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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, AlbertaT6G 2G1, CANADA
Abstract
A Banach space X 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 thestability of subprojectivity of Banach spaces under various operations, such usdirect or twisted sums, tensor products, and forming spaces of operators. Alongthe way, we obtain new classes of subprojective spaces.
Keywords: Banach space, complemented subspace, tensor product, space ofoperators.
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 space X is called subprojective if every subspace Y X containsa further subspace Z Y , complemented in X. This notion was introduced in[41], in order to study the (pre)adjoints of strictly singular operators. Recallthat an operator T P BpX,Y q is strictly singular (T P SSpX,Y q) if T is not anisomorphism on any subspace of X. In particular, it was shown that, if Y issubprojective, and, for T P BpX,Y q, T P SSpY , Xq, then T P SSpX,Y q.
Later, connections between subprojectivity and perturbation classes werediscovered. More specifically, denote by ΦpX,Y q the set of upper semi-Fredholmoperators – that is, operators with closed range, and finite dimensional kernel.If ΦpX,Y q H, we define the perturbation class
PΦpX,Y q tS P BpX,Y q : T S P ΦpX,Y q whenever T P ΦpX,Y qu.
Email addresses: [email protected] (T. Oikhberg), [email protected] (E. Spinu)
Preprint submitted to Journal of Mathematical Analysis and Applications July 24, 2014
It is known that SSpX,Y q PΦpX,Y q. In general, this inclusion is proper.However, we get SSpX,Y q PΦpX,Y q if Y is subprojective (see [1, Theorem7.51] 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) and c0 are subprojective. Common examples of non-subprojective space are L1p0, 1q (since all Hilbertian subspaces of L1 are notcomplemented), 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 ofCp∆q. By [41], Lpp0, 1q is subprojective if and only if 2 ¤ p 8. Consequently,the Hardy space Hp on the disc is subprojective for exactly the same values of p.Indeed, H8 contains the disc algebra. For 1 p 8, Hp is isomorphic to Lp.The space H1 contains isomorphic copies of Lp for 1 p ¤ 2 [42, Section 3]. Onthe other hand, VMO is subprojective ([30], see also [37] for non-commutativegeneralizations).
In this paper we examine several aspects of subprojectivity. We start bycollecting various facts needed to study subprojectivity (Section 2). Along theway, we prove that subprojectivity is stable under suitable direct sums (Proposi-tion 2.2). However, subprojectivity is not a 3-space property (Proposition 2.8).Consequently, subprojectivity is not stable under the gap metric (Proposition2.9). Considering the place of subprojective spaces in Gowers dichotomy, weobserve that each subprojective space has a subspace with an unconditionalbasis. However, we exhibit a space with an unconditional basis, but with nosubprojective subspaces (Proposition 2.11).
In Section 3, we investigate the subprojectivity of tensor products, and ofspaces of operators. A general result on tensor products (Theorem 3.1) yieldsthe subprojectivity on `pqb`q and `ppb`q for 1 ¤ p, q 8 (Corollary 3.3), as wellas of KpLp, Lqq for 1 p ¤ 2 ¤ q 8 (Corollary 3.4). We also prove thatthe space BpXq is never subprojective (Theorem 3.10), and give an example ofnon-subprojective tensor product `2rb`2 (Proposition 3.9).
Throughout Section 4, we work with CpKq spaces, with K compact metriz-able. We begin by observing that CpKq is subprojective if and only if K isscattered. Then we prove that CpK,Xq is subprojective if and only if bothCpKq and X are (Theorem 4.1). Turning to spaces of operators, we show that,for K scattered, ΠqppCpKq, `qq is subprojective (Proposition 4.4). We also studycontinuous fields on a scattered base space, proving that any scattered separableCCR C-algebra is subprojective (Corollary 4.7).
Section 5 shows that, in many cases, subprojectivity passes from a sequencespace to the associated Schatten space (Proposition 5.1).
Proceeding to Banach lattices, in Section 6 we prove that p-disjointly ho-mogeneous p-convex lattices (2 ¤ p 8) are subprojective (Proposition 6.2).
In Section 7 (Proposition 7.1), we show that the lattice Xp`pq is subprojectivewhenever X is. Consequently (Proposition 7.3), if X is a subprojective spacewith an unconditional basis and non-trivial cotype, then RadpXq is subprojec-tive.
Throughout the paper, we use the standard Banach space results and nota-
2
tion. By BpX,Y q and KpX,Y q we denote the sets of linear bounded and com-pact operators, respectively, acting between Banach spaces X and Y . UBpXqrefers to the closed unit ball of X. For p P r1,8s, we denote by p1 the “adjoint”of p (that is, 1p 1p1 1).
2. General facts about subprojectivity
We start collecting general facts about subprojectivity by observing thatsubprojectivity 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, anyseparable Banach space is a quotient of `1, which is subprojective. As noted inSection 1, there exist separable non-subprojective spaces.
However, subprojectivity does pass to direct sums.
Proposition 2.2. (a) Suppose X and Y are Banach spaces. Then the followingare equivalent:
1. Both X and Y are subprojective.
2. X ` Y is subprojective.
(b) Suppose X1, X2, . . . are Banach spaces, and E 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 view E as a space of all sequences, for which the norm E is finite.p°nXnqE refers to the space of all sequences pxnqnPN P
±nPNXn, endowed with
the norm pxnqnPN pxnXnqE . Due to the 1-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 tworesults stated below.
Proposition 2.3. Consider Banach spaces X and X 1, and T P BpX,X 1q. Sup-pose Y is a subspace of X, T |Y is an isomorphism, and T pY q is complementedin X 1. Then Y is complemented in X.
Proof. If Q is a projection from X 1 to T pY q, then T1QT is a projection fromX onto Y .
This immediately yields:
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Corollary 2.4. Suppose X and X 1 are Banach spaces, and X 1 is subprojective.Suppose, furthermore, that Y is a subspace of X, and there exists T P BpX,X 1qso that T |Y is an isomorphism. Then Y contains a subspace complemented inX.
The following version of “Principle of Small Perturbations” is folklore, andessentially contained in [5]. We include the proof for the sake of completeness.
Proposition 2.5. Suppose pxkq is a seminormalized basic sequence in a Banachspace X, and pykq is a sequence so that limk xkyk 0. Suppose, furthermore,that every subspace of spanryk : k P Ns contains a subspace complemented in X.Then spanrxk : k P Ns contains a subspace complemented in X.
Proof. Replacing xk by xkxk, we can assume that pxkq is normalized. De-note the biorthogonal functionals by xk , and set K supk xk. Passing to asubsequence, we can assume that
°k xk yk 1p2Kq. Define the operator
U P BpXq by setting Ux °k x
kpxqpykxkq. Clearly U 12, and therefore,
V IX U is invertible. Furthermore, V xk yk. If Q is a projection from Xonto a subspace W spanryk : k P Ns, then P V 1QV is a projection fromX onto a subspace Z spanrxk : k P Ns.Remark 2.6. Note that, in the proof above, the kernels and the ranges of theprojections Q and P are isomorphic, via the action of V .
Proof of Proposition 2.2. Due to Proposition 2.1, in both (a) and (b), only theimplication p2q ñ p1q needs to be established.
(a) Throughout the proof, PX and PY stand for the coordinate projectionsfrom X ` Y onto X and Y , respectively. We have to show that any subspaceE of X ` Y contains a further subspace G, complemented in X ` Y .
Show first that E contains a subspace F so that either PX |F or PY |F is anisomorphism. Indeed, suppose PX |F is not an isomorphism, for any such F .Then PX |E is strictly singular, hence there exists a subspace F E, so thatPX |F has norm less than 12. But PX PY IX`Y , hence, by the triangleinequality, PY f ¥ f PXf ¥ f2 for any f P F . Consequently, PY |F isan isomorphism.
Thus, by passing to a subspace, and relabeling if necessary, we can assumethat E contains a subspace F , so that PX |F is an isomorphism. By Corollary2.4, F contains a subspace G, complemented in X.
Set F 1 PXpF q, and let V be the inverse of PX : F Ñ F 1. By the subpro-jectivity of X, F 1 contains a subspace G1, complemented in X via a projectionQ. Then P V QPX gives a projection onto G V pG1q F .
(b) Here, we denote by Pn the coordinate projection from X p°kXkqEonto Xn. Furthermore, we set Qn
°nk1 Pk, and QK
n 1 Qn. We have toshow that any subspace Y X contains a subspace Y0, complemented in X.To this end, consider two cases.
(i) For some n, and some subspace Z Y , Qn|Z is an isomorphism. By part(a), X1` . . .`Xn QnpXq is subprojective. Apply Corollary 2.4 to obtain Y0.
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(ii) For every n, Qn|Y is not an isomorphism – that is, for every n P N,and every ε ¡ 0, there exists a norm one y P Y 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,QNi
yi, QKNi1
yi εi. By a small perturbation principle, we can assume that
Y contains norm one vectors py1iq so that QNiy1i QK
Ni1y1i 0 for every i. Write
y1i pzjqNi1
jNi1, with zj P Xj . Then Z spanrp0, . . . , 0, zj , 0, . . .q : j P Ns (zj isin j-th position) is complemented in X. Indeed, if zj 0, find zj P X
j so that
zj zj1, and xzj , zjy 1. If zj 0, set zj 0. For x pxjqjPN P X,define Rx pxzj , xjyzjqjPN. It is easy to see that R is a projection onto Z, andR does not exceed the unconditionality constant of E .
Now note that J : Z Ñ E : pα1z1, α2z2, . . .q ÞÑ pα1z1, α2z2, . . .q is anisometry. Let Y 1 spanry1i : i P Ns, and YE JpY 1q. By the subprojectivity ofE , YE contains a subspace W , which is complemented in E via a projection R1.Then J1R1JR is a projection from X onto Y0 J1pW q X.
Remark 2.7. From the last proposition it follows the (strong) p-sum of sub-projective Banach spaces is subprojective. On the other hand, the infinite weaksum of subprojective spaces need not be subprojective.
Recall that if X is a Banach space, then
`weakp pXq tx pxnq8n1 P X X X . . . : supxPX
p¸|xpxnq|pq
1p 8u.
It is known that `weakp pXq is isomorphic to Bp`p1 , Xq ( 1p 1
p1 1), see [8,
Theorem 2.2]. We show that, for X `r pr ¥ p1q, Bp`p1 , Xq contains a copyof `8, and therefore, is not subprojective. To this end, denote by peiq andpfiq the canonical bases in `r and `p1 respectively. For α pαiq P `8, defineBp`p1 , Xq Q Uα : ei ÞÑ αifi. 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 issubprojective.
Next we show that subprojectivity is not a 3-space property.
Proposition 2.8. For 1 p 8 there exists a non-subprojective Banach spaceZp, containing a subspace Xp, so that Xp and ZpXp are isomorphic to `p.
Proof. [22, Section 6] gives us a short exact sequence
0 ÝÑ `pjpÝÑ Zp
qpÝÑ `p ÝÑ 0,
where the injection jp is strictly cosingular, and the quotient map qp is strictlysingular. By [22, Theorem 6.2], Zp is not isomorphic to `p. By [22, Theorem 6.5],any non-strictly singular operator on Zp fixes a copy of Zp. Consequently, jpp`pqcontains no complemented subspaces (by [27, Theorem 2.a.3], any complementedsubspace of `p is isomorphic to `p).
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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 – thegap measure. If Y and Z are subspaces of a Banach space X, we define the gap(or opening)
ΘXpY,Zq max
supyPY,y1
distpy, Zq, supzPZ,z1
distpz, Y q(.We refer the reader to the comprehensive survey [33] for more information. Here,we note that ΘX satisfies a “weak triangle inequality”, hence it can be viewed asa measure of closeness of subspaces. The following shows that subprojectivityis not stable under ΘX .
Proposition 2.9. There exists a Banach space X with a subprojective sub-space Y so that, for every ε ¡ 0, X contains a non-subprojective space Z withΘXpY,Zq ¤ ε.
Proof. Our Y will be isomorphic to `p, where p P p1,8q is fixed. By Proposition2.8, there exists a non-subprojective Banach spaceW , containing a subspaceW0,so that both W0 and W 1 W W0 are isomorphic to `p. Denote the quotientmap W Ñ W 1 by q. Consider X W `1 W
1 and Y W0 `1 W1 E.
Furthermore, for ε ¡ 0, define Zε tεw `1 qw : w P W u. Clearly, Y isisomorphic to `p ` `p `p, hence subprojective, while Zε is 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 spaceX with a 1-unconditional basis, so that any operator on X is a strictly singularperturbation of a diagonal operator. We prove that X has no subprojectivesubspaces. In doing so, we are re-using the notation of that paper. In particular,for n P N and x P X, we define xpnq as the supremum of
°ni1 xi, where
x1, . . . , xn are successive vectors so that x °i xi. By [18, Lemma 4], for every
block subspace Y in X, every c ¡ 1, and every n P N, there exists y P Y so that1 y ¤ ypnq c. This technical result can be used to establish a remarkableproperty of X: suppose Y is a subspace of X, with a normalized block basispykq. Then any zero-diagonal (relative to the basis pykq) operator on Y is strictlysingular. Consequently, any T P BpY q can be written as T Λ S, where Λ isdiagonal, and S is zero-diagonal, hence strictly singular. This result is proved
6
in [18] for Y X, but an inspection (involving Lemma 27 and Corollary 28 ofthe cited paper) yields the generalization described above.
Suppose, for the sake of contradiction, that X contains a subprojective sub-space Y . A small perturbation argument shows we can assume Y to be a blocksubspace. Blocking further, we can assume that Y is spanned by a block basispyjq, so that 1 yj ¤ yjpjq 1 2j . We achieve the desired contradictionby showing that no subspace of Z spanry1 y2, y3 y4, . . .s is complementedin Y .
Suppose P is an infinite rank projection from Y onto a subspace of Z. WriteP Λ S, where S is a strictly singular operator with zeroes on the maindiagonal, and Λ pλjq8j1 is a diagonal operator (that is, Λyj λjyj for anyj). As supj yjpjq 8, by [18, Section 5] we have limj Syj 0. Note thatpΛSq2 ΛS, hence diag pλ2
j λjq Λ2Λ SΛSSΛS2 is strictlysingular, or equivalently, limj λjp1 λjq 0. Therefore, there exists a 0 1sequence pλ1jq so that Λ1Λ is compact (equivalenty, limjpλj λ1jq 0), whereΛ1 diag pλ1jq is a diagonal projection. Then P Λ1S1, where S1 SpΛΛ1qis strictly singular, and satisfies limj S
1yj 0. The projection P is not strictlysingular (since it is of infinite rank), hence Λ1 P S1 is not strictly singular.Consequently, the set J tj P N : λ1j 1u is infinite.
Now note that, for any j, Pyj yj ¥ 12. Indeed, Pyj P Z, hence we canwrite Pyj
°k αkpy2k1 y2kq. Let ` rj2s. By the 1-unconditionality of
our basis, yj Pyj ¥ yj α`py2`1 y2`q ¥ maxt|1 α`|, |α`|u ¥ 12. Forj P J , S1yj Pyj yj , hence S1yj ¥ 12, 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, Section2.4], any non-order continuous Banach lattice contains a subprojective subspacec0. 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 from X onto Z can be uniformly bounded. More precisely, we call aBanach space X uniformly subprojective (with constant C) if, for every sub-space Y X, there exists a subspace Z Y and a projection P : X Ñ Z withP ¤ C. The proof of [16, Proposition 2.4] essentially shows that the followingspaces are uniformly subprojective: (i) `p (1 ¤ p 8) and c0; (ii) the Lorentzsequence spaces lp,w; (iii) the Schreier space; (iv) the Tsirelson space; (v) theJames space. Additionally, Lpp0, 1q is uniformly subprojective for 2 ¤ p 8.This can be proved by combining Kadets-Pelczynski dichotomy with the resultsof [2] about the existence of “nicely complemented” copies of `2. Moreover, anyc0-saturated separable space is uniformly subprojective, since any isomorphiccopy of c0 contains a λ-isomorphic copy of c0, for any λ ¡ 1 [27, Proposition2.e.3]. By Sobczyk’s Theorem, a λ-isomorphic copy of c0 is 2λ-complementedin every separable superspace. In particular, if K is a countable metric space,then CpKq is uniformly subprojective [11, Theorem 12.30].
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However, in general, subprojectivity need not be uniform. Indeed, suppose2 p1 p2 . . . 8, and limn pn 8. By Proposition 2.2(b), X p°n Lpnp0, 1qq2 is subprojective. The span of independent Gaussian randomvariables in Lp (which we denote by Gp) is isometric to `2. Therefore, by [17,Corollary 5.7], any projection from Lp onto its subspace Gp has norm at leastc0?p, where c0 is a universal constant. Thus, X is not uniformly subprojective.
3. Subprojectivity of tensor products and spaces of operators
Throughout the paper, X b Y stands for the algebraic tensor product of Xand Y . A tensor norm rb assigns, to each pair of Banach spaces X and Y ,a norm on X b Y , in such a way that:
1. For any Banach spaces X 1 and Y 1, any T P BpX,X 1q and S P BpY, Y 1q,and any z P X b Y , we have pT b Sqzrb ¤ T Szrb.
2. For any z P X b Y , zqb ¤ zrb ¤ zpb (the left and right hand sidesrefer to the injective and projective tensor norms, respectively).
The tensor product X rbY is the completion of X b Y in the norm rb. In
particular, qb and pb stand for the injective and projective tensor products. Formore information about tensor products, the reader is referred to [7, Chapter12], or to [9].
Suppose X1, X2, . . . , Xk are Banach spaces with unconditional FDD, im-plemented by finite rank projections pP 1
1nq, pP 12nq, . . . , pP 1
knq, respectively.
That is, P 1inP
1im 0 unless n m, limN
°Nn1 P
1in IXi
point-norm, and
supN, °Nn1P 1
in 8 (this quantity is sometimes referred to as the uncon-ditional FDD constant of Xi). Let Ein ran pP 1
inq.We say that a sequence pwjq8j1 X1 bX2 b . . . bXk is block-diagonal if
there exists a sequence 0 N1 N2 . . . so that
wj P Nj1¸nNj1
E1n
b Nj1¸nNj1
E2n
b . . .b Nj1¸nNj1
Ekn.
Suppose E is an unconditional sequence space, and rb is a tensor product ofBanach spaces. The Banach space X1rbX2rb . . . rbXk is said to satisfy the E-estimate if there exists a constant C ¥ 1 so that, for any block diagonal sequencepwjqjPN in X1rbX2rb . . . rbXk, we have
C1pwjqjPNE ¤ ¸j
wj ¤ CpwjqjPNE (3.1)
Theorem 3.1. Suppose X1, X2, . . . , Xk are subprojective Banach spaces withunconditional FDD, and rb is a tensor product. Suppose, furthermore, that forany finite increasing sequence i r1 ¤ i1 . . . . . . i` ¤ ks, there existsan unconditional sequence space Ei, so that Xi1
rbXi2rb . . . rbXik satisfies the Ei-
estimate. Then X1rbX2rb . . . rbXk is subprojective.
8
Note that a tensor product need not be associative. We presume the “nat-ural” location of brackets: X1rbX2rbX3 pX1rbX2qrbX3, etc.. A differentposition of brackets is equivalent to a different ordering of X1, . . . , Xn.
A similar result for ideals of operators holds as well. We keep the notation forprojections implementing the FDD in Banach spaces X1 and X2. We say thata Banach operator ideal A is suitable (for the pair pX1, X2q) if the finite rankoperators are dense in ApX1, X2q (in its ideal norm). We say that a sequencepwjqjPN ApX1, X2q is block diagonal if there exists a sequence 0 N1 N2 . . . so that, for any j, wj pP2,Nj
P2,Nj1qwjpP1,Nj
P1,Nj1q. If E is an
unconditional sequence space, we say that ApX1, X2q satisfies the E-estimate if,for some constant C,
C1pwjqjE ¤ ¸j
wjA ¤ CpwjqjE (3.2)
holds for any finite block-diagonal sequence pwjq.Theorem 3.2. Suppose X1 and X2 are Banach spaces with unconditional FDD,so that X
1 and X2 are subprojective. Suppose, furthermore,that the ideal A issuitable for pX1, X2q, and ApX1, X2q satisfies the E-estimate for some uncondi-tional sequence E. Then ApX1, X2q is subprojective.
Before proving these theorems, we state a few consequences.
Corollary 3.3. The spaces X1qb . . . qbXn and X1pb . . . pbXn are subprojectivewhere Xi is isomorphic to either `pi p1 ¤ pi 8q, or to c0, for every 1 ¤ i ¤ n.
For n 2, this result goes back to [38] and [32] (the injective and projectivecases, respectively).
Suppose a Banach space X has an unconditional FDD implemented by pro-jections pP 1
nq – that is, P 1nP
1m 0 unless n m, supN,
°Nn1P 1
n 8,
and limN
°Nn1 P
1n IX point-norm. We say that X satisfies the lower p-
estimate if there exists a constant C so that, for any finite sequence ξj P ranP 1j ,
°j ξjp ¥ C°j ξjp. The smallest C 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, if X is an uncon-ditional sequence space, then the above definitions coincide with the standardone (see e.g. [28, Definition 1.f.4]).
Corollary 3.4. Suppose the Banach spaces X1 and X2 have unconditionalFDD, satisfy the lower and upper p-estimates respectively, and both X
1 andX2 are subprojective. Then KpX1, X2q is subprojective.
Before proceeding, we mention several instances where the above corollaryis applicable. 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 w1, . . . , wn aresuch that wj Pjwj for any j. Then
¸j
wj ¤ CAve¸j
wj ¤ CT2pXq¸
j
wj212
9
(T2pXq is the type 2 constant of X). The cotype case is handled similarly. Thus,we can state:
Corollary 3.5. Suppose the Banach spaces X1 and X2 have unconditionalFDD, cotype 2 and type 2 respectively, and both X
1 and X2 are subprojective.Then KpX1, X2q is subprojective.
This happens, for instance, if X1 Lppµq or Cp (1 p ¤ 2) and X2 Lqpµqor Cq (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. Theexistence of unconditional FDD of Cp spaces is given by [4].
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 theoremholds for a tensor product of any k1 subprojective Banach spaces that satisfyE-estimate. We will show that the statement holds for the tensor product of kBanach spaces X X1rbX2rb . . . rbXk.
For notational convenience, let Pin °nk1 P
1ik, and Ii IXi
. If A P BpXqis a projection, we use the notation AK for IX A. Furthermore, define theprojections Qn P1nbP2nb. . .bPkn and Rn PK
1nbPK2n . . .bPK
kn. Renormingall Xi’s if necessary, we can assume that their unconditional FDD constantsequal 1.
First show that, for any n, ranRKn is subprojective. To this end, write
RKn
°ki1 P
piq, where the projections P piq are defined by
P p1q P1n b I2 b . . .b Ik,
P p2q PK1n b P2n b I3 b . . .b Ik,
P p3q PK1n b PK
2n b P3n b I4 b . . .b Ik,. . . . . . . . .
P pkq PK1n b PK
2n b . . .b PKk1,n b Pkn
(note also that P piqP pjq 0 unless i j). Thus, there exists i so that P piq isan isomorphism on a subspace Y 1 Y . Now observe that the range of P piq isisomorphic to a subspace of `N8pXpiqq, where N rankPin, and
Xpiq X1rbX2rb . . . rbXi1rbXi1rb . . . rbXk.
By the induction hypothesis, Xpiq is subprojective. By Proposition 2.2, ranP piq
is subprojective for every i, hence so is ranRKn .
Now suppose Y is an infinite dimensional subspace of X. We have to showthat Y contains a subspace Z, complemented in X. If there exists n P N sothat RK
n |Y is not strictly singular, then, by Corollary 2.4, Z contains a subspacecomplemented in X.
Now suppose RKn |Z is strictly singular for any n. It is easy to see that, for any
sequence of positive numbers pεmq, one can find 0 n0 n1 n2 . . ., andnorm one elements xm P Y , so that, for any m, RK
nm1xm xmQnm
xm
10
εm. By a small perturbation, we can assume that xm RKnm1
Qnmxm. Thatis,
xm P ranpP1,nm
P1,nm1q b pP2,nm
P2,nm1q b . . .b pPk,nm
Pk,nm1q.
Let Eim ran pPi,nm Pi,nm1
q, and W spanrE1m b E2m b . . . b Ekm :m P Ns X. Applying “Tong’s trick” (see e.g. [27, p. 20]), and taking the1-unconditionality of our FDDs into account, we see that
U : X ÑW : x ÞѸm
pP1,nm P1,nm1
q b . . .b pPk,nm Pk,nm1
qxdefines a contractive projection onto W . Furthermore, Z spanrxm : m P Nsis complemented in W . Indeed, the projection Pi,nm Pi,nm1 pi,m P Nq is
contractive, hence we can identify E1mrb . . . rbEkm with pE1m b . . . b Ekmq XX. By the by Hahn-Banach Theorem, for each m there exists a contractiveprojection Um on E1mrb . . . rbE2m, with range spanrxms. By our assumption,there exists an unconditional sequence space E so that X1rb . . . rbXk satisfies theE-estimate. Then, for any finite sequence wm P E1mrb . . . rbEkm, (3.1) yields
¸k
Ukwk ¤ CpUkwkqE ¤ CpwkqE ¤ C2¸k
Ukwk.
Thus, Z is complemented in X.
Sketch of the proof of Theorem 3.2. On ApX1, X2q we define the projectionRn :ApX1, X2q Ñ ApX1, X2q : w ÞÑ PK
2nwP1n. Then the range of RKn is isomorphic
to X1 ` . . .`X
1 `X2` . . .`X2. Then proceed as in the the proof of Theorem3.1 (with k 2).
To prove Corollary 3.3, we need two auxiliary results.
Lemma 3.6. Suppose 1 pi 8 p1 ¤ i ¤ nq and X qbni1`pi .
1. If°
1pi ¡ n 1, then X satisfies the `s-estimate with 1s °1pi
pn 1q.2. If
°1pi ¤ n 1, then X satisfies the c0-estimate.
Proof. Suppose pwjq is a finite block-diagonal sequence in X. We shall showthat °j wj pwjqs, with s as in the statement of the lemma. To thisend, let pUijq be coordinate projections on `pi for every 1 ¤ i ¤ n, such thatwj U1j b . . . b Unjwj , and for each i, UikUim 0 unless k m. Lettingp1i pippi 1q, we see that
¸j
wj supξiP`p1
i,ξi¤1
x¸j
wj ,biξiy.
11
Choose biξi with ξi ¤ 1, and let ξij Uijξi. Then°j ξijp
1i ¤ 1, andx¸
j
wj ,biξiy ¤¸
j
xwj ,biξiy ¸j
xwj ,biξijy ¤¸j
wjΠni1ξij.
Now let 1r °1p1i n°
1pi. By Holder’s Inequality,
¸j
n¹i1
ξijr1r
¤n¹i1
¸j
ξijp1i
1p1i ¤ 1.
If°
1pi ¤ n 1, then r ¤ 1, hence°j Πn
i1ξij ¤ 1. Therefore, °j wj ¤maxj wj pwjqc0 . Otherwise, r ¡ 1, and
¸j
wj ¤¸
j
wjs1s¸
j
pΠni1ξijqr
1r
¤¸
j
wjs1s
pwjqs,
where 1s 1 1r °1pi n 1.
In a similar fashion, we show that °j wj ¥ pwjqs. For s 8, theinequality °j wj ¥ maxj wj is trivial. If s is finite, assume
°j wjs 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 wjsr. Then°j γ
rj 1 °
j γjwj. Further, set αij γΠlip
1lp
°nm1 Πlmp
1lq
j . An elementary calculation
shows that γj Πni1αij , and
°j α
p1iij 1. Let ξi
°j αijξij . Then ξip1i 1,
and therefore,
¸j
wj ¥ x¸j
wj ,biξiy ¸j
Πni1αijxwj ,biξijy
¸j
γjwj 1.
This establishes the desired lower estimate.
Lemma 3.7. For 1 ¤ pi ¤ 8, X `p1 pb`p2 pb . . . pb`pn satisfies the `r-estimate,where 1r °
1pi if°
1pi 1, and r 1 otherwise. Here, we interpret `8as c0.
Proof. The spaces involved all have the Contractive Projection Property (theidentity can be approximated by contractive finite rank projections). Thus, theduality between injective and projective tensor products of finite dimensionalspaces (see e.g. [9, Section 1.2.1]) shows that, for w P X,
w sup |xx,wy| : x P `p11 qb . . . qb`p1n , x ¤ 1
((here, as before, 1p1i1pi 1). Abusing the notation somewhat, we denote byPim the projection on the span of the first m basis vectors of both `pi and `p1i .
Suppose a finite sequence pwkqNk1 P X is block-diagonal, or more precisely, wk pP1,mkP1,mk1
qb . . .bpPn,mkPn,mk1
qwk for every k. Define the operator
U on X by setting Ux °Nk1
pP1,mkP1,mk1
q b . . .b pPn,mkPn,mk1
qx.
12
We also use U0 to denote the similarly defined operator on X. By “Tong’strick” (see e.g. [27, p. 20]), since X and X has an unconditional basis, U (U0)is a contractive projection onto its range W (W0). Then
¸k
wk sup |x¸
k
wk, xy| : xX ¤ 1(
sup |xUp¸
k
wkq, xy| : xX ¤ 1(
sup |x¸
k
wk, U0xy| : xX ¤ 1(.
Write U0x °Nk1 xk. By Lemma 3.6 there is an s (either 1s °
1p1i pn1q 1°
1pi or s 8) pxkqs U0x ¤ x ¤ 1. Moreover,
x¸k
wk, U0xy x¸k
wk,¸k
xky ¸k
xwk, xky,
and therefore,
¸k
wk sup ¸k
|xwk, xky| : pxkqs ¤ 1( pwkqr.
Proof of Corollary 3.3. Combine Theorem 3.1 with Lemma 3.6 and 3.7.
Proof of Corollary 3.4. To apply Theorem 3.2, we have to show that KpX1, X2qsatisfies the c0-estimate. By renorming, we can assume that the FDD constantsof X1 and X2 equal 1. Suppose pwkqNk1 is a block-diagonal sequence, withwk pP2,nk
P2,nk1qwkpP1,nk
P1,nk1q. Let w °
k wk. Then w ¥pP2,nk
P2,nk1qwpP1,nk
P1,nk1q wk, hence w ¥ maxk wk. To prove
the reverse inequality (with some constant), pick a norm one ξ P X1, and letξk pP1,nk
P1,nk1qx. Then ηk wξk satisfies pP2,nk
P2,nk1qηk ηk. Set
η wξ °k ηk. Denote by C1 (C2) lower (upper) p-estimate constants of X1
(resp. X2). Then
wξp ηp ¤ C1
¸k
ηkp ¤ C2
¸k
wkpξkp
¤ maxk
wkpC2
¸k
ξkp ¤ maxk
wkpC2C1¸k
ξkp C2C1ξp.
Taking the supremum over all ξ P UBpX1q, w ¤ pC1C2q1p maxk 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.
13
Proof. Fix 1 p 2. The Haar system forms an unconditional basis in Lpp0, 1q,hence, by [26, Theorem 3.b.1], Lpp0, 1q embeds (complmentably) into a space Ewith a symmetric basis. The space Lpp0, 1q is not subprojective (see Section 1),hence, by Proposition 2.1, neither is E .
Furthermore, a tensor product of subprojective spaces (in fact, of Hilbertspaces) need not be subprojective.
Proposition 3.9. There exists a tensor norm rb, so that `2rb`2 is not subpro-jective.
Proof. By Proposition 3.8, there exists a non-subprojective space E with 1-symmetric basis. For Banach spaces X and Y , and a P X b Y , we set arb suptpu b vqpaqEpH,Kqu, where the supremum is taken over all contractionsu : X Ñ H and v : Y Ñ K (H and K are Hilbert spaces, and EpH,Kq is theSchatten space; we identify H with its dual). Clearly rb is a norm on X b Y .It is easy to see that, for any a P X b Y , TX P BpX,X0q, and TY P BpY, Y0q,pTX b TY qpaqrb ¤ TXTY arb. Thus, to prove that rb determines a tensornorm, it suffices to establish that, for any z P X b Y , we have
zqb ¤ zrb ¤ zpb. (3.3)
To establish the left hand side of (3.3), let H be a 1-dimensional Hilbertspace (the field of scalars). For any f P UBpXq and g P UBpY q, consider thecontractions uf : X Ñ H : x ÞÑ xf, xy and vg : Y Ñ H : y ÞÑ xg, yy. Then, forz °
i xi b yi P X b Y , we have
zrb ¥ supfPUBpXq,gPUBpY q
puf b vgqpzqEpHq supfPUBpXq,gPUBpY q
¸i
xf, xiyxg, yiyEpHq zqb.
To prove the right hand side of (3.3), note that, for any two contractionsu : X Ñ H and v : Y Ñ K,
pub vqpxb yqE uxb vyE uxvy ¤ xy.
The triangle inequality establishes zrb ¤ zpb for any z.If X and Y are Hilbert spaces, then for a P XbY we have arb aEpX,Y q.
Identifying `2 with its dual, we see that E embeds into `2rb`2 as the space ofdiagonal operators. As E is not subprojective, neither is `2rb`2.
Here is another wide class of non-subprojective spaces.
Theorem 3.10. Let X be an infinite dimensional Banach space. Then BpXqis not subprojective.
14
Proof. Suppose, for the sake of contradiction, that BpXq is subprojective. Fix anorm one element x P X. For x P X define Tx P BpXq : y ÞÑ xx, yyx. ClearlyM tTx : x P Xu is a closed subspace of BpXq, isomorphic to X. Therefore, Xis subprojective. By Proposition 2.10, we can find a subspace N M with anunconditional basis. We shall deduce that BpXq contains a copy of `8, whichis 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 from N to a further subspace W , isomorphic to `1or c0, and complemented in X by a projection P . Embed BpW q isomorphicallyinto BpXq by sending T P BpW q to PTP P BpXq, where P is a projection fromX onto W . It is easy to see that BpW q contain subspaces isomorphic to `8,thus, BpXq is not subprojective.
There is only one option left: N is reflexive. Pick a subspace W N ,complemented in X. It has the Bounded Approximation Property [27, Theo-rem 1.e.13]. As in the previous paragraph, BpW q embeds isomorphically intoBpXq. Since BpW q KpW q, [12, Theorem 4(1)] shows that BpW q contains anisomorphic copy of `8. This rules out the subprojectivity of BpXq.Question 3.11. Suppose X is a subprojective Banach space. (i) Is RadpXqsubprojective? (ii) If 2 ¤ p 8, must LppXq be subprojective?
Question 3.12. Is a “classical” (injective, projective, etc.) tensor productof subprojective spaces necessarily subprojective? Note that the Fremlin tensorproduct b|π| of Banach lattices (the ordered analogue of the projective product)can destroy subprojectivity. Indeed, by [6], L2 b|π| L2 contains a copy of L1.L2 is clearly subprojective, while L1 is not (see Section 1).
4. Spaces of continuous functions
In this section we deal with CpKq spaces, mostly separable ones. It is wellknown that, if K is a compact Hausdorff set, then CpKq is separable if and onlyif K is metrizable.
Recall that a topological space is scattered (or dispersed) if every compactsubset has an isolated point. It is known that a compact set is scattered andmetrizable if and only if it is countable (in this case, CpKq, and its dual, areseparable). 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 compactHausdorff space K is scattered, then CpKq is c0-saturated. If, in addiion, Kis metrizable, then CpKq is subprojective (by Sobczyk’s Theorem, the copiesof c0 are complemented). If, on the other hand, a compact Hausdorff spaceK is not scattered, then CpKq contains a copy of Cr0, 1s, hence it cannot becomplemented. 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 if K is scattered.
15
4.1. Tensor products of CpKqIn this subsection we study the subprojectivity of projective and injective
tensor products of CpKq. Our main result is:
Theorem 4.1. Suppose K is a compact metrizable space, and X is a Banachspace. Then the following are equivalent:
1. K is scattered, and X is subprojective.
2. CpK,Xq is subprojective.
Proof. The implication p2q ñ p1q is easy. The space CpK,Xq contains copies ofCpKq and of X, hence, by Proposition 2.1, the last two spaces are subprojective.By the preceding paragraph, K must be scattered.
To prove p1q ñ p2q, first fix some notation. Suppose λ is a countable ordinal.We consider the interval r0, λs with the order topology – that is, the topologygenerated by the open intervals pα, βq, as well as r0, βq and pα, λs. Abusing thenotation slightly, we write Cpλ,Xq for Cpr0, λs, Xq.
Suppose K is scattered. By [39, Chapter 8], K is isomorphic to r0, λs, forsome countable limit ordinal λ. Fix a subprojective space X. We use inductionon λ to show that, for any countable ordinal λ,
Cpλ,Xq is subprojective. (4.1)
By Proposition 2.2, (4.1) holds for λ ¤ ω (indeed, c is isomorphic to c0, hencecpXq cqbX is isomorphic to c0pXq c0qbX). Let F denote the set of allcountable ordinals for which (4.1) fails. If F is non-empty, then it contains aminimal element, which we denote by µ. Note that µ is a limit ordinal. Indeed,otherwise it has an immediate predecessor µ 1. It is easy to see that Cpµ,Xqis isomorphic to Cpµ 1, Xq ` X, hence, by Proposition 2.2, Cpµ 1, Xq isnot subprojective. Let C0pµ,Xq tf P Cpµ,Xq : limνѵ fpνq 0u. ClearlyCpµ,Xq is isomorphic to C0pµ,Xq`X, hence we obtain the desired contradictionby showing that C0pµ,Xq is subprojective.
To do this, suppose Y is a subspace of C0pµ,Xq, so that no subspace of Yis complemented in C0pµ,Xq. For ν µ, define the projection Pν : Cpµ,Xq ÑCpν,Xq : f ÞÑ f1r0,νs. If, for some ν µ and some subspace Z Y , Pν |Z is anisomorphism, then Z contains a subspace complemented in X, by the inductionhypothesis and Corollary 2.4. Now suppose Pν |Y is strictly singular for any ν.We construct a sequence of “almost disjoint” elements of Y . To do this, take anarbitrary y1 from the unit sphere of Y . Pick ν1 µ so that y1Pν1y1 101.Now find a norm one y2 P Y so that Pν1y2 1022. Proceeding further inthe same manner, we find a sequence of ordinals 0 ν0 ν1 ν2 . . ., and asequence of norm one elements y1, y2, . . . P Y , so that yk zk 10k, wherezk pPνk Pνk1
qyk. The sequence pzkq is equivalent to the c0 basis, and thesame is true for the sequence pykq.
Moreover, spanrzk : k P Ns is complemented in Cpµ,Xq. Indeed, let ν supk νk. We claim that µ ν. If ν µ, then Pν is an isomorphism on
16
spanryk : k P Ns, contradicting our assumption. LetWk pPνkPνk1qpC0pXqq,
and find a norm one linear functional wk so that wkpzkq zk. Define
Q : C0pµ,Xq Ñ C0pµ,Xq : f ÞѸk
wkpPνk Pνk1
qfzk.Note that limk pPνk Pνk1
qf 0, hence the range of Q is precisely the spanof the elements zk. By Small Perturbation Principle, Y contains a subspacecomplemented in C0pµ,Xq.
The above theorem shows that CpKqqbX is subprojective if and only if bothCpKq and X are. We do no know whether a similar result holds for other tensorproducts. We do, however, have:
Proposition 4.2. Suppose K is a compact metrizable space, and W is either`p (1 ¤ p 8) or c0. Then CpKqpbW is subprojective if and only if K isscattered.
Proof. Clearly, if K is not scattered, then CpKq is not subprojective. So supposeK is scattered. We deal with the case of W `p, as the c0 case is handledsimilarly. As before, we can assume that K r0, λs, where λ is a countableordinal. If λ is finite, then Cpλqpb`p `N8 pb`p is clearly subprojective. Wehandle the infinite case by transfinite induction on λ. The base is easy: ifλ ω, then Cpλq c, and Cpλq is isomorphic to c0pb`p. The latter space issubprojective, by Corollary 3.3.
Suppose, for the sake of contradiction, that λ is the smallest countable or-dinal so that Cpλqpb`p is not subprojective. Reasoning as before, we concludethat λ is a limit ordinal. Furthermore, Cpλq C0pλq, hence C0pλqpb`p is notsubprojective.
Denote by Qn : `p Ñ `p the projection on the first n basis vectors in `p, andlet QK
n I Qn. For f P C0pλq and an ordinal ν λ, define Pνf χr0,νsf ,
and PKν I Pν .
Suppose X is a subspace of C0pλqpb`p which has no subspaces complementedin C0pλqpb`p. By the induction hypothesis, pPν b I`pq|Y is strictly singular forany ν λ. Furthermore, pIC0pλq b Qnq|Y must be strictly singular. Indeed,otherwise Y has a subspace Z so that pIC0pλqbQnq|Z is an isomorphism, whoserange is subprojective (the range of IC0pλq b Qn is isomorphic to the sum ofn copies of Cpλq, hence subprojective). Therefore, for any ν λ and n P N,pI PK
ν bQKn q|Y is strictly singular. Therefore we can find a normalized basis
pxiq in Y , and sequences 0 ν0 ν1 . . . λ, and 0 n0 n1 . . ., so thatxi pPK
νi1b QK
ni1qxi 103i2. By passing to a further subsequence, we
can assume that pPνi b Qniqxi 103i2. Thus, by the Small Perturbation
Principle, it suffices to show the following statement: If pyiq is a normalizedsequence is C0pλqpb`p, so that there exist non-negative integers 0 n0 n1 n2 . . ., and ordinals 0 ν0 ν1 ν2 . . . λ, with the property thatyi
pPνi Pνi1q b pQni
Qni1qyi for any i, then Y spanryi : i P Ns is
contractively complemented in CpKqpb`p.17
Denote by X the span of all x’s for which there exists an i so that x pPνi Pνi1q b pQni Qni1qx. Then Y is contractively complemented in
CpKqpb`p. In fact, we can define a contractive projection onto X as follows.
Suppose first u °Nj1 aj b bj , with bi’s having finite support in `p. Then set
Pu °8i1
pPνiPνi1qbpQni
Qni1qu. Due to our assumption on the bi’s,
there exists M so that Pu °Mi1
pPνiPνi1qbpQni
Qni1qu. To show that
Pu ¤ u, define, for ε pεiqMi1 P t1, 1uM , the operator of multiplicationby
°i1Mεiχrνi11,νis on C0pλq. The operator Vε P Bp`pq is defined similarly.
Bot Uε and Vε are contractive. Furthermore, Pu AveεpUε b Vεqu. Therefore,we can use continuity to extend P to a contractive projection from C0pλqpb`ponto X.
To construct a contractive projection from X onto Y , we need to show thatthe blocks of X satisfy the `p-estimate. That is, if xi
pPνi Pνi1q b pQni
Qni1
qxi for each i, then °i xip °i xip. To this end, use duality to
identify pC0pλqpb`pq with Bp`p, `1pr0, λqq, q. P is the “block” projection ontothe space of “block diagonal” operators which map the elements of `p supportedon pni1, nis onto the vectors in `1 supported on pνi1, νis. If T 1is are the blocks
of such an operator, then °i Tip1 °
i Tip1
, where 1p 1p1 1. (see theproof of Corollary 3.3). By duality, °i xip
°i xip.
Remark 4.3. After this paper has been completed, we were informed byE. M. Galego that, in the paper [15], currently in preparation, proves that,for compact Hausdorff spaces K1 and K2, the following are equivalent: (i) K1
and K2 are scattered; (ii) CpK1qpbCpK2q is subprojective.
4.2. Operators on CpKqProposition 4.4. Suppose K is a scattered compact metrizable space, and 1 ¤p ¤ q 8. Then the space ΠqppCpKq, `qq is subprojective.
Recall that ΠqppX,Y q stands for the space of pq, pq-summing operators – thatis, the operators for which there exists a constant C so that, for any x1, . . . , xn PX, ¸
i
Txiq1q
¤ C supxPUBpXq
¸i
|xpxiq|p1p
.
The smallest value of C is denoted by πpqpT q.Note that, if a compact Hausdorff space K is not scattered, then CpKq
contains L1 [35], hence ΠqppCpKq, `qq is not subprojective.The following lemma may be interesting in its own right.
Lemma 4.5. Suppose X is a Banach space, K is a compact metrizable scatteredspace, and 1 ¤ p ¤ q 8. Then, for any T P ΠqppCpKq, Xq, and any ε ¡ 0,there exists a finite rank operator S P ΠqppCpKq, Xq with πpqpT Sq ε.
In proving Proposition 4.4 and Lemma 4.5, we consider the cases of p qand p q separately. If p q, we are dealing with q-summing operators. By
18
Pietsch Factorization Theorem, T P BpCpKq, Xq is q-summing if and only ifthere exists a probability measure µ on K so that T factors as T j, wherej : CpKq Ñ Lqpµq is the formal identity, and T ¤ πqpT q. Moreover, µ and
T can be selected in such a way that T πqpT q. As K is scattered, thereexist distinct points k1, k2, . . . P K, and non-negative scalars α1, α2, . . ., so that°i αi 1, and µ °
i αiδki [35].Now suppose T P BpCpKq, Xq satisfies πqpT q 1. Keeping the above nota-
tion, find N P N so that p°8iN1 αiq
1q ε. Denote by u and v the operators of
multiplication by χtk1,...,kNu and χtkN1,kN2,...u, respectively, acting on Lqpµq.It is easy to see that ranku ¤ N , and vj ε. Then S T uj works in Lemma4.5.
If 1 ¤ p q, then (see e.g. [8, Chapter 10] or [40, Chaper 21]), ΠqppCpKq, Xq Πq1pCpKq, Xq, with equivalent norms. Henceforth, we set p 1. We have
a probability measure µ on K, and a factorization T T j, where j : CpKq ÑLq1pµq is the formal identity, and T : Lq1pµq Ñ X satisfies T ¤ cπq1pT q (c isa constant depending on q).
In this case, the proof of Lemma 4.5 proceeds as for q-summing operators,
except that now, we need to select N so that c°8
iN1 αi1q ε.
Proof of Proposition 4.4. Define the tensor norm rb by setting, for z °i xi b
yi P X b Y , zrb πqppzq, where z P BpX, Y q is defined by zf °ixf, xiyyi.
Recall that, for any T , πqppT q πqppTq. Furthermore, if T is weaklycompact, then TpXq Y , hence, by the injectivity of the ideal πqp,πqppT q πqppT rasq, where T ras is the astriction of T to an operator fromX to Y . If T P BpX,Y q has finite rank, write T °n
i1 xi b yi (xi P X,
and yi P Y ), and note that, by the above, πqppT q T rb.
By Lemma 4.5, we can identify ΠqppCpKq, Xq with CpKqrbX. By [35],CpKq `1 (the canonical basis in `1 corresponds to the point evaluationfunctionals), hence we can describe rb in more detail: for u °
i aibxi P `1bX,urb πqppuq, where u : `8 Ñ X is defined by ub °
ixb, aiyxi. Note thatκX u u (κX : X Ñ X is the canonical embedding), where u : c0 Ñ Xdefined via ub °
ixb, aiyxi. Thus, urb πqppuq.To finish the proof, we need to show (in light of Theorem 3.1) that `1rb`q
satisfies the `q estimate. To this end, suppose we have a block-diagonal sequencepuiqni1, and show that °i uiqrb °
i uiqrb. Abusing the notation slightly, we
identify ui with an operator from `N8 to `Nq (where N is large enough), andidentify rb with πqppq.
First show that °i uiqrb ¤ cq°i uiqrb, where c is a constant (depending
on q). We have disjoint sets pSiqni1 in t1, . . . , Nu so that uiej 0 for j R Si.Therefore there exists a probability measure µi, supported on Si, so that
uifq ¤ cq1πqppuiqqfqp8 fpLppµiq
for any f P `N8 (c1 is a constant). Now define the probability measure µ on
19
t1, . . . , Nu:µ ¸
i
πqppuiqq1 ¸
i
πqppuiqqµi.
For f P `N8, set fi fχSi. Then the vectors uifi are disjointly supported in `q,
and therefore,
p¸i
uiqfq ¸i
uifiq ¤ cq1¸i
πqppuiqqfiqp8 fipLppµiq
¤ cq1fqp8
¸i
πqppuiqqfipLppµiq.
An easy calculation shows that
fipLppµiq ¸
i
πqppuiqq1 ¸
i
πqppuiqqfipLppµq,
hencep¸i
uiqfq ¤ cq1¸i
πqppuiqqfqp8
¸i
fipLppµiq
cq1¸i
πqppuiqqfqp8 fpLppµq
.
Therefore, πqpp°i uiq ¤ c
°i πqppuiqq
1q, for some universal constant c.
Next show that °i uiqrb ¥ c1q°i uiqrb, where c1 is a constant. There exists
a probability measure µ on t1, . . . , Nu so that, for any f P `N8,
p¸u
uifqq ¥ cq2πqpp¸i
uiqqfqp8 fpLppµq
For each i let αi µ|Si`N1 , and µi µiαi (if αi 0, then clearly ui 0).
Then for any i, and any f P `N8,
uifq p¸i
uiqpχSifqq ¤ cq2πqpp
¸i
uiqqαifqp8 fpLppµiq,
hence πqppuiq ¤ c1α1qi πqpp
°i uiq (c1 is a constant). As
°i αi 1, we conclude
that°i πqppuiqq ¤ c1qπqpp
°i uiq.
4.3. Continuous fields
We refer the reader to [10, Chapter 10] for an introduction into continuousfields of Banach spaces. To set the stage, suppose K is a locally compactHausdorff space (the base space), and pXtqtPK is a family of Banach spaces (thespaces Xt are called (fibers). A vector field is an element of
±tPK Xt. A linear
subspace X of±tPK Xt is called a continuous field if the following conditions
hold:
1. For any t P K, the set txptq : x P Xu is dense in Xt.
20
2. For any x P X, the map t ÞÑ xptq is continuous, and vanishes at infinity.
3. Suppose x is a vector field so that, for any ε ¡ 0 and any t P K, thereexist an open neighborhood U Q t and y P X for which xpsq ypsq εfor any s P U . Then x P X.
Equipping X with the norm x maxt xptq, we turn it into a Banach space.In a fashion similar to Theorem 4.1, we prove:
Proposition 4.6. Suppose K is a scattered metrizable space, X is a separablecontinuous vector field on K, so that, for every t P K, the fiber Xt is subprojec-tive. Then X is subprojective.
Proof. Using one-point compactification if necessary (as in [10, 10.2.6]), we canassume that K is compact. As before, we assume that K r0, λs (λ is acountable ordinal). We denote by Xp0q the set of all x P X which vanish at λ.If ν ¤ λ, we denote by Xrνs the set of all x P Xλ which vanish outside of r0, νs.By [10, Proposition 10.1.9], xχr0,νs P X for any x P X, hence Xrνs is a Banachspace. We then define the restriction operator Pν : X Ñ Xrνs. We denote byQν : X Ñ Xν the operator of evaluation at ν.
We say that a countable ordinal λ has Property P if, whenever X is acontinuous separable vector field whose fibers are subprojective, then X is sub-projective. Using transfinite induction, we prove that any countable ordinal hasthis property.
The base of induction is easy to handle. Indeed, when λ is finite, then Xembeds into a direct sum of (finitely many) subprojective spaces Xν . Now sup-pose, for the sake of contradiction, that λ is the smallest ideal failing PropertyP. Note that λ is 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 and Xλ.
Suppose Y is a subspace of X, so that no subspace of Y is complemented inX. We shall achieve a contradiction once we show that Y contains a copy of c0.
By Proposition 2.3, Qλ is strictly singular on Y . Passing to a smaller subse-quence if necessary, we can assume that, Y has a basis pyiqiPN, so that (i) for anyfinite sequence pαiq,
°i αiyi ¥ maxi |αi|2, and (ii) for any i, Qλyi 104i.
Consequently, for any y P spanryj : j ¡ is, Qλy 104i. Indeed, we canassume that y is a norm one vector with finite support, and write y as a fi-nite sume y °
j αjyj . By the above, |αi| ¤ 2 for every i. Consequently,
Qλy ¤°j |αj |Qλyj ¤ 2
°j¡i 104j 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 onevectors, so that (i) xj P spanryi : nj ¤ i nj1s, (ii) Pνixi 104i, and(iii) Pνi1
xi 104i. 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 finda norm 1 vector x1 P spanry1, . . . , yn21s so that Pν1x1 104. We haveQλx1 104. By continuity, we can find ν2 ¡ ν1 so that Pν2x1 104.
21
Next find a norm one x2 P spanryn2 , . . . , yn31s so that Pν2x1 108. Proceedfurther in the same manner.
We claim that the sequence pxiq is equivalent to the canonical basis in c0.Indeed, for each i let x2i PνixiPνi1
xi, and x1i xix2i . Since we are workingwith the sup norm, x1i xi 1 for any i. Furthermore, the elements x1iare disjointly supported, hence, for any pαiq finite sequence of scalars pαiq,°i αix
1i maxi |αi|. By the triangle inequality,
¸i
αixi ¸i
αix1i ¤¸
i
|αi|x2i maxi|αi|
8
i1
2 204i 103 maxi|αi|,
which yields the desired result.
To state a corollary of Proposition 4.6, recall that a C-algebra A is CCR(or liminal) if, for any irreducible representation π of A on a Hilbert space H,πpAq KpHq. A C-algebra A is scattered if every positive linear functional onA is a sum of pure linear functionals (f P A is called pure if it belongs to anextreme ray of the positive cone of A). For equivalent descriptions of scatteredC-algebras, see e.g. [19, 20, 25].
Corollary 4.7. Any separable scattered CCR C-algebra is subprojective.
Proof. Suppose A is a separable scattered CCR C-algebra. As shown in [34,Sections 6.1-3], the spectrum of a separable CCR algebra is a locally compactHausdorff space. If, in addition, the algebra is scattered, then its spectrum Ais scattered as well [19, 20]. In fact, by the proof of [19, Theorem 3.1], A isseparable. It is easy to see that any separable locally compact Hausdorff spaceis metrizable.
By [10, Section 10.5], A can be represented as a vector field over A, withfibers of the form πpAq, for irreducible representations π. As A is CCR, thespaces πpAq KpHπq (Hπ being a separable Hilbert space) are subprojective.To finish the proof, apply Proposition 4.6.
The last corollary leads us to
Conjecture 4.8. A separable C-algebra is scattered if and only if it is sub-projective.
It is known ([20], see also [25]) that a scattered C-algebra is GCR. However,it need not be CCR (consider the unitization of Kp`2q).
5. Subprojectivity of Schatten spaces
In this section, we establish:
Proposition 5.1. Suppose E is a symmetric sequence space, not containing c0.Then CE is subprojective if and only if E is subprojective.
22
The assumptions of this proposition are satisfied, for instance, if E `p (1 ¤p 8), or if E is the Lorentz space lpw, pq (see [27, Proposition 4.e.3]. However,by Proposition 3.8, 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 containingc0. Suppose, furthermore, that pznq CE is a normalized sequence, so that, forevery k, limn Qkzn 0. Then, for any ε ¡ 0, CE contains sequences pznq andpz1nq, so that:
1. pznq is a subsequence of pznq.2.
°n zn z1n 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 projectionof norm not exceeding 3.
Proof. [3, Corollary 2.8] implies the existence of pznq and pz1nq, so that (1) and(2) are satisfied, and z1k a b E1k b b Ek1 ck b Ekk (k ¥ 2). Thus,z1n Z ZrZcZd, where Zr spanrabE1k : k ¥ 2s (the row component),Zc spanrb b Ek1 : k ¥ 2s (the column component), and Zd (the diagonalcomponent) contains ck b Ekk, for any k. More precisely, we can write ck ukdkvk, where uk and vk are unitaries, and dk is diagonal. Then we set Zd spanrukEiivk b Ekk : i P N, k ¥ 2s.
It remains to build contractive projections Pr, Pc, and Pd onto Zr, Zc, andZd, respectively, so that ZcYZd kerPr, ZrYZd kerPc, and ZrYZc kerPd.Indeed, then P Pr PcPd is a projection onto Zr ZcZd, and the latterspace is completely isomorphic to Z0 Zr`Zc`Zd. The spaces Zr, Zc, and Zdare either trivial (zero-dimensional), or isomorphic to `2, `2, and E , respectively.
Pd is nothing but a coordinate projection, in the appropriate basis:
PdukEijv` b Ek`
"ukEiivk b Ekk k ` ¥ 2, i j
0 otherwise
(for the sake of convenience, we set u1 v1 I`2). Next construct Pr (Pc isdealt with similarly). If a 0, just take Pr 0. Otherwise, let a1 aa, andfind f P CE so that f 1 xf, a1y. For x °
k,` xk` b Ek`, define
Prx a1 b¸`¥2
xf, x1`yE1`,
hence Prx2E °`¥2 |xf, x1`y|2. It remains to show Prx ¤ x. This inequal-
ity is obvious when Prx 0. Otherwise, set, for ` ¥ 2,
α` xf, x1`yp°`¥2 |xf, x1`y|2q12 ,
23
y I`2 b°`¥2 α`E`1, and z I`2 b E11. Then y8 °
`¥2 |α`|212 1
z8, and zxy °`¥2 α`x1` b E11. Therefore,
PrxE @f,
¸`¥2
α`x1`
D ¤ ¸`¥2
α`x1`
E
¸`¥2
α`x1` b E11
E zxyE ¤ z8xEy8 xE ,
which is what we need.
Proof of Proposition 5.1. The space CE contains an isometric copy of E , hencethe subprojectivity of CE implies that of E . To prove the converse, suppose E issubprojective, and Z0 is a subspace of CE , and show that it contains a furthersubspace Z, complemented in CE . To this end, find a normalized sequencepznq Z0, so that limn Qkzn 0 for every k. By Proposition 5.2, pznq hasa subsequence pz1nq, contained in a subspace Z1, which is complemented in CE ,and isomorphic either to E , `2, or E``2. By Proposition 2.2, Z1 is subprojective,hence spanrz1n : n P Ns contains a subspace complemented in Z1, hence also inCE .
As a consequence we obtain:
Proposition 5.3. The predual of a von Neumann algebra A is subprojective ifand only if A is atomic.
We say that A is atomic if any projection in it has an atomic subprojection.By [31, Section 1], this happens if and only if A p°iBpHiqq8.
Proof. If a von Neumann algebra A is not purely atomic, then, as explained in[31, Section 1], A contains a (complemented) copy of L1p0, 1q. This establishesthe “only if” implication of Proposition 5.3. Conversely, if A is purely atomic,then A is isometric to a (contractively complemented) subspace of C1pHq, andthe latter is subprojective.
6. p-convex and p-disjointly homogeneous Banach lattices
We say that X is p-disjointly homogeneous (p-DH for short) if every disjointnormalized 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 a p-convex p-DH Bnaach lattice. Then every sub-space, spanned by a disjoint sequence equivalent to the canonical basis of `p, iscomplemented.
24
Proof. Let pxkq X be a disjoint normalized sequence. Since X is p-DH, bypassing to a subsequence, we can assume that (xkq is an `p basic sequence. Then,in the p-concavification Xppq the disjoint sequence pxkpq is an `1 basic sequence.Therefore, there exists a functional x P rpxkpqs such that xpxkpq 1 for allk. By the Hahn-Banach Theorem x can be extended to a positive functional
in Xppq. Define a seminorm xp pxp|xp|qq 1
p on X. Denote by N thesubset of X on which this seminorm is equal to zero. Clearly, N is an ideal,therefore, the quotient space X XN is a Banach lattice, and the quotientmap Q : X Ñ X is an orthomorphism. With the defined seminorm X is anabstract Lp-space, and the disjoint sequence Qpxkq is normalized. Therefore itis an `p basic sequence that spans a complemented subspace (in particular, Q
is an isomorphism when restricted to rxks). Let P be a projection from X ontorQpxkqs. Then P Q1PQ is a projection from X onto rxks.Proposition 6.2. Let X be a p-convex, p-disjointly homogeneous Banach latticepp ¥ 2q. Then any subspace of X 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 infinitedimensional separable subspace. Then there exists a complemented order idealin X with a weak unit that contains M . Therefore, without loss of generality,we may assume that X has a weak unit. Then there exists a probability measureµ [21, p. 14] such that we have continuous embeddings
L8pµq X Lppµq L2pµq L1pµq.
Consequently, there exists a constant c1 ¡ 0 so that c1xp ¤ x for any x P X.By the proof of [28, Proposition 1.c.8], one of the following holds:
Case 1. M contains an almost disjoint bounded sequence. By Proposition 6.1M contains a copy of `p complemented in X.
Case 2. The norms and 1 are equivalent on M . Thus, there exists c2 ¡ 0so that, for any y PM ,
c2y2 ¥ c2y1 ¥ y ¥ c1yp ¥ c1y2.
In particular, M is embedded into L2pµq as a closed subspace. The orthogonalprojection from L2pµq onto M then defines a bounded projection from X ontoM .
The preceding result implies that Lorentz space Λp,W p0, 1q is subprojectivefor p ¥ 2. Indeed, Λp,W p0, 1q is p-DH and p-convex pp ¥ 1q, see [13, Theorem3] and [24]. Note that, originally, the subprojectivity of Λp,W p0, 1q (for p ¥ 2)was observed in [13, Remark 5.7].
25
7. Lattice-valued `p spaces
If X is a Banach lattice, and 1 ¤ p 8, denote by Xp`pq the completion ofthe space of all finite sequences px1, . . . , xnq (with xi P X), equipped with thenorm px1, . . . , xnq p°i |xi|pq1p, where
p¸i
|xi|pq1p sup!|¸i
αixi| :¸i
|αi|p1 ¤ 1
), with
1
p 1
p1 1.
See [28, pp. 46-48] for more information. We have:
Proposition 7.1. Suppose X is a subprojective separable space, with the lattice
structure given by an unconditional basis, and 1 ¤ p 8. Then Xp`pq issubprojective.
Proof. To show that any subspace Y Xp`pq has a further subspace Z, com-
plemented in Xp`pq, let x1, x2, . . . and e1, e2, . . . be the canonical bases in X and`p, respectively. Then the elements uij xi b ej form an unconditional basis
in Xp`pq, with
¸aijuij
¸i
p¸j
|aij |pq1pxiX ¸i
sup°
j |αj |p1¤1
|¸j
αijaij |xiX . (7.1)
Let Pn be the canonical projection onto spanruij : 0 ¤ i ¤ n, j P Ns, and setPKn I Pn. The range of Pn is isomorphic to `p, hence, if Pn|Y is not strictly
singular for some n, we are done, by Corollary 2.4. If Pn|Y is strictly singularfor every n, find a normalized sequence pyiq in Y , and 1 n1 n2 . . ., sothat Pni
yi, PKni1
yi 100i2. By small perturbation, it remains to prove
the following: if yi PKniPni1
yi, then spanryi : i P Ns contains a subspace,
complemented in Xp`pq. Further, we may assume that for each i there existsMi so that we can write
yi ¸
ni k¤ni1,1¤j¤Mi
akjukj .
For each k P rni1, ni1s (and arbitrary i P N) find a finite sequence pαkjqMij1 so
that°j |αkj |p
1 1, and |°j αkjakj | p°j |akj |pq1p. Define U : Xp`pq Ñ X :ukj ÞÑ αkjakjxk. By (7.1), U is a contraction, and U |spanryi:iPNs is an isometry.To finish the proof, recall that X is subprojective, and apply Corollary 2.4.
Remark 7.2. Using similar methods, one can prove: if K is a compact metriz-
able scattered space, and 1 ¤ p 8, then CpKqp`pq is subprojective.
Recall that, for a Banach space X, we denote by RadpXq the completion ofthe finite sums
°n rnxn (r1, r2, . . . are Rademacher functions, and x1, x2, . . . P
X) in the norm of L1pXq (equivalently, by Khintchine-Kahane Inequality, inthe norm of LppXq). If X has a unconditional basis pxiq and finite cotype, then
26
RadpXq is isomorphic to Xp`2q (here we can view X as a Banach lattice, withthe order induced by the basis pxiq). Indeed, by [28, Section 1.f], X is q-concave,for some q. An array pamnq can be identified both with an element of RadpXq(with the norm
³1
0°m
°n amnrnxm), and with an element of Xp`2q (with the
norm °mp°n |amn|2q12xm). Then
D¸m
p¸n
|amn|2q12xm ¤ ¸m
» 1
0
|¸n
amnrn|xm
» 1
0
|¸m
¸n
amnrnxm| ¤» 1
0
¸m
¸n
amnrnxm
¤ p» 1
0
¸m
¸n
amnrnxmqq1q ¤Mqp» 1
0
|¸m
¸n
amnrnxm|qq1q
¤Mq¸m
p» 1
0
|¸n
amnrn|qq1qxm ¤ CMq¸m
p¸n
|amn|2q12xm,
where Mq is a q-concavity constant, while D and C come from Khintchine’sinequality. Thus, we have proved:
Proposition 7.3. If X is a subprojective space with an unconditional basis andnon-trivial cotype, then RadpXq is 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 basisand non-trivial cotype, then RadpXq is isomorphic to X. Indeed, non-trivialcotype 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, inturn, is isomorphic to RadpXq.
Acknowledgements
The authors acknowledge the generous support of Simons Foundation, via itsTravel Grant 210060. They would also like to thank the organizers of Workshopin 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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