THE 2-CONCAVIFICATION OF A BANACH LATTICE EQUALS THE DIAGONAL OF THE FREMLIN TENSOR SQUARE

QINGYING BU, GERARD BUSKES, ALEXEY I. POPOV, ADI TCACIUC, AND VLADIMIR G. TROITSKY

Abstract. We investigate the relationship between the diagonal of the Fremlin projective tensor product of a Banach lattice E with itself and the 2-concavification of E.

1. Introduction and preliminaries

It is easy to see that the diagonal of the projective tensor product ℓ_{p}⊗_{π}ℓ_{p} is isometric to ℓ_{\frac{p}{2}} if p ⩾ 2 and to ℓ_{1} if 1 ⩽ p ⩽ 2. In this paper, we extend this fact to all Banach lattices. It turns out that the “right” tensor product for this problem is the Fremlin projective tensor product E⊗_{|π|}F of Banach lattices E and F. Given a Banach lattice E, we define (following [BB12]) the diagonal of E ⊗_{|π|}E to be the quotient of E ⊗_{|π|}E over the closed order ideal I_{oc} generated by the set {(x ⊗ y) : x ⊥ y}. We study the relationship of this diagonal with the 2-concavification of E. In the literature, it has been observed (see, e.g., [LT79]) that the p-concavification E_{(p)} of E is again a Banach lattice when E is p-convex. However, without the p-convexity assumption, E_{(p)} is only a semi-normed lattice. We show that in the case when E is 2-convex, the diagonal of E ⊗_{|π|} E is lattice isometric to E_{(2)} and that in general, the diagonal of E ⊗_{|π|} E is lattice isometric to E_{[2]}, where E_{[p]} is the completion of E_{(p)}/ ker∥·∥_{(p)}. We also show that if E satisfies the lower p-estimate then E_{[p]} is lattice isomorphic to an AL-space. In particular, if E satisfies the lower 2-estimate then the diagonal of E ⊗_{|π|}E is lattice isomorphic to an AL-space.

We consider the special case when E and F are Banach lattices with (1-unconditional) bases (e_{i}) and (f_{i}), respectively. We show that the double sequence (e_{i} ⊗ f_{j}) is an unconditional basis of E ⊗_{|π|}F (while it need not be an unconditional basis for the Banach space projective tensor product E ⊗_{π} F, see [KP70]). We also show that in this case

2010 Mathematics Subject Classification. Primary: 46B42. Secondary: 46M05, 46B40, 46B45.

Key words and phrases. Banach lattice, Fremlin projective tensor product, diagonal of tensor square, square of a Banach lattice, concavification.

The third, fourth, and fifth authors were supported by NSERC.

2 Q. BU, G. BUSKES, A. I. POPOV, A. TCACIUC, AND V. G. TROITSKY

E_{(p)} is a normed lattice and the diagonal of E ⊗_{|π|}E is lattice isometric to the completion of E_{(2)} via e_{i} ⊗ e_{i} ↦ e_{i}. Moreover, if (e_{i}) is normalized and E satisfies the lower p-estimate then the completion of E_{(p)} is lattice isomorphic to ℓ_{1}. In particular, if E satisfies the lower 2-estimate then the diagonal of E ⊗_{|π|}E is lattice isometric to ℓ_{1}.

In the rest of this section, we provide some background facts that are necessary for our exposition.

1.1. Fremlin tensor product. We refer the reader to [F72, F74] for a detailed original definition of the Fremlin tensor product E ⊗_{|π|} F of two Banach lattices E and F. However, we will only use a few facts about E ⊗_{|π|}F that we describe here.

Suppose E and F are two Banach lattices. We write E ⊗F for their algebraic tensor product; for x ∈ E and y ∈ F we write x ⊗ y for the corresponding elementary tensor in E ⊗ F. Every element of E ⊗ F is a linear combination of elementary tensors. Let G be another Banach lattice and φ: E × F → G be a bilinear map. Then φ induces a map φ̂: E ⊗ F → G such that φ̂(x ⊗ y) = φ(x, y) for all x ∈ E and y ∈ F. We say that φ is continuous if its norm, defined by

∥φ∥ = sup{∥φ(x, y)∥ : ∥x∥ ⩽ 1, ∥y∥ ⩽ 1},

is finite. We say that φ is positive if φ(x, y) ⩾ 0 whenever x, y ⩾ 0 and that φ is a lattice bimorphism if |φ(x, y)| = φ(|x|, |y|) for all x ∈ E and y ∈ F. We say that φ is orthosymmetric if φ(x, y) = 0 whenever x ⊥ y.

For u ∈ E ⊗ F, put

(1)  ∥u∥_{|π|} = sup∥φ̂(u)∥,

where the supremum is taken over all Banach lattices G and all positive bilinear maps φ from E×F to G with ∥φ∥ ⩽ 1. Theorem 1E in [F74, p. 89] proves that ∥·∥_{|π|} is a norm on E ⊗ F, and the completion of E ⊗ F with respect to this norm is again a Banach lattice. We will write E ⊗_{|π|}F for this space and call it the Fremlin tensor product of E and F. The Fremlin tensor norm is a cross norm, i.e., ∥x ⊗ y∥_{|π|} = ∥x∥ · ∥y∥ whenever x ∈ E and y ∈ F.

Remark 1. (See 1E(iii) and 1F in [F74, p. 92].) Let E, F, and G be Banach lattices. There is a one-to-one norm preserving correspondence between continuous positive bilinear maps φ: E × F → G and positive operators T: E ⊗_{|π|} F → G such that T (x ⊗ y) = φ(x, y) for all x ∈ E and y ∈ F. We will denote T = φ^{⊗}. Furthermore, φ is a lattice bimorphism if and only if T is a lattice homomorphism.

THE 2-CONCAVIFICATION IS THE DIAGONAL OF A TENSOR SQUARE 3

There is an alternative definition of E ⊗_{|π|} F, cf. [F74, 1I] and [S80, pp. 203-204]. Recall that, being a dual Banach lattice, F^{∗} is Dedekind complete by [AB06, Theorem 3.49], so that the space of regular operators L^{r}(E, F^{∗}) is a Banach lattice with respect to the regular norm ∥·∥_{r}, see [AB06, p. 255].

Proposition 2. If E and F are Banach lattices then E ⊗_{|π|}F can be identified with a closed sublattice of L^{r}(E, F^{∗})^{∗} such that ⟨x ⊗ y, T ⟩ = ⟨T x, y⟩ for x ∈ E, y ∈ F, and T ∈ L^{r}(E, F^{∗}).

Proof. Consider the map α: h ∈ (E ⊗_{|π|}F)^{∗} ↦ T ∈ L(E, F^{∗}) via ⟨T x, y⟩ = h(x ⊗ y). It is easy to see that α is one-to one and T ⩾ 0 whenever h ⩾ 0. It follows that α(h) is regular for every h.

Suppose that 0 ⩽ T: E → F^{∗}. The map φ defined by φ(x, y) = ⟨T x, y⟩ is a positive bilinear functional on E × F. Also,

∥T ∥ = sup{|⟨T x, y⟩| : ∥x∥ ⩽ 1, ∥y∥ ⩽ 1} = sup{|φ(x, y)| : ∥x∥ ⩽ 1, ∥y∥ ⩽ 1} = ∥φ∥.

By Remark 1, we can consider h = φ^{⊗}, then 0 ⩽ h ∈ (E ⊗_{|π|}F)^{∗} and ∥h∥ = ∥φ∥ = ∥T ∥. It is easy to see that T = α(h). Hence, the restriction of α to the positive cones of (E ⊗_{|π|}F)^{∗} is a bijective isometry onto the positive cone of L(E, F^{∗}). It follows by [AB06, Theorem 2.15] that α is a latice isomorphism between (E ⊗_{|π|}F)^{∗} and L^{r}(E, F^{∗}). Moreover, if T = α(h) for some h ∈ (E ⊗_{|π|}F)^{∗} then α(|h|) = |T | yields ∥h∥ = ∥|h|∥ = ∥|T |∥ = ∥T ∥_{r}. It follows that α is a lattice isometry between (E ⊗_{|π|}F)^{∗} and L^{r}(E, F^{∗}). Therefore, (E ⊗_{|π|}F)^{∗∗} is lattice isometric to L^{r}(E, F^{∗})^{∗}. Since E ⊗_{|π|}F can be viewed as a sublattice of (E ⊗_{|π|}F)^{∗∗}, it is lattice isometric to a closed sublattice of L^{r}(E, F^{∗})^{∗}. □

1.2. Functional calculus. Given x and y in a Banach lattice E, one would like to define expressions like (x^{2} + y^{2})^{\frac{1}{2}} and x^{\frac{1}{2}} y^{\frac{1}{2}} to be elements of E. This can be done point-wise if E can be represented as a function space. One could object, however, that the definition may then depend on the choice of a functional representation. Theorem 1.d.1 in [LT79] (see also [BdPvR91]) proves that there is a unique way to extend all continuous homogeneous^{1} functions from ℝ^{n} to ℝ to functions from E^{n} to E which does not depend on a particular representation of E as a function space. More precisely, for any x_{1}, …, x_{n} ∈ E there exists a unique lattice homomorphism τ from the space of all continuous homogeneous functions on ℝ^{n} to E such that if f(t_{1}, …, t_{n}) = t_{i} then τ(f) = x_{i} as i = 1, …, n. We denote τ(f) by f(x_{1}, …, x_{n}). In

1Recall that a function f: ℝ^{n} → R is called homogeneous if f(λx_{1}, …, λx_{n}) = λf(x_{1}, …, x_{n})

for any x_{1}, …, x_{n}, and λ ⩾ 0.

4 Q. BU, G. BUSKES, A. I. POPOV, A. TCACIUC, AND V. G. TROITSKY

particular, all identities and inequalities for homogeneous expressions that are valid in ℝ remain valid in E. For example,

(2)  (x_{1} + x_{2})^{\frac{1}{2}} y^{\frac{1}{2}} = ((x1^{\frac{1}{2}} y^{\frac{1}{2}} )^{2} + (x2^{\frac{1}{2}} y^{\frac{1}{2}} )^{2})^{\frac{1}{2}}

for every x_{1}, x_{2}, and y in every Banach lattice E. Note that, following convention from [LT79, p. 53], for t ∈ ℝ and p > 0, by t^{p} we mean |t|^{p} sign t. There is a certain inconsistency in notation: for example, t^{2} equals t|t|, not tt, so that (x^{2})^{\frac{1}{2}} = x while (xx)^{\frac{1}{2}} = |x|. To avoid confusion, we will distinguish xx from x^{2} throughout the paper. Note also that

(3)  x^{\frac{1}{2}} |x|^{\frac{1}{2}} = x.

In the following lemma, we collect several standard facts that we will routinely use.

Lemma 3. Given any x, y ∈ E and p > 0.

(i) |x^{\frac{1}{2}} y^{\frac{1}{2}} | = |x|^{\frac{1}{2}} |y|^{\frac{1}{2}} ; (ii) ∥x^{\frac{1}{2}} y^{\frac{1}{2}} ∥ ⩽ ∥x∥^{\frac{1}{2}} ∥y∥^{\frac{1}{2}} ; (iii) If x ⊥ y then x^{\frac{1}{2}} y^{\frac{1}{2}} = 0; (iv) If x, y ⩾ 0 then (x^{p} + y^{p})^{\frac{1}{p}} ⩾ 0;

(v) If x ∧ y = 0 then (x^{p} + y^{p})^{\frac{1}{p}} = x + y.

Proof. (i) follows from the fact that the identity holds for real numbers.

(ii) By Proposition 1.d.2(i) from [LT79], we have ∥|x|^{\frac{1}{2}} |y|^{\frac{1}{2}} ∥ ⩽ ∥x∥^{\frac{1}{2}} ∥y∥^{\frac{1}{2}} . Combining this with (i), we get the required inequality.

(iii) follows from the fact that |x^{\frac{1}{2}} y^{\frac{1}{2}} | = (|x| ∨ |y|)^{\frac{1}{2}} (|x| ∧ |y|)^{\frac{1}{2}} .

(iv) Note that (|x|^{p} + |y|^{p})^{\frac{1}{p}} ⩾ 0 for every x, y ∈ E because this inequality is true for real numbers. It follows that if x, y ⩾ 0 then (x^{p} + y^{p})^{\frac{1}{p}} = (|x|^{p} + |y|^{p})^{\frac{1}{p}} ⩾ 0.

(v) Again, for every x, y ∈ E we have |x| ∨ |y| ⩽ (|x|^{p} + |y|^{p})^{\frac{1}{p}} ⩽ |x| + |y| because this is true for real numbers. But if x ∧ y = 0 then x, y ⩾ 0 and x ∨ y = x + y. □

A Banach lattice E is said to be p-convex for some 1 ⩽ p < ∞ if there is a constant M > 0 such that ∥(∑_{i=1}^{n} x_{i}^{p})^{\frac{1}{p}} ∥ ⩽ M(∑_{i=1}^{n}∥x_{i}∥^{p})^{\frac{1}{p}} whenever x_{1}, …, x_{n} ∈ E_{+}. Similarly, E is p-concave if there is a constant M > 0 such that ∥(∑_{i=1}^{n} x_{i}^{p})^{\frac{1}{p}} ∥ ⩾ \frac{1}{M} (∑_{i=1}^{n}∥x_{i}∥^{p})^{\frac{1}{p}} whenever x_{1}, …, x_{n} ∈ E_{+}.

A Banach lattice E satisfies the upper p-estimate with constant M if ∥∑_{k=1}^{n} x_{k}∥ ⩽ M(∑_{k=1}^{n}∥x_{k}∥^{p})^{\frac{1}{p}} whenever x_{1}, …, x_{n} are disjoint. Similarly, E satisfies the lower pestimate with constant M if ∥∑_{k=1}^{n} x_{k}∥ ⩾ \frac{1}{M} (∑_{k=1}^{n}∥x_{k}∥^{p})^{\frac{1}{p}} whenever x_{1}, …, x_{n} are disjoint. It follows from Lemma 3(v) that p-convexity implies the upper p-estimate and p-concavity implies the lower p-estimate.

2. The concavification of a Banach lattice

The concavification procedure is motivated by the fact that if (x_{i}) ∈ ℓ_{r} and 1 < p < r, then the sequence (x_{i}^{p}) belongs to ℓ\frac{r}{p}.

This section is partially based on Section 1.d in [LT79]. Throughout this section, E is a Banach lattice and p ⩾ 1.

We define new vector operations on E via x ⊕ y = (x^{p} + y^{p})^{\frac{1}{p}} and α ⊙ x = α^{\frac{1}{p}} x whenever x, y ∈ E and α ∈ ℝ. (Here again, if x, y, or α are not positive then we use the convention described earlier.) Note that E endowed with these new addition and multiplication operations and the original order is again a vector lattice by Lemma 3(iv).

Define

(4)  ∥x∥_{(p)} = inf{∑_{i=1}^{n}∥v_{i}∥^{p} : |x| ⩽ v_{1} ⊕ ⋯ ⊕ v_{n}, v_{i} ⩾ 0}.

Remark 4. Note that being a vector lattice, (E, ⊕, ⊙, ⩽) satisfies the Riesz Decomposition Property (see, e.g., Theorem 1.13 in [AB06]), so that the inequality |x| ⩽ v_{1} ⊕ ⋯ ⊕ v_{n} in (4) can be replaced by equality.

It is easy to see that (4) defines a lattice semi-norm on (E, ⊕, ⊙, ⩽). This seminormed vector lattice will be denoted by E_{(p)}. It is called the p-concavification of E. As a partially ordered set, E_{(p)} coincides with E. We will see in Examples 18 and 26 that ∥·∥_{(p)} does not have to be a norm, and when it is a norm, it need not be complete.

The following fact is standard, we include the proof for completeness.

Proposition 5. If E is a p-convex Banach lattice then E_{(p)} is a Banach lattice.

Proof. Suppose that E is p-convex with constant M. Given x ∈ E. Suppose that |x| = v_{1} ⊕ ⋯ ⊕ v_{n} = (v_{1}^{p} + ⋯ + v_{n}^{p})^{\frac{1}{p}}

for some v_{i} ⩾ 0. Then ∥x∥ ⩽ M(∑_{i=1}^{n}∥v_{i}∥^{p})^{\frac{1}{p}} . It follows that \frac{1}{M^{p}} ∥x∥^{p} ⩽ ∥x∥_{(p)} ⩽ ∥x∥^{p}. This yields that ∥·∥_{(p)} is a complete norm on E_{(p)}. □

Recall that if E is a Banach lattice and x > 0, then x is an atom in E if 0 ⩽ z ⩽ x implies that z is a scalar multiple of x. We say that E is atomic or discrete if for every z > 0 there exists an atom x such that 0 < x ⩽ z.

Lemma 6. If x is an atom in a Banach lattice E then ∥x∥_{(p)} = ∥x∥^{p}

Proof. Take v_{1}, …, v_{n} ∈ E_{+} such that x = v_{1} ⊕ ⋯ ⊕ v_{n}. It follows that 0 ⩽ v_{k} ⩽ x for each k = 1, …, n, hence v_{k} = α_{k} ⊙ x = α_{k}^{1/p}x for some α_{k} ∈ ℝ_{+}. Also,

x = v_{1} ⊕ ⋯ ⊕ v_{n} = (α_{1} ⊙ x) ⊕ ⋯ ⊕ (α_{n} ⊙ x) = (α_{1} + ⋯ + α_{n}) ⊙ x,

so that ∑_{k=1}^{n} α_{k} = 1. It follows that ∑_{k=1}^{n}∥v_{k}∥^{p} = ∑_{k=1}^{n}∥αk^{\frac{1}{p}} x∥^{p} = ∥x∥^{p}, so that

∥x∥_{(p)} = ∥x∥^{p}. □

Corollary 7. If E is a discrete Banach lattice then E_{(p)} is a normed lattice.

Proof. Since we know that ∥·∥_{(p)} is a lattice semi-norm on E_{(p)}, it suffices to prove that it has trivial kernel. Suppose that y ∈ E with y ≠ 0. There is an atom x such that 0 < x ⩽ |y|. Then ∥y∥_{(p)} ⩾ ∥x∥_{(p)} = ∥x∥^{p} > 0. □

Remark 8. Thus, we know that E_{(p)} is a normed lattice in two important special cases: when E is discrete or p-convex. It would be interesting to find a general characterization of Banach lattices E for which ∥·∥_{(p)} is a norm. That is, characterize all Banach lattices E such that

inf\{∑_{i=1}^{n}∥v_{i}∥^{p} : x = (v_{1}^{p} + ⋯ + v_{n}^{p})^{\frac{1}{p}} , v_{i} > 0\} > 0

for every non-zero x ∈ E_{+}.

In general, we can only say that ∥·∥_{(p)} is a lattice seminorm on E_{(p)}. It follows that its kernel is an ideal, so that the quotient space E_{(p)}/ ker∥·∥_{(p)} is a normed lattice. Denote its completion by E_{[p]}. Clearly, E_{[p]} is a Banach lattice.

Let E be a Banach lattice. It is a standard fact (c.f., the proof of [LT79, Lemma 1.b.13]) that if there exists c > 0 such that ∥∑_{k=1}^{n} x_{k}∥ ⩾ c ∑_{k=1}^{n}∥x_{k}∥ whenever x_{1}, …, x_{n} are disjoint (that is, if E satisfies the lower 1-estimate), then E is lattice isomorphic to an AL-space. Indeed, put

|||x||| = sup{∑_{i=1}^{n}∥x_{i}∥ : x_{1}, …, x_{n} are positive and disjoint and |x| = x_{1} + ⋯ + x_{n}}.

It can be easily verified that this is an equivalent norm on E which makes E into an AL-space (with the same order).

The following lemma establishes that if E satisfies the lower p-estimate then E_{(p)} satisfies the lower 1-estimate.

THE 2-CONCAVIFICATION IS THE DIAGONAL OF A TENSOR SQUARE 7

Lemma 9. Suppose that E is a Banach lattice satisfying the lower p-estimate with constant M. Then ∥∑_{k=1}^{n} x_{k}∥_{(p)} ⩾ \frac{1}{M^{p}} ∑_{k=1}^{n}∥x_{k}∥_{(p)} whenever x_{1}, …, x_{n} are disjoint in E.

Proof. Suppose x_{1}, …, x_{n} are disjoint in E. Since |∑_{k=1}^{n} x_{k}| = ∑_{k=1}^{n}|x_{k}|, we may assume without loss of generality that x_{k} ⩾ 0 for each k. Note that ∑_{k=1}^{n} x_{k} = x_{1} ⊕ ⋯ ⊕ x_{n} by Lemma 3(v).

We will use (4) and Remark 4 to estimate ∥x_{1} ⊕ ⋯ ⊕ x_{n}∥_{(p)}. Take u_{1}, …, u_{m} in E_{+} such that x_{1} ⊕ ⋯ ⊕ x_{n} = u_{1} ⊕ ⋯ ⊕ u_{m}. Since E_{(p)} is a vector lattice, by the Riesz Decomposition Property [AB06, Theorem 1.20], for each k = 1, …, n we find v_{k,1}, …, v_{k,m} in E_{+} such that x_{k} = v_{k,1} ⊕ ⋯ ⊕ v_{k,m} and u_{i} = v_{1,i} ⊕ ⋯ ⊕ v_{n,i} for each i = 1, …, m. For each k and i we have 0 ⩽ v_{k,i} ⩽ x_{k}, so that v_{1,i}, …, v_{n,i} are disjoint for every i. It follows that u_{i} = v_{1,i} + ⋯ + v_{n,i}. By the lower p-estimate, we get ∥u_{i}∥ ⩾ \frac{1}{M} (∑_{k=1}^{n}∥v_{k,i}∥^{p})^{\frac{1}{p}} , so that M^{p}∥u_{i}∥^{p} ⩾ ∑_{k=1}^{n}∥v_{k,i}∥^{p}. For every k, we have ∥x_{k}∥_{(p)} ⩽ ∑_{i=1}^{m}∥v_{k,i}∥^{p}, so that

∑_{k=1}^{n}∥x_{k}∥_{(p)} ⩽ ∑_{k=1}^{n} ∑_{i=1}^{m}∥v_{k,i}∥^{p} ⩽ M^{p} ∑_{i=1}^{m}∥u_{i}∥^{p}.

Taking the infimum over all u_{1}, …, u_{m} in E_{+} such that x_{1} ⊕ ⋯ ⊕ x_{n} = u_{1} ⊕ ⋯ ⊕ u_{m}, we get the required inequality. □

Theorem 10. If a Banach lattice E satisfies the lower p-estimate with constant M then E_{[p]} is lattice isomorphic to an AL-space. Furthermore, if M = 1 then E_{[p]} is an AL-space.

Proof. Suppose that E satisfies a lower p-estimate with constant M. Applying Lemma 9, we have M^{p}∥∑_{k=1}^{n} x_{k}∥_{(p)} ⩾ ∑_{k=1}^{n}∥x_{k}∥_{(p)} whenever x_{1}, …, x_{n} are disjoint in E. It is easy to see that this inequality remains valid in E_{(p)}/ ker∥·∥_{(p)} and, furthermore, in E_{[p]}. □

3. Main results

Let E be a Banach lattice. Let I_{oc} be the norm closed ideal generated in E ⊗_{|π|} E by the elements of the form x ⊗ y where x ⊥ y (without loss of generality, we may also assume that x and y are positive). We can view I_{oc} as the set of all “off-diagonal” elements of E ⊗_{|π|} E. Therefore, following [BB12], we think of (E ⊗_{|π|} E)/I_{oc} as the diagonal of E ⊗_{|π|}E. We claim that this space is lattice isometric to E_{[2]}.

8 Q. BU, G. BUSKES, A. I. POPOV, A. TCACIUC, AND V. G. TROITSKY

Theorem 11. Suppose that E is a Banach lattice. Then there exists a surjective lattice isometry T: E_{[2]} → (E ⊗_{|π|}E)/I_{oc} such that T (x + ker∥·∥_{(2)}) = x ⊗ |x| + I_{oc} for each x ∈ E.

Proof. Define a map φ: E × E → E_{(2)} by φ(x, y) = x^{\frac{1}{2}} y^{\frac{1}{2}} . By the nature of the vector operations in E_{(2)}, this map is bilinear. Indeed,

φ(λx, y) = (λx)^{\frac{1}{2}} y^{\frac{1}{2}} = λ^{\frac{1}{2}} x^{\frac{1}{2}} y^{\frac{1}{2}} = λ ⊙ (x^{\frac{1}{2}} y^{\frac{1}{2}} ) = λ ⊙ φ(x, y).

Similarly, φ(x, λy) = λ ⊙ φ(x, y). Also, φ(x_{1} + x_{2}, y) = φ(x_{1}, y) ⊕ φ(x_{2}, y) by (2); we obtain φ(x, y_{1} + y_{2}) = φ(x, y_{1}) ⊕ φ(x, y_{2}) in a similar fashion. For any x, y ∈ E we have by Lemma 3(ii)

∥φ(x, y)∥_{(2)} = ∥x^{\frac{1}{2}} y^{\frac{1}{2}} ∥_{(2)} ⩽ ∥x^{\frac{1}{2}} y^{\frac{1}{2}} ∥^{2} ⩽ (∥x∥^{\frac{1}{2}} ∥y∥^{\frac{1}{2}} )^{2} = ∥x∥∥y∥,

so that ∥φ∥ ⩽ 1. Clearly, φ is a continuous lattice bimorphism; it is orthosymmetric by Lemma 3(iii).

Put N = ker∥·∥_{(2)} and let r: E_{(2)} → E_{(2)}/N be the canonical quotient map. Also, let i: E_{(2)}/N → E_{[2]} be the natural inclusion map. Consider the map (irφ)^{⊗}: E ⊗_{|π|} E → E_{[2]} as in Remark 1 (see Figure 1); then (irφ)^{⊗} is a lattice homomorphism and ∥(irφ)^{⊗}∥ ⩽ 1. Note that if x ⊥ y then (irφ)^{⊗}(x ⊗ y) = irφ(x, y) = 0. Since (irφ)^{⊗} is positive, it vanishes on I_{oc}. Consider the quotient space (E ⊗_{|π|} E)/I_{oc}; let q: E ⊗_{|π|}E → (E ⊗_{|π|}E)/I_{oc} be the canonical quotient map. Since I_{oc} ⊆ ker(irφ)^{⊗}, we can consider the induced map { \atop (irφ)^{⊗}}: (E ⊗_{|π|}E)/I_{oc} → E_{[2]} such that { \atop (irφ)^{⊗}}q = (irφ)^{⊗}.

Consider the map q⊗ from E × E to (E ⊗_{|π|} E)/I_{oc}. This map is clearly bilinear and orthosymmetric. Therefore, by Theorem 9(ii) of [BvR01], there exists a lattice homomorphism S: E_{(2)} → (E ⊗_{|π|} E)/I_{oc} such that q⊗ = Sφ. Note that for each x, y ∈ E we have

(5)  S(x^{\frac{1}{2}} y^{\frac{1}{2}} ) = Sφ(x, y) = q ⊗ (x, y) = x ⊗ y + I_{oc}.

In particular, taking y = |x|, we get Sx = x ⊗ |x| + I_{oc}.

We claim that ∥Sx∥ ⩽ ∥x∥_{(2)} for each x ∈ E. Indeed, take v_{1}, …, v_{n} ∈ E_{+} such that |x| = v_{1} ⊕ ⋯ ⊕ v_{n}. Since S is a lattice homomorphism, we have

|Sx| = S|x| = Sv_{1} + ⋯ + Sv_{n} = v_{1} ⊗ |v_{1}| + ⋯ + v_{n} ⊗ |v_{n}| + I_{oc}.

By the definition of a quotient norm,

∥Sx∥ ⩽ ∥∑_{i=1}^{n} v_{i} ⊗ |v_{i}|∥_{|π|} ⩽ ∑_{i=1}^{n}∥v_{i} ⊗ |v_{i}|∥_{|π|} = ∑_{i=1}^{n}∥v_{i}∥^{2}

because ∥·∥_{|π|} is a cross-norm. It follows now from (4) that ∥Sx∥ ⩽ ∥x∥_{(2)}.

Figure 1

[figure: E × E; φ; ⊗; E_{(2)}; S; r; E_{(2)}/N; S̃; i; E_{[2]}; T; E ⊗_{|π|}E; (irφ)^{⊗}; q; (E ⊗_{|π|}E)/I_{oc}; { \atop (irφ)^{⊗}}]

In particular, N ⊆ ker S. It follows that S induces a lattice homomorphism S̃: E_{(2)}/N → (E ⊗_{|π|}E)/I_{oc} such that S = S̃r. We now show that S̃ is an isometry. For any x ∈ E we have ∥S̃(x + N)∥ = ∥Sx∥ ⩽ ∥x∥_{(2)} = ∥x + N∥, so that ∥S̃∥ ⩽ 1. On the other hand, for every v ∈ I_{oc} we have (irφ)^{⊗}(v) = 0, so that (irφ)^{⊗}(x ⊗ |x| + v) = rφ(x, |x|) = rx = x + N by (3). Since ∥(irφ)^{⊗}∥ ⩽ 1, we get ∥x + N∥ ⩽ ∥x ⊗ |x| + v∥_{|π|}. Taking infimum over all v ∈ I_{oc}, we get ∥x + N∥ ⩽ ∥Sx∥ = ∥S̃(x + N)∥. Therefore, S̃ is an isometry. It follows that S̃ extends to a lattice isometry T: E_{[2]} → (E ⊗_{|π|}E)/I_{oc}. Note that T (x + N) = Sx = x ⊗ |x| + I_{oc} for each x ∈ E.

We claim that T is the inverse of { \atop (irφ)^{⊗}}. Indeed, for every x ∈ E we have

{ \atop (irφ)^{⊗}}T (x + N) = { \atop (irφ)^{⊗}}(x ⊗ |x| + I_{oc}) = (irφ)^{⊗}(x ⊗ |x|) = irφ(x, |x|) = irx = x + N

by (3). This means that { \atop (irφ)^{⊗}}T is the identity on E_{(2)}/N and, therefore, on E_{[2]}. On the other hand, for each x, y ∈ E it follows from (5) that

T { \atop (irφ)^{⊗}}(x ⊗ y + I_{oc}) = T (irφ)^{⊗}(x ⊗ y) = T irφ(x, y)

= T r(x^{\frac{1}{2}} y^{\frac{1}{2}} ) = S̃r(x^{\frac{1}{2}} y^{\frac{1}{2}} ) = S(x^{\frac{1}{2}} y^{\frac{1}{2}} ) = x ⊗ y + I_{oc}.

Hence, T { \atop (irφ)^{⊗}} is the identity on q(E ⊗ E). Since E ⊗ E is dense in E ⊗_{|π|} E then q(E ⊗ E) is dense in (E ⊗_{|π|} E)/I_{oc}, so that T { \atop (irφ)^{⊗}} is the identity on (E ⊗_{|π|} E)/I_{oc}. Therefore, T is the inverse of { \atop (irφ)^{⊗}}. It follows that T is onto. □

10 Q. BU, G. BUSKES, A. I. POPOV, A. TCACIUC, AND V. G. TROITSKY

Recall that if E is discrete then ∥·∥_{(2)} is a norm by Corollary 7, so that E_{(2)} is a normed lattice and E_{[2]} equals \overline{E_{(2)}}, the completion of E_{(2)}; if E is 2-convex then E_{(2)} is a Banach lattice by Proposition 5; in this case E_{[2]} = E_{(2)}.

Corollary 12. Suppose that E is Banach lattice. If E_{(2)} is a normed lattice then it is lattice isometric to a dense sublattice of (E ⊗_{|π|}E)/I_{oc} via x ∈ E_{(2)} ↦ x ⊗ |x| + I_{oc}.

Corollary 13. Suppose that E is a Banach lattice such that E_{(2)} is also a Banach lattice. Then the map T: E_{(2)} → (E ⊗_{|π|} E)/I_{oc} given by T x = x ⊗ |x| + I_{oc} is a surjective linear lattice isometry.

Remark 14. Theorem 11 provides a new characterization of the ideal I_{oc}. It was observed in the proof of Theorem 11 that (irφ)^{⊗} vanishes on I_{oc} and the induced map { \atop (irφ)^{⊗}} on (E ⊗_{|π|}E)/I_{oc} is a bijection; hence I_{oc} = ker(irφ)^{⊗}. This can be used to easily verify whether certain elements belong to I_{oc}. For example, it follows from φ(x^{\frac{1}{2}} y^{\frac{1}{2}} , x^{\frac{1}{2}} y^{\frac{1}{2}} ) = φ(x, y) that x ⊗ y − (x^{\frac{1}{2}} y^{\frac{1}{2}} ) ⊗ (x^{\frac{1}{2}} y^{\frac{1}{2}} ) is in ker(irφ)^{⊗} and, hence, in I_{oc} for every x, y ∈ E.

Similarly, one can check that x ⊗ y − y ⊗ x ∈ I_{oc} for all x, y ∈ E. Let Z be the closed sublattice of E ⊗_{|π|}E generated by vectors of the form x ⊗ y − y ⊗ x. We refer to Z as the antisymmetric part of E ⊗_{|π|}E. This yields Z ⊆ I_{oc} (this inclusion also follows from Proposition 4.33 of [L07], obtained there by very different means).

Remark 15. Suppose that E is such that E_{(2)} is a Banach lattice. Then we can identify T^{−1} in Corollary 13. Indeed, in this case, the maps i and r in the proof of Theorem 11 are just the identity maps, so that T^{−1} = φ̃^{⊗} where φ(x, y) = x^{\frac{1}{2}} y^{\frac{1}{2}} (see Figure 2). Furthermore, in this case, we have I_{oc} = ker φ^{⊗}.

Figure 2

[figure: E × E; φ; ⊗; E_{(2)}; S=T; E ⊗_{|π|}E; φ^{⊗}; q; (E ⊗_{|π|}E)/I_{oc}; φ̃^{⊗}]

THE 2-CONCAVIFICATION IS THE DIAGONAL OF A TENSOR SQUARE 11

Remark 16. Again, suppose that E is such that E_{(2)} is a Banach lattice. It follows from Corollary 13 that every equivalence class in (E ⊗_{|π|}E)/I_{oc} contains a representative of the form x ⊗ |x| for some x ∈ E. Therefore, q(E ⊗ E) = (E ⊗_{|π|} E)/I_{oc}, where q: E ⊗_{|π|}E → (E ⊗_{|π|}E)/I_{oc} is the canonical quotient map. In other words, the elements of E ⊗ E (and even elementary tensor products) are sufficient to “capture all of the diagonal” in E ⊗_{|π|}E.

As usual, one can identify q(E ⊗ E) with the quotient of E ⊗ E over I_{oc} or, more precisely, with (E ⊗ E)/((E ⊗ E) ∩ I_{oc}), where E ⊗ E is viewed as a (non-closed) subspace of E ⊗_{|π|}E. Therefore,

(6)  (E ⊗_{|π|}E)/I_{oc} = (E ⊗ E)/((E ⊗ E) ∩ I_{oc})

Combining Theorems 10 and 11, we immediately get the following.

Corollary 17. Suppose that E is a Banach lattice satisfying the lower 2-estimate with constant M. Then (E ⊗_{|π|}E)/I_{oc} is lattice isomorphic to an AL-space. If M = 1 then (E ⊗_{|π|}E)/I_{oc} is an AL-space.

4. Function spaces

In this section, we consider the case when E is a Köthe space on a σ-finite measure space (Ω, Σ, μ) as in [LT79, Definition 1.b.17]. That is, E is contained in the space L_{0}(Ω) of all measurable functions on Ω such that E contains the characteristic functions of all sets of finite measure and if f ∈ E, g ∈ L_{0}(Ω) and |g| ⩽ |f| then g ∈ E and ∥g∥ ⩽ ∥f∥.

It is easy to see that in a Köthe space, the functional calculus map τ, described in Subsection 1.2, agrees with almost everywhere pointwise operations. Indeed, fix x_{1}, …, x_{n} in E and let h: ℝ^{n} → ℝ be a homogeneous continuous function. It is easy to see that

|h(t_{1}, …, t_{n})| ⩽ M max_{1⩽i⩽n}|t_{i}|

for all t_{1}, …, t_{n} ∈ ℝ, where

M = max{|h(t_{1}, …, t_{n})| : max_{1⩽i⩽n}|t_{i}| = 1}.

It follows that

|h(x_{1}(ω), …, x_{n}(ω))| ⩽ M max_{1⩽i⩽n}|x_{i}(ω)|

for all ω ∈ Ω, so that the usual composition function h(x_{1}, …, x_{n}) defined a.e. by

h(x_{1}, …, x_{n})(ω) = h(x_{1}(ω), …, x_{n}(ω))

12 Q. BU, G. BUSKES, A. I. POPOV, A. TCACIUC, AND V. G. TROITSKY

satisfies

|h(x_{1}, …, x_{n})| ⩽ M ⋁_{1⩽i⩽n}|x_{i}| a.e.;

it follows that h(x_{1}, …, x_{n}) ∈ E. Thus, almost everywhere pointwise operations define a functional calculus on E. It follows from the uniqueness of functional calculus that this functional calculus agrees with τ ^{2}.

We proceed with a functional representation of E_{(2)} (see, e.g., [BvR01] or [JL01, p. 30]). The square of E is defined via E^{2} = {x^{2} : x ∈ E}, where, again, by x^{2} we really mean x|x| and the product is defined a.e.. Note that the map S: x ∈ E_{(2)} ↦ x^{2} ∈ E^{2} is a bijection. In view of this, we may transfer the Banach lattice structure from E_{(2)} to E^{2}. In particular, with this identification, E^{2} is a vector space. The main advantage of this approach is that addition and scalar multiplication in E^{2} are defined a.e. pointwise (the vector operations on E_{(2)} were defined exactly this way):

S(x ⊕ y) = (x ⊕ y)^{2} = ((x^{2} + y^{2})^{\frac{1}{2}} )^{2} = x^{2} + y^{2} = S(x) + S(y)

and

S(λ ⊙ x) = (λ ⊙ x)^{2} = (λ^{\frac{1}{2}} x)^{2} = λx^{2} = λS(x).

Observe, also, that if x, y ∈ E then the function xy is in E^{2}. Indeed, x^{\frac{1}{2}} y^{\frac{1}{2}} ∈ E, so that E^{2} ∋ S(x^{\frac{1}{2}} y^{\frac{1}{2}} ) = (x^{\frac{1}{2}} y^{\frac{1}{2}} )^{2} = xy.

In view of this construction, we can replace E_{(2)} with E^{2} in the preceding section. In particular, instead of the map φ: E × E → E_{(2)} defined by φ(x, y) = x^{\frac{1}{2}} y^{\frac{1}{2}} in Remark 15, we can consider the corresponding map m: E × E → E^{2} defined by m(x, y) = xy. This map is obviously a continuous orthosymmetric lattice bimorphism.

Suppose now that E^{2} is a Banach lattice (for example, E is 2-convex). Then the diagram in Figure 2 in Corollary 13 and Remark 15 becomes the diagram in Figure 3.

Figure 3

[figure: E × E; m; ⊗; E^{2}; T; E ⊗_{|π|}E; m^{⊗}; q; (E ⊗_{|π|}E)/I_{oc}; m̃^{⊗}]

2The same argument shows that on C(K)-spaces, τ agrees with the pointwise operations.

THE 2-CONCAVIFICATION IS THE DIAGONAL OF A TENSOR SQUARE 13

For x, y ∈ E, their elementary tensor product x ⊗ y can be viewed as a function on Ω^{2} via (x ⊗ y)(s, t) = x(s)y(t) for s, t ∈ Ω. This way, E ⊗ E is a subset of L_{0}(Ω^{2}). We do not know whether E ⊗_{|π|}E can still be viewed as a sublattice of L_{0}(Ω^{2}), but this is definitely the case in many important special cases.

Let D be the diagonal of Ω^{2}, that is, D = {(s, s) : s ∈ Ω}. Of course, the map s → (s, s) is a bijection between Ω and D, so that we can view D as a copy of Ω. For an arbitrary function u in L_{0}(Ω^{2}), one cannot really consider the restriction of u to D because D may have measure zero in Ω^{2}. However, such a restriction may be defined for elementary tensors via (x ⊗ y)(s, s) = x(s)y(s), which is defined a.e. on Ω. That is, the restriction of x ⊗ y to D is exactly xy = m(x, y) = m^{⊗}(x ⊗ y) (as we identify D with Ω). Extending this by linearity to E ⊗ E, we can view m^{⊗} on E ⊗ E (or even on E ⊗_{|π|}E) as the restriction to the diagonal map. Note that, in view of Remark 16 and, in particular, (6), the space E ⊗ E is sufficient to capture the diagonal part of E ⊗_{|π|}E. Furthermore, for u ∈ E ⊗ E we have u ∈ I_{oc} iff m^{⊗}(u) = 0 iff u vanishes a.e. on the diagonal. It follows that the both quotient spaces in (6) can be viewed as the space of the restrictions of the functions in E ⊗ E to D. Therefore, in the case of Köthe spaces, Corollary 13 says that the restrictions of the elements of E ⊗ E (or E ⊗_{|π|}E) to the diagonal are exactly the functions in E^{2} (again, we identify the diagonal with Ω). Moreover, the norm of the restriction (that is, the quotient norm from (6)) is the same as its E^{2} norm.

Example 18. If E = L_{p} for 1 ⩽ p < ∞ then E^{2} as a vector lattice coincides with L_{\frac{p}{2}} . In the case p ⩾ 2, E is 2-convex and hence (E⊗_{|π|}E)/I_{oc} = E_{[2]} = E_{(2)} = L_{\frac{p}{2}} . In the case 1 ⩽ p < 2, the vector lattice L_{\frac{p}{2}} (and, therefore, E_{(2)}) admits no non-trivial positive functionals by, e.g., [AB03, Theorem 5.24]. Note that every positive functional f on E_{[2]} gives rise to a positive functional f ∘ q on E_{(2)}, where q: E_{(2)} → E_{(2)}/ ker∥·∥_{(2)} is the canonical quotient map. It follows that E_{[2]}^{∗} is trivial, and so is E_{[2]}. Hence (E ⊗_{|π|}E)/I_{oc} = E_{[2]} = {0}, which is a trivial AL-space.

Example 19. Let E = C[0, 1]. In this case, E^{2} = E. Also, E ⊗_{|π|} E = C[0, 1]^{2} by Corollary 3F of [F74]. As before, we put m(x, y) = xy for x, y ∈ E. In this case, the map m^{⊗} on E ⊗ E and, therefore, on E ⊗_{|π|}E, is the restriction to the diagonal, so that I_{oc} consists of those functions that vanish on the diagonal, while (E ⊗_{|π|}E)/I_{oc} is the space of the restrictions of the functions in C[0, 1]^{2} to the diagonal, which, naturally, can again be identified with C[0, 1].

14 Q. BU, G. BUSKES, A. I. POPOV, A. TCACIUC, AND V. G. TROITSKY

5. Banach lattices with a basis

By a Banach lattice with a basis we mean a Banach lattice where the order is defined by a basis. That is, E has a (Schauder) basis (e_{i}) such that a vector x = ∑_{i=1}^{∞} x_{i}e_{i} is positive iff x_{i} ⩾ 0 for all i. It follows that the basis (e_{i}) is 1-unconditional. The converse is also true: every Banach space with a 1-unconditional basis is a Banach lattice in the induced order. It is clear that every Banach lattice with a basis is discrete.

5.1. Concavification of a Banach lattice with a basis. Since E is a Köthe space, its continuous homogeneous functional calculus in E is coordinate-wise. For example, if x = ∑_{i=1}^{∞} x_{i}e_{i} and y = ∑_{i=1}^{∞} y_{i}e_{i} then

x^{\frac{1}{2}} y^{\frac{1}{2}} = ∑_{i=1}^{∞} xi^{\frac{1}{2}} yi^{\frac{1}{2}} e_{i} and (x^{p} + y^{p})^{\frac{1}{p}} = ∑_{i=1}^{∞}(x_{i}^{p} + y_{i}^{p})^{\frac{1}{p}} e_{i}.

As before, we use the conventions t^{p} = |t|^{p} sign t here for t, p ∈ ℝ.

Next, we fix p ⩾ 1 and consider E_{(p)}. Since E is discrete, E_{(p)} is a normed lattice by Corollary 7. Hence, in this case, E_{[p]} equals \overline{E_{(p)}}, the completion of E_{(p)}. Since (e_{i}) is disjoint in E, it follows from Lemma 3(v) that x_{1}e_{1}+⋯+x_{n}e_{n} = x_{1}^{p}⊙e_{1}⊕⋯⊕x_{n}^{p}⊙e_{n}.

Lemma 20. Suppose that E is a Banach lattice with a basis (e_{i}). Then (i) ∥e_{i}∥_{(p)} = ∥e_{i}∥^{p} for each i; (ii) if x = ∑_{i=1}^{∞} x_{i}e_{i} in E then x = ⊕−∑_{i=1}^{∞} x_{i}^{p} ⊙ e_{i} in E_{(p)}; in particular, the

series converges in E_{(p)}; (iii) (e_{i}) is a 1-unconditional basis of \overline{E_{(p)}}.

Proof. (i) follows immediately from Lemma 6. To prove (ii), suppose that x = ∑_{i=1}^{∞} x_{i}e_{i} in E. For each n, we can write x = u_{n} + v_{n} = u_{n} ⊕ v_{n} where u_{n} = ∑_{i=1}^{n} x_{i}e_{i} and v_{n} = ∑_{i=n+1}^{∞} x_{i}e_{i}. Note that ∥v_{n}∥ → 0 and u_{n} = ⊕−∑_{i=1}^{n} x_{i}^{p} ⊙ e_{i}. Therefore,

∥x ⊖ (⊕−∑_{i=1}^{n} x_{i}^{p} ⊙ e_{i})∥_{(p)} = ∥x ⊖ u_{n}∥_{(p)} = ∥v_{n}∥_{(p)} ⩽ ∥v_{n}∥^{p} → 0.

This proves (ii). It follows from (ii) that the closed linear span of (e_{i}) is dense in E_{(p)} and, therefore, in \overline{E_{(p)}}. Since the sequence (e_{i}) remains disjoint in \overline{E_{(p)}}, this yields (iii). □

Proposition 21. Suppose that E is a Banach lattice with a normalized basis. If E satisfies the lower p-estimate with constant M then \overline{E_{(p)}} is lattice isomorphic (isometric if M = 1) to ℓ_{1} via (x_{i}) ∈ ℓ_{1} ↦ ∑_{i=1}^{∞} x_{i} ⊙ e_{i} ∈ E_{(p)}.

THE 2-CONCAVIFICATION IS THE DIAGONAL OF A TENSOR SQUARE 15

Proof. Let x ∈ E such that x = ∑_{i=1}^{n} x_{i} ⊙ e_{i} = ∑_{i=1}^{n} x_{i}^{1/p}e_{i}. It follows from Lemma 9 that

∥x∥_{(p)} ⩾ \frac{1}{M^{p}} ∑_{i=1}^{n}∥x_{i} ⊙ e_{i}∥_{(p)} = \frac{1}{M^{p}} ∑_{i=1}^{n}|x_{i}|.

On the other hand, by the triangle inequality, we have ∥x∥_{(p)} ⩽ ∑_{i=1}^{n}∥x_{i} ⊙ e_{i}∥_{(p)} = ∑_{i=1}^{n}|x_{i}|. □

5.2. Fremlin tensor product of Banach lattices with bases. Given Banach spaces E and F with bases (e_{i}) and (f_{i}), respectively, then the double sequence (e_{i}⊗f_{j}) is a basis for the Banach space projective tensor product E ⊗_{π} F, see [GGdL61]. However, even if these respective bases are unconditional then (e_{i} ⊗f_{j}) is not necessarily an unconditional basis for E ⊗_{π} F. Indeed, it was shown in [KP70] that the Banach space projective tensor product ℓ_{p} ⊗_{π} ℓ_{q} with 1/p + 1/q ⩽ 1 does not have an unconditional basis.

Recall that if E is a Banach lattice with a basis then the basis is automatically 1-unconditional.

Lemma 22. Suppose that E and F are Banach lattices with bases, (e_{i}) and (f_{j}), respectively. Then the double sequence (e_{i} ⊗ f_{j})_{i,j} is disjoint in E ⊗_{|π|}F. Moreover, this sequence is a 1-unconditional basis of E ⊗_{|π|}F (under any enumeration).

Proof. First, we will show that (e_{i} ⊗ f_{j}) ⊥ (e_{k} ⊗ f_{l}) provided (i, j) ≠ (k, l). Using Proposition 2, we consider E ⊗_{|π|} F as a sublattice of L^{r}(E, F^{∗})^{∗}. It suffices to show that

⟨(e_{i} ⊗ f_{j}) ∧ (e_{k} ⊗ f_{l}), T ⟩ = 0

for every positive T: E → F^{∗}. By [AB06, Theorem 3.49],

(7)  ⟨(e_{i} ⊗ f_{j}) ∧ (e_{k} ⊗ f_{l}), T ⟩ = inf_{0⩽S⩽T}{(e_{i} ⊗ f_{j})(S) + (e_{k} ⊗ f_{l})(T − S)}.

Put c = ⟨T e_{k}, f_{l}⟩ and define S: E → F^{∗} via S = ce_{k}^{∗} ⊗ f_{l}^{∗}, where e_{k}^{∗} and f_{l}^{∗} are the appropriate bi-orthogonal functionals. That is, for x ∈ E we have Sx = ce_{k}^{∗}(x)f_{l}^{∗}. Clearly, S ⩾ 0. We will show that S ⩽ T. It suffices to show that Se_{m} ⩽ T e_{m} for every m. But if m ≠ k then Se_{m} = 0 ⩽ T e_{m}. It is left to prove that Se_{k} ⩽ T e_{k}. Note that Se_{k} = cf_{l}^{∗}. It suffices to show that ⟨Se_{k}, f_{n}⟩ ⩽ ⟨T e_{k}, f_{n}⟩ for all n. But this is true because ⟨Se_{k}, f_{n}⟩ = cf_{l}^{∗}(f_{n}) = 0, when n ≠ l, and ⟨Se_{k}, f_{l}⟩ = cf_{l}^{∗}(f_{l}) = c = ⟨T e_{k}, f_{l}⟩ Now substituting this S into (7), we get

(e_{i} ⊗ f_{j})(S) + (e_{k} ⊗ f_{l})(T − S) = ce_{k}^{∗}(e_{i})f_{l}^{∗}(f_{j}) + ⟨T e_{k}, f_{l}⟩ − ⟨Se_{k}, f_{l}⟩ = 0 + c − c = 0

16 Q. BU, G. BUSKES, A. I. POPOV, A. TCACIUC, AND V. G. TROITSKY

because (i, j) ≠ (k, l).

Being a disjoint sequence in a Banach lattice, (e_{i} ⊗ f_{j})_{i,j} is a 1-unconditional basic sequence. It is left to show that its closed span is all of E ⊗_{|π|} F. Take x ∈ E and y ∈ F with ∥x∥, ∥y∥ ⩽ 1. Given any ε ∈ (0, 1), we can find basis projections P and Q on E and F, respectively, such that x_{0} = P x and y_{0} = Qy satisfy ∥x − x_{0}∥ < ε and ∥y − y_{0}∥ < ε. It follows that

∥x ⊗ y − x_{0} ⊗ y_{0}∥_{|π|} = ∥x_{0} ⊗ (y − y_{0}) + (x − x_{0}) ⊗ y_{0} + (x − x_{0}) ⊗ (y − y_{0})∥_{|π|}

⩽ ∥x_{0}∥∥y − y_{0}∥ + ∥x − x_{0}∥∥y_{0}∥ + ∥x − x_{0}∥∥y − y_{0}∥ ⩽ 3ε.

Since x_{0} ⊗ y_{0} is in span{e_{i} ⊗ f_{j} : i, j ∈ ℕ}, it follows that x ⊗ y can be approximated by elements of the span. It follows that the span is dense in E ⊗ F and, therefore, in E ⊗_{|π|}F. □

5.3. The diagonal of E ⊗_{|π|}E. Suppose that E is a Banach lattice with a basis (e_{i}). As we just observed, (e_{i} ⊗ e_{j})_{i,j} is a 1-unconditional basis in E ⊗_{|π|}E. It is easy to see that

I_{oc} = \overline{span}\{(e_{i} ⊗ e_{j}) : i ≠ j\}
(8)  (E ⊗_{|π|}E)/I_{oc} = \overline{span}\{(e_{i} ⊗ e_{i}) : i ∈ ℕ\}

In particular, we can view I_{oc} and (E ⊗_{|π|}E)/I_{oc} as two mutually complementary bands in E ⊗_{|π|}E. In view of this, our interpretation of (E ⊗_{|π|}E)/I_{oc} as the diagonal of E ⊗_{|π|}E is consistent with, e.g., Examples 2.10 and 2.23 in [R02].

It follows immediately from Corollary 12 that (E⊗_{|π|}E)/I_{oc} is lattice isometric to \overline{E_{(2)}}. Moreover, in view of (8), the map T in Corollary 12 has a particularly simple form: T: e_{i} → e_{i} ⊗ e_{i}. Thus, Corollary 12 for Banach lattices with a basis can be stated as follows.

Theorem 23. Suppose that E is a Banach lattice with a basis (e_{i}). Then the map that sends ∑_{i=1}^{∞} u_{i}e_{i} ⊗ e_{i} in E ⊗_{|π|}E into ∑_{i=1}^{∞} u_{i} ⊙ e_{i} in \overline{E_{(2)}} is a surjective lattice isometry between (E ⊗_{|π|}E)/I_{oc} and \overline{E_{(2)}}.

Combining this with Propositions 5 and 21, we get the following corollaries.

Corollary 24. Suppose that E is a Banach lattice with a basis. If E is 2-convex then (E ⊗_{|π|}E)/I_{oc} is lattice isometric to E_{(2)}.

Corollary 25. Suppose that E is a Banach lattice with a normalized basis (e_{i}), satisfying a lower 2-estimate with constant M. Then (E ⊗_{|π|}E)/I_{oc} is lattice isomorphic (isometric if M = 1) to ℓ_{1} via (x_{i}) ∈ ℓ_{1} ↦ ∑_{i=1}^{∞} x_{i}e_{i} ⊗ e_{i}.

Example 26. If E = ℓ_{p} for 1 ⩽ p < ∞ then E^{2} (and, therefore, E_{(2)}) can be identified as a vector space with ℓ_{\frac{p}{2}} . In the case p ⩾ 2, E is 2-convex and hence (E ⊗_{|π|} E)/I_{oc} = E_{[2]} = E_{(2)} = ℓ_{\frac{p}{2}} . In the case 1 ⩽ p < 2, E satisfies the lower 2estimate and hence (E ⊗_{|π|}E)/I_{oc} = E_{[2]} = ℓ_{1}. On the other hand, in the latter case, ∥·∥_{(2)} is the ℓ_{1}-norm on E_{(2)} = ℓ_{\frac{p}{2}} and we have E_{[2]} = \overline{(E_{(2)}, ∥·∥_{(2)})} = \overline{(ℓ_{\frac{p}{2}} , ∥·∥_{ℓ_{1}})} = ℓ_{1}.

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[AB06] , Positive operators, Springer, 2006.

[BB12] Q. Bu, G. Buskes, Polynomials on Banach lattices and positive tensor products, J. Math. Anal. Appl. 388 (2012), 845-862.

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(Q. Bu and G. Buskes) Department of Mathematics, University of Mississippi, University, MS 38677-1848. USA

  E-mail address: qbu@olemiss.edu, mmbuskes@olemiss.edu

(A. I. Popov) Department of Pure Mathematics, Faculty of Mathematics University of Waterloo, Waterloo, Ontario, N2L 3G1. Canada

  E-mail address: a4popov@uwaterloo.ca

(A. Tcaciuc) Mathematics and Statistics Department, Grant MacEwan University, Edmonton, AB, T5J P2P, Canada

  E-mail address: atcaciuc@ualberta.ca

(V. G. Troitsky) Department of Mathematical and Statistical Sciences, University of Alberta, Edmonton, AB, T6G 2G1. Canada

E-mail address: troitsky@ualberta.ca