RACSAM
Rev. R. Acad. Cien. Serie A. Mat.
VOL. 97 (1), 2003, pp. 33–35
An´alisis Matem´atico / Mathematical Analysis
Comunicaci´on Preliminar / Preliminary Communication
Decomposable subspaces of Banach spaces
M. Gonz ´alez and A. Martin ´ on
Abstract. We introduce and study the notion of hereditarilyA-indecomposable Banach space forA a space ideal. For a hereditarilyA-indecomposable spaceXwe show that the operators fromXinto a Banach spaceY can be written as the union of two setsAΦ+(X, Y)andA(X, Y). For some idealsA defined in terms of incomparability, the first set is open, the second set correspond to a closed operator ideal and the union is disjoint.
Subespacios descomponibles de espacios de Banach
Resumen. Introducimos y estudiamos la noci´on de hereditibilidadA-indescomponible espacio de Banach para un espacio idealA. Demostramos que para un espacioA-indescomponibleXlos operadores deX en un espacio de BanachY pueden ser escritos como la uni´on de dos conjuntosAΦ+(X, Y) yA(X;Y). Para algunos idealesAdefinidos en t´erminos de incomparabilidad, el primer conjunto es abierto, el segundo conjunto corresponde a un operador cerrado ideal y la uni ´on es disjunta.
LetAbe a space ideal in the sense of Pietsch [3]. For each Banach spaceXwe consider SA(X) :={M ⊂X :M is a subspace ofXandM /∈A}.
Definition 1 A Banach spaceX is said to beA-indecomposable if there are no subspacesM andN in SAsuch thatX=M ⊕N.
The space X is said to be hereditarily A-indecomposable (HAI) if every subspace M ofX is A- indecomposable.
Let us see some examples. Note that a nontrivial example should include aA-indecomposable space which is not inA.
Let A = F, the finite dimensional spaces. The existence of infinite dimensional, hereditarily F- indecomposable spaces has been a long-standing open problem in Banach space theory. Finally, Gowers and Maurey gave an example in [2], that we denoteXGM.
LetA = Rbe the reflexive spaces and letA = WSC be the weakly sequentially complete spaces.
James’ spaceJ is hereditarily R-indecomposable and hereditarily WSC-indecomposable space, but it is neither reflexive, nor weakly sequentially complete. The reason is thatdim(J∗∗/J) = 1.
From the previous examples we can derive new examples as follows.
Presentado por Manuel Valdivia.
Recibido: 4 de Noviembre de 2002 Aceptado: 5 de Marzo de 2003
Palabras clave / Keywords: Descomposable subespaces, totally incomparable spaces.
Mathematics Subject Classifications: 46B03, 47A53, 47B10
°c 2003 Real Academia de Ciencias, Espa˜na.
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M. Gonz´alez and A. Martin´on
LetQ:X∗∗−→X∗∗/Xdenote the quotient map. Given a closed subspaceM ofX, we can identify M∗∗/MwithQ(M⊥⊥). Thus,
Aco:={X:X∗∗/X∈A}
is a space ideal.
LetAbe one of the space idealsF,RorWSC. LetXbe a Banach space such thatX∗∗/Xis isomorphic toXGM,JorJ, respectively. The spaceXis a hereditarilyAco-indecomposable space which is not inAco. Remark 1 The HI spaces contain no unconditional basic sequence [2]. Similarly, every unconditional basic sequence in a HRI spaceXgenerates a reflexive subspace. ¥
Recall that the injection modulus of an operatorT ∈L(X, Y)is defined by j(T) := inf{kT xk:x∈X,kxk= 1}.
We will consider here the following two derived quantities
sjA(T) := sup{j(T JM) :M ∈SA(X)}and inA(T) := inf{kT JMk:M ∈SA(X)}.
Definition 2 Suppose thatSA(X)6=∅and letY be a Banach space. We define 1. ASS(X, Y) :={T ∈L(X, Y) :sjA(T) = 0}.
2. AΦ+(X, Y) :={T ∈L(X, Y) :inA(T)>0}.
ForSA(X)empty we defineASS(X, Y) =AΦ+(X, Y) =L(X, Y).
In the caseA=F, the finite dimensional spaces, the quantitiesinF andsjFwere introduced in [5]. In this caseFΦ+= Φ+, the upper semi-Fredholm operators, andFSS =SS, the strictly singular operators.
Theorem 1 For a Banach spaceX the following assertions are equivalent:
1. Xis HAI.
2. For every spaceY and everyT ∈L(X, Y),sjA(T)≤inA(T).
3. For every spaceY,L(X, Y) =AΦ+(X, Y)∪ASS(X, Y). ¤
The proof is based on the following fact thatXis HAI and ifM, N ∈SA(X), thendist(SM, SN) = 0, whereSM is the unit sphere inM.
We say that two Banach spacesXandY are totally incomparable [4] if no infinite dimensional subspace ofX is isomorphic to a subspace ofY. Given a classCof Banach spaces, the class of incomparabilityCi
was defined in [1] as follows:
Ci:={X :Xis totally incomparable with everyY ∈ C}.
The classCiis a space ideal. Moreover it is not difficult to see thatX ∈ Ciiif and only ifX has no infinite dimensional subspace inCi, and thatCiii=Ci.
Theorem 2 LetXandY be Banach spaces. Suppose thatA=AiiandSA(X)6=∅. Then AΦ+(X, Y)∩ASS(X, Y) =∅
If, additionally,Xis a HAI space, then the unionL(X, Y) =AΦ+(X, Y)∪ASS(X, Y)is disjoint. ¤
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Decomposable subspaces of Banach spaces
The proof of the previous result is based on the fact that, under the hypothesis of the statement, if M ∈SA(X), thensjA(T JM) =sjF(T JM).
Remark 2 In the caseA=Aii, we get two additional facts:
1. The componentsAΦ+(X, Y)are open and the classAΦ+is closed under products:T ∈AΦ+(X, Y) andS∈AΦ+(Y, Z)implyST ∈AΦ+(X, Z).
2. The classASSis a closed operator ideal. ¥
Acknowledgement. In this paper the first author is supported in part by DGI (Spain), Proyecto BFM2001-1147 and the second author is supported in part by Gobierno de Canarias (Spain) PI2001/039.
References
[1] ´Alvarez, T., Gonz´alez, M. and Onieva, V. M. (1987). Totally incomparable Banach spaces and three-space ideals.
Math. Nachr., 131, 83–88.
[2] Gowers, W. T. and Maurey, B. (1993). The unconditional basic sequence problem. J. Amer. Math. Soc., 6, 851–
874.
[3] Pietsch, A. (1980). Operator Ideals. North-Holland, Amsterdam.
[4] Rosenthal, H. P. (1969). On totally incomparable Banach spaces. J. Funct. Anal., 4, 167–175.
[5] Schechter, M. (1992). Quantities related to strictly singular operators, Indiana Univ. Math. J., 21, 1061–1071.
M. Gonz´alez A. Martin´on
Departamento de Matem´aticas Departamento de An´alisis Matem´atico Universidad de Cantabria Universidad de La Laguna
39071 Santander, Spain La Laguna (Tenerife), Spain [email protected] [email protected]
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