Unexpected subspaces of tensor products

Abstract

We describe complemented copies of ℓ2\ell_2 both in C(K1)⊗^πC(K2)C(K_1)\hat{\otimes}_{\pi} C(K_2) when at least one of the compact spaces KiK_i is not scattered and in L1(μ1)⊗^ϵL1(μ2)L_1(\mu_1)\hat{\otimes}_{\epsilon} L_1(\mu_2) when at least one of the measures is not atomic. The corresponding local construction gives uniformly complemented copies of the ℓ2n\ell_2^n in c0⊗^πc0c_0\hat{\otimes}_{\pi} c_0. We continue the study of c0⊗^πc0c_0\hat{\otimes}_{\pi} c_0 showing that it contains a complemented copy of Stegall's space c0(ℓ2n)c_0(\ell_2^n) and proving that (c_0\ppi c_0)'' is isomorphic to ℓ∞(ℓ∞n⊗^πℓ∞n)\ell_\infty(\ell_\infty^n\hat{\otimes}_{\pi} \ell_\infty^n), together with other results. In the last section we use Hardy spaces to find an isomorphic copy of LpL_p in the space of compact operators from LqL_q to LrL_r, where 1<p,q,r<∞1<p,q,r<\infty and 1/r=1/p+1/q.

Full text

J. London Math. Soc. (2) 74 (2006) 512–526 C
e2006 London Mathematical Society
doi:10.1112/S0024610706023118
UNEXPECTED SUBSPACES OF TENSOR PRODUCTS
F´
ELIX CABELLO S ´
ANCHEZ, DAVID P´
EREZ-GARC´
IA
and IGNACIO VILLANUEVA
Abstract
We describe complemented copies of 2both in C(K1)ˆ
⊗πC(K2)whenatleastoneofthecompact
spaces Kiis not scattered and in L1(µ1)ˆ
⊗L1(µ2) when at least one of the measures is not atomic.
The corresponding local construction gives uniformly complemented copies of the n
2in c0ˆ
⊗πc0.
We continue the study of c0ˆ
⊗πc0showing that it contains a complemented copy of Stegall’s space
c0(n
2) and proving that (c0ˆ
⊗πc0) is isomorphic to ∞(n
∞ˆ
⊗πn
∞), together with other results.
In the last section we use Hardy spaces to find an isomorphic copy of Lpin the space of compact
operators from Lqto Lr,where1<p,q,r<∞and 1/r =1/p +1/q.
Introduction
In this paper we study subspaces of tensor products of Banach spaces, with emphasis
in the ‘Varopoulos space’ C(K1)ˆ
⊗πC(K2). The contents and organization of the
article are as follows.
Section 1 contains our main result: C(K1)ˆ
⊗πC(K2) has a complemented copy of
2as long as C(K1) is infinite dimensional and K2is non-scattered (Theorem 1.2).
Actually, we give a very explicit representation of 2,namely,if(fn)isequivalent
to the standard basis of c0in C(K1)and(gn)tothatof1in C(K2), then (fn⊗gn)
spans a complemented copy of 2inside C(K1)ˆ
⊗πC(K2).
Then we ‘predualize’ Theorem 1.2 to obtain also a complemented copy of 2in
L1(µ1)ˆ
⊗L2(µ2)whenµ2is not purely atomic.
As an application, we show that the 4-fold tensor product ˆ
⊗4
πc0lacks the uniform
approximation pro perty, while ( ˆ
⊗4
πC(K))lacks the approximation property if K
is non-scattered.
In Section 2 we study the bidual of c0ˆ
⊗πc0and we prove that it is isomorphic to
∞(n
∞ˆ
⊗πn
∞), but not to ∞ˆ
⊗π∞which, as we will see, is not a direct factor in
any dual Banach space.
Finally, Section 3 deals with injective tensor product of Lebesgue spaces. We use
Hankel-like operators on the Hardy classes to exhibit a copy of Lpin the space of
compact operators K(Lq,L
r)ifwearegiven1< p,q,r< ∞such that 1/p +1/q =
1/r. Complementation, however, will not follow.
Although we have presented our main results without any mention of the
Dunford–Pettis property (DPP), our research was motivated by the study of
the DPP in tensor products.
Recall that a Banach space is said to have the DPP if every weakly compact
operator defined on it is completely continuous. (An operator is called completely
continuous if it takes weakly Cauchy sequences to norm convergent sequences,
Received 4 October 2005.
2000 Mathematics Subject Classification 46B28.
Partially supported by BMF 2001-1240 and MTM 2004-02635.

SUBSPACES OF TENSOR PRODUCTS 513
and weakly compact if it sends the unit ball into a set with weakly compact closure.)
It is a consequence of the work of Dunford and Pettis that L1(µ) spaces have
the DPP, and later Grothendieck, who first isolated this property and named it,
proved that C(K) spaces also have it. It is well known that the DPP is stable by
complemented subspaces, that reflexive (infinite-dimensional) spaces never have the
DPP and that, if Xhas the DPP, so does X. The converse is not true, as shown
in [20]. It is interesting to remark that Stegall actually constructs a Banach space X
such that Xhas the DPP but X does not: X=c0(n
2) has the DPP because
X=1(n
2) even has the Schur property (weakly convergent sequences converge in
the norm); X =∞(n
2) lacks the DPP because it contains a complemented copy
of 2.
This example was essentially unique until it was proved recently that c0ˆ
⊗πc0
has the same behaviour. The space c0ˆ
⊗πc0has the DPP because its dual 1ˆ
⊗1has
the Schur property, as in Stegall’s example. On the other hand, Bombal and the
third named author of the present article showed that C(K1)ˆ
⊗πC(K2)doesnot
have the DPP whenever at least one of the compact spaces Kiis not scattered [2].
They proved this showing an explicit example of an operator :C(K1)ˆ
⊗π
C(K2)−→ 2which is not completely continuous. Soon afterwards, the authors
of [10] pushed the same ideas further to exhibit several instances of both projective
and injective tensor products of Banach spaces lacking the DPP. In particular,
they proved that (c0ˆ
⊗πc0) lacks the DPP by extending :∞ˆ
⊗π∞−→ 2to an
operator (c0ˆ
⊗πc0) −→ 2. Actually, every weakly compact operator on ∞ˆ
⊗π∞
extends to (c0ˆ
⊗πc0) as the former space is a locally complemented subspace of the
latter, see [4]. This provided essentially the second example known to the authors
of a Banach space with the DPP whose second dual lacks it. The starting point
of our research was to find out whether there is some relation between Stegall’s
example and c0ˆ
⊗πc0. Our main result implies that the above-mentioned operator
:C(K1)ˆ
⊗πC(K2)−→ 2is in fact a projection, while the corresponding local
argument and the results of Section 2 imply that c0ˆ
⊗πc0contains a complemented
copy of c0(n
2). Thus, essentially and as far as we know, Stegall’s example remains
as the sole available example of a Banach space having the DPP and whose bidual
lacks it. To tell the truth, we should mention that the space Tconstructed by
Talagrand in [23] has the DPP, its dual has the Schur property, and T lacks the
DPP, as shown by N´u˜nez in [17]. However, a close inspection of Treveals that it
also contains a complemented copy of Stegall’s c0(n
2).
Notation. The notation and terminology used throughout the paper will be
the standard in Banach space theory, as for instance in [8,9]. We have written
Xˆ
⊗πYfor the pro jective tensor product of two given Banach spaces Xand Y,
while the injective tensor product is denoted by Xˆ
⊗Y.
If Yis a complemented subspace of X, then the (relative) projection constant
of Yin Xis the infimum of the norms of all projections from Xonto Yand it is
denoted by λ(Y,X).
If Xand Yare isomorphic, then the (multiplicative) Banach–Mazur distance
between Xand Yis
d(X, Y )=inf{TT−1with T:X−→ Yan isomorphism}.

514 F. CABELLO S ´
ANCHEZ, D. P´
EREZ-GARC´
IA AND I. VILLANUEVA
The topological dual of Xis denoted by X;thevalueofx∈Xat x∈Xis
often denoted by x,x.IfT:X−→ Yis a (linear, bounded) operator, then the
adjoint is denoted by T.WeuseK(with or without subscripts) for a compact
(Hausdorff) space, while C(K) stands for the space of all continuous functions on
K(with values in the ground field), endowed with the supremum norm.
Further notation will be introduced when needed.
1. Complemented copies of 2in c0ˆ
⊗πL∞and 1ˆ
⊗εL1
In this section we show that 2lives complemented in c0ˆ
⊗πC(K)if(andonlyif)
Kis not scattered. We use well-known weak summability arguments to show that
certain operators are bounded, the proof ultimately relying on Grothendieck’s
theorem (or on Orlicz’s theorem).
Let 1 p∞. A sequence (xn)isweaklyp-summable in Xif, for every x∈X,
the scalar sequence (x,x
n)isinp. In that case we define its p-weak norm by
(xn)ω
p=sup
x1
n|x,x
n|p1/p
.
It is clear (and easy to prove) that (xn)isweaklyp-summable if and only if the
operator p−→ Xsending ento xnis bounded (here pdenotes the conjugate of p,
that is, 1/p+1/p =1;ifp=1,thenc0should replace ∞). In fact, the norm of
that operator is equal to (xn)ω
p.
An operator T:X−→ Yis p-summing when it takes weakly p-summable
sequences into p-summable sequences. In that case we have an estimate
∞

n=1 T(xn)p1/p
K(xn)ω
p,
for some constant Kindependent of xn. The smallest possible constant in the
preceding inequality is denoted πp(T). All of this can be seen in [9, Chapter 2].
We start with the following simple technical Lemma.
Lemma 1.1. Let (fn)be a weakly 2-summable sequence in C(K1)and
(gn)a bounded sequence in C(K2).Then(fn⊗gn)is weakly 2-summable in
C(K1)ˆ
⊗πC(K2).
Proof. We may assume gn1 for all nand (fn)ω
21. Let us compute
the 2-weak norm of the sequence (fn⊗gn)inC(K1)ˆ
⊗πC(K2). Let Bbe a norm
one functional on C(K1)ˆ
⊗πC(K2)andletT:C(K1)−→ C(K2) be the associated
operator, so that
B,f ⊗g=Tf,g(f∈C(K1),g ∈C(K2)).
By Grothendieck’s inequality, Tis 2-summing, with π2(T)KG,whereKGis the
Grothendieck constant; see [9, Theorem 3.5]. Therefore,

n|B,fn⊗gn|21/2
n|Tf
n,g
n|21/2

nTf
n21/2
π2(T)(fn)nw
2KG.
Hence, (fn⊗gn)ω
2KG.

SUBSPACES OF TENSOR PRODUCTS 515
Our first result follows suit.
Theorem 1.2. Let K2be a non-scattered compact space and K1an infinite
compact space. Then C(K1)ˆ
⊗πC(K2)has a complemented copy of 2.
Proof. First of all, let us fix an isomorphic embedding :c0−→ C(K1). Such a
clearly exists: just map the unit basis of c0into a normalized sequence of functions
in C(K1) having disjoint supports. Let κ:C(K1)−→ ∞be any extension of the
inclusion map c0−→ ∞through : this can be obtained applying the Hahn–Banach
theorem to each coordinate.
Also, since K2is not scattered, there is an operator ϕ:C(K2)−→ 2mapping a
bounded sequence onto the usual basis of 2. Indeed, C(K2) contains an isomorphic
copy of 1, and therefore there even exists a surjective operator C(K2)−→ 2by
[9, Corollary 4.16].
Now, define an operator :C(K1)ˆ
⊗πC(K2)−→ 2by
(f⊗g)=κ(f)·ϕ(g),
where the product is taken coordinatewise. It is clear that is well defined and
also that κϕ.
Let us construct a bounded right inverse for . Pick a bounded sequence (gn)
such that ϕ(gn)=en(in 2)andlet(fn) be the image under of the unit basis of c0.
By Lemma 1.1 the sequence (fn⊗gn) is weakly 2-summable in C(K1)ˆ
⊗πC(K2)and
so we can define a bounded operator σ:2−→ C(K1)ˆ
⊗πC(K2)takingσ(en)=
fn⊗gn.Moreover,
(σ(en)) = (fn⊗gn)=κ(fn)·ϕ(gn)=en,
so that ◦σ=12. This shows that σ◦is a projection on C(K1)ˆ
⊗πC(K2)whose
range is isomorphic to 2(and also that is onto, even if ϕis not).
In particular, c0ˆ
⊗π∞and c0ˆ
⊗πC[0,1] do contain 2as a complemented space.
Note that if both K1and K2are scattered, then C(K1)ˆ
⊗πC(K2) has the DPP (since
the dual space is isomorphic to 1(K1)ˆ
⊗1(K2) which has the Schur property [14])
and it cannot contain a complemented reflexive subspace.
It is clear from the proof of Theorem 1.2 that the complemented copy of 2in
C(K1)ˆ
⊗πC(K2) is just the subspace spanned by the sequence (fn⊗gn). Denote it
by H.Since◦σ=12we see that the Banach–Mazur distance between Hand
2is at most σ.Sinceσ◦is a projection onto Hwe see that the relative
projection constant of Hin C(K1)ˆ
⊗πC(K2) is also bounded by σ.Ifis an
isometric embedding then =ϕ, while (in view of the proof of Lemma 1.1)
σKGM,withM=sup
ngnC(K2).
Let us consider a concrete example in detail. For the sake of clarity the action
takes place in the Cantor group ∆ = {1,−1}Nfurnished with the product topology
and Haar measure. It is convenient to regard the elements of ∆ as functions
t:N−→ {1,−1}. In this setting the Rademacher functions are just evaluations:
rn(t)=t(n)(n∈N,u∈∆).
Clearly, rnare in C(∆) and so in Lp(∆) for all p. Consider the operator ρ:2−→
L1(∆) defined by ρ(en)=rn. It is clear that ρ=1:
ρ(x)L2(∆) =x2and ρ(x)L1(∆) ρ(x)L2(∆).

516 F. CABELLO S ´
ANCHEZ, D. P´
EREZ-GARC´
IA AND I. VILLANUEVA
Actually, ρis an isomorphic embedding, according to Khinchin inequality (see [15]
or [9, p. 227]) but we will not use this fact. The adjoint ρ:L∞(∆) −→ 2(which
is a quotient map) is given by
ρ(f),e
n=

∆
rn(t)f(t)dt (f∈L∞(∆)).
However, the sequence (rn) is orthonormal in L2(∆) and so ρ(rn)=en. Hence,
the sequence (en⊗rn) spans a subspace Hisomorphic to 2and complemented
in c0ˆ
⊗πL∞(∆). Actually, His even a (necessarily complemented) subspace of the
smaller space c0ˆ
⊗πC(∆). In this case, both d(H,
2)andλ(H,c
0ˆ
⊗πL∞(∆)) are
bounded by KG.See[9, p. 29] for numerical bounds of KG.
The same ideas can be used to obtain the following local version of Theorem 1.2.
Corollary 1.3. For ev ery n,thereisHn⊂n
∞ˆ
⊗π2n
∞such that d(Hn,
n
2)KG
and λ(Hn,
n
∞ˆ
⊗π2n
∞)KG.
Proof. Let ∆(n)={1,−1}n.Weembedn
2into L1(∆(n))usingthenRademach-
ers at our disposal. Then Hnis the subspace spanned in n
∞ˆ
⊗πL∞(∆(n))bythe
system (ek⊗rk)for1kn.Ofcourse,L∞(∆(n)) is isometric to 2n
∞.
Now, we state and prove a dual version of Theorem 1.2. Consider again the
operator :c0ˆ
⊗πL∞(∆) −→ 2given by
(f⊗g)=f·ρ(g)(f∈c0,g∈L∞(∆)).
Identifying 2with its own dual let us consider the adjoint operator
:2−→ (c0ˆ
⊗πL∞(∆)).
It is obvious that G=(2)isKG-isomorphic to 2and also that it is KG-
complemented in (c0ˆ
⊗πL∞(∆)). However, takes values in 1ˆ
⊗L1(∆), which is
a subspace of (c0ˆ
⊗πL∞(∆)). Indeed we have (en)=en⊗rn∈1ˆ
⊗L1(∆) since
(en),f ⊗g=en,f ·ρ(g)=f(n)en,ρ
(g)=en,frn,g=en⊗rn,f ⊗g
for f∈c0and g∈L∞(∆). Therefore, Gis a complemented subspace of 1ˆ
⊗L1(∆).
Corollary 1.4. Let µ1and µ2be two measures. If L1(µ1)is infinite-
dimensional and µ2is not purely atomic, then L1(µ1)ˆ
⊗L1(µ2)contains a
complemented copy of 2.
Proof. The hypotheses imply that L1(µ1) contains a complemented copy of
1and L1(µ2) contains a complemented copy of L1(∆). Hence, L1(µ1)ˆ
⊗L1(µ2)
contains 1ˆ
⊗L1(∆) (hence, 2) complemented.
Our next result is a ‘formal’ consequence of Theorem 1.2 thanks to the main
result of [4]. We need the notion of a locally complemented subspace. Suppose that
Yis a closed subspace of X.WesaythatYis locally complemented in Xif for
each finite-dimensional subspace E⊂Xthere exists an operator P:E→Ysuch
that Pis the identity on Y∩E,withPMfor some Mindependent of E.

SUBSPACES OF TENSOR PRODUCTS 517
A standard compactness argument shows that for dual spaces ‘locally comple-
mented’ implies ‘complemented’. The same is true for Banach spaces which are
isomorphic to complemented subspaces of a conjugate space.
Corollary 1.5. Let K1and K2be infinite compact spaces. Then
(C(K1)ˆ
⊗πC(K2)) contains a complemented copy of 2.
Proof. The main result of [4] states that C(K1) ˆ
⊗πC(K2) is a locally com-
plemented subspace of (C(K1)ˆ
⊗πC(K2)). Since ‘to be a locally complemented
subspace of’ is a transitive property and 2is reflexive (hence, a dual space),
it suffices to find a complemented copy of 2in C(K1) ˆ
⊗πC(K2). However, the
last space obviously contains a complemented copy of ∞ˆ
⊗π∞,andTheorem1.2
applies.
Remark 1. There is an alternate way to Theorem 1.2 which gives better
estimates (although less weakly 2-summable sequences). Just use the following
result instead of Lemma 1.1.
Lemma 1.6. Let (fn)be a weakly 1-summable sequence in C(K1)and
(gn)a bounded sequence in C(K2).Then(fn⊗gn)is weakly 2-summable in
C(K1)ˆ
⊗πC(K2).
Proof. The proof is in the same way as Lemma 1.1, but we use Orlicz theorem
(instead of Grothendieck inequality) as stated in [24, Theorem 11.11]: every weakly
1-summable sequence in L1(µ) is strongly 2-summable, with

nhn21/2
KO(hn)ω
1,
where KOKGand KO=√2 in the real case. Hence, if gn1 for all nand
(fn)ω
11, we obtain (fn⊗gn)ω
2√2, taking into account that C(K2)=L1(µ)
for some measure µ.
Remark 2. There are two long time open questions related to the results of
this section. On the one hand, it is not known whether c0ˆ
⊗πc0has the uniform
approximation property (UAP), where a Banach space Xhas the UAP if there is
a constant Kand a function f:N−→ Nsuch that, given E⊂Xwith dim E=k
there is T∈L(X) such that Tis the identity on E,withTKand dim T(X)
f(k)(see[6,11]).
On the other hand, it is also not known whether c0ˆ
⊗πc0is isomorphic to
c0ˆ
⊗πc0ˆ
⊗πc0=ˆ
⊗3
πc0. We first heard of the first problem from Aleksander
Pelczy´nski (private communication), and of the second problem from Joe Diestel
(private communication). Note that it follows from the associativity of the projec-
tive tensor product that if c0ˆ
⊗πc0is isomorphic to ˆ
⊗3
πc0, then it is also isomorphic
to ˆ
⊗4
πc0. In this context the following consequence of our results maybe useful.
Corollary 1.7. The space ˆ
⊗4
πc0does not have the UAP.

518 F. CABELLO S ´
ANCHEZ, D. P´
EREZ-GARC´
IA AND I. VILLANUEVA
Proof. It follows from Theorem 1.3 that ˆ
⊗4
πc0contains uniformly complemented
copies of the trace class spaces Sn
1=n
2ˆ
⊗πn
2, and now, using the results in [22],
it follows that ˆ
⊗4
πc0does not have the UAP.
With a very similar reasoning we have the following.
Corollary 1.8. If Kis non-scattered, then (ˆ
⊗4
πC(K))lacks the approxima-
tion property (AP).
Proof. By Theorem 1.2, ˆ
⊗4
πC(K) has a complemented copy of 2ˆ
⊗π2, therefore
(ˆ
⊗4
πC(K))has a complemented copy of (2ˆ
⊗π2)=L(2), and this space does not
have the AP.
Essentially the same questions as for the c0case are, as far as we know, open for
∞:wedonotknowwhether∞ˆ
⊗π∞is isomorphic to ∞ˆ
⊗π∞ˆ
⊗π∞, and we do
not know whether (∞ˆ
⊗π∞)has the AP.
2. A description of the bidual of c0ˆ
⊗πc0
In this section we compare the spaces (c0ˆ
⊗πc0),
∞(n
∞ˆ
⊗πn
∞)and∞ˆ
⊗π∞.
Given a sequence space Sand A⊂N, we put
S[A]={x∈S:x(n)=0forn/∈A}.
The following result generalizes [3, Lemma 2]. The observation that one can
(and must) use a Riemann integral to simplify the proof is due to Klaus Floret.
Lemma 2.1. Let Xand Yhave unconditional bases and let (An)and (Bn)be
two partitions of N.LetEbe the smallest closed subspace of Xˆ
⊗πYcontaining
every X[An]⊗Y[Bn].ThenEis complemented in Xˆ
⊗πY.
Proof. We may and do assume that the bases are 1-unconditional. We have an
action of the Cantor group ∆ on Xgiven by
(t◦x)(k)=t(n)x(k)(k∈An).
Similarly,
(t◦y)(k)=t(n)y(k)(k∈Bn)
defines an action of ∆ on Y. These actions are continuous. In particular, given zin X
(respectively, in Y), the map u−→ u◦zis continuous from ∆ to X(respectively,
to Y). Thus, we can define a bilinear map B:X×Y−→ Xˆ
⊗πYthrough the
(Riemann) integral
B(x, y)=

∆
(t◦x⊗t−1◦y)dt.
It is clear that B1. Let Pdenote the linearization of B,sothatP(x⊗y)=
B(x, y). We have P=B1. It is easily seen that Pis a projection of Xˆ
⊗πY
onto E.Infact,
P(ei⊗ej)=ei⊗ejif ei⊗ej∈E
0otherwise.
This completes the proof.

SUBSPACES OF TENSOR PRODUCTS 519
Proposition 2.2. The space (c0ˆ
⊗πc0) is isomorphic to ∞(n
∞ˆ
⊗πn
∞).
Proof. We prepare the ground for Pelczy´nski decomposition method by showing
that ∞(n
∞ˆ
⊗πn
∞) embeds as a complemented subspace of (c0ˆ
⊗πc0).
Let (An)and(Bn) be two partitions of Nand let Ebe the subspace of c0ˆ
⊗πc0
defined in Lemma 2.1. We already know that Eis the range of a contractive
projection on c0ˆ
⊗πc0. We claim that Eis isometric to c0(c0[An]ˆ
⊗πc0[Bn]). It is
completely obvious that the closure of c0[An]⊗c0[Bn]inc0ˆ
⊗πc0is isometric to
c0[An]ˆ
⊗πc0[Bn], and so we treat the latter space as a subspace of c0ˆ
⊗πc0.
Now, it suffices to show that, given un∈c0[An]ˆ
⊗πc0[Bn], one has
u1+...+uk=max
1nkun.
Put u=u1+...+uk.Thatuunfor 1 nkis clear since unis the image
of uunder a contractive projection. Hence,
umax
1nkun.
To prove the reversed inequality, let Pbe the adjoint of the averaging projection
describedinLemma2.1actingon(c0ˆ
⊗πc0).Notethat
Pf, x ⊗y=

∆
f(t◦x⊗t−1◦y)dt
for all f∈(c0ˆ
⊗πc0)and x, y ∈c0. Every functional in the range of Pvanishes on
the kernel of Pand, in fact, for x∈c0[An],y ∈c0[Bm]andf∈(c0ˆ
⊗πc0)one has
Pf, x ⊗y=δnmf(x⊗y).
Now, let fbe a norm one functional attaining the norm on uand let φ=Pf.
Clearly φ(u)=f(u)andso
u=φ(u1)+...+φ(uk).
Fix ε>0andchoosexn∈c0[An]andyn∈c0[Bn]sothat
|φ(un)|<φ(xn⊗yn)+ ε
kwith xn=1andyn=un.
This can be done because the norm of a bilinear functional is equal to the norm of
its linearization on the corresponding tensor product. We obtain
u|φ(u1)|+...+|φ(uk)|
<φ(x1⊗y1)+...+φ(xk⊗yk)+ε
=φ((x1+...+xk)⊗(y1+...+yk)) + ε
φx1+...+xky1+...+yk+ε
=y1+...+yk+ε
=max
nun+ε,
and since εis arbitrary we are done.
Taki ng An=Bnsuccessive intervals consisting of nnumbers, we conclude
that c0(n
∞ˆ
⊗πn
∞) is isometric to a 1-complemented subspace of c0ˆ
⊗πc0and, pass-
ing to the biduals, that ∞(n
∞ˆ
⊗πn
∞) is isometric to a 1-complemented subspace
of (c0ˆ
⊗πc0).

520 F. CABELLO S ´
ANCHEZ, D. P´
EREZ-GARC´
IA AND I. VILLANUEVA
Next, we show that (c0ˆ
⊗πc0) is a complemented subspace of ∞(n
∞ˆ
⊗πn
∞).
We regard 1ˆ
⊗1as the conjugate of c0ˆ
⊗πc0,sothat(c0ˆ
⊗πc0) is the conjugate
of 1ˆ
⊗1.Also,wetreatn
∞ˆ
⊗πn
∞as the conjugate of n
1ˆ
⊗n
1; and the latter space
as a subspace of 1ˆ
⊗1. Using these conventions we can define an operator κ:
(c0ˆ
⊗πc0) −→ ∞(n
∞ˆ
⊗πn
∞)taking
(κ(u))n=u|n
1ˆ
⊗n
1∈n
∞ˆ
⊗πn
∞.
Quite clearly, κ1. Let us construct a projection of ∞(n
∞ˆ
⊗πn
∞)onto(c0ˆ
⊗πc0)
through κ.LetVbe a nontrivial ultrafilter on Nand define π:∞(n
∞ˆ
⊗πn
∞)−→
(c0ˆ
⊗πc0) as
π((Bn)n)=weak*−lim
V(n)Bn,
where n
∞ˆ
⊗πn
∞is treated as a subspace of c0ˆ
⊗πc0(hence, of the bidual). We check
that πis a right inverse for κ(incidentally, this will show that κis an isomorphic
embedding). One only has to show that
B=weak*−lim
n→∞ Bn(B∈(c0ˆ
⊗πc0))
where Bnis given by
Bn(A)=B(Pn(A)) (A∈1ˆ
⊗1)
and Pnis the obvious pro jection of 1ˆ
⊗1onto n
1ˆ
⊗n
1. However, this is clear since
Pn(A)convergestoAstrongly in 1ˆ
⊗1as n→∞.
Thus, each of the spaces (c0ˆ
⊗πc0) and ∞(n
∞ˆ
⊗πn
∞) is isomorphic to a com-
plemented subspace of the other and the proof will be complete if we show that
(c0ˆ
⊗πc0) is isomorphic to its ∞-sum (see [25, Theorem 24]). It clearly suffices
to see that c0ˆ
⊗πc0is isomorphic to c0(c0ˆ
⊗πc0). That c0ˆ
⊗πc0is isomorphic to
its c0-sum follows from the most elementary version of Pelczy´nski’s method
(that in [18]): both spaces are isomorphic to their squares and, in view of Lemma 2.1,
each of them contains a complemented copy of the other.
As a by-product of Corollary 1.3, Lemma 2.1 and the proof of the preceding
proposition we obtain the following.
Corollary 2.3. The space c0ˆ
⊗πc0contains a complemented copy of c0(n
2).
Thus, Stegall’s example remains as the sole known example of a Banach space
having the DDP and whose bidual lacks it.
The space ∞ˆ
⊗π∞occupies, to some extent, an intermediate position between
c0ˆ
⊗πc0and its bidual. To be more precise, there are isometries
c0ˆ
⊗πc0

−−−−→∞ˆ
⊗π∞
κ
−−−−→(c0ˆ
⊗πc0) (1)
so that κ◦is the inclusion of c0ˆ
⊗πc0in its bidual. Here, is obtained tensoriz-
ing (twice) the inclusion of c0in ∞and κis defined through the Aron–Berner
(or Davie–Gamelin) extension in a much more general setting; see [4]. For our
current purposes it suffices to identify 1ˆ
⊗1with the dual of c0ˆ
⊗πc0:inthisway
(c0ˆ
⊗πc0) is the conjugate space of 1ˆ
⊗1and ∞ˆ
⊗π∞is just the space generated
by the functionals of the form
x⊗y−→ f, xg, y(x, y ∈1,f,g∈∞),

SUBSPACES OF TENSOR PRODUCTS 521
while c0ˆ
⊗πc0is obtained with f, g ∈c0. Under these representations, the arrows
of (1) are just inclusions.
Another possibility is to identify (c0ˆ
⊗πc0) with the space of all integral operators
I(1,
∞). Then ∞ˆ
⊗π∞=N(1,
∞), the subspace of nuclear operators.
In view of Proposition 2.2 one may wonder whether ∞ˆ
⊗π∞is also isomorphic to
(c0ˆ
⊗πc0). The following result answers the last question in the negative. Its proof
shows that the results in [4] are, to some extent, optimal.
Proposition 2.4. The space ∞ˆ
⊗π∞is not complemented in any dual space.
In particular it is not isomorphic to (c0ˆ
⊗πc0).
Proof. Recall that if a Banach space is complemented in some dual space, then
it is complemented in every space containing it as a locally complemented subspace.
It is proved in [4]that∞ˆ
⊗π∞is locally complemented in (c0ˆ
⊗πc0).Thus,
if ∞ˆ
⊗π∞were complemented in some dual space, it should be complemented
in (c0ˆ
⊗πc0). We complete the proof by showing that this is not the case.
For the remainder of this section we abbreviate ∞ˆ
⊗π∞to Nand (c0ˆ
⊗πc0)
to I.
Consider the whole exact sequence
0−−−−→Nκ
−−−−→Iπ
−−−−→I/N−−−−→0
and recall that in such a diagram the subspace is complemented in the middle space
if and only if the quotient map πadmits a (linear and bounded) section, that is,
there is an operator
S:I/N−→ I
such that π◦Sis the identity on I/N. We shall see that such an Scannot exist.
Consider the operator δ:∞−→ Igiven by
δ(f)=
∞

n=1
f(n)(en⊗en)(f∈∞),
where the summation of the series is performed in the weak* topology of I=
(1ˆ
⊗1). It is clear that δis an isometric embedding. A moment’s reflection shows
that N∩δ(∞)=δ(c0). Thus, the composition π◦δfactorizes through the quotient
∞/c0and we have a commutative diagram
0−−−−→Nκ
−−−−→Iπ
−−−−→I/N−−−−→0
δ

δ




0−−−−→c0−−−−→∞−−−−→∞/c0−−−−→0
where the rows are exact and the vertical arrows are isometric embeddings.
Therefore, I/Ncontains a subspace isometric to ∞/c0. However, the latter space
contains a further subspace isometric to c0(Γ), where Γ has the power of continuum.
Since there is no one-to-one operator from c0(Γ) into the dual of a separable Banach
space (such as 1ˆ
⊗1), we conclude that πcannot have a linear and bounded
section.
Remark 3. There are a number of intriguing questions about ∞ˆ
⊗π∞.
For instance, it is not known whether ∞ˆ
⊗π∞contains a complemented subspace

522 F. CABELLO S ´
ANCHEZ, D. P´
EREZ-GARC´
IA AND I. VILLANUEVA
isomorphic to c0or not. Note that Proposition 2.4 leaves the possibility of an
affirmative answer open. We know very little about how to embed c0into ∞ˆ
⊗π∞.
It is clear that if Sis a subspace isomorphic to c0and there is another subspace
M⊂∞ˆ
⊗π∞containing Sand isomorphic to ∞,thenSis not complemented in
∞ˆ
⊗π∞. On the other hand, we know that δ(c0) is uncomplemented in ∞ˆ
⊗π∞
(this follows, for example, from [5]). However, ∞ˆ
⊗π∞is very ‘thin’ around the
diagonal, as we now claim.
Claim 2.5. No subspace of ∞ˆ
⊗π∞containing δ(c0)is isomorphic to ∞.
Proof. With the same notation as before, suppose δ(c0)⊂M⊂N,withM
isomorphic to ∞.ThenM/δ(c0) is isomorphic to a subspace of N/δ(c0). It is known
that there is only one isomorphic embedding of c0into ∞, up to automorphisms of
∞(the Lindenstrauss–Rosenthal theorem [16]; there is a remarkably simple proof
in [7]). Hence, M/δ(c0) is isomorphic to ∞/c0.
On the other hand, applying the diamond lemma (see any book on basic algebra
or just chase the diagram below) to Nand δ(∞)inIand, bearing in mind that
δ(∞)∩N=δ(c0), we get the commutative diagram
00 0







0−−−−→c0−−−−→∞−−−−→∞/c0−−−−→0
δ

δ




0−−−−→N−−−−→N+δ(∞)−−−−→(N+δ(∞))/N−−−−→0







0−−−−→N/δ(c0) (N+δ(∞))/δ(∞)−−−−→0





00
where the rows and columns are exact sequences. However, ∞is injective amongst
Banach spaces and so the middle vertical sequence splits: this implies that
N+δ(∞) (hence, I) contains an isomorphic copy of N/δ(c0) (hence, of ∞/c0).
A contradiction.
Remark 4. Proposition 2.2 cannot be predualized (twice) to obtain an iso-
morphism between c0ˆ
⊗πc0and c0(n
∞ˆ
⊗πn
∞). Actually, the latter space is (as every
c0-sum of finite-dimensional spaces) isomorphic to a subspace of c0, while the former
is not. This follows from the fact (proved by Stehle in [21]) that there are subspaces
of c0ˆ
⊗πc0failing the DPP.
3. Lpas a space of compact operators
This section only has loose connections with the preceding sections. As a motiva-
tion of our closing result, let us consider numbers p, q and rso that 1/p+1/q =1/r.

SUBSPACES OF TENSOR PRODUCTS 523
H¨older’s inequality tells us that the bilinear operator
m:Lp×Lq−→ Lr
sending (f,g) to the product f·gis continuous: actually, m=1.Itiseasyto
see that it is also surjective: indeed, if h∈Lr,andwewriteh=u|h|with u
unitary, then f=u|h|r/p belongs to Lp,g=|h|r/q belongs to Lqand h=f·g.
Moreover, hr=fpgq. Thus, the linearization of mis a quotient operator
˜m:Lpˆ
⊗πLq−→ Lr. One may wonder whether ˜mhas a right inverse so that
there is a complemented copy of Lrin Lpˆ
⊗πLq, as is the case for the operator
C(K1)ˆ
⊗πC(K2)−→ 2appearing in Section 1. All we know is the following.
•In general, ˜mdoes not have a right inverse. Indeed, take p=q=2sothat
r=1.ThenL1cannot be a subspace of L2ˆ
⊗πL2because the latter space
has the Radon–Nikod´ym property (a hereditary property) while the former
lacks it.
•However, ˜mhas a right inverse in the purely atomic case, so that pˆ
⊗πqhas
a complemented copy of r(see [1]or[3, Proposition 1]).
From now on, we assume r>1. The main result of [3] implies that there is
always a local right inverse for ˜m(explicit constructions are also available). As the
reader can imagine we say that P:X−→ Zhas a local right inverse if, for each
finite-dimensional E⊂Z, there is an operator S:E−→ Xsuch that P◦S=1E,
with SMfor some Mindependent of E.
Thus, if we consider the whole exact sequence
0−−−−→ker ˜m−−−−→Lpˆ
⊗πLq
˜m
−−−−→Lr−−−−→0,(2)
we have that ker ˜mis locally complemented in Lpˆ
⊗πLq(see, for example, [12]).
Since our hypotheses imply that Lpˆ
⊗πLqis a dual space, namely the dual of
Lpˆ
⊗Lq=K(Lp,L
q), we see that the above sequence splits if and only if ker ˜m
is complemented in some dual space. Of course this would be the case if ˜mwere
weak* continuous. However, an operator T:Lpˆ
⊗πLq−→ Lris weak* continuous if
and only if T:Lr−→ L(Lp,L
q) takes values in K(Lp,L
q)and ˜m(f)iscompact
only if f= 0. At this juncture it is not clear whether K(Lp,L
q)containsacopy
of Lrwhen 1/p +1/q =1/r. The following result answers this question in the
affirmative.
Proposition 3.1. Let 1< p,q,r < ∞be such that 1/p +1/q =1/r.
Then K(Lq,L
r)contains a copy of Lp.
Note that we have relabelled the parameters involved. We will prove the analogous
statement about the Hardy classes. Recall that the Hardy space Hp=Hp(D)
consists of those analytic functions f:D→Csuch that
fHp=sup
0<r<11
2π

2π
0|f(reiθ)|pdθ1/p
<∞.
The space Hpis a Banach space for 1 p<∞. The boundary values
f(eiθ) = lim
r→1f(reiθ)
exist almost everywhere in Tand fHp=fLp(T). This implies that Hpis
isometric to Hp(T), the subspace spanned by the functions {einθ :n0}in Lp(T).
This correspondence sends Taylor series into Fourier series: if f=∞
n=0 anzn

524 F. CABELLO S ´
ANCHEZ, D. P´
EREZ-GARC´
IA AND I. VILLANUEVA
is the Taylor series of f∈Hp, then the boundary value of fhas Fourier series
∞
n=0 aneinθ .FromnowonwetreatHpas a subspace of Lp(T). The map
∞

−∞
akeikθ −→
∞

k=0
akeikθ
is often called the Riesz projection. It is bounded on Lp=Lp(T) if and only if
1<p<∞.InthiscaseHpis a complemented subspace of Lpand, actually, the
two spaces are linearly isomorphic. Let us prove the result.
Proposition 3.2. Let 1< p,q,r < ∞satisfy 1/p +1/q =1/r.ThenHpis
isomorphic to a closed subspace of K(Hq,H
r).
Proof. Every f∈Hpinduces a Hankel-like operator H(f):Hq→Hrdefined by
H(f)(g)=R(fg)(g∈Hq),
where Ris the Riesz projection on Lr.WehaveH(f)(g)rRL(Lr)fpgr
by H¨older inequality. Hence,
H(f):Hq→HrRL(Lr)fp,
and so
H :Hp→L(Hq,H
r)RL(Lr).
Let us see that H(f) is always compact. Taking f=zkand g=zl,wehave
H(zk)(zl)=R(zk−l)=zk−lfor lk
0forl>k.
Thus, H(f) has finite rank when f=zkwith k∈N. By linearity, H(f) has finite
rank when fis a polynomial and since polynomials are dense in Hpand His
continuous we see that H(f)iscompactforeveryf∈Hp.
It remains to verify that H(f)kfpfor some constant kindependent on
f∈Hp.Takef∈Hpwith fp= 1. Identifying fwith its boundary value, there
exists g∈Lpsuch that gp=1and
g|f=

T
f¯gdθ =1.(3)
However, g|f=R(g)|fand so we can assume that the gappearing in (3)
belongs to Hp,withgpRL(Lp).Nowsince1/p=1/q +1/rthere exist a
factorization g=g1g2,withg1∈Hq,g
2∈Hrand gp=g1qg2r. Without
loss of generality we may assume g2r=1andg1q=gp.Since
1=

T
f·¯g1·¯g2dθ =

TR(f·¯g1)·¯g2dθ
we obtain H(f)(g1)Hr=R(f¯g1)Lr1 and, thus,
H(f)K(Hq,Hr)1
RL(Lp)fp,
which completes the proof.

SUBSPACES OF TENSOR PRODUCTS 525
Remark 5. A Hankel matrix A=(A[i, j])i,j 0is a matrix such that A[i, j]=
αi+jfor some sequence (αn), that is, it has the form




α0α1α2...
α1α2... ...
α2... ... ...
... ... ... ...



.
It is easily seen that the matrix of H(f) with respect to the basis (zk)k0in Hq
and Hris a Hankel matrix, with
H(f)[i, j]= 
f(i+j),
where 
f(n) denotes the nth Fourier coefficient of (the boundary value of) f.
Conversely, if the matrix of a bounded operator T:Hq→Hrwith respect to
(zk)satisfiesT[i, j ]=τi+j, for some sequence τn,then
f=
∞

k=0
τkzk
belongs to Hpand T=H(f). We leave the details to the interested reader.
We do not know whether H(Hp) is complemented in K(Hq,H
r)when
1<p<∞. It should be mentioned that the subspace of Hankel operators is
uncomplemented in K(H2), as follows from a result of Kislyakov [13]. We remark,
however, that the Hankel operators in K(H2) form a subspace isomorphic to C(T)/A,
where Ais the disk algebra (see [19]), and not to the disk algebra itself. This is
due to the unboundedness of the Riesz projection on C(T).
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526 SUBSPACES OF TENSOR PRODUCTS
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F´elix Cabello S´anchez
Departamento de Matem´aticas
Universidad de Extremadura
Avenida de Elvas
06071 Badajoz
Spain
fcabello@unex.es
Ignacio Villanueva
Departamento de An´alisis
Matem´atico
Facu lta d d e Mat em´aticas
Universidad Complutense de Madrid
28040 Madrid
Spain
ignaciov@mat.ucm.es
David P´erez-Garc´ıa
´
Area de Matem´atica Aplicada
Departamento de Matem´aticas y
F´ısica Aplicadas y Ciencias de la
Naturaleza
Escuela Superior de Ciencias
Experimentales y Tecnolog´ıa
Universidad Rey Juan Carlos
Edificio Departamental II
28933 M´ostoles (Madrid)
Spain
david.perez.garcia@urjc.es