.We have f 1,f 2 B[f 0,ε]. We must show that λf 1 +(1 λ)f 2 f 0 ε. Indeed, λf 1 +(1 λ)f 2 f 0 = λf 1 λf 0 + λf o +(1 λ)f 2 f 0 λ(f 1 f 0 ) + λ(f 0 f 2 ) + f 2 f 0 ε. Next we will show that K is a closed set. Take any g K(closure of K) sog K must be satisfied.since g K there exists {g n } K such that lim n g n g = 0. Since {g n } K for all n N, we can write g n =(T ) 1 f n, f n B[f 0,ε], so f n = T g n. We will write simply B instead of B[f 0,ε]. Since B is a closed ball, there exists an element f B such that f n f. T g n f = T g n f n + f n f T g n f n + f n f. So we have T g n f f n f.if we take limit for n, we get lim n T g n f =0. That means T g n f B. Since T is invertible and continuous (T ) 1 is also continuous due to the open mapping theorem, [3]. Thus we have g n (T ) 1 f. We know that f B and g n g so g =(T ) 1 f K. That is g K. 0 is not contained in K, otherwise if 0 K since T is one-to-one 0 would be in B but we know that 0 / K from the choice of B[f 0,ε]. Let d = inf z K z. Then d>0.here, z K =(T ) 1 B so z = sup x =1 z(x). We can find y K with y 2d, such a y will be referred to as a 2-minimal vector for f 0,ε and T. The set K B[0,d] is the set of all minimal vectors, this set maybe empty.( a minimal vector is an element of K such that z = d )If z is a minimal vector,

5 On transitive and localizing operator algebras 2959 since z K =(T ) 1 B then T z B. Since z is an element of minimal norm in K, then, in fact, T z S[f 0,ε]. T is one-to-one, we have T B[0,d] B[f 0,ε]=T (B[0,d] K) S[f 0,ε]. It follows that T B[0,d] and B(f 0,ε) are two disjoint convex sets. Indeed, if we take an element t of T B[0,d] we can write t = T f 1 for an f 1 B[0,d], so f 1 d.if t B(f 0,ε) there would be t f 0 <ε.since t = T f 1, we would have T f 1 f 0 <ε.then we would write T f 1 f 0 T. f 1 + f 0 T d + f 0 < ε which leads to a contradiction. Since one of them has nonempty interior, they can be separated by a continuous linear functional, [2]. That is, there exists a functional ˆf X with ˆf = 1 and a positive real c such that ˆf T B[0,d] c and ˆf B(f0,ε) c. By continuity, ˆf B[f0,ε] c. We say that ˆf is a minimal functional for f 0,εand T. We claim that ˆf(f 0 ) ε.indeed, for every f with f 1 we have f 0 εf f 0 = ε f ε so f 0 εf B[f 0,ε]. Then ˆf(f 0 εf) c, so that ˆf(f 0 ) ˆf(εf) c and ˆf(f 0 ) c + ε ˆf(f). If we take sup over all f with f 1, sup f 1 ˆf(f0 ) c + εsup f 1 ˆf(f). Since sup f 1 ˆf(f) = ˆf we have ˆf(f 0 ) c + ε ˆf ε, thus ˆf(f 0 ) ε. The hyperplane T ˆf = c separates K and B[0,d]. Indeed, if f B[0,d] then (T ˆf)(f) = ˆf(T f), since T f T B[0,d] we have (T ˆf)(f) c, and if f K =(T ) 1 B then T f B so (T ˆf)(f) c. For every f with f 1 we have df B[0,d]. (T ˆf)(df ) c. It follows that (T ˆf)(df )= ˆf[T (df )] = d ˆf(T f) so we can write ˆf(T f) c d sup over all f with f 1 we have and if we take sup f 1 ˆf(T f) = sup f 1 (T ˆf)(f) = T ˆf c d.

6 2960 Ö. Gök and E. Demir On the other hand, for every δ>0there exists f K with f d + δ. Since T ˆf c, d+δ (T ˆf)(f) c c d+δ f then we have T ˆf = c d. For every f K we have (T ˆf)(f) c = d T ˆf since (T ˆf)(f) c. Ify is a minimal vector we have y 2d so d 1 y. Then 2 (1) (T ˆf)(y) d T ˆf 1 2 T ˆf y. Theorem 14.([4]) Let T be a quasinilpotent operator on a Banach space X, X be the dual of X and T be the adjoint of T.If{T } c is localizing then T has a hyperinvariant subspace. Proof. Without loss of generality, T is one-to-one and has dense range. Let f 0 0 and ε (0, f 0 ) be such that B = B[f 0,ε]. For every n 1 choose a 2-minimal vector y n and a minimal functional ˆf n for f 0, ε and (T ) n. Since T is quasinilpotent, there is a subsequence {y ni } such that y n i 1 y ni 0. Indeed, otherwise there would exist δ>0such that y n 1 y n >δfor all n, so that y 1 δ y 2... δ n y n+1.since (T ) n y n+1 (T ) 1 B, we have (T ) n y n+1 d y 1 2 δn 2 y n+1. It follows that (T ) n δn 2, which contradicts the quasinilpotence of T. Since ˆf = 1 for all i, we can write that {f ni } weak*- converges to some g X due to the Alaoglu theorem, [3]. Since ˆf(f 0 ) ε for all n, then g(f 0 ) ε. In particular, g 0. Let take the sequence {(T ) n i 1 y ni 1} i=1. It is contained in B, so we can find a sequence {K i } in {T } c such that K i 1 and K i (T ) n i 1 y ni 1 converges in norm to some ω 0 due to the being a localizing algebra of {T } c. Put Y = {T } c T ω = {S T ω S {T } c }. We can easily verify that Y is a linear subspace of X. Let take S 1T ω, S 2T ω Y and scalars α, β. For this elements we will show αs 1T ω +βs 2T Y. From the linearity of S 1,S 2 and T we have, αs 1T ω + βs 2T ω =(αs 1)T ω +(βs 2)T ω =(αs 1 + βs 2)T ω.

7 On transitive and localizing operator algebras 2961 We know that {T } c is linear subspace of L(X ) then αs 1 + βs 2 {T } c. Hence Y is a linear subspace of X. In addition to this, it is clear that Y is invariant under {T } c. Also notice that Y is nontrivial because T is one-to-one and 0 T ω Y. We will show that Y kerg, so that Y is a proper T - hyperinvariant subspace. Take S {T } c ; we will show that g(s T ω) = 0. It follows from (T ˆf)(y) d T ˆf 1 2 T ˆf y that [(T ) n i ˆfni ](y ni ) 0 for every i, so that X = span(y ni ) ker((t ) n i ˆfni ). Then we can write S K i y ni 1 = α i y ni + r i, where α i is a scalar and r i ker((t ) n i ˆfni ). We claim that α i 0. Indeed, (2) ((T ) n i ˆfni )(S K i y ni 1) =α i ((T ) n i ˆfni )(y ni ), and,combining this with (1), we get (3) ((T ) n i ˆfni )(S K i y ni 1) α i 2 (T ) n i ˆfni y ni. On the other hand, (4) ((T ) n i ˆfni )(S K i y ni 1) (T ) n i ˆfni S y ni 1. It follows from (3) and (4) that so α i 2 (T ) n i ˆfni y ni (T ) n i ˆfni S y ni 1 We know that y n i 1 y ni Then (2) yields that 0 then we get α i 2 S y n i 1 y ni 0. ˆf ni ((T ) n i S K i y ni 1) = α i ˆfni ((T ) n i y ni ) α i ˆf ni (T ) n i y ni α i.1.( f 0 + ε) 0, so that ˆf ni ((T ) n i S K i y ni 1) 0. On the other hand, since S,K i (T ) c we have (T ) n i S K i y ni 1 = S T K i (T ) n i 1 y ni 1 S T ω in norm, while ˆf ni weak*- converges to g X, so that

8 2962 Ö. Gök and E. Demir Hence, ˆf ni ((T ) n i S K i y ni 1) g(s T ω). g(s T ω) = 0 that is Y ker(g). References [1] Y. A. Abromovich, C.D. Aliprantis, An Invitation To Operator Theory, American Mathematical Society, Rhode Island (2002). [2] C. D. Aliprantis, K. C. Border, Infinite Dimensional Analysis, A Hitchiker s Guide, Second Edition, Springer-Verlag, Berlin (1999). [3] J. B. Conway, A Course In Functional Analysis, Second Edition, Springer- Verlag, New York, (1990). [4] V. I. Lomonosov, H. Radjavi, V. G. Troitsky, Sesquitransitive and Localizing Operator Algebras, Integral Equations and Operator Theory, 60 (2008), [5] B. P. Rynne, M. A. Youngson, Linear Functional Analysis, Springer- Verlag, London, (2008). [6] V. G. Troitsky, Minimal vectors in arbitrary Banach spaces, Proc. American Mathematical Society, 132(2004), Received: May, 2011