Concr. Oper. 2021; 8:48–59

Research Article Open Access

Fernanda Botelho* and Richard J. Fleming The dual of the space of bounded operators on a Banach space https://doi.org/10.1515/conop-2020-0109 Received August 23, 2020; accepted January 6, 2021

Abstract: Given Banach spaces X and Y, we ask about the dual space of the L(X, Y). This paper surveys results on tensor products of Banach spaces with the main objective of describing the dual of spaces of bounded operators. In several cases and under a variety of assumptions on X and Y, the answer can best be given as the projective tensor product of X** and Y*.

Keywords: Tensor products; dual spaces; dual of spaces of operators

MSC: Primary 46B78, secondary 46B10 Dedicated to the memory of our friend Professor James E. Jamison.

1 Introduction

The space L(X, Y) of bounded linear operators from the Banach space X to the Banach space Y is one of the fundamental spaces of study in Banach space theory, and it seems natural to inquire about its dual space. A preliminary search of articles on this subject gave a rather sparse result. Hence we thought it might be of use to investigate an answer to this question. Let us begin with a very naive and elementary approach. In the case where Y is one dimensional, the answer is just X*, so what is the answer in the two dimensional case? It depends, of course, on the norm of the space. Let us consider Y = `∞(2), where X is any Banach space. ∞ Clearly T ∈ L(X, ` (2)) can be written as Tx = (φ1(x), φ2(x)) for x ∈ X and where φ1, φ2 are elements of * * X . Furthermore, kTk = max{kφ1k, kφ2k}. It follows that if ψ ∈ L(X, Y) , then ψ = (ψ1, ψ2) where ψ1, ψ2 ∈ ** 1 ** X and kψk = kψ1k + kψ2k. Thus, the dual space in this case is the space ` (2, X ). This can be extended inductively to any nite dimension. Corresponding results for other common norms can also be obtained. Notation is fairly standard. Because of a variety of uses of the letter B, we will use L(X, Y) to denote the bounded operators from X to Y and similarly, LA(X, Y) for the approximable operators (those for which the nite rank operators are dense), and LK(X, Y) for the compact operators from X to Y. The closed unit ball of a Banach space X is denoted by BX. In years gone by when researchers were discussing a particular result, someone would often say, “Well, it’s probably somewhere in Grothendieck!" That is the case here as well. A more elegant approach to the description of L(X, Y)* involves tensor products, and the result that L(X, `∞(n))* = `1(n, X**) which we de- ∞ * ** ∞ * scribed above could be expressed by L(X, ` (n)) = X ⊗ˆ π(` (n)) . Before showing how that works, let us take a little refresher course on tensor products.

*Corresponding Author: Fernanda Botelho: Department of Mathematical Sciences, University Of Memphis, Memphis TN 38152; E-mail: [email protected] Richard J. Fleming: Department of Mathematics, Central Michigan University, Mt. Pleasant, Michigan 48859; E-mail: [email protected]

Open Access. © 2020 FernandaBotelho and Richard J. Fleming, published by De Gruyter. This work is licensed under the Creative Commons Attribution alone 4.0 License. The dual of the space of bounded operators on a Banach space Ë 49

2 Tensor products

We mention here some basic and useful facts about tensor products for the benet of readers who may not be familiar with the subject. What is reported here is material taken from the wonderful book of Raymond Ryan [7]. Let B(X × Y) denote the space of bilinear functions from X × Y, where X, Y are vector spaces, to the scalar eld K (which is always the real or complex numbers). In its most elementary form, the tensor product X ⊗ Y can be thought of as a subspace of the space of linear functionals on B(X × Y). Given x ∈ X and y ∈ Y, let x ⊗ y denote the linear functional on B(X × Y) dened by

(x ⊗ y)(A) = hA, x ⊗ yi = A(x, y), where A is a bilinear form on X × Y. The tensor product X ⊗ Y is the space spanned by these elements. Thus an element of this tensor product can be written

n X u = λi xi ⊗ yi i=1 and since such representations are not unique, we may always write an element in the form

n X u = xi ⊗ yi . i=1

The projective norm, π, on the tensor product of two Banach spaces X ⊗ Y is dened by

( m m ) X X π(u) = inf kxikkyik : u = xi ⊗ yi . i=1 i=1

The inmum is taken, of course, over all representations of u. Then π is a norm with the property that π(x⊗y) = kxkkyk for x ∈ X and y ∈ Y. The tensor product X ⊗ Y with the projective norm is denoted by X ⊗π Y and its 1 1 completion is given by X⊗ˆ π Y. Some useful facts for future reference are that ` ⊗ˆ π Y = ` (Y) and this holds also for the space `1(J) where J is an arbitrary indexing set. An interesting and important result about the projective tensor product is given in the following theorem. We think it instructive to oer a proof, although we give an alternative to the interesting one given in [7].

Theorem 1. Let X and Y be Banach spaces, u ∈ X⊗ˆ π Y, and ϵ > 0. Then there exist bounded sequences {x } {y } in X Y respectively, such that the series P∞ x ⊗ y converges to u and n , n , n=1 n n

X∞ kxnkkynk < π(u) + ϵ. n=1

Proof. u ∈ X⊗ Y ϵ u Pk1 x ⊗ y π u u ϵ π u Given ˆ π and > 0 there exists 1 = i=1 i i such that ( − 1) < /8 and ( 1) ≤ Pk1 kx k ky k π u ϵ π u π u ϵ Pk1 kx k ky k π u ϵ ϵ π u ϵ i=1 i i < ( 1) + /4. Thus ( 1) < ( ) + /8 so that i=1 i i < ( ) + /8 + /4 < ( ) + /2. Without loss of generality, we may assume that, for all i ∈ {1, 2, . . . k1}, kxik = kyik. We assume this as well Pk2 xi yi u u ∈ X⊗ˆ π Y u xi ⊗yi π u u u ϵ for all , chosen below. Since − 1 there exists 2 = i=k1+1 such that ( − 1 − 2) < /16 Pk2 π u kxik kyik π u ϵ π u π u u ϵ ϵ ϵ and ( 2) ≤ i=k1+1 < ( 2) + /8. Then ( 2) ≤ ( − 1) + /16 < /8 + /16, and therefore π(u2) < ϵ/4. n+2 We continue this process to dene a sequence {un} such that π(u − u1 − u2 − · · · − un) < ϵ/2 , where Pkn Pkn n+1 un xi ⊗ yi π un kxik kyik π un ϵ π un π u u un = i=kn−1+1 , and ( ) ≤ i=kn−1+1 ≤ ( )+ /2 . Therefore ( ) ≤ ( − 1 −· · · −1)+ ϵ/2n+1 < ϵ/2n+2 + ϵ/2n+1 < ϵ/2n. It follows that

X∞ X∞ u = un = xi ⊗ yi , n=1 i=1 50 Ë Fernanda Botelho and Richard J. Fleming and ∞ ∞ X X n kxik kyik < π(u) + ϵ/2 = π(u) + ϵ. i=1 n=1 p We observe that sequences {kxik} and {kyik} are bounded by π(u) + ϵ. This completes the proof.

It follows from the above result that ( ) X∞ X∞ X∞ π(u) = inf kxnkkynk : kxnkkynk < ∞, u = xn ⊗ yn , n=1 n=1 n=1 where the inmum is taken over all such representations of u. This is another of several ways of calculating the projective norm. Dual spaces are of particular interest to us, so let us describe the dual space of the projective tensor product of two Banach spaces, X, Y. A bilinear form B on X × Y is bounded if there exists C > 0 such that |B(x, y)| ≤ Ckxkkyk for all x ∈ X and y ∈ Y. Let B(X, Y) denote the space of bounded bilinear forms on X × Y, which is a Banach space with the norm

kBk = sup{|B(x, y)| : x ∈ BX , y ∈ BY }.

Given B ∈ B(X × Y) there exists a unique linear functional B˜ on X⊗ˆ π Y satisfying B˜(x ⊗ y) = B(x, y), and it is straitforward to show that the correspondence B ↔ B˜ is an isometric isomorphism between B(X × Y) and * * (X⊗ˆ π Y) . We note also, that given B ∈ B(X × Y), we can dene an operator LB : X → Y by hy, LB(x)i = B(x, y) * and the correspondence B ↔ LB describes an isometric isomorphism between B(X × Y) and L(X, Y ). In a * similar way, the operator RB : Y → X which satises hx, RB(y)i = B(x, y) provides an isometric isomorphism * between B(X × Y) and L(Y, X ) via B ↔ RB. Hence we have

* ∼ ∼ * ∼ * (X⊗ˆ π Y) = B(X × Y) = L(X, Y ) = L(Y, X ). (1)

An element u ∈ X ⊗ Y can be considered as a bilinear form Bu on X* × Y* by

m X Bu(φ, ψ) = φ(xi)ψ(yi), i=1

u Pm x ⊗ y u X ⊗ Y B X* where = i=1 i i is any representation of . This provides a canonical embedding of into ( × Y*), and we use it to dene what is called the injective norm on X ⊗ Y, which is the norm induced by this embedding. We denote by ε(u) the injective norm of u ∈ X ⊗ Y. Hence, we have

( m ) X ε(u) = sup | φ(xi)ψ(yi)| : φ ∈ BX* , ψ ∈ BY* , i=1 where the supremum is taken over all representations of u. By consideration of the operators Lu : X* → Y and Ru : Y* → X, we have alternate forms for ε(u) given by

m X ε(u) = sup{k φ(xi)yik : φ ∈ BX* } i=1 m X = sup{k ψ(yi)xik : ψ ∈ BY* }. i=1

We will denote by X ⊗ε Y the tensor product X ⊗ Y with the injective norm and its completion (which is called the injective tensor product) by X⊗ˆ ε Y. This injective tensor product may be considered as a subspace of B(X* × Y*) or of L(X*, Y) or L(Y*, X). Of particular interest to us is that

* * * * X ⊗ˆ ε Y ⊂ B(X × Y ), L(Y , X ), or L(X, Y). (2) The dual of the space of bounded operators on a Banach space Ë 51

It is the case that ε(u) ≤ π(u) for every u ∈ X ⊗ Y and ε(x ⊗ y) = kxkkyk for every x ∈ X and y ∈ Y. It can 1 1 1 be shown that c0⊗ˆ ε X = c0(X) and ` ⊗ˆ ε X = ` [X], where ` [X] is the space of all unconditionally summable sequences in X. Furthermore, the injective tensor product of C(K) with a Banach space X, where K is a compact Hausdor space, can be identied with the space C(K, X) of continuous functions from K into X.

If we consider u in X ⊗ε Y as a continuous function on the product space BX* × BY* , which is compact * when endowed with the weak topology on each unit ball, then we can think of X⊗ˆ ε Y as embedded into

C(BX* × BY* ). A linear functional on X⊗ˆ ε Y can be extended to all of C(BX* × BY* ). Thus, if B ∈ B(X × Y), its linearization B˜ dened by B˜(x ⊗ y) = B(x, y) can extend to a bounded linear functional on C(BX* × BY* ). Since we know the description of such functionals, we have the following theorem, which describes the dual space of X⊗ˆ ε Y.

Theorem 2. Let B ∈ B(X × Y). Then B˜, as dened above, is a bounded linear functional on X⊗ˆ ε Y if and only if there exists a regular Borel measure µ on the compact space BX* × BY* such that Z B(x, y) = φ(x)ψ(y)dµ(φ, ψ) (3)

BX* ×BY* for every x ∈ X, y ∈ Y. Furthermore, the norm of B˜ is given by

kB˜k = inf kµk, where µ ranges over the set of all measures that correspond to B in this way, and this inmum is attained.

A bilinear form B on X × Y is called an integral bilinear form if B˜, as in the theorem, is a bounded linear functional on X⊗ˆ ε Y, with integral norm dened by

kBkI = inf kµk, where the inmum is taken over all the regular Borel measures on BX* × BY* that satisfy (3). The Banach space of integral bilinear forms with integral norm is denoted by BI(X × Y). Hence we have

* (X⊗ˆ ε Y) = BI(X × Y).

We mention here that a bounded bilinear form B on X × Y is said to be nuclear if and only if there exist {φ } X* {ψ } Y* P∞ kφ kkψ k bounded sequences n in and n in which satisfy n=1 n n < ∞, such that

X∞ B(x, y) = φn(x)ψn(y) n=1 x ∈ X y ∈ Y P∞ φ ⊗ ψ B for every , . An expression n=1 n n is called a nuclear representation of . The nuclear norm of B is dened by ( ) X∞ X∞ kBkN = inf kφnkkψnk : B(x, y) = φn(x)ψn(y) , n=1 n=1 the inmum taken over all the nuclear representations of B. It is clear that kBk ≤ kBkN . B x y P∞ λ φ x ψ y It is worth noting that every nuclear bilinear form is also integral. If ( , ) = n=1 n n( ) n( ), where {λn} is a summable sequence of scalars, and {φn}, {ψn} are sequences of distinct unit vectors in X*, Y* re- spectively, the sum may be interpreted as an integral like that in (3), where µ is the measure with point masses λn at the points (φn , ψn). The notions of integral and nuclear can also be applied to operators. An operator T : X → Y corresponds * to a bilinear form BT dened on X×Y by BT(x, ψ) = hTx, ψi. Then T is said to be integral if BT is integral, and the integral norm kTkI is that of BT. The space of integral operators with that norm is denoted by I(X, Y). An * operator from X to Y is said to be nuclear if it is in the range of the canonical operator N : X ⊗ˆ π Y → L(X, Y) u P∞ φ ⊗ y L L x P∞ φ x y of unit norm that associates the tensor = n=1 n n with the operator u given by u( ) = n=1 n( ) n. 52 Ë Fernanda Botelho and Richard J. Fleming

The collection of such operators is written as N(X, Y), and the nuclear norm of such an operator T is dened as ( ) X∞ X∞ kTkN = inf kφnkkynk : T(x) = φn(x)yn , n=1 n=1 where the inmum is taken over all such representations of T where {φn}, {yn} are sequences in X*, Y re- P∞ kφ kky k N spectively such that n=1 n n < ∞. We note for future reference that the map is not necessarily one-to-one. From (1) we remember that B(X × Y) is isometrically isomorphic to the space L(X, Y*), where a bilinear form B is associated with the operator T by B(x, y) = hy, Txi. Such a form B is integral if and only if the asso- ciated T is integral and the two share their integral norms. This pairing establishes an isometric isomorphism * of BI(X × Y) with I(X, Y ). This allows us to say that

* * (X⊗ˆ ε Y) = BI(X × Y) = I(X, Y ) (4) where we use “equal" signs rather than the more precise congruences.

3 Dual spaces

* In thinking about the dual space of L(X, Y), we are drawn to the fact that the injective tensor product X ⊗ˆ ε Y * can be regarded as a subspace of L(X, Y) as in (2). In fact, we can write X ⊗ˆ ε Y = LA(X, Y), the space of approximable operators. Our main theorem will be a theorem recorded by Ryan [7, Theorem 5.33] that gives a special characterization of the dual of X⊗ˆ ε Y under certain conditions on X and Y. Our proof, though put together a bit dierently, is certainly inuenced by that of Ryan. We recall that a Banach space X is said to have the approximation property if for every Banach space Y, every operator T : X → Y may be approximated on compact sets by nite rank operators. Equivalently, X has the approximation property if and only if for every Banach space Y, every compact operator from Y into X is approximable. In addition, we need to use the notion of a Banach space having the Radon-Nikodým property. A vector measure µ on a σ-algebra Σ with values in a Banach space X satises the Radon-Nikodým Theorem if µ has bounded variation and is absolutely continuous with respect to a nite positive measure λ, then µ is an indenite Bochner integral with respect to λ. We say X has the Radon-Nikodým property if every such measure as above with values in X has the Radon-Nikodým property. Before stating and proving the theorem just mentioned, we need two lemmas, also given by Ryan [7].

Lemma 3. Let Σ be a σ-algebra of subsets of Ω, M(Σ) the Banach space of scalar measures on Σ with the variation norm, and X a Banach space. Then M(Σ)⊗ˆ π X is isometrically isomorphic to the space M1(Σ, X) of vector measures with the Radon-Nikodým property, with the variation norm kµk1 = |µ|1(Ω).

Proof. u Pm µ ⊗ x M Σ ⊗ X µ ∈ M Σ X µ E Pm µ E x If = i=1 i i in ( ) , it corresponds to ( , ) by ( ) = i=1 i( ) i. It is known that π(u) = kµk1 and that the correspondence is an isometry. It remains to show that M(Σ) ⊗ X is dense R in M1(Σ, X). Let µ ∈ M1(Σ, X). There exists f ∈ L1(|µ|1, X) such that µ(E) = E fd|µ|1 for all E ∈ Σ. Since L1(|µ|1, X) = L1(|µ|1)⊗ˆ π X, there exist bounded sequences {gn}, {xn} in L1(|µ|1), X respectively such that P∞ kg kkx k f P∞ g ⊗ x µ µ E R g d|µ| n µ n=1 n n < ∞ and = n=1 n n. Let n be dened by n( ) = E n 1 for each . Then P∞ µ ⊗ x ∈ M Σ X corresponds to n=1 n n 1( , ). Before stating the next lemma, we want to dene what is called a representing measure. If T is a bounded operator on the space C(K) of continuous functions on a compact, Hausdorf space K to X, we may dene a ** ** function µT on the σ-algebra of Borel subsets of K with values in the bidual X by µT(E) = T (χ(E)). Then for each φ ∈ X* we have * hφ, µT(E)i = T φ(E), The dual of the space of bounded operators on a Banach space Ë 53

* * where T φ ∈ C(K) is a regular Borel measure on K. The measure µT is called the representing measure for T. On the other hand, given a regular vector measure µ on the Borel sets of K with values in X**, we have an associated operator Tµ dened on X* by Tµ φ = φµ. The scalar measure φµ is regular whenever µ is regular.

Lemma 4. Let T : C(K) → X be an operator with representing measure µ. Then T is a nuclear operator if and only if µ has the Radon-Nikodým property. If T is nuclear, then kTkN = kµk1.

Proof. It is known that if T is an operator on a space whose dual space has the approximation property, then T* is nuclear if and only if T is nuclear. Since T : C(K) → X in this case, we know that C(K)* is the space of regular Borel measures on the σ-algebra BK of Borel sets in K, and since this space has the approximation property, we have T is nuclear if and only if T* is nuclear. Furthermore, it is easily shown that C(K)* is com- * plemented in M(BK) by a projection of norm one. If I denotes the embedding of C(K) into M(BK), then we see that IT* is the operator associated with µ, so that IT*(φ) = φµ for all φ ∈ X*. Supppose µ has the Radon-Nikodým property. We have from Lemma 3 that M(BK)⊗ˆ π X = M1(BK , X). Therefore, µ can be thought of as a member of the projective tensor product, so we can write

X∞ µ = µn ⊗ xn . n=1

P∞ kµ kkx k S IT* S φ φµ where n=1 n n < ∞. If we write = , then ( ) = and

X∞ X∞ S(φ) = φ( µn ⊗ xn) = φ(µn ⊗ xn) n=1 n=1 X∞ X∞ = φ(xn)µn = Jxn(φ)µn , n=1 n=1

J X X** u ∈ X**⊗ M B u P∞ Jx ⊗ µ where is the canonical mapping from into . Let ˆ π ( K) be given by = n=1 n n. We N X**⊗ M B L X* M B Nu φ P∞ Jx φ µ see that the canonical map from ˆ π ( K) to ( , ( K)) dened by ( ) = n=1 n( ) n agrees with S so that S (and therefore T*) is nuclear. On the other hand, suppose T and therefore T* is nuclear. Then S = IT* is nuclear and must be dened ** as above. It follows that µ must be in M1(BK), X ) and so satises the Radon-Nikodým property. At this point we are ready to state and prove the theorem we mentioned at the beginning of this section.

Theorem 5. Let X and Y be Banach spaces for which X* has the Radon-Nikodým property and either X* or Y* has the approximation property. Then * * * (X⊗ˆ ε Y) = X ⊗ˆ π Y .

Proof. In this proof, we will assume that Y* has the approximation property. We begin by considering the * * * canonical map N from X ⊗ˆ π Y to L(Y, X ) which we have considered in the discussion of nuclear operators. If we can show that N is one-to-one, then we will establish that the space N(Y, X*) is isometrically isomorphic X*⊗ Y* u P∞ φ ⊗ ψ ∈ X*⊗ Y* φ ∈ X* ψ ∈ Y* P∞ kφ kkψ k to ˆ π . Let = n=1 n n ˆ π , where n , n , and n=1 n n < ∞. Then Nu y P∞ ψ y φ Nu ψ → P∞ kφ k ( ) = n=1 n( ) n. Suppose = 0. We may assume that n 0 and n=1 n < ∞, for if not, we can alter the representation of u so that both hold. Let T ∈ L(Y*, X**) and ϵ > 0 be given. Since {ψn} is a zero convergent sequence in Y*, it is relatively compact and because Y* has the approximation property, there is a nite rank operator S such kT(ψn) − S(ψn)k < ϵ 54 Ë Fernanda Botelho and Richard J. Fleming

S S y* Pm ψ* y* φ* ψ* ∈ Y** φ* ∈ X** i ... m for every n. The operator has the form ( ) = i=1 i ( ) i , where i , i for = 1, , . * * * * ** We have S ∈ (X ⊗ˆ π Y ) = L(Y , X ), so that X∞ X∞ hu, Si = h φn ⊗ ψn , Si = hφn ⊗ ψn , Si n=1 n=1 X∞ = hφn , Sψni n=1 ∞ m X X * * = ψi (ψn)φi (φn) n=1 i=1 m ∞ X X * * = ( ψi (ψn)φi (φn)) i=1 n=1 m ∞ X * X * = φi ( ψi (ψn)φn) i=1 n=1 = 0. P∞ ψ y φ y ∈ Y The zero occurs in the last line above since we know that n=1 ( ) n = 0 for every . This is enough P∞ y** ψ φ y** ∈ Y** {Jy y ∈ Y} w* Y** to show that n=1 ( i) n = 0 for every since : is -dense in . We conclude that hu, Si = 0. This allows us to write

X∞ X∞ |hu, Ti| = |hu, T − Si + hu, Si| ≤ kTψn − Sψnkkφnk + 0 < ϵ kφnk. n=1 n=1 P∞ kφ k ϵ hu Ti T u Since n=1 n < ∞ and is arbitrary, we must have , = 0 for all from which we conclude that = 0 and N is injective. We remark that if the assumption is that X* has the approximation property, in the above argument let T ∈ L(X*, Y**) and proceed in the same way. From (4), we see that the proof will be complete if we can show that any T ∈ I(Y, X*) is also in N(Y, X*). Given such a T, we know it corresponds to a bilinear form B ∈ BI(X ⊗ Y) by hx, Tyi = B(x, y). Because B is integral, there is a regular Borel measure ν on K = Bx* × BY* such that Z B(x, y) = φ(x)ψ(y)dν. K By the Radon-Nikodým Theorem there is a Borel measurable function h such that |h(t)| = 1 for all t ∈ K such that dν = hd|ν|, where |ν| is the total variation of ν. Let µ = |ν|, S : Y → C(K), dened by Sy(φ, ψ) = ψ(y) and R : X → L∞(µ) given by Rx(φ, ψ) = φ(x). This gives Z B(x, y) = (Sy)(Rx)dµ. K

Let V denote the restriction of R* to L1(µ) (really JL1(µ)), and let I denote the canonical embedding of C(K) into L1(µ). By the duality between L1(µ) and L∞(µ), we can see that

hRx, ISyi = hx, VISyi = hx, Tyi.

Hence we can write T = VIS. Let U : C(K) → X* denote the map U = VI. Since X* has the Radon-Nikodým U ∈ N C K X* u P∞ µ ⊗φ ∈ property, it follows from Lemma 4 that ( ( ), ). From this we know there is some = n=1 n n C K *⊗ X* µ ∈ C K * φ ∈ X* n P∞ kµ kkφ k ( ) ˆ π with n ( ) , n for each , and n=1 n n < ∞ so that X∞ Nu(f) = µn(f ) = Uf for all f ∈ C(K). n=1

Let ψn ∈ Y* be dened by ψn(y) = µn(Sy). Then

|ψn(y)| = |µn(Sy)| ≤ kµnkkSyk = kµnk sup{|Sy(φ, ψ)| :(φ, ψ) ∈ K} ≤ kµnkkyk. The dual of the space of bounded operators on a Banach space Ë 55

kψ k kµ k P∞ kψ kkφ k w P∞ ψ ⊗ φ ∈ Y*⊗ X* Thus n ≤ n , so n=1 n n < ∞. Let = n=1 n n ˆ π . We have

X∞ X∞ Nw(y) = ψn(y)φn = µn(Sy)φn n=1 n=1 = U(Sy) = Ty.

Hence, Nw = T and T ∈ N(Y, X*). We have noted before that nuclear bilinear forms are integral, and the same holds for nuclear operators. This completes the proof.

* * * * Since X⊗ˆ ε Y = Y⊗ˆ ε X and X ⊗ˆ π Y = Y ⊗ˆ π X , we have the following corollary.

Corollary 6. If either X* or Y* has the Radon-Nikodým property and either X* or Y* has the approximation property, then * * * (X⊗ˆ ε Y) = X ⊗ˆ π Y .

If both X* and Y* have the Radon-Nikodým property and either X* or Y* has the approximation property, then * * X ⊗ˆ π Y has the Radon-Nikodým property. (See [2, p.29].) We now turn to the original question. What can we say about the dual of L(X, Y)? We begin with the * nite dimensional case. We recall from (2) that X ⊗ˆ ε Y is contained in L(X, Y) and equal to LA(X, Y). Since this space is equal to L(X, Y) if either X or Y is nite dimensional, we have the following results.

Theorem 7. Suppose X and Y are Banach spaces. (i) If Y is nite dimensional, then

* * * ** * L(X, Y) = (X ⊗ˆ ε Y) = X ⊗ˆ π Y for all Banach spaces X.

(ii) If X is nite dimensional, then

* * * ** * * L(X, Y) = (X ⊗ˆ ε Y) = X ⊗ˆ π Y = X⊗ˆ π Y for all Banach spaces Y.

Proof. Since nite dimensional spaces possess both the Radon-Nikodým property and the approximation property, the results follow immediately from Corollary 6. For example, we can get results of the kind mentioned in the very beginning, such as L(X, `∞(n))* = ** 1 1 ** 1 * ** ∞ ∞ ** (X ⊗ˆ π` (n) = ` (n, X ) or L(X, ` (n)) = X ⊗ˆ π` (n) = ` (n, X ).

0 q p * q p 0 1 1 Theorem 8. If 1 < p < q < ∞, then L(` , ` ) = ` ⊗ˆ π` , where p is conjugate to p, that is, p + p0 = 1.

Proof. Since both spaces involved satisfy both the Radon-Nikodým property and the approximation property, and since every operator from `q to `p is compact (by Pitt’s Theorem), the result follows from Corollary 6.

1 1 Since every operator from c0 to ` is compact and ` has the Radon-Nikodým property, we also have

1 * 1 1 * ∞ ∞ Theorem 9. L(c0, ` ) = (` ⊗ˆ ε` ) = ` ⊗ˆ π` .

* * We note here that if either X or Y has the approximation property, then X ⊗ˆ ε Y = LK(X, Y), the compact operators from X to Y. Hence the following theorem follows from Corollary 6.

Theorem 10. (Grothendieck [3]. (See also [6].) If either X** or Y* has the Radon-Nikodým property and either has the approximation property, then * ** * LK(X, Y) = X ⊗ˆ π Y .

The three theorems previous to Theorem 10 were, of course, actually included in this last one, since in each such case, the bounded operators from X to Y were all compact, hence as we said in the beginning, “probably in Grothendieck." 56 Ë Fernanda Botelho and Richard J. Fleming

The most general statement we can make concerning duals of bounded operators is that

* * ** L(X, Y ) = (X⊗ˆ π Y) .

This is most useful when we can describe the projective tensor product of X and Y. Since we know that 1 1 1 1 1 ` ⊗ˆ π X = ` (X) and ` ⊗ˆ π` = ` , we can state the following.

Theorem 11. For any Banach space X, (i) 1 * 1 * 1 * ∞ * L(` , X ) = (` ⊗ˆ π X) = ` (X) = ` (X ). (ii) L(`1, X*)* = (`∞(X*))* = `1(X)**. (iii) 1 ∞ * 1 1 ** 1 ** ∞ * L(` , ` ) = (` ⊗ˆ π` ) = (` ) = (` ) .

Furthermore, if L1(µ) denotes the usual integration space, then

1 * 1 * 1 * (L (µ)⊗ˆ π X) = (L (µ, X)) = L(L (µ), X ), and for compact K, C(K)⊗ˆ ε X = C(K, X), so we have

Theorem 12. For a Banach space X , (i) L(L1(µ), X*)* = L1(µ, X)**. (ii) If X* has the Radon-Nikodým property, then

* * * * * ** * *** C(K, X) = (C(K)⊗ˆ ε X) = C(K) ⊗ˆ π X , so that L(C(K) , X ) = C(K, X) .

4 More on Tensor products

2 2 It was shown in Ryan’s book, [7, p. 23], that the tensor diagonal subspace of ` ⊗ˆ π` is isometrically isomorphic to `1. Actually, it is a 1-complemented subspace. The following is also true.

1 p q Theorem 13. Let 1 < p, q < ∞ with p and q conjugate. Then ` is isometrically embedded in ` ⊗ˆ π` .

Proof. The argument follows as in Ryan’s book. We denote by {en} and {fn} the standard Schauder bases for p q ` and ` respectively. We consider the subspace W spanned by en ⊗ fn. An arbitrary element u in W is of the u P∞ α e ⊗ f π u P∞ |α | B x y form = i=1 n n n. First, we have ( ) ≤ n=1 n . Towards the reversed inequality we dene ( , ) = P∞ sgn α x y |B x y | kxk kyk n=1 ( n) n n which implies that ( , ) ≤ p q , from an application of the Hölder inequality. We recall that sgn(αn) is a modulus 1 complex number such that sgn(αn)αn = |αn| if αn ≠ 0, otherwise, it is equal kBk π u hu Bi P∞ |α | π u P∞ |α | W to zero. Since ≤ 1 we have ( ) ≥ , = n=1 n . This implies that ( ) = n=1 n . Therefore is 1 1 p q 2 1 isometrically isomorphic to ` , implying that ` is isometrically embedded in ` ⊗ˆ π` . As in the ` case, ` is p q a 1-complemented subspace in ` ⊗ˆ π` .

The only linear mapping determined by the bilinear map

X∞ P(x, y) = xn yn en ⊗ fn n=1 is a norm 1 projection. Several observations can be derived from this fact: The dual of the space of bounded operators on a Banach space Ë 57

∞ p q * p p (i) ` is a 1-complemented subspace of (` ⊗ˆ π` ) = L(` , ` ). p p (ii) L1 is isomorphic to a subspace of L(` , ` ). This follows from a result by Pelczynski, as mentioned in [5]. (iii) L(`p , `p) contains a subspace isomorphic to C[0, 1]*. p p p q (iv) The unit balls in L(` , ` ) and ` ⊗ˆ π` are not weakly compact. This follows straightforwardly from The- orem 65 in [4]. p The space of sequences (xn) in a Banach space X for which the scalar sequence (φ(xn)) belongs to ` for all * w φ ∈ X is called the space of weakly p-summable sequences in X and is denoted by `p (X). If q is conjugate to p, the norm is given by

∞ w X p 1/p k(xn)kp = sup{( |φ(xn)| ) : φ ∈ BX* } n=1 X∞ = sup{( kλn xnkX : λ ∈ B`q }, 1 < p < ∞. n=1 q The norm for p = 1 is the same as above, except p is replaced by 1 and ` by c0. The following fact is stated by Ryan [7, p.134], and we sketch a proof.

Theorem 14. For any Banach space X, w q (i) if q is conjugate to p, then `p (X) = L(` , X) for 1 < p < ∞, and w (ii) for p = 1, `1 (X) = L(c0, X).

w Proof. Let 1 < p < ∞ be given and q its conjugate. For (xn) ∈ `p (X), let S((xn)) = T, where T is the operator `q X hTψ φi P∞ φ x α φ ∈ X* ψ α ∈ `q R X* → `p dened from to by , = n=1 ( n) n for and = ( n) . Let : be dened by Rφ = (φ(xn)). Then it is easy to see that

∞ X p 1/p w kRφk = ( |φ(xn)| ) ≤ kφkk(xn)kp , n=1 w so R is bounded by k(xn)k. Furthermore, given ϵ > 0, we may choose φ ∈ BX* so that k(xn)kp < P∞ |φ x |p 1/p ϵ kR φ k ϵ kRk ϵ R T ( n=1 ( n) ) + = ( ) + ≤ + . Since is dened to be the adjoint of , we conclude that w q kTk = kRk = k(xn)kp and S is an isometry. If T ∈ L(` , X), then let xn = T(en), where {en} denotes the usual q Schauder basis for ` . Then it is easy to see that S((xn)) = T. This completes the proof for part (i). The proof of the second part is similar.

q * w * * w * Thus L(` , X) = (`p (X)) and L(c0, X) = (`1 (X)) . We do not have a good characterization of the duals of the weakly p-summable spaces, although if X is a dual space with predual Y, we could write

w * q * q ** `p (X) = L(` , X) = (` ⊗ˆ π Y) , and w * * ** `1 (X) = L(c0, X) = (c0⊗ˆ π Y) . 0 In particular, if q > r, and r is conjugate to r, it follows from Corollary 4.24 in [7] that

0 0 w r * q r * q r ** q r `p (` ) = L(` , ` ) = (` ⊗ˆ π` ) = ` ⊗ˆ π` .

Of course, this also follows from Theorem 8. w w Choi and Kim [1] introduced a subspace of `p (X), denoted by `ˇp (X), which consists of sequences (xn) from X for which X p 1/p sup ( |φ(xn)| ) → 0 as m → ∞. φ∈B X* n≥m P∞ e ⊗ x `p⊗ X {e } This denition is just what is needed to show that n=1 n n converges in ˆ ε , where n is the usual p w Schauder basis for ` and (xn) ∈ `ˇp (X). We state the following interesting fact as a lemma. 58 Ë Fernanda Botelho and Richard J. Fleming

Lemma 15. If 1 ≤ p < ∞, then w p `ˇp (X) = ` ⊗ˆ ε X.

Proof. x ∈ `ˇw X S x P∞ e ⊗ x Given ( n) p ( ), let (( n)) = n=1 n n . Then ∞ ∞ X X p 1/p w ε(S((xn)) = sup{k φ(xn)enkp : φ ∈ BX* } ≤ sup ( |φ(xn)| ) = k(xn)kp . φ∈B n=1 X* n=1

On the other hand, given ϵ > 0 we may choose φ ∈ BX* such that w X p 1/p X k(xn)kp < ( |φ(xn)| ) + ϵ ≤ ε( en ⊗ xn) + ϵ. n n S u Pm z ⊗ x `p ⊗ X We conclude that is an isometry. For surjectivity, suppose = k=1 k k is an element of ε , when z z ∈ `p x ∈ X k n y Pm z x u P∞ e ⊗ y k = ( kj)j and k for each . For each let n = k=1 kn k. Then = n=1 n n and we show w that (yn) ∈ `ˇp (X). If ϵ > 0 is given, there exists for each j = 1, ... , m a positive integer nj such that

X p 1/p ϵ ( |zjn| ) < . mkxjk k≥nj

For φ ∈ BX* and n0 = max{nj : j = 1, ... , m}, we have

m m X p 1/p X X p 1/p X X p 1/p ( |φ(yn)| ) = ( |φ( zjn xj| ) ≤ ( |φ(xj|)zjn| ) n≥n0 n≥n0 j=1 n≥n0 j=1 m m X X p 1/p X X p 1/p = ( |φ(xj)zjn| ) ≤ ( |φ(xj)zjn| ) j=1 n≥n0 j=1 n≥n0 m X X p 1/p ≤ kφkkxjk( |zjn| ) < ϵ. j=1 n≥n0 w Since this holds for every φ ∈ BX* , we see that (yn) ∈ `ˇp and S((yn)) = u so that the range of S contains a p dense subspace of ` ⊗ˆ ε X and S is surjective. w With this we can describe the dual space of `ˇp (X).

w * Theorem 16. (Choi and Kim) Let 1 < p < ∞ and q its conjugate. Then `ˇp (X) consists of all linear functionals f of the form ∞ ∞ X X * f((xn)) = h(xn), f i = αjn xj (xn), j=1 n=1 q * * where zj = (αjn) ∈ ` and xj ∈ X for each j such that ∞ X * kzjkqkxj k < ∞. j=1

w Proof. It is clear that any f of the form given above is a continuous linear functional on `ˇp (X) with ∞ X * kf k ≤ kzjkqkxj k. j=1 w * p * q * * By Lemma 15 and Theorem 5, we know that `ˇp (X) = (` ⊗ˆ ε X) = ` ⊗ˆ π X , where hen ⊗ xn , zj ⊗ xj i = he z ihx x*i f ∈ `ˇw X * z α ∈ `q {x*} ∈ X* P∞ kz k kx* k n , j n , j . Hence, if p ( ) , there exist j = ( jn) and j with j=1 j q n < ∞ f P∞ z ⊗ x* ∈ `q⊗ X* such that corresponds to j=1 j j ˆ π , and ∞ ∞ ∞ ∞ ∞ X X X * X X * h(xn), f i = f ( en ⊗ xn) = hen , zjixj (xn) = αjn xj (xn). n=1 j=1 n=1 n=1 j=1 The dual of the space of bounded operators on a Banach space Ë 59

We wish to express appreciation to David Blecher who suggested the nite dimensional result and also sug- gested looking at the book of Ryan.

References

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