Subprojective Banach spaces
T. Oikhberg
Dept. of Mathematics, University of Illinois at Urbana-Champaign, Urbana IL 61801, USA
E. Spinu
Dept. of Mathematical and Statistical Sciences, University of Alberta Edmonton, Alberta T6G 2G1, CANADA
Abstract
A Banach spaceX is called subprojective if any of its infinite dimensional sub-spaces contains a further infinite dimensional subspace complemented in X. This paper is devoted to systematic study of subprojectivity. We examine the stability of subprojectivity of Banach spaces under various operations, such us direct or twisted sums, tensor products, and forming spaces of operators. Along the way, we obtain new classes of subprojective spaces.
Keywords: Banach space, complemented subspace, tensor product, space of operators.
1. Introduction and main results
Throughout this note, all Banach spaces are assumed to be infinite dimen-sional, and subspaces, infinite dimensional and closed, until specified otherwise. A Banach spaceX is calledsubprojective if every subspaceY X contains a further subspaceZ Y, complemented inX. This notion was introduced in [41], in order to study the (pre)adjoints of strictly singular operators. Recall that an operatorT PBpX, Yqisstrictly singular (T PSSpX, Yq) ifT is not an isomorphism on any subspace of X. In particular, it was shown that, if Y is subprojective, and, forT PBpX, Yq,TPSSpY, Xq, thenT PSSpX, Yq.
Later, connections between subprojectivity and perturbation classes were discovered. More specifically, denote by Φ pX, Yqthe set ofupper semi-Fredholm operators – that is, operators with closed range, and finite dimensional kernel. If Φ pX, Yq H, we define the perturbation class
PΦ pX, Yq tSPBpX, Yq:T SPΦ pX, YqwheneverT PΦ pX, Yqu.
It is known thatSSpX, Yq PΦ pX, Yq. In general, this inclusion is proper. However, we getSSpX, Yq PΦ pX, YqifY is subprojective (see [1, Theorem 7.51] for this, and for similar connections to inessential operators).
Several classes of subprojective spaces are described in [16]. For instance, the spaces`p(1¤p 8) andc0are subprojective. Common examples of
non-subprojective space are L1p0,1q (since all Hilbertian subspaces of L1 are not
complemented),Cp∆q, where ∆ is the Cantor set, or`8 (for the same reason). The disc algebra is not subprojective, since by [43, III.E.3] it contains a copy of
Cp∆q. By [41],Lpp0,1qis subprojective if and only if 2¤p 8. Consequently,
the Hardy spaceHpon the disc is subprojective for exactly the same values ofp.
Indeed,H8 contains the disc algebra. For 1 p 8,Hp is isomorphic toLp.
The spaceH1contains isomorphic copies ofLpfor 1 p¤2 [42, Section 3]. On
the other hand, VMO is subprojective ([30], see also [37] for non-commutative generalizations).
In this paper we examine several aspects of subprojectivity. We start by collecting various facts needed to study subprojectivity (Section2). Along the way, we prove that subprojectivity is stable under suitable direct sums (Proposi-tion2.2). However, subprojectivity is not a 3-space property (Proposition2.8). Consequently, subprojectivity is not stable under the gap metric (Proposition
2.9). 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 (Proposition2.11).
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 subprojectivity on`pbq`q and`pbp`q for 1¤p, q 8(Corollary3.3), as well
as ofKpLp, Lqq for 1 p¤ 2 ¤q 8 (Corollary 3.4). We also prove that
the spaceBpXqis never subprojective (Theorem3.10), and give an example of non-subprojective tensor product`2br`2 (Proposition3.9).
Throughout Section4, we work withCpKqspaces, withKcompact metriz-able. We begin by observing that CpKq is subprojective if and only if K is scattered. Then we prove that CpK, Xq is subprojective if and only if both
CpKqandX are (Theorem4.1). Turning to spaces of operators, we show that, forKscattered, ΠqppCpKq, `qqis subprojective (Proposition4.4). We also study
continuous fields on a scattered base space, proving that any scattered separable CCRC-algebra is subprojective (Corollary4.7).
Section5shows that, in many cases, subprojectivity passes from a sequence space to the associated Schatten space (Proposition5.1).
Proceeding to Banach lattices, in Section 6 we prove that p-disjointly ho-mogeneousp-convex lattices (2¤p 8) are subprojective (Proposition 6.2). In Section7 (Proposition 7.1), we show that the latticeXp`pqis subprojective
wheneverX is. Consequently (Proposition 7.3), ifX is a subprojective space with an unconditional basis and non-trivial cotype, thenRadpXqis subprojec-tive.
nota-tion. ByBpX, YqandKpX, Yqwe denote the sets of linear bounded and com-pact operators, respectively, acting between Banach spacesX and Y. UBpXq
refers to the closed unit ball ofX. ForpP r1,8s, we denote byp1the “adjoint” ofp(that is, 1{p 1{p11).
2. General facts about subprojectivity
We start collecting general facts about subprojectivity by observing that subprojectivity passes to subspaces.
Proposition 2.1. Any subspace of a subprojective Banach space is subprojec-tive.
In contrast to this, subprojectivity does not pass to quotients. Indeed, any separable Banach space is a quotient of`1, which is subprojective. As noted in
Section1, there exist separable non-subprojective spaces. However, subprojectivity does pass to direct sums.
Proposition 2.2. (a) SupposeX andY are Banach spaces. Then the following are equivalent:
1. Both X andY are subprojective. 2. X`Y is subprojective.
(b) SupposeX1, X2, . . .are Banach spaces, andE is a space with a 1-uncon-ditional basis. Then the following are equivalent:
1. The spaces E, X1, X2, . . . are subprojective. 2. p°nXnqE is subprojective.
In (b), we viewEas a space of all sequences, for which the norm}}E is finite. p°nXnqE refers to the space of all sequencespxnqnPNP
±
nPNXn, endowed with
the norm}pxnqnPN} }p}xn}Xnq}E. Due to the unconditionality (actually, 1-suppression unconditionality suffices),p°nXnqE is a Banach space.
In the proof of Proposition 2.2, and further in the paper, we will use two results stated below.
Proposition 2.3. Consider Banach spacesX andX1, andT PBpX, X1q. Sup-poseY is a subspace ofX,T|Y is an isomorphism, andTpYqis complemented
inX1. ThenY is complemented in X.
Proof. If Qis a projection from X1 to TpYq, then T1QT is a projection from
X ontoY.
Corollary 2.4. SupposeX andX1are Banach spaces, andX1 is subprojective. Suppose, furthermore, thatY is a subspace ofX, and there exists T PBpX, X1q
so thatT|Y is an isomorphism. Then Y contains a subspace complemented in
X.
The following version of “Principle of Small Perturbations” is folklore, and essentially contained in [5]. We include the proof for the sake of completeness. Proposition 2.5. Supposepxkqis a seminormalized basic sequence in a Banach
spaceX, andpykqis a sequence so thatlimk}xkyk} 0. Suppose, furthermore,
that every subspace ofspanryk:kPNscontains a subspace complemented inX.
Thenspanrxk :kPNs contains a subspace complemented inX.
Proof. Replacing xk by xk{}xk}, we can assume that pxkqis normalized.
De-note the biorthogonal functionals byxk, and setK supk}xk}. Passing to a subsequence, we can assume that°k}xkyk} 1{p2Kq. Define the operator
U PBpXqby settingU x°kxkpxqpykxkq. Clearly}U} 1{2, and therefore,
V IX U is invertible. Furthermore,V xk yk. IfQis a projection fromX
onto a subspaceW spanryk :kPNs, thenP V1QV is a projection from
X onto a subspaceZspanrxk:kPNs.
Remark 2.6. Note that, in the proof above, the kernels and the ranges of the projectionsQandP are isomorphic, via the action ofV.
Proof of Proposition2.2. Due to Proposition2.1, in both (a) and (b), only the implicationp2q ñ p1qneeds to be established.
(a) Throughout the proof,PX andPY stand for the coordinate projections
from X`Y ontoX and Y, respectively. We have to show that any subspace
E ofX`Y contains a further subspaceG, complemented inX`Y.
Show first thatE contains a subspaceF so that eitherPX|F orPY|F is an
isomorphism. Indeed, suppose PX|F is not an isomorphism, for any such F.
Then PX|E is strictly singular, hence there exists a subspaceF E, so that
PX|F has norm less than 1{2. But PX PY IX`Y, hence, by the triangle
inequality,}PYf} ¥ }f} }PXf} ¥ }f}{2 for anyf PF. Consequently,PY|F is
an isomorphism.
Thus, by passing to a subspace, and relabeling if necessary, we can assume that E contains a subspaceF, so thatPX|F is an isomorphism. By Corollary
2.4,F contains a subspaceG, complemented inX.
SetF1PXpFq, and letV be the inverse of PX :F ÑF1. By the
subpro-jectivity ofX,F1 contains a subspaceG1, complemented inX via a projection
Q. ThenP V QPX gives a projection ontoGVpG1q F.
(b) Here, we denote by Pn the coordinate projection from X p
°
kXkqE
ontoXn. Furthermore, we set Qn
°n
k1Pk, andQKn 1Qn. We have to
show that any subspaceY X contains a subspace Y0, complemented in X. To this end, consider two cases.
(i) For somen, and some subspaceZY,Qn|Z is an isomorphism. By part
(ii) For every n, Qn|Y is not an isomorphism – that is, for every n P N,
and everyε¡0, there exists a norm oney PY so that}Qny} ε. Therefore,
for every sequence of positive numbers pεiq, we can find 0 N0 N1
N2 . . ., and a sequence of norm one vectors yi P Y, so that, for every i,
}QNiyi},}QKNi 1yi} εi. By a small perturbation principle, we can assume that
Y contains norm one vectorspyi1qso thatQNiy1iQKNi 1y
1
i 0 for everyi. Write
yi1 pzjq Ni 1
jNi 1, withzjPXj. ThenZ spanrp0, . . . ,0, zj,0, . . .q:jPNs(zj is inj-th position) is complemented inX. Indeed, ifzj0, findzjPXj so that
}zj} }zj}1, and xzj, zjy 1. If zj 0, setzj 0. For x pxjqjPNP X,
defineRx pxzj, xjyzjqjPN. It is easy to see thatRis a projection ontoZ, and
}R} does not exceed the unconditionality constant ofE.
Now note that J : Z Ñ E : pα1z1, α2z2, . . .q ÞÑ pα1}z1}, α2}z2}, . . .q is an isometry. LetY1spanryi1 :iPNs, andYE JpY1q. By the subprojectivity of
E,YE contains a subspaceW, which is complemented inE via a projectionR1.
ThenJ1R
1J Ris a projection fromX ontoY0J1pWq X.
Remark 2.7. From the last proposition it follows the (strong) p-sum of sub-projective Banach spaces is subsub-projective. On the other hand, the infinite weak sum of subprojective spaces need not be subprojective.
Recall that ifX is a Banach space, then
`weakp pXq tx pxnq8n1PXXX . . .: sup
xPX
p¸|xpxnq|pq
1 p 8u.
It is known that `weak
p pXq is isomorphic toBp`p1, Xq (1p p11 1), see [8, Theorem 2.2]. We show that, for X `r pr ¥p1q, Bp`p1, Xqcontains a copy
of `8, and therefore, is not subprojective. To this end, denote by peiq and
pfiq the canonical bases in `r and `p1 respectively. For α pαiq P`8, define
Bp`p1, Xq QU α:eiÞÑαifi. Clearly,U is an isomorphism.
Note that the situation is different for r p1. Then, by Pitt’s Theorem,
Bp`p1, `rq Kp`p1, `rq. In the next section we prove that the latter space is
subprojective.
Next we show that subprojectivity is not a 3-space property.
Proposition 2.8. For1 p 8there exists a non-subprojective Banach space
Zp, containing a subspace Xp, so thatXp andZp{Xp are isomorphic to`p.
Proof. [22, Section 6] gives us a short exact sequence 0ÝÑ`p
jp
ÝÑZp qp
ÝÑ`pÝÑ0,
where the injectionjp is strictly cosingular, and the quotient mapqp is strictly
singular. By [22, Theorem 6.2],Zpis not isomorphic to`p. By [22, Theorem 6.5],
any non-strictly singular operator onZpfixes a copy ofZp. Consequently,jpp`pq
contains no complemented subspaces (by [27, Theorem 2.a.3], any complemented subspace of`p is isomorphic to`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. IfY andZ are subspaces of a Banach spaceX, we define thegap (oropening) ΘXpY, Zq max sup yPY,}y}1 distpy, Zq, sup zPZ,}z}1 distpz, Yq(.
We refer the reader to the comprehensive survey [33] for more information. Here, we note that ΘXsatisfies a “weak triangle inequality”, hence it can be viewed as
a measure of closeness of subspaces. The following shows that subprojectivity is not stable under ΘX.
Proposition 2.9. There exists a Banach space X with a subprojective sub-spaceY so that, for every ε¡0, X contains a non-subprojective spaceZ with ΘXpY, Zq ¤ε.
Proof. OurY will be isomorphic to`p, wherepP p1,8qis fixed. By Proposition
2.8, there exists a non-subprojective Banach spaceW, containing a subspaceW0, so that both W0 and W1 W{W0 are isomorphic to `p. Denote the quotient
map W Ñ W1 by q. Consider X W `1 W1 and Y W0 `1 W1 E. Furthermore, for ε ¡ 0, define Zε tεw`1 qw : w P Wu. Clearly, Y is
isomorphic to `p``p `p, hence subprojective, while Zεis isomorphic to W,
hence not subprojective. By [33, Lemma 5.9], ΘXpY, Zεq ¤ε.
Looking at subprojectivity through the lens of Gowers dichotomy and ob-serving that a subprojective Banach space does not contain hereditarily inde-composable subspaces, we immediately obtain the following.
Proposition 2.10. Every subprojective space has a subspace with an uncondi-tional basis.
The converse to the above proposition is false.
Proposition 2.11. There exists a Banach space with an unconditional basis, without subprojective subspaces.
Proof. In [18, Section 5], T. Gowers and B. Maurey construct a Banach space
X with a 1-unconditional basis, so that any operator onX 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 P N and x P X, we define }x}pnq as the supremum of
°n
i1}xi}, where
x1, . . . , xnare successive vectors so thatx
°
ixi. By [18, Lemma 4], for every
block subspaceY inX, everyc¡1, and everynPN, there existsyPY so that 1 }y} ¤ }y}pnq c. This technical result can be used to establish a remarkable
property of X: suppose Y is a subspace of X, with a normalized block basis
pykq. Then any zero-diagonal (relative to the basispykq) operator onY is strictly
singular. Consequently, anyT PBpYqcan be written asT Λ S, where Λ is diagonal, andS is zero-diagonal, hence strictly singular. This result is proved
in [18] for Y X, but an inspection (involving Lemma 27 and Corollary 28 of the cited paper) yields the generalization described above.
Suppose, for the sake of contradiction, thatX contains a subprojective sub-spaceY. A small perturbation argument shows we can assumeY to be a block subspace. Blocking further, we can assume thatY is spanned by a block basis
pyjq, so that 1 }yj} ¤ }yj}pjq 1 2j. We achieve the desired contradiction
by showing that no subspace ofZspanry1 y2, y3 y4, . . .sis complemented
inY.
SupposeP is an infinite rank projection fromY onto a subspace ofZ. Write
P Λ S, where S is a strictly singular operator with zeroes on the main diagonal, and Λ pλjq8j1 is a diagonal operator (that is, Λyj λjyj for any
j). As supj}yj}pjq 8, by [18, Section 5] we have limjSyj 0. Note that
pΛ Sq2Λ S, hence diagpλ2
jλjq Λ2ΛSΛSSΛS2is strictly
singular, or equivalently, limjλjp1λjq 0. Therefore, there exists a 01
sequencepλ1jqso that Λ1Λ is compact (equivalenty, limjpλjλ1jq 0), where
Λ1diagpλ1jqis a diagonal projection. ThenP Λ1 S1, whereS1 S pΛΛ1q is strictly singular, and satisfies limjS1yj0. The projectionP is not strictly
singular (since it is of infinite rank), hence Λ1PS1 is not strictly singular. Consequently, the setJ tjPN:λ1j1uis infinite.
Now note that, for anyj,}P yjyj} ¥1{2. Indeed,P yj PZ, hence we can
write P yj
°
kαkpy2k1 y2kq. Let ` rj{2s. By the 1-unconditionality of
our basis,}yjP yj} ¥ }yjα`py2`1 y2`q} ¥maxt|1α`|,|α`|u ¥1{2. For
jPJ,S1yjP yjyj, hence }S1yj} ¥1{2, which contradicts limj}S1yj} 0.
Remark 2.12. The preceding statement provides an example of an atomic or-der continuous Banach lattice without subprojective subspaces. By [29, Section 2.4], any non-order continuous Banach lattice contains a subprojective subspace
c0. Also, if a Banach lattice is non-atomic order continuous with an uncondi-tional basis, then it contains a subprojective subspace`2(i.e. [23, Theorem 2.3]). Finally, one might ask whether, in the definition of subprojectivity, the pro-jections fromX onto Z can be uniformly bounded. More precisely, we call a Banach space X uniformly subprojective (with constant C) if, for every sub-spaceY X, there exists a subspaceZ Y and a projectionP :X ÑZ with
}P} ¤C. The proof of [16, Proposition 2.4] essentially shows that the following spaces are uniformly subprojective: (i)`p (1¤p 8) andc0; (ii) the Lorentz
sequence spaceslp,w; (iii) the Schreier space; (iv) the Tsirelson space; (v) the
James space. Additionally,Lpp0,1q is uniformly subprojective for 2¤p 8.
This can be proved by combining Kadets-Pelczynski dichotomy with the results of [2] about the existence of “nicely complemented” copies of`2. Moreover, any
c0-saturated separable space is uniformly subprojective, since any isomorphic copy ofc0 contains a λ-isomorphic copy of c0, for any λ ¡1 [27, Proposition 2.e.3]. By Sobczyk’s Theorem, aλ-isomorphic copy of c0 is 2λ-complemented in every separable superspace. In particular, ifK is a countable metric space, thenCpKqis uniformly subprojective [11, Theorem 12.30].
However, in general, subprojectivity need not be uniform. Indeed, suppose 2 p1 p2 . . . 8, and limnpn 8. By Proposition 2.2(b), X
p°nLpnp0,1qq2 is subprojective. The span of independent Gaussian random variables inLp (which we denote byGp) is isometric to`2. Therefore, by [17,
Corollary 5.7], any projection fromLp onto its subspaceGp has norm at least
c0?p, wherec0is a universal constant. Thus,X is not uniformly subprojective.
3. Subprojectivity of tensor products and spaces of operators Throughout the paper,XbY stands for the algebraic tensor product of X
andY. Atensor norm } }br assigns, to each pair of Banach spaces X andY, a norm onXbY, in such a way that:
1. For any Banach spaces X1 and Y1, any T PBpX, X1qand S PBpY, Y1q, and anyzPXbY, we have}pTbSqz}br ¤ }T}}S}}z}br.
2. For any z P X bY, }z}bq ¤ }z}br ¤ }z}bp (the left and right hand sides refer to the injective and projective tensor norms, respectively).
The tensor product XbrY is the completion ofX bY in the norm } }br. In particular,bq andbp stand for the injective and projective tensor products. For more information about tensor products, the reader is referred to [7, Chapter 12], or to [9].
SupposeX1, X2, . . . , Xk are Banach spaces with unconditional FDD,
im-plemented by finite rank projections pP11nq, pP21nq, . . . , pPkn1 q, respectively. That is, Pin1 Pim1 0 unless n m, limN
°N
n1Pin1 IXi point-norm, and supN,}°Nn1Pin1 } 8(this quantity is sometimes referred to as the uncon-ditional FDD constant ofXi). Let EinranpPin1 q.
We say that a sequencepwjq8j1X1bX2b. . .bXk is block-diagonal if
there exists a sequence 0N1 N2 . . . so that
wj P N¸j 1 nNj 1 E1n b N¸j 1 nNj 1 E2n b. . .b N¸j 1 nNj 1 Ekn .
Suppose E is an unconditional sequence space, and br is a tensor product of Banach spaces. The Banach space X1brX2br. . .brXk is said to satisfy the
E-estimateif there exists a constantC¥1 so that, for any block diagonal sequence
pwjqjPNin X1brX2br. . .brXk, we have
C1}p}wj}qjPN}E ¤ }
¸
j
wj} ¤C}p}wj}qjPN}E (3.1)
Theorem 3.1. SupposeX1,X2, . . . ,Xk are subprojective Banach spaces with
unconditional FDD, andbr is a tensor product. Suppose, furthermore, that for any finite increasing sequence i r1 ¤ i1 . . . . . . i` ¤ ks, there exists
an unconditional sequence spaceEi, so thatXi1brXi2br. . .brXik satisfies theEi -estimate. ThenX1brX2br. . .brXk is subprojective.
Note that a tensor product need not be associative. We presume the “nat-ural” location of brackets: X1brX2brX3 pX1brX2qrbX3, etc.. A different position of brackets is equivalent to a different ordering ofX1, . . . , Xn.
A similar result for ideals of operators holds as well. We keep the notation for projections implementing the FDD in Banach spacesX1 andX2. We say that a Banach operator idealA is suitable (for the pairpX1, X2q) if the finite rank
operators are dense inApX1, X2q(in its ideal norm). We say that a sequence pwjqjPNApX1, X2qisblock diagonal if there exists a sequence 0N1 N2 . . . so that, for any j, wj pP2,Nj P2,Nj1qwjpP1,Nj P1,Nj1q. If E is an unconditional sequence space, we say thatApX1, X2qsatisfies theE-estimateif,
for some constantC,
C1}p}wj}qj}E ¤ } ¸
j
wj}A¤C}p}wj}qj}E (3.2)
holds for any finite block-diagonal sequencepwjq.
Theorem 3.2. SupposeX1andX2are Banach spaces with unconditional FDD,
so thatX1 and X2 are subprojective. Suppose, furthermore,that the ideal A is
suitable forpX1, X2q, andApX1, X2qsatisfies theE-estimate for some
uncondi-tional sequenceE. ThenApX1, X2qis subprojective.
Before proving these theorems, we state a few consequences.
Corollary 3.3. The spaces X1bq. . .bqXn and X1bp. . .bpXn are subprojective
whereXi is isomorphic to either`pi p1¤pi 8q, or toc0, for every1¤i¤n. Forn2, this result goes back to [38] and [32] (the injective and projective cases, respectively).
Suppose a Banach spaceX has an unconditional FDD implemented by pro-jections pPn1q – that is, Pn1Pm1 0 unless n m, supN,}
°N
n1Pn1} 8,
and limN
°N
n1Pn1 IX point-norm. We say that X satisfies the lower p
-estimateif there exists a constantCso that, for any finite sequenceξjPranPj1,
}°jξj}p ¥C
°
j}ξj}p. The smallestC for which the above inequality holds is
called the lower p-estimate constant. The upper p-estimate, and the upper p -estimate constant, are defined in a similar manner. Note that, ifX is an uncon-ditional sequence space, then the above definitions coincide with the standard one (see e.g. [28, Definition 1.f.4]).
Corollary 3.4. Suppose the Banach spaces X1 and X2 have unconditional FDD, satisfy the lower and upper p-estimates respectively, and both X1 and
X2 are subprojective. Then KpX1, X2qis subprojective.
Before proceeding, we mention several instances where the above corollary is applicable. Note that, ifX has type 2 (cotype 2), thenX satisfies the upper (resp. lower) 2-estimate. Indeed, suppose X has type 2, and w1, . . . , wn are
such thatwj Pjwj for anyj. Then
}¸ j wj} ¤CAve} ¸ j wj} ¤CT2pXq ¸ j }wj}2 1{2
(T2pXqis the type 2 constant ofX). The cotype case is handled similarly. Thus, we can state:
Corollary 3.5. Suppose the Banach spaces X1 and X2 have unconditional FDD, cotype 2 and type 2 respectively, and bothX1 andX2 are subprojective. ThenKpX1, X2qis subprojective.
This happens, for instance, ifX1LppµqorCp (1 p¤2) andX2Lqpµq
orCq (2¤q 8). Indeed, the type and cotype of these spaces are well known
(see e.g. [36]). The Haar system provides an unconditional basis for Lp. The
existence of unconditional FDD ofCp spaces is given by [4].
Proof of Theorem3.1. 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 anyk1 subprojective Banach spaces that satisfy E-estimate. We will show that the statement holds for the tensor product ofk
Banach spacesX X1brX2br. . .brXk.
For notational convenience, let Pin
°n
k1Pik1 , andIiIXi. If APBpXq is a projection, we use the notation AK for IX A. Furthermore, define the
projectionsQnP1nbP2nb. . .bPknandRnP1KnbP2Kn. . .bPknK. Renorming
all Xi’s if necessary, we can assume that their unconditional FDD constants
equal 1.
First show that, for any n, ranRKn is subprojective. To this end, write
RKn °ki1Ppiq, where the projectionsPpiqare defined by
Pp1q P1nbI2b. . .bIk, Pp2q PK 1nbP2nbI3b. . .bIk, Pp3q PK 1nbP2KnbP3nbI4b. . .bIk, . . . . Ppkq P1KnbP2Knb. . .bPkK1,nbPkn
(note also thatPpiqPpjq0 unless ij). Thus, there exists i so that Ppiq is
an isomorphism on a subspaceY1 Y. Now observe that the range of Ppiq is isomorphic to a subspace of`N8pXpiqq, whereN rankPin, and
XpiqX1brX2br. . .brXi1brXi 1br. . .brXk.
By the induction hypothesis,Xpiqis subprojective. By Proposition2.2, ranPpiq
is subprojective for everyi, hence so is ranRKn.
Now suppose Y is an infinite dimensional subspace ofX. We have to show that Y contains a subspace Z, complemented in X. If there existsn P N so thatRKn|Y is not strictly singular, then, by Corollary2.4,Zcontains a subspace
complemented inX.
Now supposeRKn|Zis strictly singular for anyn. It is easy to see that, for any
sequence of positive numberspεmq, one can find 0n0 n1 n2 . . ., and
εm. By a small perturbation, we can assume thatxmRKnm1Qnmxm. That is,
xmPran
pP1,nmP1,nm1q b pP2,nmP2,nm1q b. . .b pPk,nmPk,nm1q . Let Eim ranpPi,nm Pi,nm1q, and W spanrE1mbE2mb. . .bEkm :
m P Ns X. Applying “Tong’s trick” (see e.g. [27, p. 20]), and taking the
1-unconditionality of our FDDs into account, we see that
U :XÑW :xÞѸ
m
pP1,nmP1,nm1q b. . .b pPk,nmPk,nm1q
x
defines a contractive projection ontoW. Furthermore, Z spanrxm :mPNs
is complemented in W. Indeed, the projection Pi,nm Pi,nm1 pi, m P Nq is contractive, hence we can identify E1mbr. . .brEkm with pE1mb. . .bEkmq X
X. By the by Hahn-Banach Theorem, for each m there exists a contractive projection Um on E1mbr. . .brE2m, with range spanrxms. By our assumption,
there exists an unconditional sequence spaceEso thatX1br. . .brXksatisfies the
E-estimate. Then, for any finite sequencewmPE1mbr. . .brEkm, (3.1) yields
}¸ k Ukwk} ¤C}p}Ukwk}q}E ¤C}p}wk}q}E ¤C2} ¸ k Ukwk}. Thus,Z is complemented inX.
Sketch of the proof of Theorem3.2. OnApX1, X2qwe define the projectionRn:
ApX1, X2q ÑApX1, X2q:wÞÑP2KnwP1n. Then the range ofRnKis isomorphic
toX1`. . .`X1`X2`. . .`X2. Then proceed as in the the proof of Theorem
3.1(withk2).
To prove Corollary3.3, we need two auxiliary results. Lemma 3.6. Suppose1 pi 8 p1¤i¤nqandX qb
n i1`pi. 1. If °1{pi ¡ n1, then X satisfies the `s-estimate with 1{s
°
1{pi
pn1q.
2. If °1{pi¤n1, then X satisfies thec0-estimate.
Proof. Suppose pwjq is a finite block-diagonal sequence in X. We shall show
that }°jwj} }p}wj}q}s, with s as in the statement of the lemma. To this
end, letpUijq be coordinate projections on`pi for every 1 ¤i ¤n, such that
wj U1jb. . .bUnjwj, and for each i, UikUim 0 unless k m. Letting
p1ipi{ppi1q, we see that }¸ j wj} sup ξiP`p1i,}ξi}¤1 x¸ j wj,biξiy.
Choosebiξi with}ξi} ¤1, and letξij Uijξi. Then ° j}ξij}p 1 i¤1, and x¸ j wj,biξiy ¤ ¸ j xwj,biξiy ¸ j xwj,biξijy ¤ ¸ j }wj}Πni1}ξij}.
Now let 1{r°1{p1in°1{pi. By H¨older’s Inequality,
¸ j n ¹ i1 }ξij} r 1{r ¤ n ¹ i1 ¸ j }ξij}p 1 i 1{p1i ¤1. If°1{pi ¤n1, then r¤1, hence ° jΠ n i1}ξij} ¤1. Therefore, } ° jwj} ¤
maxj}wj} p}wj}qc0. Otherwise,r¡1, and
}¸ j wj} ¤ ¸ j }wj}s 1{s ¸ j pΠni1}ξij}qr 1{r ¤ ¸ j }wj}s 1{s p}wj}qs, where 1{s11{r°1{pin 1.
In a similar fashion, we show that }°jwj} ¥ p}wj}qs. For s 8, the
inequality }°jwj} ¥maxj}wj} is trivial. If sis finite, assume
°
j}wj}s 1
(we are allowed to do so by scaling). Find norm one vectors ξij P `p1i so that
ξij Uijξi, and }wj} xwj,biξijy. Let γj }wj}s{r. Then
° jγ r j 1 ° jγj}wj}. Further, setαij γ Πlip1l{p°n m1Πlmp1lq j . An elementary calculation
shows thatγjΠni1αij, and
° jα p1i ij 1. Letξi ° jαijξij. Then}ξi}p1i 1, and therefore, }¸ j wj} ¥ x ¸ j wj,biξiy ¸ j Πni1αijxwj,biξijy ¸ j γj}wj} 1.
This establishes the desired lower estimate.
Lemma 3.7. For1¤pi¤ 8,X `p1bp`p2bp. . .bp`pn satisfies the`r-estimate, where1{r°1{pi if
°
1{pi 1, and r1 otherwise. Here, we interpret `8
asc0.
Proof. The spaces involved all have the Contractive Projection Property (the identity can be approximated by contractive finite rank projections). Thus, the duality between injective and projective tensor products of finite dimensional spaces (see e.g. [9, Section 1.2.1]) shows that, for wPX,
}w} sup |xx, wy|:xP`p11bq. . .bq`p1n,}x} ¤1
(
(here, as before, 1{p1i 1{pi1). Abusing the notation somewhat, we denote by
Pim the projection on the span of the firstmbasis vectors of both`pi and`p1i. Suppose a finite sequencepwkqNk1PXis block-diagonal, or more precisely,wk
pP1,mkP1,mk1qb. . .bpPn,mkPn,mk1q
wk for everyk. Define the operator
U onX by settingU x°Nk1 pP1,mkP1,mk1q b. . .b pPn,mkPn,mk1q
x.
We also use U0 to denote the similarly defined operator on X. By “Tong’s trick” (see e.g. [27, p. 20]), sinceX andX has an unconditional basis,U (U0) is a contractive projection onto its rangeW (W0). Then
}¸ k wk} sup |x ¸ k wk, xy|:}x}X ¤1 ( sup |xUp¸ k wkq, xy|:}x}X ¤1 ( sup |x¸ k wk, U0xy|:}x}X ¤1 ( .
WriteU0x°Nk1xk. By Lemma3.6there is ans(either 1{s
° 1{p1i pn 1q 1°1{pi ors 8)}p}xk}q}s }U0x} ¤ }x} ¤1. Moreover, x¸ k wk, U0xy x ¸ k wk, ¸ k xky ¸ k xwk, xky, and therefore, }¸ k wk} sup ¸ k |xwk, xky|:}pxk}q}s¤1 ( }p}wk}q}r.
Proof of Corollary3.3. Combine Theorem3.1with Lemma3.6and3.7. Proof of Corollary3.4. To apply Theorem3.2, we have to show thatKpX1, X2q
satisfies thec0-estimate. By renorming, we can assume that the FDD constants
of X1 and X2 equal 1. Suppose pwkqNk1 is a block-diagonal sequence, with wk pP2,nk P2,nk1qwkpP1,nk P1,nk1q. Let w
°
kwk. Then }w} ¥
}pP2,nkP2,nk1qwpP1,nkP1,nk1q} }wk}, hence}w} ¥maxk}wk}. To prove
the reverse inequality (with some constant), pick a norm one ξ PX1, and let
ξk pP1,nkP1,nk1qx. Thenηk wξk satisfiespP2,nkP2,nk1qηk ηk. Set
ηwξ°kηk. Denote byC1 (C2) lower (upper)p-estimate constants ofX1
(resp. X2). Then }wξ}p }η}p¤C1¸ k }ηk}p¤C2 ¸ k }wk}p}ξk}p ¤max k }wk}pC2 ¸ k }ξk}p¤max k }wk} pC 2C1} ¸ k ξk}pC2C1}ξ}p.
Taking the supremum over allξPUBpX1q,}w} ¤ pC1C2q1{pmax k}wk}.
We do not know whether every space with a symmetric basis must contain a subprojective subspace. However, we have the followin example:
Proposition 3.8. There exists a non-subprojective Banach space with a sym-metric basis.
Proof. Fix 1 p 2. The Haar system forms an unconditional basis inLpp0,1q,
hence, by [26, Theorem 3.b.1],Lpp0,1qembeds (complmentably) into a spaceE
with a symmetric basis. The spaceLpp0,1qis not subprojective (see Section1),
hence, by Proposition2.1, neither isE.
Furthermore, a tensor product of subprojective spaces (in fact, of Hilbert spaces) need not be subprojective.
Proposition 3.9. There exists a tensor normbr, so that`2br`2 is not
subpro-jective.
Proof. By Proposition 3.8, there exists a non-subprojective space E with 1-symmetric basis. For Banach spacesX andY, andaPX bY, we set }a}br
supt}pubvqpaq}EpH,Kqu, where the supremum is taken over all contractions
u:X ÑH and v: Y ÑK (H andK are Hilbert spaces, andEpH, Kqis the Schatten space; we identifyH with its dual). Clearlybr is a norm onX bY. It is easy to see that, for anyaPX bY, TX PBpX, X0q, and TY PBpY, Y0q,
}pTXbTYqpaq}br ¤ }TX}}TY}}a}br. Thus, to prove thatbr determines a tensor
norm, it suffices to establish that, for anyzPXbY, we have
}z}bq ¤ }z}br ¤ }z}bp. (3.3) To establish the left hand side of (3.3), let H be a 1-dimensional Hilbert space (the field of scalars). For anyf PUBpXqandgPUBpYq, consider the contractionsuf :X ÑH :xÞÑ xf, xy andvg :Y ÑH :yÞÑ xg, yy. Then, for
z°ixibyiPXbY, we have }z}br ¥ sup fPUBpXq,gPUBpYq pufbvgqpzqEpHq sup fPUBpXq,gPUBpYq ¸ i xf, xiyxg, yiyEpHq }z}bq.
To prove the right hand side of (3.3), note that, for any two contractions
u:XÑH andv:Y ÑK,
}pubvqpxbyq}E }uxbvy}E }ux}}vy} ¤ }x}}y}.
The triangle inequality establishes}z}br ¤ }z}bp for anyz.
IfXandY are Hilbert spaces, then foraPXbY we have}a}br }a}EpX,Yq.
Identifying`2 with its dual, we see that E embeds into `2br`2 as the space of diagonal operators. AsE is not subprojective, neither is`2br`2.
Here is another wide class of non-subprojective spaces.
Theorem 3.10. Let X be an infinite dimensional Banach space. ThenBpXq
Proof. Suppose, for the sake of contradiction, thatBpXqis subprojective. Fix a norm one elementx PX. ForxPXdefineTxPBpXq:yÞÑ xx, yyx. Clearly
M tTx:xPXuis a closed subspace ofBpXq, isomorphic toX. Therefore,X
is subprojective. By Proposition2.10, we can find a subspaceN M with an unconditional basis. We shall deduce thatBpXqcontains a copy of `8, which is not subprojective.
If N is not reflexive, then N contains either a copy of c0 or a copy of `1,
see [27, Proposition 1.c.13]. By [27, Proposition 2.a.2], any subspace of`p (c0)
contains a further subspace isomorphic to`p(resp. c0) and complemented in`p
(resp. c0), hence we can pass fromN to a further subspaceW, isomorphic to`1
orc0, and complemented inX by a projectionP. EmbedBpWqisomorphically intoBpXqby sendingTPBpWqtoP T P PBpXq, whereP is a projection from
X onto W. It is easy to see that BpWq contain subspaces isomorphic to `8, thus,BpXqis not subprojective.
There is only one option left: N is reflexive. Pick a subspace W N, complemented inX. It has the Bounded Approximation Property [27, Theo-rem 1.e.13]. As in the previous paragraph, BpWq embeds isomorphically into
BpXq. SinceBpWq KpWq, [12, Theorem 4(1)] shows thatBpWqcontains an isomorphic copy of`8. This rules out the subprojectivity ofBpXq.
Question 3.11. Suppose X is a subprojective Banach space. (i) Is RadpXq
subprojective? (ii) If 2¤p 8, mustLppXqbe subprojective?
Question 3.12. Is a “classical” (injective, projective, etc.) tensor product of subprojective spaces necessarily subprojective? Note that the Fremlin tensor productb|π|of Banach lattices (the ordered analogue of the projective product)
can destroy subprojectivity. Indeed, by [6], L2b|π|L2 contains a copy of L1. L2 is clearly subprojective, whileL1is not (see Section1).
4. Spaces of continuous functions
In this section we deal with CpKqspaces, mostly separable ones. It is well known that, ifKis a compact Hausdorff set, thenCpKqis separable if and only ifK is metrizable.
Recall that a topological space is scattered (or dispersed) if every compact subset has an isolated point. It is known that a compact set is scattered and metrizable if and only if it is countable (in this case, CpKq, and its dual, are separable). For more information on scattered spaces, see e.g. [11, Section 12], [26, Chaper 2], [35], or [39, Sections 8.5-6].
To examine subprojectivity, recall some relevant facts from [35]. If a compact Hausdorff space K is scattered, then CpKq is c0-saturated. If, in addiion, K
is metrizable, then CpKq is subprojective (by Sobczyk’s Theorem, the copies of c0 are complemented). If, on the other hand, a compact Hausdorff space
K is not scattered, then CpKq contains a copy ofCr0,1s, hence it cannot be complemented. Furthermore, if K is scattered, then CpKq is isometric to
`1p|K|q, while, if K is not scattered, then CpKq contains a copy of L1p0,1q. Thus,CpKq is subprojective if and only ifK is scattered.
4.1. Tensor products ofCpKq
In this subsection we study the subprojectivity of projective and injective tensor products ofCpKq. Our main result is:
Theorem 4.1. SupposeK is a compact metrizable space, and X is a Banach space. Then the following are equivalent:
1. K is scattered, andX is subprojective. 2. CpK, Xqis subprojective.
Proof. The implicationp2q ñ p1qis easy. The spaceCpK, Xqcontains copies of
CpKqand ofX, hence, by Proposition2.1, the last two spaces are subprojective. By the preceding paragraph,K must be scattered.
To provep1q ñ p2q, first fix some notation. Supposeλis a countable ordinal. We consider the interval r0, λswith the order topology – that is, the topology generated by the open intervalspα, βq, as well asr0, βqandpα, λs. Abusing the notation slightly, we writeCpλ, XqforCpr0, λs, Xq.
SupposeK is scattered. By [39, Chapter 8], K is isomorphic tor0, λs, for some countable limit ordinalλ. Fix a subprojective spaceX. We use induction onλto show that, for any countable ordinalλ,
Cpλ, Xqis subprojective. (4.1) By Proposition2.2, (4.1) holds for λ¤ω (indeed, c is isomorphic toc0, hence
cpXq cbqX is isomorphic to c0pXq c0bqX). Let F denote the set of all countable ordinals for which (4.1) fails. If F is non-empty, then it contains a minimal element, which we denote byµ. Note thatµis a limit ordinal. Indeed, otherwise it has an immediate predecessorµ1. It is easy to see that Cpµ, Xq
is isomorphic to Cpµ1, Xq `X, hence, by Proposition 2.2, Cpµ1, Xq is not subprojective. LetC0pµ, Xq tf PCpµ, Xq: limνѵfpνq 0u. Clearly
Cpµ, Xqis isomorphic toC0pµ, Xq`X, hence we obtain the desired contradiction by showing thatC0pµ, Xqis subprojective.
To do this, supposeY is a subspace ofC0pµ, Xq, so that no subspace of Y
is complemented inC0pµ, Xq. Forν µ, define the projection Pν :Cpµ, Xq Ñ
Cpν, Xq:f ÞÑf1r0,νs. If, for someν µand some subspaceZ Y,Pν|Z is an
isomorphism, thenZ contains a subspace complemented inX, by the induction hypothesis and Corollary2.4. Now suppose Pν|Y is strictly singular for any ν.
We construct a sequence of “almost disjoint” elements ofY. To do this, take an arbitraryy1from the unit sphere ofY. Pickν1 µso that}y1Pν1y1} 10
1.
Now find a norm oney2PY so that }Pν1y2} 10
2{2. Proceeding further in
the same manner, we find a sequence of ordinals 0ν0 ν1 ν2 . . ., and a sequence of norm one elementsy1, y2, . . .PY, so that}ykzk} 10k, where
zk pPνkPνk1qyk. The sequencepzkqis equivalent to thec0 basis, and the same is true for the sequencepykq.
Moreover, spanrzk : k P Ns is complemented in Cpµ, Xq. Indeed, let ν
spanryk :kPNs, contradicting our assumption. LetWk pPνkPνk1qpC0pXqq, and find a norm one linear functionalwk so thatwkpzkq }zk}. Define
Q:C0pµ, Xq ÑC0pµ, Xq:f ÞѸ
k
wk pPνkPνk1qf
zk.
Note that limk}pPνkPνk1qf} 0, hence the range ofQis precisely the span of the elements zk. By Small Perturbation Principle, Y contains a subspace
complemented inC0pµ, Xq.
The above theorem shows thatCpKqqbX is subprojective if and only if both
CpKqandX are. We do no know whether a similar result holds for other tensor products. We do, however, have:
Proposition 4.2. SupposeK is a compact metrizable space, and W is either
`p (1 ¤ p 8) or c0. Then CpKqpbW is subprojective if and only if K is
scattered.
Proof. Clearly, ifKis not scattered, thenCpKqis not subprojective. So suppose
K is scattered. We deal with the case of W `p, as the c0 case is handled
similarly. As before, we can assume thatK r0, λs, where λ is a countable ordinal. If λ is finite, then Cpλqpb`p `N8bp`p is clearly subprojective. We
handle the infinite case by transfinite induction on λ. The base is easy: if
λ ω, then Cpλq c, and Cpλq is isomorphic to c0bp`p. The latter space is
subprojective, by Corollary3.3.
Suppose, for the sake of contradiction, that λis the smallest countable or-dinal so that Cpλqpb`p is not subprojective. Reasoning as before, we conclude
that λis a limit ordinal. Furthermore, Cpλq C0pλq, henceC0pλqpb`p is not
subprojective.
Denote byQn:`pÑ`p the projection on the firstnbasis vectors in`p, and
letQKn IQn. For f PC0pλq and an ordinalν λ, define Pνf χr0,νsf,
andPνKIPν.
SupposeXis a subspace ofC0pλqpb`pwhich has no subspaces complemented
in C0pλqpb`p. By the induction hypothesis,pPνbI`pq|Y is strictly singular for any ν λ. Furthermore, pIC0pλqbQnq|Y must be strictly singular. Indeed, otherwiseY has a subspaceZso thatpIC0pλqbQnq|Z is an isomorphism, whose range is subprojective (the range of IC0pλqbQn is isomorphic to the sum of
n copies of Cpλq, hence subprojective). Therefore, for anyν λ and n PN,
pIPνKbQKnq|Y is strictly singular. Therefore we can find a normalized basis
pxiqinY, and sequences 0ν0 ν1 . . . λ, and 0n0 n1 . . ., so that
}xi pPνKi1 bQnKi1qxi} 103i{2. By passing to a further subsequence, we
can assume that}pPνibQniqxi} 10
3i{2. Thus, by the Small Perturbation
Principle, it suffices to show the following statement: If pyiq is a normalized
sequence isC0pλqpb`p, so that there exist non-negative integers 0n0 n1
n2 . . ., and ordinals 0 ν0 ν1 ν2 . . . λ, with the property that
yi pPνiPνi1q b pQniQni1q
yi for anyi, then Y spanryi :i PNs is
Denote by X the span of all x’s for which there exists an i so that x pPνi Pνi1q b pQni Qni1q
x. Then Y is contractively complemented in
CpKqpb`p. In fact, we can define a contractive projection ontoX as follows.
Suppose firstu°Nj1ajbbj, withbi’s having finite support in `p. Then set
P u°8i1 pPνiPνi1qbpQniQni1q
u. Due to our assumption on thebi’s,
there existsM so thatP u°Mi1 pPνiPνi1qbpQniQni1q
u. To show that
}P u} ¤ }u}, define, for ε pεiqMi1 P t1,1uM, the operator of multiplication
by°i1M εiχrνi1 1,νis onC0pλq. The operatorVεPBp`pqis defined similarly. BotUεandVεare contractive. Furthermore,P uAveεpUεbVεqu. Therefore,
we can use continuity to extendP to a contractive projection from C0pλqpb`p
ontoX.
To construct a contractive projection fromX ontoY, we need to show that the blocks ofX satisfy the`p-estimate. That is, ifxi pPνiPνi1q b pQni
Qni1q
xi for each i, then }
°
ixi}
p °
i}xi}
p. To this end, use duality to
identifypC0pλqpb`pq with Bp`p, `1pr0, λqq,q. P is the “block” projection onto
the space of “block diagonal” operators which map the elements of`psupported
onpni1, nisonto the vectors in`1supported onpνi1, νis. IfTi1sare the blocks
of such an operator, then}°iTi}p
1
°i}Ti}p
1
, where 1{p 1{p11. (see the proof of Corollary3.3). By duality,}°ixi}p
°
i}xi}p.
Remark 4.3. After this paper has been completed, we were informed by E. M. Galego that, in the paper [15], currently in preparation, proves that, for compact Hausdorff spacesK1 and K2, the following are equivalent: (i) K1
andK2 are scattered; (ii)CpK1qpbCpK2qis subprojective. 4.2. Operators onCpKq
Proposition 4.4. SupposeK is a scattered compact metrizable space, and1¤
p¤q 8. Then the space ΠqppCpKq, `qqis subprojective.
Recall that ΠqppX, Yqstands for the space ofpq, pq-summing operators – that
is, the operators for which there exists a constantCso that, for anyx1, . . . , xnP
X, ¸ i }T xi}q 1{q ¤C sup xPUBpXq ¸ i |xpxiq|p 1{p .
The smallest value ofC is denoted byπpqpTq.
Note that, if a compact Hausdorff space K is not scattered, then CpKq
containsL1 [35], hence ΠqppCpKq, `qqis not subprojective.
The following lemma may be interesting in its own right.
Lemma 4.5. SupposeX is a Banach space,Kis a compact metrizable scattered space, and1 ¤p¤q 8. Then, for any T PΠqppCpKq, Xq, and any ε¡0,
there exists a finite rank operatorSPΠqppCpKq, Xqwith πpqpTSq ε.
In proving Proposition 4.4 and Lemma 4.5, we consider the cases ofpq
Pietsch Factorization Theorem, T P BpCpKq, Xq is q-summing if and only if there exists a probability measure µ on K so that T factors as ˜T j, where
j : CpKq Ñ Lqpµq is the formal identity, and}T} ¤ πqpTq. Moreover, µ and
˜
T can be selected in such a way that }T˜} πqpTq. As K is scattered, there
exist distinct pointsk1, k2, . . .PK, and non-negative scalarsα1, α2, . . ., so that
°
iαi1, andµ
°
iαiδki [35].
Now supposeT PBpCpKq, XqsatisfiesπqpTq 1. Keeping the above
nota-tion, findNPNso thatp°8iN 1αiq
1
q ε. Denote byuandvthe operators of multiplication byχtk1,...,kNu and χtkN 1,kN 2,...u, respectively, acting onLqpµq. It is easy to see that ranku¤N, and}vj} ε. ThenST uj˜ works in Lemma
4.5.
If 1¤p q, then (see e.g. [8, Chapter 10] or [40, Chaper 21]), ΠqppCpKq, Xq
Πq1pCpKq, Xq, with equivalent norms. Henceforth, we setp1. We have
a probability measureµonK, and a factorizationT T j˜ , wherej :CpKq Ñ Lq1pµqis the formal identity, and ˜T :Lq1pµq ÑX satisfies}T˜} ¤cπq1pTq(cis
a constant depending onq).
In this case, the proof of Lemma 4.5proceeds as for q-summing operators, except that now, we need to selectN so thatc °8iN 1αi
1{q
ε.
Proof of Proposition4.4. Define the tensor normbr by setting, forz°ixib
yiPXbY,}z}br πqppzq, where zPBpX, Yqis defined byzf
°
ixf, xiyyi.
Recall that, for any T, πqppTq πqppTq. Furthermore, if T is weakly
compact, then TpXq Y, hence, by the injectivity of the ideal πqp,
πqppTq πqppTrasq, where Tras is the astriction of T to an operator from
X to Y. IfT PBpX, Yq has finite rank, writeT °in1xi byi (xi P X,
andyiPY), and note that, by the above,πqppTq }T}br.
By Lemma 4.5, we can identify ΠqppCpKq, Xq with CpKqbrX. By [35],
CpKq `1 (the canonical basis in `1 corresponds to the point evaluation functionals), hence we can describebrin more detail: foru°iaibxiP`1bX,
}u}br πqppuq, where u: `8 ÑX is defined by ub
°
ixb, aiyxi. Note that
κXuu˜ (κX :X ÑX is the canonical embedding), where ˜u:c0 ÑX
defined via ˜ub°ixb, aiyxi. Thus,}u}br πqppu˜q.
To finish the proof, we need to show (in light of Theorem 3.1) that `1br`q
satisfies the`q estimate. To this end, suppose we have a block-diagonal sequence
puiqni1, and show that} °
iui}qbr
°
i}ui}qbr. Abusing the notation slightly, we
identify ui with an operator from `N8 to `Nq (where N is large enough), and
identify} }br withπqppq.
First show that }°iui} q r b ¤c q° i}ui} q r
b, where c is a constant (depending
onq). We have disjoint sets pSiqni1 in t1, . . . , Nuso that uiej 0 for j RSi.
Therefore there exists a probability measureµi, supported onSi, so that
}uif}q ¤cq1πqppuiqq}f}q8p}f}pL
ppµiq for any f P `N
t1, . . . , Nu: µ ¸ i πqppuiqq 1¸ i πqppuiqqµi. Forf P`N
8, setfi f χSi. Then the vectorsuifi are disjointly supported in`q, and therefore, }p¸ i uiqf}q ¸ i }uifi}q ¤c q 1 ¸ i πqppuiqq}fi}q8p}fi} p Lppµiq ¤cq1}f}q8p¸ i πqppuiqq}fi}pL ppµiq. An easy calculation shows that
}fi}pL ppµiq ¸ i πqppuiqq 1¸ i πqppuiqq}fi}pL ppµq, hence }p¸ i uiqf}q ¤cq1 ¸ i πqppuiqq }f}q8p¸ i }fi}pLppµiq cq1 ¸ i πqppuiqq }f}q8p}f}pL ppµq. Therefore,πqpp ° iuiq ¤c ° iπqppuiqq 1{q
, for some universal constantc. Next show that}°iui}qbr ¥c1q
°
i}ui}qbr, wherec1is a constant. There exists
a probability measureµont1, . . . , Nuso that, for anyf P`N8,
}p¸ u uifq}q ¥cq2πqpp ¸ i uiqq}f}q8p}f}pL ppµq
For each i letαi }µ|Si}`N1, and µi µi{αi (ifαi 0, then clearly ui 0). Then for anyi, and anyf P`N8,
}uif}q }p ¸ i uiqpχSifq} q ¤cq 2πqpp ¸ i uiqqαi}f}8qp}f}pL ppµiq, henceπqppuiq ¤c1α 1{q i πqpp ° iuiq(c1 is a constant). As ° iαi1, we conclude that°iπqppuiqq ¤c1qπqpp ° iuiq. 4.3. Continuous fields
We refer the reader to [10, Chapter 10] for an introduction into continuous fields of Banach spaces. To set the stage, suppose K is a locally compact Hausdorff space (thebase space), andpXtqtPK is a family of Banach spaces (the
spacesXtare called (fibers). Avector field is an element of
±
tPKXt. A linear
subspace X of ±tPKXt is called acontinuous field if the following conditions
hold:
2. For anyxPX, the maptÞÑ }xptq}is continuous, and vanishes at infinity. 3. Suppose xis a vector field so that, for any ε ¡0 and any t P K, there exist an open neighborhoodU Qt andyPX for which}xpsq ypsq} ε
for anysPU. ThenxPX.
EquippingX with the norm}x} maxt}xptq}, we turn it into a Banach space.
In a fashion similar to Theorem4.1, we prove:
Proposition 4.6. SupposeK is a scattered metrizable space, X is a separable continuous vector field onK, so that, for everytPK, the fiberXtis
subprojec-tive. ThenX is subprojective.
Proof. Using one-point compactification if necessary (as in [10, 10.2.6]), we can assume that K is compact. As before, we assume that K r0, λs (λ is a countable ordinal). We denote byXp0q the set of allxPX which vanish atλ.
Ifν¤λ, we denote byXrνsthe set of allxPXλ which vanish outside ofr0, νs.
By [10, Proposition 10.1.9],xχr0,νsPX for any xPX, henceXrνs is a Banach
space. We then define the restriction operatorPν :X ÑXrνs. We denote by
Qν :X ÑXν the operator of evaluation atν.
We say that a countable ordinal λ has Property P if, whenever X is a continuous separable vector field whose fibers are subprojective, thenX is sub-projective. Using transfinite induction, we prove that any countable ordinal has this property.
The base of induction is easy to handle. Indeed, when λis finite, then X
embeds into a direct sum of (finitely many) subprojective spacesXν. Now
sup-pose, for the sake of contradiction, thatλis the smallest ideal failing Property P. Note that λis a limit ordinal. Indeed, otherwise it has an immediate pre-decessor λ, and X embeds into a direct sum of two subprojective spaces – namely,Xrλs andXλ.
SupposeY is a subspace ofX, so that no subspace ofY is complemented in
X. We shall achieve a contradiction once we show thatY contains a copy ofc0. By Proposition2.3,Qλis strictly singular onY. Passing to a smaller
subse-quence if necessary, we can assume that,Y has a basispyiqiPN, so that (i) for any
finite sequencepαiq,}
°
iαiyi} ¥maxi|αi|{2, and (ii) for anyi,}Qλyi} 104i.
Consequently, for any y P spanryj : j ¡ is, }Qλy} 104i. Indeed, we can
assume that y is a norm one vector with finite support, and write y as a fi-nite sume y °jαjyj. By the above, |αi| ¤ 2 for every i. Consequently,
}Qλy} ¤ ° j|αj|}Qλyj} ¤2 ° j¡i10 4j 104i.
Now construct a sequence ν1 ν2 . . . λ of ordinals, a sequence 1
n1 n2 . . . or positive integers, and a sequence x1, x2, . . . of norm one vectors, so that (i) xj P spanryi : nj ¤ i nj 1s, (ii) }Pνixi} 10
4i, and
(iii) }Pνi 1xi} 10
4i. To this end, recall that, by Proposition 2.3 again,
Pν|Y is strictly singular for any ν λ. Pick an arbitrary ν1 λ, and find
a norm 1 vector x1 P spanry1, . . . , yn21s so that }Pν1x1} 10
4. We have }Qλx1} 104. By continuity, we can find ν2 ¡ν1 so that }Pν2x1} 10
Next find a norm onex2Pspanryn2, . . . , yn31sso that}Pν2x1} 10
8. Proceed
further in the same manner.
We claim that the sequence pxiq is equivalent to the canonical basis inc0.
Indeed, for eachiletx2i Pνixi Pνi 1xi, andx1ixix2i. Since we are working
with the sup norm, }x1i} }xi} 1 for any i. Furthermore, the elements x1i
are disjointly supported, hence, for any pαiq finite sequence of scalars pαiq,
}°iαix1i} maxi|αi|. By the triangle inequality,
}¸ i αixi} } ¸ i αix1i} ¤ ¸ i |αi|}x2i} max i |αi| 8 ¸ i1 2204i 103max i |αi|,
which yields the desired result.
To state a corollary of Proposition4.6, recall that a C-algebra Ais CCR (orliminal) if, for any irreducible representationπ ofAon a Hilbert spaceH,
πpAq KpHq. AC-algebraAisscatteredif every positive linear functional on Ais a sum of pure linear functionals (f PA is calledpure if it belongs to an extreme ray of the positive cone ofA). For equivalent descriptions of scattered
C-algebras, see e.g. [19,20,25].
Corollary 4.7. Any separable scattered CCRC-algebra is subprojective. Proof. Suppose Ais a separable scattered CCR C-algebra. As shown in [34, Sections 6.1-3], the spectrum of a separable CCR algebra is a locally compact Hausdorff space. If, in addition, the algebra is scattered, then its spectrum ˆA is scattered as well [19, 20]. In fact, by the proof of [19, Theorem 3.1], ˆA is separable. It is easy to see that any separable locally compact Hausdorff space is metrizable.
By [10, Section 10.5],A can be represented as a vector field over ˆA, with fibers of the form πpAq, for irreducible representationsπ. As A is CCR, the spacesπpAq KpHπq(Hπ being a separable Hilbert space) are subprojective.
To finish the proof, apply Proposition4.6. The last corollary leads us to
Conjecture 4.8. A separableC-algebra is scattered if and only if it is sub-projective.
It is known ([20], see also [25]) that a scatteredC-algebra is GCR. However, it need not be CCR (consider the unitization ofKp`2q).
5. Subprojectivity of Schatten spaces In this section, we establish:
Proposition 5.1. SupposeE is a symmetric sequence space, not containingc0. ThenCE is subprojective if and only if E is subprojective.
The assumptions of this proposition are satisfied, for instance, ifE`p(1¤
p 8), or ifEis the Lorentz spacelpw, pq(see [27, Proposition 4.e.3]. However, by Proposition3.8, not every symmetric sequence space is subprojective.
For the proof, we need a technical result.
Proposition 5.2. Suppose CE is a symmetric sequence space, not containing c0. Suppose, furthermore, thatpznq CE is a normalized sequence, so that, for
everyk,limn}Qkzn} 0. Then, for anyε¡0,CE contains sequencespz˜nqand
pzn1q, so that:
1. pz˜nqis a subsequence ofpznq.
2. °n}z˜nzn1} ε.
3. pz1nq lies in the subspace Z of CE, with the property that (i) Z is
3-isomorphic to either`2,E, or`2`E, and(ii)Z is the range of a projection of norm not exceeding 3.
Proof. [3, Corollary 2.8] implies the existence ofpz˜nqandpzn1q, so that (1) and
(2) are satisfied, and z1k abE1k b bEk1 ck bEkk (k ¥ 2). Thus,
z1nZZr Zc Zd, whereZrspanrabE1k :k¥2s(the row component),
Zc spanrbbEk1 : k ¥ 2s (the column component), and Zd (the diagonal
component) contains ck bEkk, for any k. More precisely, we can write ck
ukdkvk, where uk and vk are unitaries, and dk is diagonal. Then we set Zd
spanrukEiivkbEkk:iPN, k¥2s.
It remains to build contractive projectionsPr,Pc, andPd ontoZr, Zc, and
Zd, respectively, so thatZcYZdkerPr,ZrYZdkerPc, andZrYZckerPd.
Indeed, thenPPr Pc Pdis a projection ontoZr Zc Zd, and the latter
space is completely isomorphic toZ0Zr`Zc`Zd. The spacesZr,Zc, andZd
are either trivial (zero-dimensional), or isomorphic to`2,`2, andE, respectively.
Pd is nothing but a coordinate projection, in the appropriate basis:
Pd ukEijv`bEk` " ukEiivkbEkk k`¥2, ij 0 otherwise
(for the sake of convenience, we set u1 v1 I`2). Next construct Pr (Pc is dealt with similarly). Ifa0, just take Pr0. Otherwise, leta1 a{}a}, and
findf PCE so that}f} 1 xf, a1y. Forx°k,`xk`bEk`, define
Prxa1b ¸ `¥2 xf, x1`yE1`, hence}Prx}2E ° `¥2|xf, x1`y| 2. It remains to show}P rx} ¤ }x}. This
inequal-ity is obvious whenPrx0. Otherwise, set, for`¥2,
α` x
f, x1`y
p°`¥2|xf, x1`y|2q1{2
yI`2b
°
`¥2α`E`1, andz I`2bE11. Then}y}8
°
`¥2|α`|2
1{2
1
}z}8, andzxy°`¥2α`x1`bE11. Therefore,
}Prx}E @ f,¸ `¥2 α`x1` D ¤ ¸ `¥2 α`x1`E ¸ `¥2 α`x1`bE11E }zxy}E ¤ }z}8}x}E}y}8 }x}E,
which is what we need.
Proof of Proposition5.1. The spaceCE contains an isometric copy ofE, hence
the subprojectivity ofCE implies that ofE. To prove the converse, supposeE is
subprojective, andZ0 is a subspace ofCE, and show that it contains a further
subspace Z, complemented in CE. To this end, find a normalized sequence pznq Z0, so that limn}Qkzn} 0 for every k. By Proposition5.2, pznq has
a subsequencepz1nq, contained in a subspaceZ1, which is complemented in CE,
and isomorphic either toE,`2, orE``2. By Proposition2.2,Z1is subprojective, hence spanrz1n:nPNscontains a subspace complemented inZ1, hence also in
CE.
As a consequence we obtain:
Proposition 5.3. The predual of a von Neumann algebra Ais subprojective if and only ifAis atomic.
We say thatAisatomic if any projection in it has an atomic subprojection. By [31, Section 1], this happens if and only ifA p°iBpHiqq8.
Proof. If a von Neumann algebra Ais not purely atomic, then, as explained in [31, Section 1],Acontains a (complemented) copy ofL1p0,1q. This establishes the “only if” implication of Proposition5.3. Conversely, ifAis purely atomic, thenA is isometric to a (contractively complemented) subspace ofC1pHq, and the latter is subprojective.
6. p-convex andp-disjointly homogeneous Banach lattices
We say thatXisp-disjointly homogeneous(p-DH for short) if every disjoint normalized sequence contains a subsequence equivalent to the standard basis of
`p.
For the sake of completeness we present a proof of the following statement (see [14, 4.11, 4.12]).
Proposition 6.1. Let X be ap-convex p-DH Bnaach lattice. Then every sub-space, spanned by a disjoint sequence equivalent to the canonical basis of`p, is
Proof. Let pxkq X be a disjoint normalized sequence. SinceX isp-DH, by
passing to a subsequence, we can assume that (xkqis an`pbasic sequence. Then,
in thep-concavificationXppqthe disjoint sequencepxkpqis an`1basic sequence.
Therefore, there exists a functionalx P rpxkpqssuch that xpxkpq 1 for all
k. By the Hahn-Banach Theoremx can be extended to a positive functional in Xppq. Define a seminorm }x}p pxp|xp|qq
1
p on X. Denote by N the subset of X on which this seminorm is equal to zero. Clearly, N is an ideal, therefore, the quotient space ˜X X{N is a Banach lattice, and the quotient map Q : X Ñ X˜ is an orthomorphism. With the defined seminorm ˜X is an abstractLp-space, and the disjoint sequenceQpxkqis normalized. Therefore it
is an `p basic sequence that spans a complemented subspace (in particular, Q
is an isomorphism when restricted torxks). Let ˜P be a projection from ˜X onto
rQpxkqs. ThenP Q1P Q˜ is a projection fromX ontorxks.
Proposition 6.2. LetX be ap-convex,p-disjointly homogeneous Banach lattice
pp¥2q. Then any subspace ofX contains a complemented copy of either `p or
`2. Consequently,X is subprojective.
Proof. First, note that X is order continuous. Let M X be an infinite dimensional separable subspace. Then there exists a complemented order ideal in X with a weak unit that containsM. Therefore, without loss of generality, we may assume thatX has a weak unit. Then there exists a probability measure
µ[21, p. 14] such that we have continuous embeddings
L8pµq X Lppµq L2pµq L1pµq.
Consequently, there exists a constantc1¡0 so thatc1}x}p¤ }x}for anyxPX.
By the proof of [28, Proposition 1.c.8], one of the following holds:
Case 1. M contains an almost disjoint bounded sequence. By Proposition6.1 M contains a copy of`p complemented inX.
Case 2. The norms}}and}}1are equivalent onM. Thus, there existsc2¡0
so that, for anyyPM,
c2}y}2¥c2}y}1¥ }y} ¥c1}y}p¥c1}y}2.
In particular,M is embedded intoL2pµqas a closed subspace. The orthogonal projection fromL2pµqontoM then defines a bounded projection fromX onto
M.
The preceding result implies that Lorentz space Λp,Wp0,1qis subprojective
forp¥2. Indeed, Λp,Wp0,1q isp-DH andp-convexpp¥1q, see [13, Theorem
3] and [24]. Note that, originally, the subprojectivity of Λp,Wp0,1q(forp¥2)
7. Lattice-valued `p spaces
IfX is a Banach lattice, and 1¤p 8, denote byXp`pqthe completion of
the space of all finite sequencespx1, . . . , xnq(with xi PX), equipped with the
norm}px1, . . . , xnq} }p ° i|xi|pq1{p}, where p¸ i |xi|pq1{p sup ! |¸ i αixi|: ¸ i |αi|p 1 ¤1 ) , with 1 p 1 p1 1.
See [28, pp. 46-48] for more information. We have:
Proposition 7.1. SupposeX is a subprojective separable space, with the lattice structure given by an unconditional basis, and 1 ¤ p 8. Then Xp`pq is
subprojective.
Proof. To show that any subspace Y Xp`pq has a further subspace Z,
com-plemented inXp`pq, letx1, x2, . . .ande1, e2, . . .be the canonical bases inXand
`p, respectively. Then the elements uij xibej form an unconditional basis
inXp`pq, with }¸aijuij} } ¸ i p¸ j |aij|pq1{pxi}X } ¸ i sup ° j|αj|p1¤1 |¸ j αijaij| xi}X. (7.1)
LetPn be the canonical projection onto spanruij : 0 ¤i ¤n, j P Ns, and set
PnKIPn. The range ofPn is isomorphic to`p, hence, ifPn|Y is not strictly
singular for somen, we are done, by Corollary2.4. If Pn|Y is strictly singular
for every n, find a normalized sequence pyiq in Y, and 1n1 n2 . . ., so
that }Pniyi},}PnKi 1yi} 100
i{2. By small perturbation, it remains to prove
the following: if yi PnKiPni 1yi, then spanryi : i P Ns contains a subspace, complemented inXp`pq. Further, we may assume that for each i there exists
Mi so that we can write
yi
¸
ni k¤ni 1,1¤j¤Mi
akjukj.
For eachkP rni 1, ni 1s(and arbitraryiPN) find a finite sequencepαkjqMji1so
that°j|αkj|p
1
1, and|°jαkjakj| p
°
j|akj|pq1{p. DefineU :Xp`pq ÑX :
ukjÞÑαkjakjxk. By (7.1),U is a contraction, andU|spanryi:iPNs is an isometry.
To finish the proof, recall thatX is subprojective, and apply Corollary 2.4. Remark 7.2. Using similar methods, one can prove: ifKis a compact metriz-able scattered space, and 1¤p 8, then CpKqp`pqis subprojective.
Recall that, for a Banach spaceX, we denote byRadpXqthe completion of the finite sums°nrnxn (r1, r2, . . . are Rademacher functions, andx1, x2, . . .P
X) in the norm of L1pXq (equivalently, by Khintchine-Kahane Inequality, in the norm ofLppXq). IfX has a unconditional basispxiqand finite cotype, then
RadpXqis isomorphic to Xp`2q(here we can viewX as a Banach lattice, with the order induced by the basispxiq). Indeed, by [28, Section 1.f],X isq-concave,
for someq. An arraypamnqcan be identified both with an element ofRadpXq
(with the norm³10}°m°namnrnxm}), and with an element ofXp`2q(with the
norm}°mp°n|amn|2q1{2xm}). Then D}¸ m p¸ n |amn|2q1{2xm} ¤ } ¸ m »1 0 |¸ n amnrn|xm} } »1 0 |¸ m ¸ n amnrnxm|} ¤ »1 0 }¸ m ¸ n amnrnxm} ¤ p »1 0 }¸ m ¸ n amnrnxm}qq1{q ¤Mq}p »1 0 |¸ m ¸ n amnrnxm|qq1{q} ¤Mq} ¸ m p »1 0 |¸ n amnrn|qq1{qxm} ¤CMq} ¸ m p¸ n |amn|2q1{2xm},
where Mq is a q-concavity constant, while D and C come from Khintchine’s
inequality. Thus, we have proved:
Proposition 7.3. IfX is a subprojective space with an unconditional basis and non-trivial cotype, thenRadpXqis subprojective.
Remark 7.4. By [23, Theorem 2.3], if X is a non-atomic order continuous Banach lattice with an unconditional basis, then Xp`2q is isomorphic to X. Furthermore, if X is a non-atomic Banach lattice with an unconditional basis and non-trivial cotype, then RadpXq is isomorphic to X. Indeed, non-trivial cotype implies non-trivial lower estimate [28, p. 100], which, by [29, Theorem 2.4.2], implies order continuity. Therefore,X is isomorphic to Xp`2q, which, in turn, is isomorphic toRadpXq.
Acknowledgements
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 helpful suggestions.
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