COMPLEMENTABLY UNIVERSAL BANACH SPACES, II

W. B. JOHNSON AND A. SZANKOWSKI

Abstract. The two main results are:

A. If a Banach space X is complementably universal for all subspaces of c_{0} which have the bounded approximation property, then X^{∗} is non separable (and hence X does not embed into c_{0}), B. There is no separable Banach space X such that every com-

pact operator (between Banach spaces) factors through X. Theorem B solves a problem that dates from the 1970s.

1. Introduction

Given a class O of (bounded, linear) operators, it is natural to try to find a single (usually separable) Banach space U such that all the operators in O factor through U. In this case we say that O factors through U. We say that O λ-factors through U provided that for each S ∈ O there exist operators A,B such that S = BA, U is the co-domain of A and the domain of B, and ∥A∥∥B∥ ≤ λ. If there is a λ so that the class O λ-factors through U, we say that O uniformly factors through U.

These concepts were, essentially, introduced by A. Pełczyński in [14]. He used the following definition: A Banach space U is said to be complementably universal for a class B of Banach spaces provided every space in B is isomorphic to a complemented subspace of U, i.e. if for every B ∈ B, the identity on B factors through U. We shall also say that U is λ-complementably universal for the class B if for every B ∈ B, the identity on B λ-factors through U.

For B_{bas} = the class of all separable Banach spaces that have a (Schauder) basis, there is such a separable U; namely, the separable universal basis space of [14]. Kadec [11] subsequently constructed a separable Banach space with the bounded approximation property (BAP) which is complementably universal for the class of all separable Banach spaces which have the BAP. Actually, the spaces constructed by Kadec and Pełczyński are isomorphic (see [8] and [15]).

Date: January 25, 2008.

1991 Mathematics Subject Classification. Primary 46B20, 46B07; Secondary 46B99.

Key words and phrases. Universal Banach spaces, complemented subspaces, approximation property, factorization of compact operators.

Johnson was supported in part by NSF DMS-0503688, U.S.-Israel Binational Science Foundation, and a Weston Visiting Professorship at the Weizmann Institute of Science.

In part one of this paper ([9]) we proved the non existence of a separable Banach space which is complementably universal for each of the following classes of Banach spaces:

1. B_{AP} = all separable Banach spaces which have the approximation property (AP),

2. B_{p} = all subspaces of ℓ_{p} for 2 < p < ∞.

In particular, there is no separable Banach space which is complementably universal for the class of all separable Banach spaces.

In section 2 we first observe that for 1 < p < ∞, there is a subspace of ℓ_{p} which is complementably universal for the class of all subspaces of ℓ_{p} which have the AP. We also note that there is a subspace of c_{0} which is complementably universal for the class of all subspaces of c_{0} whose duals have the AP. These observations, which are very simple given results from the 1970s, were known to the authors when [9] was written and likely are known to other old timers.

The first main result of this paper is Theorem 2.2, which says that if U is a Banach space which is complementably universal for the class of subspaces of c_{0} which have BAP, then U^{∗} is non separable (and hence U cannot be isomorphic to a subspace of c_{0}). This is done by using, as was done in [9], a variation of Davie’s construction to produce a collection of subspaces of c_{0} so that there is no separable Banach space which is complementably universal for their conjugate spaces. We then use an observation of Johnson and Schechtman, contained in [7], that a subspace X of c_{0} is contained in another subspace Y of c_{0} which has the BAP (even a finite dimensional decomposition) and such that X^{∗} is isomorphic to a complemented subspace of Y^{∗}.

Pełczyński’s universal basis space U has the property that every operator that is uniformly approximable by finite rank operators factors through U. Many other spaces, including some separable reflexive spaces, [6], have the same property. All these results from the 1960s and 1970s left open the problem whether there is a separable Banach space Z so that every compact operator factors through Z. Our second main result, Theorem 2.5, is that there is no such space. We are indebted to Mariusz Wodzicki for reminding us in 1995 that whether such a space exists was still open.

We use standard Banach space theory terminology, as may be found in [12]. K denotes the class of all compact operators (between Banach spaces).

2. Results

We begin with a theorem which perhaps should be termed a “folklore result”. It is at any rate a simple consequence of results proved in the 1970s and was known to the authors when [9] was written. First we set some notation. For 1 ≤ p ≤ ∞, let {G_{n}^{p}}_{n=1}^{∞} be sequence of finite dimensional subspaces of ℓ_{p} such that for every ϵ > 0, every finite dimensional subspace of ℓ_{p} is 1 + ϵ–isomorphic to one (and hence infinitely many) of the spaces in {G_{n}^{p}}_{n=1}^{∞}. Let Y_{p} be the ℓ_{p} sum of {G_{n}^{p}}_{n=1}^{∞} and Y_{0} the c_{0} sum of {G_{n}^{0}}_{n=1}^{∞}.

Theorem 2.1. (a) The space Y_{p}, 1 < p < ∞, is complementably universal for the family of all subspaces of ℓ_{p} which have the approximation property.

(b) The space Y_{0} is complementably universal for the family of all subspaces of c_{0} whose duals have the approximation property.

Proof: To prove (a), first fix 1 < p < ∞ and let X be a subspace of ℓ_{p} which has a finite dimensional decomposition. Then by [10] (or see [12, Theorem 2.d.1]), X is isomorphic to the ℓ_{p} sum of a sequence of finite dimensional spaces. It follows from the construction of Y_{p} that X is isomorphic to a complemented subspace of Y_{p}. In the general case, where X is a subspace of ℓ_{p} which has the AP, by a theorem of Grothendieck [12, Theorem 1.e.15], X has the metric approximation property and hence the BAP . It then follows from the argument for Theorem 4 of [5] that X ⊕_{p} Y_{p} is a Π–space; that is, that there exists a sequence of finite rank projections on X ⊕_{p} Y_{p} which converges strongly to the identity operator. Since the dual of X ⊕_{p} Y_{p} also has the BAP, Theorem 1.3 in [8] yields that X ⊕_{p} Y_{p} has a finite dimensional decomposition. Therefore, by the first step of the proof, X ⊕_{p}Y_{p} is isomorphic to a complemented subspace of Y_{p}.

The proof of (b) uses the same ingredients. If X is a subspace of c_{0} which has a shrinking finite dimensional decomposition, then again by [9] (or see [12, Theorem 2.d.1]), X is isomorphic to the c_{0} sum of a sequence of finite dimensional spaces and hence is isomorphic to a complemented subspace of Y_{0}. If X is a subspace of c_{0} whose dual has the approximation property, then Grothendieck’s theorem [12, Theorem 1.e.15] implies that X^{∗} has the BAP. One then uses [5] and [8] in the same way as in the ℓ_{p} case to conclude that X is isomorphic to a complemented subspace of Y_{0}.

Theorem 2.1 suggests the following problem: What can one say about a separable Banach space which is complementably universal for the collection of all subspaces of c_{0} which have the BAP? By the results of Kadec [11] and Pełczyński [14] mentioned in the introduction, such spaces do exist. The most natural question is whether a subspace of c_{0} can have this universal property. One of the main results of this note gives a negative answer to this question:

Theorem 2.2. Let U be a Banach space which is complementably universal for the family of all subspaces of c_{0} which have the BAP. Then U^{∗} is non separable.

The main technical tool for proving Theorem 2.2 is Theorem 2.3.

Theorem 2.3. There is no separable Banach space which is complementably universal for the family D_{0} of duals to subspaces of c_{0}.

Let us observe here that, since D_{0} is closed under l_{1}-sums, by Proposition 2 in [6] it suffices to prove the following statement:

Theorem 2.4. There is no separable Banach space which is uniformly complementably universal for D_{0} .

Theorem 2.4 is proved in section 4 . The proof is similar to the proof in [9] that no separable Banach space is complementably universal for the collection of subspaces of ℓ_{p} when 2 < p < ∞, and indeed the argument in section 4 gives this result from [9]. Since the argument we need here is more involved, we chose to give a detailed, complete and streamlined proof of Theorem 2.4.

Once Theorem 2.3 is known, we complete the proof of Theorem 2.2 with the following proposition which is proved but not stated in [7, section 2].

Proposition 1. Let X be a subspace of c_{0}. Then there is a subspace Y of c_{0} which has a finite dimensional decomposition and such that X^{∗} is isomorphic to a complemented subspace of Y^{∗}.

It is obvious that Theorem 2.2 follows from the conjunction of Theorem 2.3 and Proposition 1. Here we repeat part of the discussion in [7, section 2] which yields Proposition 1 and refer to [7] for additional details. Let E_{1} ⊂ E_{2} ⊂ … be a sequence of finite dimensional subspaces of X whose union is dense in X and let Y be the subspace of the ℓ_{∞} sum (∑_{n} E_{n})_{∞} of {E_{n}}_{n=1}^{∞} consisting of sequences (e_{1}, e_{2}, …) for which lim_{n→∞} e_{n} exists in X. The space Y has a monotone finite dimensional decomposition.. Indeed, for each positive integer n define a contractive projection P_{n} on Y by setting P (e_{1}, e_{2}, …) = (e_{1}, e_{2}, …, e_{n−1}, e_{n}, e_{n}, …). It is easy to check that {P_{n}}_{n=1}^{∞} is the sequence of partial sum projections for a finite dimensional decomposition of Y. Now define an operator Q from Y into X by setting Q(e_{1}, e_{2}, …) = lim_{n} e_{n}. It is easy to check that Q is a quotient mapping from Y onto X with kernel the c_{0} sum of {E_{n}}_{n=1}^{∞}. This is a construction used to good effect by Lusky [13]. The main new points in [7] are the observations that the separable injectivity of c_{0} yields that Y is isomorphic to a subspace of c_{0}, and that, by [6], the range of Q^{∗} (which is isometric to X^{∗}) is norm one complemented in Y^{∗}. This completes the proof of Proposition 1 and hence also the proof of Theorem 2.2.

We now state our second main result.

Theorem 2.5. The class of all compact operators (between Banach spaces) does not factor through a separable Banach space.

Evidently, it is enough to prove the following

Theorem 2.6. The class of all compact operators of norm ≤ 1 does not uniformly factor through a separable Banach space.

The proofs of Theorems 2.4 and 2.6 are based on a construction which has two essential components: linear-algebraic, described in section 5 and probabilistic, described in section 6.

3. Preliminaries

We shall use the following lemma, which is a refinement of Lemma 2 in [9]:

Lemma 1. Let A be an uncountable set, let B, Z be Banach spaces and let Dε be a subspace of B, Rε be a subspace of Z for ε ∈ A. Suppose that there are δ > 0, λ > 0 and a finite dimensional space E with E ⊂ ⋂_{ε∈A} Dε, E ⊂ ⋂_{ε∈A} Rε, so that the following condition is

satisfied:

(♯_{1}) if ε, η ∈ A, ε ≠ η and T : Dε → Rη is such that

∥T_{|E} − Id_{E}∥ < δ, then ∥T ∥ > λ . Let Tε : Dε → Rε be bounded operators such that T_{ε|E} = Id_{E} for

every ε ∈ A. Then the family of operators {Tε : ε ∈ A} does not λ-factor through a separable Banach space.

Proof. Suppose U is a separable Banach space such that for every ε ∈ A, the operator Tε has a factorization Tε = QεSε with Sε : Dε → U and Qε : U → Rε so that ∥Sε∥ ≤ λ,∥Qε∥ ≤ 1. Since the space L(E, U) is separable and A is uncountable, there are ε, η ∈ A, ε ≠ η such that ∥(Sε − Sη)_{|E}∥ < δ. Let us define T = QεSη. We have (QεSε)_{|E} = Id_{E}, thus

T_{|E} − Id_{E} = Qε(Sη − Sε)_{|E}

and therefore ∥T_{|E} − Id_{E}∥ ≤ ∥Qε∥∥(Sε − Sη)_{|E}∥ < δ. Therefore, by (♯_{1}), ∥T ∥ > λ. This is a contradiction with

∥T ∥ ≤ ∥Qε∥∥Sη∥ ≤ λ.

The next lemma is a “complementably universal” version of Lemma 1:

Lemma 2. Let A be an uncountable set, let B be a Banach space and let Bε be a subspace of B for ε ∈ A. Suppose that there are δ > 0, λ > 0 and a finite dimensional space E, E ⊂ ⋂_{ε∈A} Bε so that the

following condition is satisfied:

(♯_{2}) if ε, η ∈ A, ε ≠ η and T : Bε → Bη is such that

∥T_{|E} − Id_{E}∥ < δ, then ∥T ∥ > λ . Then there is no separable Banach space which is λ-complementably

universal for the family {Bε : ε ∈ A} .

The construction of appropriate families {Bε : ε ∈ A} (for Theorems 2.2 and 2.3) and {Dε, Rε : ε ∈ A} (for Theorems 2.5 and 2.6) has two essential components: a linear-algebraic one (section 5) and a probabilistic one (section 6).

Now we would like to make explicit some tensor product notation. Given finite sets I ⊂ J, let Π_{I} denote the coordinate projection in

ℝ^{J} onto the coordinates in I.

Let E, F be vector spaces, let e_{1}, …, e_{M}; f_{1}, …, f_{N} be bases in E, F, respectively. Let I ⊂ J be finite sets and let x_{i} ∈ E, w_{i} ∈ F for i ∈ J with

x_{i} = ∑_{m=1}^{M} x(i, m)e_{m}, w_{i} = ∑_{n=1}^{N} w(i, n)f_{n}.

Then

(1)  ∑_{i∈I} w_{i} ⊗ x_{i} = ∑_{n=1}^{N} f_{n} ⊗ X^{∗}Π_{I}W (f_{n})

where, formally, X : E → ℝ^{J} and W : F → ℝ^{J} are defined by Xe_{m} = (x(i, m))_{i∈J} for m = 1, …, M and W f_{n} = (w(i, n))_{i∈J} for n = 1, …, N.

Let Z, Y be Banach spaces. For β = Σψ_{i} ⊗ z_{i} ∈ Y^{∗} ⊗ Z let us denote supp β = span {z_{i}} and β^{⊥} = ⋂ ψ_{i}^{⊥} = {y ∈ Y : ψ_{i}(y) = 0 for all i} (these notions depend on a specific representation of the tensor β but this will not interfere with our considerations).

As usual, a β ∈ Y^{∗} ⊗ Z acts on L(Z′, Y′) for any Z′ ⊃ supp β and Y′ ⊂ Y by the formula β(T) = Σψ_{i}(T z_{i}). For such a T we have

|β(T)| ≤ ∥β∥_{Y^{∗} ˆ⊗Z′}∥T ∥_{L(Z′,Y′)}.

(Recall that

∥β∥_{Y^{∗} ˆ⊗Z′} = inf{Σ∥ψ_{i}∥∥z_{i}∥ : β = Σψ_{i} ⊗ z_{i} with ψ_{i} ∈ Y^{∗}, z_{i} ∈ Z′}.)

Lemma 3. Let Z, Y be Banach spaces, let E ⊂ Z ∩ Y. Let β, γ ∈ Y^{∗} ⊗ Z. Assume that E ⊃ supp β and β(Id_{E}) = 1. Suppose that T : Z′ → Y′ where Y′ ⊂ γ^{⊥}, Z′ ⊃ supp β + supp γ. Then

∥T ∥ ≥ ∥β − γ∥_{Y^{∗} ˆ⊗Z′}^{−1}(1 − ∥β∥_{Y^{∗} ˆ⊗Z′}∥T_{|E} − Id_{E}∥).

Proof. Since the range of T is γ^{⊥}, γ(T) = 0 and thus

(β − γ)(T) = β(T) = 1 + β(T_{|E} − Id_{E}).

Hence

∥β − γ∥_{Y^{∗} ˆ⊗Z′}∥T ∥ ≥ 1 − ∥β∥_{Y^{∗} ˆ⊗Z′}∥T_{|E} − Id_{E}∥,

which proves the lemma.

4. The proofs

In this section we give a schematic outline of the proofs of Theorems 2.4 and 2.6 based on Lemmas 2 and 1. Technical details of the construction are deferred to sections 5 and 6.

Let ℝ^{∞} = {(x_{k})_{k=0}^{∞} : x_{k} ∈ ℝ^{3·2^{k}}, x_{k} ≠ 0 for finitely many k}. In ℝ^{∞} we have the natural inner product ⟨(x_{k}), (z_{k})⟩ = ∑_{k=0}^{∞}⟨x_{k}, z_{k}⟩.

Let X = (Σℓ_{1}^{3·2^{k}})_{c_{0}}, Z = (Σℓ_{∞}^{3·2^{k}})_{ℓ_{1}}.

Let e_{i}^{k} for k = 1, 2, …; i = 1, …, 3·2^{k} denote the unit vectors in ℝ^{∞}. In section 5 we shall describe vectors y_{i}^{k} ∈ ℝ^{∞} for k = 1, 2, …, i =

1, 2, …, 2^{k}. These vectors satisfy the following conditions:

(2)  ⟨y_{i}^{k}, y_{j}^{m}⟩ = 2δ_{km}δ_{ij} for all i, j, k, m,
(3)  ∥y_{i}^{k}∥_{Z} ≤ 2 · 2^{− \frac{k}{2}} , ∥y_{i}^{k}∥_{X} ≤ 3 · 2^{\frac{k}{2}} .

Let Y = span {y_{i}^{k} : k = 1, 2, …; i = 1, …, 2^{k}}. Given an x ∈ ℝ^{∞}, we identify it with the functional on Y defined by y ⇝ ⟨y, x⟩.

Let us denote for k = 1, 2, …:

I_{0}^{k} = {1, …, 2^{k−1}}, I_{1}^{k} = {2^{k−1} + 1, …, 2^{k}}.

Let us define for k = 1, 2, … and ε = 0, 1:

Y_{ε}^{k} = span {y_{i}^{k} : i ∈ I_{ε}^{k}}

and

(4)  β_{ε}^{k} = 2^{−k} ∑_{i∈I_{ε}^{k}} y_{i}^{k} ⊗ y_{i}^{k} ∈ Y^{∗} ⊗ Y_{ε}^{k} ⊂ Y^{∗} ⊗ Y.

Observe that, by (2),

(5)  β_{ε}^{k}(Id_{Y_{ε}^{k}}) = 1 for every k, ε.

Let us denote C = {0, 1}^{ℕ}. For ε ∈ C let Yε = span ⋃_{k=1}^{∞} Y_{ε(k)}^{k}. The family of spaces {Yε : ε ∈ C} will be the basis for our constructions.

In section 5 we shall obtain the following key representation for the difference β_{ε}^{k} − β_{η}^{k+1} ∈ Y^{∗} ⊗ Y:

(6)  β_{ε}^{k} − β_{η}^{k+1} = 2^{−k} ∑_{j=1}^{2^{k}} e_{j}^{k} ⊗ d_{ε,η,j}^{k}

where d_{ε,η,j}^{k} for k = 1, 2, …; j = 1, … 2^{k} and ε, η = 0, 1 satisfy the following two conditions:

(7)  ∥d_{ε,η,j}^{k}∥_{Z} ≤ Ck^{2}2^{− \frac{k}{2}} ,

the constant C being independent of k, j, ε, η, and

(8)  d_{ε,η,j}^{k} ∈ Y_{ε}^{k} + Y_{η}^{k+1}.

Proof of Theorem 2.4. With the duality generated by the inner product ⟨, ⟩, we have Z = X^{∗} (and, obviously, X ⊂ Z^{∗}).

For A ⊂ Z denote A_{⊥} = {x ∈ X : ⟨z, x⟩ = 0 for all z ∈ A} and for A ⊂ X denote A^{⊥} = {z ∈ Z : ⟨z, x⟩ = 0 for all x ∈ A}.

Let B = (Y_{⊥})^{⊥} and let Bε = ((Yε)_{⊥})^{⊥} for ε ∈ C. Evidently B and Bε are w^{∗}-closed subspaces of Z = X^{∗}, Bε ⊂ B for every ε ∈ C and we have

(Y_{1−ε(k)}^{k})^{⊥} ⊃ Bε ⊃ Y_{ε(k)}^{k} for every k.

Define β_{ε}^{k} by (4). We have ∥y_{i}^{k}∥_{Z^{∗}} = ∥y_{i}^{k}∥_{X}, thus, by (3),

∥y_{i}^{k}∥_{Z^{∗}}∥y_{i}^{k}∥_{Z} ≤ 6 for all i, k.

Hence (here Y_{ε}^{k} is equipped with the norm of Z):

(9)  ∥β_{ε}^{k}∥_{B^{∗} ˆ⊗Y_{ε}^{k}} ≤ 6 for all ε, k.

Since ∥e_{j}^{k}∥_{Z^{∗}} = 1, by the representation (6), we obtain for every k and ε ∈ C,

(10)  ∥β_{ε(k)}^{k} − β_{ε(k+1)}^{k+1}∥_{B^{∗} ˆ⊗Bε} ≤ Ck^{2}2^{− \frac{k}{2}} ,

the constant C being independent of k and ε ∈ C. Therefore for every k ≤ m,

(11)  ∥β_{ε(k)}^{k} − β_{ε(m)}^{m}∥_{B^{∗} ˆ⊗Bε} ≤ Ck^{2}2^{− \frac{k}{2}}

with another constant C.

Applying Lemma 2 and Lemma 3 we can now prove Theorem 2.4. Indeed, denote C_{k} = {ε ∈ C : ε(1) = ⋯ = ε(k) = 0}, let E = E_{k} = Y_{0}^{k} = span {y_{i}^{k} : 1 ≤ i ≤ 2^{k−1}}.

Let ε, η ∈ C_{k}, ε ≠ η, let m be the first index larger than k such that ε(m) ≠ η(m), thus Rη ⊂ β_{ε(m)}^{⊥}. Let T : Bε → Bη be such that ∥T_{|E} − Id_{E}∥ < δ = \frac{1}{12}.

Observe that, by (5), β_{ε(k)}^{k}(Id_{E}) = 1 for every ε ∈ C_{k}. By Lemma 3,

(9) and (11),

∥T ∥ ≥ ∥β_{ε(k)}^{k} − β_{ε(m)}^{m}∥_{B^{∗} ˆ⊗Bε}^{−1}(1 − ∥β_{ε(k)}^{k}∥_{B^{∗} ˆ⊗Bε}δ) > λ_{k} = \frac{1}{2C} k^{−2}2^{\frac{k}{2}} .

Since T_{ε|E} = I_{E}, by Lemma 2, there is no separable Banach space which is λ_{k}-complementably universal for the family {Bε : ε ∈ C_{k}}. (Perhaps it looks a bit strange that we can prove that λ_{1}, λ_{2}, λ_{3}, … increase while C_{1} ⊃ C_{2} ⊃ . . .. The reason is that, by the estimate of Lemma 2, λ_{k} increase together with the size of E_{k} and the spaces E_{k} do grow.) Consequently, there is no separable Banach space which is uniformly complementably universal for the family {Bε : ε ∈ C}.

To conclude, let us observe that each Bε is isomorphic to the dual of a subspace of c_{0}: being w^{∗}-closed in Z, Bε is isometric to the dual of a quotient of X and X is clearly isomorphic to a subspace of c_{0}. Finally, every subspace of a quotient of c_{0} is isomorphic to a subspace of c_{0} (see [10] or [1]).

Proof of Theorem 2.6. In Y we will define a norm ||| ||| in the following way. Let

V = {y_{i}^{k} : k = 1, 2, …; i = 1, …, 2^{k}},
W = \{2^{\frac{k}{4}} d_{ε,η,i}^{k} : k = 1, 2, …, ε, η = 0, 1, i = 1, …, 2^{k} \},

let U = absconv (V ∪ W) and let for x ∈ Y

|||x||| = inf{λ : x ∈ λU}.

Let B be the completion of Y in the norm ||| |||, let T : B → Z be the completion of Id_{Y}.

It is clear that T has norm ≤ 1. We also see that T is a compact operator, since U is dense in the unit ball of B and T (U) is the convex hull of a sequence which converges to 0 in Z (observe that, by (7), ∥2^{\frac{k}{4}} d_{ε,η,i}^{k}∥_{Z} ≤ Ck^{2}2^{− \frac{k}{4}} ).

Let Dε be the closure of Yε in B, let Rε be the closure of Yε in Z and let Tε = T_{|Dε}.

Now the argument goes like that of the previous proof:

Define β_{ε}^{k} by (4). We have ∥y_{i}^{k} ∥_{B^{∗}} = ∥y_{i}^{k}∥_{X} ≤ 2·2^{\frac{k}{2}} and ∥y_{i}^{k}∥_{Dε} ≤ 1, therefore

(12)  ∥β_{ε(k)}^{k}∥_{B^{∗} ˆ⊗Dε} ≤ 2 · 2^{\frac{k}{2}} for all ε, k.

Let us use the representation β_{ε}^{k} − β_{η}^{k+1} = 2^{−k} ∑_{j=1}^{2^{k}} e_{j}^{k} ⊗ d_{ε,η,j}^{k} (cf.(6)).

We have ∥e_{j}^{k}∥_{Z^{∗}} = 1 and, by the definition of the norm ||| |||, we have |||d_{ε(k),ε(k+1),i}^{k}||| ≤ 2^{− \frac{k}{4}} , therefore

∥β_{ε(k)}^{k} − β_{ε(k+1)}^{k+1}∥_{B^{∗} ˆ⊗Dε} ≤ 2^{− \frac{k}{4}} .

Consequently, for every m > k

(13)  ∥β_{ε(k)}^{k} − β_{ε(m)}^{m}∥_{B^{∗} ˆ⊗Dε} ≤ 6 · 2^{− \frac{k}{4}} .

Let again C_{k} = {ε ∈ C : ε(1) = ⋯ = ε(k) = 0} and E = Y_{0}^{k}. Let us take δ = \frac{1}{4}2^{− \frac{k}{2}} . Let ε, η ∈ C_{k}, ε ≠ η, let m be the first index larger than k such that ε(m) ≠ η(m), thus Rη ⊂ β_{ε(m)}^{⊥}. Let T : Dε → Rη be such that ∥T_{|E} − Id_{E}∥ < δ. By Lemma 3, (12) and (13)

∥T ∥ ≥ ∥β_{ε(k)}^{k} − β_{ε(m)}^{m}∥_{B^{∗} ˆ⊗Dε}^{−1}(1 − ∥β_{ε(k)}^{k}∥_{B^{∗} ˆ⊗Dε}δ) > 2^{\frac{k}{4} −4}.

Since T_{ε|E} = Id_{E}, by Lemma 1, the family {Tε : ε ∈ C_{k}} does not 2^{\frac{k}{4} −4}− uniformly factor through a separable Banach space, hence the family {Tε : ε ∈ C} does not uniformly factor through a separable Banach space.

5. The Construction of y_{i}^{k}

Let us denote

I_{0}^{k} = {1, …, 2^{k−1}}, I_{1}^{k} = {2^{k−1} + 1, …, 2^{k}}.

For k = 0, 1, 2, …, let u_{1}^{k}, …, u_{3·2^{k}}^{k}, be orthonormal systems in ℝ^{3·2^{k}}. For j = 1, …, 2^{k+1} denote v_{j}^{k} = u_{2^{k}+j}^{k} and let us define y_{i}^{k} ∈ ℝ^{∞} for k = 1, 2, …; i = 1, 2, …, 2^{k} by

y_{i}^{k} = v_{i}^{k−1} + u_{i}^{k}.

Due to the orthonormality of the whole system {u_{i}^{k} : k = 0, 1, 2, …; i = 1, …, 3·2^{k}}, we have

⟨y_{i}^{k}, y_{j}^{m}⟩ = 2⟨u_{i}^{k}, y_{j}^{m}⟩ = 2⟨v_{i}^{k−1}, y_{j}^{m}⟩ = 2δ_{km}δ_{ij} for all i, j, k, m,

therefore we have for every y ∈ Y (= span y_{i}^{k})

(14)  ⟨y_{i}^{k}, y⟩ = 2⟨u_{i}^{k}, y⟩ = 2⟨v_{i}^{k−1}, y⟩,

i.e. for every x ∈ Y the following tensors are identical in Y^{∗} ⊗ Y:

y_{i}^{k} ⊗ x = 2u_{i}^{k} ⊗ x = 2v_{i}^{k−1} ⊗ x.

This is a crucial observation which allows us to represent β_{ε}^{k} ∈ Y^{∗} ⊗ Y in two different ways:

β_{ε}^{k} = 2^{−k} ∑_{i∈I_{ε}^{k}} u_{i}^{k} ⊗ y_{i}^{k}

and

β_{ε}^{k} = 2^{−k} ∑_{i∈I_{ε}^{k}} v_{i}^{k−1} ⊗ y_{i}^{k}.

Now we shall transform these formulas, writing β_{ε}^{k} in terms of e_{j}^{k}, like in (1). Let

u_{j}^{k} = ∑_{j=1}^{3·2^{k}} u^{k}(i, j)e_{j}^{k} for j = 1, …, 2^{k}, v_{j}^{k} = ∑_{j=1}^{3·2^{k}} v^{k}(i, j)e_{j}^{k} for j = 1, …, 2^{k+1}.

We have for I ⊂ {1, …, 2^{k}}:

∑_{i∈I} u_{i}^{k} ⊗ y_{i}^{k} = ∑_{i∈I} u_{i}^{k} ⊗ v_{i}^{k−1} + ∑_{i∈I} u_{i}^{k} ⊗ u_{i}^{k} =

∑_{j=1}^{3·2^{k}} e_{j}^{k} ⊗ (V^{k−1})^{∗}Π_{I}U^{k}(e_{j}^{k}) + ∑_{j=1}^{3·2^{k}} e_{j}^{k} ⊗ (U^{k})^{∗}Π_{I}U^{k}(e_{j}^{k})

and

∑_{i∈I} v_{i}^{k−1} ⊗ y_{i}^{k} = ∑_{i∈I} v_{i}^{k−1} ⊗ v_{i}^{k−1} + ∑_{i∈I} v_{i}^{k−1} ⊗ u_{i}^{k} =

∑_{j=1}^{3·2^{k−1}} e_{j}^{k−1} ⊗ (V^{k−1})^{∗}Π_{I}V^{k−1}(e_{j}^{k−1}) + ∑_{j=1}^{3·2^{k−1}} e_{j}^{k} ⊗ (U^{k})^{∗}Π_{I}V^{k−1}(e_{j}^{k−1}).

where U^{k} : ℝ^{3·2^{k}} → ℝ^{2^{k}} and V^{k} : ℝ^{3·2^{k}} → ℝ^{2^{k+1}} are defined by:

(15)  U^{k}e_{j}^{k} = (u^{k}(i, j))_{i=1}^{2^{k}} and V^{k}e_{j}^{k} = (v^{k}(i, j))_{i=1}^{2^{k+1}}.

Denote now

(16)  U_{ε}^{k} = Π_{I_{ε}^{k}}U^{k}, V_{ε}^{k} = Π_{I_{ε}^{k+1}}V^{k}.

We have for ε, η = 0, 1:

β_{ε}^{k} − β_{η}^{k+1} = 2^{−k} ∑_{i∈I_{ε}^{k}} u_{i}^{k} ⊗ y_{i}^{k} − 2^{−k−1} ∑_{i∈I_{η}^{k+1}} v_{i}^{k} ⊗ y_{i}^{k+1}

= 2^{−k−1} ∑_{j=1}^{2^{k}} e_{j}^{k} ⊗ D_{ε,η}^{k}e_{j}^{k}

where the map D_{ε,η}^{k} : ℝ^{2^{k}} → Y is defined by

D_{ε,η}^{k} = 2(V_{ε}^{k−1})^{∗}U_{ε}^{k} + 2(U_{ε}^{k})^{∗}U_{ε}^{k} − (V_{η}^{k})^{∗}V_{η}^{k} − (U_{η}^{k+1})^{∗}V_{η}^{k}.

Denote

c_{1} = (V_{ε}^{k−1})^{∗}U_{ε}^{k}(e_{j}^{k}), c_{2} = [2(U_{ε}^{k})^{∗}U_{ε}^{k}−(V_{η}^{k})^{∗}V_{η}^{k}](e_{j}^{k}), c_{3} = (U_{η}^{k+1})^{∗}V_{η}^{k}(e_{j}^{k}).

Let d_{ε,η,j}^{k} = D_{ε,η}^{k}e_{j}^{k} = c_{1} + c_{2} + c_{3}. It is clear that c_{1} ∈ ℝ^{2·3^{k−1}}, c_{2} ∈ ℝ^{2·3^{k}}, c_{3} ∈ ℝ^{2·3^{k+1}}, therefore

∥d_{ε,η,j}^{k}∥_{Z} = ∥c_{1}∥_{∞} + ∥c_{2}∥_{∞} + ∥c_{3}∥_{∞}.

For a matrix Q = {q(i, j)} we denote ∥Q∥_{∞} = max |q(i, j)|. Since c_{1}, c_{2}, c_{3} are columns of the corresponding matrices, we have :

∥c_{1}∥_{∞} ≤ 2∥(V_{ε}^{k−1})^{∗}U_{ε}^{k}∥_{∞}, ∥c_{2}∥_{∞} ≤ ∥2(U_{ε}^{k})^{∗}U_{ε}^{k} − (V_{η}^{k})^{∗}V_{η}^{k}∥_{∞},
∥c_{3}∥_{∞} ≤ ∥(U_{η}^{k+1})^{∗}V_{η}^{k}∥_{∞}.

The last section of this paper is devoted to proving that there exist orthonormal systems u_{1}^{k}, …, u_{3·2^{k}}^{k} in ℝ^{3·2^{k}} so that for all k, j, ε, η we have

\begin{aligned}∥(V_{ε}^{k−1})^{∗}U_{ε}^{k}∥_{∞} & ≤ Ck^{2}2^{− \frac{k}{2}} , ∥2(U_{ε}^{k})^{∗}U_{ε}^{k} − (V_{η}^{k})^{∗}V_{η}^{k}∥_{∞} ≤ Ck^{2}2^{− \frac{k}{2}} , \\ ∥(U_{η}^{k+1})^{∗}V_{η}^{k}∥_{∞} & ≤ Ck^{2}2^{− \frac{k}{2}} , \end{aligned} ; (17)

which clearly implies our key estimate (7). The proof is probabilistic: it turns out that choosing the system u_{1}^{k}, …, u_{3·2^{k}}^{k} in ℝ^{3·2^{k}} randomly (w.r.t. the Haar measure on the orthogonal group), c_{1}, c_{2}, c_{3} will be small (i.e. of order k^{2}2^{− \frac{k}{2}} ) with large probability and therefore we can find such systems so that (17) is satisfied.

Let us observe here that the reasons for the smallness of c_{1}, c_{2}, c_{3} are somewhat different: c_{1} and c_{3} are small because the matrices U_{ε}^{k} and V_{η}^{k−1} are independent and therefore all the entries of (V_{η}^{k−1})^{∗}U_{ε}^{k} are small. On the other hand, in the matrices (U_{ε}^{k})^{∗}U_{ε}^{k} and (V_{η}^{k})^{∗}V_{η}^{k} the diagonal elements are about 1/4, respectively 1/2 and the off-diagonal elements are small (because of independence). Consequently, all the elements of 2(U_{ε}^{k})^{∗}U_{ε}^{k}−(V_{η}^{k})^{∗}V_{η}^{k} are small (the diagonal elements cancel out) and this is why c_{2} is small.

6. A probabilistic lemma

For a matrix Q = {q(i, j)} we denote ∥Q∥_{∞} = max |q(i, j)|.

Lemma 4. For j = 1, 2, let Q_{j} be an n×m_{j} matrix, with m_{1}, m_{2} ≤ 4n. Let n_{1}, n_{2}, n_{3}, n_{4} be natural numbers so that n = n_{1}+n_{2}+n_{3}+n_{4}. Then there exist disjoint sets I_{1}, I_{2}, I_{3}, I_{4} ⊂ {1, …, n} with #I_{α} = n_{α}, α = 1, 2, 3, 4, such that for α, β in {1, 2, 3, 4},

(18) ∥Q_{1}^{∗}(p_{β}Π_{I_{α}}Q_{2} − p_{α}Π_{I_{β}}Q_{2})∥_{∞} ≤ C∥Q_{1}∥_{∞}∥Q_{2}∥_{∞}n^{1/2}(log n)^{1/2},

where p_{α} = \frac{n_{α}}{n} for α = 1, 2, 3, 4.

Remark. Instead of 4 one can take here any fixed natural number.

Proof. Let X_{1}, …, X_{n} be i.i.d. variables taking values 1,2,3,4 with probabilities p_{1}, p_{2}, p_{3}, p_{4}, respectively. The random sets I_{α} for α = 1, 2, 3, 4 are defined by

I_{α} = {1 ≤ i ≤ n : X_{i} = α}

(for the time being the I_{α}’s do not satisfy the conditions #I_{α} = n_{α} ; they will be appropriately modified at the end of the proof).

Let 1 ≤ i ≤ m_{1}, 1 ≤ j ≤ m_{2} and let z_{ij} be the (i, j)-th entry of the matrix Q_{1}^{∗}(p_{β}Π_{I_{α}}Q_{2} − p_{α}Π_{I_{β}}Q_{2}). Clearly

z_{ij} = p_{β} ∑_{{k:X_{k}=α}} x(k)y(k) − p_{α} ∑_{{k:X_{k}=β}} x(k)y(k)

where (x(1), …, x(n)) is the i-th column of Q_{1} and (y(1), …, y(n)) is the j-th column of Q_{2}.

Denote M = ∥Q_{1}∥_{∞}∥Q_{2}∥_{∞}.

Claim. We have for n > 1

(19)  P [|z_{ij}| > 2M(n log n)^{1/2}] < 2n^{−4},
(20)  P [|#I_{α} − n_{α}| > 2(n log n)^{1/2}] < 2n^{−4}.

Indeed, let Y be a random variable such that

P (Y = p_{α}) = p_{β}, P (Y = −p_{β}) = p_{α}, P (Y = 0) = 1 − (p_{β} + p_{α}), let Y_{1}, Y_{2}, …, Y_{n} be independent copies of Y. It is clear that z_{ij} is equidistributed with the random variable S = ∑ x(k)y(k)Y_{k}. By Bernstein’s inequality (cf. [4],1.3.2, p.12)

P (|S| > 2M(n log n)^{1/2}) ≤ 2n^{−4} for n > 1,

which is (19). (20) is obtained analogously taking Y such that P (Y = 1) = p_{α}, P (Y = 0) = 1 − p_{α} and S = Y_{1} + ⋯ + Y_{n}. This proves the Claim.

Now we see that the probability that

(21)  |z_{ij}| ≤ 2M(n log n)^{1/2} for every 1 ≤ i ≤ m_{1}, 1 ≤ j ≤ m_{2}

and that also

(22)  |#I_{α} − p_{α}| ≤ 2(n log n)^{1/2} for α = 1, 2, 3, 4

is greater than 1−(m_{1}m_{2} +4)2n^{−4} ≥ 1−32n^{−2} −8n^{−4}. Thus for n > 7 there exist I_{α}’s so that both (21) and (22) are satisfied. By (22) it is clear that by removing from or adding to I_{α}’s fewer than 2(n log n)^{1/2} elements, we can obtain disjoint sets so that #I_{α} = n_{α} for α = 1, 2, 3, 4. This procedure will result in increasing |z_{ij}| by at most 2M(n log n)^{1/2}. Consequently, for n > 7, the Lemma is true with C = 4. By adjusting C, it remains true for all n > 1.

The next lemma obviously implies (17) and this completes our proofs.

Lemma 5. For k = 0, 1, … there exist orthonormal systems u_{1}^{k}, …, u_{3·2^{k}}^{k} in ℝ^{3·2^{k}} such that, with the notation of (16) we have

(23)  ∥u_{j}^{k}∥_{∞} ≤ 2^{− \frac{k}{2}} for j = 1, …, 3·2^{k}; k = 0, 1, …
(24) ∥2(U_{ε}^{k})^{∗}U_{ε}^{k} − (V_{η}^{k})^{∗}V_{η}^{k}∥_{∞} ≤ Ck^{2}2^{− \frac{k}{2}} for ε, η = 0, 1; k = 1, 2, …
(25)  ∥(U_{ε}^{k+1})^{∗}V_{ε}^{k}∥_{∞} ≤ Ck^{2}2^{− \frac{k}{2}} for ε = 0, 1; k = 1, 2, ….

(observe that (V_{ε}^{k−1})^{∗}U_{ε}^{k} = [(U_{ε}^{k})^{∗}V^{k−1}ε]^{∗}, thus (25) gives also

∥(V_{ε}^{k−1})^{∗}U_{ε}^{k}∥_{∞} ≤ Ck^{2}2^{− \frac{k}{2}} for ε = 0, 1; k = 2, 3, ….)

Proof. . Let W^{k} = (u^{k}(i, j))_{1≤i,j≤3·2^{k}} for k = 0, 1, 2, … be a 3·2^{k} × 3·2^{k} orthogonal matrix with ∥W^{k}∥_{∞} ≤ 2^{− \frac{k}{2}} (e.g. we can take the matrix

\frac{1}{3} ( \begin{matrix}−1 & 2 & 2 \\ 2 & −1 & 2 \\ 2 & 2 & −1 \end{matrix}) tensored with the 2^{k} × 2^{k} orthonormal Walsh ma-

trix). We denote v^{k}(i, j) = u^{k}(2^{k} + i, j) and define U^{k}, V^{k} by (15).

Let us first apply Lemma 4 to Q_{1} = Q_{2} = W^{k}, k ≥ 1 with n_{1} = n_{2} = 2^{k−1}, n_{3} = n_{4} = 2^{k}. We obtain thus disjoint sets I_{0}^{k}, I_{1}^{k}, J_{0}^{k}, J_{1}^{k} ⊂ {1, …, 3·2^{k}} such that :

#I_{0}^{k} = #I_{1}^{k} = 2^{k−1}, #J_{0}^{k} = #J_{1}^{k} = 2^{k} and
(26) ∥(W^{k})^{∗}(2Π_{I_{ε}^{k}}W^{k} − Π_{J_{η}^{k}}W^{k})∥_{∞} ≤ Ck^{2}2^{− \frac{k}{2}} for ε, η = 0, 1.

By reordering the rows of W^{k}, we can assume that

\begin{matrix}I_{0}^{k} = \{1, …, 2^{k−1}\}, I_{1}^{k} = \{2^{k−1} + 1, …, 2^{k}\}, \\ J_{0}^{k} = \{2^{k} + 1, …, 2^{k+1}\}, J_{1}^{k} = \{2^{k+1} + 1, …, 3·2^{k}\}. \end{matrix}

We see now that

(W^{k})^{∗}(2Π_{I_{ε}^{k}}W^{k} − Π_{J_{η}^{k}}W^{k}) =
=(U_{ε}^{k} + V_{η}^{k})^{∗}(2U_{ε}^{k} − V_{η}^{k}) = 2(U_{ε}^{k})^{∗}U_{ε}^{k} − (V_{η}^{k})^{∗}V_{η}^{k},

thus (26) becomes (24).

To obtain (25), we apply Lemma 4 again: Let Q_{1} be the 2^{k} × 3·2^{k} matrix consisting of the first 2^{k} rows of W^{k+1}, let Q_{2} be the 2^{k} × 3·2^{k} matrix consisting of rows numbered 2^{k} + 1, 2^{k} + 2, …, 2^{k+1} of W^{k}. Applying Lemma 4 with n_{1} = n_{2} = 2^{k−1}, n_{3} = n_{4} = 0, we obtain a set I ⊂ {1, …, 2^{k}} with #I = 2^{k−1} such that

(27)  ∥Q_{1}^{∗}(Π_{\{1,...,2^{k}\}\I}Q_{2} − Π_{I}Q_{2})∥_{∞} ≤ Ck^{2}2^{− \frac{k}{2}} .

Let us now modify W^{k+1} by multiplying the I-numbered rows of W^{k+1} by −1; the remaining rows are not changed. The modified matrix will still be called W^{k+1}. We see that then

Q_{1}^{∗}(Π_{{1,...,2^{k}}\I}Q_{2} − Π_{I}Q_{2}) = (U_{0}^{k+1})^{∗}V_{0}^{k},

thus (27) becomes (25) for ε = 0. Analogously we obtain (25) for ε = 1.

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Department of Mathematics, Texas A&M University, College Station, TX 77843–3368 U.S.A

E-mail address: johnson@math.tamu.edu

Institute of Mathematics, The Hebrew University , Jerusalem 91904 Israel

E-mail address: tomek@math.huji.ac.il