EMBEDDING `∞ INTO THE SPACE OF BOUNDED OPERATORS ON
CERTAIN BANACH SPACES
G. ANDROULAKIS, K. BEANLAND∗ , S.J. DILWORTH, F. SANACORY∗
Abstract. Sufficient conditions are given on a Banach space X which ensure that `∞ embeds in L (X), the space of all bounded linear operators on X. A basic sequence (en ) is said
to be quasisubsymmetric if for any two increasing sequences (kn ) and (`n ) of positive integers
with kn ≤ `n for all n, (ekn ) dominates (e`n ). If a Banach space X has a seminormalized
quasisubsymmetric basis then `∞ embeds in L (X).
1. Introduction
The famous open problem of whether there exists an infinite-dimensional Banach space on
which every (bounded linear) operator is a compact perturbation of a multiple of the identity,
is attributed to S. Banach (see related papers [G], [GM], and [S]). One of the reasons that
this problem has attracted a lot of attention is that if such a space X exists then by the
results of [AS] or [L], X provides a positive solution to the Invariant Subspace Problem for
Banach spaces, namely every operator on X has a non-trivial (i.e. different from zero and
the whole space) invariant subspace. Notice that if a space X satisfies the assumptions of
the above problem of Banach and if in addition X has the Approximation Property (AP)
and separable dual then L (X), the space of all operators on X endowed with the usual
operator norm, is separable (recall that a Banach space X has the AP if for every compact
subset K of X and every ε > 0 there exists a finite rank operator S ∈ L (X) such that
kx − Sxk ≤ ε for all x ∈ K). Thus if a reflexive Banach space X with a basis satisfies the
assumptions of the above problem of Banach then L (X) is separable (recall that a sequence
(en )n of non-zero elements of a Banach space is a basic sequence if there exists a constant
PM
P
B such that k N
n=1 an en k for all integers N ≤ M and all sequences of
n=1 an en k ≤ Bk
scalars (an ); the smallest such constant B is called the basis constant of (en ); the sequence
(en )n is called a basis for X if it is a basic sequence whose linear span is dense in X). In this
paper we provide sufficient conditions on a Banach space X which imply that `∞ embeds
in L (X) (which we denote by “`∞ ,→ L (X)”) and hence that L (X) is non-separable. The
question whether L (X) is non-separable for every infinite-dimensional Banach space X is a
well-known open problem.
In Section 2 we introduce the notion of a quasisubsymmetric basic sequence, give examples,
and prove our main results. We say that a basic sequence (en ) is quasisubsymmetric if for
any two increasing sequences (kn ) and (`n ) of positive integers with kn ≤ `n for all n, we
have that (ekn ) dominates (e`n ). One of our main results (Theorem 2.4) is that if a Banach
space X has a seminormalized quasisubsymmetric basis then `∞ embeds in L (X). Thus a
reflexive Banach space X with a seminormalized quasisubsymmetric basis cannot satisfy the
1991 Mathematics Subject Classification. Primary: 46B28, Secondary: 46B03.
∗
The present paper is part of the Ph.D theses of the second and fourth author which are prepared at the
University of South Carolina under the supervision of the first author.
1
2
G. Androulakis, K. Beanland, S.J. Dilworth, F. Sanacory
assumptions of the above problem of Banach. Moreover, in Proposition 2.3 we prove that
if a Banach space X has a seminormalized basis which dominates all of its subsequences
then L (X) is non-separable (we do not know whether `∞ ,→ L (X) in this case). Finally
we observe how our results provide sufficient conditions on a Banach space X which ensure
that `1 embeds complementably into the space of nuclear operators of X.
In the rest of Section 1 we review some known results about embedding `∞ into the space
of operators. If X, Y are Banach spaces then denote by L (X, Y ) the space of all bounded
operators from X to Y and K (X, Y ) the space of all compact operators from X to Y .
N.J. Kalton [K, proof of Theorem 6 (iii) ⇒ (ii)] proved that if X is an infinite-dimensional
separable Banach space and Y is any Banach space then:
(a) if c0 embeds in L (X, Y ) then `∞ embeds in L (X, Y ),
(b) if c0 embeds in Y then `∞ embeds in L (X, Y ).
Later M. Feder [F, page 201] noticed that by the result of B. Josefson [J] and A. Nissenzweig
[N], the proof of Kalton for the above results works without the assumption that X is
separable. Notice that the result (a) of Kalton as extended by Feder generalizes the result of
C. Bessaga and A. Pelczynski [BP] stating that if c0 embeds in X ∗ then `∞ embeds in X ∗ .
Recall that a sequence (en )n of non-zero elements of a Banach
an uncondiP space is calledP
tional basic sequence if there exists a constant C such that k n∈F an en k ≤ Ck n an en k for
every finitely supported sequence of scalars (an ) and
Pevery finite subset
P F of the integers, or
equivalently, there exists a constant K such that k an en k ≤ Kk bn en k for every finitely
supported sequences of scalars (an ) and (bn ) with |an | ≤ |bn | for all n. The smallestPsuch
constant K is called the unconditionality constant of (eP
n ). Also recall that a series
n xn
in a Banach space is said to converge unconditionally if n an xn converges for any bounded
sequence of scalars (an ). A basic sequence which is not unconditional is called a conditional
basic sequence. A.E. Tong and D.R. Wilken [TW] proved that if X, Y are Banach spaces, Y
has an unconditional basis and there is a noncompact operator in L (X, Y ) then `∞ embeds
in L (X, Y ). The main focus of the paper of Tong and Wilken is the well-known open problem of whether K (X, Y ) $ L (X, Y ) implies that K (X, Y ) is not complemented in L (X, Y ).
They prove that the answer is affirmative if Y has an unconditional basis.
A.E. Tong [T] proved that if X has an unconditional basis, Y is any Banach space, and
there exists a noncompact operator in L (X, Y ∗ ) then c0 embeds in L (X, Y ∗ ). This result
was generalized by Kalton [K] as follows: Let X be a Banach space with an unconditional
finite-dimensional expansion of the identity (i.e.
P there is a sequence (An ) ⊆ L (X) of finiterank operators such that for every x ∈ X, x = ∞
n=1 An x unconditionally), and let Y be any
Banach space. If there is a noncompact operator in L (X, Y ) then `∞ embeds in L (X, Y ).
Moreover, the same assumptions imply that K (X, Y ) is not complemented in L (X, Y ).
G. Emmanuele [E2] proved the following result. Assume that X and Y satisfy one of the
following two assumptions:
(i) X is a L∞ space and Y is a closed subspace of a L1 space, or
(ii) X = C[0, 1] and Y is a space with cotype 2.
If there is a noncompact operator in L (X, Y ) then `∞ embeds in L (X, Y ). Moreover, under
the same assumptions K (X, Y ) is not complemented in L (X, Y ). Related results are also
found in [E1].
Embedding `∞ into the space of all Operators on Certain Banach Spaces
3
2. Quasisubsymmetric Sequences
In this section we introduce the notion of basic quasisubsymmetric sequences and then
give examples and study their properties. The main results are Theorems 2.4 and 2.9 and
Corollary 2.14. Let c00 denote the linear space of finitely supported sequences. If (xn ) and
(yn ) are two basic sequences
P in a Banach
P space we say that (xn ) dominates (yn ) if there
exists C > 0 such that k an yn k ≤ Ck an xn k for all (an ) ∈ c00 .
Definition 2.1. A basic sequence (en )n∈N in some Banach space X is said to be quasisubsymmetric if for any two increasing sequences (kn ) and (`n ) of positive integers with kn ≤ `n
for all n, we have that (ekn ) dominates (e`n ), i.e. there exists a constant S(kn ),(`n ) > 0 such
that
X
X
an e`n ≤ S(kn ),(`n ) an ekn for all (an ) ∈ c00 .
Notice that a quasisubsymmetric sequence is automatically bounded. While in this definition the constant S(kn ),(`n ) depends on the two subsequences, the dependence can be removed
if (en ) is seminormalized (i.e. there exist positive constants d, D such that d ≤ ken k ≤ D
for all n), as the next result shows.
Lemma 2.2. Let (en ) be a seminormalized quasisubsymmetric basic sequence. Then there
exists a constant S > 0 such that for any two increasing sequences (kn ) and (`n ) of positive
integers with kn ≤ `n for all n, we have that
X
X
(1)
an ekn for all (an ) ∈ c00 .
an e`n ≤ S Proof. We will assume, to obtain a contradiction, that for all S > 0 there exist (kn ) and (`n )
increasing sequences of positive integers with kn ≤ `n for all n, and there exists (an ) ∈ c00
for which (1) is not valid.
Claim: For every S > 0 and every N ∈ N there exists (an ) ∈ c00 with an = 0 for all n < N
and there exist two increasing sequences (kn ) and (`n ) of positive integers with kn ≤ `n for
all n, such that
X
X
an e`n > S an ekn .
To prove this we fix S and N and let S̃ = S(1 + B) + (N − 1)2BD/d, where B is the basis
constant of the basic sequence (en ) and d ≤ ken k ≤ D for all n. By assumption there exists
(an ) ∈ c00 and two increasing sequences (kn ) and (`n ) of positive integers with kn ≤ `n for
all n, such that
X
X
a
e
a
e
>
S̃
n kn .
n `n P
We can assume without loss of generality that k an ekn k = 1. By the triangle inequality,
(2)
k
∞
X
an ekn k ≤ 1 + B.
n=N
Also,
(3)
|an | ≤ 2B/d for all n.
4
G. Androulakis, K. Beanland, S.J. Dilworth, F. Sanacory
Furthermore,
k
∞
X
n=N
an e`n k ≥ k
∞
X
an e`n k − k
n=1
∞
X
> S̃k
N
−1
X
an e`n k
n=1
an ekn k − (N − 1)2BD/d (by (3))
n=1
≥ S(1 + B) + (N − 1)2BD/d − (N − 1)2BD/d
∞
X
≥ Sk
an ekn k (by (2)),
n=N
which finishes the proof of the claim.
(1)
Let N1 = 1 and S1 = 2. By the claim there exists an (an ) ∈ c00 and two increasing
(1)
(1)
(1)
(1)
sequences (kn ) and (`n ) of positive integers with kn ≤ `n for all n, such that
X
X
(1)
a(1)
e
>
2
a
e
(1)
(1)
n kn = 2.
n `n
(1)
(1)
Let N2 = max(`n ) + 1 where the maximum is taken over all n in the support of (an ) and
(2)
(2)
let S2 = 4. By the claim there exists an (an ) ∈ c00 with an = 0 for all n < N2 and two
(2)
(2)
(2)
(2)
increasing sequences (kn ) and (`n ) of positive integers with kn ≤ `n for all n, such that
X
X
(2)
a(2)
e
>
4
a
e
(2)
(2)
n `n
n kn = 4.
(m−1)
Continue in this manner with Nm = max(`n
) + 1 where the maximum is taken over all n
(m−1)
(m)
(m)
m
in the support of (an
) and let Sm = 2 . Notice that Nm ≤ n ≤ kn ≤ `n < Nm+1 for
(m)
all m ∈ N and for all n in the support of (ai ). Thus we can string together the sequences
(m)
(m)
(m)
(m)
(kn ) and (`n ) and produce increasing sequences by setting an = an /2m , kn = kn and
P
(m)
(m)
an ekn converges
`n = `n for all m ∈ N and for all n in the support
of (ai ). Note that
P
and has norm at most equal to 2. However
an e`n does not converge, contradicting the
fact that (en ) is quasisubsymmetric.
The authors do not know whether every seminormalized quasisubsymmetric sequence can
be equivalently renormed so that the constant S in the statement of Lemma 2.2 can be taken
to be equal to 1. If this is true then the proof of Theorem 2.4 can be slightly simplified.
Some examples of spaces with quasisubsymmetric basic sequences follow.
(a) If any two subsequences of a basic sequence (en ) are naturally isomorphic then (en )
is quasisubsymmetric; (the examples (b) and (e) below satisfy this condition). In
particular any subsymmetric sequence is quasisubsymmetric.
(b) There exist quasisubsymmetric basic sequences which are conditional (refer to the
Introduction for the definition of conditional basic sequences). Such an example
P is the
summing basis, which is the completion of c00 equipped with the norm k an en k =
P
supN | N
n=1 an |.
(c) The sequence of the biorthogonal functionals of the basis of Schreier’s space (see e.g.
[CS]) is quasisubsymmetric. We say that a finite subset E of N is a Schreier set if
the cardinality of E is less than or equal to the minimum of E. For a = (an ) ∈ c00
Embedding `∞ into the space of all Operators on Certain Banach Spaces
5
we define the norm
kak =
X
sup
E is a Schreier set
|an |.
n∈E
Schreier’s space is the completion of c00 equipped with this norm.
(d) The sequence of the biorthogonal functionals of the basis of Tsirelson’s space (see
[FJ] and [CS]) is quasisubsymmetric. There exists a norm k · k satisfying
(
)
k
X
1
kxk = max kxkc0 , sup
kEj xk
2
j=1
for all x ∈ c00 , where k · kc0 denotes the norm of c0 and the supremum is taken over all
sequences of sets (Ej )kj=1 such that max Ej < min Ej+1 and (min Ej )kj=1 is a Schreier
set. Also Ej x denotes the natural projection of x on Ej . Tsirelson’s space is the
completion of c00 equipped with this norm.
(e) The James space has a boundedly complete basis when the norm is defined as

!2 1/2
pk+1 −1
n
X
X
xi 
kxk = sup 
k=0
i=pk
where x = (xi ) and the supremum is taken over all positive integers n and all sequences of integers such that 0 = p0 < p1 < . . . < pn+1 . Since the basis is boundedly
complete [FG, page 50] c0 6,→ J. Also it is conditional [FG, pages 49-50]. Obviously
every two subsequences of the basis are isometric. Thus our Theorem 2.4 gives an
easy way to see that `∞ ,→ L(J).
One of our main results of this section is Theorem 2.4 which gives sufficient conditions on
X for `∞ ,→ L(X). If, however, we are only interested in L(X) being non-separable, then
the assumptions of Theorem 2.4 can be weakened as the next result shows.
Proposition 2.3. If a Banach space X has a seminormalized basis which dominates its
subsequences then L (X) is non-separable.
Proof. Let (en ) be a seminormalized basis for X such that it dominates its subsequences and
let C be its basis constant (see the Introduction for the definition of the basis
If
Pconstant).
∗
e
(x)e
(kn ) is an increasing sequence in N then define T(kn ) : X → X by T(kn ) (x) = ∞
kn .
n=1 n
By our assumption each T(kn ) is bounded, i.e. T(kn ) ∈ L(X). Let (kn ), (mn ) be two different
increasing sequences in N. Then there exists some j ∈ N such that kj 6= mj . Thus
∞ X
e
e
j
j
kT(kn ) − T(mn ) k ≥ e∗n
ekn − e∗n
emn kej k
kej k
n=1
=
kekj − emj k
1
inf n ken k
kekj − emj k ≥
≥
> 0.
kej k
supn ken k
C supn ken k
We now present our first main result.
Theorem 2.4. If a Banach space X has a seminormalized quasisubsymmetric basis then
`∞ ,→ L (X).
6
G. Androulakis, K. Beanland, S.J. Dilworth, F. Sanacory
In the proof of Theorem 2.4, we will use the following remark due to Tong and Wilken
[TW, Proposition 4] which is mentioned in the Introduction.
Remark 2.5. Let (wn ) be an unconditional basic sequence in a Banach space X, and assume
that there exists a noncompact operator in L (X, [(wn )]). Then `∞ ,→ L (X).
Proof of Theorem 2.4. Let (en )n∈N be a seminormalized quasisubsymmetric basis for X.
Since (en ) is bounded, by the `1 theorem of H.P. Rosenthal [R] there exists a subsequence
(ekn )n∈N which is either equivalent to the unit vector basis of `1 or it is weakly Cauchy.
In the first case, for some K and for all (an ) ∈ c00 we have
∞
∞
∞
X
X
KX
1 X
|an | ≤ k
an ekn k ≤ k
an en k ≤ sup ken k
|an |
S
S n=1
n
n=1
n=1
where S is the constant provided by Lemma 2.2. Thus (en ) is equivalent to the unit vector
basis of `1 , hence it is an unconditional basis for X and therefore (e∗n ⊗ en )∞
n=1 is equivalent
∗
∗
∗
to the unit vector basis of c0 (where for x ∈ X and x ∈ X , x ⊗ x denotes the operator on
X defined by (x∗ ⊗ x)(y) = x∗ (y)x for every y ∈ X).
In the second case we assume that (ekn )∞
n=1 is weakly Cauchy. Define the difference sequence (zn ) of (ekn ) by zn = ek2n − ek2n−1 . Then (zn )n∈N is weakly null. Since (en ) is a
seminormalized basic sequence, we have that (zn ) is also a seminormalized basic sequence,
say 0 < d ≤ kzn k ≤ D < ∞ for all n. We claim that there exists a subsequence of (zn )n∈N
which is unconditional. In order to prove this, we use an argument of A. Brunel and L.
Sucheston [BS]. First by Mazur’s theorem [W, Corollary II.A.5] there exist intervals of the
natural numbers
j means max(Ik ) < min(Ij )) and there exist
P I1 < I2 < ... (where Ik < IP
ci ≥ 0 with i∈Ij ci = 1 for all j, such that k i∈Ij ci zi k → 0 as j → ∞.
By taking a subsequence and renaming we can assume without loss of generality that
X
1
(4)
k
ci zi k < n .
2
i∈I
n
Let (an ) ∈ c00 . Let F be a finite subset of positive integers. For n ∈ N let mn := min In . We
will show
∞
X
2C X
(5)
k
an zmn k ≤ (S +
)k
an zmn k for all (an ) ∈ c00
d
n=1
n∈N\F
where C is the basis constant of (zn ) and S is the constant provided by Lemma 2.2 for the
basic sequence (en ). First, note since (en ) is quasisubsymmetric, that for all sequences (`n )
of positive integers with `n ∈ In (for all n), and for all (an ) ∈ c00 ,
k
∞
X
an z`n k ≤ Sk
n=1
∞
X
an zmn k.
n=1
Let (an ) ∈ c00 . Assume that F = {t1 , t2 , . . . , ts }. For any (i1 , i2 , . . . , is ) with in ∈ Itn for
n = 1, . . . , s, define
s
X
X
V (i1 , i2 , . . . , is ) =
an zmn +
atn zin .
n∈N\F
n=1
Embedding `∞ into the space of all Operators on Certain Banach Spaces
7
Thus if we sum over all (i1 , i2 , . . . , is ) with in ∈ Itn for n = 1, . . . , s, we obtain
X
(6)
ci1 · · · cis V (i1 , . . . , is ) =
X
an zmn +
s
X
atn
n=1
n∈N\F
X
ci zi .
i∈Itn
Apply Lemma 2.2 for kn = mn if n ∈ N, `n = mn if n ∈ N\F and `n = in if n ∈ F , to obtain
kV (i1 , . . . , is )k ≤ Sk
∞
X
an zmn k.
n=1
Thus
X
k
X
ci1 · · · cis V (i1 , . . . , is )k ≤
in ∈Itn for n=1,...,s
ci1 · · · cis kV (i1 , . . . , is )k
in ∈Itn for n=1,...,s
X
≤
(7)
ci1 · · · cis Sk
in ∈Itn for n=1,...,s
= Sk
∞
X
∞
X
an zmn k
n=1
an zmn k.
n=1
Also note
|an |d ≤ kan zmn k = k
(8)
n
X
i=1
ai zmi −
n−1
X
ai zmi k ≤ 2Ck
i=1
∞
X
ai zmi k.
i=1
Hence by (6),
k
X
ci1 · · · cis V (i1 , . . . , is )k ≥ k
in ∈Itn for n=1,...,s
X
n∈N\F
≥k
(9)
X
n∈N\F
an zmn k −
s
X
|atn | k
n=1
2C
an zmn k −
k
d
X
ci zi k
i∈Itn
∞
X
i=1
s
X
1
ai zmi k
2tn
n=1
(by (4) and (8))
∞
X
2C X
≥k
an zmn k −
k
an zmn k.
d n=1
n∈N\F
So by (7) and (9) we obtain (5) which proves that (zmn ) is unconditional.
Let P ∈ L(X, [(zmn )]) be defined by
!
∞
∞
X
X
P
an en =
an zmn .
n=1
n=1
8
G. Androulakis, K. Beanland, S.J. Dilworth, F. Sanacory
Notice that P is bounded since
!
∞
X
P
a
e
n n ≤ ka1 ek2m1 + a2 ek2m2 + · · · k + ka1 ek2m1 −1 + a2 ekm2 −1 + · · · k
n=1
≤ 2Sk
∞
X
an en k.
n=1
Since P is a noncompact operator and (zmn ) is unconditional, by Remark 2.5, we have
`∞ ,→ L(X).
By observing that for a Banach space X the map from L (X) to L (X ∗ ) given by T 7→ T ∗ ,
where T ∗ is the adjoint operator, is an isometric embedding, Theorem 2.4 gives the following
corollary.
Corollary 2.6. If a Banach space X has a seminormalized quasisubsymmetric basis then
`∞ ,→ L(X ∗ ).
The proof of Theorem 2.4 gives the following two corollaries:
Corollary 2.7. If (en )n∈N is a seminormalized quasisubsymmetric basic sequence then either
(en )n∈N is equivalent to the unit vector basis of `1 or there exists a subsequence (ekn ) such
that (ek2n − ek2n−1 )n∈N is unconditional.
Corollary 2.8. If (en )n∈N is a seminormalized quasisubsymmetric weakly null basic sequence
then there exists a subsequence (ekn ) of (en ) which is unconditional.
Theorem 2.9. If a Banach space X has a seminormalized basis (en ) and the sequence (e∗n )
of the biorthogonal functionals is quasisubsymmetric then `∞ ,→ L (X).
Proof. By results of Kalton [K] mentioned in the Introduction, it suffices to show that c0 ,→
L (X). By Corollary 2.7 we separate two cases:
If (e∗n )n∈N is equivalent to the unit vector basis of `1 , we know that X = [(en )n∈N ] is
isomorphic to c0 . Therefore c0 ,→ L (X).
If (e∗k2n − e∗k2n−1 )n∈N is unconditional, let K be its unconditionality constant and proceed
by showing that ((e∗k2n − e∗k2n−1 ) ⊗ en )n∈N is equivalent to the unit vector basis of c0 : For
(an )n∈N ∈ c00 choose N ∈ N such that an = 0 for n > N and define the operator
T :=
N
X
an (e∗k2n − e∗k2n−1 ) ⊗ en : X → [(en )N
n=1 ].
n=1
Then
∗
T =
N
X
∗
∗
an en ⊗ (e∗k2n − e∗k2n−1 ) : [(en )N
n=1 ] → X ,
n=1
where en is considered as an element of X ∗∗ . Hence
N
X
an (e∗k2n − e∗k2n−1 ) ⊗ en = kT k = kT ∗ k.
(10)
n=1
∗
∗
∗
∗
∗ ∗
∗
∗ N
Let x ∈ [(en )N
n=1 ] with kx k = 1 and kT k ≤ 2kT x k. Let y ∈ [(en )n=1 ] be defined by
P
∗
∗
∗
∗
∗
y∗ = N
n=1 x (en )en . Then y (en ) = x (en ) for n = 1, . . . , N and ky k ≤ C where C denotes
Embedding `∞ into the space of all Operators on Certain Banach Spaces
9
the basis constant of (en ). Thus
(11)
N
X
kT ∗ k ≤ 2kT ∗ x∗ k = 2 an en (y ∗ )(e∗k2n − e∗k2n−1 ) .
n=1
(e∗k2n
e∗k2n−1 )n∈N
Now use the facts that
−
is unconditional with unconditionality constant
K (see the Introduction for the definition of the unconditionality constant) and (e∗n ) is
quasisubsymmetric to obtain:
N
N
X
X
2
an en (y ∗ )(e∗k2n − e∗k2n−1 ) ≤ 2K max |an | en (y ∗ )(e∗k2n − e∗k2n−1 )
1≤n≤N
n=1
n=1
N
!
N
X
X
≤ 2Kk(an )kc0 en (y ∗ )e∗k2n + en (y ∗ )e∗k2n−1 n=1
n=1
!
N
N
X
X
(12)
∗ ∗
∗ ∗
≤ 2Kk(an )kc0 S(n),(k2n ) en (y )en + S(n),(k2n−1 ) en (y )en n=1
n=1
∗
∗
= 2K(S(n),(k2n ) + S(n),(k2n−1 ) )ky kk(an )kc0 (since y =
N
X
en (y ∗ )e∗n )
n=1
≤ 2KC(S(n),(k2n ) + S(n),(k2n−1 ) )k(an )kc0 .
Equations (10), (11) and (12) show that ((e∗k2n − e∗k2n−1 ) ⊗ en )n∈N is dominated by the c0
basis. On the other hand, for 1 ≤ n ≤ N , T ek2n = an en . Thus
N
X
inf i kei k
k(an )kc0
an (e∗k2n − e∗k2n−1 ) ⊗ en = kT k ≥
n=1
supi kei k
which implies that ((e∗k2n − e∗k2n−1 ) ⊗ en )n∈N dominates the c0 basis. Hence c0 ,→ L (X).
Corollary 2.10. If a Banach space X has a seminormalized quasisubsymmetric basis (en )
then `∞ ,→ L([(e∗n )]), where (e∗n ) is the sequence of the biorthogonal functionals to (en ).
Proof. Consider (en ) in [(e∗n )]∗ : it is seminormalized, quasisubsymmetric and biorthogonal
to (e∗n ). Therefore by Theorem 2.9 we have `∞ ,→ L([(e∗n )]).
Finally we make some remarks related to the question of whether `1 embeds into the space
of nuclear operators of a Banach space X. If X is a Banach space let (N (X), ν(·)) denote
the space of nuclear operators on X, namely
N (X) = {T ∈ L(X) : T =
ν(T ) =
∞
X
x∗n
⊗ yn where
n=1
∞
X
inf{
kx∗n kkyn k
n=1
x∗n
∗
∈ X , yn ∈ X and
∞
X
kx∗n kkyn k < ∞}
n=1
:
∞
X
x∗n ⊗ yn is a representation of T }.
n=1
The following remark is well-known:
Remark 2.11. If X is a Banach space with the AP then N (X)∗ is isometric to L (X ∗ ).
Moreover, if T ∈ L (X ∗ ) then the action of T as a functional on N (X) gives the trace of
T S ∗ , tr (T S ∗ ), when it is applied to S ∈ N (X).
10
G. Androulakis, K. Beanland, S.J. Dilworth, F. Sanacory
Indeed, it is shown in [Ry, Proposition 4.6 (i) ⇔ (ii), cf. Corollary 4.8 (a)] that if the
Banach space X has the AP then N (X) is naturally isometric to the projective tensor
ˆ
ˆ )∗ is
product X ∗ ⊗X.
Also in [Ry, page 24] it is shown that for any Banach space Y
, (Y ∗ ⊗Y
P
isometric to L
(Y ∗ ), where for T ∈ L (Y ∗ ), the action of T as a functional on ni=1 yi∗ ⊗ yi ∈
P
ˆ gives ni=1 (T yi∗ )(yi ).
Y ∗ ⊗Y
By Remark 2.11 we have that for a Banach space X with the AP, `1 embeds complementably in N (X) if and only if `∞ embeds in L(X ∗ ).
It is easy to see (and also follows from the result of Tong [T, Theorem 1.5] mentioned in
the Introduction) that for a dual Banach space X ∗ with an unconditional basis, there exists
ˆ complementably. The
a noncompact operator in L (X ∗ ) if and only if `1 embeds in X ∗ ⊗X
next remarks give sufficient conditions on a Banach space X which ensure that `1 embeds
complementably into N (X).
Remark 2.12. Let X be a Banach space which has the AP and assume that `∞ ,→ L (X).
Then `1 ,→ N (X) complementably.
Proof. Since the map from L (X) to L (X ∗ ) given by T 7→ T ∗ is an isometric embedding, we
have that `∞ embeds in L (X ∗ ). Hence by [BP] and Remark 2.11 we obtain that `1 embeds
complementably in N (X).
Remark 2.13. Let X be a Banach space which has the AP and contains an unconditional
basic sequence (en ). Suppose that there exists a noncompact operator in L (X, [(en )]). Then
`1 ,→ N (X) complementably.
Proof. By Remark 2.5 we have that `∞ embeds in L (X). Now Remark 2.12 finishes the
proof.
Corollary 2.14. If a Banach space X has a seminormalized quasisubsymmetric basis then
`1 ,→ N (X) complementably.
Proof. Combine Theorem 2.4 and Remark 2.12.
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Department of Mathematics, University of South Carolina, Columbia, SC 29208.
[email protected], [email protected], [email protected], [email protected]