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Factorizations of natural embeddings of ln into L , II p r by T. Figiel, W. B. Johnson* and G. Schechtman ** Introduction 2 9 In this continuation of [FJS], we show that in some situations considered in [FJS], 9 1 conclusions of certain theorems can be strengthened. More explicitly, suppose that T is n an operator from some Banach space into L which factors through some L -space Z as a 1 1 J uw and normalized so that w = 1. In Corollary 12.A we show that if T is the inclusion 6 k k ] mapping from a “natural” n-dimensional Hilbertian subspace of L1 into L1, then u well- A F preserves a copy of lk with k exponential in n (where “well” and the base of the exponent 1 . h depend on u and on a quantitative measure of “naturalness”). This improves the result t k k a m of [FJS] that the same hypotheses yield that lk well-embeds into uZ. Corollary 12.B 1 [ gives a similar improvement of Corollary 1.5 in [FJS]; that is, Corollary 12.B is the same 1 v as Corollary 12.A except that the operator T is assumed to be a mapping from a space 2 0 whose dual has controlled cotype into L which acts like a quotient mapping relative to a 1 2 1 “natural” Hilbertian subspace of L . 0 1 2 Corollary20strengthenstheconclusionofProposition4.3in[FJS]inasimilarmanner; 9 / h it states that an operator from a C(K) space which well-preserves a copy of ln also well- t 2 a m preserves a copy of lk with k exponential in n (rather than just have rank which is ∞ : v exponential in n). This can be viewed as a finite dimensional analogue of a particular i X case of a result of Pe lczyn´ski [Pe1] stating that every non weakly compact operator from r a a C(K) space preserves a copy of c . 0 In Theorem 21 we apply the earlierresults in order to prove that for each m there is an m-dimensional normed space G such that any superspace of G with a good unconditional basis must contain a copy of lk with k exponential in m. ∞ We thank J. Bourgain for pointing us in the right direction on the material presented here. After proving the results in [FJS], we suggested to him that there might be a * Supported in part by NSF DMS-87-03815 ** Supported in part by the U.S.-Israel BSF 1 translation invariant operator T of bounded norm on L of the circle which is the identity 1 on the span of the first n Rademacher functions and which does not preserve lk with 1 k exponential in n. By disproving this conjecture, Bourgain started us thinking that Corollary 12.A was true. Preliminaries, a quantitative version of Rosenthal’s lemma In this section we prove, in Proposition 1 below, a quantitative version of Maurey’s formulation [M] of Rosenthal’s lemma [R] stating that an operator into L (µ) either factors 1 through an L (ν) space for some p > 1 via a change of density or is of type no better than p 1. Our approach is in fact close in spirit to Rosenthal’s original argument which (unlike some later arguments) was basically quantitative in nature. Recall that for 1 p < q ≤ ≤ ∞ and for an operator u:Z L (µ) p → C (u) = inf h h−1u:Z L p,q s q {k k k → k} where the inf is over all changes of measure h; i.e., over all 0 < h L (µ) where 1+1 = 1. ∈ s q s p In the statement of Proposition 1 and elsewhere in this paper t∗ denotes t . t−1 ∗ Proposition 1. Let 1 K there exist z ,...,z in the unit ball of Z and mutually disjoint 2+1 1 m (cid:16)2 pCkTk(cid:17) measurable sets F ,...,F such that for i = 1,...,m one has 1 m k1Fi Tzikσ1k1Fi Tzik1q−σ ≥ 2−(p1+1)K. (1) For the proof we need two lemmas. Lemma 2. Let g L with g 1. Suppose E is a µ-measurable set, 1 κ > 0. p k ∼E k Then there exists a measurable set F, F E = , such that ∩ ∅ 21/p p∗ µ(F) < , κ (cid:16) (cid:17) 2 k1F gkσ1k1F gk1q−σ > 2−p1κ. Proof. Without loss of generality we can assume that g 0. Set F = [g > γ] E, where ≥ ∼ γ > 0 is defined below. Observe that gp γp−1 g γp−1 Z ≤ Z ≤ ∼F ∼F and hence, by Hölder’s inequality, 1−t t κp γp−1 < gp g gq , − ZF ≤ (cid:16)ZF (cid:17) (cid:16)ZF (cid:17) where t = (p 1)/(q 1). Since µ(F) < 1/γ, we can fulfill both conditions of the lemma − − by choosing γ so that γp−1 = 1κp. 2 Lemma 3. Let T:Z L (µ), µ being a probability measure. Suppose 1 κ. p ∈ k ∼E k Proof. Without loss of generality we may assume that µ(E) > 0. Observe that for any measurable set A one has ∗ C (1 T:Z L (µ)) µ(A)1/p 1 T:Z L (µ) . 1,p 1 p A → ≤ k A → k The lemma follows by using this observation for A = E and A = E, because ∼ 1 T:Z L (µ) > C (1 T) C (T) C (1 T) p 1,p 1,p 1,p k ∼E → k ∼E ≥ − E ∗ ∗ K µ(E)1/p 1 T:Z L (µ) K(1 η1/p ) κ. p ≥ − k E → k ≥ −C ≥ Proof of Proposition 1. Clearly, we may assume that T = 1. Put κ = 1K , k k 2 η = (2 )−p∗, δ = 21/p p∗ and let us start with E = . Since C κ ∅ (cid:0) (cid:1) 1 T:Z L (µ) > κ, p k ∼E → k 3 using Lemma 2 we can define z and F so that µ(F ) < δ and 1 Tz satisfies (1). 1 1 1 k F1 1k1 Suppose now that, for some i 1, we have already defined z ,...,z and F ,...,F . Let 1 i 1 i ≥ E = F . As long as µ(E) η, Lemma 3 guarantees that we can use Lemma 2 again j≤i j ≤ S in order to choose z and F so that F E = , µ(F ) < δ and 1 Tz i+1 i+1 i+1 ∩ ∅ i+1 k Fi+1 i+1k1 satisfies the estimate (1). Therefore this procedure can be applied more than η/δ times. Since we have been assuming T = 1, this yields the promised lower estimate for m and k k completes the proof of the proposition. The next proposition shows that we can actually get a somewhat stronger conclusion to Proposition 1; namely, for some k proportional to m, the identity on lk can be factored 1 through T. Recall that, for a pair of linear operators T:X Y and U:X Y , the 1 1 → → factorization constant of U through T is defined to be γ (U) = inf A B : A:X X, B:Y Y , U = BTA . 1 1 T {k kk k → → } We let γ (U) = if no such factorization exists. We also put T ∞ γ (Z) = γ (id :Z Z). Z T T → Proposition 4. Let T:Z L (µ) be a bounded linear operator. Suppose that z ,...,z 1 1 m → are in the unit ball of Z and F ,...,F are mutually disjoint µ-measurable sets such that, 1 m for i = 1,...,m, 1 Tz δ T > 0. i k Fi k ≥ k k Then for some k 1δm there exist linear operators A:lk Z and B:L (µ) lk such ≥ 8 1 → 1 → 1 that BTA = id and A B T 2δ−1; i.e., γ (lk) 2δ−1 T −1 for some k 1δm. lk1 k kk kk k ≤ T 1 ≤ k k ≥ 8 For the proof we need two basically known lemmas (see [JS]). Lemma 5. Let x ,...,x be elements of L (µ) and let A ,...,A be mutually disjoint 1 m 1 1 m µ-measurable sets. If 1 < k 1m, then there exists a subset D 1,...,m with D = k ≤ 2 ⊂ { } | | such that for each i D ∈ −1 m m x dµ (2k 1) x . i i Z | | ≤ − (cid:18)2(cid:19) k k j∈DX∼{i} Aj Xi=1 4 Proof. Setting a = x dµ , for i,j = 1,...,m, and α = a , we have ij Aj | i| 1≤i6=j≤m ij R P m α x . i ≤ k k X i=1 Put s = 2k, = E 1,...,m : E = s . Write, for E , E { ⊂ { } | | } ∈ E α(E) = a . ij X X i∈Ej∈E∼{i} It is easy to see that α(E) = m−2 α = s m −1α, hence we can pick E E∈E s−2 |E| 2 2 0 ∈ E P (cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) so that α(E ) s m −1α . Let F = i E : a 1α(E ) . Then F 0 ≤ 2 2 { ∈ 0 j∈E0∼{i} ij ≥ k 0 } (cid:0) (cid:1)(cid:0) (cid:1) P has at most k elements and, clearly, each k–element subset D E F has the required 0 ⊆ ∼ property. Lemma 6. Let U:lk L (µ) be a linear operator, U 1. Suppose there exist µ- 1 → 1 k k ≤ measurable sets F ,...,F such that for i = 1,...,k 1 k 1 Ue δ, 1 Ue γ, k Fi ik ≥ k Fj ik ≤ X 1≤i6=j≤k where 0 < γ < δ. Then there exists a linear operator Q:L (µ) lk such that QU = id 1 → 1 lk1 and Q (δ γ)−1. k k ≤ − Proof. Define W:L (µ) lk by the formula Wf = ( fg dµ)k , where 1 → 1 i i=1 R g = 1 sgn(Ue ) for i = 1,...,k. i Fi i It is easy to check that W 1 and for x lk one has WUx (δ γ) x . Therefore k k ≤ ∈ 1 k k ≥ − k k the operator Q = (WU)−1W has the required properties. Proof of Proposition 4. We may assume that T = 1. Apply Lemma 5 with η = 1δ k k 2 and x = Tz for i = 1,...,m. This yields a set D 1,...,m , with D = k 1δm, i i ⊂ { } | | ≥ 8 which satisfies the assertion of Lemma 5. Writing z : i D = f ,...,f we can i 1 m { ∈ } { } define the operator A by the formula Ae = f , for i = 1,...,k. The existence of the i i operator B follows then immediately from Lemma 6. We shall combine Propositions 1 and 4 in Theorem 8 below. Before doing that we would like to state a dual version of Proposition 4. Note that, if dimX ,dimY < , then 1 1 ∞ γ (U) = γ (U∗). This follows from the principle of local reflexivity ([LT], p.33). ∗ T T 5 Corollary 7. Let V:lm X be an operator of norm 1 such that Ve δ > 0 for ∞ → k ik ≥ i = 1,...,m. Then γ (lk ) 2δ−1 for some k 1δm. V ∞ ≤ ≥ 8 Proof. Let Z = X∗. Pick norm one elements z ,...,z in Z such that z (Ve ) δ for 1 m i i ≥ i = 1,...,m. Using Proposition 4 we obtain γ (lk) 2δ−1 for some k 1δm. Since V∗ 1 ≤ ≥ 8 γ (lk ) = γ (lk) this completes the proof. V ∞ V∗ 1 The L result 1 The main results of this section are Theorem 11 and Corollary 12 below. Corollary 12.Astatesroughlythatinanygoodfactorizationofanaturalembedding ofln intoL (0,1) 2 1 (for example, the embedding sending the unit vector basis of ln to the first n Rademacher 2 functions) through an L space, the operator between the two L spaces preserves an lk 1 1 1 space with k exponential in n. We begin however with a theorem of a more general nature which is a corollary to Propositions 1 and 4. The assumptions in both this theorem and Theorem 11 are stated in terms of factorization constants of an operator into an L space 1 through L spaces via changes of densities. The relation between these constants and p factorizations of natural embeddings was one of the main tools in [FJS]. We shall return to this relation in the proof of Corollary 12. Theorem 8. Let T:Z L (µ) be a linear operator such that T = 0 and C (T) < , 1 1,q → 6 ∞ ∗ where 1 < p < q . Set σ = 1 q , ∆ = T σC (T)1−σ/C (T). Then ≤ ∞ − p∗ k k 1,q 1,p ∗ 1 C (T) p γ (lk) 2(4∆)1/σ T −1 for some k (4∆)−1/σ 1,p . T 1 ≤ k k ≥ 8 8 T (cid:16) (cid:17) k k Proof. By Maurey’s result [Ma] quoted in [FJS], for each r (1, ] there is a nonnegative ∈ ∞ function φ L (µ) such that φ dµ = 1 and r 1 r ∈ R φ−1T:Z L (φ dµ) = C (T) k r → r r k 1,r (this uses the convention 0 = 0). Set φ = 1(φ +φ ). Then for r = q and r = p 0 2 p q ∗ φ−1T:Z L (φdµ) 21/r C (T). r 1,r k → k ≤ 6 Consider the operator T = φ−1T:Z L (φdµ). Since C (T ) = C (T) for r (1, ], 1 1 1,r 1 1,r → ∈ ∞ ∗ applying Proposition 1 to the operator T , we have 21/p . We can estimate for each i 1 C ≤ ∗ 1 T z T :Z L (φdµ) 21/q C (T). k Fi 1 ikLq(φdµ) ≤ k 1 → q k ≤ 1,q Hence we obtain elements z ,...,z in Ball(Z) and sets F ,...,F so that 1 m 1 m k1Fi T1zikσL1(φdµ) ≥ 2−p1−1 2q1∗C1,q(T) σ−1C1,p(T) = 2−2∆−1kTkσ (cid:0) (cid:1) ∗ and m > C1,p(T) p . Therefore, we have 8kTk (cid:0) (cid:1) k1Fi Tzik1 = k1Fi T1zikL1(φdµ) ≥ (4∆)−σ1kTk. Now we simply apply Proposition 4 to T, z ,...,z and F ,...,F . 1 m 1 m We next state two corollaries to Theorem 8 concerning the dual situation. Corollary 9. Let U:C(K) X be a linear operator such that 0 < π (U) < , where r → ∞ 1 r < t < . Set σ = 1 r, ∆ = U σπ (U)1−σ/π (U). Then ≤ ∞ − t k k r t 1 π (U) t γ (lk ) 2(4∆)σ1 U −1 for some k (4∆)−1/σ t . U ∞ ≤ k k ≥ 8 8 U (cid:16) (cid:17) k k Proof. This follows easily from Theorem 8, because C(K)∗ is an L space, γ (lk) = 1 U∗ 1 γU(l∞k ) and C1,p(U∗) = πp∗(U) for 1 1 and π (U) = c U > 0 , then γ (lk ) ∞ → t k k U ∞ ≤ 2(4)t∗Nt∗−1 U −1 for some k 2t∗−3(c)t∗+tN1−t∗. c k k ≥ 8 Proof. This follows by using Corollary 9 with r = 1, because π (U) N U . 1 ≤ k k The next theorem and Corollary 12 below are the main results of this section. Theorem 11. Let T:Z L (µ) be a bounded linear operator and let 1 0, 1,p ≥ k k 7 for each finite rank operator u:Z L (µ) such that u = T . → 1 |Z0 |Z0 If c 25 and δ = (p 1)n, then γ (lk) < 5δ for some k > 5−δ 21/δ n. ≥ − T 1 (cid:0) (cid:1) Corollary 12.A. Let X be an n-dimensional subspace of L1 for which C1,p∗(X) C√p∗ ≤ for all 2 p∗ < . If j is the inclusion map from X into L and j = UW with 1 ≤ ∞ W:X Z, U:Z L , W 1 and Z is an L space, then γ (lk) 52D for some → → 1 k k ≤ 1 U 1 ≤ k 5−2D2n/(2D), where D = (28C U )2. ≥ k k The proof of the corollary is very similar to the proof of Corollary 1.5 in [FJS]; there are, however, some changes. Here is an outline of the proof: Let Z = WX and define 0 p by p∗ = n. If n < 2D the conclusion is obvious, so we may assume that n 2D. D ≥ Then δ = (p 1)n satisfies δ 2D, so, by Theorem 11, it is enough to prove that for any − ≤ extension V:Z L (µ) of U , one has C (V) 25 U . → 1 |Z0 1,p ≥ k k Now, for any extension U˜:Z X of U , we have id = U˜W and hence, using a → |Z0 X weak form of Grothendieck’s inequality, we obtain n1/2 = π (id ) W π (U˜) 2γ (U˜). X 2 ≤ k k 2 ≤ 2 We can now apply Theorem 1.3 in [FJS] with β = 1n1/2 to get that for any extension 2 V:Z L of U → 1 |Z0 C1,p(V) 2−3C1,p∗(X)−1n1/2. ≥ With the choice p∗ = n/D we get the desired estimate C (V) 2−3C−1D1/2 = 25 U . 1,p ≥ k k Corollary 12.B strengthens Corollary 12.A in the same way that Theorem 5.1 in [FJS] strengthens Corollary 1.5 in [FJS]. Corollary 12.B. Suppose that X L1, dimX = n, and C1,p∗(X) C√p∗ for all ⊂ ≤ 2 p∗ < . Let Y be a Banach space whose dual has finite cotype q constant C (Y∗) q ≤ ∞ and let Q:Y L be an operator for which 1 → Q(Ball(Y)) Ball(X). ⊃ 8 Let Q = UW be any factorization of Q through an L space with W 1. Then for 1 k k ≤ some absolute constant η, γ (lk) 5D for some k 5−D2n/D, where U 1 ≤ ≥ D = ηC2qC2(Y∗) U 2. q k k Sketch of proof. Follow the proof of Theorem 5.1 in [FJS] (with r replaced by p) up to the place on p. 98 where it is proved that C (U) 2. Of course, now we need and 1,p ≥ can assure that C (U˜) 25 U for any extension U˜:Z L of the restriction of U to 1,p 1 ≥ k k → U−1(X). Then apply Theorem 11. To prove Theorem 11 we need to introduce some notation and some preliminary results. Given two Banach spaces Z and W the space of all bounded linear operators between them is denoted by B(Z,W), while F(Z,W) is the set of those u B(Z,W) such ∈ that rank u < . By α we denote a norm on F(Z,W) such that α(u) u if rank u = 1. ∞ ≤ k k Given a Banach space W and numbers n,β 1, let us denote by q (n,β) the least ≥ W number k such that, whenever v:W E is a continuous linear operator with rank v n, → ≤ there exists P F(W,W) such that vP = v, P β and rankP k. (Of course, we let ∈ k k ≤ ≤ q (n,β) = , if no such k exists.) W ∞ Proposition 13. Let T B(Z,W) and let Z Z, dimT(Z ) = n < . If 1 β < 0 0 ∈ ⊆ ∞ ≤ ∞ and q (n,β) < , then there exists P F(W,W) such that P β, rankP q (n,β) W ∞ ∈ k k ≤ ≤ W and α(PT) inf α(u) : u F(Z,W),u = T . ≥ { ∈ |Z0 |Z0} Proof. Write Y = u F(Z,W) : u = T and A = inf α(u) : u Y . By { ∈ |Z0 |Z0} { ∈ } the Hahn–Banach theorem there is a norm one functional Φ on (F(Z,W),α) such that Φ(u) = A for each u Y. Observe that if S:W Z∗∗ is the linear operator defined by ∈ → (Sw)(z∗) = Φ(z∗ w), ⊗ then for all u F(Z,W) one has ∈ Φ(u) = Tr(Su) = Tr(u∗∗S). Clearly, our assumption on α yields S 1. Moreover, for all u Y one has k k ≤ ∈ (T u)∗∗S = 0. (2) − 9 Indeed, since Ker((T u)∗∗) (Ker(T u))⊥⊥ Z ⊥⊥, it suffices to verify that SW 0 − ⊇ − ⊇ ⊆ Z⊥⊥. The latter inclusion is obvious, because if w W, z∗ Z⊥ then we have (Sw)(z∗) = 0 ∈ 0 ∈ 0 0 Φ(z∗ w) = 0, since z∗ annihilates Z . Since ranku = dimTZ = n, for some u Y, 0 ⊗ 0 0 0 0 0 ∈ and since (2) implies that T∗∗S = u∗∗S for each u Y, we obtain that rankT∗∗S n. ∈ ≤ Hence, by the definition of q (n,β), there is a P F(W,W) such that T∗∗SP = T∗∗S, W ∈ P β and rankP q (n,β). Observe that, if u is any element of Y, then k k ≤ ≤ W α(PT) Φ(PT) = Tr(SPT) = Tr(T∗∗SP) = Tr(T∗∗S) = Tr(u∗∗S) = Φ(u) = A. ≥ Lemma 14. If W = L (µ) and 0 < ǫ < 1, then q (n,(1 ǫ)−1) < (2 +1)n. 1 W − ǫ Proof. Let u:W E have rank n. Write u = UQ , where Q :W W/(Ker u) is the 0 0 → → quotient map and let F = W/(Ker u). Set β = (1 ǫ)−1. Suppose first that for some k − there exists an operator Q:lk F such that Q β and Q(Ball(lk)) Ball(F). By the 1 → k k ≤ 1 ⊇ lifting property of lk there is Q :lk W such that Q Q β and Q = Q Q . By 1 1 1 → k 1k ≤ k k ≤ 0 1 the lifting property of W = L (µ) there is Q :W lk such that Q Q 1 and 1 2 → 1 k 2k ≤ k 0k ≤ Q = QQ . Let P = Q Q . Then P β, rankP k and 0 2 1 2 k k ≤ ≤ u = UQ = UQQ = UQ Q Q = uP. 0 2 0 1 2 Now the well–known volume argument shows that the unit sphere of F contains an ǫ–net (where (1 ǫ)−1 = β) of cardinality k < 1(2 +1)n. Using this fact one easily constructs − 2 ǫ the operator Q:lk F with the two properties which we have used above. 1 → Theorem11cannow beobtainedbylettingǫ = 1 andreplacingcby25 inthefollowing 2 proposition. Proposition 15. Let T:Z L (µ) satisfy the assumptions of Theorem 11, and let 1 → 0 < ǫ < 1. Then 2 p 2 n(p−1) γ (lk) < 4 +1 T −1 T 1 (1 ǫ)c ǫ k k (cid:16) (cid:17) (cid:16) (cid:17) − ∗ for some k > 4p−2 (1−ǫ)c pp 2 +1 −n(p−1). 8 ǫ (cid:0) (cid:1) (cid:0) (cid:1) Proof. Let β = (1 ǫ)−1, W = L (µ), α = C . Thanks to Lemma 14, we can apply 1 1,p − Proposition 13 which yields an operator P on L (µ) such that P < β, rank P N = 1 k k ≤ 10