Factorizations of natural embeddings of l_p^n int L_r

18 Pages·0.17 MB·English

by  Tadek Figiel

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 < p < q and σ = 1 q . Let T:Z L (µ), where µ
≤ ∞ − p∗ → 1
is a probability measure. If C (T) = K and T:Z L (µ) = K, then for some
1,p p
k → k C
∗
p
m > 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 < p < q
1
∈ k k ≤ ≤ ∞
and
1 g > 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 H¨older’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 < p < and
1
→ ∞
C (T) = K, T:Z L (µ) = K.
1,p p
k → k C
∗
If 0 < κ < K, η 0 and η1/p 1 κ/K, then µ(E) η implies that there exists
≥ C ≤ − ≤
z Ball(Z) such that 1 Tz > 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 < p ≤ ∞ (see [R]).
Corollary 10. If U:lN X , t > 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 < p < . Let
1
→ ∞
Z Z. Suppose that n = dimTZ < and that
0 0
⊆ ∞
C (u) c T > 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