COVERING A COMPACT SET IN A BANACH SPACE BY AN OPERATOR RANGE OF A BANACH SPACE WITH BASIS.

V. P. FONF, W. B. JOHNSON^{∗}, A. M. PLICHKO^{+}, AND V. V. SHEVCHYK

1. Introduction

A Banach space Y is said to have the approximation property (AP) if for every compact K ⊂ Y and every ε > 0 there exists a linear operator T of finite rank such that ∥T x − x∥ < ε, x ∈ K. If in addition, T can be chosen with ∥T ∥ ≤ λ, for some constant λ independent of K and ε, then Y is said to have the bounded approximation property (BAP). To complete the line (AP) ⟸ (BAP) denote the property of a Banach space to have a basis by (BP). Thus we have

(AP) ⟸ (BAP) ⟸ (BP).

All these properties are different (see [FJ] and [Sz]). However, the following interesting connection between the BAP and the BP was established in [P1] and [JRZ]

Theorem 1.1. A separable Banach space E has the BAP iff E is isomorphic to a complemented subspace of a Banach space with basis.

Actually the space with basis in this theorem may be chosen to be universal for all spaces with the BAP (see [P2] and [Sc]).

In this paper we establish the following connection between the AP and the BP.

Theorem 1.2. For a Banach space X the following properties are equivalent

(i) X has the AP. (ii) For each compact subset K of X there exist a Banach space Y with basis and a one-to-one linear operator T : Y → X such that T (Y) ⊃ K. (iii) There exists a reflexive Banach space R with basis and with unconditional finitedimensional decomposition such that for each compact K ⊂ B_{X} and for each γ > 0 there is a compact one-to-one operator T : R → X with T (B_{R}) ⊃ K and ∥T ∥ ≤ 1+γ.

The condition that T is one-to-one is essential. Indeed, it is not difficult to show that each compact set in every Banach space may be covered by an operator range of ℓ_{1}.

In Section 2 we prove that space R in (iii) may be chosen to be universal for all spaces with AP (see Theorem 2.1).

∗SUPPORTED IN PART BY NSF DMS-9900185, DMS-0200690, TEXAS ADVANCED RESEARCH PROGRAM 010366-0033-20013, AND THE U.S.-ISRAEL BINATIONAL SCI-ENCE FOUNDATION

+SUPPORTED IN PART BY DAAD FOUNDATION

2 V. P. FONF, W. B. JOHNSON^{∗}^{1}, A. M. PLICHKO^{+}^{2}, AND V. V. SHEVCHYK

Section 3 contains the “converse result” and a characterization of the AP in terms of the existence of an M-basis with special properties. Here we recall that a sequence {x_{n}, x_{n}^{∗}}_{n=1}^{∞} in X × X^{∗} is biorthogonal provided x_{n}^{∗}(x_{m}) = δ_{nm} for all n and m. A biorthogonal sequence {x_{n}, x_{n}^{∗}}_{n=1}^{∞} is fundamental provided the linear span of {x_{n}}_{n=1}^{∞} is dense in X and is total provided {x_{n}^{∗}}_{n=1}^{∞} separates the points of X. An M-basis is a fundamental and total biorthogonal sequence. A basis can then be thought of as a biorthogonal sequence {x_{n}, x_{n}^{∗}}_{n=1}^{∞} such that for every x in X, the series ∑ x_{n}^{∗}(x)x_{n} converges to x, so that a basis is an M-basis.

With the help of the results proved in Sections 2 and 3 we get (in Section 3) a factorization theorem for compact operators acting into spaces with the AP (see Theorem 3.3 below).

In Section 4 we deal with L_{p}-spaces. We characterize these spaces in terms of covering compact sets and in terms of the factorization of compact operators. We also classify those Banach spaces X in which every compact set is contained in the closed, convex, symmetric hull of a basic sequence which converges to zero.

We use standard Banach spaces geometry notation (see [LT]). If X is a Banach space and A ⊂ X then [A] is the closed linear span of A. Let H ⊂ X be a closed, convex, symmetric, and bounded subset of a Banach space X. Then it is well-known that the linear submanifold spanH becomes a Banach space if we introduce in it the norm with the unit ball H. We denote this later Banach space by E_{H}.

The authors wish to thank J. Lindenstrauss for a fruitful discussion and for a series of suggestions that improved the paper.

2. Covering a compact set in a Banach space with the AP

Theorem 2.1. There exists a reflexive Banach space R with basis and with unconditional finite-dimensional decomposition such that for each Banach space X with AP, for each compact K ⊂ B_{X} and for each γ > 0 there is a compact one-to-one operator T : R → X with T (B_{R}) ⊃ K and ∥T ∥ ≤ 1 + γ.

We start with some auxiliary results.

Lemma 2.2. Let {E_{i}} be a sequence of finite-dimensional subspaces of a Banach space X and {C_{i}} be a sequence of linear operators on X with C_{i}(X) ⊂ E_{i}, i = 1, 2, .... Let Y be the set of all sequences y = (x_{i})_{i=1}^{∞}, x_{i} ∈ E_{i}, such that the series ∑ x_{i} converges, and let Z be the set of all vectors x ∈ X such that x = ∑_{i=1}^{∞} C_{i}x. Then Y is a Banach space with monotone finite-dimensional decomposition (FDD) in the norm ∥y∥′ = sup{∥ ∑_{i=1}^{n} x_{i}∥ : n = 1, 2, ...} and Z is a Banach space in the norm |||x||| = sup_{n} ∥ ∑_{i=1}^{n} C_{i}x∥, x = ∑_{i=1}^{∞} C_{i}x. Moreover the operator J : Z → Y defined by Jx = (C_{i}x), x ∈ Z is an isometry.

Proof. First we prove that Y is complete. Let {y_{m}} ⊂ Y, y_{m} = (x_{i}^{m}), be such that lim_{p,q→∞} ∥y_{p} − y_{q}∥′ = 0. Clearly, for each i the sequence {x_{i}^{m}}_{m=1}^{∞} converges to some x_{i} ∈ E_{i}. First we check that the series ∑ x_{i} converges. Fix ε > 0 and find an integer s such that for each p, q ≥ s the inequality ∥y_{p} − y_{q}∥′ < ε holds. Let r be such that ∥ ∑_{i=l}^{k} x_{i}^{s}∥ < ε for each k > l > r. We have for k > l > r

∥ ∑_{i=l}^{k} x_{i}∥ = lim_{p} ∥ ∑_{i=l}^{k} x_{i}^{p}∥ ≤ lim_{p} ∥ ∑_{i=l}^{k} x_{i}^{s}∥ + lim_{p} ∥ ∑_{i=l}^{k}(x_{i}^{p} − x_{i}^{s})∥ < 3ε.

Thus y = (x_{i}) ∈ Y. Finally we show that lim y_{m} = y. Let ε and s be as above. Then for each n and for each p, q > s we have ∥ ∑_{i=1}^{n}(x_{i}^{p} − x_{i}^{q})∥ ≤ ε. By pushing q to infinity we have ∥ ∑_{i=1}^{n}(x_{i}^{p} − x_{i})∥ ≤ ε, for each n and for each p > s. By taking the supremum we have for each p > s, ∥y_{p} − y∥′ ≤ ε.

Next we prove that Z is complete. Put B_{n} = ∑_{i=1}^{n} C_{i}, n = 1, 2, ..., B_{0} = Id_{X}. It is well-known that the set W of all vectors x ∈ X such that the lim_{n} B_{n}x exists, is a Banach space in the norm

|||x||| = sup{∥B_{n}x∥ : n = 0, 1, 2, ...}.

Clearly, Z ⊂ W. We check that Z is closed in W. Let {x_{m}} ⊂ Z, |||.||| − lim x_{m} = x, x ∈ W and let ε > 0. Take m so large that |||x_{m} − x||| = sup{∥B_{n}x_{m} − B_{n}x∥ : n = 0, 1, 2, ...} < ε. Clearly, ∥x_{m} − x∥ < ε too. Next by using that x_{m} ∈ Z, find an n_{0} such that for each n > n_{0} holds ∥x_{m} − B_{n}x_{m}∥ < ε. We have for n > n_{0}

∥x − B_{n}x∥ ≤ ∥x − x_{m}∥ + ∥x_{m} − B_{n}x_{m}∥ + ∥B_{n}x_{m} − B_{n}x∥ < 3ε

which proves that lim_{n} B_{n}x = x, i.e. x ∈ Z.

The remaining part of the lemma is clear.

Remark 2.3. If for each i, dim E_{i} = 1, then the space Y has a monotone basis.

Lemma 2.4. Let E be a Banach space with monotone FDD and let K ⊂ B_{E} be compact. Then for every ε > 0 there is a compact one-to-one operator A : E → E with A(B_{E}) ⊃ K and ∥A∥ ≤ 1 + ε.

Proof. Let S_{n}, n = 1, 2, ... be the partial sum operators associated with to a given FDD E = ∑ E_{i}. Take δ > 0 with (1 + δ)^{2} < 1 + ε and let a sequence {ε_{i}}_{i=1}^{∞} be such that ∑ ε_{i} < δ/2. By using the compactness of K, find an increasing sequence {n_{i}} of integers such that

sup{∥S_{n_{i+1}}y − S_{n_{i}}y∥ : y ∈ K} < ε_{i}^{2}, i = 1, 2, ...

Put

Q_{0} = S_{n_{1}}, Q_{i} = S_{n_{i+1}} − S_{n_{i}}, i = 1, 2, ..., A_{1} = ∑_{i=0}^{∞} ε_{i}Q_{i}, ε_{0} = 1.

Clearly, A_{1} is a compact one-to-one operator with ∥A_{1}∥ ≤ 1 + δ. If y ∈ K then the series ∑_{i=0}^{∞}(1/ε_{i})Q_{i}y converges absolutely (say to z). It is not difficult to check that ∥z∥ ≤ 1 + δ and that A_{1}z = y. Thus A_{1}((1 + δ)B_{E}) ⊃ K. Put A = (1 + δ)A_{1}.

Remark 2.5. Let H = {y ∈ B_{E} : y = ∑_{i=0}^{∞} y_{i} , y_{i} ∈ Q_{i}(E), ∥y_{i}∥ ≤ ε_{i}, i = 0, 1, ...}. It is clear from the proof of the lemma that A(H) ⊃ K.

Lemma 2.6. Let L, M ⊂ Y be subspaces of a separable Banach space Y such that dim M = ∞ and L ∩ M = {0}, and let F ⊂ Y be compact. Then there is an automorphism D : Y → Y such that L ∩ D(F) ⊂ {0} and D|_{M} = Id_{M}.

Proof. Let {f_{i}} ⊂ B_{M^{⊥}} be a sequence which is total over L. Put V = cl co(B_{L} ∪ ±F). The linear manifold spanV, being the image of E_{V} under the injection of E_{V} into Y, is an operator range, and clearly codim spanV = ∞. By [Fo1], Lemma 3, there is a minimal (even basic) sequence {t_{i}} ⊂ Y with ∥t_{i}∥ < 2^{−i−1} and such that ∑ a_{i}t_{i} ∉ spanV for each non-zero bounded sequence of numbers {a_{i}}. Define an operator C : Y → Y by Cx = x+∑ f_{i}(x)t_{i}. Clearly, C is an isomorphism of Y onto Y and C|_{M} = Id_{M}. We check that C(L) ∩ F ⊂ {0}. Let u ∈ C(L) ∩ F. Then for some z ∈ L we have u = z + ∑ f_{i}(z)t_{i} ∈ F and hence ∑ f_{i}(z)t_{i} = u − z ∈ spanV. By the choice of {t_{i}} we have f_{i}(z) = 0, i = 1, 2, ..., and, since {f_{i}} is total over L, it follows that z = 0. Thus u = 0 which proves that C(L) ∩ F ⊂ {0} and hence L ∩ C^{−1}(F) ⊂ {0}. Put D = C^{−1}. This completes the proof.

The following lemma is known [P1]. We give a proof for the sake of completeness.

Lemma 2.7. For any finite-dimensional Banach space L and for any ε > 0 there is a finite-dimensional Banach space M with monotone basis which contains a (1 + ε)- isomorphic copy of L as a (1 + ε)-complemented subspace.

Proof. Let {x_{i}}_{i=1}^{n} and {x_{i}^{∗}}_{i=1}^{n} be an Auerbach system for L, n = dim L. Take an integer p with n/p < ε and define linear operators U_{i} in L as follows

U_{i}x = \frac{1}{p}x_{r}^{∗}(x)x_{r}, i = nl + r, 1 ≤ r ≤ n, l = 0, 1, ..., p − 1.

Put m = pn. Clearly, ∑_{j=1}^{m} U_{j} = Id_{L}. For i = nl + r we have the estimate

∥ ∑_{j=1}^{i} U_{j}∥ = ∥ ∑_{j=1}^{nl} U_{j} + ∑_{j=nl+1}^{nl+r} U_{j}∥ ≤ l/p + r/p ≤ (p − 1 + n)/p ≤ 1 + ε.

Let E_{i} = span{y_{i}}, ∥y_{i}∥ = 1, i = 1, ..., m, be the (1-dimensional) range of U_{i} (actually y_{i} = x_{r} for i = nl+r, 1 ≤ r ≤ n, l = 0, 1, ..., p−1). Define M as the space of all m-dimensional vectors y = (a_{i})_{i=1}^{m} with the norm ∥y∥ = max_{1≤q≤m} ∥ ∑_{i=1}^{q} a_{i}y_{i}∥. It is clear that the vectors e_{i} = (0, ..., 1, ..., 0), (1 stands on the i-th place), i = 1, ..., m, form a monotone basis of M. Define an operator A : L → M by Ax = (a_{i})_{i=1}^{m}, where a_{i} is defined by U_{i}x = a_{i}y_{i}, i = 1, ..., m. Put L_{1} = A(L). From the construction is clear that L is (1 + ε)-isomorphic to L_{1}. Finally define an operator P : M → L_{1} as follows:

P (a_{i}) = A(∑ a_{i}y_{i}), (a_{i}) ∈ M.

It is not difficult to see that P is a projection and ∥P ∥ ≤ 1 + ε.

Proof of Theorem 2.1. Take ε > 0 and a sequence of positive numbers {ε_{i}}_{i=0}^{∞}, ε_{0} = 1, with ∑_{i=1}^{∞} ε_{i} < ε.

Let K ⊂ B_{X} be a compact subset of a Banach space X with the AP. It is wellknown (see [LT], Proposition 1.e.2) that there is a sequence {x_{i}} ⊂ X with lim x_{i} = 0 and such that cl co{x_{i}} ⊃ K. By a slight modification of the proof given in [LT], Proposition 1.e.2, we can get that {x_{i}} ⊂ (1 + ε)B_{X}. Let a sequence of positive numbers {λ_{i}} be such that lim λ_{i} = ∞, {λ_{i}x_{i}} ⊂ (1 + ε)B_{X}, and lim λ_{i}x_{i} = 0. Put K_{0} = cl co{±λ_{i}x_{i}}.

Since X has the AP there is a sequence {B_{n}} of finite-dimensional operators in X such that

sup{∥B_{n}x − x∥ : x ∈ K_{0}} < ε_{n}, n = 1, 2, ...

Put C_{1} = B_{1}, C_{n+1} = B_{n+1}−B_{n}, n = 1, 2, .... Then for each x ∈ K_{0}, x = ∑_{n=1}^{∞} C_{n}x and

(2.1)  sup_{1≤n<∞} sup{∥ ∑_{i=1}^{n} C_{i}x∥ : x ∈ K_{0}} ≤ 1 + 2ε.

Next for each i put E_{i} = C_{i}(X) and define the spaces Y and Z as in Lemma 2.2. In a natural way we can consider K_{0} as a subset of Z. Thus, from 2.1 we have

sup{|||x||| : x ∈ K_{0}} ≤ 1 + 2ε.

Define a summation operator B : Y → X as follows:

B(x_{i}) = ∑_{i=1}^{∞} x_{i}, (x_{i}) ∈ Y.

Clearly, ∥B∥ = 1. Put L = KerB.

Since λ_{i} → ∞ and since K_{0} is bounded in Z, it easily follows that K considered as a subset of Z is compact. Let J : Z → Y be the natural embedding, i.e. Jx = (V_{m}x). By Lemma 2.2, J is an isometry and, in particular M = J(Z) is a (closed) subspace of Y. A simple verification shows that L ∩ M = {0}.

Now we pass to the space R. We use for R the space constructed in [J]. Recall the construction of R. Let A_{n} be the family of all n-dimensional Banach spaces with monotone basis, i.e. an n-dimensional space M ∈ A_{n} iff there is a basis {x_{i}}_{i=1}^{n} of M such that all partial sum operators have norm 1. Let {M_{n,i}}_{i=1}^{∞} be a dense (in the Banach-Mazur metric) sequence in A_{n}, n = 1, 2, .... Enumerate the set {M_{n,i}}_{n,i=1}^{∞} into a sequence {M_{n}}_{n=1}^{∞} and put

R = (∑_{n=1}^{∞} M_{n})_{ℓ_{2}}.

Clearly, R has a (monotone) basis and an unconditional finite-dimensional decomposition.

Next, by using Lemma 2.4, find a compact one-to-one operator A : Y → Y, A = ∑_{i=0}^{∞} ε_{i}Q_{i}, ∥A∥ ≤ (1 + 2ε)^{2}, such that A(B_{Y}) ⊃ J(K_{0}). Put L_{j} = Q_{j}(Y), j = 0, 1, ..., and by Lemma 2.7 find for each j a space M_{n_{j}} such that M_{n_{j}} = V_{j}+W_{j}, where V_{j} is (1 + ε_{j})-isomorphic to L_{j} and (1 + ε_{j})-complemented in M_{n_{j}}. Let T_{j} : V_{j} → L_{j} be an isomorphism satisfying ∥T_{j}∥˙∥T_{j}^{−1}∥ ≤ 1 + ε_{j}, j = 0, 1, ....

Now we define a compact 1-1 operator S : R → Y. We first define S on the subspace V = (∑_{j=0}^{∞} V_{j})_{ℓ_{2}} ⊂ R as follows: S(x_{j}) = ∑_{j=0}^{∞} ε_{j}T_{j}x_{j}, x_{j} ∈ V_{j}, j = 0, 1, .... It is not difficult to see that S|_{V} is a compact 1-1 operator, and ∥S|_{V} ∥ ≤ 1+ε (recall that ε_{0} = 1). By [Pl]; see also [Fo1]; there is a (closed) infinite-dimensional subspace E ⊂ Y with E ∩ S(V) = {0}. Next define S on the remaining part of R; i. e., on the subspace W = (∑_{n≠n_{j}} M_{n})_{ℓ_{2}} ⊕ (∑_{j=0}^{∞} W_{j})_{ℓ_{2}} in such a way that S|_{W} is a one-to-one compact operator into E with norm so small that ∥S∥ ≤ 1 + 2ε.

Put F = AS(B_{R}). It is clear from the construction of S and Remark 2.5 that F ⊃ K.

By using Lemma 2.6 find an automorphism D : Y → Y, ∥D∥ ≤ 1 + ε such that L ∩ D(F) = {0} and D_{|_{M}} = Id_{M}. Put T = BDAS. Then T : R → X is a compact one-to-one operator and T (B_{R}) ⊃ K. From the construction is clear that ∥T ∥ ≤ (1+2ε)^{3}. Clearly, for ε > 0 small enough we get ∥T ∥ ≤ 1+γ which completes the proof.

Remark 2.8. Let U be a Banach space with basis that contains (isomorphically) each Banach space with basis as a complemented subspace (for constructions of such spaces see [P2] and [Sc]). We show that such a space U may be used in Theorem 2.1 instead of R. First, by an obvious modification of the proof of Theorem 2.1, we can get that T^{−1}(K) is compact. Clearly, R ⊂ U. Let P : U → R be a projection from U onto R and A_{1} : U → U a compact one-to-one operator such that A_{1}(B_{U}) ⊃ T^{−1}(K) (see Lemma 2.4). Put F_{1} = cl A_{1}(B_{U}), L_{1} = Ker P and let D_{1} : U → U be an automorphism such that L_{1} ∩ D_{1}(F_{1}) = {0} and D_{1}|_{R} = Id_{R} (see Lemma 2.6). Then the operator T_{1} = T P DA : U → X is a one-to-one compact operator such that T_{1}(B_{U}) ⊃ K.

Remark 2.9. Let {K_{n}}_{n=1}^{∞} be a sequence of compact sets in a Banach space X with the AP. Put K = ∪_{n=1}^{∞}(n max{∥x∥ : x ∈ K_{n}})^{−1} K_{n}. Clearly, K ∪ {0} is compact and, by Theorem 2.1, there is a compact operator T : R → X with T (B_{R}) ⊃ K. In particular, T (R) ⊃ ∪_{n=1}^{∞}K_{n}.

Lemma 2.10. Let R be a reflexive Banach space with basis and T : R → X be a 1-1 linear operator from R into a separable Banach space X. Then there are a basis {y_{i}} of R and a norming M-basis {x_{i}} of X such that {T y_{i}} ⊂ {x_{i}}.

Proof. Recall that a subspace Y of X^{∗} is called norming provided the expression |||x||| := sup{f(x) : f ∈ Y, ∥f∥ ≤ 1} defines an equivalent norm on X. An M-basis {x_{n}, x_{n}^{∗}}_{n=1}^{∞} is norming provided [x_{n}^{∗}] is norming.

First we assume that [T (R)] = X. Let F ⊂ X^{∗} be a countable-dimensional norming subspace of X^{∗}. Since R is reflexive and T is 1-1, it follows that T^{∗}(F) is a countable-dimensional dense subspace of R^{∗}. Clearly, R^{∗} has a basis. Now we use the following result from [T]: Let E be a Banach space with basis and L ⊂ E a countable-dimensional dense subspace of E. Then there is a basis {z_{n}} of E with span{z_{n}} = L.

Take E = R^{∗} and L = T^{∗}(F) and let {g_{n}} be a basis of R^{∗} with span{g_{n}} = T^{∗}(F). Let {e_{n}} be the vectors in R which are biorthogonal to {g_{n}}. It is clear that x_{n} = T e_{n}, n = 1, 2, ..., is a norming M-basis of X with the desired properties.

If X_{1} = [T (R)] is a proper subspace of X we first construct as above a norming M-basis of X_{1} and then extend it (see [Fo2]) to a norming M-basis of X. The proof is complete.

Corollary 2.11. Let X be a separable Banach space with the AP. Then for each compact set K ⊂ X there is a norming M-basis {x_{i}} in X with biorthogonal functionals {x_{i}^{∗}} such that for each x ∈ K we have

x = ∑_{i=1}^{∞} x_{i}^{∗}(x)x_{i} and sup_{n} sup_{x∈K} ∥ ∑_{i=1}^{n} x_{i}^{∗}(x)x_{i}∥ < ∞.

Proof. By Theorem 2.1 there is a linear 1-1 operator T : R → X from a reflexive space R with basis into X such that T (B_{R}) ⊃ K. Apply Lemma 2.10 and complete the proof.

Remark 2.12. Let μ be a probability measure in a separable Banach space X. An M-basis {x_{i}} for X (with biorthogonal functionals {x_{i}^{∗}}) is called a stochastic basis for (X, μ) if μ {x ∈ X : x = ∑ x^{∗}(x)x_{i}} = 1, (see [He]). It follows from Remark 2.9 and Corollary 2.11 that if a Banach space X has the AP then for each probability measure μ, (X, μ) has a stochastic basis.

Remark 2.13. In view of Theorem 2.1 the following question is natural: is it possible to substitute for the space R in Theorem 2.1 a space with an unconditional basis? The following example shows that the answer is negative.

By [GL], for each integer n there is a finite-dimensional space E_{n} such that for each Banach space Y with 1-unconditional basis and for any two operators A : E_{n} → Y and B : Y → E_{n} with BA = Id_{E_{n}}, the inequality ∥A∥∥B∥ ≥ n holds. Put X = (∑ E_{n})_{ℓ_{2}} and let X_{1} be a copy of X. Let T : X → X_{1} be a linear operator such that T |_{E_{n}} = Id_{E_{n}}/\sqrt{n}, n = 1, 2, .... Set K = T (B_{X}). Clearly, K is a compact set. We check that K cannot be covered by a one-to-one operator range of a Banach space with unconditional basis. Suppose the contrary, i.e., for some Banach space Y with unconditional basis and for some operator B : Y → X_{1}, K is a subset of B(Y). Consider the operator A = B^{−1}T : X → Y. A simple verification shows that A is a closed operator, so A is bounded by the closed graph theorem. Clearly, T = BA and \sqrt{n}T |_{E_{n}} = Id|_{E_{n}} = \sqrt{n}BA|_{E_{n}} for each n. By taking into account the property of the space E_{n} we deduce that \sqrt{n}∥A|_{E_{n}}∥∥B∥ ≥ n. In particular, ∥A∥ ≥ \sqrt{n}/∥B∥ for each n which contradicts the boundedness of A.

3. The converse result

Theorem 3.1. Let X be a Banach space such that for each compact H ⊂ X there are a Banach space Y with the AP and a one-to-one operator T : Y → X such that T Y ⊃ H. Then X has the AP.

Proof. Fix a compact K ⊂ X and ε > 0. We shall construct a finite-dimensional operator C in X such that sup{∥Cx − x∥ : x ∈ K} < ε.

By a well-known result (see [LT], Proposition 1.e.2) there is a sequence {x_{i}} ⊂ X such that lim x_{i} = 0 and that cl co{x_{i}} ⊃ K. Take a sequence of numbers {λ_{i}} with lim λ_{i} = ∞ and such that lim λ_{i}x_{i} = 0 and put H = cl co{±λ_{i}x_{i}}. Let A : E_{H} → X be the natural injection (recall that E_{H} is the Banach space spanH with the unit ball H). Let T : Y → X be a one-to-one linear operator from a Banach space Y with the AP such that T (Y) ⊃ H. A simple verification shows that the operator T^{−1}A : E_{H} → Y is closed and by the closed graph theorem it is bounded. In particular, the set T^{−1}(H) is bounded. From lim λ_{i} = ∞ it easily follows that the set F = T^{−1}(K) is compact. Next by using the AP of Y find a finite-dimensional operator B : Y → Y,

By = ∑_{i=1}^{n} f_{i}(y)e_{i}, f_{i} ∈ Y^{∗}, e_{i} ∈ Y, y ∈ Y

such that sup{∥By − y∥ : y ∈ F } < ε/2∥T ∥.

Recall that the bw^{∗}-topology on Y^{∗} is just the topology of uniform convergence on the compact subsets of Y. By the Krein-Smulian theorem the bw^{∗}-closure of each convex subset of Y^{∗} coincides with its w^{∗}-closure. In particular, bw^{∗} − clT^{∗}(X^{∗}) = w^{∗} − clT^{∗}(X^{∗}). Since T is one-to-one it follows that w^{∗} − clT^{∗}(X^{∗}) = Y^{∗} and hence bw^{∗} − clT^{∗}(X^{∗}) = Y^{∗}. The last equality allows us to find for each i, 1 ≤ i ≤ n, a linear functional g_{i} ∈ T^{∗}(X^{∗}) such that

sup{|g_{i}(y) − f_{i}(y)| : y ∈ F } < ε/(2n∥e_{i}∥∥T ∥).

Take h_{i} ∈ X^{∗} such that T^{∗}h_{i} = g_{i}, i = 1, ..., n, and define an operator C : X → X by

Cx = ∑_{i=1}^{n} h_{i}(x)T e_{i}, x ∈ X.

We check that C is as desired. Take x ∈ K, put y = T^{−1}x ∈ F, and write

∥Cx − x∥ = ∥ ∑_{i=1}^{n} h_{i}(x)T e_{i} − x∥ = ∥T (∑_{i=1}^{n} h_{i}(x)e_{i} − T^{−1}x)∥ ≤ ∥T ∥∥ ∑_{i=1}^{n} g_{i}(y) − y∥ =

∥T ∥∥ ∑_{i=1}^{n} f_{i}(y)−y+∑_{i=1}^{n}(g_{i}(y)−f_{i}(y))e_{i}∥ ≤ ∥T ∥(ε/2∥T ∥+∑_{i=1}^{n} |g_{i}(y)−f_{i}(y)|∥e_{i}∥) < ε.

The proof is complete.

Alternate proof. By a theorem of Grothendieck (see Theorem 1.e.4 in [LT]), it is enough to verify that an arbitrary compact operator S from an arbitrary Banach space Z into X is (uniformly) approximable by finite dimensional operators. The operator S factors compactly through some Banach space W (W can even be taken reflexive; see e. g. [F]); that is, there are compact operators A : Z → W, D : W → X so that DA = S. Let T be a one-to-one operator from some space Y which has the AP into X so that T Y contains the closure of D(B_{W}). The linear mapping T^{−1}D is closed, hence is a bounded linear operator. Moreover, T^{−1}DA : Z → Y is a compact operator into a space with the AP, hence is approximable by finite dimensional operators, whence S = T (T^{−1}DA) is also approximable by finite dimensional operators.

Corollary 3.2. For a Banach space X the following assertions are equivalent:

(i) X has the AP. (ii) Any compact set K ⊂ X may be covered by a 1-1 operator range of a Banach space with basis. (iii) For any compact set K ⊂ X there is an M-basic sequence {x_{i}} in X (with biorthogonal functionals {x_{i}^{∗}}) such that x = ∑ x_{i}^{∗}(x)x_{i} for each x ∈ K.

With the help of the above results we get the following factorization

Theorem 3.3. For a Banach space X the following assertions are equivalent:

(i) X has the AP. (ii) For each Banach space Z and for each compact operator A : Z → X there are compact operators B : Z → R and T : R → X such that T is one-to-one and that

A = T B. (Here R = ( ∑_{n=1}^{∞} M_{n})_{ℓ_{2}} is the space from Theorem 2.1.)

(iii) For each Banach space Z and for each compact operator A : Z → X there are a Banach space Y with AP and compact operators B : Z → Y and T : Y → X such that A = T B.

Proof. (i) ⟹ (ii). Put K = clA(B_{Z}) and as it was done in the proofs of Theorems 2.1 and 3.1, find {x_{i}}_{i=1}^{∞} ⊂ X, lim_{i} x_{i} = 0 with cl co{x_{i}}_{i=1}^{∞} ⊃ K and a sequence of numbers {λ_{i}}_{i=1}^{∞}, lim_{i} λ_{i} = ∞ such that lim_{i} λ_{i}x_{i} = 0. Put K_{1} = cl co{λ_{i}x_{i}}_{i=1}^{∞} and by using Theorem 2.1 find a one-to-one compact operator T : R → X T (aB_{R}) ⊃ K_{1} for some a > 0. Put B = T^{−1}A and check that B is a compact operator. Indeed, since T^{−1}(K_{1}) is bounded it follows that T^{−1}(K) is compact. But B(B_{Z}) = T^{−1}A(B_{Z}) ⊂ T^{−1}(K) which proves that B is a compact operator and completes the proof of (i) ⟹ (ii). (ii) ⟹ (iii). Obvious. (iii) ⟹ (i). This is immediate from the result of Grothendieck (Theorem 1.e.4 in [LT]) used in the alternate proof of Theorem 3.1.

4. Covering compact sets in L_{p}-spaces and factorizing compact operators into L_{p}-spaces

As we already mentioned in Introduction, any compact subset of any Banach space may be covered by an operator range of ℓ_{1}. In this section we prove that every compact subset of a Banach space X may be covered by a one-to-one range of ℓ_{1} if and only if X is an L_{1}-space (actually we prove a much stronger result, see Theorem 4.7 below). Also we get a characterization of L_{p}-spaces in terms of covering compact sets and in terms of the factorization of compact operators.

We start with with two auxiliary results.

Lemma 4.1. For each natural number n there is m(n) < ∞ so that if E is an n-dimensional space which is K-isomorphic to a K-complemented subspace of L_{p}, 1 ≤ p ≤ ∞, and E has a basis with constant K, then E ⊕ ℓ_{p}^{m(n)} is f(K)-isomorphic to ℓ_{p}^{n+m(n)}.

Proof. By [PR], E is 2K-isomorphic to a 2K-complemented subspace of ℓ_{p}^{m}, where m depends only on n. Then by [BDGJN], E ⊕_{p} ℓ_{p}^{mn} is f(K)-isomorphic to ℓ_{p}^{(m+1)n}.

The next lemma very slightly improves the results in [JRZ] and [NW] that a separable L_{p} space has a basis with certain desirable properties.

Lemma 4.2. Let X be a separable L_{p} space, 1 ≤ p ≤ ∞. Then X has a basis {x_{n}}_{n=1}^{∞} which satisfies

sup_{1≤n<m<∞} d(ℓ_{p}^{m−n}, span {x_{n}, x_{n+1}, …, x_{m−1}}) < ∞.

Proof. Note that by Lemma 4.1 it is enough to show that for any sequence 0 = n_{1} < n_{2} < …, there is a basis {x_{n}}_{n=1}^{∞} for X and a constant K so that for all k = 1, 2, …, {x_{i}}_{i=n_{k}+1}^{n_{k+1}} is K-equivalent to the unit vector basis for ℓ_{p}^{n_{k+1}−n_{k}}.

First we indicate how to get such a basis when p < ∞. In this case X has a complemented subspace which is isomorphic to ℓ_{p}, and hence X is isomorphic to X ⊕ ℓ_{p} since for some Y we have X ∼ Y ⊕ ℓ_{p} ∼ Y ⊕ ℓ_{p} ⊕ ℓ_{p} ∼ X ⊕ ℓ_{p}. Let {y_{n}}_{n=1}^{∞} be a basis for X and let {e_{n}}_{n=1}^{∞} be the unit vector basis for ℓ_{p}. Make a basis for X ⊕ℓ_{p} by interlacing these bases for X and ℓ_{p}; i. e., use {y_{1} ⊕0, 0⊕e_{1}, 0⊕e_{2}, …, 0⊕ e_{m_{1}}, y_{2} ⊕ 0, 0 ⊕ e_{m_{1}+1}, 0 ⊕ e_{m_{1}+2}, …, 0 ⊕ e_{m_{2}}, … }. As long as m_{k} grows fast enough, this will be a basis that has the desired property.

The same argument works when X is a separable L_{∞} space which contains a (necessarily complemented) isomorphic copy of c_{0}, but to do the general case, one needs to “localize” the construction. Again, let {y_{n}}_{n=1}^{∞} be a basis for X. Let ϵ > 0; ϵ = 1 is fine for our purposes. For appropriate m_{1}, take a sequence {e_{i}^{1}}_{i=1}^{m_{1}} which is 1 + ϵ-equivalent to the unit vector basis of ℓ_{∞}^{m_{1}} and which lies in the linear span of {y_{n}}_{n=1}^{∞}; say, in span {y_{1}, …, y_{s_{1}}}. Let E_{1} be the linear span of {e_{I}^{1}}_{i=1}^{m_{1}} and write span {y_{1}, …, y_{s_{1}}} = E_{1} + F_{1}, where F_{1} is the kernel of a norm 1 + ϵ projection from span {y_{1}, …, y_{s_{1}}} onto E_{1}. The vectors e_{1}^{1}, …, e_{m_{1}}^{1} form the first m_{1} terms of our basis.

For the construction of the next m_{2} basis vectors (where m_{2} is appropriately large, depending on s_{1} as well as the sequence {n_{k}}_{k=1}^{∞}), take a subspace E_{2} of the linear span of {y_{i}}_{i=s_{1}+1}^{s_{2}} for some s_{2} so that E_{2} is 1 + ϵ-isomorphic to the ℓ_{∞} space of its dimension. As long as its dimension is at least m_{1}, the space F_{1}+E_{2} is isomorphic to the ℓ_{∞} space of its dimension (which we take to be m_{2}) with isomorphism constant depending only on the basis constant of {y_{n}}_{n=1}^{∞}. Let {e_{i}^{2}}_{i=1}^{m_{2}} be a basis for F_{1} + E_{2} which is d(F_{1} + E_{2}, ℓ_{∞}^{m_{2}})-equivalent to the unit vector basis for ℓ_{∞}^{m_{2}}; these vectors form the next m_{2} terms of the desired basis for X.

Now just iterate this construction. Let F_{2} be the kernel of a projection of norm d(F_{1} + E_{2}, ℓ_{∞}^{m_{2}}) from the linear span of {y_{i}}_{i=s_{1}+1}^{s_{2}} onto the linear span of {e_{i}^{2}}_{i=1}^{m_{2}}. Then select a subspace E_{3} of the linear span of {y_{i}}_{i=s_{2}+1}^{s_{3}} for some s_{3} so that E_{3} is 1 + ϵ-isomorphic to the ℓ_{∞} space of its (appropriately large) dimension and continue as in the previous step. It is evident that for any {n_{k}}_{k=1}^{∞} this construction can produce a basis {x_{n}}_{n=1}^{∞} which has the property mentioned in the first paragraph of the proof.

Theorem 4.3. The following are equivalent for a Banach space X.

1. X is a L_{p} space.

2. Every compact operator into X factors through some L_{p} space.

3. If T is a compact operator from some Banach space Z into X and Y is a separable L_{p} space then there are compact operators A from Z into Y and B from Y into X so that T = BA and B is one-to-one.

Remark 4.4. The equivalence of (1) with (2) is known; see [J].

Proof. We show that (1) implies (3). First assume that X is a separable L_{p} space and let T : Z → X be a compact operator; without loss of generality ∥T ∥ ≤ 1. Take a basis {x_{n}}_{n=1}^{∞} for X which has the property given in Lemma 4.2. Let {S_{n}}_{n=1}^{∞} be the partial sum projections associated with {x_{n}}_{n=1}^{∞} and set S_{0} = 0. Take 0 = n_{0} < n_{1} < … so that ∥(S_{n_{k}} − S_{n_{k−1}})T ∥ < 4^{−k} for each k > 1. Now let Y be any separable L_{p} space and let {y_{n}}_{n=1}^{∞} be a basis for Y which also has the property given in Lemma 4.2. Thus for each k, the spaces span y_{n_{k−1}+1}, …, y_{n_{k}} and span span x_{n_{k−1}+1}, …, x_{n_{k}} are isomorphic with isomorphism constant independent of k. Next we use an idea similar to one used in the proof of Lemma 2.4. Take a norm one isomorphism U_{k} from y_{n_{k−1}+1}, …, y_{n_{k}} onto span x_{n_{k−1}+1}, …, x_{n_{k}} so that

sup_{k} ∥U_{k}∥ < ∞. Define A : Z → Y by Az = ∑_{k=1}^{∞} 2^{k−1}U_{k}^{−1}(S_{n_{k}} − S_{n_{k−1}})T. Let

{P_{n}}_{n=1}^{∞} be the partial sum projections associated with the basis {y_{n}}_{n=1}^{∞}. Define

B : Y → X by By = ∑_{k=1}^{∞} 2^{−k+1}U_{k}(S_{n_{k}} − S_{n_{k−1}})y. This does it.

If X is a non separable L_{p} space, we just replace X with any separable L_{p} subspace which contains the range of T and proceed as above.

The implication (3) implies (2) is trivial, so we turn to the implication (2) implies (1). First assume 1 < p < ∞, so that every separable L_{p} space is isomorphic to a complemented subspace of L_{p}(0, 1) by [LP]. Thus condition (2) implies that every compact operator into X factors through L_{p}(0, 1). Now the operators from a space Z into X which factor through L_{p}(0, 1) forms a Banach space under the factorization norm !T! = inf ∥A∥ · ∥B∥, where the infimum is over all factorizations T = BA with A : Z → L_{p}(0, 1), B : L_{p}(0, 1) → X (use the fact that L_{p}(0, 1) is isometric to the ℓ_{p} sum of itself). Thus the uniform boundedness principle yields that there is a constant K = K_{Z} so that !T! ≤ K∥T ∥ for all compact T from Z to X. Applying this with Z the ℓ_{1} sum of all finite dimensional subspaces of X, we conclude that there is a constant K so that the inclusion from every finite dimensional subspace of X into X factors through L_{p}(0, 1) with factorization constant at most K. This implies that X is a L_{p} space.

This argument does not quite work for p = 1 or p = ∞ since not every separable L_{1} space is isomorphic to a complemented subspace of L_{1}(0, 1) and no infinite dimensional separable L_{∞} space is isomorphic to a complemented subspace of L_{∞}[0, 1]. Here is how to get around that annoyance. The argument in [LP] yields that the injection J_{Y} from a separable L_{p} space into its second dual factors through L_{p}(0, 1) in the extreme case p = 1. When p = ∞, the same is true by the injectivity of the second dual of a L_{∞} space. The argument in the preceding paragragh then yields that the inclusion from every finite dimensional subspace of X into X^{∗∗} factors through L_{p}(0, 1) with factorization constant at most K. This is enough, by [LR], to guarantee that X is a L_{p} space.

Remark 4.5. It is evident that condition (3) in Theorem 4.3 is equivalent to:

4. For every compact subset K of X and every separable L_{p} space Y there is a one-to-one compact operator S from Y to X so that S(B_{Y}) ⊃ K.

That the one-to-oneness is essential in Remark 4.5 is given by the following proposition.

Proposition 4.6. Let 1 ≤ p < ∞. The following are equivalent for a Banach space X.

5. For every compact subset K of X and every L_{p} space Y there is a compact operator S from Y to X so that S(B_{Y}) ⊃ K.

6. For every compact subset K of X there is a L_{p} space Y and an operator S from Y to X so that S(B_{Y}) ⊃ K.

7. X is isomorphic to a quotient of L_{p}(μ) for some measure μ.

Proof. The case p = 1 is easy because every Banach space is a quotient of ℓ_{1}(Γ) for some set Γ and every L_{1} space contains a complemented copy of ℓ_{1}. So assume that 1 < p < ∞. If condition (6) holds, then the space Y can always be taken separable and hence Y can be taken to be L_{p}(0, 1) because every separable L_{p} space is isomorphic to a complemented subspace of L_{p}(0, 1).

We claim that there is a constant C so that every compact subset of the unit ball of X is covered by S(B_{L_{p}(0,1)}) for some operator S from L_{p}(0, 1) into X for which ∥S∥ ≤ C. Indeed, otherwise there would be compact subsets K_{n} ⊂ B_{X} so that if

K_{n} ⊂ S(B_{L_{p}(0,1)}) with S : L_{p}(0, 1) → X, then ∥S∥ > n^{2}. Set K = sup_{n=1}^{∞} n^{−1}K_{n} ∪ {0}.

Then K is a compact subset of B_{X} and so there is an operator S : L_{p}(0, 1) → X so that K ⊂ S(B_{L_{p}(0,1)}). Then (nS)(B_{L_{p}(0,1)}) ⊃ K_{n} and so ∥nS∥ ≥ n^{2} for all n. This proves the claim.

By the claim, there is a C so that for each finite dimensional subspace E of X there is an operator S_{E} : L_{p}(0, 1) → X so that S_{E}B_{L_{p}(0,1)} ⊃ B_{E} and ∥S_{E}∥ ≤ C. Direct the finite dimensional subspaces of X by inclusion. Then any weak-operator cluster point of S_{E}^{∗} defines an isomorphism from X^{∗} into L_{q}(0, 1), where 1/p + 1/q = 1. We now know X is reflexive, so that X is a quotient of L_{p}(0, 1). This completes the proof that condition (6) implies condition (7).

Finally we assume that (7) is satisfied. Let Q be a quotient map from L_{p}(μ) onto X. If K is a compact subset of X, then it is easy too see that some superspace of K in X is the image of some separable L_{p} subspace of L_{p}(μ) and hence of L_{p}(0, 1). That is, without loss of generality we can assume that X is separable and that L_{p}(μ) is L_{p}(0, 1). Now by Michael’s selection theorem (or the Bartle-Graves theorem) there is a compact subset K_{1} of L_{p}(0, 1) so that QK_{1} = K. By Lemma 2.4, there is a compact operator T on L_{p}(0, 1) so that T B_{L_{p}(0,1)} ⊃ K_{1}. Then S := QT is a compact operator from L_{p}(0, 1) into X for which SB_{L_{p}(0,1)} ⊃ K. This completes the proof of Proposition 4.6.

We have frequently used the fact that every compact subset of a Banach space is contained in the closed convex symmetric hull of a sequence which converges to zero.

What are the Banach spaces for which the null sequence can always be chosen to have some extra property such as being basic or at least countably linearly independent? It is a good exercise for students to show that Hilbert space does not have one of these stronger properties. A complete answer to the question is given by the next theorem.

Theorem 4.7. The following are equivalent for a Banach space X.

1. X is a L_{1} space.

2. For each compact subset K of X there is a basic sequence which converges to zero and whose closed convex symmetric hull contains K.

3. For each compact subset K of X there is a sequence {y_{n}}_{n=1}^{∞} which converges to

zero and whose closed convex symmetric hull contains K and such that 0 ≠ ∑_{n=1}^{∞} λ_{n}y_{n}

whenever 0 < ∑_{n=1}^{∞} |λ_{n}| < ∞.

If X is separable, the preceding conditions are equivalent to

4. For each compact subset K of X there is a basis for X which converges to zero and whose closed convex symmetric hull contains K.

Proof. Assume that (1) holds and K is a compact subset of B_{X}. Let Y be a separable L_{1} subspace of X which contains K and let {x_{n}}_{n=1}^{∞} be a basis for X_{0} which satisfies the conclusion of Lemma 4.3. Let {S_{n}}_{n=1}^{∞} be the partial sum projections associated with {x_{n}}_{n=1}^{∞} and set S_{0} = 0. From Lemma 2.4 we get 0 = n_{0} < n_{1} < …

so that image of B_{X} under the operator T := ∑_{k=1}^{∞} 4^{−k+1}(S_{n_{k}} − S_{n_{k−1}}) contains K.

By the property of the sequence {x_{n}}_{n=1}^{∞}, for each k there is a basis z_{n_{k−1}}, …, z_{n_{k}} for the linear span E_{k} of x_{n_{k−1}}, …, x_{n_{k}} so that the unit ball of E_{k} is contained in the convex symmetric hull of z_{n_{k−1}}, …, z_{n_{k}} and such that sup_{1≤n<∞} ∥z_{n}∥ := C < ∞. Let C_{1} := sup_{k} ∥S_{n_{k}} − S_{n_{k−1}}∥. From the property of the operator T we deduce that K is contained in the closed convex symmetric hull of the sequence y_{n}, where y_{j} = 2CC_{1}2^{−k+1}z_{j} if n_{k−1} < j ≤ n_{k}. This completes the proof that (1) implies (2) and that (1) implies (4) when X is separable. The only remaining non trivial implication is that (3) implies (1). But (3) implies that every compact set in X can be covered by the image of the unit ball of ℓ_{1} under some one-to-one operator from ℓ_{1} into X, so X is a L_{1} space by condition (4) in the remark after Theorem 4.3.

References

[BDGJN] G. Bennett, L. E. Dor, V. Goodman, W. B. Johnson, and C.M. Newman, “On uncomplemented subspaces of L_{p}, 1 < p < 2,” Israel J. Math. 26 (1977), 178–187.

[F] T. Figiel, “Factorization of compact operators and applications to the approximation problem,” Studia Math. 45 (1973), 191–210.

[FJ] T. Figiel and W. B. Johnson, “The approximation property does not imply the bounded approximation property,” Proc. Amer. Math. Soc. 41 (1973). 197–200.

[Fo1] V. P. Fonf, “One property of families of embedded Banach spaces,” J. Soviet Math. 59 (1992), 687–690.

[Fo2] V. P. Fonf, “On the extension of operator bases in Banach spaces,” Teor. Funkt., Funkt. Anal. i Prilozh. 54 (1990), 37–42 (Russian), English transl. in Soviet Math. 1991, 319–322.

[GL] Y. Gordon and D. R. Lewis, “Absolutely summing operators and local unconditional structures,” Acta Math. 133 (1974), 27–48.

[He] W. Herer, “Stochastic bases in Frechet spaces,” Demonstratio Math. 14 (1981), 719–724.

[J] W. B. Johnson, “Factoring compact operators,” Israel J. Math. 9 (1971), 337–345.

[JRZ] W. B. Johnson, H. P. Rosenthal and M. Zippin, “On bases, finite-dimensional decompositions and weaker structures in Banach spaces,” Studia Math. 43 (1971), 77–92.

[LT] J. Lindenstrauss and L. Tzafriri, “Classical Banach Spaces I,” Springer - Verlag, 1977.

[NW] N. J. Nielsen and P. Wojtaszczyk, “A remark on bases in L_{p}-spaces with an application to complementably universal L_{∞}-spaces,” Bull. Acad. Polon. Sci., S’er. Sci. Math. Astronom. Phys. 21 (1973), 249–254.

[P1] A. Pełczyński, “Any Banach space with the bounded approximation property is a complemented subspace of a Banach space with a basis,” Studia Math. 40 (1971), 239–242.

[P2] A. Pełczyński, “Universal bases,” Studia Math. 32 (1969), 247–268.

[PR] A. Pełczyński and H. P. Rosenthal, “Localization techniques in L_{p} spaces,” Studia Math. 52 (1974/75), 263–289.

[Pl] A. M. Plichko, “The choice of subspaces with special properties in a Banach space and some properties of quasi-complements,” Functional Anal. and Appl. 15 (1981), 88–89 (Transl. from Russian).

[Sc] G. Schechtman, “On Pełczyński’s paper “Universal bases,”” Israel J. Math. 22 (1975), 181– 184.

[Sz] S. J. Szarek, “A Banach space without a basis which has the bounded approximation property,” Acta Math. 159 (1987), 81–98.

[T] P. Terenzi, “A complement to Krein-Milman-Rutman theorem with applications,” Instituto Lombardo (Rend. Sc.) A 113 (1979), 341–353.

V. P. Fonf

Department of Mathematics

Ben-Gurion University of the Negev P.O.Box 653 Beer-Sheva 84105 Israel e-mail: fonf@black.bgu.ac.il and Department of Mathematics Texas A&M University College Station Texas, 77843 USA W. B. Johnson Department of Mathematics Texas A&M University College Station Texas, 77843 USA e-mail: johnson@math.tamu.edu

A. M. Plichko

Department of Mathematics

Pedagogical University

Shevchenko str. 1

Kirovograd 25006

Ukraine

e-mail: aplichko@kspu.kr.ua V. V. Shevchik Sebastian-Kneipp Gasse, 7 Augsburg 86152 Germany e-mail: Shevchyk@t-online.de