COMPLEMENTED SUBSPACES OF BOUNDED LINEAR OPERATORS. Manijeh Bahreini Esfahani, B.S., M.S.. Dissertation Prepared for the Degree of.
COMPLEMENTED SUBSPACES OF BOUNDED LINEAR OPERATORS Manijeh Bahreini Esfahani, B.S., M.S.
Dissertation Prepared for the Degree of DOCTOR OF PHILOSOPHY
UNIVERSITY OF NORTH TEXAS August 2003
APPROVED: Elizabeth Bator, Major Professor Paul Lewis, Committee Member Joseph Iaia, Committee Member Neal Brand, Chair of the Department of Mathematics C. Neal Tate, Dean of the Robert B. Toulouse School of Graduate Studies
Bahreini Esfahani, Manijeh. Complemented Subspaces of Bounded Linear Operators. Doctor of Philosophy (Mathematics), August 2003, 66 pp., bibliography, 17 titles. For many years mathematicians have been interested in the problem of whether an operator ideal is complemented in the space of all bounded linear operators. In this dissertation the complementation of various classes of operators in the space of all bounded linear operators is considered. This paper begins with a preliminary discussion of linear bounded operators as well as operator ideals. Let L(X, Y ) be a Banach space of all bounded linear operator between Banach spaces X and Y , K(X, Y ) be the space of all compact operators, and W(X, Y ) be the space of all weakly compact operators. We denote space all operator ideals by O. Work by Bator, Lewis, and Kalton on the space of compact and weakly compact operators motivates much of this paper. Conditions that make K(X, Y ) and W(X, Y ) complemented in L(X, Y ) are generalized to the space of operator ideals. Also an extension will be given from the previous results that yield copies of co as well as l ∞ in L(X, Y ) to the space O(X, Y ). For instance, if X is an infinite-dimensional space and co embeds in L(X, Y ), then l ∞ embeds in L(X, Y ). This notion is generalized to the case when X is separable, Y contain a copy of co, and O(X, Y ) is complemented in L(X, Y ). Then l ∞ embeds isomorphically in O(X, Y ) if and only if co embeds isomorphically in O(X, Y ). The results concerning operator ideals are applied to the study of CC(X, Y ), the Banach space of completely continuous operators. For example, suppose O is a separably determined closed operator ideal that has property (*). Then O(X, co) = L(X, co) if and only if O(X, co) is complemented in L(X, co). It is immediate that if X be a Banach space, then CC(X, co) = L(X, co) if and only if CC(X, co) is complemented in L(X, co).
Copyright 2003 by Manijeh Bahreini Esfahani, B.S., M.S.
ii
ACKNOWLEDGEMENTS I am grateful to many people for their support and encouragement over the years it has taken to earn this degree. I wish to express my deepest gratitude to my major professor, Dr. Elizabeth Bator, for her guidance, encouragement, and never ending patience. I especially would like to thank my committee memebers, Dr. Paul Lewis and Dr. Joseph Iaia, for their participation, many helpful suggestions and much appreciated advice. In particular, I wish to express my appreciation to all the wonderful people in the math department especially, Dr. Allen, Dr. Brand, Dr. Kallman, Dr. Zafarani, Belinda, Beth, Inez, Jenny, Kiko, Maryam, Ross, Shemsi for their help, cooperation, and the support given to me while I pursued my studies. This study is dedicated to my dear mother(Zahra Motaei), my loving husband(Mehdi Taghvaei), my beautiful children(Zainab and Fatemeh), and to my family in Iran for their encouragement, support, and con.dence in me to complete my research and accomplish my goal through my life.
iii
CONTENTS ACKNOWLEDGEMENTS .......................................................................................................... iii Chapters 1 INTRODUCTION ....................................................................................................................... 1 2 COMPACT OPERATORS ........................................................................................................14 3 WEAKLY COMPACT AND UNCONDITIONALLY CONVERGING OPERATORS..........30 4 SEPARABLY DETERMINED OPERATOR IDEALS.............................................................50 BIBLIOGRAPHY..........................................................................................................................65
iv
CHAPTER 1 INTRODUCTION Numerous authors have dealt with the question of whether an operator ideal is complemented in the space of all bounded linear operators. In this dissertation the complementation of various classes of operators in the space of all bounded linear operators is studied. Work by Bator, Lewis, and Kalton on the space of compact and weakly compact operators motivates much of this paper. This paper begins with a preliminary discussion of linear bounded operators as well as operator ideals. We collect some known results which will be needed for our investigation throughout this paper. First we start with some theorems due to Kalton. Then a precise definition of the space of operator ideals is given. It will be shown that all finite rank operators belong to the operator ideal. Last result in Chapter 1 is the generalization of Lemma 2 in [K] for compact operators to operator ideals. Chapter 2 is devoted to studying compact operators. The first main result of this chapter is due to Kalton. This result shows that K(X, Y ) contains a copy of l∞ if and only if either X contains a complemented copy of l1 or Y contains a copy of l∞ . Building on work of Kalton and Feder, Emmanuele and John showed that K(X, Y ) is not complemented in L(X, Y ) if co embeds in K(X, Y ). An alternate proof of this theorem has been given by Bator and Lewis which is presented in this chapter. 1
In Chapter 3 particular attention has been paid to the case in which operator ideal is the space of all weakly compact operators as well as the space of all unconditionally converging operators. Our first result, which is very useful for the spaces of weakly compact operators, is a direct consequence of Lemma 1.4 in Chapter 1. In Theorem 2 of [E1, p.126], Emmanuele showed that if X has the Gelfand-Phillips property and the Dunford-Pettis property and does not have the Schur property, then W (X, Y ) is not complemented in L(X, Y ) whenever Y contains a copy of co . Later Bator and Lweis improved this result by only assuming that X is not a Grothendick space. However, it is not necessarily the case that if X is a Grothendick space and Y contains a copy of co , then W (X, Y ) = L(X, Y ). We will show that a sufficient condition to establish W (X, Y ) = L(X, Y ) is the separability of X. Hence different conditions will be needed in the non-separable case. We establish a complete characterization of Banach spaces X so that W (X, Y ) is complemented in L(X, Y ) whenever Y contains a copy of l∞ . Our last results in this chapter will be an investigation of the space U(X, Y ). It will be shown that if X is separable, then co does not embed in X or Y if and only if U(X, Y ) is complemented in L(X, l∞ ). In the non-separable case, co does not embed in X if and only if U(X, l∞ ) = L(X, l∞ ).
2
An investigation of the subspaces K(X, Y ), W (X, Y ), and U(X, Y ) of L(X, Y ) was made in the previous chapters. In Chapter 4 we generalize some of these results to arbitrary operator ideals. First an extension will be given from the previous results that yield copies of co as well as l∞ in L(X, Y ) to the space O(X, Y ). For instance, if X is an infinite-dimensional space and co embeds in L(X, Y ), then l∞ embeds in L(X, Y ). This notion is generalized to the case when X is separable, Y contains a copy of co , and O(X, Y ) is complemented in L(X, Y ). Then l∞ embeds isomorphically in O(X, Y ) if and only if co embeds isomorphically in O(X, Y ). Also assumptions that make K(X, Y ) and W (X, Y ) complemented in L(X, Y ) are generalized to the space of operator ideals. Finally, the results concerning operator ideals are applied to the study of CC(X, Y ), the Banach space of completely continuous operators. For example, suppose O is a separably determined closed operator ideal that has property (*). Then O(X, co ) = L(X, co ) if and only if O(X, co ) is complemented in L(X, co ). It is immediate that if X is a Banach space, then CC(X, co ) = L(X, co ) if and only if CC(X, co ) is complemented in L(X, co ). Throughout this dissertation X and Y will always denote infinite dimentional Banach spaces and X ∗ is the dual of X. L(X, Y ) denotes the space of bounded linear operators between X and Y equipped by the usual operator norm. The following topologies on L(X, Y ) are used in our investigation. The weak-operator topology w is defined by the linear functional T → y ∗(T x) where y ∗ ∈ Y ∗ and x ∈ X. The dual
3
weak-operator topology w is defined by the linear functional T → x∗∗ (T ∗ y ∗) where y ∗ ∈ Y ∗ , x∗∗ ∈ X ∗∗ . Also we say that Tα → T in the strong-operator topology if Tα (x) → T (x) for all x ∈ X. It is readily seen that w w; that is, w-operator topology is weaker than w -operator topology. Also, if X is reflexive then w = w. We say that X embeds in Y if there is a linear homeomorphism from X to Y , i.e. there is an isomorphic embedding T : X → Y . A projection P is an idempotent (P 2 = P ) operator from a Banach space X to itself. A subspace Y of X is complemented if there exists a projection P from X to Y . If A ⊆ X, then [A] denotes the closed linear span of A. The canonical unit vector basis of co is denoted by en , and the canonical unit vector basis of l1 is denoted by (e∗n ). We denote the nth coordinate map ξ −→ ξn in l∞ by ξ −→ πn (ξ). Also for M a subset of the positive integers N , l∞ (M) is defined to be the set of all ξ ∈ l∞ with ξk = 0 for k ∈ M. A series
n
xn is said to be weakly unconditionally Cauchy (w.u.C) if, given any
permutation π of the natural numbers, ( nk=1 xπ(k) ) is a weakly Cauchy sequence. In another words, a series Recall that
n
xn is wuC if and only if
n
xn is weakly subseries convergent if
n
|x∗ xn | < ∞ for each x∗ ∈ X ∗ .
n
xn converges weakly to some
element of X for every increasing sequence of integers. This implies, by the Orliczpettis Theorem [D], that
n
xn is subseries convergent in norm, hence unconditionally
convergent in norm.
4
The reader is referred to Diestel [D] or Dunford-Schwartz [DS] for undefined notation and terminology. First results we require are due to Bessaga and Pelczynski ([BP1],[BP2]). Proposition 1.1. (i) Let X be a Banach space containing no copy of c0 . Then every weakly unconditionally Cauchy series converges to a point in X. (ii) Let X be a Banach space such that X ∗ contains a copy of c0 . Then X ∗ contains a copy of l∞ and X contains a complemented copy of l1 . (iii) Let X be a Banach space containing no complemented copy of l1 . If is weak ∗ -unconditionally convergent in X ∗ , then
∞
∗ i=1 en
∞
∗ n=1 en
converges in norm.
Proof. (i). It is immediate from Bessaga-Pelczynski Theorem [DU, p. 22 or D, p. 45]. (ii). This is corollary 6 [DU, p. 23] or Theorem 10 [D, p. 48]. (iii). It follows from Corollary 7 [DU, p. 23] and (ii). The following lemma is well-known. We include the proof as we frequently use the result. Lemma 1.2. If Xo is separable, then there is an isometry from Xo into l∞ . In particular, if Xo ⊆ X and Xo is separable, then there exists T : X → l∞ such that T = 1 and T|Xo is an isometry.
5
Proof. Let {xn } be dense in BXo . Note that for each xn , there exits x∗n ∈ Xo∗ with x∗n = 1 and x∗n (xn ) = xn . Now define J : Xo −→ l∞ by J(x) = (x∗n x)∞ n=1 . Observe that J is an isometry. To see this, let x ∈ BXo . Then
||J(x)|| = ||(x∗n x)∞ n=1 || = sup |x∗n x| n
≤ ||x||.
Now let (xnk ) be a subsequence of (xn ) such that xnk → x. It follows that ||xnk || → ||x||. Note that xnk ∗ (x) = xnk ∗ (xnk ) − xnk ∗ (x − xnk ) and x − xnK → 0. Therefore, xnk ∗ (x − xnk ) → 0 and xnk ∗ (xnk ) → ||xnk || → ||x||. Hence |xnk ∗ (x)|k→∞ → ||x|| and
∗ ∗ ||J(x)|| = ||(x∗n x)∞ n=1 || = sup |xn x| ≥ sup |xnk x|. n
k
Now since supk |x∗nk x| → ||x||, we have ||J(x)|| ≥ ||x|| and by the previous argument ||J(x|| = ||x||. Also ||J|| = supx∈BX ||J(x)|| = supx∈BXo ||x|| = 1 , so ||J|| = 1. Next let x ∈ Xo . Then J(x) = (x∗1 (x), x∗2 (x), ...). Note that |x∗n x| ≤ 1 for each n. Using the Hahn-Banach Theorem for each component, x∗n may be extended to yn∗ . It follows that |yn∗ x| ≤ 1 for each x ∈ X. Thus J : X0 −→ l∞ may be extended to a bounded linear operator T : X −→ l∞ . Note that ||T || ≤ 1.
6
The next proposition is very closely to results of Rosenthal [Ro] and could be used to prove the results of Rosenthal in the countable case. Proposition 1.3. Let A : l∞ → l∞ be a bounded linear map. Suppose A(en ) = 0 for all n. Then there exists an infinite subset M of N such that A(ξ) = 0 for ξ ∈ l∞ (M). Proof. We may choose an uncountable collection (Nα ; α ∈ A) of infinite subsets of N such that for α = β, Nα ∩ Nβ is finite. To see this, let I be the rationals in (0, 1) and A be the irrationals in (0, 1). Then for α ∈ A, let Nα be a sequence in I converging to α. Now suppose the proposition is false. Then for each α, there exists ξ (α) ∈ l∞ (Nα ) such that ξ (α) = 1 and A(ξ (α) ) = 0. Let φ be a finite subset of A. Then it will be shown that
ξ (α) = η + ζ
α∈φ
where η ≤ 1 and ζ is in the linear span of (e1 , e2 , . . .). To verify this, first note that ξ α ∈ l∞ (Nα ) = {ξ : ξk = 0
7
∀k ∈ Nα } and
Nα ∩ Nβ is finite for α = β. Then we have the folowing,
ξ (α) = ξ 1 + ξ 2 + . . . + ξ (α) + . . . + ξ m
α∈φ
= (ξ11 , ξ21, ξ31 , . . . , ξn1 , . . .) + (ξ12 , ξ22, ξ32 , . . . , ξn2 , . . .) + (ξ13 , ξ23, ξ33 , . . . , ξn3 , . . .) + . . . + (ξ1m , ξ2m, ξ3m , . . . , ξnm, . . .). = (
m
ξ1i , ξ2β , ξ3γ , . . . ,
i=1
m
ξni , . . .)
i=1
m m ξ1i , 0, 0, . . . , ξni , . . .) = (0, ξ2β , ξ3γ , . . . , 0, . . .) + (
=
(0, ξ2β , ξ3γ , . . . , 0, . . .)
i=1 m
+ {(
i=1
ξ1i )e1
i=1
+ 0 · e2 + 0 · e3 + . . . , (
m
ξni )em , . . .}
i=1
= η + ζ.
To justify the third equation, note that finitely many components have more than one nonzero numbers. Therefore, in the other components there is only one nonzero
8
number. It follows that, (α) A(ξ ) = A( (ξ (α) )) α∈φ
α∈φ
= A(η + ξ) = A(η) + A(ξ) = A(η) + A(
n
βi ei )
i=1
= A(η) +
n
βi A(ei )
i=1
= A(η) + 0 = A(η) ≤ A · η = A .
Hence |
(A(ξ(α) ))m | = |(
α∈φ
where φ ⊂ A. Thus |
α∈φ
A(ξ(α) ))m | ≤ A
α∈φ
πm (A(ξ(α) ))| < ∞.
Therefore the set Am = {α; πm (A(ξ (α) )) = 0} is countable. Hence ∪m Am is also countable. Now if α ∈ ∪Am , then πm (A(ξ (α) )) = 0 for each m. It follows that Aξ (α) = 0 for all α ∈ A − (∪Am ). But the set A − (∪Am ) is uncountable. This contradiction completes the proof. We next apply proposition 1.3 to obtain the form required for studing spaces of bounded linear operators. 9
Proposition 1.4. Let X be a separable Banach space and suppose Φ : l∞ → L(X, l∞ ) is a bounded linear operator with Φ(en ) = 0 for all n. Then there is an infinite subset M of N such that for ξ ∈ l∞ (M), Φ(ξ) = 0. Proof. First note that the unit ball of X is separable since a subspace of a separable metric space is separable. Now let {xn } be a countable dense subset of the BX and define θ : L(X, l∞ ) → l∞
πm (θ(T )) = πα1 (m) (T xα2 (m) )
by
where α : N → N × N is some bijection. It will be shown that θ is an isometric embedding. To verify this, let T ∈ L(X, l∞ ). Then for x ∈ X,
T (x) = (T1 (x), T2 (x), . . . , Tn (x), . . .).
Note that {xn } is dense in BX , that is, for each x ∈ BX there is {xm } ⊆ {xn } such that (xm ) → x. Recall that α : N → N × N is a bijection. It follows that for each (n, m) ∈ N × N there exists an uniqe k ∈ N such that α(k) = (n, m) = (α1 (k), α2 (k)). Then
T =
sup T (x) = sup T (xm ) m
x∈BX
= sup sup |πn (T xm )| m
n
= sup |πα1 (k) (T xα2 (k) )|. k
10
(1)
On the other hand, θ : L(X, l∞ ) → l∞ . Thus πm (θ(T )) = πα1 (m) (T xα2 (m) ) and
θ(T ) = (πα1 (1) (T xα2 (1) ), πα1 (2) (T xα2 (2) ), . . . , πα1 (k) (T xα2 (k) ), . . .).
Therefore, θ(T ) = sup πα1 (k) (T xα2 (k) ) . k
(2)
Now (1) and (2) imply that T = θ(T ) . Thus θ is an isometric embedding and L(X, l∞ ) → l∞ . It follows that Φ : l∞ → l∞ is a bounded linear operator with Φ(en ) = 0 for all n. Now using the previous Proposition, there is an infinite subset M of N such that Φ(ξ) = 0 for ξ ∈ l∞ (M). Definition 1.5. O is said to be an operator ideal if for all Banach spaces X, Y , Z, and W we have the following. (i) O(X, Y ) is a subspace of L(X, Y ) and (ii) If S ∈ L(Z, X), T ∈ O(X, Y ), and R ∈ L(Y, W ), then we have RST ∈ O(Z, W ). Lemma 1.6. Let O be a non trivial operator ideal and X and Y be Banach spaces. Then every finite rank operator from X to Y is in O(X, Y ). Proof. It suffises to show that if x∗ ∈ X ∗ and y ∈ Y , then x∗ ⊗ y ∈ O(X, Y ) where (x∗ ⊗ y)(x) = x∗ (x)y. Fix x∗o ∈ X ∗ and yo ∈ Y . Let ϕ = 0 in O(X, Y ). Then choose 11
xˆ ∈ X such that ϕ(ˆ x) = zˆ = 0. Now define T : X → X by T (x) = x∗o (x)ˆ x. It follows that ϕ(T (x)) = x∗o (x)ˆ z . Then choose yo∗ ∈ BY∗ such that yo∗ (ˆ z ) = 1. Next define S : Y → Y by S(y) = yo∗ (y)yo and observe that S(ˆ z ) = yo . Note that SϕT (x) = x∗o (x)yo = (x∗o ⊗ yo )(y). As ϕ ∈ O(X, Y ) we have x∗o ⊗ yo ∈ O(X, Y ). Theorem 1.7. Let X be a separable space and O(X, Y ) be complemented in L(X, Y ). Suppose Φ : l∞ → L(X, Y ) has the following properties: (i) Φ(en ) ∈ O(X, Y ) for each n, (ii) {Φ(ξ)(x), ξ ∈ l∞ , x ∈ X} is separable. Then for every infinite subset M of N, there exists an infinite subset M0 of M with Φ(ξ) ∈ O(X, Y ) for ξ ∈ l∞ (M0 ). Proof. Let Yo be the linear span of {Φ(ξ)x; ξ ∈ l∞ , x ∈ X}. Then by (ii), Yo is a separable subspace of Y . Apply Lemma 1.2 and select S : Y −→ l∞ so that ||S|| = 1 and S|Yo is an isometry. Suppose Γ : L(E, F ) −→ O(E, F ) is a bounded projection. It is clearly sufficient to establish the result when M = N. Define maps Ψ : l∞ −→ L(X, l∞ ) and Ψ1 : l∞ −→ O(X, l∞ ) by
ψ(ξ)x = SΦ(ξ)x
and ψ1 (ξ)x = SΓ(Φ(ξ))x. 12
where x ∈ X and ξ ∈ l∞ . Then both Ψ and Ψ1 are bounded and linear. As Φ(en ) ∈ O(X, Y ) for each n, we have Γ(Φ(en )) = Φ(en ). Therefore,
Ψ(en ) = SΦ(en ) = SΓ[Φ(en )] = Ψ1 (en )
for all n. Let η = Ψ − Ψ1 . Then η : l∞ −→ L(X, Y ) is a bounded linear operator with η(en ) = 0 for all n. Now since X is separable we may apply Proposition 1.4. Hence there is an infinite subset Mo of N such that η(ξ) = 0 for all ξ ∈ l∞ (Mo ). It follows that Ψ(ξ) = Ψ1 (ξ) for all ξ ∈ l∞ (Mo ). In particular, Ψ(ξ) ∈ O(X, l∞ ) for all ξ ∈ l∞ (Mo ). Now note that S is an isometry on Yo and the range of Ψ(ξ) lies in S(Yo ). Hence Φ(ξ) ∈ O(X, Y ) for all ξ ∈ l∞ (Mo ). .
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CHAPTER 2 COMPACT OPERATORS In this Chapter we study the structure of the Banach space K(X, Y ) of all compact linear operators between two Banach spaces X and Y . We investigate two distinct problems, subspaces isomorphic to l∞ and comlementation of K(X, Y ) in L(X, Y ). We start with collecting together some known results which will be used in this chapter. The first result is essentially due to Kalton [K]. Theorem 2.1. Assume A is a subset of K(X, Y ); then A is a weakly compact subset if and only if A is a w -compact subset. Corollary 2.2. Let Tn be a sequence of compact operators such that Tn → T in w topology where T is a compact operator. Then Tn → T weakly and there is a sequence Sn of convex combinations {Tn ; n = 1, 2, . . .} with Tn − T → 0. Proof. Let A = {Tn , T }. By assumption A is a W -compact subset in K(X, Y ). Then Theorem 1.1 implies that A is a weakly compact subset. Thus Tn → T weakly. The second part of the theorem is immediate [DS,P.422]. The next result is an easy consequence of the Orlicz-Pettis Theorem:“Let be a series in X such that every subseries of of X. Then
n
n
n
xn
xn is weakly convergent to a point
xn is unconditionally convergent in norm.” 14
Proposition 2.3. Suppose Y is a Banach space and T : l∞ → Y is weakly compact operator. Then the series
∞
n=1 ξn T en
converges in norm for each ξ = (ξn ) ∈ l∞ .
Proof. Using the Orlicz-Pettis Theorem, it is enough to show that every subseries of ∞
n=1 ξn T (en )
in Y . Suppose
is weakly convergent. Notice that T (Bl∞ ) is a weakly compact subset ∞
k=1 ξnk T (enk )
is a subseries of
and assume that ξ ≤ 1. Then Sn =
∞
n=1 ξn T (en ).
n
i=1 ξi T (ei )
= T(
Set Sn =
n
i=1 ξi ei )
n
i=1 ξi T (ei )
⊆ T (Bl∞ ). It
follows that Sn has a weakly convergent subsequence. Now by passing to any weakly convergent subseqence of Sn we see that
(Snk )k = (
k
ξni T (eni )k →
∞
ξn T (en ),
n=1
i=1
in the weak topology. Now use Orlicz-Pettis theorem. The next result is due to Rosenthal [R]. Proposition 2.4. Assume Y is a Banach space containing no isomorphic copy of l∞ . Then every bounded linear map T : l∞ → Y is weakly compact. Now we study conditions that yields copy of l∞ in K(X, Y ). The next theorem is due to Kalton [K].
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Theorem 2.5. Let X and Y be Banach spaces. The following are equivalent: (i) K(X, Y ) contains a copy of l∞ . (ii) Either X contains a complemented copy of l1 or Y contains a copy of l∞ . Proof. (ii) ⇒ (i). Suppose X contains a complemented copy of l1 . Then Proposition 1.1 ensures a copy of l∞ in X ∗ . Let S : l∞ → X ∗ be a linear embedding. Fix y ∈ Y with y = 1. Next define Φ : l∞ → K(X, Y ) by
Φ(ξ)(x) = S(ξ)(x)y
for all x ∈ X. Note that,
Φ(ξ) = = =
sup Φ(ξ)x
x∈BX
sup S(ξ)(x)y
x∈BX
sup S(ξ)(x) · y
x∈BX
= S(ξ)
Thus Φ is a linear embedding and l∞ → K(X, Y ). Now assume that Y contains a copy of l∞ and S : l∞ → Y is a linear embedding.
16
Then fix x∗ ∈ X ∗ with x∗ = 1. Define Φ : l∞ → K(X, Y ) by
Φ(ξ)(x) = x∗ (x)S(ξ)
for all x ∈ X and observe that,
Φ(ξ) = = =
sup Φ(ξ)x
x∈BX
sup x∗ (x)S(ξ)
x∈BX
sup |x∗ (x)|. S(ξ)
x∈BX
= x∗ . S(ξ) = S(ξ) .
Hence Φ is a linear embedding and l∞ → K(X, Y ). (i) =⇒ (ii). To prove the converse, suppose (ii) is false. Use the hypotheses and let Φ : l∞ → K(X, Y ) be a linear embedding. Set Φ(en ) = Tn in K(X, Y ). Then for x ∈ X we define Λx : l∞ → Y by
Λx ξ = Φ(ξ)x.
Although it is straightforward to check that Λx is linear and continuous, the proofs
17
are included here for the convenience of the reader. To see the Λx is linear, note that
Λαx1 +x2 ξ = Φ(ξ)(αx1 + x2 ) = αΦ(ξ)x1 + Φ(ξ)x2 = αΛx1 ξ + Λx2 ξ.
Now suppose ξm → ξ in l∞ . Since Φ is an isomorphism, we have Φ(ξm ) → Φ(ξ) in X. Therefore, Φ(ξm )x → Φ(ξ)x for each x ∈ X. It follows that Λx ξm → Λx ξ. Thus Λx is a continuous map. Next as Y contains no copy of l∞ , by Proposition 2.4 every bounded linear map Λx : l∞ → Y is weakly compact. Then proposition 2.3 yields that the series ∞
n=1 ξn Λx en
converges in norm in Y for each ξ = (ξn ) ∈ l∞ . Obseve that ∞
ξn Λx en =
n=1
It follows that
∞
n=1 ξn Tn
∞
ξn Φ(en )x =
n=1
∞
ξn Tn (x).
n=1
converges in the strong operator topology in L(X, Y ) and
hence in the topology w. Next define Ψ : l∞ → L(X, Y ) by
Ψ(ξ) =
∞
ξn Tn .
n=1
It will be shown that Ψ is a bounded linear operator. In fact, it is enough to show that Φ is weak ∗ − weak continuous and then apply the Closed Graph Theorem. To get started, recall that the series
∞
n=1 ξn Tn
18
converges in topology w for all
ξ = (ξn ) ∈ l∞ . It follows that for fixed x ∈ X and y ∗ ∈ Y ∗ ,
∗
y (
∞
ξn Tn )x =
n=1
∞
y ∗ (ξn Tn (x))
n=1
converges for each ξ ∈ l∞ . Then choose ξn = ±1 so that
∞
n=1
|y ∗(Tn (x))| < ∞. Now
suppose ξ m → 0 in weak ∗ topology, it will be shown that Ψ(ξ m ) → 0 in w topology. Next, without loss of generality, let ξ m ≤ 1. Fix y ∗ ∈ Y ∗ , x ∈ X and let > 0. Choose N ∈ N such that ∞
|y ∗(Tn (x))| < . 2 n=N +1 Then choose M ∈ N so that for m ≥ M we have,
|ξnm|
0.
n=m
Next define Φ : l∞ → L(l1 , Y ) by Φ(ξ)(λ) = that Φ(en )(λ) =
∞
m m=1 en λm ym
∞
n=1 ξn λn yn
where λ = (λn ) ∈ l1 . Note
= λn yn . It follows that Φ(en ) is a rank one operator
and is compact. Also for each ξ ∈ l∞ and λ ∈ l1 , Φ(ξ)(λ) =
∞
n=1 ξn λn yn
where
|ξn λn | < ∞. That is, Φ(ξ)(λ) is an infinite sum of {yn } for all ξ ∈ l∞ and λ ∈ l1 . Therefore {Φ(ξ)(x) : ξ ∈ l∞ , x ∈ l1 } ⊆ [yn : n ∈ N] and is separable. Now apply lemma 1.6 and select an infinite subset M of N with Φ(χM ) ∈ K(l1 , F ). However, Φ(χM )(en ) =
m
χM (m)em n yi = yn for each n ∈ M.
This implies that the sequence {Φ(χM )(em ); m ∈ M} has no cluster point. This contradiction concludes the proof. Building on work of Kalton [K] and Feder [F1], [F2], Emmanuele [4] and John [J] independently established the following result. Next theorem has an alternate proof which was presented by Bator and Lewis [BL1]. Theorem 2.7. If X and Y are infinite dimensional Banach spaces and c0 embeds isomorphically in K(X, Y ), then K(X, Y ) is not complemented in L(X, Y ).
26
Proof. The proof is divided into three cases: (i) l1 is complemented in X; (ii) co embeds isomorphically in Y ; (iii) c0 does not embed in Y , l1 is not complemented in X, and yet c0 does embed in K(E, F ). (i). Lemma 2.6, explicitly deals with (i). (ii). Suppose then that c0 → Y , and let (yn ) be a bounded sequence in Y so that (yn ) is equivalent to (en ), i.e. (yn ) ∼ (en ). Let (x∗n ) be a weak ∗ -null and normalized sequence in X ∗ [D, Chp. XII], and define φ : l∞ → L(X, co ) by φ(ξ)(x) = (ξn x∗n (x)) for ξ = (ξn ) ∈ l∞ and x ∈ X. It is easy to check that φ is an isomorphism; in fact φ is an isometry. Moreover, if M is a non-empty subset of N, then (φ(χM ))∗ (e∗n ) = x∗n for each n ∈ M. Hence φ(χM ) is not a compact operator if M is any infinite subset of N. However, φ(en )(x) = x∗n (x)en , i.e., φ(en ) is a rank one ( and therefore compact ) operator. Next we let Xo be a separable subspace of X which norms {x∗n : n ∈ N}. Let R : L(X, co ) → L(Xo , co ) be the natural restriction map, and define φˆ : l∞ → L(Xo , co ) to be R ◦ φ. Since φ is an isometry and Xo norms x∗n for each n, φˆ is an isometry. ˆ M ) is not a compact Further, the argument in the preceding paragraph shows that φ(χ operator if M is an infinite subset of N. Now use the injectivity of l∞ [D, p. 71], and select an operator S : Y → l∞ so that
27
ˆ )(x) = S(T (x)) for S(yn ) = en for each n. Define Sˆ : L(Xo , Y ) → L(Xo , l∞ ) by (ST x ∈ Xo and T ∈ L(Xo , Y ). Also, let R1 : L(X, Y ) → L(Xo , Y ) be the restriction map and let i : L(X, co ) → L(X, Y ) be the natural map obtained by identifying c0 with its isomorphic image in Y . Use R1 and i to define an operator ψ : l∞ → L(Xo , l∞ ) : ˆ 1 iφ(ξ). Note that ψ(ξ)(xo ) = φ(ξ)(xo ) for ξ ∈ l∞ and x0 ∈ X0 , i.e., ψ(ξ) = SR ˆ ψ(ξ) = φ(ξ) for each ξ ∈ l∞ . Suppose that K(X, Y ) is complemented in L(X, Y ) and P : L(X, Y ) → K(X, Y ) ˆ 1 P iφ(ξ), and observe that is a projection. Define Γ : l∞ → K(Xo , l∞ ) by Γ(ξ) = SR Γ(en ) = ψ(en ) for all n. An application of Proposition 1.4 produces an infinite set M contained in N so that Γ(ξ) = ψ(ξ) for all ξ = (ξn ) ∈ l∞ with ξn = 0 if n ∈ M. In particular, Γ(χM ) = ψ(χM ). But this is a contradiction since Γ(χM ) is a compact operator and ψ(χM ) is not compact. (iii) Suppose that co does not embed in Y , l1 is not complemented in X, and co → K(X, Y ). As above, suppose that P : L(X, Y ) → K(X, Y ) is a projection, and let ψ : co → K(X, Y ) be an isomorphism. Define φ : l∞ → L(X, Y ) by φ(ξ)(x) =
ξn ψ(en )(x), i.e.,
φ(ξ) =
ξn ψ(en ) (strong operator topology).
28
Since l∞ does not embed in K(X, Y ) by Theorem 2.5, an application of the Proposition 2.4 guarantees that P ◦ φ is weakly compact. But any weakly compact operator is unconditionally converging [D]. This is a clear contradiction since ψ is an isomorphism.
29
CHAPTER 3
WEAKLY COMPACT AND UNCONDITIONALLY CONVERGING OPERATORS This chapter is devoted to studing spaces of weakly compact operators and spaces of unconditionally converging operators. Before giving our results let us recall a few definitions. If X and Y are Banach spaces, then W (X, Y ) denotes the space of all weakly compact operators and U(X, Y ) denotes the space of all unconditionally converging operators. Let X and Y be Banach spaces and T : X −→ Y be a bounded linear operator. The operator T is said to be weakly compact if T (BX ) is compact in Y when Y is given the weak topology. Further, T is an unconditionally converging operator if T sends weakly unconditionally converging series into unconditionally series. Note that a series weakly unconditionally convergent series if
∞ n
∞ n
xn in X is
|x∗ (xn )| < ∞ for each x∗ ∈ X ∗ .
Our first result, which is very useful for the spaces of weakly compact operators, is a direct consequence of Lemma 1.6 in Chapter 1. It should be noted that spaces of weakly compact operators form an operator ideal.
30
Lemma 3.1. If X is a separable Banach space, W (X, Y ) is complemented in L(X, Y ), and φ : l∞ → L(X, Y ) is an operator so that (i) φ(en ) ∈ W (X, Y ) for each n and (ii) {φ(b)(x) : b ∈ l∞ , x ∈ X} is separable, then every infinite subset M of N has an infinite subset M0 so that φ(b) ∈ W (X, Y ) for each b ∈ l∞ (M0 ). We now establish a complete characterization of Banach spaces X so that W (X, co ) is complemented in L(X, co ). A Banach space X is called a Grothendieck space if every weak ∗ convergent sequence in X ∗ converges weakly in X ∗ . Theorem 3.2. Let X be a Banach space. Then the following are equivalent: (i) X is a Grothendieck space. (ii) W (X, co ) = L(X, co ). (iii) W (X, co) is complemented in L(X, co ). Proof. (i) ⇒ (ii). Since X is a Grothendieck space every bounded linear operator T : X → co is weakly compact [DU, p.179]. (ii) ⇒ (iii). It is trivial. (iii) ⇒ (i). Suppose X is not a Grothendieck space. Then there is a weak ∗ null normalized sequence (x∗n ) in X ∗ with no weakly null subsequence. Hence S : X → co
31
defined by S(x) =
(x∗n (x))
=
∞
x∗n (x)en
n=1
is not a weakly compact operator. To see this, it will be shown that S ∗ is not a weakly compact operator. Observe that
< S ∗ (e∗n ), x > = < e∗n , S(x) > =
m=1
=
∞
x∗m (x)e∗n (em )
m=1
= x∗n (x).
Thus S ∗ (e∗n ) = x∗n . It follows that S ∗ is not weakly compact and neither is S. Choose a bounded sequence (xn ) in BX such that S(xn ) has no weakly convergent subsequence. Let Xo = [xn ]. It follows that Xo is a separable subspace of X so that S|Xo is not a weakly compact operator. Letting wn∗ = x∗n |Xo , and by passing to a subsequence if necessay, we have (wn∗ ) is weak ∗ null and has no weakly null subsequence. Now define T : l∞ → L(X, co ) by
T (ξ)(x) =
n
32
ξn x∗n (x)en .
Note that if x ∈ X and ξ ∈ l∞ , then
T (ξ)(x) =
∞
ξn x∗n (x)en
n=1
= sup |ξn x∗n (x)| n
≤ ξ sup x∗n x . n
Thus T is a bounded linear transformation. Next let R : L(X, co ) → L(Xo , co ) be the natural restricton map. Define ψ : l∞ → L(Xo , co) by ψ(ξ)(x) = RT (ξ)(x), for ξ ∈ l∞ and x ∈ Xo . Observe that for all n ∈ N and x ∈ X,
T (en )(x) =
∗ ∗ em n xm (x)em = xn (x)en .
m
Thus T (en ) is a compact, and hence weakly compact, operator for all n ∈ N. It follows that ψ(en ) ∈ W (Xo , co ) for each n. Also note that {ψ(ξ)(x), ξ ∈ l∞ , x ∈ Xo } ⊆ co . Hence conditions (i) and (ii) of Lemma 1.6 are satisfied. Select an infinite subset M
33
of N so that ψ(ξ) ∈ W (Xo , co) for each ξ ∈ l∞ (M). Thus ψ(χM ) ∈ W (Xo, co ). But
< (T (χM ))∗ (e∗n ), x > = < e∗n , (T (χM ))x > =
m∈M
x∗m (x)e∗n (em )
m∈M
= x∗n (x),
for all x ∈ X. Hence (RT (χM ))∗ (e∗n ) = x∗n |Xo = wn∗ . But (wn∗ ) has no weakly null subsequence. It follows that RT (χM ) is not a weakly compact operator. This contradiction concludes the proof. In Theorem 2 of [E1, p.126], Emmanuele showed that if X has the Gelfand-Phillips property and the Dunford-Pettis property and does not have the Schur property, then W (X, Y ) is not complemented in L(X, Y ) whenever Y contains a copy of co . The next theorem which is due to Bator and Lewis [BL2] improves this result by only assuming that X is not a Grothendieck space.
34
Theorem 3.3. Assume X is not a Grothendieck space and co embeds isomorphically in Y , then W (X, Y ) is not complemented in L(X, Y ). Proof. First choose a weak ∗ null normalized sequence (x∗n ) in X ∗ with no weakly null subsequence. Define φ : X → co by φ(x) =
∞
∗ n=1 xn (x)en .
As it was shown in the
previous theorem φ is not weakly compact. Hence there is a separable subspace Xo of X so that φ|Xo is not weakly comact. It follows that wn∗ = x∗n |Xo is weak ∗ null sequence with no weakly null subsequence. Now define T : l∞ → L(X, co ) by
T (x)(ξ) =
ξn x∗n (x)en
where x ∈ X and ξ = (ξn ) ∈ l∞ . Certainly T is a linear bounded transformation. Next let ˆi : co → Y be an isomorphism and ˆi(en ) = fn for each n ∈ N. Use the ˆ n ) = en for each injectivity of l∞ and select an operator Sˆ : Y → l∞ so that S(f ˆ where x ∈ Xo n ∈ N. Then define S : L(Xo , Y ) → L(Xo , l∞ ) by (SA)(x) = S(A(x)) and A ∈ L(Xo , Y ). Also define the natural embedding i : L(X, co ) → L(X, Y ) by (iB)(x) = ˆi(B(x)) for x ∈ X and B ∈ L(X, l∞ ). Let R : L(X, co ) → L(Xo , co ) be the natural restriction map. Now use S, R, i, and T to define ψ : Xo → l∞ by
ψ(ξ)(x) = SRiT (ξ)(x).
Last, suppose W (X, Y ) is complemented in L(X, Y ) and let P : L(X, Y ) → 35
W (X, Y ) be the projection. Then define Γ : l∞ → W (Xo , l∞ ) by
Γ(ξ)(x) = SRP iT (ξ)(x).
and observe that Γ(en ) = ψ(en ) for all n ∈ N. To see this, let x ∈ X. Then
(iT (en ))(x) = ˆi(T (en )) = ˆi(
∗ ∗ ˆ ∗ em n xm (x).em ) = i(xn (x).en ) = xn (x).fn .
m
Thus iT (en ) is a weakly compact operator and P (iT (en )) = iT (en ). It follows that Γ(en ) = SRP (iT (en )) = SRiT (en ) = ψ(en ) for each n. Proposition 1.4 implies that there exists an infinite subset M ⊆ N so that Γ(ξ) = ψ(ξ) for all ξ ∈ l∞ (M). Thus ψ(χM ) = RSiT (χM ) ∈ W (Xo , l∞ ). Note that RSiT (χM ) = RT (χm ) since (Si)en = en for all n ∈ N. It folows that RT (χM ) ∈ W (Xo , l∞ ). However, as it was shown in the previous theorem, RT (χM ) is not a weakly compact operator. Remark. It follows immediately from the above that whenever X is not a Grothendieck space and co embeds isomorphically in Y , then W (X, Y ) = L(X, Y ). However, it is not necessarily the case that if X is a Grothendieck space and Y containes a copy of co , then W (X, Y ) = L(X, Y ). For instance, it is well known that l∞ is a Grothendieck but not a reflexive space. Thus W (l∞ , l∞ ) = L(l∞ , l∞ ). However, if X is a separable, Grothendieck space it is readily seen that X is reflexive and W (X, Y ) = L(X, Y ).
36
Thus we have the following. Corollary 3.4. Suppose X is a separable Banach space and co embeds isomorphically in Y . Then the following are equivalent: (i) X is a Grothendieck space. (ii) W (X, Y ) = L(X, Y ). (iii) W (X, Y ) is complemented in L(X, Y ). It would be interesting to know if the assumption ”X is a separable” in the above corollary can be eliminated. However, the above remark shows that Corollary 3.4 can not be simply generalaized by deleting ”separable”. Hence, different conditions must be added in the non-separable case. In order to start, we need the following. Theorem 3.5. If X is not a reflexive Banach space, then W (X, l∞ ) is not complemented in L(X, l∞ ). Proof. Since X is not a reflexive space, there is a sequence (xn ) ⊂ X with no weakly convergent subsequence. Let Xo = [xn ]. Then Xo is a separable and not a reflexive space. Therefore Xo is not a Grothendieck space. Thus there is a weak ∗ null sequence (yn∗ ) in Xo∗ so that no subsequence of (yn∗ ) converges weakly to a point in Xo∗ . Now ˆ define Sˆ : Xo → co by S(x) = (yn∗ (x)). As it was shown in the proof of the Theorem 1.2, Sˆ : Xo → co → l∞ is not a weakly compact operator. Now for each n ∈ N, let
37
x∗n be a Hahn Banach extension of yn∗ . Define S : X → l∞ by
S(x) = (x∗n (x))
for all x ∈ X. It is easy to see that S is not a weakly compact operator.
∗ n ξn (xn (x))en
Next it will be shown that the series
is weak ∗ convergent series
in l∞ for each ξ ∈ l∞ and x ∈ X. Fix ξ = (ξn ) ∈ l∞ , x ∈ X, and let λ = (λn ) ∈ l1 . Note that λ =
m
=
m
ξn x∗n (x)en )
n
ξn x∗n (x)e∗m (en ))
n
λm ξm x∗m (x),
m
and observe that
|
λm ξm x∗m (x)| ≤
m
|λm ||ξm||x∗m (x)|
m
≤ ξ x∗m x
|λm |
m
< ∞.
It follows that the series
∗ n ξn (xn (x)).en
is weak ∗ absolutely summable in l∞ . 38
Now define T : l∞ → L(X, l∞ ) by
T (ξ)(x) =
ξn (x∗n (x)).en
(weak ∗ topology ),
n
where ξ = (ξn ) ∈ l∞ , x ∈ X, and x∗n ∈ X ∗ . The above argument shows that T is a bounded linear transformation. Use T and the restriction map R : L(X, l∞ ) → L(Xo , l∞ ) to define ψ : l∞ → L(Xo , l∞ ) by
ψ(ξ)(x) = RT (ξ)(x).
Next suppose that W (X, l∞ ) is complemented in L(X, l∞ ). Let P : L(X, l∞ ) → W (X, l∞ ) be a projection. Define ϕ : l∞ → W (Xo , l∞ ) by ϕ = RP T . Note that as T (en ) is a rank one operator we have T (en ) ∈ W (X, l∞ ). Hence
ϕ(en ) = RP T (en ) = RT (en ) = ψ(en )
for all n ∈ N. Proposition 1.4 ensures that there is an infinite set M ⊆ N so that ϕ(ξ) = ψ(ξ) for all ξ ∈ l∞ (M). Thus ψ(χM ) is a weakly compact operator. However,
39
if x ∈ Xo , we have
< (ψ(χM ))∗ (e∗n ), x > = < e∗n , (ψ(χM ))x > =
m∈M
x∗m (x).e∗n (em )
m∈M
= x∗n (x),
Therefore (ψ(χm ))∗ (e∗n ) = x∗n |Eo = yn∗ , ∀n ∈ M. Thus (ψ(χm ))∗ is not a weakly compact operator and neither is ψ(χm ). This contradiction ends the proof. We now establish a complete characterization of Banach spaces X so that W (X, l∞ ) is complemented in L(X, l∞ ). Corollary 3.6. Let X be a Banach space. Then the following are equivalent: (i) X is a reflexive space. (ii) W (X, l∞ ) = L(X, l∞ ). (iii) W (X, l∞ ) is complemented in L(X, l∞ ). The next result improves Theorem 3.5 to spaces Y which contain a copy of l∞ . Theorem 3.7. Assume X is not a reflexive space and l∞ embeds isomorphically in Y ; then W (X, Y ) is not complemented in L(X, Y ). 40
Proof. Let (xn ) be a sequence in X with no weakly convergent subsequence. Set Xo = [xn ]. Note that Xo is not a Grothendieck space since Xo is a separable and not a reflexive space. Thus there is a weak ∗ null sequence (yn∗ ) in Xo∗ so that no subsequence of (yn∗ ) converges weakly to a point in Xo∗ . Then Sˆ : Xo → co → l∞ ˆ defined by S(x) = (yn∗ (x)) is not a weakly compact operator. Now define S : X → l∞ by S(x) = (x∗n (x)) where x∗n is a Hahn Banach extension of yn∗ for each n ∈ N. It is clear that S is not a weakly compact operator. Next note that the series
∗ n ξn (xn (x))en
is a weak ∗ absolutely summable in l∞
for each ξ ∈ l∞ and x ∈ X. Define T : l∞ → L(X, l∞ ) by
T (ξ)(x) =
ξn (x∗n (x))en
(weak ∗ topology ),
n
for each ξ = (ξn ) ∈ l∞ , x ∈ X, and x∗n ∈ X ∗ . The above argument shows that T is a bounded linear transformation. Now let ˆi : l∞ → Y be an isomorphism and ˆi(en ) = fn for each n ∈ N. Use the injectivity of l∞ and select an operator Sˆ : Y → l∞ so that ˆ n ) = en for each n ∈ N. Then define S : L(Xo , Y ) → L(Xo , l∞ ) by S(f
ˆ (SA)(x) = S(A(x))
where x ∈ Xo and A ∈ L(Xo , Y ). Suppose R : L(X, l∞ ) → L(Xo , l∞ ) is the natural
41
restriction map. Define the natural embedding i : L(X, l∞ ) → L(X, Y ) by
(iB)(x) = ˆi(B(x))
for x ∈ X and B ∈ L(X, l∞ ). Use S, R, i, and T to define ψ : Xo → l∞ by
ψ(ξ)(x) = SRiT (ξ)(x).
Now suppose W (X, Y ) is complemented in L(X, Y ) and let P : L(X, Y ) → W (X, Y ) be the projection. Then define Γ : l∞ → W (Xo , l∞ ) by
Γ(ξ)(x) = SRP iT (ξ)(x).
and observe that Γ(en ) = ψ(en ) for all n ∈ N. As in the proof of the previous theorem, there is M ∈ P∞ (N) such that ψ(χM ) ∈ W (Xo , l∞ ). However, the proof of the previous theorem shows that this is not true. The next corollary has a different nature, because we do not require the separability of the Banach space X. Corollary 3.8. Let X be a Banach space and Y contains a copy of l∞ . Then the following are equivalent. (i) X is a reflexive space . 42
(ii) W (X, Y ) = L(X, Y ). (iii) W (X, Y ) is complemented in L(X, Y ). The next theorem is due to Bator and Lewis [BL2]. Theorem 3.9. Suppose X is a Banach space containing a complemented copy of l1 and Y is a non-reflexive Banach space. Then W (X, Y ) is not complemented in L(X, Y ). Proof. By the way of contradiction suppose that W (X, Y ) is complemented in L(X, Y ) and Γ : L(X, Y ) → W (X, Y ) is a projection. Select a subspace H which is complemented in X and isomorphic to l1 . let P : X → H be a projection. Now if T : H → Y is an operator, then it is readily seen that ΓT P ∈ W (X, Y ). Further, if R : L(X, Y ) → L(H, Y ) is the natural restriction map, then RΓT P ∈ W (H, Y ). In fact, T → RΓT P defines a projection from L(H, Y ) onto W (H, Y ). Use the hypotheses and choose a bounded sequence (yn ) in BY so that no subsequence of (yn ) converges weakly to a point of Y . It follows that {yn : n ∈ M} is not relatively weakly compact. Next define φ : ∞ → L(1 , Y ) by
φ(ξ)(λ) =
∞
ξ n λn y n ,
n=1
where ξ = (ξn ) ∈ l∞ and λ = (λn ) ∈ l1 . It should be noted that φ(en )(λ) = λn yn . Thus φ(en ) is a weakly compact operator. Further, {φ(ξ)(x) : ξ ∈ l∞ , x ∈ 1 } ⊆ [yn : 43
n ∈ N ]. Now use Lemma 1.6 and select an infinite set M ⊆ N so that φ(χm ) ∈ W (1 , Y ). But φ(χM )(e∗n ) = yn for each n ∈ M and we have a contradiction. Our final part in this Chapter will be an investigation of the space U(X, Y ). It will be shown that if X is separable, then co does not embed in X or Y if and only if U(X, Y ) is complemented in L(X, l∞ ). In the non-separable case, co does not embed in X if and only if U(X, l∞ ) = L(X, l∞ ).
44
The next theorem is due to Bator and Lewis [BL2] Theorem 3.10. If X is separable and U(X, Y ) = L(X, Y ), then U(X, Y ) is not complemented in L(X, Y). Now we present another version of the previos theorem. Theorem 3.11. Let X be a separable Banach space and Y be a Banach space. Then the following are equivalent. (i) co deos not embed in X or Y . (ii) U(X, Y ) = L(X, Y ). (iii) U(X, Y ) is complemented in L(X, Y ). Proof. (i) ⇒ (ii). Suppose U(X, Y ) = L(X, Y ). Let A : X → Y be an operator such that A is not unconditionally converging. Then there is a subspace H of X isomorphic to co such that restriction A|H is an isomorphism [D, p. 54]. This is a contradiction with (i). (ii) ⇒ (iii). It is obvious. (iii) ⇒ (i). Suppose (i) false. Then X and Y contain a copy of co . Now by Sobczyk’s theorem there is a complemented subspace Xo of X such that Xo is isomorphic to co . Let P˜ : X → XO be a projection and I : Xo → co be the isometry from Xo to co with I(xn ) = (en ) for each n. Let A = I P˜ . Then T : l∞ → L(X, co )
45
defined by T (ξ)(x) =
ξe∗n (A(x)).en
n
is a bounded and linear transformation. Let ˜i : co → Y be a linear embedding and ˜i(en ) = yn for each n. Now we use injectivity of l∞ and select an operator S˜ : Y → l∞ ˜ n ) = en for each n. Then define operators S˜ : L(X, Y ) → L(X, l∞ ) and so that S(y the natural embedding i : L(X, co ) → L(X, Y ) by
˜ (SB)(x) = S(B(x))
and (iD)(x) = ˜i(D(x)) for x ∈ X, B ∈ L(X, Y ), and D ∈ L(X, co ). Now define ψ : l∞ → L(X, l∞ ) by
ψ(ξ)(x) = SiT (ξ)(x)
. Next let P : L(X, Y ) → U(X, Y ) be the projection map and define Γ : l∞ → U(X, l∞ ) by Γ(ξ)(x) = SP iT (ξ)(x).
46
Observe that Γ(en ) = ψ(en ) for all n ∈ N. Using proposition 1.4, there is an infinite subset M of N such that ψ(ξ) = Γ(x) for each ξ ∈ l∞ (M). Hence ψ(χM ) is an unconditionally converging operator. But ψ(χM )(xn ) = en for each n. Finally, note that the series
n en
n
xn is weakly unconditionally converging but
n
ψ(χM )(xn ) =
does not converge unconditionally. It follows that ψ(χM ) is not unconditionally
converging. This contradiction completes the proof. Theorem 3.12. Let X be a Banach space. Then the following are equivalent. (i) X does not contain a copy of co . (ii) U(X, l∞ ) = L(X, l∞ ). (iii) U(X, l∞ ) is complemented in L(X, l∞ ). Proof. (i) ⇒ (ii). Suppose U(X, l∞ ) = L(X, l∞ ). Let A : X → l∞ be an operator such that A is not unconditionally converging. Then there is a subspace H of X isomorphic to co such that restriction A|H is an isomorphism [D]. This is a contradiction with (i). (ii) ⇒ (i). Suppose U(X, l∞ ) = L(X, l∞ ) and co embeds isomorphically in X. Let i : co → l∞ be the natural embedding map. Using injectivity of l∞ there is an extension A : X → l∞ so that A|co = i. It is clear to see that A is an isomorphism on co . Thus A ∈ U(X, l∞ ) and we have a contradiction. (ii) ⇒ (iii). It is obvious. (iii) ⇒ (ii). Suppose (ii) false. Select an operator A : X → l∞ so that A is 47
not unconditionally converging. Thus there is a bounded sequence (hn ) ∈ X such that (hn ) ∼ (en ) and A(hn ) = en for each n ∈ N. Set H = [hn ] and Aˆ = A|H . Note that A(x) =
∗ n en (x)en
in the weak ∗ topology for each x ∈ X. Now define
T : l∞ → L(X, l∞ ) by
T (ξ)(x) =
ξn e∗n (x)en
weak ∗ topology
n
where ξ = (ξn ) ∈ l∞ and x ∈ X. T is a bounded linear transformation. Now use the restriction map R : L(X, l∞ ) → L(H, l∞ ) to define ψ : l∞ → L(H, l∞ ) by
ψ(ξ)(x) = RT (ξ)(x)
for ξ ∈ l∞ and x ∈ H. Next, let P : L(X, l∞ ) → U(X, l∞ ) be a projection. Use operators T , R, and P to define ϕ : l∞ → U(H, l∞ ) by
ϕ(ξ)(x) = RP T (ξ)(x)
and observe that ϕ(en ) = ψ(en ) for each n ∈ N. Now by the Proposition 1.4 select M ∈ P ∞ (N) so that ϕ(ξ) = ψ(ξ) for all ξ ∈ l∞ (M). Thus ψ(χM ) is an unconditionally converging operator. However, ψ(χM )(hn ) = en for each n. Hence ψ(χM ) is an
48
isomorphism on H ∼ = co . It follows that ψ(χM ) is not unconditionally converging. This contradiction finishes the proof.
49
CHAPTER 4
SEPARABLY DETERMINED OPERATOR IDEALS An investigation of the subspaces K(X, Y ), W (X, Y ), and U(X, Y ) of L(X, Y ) was made in the previous chapters. In this chapter we generalize some of these results to arbitrary operator ideals. Let recall the Diestel-Faires Theorem [DU] which will be needed in this chapter. Let F be an algebra of subsets of the set Ω and G : F → X be a bounded vector measure. If G is not strongly additive, then there is a topological isomorphism T : co → X and a sequence (En ) of disjoint members of F such that T (en ) = G(En ). If in addition F is a σ-algebra, then the above statement remains true if the space co is replaced by the space l∞ . We first consider conditions that yield copies of co as well as l∞ in O(X, Y ). The next three results are due to Lewis [L]. Lemma 4.1. If l∞ embeds in X ∗ or in Y , then l∞ embeds in O(X, Y ). Proof. First suppose l∞ → X ∗ and let y ∈ Y such that y = 1. For x∗ ∈ l∞ → X ∗ , define x∗ ⊗ y : X → Y by (x∗ ⊗ y)(x) = x∗ (x)y.
50
As x∗ ⊗ y is a rank one operator, x∗ ⊗ y ∈ O(X, Y ). Also it is easy to see that x∗ ⊗ y = x∗ . It follows that ϕ : l∞ → O(X, Y ) defined by
ϕ(x∗ )(x) = (x∗ ⊗ y)(x)
for x ∈ X is an isomorphic embedding. Next, let l∞ → Y . Then choose x∗ ∈ X ∗ with x∗ = 1. Define ϕ : l∞ → O(X, Y ) by ϕ(y)(x) = (x∗ ⊗ y)(x) where y ∈ l∞ → Y and x ∈ X. With the same argument as above, ϕ is an isomorphic embedding of l∞ into O(X, Y ). Theorem 4.2. co embeds in O(X, Y ) if and only if there exists a non-null sequence (Tn ) in O(X, Y ) so that
n
Tn (x) is weakly unconditionally convergent in Y for each
x ∈ X. Proof. Suppose T : co → O(X, Y ) is an isomorphism. Let Tn = T (en ). Then Tn ∈ O(X, Y ) and
n Tn (x) =
Y for each x ∈ X since
n
n
T (en )(x) is weakly unconditionally convergent in
T (en ) is weakly unconditionally convergent in O(X, Y ).
Conversely, suppose (Tn ) is as in the hypothesis. Since conditionally convergent in Y for each x ∈ X, {
n∈A
n
Tn (x) is weakly un-
Tn (x) : A ⊆ N} is bounded
for each x ∈ X by the Uniform Bounded Principle. Now let F be the algebra of 51
subsets of N consisting of the finite sets and their complements. Then the measure µ : F → O(X, Y ) defined by
µ(A) =
n∈A
Tn
if A is f inite
− n∈A Tn
if N \ A is f inite
is bounded and finitely additive. However, the measure µ is not strongly additive. Finally, the Diestel-Faires Theorem ensures that there is a copy of co in O(X, Y ). Now we generalize the previous results that yield copies of co as well as l∞ in L(X, Y ) to the space O(X, Y ). Theorem 4.3. Let X be separable, Y contain an isomorphic copy of co , and O(X, Y ) be complemented in L(X, Y ). Then l∞ embeds isomorphically in O(X, Y ) if and only if co embeds isomorphically in O(X, Y ) Proof. If l∞ → O(X, Y ), then obviously co → O(X, Y ). Suppose co → O(X, Y ). The Josefson-Nissenzweig theorem yields a weak ∗ null normalized sequence (x∗n ) in X ∗ . Define a map T : l∞ → L(X, co ) → L(X, Y ) by
T (ξ)(x) = (ξn x∗n (x)).
Note that T is an embedding of l∞ into L(X, Y ). If x ∈ X, then T (en )(x) = x∗n (x) for all n ∈ N. Hence T (en ) ∈ O(X, Y ). Also {T (ξ)(x); ξ ∈ l∞ , x ∈ X} ⊆ co and is 52
separable. Therefore conditions (i) and (ii) of Lemma 1.6 hold. Hence there is an infinite subset M of N such that T (ξ) ∈ O(X, Y ) for each ξ ∈ l∞ (M). Thus T is an embedding of l∞ into O(X, Y ) and l∞ → O(X, Y ). Theorem 4.4. Let X, Y be separable, and O(X, Y ) be complemented in L(X, Y ). Then l∞ embeds isomorphically in O(X, Y ) if and only if co embeds isomorphically in O(X, Y ). Proof. If l∞ → O(X, Y ), then obviously co → O(X, Y ). Conversely, suppose co → O(X, Y ). Then the case co → Y follows directly by the previous theorem. Thus we assume co → Y . Let A : co → O(X, Y ) be an isomorphism. Then for x ∈ X the series
n
A(en )x converges unconditionally in Y . To see this, note that
| < A(en )(x), y ∗ ) > | =
n
Thus
n
|(x ⊗ y ∗ )(A(en ))| < ∞.
n
A(en )x is weakly unconditionally convergent in Y and co → Y . Now let F
be the σ-algebra of subsets of N. Define G : F → L(X, Y ) by G(M) =
n∈M
A(en ) in
the strong operator topology for all M ⊆ N. It is clear to see that G is finitely additive but not strongly additive. To see this, let En = {n} be disjoint members of F . Then the series
∞
n=1
G(En ) does not converge in norm. Also G has bounded range by
the Uniform Bounded Principle. Now by Diestel-Fairs theorem l∞ → L(X, Y ). Let 53
T : l∞ → L(X, Y ) be an embedding. Then T (en ) = G(En ) = G(n). But G(n) = A(en ) for all n ∈ N. Hence T (en ) ∈ O(X, Y ), for all n ∈ N. Also, {T (ξ)(x); ξ ∈ l∞ , x ∈ X} ⊆ Y and is separable. Therefore Lemma 1.6 ensures that there is an infinite subset M of N such that T (ξ) ∈ O(X, Y ) for each ξ ∈ l∞ (M). Thus T is an embedding of l∞ into O(X, Y ) and l∞ → O(X, Y ). We will use the following notation. Let A : X → l∞ and M be a nonempty subset of N. We define the formal series AM : X → l∞ by AM (x) =
m∈M
e∗m (A(x)).em .
Definition 4.5. Suppose O is a closed operator ideal. We say O has the property (∗) if whenever X is a Banach space and A ∈ O(X, l∞ ) then there is an infinite subset Mo ⊆ N such that AM ∈ O(X, l∞ ) for all infinite subsets M ⊆ Mo . Theorem 4.6. let O be a separably determined closed operator ideal that has property (*). Then the following are equivalent: (i) O(X, co ) = L(X, co ). (ii) O(X, co ) is complemented in L(X, co ). Proof. (i) ⇒ (ii) is obvious. (ii) ⇒ (i). Suppose (i) is false. Let A : X → co such that A ∈ O(X, co ). Hence there is a separable subspace Xo of X such that Ao = A|Xo : Xo → co is not in O(Xo , co ). Choose Mo ∈ P∞ (N) such that AoM ∈ O(Xo , co ) for each M ∈ P∞ (Mo ).
54
Now define T : l∞ → L(X, co ) by
T (ξ)(x) =
ξn e∗n (A(x))en
n
for ξ = (ξn ) ∈ l∞ and x ∈ X. Let R : L(X, co ) → L(Xo , co ) be the restriction map. Define ϕ : l∞ → L(Xo , co ) by
ϕ(ξ)(x) = RT (ξ)(x)
where ξ ∈ l∞ and x ∈ Xo . Now we verify conditions (i) and (ii) in lemma 1.6. Let x ∈ X. Then T (en )(x) =
∗ ∗ em n em (A(X))em = en (A(x)).en .
m
Note that T (en ) ∈ O(X, co ) for all n ∈ N, as it is a rank one operator. Thus ϕ(en ) = RT (en ) is in O(Xo , co ) for each n ∈ N and (i) of Lemma 1.6 is satisfied. Now {ϕ(ξ)(x), ξ ∈ l∞ , x ∈ Xo } ⊆ co is separable. Therefore, there is an infinite subset M of Mo so that ϕ(ξ) ∈ O(Xo , co ) for each ξ ∈ l∞ (M). It follows that φ(χM ) ∈ O(Xo , co ).
55
Hence R(T (χM )) is in O(Xo , co ). But for x ∈ Xo ,
RT (χM )(x) = T (χM )(x) =
∞
χM (j)e∗j (A(x))ej
j=1
=
e∗j (A(x)ej
j∈M
= AoM (x).
Since AoM = φ(χM ) ∈ O(Xo , co ) we have a contradiction. The next corollary follows immediately from the above. Corollary 4.7. let X be separable and O be an operator ideal with property (∗) . Then the following are equivalent: (i) O(X, co ) = L(X, co ). (ii) O(X, co ) is complemented in L(X, co ). Theorem 4.8. Let O be a separably determined ideal with property (∗). If O(X, co ) = L(X, co ) and co → Y , then O(X, Y ) is not complemented in L(X, Y ). Proof. Let A : X → co such that A ∈ O(X, co ). Hence there is a separable subspace Xo of X such that Ao = A|Xo ∈ O(Xo , co ). Choose Mo ∈ P∞ (N) such that AoM ∈
56
O(Xo , co ) for each M ∈ P∞ (Mo ). Define T : l∞ → L(X, co ) by
T (ξ)(x) =
ξn e∗n (A(x)).en
n
where ξ = (ξn ) ∈ l∞ and x ∈ X. Now let ˆi : co → Y be an isomorphism defined by ˆi(en ) = yn . Let i1 : co → l∞ be the injection map. Use the injectivity of l∞ to ˆ n ) = en for each n. Then define operators select an operator Sˆ : Y → l∞ so that S(y S : L(X, Y ) → L(X, l∞ ) and i : L(X, co ) → L(X, Y ) by
ˆ (S(B))(x) = S(B(x))
and
(i(D))(x) = ˆi(D(x)) for x ∈ X, B ∈ L(X, Y ), and D ∈ L(X, co ). Let R : L(X, co ) → L(Xo , co ) be the restriction map. Define ϕ : l∞ → L(Xo , l∞ ) by
ϕ(ξ)(x) = RSiT (ξ)(x).
Next suppose O(X, Y ) is complemented in L(X, Y ) and P : L(X, Y ) → O(X, Y ) be a projection map. Define Γ : l∞ → O(Xo , l∞ ) by
Γ(ξ)(x) = RSP iT (ξ)(x). 57
Since T (en ) is a rank one operator for each n ∈ N, we have Γ(en ) = ϕ(en ). Using Proposition 1.4, there is an infinite subset M of Mo such that ϕ(ξ) ∈ O(Xo, l∞ ) for each ξ ∈ l∞ (M). It follows that ϕ(χM ) ∈ O(Xo , l∞ ). Thus RSiT (χM ) is in O(Xo , l∞ ). Since (Si)en = en for each n, we have RSiT (χM ) = RT (χM ). If x ∈ Xo , then
RT (χM )(x) = T (χM )(x) =
∞
χM (j)e∗j (A(x))ej
j=1
=
e∗j (A(x))ej
j∈M
= AoM (x).
Hence AoM = φ(χM ) ∈ O(Xo , co ). This contradiction completes the proof. Theorem 4.9. Let O be a separably determined ideal with property (∗). Then the following are equivalent: (i) O(X, l∞ ) = L(X, l∞ ). (ii) O(X, l∞ ) is complemented in L(X, l∞ ). Proof. (i) ⇒ (ii) is obvious. (ii) ⇒ (i). Suppose (i) is false. Let A : X → l∞ such that A ∈ O(X, l∞ ). Then there is a separable subspace Xo ⊆ X such that Ao = A|Xo ∈ O(Xo , l∞ ). Choose Mo ∈ P∞ (N) so that AoM ∈ O(Xo , l∞ ) for each M ∈ P∞ (Mo ). Note that the operator 58
A must be of the form A(x) = (x∗n x) for some bounded sequence (x∗n ) in X. Next observe that
∞
∗ n=1 ξn (xn (x)).en
is weak ∗ absolutely summable in l∞ for each ξ ∈ l∞
and x ∈ X. Now define T : l∞ → L(X, l∞ ) by
T (ξ)(x) =
∞
ξn (x∗n (x))en
(weak ∗ topology)
n=1
where ξ = (ξn ) ∈ l∞ , x ∈ X, and x∗n ∈ X ∗ . It is easy to see that T is a bounded linear transformation. Now use T and the restriction map R : L(X, l∞ ) → L(Xo , l∞ ) to define ϕ : l∞ → L(Xo , l∞ ) by
ϕ(ξ)(x) = RT (ξ)(x).
Next suppose O(X, l∞ ) is complemented in L(X, l∞ ) and P : L(X, l∞ ) → O(X, l∞ ) be a projection map. Then Γ : l∞ → O(Xo , l∞ ) defined by
Γ(ξ)(x) = RP T (ξ)(x)
is an operator so that Γ(en ) = ϕ(en ) for all n ∈ N. Use Proposition 1.4 to select an infinite subset M of Mo such that Γ(ξ) = ϕ(ξ) for all ξ ∈ l∞ (M). Thus ϕ(ξ) = RT (ξ) is in O(Xo , l∞ ) for each ξ ∈ l∞ (M). To complete the proof, observe that T (χM ) is
59
not in O(Xo , l∞ ). If x ∈ Xo ,
T (χM )(x) =
∞
χM (j)(x∗j (x))ej
j=1
=
(x∗j (x))ej
j∈M
= AoM (x).
Since T (χM ) = AoM ∈ O(Xo, l∞ ) we have a contradiction. We now apply the results of this chapter to the study of CC(X, Y ), the Banach space of completely continuous operators from the Banach space X to the Banach space Y . Recall that T ∈ L(X, Y ) is called completely continuous if T maps weakly convergent sequences into norm convergent sequences. A Banach space X is a Schur space if every weakly null sequence in X is norm null. Lemma 4.10. let X be a Schur space and Y be a Banach space. Then CC(X, Y ) = L(X, Y ). Proof. Let T : X → Y be in L(X, Y ) and (xn ) be a weakly null sequence in X. Since X is a Schur space, then (xn ) also is a norm null sequence in X. Clearly (T (xn )) is a norm null sequence in Y . Hence T is in CC(X, Y ) and CC(X, Y ) = L(X, Y ). It is not necessarily the case that if X is not Schur then there is a T ∈ L(X, co ) such that T ∈ CC(X, co ). In particular, it is well known that any operator T : l∞ → co 60
is weakly compact. As l∞ has the Dunford-Pettis property, T is also completely continuous. However, we do have this result whenever X is separable. Theorem 4.11. Let X be a separable Banach space. Then the following are equivalent: (i) X is a Schur space. (ii) CC(X, co ) = L(X, co ). Proof. (i) ⇒ (ii) is obvious by previous lemma. (ii) ⇒ (i). Suppose X is not a Schur space. Thus there exists a normalized sequence (xn ) in X such that (xn ) is weakly null. Now for each n, choose x∗n in BX ∗ so that x∗n xn = xn = 1. But BX ∗ is weak ∗ compact and metrizable since X is separable. Therefore (x∗n ) has a weak ∗ convergent subsequence. By passing to such subsequence if necesssary, we may assume that there is a x∗ ∈ X ∗ such that (x∗n ) converges weak ∗ to x∗ . Consider the weak ∗ null sequence yn∗ = x∗n − x∗ . Define T : X → co by T (x) = (yn∗ (x)). We will show that T is not completely continuous. Since (xn ) is weakly null in X, it is enough to show that T (xn ) is not norm null. Note that the j th coordinate of T (xj ) is yj∗(xj ) = (x∗n − x∗ )(xj ) = x∗j (xj ) − x∗ (xj ) = 1 − x∗ (xj ). 61
As (xj ) is weakly null, there exists J ∈ N such that if j ≥ J, then x∗ (xj ) < 12 , and therefore yj∗(xj ) > 12 . Thus T (xn ) is not norm null. This concludes that T is not in CC(X, co ) and contradicts (i). Theorem 4.12. Let X be a Banach space. Then the following are equivalent: (i) CC(X, co ) = L(X, co ). (ii) CC(X, co ) is complemented in L(X, co ). If X is separable, then (i) and (ii) are equivalent to (iii) X is a Schur space. Proof. Using Theorem 4.6 it is enough to show that the operator ideal of completely continuous operators from X to co has the property (∗). Let A : X → co be an operator which is not completely continuous. Let (xn ) be a weakly null sequence in X and δ > 0 such that A(xn ) > δ. It is clear that (A(xn )) is weakly nulll. Thus for n1 = 1 there exists N1 ∈ N such that A(xn1 )N1 > δ. Now since (A(xn )) is weakly null in co , lim e∗N1 (A(xn )) = 0.
n→∞
Hence there is n2 > n1 such that e∗i (A(xn )) < δ for 1 ≤ i ≤ N1 and for all n ≥ n2 . It then follows that there exists N2 > N1 such that e∗N2 (A(xn2 )) > δ. Continue this process to obtain a subsequence (xni ) of (xn ) and an increasing sequence (Ni ) of N
62
such that e∗Ni (A(xni )) > δ for all i ∈ N. Now M = {Ni , i = 1, 2, ...} is an infinite subset of N and let Xo = [xn ]. The following is a direct result of Theorem 4.7. Corollary 4.13. Let X be a Banach space and co → Y . Then the following are equivalent: (i) CC(X, Y ) = L(X, Y ). (ii) CC(X, Y ) is complemented in L(X, Y ). If X is separable, then (i) and (ii) are equivalent to (iii) X is a Schur space. Theorem 4.14. Let X be a Banach space. Then the following are equivalent: (i) X is a Schur space. (ii) CC(X, l∞ ) = L(X, l∞ ). Proof. (i) ⇒ (ii). It is trivial. (ii) ⇒ (i). Suppose X is not a Schur space. Thus there exists a normalized ∗ sequence (xn ) in X such that (xn ) is weakly null. Now for each n, choose x∗n in BX
63
so that x∗n xn = xn = 1. Define T : X → l∞ by
T (x) = (x∗n ).
Observe that (xn ) is weakly null in X but T (xn ) = sup x∗n (x) = 1. Therefore T is not completely continuous and this contradicts (i). Corollary 4.15. Let X be a Banach space and l∞ → Y . Then the following are equivalent: (i) X is a Schur space. (ii) CC(X, Y ) = L(X, Y ). (iii) CC(X, Y ) is complemented in L(X, Y ).
64
BIBLIOGRAPHY
[BL1] Bator, Elizabeth M. and Lewis, Paul, Complemented Spaces of Operators, (preprint) [BL2] Bator, Elizabeth M. and Lewis, Paul, Complemented Spaces of Operators, Bull. Polish Acad. Sci. vol. 50, No.4, (2002), 413-416. [BP1] Bessaga, C. and Pelczynski, A., On bases and unconditional convergence of series in Banach spaces, Studia Math. 17, (1958), 151-174. [BP2] Bessaga, C. and Pelczynski, A., Some remarks on conjugate spaces containing a subspace isomorphic to the space co , Bull. Acad. Polon. Sci. 6, (1958), 249250 [D]
Diestel, J., Sequences and Series in Banach spaces, Grad. Texts in Math. no. 92, Springer-Verlag, New York, (1984).
[DU]
Diestel, J. and Uhl, J.J., Jr., Vector Measures, American Mathematical Society, Providence, Rhode Island (1977).
[DS]
Dunford, Nelson and Schwartz, Jacob T., Linear Operators: Part I, John Wiley and Sons, New York (1957).
[E1]
Emmanuele, G., Remarks on the uncomplemented subspace W (E, F ), J. Funct. Analysis 99(1991), 125-130. 65
[E2]
Emmanuele, G., A remark on the containment of c0 in spaces of compact operators, Math. Proc. Cambridge Phil. Soc. 111(1992), 331-335.
[EJ]
Emmanuele, G. and John, K., Uncomplementability of spaces of compact operators in larger spaces of operators, Czechoslovak Math. J. 47(1997), 19-31.
[F1]
Feder, M., On subspaces of spaces with an unconditional basis and spaces of operators, Illinois J. Math. 24(1980), 196-205.
[F2]
Feder, M., On the non-existence of a projection onto the spaces of compact operators, Canad. Math. Bull. 25(1982), 78-81.
[J]
ohn, K. On the uncomplemented subspace K(X, Y ), Czechoslovak Math. J. 42(1992), 167-173.
[K]
Kalton, N., Spaces of compact operators, Math. Ann. 208(1974), 267-278.
[L]
Lewis, P., Spaces of operators and co, Studia Mathematica, 145 (3) (2001), 213-218.
[LT]
Lindenstrauss, J., and Tzafriri, L., Classical Banach spaces I, Springer-Verlag, Berlin, 1977.
[R]
osenthal, H.P., On relatively disjoint families of measures with some applications to Banach space theorv, Studia Math, 37, (1971)13-36.
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