p-Summing Operators on Injective Tensor Products of Spaces

by

Stephen Montgomery-Smith^{(∗)} and Paulette Saab^{(∗∗)}

Abstract Let X, Y and Z be Banach spaces, and let ∏_{p}(Y, Z) (1 ≤ p < ∞) denote the space of p-summing operators from Y to Z. We show that, if X is a £_{∞}-space, then a bounded linear operator T : X ˆ⊗_{ϵ}Y ⟶ Z is 1-summing if and only if a naturally associated operator T^{#} : X ⟶ ∏_{1}(Y, Z) is 1-summing. This result need not be true if X is not a £_{∞}-space. For p > 1, several examples are given with X = C[0, 1] to show that T^{#} can be p-summing without T being psumming. Indeed, there is an operator T on C[0, 1] ˆ⊗_{ϵ}ℓ_{1} whose associated operator T^{#} is 2-summing, but for all N ∈ N, there exists an N-dimensional subspace U of C[0, 1] ˆ⊗_{ϵ}ℓ_{1} such that T restricted to U is equivalent to the identity operator on ℓ_{∞}^{N}. Finally, we show that there is a compact Hausdorff space K and a bounded linear operator T : C(K) ˆ⊗_{ϵ}ℓ_{1} ⟶ ℓ_{2} for which T^{#} : C(K) ⟶ ∏_{1}(ℓ_{1}, ℓ_{2}) is not 2-summing.

(∗) Research supported in part by an NSF Grant DMS 9001796 (∗∗) Research supported in part by an NSF Grant DMS 87500750 A.M.S. (1980) subject classification: 46B99 Introduction Let X and Y be Banach spaces, and let X ˆ⊗_{ϵ}Y denote their injective tensor product. In this paper, we shall study the behavior of those operators on X ˆ⊗_{ϵ}Y that are p-summing.

If X, Y and Z are Banach spaces, then every p-summing operator T : X ˆ⊗_{ϵ}Y ⟶ Z induces a p-summing linear operator T^{#} : X ⟶ ∏_{p}(Y, Z). This raises the following question: given two Banach spaces Y and Z, and 1 ≤ p < ∞, for what Banach spaces X is it true that a bounded linear operator T : X ˆ⊗_{ϵ}Y ⟶ Z is p-summing whenever T^{#} : X ⟶ ∏_{p}(Y, Z) is p-summing?

In [11], it was shown that whenever X = C(Ω) is a space of all continuous functions on a compact Hausdorff space Ω, then T : C(Ω) ˆ⊗_{ϵ}Y ⟶ Z is 1-summing if and only if T^{#} : C(Ω) ⟶ ∏_{1}(Y, Z) is 1-summing. We will extend this result by showing that this result still remains true if X is any £_{∞}-space. We will also give an example to show that the result need not be true if X is not a £_{∞}-space. For this, we shall exhibit a 2-summing operator T on ℓ_{2} ˆ⊗_{ϵ}ℓ_{2} that is not 1-summing, but such that the associated operator T^{#} is 1-summing.

The case p > 1 turns out to be quite different. Here, the £_{∞}-spaces do not seem to play any important role. We show that for each 1 < p < ∞, there exists a bounded linear operator T : C[0, 1] ˆ⊗_{ϵ}ℓ_{2} ⟶ ℓ_{2} such that T^{#} : C[0, 1] ⟶ ∏_{p}(ℓ_{2}, ℓ_{2}) is p-summing, but such that T is not p-summing. We will also give an example that shows that, in general, the condition on T^{#} to be 2-summing is too weak to imply any good properties for the operator T at all. To illustrate this, we shall exhibit a bounded linear operator T on C[0, 1] ˆ⊗_{ϵ}ℓ_{1} with values in a certain Banach space Z, such that T^{#} : C[0, 1] ⟶ ∏_{2}(ℓ_{1}, Z) is 2-summing, but for any given N ∈ N, there exists a subspace U of C[0, 1] ˆ⊗_{ϵ}ℓ_{1}, with dim U = N, such that T restricted to U is equivalent to the identity operator on ℓ_{∞}^{N}.

Finally, we show that there is a compact Hausdorff space K and a bounded linear operator T : C(K) ˆ⊗_{ϵ}ℓ_{1} ⟶ ℓ_{2} for which T^{#} : C(K) ⟶ ∏_{1}(ℓ_{1}, ℓ_{2}) is not 2-summing.

I - Definitions and Preliminaries

Let E and F be Banach spaces, and let 1 ≤ q ≤ p < ∞. An operator T : E ⟶ F is said to be (p, q)-summing if there exists a constant C ≥ 0 such that for any finite sequence e_{1}, e_{2}, …, e_{n} in E, we have

(∑_{i=1}^{n} ∥ T (e_{i}) ∥^{p})^{\frac{1}{p}} ≤ C sup \{(∑_{i=1}^{n} |e^{∗}(e_{i})|^{q})^{\frac{1}{q}} : e^{∗} ∈ E^{∗}, ∥ e^{∗} ∥≤ 1\} .

We let π_{p,q}(T) denote the smallest constant C such that the above inequality holds, and let ∏_{p,q}(E, F) be the space of all (p, q)-summing operators from E to F with the norm π_{p,q}. It is easy to check that ∏_{p,q}(E, F) is a Banach space. In the case p = q, we will simply write ∏_{p}(E, F) and π_{p}. We will use the fact that T ∈ ∏_{p,q}(E, F) if and only if ∑_{n} ∥T e_{n}∥^{p} < ∞ for every infinite sequence (e_{n}) in E with ∑_{n} |e^{∗}(e_{n})|^{q} < ∞ for each e^{∗} ∈ E^{∗}. That is to say, T is in ∏_{p,q}(E, F) if and only if T sends all weakly ℓ_{q}-summable sequences into strongly ℓ_{p}-summable sequences. In what follows we shall mainly be interested in the case where p = q and p = 1 or 2.

Given two Banach spaces E and F, we will let E ˆ⊗_{ϵ}F denote their injective tensor product, that is, the completion of the algebraic tensor product E ⊗ F under the cross norm ∥ · ∥_{ϵ} given by the following formula. If ∑_{i=1}^{n} e_{i} ⊗ x_{i} ∈ E ⊗ F, then

∥ ∑_{i=1}^{n} e_{i} ⊗ x_{i} ∥_{ϵ}= sup {| ∑_{i=1}^{n} e^{∗}(e_{i})x^{∗}(x_{i})| : ∥ e^{∗} ∥≤ 1, ∥ x^{∗} ∥≤ 1, e^{∗} ∈ E^{∗}, x^{∗} ∈ F^{∗}} .

We will say that a bounded linear operator T between two Banach spaces E and F is called an integral operator if the bilinear form τ defines an element of (E ˆ⊗_{ϵ}F^{∗})^{∗}, where τ is induced by T according to the formula τ(e, x^{∗}) = x^{∗}(T e) (e ∈ E, x^{∗} ∈ F^{∗}). We will define the integral norm of T, denoted by ∥ T ∥_{int}, by

∥ T ∥_{int} = sup {| ∑_{i=1}^{n} x_{i}^{∗}(T e_{i})| : ∥ ∑_{i=1}^{n} e_{i} ⊗ x_{i}^{∗} ∥_{ϵ}≤ 1} .

The space of all integral operators from a Banach space E into a Banach space F will be denoted by I(E, F). We note that I(E, F) is a Banach space under the integral norm ∥ ∥_{int}.

We will say that a Banach space X is a £_{∞}-space if, for some λ > 1, we have that for every finite dimensional subspace B of X, there exists a finite dimensional subspace E of X containing B, and an invertible bounded linear operator T : E ⟶ ℓ_{∞}^{dim E} such that ∥ T ∥ ∥ T^{−1} ∥≤ λ.

It is well known that for any Banach spaces E and F, if T is in I(E, F), then it is also in ∏_{1}(E, F), with π_{1}(T) ≤∥ T ∥_{int}. But I(E, F) is strictly included in ∏_{1}(E, F). It was shown in [12, p. 477] that a Banach space E is a £_{∞}-space if and only if for any Banach space F, we have that I(E, F) = ∏_{1}(E, F). We will use this characterization of £_{∞}-spaces in the sequel.

Finally, we note the following characterization of 1-summing operators (called right semi-integral by Grothendieck in [5]), which will be used later. Proposition 1 Let E and F be Banach spaces. Then the following properties about a bounded linear operator T from E to F are equivalent:

(i) T is 1-summing; (ii) There exists a Banach space F_{1}, and an isometric injection φ : F ⟶ F_{1}, such that

φ ∘ T : E ⟶ F_{1} is an integral operator.

For all other undefined notions we shall refer the reader to either [3], [7] or [10].

II 1-Summing and Integral Operators

Let X and Y be Banach spaces with injective tensor product X ˆ⊗_{ϵ}Y. For a Banach space Z, any bounded linear operator T : X ˆ⊗_{ϵ}Y ⟶ Z induces a linear operator T^{#} on X by

T^{#}x(y) = T (x ⊗ y) (y ∈ Y).

It is clear that the range of T^{#} is the space £(Y, Z) of bounded linear operators from Y into Z, and that T^{#} is a bounded linear operator.

In this section, we are going to investigate the 1-summing operators, and the integral operators, on X ˆ⊗_{ϵ}Y. We will use Proposition 1 to relate these two ideas together. First of all, we have the following result. Theorem 2 Let X, Y and Z be Banach spaces, and let T : X ˆ⊗_{ϵ}Y ⟶ Z be a bounded linear operator. Denote by i : Z ⟶ Z^{∗∗} the isometric embedding of Z into Z^{∗∗}. Then the following two properties are equivalent:

(i) T ∈ I(X ˆ⊗_{ϵ}Y, Z); (ii) î ∘ T ∈ I(X, I(Y, Z^{∗∗})), where î: I(Y, Z) ⟶ I(Y, Z^{∗∗}) is defined by î(U) = i ∘ U for

each U ∈ I(Y, Z). In particular, if T^{#} ∈ I(X, I(Y, Z)), then T ∈ I(X ˆ⊗_{ϵ}Y, Z).

Proof: First, we show that (X ˆ⊗_{ϵ}Y) ˆ⊗_{ϵ}Z^{∗} and X ˆ⊗_{ϵ}(Y ˆ⊗_{ϵ}Z^{∗}) are isometrically isomorphic to one another. To see this, note that the algebraic tensor product is an associative operation, that is, (X ⊗ Y) ⊗ Z^{∗} and X ⊗ (Y ⊗ Z^{∗}) are algebraically isomorphic. Also, they are both generated by elements of the form ∑_{i=1}^{n} x_{i} ⊗y_{i} ⊗z_{i}^{∗}, where x_{i} ∈ X, y_{i} ∈ Y and z_{i}^{∗} ∈ Z^{∗}. Now, if we let B(X^{∗}), B(Y^{∗}) and B(Z^{∗∗}) denote the dual unit balls of X^{∗}, Y^{∗} and Z^{∗∗} equipped with their respective weak^{∗} topologies, then the spaces (X ⊗_{ϵ} Y) ⊗_{ϵ} Z^{∗} and X ⊗_{ϵ} (Y ⊗_{ϵ} Z^{∗}) embed isometrically into C (B(X^{∗}) × B(Y^{∗}) × B(Z^{∗∗})) in a natural way, by

⟨∑_{i=1}^{n} x_{i} ⊗ y_{i} ⊗ z_{i}^{∗}, (x^{∗}, y^{∗}, z^{∗∗})⟩ = ∑_{i=1}^{n} x^{∗}(x_{i})y^{∗}(y_{i})z^{∗∗}(z_{i}^{∗}),

where ∑_{i=1}^{n} x_{i} ⊗ y_{i} ⊗ z_{i}^{∗} is in (X ⊗_{ϵ} Y) ⊗_{ϵ} Z^{∗} or X ⊗_{ϵ} (Y ⊗_{ϵ} Z^{∗}), and (x^{∗}, y^{∗}, z^{∗∗}) is in the

compact set B(X^{∗})×B(Y^{∗})×B(Z^{∗∗}). Thus both spaces (X ˆ⊗_{ϵ}Y) ˆ⊗_{ϵ}Z^{∗} and X ˆ⊗_{ϵ}(Y ˆ⊗_{ϵ}Z^{∗}) can be thought of as the closure in C (B(X^{∗}) × B(Y^{∗}) × B(Z^{∗∗})) of the algebraic tensor product of X, Y and Z^{∗}.

Now let us assume that T : X ˆ⊗_{ϵ}Y ⟶ Z is an integral operator. Then the bilinear map τ on X ˆ⊗_{ϵ}Y × Z^{∗}, given by τ(u, z^{∗}) = z^{∗}(T u) for u ∈ X ˆ⊗_{ϵ}Y and z^{∗} ∈ Z^{∗}, defines an element of (X ˆ⊗_{ϵ}Y ˆ⊗_{ϵ}Z^{∗})^{∗}, that is,

(*)  ∥ T ∥_{int}= sup {| ∑_{i=1}^{n} z_{i}^{∗} (T (x_{i} ⊗ y_{i})) : ∥ ∑_{i=1}^{n} x_{i} ⊗ y_{i} ⊗ z_{i}^{∗} ∥_{ϵ}≤ 1} .

To show that for every x in X the operator T^{#}x is in I(Y, Z), with

∥ T^{#}x ∥_{int}≤∥ x ∥ ∥ T ∥_{int},

is easy. This is because, for each x ∈ X, the operator T^{#}x is the composition of T with the bounded linear operator from Y to X ˆ⊗_{ϵ}Y, which to each y in Y gives the element x ⊗ y.

If i : Z ⟶ Z^{∗∗} denotes the isometric embedding of Z into Z^{∗∗}, it induces a bounded linear operator î : I(Y, Z) ⟶ I(Y, Z^{∗∗}) given by î(U) = i ∘ U for all U ∈ I(Y, Z). It is immediate that î is an isometry. We will now show that the operator î ∘ T^{#} : X ⟶ I(Y, Z^{∗∗}) is integral. It is well known (see [3, p. 237]) that the space I(Y, Z^{∗∗}) is isometrically isomorphic to the dual space (Y ˆ⊗_{ϵ}Z^{∗})^{∗}. Thus to show that î ∘ T^{#} : X ⟶ (Y ˆ⊗_{ϵ}Z^{∗})^{∗} is an integral operator, we need to show that it induces an element of (X ˆ⊗_{ϵ}(Y ˆ⊗_{ϵ}Z^{∗}))^{∗}. For this, it is enough to note that, by our discussion concerning the isometry of (X ˆ⊗_{ϵ}Y) ˆ⊗_{ϵ}Z^{∗} and X ˆ⊗_{ϵ}(Y ˆ⊗_{ϵ}Z^{∗}), that

(**)  ∥ î ∘ T^{#} ∥_{int}= sup {| ∑_{i=1}^{n} î ∘ T^{#}x_{i}, y_{i} ⊗ z_{i}^{∗}| :∥ ∑_{i=1}^{n} x_{i} ⊗ y_{i} ⊗ z_{i}^{∗} ∥_{ϵ}≤ 1} .

But for each x ∈ X, y ∈ Y and z^{∗} ∈ Z^{∗}, we have

⟨î ∘ T^{#}x, y ⊗ z^{∗}⟩ = ⟨T (x ⊗ y), z^{∗}⟩.

Hence, from (*) and (**), it follows that

∥ î ∘ T ∥_{int}=∥ T ∥_{int} .

Thus we have shown that (i) ⇒ (ii). The proof of (ii) ⇒ (i) follows in a similar way. If î∘T^{#} : X ⟶ I(Y, Z^{∗∗}) is an integral operator, then one can show that i∘T : X ˆ⊗_{ϵ}Y ⟶ Z^{∗∗} is integral, which in turn implies that T itself is integral (see [3, p. 233]).

Finally, the last assertion follows easily, since if T^{#} : X ⟶ I(Y, Z) is integral, then î ∘ T is integral (see [3, p. 232]).

Since the mapping î : I(Y, Z) ⟶ I(Y, Z^{∗∗}) is an isometry, Proposition 1 coupled with Theorem 2 implies that, if T : X ˆ⊗_{ϵ}Y ⟶ Z is an integral operator, then T^{#} : X ⟶ I(Y, Z) is 1-summing. This result can be shown directly from the definitions. In what follows we shall present a sketch of that alternative approach. Theorem 3 Let X, Y and Z be Banach spaces, and let T : X ˆ⊗_{ϵ}Y ⟶ Z be a bounded linear operator. If T is integral, then T^{#} : X ⟶ I(Y, Z) is 1-summing. If in addition X is a £_{∞}-space, then T : X ˆ⊗_{ϵ}Y ⟶ Z is integral if and only if T^{#} : X ⟶ I(Y, Z) is integral. Proof: First, we will show that, if T : X ˆ⊗_{ϵ}Y ⟶ Z is an integral operator, then T^{#} is in ∏_{1} (X, I(Y, Z)) with π_{1}(T^{#}) ≤∥ T ∥_{int}. Let x_{1}, x_{2}, …, x_{n} be in X, and fix ϵ > 0. For each i ≤ n, there exists n_{i} ∈ N, (y_{ij})_{j≤n_{i}} in Y, and (z_{ij}^{∗})_{j≤n_{i}} in Z^{∗}, such that

∥ ∑_{j=1}^{ni} y_{ij} ⊗ z_{ij}^{∗} ∥_{ϵ}≤ 1, and

∥ T^{#}x_{i} ∥_{int}≤ ∑_{j=1}^{n_{i}} z_{ij}^{∗} (T (x_{i} ⊗ y_{ij})) + \frac{ϵ}{2^{i}} .

Since T is an integral operator, and

∥ ∑_{i=1}^{n} ∑_{j=1}^{n_{i}} x_{i} ⊗ y_{ij} ⊗ z_{ij}^{∗} ∥_{ϵ}≤ sup {∑_{i=1}^{n} |x^{∗}(x_{i})| : ∥ x^{∗} ∥≤ 1, x^{∗} ∈ X^{∗}} ,

it follows that

∑_{i=1}^{n} ∑_{j=1}^{n_{i}} z_{ij}^{∗} (T (x_{i} ⊗ y_{ij})) ≤∥ T ∥_{int} sup {∑_{i=1}^{n} |x^{∗}(x_{i})| : ∥ x^{∗} ∥≤ 1, x^{∗} ∈ X^{∗}} .

Therefore

∑_{i=1}^{n} ∥ T^{#}x_{i} ∥_{int} ≤∥ T ∥_{int} sup {∑_{i=1}^{n} |x^{∗}(x_{i})| : x^{∗} ∈ X^{∗}, ∥ x^{∗} ∥≤ 1} + ϵ.

Now, if in addition X is a £_{∞}-space, then by [12, p. 477], the operator T^{#} is indeed integral.

Remark 4 If X = C(Ω) is a space of continuous functions defined on a compact Hausdorff space Ω, one can deduce a similar result to Theorem 3 from the main result of [13].

Our next result extends a result of [16] to £_{∞}-spaces, where it was shown that whenever X = C(Ω), a space of all continuous functions on a compact Hausdorff space Ω, then a bounded linear operator T : C(Ω) ˆ⊗_{ϵ}Y ⟶ Z is 1-summing if and only if T^{#} : C(Ω) ⟶ ∏_{1}(Y, Z) is 1-summing. This also extends a result of [14] where similar conclusions were shown to be true for X = A(K), a space of continuous affine functions on a Choquet simplex K (see [2]).

We note that one implication follows with no restriction on X. If X, Y and Z are Banach spaces, and T : X ˆ⊗_{ϵ}Y ⟶ Z is a 1-summing operator, then T^{#} takes its values in ∏_{1}(Y, Z). This follows from the fact that for each x ∈ X, the operator T^{#}x is the composition of T with the bounded linear operator from Y into X ˆ⊗_{ϵ}Y which to each y in Y gives the element x ⊗ y in X ˆ⊗_{ϵ}Y, and hence

π_{1}(T^{#}x) ≤∥ x ∥ π_{1}(T).

Moreover, one can proceed as in [16] to show that T^{#} : X ⟶ ∏_{1}(Y, Z) is 1-summing. Theorem 5 If X is a £_{∞} space, then for any Banach spaces Y and Z, a bounded linear operator T : X ˆ⊗_{ϵ}Y ⟶ Z is 1-summing if and only if T^{#} : X ⟶ ∏_{1}(Y, Z) is 1-summing. Proof: Let T : X ˆ⊗_{ϵ}Y ⟶ Z be such that T^{#} : X ⟶ ∏_{1}(Y, Z) is 1-summing. Since X is a £_{∞}-space, it follows from [14, p. 477] that T^{#} : X ⟶ ∏_{1}(Y, Z) is an integral operator. Let φ denote the isometric embedding of Z into C (B(Z^{∗})), the space of all continuous scaler functions on the unit ball B(Z^{∗}) of Z^{∗} with its weak^{∗}-topology. This induces an isometry

φ̂ : ∏_{1}(Y, Z) ⟶ ∏_{1} ((Y, C(B(Z^{∗}))) ,
φ̂(U) = φ ∘ U for all U ∈ ∏_{1}(Y, Z).

Now, it follows from [15, p. 301], that ∏_{1} (Y, C(B(Z^{∗}))) is isometric to I (Y, C(B(Z^{∗}))). Hence we may assume that φ̂ ∘ T^{#} : X ⟶ I (Y, C(B(Z^{∗}))) is an integral operator. Moreover, it is easy to check that (φ ∘ T)^{#} = φ̂ ∘ T^{#}. By Theorem 2 the operator φ ∘ T : X ˆ⊗_{ϵ}Y ⟶ C(B(Z^{∗})) is an integral operator, and hence T is in ∏_{1} (X ˆ⊗_{ϵ}Y, Z) by Proposition 1.

In the following section we shall, among other things, exhibit an example that illustrates that it is crucial for the space X to be a £_{∞}-space if the conclusion of Theorem 5 is to be valid.

III 2-summing Operators and some Counter-examples.

In this section we shall study the behavior of 2-summing operators on injective tensor product spaces. As we shall soon see, the behavior of such operators when p = 2 is quite different from when p = 1. For instance, unlike the case p = 1, the £_{∞}-spaces don’t seem to play any particular role. In fact, we shall exhibit operators T on C[0, 1] ˆ⊗_{ϵ}ℓ_{2} which are not 2-summing, yet their corresponding operators T^{#} are. We will also give other interesting examples that answer some other natural questions.

We will present the next theorem for p = 2, but the same result is true for any 1 ≤ p < ∞, with only minor changes.

Theorem 6 Let X, Y and Z be Banach spaces. If T : X ˆ⊗_{ϵ}Y ⟶ Z is a 2-summing operator, then T^{#} : X ⟶ ∏_{2}(Y, Z) is a 2-summing operator.

Proof: If T : X ˆ⊗_{ϵ}Y ⟶ Z is 2-summing, then using the same kind of arguments that we have given above, it can easily be shown that for each x ∈ X, that T^{#}x ∈ ∏_{2}(Y, Z), with π_{2}(T^{#}x) ≤ π_{2}(T) ∥ x ∥.

Now we will show that T^{#} : X ⟶ ∏_{2}(Y, Z) is 2-summing. Let (x_{n}) be in X such that ∑_{n} |x^{∗}(x_{n})|^{2} < ∞ for each x^{∗} in X^{∗}. Fix ϵ > 0. For each n ≥ 1, let (y_{nm}) be a sequence in Y such that

sup {( ∑_{m=1}^{∞} |y^{∗}(y_{nm})|^{2})^{1/2} : ∥ y^{∗} ∥≤ 1, y^{∗} ∈ Y^{∗}} ≤ 1,

and

π_{2} (T^{#}x_{n}) ≤ ( ∑_{m=1}^{∞} ∥ T (x_{n} ⊗ y_{nm}) ∥^{2})^{1/2} + \frac{ϵ}{2^{n}} .

Then

[π_{2} (T^{#}x_{n})]^{2} ≤ ∑_{m=1}^{∞} ∥ T (x_{n} ⊗ y_{nm}) ∥^{2} +\frac{ϵ}{2^{n−1}} ( ∑_{m=1}^{∞} ∥ T (x_{n} ⊗ y_{nm}) ∥^{2})^{1/2} + \frac{ϵ^{2}}{2^{2n}} .

Now, consider the sequence (x_{n} ⊗ y_{nm}) in X ˆ⊗_{ϵ}Y. For each ξ ∈ (X ˆ⊗_{ϵ}Y )^{∗} ≃ I(X, Y^{∗}) we have that

\begin{aligned}∑_{m,n} |ξ(x_{n})(y_{nm})|^{2} & = ∑_{n=1}^{∞} ∑_{m=1}^{∞} |ξ(x_{n})(y_{nm})|^{2} \\ & ≤ ∑_{n=1}^{∞} ∥ ξ(x_{n}) ∥^{2} . \end{aligned}

Since ξ ∈ I(X, Y^{∗}), it follows that ξ ∈ ∏_{2}(X, Y^{∗}), and so

∑_{n=1}^{∞} ∥ ξ(x_{n}) ∥^{2}< ∞.

Hence we have shown that for all ξ ∈ (X ˆ⊗_{ϵ}Y)^{∗},

∑_{m,n} |ξ(x_{n})(y_{nm})|^{2} < ∞.

Since T ∈ ∏_{2} (X ˆ⊗_{ϵ}Y, Z), we have that

∑_{m,n} ∥ T (x_{n} ⊗ y_{nm}) ∥^{2}< ∞,

and therefore

∑_{n} [π_{2} (T^{#}x_{n})]^{2} < ∞.

Remark 7 The above result extends a result of [1], where it was shown that if T : X ˆ⊗_{ϵ}Y ⟶ Z is p-summing for 1 ≤ p < ∞, then T^{#} : X ⟶ £(Y, Z) is p-summing.

Now we shall give the example that we promised at the end of section II. Theorem 8 There exists a bounded linear operator T : ℓ_{2} ˆ⊗_{ϵ}ℓ_{2} ⟶ ℓ_{2} such that T is not 1-summing, yet T^{#} : ℓ_{2} ⟶ π_{1}(ℓ_{2}, ℓ_{2}) is 1-summing. Proof: First, we note the well known fact that ℓ_{2} ˆ⊗_{ϵ}ℓ_{2} = K(ℓ_{2}, ℓ_{2}), the space of all compact operators from ℓ_{2} to ℓ_{2}. Now we define T as the composition of two operators.

Let P : K(ℓ_{2}, ℓ_{2}) ⟶ c_{0} be the operator defined so that for each K ∈ K(ℓ_{2}, ℓ_{2}),

P (K) = (K(e_{n})(e_{n})) ,

where (e_{n}) is the standard basis of ℓ_{2}. It is well known [10, p.145] that the sequence (e_{n}⊗e_{n}) in ℓ_{2} ˆ⊗_{ϵ}ℓ_{2} is equivalent to the c_{0}-basis, and that the operator P defines a bounded linear projection of K(ℓ_{2}, ℓ_{2}) onto c_{0}.

Let S : c_{0} ⟶ ℓ_{2} be the bounded linear operator such that for each (α_{n}) ∈ c_{0}

S(α_{n}) = (\frac{α_{n}}{n} ) .

It is easily checked [7, p. 39] that S is a 2-summing operator that is not 1-summing.

Now we define T : K(ℓ_{2}, ℓ_{2}) ⟶ ℓ_{2} to be T = S ∘ P. Thus T is 2-summing but not 1-summing. It follows from Theorem 6 that the induced operator T^{#} : ℓ_{2} ⟶ ∏_{2}(ℓ_{2}, ℓ_{2}) is 2-summing. Since ℓ_{2} is of cotype 2, it follows from [10, p. 62], that for any Banach space E, we have ∏_{2}(ℓ_{2}, E) = ∏_{1}(ℓ_{2}, E), and that there exists a constant C > 0 such that for all U ∈ ∏_{2}(ℓ_{2}, E) we have

π_{1}(U) ≤ Cπ_{2}(U).

This implies that T^{#} is 1-summing as an operator taking its values in ∏_{1}(ℓ_{2}, ℓ_{2}).

Remark 9 We do not need to use Theorem 6 to show that T^{#} is 1-summing in the example above. Instead, we can use the following argument. First note that T^{#} factors as follows:

ℓ_{2} ⟶^{T^{#}} π_{1}(ℓ_{2}, ℓ_{2})
A↓
ℓ_{2} ↗ B

Here A : ℓ_{2} → ℓ_{2} is the 1-summing operator defined by

A(α_{n}) = (\frac{α_{n}}{n} ) ,

for each (α_{n}) ∈ ℓ_{2}, and B : ℓ_{2} ⟶ π_{1}(ℓ_{2}, ℓ_{2}) is the natural embedding of ℓ_{2} into the space π_{1}(ℓ_{2}, ℓ_{2}) defined by

B(β_{n})(γ_{n}) = (β_{n}γ_{n})

for each (β_{n}), (γ_{n}) ∈ ℓ_{2}.

Now we will give two examples concerning the case when p > 1. We will show that we do not have a converse to Theorem 8, even when the underlying space X is a £_{∞}-space.

First, let us fix some notation. In what follows we shall denote the space ℓ_{p}(Z) by ℓ_{p}, and call its standard basis {e_{n} : n ∈ Z}. Thus if x = (x(n)) ∈ ℓ_{p}, then x(n) = ⟨x, e_{n}⟩, and

∥ x ∥_{ℓ_{p}}= (∑_{n=1}^{∞} |⟨x, e_{n}⟩|^{p}⟩)^{\frac{1}{p}} .

For f ∈ L_{p}[0, 1], we let

∥ f ∥_{L_{p}}= (∫_{0}^{1} |f(t)|^{p}dt)^{\frac{1}{p}} .

If Ω is a compact Hausdorff space, and Y is a Banach space, then C(Ω, Y) = C(Ω) ˆ⊗_{ϵ}Y will denote the Banach space of continuous Y -valued functions on Ω under the supremum norm.

We recall that since ℓ_{2} is of cotype 2, we have that ∏_{2}(ℓ_{2}, ℓ_{2}) = ∏_{1}(ℓ_{2}, ℓ_{2}). We also recall that, if u = ∑_{n=1}^{∞} α_{n}e_{n} ⊗ e_{n} is a diagonal operator in ∏_{2}(ℓ_{2}, ℓ_{2}), then

π_{2}(u) = (∑_{n=1}^{∞} |α_{n}|^{2})^{\frac{1}{2}} = the Hilbert-Schmidt norm of u.

Theorem 10 For each 1 < p < ∞, there is a bounded linear operator T : C([0, 1], ℓ_{2}) → ℓ_{2} that is not p-summing, but such that T^{#} : C[0, 1] ⟶ Π_{p}(ℓ_{2}, ℓ_{2}) is p-summing.

Proof: We present the proof for p ≤ 2. The case where p > 2 follows by the same argument. For each n ∈ Z, let ϵ_{n}(t) : [0, 1] → C, ϵ_{n}(t) = e^{2π int} denote the standard trigonometric basis of L_{2}[0, 1]. If f ∈ L_{1}[0, 1], let f̂(n) = ∫_{0}^{1} f(t)ϵ_{n}(t)dt denote the usual Fourier coefficient of f. For each λ = (λ_{n}), where |λ_{n}| ≤ 1 for all n ∈ Z, define the operator

T_{λ} : C ([0, 1], ℓ_{2}) ⟶ ℓ_{2}

such that for φ ∈ C ([0, 1], ℓ_{2}) we have

T_{λ}φ = (λ_{n} ⟨φ̂(n), e_{n}⟩ ) .

Here φ̂(n) = Bochner – ∫_{0}^{1} φ(t)ϵ_{n}(t)dt.

The operator T_{λ} is a bounded linear operator, with ∥ T_{λ}φ ∥_{ℓ_{2}}≤∥ φ ∥. To see this,

note that for φ ∈ C ([0, 1], ℓ_{2}) we have

∥ T_{λ}φ ∥_{ℓ_{2}}^{2} = ∑_{n} |λ_{n}|^{2}| ⟨φ̂(n), e_{n}⟩ |^{2}

≤ ∑_{n} | ⟨φ̂(n), e_{n}⟩ |^{2}

≤ ∑_{n} ∫_{0}^{1} | ⟨φ(t), e_{n}⟩ |^{2}dt

= ∫_{0}^{1} ∥ φ(t) ∥_{ℓ_{2}}^{2} dt

≤ sup_{t} ∥ φ(t) ∥_{ℓ_{2}}^{2} .

Now, note that if f ∈ C ([0, 1]), and x ∈ ℓ_{2}, then

T_{λ}(f ⊗ x) = (λ_{n}f̂(n)⟨x, e_{n}⟩) ,

and hence the operator T_{λ}^{#} : C[0, 1] →£(ℓ_{2}, ℓ_{2}) is such that

T_{λ}^{#}f(x) = (λ_{n}f̂(n)⟨x, e_{n}⟩) .

Thus

π_{2}(T_{λ}^{#}f) = (∑_{n} |λ_{n}|^{2}|f̂(n)|^{2})^{\frac{1}{2}} .

Hence, by Hölder’s inequality,

π_{2}(T_{λ}^{#}f) ≤∥ (λ_{n}) ∥_{ℓ_{r}}∥ (f̂(n)) ∥_{ℓ_{q}},

where \frac{1}{r} + \frac{1}{q} = \frac{1}{2}. By the Hausdorff-Young inequality, we have that

∥ (f̂(n)) ∥_{ℓ_{q}}≤∥ f ∥_{L_{p}},

where 1 ≤ p ≤ 2 and \frac{1}{p} + \frac{1}{q} = 1. Thus

π_{2}(T_{λ}^{#}f) ≤∥ (λ_{n}) ∥_{ℓ_{r}} ∥ f ∥_{L_{p}},

for 1 ≤ p ≤ 2, 2 ≤ r ≤ ∞ and \frac{1}{p} = \frac{1}{r} + \frac{1}{2}. This shows that if ∥ (λ_{n}) ∥_{ℓ_{r}}< ∞, then

(1) T_{λ}^{#} (C[0, 1]) ⊆ π_{2}(ℓ_{2}, ℓ_{2}) = π_{p}(ℓ_{2}, ℓ_{2});

(2) T_{λ}^{#} : C[0, 1] ⟶ π_{p}(ℓ_{2}, ℓ_{2}) is p-summing. Now, let U ⊂ C ([0, 1], ℓ_{2}) be the closed linear span of {ϵ_{i} ⊗ e_{i}, a_{i} ∈ Z}. Then U is

isometrically isomorphic to ℓ_{2}. This is because

\begin{aligned}∥ ∑_{i} μ_{i}ϵ_{i} ⊗ e_{i} ∥ & = sup_{t∈[0,1]} ∥ (μ_{n}ϵ_{n}(t)) ∥_{ℓ_{2}} \\ & =∥ (μ_{i}ϵ_{i}(t_{0})) ∥_{ℓ_{2}}, \end{aligned}

for some t_{0} ∈ [0, 1], and hence

∥ ∑_{i} μ_{i}ϵ_{i} ⊗ e_{i} ∥= (∑_{i} |μ_{i}|^{2})^{\frac{1}{2}} .

Moreover

T_{λ}(ϵ_{i} ⊗ e_{i}) = λ_{i}e_{i} for all i ∈ Z,

Therefore, we have the following commuting diagram

U ⟶^{T_{λ|U}} ℓ_{2}
Q↓  ↗ S_{λ}
ℓ_{2}

where Q : U → ℓ_{2} is the isomorphism from U onto ℓ_{2} such that Q(ϵ_{n} ⊗ e_{n}) = e_{n} for all n ∈ Z, and S_{λ} : ℓ_{2} ⟶ ℓ_{2} is the operator given by S_{λ}(e_{n}) = λ_{n}e_{n}. So to show that T_{λ} is not p-summing, it is sufficient to show that one can pick λ = (λ_{n}) such that S_{λ} is not p-summing. To do this, we consider two cases. If p = 2, we take λ_{n} = 1 for all n ∈ Z. Then the map S_{λ} induced on ℓ_{2} is the identity map which is not s-summing for any s < ∞. If 1 < p < 2, let λ_{n} = \frac{1}{|n + 1|^{\frac{1}{r}} log |n + 1|}, so that ∥ (λ_{n}) ∥_{ℓ_{r}}< ∞. Then the map S_{λ} : ℓ_{2} ⟶ ℓ_{2} is not s-summing for any s < r. To show this, we may assume, without loss of generality, that s ≥ 2. Let x_{n} = e_{n} for all n ≥ 1, and note that

sup_{x^{∗}∈B(ℓ_{2})} (∑_{n} |x^{∗}(x_{n})|^{s})^{\frac{1}{s}} ≤∥ x^{∗} ∥_{ℓ_{2}}≤ 1,

whilst

(∑_{n} ∥ λ_{n}x_{n} ∥^{s})^{\frac{1}{s}} = ∞.

While the operators T_{λ} in the previous example failed to be p-summing, they were all (2,1)-summing. This suggests the following question: suppose T : C ([0, 1], Y) ⟶ Z is a bounded linear operator such that T^{#} : C[0, 1] ⟶ ∏_{2}(Y, Z) is 2-summing. What can we say about T? Is T (2, 1)-summing? The following example shows that T can be very bad.

Theorem 11 There exists a Banach space Z, and a bounded linear operator T : C ([0, 1], ℓ_{1}) → Z such that T^{#} : C[0, 1] → ∏_{2}(ℓ_{1}, Z) is 2-summing, with the property that, for any N ∈ N, there exists a subspace U of C ([0, 1], ℓ_{1}) with dim U = N, such that T restricted to U behaves like the identity operator on ℓ_{∞}^{N}. In particular T is not (2,1)-summing.

Proof: If X and Y are Banach spaces, we denote by X ˆ⊗_{π}Y the projective tensor product, that is, the completion of the algebraic tensor product of X and Y under the norm

∥ u ∥_{π}= inf{∑_{i=1}^{n} ∥ x_{i} ∥∥ y_{i} ∥, u = ∑_{i=1}^{n} x_{i} ⊗ y_{i}}.

It is well known that (X ˆ⊗_{π}Y)^{∗} is isometrically isomorphic to the space £(X, Y^{∗}) of all bounded linear operators from X to Y^{∗}.

Let Z = C ([0, 1], ℓ_{1}) + L_{2}[0, 1] ˆ⊗_{π}ℓ_{2} be the Banach space with the norm

∥ x ∥_{Z}= inf{∥ x′ ∥_{ϵ} + ∥ x′′ ∥_{π}: x = x′ + x′′},

where ∥ ∥_{ϵ} denotes the sup norm in C ([0, 1], ℓ_{1}), and ∥ ∥_{π} denotes the norm of the projective tensor product L_{2}[0, 1] ˆ⊗_{π}ℓ_{2}. Let

T : C ([0, 1], ℓ_{1}) ⟶ Z

be the identity operator.

We first see that for each f ∈ C[0, 1], the operator T^{#}f : ℓ_{1} → Z is 2-summing with

π_{2}(T^{#}f) ≤ π_{2}(I) ∥ T^{#}f ∥_{£(ℓ_{2},Z)},

where I : ℓ_{1} ⟶ ℓ_{2} is the natural mapping. This is because, for each f ∈ C[0, 1], and each x ∈ ℓ_{1}, we have that

∥ T (f ⊗ x) ∥≤∥ f ⊗ x ∥_{L_{2} ˆ⊗_{π}ℓ_{2}}≤∥ f ∥_{L_{2}} ∥ x ∥_{ℓ_{2}} .

To see that T^{#} : C[0, 1] ⟶ ∏_{2}(ℓ_{1}, X) is 2-summing, note that ∥ T^{#}f ∥_{£(ℓ_{2},Z)}≤∥ f ∥_{L_{2}}, and hence if f_{1}, …, f_{n} ∈ C[0, 1], then

\begin{aligned}(∑_{k=1}^{n} [π_{2}(T^{#}f_{k})]^{2})^{\frac{1}{2}} & ≤ π_{2}(I) (∑_{k=1}^{n} ∥ f_{k} ∥_{L_{2}}^{2})^{\frac{1}{2}} \\ & ≤ π_{2}(I)π_{2}(J) sup_{t∈[0,1]} ∥( ∑_{K=1}^{n} |f_{k}(t)|^{2})^{\frac{1}{2}} ∥ . \end{aligned}

Here J : C[0, 1] ⟶ L_{2}[0, 1] denotes the natural mapping.

Now we define the space U, a closed linear subspace of C ([0, 1], ℓ_{1}). Let {f_{ij} : 1 ≤ i, j ≤ N} be disjoint functions in C[0, 1], for which 0 ≤ f_{ij} ≤ 1, ∥ f_{ij} ∥= 1, each f_{ij} is supported in an interval of length \frac{1}{N^{2}} , and

∫_{0}^{1} f_{ij}dt = \frac{1}{2N^{2}} and ∫_{0}^{1} f_{ij}^{2}dt = \frac{1}{3N^{2}} .

Let {e_{ij} : 1 ≤ i, j ≤ N} be distinct unit vectors in ℓ_{1}. We let U = {∑_{i,j} λ_{i}f_{ij} ⊗e_{ij}, λ_{i} ∈ R}. Now we consider T restricted to U. If ∑_{i,j} λ_{i}f_{ij} ⊗ e_{ij} ∈ U, then

∥ ∑_{i,j} λ_{i}f_{ij} ⊗ e_{ij} ∥_{ϵ}≤ sup_{i} |λ_{i}|,

and hence

∥ ∑_{i,j} λ_{i}f_{ij} ⊗ e_{ij} ∥_{Z}≤ sup_{i} |λ_{i}|.

Let y_{i}^{∗} = N ∑_{j} f_{ij} ⊗ e_{ij}, and set x = ∑_{i,j} λ_{i}f_{ij} ⊗ e_{ij}. Then whenever x = x′ + x′′, with

x′ ∈ C ([0, 1], ℓ_{1}) and x′′ ∈ L_{2}[0, 1] ˆ⊗_{π}ℓ_{2}, we know that

|y_{i}^{∗}(x)| ≤ |y_{i}^{∗}(x′)| + |y_{i}^{∗}(x′′)|.

Hence

|y_{i}^{∗}(x)| ≤∥ y_{i}^{∗} ∥_{C([0,1],ℓ_{1})^{∗}} ∥ x′ ∥_{ϵ} + ∥ y_{i}^{∗} ∥_{(L_{2}[0,1] ˆ⊗_{π}ℓ_{2})^{∗}} ∥ x′′ ∥_{π} .

But

\begin{aligned}∥ y_{i}^{∗} ∥_{C([0,1],ℓ_{1})^{∗}} & = N ∑_{i=1}^{N} ∫_{supp f_{ij}} |f_{ij}|dt \\ & = N · \frac{N}{2N^{2}} = \frac{1}{2}, \end{aligned}

and, since (L_{2}[0, 1] ˆ⊗_{π}ℓ_{2})^{∗} is isometric to £(L_{2}[0, 1], ℓ_{2}),

∥ y_{i}^{∗} ∥_{(L_{2}[0,1] ˆ⊗_{π}ℓ_{2})^{∗}} = sup \{[∑_{j=1}^{N}(N ∫_{0}^{1} f_{ij}gdt)^{2}]^{\frac{1}{2}} : ∥ g ∥_{L_{2}}≤ 1\}

≤ sup \{N[∑_{j=1}^{N} ∫_{0}^{1} f_{ij}^{2}dt · ∫_{supp f_{ij}} |g|^{2}dt]^{\frac{1}{2}} :∥ g ∥_{L_{2}}≤ 1\}

= \frac{1}{\sqrt{3}} \{(∑_{j=1}^{N} ∫_{supp f_{ij}} |g|^{2}dt)^{\frac{1}{2}} :∥ g ∥_{2}≤ 1\}

= \frac{1}{\sqrt{3}}.

Therefore

|y_{i}^{∗}(x)| ≤ \frac{1}{2} ∥ x′ ∥_{ϵ} +\frac{1}{\sqrt{3}} ∥ x′′ ∥_{π}, ≤ \frac{1}{\sqrt{3}} ∥ x ∥ .

However,

\begin{aligned}y_{i}^{∗}(x) & = N ∑_{j=1}^{N} λ_{i} ∫_{0}^{1} f_{ij}^{2}dt \\ & = N^{2}λ_{i} \frac{1}{3N^{2}} = \frac{λ_{i}}{3} . \end{aligned}

Therefore

\begin{aligned}∥ ∑_{i,j} λ_{i}f_{ij} ⊗ e_{ij} ∥_{Z} & ≥ \sqrt{3} sup_{i} |y_{i}^{∗}(x)| \\ & ≥ \frac{1}{\sqrt{3}} sup |λ_{i}|. \end{aligned}

Thus the space U is isomorphic to ℓ_{∞}^{N}, and we have the commuting diagram

U ⟶^{T_{|U}} T (U)
A↓ ↑_{A^{−1}}
ℓ_{∞}^{N} ⟶^{id_{ℓ^{N}_{∞}}} ℓ_{∞}^{N}

where A : U → ℓ_{∞}^{N} is the isomorphism between U and ℓ_{∞}^{N}.

IV Operators that factor through a Hilbert space

It is well known that £(X, ℓ_{2}) = ∏_{2}(X, ℓ_{2}) whenever X is C(K) or ℓ_{1}. One might ask whether this is true when X = C(K, ℓ_{1}). Indeed one could ask the weaker question: if T : C(K, ℓ_{1}) ⟶ ℓ_{2} is bounded, does it follow that the induced operator T^{#} is 2-summing? We answer this question in the negative.

Theorem 12 There is a compact Hausdorff space K and a bounded linear operator T : C(K, ℓ_{1}) ⟶ ℓ_{2} for which T^{#} : C(K) ⟶ ∏_{1}(ℓ_{1}, ℓ_{2}) is not 2-summing. Proof: First, we show that there is a compact Hausdorff space K, and an operator R : C(K) ⟶ ℓ_{∞} that is (2,1)-summing but not 2-summing. To see this, let K = [0, 1], and consider the natural embedding C[0, 1] ⟶ L_{2,1}[0, 1], where L_{2,1}[0, 1] is the Lorentz space on [0, 1] with the Lebesque measure (see [6]). By [11], it follows that this map is (2,1)-summing. To show that this map is not 2-summing, we argue in a similar fashion to [8]. For n ∈ N, consider the functions e_{i}(t) = f(t + \frac{1}{i} mod 1) (1 ≤ i ≤ n), where f(t) = \frac{1}{\sqrt{t}} if t ≥ \frac{1}{n} and \sqrt{n} otherwise. Then it is an easy matter to verify that for some constant C > 0,

(∑_{i=1}^{n} |e^{∗}(e_{i})|^{2})^{\frac{1}{2}} ≤ C\sqrt{log n}

for every e^{∗} in the unit ball of C[0, 1]^{∗}, whereas

(∑_{i=1}^{n} ∥e_{i}∥_{L_{2,1}[0,1]}^{2})^{\frac{1}{2}} ≥ C^{−1} log n.

Finally, since L_{2,1}[0, 1] is separable, it embeds isometrically into ℓ_{∞}.

Define T : C(K, ℓ_{1}) → ℓ_{2} as follows: for φ = (f_{n}) ∈ C(K, ℓ_{1}), let

T (f_{n}) = ∑_{n} Rf_{n}(n)e_{n}.

Then T is bounded, for

∥ T (f_{n}) ∥_{2} = (∑_{n} |Rf_{n}(n)|^{2})^{\frac{1}{2}}

≤ (∑_{n} ∥ Rf_{n} ∥_{ℓ_{∞}}^{2})^{\frac{1}{2}}

≤ π_{2,1}(R) sup_{t∈K} ∑_{n} |f_{n}(t)|.

Thus

∥ T ∥≤ π_{2,1}(R).

But T^{#} : C(K) ⟶ £(ℓ_{1}, ℓ_{2}) is not 2-summing, because for each f ∈ C(K), the operator T^{#}f : ℓ_{1} ⟶ ℓ_{2} is the diagonal operator ∑_{n} Rf(n)e_{n} ⊗ e_{n}. Hence the strong operator norm of T^{#}f is

∥ T^{#}f ∥= sup_{n} |Rf(n)| =∥ Rf ∥_{ℓ_{∞}} .

Thus T^{#} : C(K) ⟶ £(ℓ_{1}, ℓ_{2}) is not 2-summing, because R : C(K) ⟶ ℓ_{∞} is not 2-summing.

Discussions and concluding remarks Remark 13 Theorem 12 shows that if X and Y are Banach spaces such that £(X, ℓ_{2}) = ∏_{2}(X, ℓ_{2}) and £(Y, ℓ_{2}) = ∏_{2}(X, ℓ_{2}), then X ˆ⊗_{ϵ}Y need not share this property. This observation could also be deduced from arguments presented in [4] (use Example 3.5 and the proof of Proposition 3.6 to show that there is a bounded operator T : (ℓ_{1} ⊕ ℓ_{1} ⊕ … ⊕ ℓ_{1})_{ℓ_{∞}} ⟶ ℓ_{2} that is not p-summing for any p < ∞).

Remark 14 In the proof of Theorem 2 we showed that the injective tensor product is an associative operation, that is, if X, Y and Z are Banach spaces, then (X ˆ⊗_{ϵ}Y) ˆ⊗_{ϵ}Z is isometrically isomorphic to X ˆ⊗_{ϵ}(Y ˆ⊗_{ϵ}Z). It is not hard to see that the same is true for the projective tensor product. However, we can conclude from Theorem 12 that what is known as the γ_{2}^{∗}-tensor product is not an associative operation.

If E and F are Banach spaces, and T : E ⟶ F is a bounded linear operator, following [10], we say that T factors through a Hilbert space if there is a Hilbert space H, and operators B : E ⟶ H and A : H ⟶ F such that T = A ∘ B. We let γ_{2}(T) = inf{∥ A ∥ ∥ B ∥}, where the infimum runs over all possible factorization of T, and denote the space of all operators T : E ⟶ F that factor through a Hilbert space by Γ_{2}(E, F). It is not hard to check that γ_{2} defines a norm on Γ_{2}(E, F), making Γ_{2}(E, F) a Banach space. We define the γ_{2}^{∗}-norm ∥ ∥_{∗} on E ⊗ F (see [9] or [10]) in which the dual of E ⊗F is identified with Γ_{2}(E, F^{∗}), and let E ˆ⊗_{γ_{2}^{∗}}F denote the completion of (E ⊗F, ∥ ∥_{∗}).

The operator T : C(K) ˆ⊗_{γ_{2}^{∗}}ℓ_{1} ⟶ ℓ_{2} exhibited in Theorem 12, induces a bounded linear functional on [(C(K) ˆ⊗_{γ_{2}^{∗}}ℓ_{1}) ˆ⊗_{γ_{2}^{∗}}ℓ_{2}]^{∗}. Now we see that if C(K) ˆ⊗_{γ_{2}^{∗}}(ℓ_{1} ˆ⊗_{γ_{2}^{∗}}ℓ_{2}) were isometrically isomorphic to (C(K) ˆ⊗_{γ_{2}^{∗}}ℓ_{1}) ˆ⊗_{γ_{2}^{∗}}ℓ_{2}, then the operator T^{#} : C(K) →£(ℓ_{1}, ℓ_{2}) would induce a bounded linear functional on [C(K) ˆ⊗_{γ_{2}^{∗}}(ℓ_{1} ˆ⊗_{γ_{2}^{∗}}ℓ_{2})]^{∗}, showing that T^{#} ∈ Γ_{2} (C(K), £(ℓ_{1}, ℓ_{2})), implying that T^{#} would be 2-summing [10, p. 62]. This contradiction shows that C(K) ˆ⊗_{γ_{2}^{∗}}(ℓ_{1} ˆ⊗_{γ_{2}^{∗}}ℓ_{2}) and (C(K) ˆ⊗_{γ_{2}^{∗}}ℓ_{1}) ˆ⊗_{γ_{2}^{∗}}ℓ_{2} cannot be isometrically isomorphic.

Another example showing that the γ_{2}^{∗}-tensor product is not associative was given by Pisier (private communication).

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University of Missouri Dept. of Math. Columbia, MO 65211