Subprojective Banach spaces

0.3 MB·

by  Timur Oikhberg

SUBPROJECTIVE BANACH SPACES
TIMUR OIKHBERGAND EUGENIU SPINU
4
1 Abstract. ABanachspaceX iscalledsubprojectiveifanyofitsinfinitedimen-
0 sionalsubspacesY containsafurtherinfinitedimensionalsubspacecomplemented
2 in X. This paper is devoted to systematic study of subprojectivity. We examine

n thestabilityofsubprojectivityofBanachspacesundervariousoperations,suchus
a direct or twisted sums, tensor products, and forming spaces of operators. Along
J the way,we obtain new classes of subprojective spaces.

7
1

]
A 1. Introduction and main results
F
We examine various aspects of subprojectivity. Throughout this note, all Banach
.
h spaces are assumed to be infinite dimensional, and subspaces, infinite dimensional
t
a and closed, until specified otherwise.
m A Banach space X is called subprojective if every subspace Y X contains a
⊂
[ further subspace Z Y, complemented in X. This notion was introduced in [40], in
⊂
ordertostudythe(pre)adjointsofstrictlysingularoperators. Recallthatanoperator
1
v T B(X,Y) is strictly singular (T (X,Y)) if T is not an isomorphism on any
1 sub∈space of X. In particular, it w∈asSsShown that, if Y is subprojective, and, for
3
T B(X,Y), T∗ (Y∗,X∗), then T (X,Y).
2
∈ ∈ SS ∈ SS
4 Later, connections between subprojectivity and perturbation classes were discov-
. ered. Morespecifically, denotebyΦ (X,Y)thesetofupper semi-Fredholm operators
1 +
0 –thatis,operatorswithclosedrange,andfinitedimensionalkernel. IfΦ (X,Y) = ,
+
6 ∅
4 we define the perturbation class
1
: PΦ (X,Y) = S B(X,Y) :T +S Φ (X,Y) whenever T Φ (X,Y) .
v + { ∈ ∈ + ∈ + }
Xi It is known that (X,Y) PΦ+(X,Y). In general, this inclusion is proper.
SS ⊂
However, we get (X,Y) = PΦ (X,Y) if Y is subprojective (see [1, Theorem
r +
SS
a 7.51] for this, and for similar connections to inessential operators).
Several classes of subprojective spaces are described in [16]. Common examples
of non-subprojective space are L (0,1) (since all Hilbertian subspaces of L are not
1 1
complemented), C(∆), where ∆ is the Cantor set, or ℓ (for the same reason). The
∞
disc algebra is not subprojective, since by e.g. [41, III.E.3] it contains a copy of
C(∆). By [40], L (0,1) is subprojective if and only if 26 p < . Consequently, the
p
∞
Hardyspace H on thediscis subprojectivefor exactly the samevalues of p. Indeed,
p
H contains thediscalgebra. For 1 < p < , H isisomorphictoL . ThespaceH
∞ p p 1
∞
2010 Mathematics Subject Classification. Primary: 46B20, 46B25; Secondary: 46B28, 46B42.
The authors acknowledge the generous support of Simons Foundation, via its Travel Grant
210060. They would also like to thank the organizers of Workshop in Linear Analysis at Texas
A&M, where part of this work was carried out. Last but not least, they express their gratitude to
L.Bunce, T. Schlumprecht,B. Sari, and N.-Ch. Wong for many helpfulsuggestions.
1
2 T.OIKHBERGANDE.SPINU
contains isomorphic copies of L for 1 < p 6 2 [42, Section 3]. On the other hand,
p
VMO is subprojective ([29], see also [36] for non-commutative generalizations).
Westartourpaperbycollectingvariousfactsneededtostudysubprojectivity(Sec-
tion 2). Along the way, we prove that subprojectivity is stable under suitable direct
sums (Proposition 2.1). However, subprojectivity is not a 3-space property (Propo-
sition 2.7). Consequently, subprojectivity is not stable underthe gap metric (Propo-
sition 2.8). Considering the place of subprojective spaces in Gowers dichotomy, we
observe that each subprojective space has a subspace with an unconditional basis.
However, we exhibit a space with an unconditional basis, but with no subprojective
subspaces (Proposition 2.10).
In Section 3, we investigate the subprojectivity of tensor products, and of spaces
of operators. A general result on tensor products (Theorem 3.1) yields the subpro-
jectivity on ℓ ˇℓ and ℓ ˆℓ for 1 6 p,q < (Corollary 3.3), as well as of (L ,L )
p q p q p q
⊗ ⊗ ∞ K
for 1 < p 6 2 6 q < (Corollary 3.4). We also prove that the space B(X) is
∞
never subprojective (Theorem 3.9), and give an example of non-subprojective tensor
product ℓ ℓ (Proposition 3.8).
2 α 2
⊗
Throughout Section 4, we work with C(K) spaces, with K compact metrizable.
WebeginbyobservingthatC(K)issubprojectiveifandonlyifK isscattered. Then
weprovethatC(K,X)issubprojectiveifandonlyifbothC(K)andX are(Theorem
4.1). Turning to spaces of operators, we show that, for K scattered, Π (C(K),ℓ )
qp q
is subprojective (Proposition 4.4). Then we study continuous fields on a scattered
base space, proving that any scattered separable CCR C∗-algebra is subprojective
(Corollary 4.7).
Section 5 shows that, in many cases, subprojectivity passes from a sequence space
to the associated Schatten spaces (Proposition 5.1).
ProceedingtoBanachlattices,inSection6weprovethatp-disjointlyhomogeneous
p-convex lattices (2 6 p < ) are subprojective (Proposition 6.2). In Section 7
∞
^
(Proposition 7.1), we show that the lattice X(ℓ ) is subprojective whenever X is.
p
Consequently (Proposition 7.3), if X is a subprojective space with an unconditional
basis and non-trivial cotype, then Rad(X) is subprojective.
Throughoutthepaper,weusethestandardBanachspaceresultsandnotation. By
B(X,Y) and (X,Y) we denote the sets of linear bounded and compact operators,
K
respectively, acting between Banach spaces X and Y. B(X) refers to the closed unit
ball of X. For p [1, ], we denote by p′ the “adjoint” of p (that is, 1/p+1/p′ = 1).
∈ ∞
2. General facts about subprojectivity
We begin this section by showing that subprojectivity passes to direct sums.
Proposition 2.1. (a) Suppose X and Y are Banach spaces. Then the following are
equivalent:
(1) Both X and Y are subprojective.
(2) X Y is subprojective.
⊕
(b) Suppose X ,X ,... are Banach spaces, and is a space with a 1-unconditional
1 2
E
basis. Then the following are equivalent:
(1) The spaces ,X ,X ,... are subprojective.
1 2
E
SUBPROJECTIVE BANACH SPACES 3
(2) ( X ) is subprojective.
n n E
P
In (b), we view as a space of sequences of scalars, equipped with the norm
E
k·kE. ( nXn)E refers to the space of all sequences (xn)n∈N ∈ n∈NXn, endowed
with thePnorm k(xn)n∈Nk = k(kxnkXn)kE. Due to the 1-uncondQitionality (actually,
1-suppression unconditionality suffices), ( X ) is a Banach space.
n n E
We begin by making two simple observaPtions, to be usedseveral times throughout
this paper.
Proposition 2.2. Consider Banach spaces X and X′, and T B(X,X′). Suppose
∈
Y is a subspace of X, T is an isomorphism, and T(Y) is complemented in X′.
Y
|
Then Y is complemented in X.
Proof. If Q is a projection from X′ to T(Y), then T−1QT is a projection from X
onto Y.
This immediately yields:
Corollary 2.3. Suppose X and X′ are Banach spaces, and X′ is subprojective.
Suppose, furthermore, that Y is a subspace of X, and there exists T B(X,X′) so
∈
that T is an isomorphism. Then Y contains a subspace complemented in X.
Y
|
The following version of “Principle of Small Perturbations” is folklore, and essen-
tially contained in [5]. We include the proof for the sake of completeness.
Proposition 2.4. Suppose (x ) is a seminormalized basic sequence in a Banach
k
space X, and (y ) is a sequence so that lim x y = 0. Suppose, furthermore,
k k k k
that every subspace of span[y : k N] contkains−a skubspace complemented in X.
k
Then span[x :k N] contains a sub∈space complemented in X.
k
∈
Proof. Replacing x by x / x , we can assume that (x ) normalized. Denote the
k k k k
k k
biorthogonal functionals by x∗, and set K = sup x∗ . Passing to a subsequence,
k kk kk
we can assume that x y < 1/(2K). Define the operator U B(X) by
kk k − kk ∈
setting Ux = kx∗k(Px)(yk −xk). Clearly kUk < 1/2, and therefore, V = IX +U
is invertible. FPurthermore, Vxk = yk. If Q is a projection from X onto a subspace
W span[y : k N], then P = V−1QV is a projection from X onto a subspace
k
Z ⊂span[x :k ∈N].
k
⊂ ∈
Remark 2.5. Note that, in the proof above, the kernels and the ranges of the
projections Q and P are isomorphic, via the action of V.
Proof of Proposition 2.1. It is easy to see that subprojectivity is inherited by sub-
spaces. Thus, in both (a) and (b), only the implication (2) (1) needs to be
⇒
established.
(a) Throughout the proof, P and P stand for the coordinate projections from
X Y
X Y onto X and Y, respectively. We have to show that any subspace E of X Y
⊕ ⊕
contains a further subspace G, complemented in X Y.
⊕
Show first that E contains a subspace F so that either P or P is an iso-
X F Y F
| |
morphism. Indeed, suppose P is not an isomorphism, for any such F. Then
X F
|
P is strictly singular, hence there exists a subspace F E, so that P has
X E X F
| ⊂ |
norm less than 1/2. But P + P = I , hence, by the triangle inequality,
X Y X⊕Y
4 T.OIKHBERGANDE.SPINU
P f > f P f > f /2 for any f F. Consequently, P is an iso-
Y X Y F
k k k k − k k k k ∈ |
morphism.
Thus, by passing to a subspace, and relabeling if necessary, we can assume that
E contains a subspace F, so that P is an isomorphism. By Corollary 2.3, F
X F
|
contains a subspace G, complemented in X.
Set F′ = P (F), and let V be the inverse of P :F F′. By the subprojectivity
X X
→
of X, F′ contains a subspace G′, complemented in X via a projection Q. Then
P = VQP gives a projection onto G = V(G′) F.
X
⊂
(b) Here, we denote by P the coordinate projection from X = ( X ) onto
n k k E
Xn. Furthermore, we set Qn = nk=1Pk, and Q⊥n = 1 − Qn. WePhave to show
that any subspace Y X containsPa subspace Y0, complemented in X. To this end,
⊂
consider two cases.
(i) For some n, and some subspace Z Y, Q is an isomorphism. By part (a),
n Z
⊂ |
X ... X = Q (X) is subprojective. Apply Corollary 2.3 to obtain Y .
1 n n 0
(⊕ii) For⊕every n, Q is not an isomorphism – that is, for every n N, and every
n Y
| ∈
ε > 0, there exists a norm one y Y so that Q y < ε. Therefore, for every
n
∈ k k
sequence of positive numbers (ε ), we can find 0 = N < N < N < ..., and a
i 0 1 2
sequence of norm one vectors y Y, so that, for every i, Q y , Q⊥ y < ε .
i ∈ k Ni ik k Ni+1 ik i
By a small perturbation principle, we can assume that Y contains norm one vectors
(y′) so that Q y′ = Q⊥ y′ = 0 for every i. Write y′ = (z )Ni+1 , with z X .
i Ni i Ni+1 i i j j=Ni+1 j ∈ j
Then Z = span[(0,...,0,z ,0,...) : j N] (z is in j-th position) is complemented
j j
∈
in X. Indeed, if z = 0, find z∗ X∗ so that z∗ = z −1, and z∗,z = 1. If
j 6 j ∈ j k jk k jk h j ji
zj = 0, set zj∗ = 0. For x = (xj)j∈N ∈ X, define Rx = (hzj∗,xjizj)j∈N. It is easy
to see that R is a projection onto Z, and R does not exceed the unconditionality
k k
constant of .
E
Now note that J : Z : (α z ,α z ,...) (α z ,α z ,...) is an isometry.
1 1 2 2 1 1 2 2
Let Y′ = span[y′ : i N→], aEnd Y = J(Y′). By7→the sukbpkrojecktivkity of , Y contains
i ∈ E E E
a subspace W, which is complemented in via a projection R . Then J−1R JR is
1 1
E
a projection from X onto Y = J−1(W) X.
0
⊂
Remark 2.6. From the last proposition it follows the (strong) p-sum of subprojec-
tive Banach spaces is subprojective. On the other hand, the infinite weak sum of
subprojective spaces need not be subprojective.
Recall that if X is a Banach space, then
1
ℓwpeak(X) = {x = (xn)∞n=1 ∈ X ×X ×X... :x∗s∈uXp∗(X|x∗(xn)|p)p < ∞}.
It is known that ℓweak(X) is isomorphic to B(ℓ ,X) (1 + 1 = 1), see [9, Theo-
p p′ p p′
rem 2.2]. We show that, for X = ℓ (r p′), B(ℓ ,X) contains a copy of ℓ , and
r p′ ∞
≥
therefore, is not subprojective. To this end, denote by (e ) and (f ) the canonical
i i
bases in ℓ and ℓ respectively. For α = (α ) ℓ , define B(ℓ ,X) Uα : e
r p′ i ∞ p′ i
∈ ∋ 7→
α f . Clearly, U is an isomorphism.
i i
Notethatthesituationisdifferentforr < p′. Then,byPitt’sTheorem,B(ℓ ,ℓ )=
p′ r
(ℓ ,ℓ ). In the next section we prove that the latter space is subprojective.
p′ r
K
Next we show that subprojectivity is not a 3-space property.
SUBPROJECTIVE BANACH SPACES 5
Proposition 2.7. For 1< p < there exists a non-subprojective Banach space Z ,
p
∞
containing a subspace X , so that X and Z /X are isomorphic to ℓ .
p p p p p
Proof. [23, Section 6] gives us a short exact sequence
0 ℓ jp Z qp ℓ 0,
p p p
−→ −→ −→ −→
where the injection j is strictly cosingular, and the quotient map q is strictly
p p
singular. By [23, Theorem 6.2], Z is not isomorphic to ℓ . By [23, Theorem 6.5],
p p
any non-strictly singular operator on Z fixes a copy of Z . Consequently, j (ℓ )
p p p p
contains no complemented subspaces (by [26, Theorem 2.a.3], any complemented
subspace of ℓ is isomorphic to ℓ ).
p p
It is easy to see that subprojectivity is stable under isomorphisms. However, it
is not stable under a rougher measure of “closeness” of Banach spaces – the gap
measure. If Y and Z are subspaces of a Banach space X, we define the gap (or
opening)
Θ (Y,Z) = max sup dist(y,Z), sup dist(z,Y) .
X
(cid:8)y∈Y,kyk=1 z∈Z,kzk=1 (cid:9)
We refer the reader to the comprehensive survey [32] for more information. Here,
we note that Θ satisfies a “weak triangle inequality”, hence it can be viewed as a
X
measure of closeness of subspaces. The following shows that subprojectivity is not
stable under Θ .
X
Proposition 2.8. There exists a Banach space X with a subprojective subspace Y
so that, for every ε> 0, X contains a non-subprojective space Z with Θ (Y,Z) 6 ε.
X
Proof. Our Y will be isomorphic to ℓ , where p (1, ) is fixed. By Proposition
p
∈ ∞
2.7, there exists a non-subprojective Banach space W, containing a subspace W ,
0
so that both W and W′ = W/W are isomorphic to ℓ . Denote the quotient map
0 0 p
W W′ by q. Consider X = W W′ and Y = W W′ E. Furthermore, for
1 0 1
→ ⊕ ⊕ ⊂
ε > 0, define Z = εw qw : w W . Clearly, Y is isomorphic to ℓ ℓ ℓ ,
ε 1 p p p
{ ⊕ ∈ } ⊕ ∼
hence subprojective, while Z is isomorphic to W, hence not subprojective. By [32,
ε
Lemma 5.9], Θ (Y,Z ) 6 ε.
X ε
Looking at subprojectivity through the lens of Gowers dichotomy and observing
that a subprojective Banach space does not contain hereditarily indecomposable
subspaces, we immediately obtain the following.
Proposition 2.9. Every subprojective space has a subspace with an unconditional
basis.
The converse to the above proposition is false.
Proposition 2.10. There existsaBanachspace with anunconditional basis, without
subprojective subspaces.
Proof. In[18,Section 5], T.Gowers andB.MaureyconstructaBanach spaceX with
a1-unconditional basis, so that any operator on X is a strictly singular perturbation
of a diagonal operator. We prove that X has no subprojective subspaces. In doing
so, we are re-using the notation of that paper. In particular, for n N and x X,
∈ ∈
6 T.OIKHBERGANDE.SPINU
we define x as the supremum of n x , where x ,...,x are successive
k k(n) i=1k ik 1 n
vectors so that x = ixi. It is known tPhat, for every block subspace Y in X, every
c > 1, and every nP N, there exists y Y so that 1 = y 6 y (n) < c. This
∈ ∈ k k k k
technical result can be used to establish a remarkable property of X: suppose Y is a
subspace of X, with a normalized block basis (y ). Then any zero-diagonal (relative
k
to the basis (y )) operator on Y is strictly singular. Consequently, any T B(Y)
k
∈
can be written as T = Λ+S, where Λ is diagonal, and S is zero-diagonal, hence
strictly singular. This result is proved in [18] for Y = X, but an inspection yields
the generalization described above.
Suppose, for the sake of contradiction, that X contains a subprojective subspace
Y. A small perturbation argument shows we can assume Y to be a block subspace.
Blocking further, we can assume that Y is spanned by a block basis (y ), so that
j
1= y 6 y < 1+2−j. We achieve the desired contradiction by showing that
j j (j)
k k k k
no subspace of Z = span[y +y ,y +y ,...] is complemented in Y.
1 2 3 4
Suppose P is an infinite rank projection from Y onto a subspace of Z. Write P =
Λ+S, where S is a strictly singular operator with zeroes on the main diagonal, and
Λ= (λ )∞ is a diagonal operator (that is, Λy =λ y for any j). As sup y <
j j=1 j j j jk jk(j)
, by [18, Section 5] we have lim Sy = 0. Note that (Λ+ S)2 = Λ+ S, hence
j j
∞
diag(λ2 λ ) = Λ2 Λ = S ΛS SΛ S2 is strictly singular, or equivalently,
j − j − − − −
lim λ (1 λ ) = 0. Therefore, there exists a 0 1 sequence (λ′) so that Λ′ Λ
j j − j − j −
is compact (equivalenty, lim (λ λ′) = 0), where Λ′ = diag(λ′) is a diagonal
j j − j j
projection. ThenP = Λ′+S′,whereS′ = S+(Λ Λ′)isstrictlysingular,andsatisfies
−
lim S′y = 0. The projection P is not strictly singular (since it is of infinite rank),
j j
henceΛ′ = P S′ isnotstrictly singular. Consequently, thesetJ = j N : λ′ =1
− { ∈ j }
is infinite.
Now note that, for any j, Py y > 1/2. Indeed, Py Z, hence we can
j j j
k − k ∈
write Py = α (y +y ). Let ℓ = j/2 . By the 1-unconditionality of our
j k k 2k−1 2k ⌈ ⌉
basis, yj PPyj > yj αℓ(y2ℓ−1 +y2ℓ) > max 1 αℓ , αℓ > 1/2. For j J,
k − k k − k {| − | | |} ∈
S′y = Py y , hence S′y > 1/2, which contradicts lim S′y = 0.
j j j j j j
− k k k k
Remark 2.11. The preceding statement provides an example of an atomic order
continuous Banach lattice without subprojective subspaces. One can also observe
that if a Banach lattice is not order continuous, then it contains a subprojective
subspacec . Also, if a Banach lattice is non-atomic order continuous with an uncon-
0
ditional basis, then it contains a subprojective subspace ℓ (i.e. [24, Theorem 2.3]).
2
Finally, onemightaskwhether,inthedefinitionofsubprojectivity,theprojections
from X onto Z can be uniformly bounded. More precisely, we call a Banach space
X uniformly subprojective (with constant C) if, for every subspace Y X, there
⊂
exists a subspace Z Y and a projection P : X Z with P 6 C. The proof
⊂ → k k
of [16, Proposition 2.4] essentially shows that the following spaces are uniformly
subprojective: (i) ℓ (1 6 p < ) and c ; (ii) the Lorentz sequence spaces l ;
p 0 p,w
∞
(iii) the Schreier space; (iv) the Tsirelson space; (v) the James space. Additionally,
L (0,1) is uniformly subprojective for 2 6 p < . This can be proved by combining
p
∞
Kadets-Pelczynski dichotomy with the results of [2] about the existence of “nicely
complemented”copiesofℓ . Moreover, anyc -saturatedseparablespaceisuniformly
2 0
subprojective, sinceany isomorphiccopy ofc contains aλ-isomorphiccopy ofc , for
0 0
SUBPROJECTIVE BANACH SPACES 7
any λ > 1 [26, Proposition 2.e.3]. By Sobczyk’s Theorem, a λ-isomorphic copy of c
0
is 2λ-complemented in every separable superspace. In particular, if K is a countable
metric space, then C(K) is uniformly subprojective [12, Theorem 12.30].
However, in general, subprojectivity need not be uniform. Indeed, suppose 2 <
p < p < ... < , and lim p = . By Proposition 2.1(b), X = ( L (0,1))
1 2 ∞ n n ∞ n pn 2
is subprojective. The span of independent Gaussian random variablePs in Lp (which
we denote by G ) is isometric to ℓ . Therefore, by [17, Corollary 5.7], any projection
p 2
from Lp onto Gp has norm at least c0√p, where c0 is a universal constant. Thus, X
is not uniformly subprojective.
3. Subprojectivity of tensor products and spaces of operators
SupposeX ,X ,...,X areBanachspaceswithunconditionalFDD,implemented
1 2 k
by finite rank projections (P′ ), (P′ ), ..., (P′ ), respectively. That is, P′ P′ = 0
1n 2n kn in im
unless n = m, lim N P′ = I point-norm, and sup N P′ <
N n=1 in Xi N,±k n=1± ink ∞
(this quantity is somPetimes referred to as the FDD constant ofPXi). Let Ein =
ran(P′ ).
in
We say that a sequence (w )∞ X X ... X is block-diagonal if there
j j=1 ⊂ 1 ⊗ 2 ⊗ ⊗ k
exists a sequence 0 = N < N < ... so that
1 2
Nj+1 Nj+1 Nj+1
w E E ... E .
j 1n 2n kn
∈ ⊗ ⊗ ⊗
(cid:0)n=XNj+1 (cid:1) (cid:0)n=XNj+1 (cid:1) (cid:0)n=XNj+1 (cid:1)
Suppose is an unconditional sequence space, and ˜ is a tensor product of Banach
E ⊗
spaces. The Banach space X ˜X ˜ ... ˜X is said to satisfy the -estimate if
1 2 k
⊗ ⊗ ⊗ E
there exists a constant C > 1 so that, for any block diagonal sequence (wj)j∈N in
X ˜X ˜ ... ˜X , we have
1 2 k
⊗ ⊗ ⊗
(3.1) C−1 ( wj )j∈N E 6 wj 6 C ( wj )j∈N E
k k k k k k k k k k
Xj
Theorem 3.1. Suppose X , X , ..., X are subprojective Banach spaces with un-
1 2 k
conditional FDD, and ˜ is a tensor product. Suppose, furthermore, that for any
⊗
finite increasing sequence i = [1 6 i < ... < ...i 6 k], there exists an uncondi-
1 ℓ
tional sequence space Ei, so that Xi1⊗˜Xi2⊗˜ ...⊗˜Xik satisfies the Ei-estimate. Then
X ˜X ˜ ... ˜X is subprojective.
1 2 k
⊗ ⊗ ⊗
A similar result for ideals of operators holds as well. We keep the notation for
projections implementing the FDD in Banach spaces X and X . We say that a
1 2
Banachoperatorideal issuitable(forthepair(X ,X ))ifthefiniterankoperators
1 2
A
are dense in (X1,X2) (in its ideal norm). We say that a sequence (wj)j∈N
A ⊂
(X ,X ) is block diagonal if there exists a sequence 0= N < N < ... so that, for
1 2 1 2
A
any j, w = (P P )w (P P ). If is an unconditional sequence
j 2,Nj − 2,Nj−1 j 1,Nj − 1,Nj−1 E
space, we say that (X ,X ) satisfies the -estimate if, for some constant C,
1 2
K E
(3.2) C−1 ( w ) 6 w 6 C ( w )
j j E j A j j E
k k k k k k k k k k
Xj
holds for any finite block-diagonal sequence (w ).
j
8 T.OIKHBERGANDE.SPINU
Theorem 3.2. Suppose X and X are Banach spaces with unconditional FDD, so
1 2
that X∗ and X are subprojective. Suppose, furthermore,that the ideal is suitable
1 2 A
for(X ,X ), and (X ,X ) satisfies the -estimate for some unconditional sequence
1 2 1 2
A E
. Then (X ,X ) is subprojective.
1 2
E A
Before proving these theorems, we state a few consequences.
Corollary 3.3. The spaces X ˇ ... ˇX and X ˆ ... ˆX are subprojective where
1 n 1 n
⊗ ⊗ ⊗ ⊗
X is ether isomorphic to ℓ (1 p < ) or c for every 1 i n.
i pi ≤ i ∞ 0 ≤ ≤
For n = 2, this resultgoes back to [37]and [31] (the injective and projective cases,
respectively).
SupposeaBanachspaceX hasanFDDimplementedbyprojections(P′)–thatis,
n
P′P′ = 0 unless n = m, sup N P < , and lim N P = I point-
n m N,±k n=1± nk ∞ N n=1 n X
norm. We say that X satisfies thePlower p-estimate if there exPists a constant C so
that, for any finitesequence ξ ranP , ξ p >C ξ p. ThesmallestC for
j ∈ j k j jk jk jk
which the above inequality holds is calledPthe lower p-ePstimate constant. The upper
p-estimate, and the upper p-estimate constant, are defined in a similar manner. Note
that, if X is an unconditional sequence space, then the above definitions coincide
with the standard one (see e.g. [27, Definition 1.f.4]).
Corollary 3.4. Suppose the Banach spaces X and X have unconditional FDD,
1 2
satisfy the lower and upper p-estimates respectively, and both X∗ and X are sub-
1 2
projective. Then (X ,X ) is subprojective.
1 2
K
Before proceeding, we mention several instances where the above corollary is ap-
plicable. Note that, if X has type 2 (cotype 2), then X satisfies the upper (resp.
lower) 2-estimate. Indeed, suppose X has type 2, and w ,...,w are such that
1 n
w = P w for any j. Then
j j j
1/2
w 6 CAve w 6 CT (X) w 2
j ± j 2 j
k k k ± k k k
Xj Xj (cid:16)Xj (cid:17)
(T (X) is the type 2 constant of X). The cotype case is handled similarly. Thus, we
2
can state:
Corollary 3.5. Suppose the Banach spaces X and X have unconditional FDD, co-
1 2
type2andtype 2respectively, andbothX∗ andX are subprojective. Then (X ,X )
1 2 K 1 2
is subprojective.
This happens, for instance, if X = L (µ) or C (1 < p 6 2) and X = L (µ) or
1 p p 2 q
C (2 6 q < ). Indeed, the type and cotype of these spaces are well known (see
q
∞
e.g. [35]). The Haar system provides an unconditional basis for L . The existence
p
of unconditional FDD of C spaces is given by [4].
p
Proof of Theorem 3.1. We will prove the theorem by induction on k. Clearly, we
can take k = 1 as the basic case. Suppose the statement of the theorem holds for
a tensor product of any k 1 subprojective Banach spaces that satisfy -estimate.
− E
We will show that the statement holds for the tensor product of k Banach spaces
X = X ˜X ˜ ... ˜X .
1 2 k
⊗ ⊗ ⊗
SUBPROJECTIVE BANACH SPACES 9
For notational convenience, let P = n P′ , and I = I . If A B(X) is a
in k=1 ik i Xi ∈
projection, we use the notation A⊥ for IXP A. Furthermore, define the projections
−
Q = P P ... P and R = P⊥ P⊥ ... P⊥. Renorming all X ’s if
n 1n ⊗ 2n ⊗ ⊗ kn n 1n ⊗ 2n ⊗ kn i
necessary, we can assume that their unconditional FDD constants equal 1.
First show that, for any n, ranR⊥ is subprojective. To this end, write R⊥ =
n n
k P(i), where the projections P(i) are defined by
i=1
P P(1) = P I ... I ,
1n 2 k
⊗ ⊗ ⊗
P(2) = P⊥ P I ... I ,
1n ⊗ 2n⊗ 3⊗ ⊗ k
P(3) = P⊥ P⊥ P I ... I ,
1n ⊗ 2n⊗ 3n ⊗ 4⊗ ⊗ k
... ... ...
P(k) = P⊥ P⊥ ... P⊥ P
1n ⊗ 2n⊗ ⊗ k−1,n⊗ kn
(note also that P(i)P(j) = 0 unless i = j). Thus, there exists i so that P(i) is an
isomorphismonasubspaceY′ Y. NowobservethattherangeofP(i) isisomorphic
⊂
to a subspace of ℓN(X(i)), where N = rankP , and
∞ in
X(i) = X ˜X ˜ ... ˜X ˜X ˜ ... ˜X .
1 2 i−1 i+1 k
⊗ ⊗ ⊗ ⊗ ⊗ ⊗
By the induction hypothesis, X(i) is subprojective. By Proposition 2.1, ranP(i) is
subprojective for every i, hence so is ranR⊥.
n
Now suppose Y is an infinite dimensional subspace of X. We have to show that
Y contains a subspace Z, complemented in X. If there exists n N so that R⊥
∈ n|Y
is not strictly singular, then, by Corollary 2.3, Z contains a subspace complemented
in X.
Now suppose R⊥ is strictly singular for any n. It is easy to see that, for any
n|Z
sequence of positive numbers (ε ), one can find 0 = n < n < n < ..., and norm
m 0 1 2
one elements x Y, so that, for any m, R⊥ x + x Q x < ε . By
m ∈ k nm−1 mk k m − nm mk m
a small perturbation, we can assume that x = R⊥ Q x . That is,
m nm−1 nm m
x ran (P P ) (P P ) ... (P P ) .
m ∈ 1,nm − 1,nm−1 ⊗ 2,nm − 2,nm−1 ⊗ ⊗ k,nm − k,nm−1
(cid:16) (cid:17)
LetE = ran(P P ),andW = span[E E ... E : m N] X.
im i,nm− i,nm−1 1m⊗ 2m⊗ ⊗ km ∈ ⊂
Applying “Tong’s trick” (see e.g. [26, p. 20]), and taking the 1-unconditionality of
our FDDs into account, we see that
U : X W :x (P P ) ... (P P ) x
→ 7→ 1,nm − 1,nm−1 ⊗ ⊗ k,nm − k,nm−1
Xm (cid:0) (cid:1)
defines a contractive projection onto W. Furthermore, Z = span[x : m N] is
m
complemented in W. Indeed, the projection P P (i,m N) is c∈ontrac-
i,nm − i,nm−1 ∈
tive, hence we can identify E ˜ ... ˜E with (E ... E ) X. By the
1m km 1m km
⊗ ⊗ ⊗ ⊗ ∩
by Hahn-Banach Theorem, for each m there exists a contractive projection U on
m
E ˜ ... ˜E , with range span[x ]. By our assumption, there exists an uncondi-
1m 2m m
⊗ ⊗
tional sequence space so that X ˜ ... ˜X satisfies the -estimate. Then, for any
1 k
E ⊗ ⊗ E
finite sequence w E ˜ ... ˜E , (3.1) yields
m 1m km
∈ ⊗ ⊗
U w 6 C ( U w ) 6 C ( w ) 6 C2 U w .
k k k k E k E k k
k k k k k k k k k k k k
Xk Xk
Thus, Z is complemented in X.
10 T.OIKHBERGANDE.SPINU
Sketch of the proof of Theorem 3.2. On (X ,X ) we define the projection R :
1 2 n
A
(X ,X ) (X ,X ) : w P⊥wP . Then the range of R⊥ is isomorphic
A 1 2 → A 1 2 7→ 2n 1n n
to X∗ ... X∗ X ... X . Then proceed as in the the proof of Theorem 3.1
1 ⊕ ⊕ 1 ⊕ 2⊕ ⊕ 2
(with k = 2).
To prove Corollary 3.3, we need two auxiliary results.
Lemma 3.6. Suppose 1< p < (1 i n) and X = ˇn ℓ .
i ∞ ≤ ≤ ⊗i=1 pi
(1) If 1/p > n 1, thenX satisfies theℓ -estimatewith 1/s = 1/p (n 1).
i s i
− − −
(2) IfP1/pi 6 n 1, then X satisfies the c0-estimate. P
−
P
Proof. Suppose (w ) is a finite block-diagonal sequence in X. We shall show that
j
w = ( w ) ,withsasinthestatementofthelemma. Tothisend,let(U )
k j jk k k jk ks ij
bePcoordinateprojectionsonℓpi forevery1 ≤ i≤ n,suchthatwj = U1j⊗...⊗Unjwj,
and for each i, U U =0 unless k = m. Letting p′ = p /(p 1), we see that
ik im i i i −
w = sup w , ξ .
j j i i
kXj k ξi∈ℓp′i,kξik61(cid:12)(cid:12)hXj ⊗ i(cid:12)(cid:12)
(cid:12) (cid:12)
Choose ⊗iξi with kξik ≤ 1, and let ξij = Uijξi. Then jkξijkp′i 6 1, and
P
w , ξ 6 w , ξ = w , ξ 6 w Πn ξ .
(cid:12)(cid:12)hXj j ⊗i ii(cid:12)(cid:12) Xj (cid:12)h j ⊗i ii(cid:12) Xj (cid:12)h j ⊗i iji(cid:12) Xj k jk i=1k ijk
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12)
Now let 1/r = 1/p′ = n 1/p . By H¨older’s Inequality,
i − i
P n P 1/r n 1/p′
ξij r 6 ξij p′i i 6 1.
k k k k
(cid:16)Xj (cid:0)Yi=1 (cid:1) (cid:17) Yi=1(cid:16)Xj (cid:17)
If 1/p 6 n 1, then r 1, hence Πn ξ 6 1. Therefore, w 6
i − ≤ j i=1k ijk k j jk
maPxjkwjk = (kwjk)c0. Otherwise, r > 1P, and P
1/s 1/r 1/s
w 6 w s (Πn ξ )r 6 w s = ( w ) ,
k jk k jk i=1k ijk k jk k jk s
Xj (cid:16)Xj (cid:17) (cid:16)Xj (cid:17) (cid:16)Xj (cid:17)
where 1/s = 1 1/r = 1/p n+1.
i
− −
In a similar fashion, wPe show that k jwjk > (kwjk)s. For s = ∞, the inequality
k jwjk > maxjkwjk is trivial. If s isPfinite, assume jkwjks = 1 (we are allowed
toPdo so by scaling). Find norm one vectors ξij Pℓp′ so that ξij = Uijξi, and
∈ i
w = w , ξ . Let γ = w s/r. Then γr = 1 = γ w . Further,
k jk h j ⊗i iji j k jk j j j jk jk
set α = γΠl=6 ip′l/(Pnm=1Πl6=mp′l). An elementary cPalculation showPs that γ = Πn α ,
ij j j i=1 ij
p′
and α i = 1. Let ξ = α ξ . Then ξ =, and therefore,
j ij i j ij ij k kp′
P P
w > w , ξ = Πn α w , ξ = γ w = 1.
k jk h j ⊗i ii i=1 ijh j ⊗i iji jk jk
Xj Xj Xj Xj
This establishes the desired lower estimate.
Lemma 3.7. For 1 6 p 6 , X = ℓ ˆℓ ˆ ... ˆℓ satisfies the ℓ -estimate,
i ∞ p1⊗ p2⊗ ⊗ pn r
where 1/r = 1/p if 1/p < 1, and r = 1 otherwise. Here, we interpret ℓ as
i i ∞
c0. P P