Interpolator Symmetries and New Kalton-Peck Spaces

Abstract

We study the six diagrams generated by the first three Schechter interpolators $\Delta _2(f)= f''(1/2)/2!, \Delta _1(f)= f'(1/2), \Delta _0(f)=f(1/2)$ acting on the Calderón space associated to the pair $(\ell _\infty , \ell _1)$. We will study the remarkable and somehow unexpected properties of all the spaces appearing in those diagrams: two new spaces (and their duals), two Orlicz spaces (and their duals) in addition to the third order Rochberg space, the standard Kalton-Peck space $Z_2$ and, of course, the Hilbert space $\ell _2$. We will also deal with a nice test case: that of weighted $\ell _2$ spaces, in which case all involved spaces are Hilbert spaces.

1 Introduction

The aim of this paper is to present the seven natural Banach spaces generated by the first three interpolators of the complex interpolation method when applied to the couple $(\ell _\infty , \ell _1)$ at 1/2. They are three Rochberg spaces $\ell _2$, $Z_2$ and $Z_3$, two Orlicz spaces $\ell _f, \ell _g$ generated by the Orlicz functions $f(t)=t^2 \log t^2, g(t)=t^2 \log ^4 t$; and two new spaces $\wedge , \bigcirc$. We present the basic diagram generated by these three interpolators and the six possible diagrams they generate, which produce the seven spaces just mentioned, their duals, and nothing more. And it is so by virtue of the symmetries of the six diagrams: some are overt (described in Sect. 2.3), but some are deeply concealed and unexpected (like those described between Proposition 3.2 and 3.7).

Following [7], we will work with a variant ${\mathcal {C}}$ of the Calderón space considered in [3, Section 4.1] when working with the pair $(\ell _\infty , \ell _1)$: if ${\mathbb {S}}$ is the open strip $\{z\in {\mathbb {C}}: 0<\textrm{Re}(z)<1\}$ in the complex plane, ${\mathcal {C}}$ will be the space of continuous bounded functions on ${\mathbb {S}}$ that are also weak$^*$-continuous as functions $f:\overline{{\mathbb {S}}}\longrightarrow \ell _\infty$ and that moreover are holomorphic on ${\mathbb {S}}$ and satisfy the boundary condition $f(k+it)\in X_k$ for each $t\in \mathbb {R}$ and $\sup _t\Vert f(k+it)\Vert _{X_k}<\infty$, valid for $k=0,1$. The Calderón space ${\mathcal {C}}$ is complete with the norm $\Vert f\Vert = \sup \{\Vert f(k+it)\Vert _{k}: k=0,1; t\in \mathbb {R}\}$. The evaluation maps $\delta _z: {\mathcal {C}}\longrightarrow \ell _\infty$ are continuous for all $z\in {\mathbb {S}}$, and given $\theta \in (0,1)$ and $p=\theta ^{-1}$ one obtains $\ell _p = \left\{ f(\theta ): f\in {\mathcal {C}}\right\}$ with the standard norm in $\ell _p$ equal to the quotient norm in $\Vert x\Vert _{\theta }= \inf \big \{\Vert f\Vert : x=f(\theta ), f\in \mathcal C\big \}$. See [3, Lemma 4.1.1] and [5, Section 10.8] for details.

For the rest of the paper we will focus on the Hilbert space case: $\theta =1/2$; $p=2$. We consider the interpolators $\Delta _k:{\mathcal {C}} \rightarrow \ell _\infty$ defined by $\Delta _k(f) =f^{(k)}(1/2)/k!$ for $k=0,1,2,\ldots$ Following Rochberg [29] (see also [6, 7]), the $n^{th}$ Rochberg space is defined as ${\mathfrak {R}}_n =\{(\Delta _{n-1}(f),\dots , \Delta _0(f)): f\in {\mathcal {C}}\}$ endowed with its natural quotient norm. This yields ${\mathfrak {R}}_1= \ell _2$ and ${\mathfrak {R}}_2= Z_2$, the Kalton-Peck space [24]. We will denote $\mathfrak R_3$ with the more friendly name $Z_3$. Among the distinguished subspaces of $Z_3$ we will encounter the three Orlicz spaces $\ell _2=\{(w,0,0) \in Z_3\}$, $\ell _f=\{(0,x,0) \in Z_3\}$ and $\ell _g=\{(0,0,y) \in Z_3\}$, and the three spaces $Z_2=\{(w,x,0) \in Z_3\}$, $\wedge =\{(w,0,y) \in Z_3\}$ and $\bigcirc =\{(0,x,y) \in Z_3\}$.

Let us now aim at diagrams: It is a fact uncovered through [4, 11, 24] that $Z_2$ admits two natural representations $0\rightarrow \ell _2\rightarrow Z_2\rightarrow \ell _2\rightarrow 0$ and $0\rightarrow \ell _f\rightarrow Z_2\rightarrow \ell _f^*\rightarrow 0$ as a non-trivial twisted sum that are associated to the two permutations $(\Delta _1,\Delta _0)$ and $(\Delta _0,\Delta _1)$. In the same way, we will show (Sect. 2) that $Z_3$ admits six natural representations as a twisted sum space associated with the six diagrams generated by the six permutations of the three interpolators $(\Delta _2, \Delta _1, \Delta _0)$. Indeed, if we denote [abc] the diagram obtained from the permutation $(\Delta _a, \Delta _b, \Delta _c)$, the six diagrams are (we will omit the arrow $0\rightarrow$ at the beginning and $\rightarrow 0$ at the end of the exact sequences forming the rows and columns):

We will prove:

• Properties shared by all spaces/sequences

  • (1) All the spaces in the diagrams are hereditarily $\ell _2$ (Proposition 5.1) and have basis.

  • (2) All the exact sequences are nontrivial (Corollary 5.7).

  • (3) All quotient maps, except perhaps $q_{1,2}$ and $q_{2,1}$ (see below), are strictly singular (Proposition 5.14).

• Properties similar to those of $Z_2$

  • (1) The spaces $\bigcirc$, $\wedge$, $\bigcirc ^*$ and $\wedge ^*$ admit a symmetric two-dimensional decomposition.

  • (2) $Z_3$ admits a symmetric three-dimensional decomposition (Proposition 3.1) and it is isomorphic to its dual [6, Prop. 5.5 and Cor. 5.7].

  • (3) Every infinite dimensional complemented subspace of $Z_3$ contains a copy of $Z_3$ complemented in the whole space (Proposition 5.5).

  • (4) The spaces $Z_3$, $\wedge$ and $\wedge ^*$ contain no complemented copies of $\ell _2$ and admit no unconditional basis (Proposition 5.12).

  • (5) Every basic sequence in $Z_3$ contains a subsequence equivalent to the canonical basis of one of the spaces $\ell _2, \ell _f, \ell _g$ (Theorem 5.8).

• Properties different from those of $Z_2$

  • (1) None of the spaces $\bigcirc$, $\wedge$, $\bigcirc ^*$ and $\wedge ^*$ is isomorphic to a subspace or a quotient of $Z_2$ (Proposition 5.4).

  • (2) $\wedge$ and $\bigcirc$ are not isomorphic to their duals (Proposition 5.10).

  • (3) Neither of the spaces $\wedge$ and $\wedge ^*$ is isomorphic to either $\bigcirc$ or $\bigcirc ^*$ (Proposition 5.11).

• Open questions

  • (1) We have been unable to show that $\bigcirc$ (hence $\bigcirc ^*$ also) contains no complemented copies of $\ell _2$. From that it would follow also $q_{1,2}$ and $q_{2,1}$ are strictly singular, hence that $\bigcirc$ and $\bigcirc ^*$ do not have an unconditional basis (Remark 5.15), which would complete our scheme.

  • (2) We could not cover in this paper the case of interpolation at an arbitrary $\theta \ne 1/2$. In that case, the first thing one loses is duality and its associated symmetries: $Z_p$ is no longer isomorphic to $Z_p^*$. The same is valid for weighted $\ell _p$-spaces or weighted versions of a given space with an unconditional basis.

2 The Six Diagrams Generated by the Three Interpolators $\Delta _2, \Delta _1, \Delta _0$

A Banach space space Z is a twisted sum of Y and X if there exists an exact sequence $0\rightarrow Y \rightarrow Z \rightarrow X \rightarrow 0$ (namely, a diagram formed by Banach spaces and continuous operators so that the kernel of each of them coincides with the image of the previous one). Twisted sums of Y and X correspond to a special type of maps $X\longrightarrow Y$, called quasi-linear maps [5, 24]. We need to widen this notion as in [12, 15] assuming that Y is continuously embedded in an “ambient” Hausdorff topological vector space Banach space $\Sigma$ which, for us, will be a Banach or quasi-Banach space. There are indeed natural situations in which these “generalized” quasi-linear maps appear: centralizers between quasi-Banach function spaces [22]; differentials generated by two interpolators [11]; or G-actions on twisted sums [12].

Definition 2.1

A quasi-linear map $\Omega : X \curvearrowright Y$ with ambient space $\Sigma$ is a homogeneous map $\Omega : X\longrightarrow \Sigma$ for which there exists a constant C such that for $x_1,x_2 \in X$,

• $\Omega (x_1+x_2)- \Omega (x_1)- \Omega (x_2)\in Y$ and

• $\Vert \Omega (x_1+x_2)- \Omega (x_1)- \Omega (x_2)\Vert _Y \le C (\Vert x_1\Vert _X+ \Vert x_2\Vert _X)$.

A quasi-linear map $\Omega$ as above defines a twisted sum $Y\oplus _\Omega X= \{(\beta ,x)\in \Sigma \times X: \beta -\Omega (x)\in Y\}$ endowed with the quasinorm $\Vert (\beta ,x)\Vert _\Omega =\Vert \beta - \Omega (x)\Vert _Y+\Vert x\Vert _X$; the embedding $\jmath : Y\longrightarrow Y\oplus _\Omega X$ given by $j(y)=(y,0)$ is isometric and the quotient map $\pi : Y\oplus _\Omega X\longrightarrow X$ is given by $\pi (\beta ,x)=x$. They define the exact sequence , that shall be referred to as the exact sequence generated by $\Omega$. Since X and Y are complete, $(Y\oplus _\Omega X,\Vert (\cdot , \cdot ) \Vert _{\Omega })$ is a quasi-Banach space [14, Lemma 1.5.b]. When Y and X are B-convex Banach spaces, the quasi-norm in $Y\oplus _\Omega X$ is equivalent to a norm [20, Theorem 2.6]. This is the case for the spaces we consider in this paper.

Definition 2.2

A quasi-linear map $\Omega :X\curvearrowright Y$ with ambient space $\Sigma$ is bounded if there exists a constant D so that $\Omega x \in Y$ and $\Vert \Omega x\Vert _Y\le D\Vert x\Vert _X$ for each $x\in X$. It is trivial if there exists a linear map $L:X\longrightarrow \Sigma$ so that $\Omega - L: X\longrightarrow Y$ is bounded. Two quasilinear maps $\Omega _1, \Omega _2$ $X \curvearrowright Y$ with ambient space $\Sigma$ are boundedly equivalent if $\Omega _1 -\Omega _2: X \rightarrow Y$ is bounded. This implies that $\Vert (\cdot , \cdot )\Vert _{\Omega _1}$ and $\Vert (\cdot ,\cdot )\Vert _{\Omega _2}$ are equivalent quasi-norms. The quasilinear maps $\Omega _1: X_1\curvearrowright Y_1$ and $\Omega _2: X_2\curvearrowright Y_2$ are isomorphically equivalent, denoted $\Omega _1\simeq \Omega _2$, if there exist three isomorphisms S, T, U forming a commutative diagram

(1)

The following notions of domain and range generalize the classical domain and range for $\Omega$-operators obtained from an interpolation process [8, 17, 18], for centralizers on function spaces [4] or for G-centralizers in suitable G-Banach spaces [12].

Definition 2.3

Let $\Omega : X\curvearrowright Y$ be a quasi-linear map with ambient space $\Sigma$. The domain of $\Omega$ is the set $\textrm{Dom}\,\Omega = \{ x\in X: \Omega x \in Y\}$, and the range of $\Omega$ is the set $\textrm{Ran}\,\Omega = \{\beta \in \Sigma :\exists x\in X: \beta -\Omega x\in Y\}$.

Since $\Omega$ is quasi-linear, $\textrm{Dom}\,\Omega$ is a linear subspace of X as well as $\textrm{Ran}\,\Omega$. The space $\textrm{Dom}\,\omega$ can be endowed with the quasi-norm $\Vert x\Vert _D = \Vert \Omega x \Vert + \Vert x\Vert$ so that it is isometric to the subspace $\{ (0, x)\in Y\oplus _\Omega X$. The space $\textrm{Ran}\,\Omega$ can be endowed with the quasi-norm $\Vert \beta \Vert _R = \inf \{\Vert \beta -\Omega x\Vert +\Vert x\Vert \}$ where the infimum is taken over all $x\in X: \beta -\Omega x \in Y$. In this way $\textrm{Ran}\,\Omega$ can be identified with the quotient $(Y\oplus _\Omega X)/\textrm{Dom}\,\Omega$ with quotient map $(\beta , x) \rightarrow \beta$. What is not guaranteed is that either $\textrm{Dom}\,\Omega$ is a closed subspace of $Y\oplus _\Omega X$ or, equivalently, that $\textrm{Ran}\,\Omega$ is Hausdorff. Now, if $\Omega : X\rightarrow \Sigma$ is continuous at 0 for some choice of the ambient space $\Sigma$ then $\textrm{Dom}\,\Omega$ is closed. Indeed, if $(0,x_n)\rightarrow (z, x)$ then $\Vert z- \Omega (x_n-x)\Vert + \Vert x_n-x\Vert \rightarrow 0$. Thus $x_n-x\rightarrow 0$ and, by continuity, $\Omega (x_n-x)\rightarrow 0$ in $\Sigma$; thus $\Vert z\Vert _\Sigma \le \Vert z- \Omega (x_n-x)\Vert _\Sigma + \Vert \Omega (x_n-x)\Vert _\Sigma \rightarrow 0$, which means that $\Vert z\Vert _\Sigma =0$ and, by Hausdorffness, $z=0$. In fact, if $\textrm{Ran}\,\Omega$ is Hausdorff then we could choose it as ambient space: the formal identity establishes a continuous inclusion $Y\rightarrow \textrm{Ran}\,\Omega$ since $\Vert y\Vert _R\le \Vert y\Vert$ (with the choice $x=0$) and $\Omega : X\rightarrow \textrm{Ran}\,\omega$.

Given an interpolation pair of Banach spaces $(X_0,X_1)$ with ambient space $\Sigma$ and associated Calderón space ${\mathcal {C}}$, we fix the following terminology:

Operator acting on the pair:

An operator $T:\Sigma \rightarrow \Sigma$ is said to act on the pair $(X_0,X_1)$ if $T[X_i]\subset X_i$ for $i=0,1$

Interpolator:

An operator $\Delta : {\mathcal {C}} \rightarrow \Sigma$ is an interpolator if every T acting on the pair admits an operator $T_{{\mathcal {C}}}: {\mathcal {C}}\rightarrow {\mathcal {C}}$ such that $\Delta T_{{\mathcal {C}}} = T \Delta$.

Consistent family of interpolators:

A family $\{\Delta _i: i\in I\}$ of interpolators on ${\mathcal {C}}$ is said to be consistent if for each operator T acting on the pair $(X_0,X_1)$ there exists an operator $T_{{\mathcal {C}}}$ on $\mathcal C$ such that $T \Delta _i= \Delta _i T_{{\mathcal {C}}}$ for every $i\in I$.

Given a finite sequence $\{\Delta _i: i=0,\ldots ,n+k\}$ of interpolators we will consider the pair $(\Psi ,\Phi )$ of interpolators $\Psi =\langle \Delta _{k+n-1},\ldots \Delta _{k}\rangle : {\mathcal {C}}\rightarrow \Sigma ^n$ and $\Phi =\langle \Delta _{k-1},\ldots \Delta _0 \rangle :{\mathcal {C}} \rightarrow \Sigma ^k,$ given by $\Psi (f) = (\Delta _{k+n-1} f,\ldots \Delta _{k} f)$ and $\Phi (f)=(\Delta _{k-1} f,\ldots \Delta _0 f)$. Proceeding in the standard way, see [7] and [11], we obtain the following commutative diagram with exact rows and columns:

(2)

in which $X_\Phi = \Phi ({\mathcal {C}})$, $X_\Psi = \Psi ({\mathcal {C}})$, $X_{\langle \Psi ,\Phi \rangle }= \langle \Psi ,\Phi \rangle (\mathcal C)$ and all the spaces are endowed with their natural quotient norms. The maps $\imath$ and $\rho$ are defined by $\imath \Psi g=(\Psi g,0)$ and $\rho (\Psi f,\Phi f)=\Phi f$. If $B_\Phi : X_\Phi \rightarrow {\mathcal {C}}$ denotes an homogeneous bounded selection for the quotient map $\Phi :{\mathcal {C}}\rightarrow X_\Phi$ then the differential associated to $(\Psi ,\Phi )$ is the map $\Omega _{\Psi ,\Phi }: X_\Phi \rightarrow \Sigma ^n$ given by $\Omega _{\Psi ,\Phi } = \Psi \circ B_\Phi$. We have that $\Omega _{\Psi ,\Phi }: X_\Phi \curvearrowright \Psi (\ker \Phi )$ is a quasilinear map with ambient space $\Sigma ^n$. The differential $\Omega _{\Psi , \Phi }$ is continuous at 0 and, consequently, the domain of $\Omega _{\Psi ,\Phi }$ is closed, its range is Hausdorff, and one also has the inverse exact sequence . Moreover [11, Proposition 3.8]:

Proposition 2.4

The following identities, with equivalence of norms in (1) and (2), hold:

• (1) $\textrm{Dom}\,\Omega _{\Psi ,\Phi } =\Phi (\ker \Psi )$.

• (2) $\textrm{Ran}\,\Omega _{\Psi ,\Phi } = X_\Psi$.

• (3) $\Omega _{\Phi ,\Psi } = (\Omega _{\Psi ,\Phi })^{-1}$.

From now on we will focus on the pair $(\ell _\infty , \ell _1)$ and the sequence of interpolators $\Delta _k: {\mathcal {C}}\rightarrow \ell _\infty$ given by $\Delta _k(f)=f^{(k)}(1/2)/k!$. These are interpolators because the evaluation map of the $n^{th}$- derivative $\delta ^{(n)}_z:{\mathcal {C}}\rightarrow \ell _\infty$ at an interior z is continuous [6, Lemma 2.4] for each $n\in {\mathbb {N}}$. Moreover, each finite sequence $\{\Delta _k: n\le k \le m\}$ is consistent. More specifically, we will focus on diagram (2) obtained from the first three interpolators $\Delta _2, \Delta _1, \Delta _0$. There are six possible permutations of these interpolators, and therefore six different diagrams.

2.1 The Diagram [abc].

Let (a, b, c) be a permutation of (0, 1, 2). Observe that $\ker \langle \Delta _b, \Delta _c\rangle = \ker \Delta _b \cap \ker \Delta _c$. We denote by [abc] the diagram generated by the triple $(\Delta _a,\Delta _b,\Delta _c)$:

where the maps are given by

• $j(\Delta _a h)=(\Delta _a h,0)$,   $k(\Delta _a h)=(\Delta _a h,0,0)$,   $h\in \ker \Delta _b\cap \ker \Delta _c$;

• $l(\Delta _a g,\Delta _b g)=(\Delta _a g,\Delta _b g,0)$,   $q(\Delta _a g,\Delta _b g)=\Delta _b g$,   $i(\Delta _b g)=(\Delta _b g,0)$,   $g\in \ker \Delta _c$;

• $s(\Delta _a f,\Delta _b f,\Delta _c f)=\Delta _c f$,  $r(\Delta _a f,\Delta _b f,\Delta _c f)=(\Delta _b f,\Delta _c f)$,  $p(\Delta _b f,\Delta _c f)=\Delta _c f$,   $f\in {\mathcal {C}}$.

2.2 The Quasi-Linear Maps

We simplify the notation for the quasi-linear maps as follows:

$$\begin{aligned} \Omega _{a,b}=\Omega _{\Delta _a,\Delta _b}; \quad \Omega _{a,\langle b,c\rangle }= \Omega _{\Delta _a,\langle \Delta _b, \Delta _c\rangle } \quad \text {and}\quad \Omega _{\langle a,b\rangle ,c}= \Omega _{\langle \Delta _a, \Delta _b \rangle ,\Delta _c}. \end{aligned}$$

It follows from Proposition 2.4 that

• (1) the central column of [abc] is generated by $\Omega _{a,\langle b,c\rangle }$,

• (2) the central row of [abc] is generated by $\Omega _{\langle a,b\rangle ,c}$,

• (3) the lower row of [abc] is generated by $q\circ \Omega _{\langle a,b\rangle ,c}\simeq \Omega _{b,c}$, since $q\circ \langle \Delta _a,\Delta _b\rangle = \Delta _b$.

• (4) the left column of [abc] is generated by $\Omega _{a,\langle b,c\rangle }\circ i$.

2.3 Elementary Symmetries

The following equivalences are obvious, or can be derived from Proposition 2.4:

$$\begin{aligned} \Omega _{\langle b,c\rangle ,a}\simeq & {} \Omega _{\langle c,b\rangle ,a}, \quad \Omega _{a,\langle b,c\rangle } \simeq \Omega _{a,\langle c,b\rangle }\\ (\Omega _{a,\langle b,c\rangle })^{-1}\simeq & {} \Omega _{\langle b,c \rangle ,a}, \quad (\Omega _{\langle a,b\rangle ,c})^{-1} \simeq \Omega _{c,\langle a,b\rangle }, \quad (\Omega _{a,b})^{-1}\simeq \Omega _{b,a}. \end{aligned}$$

3 Determination of the Spaces in the Diagrams

We will show that the six diagrams [abc] corresponding to the permutations of (0, 1, 2) can be drawn (with equivalence of norms) with the self-dual spaces ${\mathfrak {R}}_1= \Delta _0({\mathcal {C}})= \ell _2$; ${\mathfrak {R}}_2 = \langle \Delta _1, \Delta _0\rangle (\mathcal C)= Z_2$, [7, 24] and ${\mathfrak {R}}_3= \langle \Delta _2, \Delta _1, \Delta _0\rangle ({\mathcal {C}})$ from now on denoted $Z_3$; the Orlicz spaces $\ell _f$ and $\ell _g$ and their duals, and the new spaces $\wedge$ and $\bigcirc$ and their duals. The properties of these spaces will be considered in Sect. 5. We begin showing that the spaces in the diagrams admit symmmetric Schauder decompositions and bases:

Proposition 3.1

The unit vector basis $(e_n)$ is a symmetric basis for the three Banach spaces $\Delta _c ({\mathcal {C}})$, $\Delta _b(\ker \Delta _c)$ and $\Delta _a(\ker \Delta _b\cap \ker \Delta _c)$. Similarly, $\langle \Delta _a,\Delta _b\rangle (\ker \Delta _c)$ and $\langle \Delta _a,\Delta _b \rangle ({\mathcal {C}})$ have a symmetric two-dimensional decomposition and $\langle \Delta _a,\Delta _b,\Delta _c\rangle ({\mathcal {C}})$ has a symmetric three-dimensional decomposition. Moreover, all the spaces in the diagrams admit a basis.

Proof

Observe that since the family $\{\Delta _{n+k}, \ldots , \Delta _1\}$ is consistent, given an operator $T:\Sigma \rightarrow \Sigma$ acting on the pair the induced operator $T (\Delta _k f)= \Delta _k (T_{\mathcal C}f)$ defines an operator $\tau _k$ on $X_{\Delta _k}= \Delta _k({\mathcal {C}})$ in the form $\tau _k(\Delta _k f) = \Delta _k (T_{{\mathcal {C}}}f)$: Indeed, if $B_k$ is a homogeneous bounded selection for $\Delta _k$ then $\Vert \tau _k(\Delta _k f)\Vert _{X_{\Delta _k}} = \Vert \tau _k(\Delta _k B_k \Delta _k f)\Vert _{X_{\Delta _k}} = \Vert \Delta _k(T_{{\mathcal {C}}} B_k \Delta _k f)\Vert _{X_{\Delta _k}} \le \Vert \Delta _k\Vert \Vert T_{{\mathcal {C}}}\Vert \Vert B_k\Vert \Vert \Delta _k f\Vert _{X_{\Delta _k}}$. Let now X be any of the first three spaces in the statement and let $P_n$ denote the natural projection onto the subspace generated by $\{e_1,\ldots , e_n\}$. Since $P_n$ is a norm-one operator on $\ell _\infty$ and $\ell _1$, $(P_n)$ is a bounded sequence of operators on X by the argument above. Clearly $(e_n)$ is contained in X and generates a dense subspace. Since for each $x\in \text {span}\{e_n: n\in \mathbb {N}\}$, $P_n x$ converges to x in X, it does the same for each $x\in X$. Thus $(e_n)$ is a Schauder basis for X, and considering the operators associated to permutations of the basis. The argument at the beginning of the proof shows that the basis is symmetric. The remaining results on FDD’s are proved in a similar way, using the operators induced by $P_n$ in each of the spaces.

All the spaces have a basis because if $(E_n)$ is a FDD for X with FDD-constant K and each $E_n$ has a basis $(x^n_i)_{i=1}^{k_n}$ with basis constant $\le M$ then $\left( (x^n_i)_{i=1}^{k_n}\right) _{n=1}^\infty$ is a basis for X with basis constant $\le KM$ [9, Proposition 6.5].

$\square$

The next result shows that some of the spaces in the diagrams coincide. Note that algebraic equality implies isomorphism because if $\tau _1: X_1\rightarrow Y$ and $\tau _2: X_2\rightarrow Y$ are operators between Banach spaces with $\tau _1(X_1)= \tau _2(X_2)$ then the quotients $X_1/\ker \tau _1$ and $X_2/\ker \tau _2$ are isomorphic: if $T_i: X_i/\ker \tau _i\rightarrow Y$ denotes the injective operator induced by $\tau _i$ then $T_2^{-1}\circ T_1: X_1/\ker \tau _1\rightarrow X_2/\ker \tau _2$ is a closed bijective operator, which is continuous by the closed graph theorem.

Proposition 3.2

The following equalities hold:

• (1) $\Delta _2(\ker \Delta _1\cap \ker \Delta _0)= \Delta _1(\ker \Delta _0)= \Delta _0({\mathcal {C}})$,

• (2) $\langle \Delta _2, \Delta _1\rangle (\ker \Delta _0)=\langle \Delta _1, \Delta _0\rangle ({\mathcal {C}})$,

• (3) $\Delta _1(\ker \langle \Delta _0, \Delta _2 \rangle )=\Delta _0(\ker \Delta _1)$.

Proof

Let $\varphi :\mathbb {S}\rightarrow \mathbb {D}$ be a conformal equivalence such that $\varphi (1/2) =0$. Since $\varphi '(1/2)\ne 0$, we can define $\phi = \varphi '(1/2)^{-1}\varphi$. (1) For each $g\in \ker \Delta _0$ there is $f\in {\mathcal {C}}$ such that $g=\phi \cdot f$, hence $\Delta _1 g= \Delta _0 f$, and we get $\Delta _1(\ker \Delta _0)\subset \Delta _0 ({\mathcal {C}})$. Conversely, if $f\in {\mathcal {C}}$ then $g=\phi \cdot f\in \ker \Delta _0$ and $\Delta _0 f= \Delta _1 g$, so the second equality is proved. The first equality can be proved in a similar way. It was proved in [7, Theorem 4] that $j(x_1,x_0)= (x_1,x_0,0)$ and $q(y_2,y_1,y_0)= y_0$ define an exact sequence

and (2) follows from $\langle \Delta _2, \Delta _1, \Delta _0 \rangle (\ker \Delta _0)= \ker q$ and $\langle \Delta _1, \Delta _0,0 \rangle ({\mathcal {C}}) = \textrm{Im}\,j$. (3) Note that $y\in \Delta _0 (\ker \Delta _1)$ if and only if $(0,y)\in \langle \Delta _1,\Delta _0 \rangle (\mathcal C)= \langle \Delta _2, \Delta _1 \rangle (\ker \Delta _0)$; equivalently, $y\in \Delta _1(\ker \Delta _0\cap \ker \Delta _2)= \Delta _1(\ker \langle \Delta _0, \Delta _2 \rangle )$.$\square$

Next we identify the corner spaces as Orlicz sequence spaces. Let us consider the Orlicz functions $f(t)=t^2(\log t)^2$ and $g(t)=t^2 (\log t)^4$.

Proposition 3.3

$\Delta _0(\ker \Delta _1)=\ell _f$ and $\Delta _0(\ker \Delta _1\cap \ker \Delta _2)=\ell _g$.

Proof

The first equality was essentially proved in [24, Lemma 5.3]. With our notation,

$$\begin{aligned} \Delta _0(\ker \Delta _1)= \textrm{Dom}\,\Omega _{1,0}=\{x\in \ell _2: \Omega _{1,0}x\in \ell _2\} \end{aligned}$$

and $\Omega _{1,0}:\ell _2\rightarrow \ell _\infty$ is given by $\Omega _{1,0}= 2x\log (|x|/\Vert x\Vert _2)$. Thus

$$\begin{aligned} \Delta _0(\ker \Delta _1)= \{x\in \ell _2: x\log |x|\in \ell _2\}=\ell _f. \end{aligned}$$

Similarly, since $\Delta _0 (\ker \Delta _1 \cap \ker \Delta _2)= \textrm{Dom}\,\Omega _{\langle 2,1 \rangle , 0}$ and $\Omega _{\langle 2,1 \rangle , 0}:\ell _2\rightarrow \ell _\infty \times \ell _\infty$ is given by

$$\begin{aligned} \Omega _{\langle 2,1\rangle ,0}x= \left( 2x \log ^2\frac{|x|}{\Vert x\Vert _2}, 2x\log \frac{|x|}{\Vert x\Vert _2}\right) \end{aligned}$$

(see [7]), we have $\Delta _0 (\ker \Delta _1 \cap \ker \Delta _2) = \{x\in \ell _2: (2x\log ^2|x|,2x\log |x|)\in Z_2\}$. Therefore $x\in \Delta _0 (\ker \Delta _1 \cap \ker \Delta _2)$ if and only if $x\in \ell _2$, $2x\log |x|\in \ell _2$ and

$$\begin{aligned} 2x\log ^2|x|-\Omega _{1,0}(2x\log |x|)= 2x\log ^2|x|- 4x\log |x| \log \frac{|x\log |x||}{\Vert 2x\log |x|\Vert _2} \in \ell _2. \end{aligned}$$

Since $\log |x\log |x||=\log |x|+ \log |\log |x||$, we conclude that $\Delta _0 (\ker \Delta _1 \cap \ker \Delta _2)= \{x\in \ell _2: x\log ^2|x| \in \ell _2\} =\ell _g$. $\square$

The second equality in the following result appears observed in [4, Example after Corollary 3].

Proposition 3.4

$\Delta _2(\ker \Delta _0)= \Delta _1({\mathcal {C}}) = \ell _f^*$.

Proof

For the first equality, $\langle \Delta _1,\Delta _0 \rangle (\mathcal C)= \langle \Delta _2, \Delta _1\rangle (\ker \Delta _0)$ by Proposition 3.2. Thus

$$\begin{aligned} x\in \Delta _1({\mathcal {C}})\Leftrightarrow & {} (x,f(1/2))=(f'(1/2),f(1/2)) \text { for some } f\in {\mathcal {C}}\\\Leftrightarrow & {} (x,g'(1/2))=(g''(1/2),g'(1/2)) \text { for some } g\in \ker \Delta _0\\\Leftrightarrow & {} x\in \Delta _2(\ker \Delta _0). \end{aligned}$$

For the second equality, since $Z_2 = \langle \Delta _1, \Delta _0\rangle ({\mathcal {C}})$, we have a natural exact sequence

(3)

with $i(x) =(0,x)$ and $p(y,x)=y$. Moreover (see [24, Section 5]), the expression $\langle U_2(y,x),(b,a)\rangle = \langle -x,b\rangle + \langle y,a\rangle$ defines a bijective isomorphism $U_2:Z_2\rightarrow Z_2^*$, where $\langle \cdot ,\cdot \rangle$ denotes the Riesz product. Since $i^*U_2= p$, we get $\Delta _1({\mathcal {C}})= \ell _f^*$.

$\square$

The following three results were unexpected for us since, at first glance, the first two spaces seem to be incomparable.

Proposition 3.5

$\Delta _0(\ker \Delta _2)=\Delta _0(\ker \Delta _1) =\ell _f$.

Proof

The second equality is proved in Proposition 3.3. Moreover, the map $\Omega _{2,0}:\ell _2\rightarrow \ell _\infty$ is given by $\Omega _{2,0}= 2x\log ^2 (|x|/\Vert x\Vert )$. Thus

$$\begin{aligned} \Delta _0(\ker \Delta _2)= \textrm{Dom}\,\Omega _{2,0} =\{x\in \ell _2:x\log ^2|x|\in \Delta _2(\ker \Delta _0)= \ell _f^* \}. \end{aligned}$$

Since $\ell _f= \{x\in \ell _2: x\log |x|\in \ell _2\}$, $\ell _f^*= \{x\in \ell _\infty : x\log ^{-1} |x|\in \ell _2\}$ [27, Example 4.c.1]. Then

$$\begin{aligned} x\in \Delta _0(\ker \Delta _2)\Leftrightarrow x\in \ell _2\; \text { and }\; \frac{x\log ^2|x|}{\log (|x|\log ^2|x|)}= \frac{x\log ^2|x|}{\log |x| +2\log |\log x|} \in \ell _2. \end{aligned}$$

Thus $x\in \Delta _0(\ker \Delta _2)$ if and only if $x\log |x|\in \ell _2$; equivalently $x\in \ell _f$. $\square$

Like in the proof of Proposition 3.4, it was proved in [6, Proposition 5.1] that the expression

$$\begin{aligned} \langle U_3(x_2,x_1,x_0),(y_2,y_1,y_0) \rangle = \langle x_0,y_2\rangle -\langle x_1,y_1\rangle +\langle x_2,y_0\rangle \end{aligned}$$

defines a bijective isomorphism $U_3:Z_3\rightarrow Z_3^*$ given by $U_3(x_2,x_1,x_0)= (x_0,-x_1,x_2)$. This fact will be a tool to prove the next result.

Proposition 3.6

$\Delta _2(\ker \Delta _1)=\Delta _2(\ker \Delta _0) =\ell _f^*$.

Proof

The second equality is proved in Proposition 3.4, and we derive the first equality from Proposition 3.5 by constructing an isomorphism from $\Delta _2(\ker \Delta _1)$ onto $\Delta _0(\ker \Delta _2)^*$ that takes $e_n$ to $e_n$ for every $n\in \mathbb {N}$. Recall that if M and N are closed subspaces of X with $N\subset M$ then $(M/N)^* \simeq N^\perp /M^\perp$. Thus, with the natural identifications we get

$$\begin{aligned} \Delta _0(\ker \Delta _2)\simeq \frac{\langle \Delta _1,\Delta _0 \rangle (\ker \Delta _2)}{\Delta _1(\ker \Delta _0 \cap \ker \Delta _2)}\; \Longrightarrow \; \Delta _0(\ker \Delta _2)^* \simeq \frac{\left( \Delta _1(\ker \Delta _0\cap \ker \Delta _2) \right) ^\perp }{\left( \langle \Delta _1, \Delta _0\rangle (\ker \Delta _2) \right) ^\perp } \end{aligned}$$

and

$$\begin{aligned} \Delta _2(\ker \Delta _1)\simeq \frac{\langle \Delta _0,\Delta _2\rangle (\ker \Delta _1)}{\Delta _0(\ker \Delta _2 \cap \ker \Delta _1)}, \end{aligned}$$

and we conclude that $U_3$ induces an isomorphism from $\Delta _2 (\ker \Delta _1)$ onto $\Delta _0(\ker \Delta _2)^*$ by showing that $U_3$ takes $\langle \Delta _0, \Delta _2 \rangle (\ker \Delta _1)$ onto $\left( \Delta _1 (\ker \Delta _0\cap \ker \Delta _2) \right) ^\perp$ and $\Delta _0(\ker \Delta _2 \cap \ker \Delta _1)$ onto $\left( \langle \Delta _1,\Delta _0 \rangle (\ker \Delta _2) \right) ^\perp$. Indeed, $\Delta _1 (\ker \Delta _0\cap \ker \Delta _2)$ can be identified with the subspace of the vectors (0, y, 0) in $Z_3$. Then $\left( \Delta _1 (\ker \right. \left. \Delta _0\cap \ker \Delta _2) \right) ^\perp$ is the subspace of the vectors (x, 0, z) in $Z_3^*$, which coincides with $U_3\left( \langle \Delta _0, \Delta _2\rangle (\ker \Delta _1)\right)$, and similarly $\langle \Delta _1, \Delta _0 \rangle (\ker \Delta _2)^\perp = U_3\left( \Delta _0(\ker \Delta _2\right. \left. \cap \ker \Delta _1)\right)$, and it is clear that the induced isomorphism takes $e_n$ to $e_n$ for every $n\in \mathbb {N}$. $\square$

What follows is perhaps the most surprising symmetry:

Proposition 3.7

$\Delta _1(\ker \Delta _2)=\Delta _1 (\ker \Delta _0) =\ell _2$.

Proof

Proposition 3.5 implies $\ker \Delta _0 +\ker \Delta _1 = \ker \Delta _0+ \ker \Delta _2$, from which we get

$$\begin{aligned} \Delta _1 (\ker \Delta _0)=\Delta _1 (\ker \Delta _0 + \ker \Delta _2)\supset \Delta _1 (\ker \Delta _2), \end{aligned}$$

while Proposition 3.6 implies $\ker \Delta _2 +\ker \Delta _0 = \ker \Delta _2+ \ker \Delta _1$. Thus

$$\begin{aligned} \Delta _1 (\ker \Delta _2)=\Delta _1 (\ker \Delta _2+ \ker \Delta _0)\supset \Delta _1 (\ker \Delta _0). \end{aligned}$$

$\square$

A rich theory [1, 30], see also [5, Section 10.8], contemplates $Z_2$ as a Fenchel-Orlicz space, with the meaning described next. A function $\varphi : \mathbb {C}^n \rightarrow [0, \infty )$ is a (quasi) Young function if it is (quasi) convex, $\varphi (0) = 0$, $\lim _{t \rightarrow \infty } \varphi (tx) = \infty$ and $\varphi (z x) = \varphi (x)$ for every $z \in \mathbb {C}$ with $|z|=1$ and every $x \ne 0$. If we call two positive functions $\phi , \psi$ equivalent when $\phi /\psi$ is both upper and lower bounded, a quasi-convex function $\phi$ on $\mathbb {R}^n$ is equivalent to its convex hull $co\phi (x) = \inf \{ \sum \theta _i \phi (x_i): x=\sum \theta _i x_i, \sum \theta _i=1, \theta _i\ge 0\}$. A Young function $\varphi$ generates the Fenchel-Orlicz space

$$\begin{aligned} \ell _{\varphi } = \left\{ (x^j)_{j \ge 1} \subset \mathbb {C}^n: \exists \rho > 0 \text { such that } \sum \varphi \Big (\frac{1}{\rho } x^j\Big ) < \infty \right\} \end{aligned}$$

endowed with the norm $\Vert (x^j)_{j \ge 1}\Vert _{\varphi } = \inf \{\rho > 0: \sum \varphi (\frac{1}{\rho } x^j) \le 1\}$. The case $n = 1$ correspond to Orlicz spaces. We will say that a quasi-Young function $\phi$ generates the Fenchel-Orlicz space $\ell _{\varphi }$ when $co \phi$ is equivalent to $\varphi$.

The Rochberg spaces associated to the scale of $\ell _p$-spaces are Fenchel-Orlicz spaces in a natural way (see [16]). Indeed, given $\theta \in (0,1)$ and $n \ge 2$ there is a Young function $\varphi _n: \mathbb {C}^{n} \rightarrow [0, \infty )$ such that $Z_n=\ell _{\varphi _n}$. More precisely:

• $\ell _2$ is $\ell _{\varphi _1}$, the Orlicz space generated by the Orlicz function $\varphi _1(x)= |x_0|^2.$

• $Z_2$ is $\ell _{\varphi _2}$, the Fenchel-Orlicz space generated by the quasi-Young function

  $$\begin{aligned} \varphi _2(x_1,x_0)= |x_1 - x_0\log |x_0||^2 + |x_0|^2. \end{aligned}$$

  Keep track that $\varphi _2(x_1,0) = |x_1 |^2$, so $\ell _2 = \{(x, y) \in \ell _{\varphi _2}: y = 0\}$; while $\varphi _2(0,x_0) = |x_0\log |x_0||^2 + |x_0|^2 \sim f$, so $\ell _{f} = \textrm{Dom}\,\textsf{K}\hspace{-1pt}\textsf{P}= \{(x, y) \in \ell _{\varphi _2}: x = 0\}$.

• $Z_3$ is $\ell _{\varphi _3}$, the Fenchel-Orlicz space generated by the quasi-Young function

  $$\begin{aligned} \varphi _3(x_2,x_1,x_0) = \varphi _2(x_1, x_0) + \varphi _1(x_2 - g_{(x_1, x_0)}[2]) \end{aligned}$$

  where f[i] stands for $\frac{f^{(i)}(1/2)}{i!}$ and $g_x(z)= |x|^{2z-1}x$, so that $g_x[1]= 2x\log |x|$. Now, we set $g_{(x_1, x_0)} = g_{x_0} + \frac{\varphi }{k_2} g_{x_1 - g_{x_0}[1]}$, with $\varphi : \mathbb {S} \rightarrow \mathbb {D}$ a conformal map such that $\varphi (\frac{1}{2}) = 0$ and $k_2$ is adjusted so that $g_{(x_1,x_0)}[1]=x_1$. One therefore has $g_{(x_1, x_0)}(z) = g_{x_0}(z) + \frac{\varphi (z)}{k_2} g_{x_1 - 2x_0\log |x_0|}(z)$ $= |x_0|^{2z-1}x_0 + \frac{\varphi (z)}{k_2} |x_1 - 2x_0\log |x_0||^{2z-1}(x_1 - 2x_0\log |x_0|)$ to get, after a few tedious computations,

  $$\begin{aligned} g_{(x_1, x_0)}[2]= & {} 2x_0\log ^2|x_0| +\frac{\varphi '(1/2)}{k_2} 2(x_1 - 2x_0\log |x_0|) \log (\left| x_1 - 2x_0\log |x_0|\right| )\\{} & {} + \frac{\varphi ''(1/2)}{2k_2} (x_1 - 2x_0\log |x_0|). \end{aligned}$$

• $\wedge$ is generated by $\varphi _3(x_2, 0, x_0) = \varphi _2(0, x_0) + |x_2 - g_{(0, x_0)}[2]|^2$.

• $\bigcirc$ is generated by $\varphi _3(0, x_1, x_0) = \varphi _2 (x_1, x_0) + |g_{(x_1, x_0)}[2]|^2$.

4 Construction of the Diagrams

As we said before, $\langle \Delta _a, \Delta _b, \Delta _c\rangle ({\mathcal {C}})\simeq Z_3$ for each permutation (a, b, c) of (2, 1, 0).

Diagram [210]: By Proposition 3.2, $\Delta _2(\ker \Delta _1 \cap \ker \Delta _0)=\Delta _1(\ker \Delta _0)= \Delta _0({\mathcal {C}})\simeq \ell _2$ and $\langle \Delta _2,\Delta _1\rangle (\ker \Delta _0)= \langle \Delta _1,\Delta _0\rangle ({\mathcal {C}})\simeq Z_2$. We thus get

The two quasilinear maps generating the two middle sequences are $\Omega _{\langle 2,1\rangle ,0}$ and $\Omega _{2, \langle 1,0\rangle }$; both can be found explicitly in [7] (and implicit in [29]) and also at the appropriate places in this paper.

Diagram [012]: By Propositions 3.3 and 3.7, $\Delta _0 (\ker \Delta _1 \cap \ker \Delta _2)=\ell _g$ and $\Delta _1(\ker \Delta _2)=\ell _2$. So we have the spaces in the left column. The next result provides the spaces in the lower row.

Proposition 4.1

(a) $\Delta _2({\mathcal {C}})$ is isomorphic to $\ell _g^*$. (b) $\langle \Delta _1, \Delta _2\rangle ({\mathcal {C}})$ is isomorphic to $\bigcirc ^*$.

Proof

(a) By Proposition 3.3, $\ell _g=\Delta _0 (\ker \Delta _1 \cap \ker \Delta _2)$ which is isomorphic to a closed subspace of $Z_3$, namely $\{(x_2,x_1,x_0)\in Z_3: x_2=x_1=0\}$. Hence $\ell _g^*\simeq Z_3^*/\left( \Delta _0(\ker \Delta _1 \cap \ker \Delta _2) \right) ^\perp$. Since $\left( \Delta _0(\ker \Delta _1 \cap \ker \Delta _2) \right) ^\perp = U_3\left( \langle \Delta _0, \Delta _1\rangle (\ker \Delta _2) \right)$ then

$$\begin{aligned} \Delta _2(\mathcal {C})\simeq \frac{Z_3}{\langle \Delta _0, \Delta _1 \rangle (\ker \Delta _2)}\simeq \ell _g^*. \end{aligned}$$

(b) The space $\bigcirc =\langle \Delta _0, \Delta _1\rangle (\ker \Delta _2)$ is isomorphic to $\{(x_2,x_1,x_0)\in Z_3: x_2=0\}$, a closed subspace of $Z_3$. Hence $\bigcirc ^*\simeq Z_3^*/\left( \langle \Delta _0, \Delta _1\rangle (\ker \Delta _2) \right) ^\perp$. Since

$$\begin{aligned} \left( \langle \Delta _0, \Delta _1 \rangle (\ker \Delta _2) \right) ^\perp = U_3\left( \Delta _0 (\ker \Delta _1 \cap \ker \Delta _2)\right) \end{aligned}$$

we get $\langle \Delta _1, \Delta _2\rangle ({\mathcal {C}})\simeq Z_3/ (\Delta _0(\ker \Delta _1 \cap \ker \Delta _2))\simeq \bigcirc ^*$. $\square$

Thus we obtain the diagram:

Diagram [201]: $\Omega _{2,\langle 0,1\rangle } \simeq \Omega _{2,\langle 1,0 \rangle }$ gives the central column (coincides with that of [210]), and Propositions 3.4 and 3.5 give the lower row. Thus, we get

Arguing as in the proof of Proposition 4.1, we get ($U_3$ appeared before Proposition 3.6):

Proposition 4.2

$\langle \Delta _2, \Delta _0\rangle (\mathcal {C})$ is isomorphic to $\wedge ^*=\langle \Delta _2,\Delta _0 \rangle (\ker \Delta _1)^*$.

Proof

Since the space $\wedge =\langle \Delta _2, \Delta _0\rangle (\ker \Delta _1)$ is isomorphic to a subspace of $Z_3$, we get $\wedge ^*\simeq Z_3^*/ \left( \langle \Delta _2,\Delta _0\rangle (\ker \Delta _1)\right) ^\perp$. Moreover $\left( \langle \Delta _2,\Delta _0 \rangle (\ker \Delta _1)\right) ^\perp = U_3 \left( \Delta _1(\ker \Delta _2\cap \ker \Delta _0)\right)$, and therefore $\langle \Delta _2,\Delta _0\rangle (\mathcal {C})\simeq Z_3/\left( \Delta _1(\ker \Delta _2\cap \right. \left. \ker \Delta _0)\right) \simeq \wedge ^*$. $\square$

Diagram [120]: $\Omega _{\langle 1,2\rangle ,0}\simeq \Omega _{\langle 2,1\rangle ,0}$ gives the central row, and $\Delta _1 (\ker \Delta _2 \cap \ker \Delta _0)= \ell _f$ and $\Delta _2(\ker \Delta _0)= \ell _f^*$ by Propositions 3.2, 3.3 and 3.6. Since $\wedge ^*\simeq \langle \Delta _2, \Delta _0 \rangle (\mathcal {C})$ by Proposition 4.2 and $\Delta _0(\mathcal {C}) =\ell _2$, we get

Diagram [021]: $\Omega _{0,\langle 2,1\rangle }\simeq \Omega _{0,\langle 1,2\rangle }$ gives the central column and $\Omega _{\langle 0,2\rangle ,1}\simeq \Omega _{\langle 2,0\rangle ,1}$ gives the central row. Since $\Delta _2(\ker \Delta _1)=\ell _f^*$ by Proposition 3.6, we get

Diagram [102]: $\Omega _{1,\langle 0,2\rangle }\simeq \Omega _{1,\langle 2,0\rangle }$ gives the central column, and $\Omega _{\langle 1,0\rangle ,2}\simeq \Omega _{\langle 0,1\rangle ,2}$ gives the central row. Moreover, $\Delta _0(\ker \Delta _2) \simeq \ell _f$ by Proposition 3.5. So we get

5 Properties of the Spaces

Here we describe some isomorphic properties of the spaces in the diagrams. Recall that a Banach space X is hereditarily $\ell _2$ if every closed infinite dimensional subspace of X contains a subspace isomorphic to $\ell _2$. Being hereditarily $\ell _2$ is inherited by subspaces, but not by quotients since every separable reflexive space is a quotient of a reflexive hereditarily $\ell _2$ space [2, Theorem 6.2]. To be hereditarily $\ell _2$ is a three-space property [14, Theorem 3.2.d].

Proposition 5.1

All the spaces appearing in the diagrams [abc] are hereditarily $\ell _2$.

Proof

Each infinite dimensional subspace of a reflexive Orlicz sequence space contains a copy of $\ell _p$ for $p\in [\alpha , \beta ]$, being $\alpha$ (resp. $\beta$) the lower (resp. upper) Boyd index of the space [26, Proposition I.4.3, Theorem I.4.6]. Since $Z_3$ has type $2-\varepsilon$ and cotype $2+\varepsilon$ for each $\varepsilon >0$, the same happens with $\ell _f$ and $\ell _g$ and their dual spaces, hence their Boyd indices are 2 and these spaces are hereditarily $\ell _2$. Apply the 3-space property for all the other spaces. $\square$

Recall from [25, Corollary 13] that if M is an Orlicz function satisfying the $\Delta _2$-condition and $2\le q <\infty$ then the space $\ell _M$ has cotype q if and only if there exists $K>0$ such that $M(tx)\ge Kt^qM(x)$ for all $0\le t,x\le 1$. Consequently, the spaces $\ell _f$ and $\ell _g$ have cotype 2 and $\ell _f^*$ and $\ell _g^*$ have type 2. We need one more technical result:

Proposition 5.2

Let X be a Banach space.

• (1) If X has type 2 then every subspace isomorphic to $\ell _2$ is complemented.

• (2) If X has an unconditional basis and cotype 2 then every subspace of X isomorphic to $\ell _2$ contains an infinite dimensional subspace complemented in X.

Proof

(a) is a consequence of Maurey’s extension theorem; see [19, Corollary 12.24]. (b) The following argument is similar to the proof of [28, Theorem 3.1] for subspaces of $L_p$, $1<p<2$, with an unconditional basis. Let $(e_n)$ be an unconditional basis of X, let $(x_k)$ be a normalized block basis of $(e_n)$, and take a sequence $(c_j)$ of scalars and a successive sequence $(B_k)$ of intervals of integers so that $x_k = \sum _{i\in B_k} c_i e_i$. We consider the sequence of projections $(P_k)$ in X defined by $P_ke_j=e_j$ if $j\in B_k$, and $P_ke_j=0$ otherwise. Let $Q_k$ be a norm-one projection on $\text {span}\{e_j: j\in B_k\}$ onto the one-dimensional subspace generated by $x_k$. We claim that $Px=\sum _{k=1}^\infty Q_kP_k x$ defines a projection on X onto the closed subspace generated by $(x_k)$. If $x\in X$ then $\sum _{k=1}^\infty P_k x$ is unconditionally converging and $\Vert \sum _{k=1}^\infty P_k x\Vert \le D\Vert x\Vert$ for some $D>0$. Moreover, since X has cotype 2, $\left( \sum _{k=1}^\infty \Vert P_k x\Vert ^2 \right) ^{1/2}\le E \Vert \sum _{k=1}^\infty P_k x\Vert$ for some $E>0$. We write $Q_kP_kx = s_k x_k$ for each k. Then

$$\begin{aligned} \left( \sum _{k=1}^\infty |s_k|^2 \right) ^{1/2} \le \left( \sum _{k=1}^\infty \Vert P_k x\Vert ^2 \right) ^{1/2}\le E\cdot D \Vert x\Vert . \end{aligned}$$

Hence $\sum _{k=1}^\infty Q_kP_k x$ converges, and it is easy to check that P is the required projection. $\square$

Corollary 5.3

Each infinite dimensional subspace of one of the spaces $\ell _f$, $\ell _g$, $\ell _f^*$ and $\ell _g^*$ contains a complemented copy of $\ell _2$.

Since $Z_2 \simeq Z_2^*$ [24], a space X is (isomorphic to) a subspace (resp. a quotient) of $Z_2$ if and only if $X^*$ is a quotient (resp. a subspace) of $Z_2$.

Proposition 5.4

None of the spaces $\bigcirc$, $\bigcirc ^*$, $\wedge$ and $\wedge ^*$ is (isomorphic to) a subspace or a quotient of $Z_2$.

Proof

It was proved in [24, Theorem 5.4] that every normalized basic sequence in $Z_2$ has a subsequence equivalent to the basis of one of the spaces $\ell _2$ or $\ell _f$. Thus none of the four spaces is a subspace of $Z_2$ because $\bigcirc$ and $\wedge$ contain a copy of $\ell _g$ and $\bigcirc ^*$ and $\wedge ^*$ contain a copy of $\ell _f^*$, as we can see in the diagrams.

$\square$

Next we extend to $Z_3$ some fundamental results about $Z_2$. The following one is in [21] for $Z_2$ and the proof we present is similar to that in [5, Proposition 10.9.1] for $Z_2$.

Proposition 5.5

An operator $\tau : Z_3\rightarrow X$ either is strictly singular or an isomorphism on a complemented copy of $Z_3$.

Proof

Since the quotient map in the sequence $0\rightarrow \ell _2\rightarrow Z_3 \rightarrow Z_2 \rightarrow 0$ is strictly singular (see [7]) an operator $\tau : Z_3 \rightarrow X$ is strictly singular if and only if $\tau |_{\ell _2}$ is strictly singular. So, let $\tau$ be a non-strictly singular operator. Let us assume first that $\tau |_{\ell _2}$ is an embedding so that we can assume that $\Vert \tau (y,0)\Vert \ge \Vert y\Vert$ for all $y\in \ell _2$. Observe the commutative diagram:

• The composition $Q\,(\tau ,\textbf{id})$ is strictly singular since it factors through $\pi$.

• $Q\,(\tau ,\textbf{id})=Q(\tau ,0)+Q\,(0,\textbf{id})$.

• $Q\,(0, \textbf{id})$ is an embedding since

  $$\begin{aligned} \Vert Q(0,z)\Vert= & {} \inf _{y\in \ell _2} \Vert (0,z) - (\tau , \imath )(y) \Vert = \inf _{y\in \ell _2} \Vert (-\tau y, z- y) \Vert \\= & {} \inf _{y\in \ell _2} \big \{\Vert \tau (y,0)\Vert + \Vert z-y\Vert \big \}\ge \Vert y\Vert + \Vert z\Vert - \Vert y\Vert = \Vert z\Vert . \end{aligned}$$

Thus, $Q(\tau ,0)$, being the difference (or sum) between a strictly singular operator and an embedding, has to have closed range and finite dimensional kernel [27, Proposition 2.c.10] and therefore it must be an isomorphism on some finite codimensional subspace of $Z_3$, and the same happens to $\tau$. All subspaces of $Z_3$ with codimension 3 are isomorphic to $Z_3$ and thus we are done.

In the general case, if $\tau$ is not strictly singular, then $\tau |_{U}$ is an embedding for some subspace U of $\ell _2$ generated by a normalized block basis $(u_n)$ of the canonical basis. We consider the operator $\tau _U: \ell _2 \rightarrow \ell _2$ given by $\tau _U(e_n)=u_n$, which acts on the pair. It was shown by Kalton [21] that the operator $S_U: Z_2 \longrightarrow Z_2$ defined by $S_U(e_n, 0)= (u_n, 0)$ and $S_U(0,e_n)=(\Omega _{1,0} u_n, u_n)$ is continuous and makes commutative the following diagram:

(4)

The operator $S_U$ can be described by the matrix $S_U = \left( \begin{array}{cc} u &{} 2u\log u \\ 0 &{} u \\ \end{array} \right)$. The theory developed in [12, Proposition 7.1] explains why the upper-right entry of the matrix has to be $2u\log u$ and why there is also a commutative diagram

(5)

in which

$$\begin{aligned} R_U = \left( \begin{array}{ccc} u &{} 2u\log u &{} 2u\log ^2 u \\ 0 &{} u &{} 2u\log u \\ 0 &{} 0 &{} u \\ \end{array} \right) \end{aligned}$$

Since $\tau _U$ is an into isometry, so are $S_U$ and $R_U$. Thus, $R_U(Z_3)$ is an isometric copy of $Z_3$. Let us show it is complemented. With that purpose, consider $Z_3^{U}$ the space $Z_3$ constructed with each block $u_n$ in place of $e_n$; namely, $Z_2^U$ is the twisted sum space $U\oplus _{\Omega _{1,0}^U} U$ constructed with $\Omega _{1,0}^U(u ) = 2\sum \lambda _n \log \frac{|u|}{\Vert u\Vert }$ for $u\in U$ and then $Z_3^U$ is the space $Z_2^U \oplus _{\Omega _{\langle 2,1\rangle , 0}^U} U$ with the corresponding definition for $\Omega _{\langle 2,1\rangle , 0}^U$. We can in this way understand $R_U$ as an operator $R_U': Z_3^U \rightarrow Z_3$ in the obvious form: $R_U'(u_n,0,0)=R_U(e_n,0,0)$, $R_U'(0,u_n,0)=R_U(0, e_n,0)$ and $R_U'(0,0, u_n)=R_U(0, 0, e_n)$. Consider the diagram

Here $D_U$ is the obvious isomorphism between $Z_3^U$ and $(Z_3^U)^*$ induced by D. The diagram is commutative: for normalized blocks $u_i, u_j, u_k, u_l, u_m, u_n$ one has

$$\begin{aligned} R_U' (u_i, u_j, u_k) = (u_i + 2u_j\log u_j + 2 u_k \log ^2 u_k, u_j + 2u_k\log u_k, u_k ) \end{aligned}$$

while the action of $D\left( u_i + 2u_j\log u_j + 2 u_k \log ^2 u_k, u_j + 2u_k\log u_k, u_k \right)$ over$\left( u_l + 2u_m\log u_m + 2u_n\log ^2 u_n, u_m +2u_n\log u_n, u_n\right)$ gives

$$\begin{aligned}{} & {} (u_i + 2u_j\log u_j + 2 u_k \log ^2 u_k)u_n - (u_j + 2u_k\log u_k)(u_m +2u_n\log u_n) \\{} & {} \quad + u_k(u_l + 2u_m\log u_m + 2u_n\log ^2 u_n); \end{aligned}$$

namely

$$\begin{aligned}{} & {} \delta _{in} + 2\delta _{jn}\log u + 2 \delta _{kn}\log ^2 u - \delta _{jm} - 2\delta _{jn}\log u - 2\delta _{km}\log u - 4\delta _{kn}\log ^2 u \\{} & {} \quad + \delta _{kl} + 2\delta _{km}\log u + 2 \delta _{kn}\log ^2 u \end{aligned}$$

which is $\delta _{in}- \delta _{jm}+ \delta _{kl}$. Thus

$$\begin{aligned} {(R_U')}^* D R_U' (u_i, u_j, u_k) (u_l, u_m, u_n)= & {} DR_U'(u_i, u_j, u_k) \left( R_U'(u_l, u_m, u_n)\right) \\= & {} \langle R_U'(u_i, u_j, u_k), R_U'(u_l, u_m, u_n)\rangle \\= & {} \delta _{in} - \delta _{jm} + \delta _{kl}\\= & {} D_U(u_i, u_j, u_k) (u_l, u_m, u_n) \end{aligned}$$

Therefore, $D_U^{-1}{(R_U')}^* D$ is a projection onto the range of $R_U$, as desired, and one can repeat the same argument as before working now with $\tau |_{U}$ instead of $\tau |_{\ell _2}$. $\square$

Corollary 5.6

Every operator from $Z_3$ into a twisted Hilbert space is strictly singular. In particular, $Z_3$ does not contain complemented copies of either $Z_2$ or $\ell _2$.

Proof

That $Z_3$ cannot be a subspace of a twisted Hilbert space was proved in [7, Prop. 12]. $\square$

Corollary 5.7

The six representations of $Z_3$ as a twisted sum in the diagrams are non-trivial.

Proof

Since $Z_3$ contains no complemented copy of $\ell _2$ and $Z_3\simeq Z_3^*$ [6, Prop. 5.5 and Cor. 5.7], by Corollary 5.3 the exact sequences   $Z_2\rightarrow Z_3 \rightarrow \ell _2$,   $\wedge \rightarrow Z_3\rightarrow \ell _f^*$   and   $\bigcirc \rightarrow Z_3 \rightarrow \ell _g^*$   have strictly singular quotient map, while   $\ell _2\rightarrow Z_3 \rightarrow Z_2$,   $\ell _f\rightarrow Z_3\rightarrow \wedge ^*$   and   $\ell _g\rightarrow Z_3 \rightarrow \bigcirc ^*$ have strictly cosingular embedding. Of course, the second part is a dual result of the first one. $\square$

In [24, Theorem 5.4] it is proved that every normalized basic sequence in $Z_2$ admits a subsequence equivalent to the basis of one of the spaces $\ell _2$ or $\ell _{f}$. For $Z_3$ we have:

Theorem 5.8

Every normalized basic sequence in $Z_3$ admits a subsequence equivalent to the basis of one of the spaces $\ell _2, \ell _f, \ell _g$.

Proof

Let $(y_n, x_n, z_n)_n$ be a normalized basic sequence in $Z_3$. If $\Vert z_n\Vert \rightarrow 0$ as $n\rightarrow \infty$, we can assume that $\sum \Vert z_n\Vert <\infty$ and thus that, up to a perturbation, $(y_n, x_n,z_n)$ is a basic sequence in $Z_2$; therefore it admits a subsequence equivalent to the basis of either $\ell _2$ or $\ell _{f}$ [24, Theorem 5.4].

If $\Vert z_n\Vert \ge \varepsilon$ then we can assume after perturbation that there is a block basic sequence $(u_n)$ in $\ell _2$ such that $\sum \Vert z_n - u_n\Vert < \infty$. Since

$$\begin{aligned} (y_n, x_n, z_n)= & {} (y_n, x_n, z_n) - ( \Omega _{\langle 2,1\rangle , 0}u_n, u_n) + ( \Omega _{\langle 2,1\rangle , 0}u_n, u_n)\\= & {} ((y_n, x_n)- \Omega _{\langle 2,1\rangle , 0}u_n, z_n - u_n) + ( \Omega _{\langle 2,1\rangle , 0}u_n, u_n) \end{aligned}$$

and $z_n - u_n\rightarrow 0$ we can assume that $((y_n, x_n)- \Omega _{\langle 2,1\rangle , 0}u_n, z_n - u_n)$ admits a subsequence equivalent to the basis of either $\ell _2$ or $\ell _{f}$. We conclude showing that $( \Omega _{\langle 2,1\rangle , 0}u_n, u_n)$ is equivalent to the canonical basis of $\ell _{g}$. And thus the plan is to show that $\sum (x_n \Omega _{\langle 2,1\rangle , 0}u_n, \sum x_n u_n)$ converges in $Z_3$ if and only if $(x_n)\in \ell _{g}$. In order to show that, we simplify the notation: let x be a scalar sequence, let $u=(u_n)$ be the sequence of blocks and let us denote $xu = \sum x_n u_n$. Showing that $(x\Omega _{\langle 2,1\rangle , 0} u, xu)$ converges in $Z_3$ is the same as showing that its norm is finite. Recall that for a positive normalized z one has $\Omega _{\langle 2,1\rangle , 0}(z)= (2z\log ^2 z, 2z \log z)$. Since

$$\begin{aligned} \Vert (x\Omega _{\langle 2,1\rangle , 0} u, xu)\Vert _{Z_3}= & {} \Vert (x\Omega _{\langle 2,1\rangle , 0} u - \Omega _{\langle 2,1\rangle , 0}( xu)\Vert _{Z_2} + \Vert xu\Vert _2\\= & {} \Vert (x\Omega _{\langle 2,1\rangle , 0} u - \Omega _{\langle 2,1\rangle , 0}( xu)\Vert _{Z_2} + \Vert xu\Vert _2, \end{aligned}$$

assuming $\Vert u_n\Vert =1$ for all n and $\Vert xu\Vert =1$, one gets

$$\begin{aligned} x\Omega _{\langle 2,1\rangle , 0} u - \Omega _{\langle 2,1\rangle , 0}( xu)= & {} \left( x2u\log ^2u, 2x\log u\right) - \left( 2xu\log ^2(xu), 2xu\log (xu)\right) \\= & {} \left( 2xu(\log ^2 u - \log ^2 xu), 2xu(\log u - \log (ux))\right) \\= & {} \left( 2xu(\log ^2 u - (\log ^2 x + \log ^2 u + 2\log x\log u), -2xu \log x)\right) \\= & {} \left( -2xu(\log ^2 x + 2\log x\log u), -2xu \log x)\right) \end{aligned}$$

and therefore

$$\begin{aligned}{} & {} \Vert (x\Omega _{\langle 2,1\rangle , 0} u - \Omega _{\langle 2,1\rangle , 0}( xu)\Vert _{Z_2} =\Vert \left( -2xu(\log ^2 x + 2\log x\log u), -2xu \log x)\right) \Vert _{Z_2}\\{} & {} \quad = \Vert -2xu(\log ^2 x + 2\log x\log u) + 4xu \log x \log \left( 2xu \log x\right) \Vert _2 + \Vert 2xu \log x\Vert _2\\{} & {} \quad = \Vert 2xu \left( \log ^2 x + 2\log 2\log x + 2 \log x \log \log x\right) \Vert _2 + \Vert 2xu \log x\Vert _2. \end{aligned}$$

That means that the sequence x satisfies $x(\log ^2|x|) \in \ell _2$; namely, $x\in \ell _{g}$.$\square$

This result has consequences for the structure of the spaces $Z_3$, $\wedge$ and $\bigcirc$.

Proposition 5.9

$Z_3$ has no complemented subspace with an unconditional basis.

Proof

If $(x_n)$ were an unconditional basic sequence in $Z_3$ generating a complemented subspace, it would admit a subsequence $(x_{n_k})$ equivalent to the basis of one of the spaces $\ell _2, \ell _f, \ell _g$ by Theorem 5.8. Since this subsequence would generate a complemented subspace of $Z_3$, we would conclude that $Z_3$ contains a complemented copy of $\ell _2$, by Corollary 5.3, which cannot happen. $\square$

Proposition 5.10

The spaces $\wedge$ and $\bigcirc$ are not isomorphic to their dual spaces.

Proof

Both $\wedge$ and $\bigcirc$ are subspaces of $Z_3$, hence Theorem 5.8 applies. But $\wedge ^*$ and $\bigcirc ^*$ contain a copy of $\ell _f^*$, as we can see in the diagrams, while the canonical basis of $\ell _f^*$ (or any of its subsequences) is not equivalent to those of $\ell _2, \ell _f$ or $\ell _g$. $\square$

Proposition 5.11

The space $\wedge$ (hence $\wedge ^*$ also) is not isomorphic to either $\bigcirc$ or $\bigcirc ^*$.

Proof

The idea for the proof is to show that every weakly null sequence in $\wedge$ contains a subsequence equivalent to the canonical basis of either $\ell _2$ or $\ell _g$, so that $\wedge$ cannot contain either $\ell _f$ or $\ell _f^*$ and therefore it cannot be isomorphic to either $\bigcirc$ or $\bigcirc ^*$. Why it is so is essentially contained in the proof of Theorem 5.8, taking into account that the elements of $\wedge$ have the form (y, 0, z). Our interest lies now in showing that when $(u_n)$ are blocks in $\ell _2$ (actually in $\ell _f$) and $\sum (x_ny_n, 0, u_n)$ converges in $Z_3$ then $x=(x_n)$ is in either $\ell _2$ or $\ell _g$. Using the same notation as then, since $\left\| (xy, 0, xu)\right\| _{Z_3} = \left\| (xy, 0) - \Omega _{\langle 2,1\rangle , 0}(xu)\right\| _{Z_2} + \Vert xu\Vert _{\ell _2}$, and since (xy, 0) and xu converge when $x\in \ell _2$, our only concern is when $\Omega _{\langle 2,1\rangle , 0}(xu)$ converges in $Z_2$. But this means that $x\in \textrm{Dom}\,\Omega _{\langle 2,1\rangle , 0}=\ell _g$.$\square$

Proposition 5.12

The spaces $\wedge$ and $\wedge ^*$ do not contain $\ell _2$ complemented. Consequently, they do not have an unconditional basis.

Proof

Consider the diagram [120]. Its lower sequence comes defined by $\bigtriangleup (x) = x\log ^2 x$, obtained from the composition $\Omega _{\langle 2,1\rangle , 0}x=(x\log ^2 x, x \log x)$ with the projection onto the first coordinate. Let u be a sequence of disjoint blocks of the canonical basis of $\ell _2$ and let $x\in \ell _2$.

$$\begin{aligned} \bigtriangleup (xu)= & {} xu\log ^2(xu)= xu\left( \log x+ \log u)^2\right) \\= & {} xu\left( \log ^2 x + \log ^2 u + 2\log x \log u\right) \\= & {} xu\log ^2 x + xu\log ^2 u + 2xu\log x \log u \end{aligned}$$

Observe that the second term $x\rightarrow xu\log ^2 u$ is linear while the third term $x\rightarrow 2x \log x u \log u$ is $x \rightarrow \Omega _{1,0}(x)$, according to [5, Lemma 9.3.10] and up to a weight and a linear map. This map is bounded when considered with values in its range $\ell _f^*$, which yields that the restriction $\bigtriangleup |_{[u]}$ is, up to a linear plus a bounded map, $\bigtriangleup$ once again. Therefore, the quotient map Q in $0\rightarrow \ell _f^*\rightarrow \wedge ^* {\mathop {\rightarrow }\limits ^{Q}}\ell _2\rightarrow 0$ is strictly singular; hence $Q^*$, the embedding in its dual sequence $0\rightarrow \ell _2 {\mathop {\rightarrow }\limits ^{Q^*}}\wedge \rightarrow \ell _{f}\rightarrow 0$, which is the left column in diagram [201], is strictly cosingular.

The rest is similar to [6, Prop. 15]: Assume that $\wedge ^*$ contains a subspace A isomorphic to $\ell _2$ complemented by some projection P. Since Q is strictly singular, there exist an infinite dimensional subspace $A'\subset \ell _2$ and a nuclear operator $K:A'\rightarrow \wedge ^*$ nuclear norm $\Vert K\Vert _n<1$ such that $I-K: A'\rightarrow A$ is a bijective isomorphism. Let N be a nuclear operator on $\wedge ^*$ extending K with $\Vert N\Vert _n<1$. Then $I_{\wedge ^*}-N$ is invertible, where $I_{\wedge ^*}$ is the identity on $\wedge ^*$, $(I_{\wedge ^*}-N)^{-1}= \sum _{k\ge 0}N^k$, and $(I_{\wedge ^*}-N)\circ P\circ (I_{\wedge ^*}-N)^{-1}$ is a projection on $\wedge ^*$ onto $A'$. This cannot be since the embedding map $Q^*$ is strictly cosingular. Since $\wedge$ is reflexive, it cannot contain $\ell _2$ complemented also. As for the second part, since $\wedge$ is a subspace of $Z_3$, the argument in the proof of Corollary 5.9 also proves the result. $\square$

Corollary 5.13

All the exact sequences appearing in the six diagrams are non-trivial.

Proof

Corollary 5.7 showed that the sequences passing through $Z_2$ are nontrivial. The non-triviality for those passing through $\wedge$ and $\wedge ^*$ follows from the fact that these spaces do not admit an unconditional basis (Proposition 5.12); for those passing through $\bigcirc$ follows from the fact that $\ell _f\oplus \ell _f\simeq \ell _f$ does not contain copies of $\ell _g$ and $\ell _g\oplus \ell _2\simeq \ell _g$ does not contain copies of $\ell _f$; and for those passing through $\bigcirc ^*$ we can argue as for $\bigcirc$.$\square$

This corollary can be improved.

Proposition 5.14

The following maps:

• (1) $Q_0$, $Q_1$, $Q_2$, $Q_{1,0}$, $Q_{0,1}$, $Q_{2,0}$, $Q_{0,2}$, $Q_{1,2}$, $Q_{2,1}$;

• (2) $p_{1,0}$, $p_{0,1}$, $p_{2,0}$, $p_{0,2}$, $p_{2,1}$, $p_{1,2}$; and

• (3) $q_{1,0}$, $q_{0,1}$, $q_{2,0}$, $q_{0,2}$

are strictly singular.

Proof

(1) That $Q_0$, $Q_1$, $Q_2$, $Q_{1,0}$ and $Q_{0,1}$ are strictly singular is a consequence of Proposition 5.5, because $\ell _2$, $\ell _f^*$, $\ell _g^*$ and $Z_2$ do not contain $Z_3$. The lower part in the diagram [120]

plus the technique used before shows that $Q_{2,0}$, hence $Q_{0,2}$, is strictly singular. Therefore, its restrictions $p_{2,0}$ and $p_{0,2}$ are strictly singular too. The restriction of $p_{1,2}$ to $\ell _f$ is the canonical inclusion of $\ell _f$ into $\ell _2$, which is strictly singular due to the criterion [27, Theorem 4.a.10] asserting that given two Orlicz spaces $\ell _M$, $\ell _N$ for which the canonical inclusion $\jmath : \ell _M\rightarrow \ell _N$ is continuous then $\jmath$ is strictly singular if and only if for each $B>0$ there is a sequence $\tau _1, \dots , \tau _n$ in (0, 1] such that $\sum M(\tau _i t) \ge B \sum N(\tau _i t)$ for all $t \in [0,1]$. Straightforward calculations yield that the canonical inclusions $\ell _{g} \rightarrow \ell _{f}$ and $\ell _{f}\rightarrow \ell _2$ are strictly singular. Thus, also $p_{0,2}$ is strictly singular and consequently the lower part of diagram [102]

yields that $Q_{0,2}$, hence $Q_{2,0}$ too, is strictly singular. (2) the maps are restrictions of $Q_{1,0}$, $Q_{0,1}$, $Q_{2,0}$ and $Q_{0,2}$. (3) follows from Corollary 5.3 because $Z_2$ and $\wedge ^*$ contain no complemented copy of $\ell _2$.$\square$

Remark 5.15

We have been unable to prove that $q_{1,2}$ and $q_{2,1}$ are strictly singular, from where it would follow that $\bigcirc$ and $\bigcirc ^*$ do not have an unconditional basis.

6 The Case of Weighted Hilbert Spaces

This is an interesting test case by its simplicity (all exact sequences are trivial and all spaces are isomorphic to Hilbert spaces), and provides some insight about what occurs in other situations. Let $w=(w_n)$ be a weight sequence (a non-increasing sequence of positive numbers such that $\lim w_n=0$ and $\sum w_n=\infty$) and let $w^{-1}=(w_n^{-1})$. Note that $\ell _2(w)^*$ is isometric to $\ell _2(w^{-1})$.

If ${\mathcal {C}}$ is the Calderón spaces for the couple $(\ell _2(w^{-1}), \ell _2(w))$, an homogeneous bounded selector for the interpolator $\Delta _0:{\mathcal {C}}\rightarrow \Sigma$ is $B(x)(z) = w^{2z-1}x$. Therefore $B(x)'(z) = 2w^{2z-1}\log w \cdot x$ and $\Omega _{1,0} x =\Delta _1 Bx = 2\log w \cdot x$. The Rochberg space ${\mathcal {R}}_2$ will be

$$\begin{aligned} Z_2(w)= \{(y, x): x\in \ell _2, \quad y - 2\log w \cdot x\in \ell _2\} \end{aligned}$$

from where $\textrm{Dom}\,\Omega _{1,0} = \{x\in \ell _2: 2\log w \cdot x\in \ell _2\} = \ell _2(\log w) = \{(0,x)\in Z_2(w)\}$ and $\textrm{Ran}\,\Omega _{1,0} = \ell _2((\log w)^{-1})$ so that $(\Omega _{1,0})^{-1}x = \frac{1}{2\log w} x$; thus $\textrm{Dom}\, (\Omega _{1,0})^{-1} = \{x\in \ell _2((\log w)^{-1}): (\log w)^{-1} \cdot x\in \ell _2(\log w)\}=\ell _2= \textrm{Ran}\,(\Omega _{1,0})^{-1}$, as we already know.

Next, $B(x)''(z) = 4w^{2z-1}\log ^2 w \cdot x$, and thus $\Delta _2 B(x) = 2\log ^2 w \cdot x$. Therefore

$$\begin{aligned} \Omega _{\langle 2,1\rangle , 0}(x) = \left( \Delta _2 B(x), \Delta _1 B(x)\right) = \left( 2\log ^2 w\cdot x, 2\log w \cdot x\right) \end{aligned}$$

defines a linear map with domain $\textrm{Dom}\,\Omega _{\langle 2,1\rangle , 0} = \{x\in \ell _2: (2\log ^2 w\cdot x, 2\log w \cdot x) \in Z_2(w)\} = \ell _2(\log ^2 w)$ since one must have $2\log w \cdot x \in \ell _2$ and $2\log ^2 w\cdot x - 4\log ^2\cdot w = - 2\log ^2 w \cdot x \in \ell _2$. Therefore we have some parts of the first two diagrams [210] and [012]

We need to know now who are $\bigcirc = \textrm{Dom}\,\Omega _{2, \langle 1, 0\rangle }$ and $\blacksquare = \bigcirc /\ell _2(\log ^2 w)$. To get the first of those spaces we need to know $\Omega _{2, \langle 1, 0\rangle }$. Recall from the standard diagram

that if A, B are homogeneous bounded selectors for a and b then

$$\begin{aligned} W (y,x) = B(y - \Omega _{b,a} x) + A x \end{aligned}$$

is a selector for (b, a) and therefore $\Omega _{c, (b,a)} = c W$. With this info at hand, we need a selector W for $\langle \Delta _1,\Delta _0\rangle$ to then obtain $\Omega _{2, \langle 1, 0\rangle } = \Delta _2 W$. Now, the selector for $\Delta _0$ is $Bx(z)=w^{2z-1} x$ as we already know, and the selector for $\Delta _1: \ker \delta _0 \rightarrow \ell _2$ is $\frac{1}{\varphi '(1/2)}\varphi B$ where $\varphi$ is a conformal mapping with $\varphi (1/2)=0$. Thus, $W(y,x)= \frac{\varphi }{\varphi '(1/2)} B(y - \Omega _{1,0} x) + Bx$, and elementary calculations yield

$$\begin{aligned} \Omega _{2, \langle 1, 0\rangle }(y,x)= & {} \frac{1}{2}W(y,x)''(1/2) = \Omega _{1,0}(y-\Omega _{1,0} x) \\{} & {} + \frac{\varphi ''(1/2)}{2\varphi '(1/2)}(y-\Omega _{1,0} x) + \frac{1}{2}Bx''(1/2)\\= & {} 2\log w \cdot (y-2\log w \cdot x) + \frac{\varphi ''(1/2)}{2\varphi '(1/2)}(y- 2\log w \cdot x) + 2\log ^2 w \cdot x. \end{aligned}$$

Setting $d= \frac{\varphi ''(1/2)}{2\varphi '(1/2)}$ one gets $\Omega _{2, \langle 1, 0\rangle }(y,x) = (2\log w +d) y - (2\log ^2 w + 2d\log w) x$. This yields $\textrm{Dom}\,\Omega _{2, \langle 1, 0\rangle } = \{(y,x)\in Z_2(w): (2\log w +d) y - (2\log ^2 w + 2d\log w) x\in \ell _2\}$ and then $\textrm{Dom}\,\Omega _{2, \langle 1, 0\rangle }|_{\textrm{Dom}\,\Omega _{1,0}} = \{(0,x)\in Z_2(w): (2\log ^2 w + 2d\log w) x\in \ell _2\} = \ell _2(\log ^2 w)$. And since $d y - 2d\log w x\in \ell _2$ when $(y,x)\in Z_2(w)$ one gets

$$\begin{aligned} \bigcirc= & {} \{(y,x)\in Z_2(w): (2\log w +d) y - (2\log ^2 w + 2d\log w) x\in \ell _2\}\\= & {} \{(y,x)\in Z_2(w): \log w y - \log ^2 w x\in \ell _2\}\\= & {} \{(y,x)\in Z_2(w): \log w (y - \log w x)\in \ell _2\}\\= & {} \{(y,x): x\in \ell _2 \quad \textrm{and}\quad y - \log w x \in \ell _2(\log w)\}. \end{aligned}$$

By obvious reasons we will call this space $\bigcirc = Z_{\ell _2(\log w)}(w)$. It is clear that $\bigcirc$ is a twisting $0\longrightarrow \ell _2(\log w) \longrightarrow Z_{\ell _2(\log w)}(w) \longrightarrow \ell _2(\log w) \longrightarrow 0$ of $\ell _2(\log w)$ obtained with the same quasilinear map $\Omega x = 2\log w x$. This is a bonus effect of working with weighted spaces in which all maps are linear. On the other hand, $\blacksquare$ is the domain of $\Delta _2 \Omega _{2, \langle 1, 0\rangle }^{-1}$. We showed in Proposition 3.6 that $\Delta _0(\ker \Delta _2)=\Delta _0(\ker \Delta _1) \Longrightarrow \Delta _1(\ker \Delta _2)= \Delta _1(\ker \Delta _0)$, which in this case yields $\textrm{Dom}\,(\Omega ) = \ell _2(\log w) \Longrightarrow \blacksquare = \ell _2$. Thus, giving the analogous meaning as before to the space $Z_{\ell _2((\log w)^{-1})}(w)$, diagrams [210] and [012] are

The other relevant new space appears in [201]

that we can identify as the pullback space $\wedge = \{(y,0,x)\in Z_3\}$ generated with the map $\Omega _{2, \langle 1, 0\rangle }|_{\textrm{Dom}\,\Omega _{1,0}} x = -(2\log ^2 w + 2d\log w) x$. We thus get that [102] and [201] are

The vertical sequence on the left is defined by $\Omega x = 2\log w x$ because this is the derivation associated to the interpolation couple $\left( \ell _2(w^{-1}\log w), \ell _2(w\right. \left. \log w)\right) _{1/2} = \ell _2(\log w)$. Since $\textrm{Dom}\,\Omega = \{ x \in \ell _2(\log w): \log w x \in \ell _2(\log w ) \}= \{ x \in \ell _2(\log w): \log ^2 w x \in \ell _2 \}= \ell _2(\log ^2 w)$ one gets that [021] and [120] are