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On hereditarily indecomposable Banach spaces
Annals of Pure and Applied Logic 126 (2004) 293 – 299 www.elsevier.com/locate/apal
On hereditarily indecomposable Banach spaces Tadeusz Figiela; b; ∗ , Ryszard Frankiewicza; b , Ryszard Komorowskia; b;1 , Czes law Ryll-Nardzewskia; b a Institute
of Mathematics, Polish Academy of Sciences, Warsaw, Poland of Mathematics, Technical University, Wroclaw, Poland
b Institute
Abstract A set-theoretical proof of Gowers’ Dichotomy Theorem is presented together with its application to another dichotomy concerning asymptotic l2 basic sequences. c 2003 Published by Elsevier B.V. MSC: 03E02; 03E05; 46E27 Keywords: Ramsey set; Banach space; Hereditarily indecomposable Banach space; Unconditional basis
1. Notation and preliminaries Gowers’ dichotomy theorem states that every Banach space X contains a subspace with an unconditional basis or a subspace X0 which is hereditarily indecomposable. Recall that X0 is said to be hereditarily indecomposable if no pair of in=nite-dimensional subspaces U and V of X0 forms an unconditional sum. The existence of hereditarily indecomposable spaces (HI ) was used by Gowers [5] (see also [9,12]) in his solution to the homogeneous Banach space problem. In this paper we emphasize the set theoretical, or combinatorial, character of the dichotomy theorem. More precisely, the theorem is similar to the well known theorem (Silver, [3]) stating that every analytic set on the real line is a Ramsey set. We skip the de=nition of a Ramsey set, as we do not need it in this paper. Similarly to the
Supported by KBN Grant 5-P03A-037-20. Corresponding author. E-mail addresses: [email protected] (T. Figiel), [email protected] (R. Frankiewicz), [email protected] (C. Ryll-Nardzewski). 1 1955 –2003. ∗
c 2003 Published by Elsevier B.V. 0168-0072/$ - see front matter doi:10.1016/j.apal.2003.11.005
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T. Figiel et al. / Annals of Pure and Applied Logic 126 (2004) 293 – 299
Silver theorem, the dichotomy theorem states that there are two possibilities: either there exists an object with “nice” properties, or the complete opposite, an object with extremely “bad” properties. The main idea of our proof is also somewhat related to the Baire category argument applied to Ellentuck topology [3], the main part of the proof of Silver theorem. In addition, we work on a countable structure, instead of a whole Banach space. This is analogous to the LLowenheim–Skolem–Tarski theorem [2], well known to model theorists. Let us recall two results on HI spaces. Theorem 1.1 (Gowers and Maurey [6]). A hereditarily indecomposable (HI ) Banach space is not isomorphic to any of its proper subspaces. Theorem 1.2 (Gowers [5]). Every in4nite-dimensional Banach space contains either a subspace with an unconditional basis or a subspace with the HI property. In fact, the authors of [6] proved that the space of operators on every HI space is small. Namely, every bounded linear operator on X is of the form S + I , where S is a strictly singular operator. Let us also mention that a similar result was obtained by Shelah [11] who, under the assumption that V = L, constructed a Banach space with density !1 , such that every bounded operator T : X → X is of the form S + I , where S has a separable range. In this paper we consider normed linear spaces of two categories: separable real Banach spaces and countable spaces over the =eld Q of rational numbers. Standard notation from Banach space theory together with fundamental facts and terminology can be found in [7,8]. We will also use the following notation. The completion of a normed linear space X will be denoted by XO . We write either Y ¡X or X ¿Y to denote that Y is a linear subspace of X and YO is in=nite dimensional. Recall that a sequence (ej )∞ (or simply basis), if every j = 1 in a Banach space X is a Schauder basis ∞ vector x ∈ X has a unique representation of the form x = j = 1 tj ej for some scalars ∞ t1 ; t2 ; : : :,. A basis (ej )j = 1 is called an unconditional basis, if there is a constant C such that ∞ ∞ j tj ej 6 C tj ej ; j=1 j=1 ∞ for all x = j=1 tj ej ∈ X and j ∈ {−1; 1}. For a given C such a basis is called Cunconditional. The in=mum of such constants C is denoted by unc((ej )∞ j = 1 ). A pair (U; V ) of subspaces of X is called C-unconditional if ∀ (u; v) ∈ U × V
u − v 6 Cu + v:
As in the de=nition of unc((ej )∞ j=1 ), let unc(U; V ) denote the in=mum of such constants C. If there is no such constant, let unc(U; V ) = ∞. Clearly, X is hereditarily indecomposable if and only if unc(U; V ) = ∞ for all U; V ¡X .
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In Section 3 we formulate a result which can be obtained by adapting the method of Section 2 to the situation, where one considers all =nite partitions of the set of positive integers (not only partitions into two subsets as in Section 2). We =nd a natural condition on the given Banach space X (that condition is essentially necessary), which allows one to prove that every subspace Y of X contains an unconditional basic sequence which resembles the usual basis of the well-known space T2 (the 2-convexi=ed Tsirelson’s space).
2. Proof of Gowers’ theorem In this section we shall present a proof of Theorem 1.2 similar to that given by Maurey [9] (see also [12]). Our approach eliminates certain lengthy technical steps of the original construction, and replaces them by some elementary observations. More precisely, the proof presented below does not require -nets in each induction step in the construction of an unconditional basic sequence. Throughout this section we assume that X is a countable normed linear space over the =eld Q and XO is in=nite-dimensional. At the end we shall show how this leads to a proof of Theorem 1.2 for the Banach space XO . Let B denote the set of all =nite subsets of the set of positive integers N . For K¡X let AK denote the set of all =nite sequences (x1 ; : : : ; xn ) with n¿0 and xi ∈ K\{0} for i = 1; : : : ; n. Note that ∅ ∈ AK (this is a unique sequence of length 0). Of course, the product AK × B is countable. For X1 ; X2 ⊆ X , xO = (x1 ; : : : ; xn ) and (x; O J ) ∈ AK × B let X1 + X2 = span{X1 ∪ X2 };
x| O J = span{xi : i ∈ J };
x| O J = span{xi : i ∈= J }:
Remark 2.1. Fix xO = (x1 ; : : : ; xn ). If ∀J ∈ B unc(x| O J ; x| O J )6C, then the sequence x1 ; : : : ; xn is C-unconditional. In order to prove Theorem 1.2, we need a technical de=nition and some lemmas. Suppose that X ¡X and C ¿ 1. If (x; O J ) ∈ AX × B we shall say that (x; O J ) is (X ; C)-good if ∀Y ¡ X ∃U; V ¡ Y
unc(x| O J + U; x| O J + V ) ¡ C:
We shall say that (x; O J ) is (X ; C)-bad if ∀U; V ¡ X
unc(x| O J + U; x| O J + V ) ¿ C:
If every element (x; O J ) ∈ AX × B is (X ; C)-good or it is (X ; C)-bad then the subspace X ¡X shall be called C-dichotomous. Lemma 2.2. ∀C¿1 ∃ X ¡X such that X is C-dichotomous.
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Proof. Let us =x an enumeration ((xOn ; Jn ))∞ n=0 of the set AX × B. For any =xed C¿1 the following statement holds. ∀K ¡ X ∀(x; O J ) ∈ AK × B
(∗)
one of the following is true: ∀Y ¡ K∃U; V ¡ Y
unc(x| O J + U; x| O J + V )¡C;
(2.1)
∃Y ¡ K∀U; V ¡ Y
unc(x| O J + U; x| O J + V ) ¿ C:
(2.2)
Let us de=ne a sequence (Kn )∞ n = 1 (Kn+1 ¡Kn ) of subspaces of X we need for the construction of the space X . Set K0 = X . Assume that the sequence of subspaces K0 ¿ K1 ¿ · · · ¿Kn has been constructed. We now de=ne Kn+1 . If condition (2.1) of (*) holds with (x; O J ) = (xOn ; Jn ) and K = Kn , then set Kn+1 = Kn . Otherwise, (xOn ; Jn ) and K = Kn satisfy (2.2) of (*) and then we set Kn+1 = Y ¡Kn . We choose a sequence (zn )∞ n = 1 with zn ∈ Kn such that the zn ’s are linearly indepen dent in the completion of X . We set X = span{(zn )∞ n = 1 }. Clearly, X ¡X . We claim that X is C-dichotomous. To show this, we =x an element (xOk ; Jk ) ∈ AX × B. Now, if K = X and (x; O J ) = (xOk ; Jk ) in (*), then by (*) two cases are possible. If (2.1) holds, then (xOk ; Jk ) is (X ; C)-good. Assume now that (2.1) does not hold. Fix U; V ¡X . Recall that in our inductive construction Kn+1 has been chosen so that ∀U ; V ¡ Kn+1
unc(xOk |Jk + U ; xOk |Jk + V ) ¿ C:
Set U0 = U ∩ Kn+1 and V0 = V ∩ Kn+1 . Observe that U0 (resp. V0 ) is of =nite codimension in U (resp. in V ), because Kn+1 ∩ X ⊇ span{zn+1 ; zn+2 ; : : :} is a space of =nite codimension in X . It follows that U0 , V0 ¡Kn+1 . Setting U = U0 and V = V0 , we obtain the estimate unc(xOk |Jk + U; xOk |Jk + V ) ¿ unc(xOk |Jk + U ; xOk |Jk + V ) ¿ C: Since U; V ¡X can be arbitrary, this means that in this case (xOk ; Jk ) is (X ; C)-bad. This completes the proof of the lemma. Lemma 2.3. Fix C¿1 and suppose that X ¡X is C-dichotomous. If ((x1 ; : : : ; xn ); J ) ∈ AX × B is (X ; C)-good, then ∀Y ¡ X ∃W ¡ Y ∀w ∈ W
((x1 ; : : : ; xn ; w); J ) is (X ; C)-good:
Proof. Let us assume that n + 1 ∈ J , the second case (n + 1 ∈= J ) being analogous. Suppose to the contrary that ∃Y ¡ X ∀W ¡ Y ∃w ∈ W
((x1 ; : : : ; xn ; w); J ) is not (X ; C)-good:
Since X is C-dichotomous, this means exactly that ∃Y ¡ X ∀W ¡ Y ∃w ∈ W ∀U; V ¡X unc((x1 ; : : : ; xn ; w)|J + U; (x1 ; : : : ; xn ; w)|J + V ) ¿ C:
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Changing the order of the quanti=ers ∃ w ∈ W and ∀U; V ¡X in the latter formula, we obtain that ∃Y ¡ X ∀U; V; W ¡ Y ∃w ∈ W; unc((x1 ; : : : ; xn ; w)|J + U; (x1 ; : : : ; xn ; w)|J + V ) ¿ C: Setting W = U , we have for any w ∈ W (x1 ; : : : ; xn ; w)|J = (x1 ; : : : ; xn )|J and (x1 ; : : : ; xn ; w)|J + U = (x1 ; : : : ; xn )|J + U: Hence, we obtain ∃Y ¡ X ∀U; V ¡ Y
unc((x1 ; : : : ; xn )|J + U; (x1 ; : : : ; xn )|J + V ) ¿ C:
The latter formula means that ((x1 ; : : : ; xn ); J ) is not (X ; C)-good and this contradicts the assumption of the lemma. Lemma 2.4. Fix C¿1 and suppose that X ¡X is C-dichotomous. If xO = (x1 ; : : : ; xn ) ∈ AX and for each J ∈ B (x; O J ) is (X ; C)-good then ∀Y ¡ X ∃W ¡ Y ∀w ∈ W ∀J ∈ B
((x1 ; : : : ; xn ; w); J ) is (X ; C)-good:
Proof. Let J1 ; : : : ; Jm be any enumeration of the subsets of {1; 2; : : : ; n}. Fix Y ¡X . By Lemma 2.3, for ((x1 ; : : : ; xn ); J1 ) there exists a subspace W1 ¡Y such that ∀w ∈ W1 ((x1 ; : : : ; xn ; w); J1 ) is (X ; C)-good. Similarly, for ((x1 ; : : : ; xn ); J2 ) we have W2 ¡W1 ¡Y such that for each w ∈ W2 both pairs ((x1 ; : : : ; xn ; w); J2 ) and ((x1 ; : : : ; xn ; w); J1 ) are (X ; C)-good. After m such steps we have Y ¿W1 · · · ¿Wm and ∀ w ∈ Wm ∀Ji
(i = 1; : : : ; m)
((x1 ; : : : ; xn ; w); Ji ) is (X ; C)-good:
Finally, let W = Wm . Lemma 2.5. If (∅; ∅) ∈ AX × B is (X; C)-good then X contains a C-unconditional basic sequence. Proof. By Lemma 2.2 there is X ¡X such that X is C-dichotomous. It is easy to verify that (∅; J ) is (X ; C)-good for each J ∈ B. Now using Lemma 2.4 we can de=ne a sequence (xi )∞ i = 1 in X so that the xi ’s are linearly independent in the completion of X and for each natural n and each J ∈ B the pair ((x1 ; : : : ; xn ); J ) is (X ; C)-good. Using Remark 2.1 we conclude that the sequence (xi )∞ i = 1 is C-unconditional. Lemma 2.6. Let Y be an in4nite-dimensional Banach space. Let X ⊆ Y be a countable subset of Y that is a linear subspace over Q such that XO ¡Y . Suppose that C¿1
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and (if W , U , V denote closed subspaces of Y ) ∀W ¡ Y ∃U; V ¡ W
unc(U; V )¡C:
Then (∅; ∅) ∈ AX × B is (X; C )-good for each C ¿ C.
Proof. Fix any X ¡X and choose two subspaces U; V ¡XO such that unc(U; V )¡C. Pick an )¿1 so that )2 ¡C =C. Using a standard Banach space argument, one can construct a linear isomorphism T : X → X such that T ¡) and T −1 ¡) and there exist U , V ¡X such that T −1 (U ) ⊂ U and T −1 (V ) ⊂ V . It is then easy to verify that unc(U ; V ) 6 T T −1 unc(U; V ) 6 )2 C ¡ C : This completes the proof of the lemma. Proof of Theorem 1.2. In this proof we are dealing directly only with Banach spaces. Let Z be an in=nite-dimensional Banach space. Consider the following condition: ∃C ¿ 1∃Z ¡ Z∀W ¡ Z
∃U; V ¡ W
unc(U; V )¡C:
If Z satis=es this condition then, by Lemmas 2.5 and 2.6, Z contains an unconditional basic sequence. If Z fails this condition then a well-known diagonal argument shows that Z has a HI subspace (cf. the proof of Theorem C in [12]). 3. A result on asymptotic l2 basic sequences In this section we shall indicate an application of the approach presented in Section 2 to construction of so called asymptotic l2 basic sequences in a suitable class of Banach spaces. Similar arguments are applicable to asymptotic lp basic sequences, 16p¡∞, and to asymptotic c0 basic sequences. We shall only state the main theorem without proving it here. A complete proof will be presented in a more general setting in [4]. Before stating this result we need some de=nitions. De&nition 3.1. Let X be a Banach space and 16C¡∞. We say that a sequence (U1 ; : : : ; Un ) of subspaces of X is C-l2 , if ∀(u1 ; : : : ; un ) ∈ U1 × · · · × Un ; 1=2 1=2 n n n √ 1 √ uj 2 6 uj uj 2 : 6 C C j=1 j=1 j=1 We say that X contains ln2 -sums C-isomorphically if for each n¿2 there exists a C-l2 n-tuple (U1 ; : : : ; Un ) with Ui ¡X for i = 1; : : : ; n.
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A basic sequence (xi ) in X is said to be asymptotic C-l2 if for each n¿2 and each sequence A1 ; : : : ; An of mutually disjoint subsets of the set {m ∈ N : m¿n} the n-tuple (span{xj : j ∈ Ai })ni= 1 is C-l2 . Clearly, an orthonormal basis in a Hilbert space is asymptotic C-l2 , where C = 1. Another example is the usual basis of T2 , the 2-convexi=ed Tsirelson’s space (see [1,10]), which is asymptotic C-l2 with C = 2. (Recall that the space T2 contains no isomorphic copy of any lp space or c0 .) A standard argument proves that (every subspace of) a Banach space with an asymptotic C-l2 basis contains l2 -sums C uniformly for each C ¿C. Conversely, by adapting the method used in Section 2 one can prove the following. Theorem 3.2. Let X be a Banach space. If each subspace Y ¡X contains l2 -sums C-isomorphically for some C¡∞, then there is X0 ¡X and C0 ¡∞ such that each subspace Y ¡X0 contains l2 -sums C0 -isomorphically. Moreover, if C0 ¡C ¡∞ then each subspace Y ¡X0 contains an asymptotic C -l2 basic sequence. References [1] P.G. Casazza, T. Shura, Tsirelson’s Space, in: Lecture Notes in Mathematics, Vol. 1363, Springer, Berlin, 1989. [2] C.C. Chang, H.J. Keisler, Model Theory, in: Studies in Logic and Foundation of Mathematics, Vol. 73, North-Holland, Amsterdam, 1977. [3] D. Ellentuck, A new proof that analytic sets are Ramsey, J. Symbolic Logic 39 (1974) 163–165. [4] T. Figiel, R. Frankiewicz, R. Komorowski, C. Ryll-Nardzewski, Selecting basic sequences in ’-stable Banach spaces, Studia Math. 159 (2003) 499–515. [5] W.T. Gowers, A new dichotomy for Banach spaces, Geom. Funct. Anal. 6 (1996) 1083–1093. [6] W.T. Gowers, B. Maurey, The unconditional basic sequence problem, J. Math. Soc. 6 (1993) 851–874. [7] J. Lindenstrauss, L. Tzafriri, Classical Banach Spaces I, Springer, Berlin, 1977. [8] J. Lindenstrauss, L. Tzafriri, Classical Banach Spaces II, Springer, Berlin, 1979. [9] B. Maurey, Quelques progrSes dans la comprTehension de la dimension in=nie, in: Espaces de Banach classiques et quantiques, JournTee Annuelle, Soc. Math. de France, 1994, 1–29. [10] G. Pisier, The Volume of Convex Bodies and Banach Space Geometry, Texas A&M University, UniversitTe Paris VI, Texas, 1989. [11] S. Shelah, A Banach space with few operators, Israel J. Math. 30 (1978) 181–191. [12] N. Tomczak-Jaegermann, A solution of homogeneous Banach space problem, Canad. Math. Soc. 3 (1996) 267–286.
On hereditarily indecomposable Banach spaces Tadeusz Figiela; b; ∗ , Ryszard Frankiewicza; b , Ryszard Komorowskia; b;1 , Czes law Ryll-Nardzewskia; b a Institute
of Mathematics, Polish Academy of Sciences, Warsaw, Poland of Mathematics, Technical University, Wroclaw, Poland
b Institute
Abstract A set-theoretical proof of Gowers’ Dichotomy Theorem is presented together with its application to another dichotomy concerning asymptotic l2 basic sequences. c 2003 Published by Elsevier B.V. MSC: 03E02; 03E05; 46E27 Keywords: Ramsey set; Banach space; Hereditarily indecomposable Banach space; Unconditional basis
1. Notation and preliminaries Gowers’ dichotomy theorem states that every Banach space X contains a subspace with an unconditional basis or a subspace X0 which is hereditarily indecomposable. Recall that X0 is said to be hereditarily indecomposable if no pair of in=nite-dimensional subspaces U and V of X0 forms an unconditional sum. The existence of hereditarily indecomposable spaces (HI ) was used by Gowers [5] (see also [9,12]) in his solution to the homogeneous Banach space problem. In this paper we emphasize the set theoretical, or combinatorial, character of the dichotomy theorem. More precisely, the theorem is similar to the well known theorem (Silver, [3]) stating that every analytic set on the real line is a Ramsey set. We skip the de=nition of a Ramsey set, as we do not need it in this paper. Similarly to the
Supported by KBN Grant 5-P03A-037-20. Corresponding author. E-mail addresses: [email protected] (T. Figiel), [email protected] (R. Frankiewicz), [email protected] (C. Ryll-Nardzewski). 1 1955 –2003. ∗
c 2003 Published by Elsevier B.V. 0168-0072/$ - see front matter doi:10.1016/j.apal.2003.11.005
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T. Figiel et al. / Annals of Pure and Applied Logic 126 (2004) 293 – 299
Silver theorem, the dichotomy theorem states that there are two possibilities: either there exists an object with “nice” properties, or the complete opposite, an object with extremely “bad” properties. The main idea of our proof is also somewhat related to the Baire category argument applied to Ellentuck topology [3], the main part of the proof of Silver theorem. In addition, we work on a countable structure, instead of a whole Banach space. This is analogous to the LLowenheim–Skolem–Tarski theorem [2], well known to model theorists. Let us recall two results on HI spaces. Theorem 1.1 (Gowers and Maurey [6]). A hereditarily indecomposable (HI ) Banach space is not isomorphic to any of its proper subspaces. Theorem 1.2 (Gowers [5]). Every in4nite-dimensional Banach space contains either a subspace with an unconditional basis or a subspace with the HI property. In fact, the authors of [6] proved that the space of operators on every HI space is small. Namely, every bounded linear operator on X is of the form S + I , where S is a strictly singular operator. Let us also mention that a similar result was obtained by Shelah [11] who, under the assumption that V = L, constructed a Banach space with density !1 , such that every bounded operator T : X → X is of the form S + I , where S has a separable range. In this paper we consider normed linear spaces of two categories: separable real Banach spaces and countable spaces over the =eld Q of rational numbers. Standard notation from Banach space theory together with fundamental facts and terminology can be found in [7,8]. We will also use the following notation. The completion of a normed linear space X will be denoted by XO . We write either Y ¡X or X ¿Y to denote that Y is a linear subspace of X and YO is in=nite dimensional. Recall that a sequence (ej )∞ (or simply basis), if every j = 1 in a Banach space X is a Schauder basis ∞ vector x ∈ X has a unique representation of the form x = j = 1 tj ej for some scalars ∞ t1 ; t2 ; : : :,. A basis (ej )j = 1 is called an unconditional basis, if there is a constant C such that ∞ ∞ j tj ej 6 C tj ej ; j=1 j=1 ∞ for all x = j=1 tj ej ∈ X and j ∈ {−1; 1}. For a given C such a basis is called Cunconditional. The in=mum of such constants C is denoted by unc((ej )∞ j = 1 ). A pair (U; V ) of subspaces of X is called C-unconditional if ∀ (u; v) ∈ U × V
u − v 6 Cu + v:
As in the de=nition of unc((ej )∞ j=1 ), let unc(U; V ) denote the in=mum of such constants C. If there is no such constant, let unc(U; V ) = ∞. Clearly, X is hereditarily indecomposable if and only if unc(U; V ) = ∞ for all U; V ¡X .
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In Section 3 we formulate a result which can be obtained by adapting the method of Section 2 to the situation, where one considers all =nite partitions of the set of positive integers (not only partitions into two subsets as in Section 2). We =nd a natural condition on the given Banach space X (that condition is essentially necessary), which allows one to prove that every subspace Y of X contains an unconditional basic sequence which resembles the usual basis of the well-known space T2 (the 2-convexi=ed Tsirelson’s space).
2. Proof of Gowers’ theorem In this section we shall present a proof of Theorem 1.2 similar to that given by Maurey [9] (see also [12]). Our approach eliminates certain lengthy technical steps of the original construction, and replaces them by some elementary observations. More precisely, the proof presented below does not require -nets in each induction step in the construction of an unconditional basic sequence. Throughout this section we assume that X is a countable normed linear space over the =eld Q and XO is in=nite-dimensional. At the end we shall show how this leads to a proof of Theorem 1.2 for the Banach space XO . Let B denote the set of all =nite subsets of the set of positive integers N . For K¡X let AK denote the set of all =nite sequences (x1 ; : : : ; xn ) with n¿0 and xi ∈ K\{0} for i = 1; : : : ; n. Note that ∅ ∈ AK (this is a unique sequence of length 0). Of course, the product AK × B is countable. For X1 ; X2 ⊆ X , xO = (x1 ; : : : ; xn ) and (x; O J ) ∈ AK × B let X1 + X2 = span{X1 ∪ X2 };
x| O J = span{xi : i ∈ J };
x| O J = span{xi : i ∈= J }:
Remark 2.1. Fix xO = (x1 ; : : : ; xn ). If ∀J ∈ B unc(x| O J ; x| O J )6C, then the sequence x1 ; : : : ; xn is C-unconditional. In order to prove Theorem 1.2, we need a technical de=nition and some lemmas. Suppose that X ¡X and C ¿ 1. If (x; O J ) ∈ AX × B we shall say that (x; O J ) is (X ; C)-good if ∀Y ¡ X ∃U; V ¡ Y
unc(x| O J + U; x| O J + V ) ¡ C:
We shall say that (x; O J ) is (X ; C)-bad if ∀U; V ¡ X
unc(x| O J + U; x| O J + V ) ¿ C:
If every element (x; O J ) ∈ AX × B is (X ; C)-good or it is (X ; C)-bad then the subspace X ¡X shall be called C-dichotomous. Lemma 2.2. ∀C¿1 ∃ X ¡X such that X is C-dichotomous.
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Proof. Let us =x an enumeration ((xOn ; Jn ))∞ n=0 of the set AX × B. For any =xed C¿1 the following statement holds. ∀K ¡ X ∀(x; O J ) ∈ AK × B
(∗)
one of the following is true: ∀Y ¡ K∃U; V ¡ Y
unc(x| O J + U; x| O J + V )¡C;
(2.1)
∃Y ¡ K∀U; V ¡ Y
unc(x| O J + U; x| O J + V ) ¿ C:
(2.2)
Let us de=ne a sequence (Kn )∞ n = 1 (Kn+1 ¡Kn ) of subspaces of X we need for the construction of the space X . Set K0 = X . Assume that the sequence of subspaces K0 ¿ K1 ¿ · · · ¿Kn has been constructed. We now de=ne Kn+1 . If condition (2.1) of (*) holds with (x; O J ) = (xOn ; Jn ) and K = Kn , then set Kn+1 = Kn . Otherwise, (xOn ; Jn ) and K = Kn satisfy (2.2) of (*) and then we set Kn+1 = Y ¡Kn . We choose a sequence (zn )∞ n = 1 with zn ∈ Kn such that the zn ’s are linearly indepen dent in the completion of X . We set X = span{(zn )∞ n = 1 }. Clearly, X ¡X . We claim that X is C-dichotomous. To show this, we =x an element (xOk ; Jk ) ∈ AX × B. Now, if K = X and (x; O J ) = (xOk ; Jk ) in (*), then by (*) two cases are possible. If (2.1) holds, then (xOk ; Jk ) is (X ; C)-good. Assume now that (2.1) does not hold. Fix U; V ¡X . Recall that in our inductive construction Kn+1 has been chosen so that ∀U ; V ¡ Kn+1
unc(xOk |Jk + U ; xOk |Jk + V ) ¿ C:
Set U0 = U ∩ Kn+1 and V0 = V ∩ Kn+1 . Observe that U0 (resp. V0 ) is of =nite codimension in U (resp. in V ), because Kn+1 ∩ X ⊇ span{zn+1 ; zn+2 ; : : :} is a space of =nite codimension in X . It follows that U0 , V0 ¡Kn+1 . Setting U = U0 and V = V0 , we obtain the estimate unc(xOk |Jk + U; xOk |Jk + V ) ¿ unc(xOk |Jk + U ; xOk |Jk + V ) ¿ C: Since U; V ¡X can be arbitrary, this means that in this case (xOk ; Jk ) is (X ; C)-bad. This completes the proof of the lemma. Lemma 2.3. Fix C¿1 and suppose that X ¡X is C-dichotomous. If ((x1 ; : : : ; xn ); J ) ∈ AX × B is (X ; C)-good, then ∀Y ¡ X ∃W ¡ Y ∀w ∈ W
((x1 ; : : : ; xn ; w); J ) is (X ; C)-good:
Proof. Let us assume that n + 1 ∈ J , the second case (n + 1 ∈= J ) being analogous. Suppose to the contrary that ∃Y ¡ X ∀W ¡ Y ∃w ∈ W
((x1 ; : : : ; xn ; w); J ) is not (X ; C)-good:
Since X is C-dichotomous, this means exactly that ∃Y ¡ X ∀W ¡ Y ∃w ∈ W ∀U; V ¡X unc((x1 ; : : : ; xn ; w)|J + U; (x1 ; : : : ; xn ; w)|J + V ) ¿ C:
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Changing the order of the quanti=ers ∃ w ∈ W and ∀U; V ¡X in the latter formula, we obtain that ∃Y ¡ X ∀U; V; W ¡ Y ∃w ∈ W; unc((x1 ; : : : ; xn ; w)|J + U; (x1 ; : : : ; xn ; w)|J + V ) ¿ C: Setting W = U , we have for any w ∈ W (x1 ; : : : ; xn ; w)|J = (x1 ; : : : ; xn )|J and (x1 ; : : : ; xn ; w)|J + U = (x1 ; : : : ; xn )|J + U: Hence, we obtain ∃Y ¡ X ∀U; V ¡ Y
unc((x1 ; : : : ; xn )|J + U; (x1 ; : : : ; xn )|J + V ) ¿ C:
The latter formula means that ((x1 ; : : : ; xn ); J ) is not (X ; C)-good and this contradicts the assumption of the lemma. Lemma 2.4. Fix C¿1 and suppose that X ¡X is C-dichotomous. If xO = (x1 ; : : : ; xn ) ∈ AX and for each J ∈ B (x; O J ) is (X ; C)-good then ∀Y ¡ X ∃W ¡ Y ∀w ∈ W ∀J ∈ B
((x1 ; : : : ; xn ; w); J ) is (X ; C)-good:
Proof. Let J1 ; : : : ; Jm be any enumeration of the subsets of {1; 2; : : : ; n}. Fix Y ¡X . By Lemma 2.3, for ((x1 ; : : : ; xn ); J1 ) there exists a subspace W1 ¡Y such that ∀w ∈ W1 ((x1 ; : : : ; xn ; w); J1 ) is (X ; C)-good. Similarly, for ((x1 ; : : : ; xn ); J2 ) we have W2 ¡W1 ¡Y such that for each w ∈ W2 both pairs ((x1 ; : : : ; xn ; w); J2 ) and ((x1 ; : : : ; xn ; w); J1 ) are (X ; C)-good. After m such steps we have Y ¿W1 · · · ¿Wm and ∀ w ∈ Wm ∀Ji
(i = 1; : : : ; m)
((x1 ; : : : ; xn ; w); Ji ) is (X ; C)-good:
Finally, let W = Wm . Lemma 2.5. If (∅; ∅) ∈ AX × B is (X; C)-good then X contains a C-unconditional basic sequence. Proof. By Lemma 2.2 there is X ¡X such that X is C-dichotomous. It is easy to verify that (∅; J ) is (X ; C)-good for each J ∈ B. Now using Lemma 2.4 we can de=ne a sequence (xi )∞ i = 1 in X so that the xi ’s are linearly independent in the completion of X and for each natural n and each J ∈ B the pair ((x1 ; : : : ; xn ); J ) is (X ; C)-good. Using Remark 2.1 we conclude that the sequence (xi )∞ i = 1 is C-unconditional. Lemma 2.6. Let Y be an in4nite-dimensional Banach space. Let X ⊆ Y be a countable subset of Y that is a linear subspace over Q such that XO ¡Y . Suppose that C¿1
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and (if W , U , V denote closed subspaces of Y ) ∀W ¡ Y ∃U; V ¡ W
unc(U; V )¡C:
Then (∅; ∅) ∈ AX × B is (X; C )-good for each C ¿ C.
Proof. Fix any X ¡X and choose two subspaces U; V ¡XO such that unc(U; V )¡C. Pick an )¿1 so that )2 ¡C =C. Using a standard Banach space argument, one can construct a linear isomorphism T : X → X such that T ¡) and T −1 ¡) and there exist U , V ¡X such that T −1 (U ) ⊂ U and T −1 (V ) ⊂ V . It is then easy to verify that unc(U ; V ) 6 T T −1 unc(U; V ) 6 )2 C ¡ C : This completes the proof of the lemma. Proof of Theorem 1.2. In this proof we are dealing directly only with Banach spaces. Let Z be an in=nite-dimensional Banach space. Consider the following condition: ∃C ¿ 1∃Z ¡ Z∀W ¡ Z
∃U; V ¡ W
unc(U; V )¡C:
If Z satis=es this condition then, by Lemmas 2.5 and 2.6, Z contains an unconditional basic sequence. If Z fails this condition then a well-known diagonal argument shows that Z has a HI subspace (cf. the proof of Theorem C in [12]). 3. A result on asymptotic l2 basic sequences In this section we shall indicate an application of the approach presented in Section 2 to construction of so called asymptotic l2 basic sequences in a suitable class of Banach spaces. Similar arguments are applicable to asymptotic lp basic sequences, 16p¡∞, and to asymptotic c0 basic sequences. We shall only state the main theorem without proving it here. A complete proof will be presented in a more general setting in [4]. Before stating this result we need some de=nitions. De&nition 3.1. Let X be a Banach space and 16C¡∞. We say that a sequence (U1 ; : : : ; Un ) of subspaces of X is C-l2 , if ∀(u1 ; : : : ; un ) ∈ U1 × · · · × Un ; 1=2 1=2 n n n √ 1 √ uj 2 6 uj uj 2 : 6 C C j=1 j=1 j=1 We say that X contains ln2 -sums C-isomorphically if for each n¿2 there exists a C-l2 n-tuple (U1 ; : : : ; Un ) with Ui ¡X for i = 1; : : : ; n.
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A basic sequence (xi ) in X is said to be asymptotic C-l2 if for each n¿2 and each sequence A1 ; : : : ; An of mutually disjoint subsets of the set {m ∈ N : m¿n} the n-tuple (span{xj : j ∈ Ai })ni= 1 is C-l2 . Clearly, an orthonormal basis in a Hilbert space is asymptotic C-l2 , where C = 1. Another example is the usual basis of T2 , the 2-convexi=ed Tsirelson’s space (see [1,10]), which is asymptotic C-l2 with C = 2. (Recall that the space T2 contains no isomorphic copy of any lp space or c0 .) A standard argument proves that (every subspace of) a Banach space with an asymptotic C-l2 basis contains l2 -sums C uniformly for each C ¿C. Conversely, by adapting the method used in Section 2 one can prove the following. Theorem 3.2. Let X be a Banach space. If each subspace Y ¡X contains l2 -sums C-isomorphically for some C¡∞, then there is X0 ¡X and C0 ¡∞ such that each subspace Y ¡X0 contains l2 -sums C0 -isomorphically. Moreover, if C0 ¡C ¡∞ then each subspace Y ¡X0 contains an asymptotic C -l2 basic sequence. References [1] P.G. Casazza, T. Shura, Tsirelson’s Space, in: Lecture Notes in Mathematics, Vol. 1363, Springer, Berlin, 1989. [2] C.C. Chang, H.J. Keisler, Model Theory, in: Studies in Logic and Foundation of Mathematics, Vol. 73, North-Holland, Amsterdam, 1977. [3] D. Ellentuck, A new proof that analytic sets are Ramsey, J. Symbolic Logic 39 (1974) 163–165. [4] T. Figiel, R. Frankiewicz, R. Komorowski, C. Ryll-Nardzewski, Selecting basic sequences in ’-stable Banach spaces, Studia Math. 159 (2003) 499–515. [5] W.T. Gowers, A new dichotomy for Banach spaces, Geom. Funct. Anal. 6 (1996) 1083–1093. [6] W.T. Gowers, B. Maurey, The unconditional basic sequence problem, J. Math. Soc. 6 (1993) 851–874. [7] J. Lindenstrauss, L. Tzafriri, Classical Banach Spaces I, Springer, Berlin, 1977. [8] J. Lindenstrauss, L. Tzafriri, Classical Banach Spaces II, Springer, Berlin, 1979. [9] B. Maurey, Quelques progrSes dans la comprTehension de la dimension in=nie, in: Espaces de Banach classiques et quantiques, JournTee Annuelle, Soc. Math. de France, 1994, 1–29. [10] G. Pisier, The Volume of Convex Bodies and Banach Space Geometry, Texas A&M University, UniversitTe Paris VI, Texas, 1989. [11] S. Shelah, A Banach space with few operators, Israel J. Math. 30 (1978) 181–191. [12] N. Tomczak-Jaegermann, A solution of homogeneous Banach space problem, Canad. Math. Soc. 3 (1996) 267–286.