Extendibility of Bilinear Forms on Banach Sequence Spaces

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This theorem will follow as a consequence of Theorem 2.2 below. We also characterize the Banach sequence spaces satisfying a bilinear Hahn-Banach theorem as those “between” c0 and ℓ∞ (see Corollary 2.4). As a byproduct, we obtain a partial answer to the following open problem: if a sequence Xn of n-dimensional Banach spaces is uniformly complemented L n in some ∞, must these spaces be uniformly isomorphic to ℓ∞? Corollary 2.3 gives a positive answer for sections of a Banach sequence space (see Proposition 2.5).

1.1. Preliminaries. We briefly collect here some basic definitions that will be used through- out the paper. We will consider real or complex Banach spaces, that will be denoted X, Y,.... Unless otherwise stated they will be assumed to be infinite dimensional. The duals will be denoted by X∗, Y ∗, . . .. Given two Banach spaces X and Y , we write X Y if they 1 ≈ are isomorphic and X Y if they are isometrically isomorphic. We refer to [1, 25] for basic concepts and notations≈ on Banach spaces. We denote L 2(X,Y ) for the Banach space of all scalar valued, continuous, bilinear mappings (in short bilinear forms) on X Y . We write L 2(X) whenever Y = X. × The space of extendible bilinear forms is denoted by E 2(X,Y ). The extendible norm

T E = T E 2 := inf c> 0 : for all W X,Z Y there is an extension of T k k k k (X,Y ) { ⊇ ⊇ to W Z with norm c × ≤ } 2 makes E (X,Y ) a Banach space. Since every ℓ∞(I) space is injective (in fact, has the metric extension property), every bilinear form on such spaces is extendible and the extendible and uniform norm coincide. Moreover, a bilinear form T on X Y is extendible if and only if it × extends to ℓ∞(I) ℓ∞(J), for some ℓ∞(I) X and ℓ∞(J) Y . The supremum defining the extendible norm can× be taken only over the⊃ extensions to ℓ⊃(I) ℓ (J). ∞ × ∞ We write L (X; Y ) for the space of all (continuous, linear) operators u : X Y . We denote by Π (X; Y ) the space of absolutely summing operators, Γ (X; Y ) for the →-factorable and 1 ∞ ∞ ∆2(X; Y ) for the 2-dominated. Their corresponding norms are, respectively, π1, γ∞ and δ2 (see [13, 15] for definitions and basic properties). We are going to use the theory tensor products and operator ideals as presented in [13]. We recall some notation and definitions for completeness. The projective tensor norm π is defined, for a z in the tensor product X Y , by ⊗ r π(z) = inf  x y  , k jk k jk Xj=1  where the infimum is taken over all the representations of zof the form z = r x y . The j=1 j ⊗ j right-injective associate of π is denoted by π . This tensor norm is greatestP right-injective tensor norm and makes the following inclusion\ an isometry:

1 ,( ∗ X Y ֒ X ℓ (B ⊗π\ → ⊗π ∞ Y ∗ where BY ∗ is the unit ball of Y (see [13, Theorem 20.7.] for details). Likewise, the injective associate /π is the largest injective tensor norm and is induced by the isometric inclusion \ 1 .( ∗ X Y ֒ ℓ (B ∗ ) ℓ (B ⊗/π\ → ∞ X ⊗π ∞ Y BILINEARFORMSONBANACHSEQUENCESPACES 3

The metric extension property of ℓ∞(I) spaces implies that extendible bilinear forms are precisely the /π -continuous ones: \ 1 E 2(X,Y ) = X Y ∗ . ⊗/π\ We refer to [13, 15] for all the basic (and not so basic) facts and any undeﬁned notation on tensor norms and operator ideals.

Given a family Xn n of Banach spaces where dim Xn = n, we say that Xn are K-uniformly complemented in {X if} for each n we have a mapping i : X X and q : X X such n n → n → n that qn in is the identity on Xn and in qn K. In this case we also say that X contains ◦ k k k k≤ n Kn Xn uniformly complemented. We note that if X contains uniform copies of ℓ∞ (i.e., with n the sup norm), then the ℓ∞ are uniformly complemented since they are injective spaces. 1.2. Banach sequence spaces. By a Banach sequence space (also known as Köthe sequence N space) we will mean a Banach space X K of sequences in K such that ℓ1 X ℓ∞ with ⊆ N ⊆ ⊆ norm one inclusions satisfying that if x K and y X are such that xn yn for all n N, then x belongs to X and x y∈. ∈ | | ≤ | | ∈ k k ≤ k k If X is a Banach sequence space, we denote by en n the sequence of canonical vectors, which is always a 1-unconditional basic sequence.{ We} define X = span e ,...,e and N { 1 N } X0 = span en n. This last space is usually referred to as the minimal kernel of X. Given x X we{ write} xN = (x ,...,x ). There are inclusions iX : X ֒ X and projections ∈ 1 N N N → πX : X X given by iX (x ,...,x ) = (x ,...,x , 0, 0,...) and πX (x) = xN . The N → N N 1 N 1 N N inclusions are isometric and the projections have norm 1. For the case X = ℓp (1 p ), N ≤ ≤∞ we write ℓp for XN . Given a Banach sequence space X, its Köthe dual is defined as N X× = (z ) K : z x < for all x X . { n n ∈ n | n n| ∞ ∈ } P With the norm z × = sup z x it is again a Banach sequence space. k kX kxkX ≤1 n | n n| Following [24, 1.d], a Banach sequenceP space X is said to be r-convex (with 1 r< ) if there exists a constant κ> 0 such that for any choice x ,...,x X we have ≤ ∞ 1 N ∈ N N 1/r ∞ 1/r x (k) r κ x r . | j | ≤ k jkX Xj=1 k=1 X Xj=1

On the other hand, X is s-concave (with 1 s< ) if there is a constant κ> 0 such that ≤ ∞ N N 1/s 1/s ∞ x s κ x (k) s k jkX ≤ | j | Xj=1 Xj=1 k=1 X

for all x1,...,xN X. It is well known∈ that ℓ is r convex for 1 r p and s-concave for p s< . p ≤ ≤ ≤ ∞ The following result is probably known. However we were not able to ﬁnd a proper reference of this fact and we include here a short proof. It is modelled along the same lines as the proof of the fact that if the canonical vectors form a basis of X then both duals coincide. Proposition 1.2. If X is a Banach sequence space, its K¨othe dual X× is a 1-complemented subspace of the usual dual X∗. 4 DANIELCARANDOANDPABLOSEVILLA-PERIS

Proof. Let us see first that the mapping i : X× X∗ defined by i(z) = ϕ : X K, with → z → ϕz(x)= n znxn, is an isometry. It is clearly well defined; moreover P ϕz = sup znxn sup znxn = z X× . k k x∈BX ≤ x∈BX | | k k Xn Xn

To see the reverse inequality, for any t,s K we take ε(t,s) K with ε(t,s) = 1 such that st = ε(t,s)st; then for every x B and∈ every z X× we have∈ | | | | ∈ X ∈

znxn = ε(zn,xn)znxn = ε(zn,xn)znxn sup znan = ϕz , | | ≤ a∈BX k k Xn Xn Xn Xn

which gives z X× ϕz . On the otherk k hand,≤ kthek mapping q : X∗ X× given by q(ϕ)= ϕ(e ) defines a norm-one → n n projection. Indeed, given x X and fixed N we have ∈ N N N x ϕ(e ) = ε(x , ϕ(e ))x ϕ(e )= ϕ ε(x , ϕ(e ))x e | n n | n n n n n n n n nX=1 nX=1 nX=1 N ϕ ε(xn, ϕ(en))xnen ϕ x . ≤ k k X ≤ k k k k Xn=1 ∞ This shows that n=1 xnϕ(en) ϕ x , which gives that q is well defined and q(ϕ) | | ≤ k k k k k k≤ ϕ . Furthermore,Pq is a projection, since clearly q i(z)= q(ϕz) = (zn)n = z. k k ◦ 2. Extension of bilinear forms on Banach sequence spaces

In what follows KG denotes the Grothendieck’s constant. We begin by proving the following known fact, which was stated as Theorem 3.4 in [20] without the estimates for the norms (see [11, Lemma 2.4] for a result in the same spirit).

Proposition 2.1. If every bilinear form B : X Y K is extendible with B E K B , then every operator u : X∗ ℓ is absolutely 1-summing× → and π (u) K K ku .k ≤ k k → 2 1 ≤ G k k Proof. We first note that, by definition of the tensor norm /π (see [13, Section 20.7]), E 2(X,Y ) is isometrically the dual of X Y . Then, our hypothesis\ is equivalent to the ⊗/π\ inequality π K/π on X Y and, as a consequence, we also have π K/π on X Y . Since both π≤ and /π\ are⊗ right-injective, an application of Dvoretzky’s\ ≤ theore\m [15,⊗ 19.1] and the previous\ inequality\ gives an isomorphism (1) X ℓN X ℓN ⊗/π\ 2 −→ ⊗π\ 2 N L N ∗ N with norm at most K. Since ℓ2 is finite dimensional, (X; ℓ2 ) and X ℓ2 coincide as sets. Then, the embedding in [13, Section 17.6] is actually surjective and [13⊗, Sections 21.5 and 1 27.2] give X∗ ℓN Γ (X; ℓN ) (see [15, Chapter 9] or [13, Section 18] for the definition ⊗w∞ 2 ≈ ∞ 2 of Γ∞(X; Y )). Therefore, 1 1 1 (2) X ℓN ∗∗ Γ (X,ℓN ) ∗ X∗ ℓN ∗ Π (X∗; ℓN ) . ⊗π\ 2 ≈ ∞ 2 ≈ ⊗w∞ 2 ≈ 1 2 Now, by [13, Sections 17.12 and 27.2], the operator ideal ∆2 is associated to w2 and Π1 is associated to π . On the other hand Grothendieck’s inequality [13, Section 20.17] states that \ BILINEARFORMSONBANACHSEQUENCESPACES 5

L ∗ N 1 ∗ N w2 KG/π and clearly we have (X ; ℓ2 ) ∆2(X ; ℓ2 ). Using (2) to take biduals in (1) we≥ have an\ isomorphism ≈ 1 (3) L (X∗; ℓN ) X ℓN ∗∗ X∗∗ ℓN ∗∗ Π (X∗; ℓN ), 2 −→ ⊗/π\ 2 −→ ⊗π\ 2 ≈ 1 2 where the ﬁrst mapping has norm bounded by KG and the second one by K. Since both L N and Π1 are maximal operator ideals, the same holds if we put ℓ2 instead of ℓ2 .

With this result we can now prove the following one, from which Theorem 1.1 follows as an immediate consequence. Theorem 2.2. Let X be a Banach space with an unconditional basis and Y be any infinite dimensional Banach space such that every bilinear form on X Y is extendible. Then X c . × ≈ 0 Proof. Let us see first that, under our assumptions, the basis of X must be shrinking. Sup- pose it is not. Since it is unconditional, by James theorem [19, Corollary 2] (see also [1, Theorem 3.3.1]) X must contain a complemented copy of ℓ1. Since the property of all bilinear forms being extendible is inherited by complemented subspaces, it follows that every bilinear form on ℓ Y is extendible. This implies [22, Lemma 6] that every continuous linear op- 1 × erator from Y to ℓ∞ is absolutely 2-summing. By the so called Lp-Local Technique Lemma for Operator Ideals [13, Section 23.1], the same holds for every operator from Y to ℓ∞(I), for any index set I. But this is not possible, since there exists an isometric embedding from Y into some ℓ∞(I), and this cannot be absolutely 2-summing (otherwise, the identity on Y would be so, but Y is infinite dimensional). This means that the canonical basis of X must be shrinking. We can assume that the basis is 1-unconditional, so that the coordinate basis is an 1-unconditional basis of X∗. We also ∗ know from Proposition 2.1 that all operators from X to ℓ2 are absolutely 1-summing. By ∗ 2 [23] (see also [25, Theorem 8.21]) this implies that the basis of X is (KGK) -equivalent to the basis of ℓ1. Although ℓ1 has many non-isomorphic preduals, if the coordinate basis is equivalent to that of ℓ1, a standard computation shows that the canonical basis on X must 2 be (KGK) -equivalent to the basis of c0.

Note that the proof not only shows that X must be isomorphic to c0, but also gives an esti- mation of the Banach-Mazur distance between X and c0 whenever X has an 1-unconditional basis. As a consequence, we can also characterize the pairs of Banach sequence spaces on which every bilinear form is extendible. Corollary 2.3. If X and Y are Banach sequence spaces, then the following are equivalent. (i) L 2(X,Y )= E 2(X,Y ) and B E K B for all bilinear form B on X Y . k k ≤ 1k k × (ii) The canonical basic sequence of X and Y are K2-equivalent to the canonical basis of c0. N N N (iii) K3 := sup d(XN ,ℓ∞), d(YN ,ℓ∞) : N is ﬁnite (where d denotes the Banach- Mazur distance).{ ∈ } (iv) The spaces X and Y (N N) are K -uniformly complemented in some L (µ). N N ∈ 4 ∞ Moreover, we have K K K (K K )2 and K K2 K4 K8. 4 ≤ 3 ≤ 2 ≤ G 1 1 ≤ 2 ≤ G 4 Proof. If (i) holds on X Y , then the same holds for XN YN for any N and, by the density lemma [13, Section 13.4],× for X Y . By Theorem 2.2,× both bases are (K K)2-equivalent 0 × 0 G to the basis of c0. 6 DANIELCARANDOANDPABLOSEVILLA-PERIS

The implications (ii) (iii) (iv) are immediate, as well as the inequalities K K ⇒ ⇒ 4 ≤ 3 ≤ K2. 2 If (iv) holds, bilinear forms on XN YN are extendible with E K4 and, as before, the same holds for X Y . By Theorem× 2.2 their canonical basesk · k ≤ are Kk2 ·K k 4-equivalent to 0 × 0 G 4 the canonical basis of c0, which is (ii). Now suppose (ii) holds and take a bilinear form B : X Y K. We know from Proposi- tion 1.2 that X× is 1-complemented in X∗. Then (X×)∗×is isometrically→ a (complemented) ∗∗ × × ∗ ×× subspace of X . Since (ii) implies X = ℓ1, we also have (X ) = X = ℓ∞. The same holds for Y , so we obtain the following diagrams:

i1 / ×× i2 / ∗∗ j1 / ×× j2 / ∗∗ X X O X Y Y O Y , u v

ℓ∞ ℓ∞ where i1, i2, j1 and j2 are isometric injections and u and v are isomorphisms with (4) u u−1 K and v v−1 K . k k k k≤ 2 k k k k≤ 2 We can extend B (in the canonical way) to a bilinear form B˜ : X∗∗ Y ∗∗ K with the × → same norm as B, and then define a bilinear form B on ℓ ℓ by B = B˜ (i u, j v). ∞ × ∞ ◦ 2 ◦ 2 ◦ We have obtained the factorization B = B (u−1 i , v−1 j ). Since on ℓ ℓ every b 1 1 b ∞ ∞ bilinear form is extendible (with the extendible◦ norm◦ equal to◦ the usual norm), from× the ideal b property of extendible bilinear forms and inequalities (4) we conclude that B is extendible and B E K2 B . k k ≤ 2 k k It follows from the previous corollary (and its proof) that a Banach sequence space on which every bilinear form is extendible must satisfy the sublattice inclusions (5) c X ℓ . 0 ⊂ ⊂ ∞ Conversely, let X be a Banach sequence space satisfying (5). By a closed graph argument, both inclusions are continuous and it is easy to check that X satisfies the equivalent conditions of Corollary 2.3. Note also that a Banach sequence space satisfies (5) if and only if its Köthe dual is ℓ1. As a consequence, we have the following version of Theorem 1.1 for Banach sequence spaces. Corollary 2.4. The Banach sequence spaces X on which every bilinear form is extendible × are those satisfying (5). Also, this happens if and only if X = ℓ1.

Examples of such spaces are c0 ℓ∞, c0(ℓ∞) and ℓ∞(c0). It is not hard to see that these spaces are mutually non-isomorphic⊕ Banach sequence spaces (see also [12], where the authors show that c0(ℓ∞) and ℓ∞(c0) are not isomorphic even as Banach spaces). If a sequence Xn of n-dimensional Banach spaces is uniformly complemented in some L∞, n it is an open problem if these spaces have to be uniformly isomorphic to ℓ∞. Taking X = Y in Corollary 2.3, the implication (iv) (iii) gives the following partial answer. ⇒ Proposition 2.5. If the N-dimensional sections XN of a Banach sequence space X are N uniformly complemented in some L∞(µ), then they must be uniformly isomorphic to ℓ∞.

Note also that, since ℓ∞ and c0 are the only symmetric Banach sequence spaces satisfying (ii) of Corollary 2.3, these two are the only symmetric Banach sequence spaces on which every bilinear form is extendible. BILINEARFORMSONBANACHSEQUENCESPACES 7

If X ,...,X are Banach spaces such that every n-linear form on X X is extendible, 1 n 1×···× n then it is known (and easy to see) that so is every bilinear form on Xi Xj for each pair i = j. Indeed, given B L 2(X X ), we can multiply it by linear functionals× to obtain a 6 ∈ i × j n-linear form on X1 Xn. This is extendible by our hypothesis. From this, it is rather immediate to conclude×···× that B is extendible. As a consequence, multilinear versions of our results follow directly from the bilinear ones. If X and Y are Banach sequence spaces such that every bilinear form on X Y is extendible, N × then we know from Theorem 2.3 (iii) that both X and Y contain the ℓ∞ uniformly. We can extend this statement to subspaces of Banach lattices.

Proposition 2.6. Let X1, X2 be subspaces of Banach lattices such that every n-linear form on X1, X2 is extendible. Then every inﬁnite dimensional complemented subspace of each Xj N contains the ℓ∞ uniformly.

Proof. Suppose that there exists a complemented subspace E of X1 that does not contain the N ℓ∞ uniformly. By [21, Corollary 1], E must contain uniformly complemented N-dimensional subspaces E such that sup d(E ,ℓN ) < for p = 1 or 2. Since E is complemented N N N p ∞ in X1, the EN are also uniformly complemented in X1. On the other hand, again by [21, Corollary 1], X2 must contain uniformly complemented N-dimensional subspaces FN such that sup d(F ,ℓN ) < for q = 1, 2 or . Our hypotheses ensure that bilinear forms on N N q ∞ ∞ X1 X2 are extendible. Since EN and FN are uniformly complemented in X1 and X2 and they × N N are (uniformly) isomorphic to ℓ and ℓ , there must exist K > 0 such that B E 2 N N p q k k (ℓp ,ℓq ) ≤ K B L 2 N N for all N. Now, the density lemma [13, Section 13.4] implies that every k k (ℓp ,ℓq ) bilinear form on ℓ ℓ (or ℓ c ) must be extendible, which contradicts Theorem 2.2. p × q p × 0 The converse of Proposition 2.6 does not hold. For example, the Schreier space is c0- saturated and there are non-extendible bilinear forms on it (since there are bilinear forms which are not weakly sequentially continuous). Another conterexample of the converse is d∗(w, 1), the predual of the Lorentz sequence space d(w, 1) (see [27] or [16] for a descrip- tion of the predual). Since these examples are Banach sequence spaces, they also show that N assertion (iii) in Theorem 2.3 is strictly stronger than containing ℓ∞ uniformly. In Banach sequence spaces, diagonal bilinear forms are the simplest ones. These are the bilinear forms Tα : X1 X2 C given by × → ∞ 1 2 1 2 Tα(x ,x )= αkxkxk , Xk=1 for some sequence (αk)k of scalars. We end this note showing, under some assumptions, which are the spaces on which all diagonal bilinear forms are extendible. Following standard notation, given a symmetric Banach sequence space X we consider the fundamental function of X, given by λ (N) := N e for N N. X k=1 k X ∈ Given two sequence of real numbers (an)n and P (bn)n we write an bn whenever there is a universal constant C > 0 such that an Cbn for every n. If an bn and bn an, we write a b . ≤ n ≍ n Theorem 2.7. Let X and Y be symmetric Banach sequence spaces, each being 2-convex or 2-concave. Then all diagonal bilinear forms on X Y are extendible if and only if either X = Y = ℓ or X,Y c ,ℓ × 1 ∈ { 0 ∞} 8 DANIELCARANDOANDPABLOSEVILLA-PERIS

Proof. The if part is clear: by [8, Proposition 2.3] (see also [9, Proposition 1.2]) on ℓ1 diagonal bilinear forms are integral (and, therefore, extendible), and in the other cases all bilinear forms are extendible. N For the converse, we consider the diagonal bilinear form given by φN (x,y) = i=1 xiyi. It is easily computed that φN L 2 N = 1; on the other hand, by [8, PropositionP 1.1] or k k (ℓ2 ) N N [11, Proposition 2.5] we have φ E 2 N = φ N 2 N = N. Let now id : ℓ X and k N k (ℓ2 ) k N k (ℓ2 ) X 2 → N idN : ℓN Y be the identity mappings. Comparing the usual and extendible norms of the Y 2 → N bilinear forms φ and φ ((idN )−1, (idN )−1), we get N N ◦ X Y N idN idN (idN )−1 (idN )−1 . k X kk Y kk X k Y k By [28, 16.4] (see also [14, page 138]), since X is a symmetric Banach sequence space we have d(ℓN , X )= idN (idN )−1 (and the same for Y ). Therefore, 2 N k X k k X k N N idN idN (idN )−1 (idN )−1 = d(ℓN , X )d(ℓN ,Y ) . k X kk Y kk X k Y k 2 N 2 N N N Since we always have d(ℓ2 , XN ) √N and d(ℓ2 ,YN ) √N, we can conclude that √N N N N −1 ≤ ≤ ≍ d(ℓ2 , XN )= idX (idX ) (and the same for YN ). We now apply [14, Lemma 1 (i)] and get: k k k k 1 λ (N) max , X 1. λX (N) N ≍ From this we can conclude that X must be ℓ1, c0 or ℓ∞. Indeed, suppose we split the natural numbers N = I J, so that 1 1 and λX (N) 1. We have then that ∪ λX (N) N∈I ≍ N N∈J ≍ (λ (N)) is bounded and (N ) (λ (N)) . Since (λ (N)) N is non-decreasing, X N∈I N∈J X N∈J X N∈ either I or J must be finite. If J is finite, then (λX (N))N∈N is bounded and then the norm in X is equivalent to the sup norm and X is c or ℓ . If I is finite, then N 0 ∞ (λX (N))N∈N. Although the fundamental sequence of a symmetric Banach sequence space does not characterize the norm, for this extreme case it is possible to prove that the norm on X must be isomorphic to ℓ1: from the estimate N (λX (N))N∈N we easily obtain × λ × (N) 1 and, by the previous case, X must be ℓ . Then we have X = ℓ . Proceeding X ≍ ∞ 1 in the same way, Y has to be either ℓ1, c0 or ℓ∞. It remains to show that on c0 ℓ1 and on ℓ1 ℓ∞ there are non-extendible diagonal bilinear forms. The mapping c ℓ given× by (x,x′) ×x′(x) is the diagonal bilinear form induced by 0 × 1 7→ the formal identity. An extension of this mapping to c0 ℓ∞ would give a projection from ℓ to ℓ (see [13, 1.5]), which does not exist. For ℓ ℓ ×we can reason in a similar way. ∞ 1 1 × ∞ Both assumptions on symmetry and concavity/convexity in “only if” part of the previous theorem cannot be omitted. Indeed, if we take c0 ℓ1 (that seen as a sequence space is neither symmetric nor 2-concave or 2-convex), then every⊕ diagonal bilinear form on (c ℓ ) (c ℓ ) 0 ⊕ 1 × 0 ⊕ 1 is the sum of a diagonal bilinear form on c0 c0 and a diagonal bilinear form on ℓ1 ℓ1, and is therefore extendible. × ×

Acknowledgements We wish to thank Andreas Defant for valuable comments and useful conversations that improved the paper. We also wish to thank the referees, for their suggestions which improved the ﬁnal shape of the paper and for pointing to us some questions and remarks which lead to Corollary 2.4 and Proposition 2.5. BILINEARFORMSONBANACHSEQUENCESPACES 9

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Departamento de Matematica´ - Pab I, Facultad de Cs. Exactas y Naturales, Universidad de Buenos Aires, (1428) Buenos Aires, Argentina and IMAS - CONICET E-mail address: [email protected]

Instituto Universitario de Matematica´ Pura y Aplicada and DMA, ETSIAMN, Universitat Politecnica` de Valencia,` Valencia, Spain E-mail address: [email protected]