Subprojective Banach Spaces

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 space X is called subprojective if any of its inﬁnite dimensional subspaces 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 dimensional, and subspaces, infinite dimensional and closed, until specified otherwise. A Banach space X is called subprojective if every subspace Y X contains a further subspace Z Y , complemented in X. This notion was introduced in [41], in order to study the (pre)adjoints of strictly singular operators. Recall that an operator T P BpX,Y q is strictly singular (T P SSpX,Y q) if T is not an isomorphism on any subspace of X. In particular, it was shown that, if Y is subprojective, and, for T P BpX,Y q, T ¦ P SSpY ¦,X¦q, then T P SSpX,Y q. Later, connections between subprojectivity and perturbation classes were discovered. More specifically, denote by Φ pX,Y q the set of upper semi-Fredholm operators – that is, operators with closed range, and finite dimensional kernel. If Φ pX,Y q H, we define the perturbation class

P Φ pX,Y q tS P BpX,Y q : T S P Φ pX,Y q whenever T P Φ pX,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 P Φ pX,Y q. In general, this inclusion is proper. However, we get SSpX,Y q P Φ pX,Y q if Y 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) and c0 are 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, 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]. 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- tion 2.2). However, subprojectivity is not a 3-space property (Proposition 2.8). 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 (Proposition 2.11). In Section3, we investigate the subprojectivity of tensor products, and of spaces of operators. A general result on tensor products (Theorem 3.1) yields q p the subprojectivity on `pb`q and `pb`q for 1 ¤ p, q 8 (Corollary 3.3), as well as of KpLp,Lqq for 1 p ¤ 2 ¤ q 8 (Corollary 3.4). We also prove that the space BpXq is never subprojective (Theorem 3.10), and give an example of r non-subprojective tensor product `2b`2 (Proposition 3.9). Throughout Section4, we work with CpKq spaces, with K compact metrizable. 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 CpKq 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 study continuous ﬁelds on a scattered base space, proving that any scattered separable CCR C¦-algebra is subprojective (Corollary 4.7). Section5 shows that, in many cases, subprojectivity passes from a sequence space to the associated Schatten space (Proposition 5.1). Proceeding to Banach lattices, in Section6 we prove that p-disjointly homogeneous p-convex lattices (2 ¤ p 8) are subprojective (Proposition 6.2). In Section7 (Proposition 7.1), we show that the lattice Xp`pq is subprojective whenever X is. Consequently (Proposition 7.3), if X is a subprojective space with an unconditional basis and non-trivial cotype, then RadpXq is subprojective. 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 compact operators, respectively, acting between Banach spaces X and Y . UBpXq refers to the closed unit ball of X. For p P r1, 8s, we denote by p1 the “adjoint” of p (that is, 1{p 1{p1 1).

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 subprojective.

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) Suppose X and Y are Banach spaces. Then the following are equivalent: 1. Both X and Y are subprojective. 2. X ` Y is subprojective.

(b) Suppose X1,X2,... are Banach spaces, and E is a space with a 1-unconditional basis. Then the following are equivalent:

1. The spaces E,X1,X2,... are subprojective. ° p q 2. n Xn E is subprojective. }¤} ° In (b), we view E as a space of all sequences, for which± the norm E is ﬁnite. p XnqE refers to the space of all sequences pxnqnP P P Xn, endowed with n N n N }p q P } }p} } q} the norm xn n N xn Xn E . Due° to the 1-unconditionality (actually, 1- p q suppression unconditionality suﬃces), n Xn E 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 spaces X and X1, and T P BpX,X1q. Sup- pose Y is a subspace of X, T |Y is an isomorphism, and T pY q is complemented in X1. Then Y is complemented in X. Proof. If Q is a projection from X1 to T pY q, then T ¡1QT is a projection from X onto Y .

This immediately yields:

3 Corollary 2.4. Suppose X and X1 are Banach spaces, and X1 is subprojective. Suppose, furthermore, that Y is a subspace of X, and there exists T P BpX,X1q so that T |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. Suppose pxkq is a seminormalized basic sequence in a Banach space X, and pykq is a sequence so that limk }xk¡yk} 0. Suppose, furthermore, r P s that every subspace of span yk : k N contains a subspace complemented in X. r P s Then span xk : k N contains a subspace complemented in X.

Proof. Replacing xk by xk{}xk}, we can assume that pxkq is normalized. De- ¦ } ¦} note the biorthogonal functionals° by xk , and set K supk xk . Passing to a } ¡ } {p q subsequence, we can assume° that k xk yk 1 2K . Deﬁne the operator P p q ¦p qp ¡ q } } { U B X by setting Ux k xk x yk xk . Clearly U 1 2, and therefore, V IX U is invertible. Furthermore, V xk yk. If Q is a projection from X r P s ¡1 onto a subspace W span yk : k N , then P V QV is a projection from r P s X onto a subspace Z span xk : k N . Remark 2.6. Note that, in the proof above, the kernels and the ranges of the projections Q and P are isomorphic, via the action of V . Proof of Proposition 2.2. Due to Proposition 2.1, in both (a) and (b), only the implication p2q ñ p1q needs to be established. (a) Throughout the proof, PX and PY stand for the coordinate projections from X ` Y onto X and Y , respectively. We have to show that any subspace E of X ` Y contains a further subspace G, complemented in X ` Y . Show ﬁrst that E contains a subspace F so that either PX |F or PY |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 subspace F E, so that PX |F has norm less than 1{2. But PX PY IX`Y , hence, by the triangle inequality, }PY f} ¥ }f}¡}PX f} ¥ }f}{2 for any f P F . Consequently, PY |F is an isomorphism. Thus, by passing to a subspace, and relabeling if necessary, we can assume that E contains a subspace F , so that PX |F is an isomorphism. By Corollary 2.4, F contains a subspace G, complemented in X. 1 1 Set F PX pF q, and let V be the inverse of PX : F Ñ F . By the subprojectivity of X, F 1 contains a subspace G1, complemented in X via a projection p 1q Q. Then P V QPX gives a projection onto G V G F . ° p q (b) Here, we denote by Pn the coordinate° projection from X k Xk E n K ¡ onto Xn. Furthermore, we set Qn k1 Pk, and Qn 1 Qn. We have to show 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.

4 | P (ii) For every n, Qn Y is not an isomorphism – that is, for every n 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, }Q y }, }QK y } ε . By a small perturbation principle, we can assume that Ni i Ni 1 i i Y contains norm one vectors py1q so that Q y1 QK y1 0 for every i. Write i Ni i Ni 1 i 1 p qNi 1 P rp q P s yi zj jN 1, with zj Xj. Then Z span 0,..., 0, zj, 0,... : j N (zj is i ¦ P ¦ in j-th position) is complemented in X. Indeed, if zj 0, find zj Xj so that } ¦} } }¡1 x ¦ y ¦ p q P zj zj , and zj , zj 1. If zj 0, set zj 0. For x xj jP X, ¦ N px y q P define Rx zj , xj zj j N. It is easy to see that R is a projection onto Z, and }R} does not exceed the unconditionality constant of E. Now note that J : Z Ñ E : pα1z1, α2z2,...q ÞÑ pα1}z1}, α2}z2},...q is an 1 r 1 P s p 1q isometry. Let Y span yi : i N , and YE J Y . By the subprojectivity of E, YE contains a subspace W , which is complemented in E via a projection R1. ¡1 ¡1 Then J R1JR is a projection from X onto Y0 J pW q X. Remark 2.7. From the last proposition it follows the (strong) p-sum of subprojective Banach spaces is subprojective. On the other hand, the infinite weak sum of subprojective spaces need not be subprojective. Recall that if X is a Banach space, then ¸ 8 ¦ 1 weakp q t p q P ¢ ¢ p | p q|pq p 8u `p X x xn n1 X X X... : sup x xn . x¦PX¦

weakp q p 1 q 1 1 It is known that `p X is isomorphic to B `p ,X ( p p1 1), see [8, 1 Theorem 2.2]. We show that, for X `r pr ¥ p q, Bp`p1 ,Xq contains 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 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 is subprojective. Next we show that subprojectivity is not a 3-space property. Proposition 2.8. For 1 p 8 there exists a non-subprojective Banach space Zp, containing a subspace Xp, so that Xp and Zp{Xp are isomorphic to `p. Proof. [22, Section 6] gives us a short exact sequence

jp qp 0 ÝÑ `p ÝÑ Zp ÝÑ `p ÝÑ 0, where the injection jp is strictly cosingular, and the quotient map qp is strictly singular. 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`pq contains no complemented subspaces (by [27, Theorem 2.a.3], any complemented subspace of `p is isomorphic to `p).

5 It is easy to see that subprojectivity is stable under isomorphisms. However, it is not stable under a rougher measure of “closeness” of Banach spaces – the gap measure. If Y and Z are subspaces of a Banach space X, we deﬁne the gap (or opening) ( ΘX pY,Zq max sup distpy, Zq, sup distpz, Y q . yPY,}y}1 zPZ,}z}1

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 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 subspace Y so that, for every ε ¡ 0, X contains a non-subprojective space Z with ΘX pY,Zq ¤ ε.

Proof. Our Y will be isomorphic to `p, where p P p1, 8q is fixed. By Proposition 2.8, there exists a non-subprojective Banach space W , containing a subspace W0, 1 so that both W0 and W W {W0 are isomorphic to `p. Denote the quotient 1 1 1 map W Ñ W by q. Consider X W `1 W and Y W0 `1 W E. Furthermore, for ε ¡ 0, define Zε tεw `1 qw : w P W u. Clearly, Y is isomorphic to `p ` `p `p, hence subprojective, while Zε is isomorphic to W , hence not subprojective. By [33, Lemma 5.9], ΘX pY,Zεq ¤ ε. Looking at subprojectivity through the lens of Gowers dichotomy and observing 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 unconditional 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 on X is a strictly singular perturbation of a diagonal operator. We prove that X has no subprojective subspaces. In doing so, we are re-using the notation of that paper.° In particular, P P } } n } } for n N and x X, we define x pnq as the° supremum of i1 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 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 basis pykq) operator on Y is strictly singular. Consequently, any T P BpY q can be written as T Λ S, where Λ is diagonal, 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 of the cited paper) yields the generalization described above. Suppose, for the sake of contradiction, that X contains a subprojective subspace Y . A small perturbation argument shows we can assume Y to be a block subspace. Blocking further, we can assume that Y is spanned by a block basis ¡j pyjq, so that 1 }yj} ¤ }yj}pjq 1 2 . We achieve the desired contradiction by showing that no subspace of Z spanry1 y2, y3 y4,...s is complemented in Y . Suppose P is an infinite rank projection from Y onto a subspace of Z. Write P Λ S, where S is a strictly singular operator with zeroes on the main p q8 diagonal, and Λ λj j1 is a diagonal operator (that is, Λyj λjyj for any } } 8 j). As supj yj pjq , by [18, Section 5] we have limj Syj 0. Note that p q2 p 2 ¡ q 2 ¡ ¡ ¡ ¡ 2 Λ S Λ S, hence diag λj λj Λ Λ S ΛS SΛ S is strictly singular, or equivalently, limj λjp1 ¡ λjq 0. Therefore, there exists a 0 ¡ 1 p 1 q 1 ¡ p ¡ 1 q sequence λj so that Λ Λ is compact (equivalenty, limj λj λj 0), where 1 p 1 q 1 1 1 p ¡ 1q Λ diag λj is a diagonal projection. Then P Λ S , where S S Λ Λ 1 is strictly singular, and satisfies limj S yj 0. The projection P is not strictly singular (since it is of infinite rank), hence Λ1 P ¡ S1 is not strictly singular. t P 1 u Consequently, the set J j N : λj 1 is infinite. } ¡ } ¥ { P Now note° that, for any j, P yj yj 1 2. Indeed, P yj Z, hence we can p q r { s write P yj k αk y2k¡1 y2k . Let ` j 2 . By the 1-unconditionality of our basis, }yj ¡ P yj} ¥ }yj ¡ α`py2`¡1 y2`q} ¥ maxt|1 ¡ α`|, |α`|u ¥ 1{2. For 1 1 1 j P J, S yj P yj ¡ yj, hence }S yj} ¥ 1{2, which contradicts limj }S yj} 0.

Remark 2.12. The preceding statement provides an example of an atomic order 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 unconditional 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 projections from X onto Z can be uniformly bounded. More precisely, we call a Banach space X uniformly subprojective (with constant C) if, for every subspace Y X, there exists a subspace Z Y and a projection P : X Ñ Z with }P } ¤ C. The proof of [16, Proposition 2.4] essentially shows that the following spaces are uniformly subprojective: (i) `p (1 ¤ p 8) and c0; (ii) the Lorentz sequence spaces lp,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 of c0 contains a λ-isomorphic copy of c0, for any λ ¡ 1 [27, Proposition 2.e.3]. By Sobczyk’s Theorem, a λ-isomorphic copy of c0 is 2λ-complemented in every separable superspace. In particular, if K is a countable metric space, then CpKq is uniformly subprojective [11, Theorem 12.30].

7 However, in general, subprojectivity need not be uniform. Indeed, suppose 8 8 2° p1 p2 ... , and limn pn . By Proposition 2.2(b), X p p qq n Lpn 0, 1 2 is subprojective. The span of independent Gaussian random variables 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 least c0 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 X and Y .A tensor norm } ¤ }br 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 X1 and Y 1, any T P BpX,X1q and S P BpY,Y 1q, and any z P X b Y , we have }pT b Sqz}br ¤ }T }}S}}z}br .

2. For any z P X b Y , }z}bq ¤ }z}br ¤ }z}bp (the left and right hand sides refer to the injective and projective tensor norms, respectively). r The tensor product XbY is the completion of X b Y in the norm } ¤ }br . In particular, bq and bp 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]. Suppose X1, X2,..., Xk are Banach spaces with unconditional FDD, im- p 1 q p 1 q p 1 q plemented by ﬁnite rank projections P1n °, P2n ,..., Pkn , respectively. That is, P 1 P 1 0 unless n m, lim N P 1 I point-norm, and °in im N n1 in Xi } N ¨ 1 } 8 supN,¨ n1 Pin (this quantity is sometimes referred to as the uncon- p 1 q ditional FDD constant of Xi). Let Ein ran Pin . p q8 b b b We say that a sequence wj j1 X1 X2 ... Xk is block-diagonal if there exists a sequence 0 N1 N2 ... so that

N¸j 1 ¨ N¸j 1 ¨ N¸j 1 ¨ wj P E1n b E2n b ... b Ekn .

nNj 1 nNj 1 nNj 1

Suppose E is an unconditional sequence space, and br is a tensor product of r r r Banach spaces. The Banach space X1bX2b ... bXk is said to satisfy the E- estimate if there exists a constant C ¥ 1 so that, for any block diagonal sequence p q P br br br wj j N in X1 X2 ... Xk, we have ¸ ¡1 }p} }q P } ¤ } } ¤ }p} }q P } C wj j N E wj C wj j N E (3.1) j

Theorem 3.1. Suppose X1, X2,..., Xk are subprojective Banach spaces with unconditional FDD, and br is a tensor product. Suppose, furthermore, that for any ﬁnite increasing sequence i r1 ¤ i1 ...... i` ¤ ks, there exists r r r an unconditional sequence space Ei, so that Xi bXi b ... bXi satisﬁes the Ei- r r r 1 2 k estimate. Then X1bX2b ... bXk is subprojective.

8 Note that a tensor product need not be associative. We presume the “nat- r r r r ural” location of brackets: X1bX2bX3 pX1bX2qbX3, etc.. A different position 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 for projections implementing the FDD in Banach spaces X1 and X2. We say that a Banach operator ideal A is suitable (for the pair pX1,X2q) if the finite rank operators are dense in ApX1,X2q (in its ideal norm). We say that a sequence p q P p q wj j N A X1,X2 is block diagonal if there exists a sequence 0 N1 N2 p ¡ q p ¡ q ... so that, for any j, wj P2,Nj P2,Nj¡1 wj P1,Nj P1,Nj¡1 . If E is an unconditional sequence space, we say that ApX1,X2q satisfies the E-estimate if, for some constant C, ¸ ¡1 C }p}wj}qj}E ¤ } wj}A ¤ C}p}wj}qj}E (3.2) j holds for any finite block-diagonal sequence pwjq.

Theorem 3.2. Suppose X1 and X2 are Banach spaces with unconditional FDD, ¦ so that X1 and X2 are subprojective. Suppose, furthermore,that the ideal A is suitable for pX1,X2q, and ApX1,X2q satisfies the E-estimate for some unconditional sequence E. Then ApX1,X2q is subprojective. Before proving these theorems, we state a few consequences. q q p p Corollary 3.3. The spaces X1b ... bXn and X1b ... bXn are subprojective p ¤ 8q ¤ ¤ where Xi is isomorphic to either `pi 1 pi , or to c0, for every 1 i n. For n 2, this result goes back to [38] and [32] (the injective and projective cases, respectively). Suppose a Banach space X has an unconditional FDD implemented° by projections pP 1 q – that is, P 1 P 1 0 unless n m, sup } N ¨P 1 } 8, °n n m N,¨ n1 n N 1 and limN n1 Pn IX point-norm. We say that X satisfies the lower p- P 1 estimate° if there° exists a constant C so that, for any finite sequence ξj ran Pj, } }p ¥ } }p j ξj C j ξj . 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 unconditional 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,X2q is subprojective. Before proceeding, we mention several instances where the above corollary is 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 are such that wj Pjwj for any j. Then ¡ © ¸ ¸ ¸ 1{2 2 } wj} ¤ CAve¨} ¨wj} ¤ CT2pXq }wj} j j j

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 unconditional ¦ FDD, cotype 2 and type 2 respectively, and both X1 and X2 are subprojective. Then KpX1,X2q is subprojective.

This happens, for instance, if X1 Lppµq or Cp (1 p ¤ 2) and X2 Lqpµq or 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. The existence 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 theorem holds for a tensor product of any k ¡ 1 subprojective Banach spaces that satisfy E-estimate. We will show that the statement holds for the tensor product of k br br br Banach spaces X X1 X2 ... Xk. ° n 1 P p q For notational convenience, let Pin k1 Pik, and Ii IXi . If A B X K is a projection, we use the notation A for IX ¡ A. Furthermore, deﬁne the b b b K b K b K projections Qn P1n P2n ... Pkn and Rn P1n P2n ... Pkn. Renorming all Xi’s if necessary, we can assume that their unconditional FDD constants equal 1. K First° show that, for any n, ran Rn is subprojective. To this end, write K k piq piq Rn i1 P , where the projections P are deﬁned by

p1q P P1n b I2 b ... b Ik, p2q K b b b b P P1n P2n I3 ... Ik, p3q K b K b b b b P P1n P2n P3n I4 ... Ik, ...... pkq K b K b b K b P P1n P2n ... Pk¡1,n Pkn

(note also that P piqP pjq 0 unless i j). Thus, there exists i so that P piq is an isomorphism on a subspace Y 1 Y . Now observe that the range of P piq is N piq isomorphic to a subspace of `8pX q, where N rank Pin, and

piq r r r r r r X X1bX2b ... bXi¡1bXi 1b ... bXk.

By the induction hypothesis, Xpiq is subprojective. By Proposition 2.2, ran P piq K is subprojective for every i, hence so is ran Rn . Now suppose Y is an inﬁnite dimensional subspace of X. We have to show that Y contains a subspace Z, complemented in X. If there exists n P N so K| that Rn Y is not strictly singular, then, by Corollary 2.4, Z contains a subspace complemented in X. K| Now suppose Rn 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 ..., and norm one elements x P Y , so that, for any m, }RK x } }x ¡ Q x } m nm¡1 m m nm m

10 ε . By a small perturbation, we can assume that x RK Q x . That m m nm¡1 nm m is, ¡ © P p ¡ q b p ¡ q b b p ¡ q xm ran P1,nm P1,nm¡1 P2,nm P2,nm¡1 ... Pk,nm Pk,nm¡1 .

p ¡ q r b b b Let Eim ran Pi,nm Pi,nm¡1 , and W span E1m E2m ... Ekm : 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 ¸ ¨ Ñ ÞÑ p ¡ q b b p ¡ q U : X W : x P1,nm P1,nm¡1 ... Pk,nm Pk,nm¡1 x m r P s defines a contractive projection onto W . Furthermore, Z span xm : m N ¡ p P q is complemented in W . Indeed, the projection Pi,nm Pi,nm¡1 i, m N is r r contractive, hence we can identify E1mb ... bEkm with pE1m b ... b Ekmq X X. By the by Hahn-Banach Theorem, for each m there exists a contractive r r projection Um on E1mb ... bE2m, with range spanrxms. By our assumption, r r there exists an unconditional sequence space E so that X1b ... bXk satisfies the r r E-estimate. Then, for any finite sequence wm P E1mb ... bEkm,(3.1) yields ¸ ¸ 2 } Ukwk} ¤ C}p}Ukwk}q}E ¤ C}p}wk}q}E ¤ C } Ukwk}. k k Thus, Z is complemented in X.

Sketch of the proof of Theorem 3.2. On ApX1,X2q we define the projection Rn : p q Ñ p q ÞÑ K K A X1,X2 A X1,X2 : w P2nwP1n. Then the range of Rn is isomorphic ¦ ` ` ¦ ` ` ` to X1 ... X1 X2 ... X2. Then proceed as in the the proof of Theorem 3.1 (with k 2). To prove Corollary 3.3, we need two auxiliary results. 8 p ¤ ¤ q bq n Lemma 3.6. Suppose 1 pi 1 i n and X i1`pi . ° ° 1. If 1{pi ¡ n ¡ 1, then X satisfies the `s-estimate with 1{s 1{pi ¡ pn ¡ 1q. ° 2. If 1{pi ¤ n ¡ 1, then X satisfies the c0-estimate. p q Proof. °Suppose wj is a finite block-diagonal sequence in X. We shall show } } }p} }q} that j wj wj s, with s as in the statement of the lemma. To this p q ¤ ¤ end, let Uij be coordinate projections on `pi for every 1 i n, such that wj U1j b ... b Unjwj, and for each i, UikUim 0 unless k m. Letting 1 {p ¡ q pi pi pi 1 , we see that § § ¸ § ¸ § } wj} sup §x wj, biξiy§. P 1 } }¤ ξi `p , ξi 1 j i j

11 ° 1 b } } ¤ } }pi ¤ Choose iξi with ξi 1, and let ξij Uijξi. Then j ξij 1, and § § § ¸ § ¸ § § ¸ § § ¸ x b y ¤ §x b y§ §x b y§ ¤ } } n } } § wj, iξi § wj, iξi wj, iξij wj Πi1 ξij . j j j j ° ° { { 1 ¡ { Now let 1 r 1 pi n 1 pi. By H¨older’sInequality,

¡ © ¡ © 1 ¸ ¹n ¨ 1{r ¹n ¸ 1{p r p1 i }ξij} ¤ }ξij} i ¤ 1. j i1 i1 j ° ° ° { ¤ ¡ ¤ n } } ¤ } } ¤ If 1 pi n 1, then r 1, hence j Πi1 ξij 1. Therefore, j wj } } p} }q ¡ maxj wj wj c0 . Otherwise, r 1, and ¡ © ¡ © ¡ © ¸ ¸ 1{s ¸ 1{r ¸ 1{s } } ¤ } }s p n } }qr ¤ } }s p} }q wj wj Πi1 ξij wj wj s, j j j j ° { ¡ { { ¡ where 1 s 1 1 r 1 pi n 1. ° } } ¥ p} }q 8 In a similar° fashion, we show that j wj wj s. For°s , the } } ¥ } } } }s inequality j wj maxj wj is trivial. If s is ﬁnite, assume j wj 1 (we are allowed to do so by scaling). Find norm one vectors ξij P `p1 so that ° i ξ U ξ , and }w } xw , b ξ y. Let γ }w }s{r. Then γr 1 ij ij i j j i ij ° j j j j ° 1 n 1 Π p {p Π p q } } l i l m1 l m l j γj wj . Further, set αij γj . An elementary calculation ° 1 ° n pi shows that γ Π α , and α 1. Let ξ α ξ . Then }ξ } 1 1, j i1 ij j ij i j ij ij i pi and therefore, ¸ ¸ ¸ ¸ } } ¥ x b y n x b y } } wj wj, iξi Πi1αij wj, iξij γj wj 1. j j j j

This establishes the desired lower estimate. ¤ ¤ 8 bp bp bp Lemma 3.7.°For 1 °pi , X `p1 `p2 ... `pn satisfies the `r-estimate, where 1{r 1{pi if 1{pi 1, and r 1 otherwise. Here, we interpret `8 as c0. 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 w P X, ( }w} sup |xx, wy| : x P ` 1 bq ... bq` 1 , }x} ¤ 1 p1 pn { 1 { (here, as before, 1 pi 1 pi 1). Abusing the notation somewhat, we denote by P the projection on the span of the first m basis vectors of both ` and ` 1 . im pi pi p qN P Suppose a finite sequence wk k1 X is block-diagonal,¨ or more precisely, wk pP ¡P qb...bpP ¡P q w for every k. Define the operator 1,mk 1,mk¡1 ° n,mk n,mk¡1 k ¨ N p ¡ q b b p ¡ q U on X by setting Ux k1 P1,mk P1,mk¡1 ... Pn,mk Pn,mk¡1 x.

12 ¦ We also use U0 to denote the similarly deﬁned operator on X . By “Tong’s ¦ trick” (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 ¸ ¸ ( } wk} sup |x wk, xy| : }x}X¦ ¤ 1 k k ¸ ( sup |xUp wkq, xy| : }x}X¦ ¤ 1 ¸ k ( sup |x wk,U0xy| : }x}X¦ ¤ 1 . k ° ° N { { 1 ¡ p ¡ Write U0x° k1 xk. By Lemma 3.6 there is an s (either 1 s 1 pi n 1q 1 ¡ 1{pi or s 8) }p}xk}q}s }U0x} ¤ }x} ¤ 1. Moreover, ¸ ¸ ¸ ¸ x wk,U0xy x wk, xky xwk, xky, k k k k and therefore, ¸ ¸ ( } wk} sup |xwk, xky| : }pxk}q}s ¤ 1 }p}wk}q}r. k k

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,X2q satisﬁes the c0-estimate. By renorming, we can assume that the FDD constants p qN of X1 and X2 equal 1. Suppose wk k1 is a block-diagonal° sequence, with p ¡ q p ¡ q } } ¥ wk P2,nk P2,nk¡1 wk P1,nk P1,nk¡1 . Let w k wk. Then w }p ¡ q p ¡ q} } } } } ¥ } } P2,nk P2,nk¡1 w P1,nk P1,nk¡1 wk , hence w maxk wk . To prove the reverse inequality (with some constant), pick a norm one ξ P X1, and let p ¡ q p ¡ q ξk P1,nk° P1,nk¡1 x. Then ηk wξk satisﬁes P2,nk P2,nk¡1 ηk ηk. Set η wξ k ηk. Denote by C1 (C2) lower (upper) p-estimate constants of X1 (resp. X2). Then ¸ ¸ p p p p p }wξ} }η} ¤ C1 }ηk} ¤ C2 }wk} }ξk} k ¸ k ¸ p p p p ¤ max }wk}pC2 }ξk} ¤ max }wk} C2C1} ξk} C2C1}ξ} . k k k k

1{p Taking the supremum over all ξ P UBpX1q, }w} ¤ pC1C2q 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 symmetric 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 E with a symmetric basis. The space Lpp0, 1q is not subprojective (see Section1), hence, by Proposition 2.1, neither is E. Furthermore, a tensor product of subprojective spaces (in fact, of Hilbert spaces) need not be subprojective. r r Proposition 3.9. There exists a tensor norm b, so that `2b`2 is not subprojective. 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 }a}br supt}pu b vqpaq}EpH,Kqu, where the supremum is taken over all contractions u : X Ñ H and v : Y Ñ K (H and K are Hilbert spaces, and EpH,Kq is the Schatten space; we identify H with its dual). Clearly br 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, r }pTX b TY qpaq}br ¤ }TX }}TY }}a}br . Thus, to prove that b determines a tensor norm, it suﬃces to establish that, for any z P X b Y , 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 any f P UBpX¦q and g P UBpY ¦q, consider the Ñ ÞÑ x y Ñ ÞÑ x y contractions° uf : X H : x f, x and vg : Y H : y g, y . Then, for b P b z i xi yi X Y , we have } } r ¥ p b qp q z b sup uf vg z EpHq fPUBpX¦q,gPUBpY ¦q § ¸ § § § x yx y } } q sup f, xi g, yi EpHq z b. P p ¦q P p ¦q f UB X ,g UB Y i

To prove the right hand side of (3.3), note that, for any two contractions u : X Ñ H and v : Y Ñ K,

}pu b vqpx b yq}E }ux b vy}E }ux}}vy} ¤ }x}}y}.

The triangle inequality establishes }z}br ¤ }z}bp for any z. If X and Y are Hilbert spaces, then for a P XbY we have }a}br }a}EpX¦,Y q. r Identifying `2 with its dual, we see that E embeds into `2b`2 as the space of r diagonal operators. As E is not subprojective, neither is `2b`2. Here is another wide class of non-subprojective spaces. Theorem 3.10. Let X be an inﬁnite dimensional Banach space. Then BpXq is not subprojective.

14 Proof. Suppose, for the sake of contradiction, that BpXq is subprojective. Fix a ¦ ¦ ¦ norm one element x P X . For x P X define Tx P BpXq : y ÞÑ xx , yyx. Clearly M tTx : x P Xu is a closed subspace of BpXq, isomorphic to X. Therefore, X is subprojective. By Proposition 2.10, we can find a subspace N M with an unconditional basis. We shall deduce that BpXq contains 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 from N to a further subspace W , isomorphic to `1 or c0, and complemented in X by a projection P . Embed BpW q isomorphically into BpXq by sending T P BpW q to PTP P BpXq, where P is a projection from X 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 into BpXq. Since BpW q KpW q,[12, Theorem 4(1)] shows that BpW q contains an isomorphic copy of `8. This rules out the subprojectivity of BpXq. Question 3.11. Suppose X is a subprojective Banach space. (i) Is RadpXq subprojective? (ii) If 2 ¤ p 8, must LppXq be subprojective? Question 3.12. Is a “classical” (injective, projective, etc.) tensor product of subprojective spaces necessarily subprojective? Note that the Fremlin tensor product 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 Section1).

4. Spaces of continuous functions In this section we deal with CpKq spaces, mostly separable ones. It is well known that, if K is a compact Hausdorff set, then CpKq is separable if and only if K 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 of Cr0, 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 if K is scattered.

15 4.1. Tensor products of CpKq In 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 Banach space. 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 of CpKq 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 topology generated by the open intervals pα, βq, as well as r0, βq and pα, λs. Abusing the notation slightly, we write Cpλ, Xq for Cpr0, λs,Xq. Suppose K is scattered. By [39, Chapter 8], K is isomorphic to r0, λs, for some countable limit ordinal λ. Fix a subprojective space X. We use induction on 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, hence q q cpXq cbX is isomorphic to c0pXq c0bX). 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 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, Xq is isomorphic to Cpµ ¡ 1,Xq ` X, hence, by Proposition 2.2, Cpµ ¡ 1,Xq is not subprojective. Let C0pµ, Xq tf P Cpµ, Xq : limνÑµ fpνq 0u. Clearly Cpµ, Xq is isomorphic to C0pµ, Xq`X, hence we obtain the desired contradiction by showing that C0pµ, Xq is subprojective. To do this, suppose Y is a subspace of C0pµ, Xq, so that no subspace of Y is complemented in C0pµ, Xq. For ν µ, define the projection Pν : Cpµ, Xq Ñ Cpν, Xq : f ÞÑ f1r0,νs. If, for some ν µ and some subspace Z Y , Pν |Z is an isomorphism, then Z contains a subspace complemented in X, by the induction hypothesis 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 an } ¡ } ¡1 arbitrary y1 from the unit sphere of Y . Pick ν1 µ so that y1 Pν1 y1 10 . P } } ¡2{ Now find a norm one y2 Y so that Pν1 y2 10 2. Proceeding further in the same manner, we find a sequence of ordinals 0 ν0 ν1 ν2 ..., and a ¡k sequence of norm one elements y1, y2,... P Y , so that }yk ¡ zk} 10 , where p ¡ q p q zk Pνk Pνk¡1 yk. The sequence zk is equivalent to the c0 basis, and the same is true for the sequence pykq. r P s p q Moreover, span zk : k N is complemented in C µ, X . Indeed, let ν supk νk. We claim that µ ν. If ν µ, then Pν is an isomorphism on

16 r P s p ¡ qp p qq span yk : k N , contradicting our assumption. Let Wk Pνk Pνk¡1 C0 X , and ﬁnd a norm one linear functional wk so that wkpzkq }zk}. Deﬁne ¸ ¨ p q Ñ p q ÞÑ p ¡ q Q : C0 µ, X C0 µ, X : f wk Pνk Pνk¡1 f zk. k }p ¡ q } Note that limk Pνk Pνk¡1 f 0, hence the range of Q is precisely the span of the elements zk. By Small Perturbation Principle, Y contains a subspace complemented in C0pµ, Xq. The above theorem shows that CpKqbqX is subprojective if and only if both CpKq and X are. We do no know whether a similar result holds for other tensor products. We do, however, have:

Proposition 4.2. Suppose K is a compact metrizable space, and W is either p `p (1 ¤ p 8) or c0. Then CpKqbW is subprojective if and only if K is scattered. Proof. Clearly, if K is not scattered, then CpKq is 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 that K r0, λs, where λ is a countable p N p ordinal. If λ is finite, then Cpλqb`p `8b`p is clearly subprojective. We handle the infinite case by transfinite induction on λ. The base is easy: if p λ ω, then Cpλq c, and Cpλq is isomorphic to c0b`p. The latter space is subprojective, by Corollary 3.3. Suppose, for the sake of contradiction, that λ is the smallest countable or- p dinal so that Cpλqb`p is not subprojective. Reasoning as before, we conclude p that λ is a limit ordinal. Furthermore, Cpλq C0pλq, hence C0pλqb`p is not subprojective. Denote by Qn : `p Ñ `p the projection on the first n basis vectors in `p, and K ¡ P p q let Qn I Qn. For f C0 λ and an ordinal ν λ, define Pν f χr0,νsf, K ¡ and Pν I Pν . p Suppose X is a subspace of C0pλqb`p which has no subspaces complemented p qbp p b q| in C0 λ `p. By the induction hypothesis, Pν I`p Y is strictly singular for p b q| any ν λ. Furthermore, IC0pλq Qn Y must be strictly singular. Indeed, p b q| otherwise Y has a subspace Z so that IC0pλq Qn Z is an isomorphism, whose b range is subprojective (the range of IC0pλq Qn is isomorphic to the sum of n copies of Cpλq, hence subprojective). Therefore, for any ν λ and n P N, p ¡ K b Kq| I Pν Qn 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 that }x ¡ pP K b QK qx } 10¡3i{2. By passing to a further subsequence, we i νi¡1 ni¡1 i }p b q } ¡3i{ can assume that Pνi Qni xi 10 2. Thus, by the Small Perturbation Principle, it suffices to show the following statement: If pyiq is a normalized p sequence is C0pλqb`p, so that there exist non-negative integers 0 n0 n1 n2 ..., and ordinals 0 ν0 ν1¨ ν2 ... λ, with the property that p ¡ q b p ¡ q r P s yi Pνi Pνi¡1 Qni Qni¡1 yi for any i, then Y span yi : i N is p contractively complemented in CpKqb`p.

17 Denote by X the span of all¨ x’s for which there exists an i so that x p ¡ q b p ¡ q Pνi Pνi¡1 Qni Qni¡1 x. Then Y is contractively complemented in p qbp C K `p. In fact,° we can define a contractive projection onto X as follows. N b Suppose first u j1 aj bj, with bi’s having finite support in `p. Then set °8 ¨ P u pP ¡P qbpQ ¡Q q u. Due to our assumption on the b ’s, i1 νi νi¡1 °ni ni¡1 ¨ i M p ¡ qbp ¡ q there exists M so that P u i1 Pνi Pνi¡1 Qni Qni¡1 u. To show that } } ¤ } } p qM P t¡ uM P u° u , define, for ε εi i1 1, 1 , the operator of multiplication p q P p q by i1 Mεiχrνi¡1 1,νis on C0 λ . The operator Vε B `p is defined similarly. Bot Uε and Vε are contractive. Furthermore, P u AveεpUε b Vεqu. Therefore, p we can use continuity to extend P to a contractive projection from C0pλqb`p onto X. To construct a contractive projection from X onto Y , we need to show that the blocks of X satisfy the ` -estimate. That is, if x pP ¡ P q b pQ ¡ ¨ p ° ° i νi νi¡1 ni q } }p } }p Qni¡1 xi for each i, then i xi i xi . To this end, use duality to p ¦ ¦ identify pC0pλqb`pq with Bp`p, `1pr0, λqq, q. P is the “block” projection onto the space of “block diagonal” operators which map the elements of `p supported p s p s 1 on ni¡1, ni onto the vectors° in `1 supported° on νi¡1, νi . If Ti s are the blocks } }p1 } }p1 { { 1 of such an operator, then i Ti ° i Ti ,° where 1 p 1 p 1. (see the } }p } }p proof of Corollary 3.3). By duality, i xi i xi . 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 spaces K1 and K2, the following are equivalent: (i) K1 p and K2 are scattered; (ii) CpK1qbCpK2q is subprojective.

4.2. Operators on CpKq Proposition 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 – that is, the operators for which there exists a constant C so that, for any x1, . . . , xn P X, ¡ © ¡ © ¸ 1{q ¸ 1{p q ¦ p }T xi} ¤ C sup |x pxiq| . ¦P p ¦q i x UB X i

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 scattered space, 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 q and 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 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 } ¤ πqpT q. Moreover, µ and T˜ can be selected in such a way that }T˜} πqpT q. As K is scattered, there P °exist distinct points°k1, k2,... 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- °8 1 P p q q tion, find N N so that iN 1 αi ε. Denote by u and v the operators of p q multiplication by χtk1,...,kN u and χtkN 1,kN 2,...u, respectively, acting on Lq µ . It is easy to see that rank u ¤ N, and }vj} ε. Then S T˜ 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 set p 1. We have a probability measure µ on K, and a factorization T T˜ j, where j : CpKq Ñ Lq1pµq is the formal identity, and T˜ : Lq1pµq Ñ X satisfies }T˜} ¤ cπq1pT q (c is a constant depending on q). In this case, the proof of Lemma 4.5 proceeds ° as for q-summing¨ operators, 8 1{q except that now, we need to select N so that c iN 1 αi ε. ° br b Proof of Proposition 4.4. Define the tensor norm by setting, for z° i xi P b } } p q P p ¦ q x y yi X Y , z br πqp z , where z B X ,Y is defined by zf i f, xi yi. ¦¦ Recall that, for any T , πqppT q πqppT q. Furthermore, if T is weakly ¦¦ ¦¦ compact, then T pX q Y , hence, by the injectivity of the ideal πqp, p q p rasq ras ¦¦ πqp T πqp T , where T is the astriction of T ° to an operator from ¦¦ P p q n ¦ b ¦ P ¦ X to Y . If T B X,Y has finite rank, write T i1 xi yi (xi X , and yi P Y ), and note that, by the above, πqppT q }T }br . ¦ r By Lemma 4.5, we can identify ΠqppCpKq,Xq with CpKq bX. By [35], p q¦ C K `1 (the canonical basis in `1 corresponds to the° point evaluation br b P b functionals), hence we can describe in more detail: for u° i ai xi `1 X, } } p q Ñ x y u br πqp u , where u : `8 X is defined by ub i b, ai xi. Note that ¥ ¦¦ Ñ ¦¦ Ñ κX u u˜ (κX°: X X is the canonical embedding), whereu ˜ : c0 X x y } } p q defined viaub ˜ i b, ai xi. Thus, u br πqp u˜ . r To finish the proof, we need to show (in light of Theorem 3.1) that `1b`q satisfies the `q estimate. To° this end,° suppose we have a block-diagonal sequence p qn } }q } }q ui i1, and show that i ui br i ui br . Abusing the notation slightly, we N N identify ui with an operator from `8 to `q (where N is large enough), and } ¤ } p¤q identify br with πqp° . ° } }q ¤ q } }q First show that i ui br c i ui br , where c is a constant (depending p qn t u R on q). We have disjoint sets Si i1 in 1,...,N so that uiej 0 for j Si. Therefore there exists a probability measure µi, supported on Si, so that

q q q q¡p p }uif} ¤ c πqppuiq }f}8 }f} 1 Lppµiq

N for any f P `8 (c1 is a constant). Now deﬁne the probability measure µ on

19 t u 1,...,N : ¸ ¨ ¸ q ¡1 q µ πqppuiq πqppuiq µi. i i P N For f `8, set fi fχSi . Then the vectors uifi are disjointly supported in `q, and therefore, ¸ ¸ ¸ q q q q q¡p p }p uiqf} }uifi} ¤ c πqppuiq }fi}8 }fi} 1 Lppµiq i i ¸ i q q¡p q p ¤ c }f}8 πqppuiq }fi} . 1 Lppµiq i An easy calculation shows that ¸ ¨ ¸ p q ¡1 q p }fi} πqppuiq πqppuiq }fi} , Lppµiq Lppµq i i hence ¸ ¸ ¨ ¸ q q q q¡p p }p uiqf} ¤ c πqppuiq }f}8 }fi} 1 Lppµiq i ¸i ¨ i q q q¡p p c πqppuiq }f}8 }f} . 1 Lppµq i ° ° ¨ { p q ¤ p qq 1 q Therefore, πqp i ui ° c i πqp u°i , for some universal constant c. } }q ¥ 1q } }q 1 Next show that i ui br c i ui br , where c is a constant. There exists N a probability measure µ on t1,...,Nu so that, for any f P `8, ¸ ¸ q q q q¡p p }p uifq} ¥ c πqpp uiq }f}8 }f} 2 Lppµq u i

For each i let αi }µ|S } N , and µi µi{αi (if αi 0, then clearly ui 0). i `1 N Then for any i, and any f P `8, ¸ ¸ q q q q q¡p p }uif} }p uiqpχS fq} ¤ c πqpp uiq αi}f}8 }f} , i 2 Lppµiq i i ° ° p q ¤ 1 1{q p q 1 hence°πqp ui c αi πqp ° i ui (c is a constant). As i αi 1, we conclude p qq ¤ 1q p q that i πqp ui c πqp i ui .

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 p q Hausdorff space (the base space), and Xt tPK is a family of Banach± spaces (the spaces 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, there exist an open neighborhood U Q t and y P X for which }xpsq ¡ ypsq} ε for 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 separable continuous vector field on K, so that, for every t P K, the fiber Xt is subprojective. Then X 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 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 Banach space. We then define the restriction operator Pν : X Ñ Xrνs. We denote by Qν : X Ñ Xν the operator of evaluation at ν. We say that a countable ordinal λ has Property P if, whenever X is a continuous separable vector field whose fibers are subprojective, then X is subprojective. Using transfinite induction, we prove that any countable ordinal has this property. The base of induction is easy to handle. Indeed, when λ is finite, then X embeds into a direct sum of (finitely many) subprojective spaces Xν . Now suppose, for the sake of contradiction, that λ is the smallest ideal failing Property P. Note that λ is a limit ordinal. Indeed, otherwise it has an immediate predecessor λ¡, and X embeds into a direct sum of two subprojective spaces – namely, Xrλ¡s and Xλ. Suppose Y is a subspace of X, so that no subspace of Y is complemented in X. 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- p q P quence if necessary, we° can assume that, Y has a basis yi i N, so that (i) for any p q } } ¥ | |{ } } ¡4i finite sequence αi , i αiyi maxi αi 2, and (ii) for any i, Qλyi 10 . ¡4i Consequently, for any y P spanryj : j ¡ is, }Qλy} 10 . Indeed, we can assume 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, } } ¤ | |} } ¤ ¡4j ¡4i Qλy j αj Qλyj 2 j¡i 10 10 . 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 P r ¤ s } } ¡4i vectors, so that (i) xj span yi : nj i nj 1 , (ii) Pνi xi 10 , and } } ¡4i (iii) Pνi 1 xi 10 . 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 P r s } } ¡4 a norm 1 vector x1 span y1, . . . , yn2¡1 so that Pν1 x1 10 . We have } } ¡4 ¡ } } ¡4 Qλx1 10 . By continuity, we can find ν2 ν1 so that Pν2 x1 10 .

21 P r s } } ¡8 Next ﬁnd a norm one x2 span yn2 , . . . , yn3¡1 so that Pν2 x1 10 . Proceed further in the same manner. We claim that the sequence pxiq is equivalent to the canonical basis in c0. 2 1 ¡ 2 Indeed, for each i let xi Pνi xi Pνi 1 xi, and xi xi xi . Since we are working } 1 } } } 1 with the sup norm, xi xi 1 for any i. Furthermore, the elements xi p q p q are° disjointly supported, hence, for any αi ﬁnite sequence of scalars αi , } 1 } | | i αixi maxi αi . By the triangle inequality, § § 8 § ¸ ¸ § ¸ ¸ } }¡} 1 } ¤ | |} 2} | | ¤ ¡4i ¡3 | | § αixi αixi § αi xi max αi 2 20 10 max αi , i i i i i i1 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 on A is a sum of pure linear functionals (f P A¦ is called pure if it belongs to an extreme ray of the positive cone of A¦). For equivalent descriptions of scattered C¦-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 compact Hausdorff space. If, in addition, the algebra is scattered, then its spectrum Aˆ is scattered as well [19, 20]. In fact, by the proof of [19, Theorem 3.1], Aˆ 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 Aˆ, with fibers of the form πpAq, for irreducible representations π. As A is CCR, the spaces 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 subprojective. 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 containing c0. Suppose, furthermore, that pznq CE is a normalized sequence, so that, for every k, limn }Qkzn} 0. Then, for any ε ¡ 0, CE contains sequences pz˜nq and p 1 q zn , so that:

1. pzñq is a subsequence of pznq. ° } ¡ 1 } 2. n zñ zn ε. p 1 q 3. zn lies in the subspace Z of CE , with the property that (i) Z is 3- isomorphic to either `2, E, or `2 È, and (ii) Z is the range of a projection of norm not exceeding 3.

p q p 1 q Proof. [3, Corollary 2.8] implies the existence of z˜n and zn , so that (1) and 1 b b b ¥ (2) are satisﬁed, and zk a E1k b Ek1 ck Ekk (k 2). Thus, 1 r b ¥ s zn Z Zr Zc Zd, where Zr span a E1k : k 2 (the row component), Zc spanrb b Ek1 : k ¥ 2s (the column component), and Zd (the diagonal component) 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 r b P ¥ s span ukEiivk Ekk : i N, k 2 . It remains to build contractive projections Pr, Pc, and Pd onto Zr, Zc, and Zd, respectively, so that ZcYZd ker Pr, ZrYZd ker Pc, and ZrYZc ker Pd. Indeed, then P Pr Pc Pd is a projection onto Zr Zc Zd, and the latter space is completely isomorphic to Z0 Zr `Zc `Zd. The spaces Zr, Zc, and Zd are either trivial (zero-dimensional), or isomorphic to `2, `2, and E, respectively. Pd is nothing but a coordinate projection, in the appropriate basis: " ¨ u E v b E k ` ¥ 2, i j P u E v b E k ii k kk d k ij ` k` 0 otherwise

(for the sake of convenience, we set u1 v1 I`2 ). Next construct Pr (Pc is 1 {} } dealt with similarly). If a 0, just take Pr 0.° Otherwise, let a a a , and P ¦ } } x 1y b ﬁnd f CE so that f 1 f, a . For x k,` xk` Ek`, deﬁne ¸ 1 Prx a b xf, x1`yE1`, `¥2 ° } }2 |x y|2 } } ¤ } } hence Prx E `¥2 f, x1` . It remains to show Prx x . This inequality is obvious when Prx 0. Otherwise, set, for ` ¥ 2,

xf, x y α ° 1` , ` p |x y|2q1{2 `¥2 f, x1`

23 ° ° ¨ { b b } } | |2 1 2 y I`2 `¥2 α`°E`1, and z I`2 E11. Then y 8 `¥2 α` 1 } } b z 8, and zxy `¥2 α`x1` E11. Therefore, @ ¸ D ¸ } } ¤ Prx E f, α`x1` α`x1` E ¥ ¥ ¸` 2 ` 2 b } } ¤ } }8} } } }8 } } α`x1` E11 E zxy E z x E y x E , `¥2 which is what we need.

Proof of Proposition 5.1. The space CE contains an isometric copy of E, hence the subprojectivity of CE implies that of E. To prove the converse, suppose E is subprojective, and Z0 is a subspace of CE , 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 Proposition 5.2, pznq has p 1 q a subsequence zn , 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, r 1 P s hence span zn : n N contains a subspace complemented in Z1, hence also in CE . As a consequence we obtain: Proposition 5.3. The predual of a von Neumann algebra A is subprojective if and only if A is atomic.

We say that A is atomic if any projection in it has° an atomic subprojection. p p qq By [31, Section 1], this happens if and only if A i B Hi 8. 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 establishes the “only if” implication of Proposition 5.3. Conversely, if A is purely atomic, then A¦ is isometric to a (contractively complemented) subspace of C1pHq, and the 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 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 a p-convex p-DH Bnaach lattice. Then every subspace, spanned by a disjoint sequence equivalent to the canonical basis of `p, is complemented.

24 Proof. Let pxkq X be a disjoint normalized sequence. Since X is p-DH, by passing to a subsequence, we can assume that (xkq is an `p basic sequence. Then, p in the p-concavification Xppq the disjoint sequence pxk q is an `1 basic sequence. ¦ p ¦ p Therefore, there exists a functional x P rpxk qs such that x pxk q 1 for all k. By the Hahn-Banach Theorem x¦ can be extended to a positive functional ¦ ¦ p 1 in Xppq . Define a seminorm }x}p px p|x |qq 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 abstract Lp-space, and the disjoint sequence Qpxkq is normalized. Therefore it is 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˜ onto ¡1 rQpxkqs. Then P Q PQ˜ is a projection from X onto rxks. Proposition 6.2. Let X be a p-convex, p-disjointly homogeneous Banach lattice pp ¥ 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 infinite dimensional separable subspace. Then there exists a complemented order ideal in 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 measure µ [21, p. 14] such that we have continuous embeddings

L8pµq X Lppµq L2pµq L1pµq.

Consequently, there exists a constant c1 ¡ 0 so that c1}x}p ¤ }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 6.1 M contains a copy of `p complemented in X.

Case 2. The norms }¤} and }¤}1 are equivalent on M. Thus, there exists c2 ¡ 0 so that, for any y P M,

c2}y}2 ¥ c2}y}1 ¥ }y} ¥ c1}y}p ¥ c1}y}2.

In particular, M is embedded into L2pµq as a closed subspace. The orthogonal projection from L2pµq onto M then deﬁnes a bounded projection from X onto M.

The preceding result implies that Lorentz space Λp,W p0, 1q is subprojective for p ¥ 2. Indeed, Λp,W p0, 1q is p-DH and p-convex pp ¥ 1q, see [13, Theorem 3] 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 of p q P the space of all ﬁnite sequences° x1, . . . , xn (with xi X), equipped with the }p q} }p | |pq1{p} norm x1, . . . , xn i xi , where ¸ ! ¸ ¸ ) 1 1 1 p |x |pq1{p sup | α x | : |α |p ¤ 1 , with 1. i i i i p p1 i i i 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 is subprojective. Proof. To show that any subspace Y Xp`pq has a further subspace Z, complemented 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 ¸ ¸ ¸ ¸ ¸ ¨ } } } p | |pq1{p } } | | } aijuij aij xi X ° sup αijaij xi X . (7.1) | |p1 ¤ i j i j αj 1 j r ¤ ¤ P s Let Pn be the canonical projection onto span uij : 0 i n, j N , and set K ¡ | Pn 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 singular for every n, ﬁnd a normalized sequence pyiq in Y , and 1 n1 n2 ..., so that }P y }, }P K y } 100¡i{2. By small perturbation, it remains to prove ni i ni 1 i the following: if y P K P y , then spanry : i P s contains a subspace, i ni ni 1 i i N complemented in Xp`pq. Further, we may assume that for each i there exists Mi so that we can write ¸ yi akjukj.

ni k¤ni 1,1¤j¤Mi

M For each k P rn 1, n s (and arbitrary i P ) find a finite sequence pα q i so ° i i 1 ° ° N kj j1 | |p1 | | p | |pq1{p p q Ñ that j αkj 1, and j αkjakj j akj . Define U : X `p X : u ÞÑ α a x . By (7.1), U is a contraction, and U| r P s is an isometry. kj kj kj k span yi:i N 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 metrizable scattered space, and 1 ¤ p 8, then CpKqp`pq is subprojective.

Recall that,° for a Banach space X, we denote by RadpXq the completion of P the ﬁnite sums n rnxn (r1, r2,... are Rademacher functions, and x1, x2,... X) in the norm of L1pXq (equivalently, by Khintchine-Kahane Inequality, in the 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, with the order induced by the basis pxiq). Indeed, by [28, Section 1.f], X is q-concave, p q p q for some q. An³ array° a°mn can be identiﬁed both with an element of Rad X 1 (with the norm } a r x }), and with an element of Xp` q (with the ° ° 0 m n mn n m 2 } p | |2q1{2 } norm m n amn xm ). Then » ¸ ¸ ¸ 1 ¸ 2 1{2 D} p |amn| q xm} ¤ } | amnrn|xm} 0 m» n »m n 1 ¸ ¸ 1 ¸ ¸ } | amnrnxm|} ¤ } amnrnxm} 0 0 » m n m n» 1 ¸ ¸ 1 ¸ ¸ q 1{q q 1{q ¤ p } amnrnxm} q ¤ Mq}p | amnrnxm| q } 0 0 m »n m n ¸ 1 ¸ ¸ ¸ q 1{q 2 1{2 ¤ Mq} p | amnrn| q xm} ¤ CMq} p |amn| q xm}, m 0 n m n where Mq is a q-concavity constant, while D and C come from Khintchine’s inequality. Thus, we have proved: Proposition 7.3. If X is a subprojective space with an unconditional basis and non-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 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 to RadpXq.

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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