On the Bounded Approximation Property in Banach spaces
Abstract.
We prove that the kernel of a quotient operator from an space onto a Banach space with the Bounded Approximation Property (BAP) has the BAP. This completes earlier results of Lusky –case – and Figiel, Johnson and Pełczyński –case separable. Given a Banach space , we show that if the kernel of a quotient map from some space onto has the BAP then every kernel of every quotient map from any space onto has the BAP. The dual result for spaces also hold: if for some space some quotient has the BAP then for every space every quotient has the BAP.
1. Preliminaries
An exact sequence in the category of Banach spaces and bounded linear operators is a diagram in which the kernel of each arrow coincides with the image of the preceding; the middle space is also called a twisted sum of and . By the open mapping theorem this means that is isomorphic to a subspace of and is isomorphic to the corresponding quotient. Two exact sequences and are said to be equivalent if there exists an operator making commutative the diagram
The classical 3lemma (see [4, p. 3]) shows that must be an isomorphism. An exact sequence is said to split if it is equivalent to the trivial sequence . There is a correspondence (see [4, Thm. 1.5.c, Section 1.6]) between exact sequences of Banach spaces and the socalled linear maps which are homogeneous maps (we use this notation to stress the fact that these are not linear maps) with the property that there exists some constant such that for all finite sets one has . The infimum of those constants is called the linearity constant of and denoted .
The process to obtain a linear map out from an exact sequence is the following: get a homogeneous bounded selection for the quotient map , and then a linear selection for the quotient map. Then is a linear map. A linear map induces the exact sequence of Banach spaces in which means the completion of the vector space endowed with the quasinorm . The linearity of makes this quasinorm equivalent to a norm (see [3]). The embedding is while the quotient map is . The exact sequences
are equivalent setting as the
operator . We will use the notation
to
mean that is a linear map associated to that exact
sequence. Two linear maps are said
to be equivalent, and we write , if the
induced exact sequences are equivalent. Two maps are equivalent if and only if the difference can be written as , where is a
homogeneous bounded map and is a linear map.
Recall from [7] the definition and basic properties of the dual of a Banach space . We define the dual of as the Banach space of linear maps such that for a prefixed Hamel basis of . The space is defined as the closed linear span in of the evaluation functionals given by . Given a linear map and a Hamel basis we denote the socalled canonical form of , namely, the map
where . The two basic properties of are displayed in the following proposition; the proof can be found in [7, Prop. 3.2]:
Proposition 1.1.
There is a linear map with the property that given a linear map then there exists an operator such that and .
It is easy to give examples of linear maps on finite dimensional spaces with infinitedimensional range: set defined by , , and by homogeneity on the rest. It is however possible to modify a linear map defined on a finitedimensional Banach space so that the image is finite dimensional [7, Lemma 2.3]. We need an improvement of that result:
Lemma 1.1.
Let be a linear map and let be a finite dimensional subspace of . Given a finite set of points in the unit sphere of so that for and , there is a linear map verifying:

for all .

.


The image of by spans a finite dimensional space.
Proof.
Let be a Hamel basis for formed by norm one vectors. Assume that . Fix a finite set of elements of norm at most such that the unit ball of is contained in the convex hull of . Be sure that the set contains all and that for . Modify now as follows: for all ; if is a norm one element of and then . It is necessary to establish some order between the convex combinations of the to have the value of univocally defined and equal to at the points of , but this can be easily done in many ways. Condition (1) is therefore fulfilled. We show (2): if is a norm one element of then
If is a point not in then , and thus (2) is proved. Observe that if is a bounded homogeneous map then . This and the previous estimate yield (3):
Condition (4) is clear. ∎
A few facts about the connections between linear maps and the associated exact sequences will be needed in this paper. Given an exact sequence and an operator , there is a commutative diagram
in which . Moreover,
Lemma 1.2.
If one has a commutative diagram
(1) 
then
i.e., is a linear map associated with the lower sequence in (1).
Proof.
What one has to check is that the two sequences
are equivalent; and this happens via the map . Indeed, that the map makes the diagram commutative is clear; its continuity follows from the estimate:
∎
Analogously, given an exact sequence and an operator , there is a commutative diagram
in which . Moreover,
Lemma 1.3.
If one has a commutative diagram
(2) 
then
i.e., is a linear map associated with the lower sequence in (2).
Proof.
One has to check that the two sequences
are equivalent; and this happens via the map , where is a linear selection for such that for some bounded selection for one has . Which means that is a linear map associated with the lower sequence. The commutativity of the diagram is clear and the continuity of follows from the estimate:
∎
As an immediate consequence we have.
Lemma 1.4.
Proof.
Following [12], an exact sequence is said to locally split if its dual sequence splits.
Lemma 1.5.
An exact sequence (locally) splits if and only if for every operator and the sequence (locally) splits.
Proof.
The sufficiency is obvious in both cases. To prove the necessity is simple for the splitting: if with homogeneous bounded and linear then with homogeneous bounded and linear. The case of local spliting follows from this and the observation that if one has the commutative diagram
then the biduals form also a commutative diagram
Kalton shows in [12, Thm. 3.5] that an exact sequence locally splits if and only if its bidual sequence splits. Thus, since splits, so does . ∎
Definition 1.1.
A Banach space has the BAP if for each finite dimensional subspace there is a finite rank operator such that and for each .
2. Results for spaces
We assume in what follows that denotes an arbitrary space. In [16, 17] Lusky shows that when is separable and has the BAP then the kernel of every quotient map has the BAP. Theorem 2.1 (b) of [8] asserts that if is a quotient map and has the BAP then both and have the BAP. It is wellknown that when has the BAP then also has the BAP, but the converse fails since there exist spaces with basis whose dual do not have AP [15, 1.e.7(b)]. Therefore, the missing case is to show that the kernel of an arbitrary quotient map has the BAP when has the BAP.
Lemma 2.1.
Let be a Banach space with the BAP. Then has the BAP.
Proof.
Fix a Hamel basis for and let the universal map appearing in Proposition 1.1 verifying for all . Let be a finite dimensional subspace of . We can assume without loss of generality that . Take and let be a finite set of for which . Let be a finite rank operator fixing . Let be the version of verifying that the image of is finitedimensional, which has moreover been done so that and for all and all . By the properties of there is an operator such that
Given any , if then
and therefore the image of spans a finite dimensional space. This means that also the range of is finite dimensional. Moreover, fixes since for one has that if with one has
Finally,
∎
Theorem 2.1.
Let be an exact sequence in which has the BAP. Then has the BAP.
Proof.
The universal property of the construction mentioned in Proposition 1.1 yields a commutative diagram
Therefore, by virtue of Lemma 1.4 (1) there is an exact sequence
This sequence locally splits: the dual sequence splits since the dual of an space is injective. Thus, has the BAP, as well as both and . ∎
Question. Does a similar result hold for other
wellknown variations of the BAP such as the commuting
bounded approximation property (CBAP), the uniform
approximation property (UAP) or the existence of finite
dimensional decomposition (FDD)? The previous proof cannot be
translated to cover the case of the CBAP or FDD since these
properties do not pass to complemented subspaces [2];
also, the result cannot be translated to the case of the UAP since
Lemma 1.1 spoils the estimate on the dimension of the
required finite dimensional operator.
The phenomenon described in the proposition –all the kernels of all the quotient maps from an space onto a space with the BAP have the BAP– is part of a general stability result
Proposition 2.1.
Given two exact sequences
(3) 
Then has the BAP if and only if has the BAP.
Proof.
Observe that if one has a commutative diagram formed by short exact sequences
in which we assume that:

and locally split.

has the BAP

has the BAP.
then also has the BAP: Since locally split and both have the BAP, then has the BAP [10], see also [4, Thm. 7.3.e]. Since locally splits, then has the BAP. Apply now this schema to diagram (3) depicted in the form
and recall that both and locally split since the quotient is an space.∎
The following lemma, rather its dual version, will be required to work with spaces; we include its proof for the sake of co