Quasicomplemented Subspaces of Banach Spaces and Separable Quotients

Abstract

We present a result which generalizes the theorems of Gurariy-Kadets and Lindenstrauss-Rosenthal on quasicomplemented subspaces of Banach spaces. We also provide a solution in wide classes of Banach spaces of a problem posed by Ivan Singer concerning the existence of a $w^*$-basic sequence in the dual of a Banach space which is biorthogonal to a given basic sequence (or more generally, a given minimal sequence) in that space, and other related problems.

1 Introduction

A closed subspace X of a Banach space E is said to be quasicomplemented if there exists a closed subspace $Y\subset E$ such that $X\cap Y = \{0\}$ and $X+Y$ is dense in E. In such case, the subspace Y is called a quasicomplement of X. This concept was coined by Murray [16], who proved that every closed subspace of a separable reflexive space is quasicomplemented. Shortly thereafter, Mackey [15] showed that the same assertion holds true in any separable Banach space. This result, known as the Murray–Mackey theorem, has been generalized in several ways. Gurariy and Kadets [9] proved that if X is a closed infinite-dimensional and infinite-codimensional subspace of a separable Banach space E, then for any $\varepsilon > 0$ there exists a quasicomplement Y of X which is isomorphic and $(1+\varepsilon )$-isometric to X. On the other hand, Fonf and Shevchik [8] showed that, under the same hypothesis, the subspace X admits a quasicomplement Y with the property that the restriction to Y of the canonical quotient map $Q: E\rightarrow E/X$ is compact, thus, $Q|_Y:Y\rightarrow E/X$ is a one-to-one, compact and dense-range operator (in particular, $X+Y_1$ is not closed in E for any closed infinite-dimensional subspace $Y_1\subset Y$). Following the terminology of [8], we say that the subspace Y is a compactly adjacent quasicomplement of X.

A nonseparable extension of Murray–Mackey theorem was provided by Rosenthal [17, 18], who proved that a closed subspace X of a Banach space E is quasicomplemented whenever $X^*$ is $w^*$-separable and $X^{\perp }$ (the annihilator of X in $E^*$) contains a reflexive subspace. In [18] (see e.g. [10, Theorem 5.83]), it was shown that the latter condition is fulfilled by every closed infinite-codimensional subspace of $\ell ^{\infty }(\Gamma )$, for any infinite set $\Gamma$. Thus, if X is a closed subspace of $\ell ^{\infty }(\Gamma )$ with $w^*$-separable dual (in particular, if X is a closed subspace of $\ell ^{\infty }$) then X is quasicomplemented. The $w^*$-separability of $X^*$ is essential in this result: as Lindenstrauss pointed out in [14] (see also [10, Corollary 5.89]), the space $c_0(\Gamma )$ admits no quasicomplement in $\ell ^{\infty }(\Gamma )$ if the set $\Gamma$ is uncountable. In an unpublished manuscript, Lindenstrauss and Rosenthal gave a generalization of the criterion above, which asserts that a closed subspace of a Banach space E is quasicomplemented whenever $X^*$ is $w^*$-separable and E/X has a separable infinite-dimensional quotient (that is, there is a closed subspace $Y\subset E$ containing X such that E/Y is separable and infinite-dimensional). A proof of this result was supplied by Johnson [12]. His argument relies on a well-known result by himself and Rosenthal [13] (see e.g. [20, Theorem 1.5]), which will play also an important role in this paper, and guarantees the existence of a $w^*$-basic sequence in the dual of every separable space. Recall that a sequence $\{f_n\}_n$ in the dual of a Banach space E is said to be $w^*$-basic if there exists a sequence $\{x_n\}_n\subset E$ such that $\{x_n, f_n\}_n$ is a biorthogonal system and, for every $f\in \overline{\textrm{span} \{f_n\}_n}^{w^*}$, the series $\sum _{n\ge 1} f(x_n)f_n$ converges to f in the $w^*$-topology. We refer to [20] for the basics on such sequences, and to [10, Chapter 5.7] for more information on quasicomplements in the nonseparable setting.

Some other results, concerning the existence of quasicomplemented subspaces of separable spaces with a special behaviour with respect to operator ranges in those spaces, have been recently established in [7]. Recall that a linear subspace R of a Banach space E is an operator range in E if there exist a Banach space Z and a bounded linear operator $T:Z\rightarrow E$ such that $R = T(Z)$. These subspaces play an important role in the Geometry of Banach spaces. In this respect, we mention a result of Saxon and Wilansky [19], which asserts that a Banach space E has a separable infinite-dimensional quotient if and only if E contains a proper dense operator range. We refer to [4, 6] for detailed accounts on operator ranges and their applications in the settings of Banach and Hilbert spaces, respectively. Corollary 2.1 in [7] yields that if X is a closed infinite-dimensional subspace of a separable Banach space E and R is an infinite-codimensional operator range in E with $X\subset R$, then there exists a quasicomplement Y of X which is essentially disjoint with respect to R, that is, $Y\cap R = \{0\}$.

This paper consists in two sections apart from this Introduction. In the next one, we establish a criterion for quasicomplementation, which provides a simultaneous extension of the theorem of Lindenstrauss and Rosenthal and the aforesaid results in [7,8,9]. To state it, we introduce the following concept, motivated by [8] (recall that a bounded linear operator $T:E\rightarrow Y$ between Banach spaces is said to be nuclear if there exist sequences $\{y_n\}_n\subset Y$ and $\{f_n\}_n\subset E^*$ such that $\sum _{n\ge 1} \Vert f_n\Vert \Vert y_n\Vert < \infty$ and $T(x) = \sum _{n\ge 1} f_n(x)y_n$ for all $x\in E$).

Definition 1.1

Let X be a closed subspace of a Banach space E. A closed subspace $Y\subset E$ is a nuclearly adjacent quasicomplement of X if the restriction to Y of the quotient map $Q:E\rightarrow E/X$ is a one-to-one, dense-range and nuclear operator.

It is well-known that every nuclear operator is compact, thus nuclearly adjacent quasicomplements are compactly adjacent. Our main result in Sect. 2 implies the following two statements for a closed subspace X of a Banach space E with $\dim (X) = \textrm{codim}\, (X)=\infty$, which yield a nonseparable extension of the theorems of Gurariy-Kadets and Fonf-Shevchik, and a refinement of the result of Lindenstrauss-Rosenthal, respectively:

(A):

If E/X is separable then for any $\varepsilon > 0$ and any closed subspace $Y\subset X$ with $w^*$-separable dual there exists a nuclearly adjacent quasicomplement Z of X wich is isomorphic and $(1+\varepsilon )$-isometric to Y.

(B):

If $X^*$ is $w^*$-separable and E/X has a separable (infinite-dimensional) quotient, then for any $\varepsilon >0$ and any closed subspace $Y\subset E$ such that $X\subset Y$ and E/Y is separable, there exists a subspace $Z\subset E$ isomorphic and $(1+\varepsilon )$-isometric to Y such that X is a nuclearly adjacent quasicomplement of Z.

Our criterion yields as well certain disjointness properties of the quasicomplements with respect to operator ranges (or more generally, countable unions of operator ranges) in E and/or $E^*$, which provide some nonseparable extensions of the aforesaid result in [7]. We also establish a characterization of the closed subspaces of a Banach space E that admit a nuclearly adjacent quasicomplement, and those subspaces of E that are nuclearly adjacent quasicomplements of some subspace of that space, which provides a converse of the theorem of Lindenstrauss-Rosenthal. These results lead in addition to several characterizations of the classes of separable Banach spaces and Banach spaces with a separable quotient and $w^*$-separable dual in terms of quasicomplements.

In Sect. 3 we provide several further strengthenings of the theorem of Lindenstrauss-Rosenthal, which give solution to some questions regarding $w^*$-basic sequences. Our first result there establishes the equivalence of, among others, the following assertions for a closed infinite-dimensional and infinite-codimensional subspace X with $w^*$-separable dual of a Banach space E:

• (a) E/X has a separable infinite-dimensional quotient.

• (b) X has a proper quasicomplement $Y\subset E$ (that is, such that $X+Y\ne E$).

• (c) Every sequence $\{x_n^*\}_n\subset X^*$ admits a $w^*$-basic extension $\{f_n\}_n\subset E^*$ (that is, a $w^*$-basic sequence $\{f_n\}_n\subset E^*$ such that $f_n|_{X} = x_n^*$ for all n).

• (d) Every total sequence $\{x_n^*\}_n\subset X^*$ (that is, every sequence $\{x_n^*\}_n$ that separates the points of X) admits a $w^*$-basic extension $\{f_n\}_n\subset E^*$.

We also prove that assertions (a) and (c) are equivalent with no assumption on the $w^*$-separability of $X^*$. It is worth mentioning that every $w^*$-basic sequence is basic (see e.g. [20, Proposition 1.8]), and that the analogue of property (c) requiring that $\{f_n\}_n$ is basic always holds true [20, Appendix (3)]. Assertion (d) is closely related with a question posed by Singer [20, Problem 1.6.b], which asks if for a given basic sequence $\{x_n\}_n$ in a Banach space E such that $[\{x_n\}_n]$ (the closed linear span of $\{x_n\}_n$) is infinite-codimensional, there is a $w^*$-basic sequence $\{f_n\}_n\subset E^*$ such that $\{x_n, f_n\}_n$ is a biorthogonal system. We show that if the subspace X is separable then conditions $(a)-(d)$ are fulfilled if and only if for every minimal sequence $\{x_n\}_n$ in X there is a $w^*$-basic sequence $\{f_n\}_n\subset E^*$ such that $\{x_n, f_n\}_n$ is a biorthogonal system (recall that a sequence $\{x_n\}_n$ in a Banach space X is minimal provided that there is a sequence $\{x_n^*\}_n\subset X^*$ such that $\{x_n, x_n^*\}_n$ is biorthogonal). We also prove that that if $\{x_n\}_n$ is a basic sequence in E with $\textrm{codim}\, [\{x_n\}_n] = \infty$ then, $\{x_n\}_n$ admits a $w^*$-basic sequence $\{f_n\}_n\subset E^*$ of biorthogonal functionals if and only if the subspace $[\{x_n\}_n]$ is quasicomplemented in E. These facts lead to an affirmative answer of Singer’s problem (even in its stronger form replacing basic sequences with minimal ones) in wide classes of Banach spaces, like the one of weakly Lindelöf determined. Finally, we show that any minimal sequence $\{x_n\}_n$ in a Banach space E with $\textrm{codim}\, [\{x_n\}_n]= \infty$ admits a $w^*$-basic sequence $\{f_n\}_n$ in $E^{***}$ such that $\{x_n, f_n\}_n$ is a biorthogonal system.

Let us fix some notations and terminology. All the spaces considered in this paper are over the real field. Given a normed space E, we denote by $B_E$ and $I_E$ the closed unit ball of E and the identity operator on E, respectively, and by ${\mathcal {R}}(E)$ the class made up of all infinite-codimensional operator ranges in that space. It is worth to mention that ${\mathcal {R}}(E)$ contains all closed infinite-codimensional subspaces of E (see e.g. [4]) and proper dense operator ranges in E (see e.g. [1, Corollary 2.17]). As we already mentioned, if A is a subset of E, then the symbol $A^{\perp }$ stands for the annihilator subspace of A, that is, $A^{\perp } = \{f\in E^*:\, f(x)=0\,\, \text {for all} \,\, x\in A\}$. Analogously, given a set $B\subset E^*$ we write $B_{\perp } = \{x\in E:\, f(x)=0\,\, \text {for all} \,\, f\in B\}$. Some of our results or their proofs involve minimal sequences in a Banach space E with sequences of biorthogonal functionals lying in some specific subspace of $E^*$. A sequence $\{x_n\}_n\subset E$ is said to be B-minimal, for some set $B\subset E^*$, if there exists a sequence $\{f_n\}_n\subset B$ such that $\{x_n, f_n\}_n$ is a biorthogonal system. Analogously, if $A\subset E$, then a sequence $\{f_n\}_n\subset E^*$ is called A-minimal if there is a sequence $\{x_n\}_n\subset A$ such that $\{x_n, f_n\}_n$ is biorthogonal. In the case that $A = E$, we say that the sequence $\{f_n\}_n$ is $w^*$-minimal. In the sequel, otherwise stated, the words “subspace” and “operator” refer respectively to a closed infinite-dimensional and infinite-codimensional subspace of a Banach space and a bounded linear operator between Banach spaces, “separable quotient” means separable infinite-dimensional quotient, and the term “basis” stands for Schauder basis.

2 Nuclearly Adjacent Quasicomplements

The first result of this section is the aforementioned criterion for the existence of quasicomplements.

Theorem 2.1

Let X and Y be subspaces of a Banach space E with $X\subset Y$, and let $\{R_k\}_{k\ge 1}\subset {\mathcal {R}}(E)$ and $\{S_k\}_{k\ge 1}\subset {\mathcal {R}}(E^*)$ be countable sets of operator ranges such that

$$\begin{aligned} X\subset \bigcap _{k\ge 1} R_k \quad \, \, \text {and} \, \, \quad Y^{\perp }\subset \bigcap _{k\ge 1} S_k. \end{aligned}$$

If $X^*$ is $w^*$-separable and E/Y is separable then, for every $\varepsilon > 0$ there exists an isomorphism $\varphi :E\rightarrow E$ with $\Vert \varphi -I_E\Vert < \varepsilon$ satisfying the following properties:

• (1) $\varphi (X)$ is a nuclearly adjacent quasicomplement of Y and $\varphi (X)\cap \left( \bigcup _{k\ge 1} R_k\right) = \{0\}.$

• (2) X is a nuclearly adjacent quasicomplement of $\varphi (Y)$ and $\varphi (Y)^{\perp }\cap \left( \bigcup _{k\ge 1} S_k\right) = \{0\}.$

An important ingredient in the proof of Theorem 2.1 is the next result from [11], which ensures the existence of minimal sequences in a Banach space E with a disjoint behaviour with respect to a given countable union of operator ranges in that space. Following the terminology of that paper, we say that a sequence $\{x_n\}_n\subset E$ has property $(*)$ with respect to a set $V\subset E$ if the conditions $\{a_n\}_n\in \ell ^1$ and $\sum _{n\ge 1} a_n x_n\in V$ imply $a_n=0$ for all n.

Lemma 2.2

([11, Lemma 2.1]). Let E be a Banach space, let $E_0$ and F be closed infinite-dimensional (not necessarily infinite-codimensional) subspaces of E and $E^*$ respectively, and let $\{R_k\}_{k\ge 1}$ be a countable subset of ${\mathcal {R}}(E_0)$. If $\{u_n\}_n\subset E_0$ is an F-minimal sequence, then there exist an isomorphism $\varphi :E\rightarrow E$ and an F-minimal sequence $\{v_n\}_n\subset B_{E_0}$ such that:

• (1) $\varphi (u_n)$ and $v_n$ are collinear for each n.

• (2) $\varphi (E_0) = E_0$.

• (3) $\varphi ^*(F) = F$.

• (4) $\{v_n\}_n$ satisfies property $(*)$ with respect to the set $\bigcup _{k\ge 1} R_k$.

We shall also need the following easy fact.

Lemma 2.3

Let X be a subspace of a Banach space E, and let T and N be endormophisms on E such that N is nuclear and $T(X)\subset X$. If the operator $\psi = T+N$ is an isomorphism on E, then the restriction to X of the quotient map $E\rightarrow E/\psi ^{-1}(X)$ is nuclear.

Proof

Let us write $X_0:= \psi ^{-1}(X)$, and let $Q:E\rightarrow E/X$ and $S:E\rightarrow E/X_0$ be the quotient maps from E onto E/X and $E/X_0$, respectively. Since $\psi$ is an isomorphism, the formula

$$\begin{aligned} \Psi (S(u)) = Q(\psi (u)), \quad S(u)\in E/X_0 \end{aligned}$$

defines an isomorphism $\Psi :E/X_0\rightarrow E/X$. On the other hand, for each $x\in X$ we have

$$\begin{aligned} Q(\psi (x)) = Q(T(x)+N(x)) = Q(N(x)). \end{aligned}$$

Therefore $S(x) = \Psi ^{-1}(Q(N(x)))$ for each $x\in X$, that is, $S|_{ X} = (\Psi ^{-1}\circ Q) \circ (N|_{ X})$, and bearing in mind that the operator $N|_{ X}$ is nuclear (as N is) we deduce that $S|_{ X}$ is nuclear as well. $\square$

Apart from the previous lemmas, in the proof of Theorem 2.1 we shall use a well-known result of Markushevich which ensures the existence of an M-basis in every separable Banach space (see e.g. [10, Theorem 1.22]). Recall that a sequence $\{x_n\}_n$ in a (separable) Banach space X is said to be an M-basis of X if $\{x_n\}_n$ is minimal, $[\{x_n\}_n] =X$, and the (unique) sequence $\{x_n^*\}_n$ in $X^*$ of biorthogonal functionals associated to $\{x_n\}_n$ is total.

Proof of Theorem 2.1

Let Q be the canonical quotient map from E onto E/Y, and set $R_0=Y$ and $S_0 = X^{\perp }$. The construction of the isomorphism $\varphi$ relies on the following two statements.

Claim 1. There is a sequence $\{e_n\}_n\subset B_E$ satisfying property $(*)$ with respect to $\bigcup _{k= 0}^{\infty } R_k$ such that $\{Q(e_n)\}_n$ is an M-basis of E/Y.

Indeed, since E/Y is separable, according to the aforementioned result of Markushevich, there is a sequence $\{u_n\}_n\subset E$ such that $\{Q(u_n)\}_n$ is an M-basis of E/Y. In particular, $\{u_n\}_n$ is a $Y^{\perp }$-minimal sequence. As $R_0 = Y\in {\mathcal {R}}(E)$, Lemma 2.2 (applied to the sequence $\{u_n\}_n$, the subspaces $E_0=E$ and $F= Y^{\perp }\subset E^*$, and the set of operator ranges $\{R_k\}_{k\ge 0}$) guarantees the existence of a $Y^{\perp }$-minimal sequence $\{e_n\}_n\subset B_E$, a sequence of scalars $\{\lambda _n\}_n$ and an isomorphism $\phi :E\rightarrow E$ such that $\{e_n\}_n$ satisfies property $(*)$ with respect to the set $\bigcup _{k\ge 0} R_k$, $\phi (u_n) = \lambda _n e_n$ for each n and $\phi ^*(Y^{\perp }) = Y^{\perp }$. Evidently $\lambda _n\ne 0$ for all n. Observe also that $\phi ^*(Y^{\perp }) = [\phi ^{-1}(Y)]^{\perp }$, so $Y = [\phi ^*(Y^{\perp })]_{\perp } = \phi ^{-1}(Y)$, and hence $Y = \phi (Y)$. This allows us to consider the operator $\Phi :E/Y\rightarrow E/Y$ defined for each $Q(u)\in E/Y$ as

$$\begin{aligned} \Phi \left( Q(u)\right) = Q(\phi (u)). \end{aligned}$$

It is clear that $\Phi$ is an isomorphism. Consequently, the sequence $\{\Phi \left( Q(u_n)\right) \}_n$ is an M-basis of E/Y, and bearing in mind that $\Phi \left( Q(u_n)\right) = Q\left( \phi (u_n)\right) = \lambda _n Q(e_n)$ for each n, it follows that $\{Q(e_n)\}_n$ is also an M-basis of E/Y.

Claim 2. There is an X-minimal sequence $\{z_n^*\}_n\subset B_{E^*}$ which is total over X (i.e. $\{z_n^*|_X\}_n$ separates the points of X) and satisfies property $(*)$ with respect to $\bigcup _{k=0}^{\infty } S_k$.

Indeed, the $w^*$-separability of $X^*$ yields a total X-minimal sequence $\{x_n^*\}_n\subset X^*$ (see e.g. [10, Lemma 1.21]). Let $\{h_n\}_n\subset E^*$ be a sequence such that $h_n|_X = x_n^*$ for all n and let $\pi :E\rightarrow E^{**}$ denote the canonical isometry. Since $S_0= X^{\perp }\in {\mathcal {R}}(E^*)$, according to Lemma 2.2 (applied in $E^*$, to the sequence $\{h_n\}_n$, the subspaces $E_0=E^*$ and $F= \pi (X)\subset E^{**}$, and the set of operator ranges $\{S_k\}_{k\ge 0}\subset {\mathcal {R}}(E^*)$), we deduce the existence of an X-minimal sequence $\{z_n^*\}_n\subset B_{E^*}$, a sequence of non-zero scalars $\{\lambda _n'\}_n$ and an isomorphism $\zeta :E^*\rightarrow E^*$ such that $\{z_n^*\}_n$ satisfies property $(*)$ with respect to $\bigcup _{k\ge 0} S_k$, $\zeta (h_n) = \lambda _n' z_n^*$ for all n and $\zeta ^*\left( \pi (X)\right) = \pi (X)$. We only need to check that $\{z_n^*\}_n$ is total over X. Pick $x\in X$ such that $z_n^*(x)=0$ for all n. Since

$$\begin{aligned} \langle \zeta ^*(\pi (x)), \, h_n \rangle = \langle \pi (x),\, \zeta (h_n)\rangle = \langle \zeta (h_n), \, x\rangle = \lambda _n' z_n^*(x) \end{aligned}$$

we get $\langle \zeta ^*\left( \pi (x)\right) ,\, h_n\rangle = 0$ for all n. Taking into account that $\xi ^*(\pi (x)) \in \pi (X)$ and the sequence $\{h_n\}_n$ is total over X it follows that $\xi ^*(\pi (x))=0$, hence $x=0$ and the claim is proved.

Now, choose numbers $a_n> 0$ such that $\sum _{n\ge 1} a_n<\varepsilon /(1+\varepsilon )$, and define, for each $u\in E$,

$$\begin{aligned} \varphi (u) = u + \sum _{n\ge 1} a_n z_n^*(u)e_n. \end{aligned}$$

(2.1)

Then, $\varphi$ is an isomorphism on E with $\Vert \varphi -I_E\Vert < \varepsilon$. We shall show that $\varphi$ satisfies properties (1) and (2). Before proceeding, we prove the next statement.

Claim 3. For each subspace $Z\subset E$, the restriction to $\varphi (Z)$ of the quotient map $E\rightarrow E/Z$ and the restriction to Z of the quotient map $E\rightarrow E/\varphi (Z)$ are nuclear operators.

In fact, since $\varphi$ is the sum of the identity operator and a nuclear endomorphism on E, the first assertion follows applying Lemma 2.3 to the subspace $\varphi (Z)$. Now, denote by S the quotient map $E\rightarrow E/Z$, let us write $\psi = \varphi ^{-1}$ and define, for each n, $v_n = \psi (e_n)$. For every $u\in E$ we have

$$\begin{aligned} \varphi \left( u-\sum _{n\ge 1} a_n z_n^*(u)v_n\right){} & {} = \varphi (u)-\sum _{n\ge 1} a_n z_n^*(u)\varphi (v_n)\\{} & {} = \varphi (u)-\sum _{n\ge 1} a_n z_n^*(u)e_n= u, \end{aligned}$$

therefore

$$\begin{aligned} \psi (u) = u - \sum _{n\ge 1} a_n z_n^*(u)v_n. \end{aligned}$$

Since $\sum _{n\ge 1} a_n \Vert z_n^*\Vert \Vert v_n\Vert \le \Vert \psi \Vert \sum _{n\ge 1} a_n \Vert z_n^*\Vert \Vert e_n\Vert < \infty$ we have that $\psi$ is the sum of the identity and a nuclear endomorphism on E, and a new appeal to Lemma 2.3, applied to the isomorphism $\psi$ and the subspace Z, yields that the operator $S|_{Z}:Z\rightarrow E/\varphi (Z)$ is nuclear.

Now, we shall show that

$$\begin{aligned} \varphi (X)\cap \left( \bigcup _{k\ge 0} R_k\right) = \{0\}. \end{aligned}$$

(2.2)

Indeed, pick $x\in X$ such that $\varphi (x) \in \left( \bigcup _{k\ge 0} R_k\right)$, and take an integer $k\ge 0$ with $u\in R_k$. Since $X\subset R_k$ we have $\varphi (x)-x\in R_k$, hence

$$\begin{aligned} \sum _{n\ge 1} a_n z_n^*(x)e_n\in R_k. \end{aligned}$$

It is clear that $\{a_n z_n^*(x)\}_n\in \ell ^1$. As the sequence $\{e_n\}_n$ satisfies property $(*)$ with respect to $\bigcup _{k\ge 0} R_k$ we obtain $z_n^*(x)=0$ for all n. Taking into account that the sequence $\{z_n^*\}_n$ is total over X we deduce that $x=0$, and (2.2) is proved.

Proceeding in a similar way we get

$$\begin{aligned} X\cap \varphi (Y)=\{0\}. \end{aligned}$$

(2.3)

Indeed, choose $y\in Y$ with $\varphi (y)\in X$. Since $X\subset Y$ we have $\varphi (y)-y\in Y$. Let $\{f_n\}_n\subset Y^{\perp }$ denote the sequence of biorthogonal functionals associated to $\{Q(e_n)\}_n$ through the canonical identification between $Y^{\perp }$ and $(E/Y)^{*}$. As $\varphi (y)-y\in Y$ and $\{e_n, f_n\}_n$ is a biorthogonal system, for each n we have

$$\begin{aligned} 0 = f_n(\varphi (y)-y) = a_n z_n^*(y), \end{aligned}$$

that is, $z_n^*(y)=0$. Therefore $\varphi (y) = y$, hence $y\in X$, and using again the fact that $\{z_n^*\}_n$ is total over X we obtain $y=0$.

Next, we shall prove that

$$\begin{aligned} \varphi (Y)^{\perp }\cap \left( \bigcup _{k\ge 0} S_k\right) = \{0\}. \end{aligned}$$

(2.4)

Choose $g\in \varphi (Y)^{\perp }\cap \left( \bigcup _{k\ge 0} S_k\right)$. Since $\varphi (Y)^{\perp } = (\varphi ^*)^{-1}(Y^{\perp })$ we have $\varphi ^*(g)\in Y^{\perp }$. Pick an integer $k\ge 0$ such that $g\in S_k$. As $Y^{\perp }\subset S_k$ we get $\varphi ^*(g)-g\in S_k$. Notice that $\varphi ^*(g) = g + \sum _{n\ge 1} a_n g(e_n)z_n^*$. Therefore $\sum _{n\ge 1} a_n g(e_n)z_n^*\in S_k.$ Bearing in mind that $\{a_n g(e_n)\}_n\in \ell ^1$ and the sequence $\{z_n^*\}_n$ satisfies property $(*)$ with respect to the set $\bigcup _{k\ge 0} S_k$ we obtain $g(e_n)=0$ for all n. Taking into account that g identifies with a functional from $(E/Y)^*$ and the sequence $\{Q(e_n)\}_n$ is linearly dense in E/Y it follows that $g=0$, and (2.4) is checked.

A similar argument, using the fact that $\{z_n^*\}_n$ is an X-minimal sequence, yields that

$$\begin{aligned} \varphi (X)^{\perp }\cap Y^{\perp } = \{0\}. \end{aligned}$$

Since $\varphi (X)^{\perp }\cap Y^{\perp } = \left( \varphi (X)+Y\right) ^{\perp }$ we have that $\varphi (X)+Y$ is dense in E. This and (2.2) entail that $\varphi (X)$ is a quasicomplement of Y satisfying $\varphi (X)\cap \left( \bigcup _{k\ge 1} R_k\right) =\{0\}$. Moreover, thanks to Claim 3 we deduce that the restriction to $\varphi (Y)$ of the quotient map $Q:E\rightarrow E/Y$ is a nuclear operator, and bearing in mind that $\varphi (X)\subset \varphi (Y)$, it follows that the restriction of Q to the subspace $\varphi (X)$ is nuclear as well. Thus, $\varphi (X)$ is a nuclearly adjacent quasicomplement of Y.

Analogously, from (2.3) and (2.4) we infer that $\varphi (Y)$ is a quasicomplement of X such that $\varphi (Y)^{\perp }\cap \left( \bigcup _{k\ge 1} S_k\right) =0$. On the other hand, using again Claim 3 we deduce that the restriction to Y of the quotient map $S: E\rightarrow E/\varphi (Y)$ is nuclear. Since $X\subset Y$ it follows that $S|_X:X\rightarrow E/\varphi (Y)$ is also a nuclear operator. Consequently, X is a nuclearly adjacent quasicomplement of $\varphi (Y)$. $\square$

Remark 2.4

In Theorem 2.1, the operator ranges $R_k$ and $S_k$ can not be replaced by arbitrary (infinite-dimensional and infinite-codimensional) linear subspaces. In [5], Drewnowski proved that every infinite-dimensional Banach space E contains a dense linear infinite-codimensional subspace V such that, if M is a closed subspace of E with $M\cap V = \{0\}$, then $\dim (M)< \infty$.

Next, we shall give some applications of Theorem 2.1 to the study of quasicomplements. Prior to this, we establish a couple of easy consequences of the proof of that result, which have independent interest. Recall that a subset M of a Banach space E is said to be spaceable if M contains a closed infinite-dimensional subspace of E. For a detailed account on spaceability and other related topics we refer to the book [2].

Proposition 2.5

Let E be a Banach space and $\{R_k\}_k$ be a countable subset of ${\mathcal {R}}(E)$. If $\bigcap _{k\ge 1} R_k$ is spaceable, then the set $\left( E\setminus \bigcup _{k\ge 1} R_k\right) \cup \{0\}$ is spaceable as well. More precisely, if X is a subspace of E with $w^*$-separable dual such that $X\subset \bigcap _{k\ge 1} R_k$ then, for any $\varepsilon > 0$ there exists an isomorphism $\varphi :E\rightarrow E$ such that $\Vert \varphi -I_E\Vert < \varepsilon$ and

$$\begin{aligned} \varphi (X) \subset \left( E\setminus \bigcup _{k\ge 1} R_k\right) \cup \{0\}. \end{aligned}$$

Proof

Let X be a subspace of E such that $X^*$ is $w^*$-separable and $X\subset \bigcap _{k\ge 1} R_k$, and fix $\varepsilon > 0$. Because of the $w^*$-separability of $X^*$, there is a sequence $\{z_n^*\}_n$ in $B_{E^*}$ which is total over X. On the other hand, according to Lemma 2.2 (applied to any minimal sequence in E), we can find a minimal sequence $\{e_n\}_n\subset B_E$ satisfying property $(*)$ with respect to $\bigcup _{k\ge 1} R_k$.

Now, let $\{a_n\}_n$ be a sequence of positive scalars such that $\sum _{n\ge 1} a_n< \varepsilon /(1+\varepsilon )$, and define, for each $u\in E$,

$$\begin{aligned} \varphi (u) = u + \sum _{n\ge 1} a_n z_n^*(u) e_n. \end{aligned}$$

Then, $\varphi$ is an isomorphism on E with $\Vert \varphi -I_E\Vert < \varepsilon$. Moreover, if a vector $x\in X$ satisfies $\varphi (x) \in \bigcup _{k\ge 1} R_k$ then $\sum _{n\ge 1} a_n z_n^*(x) e_n \in \bigcup _{k\ge 1} R_k$. Since $\{e_n\}_n$ has property $(*)$ with respect to $\bigcup _{k\ge 1} R_k$ it follows that $z_n^*(x)=0$ for all n, and taking into account that $\{z_n^*\}_n$ is total over X we get $x=0$. $\square$

Proposition 2.6

Let X be a subspace of a Banach space E. If E/X has a separable quotient then, for any subspace $Y\subset E$ such that $X\subset Y$ and E/Y is separable and any $\varepsilon > 0$ there exists an isomorphism $\varphi :E\rightarrow E$ with $\Vert \varphi -I_E\Vert < \varepsilon$ satisfying the following properties:

• (1) $X+\varphi (Y)$ is dense in E and $X+\varphi (Y)\ne E$.

• (2) $\varphi (X)+Y$ is dense in E and $\varphi (X) + Y \ne E$.

Proof

Let Q be the quotient map from E onto E/Y. By Markushevich’s theorem, there is a sequence $\{e_n\}_n\subset B_E$ such that $\{Q(e_n)\}_n$ is an M-basis of E/Y. On the other hand, Lemma 2.2 (applied to any given X-minimal sequence $\{h_n\}_n\subset E^*$) ensures the existence of an X-mininal sequence $\{z_n^*\}_n\subset B_{E^*}$ satisfying property $(*)$ with respect to $X^{\perp }$.

Now, let $\{a_n\}_n$ be a sequence of positive numbers such that $\sum _{n\ge 1} a_n < \varepsilon /(1+\varepsilon )$, and consider the isomorphism $\varphi :E\rightarrow E$ defined for each $u\in E$ by the formula

$$\begin{aligned} \varphi (u) = u+\sum _{n\ge 1} a_n z_n^*(u)e_n. \end{aligned}$$

The arguments employed in the proof of Theorem 2.1 (with the exception of equalities (2.2) and (2.3)) yield that $\varphi$ satisfies the required properties. $\square$

Now, we shall establish the promised results on quasicomplements. The first one provides a simultaneous nonseparable generalization of the aforementioned results in [7,8,9].

Corollary 2.7

If X is a subspace of a Banach space E then the following assertions are equivalent:

• (1) E/X is separable.

• (2) X admits a nuclearly adjacent quasicomplement.

• (3) X admits a compactly adjacent quasicomplement.

• (4) For any subspace $X_0\subset E$ with $w^*$-separable dual such that $X_0\subset X$, any $\varepsilon > 0$ and any countable set $\{R_k\}_k\subset {\mathcal {R}}(E)$ with

  $$\begin{aligned} X_0\subset \bigcap _{k\ge 1} R_k \end{aligned}$$

  there is an isomorphism $\varphi :E\rightarrow E$ such that $\Vert \varphi -I_E\Vert < \varepsilon$ and if $Z = \varphi (X_0)$, then Z is a nuclearly adjacent quasicomplement of X satisfying

  $$\begin{aligned} Z\cap \left( \bigcup _{k\ge 1} R_k\right) = \{0\}. \end{aligned}$$

Proof

The implication $(1)\Rightarrow (4)$ follows applying Theorem 2.1 (with $Y=X$, $X=X_0$ and $S_k = X^{\perp }\subset E^*$ for all k), and $(4)\Rightarrow (2)\Rightarrow (3)$ are obvious. Assume that property (3) is satisfied, let Y be a compactly adjacent quasicomplement of X and let Q be the quotient map from E onto E/X. Then, $Q|_{Y}$ is compact and $\overline{Q(Y)} = E/X$. Since compact operators have separable range it follows that E/X is separable, thus $(3)\Rightarrow (1)$. $\square$

As an application of the previous result we obtain the following corollary.

Corollary 2.8

Let $\{X_k\}_{k\ge 1}$ be a countable set of closed infinite-codimensional subspaces of a Banach space E, and let $X = \bigcap _{k\ge 1} X_k$. If $\dim (X)=\infty$ and E/X is infinite-dimensional and separable, then there exists a subspace $Z\subset E$ which is a nuclearly adjacent quasicomplement of $X_k$ for each $k\ge 1$.

Proof

The former corollary, applied to the set of operator ranges $\{X_k\}_k\subset {\mathcal {R}}(E)$, ensures the existence of a subspace $Z\subset E$ which is a nuclearly adjacent quasicomplement of X and satisfies $Z \cap \left( \bigcup _{k\ge 1} X_k\right) = \{0\}$. Choose $k\ge 1$. Then $X_k\cap Z = \{0\}$ and since $X+Z$ is dense in E, so is $X_k+Z$. Denote by $Q:E\rightarrow E/X$ and $Q_k:E\rightarrow E/X_k$ the quotient maps from E onto E/X and $E/X_{k}$, respectively. As $X\subset X_k$, the formula

$$\begin{aligned} j_k (u+X) = u+ X_k \end{aligned}$$

defines a bounded linear operator $j_k:E/X\rightarrow E/X_k$. Since $Q_k = j_k \circ Q$ and the operator $Q|_Z$ is nuclear it follows that $Q_k|_Z = j_k \circ Q|_Z$ is nuclear as well. Therefore, Z a nuclearly adjacent quasicomplement of $X_k$. $\square$

Notice that the property of being nuclearly adjacent for a couple of quasicomplemented subspaces is not symmetric. Indeed, if E is any nonseparable Banach space with a separable quotient and X is a subspace of E such that E/X is separable then, according to Corollary 2.7 we have that X admits a separable nuclearly adjacent quasicomplement, say Y. But, since the space E/Y is nonseparable, a new appeal to that corollary yields that Y does not admit any compactly adjacent quasicomplement, thus X is not a compactly adjacent quasicomplement of Y.

Next, we establish a refinement of the theorem of Lindenstrauss-Rosenthal and a converse of it, which yields a characterization of the subspaces of a Banach space E that are nuclearly adjacent quasicomplements of a subspace of E.

Corollary 2.9

If X is a subspace of a Banach space E then the following assertions are equivalent:

• (1) E/X has a separable quotient and $X^*$ is $w^*$-separable.

• (2) X is a nuclearly adjacent quasicomplement of some subspace $Z\subset E$.

• (3) X is a compactly adjacent quasicomplement of some subspace $Z\subset E$.

• (4) X has a proper quasicomplement $Z\subset E$ such that E/Z is separable.

• (5) E/X has a separable quotient and, for every subspace $Y\subset E$ such that E/Y is separable and $X\subset Y$, any $\varepsilon >0$ and any countable subset $\{S_k\}_k\subset {\mathcal {R}}(E^*)$ with

  $$\begin{aligned} Y^{\perp }\subset \bigcup _{k\ge 1} S_k \end{aligned}$$

  there is an isomorphism $\varphi :E\rightarrow E$ such that $\Vert \varphi -I_E\Vert < \varepsilon$ and if $Z = \varphi (Y)$, then X is a nuclearly adjacent quasicomplement of Z satisfying

  $$\begin{aligned} Z^{\perp }\cap \left( \bigcup _{k\ge 1} S_k\right) = \{0\}. \end{aligned}$$

Proof

The implication $(1)\Rightarrow (5)$ follows immediately from Theorem 2.1, and $(5)\Rightarrow (2)\Rightarrow (3)$ are obvious. Assume that X is a compactly adjacent quasicomplement of some subspace $Z\subset E$. Then, the restriction to X of the quotient map $T:E\rightarrow E/Z$ is a compact operator with $\overline{T(X)}=E/Z$. Hence E/Z is separable, and since E/Z is infinite-dimensional it follows that

$$\begin{aligned} T(X)\ne E/Z. \end{aligned}$$

In particular Z is a proper quasicomplement of X, and $(3)\Rightarrow (4)$ is proved.

Now, let Z be a subspace of E satisfying property (4). Then, the restriction to Z of the quotient map $Q:E\rightarrow E/X$ has a proper dense range, thus, according to the result of Saxon and Wilansky [19] mentioned in the Introduction, the space E/X has a separable quotient. On the other hand, since $Z^{\perp }$ identifies with $(E/Z)^*$ and E/Z is separable, we have that $Z^{\perp }$ is $w^*$-separable. Moreover, as $X\cap Z = \{0\}$ it follows that the restriction to X of the quotient map $T:E\rightarrow E/Z$ is one-to-one. Since its adjoint $(T_{| X})^*$ identifies with the restriction to $Z^{\perp }$ of the restriction operator $q^*:E^*\rightarrow X^*$ we deduce that

$$\begin{aligned} \overline{q^*(Z^{\perp })}^{w^*} = X^*. \end{aligned}$$

Therefore, $X^*$ is $w^*$-separable, and hence, (4) implies (1). $\square$

Remark 2.10

In view of the disjointness properties of the quasicomplement Z in Corollaries 2.7 and 2.9 it is natural to ask if:

• (a) for a given subspace X of a Banach space E such that E/X is separable, and a given countable set $\{S_k\}_k\subset {\mathcal {R}}(E^*)$ with $X^{\perp }\subset \bigcap _{k\ge 1} S_k$, there exists a quasicomplement Z of X satisfying

  $$\begin{aligned} Z^{\perp }\cap \left( \bigcup _{k\ge 1} S_k\right) = \{0\}. \end{aligned}$$

• (b) for a given subspace X of a Banach space E such that E/X has a separable quotient and $X^*$ is $w^*$-separable, and a given countable set $\{R_k\}_k\subset {\mathcal {R}}(E)$ with $X\subset \bigcap _{k\ge 1} R_k$, there exists a quasicomplement Z of X satisfying

  $$\begin{aligned} Z\cap \left( \bigcup _{k\ge 1} R_k\right) = \{0\}. \end{aligned}$$

The answer to both questions is negative, even if the sets $\{R_k\}_k$ and $\{S_k\}_k$ reduce to only one operator range. Indeed, let $\Gamma$ be any uncountable set, consider the Banach space $E = \ell ^{\infty }(\Gamma )$, and let S be the annihilator subspace of $c_0(\Gamma )$ in $E^*$. By Rosenthal’s theorem mentioned in the Introduction, S contains a reflexive infinite-dimensional subspace, say V. Then, V is necessarily $w^*$-closed, and we can assume that V is separable. Set $X:=V_{\perp }$. Evidently, $(E/X)^*$ is separable, so E/X is separable as well (consequently X is quasicomplemented in E), and $X^{\perp }\subset S$. However, if $Z\subset E$ is any quasicomplement of X in E we have $Z^{\perp }\cap S \ne \{0\}.$ Indeed, otherwise Z would be a quasicomplement of $c_0(\Gamma )$ in E, which contradicts Lindenstrauss’ theorem in [14]. Thus, question (a) has a negative answer, and the disjointness property in Corollary 2.7 is essentially optimal. As regards question (b), consider again the Banach space $E = \ell ^{\infty }(\Gamma )$ (with $\Gamma$ uncountable), set $R = c_0(\Gamma )$ and let X be any separable subspace of $c_0(\Gamma )$. Then, X is quasicomplemented in E, but a new appeal to Lindenstrauss’ theorem yields that if $Z\subset E$ is a quasicomplement of X then $Z\cap R \ne \{0\}.$

The lack of symmetry of the property of being nuclearly adjacent for a couple of quasicomplemented subspaces of a Banach space leads to the following concept.

Definition 2.11

If X and Y are quasicomplemented subspaces of a Banach space E, then we say that X and Y are mutually nuclearly adjacent (mutually compactly adjacent) if X is nuclearly (compactly) adjacent to Y and Y is nuclearly (compactly) adjacent to X.

Next, we establish a characterization of the subspaces of a Banach space that admit a mutually nuclearly adjacent quasicomplemented subspace, which enhances also the disjointness properties of Corollaries 2.7 and 2.9.

Corollary 2.12

If X is a subspace of a Banach space E then the following assertions are equivalent:

• (1) E/X is separable and $X^*$ is $w^*$-separable.

• (2) X admits a mutually nuclearly adjacent quasicomplement.

• (3) X admits a mutually compactly adjacent quasicomplement.

• (4) For any $\varepsilon > 0$ and any two countable sets $\{R_k\}_k\subset {\mathcal {R}}(E)$ and $\{S_k\}_k\subset {\mathcal {R}}(E^*)$ with

  $$\begin{aligned} X\subset \bigcap _{k\ge 1} R_k \quad \text { and } \quad X^{\perp }\subset \bigcap _{k\ge 1} S_k \end{aligned}$$

  there exists an isomorphism $\varphi :E\rightarrow E$ such that $\Vert \varphi -I_E\Vert < \varepsilon$, and if $Z = \varphi (X)$ then Z is a mutually nuclearly adjacent quasicomplement of X satisfying

  $$\begin{aligned} Z\cap \left( \bigcup _{k\ge 1} R_k\right) = \{0\} \quad \text {and} \quad Z^{\perp } \cap \left( \bigcup _{k\ge 1} S_k\right) = \{0\}. \end{aligned}$$

Proof

The implication $(1)\Rightarrow (4)$ follows applying Theorem 2.1 with $Y=X$, and $(4)\Rightarrow (2)\Rightarrow (3)$ are trivial. On the other hand, if X has a mutually compactly adjacent quasicomplement then, thanks to Corollary 2.7 we have that E/X is separable, and Corollary 2.9 yields that $X^*$ is $w^*$-separable, thus (3) implies (1). $\square$

We end this section by applying the previous results to obtain several characterizations of separable Banach spaces and spaces with $w^*$-separable dual in terms of nuclearly adjacent quasicomplements.

Corollary 2.13

If E is a Banach space then the following assertions are equivalent:

• (1) E is separable.

• (2) For every subspace $X\subset E$ and every $\varepsilon > 0$ there is an isomorphism $\varphi :E\rightarrow E$ such that $\Vert \varphi -I_E\Vert < \varepsilon$ and $\varphi (X)$ is a mutually nuclearly adjacent quasicomplement of X.

• (3) Every subspace of E has a mutually nuclearly adjacent quasicomplement.

• (4) E has a separable subspace which admits a nuclearly adjacent quasicomplement.

Proof

The implication $(1)\Rightarrow (2)$ follows from Corollary 2.12, and $(2)\Rightarrow (3)\Rightarrow (4)$ are obvious. On the other hand, if X is a subspace of E satisfying (4), then E/X is separable by Corollary 2.7. Since X is separable, E is separable as well, thus (4) implies (1). $\square$

Corollary 2.14

If E is a Banach space then the following assertions are equivalent:

• (1) E has a separable quotient and $E^*$ is $w^*$-separable.

• (2) There exists a subspace $X\subset E$ with the following property: for any $\varepsilon >0$ there is an isomorphism $\varphi :E\rightarrow E$ such that $\Vert \varphi -I_E\Vert < \varepsilon$ and $\varphi (X)$ is a mutually nuclearly adjacent quasicomplement of X.

• (3) E has a subspace with a mutually nuclearly adjacent quasicomplement.

• (4) E has a subspace with $w^*$-separable dual that admits a nuclearly adjacent quasicomplement.

Proof

Assume that E satisfies (1), and let X be a subspace of E such that E/X is separable. Bearing in mind that $X^*$ is separable (as $E^*$ is), thanks to Corollary 2.12 we have that X satisfies (2). The implication $(2)\Rightarrow (3)$ is trivial. On the other hand, if X is a subspace of E that admits a mutually nuclearly adjacent quasicomplement then, thanks to Corollary 2.9 it follows that $X^*$ is $w^*$-separable, therefore $(3)\Rightarrow (4)$. Now, let X be a subspace of E with $w^*$-separable dual that admit a nuclearly adjacent quasicomplement. Then E/X is separable by Corollary 2.7. Moreover, since $X^*$ is $w^*$-separable and $(E/X)^*$ are $w^*$-separable, $E^*$ is $w^*$-separable as well. In fact, according to a result of Singer [20, Lemma 7.5], for every total sequence $\{x_n^*\}_n\subset X^*$ there is a total sequence $\{f_n\}_n\subset E^*$ such that $f_n|_{X}=x_n^*$ for all n, thus $(4)\Rightarrow (1)$. $\square$

Notice that the assumption “E has a separable quotient” in assertion (1) from the previous corollary is necessary to achieve the implication $(1)\Rightarrow (2)$. Indeed, as it is well-known, the existence of a Banach space with no separable quotient is equivalent to the existence of a Banach space with $w^*$-separable dual and no separable quotient.

3 The Theorem of Lindenstrauss–Rosenthal and $w^*$-Basic Sequences

In this section, we establish some relationships between the existence of a quasicomplement of a subspace X of a Banach space E and the possibility of finding $w^*$-basic extensions to E of a given sequence of functionals $\{x_n^*\}_n\subset X^*$, as well as the related problem of Singer which wonders on the existence of $w^*$-basic sequences of biorthogonal functionals in $E^*$ associated to a given basic sequence (or more generally, a given minimal sequence) $\{x_n\}_n\subset E$. The first result in this direction is the following strengthening of the theorem of Lindenstrauss and Rosenthal. As regards assertion (8) in this theorem, recall that the dual of every Banach space with $w^*$-separable dual contains a total sequence which is $w^*$-minimal (see e.g. [10, Lemma 1.21]). The second property is motivated by question (b) in Remark 2.10. That property implies that if E is a Banach space, and $U\subset E$ is a subspace with $w^*$-separable dual, then for each subspace $X\subset E$ such that E/X has a separable quotient, $U\subset X$ and X/U is not too big so that $(X/U)^*$ is $w^*$-separable, then U admits a quasicomplement $Y\subset E$ satisfying the disjointness property $Y\cap X = \{0\}$.

Theorem 3.1

Let X be a subspace of a Banach space E. If $X^*$ is $w^*$-separable then the following assertions are equivalent:

• (1) E/X has a separable quotient.

• (2) For every closed infinite-dimensional subspace $U\subset X$ (not necessarily of infinite codimension in X) such that $(X/U)^*$ is $w^*$-separable, there exists a subspace $Y\subset E$ such that both X and U are nuclearly adjacent quasicomplements of Y.

• (3) X is a nuclearly adjacent quasicomplement of some subspace $Y\subset E$.

• (4) X has a proper quasicomplement.

• (5) There exists a subspace $Y\subset E$ such that $X+Y$ is dense in E and $X+Y\ne E$.

• (6) Every sequence $\{x_n^*\}_n\subset X^*$ has a $w^*$-basic extension $\{f_n\}_n\subset E^*$.

• (7) Every $w^*$-minimal sequence $\{x_n^*\}_n\subset X^*$ has a $w^*$-basic extension $\{f_n\}_n\subset E^*$.

• (8) Every total $w^*$-minimal sequence $\{x_n^*\}_n\subset X^*$ has a $w^*$-basic extension $\{f_n\}_n\subset E^*$.

In the proof of Theorem 3.1, we shall use the following lemmas, the first of which enhances some ideas of Johnson [12].

Lemma 3.2

Let X be a subspace of a Banach space E. If E/X has a separable quotient then, for every sequence $\{x_n^*\}_n\subset X^*$ there exists a $w^*$-basic extension $\{f_n\}_n\subset E^*$ such that the restriction to X of the canonical quotient map $Q:E\rightarrow E/[\{f_n\}_n]_{\perp }$ is nuclear.

Proof

Since E/X has a separable quotient, according to the aforementioned result of Johnson and Rosenthal [13], $X^{\perp }$ contains a $w^*$-basic sequence $\{g_n\}_n$. Let $\{u_n\}_n$ be a sequence in E such that $\{u_n, g_n\}_n$ is a biorthogonal system, let $\{z_n^*\}_n\subset E^*$ be any extension of $\{x_n^*\}_n$ and $\{a_n\}_n\subset (0,\infty )$ be a sequence with $\sum _{n\ge 1} a_n \Vert u_n\Vert \Vert z_n^*\Vert < 1$. Consider the isomorphism $\psi :E\rightarrow E$ defined for each $u\in E$ as $\psi (u) = u + \sum _{n\ge 1} a_n z_n^*(u)u_n$, and set, for each n,

$$\begin{aligned} f_n = a_n^{-1} g_n + z_n^*. \end{aligned}$$

Since $\{g_n\}_n\subset X^{\perp }$ we have that the sequence $\{f_n\}_n$ is an extension of $\{x_n^*\}_n$ to E. Notice that $\psi ^*(f) = f + \sum _{n\ge 1} a_n f(u_n)z_n^*$ for all $f\in E^*$. Thus, as $\{u_n, g_n\}_n$ is a biorthogonal system we get $\psi ^*(g_n)= a_n f_n$ for all n. Bearing in mind that $\psi ^*$ is a $w^*$-isomorphism on $E^*$ and $\{g_n\}_n$ is $w^*$-basic, we have that $\{a_n f_n\}_n$ is $w^*$-basic, hence $\{f_n\}_n$ is $w^*$-basic as well.

Now, we shall show that the restriction to X of the quotient map $Q:E\rightarrow E/[\{f_n\}_n]_{\perp }$ is nuclear. For each n we define

$$\begin{aligned} v_n = a_n \psi ^{-1}(u_n). \end{aligned}$$

It is clear that $\{v_n, f_n\}_n$ is a biorthogonal system. As $\{f_n\}_n$ is a $w^*$-basic sequence, thanks to [20, Proposition 1.8] we have that $\{Q(v_n)\}_n$ is a basis of $E/[\{f_n\}_n]_{\perp }$. Bearing in mind that $(E/[\{f_n\}_n]_{\perp })^*$ identifies with $\overline{[ \{f_n\}_n]}^{w^*}$ and the sequence $\{f_n\}_n$ is an extension of $\{x_n^*\}_n\subset X^*$ we get

$$\begin{aligned} Q(x) = \sum _{n\ge 1} x_n^*(x)Q(v_n) \quad \text {for all} \quad x\in X. \end{aligned}$$

Since $\sum _{n\ge 1} a_n \Vert u_n\Vert \Vert z_n^*\Vert < 1$,   $\Vert x_n^*\Vert \le \Vert z_n^*\Vert$ and $\Vert Q(v_n)\Vert =\left\| Q(a_n\psi ^{-1}(u_n))\right\| \le a_n \Vert \psi ^{-1}\Vert \Vert u_n\Vert$ for all n we have $\sum _{n\ge 1} \Vert x_n^*\Vert \Vert Q(v_n)\Vert \le \Vert \psi ^{-1}\Vert \sum _{n\ge 1} a_n \Vert u_n\Vert \Vert z_n^*\Vert < \infty .$ Consequently, $Q|_{ X}$ is a nuclear operator. $\square$

Lemma 3.3

Let X be a subspace of a Banach space E. If $\{f_n\}_n$ is a $w^*$-basic X-minimal sequence in $E^*$ then $X+[\{f_n\}_n]_{\perp }$ is dense in E. Moreover, if $X+[\{f_n\}_n]_{\perp }=E$ then $\{f_n|_{X}\}_n$ is a $w^*$-basic sequence in $X^*$.

Proof

Let us write $Y = [\{f_n\}_n]_{\perp }$, let Q denote the quotient map from E onto E/Y, and let $\{x_n\}_n$ be a sequence in X such that $\{x_n, f_n\}_n$ is a biorthogonal system. Since $\{f_n\}_n$ is $w^*$-basic and $Y^{\perp } = \overline{\textrm{span} \{f_n\}_n}^{w^*}$ we have $[\{x_n\}_n]^{\perp }\cap Y^{\perp } = \{0\}$, and therefore $X^{\perp }\cap Y^{\perp } = \{0\}$. Thus, the restriction to $Y^{\perp }$ of the restriction operator $q^*:E^*\rightarrow X^*$ is one-to-one. Bearing in mind that $q^*|_{Y^{\perp }}$ identifies with the adjoint operator of $Q|_{X}$ it follows that $\overline{Q(X)} = E/Y$, which entails that $X+Y$ is dense in E.

Now, assume that $X+Y= E$. Then $Q(X) = E/Y$, therefore Q induces an isomorphism ${\widehat{Q}}:X/(X\cap Y)\rightarrow E/Y$. Notice that the adjoint operator of ${\widehat{Q}}$ identifies with the restriction to $Y^{\perp }$ of the restriction map $q^*:E^*\rightarrow Z$, where $Z = \{f|_{X}: \, f\in (X\cap Y)^{\perp }\}\subset X^*$. Thus $q^*|_{Y^{\perp }}$ is an isomorphism from $Y^{\perp }$ onto Z for the weak$^*$ topologies, and bearing in mind that $\{f_n\}_n$ is $w^*$-basic it follows that $\{f_n|_{X}\}_n$ is $w^*$-basic as well. $\square$

Proof of Theorem 3.1

The implications $(2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (5)$ and $(6)\Rightarrow (7)\Rightarrow (8)$ are obvious, and $(1)\Rightarrow (6)$ is a consequence of Lemma 3.2. Thus, it is enough to prove that $(1)\Rightarrow (2)$ and $(8)\Rightarrow (5) \Rightarrow (1)$.

$(1)\Rightarrow (2)$ Let U be a closed infinite-dimensional subspace of X such that $(X/U)^*$ is $w^*$-separable. Since $X^*$ is $w^*$-separable, so is $U^*$. Thus, there is a $w^*$-minimal total sequence $\{u_n^*\}_n\subset U^*$, and using [20, Lemma 7.5] we deduce the existence of a sequence $\{x_n^*\}_n\subset X^*$ which is total and satisfies $x_n^*|_{U} = u_n^*$ for all n. Taking into account that E/X has a separable quotient, thanks to Lemma 3.2 there is a $w^*$-basic sequence $\{f_n\}_n\subset E^*$ such that $f_n|_{X}= x_n^*$ for all n and the restriction to X of the quotient operator $Q: E\rightarrow E/[\{f_n\}_n]_{\perp }$ is nuclear. We shall show that the subspace $Y = [\{f_n\}_n]_{\perp }$ satisfies the required properties. Evidently, $Q|_{U}$ is a nuclear operator. Therefore, it suffices to show that

• (a) $X\cap Y = \{0\}$ and

• (b) $U+Y$ is dense in E.

If $x\in X\cap Y$ then $f_n(x)=0$ for all n. As $\{f_n\}_n$ is an extension of $\{x_n^*\}_n\subset X^*$ we get $x_n^*(x)=0$ for all n, and bearing in mind that $\{x_n^*\}_n$ is total over X we obtain $x=0$. On the other hand, since $\{f_n\}_n$ is an extension of $\{u_n^*\}_n$ and this sequence is U-minimal, we have that $\{f_n\}_n$ is a $w^*$-basic U-minimal sequence, and an appeal to Lemma 3.3 entails that $X+U$ is dense in E.

$(8) \Rightarrow (5)$ First, we observe that $X^*$ contains a total $w^*$-minimal sequence $\{x_n^*\}_n$ which is not $w^*$-basic. Indeed, if X nonseparable, this property is satisfied by any total $w^*$-minimal sequence $\{x_n^*\}_n$ in $X^*$ (otherwise any sequence $\{x_n\}_n\subset X$ such that $\{x_n, x_n^*\}_n$ is a biorthogonal system should be a basis of X by [20, Proposition 1.8]). If X is separable then, according to [20, Proposition 8.12], X has an M-basis $\{x_n\}_n$ which is not a basis, and a new appeal to [20, Proposition 1.8] yields that the sequence of associated biorthogonal functionals $\{x_n^*\}_n\subset X^*$ is not $w^*$-basic. Now, applying assertion (8) to the (non $w^*$-basic) sequence $\{x_n^*\}_n$, we obtain a $w^*$-basic extension $\{f_n\}_n\subset E^*$ of that sequence, and using Lemma 3.3 it follows that the subspace $Y = [\{f_n\}_n]_{\perp }$ satisfies (5).

$(5) \Rightarrow (1)$ Let Y be a subspace of E satisfying (5) and consider the Banach space $F = X\times Y$ with a product norm (e. g. $|||(x,y)|||=\Vert x\Vert +\Vert y\Vert$ for every $(x,y)\in X\times Y$, being $\Vert \cdot \Vert$ the inherited norm of E in X and Y). Also consider the operator $T:F\rightarrow E$ defined by the formula

$$\begin{aligned} T(x,y) = x+y, \quad (x,y)\in F. \end{aligned}$$

Clearly, T is bounded and T(F) is a proper and dense linear subspace of E which contains X. Thus, if ${\widetilde{Q}}:E\rightarrow E/X$ denotes the quotient map, then the operator $S:= {\widetilde{Q}}\circ T:F\rightarrow E/X$ satisfies $\overline{S(F)} = E/X$ and $S(F)\ne E/X$, that is, the space E/X contains a proper dense operator range, and an appeal to the aforementioned result of Saxon and Wilansky [19] yields that E/X has a separable infinite-dimensional quotient. $\square$

Remark 3.4

The proof of the implication $(8)\Rightarrow (5)$ in Theorem 3.1 yields that if the subspace $X\subset E$ is does not have basis (in particular if X is nonseparable), then assertions $(1)-(8)$ are equivalent to the next one:

“There exists a total $w^*$-minimal sequence $\{x_n^*\}_n\subset X^*$ which admits a $w^*$-basic extension $\{f_n\}_n\subset E^*$”.

It is worth mentioning that the $w^*$-separability of $X^*$ is not essential in several implications of the proof of Theorem 3.1. A slight variant of the arguments employed there leads to the following characterization of those subspaces X of a Banach space E with the property that every sequence $\{x_n^*\}_n\subset X^*$ admits a $w^*$-basic extension $\{f_n\}_n\subset E^*$.

Proposition 3.5

If X is a subspace of a Banach space E then the following assertions are equivalent:

• (1) E/X has a separable quotient.

• (2) There exists a subspace $Y\subset E$ such that $X+Y$ is dense in E and $X+Y\not =E$.

• (3) Every sequence $\{x_n^*\}\subset X^*$ has a $w^*$-basic extension $\{f_n\}_n\subset E^*$.

• (4) Every $w^*$-minimal sequence $\{x_n^*\}_n\subset X^*$ has a $w^*$-basic extension $\{f_n\}_n\subset E^*$.

Moreover, if E/X has a separable quotient, then for every $\varepsilon > 0$ the subspace Y in assertion (2) can be chosen to be isomorphic and $(1+\varepsilon )$-isometric to any given subspace $Z\subset E$ such that $X\subset Z$ and E/Z is separable.

Proof

The implication $(1)\Rightarrow (3)$ follows from Lemma 3.2, $(3)\Rightarrow (4)$ is obvious, and $(2)\Rightarrow (1)$ can be achieved using the argument employed in the proof of implication $(5)\Rightarrow (1)$ from the previous theorem. Thus, it is enough to prove that (4) implies (2). According to [20, Proposition 1.12] there exist a basic sequence $\{x_n\}_n$ in X and a non-basic sequence $\{x_n^*\}_n\subset X^*$ such that $\{x_n, x_n^*\}_n$ is a biorthogonal system. In particular, $\{x_n^*\}_n$ is $w^*$-minimal. Thus, assuming (4), we can find a $w^*$-basic sequence $\{f_n\}_n\subset E^*$ such that $f_n|_{X} = x_n^*$ for all n. Set $Y = [\{f_n\}_n]_{\perp }$. As $\{f_n\}_n$ is $w^*$-basic and $\{x_n^*\}_n$ is not basic, thanks to Lemma 3.3 it follows that $X+Y$ is a proper dense linear subspace of E. Therefore, $(4)\Rightarrow (2)$. The statement after assertion (4) is an immediate consequence of Proposition 2.6. $\square$

A well-known result of Argyros et al. [3] asserts that every dual Banach space has a separable quotient. As an application of this result and the former proposition we obtain the following corollary.

Corollary 3.6

Let E be a Banach space and Z be a subspace of $E^*$. If Z is $w^*$-closed then, for every sequence $\{g_n\}_n\subset E^{**}$ there exists a $w^*$-basic sequence $\{f_n\}_n\subset E^{**}$ such that

$$\begin{aligned} f_n(z) = g_n(z) \quad \text {for all} \quad n\ge 1 \,\, \text {and} \,\, z\in Z. \end{aligned}$$

Proof

For each $n\ge 1$ and $z\in Z$ we set $x_n^*(z) = g_n(z)$. It is clear that $x_n^*\in Z^*$. Since Z is $w^*$-closed, $E^*/Z$ identifies with the dual space of $Z_{\perp }$. Thus, by the aforesaid result in [3], $E^*/Z$ has a separable quotient, and using the previous proposition we deduce the existence of a $w^*$-basic sequence $\{f_n\}_n\subset E^{**}$ such that $f_{n}|_{Z} = x_n^* = g_n|_{Z}$ for all $n\ge 1$. $\square$

As we already mentioned, the weak version of property (3) in Proposition 3.5 replacing $w^*$-basic extensions with basic ones always holds true, that is, if X is a subspace of a Banach space E, then for every sequence $\{x_n^*\}_n\subset X^*$ there exists a basic sequence $\{f_n\}_n\subset E^*$ such that $f_n|_{X} = x_n^*$ for all n. This implies that any minimal sequence $\{x_n\}_n$ in E admits a basic sequence $\{f_n\}_n\subset E^*$ of biorthogonal functionals. In the rest of this section we treat with the problem of Singer concerning the existence of $w^*$-basic sequence of biorthogonal functionals associated to basic sequences (or more generally, minimal sequences) in a Banach space. The next corollary of Theorem 3.1 characterizes the separable subspaces of a Banach space with the property that every minimal sequence in that subspace admits a $w^*$-basic sequence of biorthogonal functionals.

Corollary 3.7

If X is a separable subspace of a Banach space E, then the following assertions are equivalent:

• (1) E/X has a separable quotient.

• (2) X has a proper quasicomplement.

• (3) For every minimal sequence $\{x_n\}_n$ in X there exists a $w^*$-basic sequence $\{f_n\}_n\subset E^*$ such that $\{x_n, f_n\}_n$ is a biorthogonal system.

• (4) For every M-basis $\{x_n\}_n$ of X there exists a $w^*$-basic sequence $\{f_n\}_n\subset E^*$ such that $\{x_n, f_n\}_n$ is a biorthogonal system.

Proof

The equivalence between (1) and (2) follows from Theorem 3.1, and $(3)\Rightarrow (4)$ is trivial. On the other hand, again by Theorem 3.1, if property (2) is satisfied and $\{x_n\}_n$ is a minimal sequence in X, then for every sequence $\{x_n^*\}_n\subset X^*$ such that $\{x_n, x_n^*\}_n$ is a biorthogonal system there is a $w^*$-basic sequence $\{f_n\}_n\subset E^*$ such that $f_n|_{X} = x_n^*$ for all n, so $\{x_n, f_n\}_n$ is a biorthogonal system, hence $(2)\Rightarrow (3)$. Now, assume that property (4) holds. According to [20, Proposition 8.12], X has a non-basic M-basis $\{x_n\}_n$. Let $\{x_n^*\}_n\subset X^*$ be the corresponding sequence of biorthogonal functionals. Because of the assumption, there exists a $w^*$-basic sequence $\{f_n\}_n\subset E^*$ such that $\{x_n, f_n\}_n$ is a biorthogonal system. Set $Y = [\{f_n\}_n]_{\perp }$. According to Lemma 3.3, $X+Y$ is dense in E. On the other hand, since $[\{x_n\}_n]= X$ we get $f_n|_{X} = x_n^*$ for all n. As $\{x_n\}_n$ is not basic, thanks to [20, Proposition 1.8] we have that $\{x_n^*\}_n$ is not $w^*$-basic, and a new appeal to Lemma 3.3 guarantees that $X+Y\ne E$. Moreover, since $\{f_n\}_n$ is total over X (being an extension of the total sequence $\{x_n^*\}_n$) we get $[\{f_n\}_n]_{\perp }\cap X = \{0\}$, hence $(4)\Rightarrow (2)$. $\square$

In the case that the subspace X has a basis, we have the following variant of the previous result, which yields an affirmative answer to Singer’s problem in wide classes of Banach spaces.

Theorem 3.8

Let E be a Banach space. If X is a subspace of E with a basis, then the following assertions are equivalent:

• (1) X is quasicomplemented.

• (2) Every basis of X has a $w^*$-basic sequence $\{f_n\}_n\subset E^*$ of biorthogonal functionals.

• (3) There exists a basis of X with a $w^*$-basic sequence $\{f_n\}_n\subset E^*$ of biorthogonal functionals.

• (4) There exists a total $w^*$-minimal sequence $\{x_n^*\}_n\subset X^*$ which admits a $w^*$-basic extension $\{f_n\}_n\subset E^*$.

Proof

We only need to show that $(1)\Rightarrow (2)$ and $(3)\Rightarrow (4)\Rightarrow (1)$. Assume that X is quasicomplemented and let Y be a quasicomplement of X. Let us fix a basis $\{x_n\}_n$ of X, and denote by $\{x_n^*\}_n\subset X^*$ the corresponding sequence of biorthogonal functionals. If $X+Y\ne E$, then Corollary 3.7 yields a $w^*$-basic sequence $\{f_n\}_n$ such that $\{x_n,f_n\}_n$ is a biorthogonal system. If $X+Y=E$ then $E=X\oplus Y$ (i.e. E is the direct sum of X and Y) and $E^* = X^{\perp }\oplus Y^{\perp }$. Thus, the restriction to $Y^{\perp }$ of the restriction operator $q^*:E^*\rightarrow X^*$ is an isomorphism for the norm topologies as well as the weak$^*$ topologies. Let $\rho = q^*|_{Y^{\perp }}$ and define, for each $n\ge 1$,

$$\begin{aligned} f_n = \rho ^{-1}(x_n^*). \end{aligned}$$

As $\{x_n\}_n$ is a basis of X we have that $\{x_n^*\}_n$ is a $w^*$-basis of $X^*$, that is, $\{x_n^*\}_n$ is a $w^*$-basic sequence in $X^*$ satisfying $\overline{[\{x_n^*\}_n]}^{w^*} = X^*$. Therefore, $\{f_n\}_n$ is a $w^*$-basic sequence in $E^*$. Moreover, since $\{x_n, x_n^*\}_n$ is a biorthogonal system in X and $\rho (f_n) = x_n^*$ for each n, we have that $\{x_n, f_n\}_n$ is a biorthogonal system in E, thus $(1)\Rightarrow (2)$. Now, assume that there exist a basis $\{x_n\}_n$ of X and a $w^*$-basic sequence $\{f_n\}_n$ in $E^*$ such that $\{x_n, f_n\}_n$ is a biorthogonal system, and let $\{x_n^*\}_n$ be the sequence of biorthogonal functionals in $X^*$ associated to $\{x_n\}_n$. Since $[\{x_n\}_n] = X$ it follows that $f_n|_{X} = x_n^*$ for all n, therefore the sequence $\{x_n^*\}_n$ satisfies (4). To finish, suppose that there exists an X-minimal $w^*$-basic sequence $\{f_n\}_n\subset E^*$ such that $\{f_n|_{X}\}_n$ is total over X, and set $Y = [\{f_n\}_n]_{\perp }$. According to Lemma 3.3 we have that $X+ Y$ is dense in E. On the other hand, since $\{f_n\}_n$ is total over X we get $Y\cap X = \{0\}$, therefore $(4)\Rightarrow (1)$. $\square$

A Banach space E is called weakly Lindelöf determined (for short, WLD) if there exist a set $\Gamma$ and a bounded linear one-to-one ($w^*$, $\tau _p$)-continuous operator $T:E^*\rightarrow \ell ^{\infty }_{c}(\Gamma )$, where $\ell ^{\infty }_{c}(\Gamma )$ is the space made up of all $x\in \ell ^{\infty }(\Gamma )$ with countable support, and $\tau _p$ denotes the topology of pointwise convergence on that space. We refer to [10, Chapter 5] for several characterizations and the general properties of these spaces. It is well-known (see [10, Corollaries 5.43 and 5.74]) that every subspace of a WLD space is WLD and quasicomplemented. Consequently, by Theorem 3.8, if E is a WLD space, then every basic sequence $\{x_n\}_n\subset E$ such that $\textrm{codim}\, [\{x_n\}_n] = \infty$ has a $w^*$-basic sequence $\{f_n\}_n\subset E^*$ of biorthogonal functionals. Actually, the same result holds replacing basic sequences with minimal ones as it is stated in the next corollary.

Corollary 3.9

If E is a WLD Banach space, then for every minimal sequence $\{x_n\}_n$ such that $\textrm{codim}\, [\{x_n\}_n] = \infty$ there exists a $w^*$-basic sequence $\{f_n\}_n\subset E^*$ such that $\{x_n, f_n\}_n$ is a biorthogonal system.

Proof

Since $(E/[\{x_n\}_n])^*$ identifies with the annihilator subspace of $[\{x_n\}_n]$, which is a $w^*$-closed subspace of $E^*$, thanks to the definition of WLD spaces we have that $E/[\{x_n\}_n]$ is WLD as well. Therefore every subspace of $E/[\{x_n\}_n]$ is quasicomplemented. In particular, $E/[\{x_n\}_n]$ has a separable quotient (see e.g. [19]), and an appeal to Corollary 3.7, applied to the subspace $X = [\{x_n\}_n]$, yields a $w^*$-basic sequence $\{f_n\}_n\subset E^*$ such that $\{x_n, f_n\}_n$ is a biorthogonal system. $\square$

The conclusion of Corollary 3.9 also holds true for the non WLD space $\ell ^{\infty }(\Gamma )$. Indeed, if X is a subspace of $\ell ^{\infty }(\Gamma )$ then, according to Rosenthal’s result in [18] mentioned in the Introduction, $\ell ^{\infty }(\Gamma )/X$ has a separable quotient. Thus, if $\{x_n\}_n$ is a minimal sequence in $\ell ^{\infty }(\Gamma )$, then $\ell ^{\infty }(\Gamma )/[\{x_n\}_n]$ has a separable quotient. Hence, $\{x_n\}_n$ admits a $w^*$-basic sequence of biorthogonal functionals. In the case that the set $\Gamma$ is uncountable, this statement follows as well from the next result.

Corollary 3.10

Let E be a Banach space and let $\{z_n\}_n$ be a minimal sequence in $E^*$. If $[\{z_n\}]_{\perp }$ is infinite-dimensional, then there exists a $w^*$-basic sequence $\{f_n\}_n\subset E^{**}$ such that $\{z_n, f_n\}_n$ is a biorthogonal system.

Proof

Let us write $X:=[\{z_n\}]_{\perp }$. Since $E^*/X^{\perp }$ identifies with $X^*$ (and it is infinite-dimensional), according to the aforementioned result in [3] we have that $E^*/X^{\perp }$ has a separable quotient. Equivalently, there is a subspace $Y\subset E^*$ such that $Y\supset X^{\perp }=\overline{[\{z_n\}_n]}^{w^*}$ and $E^*/Y$ is separable. Since $Y\supset [\{z_n\}_n]$, we have that $E^*/[\{z_n\}_n]$ has also a separable quotient, and we can apply Corollary 3.7 (in $E^*$). $\square$

As a particular case of this result, we obtain the following weak answer to Singer’s problem for minimal sequences.

Corollary 3.11

Let $\{x_n\}_n$ be a minimal sequence in a Banach space E. If $[\{x_n\}_n]$ is infinite-codimensional, then there is a $w^*$-basic sequence $\{f_n\}_n$ in $E^{***}$ such that $\{x_n, f_n\}_n$ is a biorthogonal system.

Proof

For each n we write $z_n = \pi (x_n)$, where $\pi :E\rightarrow E^{**}$ denotes the canonical isometry. Then, $\{z_n\}_n$ is a minimal sequence in $E^{**}$. Moreover, since $\textrm{codim}\, [\{x_n\}_n]= \infty$ we get $\dim \, ([\{z_n\}]_{\perp })= \dim \, ([\{x_n\}_n]^{\perp }) = \dim \, (E/[\{x_n\}_n])^*= \infty$, and the former corollary applies. $\square$