General Mathematics Vol. 16, No. 1 (2008), A. P. Madrid, C. C. Peña

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1 General Mathematics Vol. 16, No. 1 (2008), On X - Hadamard and B- derivations 1 A. P. Madrid, C. C. Peña Abstract Let F be an infinite dimensional complex Banach space endowed with a bounded shrinking basis X. We seek conditions to relate X- Hadamard derivations and B-derivations supported on multiplier operators of F relative to X. It is seeing that in general the former class is larger than the first and some facts on basis problems are also considered Mathematics Subject Classification: 47B39 Key words: Bounded shrinking, sequence and unconditional basis, Associated sequence of coefficient functionals (or a.s.c.f.) of a basis, Bounded and Hadamard derivations. 1 Introduction Throughout this article by F we will denote a complex infinite dimensional Banach space endowed with a bounded shrinking basis X = {x n }. Let F F be the tensor product Banach space of F and F, i.e. the completion of the usual algebraic tensor product with respect to the following cross norm defined for u F F as { n } n u n = inf x j x j : u = x j x j. j=1 j=1 1 Received 3 August 2007 Accepted for publication (in revised form) 10 December

2 42 A. P. Madrid, C. C. Peña The space F F is indeed a Banach algebra under the product so that (x x )(y y ) = y,x (x x ) if x,y F,x,y F. Then F F is isometric isomorphic to the Banach algebra N F (F) of nuclear operators on F (cf. [5], Th. C.1.5, p. 256). This fact allows the transference of the investigation of properties and structure of bounded derivations to a more tractable frame which has essentially the same profile. For previous researches on this matter in a purely algebraic setting, in the frame of Hilbert spaces or on certain Banach algebras of operators the reader can see [1], [2], [3]. The class of bounded derivations on F F denoted as D(F F ) becomes a closed subspace of B(F F ). Example 1. If ad v (u) = u v v u for u,v F F then ad v D(F F ). As usual, {ad v } v F b F is the set of inner derivations on F F. Example 2. Let δ F : B(F) B(F F ),δ F (T) δ T where δ T is the unique linear bounded operator on F F so that δ T (x x ) = T(x) x x T (x ) for all basic tensor x x F F. By the universal property on tensor products δ F is well defined. Indeed, R(δ F ) D(F F ) and δ F B(B(F),B(F F )). Example 3. ad x x = δ x x, where as usual x x B(F) denotes the finite rank operator (x x )(y) = y,x x, with x,y F and x F. Proposition 1. (cf. [6], [7]) Let F be a Banach space, {x n } be a shrinking basis of F and let {x n} be its a.s.c.f.. The system of all basic tensor products x x m is basis of F F, arranged into a single sequence as follows: If m N let n N so that (n 1) 2 < m n 2 and then let s write x m = x σ1 (m) x σ 2 (m), with { (m (n 1) 2,n) if (n 1 2 ) + 1 m (n 1) 2 + n, σ(m) = (n,n 2 m + 1) if (n 1) 2 + n m n 2.

3 On X - Hadamard and B- derivations 43 Remark 1. In particular, σ : N N N becomes a bijective function. Since F F (F F ) we will also write z m = x σ 1 (m) x σ 2 (m), m N. Thus {z m} becomes the a.sc.f. of {z m }. Theorem 1. (cf. [4]) Let F be an infinite dimensional Banach space with a shrinking basis {x n}. Given δ D(F F ) there are unique sequences {h n } n N and {g v u} u,v N so that if u,v N then δ(z σ 1 (u,v)) = (h u h v )z σ 1 (u,v) + (gu n z σ 1 (u,v) gn v z σ 1 (u,n)). Indeed, h = h[δ] = ( δ(z n 2),z n 2 ) n N and η = η[δ] = (η m n ) n,m=1, with η m n = h σ 1 (m,1) n,1 = h m2 σ(n 2 ) = { δ(z n 2),z m 2 if n m 0, if n = m. In the sequel we will say that they are the h and g sequences of δ. Example 4. Let {v n } C N so that v = v n z n is a well defined element of F F. Then h [adv] = {v n 2 n+1 v 1 },η m [ad v ] = v m 2 n+1 if 1 n < m and η m n = v (n 1) 2 +m if n > m. Definition 1. A derivation δ D(F F ) is said to be an X-Hadamard derivation if its g- sequence is null. We will denote the set of all those derivations as D X (F F ). In [4] it is proved that the former is a complementary Banach subspace of D(F F ). Definition 2. An operator δ D(F F ) will be called a B-derivation if there exists T B(F) so that δ = δ T according to the notation of Example 2. We will denote the class of such derivations as D B (F F ). Remark 2. Any B-derivations is infinitely supported because δ T = δ T+λIdF if T F and λ C. More precisely, ker(δ F = C Id F )( see Lema 1 below ).

4 44 A. P. Madrid, C. C. Peña In Th. 2 will prove that any X-Hadamard derivation is a B-derivation. In Proposition 2 and Proposition 3 we will analize necessary and sufficient conditions under which certain natural series of Hadamard derivations are realized as B derivations. It ll then be clear how h-sequences determine their structures since the corresponding supports become multiplier operators included by them. 2 X-Hadamard and B-derivations Lemma 1. ker(δ F ) = C Id F. Proof. The inclusion is evident. Let T B(F) so that δ T = 0 and let λ σt. If λ belongs to the compression spectrum of T let x F {0} so that x R(T λidf 0. For all x F we have ) x,t (x ) = T(x),x = λx,x = x,λx, i.e. (t λid F )(x ) = 0. Moreover, since (T(F) λx) x = x (T (x ) λx ) = 0, the projective norm is a cross-norm and x 0 then T = λid F. If λ σ ap (T) we choose a sequence {y n } of unit vectors of F so that T(y n ) λy n 0. If y F then 0 = lim n (T(y n ) λy n ) y π = lim n y n (T (y ) λy ) π = T (y ) λy. Reasoning as above we conclude that T = λid F. Lemma 2. (i) If r,s N then {0, 1, 1,...} if r = s = 1, (1) h[δ xr x ] = {0, 0,...} if r s, s e r if r = s > 1. (ii) If r s then η[δ xr x ] = s er s is the zero matrix elsewhere and has a one in the (s,r) entry. All derivations δ xn x with n N are of Hadamard type. n

5 On X - Hadamard and B- derivations 45 Proof. (i) If r,s,n N we get (2) δx r x s(x n x m) = (x r x s)(x n ) x m x n (x r x s) (x m) = [ x n,x s x r ] x m x n [ x r,x m x s] = δ n s (x r x m) δ m r (x n x s). (3) Letting m = 1 in (2) then δ xr x s (x n x 1) = h p n,1 z p p=1 = δ n s (x r x 1) δ 1 r (x n x s) = δ n s z r 2 δ 1 r z σ 1 (n,s). If r = s = 1 by (3) is and the first assertion follows. If r = s > 1 by (3) is δ xr x r (x n x 1) = δ n r z r 2 and our thirth claim follows. Finally, if r s = n then (3) becomes δ xr x s (x s x 1) = z r 2 δ r 1 z s 2 s+1 and clearly h s [δ xr x ] = 0. If s {r,n} then s δ xr x s (x n x 1) = δ r 1 z σ 1 (n,s). But σ 1 (n,s) = n 2 if and only if s = 1 and as r s then h n [δ xr x s ] = 0. (ii) If r,s,n,m N and n m then (4) η m n [δ xr x s ] = δ X(x r x s)(z n 2),z m 2 = δ X x r x s)(x n x 1),x m x 1 = δ n s (x r x 1) (x n ) (x r x s) (x 1),x m x 1 = δ n s δ r m. The conclusion is now clear. Remark 3. Given an elementary tensor x x X X and m N we have ( m ) m δ xn x (x x ) = [ x,x n n (x n x ) x n,x (x x n)] ( m ) ( m ) = x,x n x n x x x n,x x n

6 46 A. P. Madrid, C. C. Peña and so lim m 0 ( m δ xn x n ) (x x ) 0. Since the basis X is assumed to be bounded then ρ = inf x n x p is positive (cf. [7], Corollary 3.1, p.20). n,p N Consequently, if n,m,p N and n p then (5) δ x n x m i.e. the series δ x n x n δ xn x m ( xm x m x p π xp ) = x n x m inf x n / sup x m > 0, n N m N is not convergent. Remark 4. The set {δ xn x n } is linearly dependent. For, let {c n } be a sequence of scalars so that c n δ xn x 0. In particular, by (4) is n {c n } c 0. If r,s are two positive integers then [ ] c n δ xn x (x n r x s) = (c r c s )(x r x s) = 0, i.e. c r = c s. Hence {c n } becomes the constant zero sequence and the assertion follows. Theorem 2. Every X-Hadamard derivation is a B-derivation. Proof. If δ D(F F ) and x F the series x,x n h n [δ] x n converges. For, if p,q N then x,x n h n [δ] x n = n=p δ [( ) x,x n x n n=p δ x,x n x n, n=p x 1 x 1 ] i.e. the sequence of corresponding partial sums is a Cauchy sequence. So, it is defined a linear operator M h[δ] : x x,x n h n [δ] x n that π

7 On X - Hadamard and B- derivations 47 is bounded as a consequence of the Banach-Steinhauss theorem. Hence h[δ] M(f,X), i.e. h[δ] is a multiplier of F relative to the basis X. Analogously, if x F the series x m,x h m [δ] x m also converges because if p,q N we get x m,x h m [δ] x m = x 1 x m=p 1 x m,x h m [δ] x m m=p π ( = δ x 1 x 1 x m,x xm) m=p δ x m,x x m. It is immediate that M h[δ] (x ) = n=p x m,x h m [δ] x m for all x F and h[δ] is also realizes as a multiplier of F relative to the basis X. Now, if x x is a fixed basic tensor in F F we can write δ(x x ) = x,x n x m,x (h n [δ] h m [δ])(x n x m ) = M h[δ] (x) x x M h[δ] (x ) and definitely δ = δ Mh[δ]. Proposition 2. Let {ζ} C N so that δ = π ζ n δ xn x is a well defined n Hadamard derivation. (i) h[δ] c, {ζ} c 0 and ζ m = h m [δ] lim n h n [δ] if m N. (ii) δ = δ S where S B(F) is defined for x F as S(x) = h n [δ] x,x n x n x lim n h n [δ]. Proof. (i) If m N it is readily seeing that δ(x m x 1) = (ζ m ζ 1 ) z m 2. Thus h 1 [δ] = 0 and h m [δ] = ζ m ζ 1 if m > 1. By (4) we have that {ζ} c 0 and we get (ii).

8 48 A. P. Madrid, C. C. Peña (ii) Let S n = n k=1 (h k [δ] lim n x k x k),n N. By the uniform boundedness principle and (ii) the sequence {S n } n N is bounded in B(F). Whence, since S n (x) S(x) if x F then S B(F). Indeed, if n,m N we get δ S (x n x m)= = ((h n [δ] lim k h k [δ])x n ) x m x n ((h m [δ] lim k h k [δ])x m) = (h n [δ] h m [δ]) (x n x m) = δ(x n x m), i.e. δ = δ S Proposition 3. Let {h n } M(F, {x n } ) M(F, {x n} ) c such that h 1 = 0. On writing h 0 lim n h n the series (h n h 0 ) δ xn x n converges to a Hadamard derivation δ on ˆ F so that h[δ] = {h n }. Proof. If S = Let S n = (h k h 0 ) x k x k then S B(F) and k=1 S {h n } M(F,{xn} ) + h 0. n (h k h 0 ) x k x k 6,nN. Given x F the sequence {S n (x)} k=1 converges because {h n } is a multiplier of F and ({x n } ) is an F complete biorthogonal system. Therefore { S n } becomes bounded. Now if we fix an elementary tensor y y F ˆ F and n N then (6) (δ Sn δ S )(y y ) π = (h k h 0 ) y,x k x k y k>n y k h 0 ) x k,y k>n(h x k π (h k h 0 ) y,x k x k y k>n + y (h k h 0 ) x k,y x k. k>n

9 On X - Hadamard and B- derivations 49 Since {h n } M(F, {x n } ) M(F, {x n} ) and {x n } is a shrinking basis by (5) we see that lim n (δ Sn δ S )(y y ) = 0. Indeed, as F F is dense in F ˆ F, {S n } is bounded and δ T 2 T for all T B(F) then δ S = (h n h 0 ) δ xn x n. Problem 1 Giving T B(F) then η[t] = { T(x n ),x m } n,m=1. So it is obvious that D X (F ˆ F ). It would be desirable to decide if D B (F ˆ F ) is a Banach space. Remark 5. Is {δ xn x n } a basis of D X (F ˆ F )?- In general this is not the case. For instance, let F = l p (N) with 1 < p < and let X = {e n }, where e n = {δ n,m } m=1 and δ n,m the current Kronecker symbol if n,m N. Then X is not only a shrinking basis, it is further an unconditional basis of F. Consequently, if T(x) = x,e 2n for x F then T B(F). It is readily seeing that δ T is an X-Hadamard derivation. Since h[δ T ] = {0, 1, 0, 1,...} by Prop. 2 {δ en e n } can not be a basis of D X (F ˆ F ). Problem 2 Is {δ xn x n } a sequence basis?- Can be be constructed a basis of D X ((F ˆ F )?- References [1] A.L. Barrenechea and C.C. Peña, On derivation over rings of triangular matrices, Bulletin CXXXI de l Académie Serbe des Sciences Mathematiques, 30, 77-84,(2005). [2] A.L. Barrenechea and C.C. Peña, Some remarks about bounded derivations on the Hilbert algebra of square summable matrices, Matematicki Vesinik, 57, No.4,78-95,(2005). [3] A.L. Barrenechea and C.C. Peña, On innerness of derivations on S(H), Lobachevskii L. of Math., Vol.18,21-32,(2005). [4] A.L. Barrenechea and C.C. Peña,On the structure of derivations on certain non-amenable nuclear Banach algebras, Preprint, (2007).

10 50 A. P. Madrid, C. C. Peña [5] V. Runde, Lectures on amenability, Springer-Verlag,Berlin, Heidelberg, N.Y.,(2002). [6] R. Schatten, A theory of cross spaces, Ann. of Math. Studies 26, Princeton university Press, (1950). [7] I. Singer, Bases in Banach spaces, I. Springer-Verlag, Berlin-Heidelberg- N.Y.,(1970). A.P.Madrid Comisión de Investivaciones Científicas de la Pcia. de Buenos Aires (CIC) - UNCPBA, Departamento de Matemáticas-NUCOMPA, apmadrid8@hotmail.com C. C. Peña UNCPBA, Departamento de Matemáticas-NUCOMPA, ccpenia@gmail.com