WEAKLY NULL SEQUENCES IN L 1

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1 JOURNAL OF THE AMERICAN MATHEMATICAL SOCIETY Volume 0, Number 1, January 007, Pages 5 36 S (06) Article electronically published on September 19, 006 WEAKLY NULL SEQUENCES IN L 1 WILLIAM B JOHNSON, BERNARD MAUREY, AND GIDEON SCHECHTMAN 1 Introduction In [MR], the second author H P Rosenthal constructed the first examples of weakly null normalized sequences which do not have any unconditionally basic subsequences Much more recently, the second author W T Gowers [GM] constructed infinite dimensional Banach spaces which do not contain any unconditionally basic sequences These later examples are of a different character than the examples in [MR] For example, in [MR], it is shown that if K is a sufficiently complex countable compact metric space, then the space C(K) contains a weakly null normalized sequence which does not have an unconditionally basic subsequence, it is known [PS] that every infinite dimensional subspace of such a C(K) space contains an unconditionally basic sequence, in fact, a sequence which is equivalent to the unit vector basis of c 0 In [MR] it was asked if a similar example exists in L 1 The space L 1 is similar to C(K) in that every infinite dimensional subspace contains an unconditionally basic sequence Indeed, if the subspace is not reflexive, then l 1 embeds into the subspace [KP] If a subspace X of L 1 is reflexive, then it embeds into L p for some 1 <p [R] SinceL p has an unconditional basis, every weakly null normalized sequence in X has an unconditionally basic subsequence The main result in this paper, Theorem 1, is that there is a weakly null normalized sequence {f i } in L 1 which has no unconditionally basic subsequence In fact, the sequence {f i } has the stronger property that for every ε>0, the (conditional) Haar basis is (1 + ε)-equivalent to a block basis of every subsequence of {f i } This is analogous to the result in [MR] that if K is a sufficiently complex countable compact metric space, then the space C(K) contains a weakly null normalized sequence {x n } n=1 so that every initial segment of the (conditional) summing basis is (1 + ε)-equivalent to a block basis of every subsequence of {x n } n=1 Theorem 1 can also be compared to the result of [MS] that for 1 <p<, there is a 1-symmetric basic sequence {g n } n=1 in L p so that the Haar basis is equivalent to a block basis of every subsequence of {g n } n=1 In this result, there is a lower Received by the editors June 8, Mathematics Subject Classification Primary 46B15, 46E30 Key words phrases L 1, Haar basis, unconditional basic sequence, weakly null The first author was supported in part by NSF grant DMS NSF grant DMS , Texas Advanced Research Program the US-Israel Binational Science Foundation The last author was supported in part by the Israel Science Foundation the US-Israel Binational Science Foundation was a participant in the NSF Workshop in Linear Analysis Probability, Texas A&M University 5 c 006 by the authors

2 6 WILLIAM B JOHNSON, BERNARD MAUREY, AND GIDEON SCHECHTMAN bound C p > 1 to the constant of equivalence to the Haar basis because every block basis of {g n } n=1 is monotonely unconditional the Haar basis for L p is not The approach to proving Theorem 1 also yields information about other rearrangement invariant function spaces In Theorem we prove that if X is a χ rearrangement invariant function space on [0, 1] such that inf [0,t] t>0 =0 t 1/ 1 0 χ dt [0,t] t <, then there is a weakly null normalized sequence {f i } in X such that for each ε>0, the Haar basis is (1 + ε)-equivalent to a block basis of every subsequence of {f i } The first condition says that, in some sense, X is not to the right of L Some such condition is needed because the theorem is false in L p for p>by the results of [KP] The second condition in Theorem is technical probably can be weakened; however, in order for anything resembling our construction to work, the Rademacher functions in X must be equivalent to an orthonormal sequence in a Hilbert space We use stard Banach space theory terminology as can be found in [LT] By a Haar system we mean here any sequence of functions distributionally equivalent to the (L normalized, mean zero) traditional Haar functions, ie, any sequence of functions {k i,n }, n n=0, with k i,n being a characteristic function of a set A i,n of measure n, k i,n equal to 1 on A i 1,n+1 to 1 ona i,n+1 (In particular A i,n+1 A i 1,n+1 are disjoint their union is A i,n ) A peculiar weakly null sequence in L 1 In proving Theorem 1, we use the well-known fact that j independent sets each having measure less than 1/j are essentially disjoint Rather than hunt for a reference that states this in a form suitable for our use, we formulate what we need as a lemma Thanks are due to S Kwapień for simplifying the proof Lemma 1 Let 0 < θ < 1 let j be a positive integer Let (Ω, F,µ) be a probability space let h 1,h,,h j be independent symmetric rom variables on Ω so that for each i, h i is the indicator function of a set having probability θ Then there are disjoint sets A B in the algebra generated by h 1,h,,h j so that (1) µ(a) =µ(b) θj (1 θ(j 1)) 1 () h i χ A θ j(j 1) Proof Since j h i = θj [ j ] (3) µ h i =1 = (4) [ j h i ] j h i [ j h i =1] h i = θj(1 θ) j 1, = θj [ j h i =1] = θj θj(1 θ) j 1 θ j(j 1) j h i

3 WEAKLY NULL SEQUENCES IN L 1 7 Set Then, by (3), by (4) [ j A := h i =1 [ j B := h i =1 ] h i =1, ] h i = 1 µ(a) =µ(b) θj (1 θ(j 1)) 1 h i χ A θ j(j 1) Theorem 1 There is a weakly null normalized sequence {f i } in L 1 such that for each ε>0, the Haar basis is (1 + ε)-equivalent to a block basis of every subsequence of {f i } Consequently, {f i} has no unconditionally basic subsequence Proof Let A be the algebra generated by the dyadic subintervals of (0, 1) let {E n } n=1 be an ordering of the nonempty elements of A so that each element of A appears infinitely many times in the sequence {E n } n=1 We will define by recursion an increasing sequence {a n } n=1 of positive numbers a rapidly growing sequence {k n } n=1 of powers of two to satisfy certain conditions to be specified later We then define, for each n, a sequence {h i,n } of functions on (0, 1) so that (i) h i,n = χ Ai,n with A i,n E n, (ii) h i,n =0, (iii) λ(a i,n )=λ(e n )/k n (λ is Lebesgue measure), (iv) h i,n is A-measurable, (v) {h i,n } are independent relative to E n with normalized Lebesgue measure on E n Having done this, we define the desired sequence {f i } by (5) f i = a n h i,n n=1 The a n s are going to be chosen so that n=1 a n h i,n 1 = n=1 a nλ(e n )/k n converges, since each sequence {h i,n } is clearly weakly null, {f i} is also Since the proof that {f i } has the other desired properties is a bit technical, we first describe in words the main part of the construction The sequence {f i } itself of course cannot be symmetric if it has the properties we claim However, notice that for each n, the sequence {a n h i,n } is a 1-symmetric basic sequence which is equivalent (with constant depending on n) to the unit vector basis of l It turns out that this allows all estimates we make when summing the f i s to depend only on the number of terms in the sums Suppose that we want to build a Haar function whose support is a set E in A; that is, we want a linear combination of {f i } (for i in a given infinite set of natural numbers) which approximates a mean zero function whose absolute value is approximately χ E The definitions of {a n } n=1 {k n } n=1 guarantee that for an appropriate m n, if we sum m n different f i (say, for i in B) normalize

4 8 WILLIAM B JOHNSON, BERNARD MAUREY, AND GIDEON SCHECHTMAN appropriately, the resulting vector a 1 n i B f i is a small perturbation of i B h i,n, this latter vector is (by Lemma 1) a small perturbation of an A-measurable function, h B, which takes on only the values { 1, 0, 1} Moreover, the measure of the support S(B) ofh B will be a percentage of the measure of the set E n Ifthat percentage were 100%, we would have achieved our goal since there are arbitrarily large n with E n = E We cannot make the percentage 100%, but we can repeat the process, replacing E by E \ S(B), get another linear combination of {f i } whose absolute value is approximately χ F with F the same percentage of E \ S(B) By iterating this process, we obtain a linear combination of {f i } whose absolute value is approximately χ E We turn now to the actual proof of Theorem 1 Suppose we have defined {a 1,a,,a N } {k 1,k,,k N } Since for each n, the sequence {a n h i,n } is equivalent to the unit vector basis of l, there is a constant M N so that for all finite sets σ of natural numbers we have N (6) a n h i,n < σ 1/ M N i σ n=1 We want to define k N+1 a N+1 so that if we add up εk N+1 (with ε> N, say) of the f i s, then the terms involving a N+1 h i,n+1 are dominant To guarantee that these terms dominate the terms involving a n h i,n with n N, it suffices, as we shall see, to have (7) N k 1/ N+1 M N a N+1 λ(e N+1 ) We shall also see below that in order to guarantee that the terms involving a N+1 h i,n+1 are negligible with respect to the terms involving a n h i,n with n N ifweaddupfewerthank N of the f i s, it suffices (in fact, is over kill ) to have a N+1 λ (E N+1 ) (8) min{λ (E n):1 n N} k N+1 N+1 k N There is of course no difficulty in achieving (7) (8) simultaneously For example, let a N+1 = k 3/4 N+1 with k N+1 a sufficiently large power of This completes the description of how to define the sequence {f i } To see that {f i } satisfies the conclusion of Theorem 1, we need the following: Claim 1 Let τ>0, let F 1 be a nonempty set in A, suppose that M is an infinite subset of N Then there is an f in the linear span of {f i } i M disjoint subsets A A, B Aof F 1 with λ(a) =λ(b) >λ(f 1 )/ τ so that f χ A 1 <τ Once we have the claim, it is of course easy to get the stronger conclusion with A B = F 1 keeping the second approximation conclusion as is Just divide F 1 \ (A B) into two disjoint sets of A of the same measure; add one of them to A the other to B It is now evident from Claim 1 that for any sequence {ε n } n=1 of positive numbers any infinite subset M of N we can build a Haar system {g n } n=1 a block basis {u n } n=1 of {f i } i M so that for each n, g n u n 1 <ε n This implies that {f i } satisfies the conclusion of Theorem 1 Proof of Claim 1 Let ε 1 be an appropriately large power of ; say, ε 1 = m We choose an appropriately large N 1 with E N1 = F 1 let σ 1 be a subset of M which has cardinality εk N1 By conditions (i) (v) Lemma 1 (applied in the

5 probability space (F 1, A 1 F 1, B 1 F 1 in A so that WEAKLY NULL SEQUENCES IN L 1 9 λ λ(f 1 ) )withθ = k 1 N 1 λ(a 1 )=λ(b 1 ) ε (1 ε)λ(f 1) j = εk N1 ), there are disjoint sets h i,kn1 χ A1 B1 <ε 1 i σ 1 Set F := F 1 \ (A 1 B 1 )(sothatλ(f ) [1 ε(1 ε)]λ(f 1 )) repeat the construction, replacing F 1 by F Weusethesameε choose an appropriately large N >N 1 with E N = F let σ be a subset of M of cardinality εk N with max σ 1 < min σ ThistimewegetdisjointsetsA F, B F in A so that λ(a )=λ(b ) ε (1 ε)λ(f ) h i,kn χ A B <ε 1 i σ Next set F 3 := F \ (A A 3 )(sothatλ(f 3 ) [1 ε(1 ε)] λ(f 1 )) repeat Continue in this way mε 1 = m m steps, thereby obtaining k 1 <k < <k m m (which can grow as fast as we like), disjoint subsets A 1,A,,A m m,b 1,B, B m m of F 1 which are in A, subsets σ 1,σ,,σ m m of M, with maxσ j 1 < min σ j σ j = εk Nj so that λ(a j )=λ(b j ), n (9) λ(f 1 \ (A j B j )) [1 ε(1 ε)], (10) i σ j h i,knj χ Aj Bj 1 <ε Set A := m m A j, B := m m B j,σ := m m σ j Then from (9) we have that (11) λ(a) =λ(b) while (10) gives (1) m m h i,knj i σ j We need to verify that m m From (6), (7), (8) we get i σ j (a 1 N j f i h i,knj ) 1 1 [1 ε(1 ε)]m/ε λ(f 1 ) χ A <mε 1 i σ j h i,knj is close to the linear span of {f i } i M a 1 N j ((εk Nj ) 1/ ) a M Nj 1 + εk n λ(e n ) Nj n=n j +1 k n Nj+1 ε 1/ λ(e Nj )+ε Nj 1 λ(e Nj ):=( ), we can assume that ( ) j 1 ε since ε is specified before N 1 From this (1) we get that m m (13) a 1 N j f i χ A 1 < (m +1)ε = ε(1 log ε) i σ j

6 30 WILLIAM B JOHNSON, BERNARD MAUREY, AND GIDEON SCHECHTMAN From (11) (13) it is clear that if ε is sufficiently small, the claim is satisfied if we set f := m m a 1 N j i σ j f i 3 Rearrangement invariant function spaces Here we generalize the first statement of Theorem 1 to the case of rearrangement invariant function spaces It is clear that this statement does not hold for every rearrangement invariant function spaces, not even for nice ones Indeed, for p<, every weakly null normalized sequence in L p contains a subsequence equivalent to the unit vector basis of either l p or l [KP] It is natural to conjecture that the theorem may still hold for spaces which are strictly to the left of L,insome sense Actually a lot more is true: Theorem Let (X, ) be a rearrangement invariant function space on [0, 1] such that χ [0,t] (14) inf =0 t>0 t 1/ 1 (15) χ [0,t] dt 0 t < Then there is a weakly null normalized sequence {f i } in X such that for each ε>0, the Haar basis is (1 + ε)-equivalent to a block basis of every subsequence of {f i } We begin with a lemma which replaces Lemma 1 The main difference is that we give a lattice, rather than norm, estimate for the error Lemma Let 0 < θ < 1 let j be a positive integer Let (Ω, F,µ) be a probability space let h 1,h,,h j be independent symmetric rom variables on Ω so that for each i, h i is the indicator function of a set having probability θ Denote by S the support of j h i Then there are disjoint subsets A B of S in the algebra generated by h 1,h,,h j, for each s =, 3,,asetC s S of measure at most (θj) s 1 exp( θj 1 θ ) µ(s) so that (16) µ(a) =µ(b) θj ( θ(j 1) ) exp µ(s) exp 1 θ (17) h i χ A χc Cs s=3 ( θ(j 1) ) 1 θ Remark 1 In particular, if (Ω, F, µ) is a subinterval of [0, 1] with normalized Lebesgue measure is a rearrangement invariant norm on [0, 1], then, putting σ = µ(s)exp ( ) θj 1 θ, (18) j 1 h i χ A χ[0,θjσ/] + χ [0,(θj) t σ/(t+1)!] t=

7 WEAKLY NULL SEQUENCES IN L 1 31 However, this norm estimate of the error does not seem to be enough to deduce Theorem we will have to use the lattice estimate given in the lemma Proof As in the proof of Lemma 1 set [ j A := h i =1 For s =, 3, set Then clearly h i χ A [ j B := h i =1 [ j ] C s := h i s r= ] h i =1 ] h i = 1 ( j ) h i χ χ [ j + h i =r] C For s j the measure of C s is ( ) j θ r (1 θ) j r (θj) r r r! r=s r=s The measure of S is 1 (1 θ) j θj, while A B both have measure θj(1 θ) j 1 (θj)s e θj θj ( θ(j 1) exp 1 θ ) χ Cs Since µ(s) µ(a)+µ(b) θj exp ( θ(j 1) ) 1 θ,wealsogetthat µ(c s ) (θj)s 1 exp(θj + θ(j 1) 1 θ ) µ(s) (θj)s 1 exp( θj 1 θ ) µ(s) The proof of Theorem is very similar to that of Theorem 1 we shall only sketch it Proof of Theorem We are going to define inductively an increasing sequence of positive numbers {a n }, a sequence {k n } n=1 of powers of two a double sequence {h i,n } of functions on [0, 1] Given k n, {h i,n } is defined in exactly the same way as in the beginning of the proof of Theorem 1 to satisfy conditions (i) (v) there We now describe how, given a 1,,a N k 1,,k N, to define a N+1 k N+1 We note first that (15) implies in particular that for each n, {h i,n } is equivalent to an orthonormal sequence We delay the proof of that to Lemma 3 below Once we know this, we deduce, as in the L 1 case, that there is a constant M N so that for all finite sets σ of natural numbers we have N (19) a n h i,n < σ 1/ M N n=1 i σ s=3

8 3 WILLIAM B JOHNSON, BERNARD MAUREY, AND GIDEON SCHECHTMAN We want to define k N+1 >k N ( a power of two) a N+1 >a N so as to satisfy (0) N k 1/ N+1 M N a N+1 χ EN+1 (1) a N+1 χ min{ χ :1 n N} E n λ(e [0, N+1 ) k ] N+1 N+1 k N We can choose k N+1 a N+1 to satisfy (0) (1) simultaneously since, by condition (14) on the space, we clearly can build increasing positive sequences {a n } {k n } tending to infinity with a n /kn 1/ a n χ [0,k 1 n ] 0 By passing to a subsequence, the convergence to these limits can be made arbitrarily fast Condition (1) implies in particular that n=1 a n h i,n < So we can define f i = n=1 a nh i,n as before Lemma 3 guarantees that {f i } is weakly null The main part of the proof is contained in the following claim, replacing Claim 1, which is clearly enough to finish the proof of the theorem Note that since the conditions on the space ensure that it is not L, the measure condition on the sets A B in the claim imply that χ A χ B approximates a Haar function supported on F 1 to an arbitrary degree of approximation (see the comment after the statement of Claim 1) Claim Let τ > 0, let F 1 be a nonempty set in A, suppose that M is an infinite subset of N Then there are f in the linear span of {f i } i M disjoint subsets A, B Aof F 1 with λ(a) =λ(b) >λ(f 1 )/ τ so that f χ A +χ B <τ Proof of Claim Let ε 1 be an appropriately large power of, say, ε 1 = m We choose an appropriately large N 1 with E N1 = F 1 let σ 1 be a subset of M which has cardinality εk N1 By conditions (i) (v) Lemma (applied in the λ probability space (F 1, λ(f 1 ))withθ = k 1 N 1 j = εk N1 ), there are disjoint sets A 1,B 1 in A another set S 1 in A such that A 1,B 1 S 1 F 1, for each s =, 3,,asetC 1,s S 1 of measure at most ε s 1 e 3ε λ(s 1 ) so that λ(a 1 )=λ(b 1 ) εe ε λ(f 1 ) e ε λ(s 1) i σ 1 h i,kn1 χ A1 B1 χc1, + χ C1,s Set F := F 1 \ S 1 repeat the construction, replacing F 1 by F We use the same ε choose an appropriately large N >N 1 with E N = F let σ be a subset of M of cardinality εk N with max σ 1 < min σ Thistimeweget sets A,B S F in A with A,B disjoint, for each s =, 3,,aset C,s S of measure at most ε s 1 e 3ε λ(s ) so that λ(a )=λ(b ) εe ε λ(f ) e ε λ(s ) s=3

9 WEAKLY NULL SEQUENCES IN L 1 33 i σ h i,kn χ A B χc, + χ C,s We continue in an obvious way, setting F 3 = F 1 \ (S 1 S ), getting subsets σ 1,,σ l of M with max σ i < min σ i+1, disjoint subsets S 1,S,,S l of F 1,a disjoint couple of sets A j,b j S j, j =1,,,l, all in A, a double sequence of sets C j,s, j =1,,,l, s=1,,, satisfying for all j =1,,,l, λ(c j,s ) εs 1 e 3ε λ(s j ), λ(a j )=λ(b j ) εe ε λ(f 1 \ j 1 r=1 i σ j h i,knj χ Aj Bj χcj, + s=3 S r ) e ε λ(s j) χ Cj,s We choose l so that λ(f 1 \ l r=1 S r) <ελ(f 1 ) This clearly can be done since S j eats at least an εe ε portion of F 1 \ j 1 r=1 S rset l l A = A j,b= for all s =, 3, Then () C s = l C j,s B j λ(c s ) εs 1 e 3ε λ(f 1 ), λ(a) =λ(b) e ε λ( l Consequently, l (3) l h i,knj i σ j h i,knj i σ j r=1 s=3 S r ) e ε ε λ(f 1 ), χ A χ C Cs s=3 χ A χ + [0,εe χ 3ε /] [0,ε t e 3ε /(t+1)!] For ε small enough the last expression is smaller than ε 10 χ [0,t] dt 0 t condition (15) implies that the last quantity is smaller than τ/ifε is small enough Also, (3) implies that, if ε is small enough, then λ(a),λ(b) >λ(f 1 )/ τ t=

10 34 WILLIAM B JOHNSON, BERNARD MAUREY, AND GIDEON SCHECHTMAN The rest of the proof is very similar to that of the L 1 case We need to verify that l i σ j h i,knj is close to the linear span of {f i } i M From (19), (0), (1) we get (a 1 N j f i h i,knj ) a 1 N j (εk Nj ) 1/ M Nj 1 + εk Nj a n χ [0, λ(en) ] kn i σ j n=n j +1 N j+1 ε 1/ χ ENj + ε N j χ ENj := ( ), we can assume that ( ) j 1 ε since ε is specified before N 1 From this (3) we get that (4) l a 1 N j f i χ A 1 <ε+ τ/ <τ i σ j for small enough ε Thus the claim is satisfied for f := l a 1 N j i σ j f i Lemma 3 Assume (X, ) is a rearrangement invariant function space on [0, 1] 1 0 χ dt [0,t] t < Let {r i } be a sequence of three valued symmetric rom variables whose absolute values are characteristic functions of sets {A i } of equal measure, all contained in one set A, assume the r i s are independent as rom variables in the probability space A with the normalized Lebesgue measure Then, {r i } is equivalent to an orthonormal sequence Proof By a simple classical computation in the space L 4 (A, a j r j 4 4 = m a 4 j +6m ( ) a ja k 3m a j j<k dt λ(a) )weget where m = λ(a j )/λ(a), hence the L L 4 norms are equivalent, this yields that the L 1 L norms are also equivalent By the fact that the norm in X dominates the L 1 norm, we get that {r i }, as a sequence of elements of X, dominates dt λ(a) the same sequence considered as a sequence in L 1 (A, ) The latter is equivalent to an orthogonal sequence by the preceding argument For the other inequality we first assume as we may that A =[0, 1] For each sequence of coefficients {a i } with a i = 1 each t>0wehavethewell-known inequality (see, for example, [LeTa, p 90]) Since a i r i λ( a i r i >t) e t / jχ = (j 1< ai χ, r i j) ( ai r i >t) t=0

11 WEAKLY NULL SEQUENCES IN L 1 35 we get that a i r i is dominated by χ ( ai χ r i >t) [0,e t / ] The last sum is comparable to t=0 t=0 1 χ [0,e t dt = ds χ / ] [0,s] s 0 0 log 1 s, which, by (15), is finite Remark The proofs of Theorems 1 Theorem actually show that the weakly null sequence obtained has the stronger property that every subsequence has a block basis which is an arbitrarily small perturbation of a Haar system Under some mild extra condition on the rearrangement invariant space (separability, for example, is sufficient), the closed linear span of a Haar system is norm one complemented (via a conditional expectation), which implies that the corresponding block basis of the weakly null sequence spans a complemented subspace Remark 3 One of the referees asked whether the sequence {f i } in Theorem 1 can be constructed so that it all of its subsequences span a space isomorphic to L 1 We cannot answer this question However, there does not exist a normalized weakly null sequence in L p,1<p<, so that every subsequence has a further subsequence which spans a subspace isomorphic to L p Indeed, by passing to an appropriate subsequence which is a small perturbation of a block basis for the Haar system, we can assume that such a sequence {f i } is an unconditional basis for L p But by using the Kadec-Pe lczyński [KP] dichotomy for unconditional sequences in L q,1/p +1/q = 1 dualizing, we see that {f i } must have a subsequence which is equivalent to the unit vector basis of either l p or of l,sonofurther subsequence can span an isomorph of L p References [GM] Gowers, W T Maurey, B, The unconditional basic sequence problem J Amer Math Soc 6 (1993), no 4, MR10138 (94k:4601) [KP] Kadec, M I Pe lczyński, A, Bases, lacunary sequences complemented subspaces in the spaces L p Studia Math 1 (196), MR (7:851) [LeTa] Ledoux, M Talagr, M, Probability in Banach spaces Isoperimetry processes Ergebnisse der Mathematik und ihrer Grenzgebiete (3) [Results in Mathematics Related Areas (3)], 3 Springer-Verlag, Berlin, 1991 MR (93c:60001) [LT] Lindenstrauss, J Tzafriri, L, Classical Banach spaces II Function spaces Ergebnisse der Mathematik und ihrer Grenzgebiete [Results in Mathematics Related Areas], 97 Springer-Verlag, Berlin-New York, 1979 MR (81c:46001) [MR] Maurey, B Rosenthal, H P, Normalized weakly null sequence with no unconditional subsequence Studia Math 61 (1977), no 1, MR (55:11010) [MS] Maurey, B Schechtman, G, Some remarks on symmetric basic sequences in L 1 Compositio Math 38 (1979), no 1, MR05364 (80c:46039) [PS] Pe lczyński, A Semadeni, Z, Spaces of continuous functions (III) (Spaces C(Ω) for Ω without perfect subsets) Studia Math 18 (1959), 11 MR (1:658) [R] Rosenthal, H P, On subspaces of L p Ann of Math 97 (1973), MR031 (47:784)

12 36 WILLIAM B JOHNSON, BERNARD MAUREY, AND GIDEON SCHECHTMAN Department of Mathematics, Texas A&M University, College Station, Texas address: johnson@mathtamuedu Laboratoire d Analyse et de Mathématiques Appliquées, UMR CNRS 8050, Université de Marne la Vallée, Champs-sur-Marne, France address: maurey@univ-mlvfr Department of Mathematics, Weizmann Institute of Science, Rehovot, Israel address: gideonschechtman@weizmannacil