On Garling sequence spaces

TL;DR

This paper introduces a new class of Banach spaces inspired by Garling's example, demonstrating unique subsymmetric bases and properties like $ ext{l}_p$-saturation and complementable homogeneity, thus providing novel examples in Banach space theory.

Contribution

It constructs and analyzes a new class of Banach spaces with unique subsymmetric bases and specific structural properties, filling a gap in existing examples.

Findings

Existence of Banach spaces with unique subsymmetric bases

Spaces are $ ext{l}_p$-saturated and complementably homogeneous

First known examples with both properties coexist

Abstract

The aim of this paper is to introduce and investigate a new class of separable Banach spaces modeled after an example of Garling from 1968. For each $1\leqslant p<\infty$ and each nonincreasing weight $\textbf{w}\in c_0\setminus\ell_1$ we exhibit an $\ell_p$-saturated, complementably homogeneous, and uniformly subprojective Banach space $g(\textbf{w},p)$. We also show that $g(\textbf{w},p)$ admits a unique subsymmetric basis despite the fact that for a wide class of weights it does not admit a symmetric basis. This provides the first known examples of Banach spaces where those two properties coexist.

Equations

$$\Vert f{\Vert }_{g(\mathbf{w},p)}=\underset{\varphi \in \mathcal{O}}{\sup }{\left(\sum _{n=1}^{\infty }|a_{\varphi (n)}{|}^p{w}_n\right)}^{1/p}$$

$$g(\mathbf{w},p):=\{f=(a_n{)}_{n=1}^{\infty }:\Vert f{\Vert }_{g(\mathbf{w},p)}<\infty \}.$$

$$\mathcal{W}:=\left\{(w_n{)}_{n=1}^{\infty }\in c_0\setminus {\ell }_1:1=w_1\ge w_2>\cdots w_n\ge w_{n+1}\ge \cdots >0\right\}.$$

$$d(\mathbf{w},p)=\{f=(a_n{)}_{n=1}^{\infty }:\Vert f{\Vert }_{d(\mathbf{w},p)}<\infty \},$$

$$\Vert f{\Vert }_{d(\mathbf{w},p)}=\underset{\sigma \in \Pi }{\sup }{\left(\sum _{n=1}^{\infty }|a_{\sigma (n)}{|}^p{w}_n\right)}^{1/p}.$$

$${\Vert \sum _{n=1}^{\infty }{a}_n{\mathbf{x}}_n\Vert }_X\le C{\Vert \sum _{n=1}^{\infty }{a}_n{\mathbf{y}}_n\Vert }_Y,(a_n{)}_{n=1}^{\infty }\in c_{00},$$

$$\Vert f{\Vert }_g^p=\underset{\varphi \in {\mathcal{O}}_f}{\sup }\sum _{n=1}^{r(\varphi )}|a_{\varphi (n)}{|}^p{w}_n.$$

$$\Vert f{\Vert }_{d_{\infty }(\mathbf{w},p)}=\underset{n}{\sup }{\left(\sum _{k=1}^n{w}_k\right)}^{1/p}{a}_n^{\ast }<\infty ,$$

$${\ell }_p⊊d(\mathbf{w},p)\subseteq g(\mathbf{w},p)\subseteq d_{\infty }(\mathbf{w},p),$$

$$\sum _{n=1}^{r(\varphi )}|a_{\varphi (n)}{|}^p{w}_n=\sum _{n=1}^{r(\varphi )}|a_{\sigma (n)}{|}^p{w}_n\le \sum _{n=1}^{\infty }|a_{\sigma (n)}{|}^p{w}_n\le \Vert f{\Vert }_{d(\mathbf{w},p)}^p.$$

$$a_n^{\ast }{\left(\sum _{k=1}^n{w}_k\right)}^{1/p}\le {\left(\sum _{k=1}^n|a_{\varphi (k)}{|}^p{w}_k\right)}^{1/p}\le \Vert f{\Vert }_{d(\mathbf{w},p)},$$

$$M_{ϵ}((a_n{)}_{n=1}^{\infty })=({ϵ}_n{a}_n{)}_{n=1}^{\infty }.$$

$$P_A:{\mathbb{F}}^{\mathbb{N}}\to {\mathbb{F}}^{\mathbb{N}},P_A((a_n{)}_{n=1}^{\infty })=({\lambda }_n{a}_n{)}_{n=1}^{\infty },$$

$$b_n=\{\begin{array}{ll}{a}_k& \text{ if }n=\varphi (k),\\ 0& \text{ if }n\notin R(\varphi ).\end{array}$$

$$\Vert V_{\varphi }(f){\Vert }_g^p=\underset{\psi \in \mathcal{O}}{\sup }\sum _{n=1}^{\infty }|a_{\varphi (\psi (n))}{|}^p{w}_n\le \Vert f{\Vert }_g.$$

$$\sum _{n=1}^{\infty }|b_{\psi (n)}{|}^p{w}_n$$

$$\le \sum _{n\in J}|a_{\rho (n)}{|}^p{w}_n\le \Vert f{\Vert }_g^p.$$

$$\sum _{n=1}^{|B|}{w}_n=\Vert f{\Vert }_g={\Vert \sum _{n\in A}{\mathbf{e}}_n\Vert }_g=\sum _{n=1}^{|A|}{w}_n.$$

$$\Vert f{\Vert }_g={\left(\sum _{n=1}^{r(\psi )}|a_{\psi (n)}{|}^p{w}_n\right)}^{1/p},$$

$$\sum _{n=1}^{r(\varphi )}|a_{\varphi (n)}{|}^p{w}_n=\sum _{n\in R(\psi )}|a_n{|}^p{w}_{\gamma (n)}\le \sum _{n\in R(\psi )}|a_n{|}^p{w}_{{\psi }^{-1}(n)}=\sum _{n=1}^{r(\psi )}|a_{\psi (n)}{|}^p{w}_n.$$

$$\Vert f{\Vert }_g^p=\sum _{n=1}^{r(\psi )}|a_{\psi (n)}{|}^p{w}_n\le \Vert h{\Vert }_g^p,$$

$${\mathbf{y}}_n=\sum _{i=p_n}^{p_{n+1}-1}{a}_i{\mathbf{x}}_i$$

$$\underset{n\in \mathbb{N}}{\inf }\underset{p_n\le i\le p_{n+1}-1}{\sup }|a_i|>0.$$

$$C_n=\underset{p_n\le i\le p_{n+1}-1}{\max }|a_i|,C=\underset{n}{\sup }\Vert {\mathbf{x}}_n\Vert ,\text{and}B=\underset{n}{\inf }\Vert y_n\Vert .$$

$${\mathbf{y}}_n=\sum _{i=p_n}^{p_{n+1}-1}{a}_i{\mathbf{g}}_i,n\in \mathbb{N},$$

$$\sum _{n=1}^{\infty }|c_{\varphi (n)}{|}^p{w}_n$$

$$=\sum _{n=1}^{\infty }|b_n{|}^p\sum _{j=1}^{r_n}|a_{\varphi (j+q_n)}{|}^p{w}_{j+q_n}$$

$$\le \sum _{n=1}^{\infty }|b_n{|}^p\sum _{j=1}^{r_n}|a_{\varphi (j+q_n)}{|}^p{w}_j$$

$$\le \sum _{n=1}^{\infty }|b_n{|}^p\Vert {\mathbf{y}}_n{\Vert }_g^p$$

$$=\sum _{n=1}^{\infty }|b_n{|}^p.$$

$${\mathbf{y}}_n=\sum _{i=p_n}^{p_{n+1}-1}{a}_i{\mathbf{x}}_i,n\in \mathbb{N},$$

$$\hat{}{\mathbf{y}}_n=\sum _{i={\hat{p}}_n}^{{\hat{p}}_{n+1}-1}{a}_{\varphi (i)}{\mathbf{x}}_i,n\in \mathbb{N},$$

$$\hat{}{\mathbf{y}}_n=\sum _{i=p_n}^{p_{n+1}-1}{a}_i{\mathbf{g}}_i,n\in \mathbb{N},$$

Full text

F. Albiac acknowledges the support of the Spanish Ministry for Economy and Competitivity Grants MTM2014-53009-P for Análisis Vectorial, Multilineal y Aplicaciones, and MTM2016-76808-P for * Operators, lattices, and structure of Banach spaces*. J.L. Ansorena acknowledges the support of the Spanish Ministry for Economy and Competitivity Grant MTM2014-53009-P for Análisis Vectorial, Multilineal y Aplicaciones

1. Introduction, notation, and terminology

Let $1\leq p<\infty$ and let $\mathbf{w}=(w_{n})_{n=1}^{\infty}$ be a weight, i.e., a sequence of positive scalars. Assume that $\mathbf{w}$ is decreasing. Given a sequence of scalars $f=(a_{n})_{n=1}^{\infty}$ we put

[TABLE]

where $\mathcal{O}$ denotes the set of increasing functions from $\mathbb{N}$ to $\mathbb{N}$. We consider the normed space

[TABLE]

A straightforward application of Fatou’s lemma gives that $g(\mathbf{w},p)$ is a Banach space. Imposing the further conditions $\mathbf{w}\in c_{0}$ and $\mathbf{w}\notin\ell_{1}$ will prevent us, respectively, from having $g(\mathbf{w},p)=\ell_{p}$ or $g(\mathbf{w},p)=\ell_{\infty}$. We will assume as well that $\mathbf{w}$ is normalized, i.e., $w_{1}=1$. Thus, we put

[TABLE]

The germ of this family of spaces goes back to [Ga68], where Garling showed that for $\mathbf{w}=(n^{-1/2})_{n=1}^{\infty}$ the canonical unit vectors of $g(\mathbf{w},1)$ form a subsymmetric basic sequence that is not symmetric. Indeed, until then these two concepts were believed to be the same. Other Garling sequence spaces have appeared in past literature since then. In [Pu76], Pujara investigated a variation of $g(\mathbf{w},2)$ for $\mathbf{w}=(n^{-1/2})_{n=1}^{\infty}$. He proved that those spaces are uniformly convex, and that their canonical basis is subsymmetric but not symmetric. Dilworth el al. [DOSZ11] studied derived forms of $g(\mathbf{w},1)$ for various choices of weights $\mathbf{w}$ that yield spaces in which the canonical basis is 1-greedy and subsymmetric but not symmetric.

Of course, all these spaces are inspired in sequence Lorentz spaces defined as

[TABLE]

where, if $\Pi$ is the set of permutations on $\mathbb{N}$,

[TABLE]

The main difference between Lorentz sequence spaces and Garling sequence spaces is that, unlike for $d(\mathbf{w},p)$, the canonical basis in $g(\mathbf{w},p)$ is not symmetric for a wide class of weights. We will tackle this issue in Section 5. The canonical basis is subsymmetric, though. Moreover, while $d(\mathbf{w},p)$ is known to admit a unique symmetric basis, in Section 4 we prove that $g(\mathbf{w},p)$ has a unique subsymmetric basis. Along the way to proving the uniqueness of subsymmetric basis, we will show in Section 3 that many structural results of $d(\mathbf{w},p)$ carry over to $g(\mathbf{w},p)$. To be precise, we will show that $g(\mathbf{w},p)$ is complementably homogeneous and uniformly complementably $\ell_{p}$-saturated. As a consequence of the latter, $g(\mathbf{w},p)$ is uniformly subprojective. Both subprojectivity and complementable homogeneity have been used in the study of the closed ideal structure of the algebra of endomorphisms of a Banach space. Subprojectivity was introduced in [Wh64] in order to study the closed operator ideal of strictly singular operators. For instance, if $X$ and $Y$ are Banach spaces, $Y$ is subprojective and $T\in\mathcal{L}(X,Y)$, then the dual operator $T^{*}$ is strictly singular only if $T$ is. If furthermore $X=Y$ and is reflexive, the converse holds, i.e., if $X$ is reflexive then $T\in\mathcal{L}(X)$ is strictly singular if and only if $T^{*}$ is. See [OS15] for further discussion of subprojectivity. Complementable homogeneity, meanwhile, has been used in several recent papers (cf., e.g., [CJZ11, LSZ14, Zh14]) to show the existence of a unique maximal ideal in $\mathcal{L}(X)$. In particular, the set of all $X$-strictly singular operators—those for which the restrictions to subspaces isomorphic to $X$ are never bounded below—acting on a complementably homogeneous Banach space $X$ forms the unique maximal ideal in $\mathcal{L}(X)$ whenever it is closed under addition. By an argument in [KPSTT12, Corollary 5.2] and the second paragraph of the proof of [KPSTT12, Theorem 5.3], it follows that the set of $g(\mathbf{w},p)$-strictly singular operators forms the unique maximal ideal in $\mathcal{L}(g(\mathbf{w},p))$.

We close this introductory section by setting the notation and recalling the terminology that will be most heavily used. We will write $\mathbb{F}$ for the real or complex field and denote by $(\mathbf{e}_{n})_{n=1}^{\infty}$ the canonical basis of $\mathbb{F}^{\mathbb{N}}$, i.e., $\mathbf{e}_{n}=(\delta_{k,n})_{k=1}^{\infty}$, were $\delta_{k,n}=1$ if $n=k$ and $\delta_{k,n}=0$ otherwise. On occasion we will need to compare linear combinations of the vectors $(\mathbf{e}_{n})_{n=1}^{\infty}$ with respect to different norms; thus to avoid confusion we may denote by $(\mathbf{d}_{n})_{n=1}^{\infty}$, $(\mathbf{f}_{n})_{n=1}^{\infty}$, and $(\mathbf{g}_{n})_{n=1}^{\infty}$ the vectors $(\mathbf{e}_{n})_{n=1}^{\infty}$ when seen, respectively, inside the spaces $d(\mathbf{w},p)$, $\ell_{p}$, or $g(\mathbf{w},p)$. Also for simplicity, when $p$ and $\mathbf{w}$ are clear from context the norms in the spaces $d(\mathbf{w},p)$ and $g(\mathbf{w},p)$ will be denoted by $\|\cdot\|_{d}$ and $\|\cdot\|_{g}$, respectively. As it customary, we write $c_{00}$ for the space of all scalar sequences with finitely many nonzero entries. If $(\mathbf{x}_{n})_{n=1}^{\infty}$ is a basis for a Banach space $X$, and $f=\sum_{n=1}^{\infty}a_{n}\mathbf{x}_{n}\in X$, then the support of $f$ is the set $\operatorname{supp}(f):=\{n\in\mathbb{N}:a_{n}\neq 0\}$ of indices corresponding to its nonzero entries. Given a basic sequence $(\mathbf{z}_{n})_{n=1}^{\infty}$ in $X$, $\langle\mathbf{z}_{n}\rangle_{n=1}^{\infty}$ denotes its linear span and $[\mathbf{z}_{n}]_{n=1}^{\infty}$ will be its closed linear span.

Given families of positive real numbers $(\alpha_{i})_{i\in I}$ and $(\beta_{i})_{i\in I}$, the symbol $\alpha_{i}\lesssim\beta_{i}$ for $i\in I$ means that $\sup_{i\in I}\alpha_{i}/\beta_{i}<\infty$, while $\alpha_{i}\approx\beta_{i}$ for $i\in I$ means that $\alpha_{i}\lesssim\beta_{i}$ and $\beta_{i}\lesssim\alpha_{i}$ for $i\in I$. Now suppose $(\mathbf{x}_{n})_{n=1}^{\infty}$ and $(\mathbf{y}_{n})_{n=1}^{\infty}$ are basic sequences in $X$ and $Y$, respectively. If for some positive $C$,

[TABLE]

we say that $(\mathbf{y}_{n})_{n=1}^{\infty}$ $C$-dominates $(\mathbf{x}_{n})_{n=1}^{\infty}$, and write $(\mathbf{x}_{n})_{n=1}^{\infty}\lesssim_{C}(\mathbf{y}_{n})_{n=1}^{\infty}$. When the value of the constant is irrelevant we will simply say that $(\mathbf{y}_{n})_{n=1}^{\infty}$ dominates $(\mathbf{x}_{n})_{n=1}^{\infty}$, and write $(\mathbf{x}_{n})_{n=1}^{\infty}\lesssim(\mathbf{y}_{n})_{n=1}^{\infty}$. Whenever $(\mathbf{x}_{n})_{n=1}^{\infty}\lesssim_{C}(\mathbf{y}_{n})_{n=1}^{\infty}$ and $(\mathbf{y}_{n})_{n=1}^{\infty}\lesssim_{C}(\mathbf{x}_{n})_{n=1}^{\infty}$, we say that $(\mathbf{x}_{n})_{n=1}^{\infty}$ and $(\mathbf{y}_{n})_{n=1}^{\infty}$ are $C$-equivalent, and write $(\mathbf{x}_{n})_{n=1}^{\infty}\approx_{C}(\mathbf{y}_{n})_{n=1}^{\infty}$ or, simply, $(\mathbf{x}_{n})_{n=1}^{\infty}\approx(\mathbf{y}_{n})_{n=1}^{\infty}$.

By a sign we mean a scalar of modulus one. A basic sequence $(\mathbf{x}_{n})_{n=1}^{\infty}$ is called unconditional if $(\mathbf{x}_{\sigma(n)})_{n=1}^{\infty}$ is a basic sequence for any $\sigma\in\Pi$. It is well known that a basic sequence $(\mathbf{x}_{n})_{n=1}^{\infty}$ is unconditional if and only if there exists a constant $C\geq 1$ so that $(\mathbf{x}_{n})_{n=1}^{\infty}\approx_{C}(\epsilon_{n}\mathbf{x}_{n})_{n=1}^{\infty}$ for any choice of signs $(\epsilon_{n})_{n=1}^{\infty}$. A basic sequence $(\mathbf{x}_{n})_{n=1}^{\infty}$ is called subsymmetric if it is unconditional and equivalent to all its subsequences. It is called symmetric whenever it is equivalent to each of its permutations. If $(\mathbf{x}_{n})_{n=1}^{\infty}$ is a subsymmetric basic sequence then there is a constant $C\geq 1$ such that $(\mathbf{x}_{n})_{n=1}^{\infty}\approx_{C}(\epsilon_{n}\mathbf{x}_{\phi(n)})_{n=1}^{\infty}$ for any $\phi\in\mathcal{O}$ and any choice of signs $(\epsilon_{n})_{n=1}^{\infty}$. In this case we say that $(\mathbf{x}_{n})_{n=1}^{\infty}$ is $C$-subsymmetric. Similarly, if $(\mathbf{x}_{n})_{n=1}^{\infty}$ is symmetric then there is $C\geq 1$ such that $(\mathbf{x}_{n})_{n=1}^{\infty}\approx_{C}(\epsilon_{n}\mathbf{x}_{\sigma(n)})_{n=1}^{\infty}$ for any choice of signs $(\epsilon_{n})_{n=1}^{\infty}$ and any $\sigma\in\Pi$, in which case we say that $(\mathbf{x}_{n})_{n=1}^{\infty}$ is $C$-symmetric. Note that $C$-symmetry implies $C$-subsymmetry, which in turn implies $C$-unconditionality. Note also that every subsymmetric basis $(\mathbf{x}_{n})_{n=1}^{\infty}$ is semi-normalized, i.e., $\|\mathbf{x}_{n}\|\approx 1$ for $n\in\mathbb{N}$.

Given a function $\phi$ we denote by $\operatorname{R}(\phi)$ its range. Let $\mathcal{O}_{f}$ be the set of increasing functions from an integer interval $[1,\dots,r]\cap\mathbb{N}$ into $\mathbb{N}$. Given $\phi\in\mathcal{O}_{f}$ we denote by $r(\phi)$ the largest integer in its domain. A function in $\mathcal{O}_{f}$ is univocally determined by its range. For $\mathbf{w}=(w_{n})_{n=1}^{\infty}$, $1\leq p<\infty$, and $f=(a_{n})_{n=1}^{\infty}\in\mathbb{F}^{\mathbb{N}}$ we have

[TABLE]

A Banach space with an unconditional (respectively, subsymmetric or symmetric) basis is said to have a unique unconditional (respectively, unique subsymmetric or unique symmetric basis) if any two semi-normalized unconditional (respectively, subsymmetric or symmetric) bases of $X$ are equivalent.

Given infinite-dimensional Banach spaces $X$ and $Y$, $\mathcal{L}(X,Y)$ denotes the space of bounded linear operators from $X$ into $Y$, and put $\mathcal{L}(X)=\mathcal{L}(X,X)$. The symbol $X\approx Y$ means that $X$ and $Y$ are isomorphic. The space $X$ is said to be $Y$-saturated (or hereditarily $Y$) if every infinite-dimensional closed subspace of $X$ admits a further subspace $Z\approx Y$. If, in addition, there is a constant $C$ such that $Z$ can be chosen $C$-complemented in $X$ we say that $X$ is uniformly complementably $Y$-saturated. A Banach space $X$ is said to be complementably homogeneous whenever for every closed subspace $Y$ of $X$ such that $Y\approx X$, there exists another closed subspace $Z$ such that $Z$ is complemented in $X$, $Z\approx X$, and $Z\subseteq Y$. A Banach space $X$ is said to be uniformly subprojective (see [OS15]) if there is a constant $C$ such that for every infinite-dimensional subspace $Y\subseteq X$ there is a further subspace $Z\subseteq Y$ that is $C$-complemented in $X$.

Most other terminology is standard, such as might appear, for instance, in [AK2016].

2. Preliminaries

In order to state some embeddings involving Garling sequence spaces we will invoke the weak Lorentz sequence space $d_{\infty}(\mathbf{w},p)$ for $\mathbf{w}\in\mathcal{W}$ and $1\leq<\infty$, consisting of all sequences $f=(a_{n})_{n=1}^{\infty}\in c_{0}$ so that

[TABLE]

where $(a_{n}^{*})_{n=1}^{\infty}$ denotes the decreasing rearrangement of $f$. Notice that $d_{\infty}(\mathbf{w},p)$ is a quasi-Banach space (see [CRS2007]).

Lemma 2.1**.**

Suppose $1\leq p<\infty$ and $\mathbf{w}=(w_{n})_{n=1}^{\infty}\in\mathcal{W}$. A sequence $f=(a_{n})_{n=1}^{\infty}\in\mathbb{F}^{\mathbb{N}}$ belongs to the space $d_{\infty}(\mathbf{w},p)$ if and only if $D_{f}:=\sup_{n}\|\sum_{k=1}^{n}a_{k}\mathbf{e}_{k}\|_{d_{\infty}(\mathbf{w},p)}<\infty$, in which case $\|f\|_{d_{\infty}(\mathbf{w},p)}=D_{f}$.

Proof.

We may suppose that $\{n\in\mathbb{N}\colon a_{n}\not=0\}$ is infinite (otherwise there is nothing to prove). Put $s_{n}=(\sum_{k=1}^{n}w_{k})^{1/p}$. Assume that $D_{f}<\infty$. In order to show that $f\in c_{0}$ it suffices to see that for each $\varepsilon>0$ the set $\{k\in\mathbb{N}\colon|a_{k}|\geq\varepsilon\}$ is finite. Let $m\in\mathbb{N}$ be such that $D_{f}<\varepsilon s_{m}$. For a given $j$, denote by $(b_{n}^{*})_{n=1}^{\infty}$ the decreasing rearrangement of $\sum_{n=1}^{j}a_{n}\mathbf{e}_{n}$. We have $b_{m}^{*}<\varepsilon$ and so $|\{n\leq j\colon|a_{n}|\geq\varepsilon\}|\leq m-1$. Since $j$ is arbitrary, we are done.

Now we consider the decreasing rearrangement $(a_{n}^{*})_{n=1}^{\infty}$ of $f$. Given $n\in\mathbb{N}$, pick $r\in\mathbb{N}$ such that $|a_{k}|<a_{n}^{*}$ for $k>r$. We have $a_{n}^{*}=c_{n}^{*}$, where $(c_{n}^{*})_{n=1}^{\infty}$ is the decreasing rearrangement of $\sum_{n=1}^{r}a_{n}\mathbf{e}_{n}$. Therefore $a_{n}^{*}s_{n}\leq D_{f}$. In other words, $f\in d_{\infty}(\mathbf{w},p)$ and $\|f\|_{d_{\infty}(\mathbf{w},p)}\leq D_{f}$.

The reverse inequality and the converse implication are obvious. ∎

Proposition 2.2**.**

Let $1\leq p<\infty$ and $\mathbf{w}\in\mathcal{W}$. Then

[TABLE]

with norm-one inclusions.

Proof.

The embedding $\ell_{p}\subsetneq d(\mathbf{w},p)$ is clear and well known.

Let $f=(a_{n})_{n=1}^{\infty}\in\mathbb{F}^{\mathbb{N}}$ and $\phi\in\mathcal{O}_{f}$. Pick $\sigma\in\Pi$ extending $\phi$. Then

[TABLE]

Taking the supremum on $\phi$ we get $\|f\|_{g(\mathbf{w},p)}\leq\|f\|_{d(\mathbf{w},p)}$.

Now let $f=(a_{n})_{n=1}^{\infty}\in c_{00}$. For a given $n\in\mathbb{N}$ there is $\phi\in\mathcal{O}_{f}$ with $r(\phi)=n$ and $|a_{\phi(k)}|\geq a_{n}^{*}$ for $1\leq k\leq r(\phi)=n$. We have

[TABLE]

so that $\|f\|_{d_{\infty}(\mathbf{w},p)}\leq\|f\|_{g(\mathbf{w},p)}$. An appeal to Lemma 2.1 concludes the proof. ∎

Before proceeding with the proof of the aforementioned subsymmetry of the canonical basis of Garling sequence spaces, let us introduce some linear maps that will be handy.

$\bullet$ Given a sequence of signs $\epsilon=(\epsilon_{n})_{n=1}^{\infty}$ we define $M_{\epsilon}\colon\mathbb{F}^{\mathbb{N}}\to\mathbb{F}^{\mathbb{N}}$ by

[TABLE]

$\bullet$ The coordinate projection on $A\subseteq\mathbb{N}$ is defined by

[TABLE]

where $\lambda_{n}=1$ if $n\in A$ and $\lambda_{n}=0$ otherwise.

$\bullet$ Given $\phi\in\mathcal{O}$ we define $V_{\phi}\colon\mathbb{F}^{\mathbb{N}}\to\mathbb{F}^{\mathbb{N}}$ by $V_{\phi}((a_{n})_{n=1}^{\infty})=(a_{\phi(n)})_{n=1}^{\infty}$,

and

$\bullet$ $U_{\phi}\colon\mathbb{F}^{\mathbb{N}}\to\mathbb{F}^{\mathbb{N}}$ by $U_{\phi}((a_{n})_{n=1}^{\infty})=(b_{n})_{n=1}^{\infty}$, where

[TABLE]

Remark 2.3*.*

Given $\phi\in\mathcal{O}$ and $n\in\mathbb{N}$, We have $U_{\phi}\circ V_{\phi}=\mathrm{Id}_{\mathbb{F}^{\mathbb{N}}}$ and $V_{\phi}(\mathbf{e}_{n})=\mathbf{e}_{\phi(n)}$.

Proposition 2.4**.**

Let $\mathbf{w}\in\mathcal{W}$ and $1\leq p<\infty$. Let $\epsilon=(\epsilon_{n})_{n=1}^{\infty}$ be a sequence of signs, let $A$ be a subset of $\mathbb{N}$, and let $\phi$ a map in $\mathcal{O}$.

• (i)

$M_{\epsilon}$* and $P_{A}$ are norm-one operators from $g(\mathbf{w},p)$ into $g(\mathbf{w},p)$.*

• (ii)

$V_{\phi}$* and $U_{\phi}$ are norm-one operators from $g(\mathbf{w},p)$ into $g(\mathbf{w},p)$.*

• (iii)

$V_{\phi}$* is an isometric embedding from $g(\mathbf{w},p)$ into $g(\mathbf{w},p)$.*

• (iv)

The standard unit vectors form a $1$-subsymmetric basic sequence in $g(\mathbf{w},p)$.

Proof.

(i) is clear; (iii) is a consequence of (ii) and Remark 2.3; (iv) is a consequence of (i), (iii), and Remark 2.3. Thus we must only care to show (ii). Let $f=(a_{n})_{n=1}^{\infty}\in\mathbb{F}^{\mathbb{N}}$. It is obvious that $\|M_{\epsilon}(f)\|_{g}=\|f\|_{g}$ and that $\|P_{A}(f)\|_{g}\leq\|f\|_{g}$. Since $\phi\circ\psi\in\mathcal{O}$ for all $\psi\in\mathcal{O}$,

[TABLE]

Let $U_{\phi}(f)=(b_{n})_{n=1}^{\infty}$. Pick $\psi\in\mathcal{O}$. The function $\phi^{-1}\circ\psi$ is an increasing map from a set $A\subseteq\mathbb{N}$ to $\mathbb{N}$. Put $J=\{n\in\mathbb{N}\colon n\leq|A|\}$ and choose $\gamma\colon J\to A$ increasing and bijective. Since $\rho:=\phi^{-1}\circ\psi\circ\gamma\in\mathcal{O}\cup\mathcal{O}_{f}$ and $n\leq\gamma(n)$ for all $n\in J$,

[TABLE]

Taking the supremum on $\psi$ we get $\|U_{\phi}(f)\|_{g}\leq\|f\|_{g}$. ∎

Definition 2.5**.**

Let $(\mathbf{x}_{n})_{n=1}^{\infty}$ be a basis for a Banach space $X$ and let $g=\sum_{n=1}^{\infty}b_{n}\mathbf{x}_{n}$ be a vector in $X$. Given another $f=\sum_{n=1}^{\infty}a_{n}\mathbf{x}_{n}\in X$ we write $f\prec g$ (with respect to the basis $(\mathbf{x}_{n})_{n=1}^{\infty}$) if

• •

$\operatorname{supp}(f)\subseteq\operatorname{supp}(g)$,

• •

$a_{n}=b_{n}$ for all $n\in\operatorname{supp}(f)$, and

• •

$\|f\|=\|g\|$.

We say that $g$ is minimal in $X$ (with respect to $(\mathbf{x}_{n})_{n=1}^{\infty}$) if it is minimal in $X$ equipped with the partial order $\prec$, i.e., if $f\prec g$ implies $f=g$.

Remark 2.6*.*

Note that if $g$ is finitely supported with respect to a basis $(\mathbf{x}_{n})_{n=1}^{\infty}$ then there exists a minimal $f\prec g$.

Remark 2.7*.*

Every sequence in $d(\mathbf{w},p)$, $1\leq p<\infty$, is minimal. The situation is quite different for $g(\mathbf{w},p)$. For instance, if $(1-w_{2})t^{p}\geq 1$, then the vector $\mathbf{g}_{1}+t\mathbf{g}_{2}$ is not minimal.

Lemma 2.8**.**

Let $\mathbf{w}=(w_{n})_{n=1}^{\infty}\in\mathcal{W}$ and $1\leq p<\infty$. For any $A\subseteq\mathbb{N}$ finite we have:

• (i)

$\|\sum_{n\in A}\mathbf{e}_{n}\|_{g}=\sum_{n=1}^{|A|}w_{n}$.

• (ii)

The vector $\sum_{n\in A}\mathbf{e}_{n}$ is minimal in $g(\mathbf{w},p)$ (with respect to its canonical basis).

Proof.

Part (i) can be obtained, for example, from Proposition 2.2. To prove (ii), let $f\prec\sum_{n\in A}\mathbf{e}_{n}$. Then $f=\sum_{n\in B}\mathbf{e}_{n}$ for some $B\subseteq A$, and

[TABLE]

Consequently $|B|=|A|$ and so $B=A$. ∎

Lemma 2.9**.**

Let $\mathbf{w}\in\mathcal{W}$ and $1\leq p<\infty$. Suppose $f\in c_{00}$ is minimal in $g(\mathbf{w},p)$ with respect to the canonical basis. Then

[TABLE]

where $\psi\in\mathcal{O}_{f}$ is determined by $\operatorname{R}(\psi)=\operatorname{supp}f$.

Proof.

Given $\phi\in\mathcal{O}_{f}$, let $\psi$ be the map in $\mathcal{O}_{f}$ determined by $\operatorname{R}(\psi)=\operatorname{R}(\phi)\cap\operatorname{supp}f$. Let $\gamma$ by the inverse function of $\phi$ restricted to $\operatorname{R}(\psi)$. We have

[TABLE]

Therefore in order to compute the supremum in (1.1) we can restrict ourselves to $\{\phi\in\mathcal{O}_{f}\colon\operatorname{R}(\phi)\subseteq\operatorname{supp}f\}$, so that this supremum is attained. Let $\psi$ be the element in $\mathcal{O}_{f}$ with $\operatorname{R}(\psi)\subseteq\operatorname{supp}(f)$ where the supremum in (1.1) is attained. Let $h$ be the projection of $f$ onto $\operatorname{R}(\psi)$. Then

[TABLE]

hence $h=f$ and $\operatorname{R}(\psi)=\operatorname{supp}f$. ∎

Recall that a block basic sequence of a basic sequence $(\mathbf{x}_{n})_{n=1}^{\infty}$ in a Banach space $X$ is a sequence $(\mathbf{y}_{n})_{n=1}^{\infty}$ of non-zero vectors of the form

[TABLE]

for some (unique) $(a_{n})_{n=1}^{\infty}\in\mathbb{F}^{\mathbb{N}}$ and some increasing sequence of integers $(p_{n})_{n=1}^{\infty}$ with $p_{1}=1$.

Definition 2.10**.**

Let $\mathbf{y}_{n}=\sum_{i=p_{n}}^{p_{n+1}-1}a_{i}\mathbf{x}_{i}$, $n\in\mathbb{N},$ be a block basic sequence of a basic sequence $(\mathbf{x}_{n})_{n=1}^{\infty}$ in a Banach space $X$. Then:

1. (a)

$(\mathbf{y}_{n})_{n=1}^{\infty}$ is said to be uniformly null if $\lim_{n}a_{n}=0$. 2. (b)

$(\mathbf{y}_{n})_{n=1}^{\infty}$ is said to verify the gliding hump property if

[TABLE]

Remark 2.11*.*

A subsequence of a block basic sequence also is a block basic sequence. Moreover, we have the following dichotomy: a block basic sequence is either uniformly null or has a subsequence verifying the gliding hump property.

Lemma 2.12**.**

Let $(\mathbf{x}_{n})_{n=1}^{\infty}$ be a semi-normalized basis in a Banach space $X$. If $(\mathbf{y}_{n})_{n=1}^{\infty}$ is a semi-normalized uniformly null block basic sequence of $(\mathbf{x}_{n})_{n=1}^{\infty}$ then $\lim_{n}|\operatorname{supp}(\mathbf{y}_{n})|=\infty$.

Proof.

Assume the claim fails. Then there is an infinite subset $A\subseteq\mathbb{N}$ with $D=\sup_{n\in A}|\operatorname{supp}(\mathbf{y}_{n})|<\infty$. Write $\mathbf{y}_{n}=\sum_{i=p_{n}}^{p_{n+1}-1}a_{i}\mathbf{x}_{i}$ and put

[TABLE]

We have $B\leq CDC_{n}$ for all $n\in A$. Letting $n$ tend to infinity trough $A$ we obtain $B\leq 0$, an absurdity. ∎

We shall also need two properties on block basic sequences of subsymmetric bases that we gather in the following proposition.

Proposition 2.13**.**

Let $X$ be a Banach space with a subsymmetric basis $(\mathbf{x}_{n})_{n=1}^{\infty}$ and suppose $(\mathbf{y}_{n})_{n=1}^{\infty}$ is a semi-normalized block basic sequence of $(\mathbf{x}_{n})_{n=1}^{\infty}$.

1. (i)

If $\sup_{n}|\operatorname{supp}(\mathbf{y}_{n})|<\infty$ then $(\mathbf{y}_{n})_{n=1}^{\infty}$ is equivalent to $(\mathbf{x}_{n})_{n=1}^{\infty}$. 2. (ii)

If $(\mathbf{y}_{n})_{n=1}^{\infty}$ verifies the gliding hump property then $(\mathbf{x}_{n})_{n=1}^{\infty}\lesssim(\mathbf{y}_{n})_{n=1}^{\infty}$.

Proof.

The proofs follow the steps of [ACL73]*Proposition 3 and [ACL73]*Proposition 4, respectively, and so we leave them out for the reader. ∎

To make this paper as self-contained as possible we record the following well known fact.

Proposition 2.14** (cf. [ACL73]*Proposition 1).**

Every subsymmetric basic sequence in a Banach space is either weakly null or else equivalent to the canonical basis of $\ell_{1}$.

Corollary 2.15**.**

Let $(\mathbf{x}_{n})_{n=1}^{\infty}$ be a basic sequence in a Banach space $X$. If $(\mathbf{x}_{n})_{n=1}^{\infty}$ is equivalent to the unit vector basis of $g(\mathbf{w},p)$ for $\mathbf{w}\in\mathcal{W}$ and $1\leq p<\infty$, then it is weakly null.

Proof.

It is straightforward from Proposition 2.2, Proposition 2.4, and Proposition 2.14. ∎

We put an end to this preliminary section recalling a precise version of Bessaga-Pelczyński Selection Principle that we will need below.

Theorem 2.16** (cf. [AK2016]*Proposition 1.3.10).**

Let $X$ be a Banach space with a basis $(\mathbf{x}_{n})_{n=1}^{\infty}$. Let $(\mathbf{y}_{n})_{n=1}^{\infty}$ be a normalized weakly null sequence in $X$ and let $\epsilon>0$. Then there exists a subsequence $(\mathbf{y}_{n_{k}})_{n=1}^{\infty}$ of $(\mathbf{y}_{n})_{n=1}^{\infty}$ that is $(1+\epsilon)$-equivalent to a normalized block basic sequence $(\mathbf{z}_{k})_{k=1}^{\infty}$ of $(\mathbf{x}_{n})_{n=1}^{\infty}$. Moreover, if $T\colon[\mathbf{z}_{k}]_{k=1}^{\infty}\to[\mathbf{y}_{n_{k}}]_{n=1}^{\infty}$ is the isomorphism given by $T(\mathbf{z}_{k})=\mathbf{y}_{n_{k}}$, we have that whenever $Z\subseteq[\mathbf{z}_{k}]_{k=1}^{\infty}$ is $C$-complemented in $X$ then $T(Z)$ is $C(1+\epsilon)$-complemented in $X$.

3. Geometric properties of Garling sequence spaces

Our ultimate goal in this section is to establish the following set of structural results about the spaces $g(\mathbf{w},p)$.

Theorem 3.1**.**

Let $1\leq p<\infty$ and $\mathbf{w}=(w_{n})_{n=1}^{\infty}\in\mathcal{W}$.

• (i)

The unit vectors in $g(\mathbf{w},p)$ span the entire space and form a boundedly complete basis.

• (ii)

Every basic sequence in $g(\mathbf{w},p)$ equivalent to its canonical basis admits a subsequence spanning a subspace complemented in $g(\mathbf{w},p)$.

• (iii)

If $1<p<\infty$ then $g(\mathbf{w},p)$ is reflexive, but if $p=1$ then $g(\mathbf{w},1)$ is nonreflexive.

• (iv)

No subspace of $\ell_{p}$ is isomorphic to $g(\mathbf{w},p)$.

• (v)

For every $\epsilon>0$ and every infinite-dimensional closed subspace $Y$ of $g(\mathbf{w},p)$ there exists a further subspace $Z\subseteq Y$ that is $(1+\epsilon)$-isomorphic to $\ell_{p}$ and $(1+\epsilon)$-complemented in $g(w,p)$.

• (vi)

The identity operator on $\ell_{p}$ factors through the inclusion map $I_{d,g}\colon d(\mathbf{w},p)\to g(\mathbf{w},p)$.

To that end, let us begin with the following Proposition.

Proposition 3.2**.**

Let $1\leq p<\infty$ and $\mathbf{w}=(w_{n})_{n=1}^{\infty}\in\mathcal{W}$. Every normalized block basic sequence of the canonical basis of $g(\mathbf{w},p)$ is 1-dominated by the canonical basis of $\ell_{p}$.

Proof.

Let

[TABLE]

be a normalized block basic sequence. For $i\in\mathbb{N}$ define $n=n(i)\in\mathbb{N}$ by $p_{n}\leq i\leq p_{n+1}-1$. Let $(b_{n})_{n=1}^{\infty}\in c_{00}$. Denote $(c_{i})_{i=1}^{\infty}:=\sum_{n=1}^{\infty}b_{n}\mathbf{y}_{n}$. We have $c_{i}=b_{n(i)}a_{i}$ for all $i\in\mathbb{N}$. Let $\phi\in\mathcal{O}$. For each $n\in\mathbb{N}$, $J_{n}:=\{i\in\mathbb{N}\colon p_{n}\leq\phi(i)\leq p_{n+1}-1\}$ is an integer interval and, then, for some $q_{n}$, $r_{n}\in\mathbb{N}\cup\{0\}$, $J_{n}=\{j\in\mathbb{N}\colon 1+q_{n}\leq j\leq q_{n}+r_{n}\}$. We have

[TABLE]

Taking the supremum on $\phi$ yields $\left\|\sum_{n=1}^{\infty}b_{n}\mathbf{y}_{n}\right\|^{p}\leq\sum_{n=1}^{\infty}|b_{n}|^{p}.$ ∎

We are now able to state a powerful result. In its proof we will make use of the following construction.

Definition 3.3**.**

Let $(\mathbf{x}_{n})_{n=1}^{\infty}$ be a basis in a Banach space $X$. Given a block basic sequence $(\mathbf{y}_{n})_{n=1}^{\infty}$ of $(\mathbf{x}_{n})_{n=1}^{\infty}$,

[TABLE]

its left-shifted block basic sequence $(\hat{}\mathbf{y}_{n})_{n=1}^{\infty}$ is the block basic sequence of $(\mathbf{x}_{n})_{n=1}^{\infty}$ constructed from $(\mathbf{y}_{n})_{n=1}^{\infty}$ as follows: Consider $\phi\in\mathcal{O}$ with range $\{i\in\mathbb{N}\colon a_{i}\not=0\}$ and put

[TABLE]

where $\hat{p}_{1}=1$, and $\hat{p}_{n+1}-\hat{p}_{n}=|\operatorname{supp}\mathbf{y}_{n}|$ for all $n\in\mathbb{N}$.

Theorem 3.4**.**

Let $1\leq p<\infty$ and $\mathbf{w}\in\mathcal{W}$. Suppose $(\mathbf{y}_{n})_{n=1}^{\infty}$ is a uniformly null normalized block basic sequence of the canonical basis of $g(\mathbf{w},p)$. Then for each $\epsilon>0$ there exists a subsequence $(\mathbf{y}_{n_{k}})_{k=1}^{\infty}$ of $(\mathbf{y}_{n})_{n=1}^{\infty}$ that is $(1+\epsilon)$-equivalent to the canonical basis of $\ell_{p}$ and such that $[{\mathbf{y}}_{n_{k}}]_{k=1}^{\infty}$ is $(1+\epsilon)$-complemented in $g(\mathbf{w},p)$.

Proof.

By Proposition 2.4 (i), without loss of generality we can assume that each $\mathbf{y}_{n}$ has nonnegative coefficients with respect to $(\mathbf{g}_{i})_{i=1}^{\infty}$. For each $n\in\mathbb{N}$ pick a minimal $\tilde{}\mathbf{y}_{n}\prec\mathbf{y}_{n}$. Let

[TABLE]

be the left-shifted block basic sequence of $({\tilde{}\mathbf{y}}_{n})_{n=1}^{\infty}$, where $(a_{i})_{i=1}^{\infty}$ is a sequence of positive scalars converging to zero and $(p_{n})_{n=1}^{\infty}$ is an increasing sequence of integers with $p_{1}=1$. By Lemma 2.12 we have $\lim_{n}(p_{n+1}-p_{n})=\infty$. Let $\mathbf{w}=(w_{n})_{n=1}^{\infty}$ and pick $\alpha=(1+\epsilon)^{-p}\in(0,1)$. We claim that there exist increasing sequences of positive integers $(q_{k})_{k=1}^{\infty}$ and $(n_{k})_{k=1}^{\infty}$ with $q_{1}=1$ so that for each $k\in\mathbb{N}$:

• •

$q_{k+1}-q_{k}=p_{1+n_{k}}-p_{n_{k}}$,

• •

$A_{k}:=\sum_{i=q_{k}}^{q_{k+1}-1}a_{i+p_{n_{k}}-q_{k}}^{p}w_{i}\geq\alpha$,

To prove the claim we proceed recursively. Suppose that $q_{k}$ and $n_{k-1}$ have been constructed (for the case $k=1$ we put $n_{0}=0$). Since $\mathbf{w}\in c_{0}$ we can find $L\in\mathbb{N}$ so that

[TABLE]

Then we choose $M\in\mathbb{N}$ such that for any $i\geq M$

[TABLE]

Select $j=n_{k}\in\mathbb{N}$ so that $j>n_{k-1}$, $p_{j}\geq M$ and $p_{j+1}-p_{j}>L+q_{k}$. Set $q_{k+1}=p_{j+1}-p_{j}+q_{k}$. Let $\phi\in\mathcal{O}$ determined by

[TABLE]

We have $\tilde{}\mathbf{y}_{j}=V_{\phi}(\hat{}\mathbf{y}_{j})$. Hence, by Lemma 2.4 (iii), $\hat{}\mathbf{y}_{j}$ is minimal. Then, by Lemma 2.9,

[TABLE]

and so

[TABLE]

This completes the proof of our claim.

Consider the linear map $S\colon\mathbb{F}^{\mathbb{N}}\to\mathbb{F}^{\mathbb{N}}$ given by

[TABLE]

Let $f=(c_{i})_{i=1}^{\infty}\in\mathbb{F}^{\mathbb{N}}$. Hölder’s inequality gives

[TABLE]

That is, $S$ is an $\alpha^{-1/p}$-bounded operator from $g(\mathbf{w},p)$ into $\ell_{p}$.

The left-shifted block basic sequence of $(\hat{}\mathbf{y}_{n_{k}})_{k=1}^{\infty}$, which we denote by $(\mathbf{z}_{k})_{k=1}^{\infty}$, is given by the formula

[TABLE]

Since the left-shifted block basic sequences of $(\tilde{}\mathbf{y}_{n_{k}})_{n=1}^{\infty}$ and $(\hat{\mathbf{y}}_{n_{k}})_{k=1}^{\infty}$ coincide, there is $\psi\in\mathcal{O}$ such that $V_{\psi}(\mathbf{z}_{k})=\tilde{}\mathbf{y}_{n_{k}}$ for all $k\in\mathbb{N}$. Let $T=S\circ U_{\psi}\circ P_{A}$. By Proposition 2.4 (i) and (ii), $T$ is a $(1+\epsilon)$-bounded operator from $g(\mathbf{w},p)$ into $\ell_{p}$.

By Proposition 3.2, $(\mathbf{y}_{n_{k}})_{k=1}^{\infty}\lesssim_{1}(\mathbf{f}_{k})_{k=1}^{\infty}$. In other words, there is a norm-one operator $R\colon\ell_{p}\to g(\mathbf{w},p)$ with $R(\mathbf{f}_{k})=\mathbf{y}_{n_{k}}$ for all $k\in\mathbb{N}$. For every $k\in\mathbb{N}$ we have

[TABLE]

Consequently, $T\circ R=\mathrm{Id}_{\ell_{p}}$. We infer that $R\circ T$ is a projection from $g(\mathbf{w},p)$ onto $[\mathbf{y}_{n_{k}}]_{k=1}^{\infty}$ and that $(\mathbf{f}_{k})_{k=1}^{\infty}\lesssim_{1+\epsilon}(\mathbf{y}_{n_{k}})_{k=1}^{\infty}$. ∎

Remark 3.5*.*

If we consider the natural lattice structure both in $g(\mathbf{w},p)$ and in $\ell_{p}$, the maps $R$ and $T$ constructed in the proof of Theorem 3.4 are lattice homomorphisms.

Corollary 3.6**.**

Every normalized block basic sequence of $(\mathbf{g}_{n})_{n=1}^{\infty}$ admits a subsequence that dominates $(\mathbf{g}_{n})_{n=1}^{\infty}$.

Proof.

This is an immediate consequence of Remark 2.11, Proposition 2.13 (ii), Theorem 3.4, and the the embedding $\ell_{p}\subseteq g(\mathbf{w},p)$ provided by Proposition 2.2, which reads as $(\mathbf{g}_{n})_{n=1}^{\infty}\lesssim(\mathbf{f}_{n})_{n=1}^{\infty}$. ∎

Before proving Theorem 3.1 we need a couple more results.

Proposition 3.7**.**

Let $1\leq p<\infty$ and $\mathbf{w}\in\mathcal{W}$. If $(\mathbf{y}_{n})_{n=1}^{\infty}$ is a semi-normalized block basic sequence of the canonical basis of $g(\mathbf{w},p)$ then

[TABLE]

Proof.

From Remark 2.11 in combination with Proposition 2.13 (ii) and Theorem 3.4 we infer that we can find a subsequence $(\mathbf{y}_{n_{k}})_{k=1}^{\infty}$ that either dominates $(\mathbf{g}_{n})_{n=1}^{\infty}$ or else is equivalent to $(\mathbf{f}_{n})_{n=1}^{\infty}$. In either case, since, by Proposition 2.2, $(\mathbf{f}_{n})_{n=1}^{\infty}$ dominates $(\mathbf{g}_{n})_{n=1}^{\infty}$, we have $(\mathbf{g}_{n})_{n=1}^{\infty}\lesssim(\mathbf{y}_{n_{k}})_{k=1}^{\infty}$. Put $\mathbf{w}=(w_{n})_{n=1}^{\infty}$. Appealing to the $1$-unconditionality of $(\mathbf{g}_{n})_{n=1}^{\infty}$ and to Lemma 2.8,

[TABLE]

Since $\mathbf{w}\notin\ell_{1}$ we are done. ∎

Proposition 3.8**.**

Let $1\leq p<\infty$ and $\mathbf{w}=(w_{n})_{n=1}^{\infty}\in\mathcal{W}$. Suppose $(\mathbf{y}_{n})_{n=1}^{\infty}$ is a block basic sequence of the canonical basis of $g(\mathbf{w},p)$. Then for any $\epsilon>0$ there exists a further block basic sequence of $(\mathbf{y}_{n})_{n=1}^{\infty}$ that is $(1+\epsilon)$-equivalent to the canonical basis of $\ell_{p}$ and $(1+\epsilon)$-complemented in $g(\mathbf{w},p)$.

Proof.

Without loss of generality we assume that $(\mathbf{y}_{n})_{n=1}^{\infty}$ is normalized. By Proposition 3.7 we can recursively construct an increasing sequence $(p_{n})_{n=1}^{\infty}$ of positive integers such that $\left\|\sum_{i=p_{n}}^{p_{n+1}-1}\mathbf{y}_{i}\right\|_{g}\geq n$ for all $n\in\mathbb{N}$. Then the block basic sequence

[TABLE]

is uniformly null, and so by Theorem 3.4 we can find a subsequence $(\mathbf{z}_{n_{k}})_{k=1}^{\infty}$ that is $(1+\epsilon)$-equivalent to the canonical basis of $\ell_{p}$ and $(1+\epsilon)$-complemented in $g(\mathbf{w},p)$.∎

Completion of the Proof of Theorem 3.1.

(i) By Proposition 3.7, no block basic sequence of $(\mathbf{g}_{n})_{n=1}^{\infty}$ is equivalent to the canonical basis of $c_{0}$. Appealing to [AK2016, Theorem 3.3.2]) we infer that $(\mathbf{g}_{n})_{n=1}^{\infty}$ is a boundedly complete basic sequence.

Let us prove that $[\mathbf{g}_{n}]_{n=1}^{\infty}=g(\mathbf{w},p)$. Take $f=(a_{n})_{n=1}^{\infty}\in g(\mathbf{w},p)$. Then $\sup_{m}\|\sum_{n=1}^{m}a_{n}\mathbf{g}_{n}\|_{g}<\infty.$ Since $(\mathbf{g}_{n})_{n=1}^{\infty}$ is boundedly complete, the series $\sum_{n=1}^{\infty}a_{n}\mathbf{g}_{n}$ converges to some $h\in g(\mathbf{w},p)$. Obviously, $f=h$.

(ii) Let $(\mathbf{y}_{n})_{n=1}^{\infty}$ be a basic sequence in $g(\mathbf{w},p)$ equivalent to its canonical basis. By Corollary 2.15, $(\mathbf{y}_{n})_{n=1}^{\infty}$ is weakly null. Then, applying Theorem 2.16 and passing to a subsequence, we can assume that $(\mathbf{y}_{n})_{n=1}^{\infty}$ is a block basic sequence of $(\mathbf{g}_{n})_{n=1}^{\infty}$. By Theorem 3.4 and Remark 2.11, since $(\mathbf{g}_{n})_{n=1}^{\infty}$ has no subsequence equivalent to the $\ell_{p}$-basis, passing to a further subsequence we have that $(\mathbf{y}_{n})_{n=1}^{\infty}$ has the gliding hump property. Let $K\geq 1$ be such that $(\mathbf{y}_{n})_{n=1}^{\infty}\lesssim_{K}(\mathbf{g}_{n})_{n=1}^{\infty}$. Write

[TABLE]

for some $(a_{n})_{n=1}^{\infty}\in\mathbb{F}^{\mathbb{N}}$ and some increasing sequence $(p_{n})_{n=1}^{\infty}$ with $p_{1}=1$. For each $n\in\mathbb{N}$ pick an integer sequence $({i_{n}})_{n=1}^{\infty}$ with $p_{n}\leq i_{n}<p_{n+1}$ and so that $\inf_{n\in\mathbb{N}}|a_{i_{n}}|:=c>0$. Observe that for any $(b_{n})_{n=1}^{\infty}\in c_{00}$, by 1-subsymmetry of $(\mathbf{g}_{n})_{n=1}^{\infty}$ we have

[TABLE]

Thus we can define a bounded linear map $P\colon g(\mathbf{w},p)\to[\mathbf{y}_{n}]_{n=1}^{\infty}$ by the rule

[TABLE]

Since $P(\mathbf{y}_{n})=\mathbf{y}_{n}$ for all $n\in\mathbb{N}$ we are done.

(iii) Let us suppose that $(\mathbf{g}_{n})_{n=1}^{\infty}$ fails to be shrinking. Then due to unconditionality of $(\mathbf{g}_{n})_{n=1}^{\infty}$, there exists a block basic sequence $(\mathbf{y}_{n})_{n=1}^{\infty}$ of $(\mathbf{g}_{n})_{n=1}^{\infty}$ that is equivalent to the canonical $\ell_{1}$-basis (see, e.g., [AK2016, Theorem 3.3.1]). Now, by Proposition 3.8, we can find a further block basic sequence equivalent to the canonical $\ell_{p}$-basis. However, this is impossible if $p\neq 1$. Thus, for $p>1$, $(\mathbf{g}_{n})_{n=1}^{\infty}$ is shrinking, and since it is also boundedly complete by part (i), and unconditional, it must span a reflexive space by a theorem of James (see, e.g., [AK2016, Theorem 3.2.19]). This proves the first part of (iii). The second part about nonreflexivity in case $p=1$ follows since $g(\mathbf{w},1)$ contains a (nonreflexive) copy of $\ell_{1}$ by Proposition 3.8.

(iv) Assume that $g(\mathbf{w},p)$ embeds into $\ell_{p}$. Then $\ell_{p}$ contains a basic sequence $(\mathbf{y}_{n})_{n=1}^{\infty}\approx(\mathbf{g}_{n})_{n=1}^{\infty}$. By Corollary 2.15, $(\mathbf{y}_{n})_{n=1}^{\infty}$ is weakly null. Hence (see, e.g., [AK2016, Proposition 2.1.3]) $(\mathbf{y}_{n})_{n=1}^{\infty}$ has a sequence equivalent to the canonical basis $(\mathbf{f}_{n})_{n=1}^{\infty}$ of $\ell_{p}$. By Proposition 2.4, $(\mathbf{g}_{n})_{n=1}^{\infty}\approx(\mathbf{f}_{n})_{n=1}^{\infty}$, which it false by Proposition 2.2.

(v) Fix $\epsilon>0$, and let $Y$ be any infinite-dimensional closed subspace of $g(\mathbf{w},p)$. By a classical result of Mazur (see, e.g., [AK2016, Theorem 1.4.5]), we can find a normalized basic sequence $(\mathbf{y}_{n})_{n=1}^{\infty}$ in $Y$. In case $p>1$, this basic sequence must be weakly null by the reflexivity of $g(\mathbf{w},p)$. For the case $p=1$, we can assume by Rosenthal’s $\ell_{1}$ Theorem (see, e.g., [AK2016, Theorem 11.2.1]) together with James’ $\ell_{1}$ Distortion Theorem (see, e.g., [AK2016, Theorem 11.3.1]) that $(\mathbf{y}_{n})_{n=1}^{\infty}$ admits a weakly Cauchy subsequence. By passing to a further subsequence if necessary, the sequence

[TABLE]

is normalized and weakly null. Let us relabel if necessary so that in all cases $1\leq p<\infty$ we have found a normalized and weakly null basic sequence $(\mathbf{y}_{n})_{n=1}^{\infty}$ in $Y$. By Theorem 2.16 we can pass to a subsequence if necessary so that $(\mathbf{y}_{n})_{n=1}^{\infty}$ is $(\sqrt{1+\epsilon})$-equivalent to a normalized block basic sequence $(\mathbf{y}^{\prime}_{n})_{n=1}^{\infty}$ of the canonical basis of $g(\mathbf{w},p)$. Moreover, if $Z\subseteq[\mathbf{y}_{n}^{\prime}]_{n=1}^{\infty}$ is $C$-complemented in $g(\mathbf{w},p)$ and $T$ is the operator defined by $T(\mathbf{y}^{\prime}_{n})=\mathbf{y}_{n}$, then $T(Z)$ is $(\sqrt{1+\epsilon})C$-complemented in $g(\mathbf{w},p)$. Now we apply Proposition 3.8 to find a block basic sequence $(\mathbf{z}^{\prime}_{n})_{n=1}^{\infty}$ of $(\mathbf{y}^{\prime}_{n})_{n=1}^{\infty}$ which is $(\sqrt{1+\epsilon})$-equivalent to the canonical basis $(\mathbf{f}_{n})_{n=1}^{\infty}$ of $\ell_{p}$ such that $[\mathbf{z}^{\prime}_{n}]_{n=1}^{\infty}$ is $(\sqrt{1+\epsilon})$-complemented in $g(\mathbf{w},p)$. For $n\in\mathbb{N}$ let $\mathbf{z}_{n}=T(\mathbf{z}^{\prime}_{n})$ It is clear that $(\mathbf{z}_{n})_{n=1}^{\infty}$ is $(\sqrt{1+\epsilon})$-equivalent to $(\mathbf{z}^{\prime}_{n})_{n=1}^{\infty}$, and hence $(1+\epsilon)$-equivalent to $(\mathbf{f}_{n})_{n=1}^{\infty}$. It follows that $[\mathbf{z}_{n}]_{n=1}^{\infty}$ is a subspace of $Y$ that is $(1+\epsilon)$-isomorphic to $\ell_{p}$ and $(1+\epsilon)$-complemented in $g(\mathbf{w},p)$.

(vi) The block basic sequence

[TABLE]

is normalized and uniformly null. By Lemma 2.8, each $\mathbf{y}_{n}$ is minimal. Thus, given $\epsilon>0$, by Theorem 3.4 we can find a subsequence $(\mathbf{y}_{n_{k}})_{k=1}^{\infty}$ of $(\mathbf{y}_{n})_{n=1}^{\infty}$ that is $(1+\epsilon)$-equivalent to the canonical $\ell_{p}$-basis and such that $[\mathbf{y}_{n_{k}}]_{k=1}^{\infty}$ is $(1+\epsilon)$-complemented in $g(\mathbf{w},p)$. Let $P\colon g(\mathbf{w},p)\to[\mathbf{y}_{n_{k}}]_{k=1}^{\infty}$ be a projection with $\|P\|\leq 1+\epsilon$ and let $S\colon[\mathbf{y}_{n_{k}}]_{k=1}^{\infty}\to\ell_{p}$ be given by $S(\mathbf{y}_{n_{k}})=\mathbf{f}_{k}$. We have $\|S\|\leq 1+\epsilon$. Pick $(\mathbf{z}_{n})_{n=1}^{\infty}$ in $d(\mathbf{w},p)$ so that $\mathbf{y}_{n}=I_{d,g}(\mathbf{z}_{n})$. The sequence $(\mathbf{z}_{n})_{n=1}^{\infty}$ is also normalized. Then, by the $d(\mathbf{w},p)$-analog of Proposition 3.2 (see [ACL73]*Proposition 5) $(\mathbf{z}_{n_{k}})_{n=1}^{\infty}$ is $1$-dominated by the canonical basis of $\ell_{p}$, that is, the linear operator $T\in\mathcal{L}(\ell_{p},d(\mathbf{w},p))$ given by $T(\mathbf{f}_{k})=\mathbf{z}_{n_{k}}$ for $k\in\mathbb{N}$ verifies $\|T\|\leq 1$. The composition operator $S\circ P\circ I_{d,g}\circ T$ is the identity map on $\ell_{p}$. ∎

The longed for features of Garling sequence spaces that were advertised in Section 1 follow now readily from Theorem 3.1.

Corollary 3.9**.**

Let $\mathbf{w}\in\mathcal{W}$ and $1\leq p<\infty$. The space $g(\mathbf{w},p)$ is complementably homogeneous and uniformly complementably $\ell_{p}$-saturated.

Remark 3.10*.*

There are $d(\mathbf{w},p)$-analogs of Theorem 3.1 (ii) and Theorem 3.1 (v) (cf. [CL74]*Corollary 12 and [ACL73]*Theorem 1, respectively). Then, for $\mathbf{w}\in\mathcal{W}$ and $1\leq p<\infty$, $d(\mathbf{w},p)$ is also complementably homogeneous and uniformly complementably $\ell_{p}$-saturated.

4. Uniqueness of subsymmetric basis in $g(\mathbf{w},p)$

The spaces $g(\mathbf{w},p)$ do not have a unique unconditional basis since they are not isomorphic to any of the only three spaces that enjoy that property, namely $c_{0}$, $\ell_{1}$, or $\ell_{2}$ (see, e.g., [AK2016]*Theorem 9.3.1). The aim of this section is to show that they do have a unique subsymmetric basis.

Theorem 4.1**.**

Let $1\leq p<\infty$ and $\mathbf{w}\in\mathcal{W}$. Suppose $(\mathbf{y}_{n})_{n=1}^{\infty}$ is a subsymmetric basic sequence in $g(\mathbf{w},p)$. Then every subsymmetric basis for $Y=[\mathbf{y}_{n}]_{n=1}^{\infty}$ is equivalent to $(\mathbf{y}_{n})_{n=1}^{\infty}$. In particular, every subsymmetric basis in $g(\mathbf{w},p)$ is equivalent to its canonical basis.

Proof.

Assume $Y$ is not isomorphic to $\ell_{p}$, else we are done. Now let $(\mathbf{u}_{n})_{n=1}^{\infty}$ be any subsymmetric basis for $Y$. Since neither $(\mathbf{y}_{n})_{n=1}^{\infty}$ nor $(\mathbf{u}_{n})_{n=1}^{\infty}$ are equivalent to the canonical basis of $\ell_{1}$, Proposition 2.14 gives that both sequences are weakly null. Thus, we may apply Theorem 2.16 to obtain $(\mathbf{u}^{\prime}_{n})_{n=1}^{\infty}$ a normalized block basic sequence of $(\mathbf{y}_{n})_{n=1}^{\infty}$ and $(\mathbf{y}^{\prime}_{n})_{n=1}^{\infty}$ a normalized block basic sequence of $(\mathbf{u}_{n})_{n=1}^{\infty}$ with $(\mathbf{u}^{\prime}_{n})_{n=1}^{\infty}\approx(\mathbf{u}_{n})_{n=1}^{\infty}$ and $(\mathbf{y}^{\prime}_{n})_{n=1}^{\infty}\approx(\mathbf{y}_{n})_{n=1}^{\infty}$. By Theorem 3.4 neither $(\mathbf{u}^{\prime}_{n})_{n=1}^{\infty}$ nor $(\mathbf{y}^{\prime}_{n})_{n=1}^{\infty}$ are uniformly null since, otherwise, either $(\mathbf{u}_{n})_{n=1}^{\infty}$ or $(\mathbf{y}_{n})_{n=1}^{\infty}$ would be equivalent to the canonical basis of $\ell_{p}$. Combining subsymmetry with Remark 2.11 it follows that, passing to subsequences if necessary, we may assume that both $(\mathbf{u}^{\prime}_{n})_{n=1}^{\infty}$ and $(\mathbf{y}^{\prime}_{n})_{n=1}^{\infty}$ verify the gliding hump property. Proposition 2.13 (ii) yields

[TABLE]

∎

Remark 4.2*.*

It is also easy to check that Theorem 4.1 holds as well for $d(\mathbf{w},p)$ in place of $g(\mathbf{w},p)$ and so $d(\mathbf{w},p)$ admits a unique subsymmetric basis (which is also symmetric).

5. Nonexistence of symmetric basis in $g(\mathbf{w},p)$

In 2004, B. Sari [Sa04, §6] proved that there exist Banach spaces, called Tirilman spaces, admitting a subymmetric basis but failing to admit a symmetric one (in fact, those spaces do not even contain symmetric basic sequences!). The aim of this section is, on the one hand, to determine when $g(\mathbf{w},p)$ admits a symmetric basis, and on the other hand, to show that for a wide class of weights $\mathbf{w}\in\mathcal{W}$, the space $g(\mathbf{w},p)$ does not have such a basis despite having a unique subsymmetric basis.

Theorem 5.1**.**

Let $1\leq p<\infty$ and $\mathbf{w}=(w_{n})_{n=1}^{\infty}\in\mathcal{W}$. The following are equivalent.

1. (i)

The canonical basis is a symmetric basis in $g(\mathbf{w},p)$. 2. (ii)

$g(\mathbf{w},p)$* admits a symmetric basis.* 3. (iii)

The canonical bases of $g(\mathbf{w},p)$ and $d(\mathbf{w},p)$ are equivalent, that is $g(\mathbf{w}.p)=d(\mathbf{w},p)$ up to an equivalent norm. 4. (iv)

$g(\mathbf{w},p)$* and $d(\mathbf{w},p)$ are isomorphic.* 5. (v)

$d(\mathbf{w},p)$* contains a subspace isomorphic to $g(\mathbf{w},p)$.* 6. (vi)

The inclusion map $I_{d,g}\colon d(\mathbf{w},p)\to g(\mathbf{w},p)$ preserves a copy of $d(\mathbf{w},p)$ that is complemented in $g(\mathbf{w},p)$. More precisely, there exists a subspace $Y$ of $d(\mathbf{w},p)$ such that $Y\approx d(\mathbf{w},p)$, $I_{d,g}|_{Y}$ is bounded below, and $I_{d,g}(Y)$ is complemented in $g(\mathbf{w},p)$. 7. (vii)

There is an operator $T\in\mathcal{L}(d(\mathbf{w},p),g(\mathbf{w},p))$ such that $T\circ I_{d,g}$ preserves a copy of $d(\mathbf{w},p)$.

Proof.

The implications (i) $\Rightarrow$ (ii), (iii) $\Rightarrow$ (i), (iii) $\Rightarrow$ (iv), (iv) $\Rightarrow$ (v), and (iii) $\Rightarrow$ (vi) are all trivial.

(ii) $\Rightarrow$ (i) A symmetric basis for $g(\mathbf{w},p)$ is also subsymmetric and so, by Theorem 4.1, equivalent to $(\mathbf{g}_{n})_{n=1}^{\infty}$. Hence $(\mathbf{g}_{n})_{n=1}^{\infty}$ is symmetric.

(i) $\Rightarrow$ (iii) There is uniform constant $K$ such that $(\mathbf{g}_{\sigma(n)})_{n=1}^{\infty}\approx_{K}(\mathbf{g}_{n})_{n=1}^{\infty}$ for all $\sigma\in\Pi$. Given $f=(a_{n})_{n=1}^{\infty}\in c_{0}$ and $\sigma\in\Pi$ we have

[TABLE]

Taking the supremum in $\sigma$ we get $\|f\|_{d}\leq\|f\|_{g}$. In other words, $(\mathbf{d}_{n})_{n=1}^{\infty}\lesssim_{K}(\mathbf{g}_{n})_{n=1}^{\infty}$. The estimate $(\mathbf{g}_{n})_{n=1}^{\infty}\lesssim_{1}(\mathbf{d}_{n})_{n=1}^{\infty}$ is a consequence of Proposition 2.2.

(v) $\Rightarrow$ (iii) Let $(\mathbf{y}_{n})_{n=1}^{\infty}$ be a basic sequence in $d(\mathbf{w},p)$ equivalent to the canonical basis of $g(\mathbf{w},p)$. By Corollary 2.15 $(\mathbf{y}_{n})_{n=1}^{\infty}$ is weakly null. Then, via Theorem 2.16 and passing to a subsequence, we obtain that $(\mathbf{y}_{n})_{n=1}^{\infty}$ is equivalent to a (semi-normalized) block basic sequence of $(\mathbf{d}_{n})_{n=1}^{\infty}$. By [ACL73, Corollary 2] and passing to a further subsequence, we have $(\mathbf{d}_{n})_{n=1}^{\infty}\lesssim(\mathbf{y}_{n})_{n=1}^{\infty}$. Since $(\mathbf{g}_{n})_{n=1}^{\infty}\lesssim_{1}(\mathbf{d}_{n})_{n=1}^{\infty}$ we are done.

(vi) $\Rightarrow$ (vii) Let $P$ be a projection from $g(\mathbf{w},p)$ onto $I_{d,g}(Y)$ and $S$ be the inverse map of $I_{d,g}|_{Y}$. We have that $S\circ P\circ I_{d,g}$ preserves a copy of $d(\mathbf{w},p)$.

(vii) $\Rightarrow$ (iii) By [KPSTT12, Theorem 5.5], the sequence $((T\circ I_{d,g})(\mathbf{d}_{n}))_{n=1}^{\infty}$ does not converges to zero in norm. This means that we can find a subsequence $(T(\mathbf{g}_{n_{k}}))_{k=1}^{\infty}$ that is semi-normalized. As $(\mathbf{g}_{n})_{n=1}^{\infty}$ is weakly null by Corollary 2.15, so is $(T(\mathbf{g}_{n_{k}}))_{k=1}^{\infty}$. Pass to a further subsequence if necessary so that, by applying Theorem 2.16, $(T(\mathbf{g}_{n_{k}}))_{k=1}^{\infty}$ is (semi-normalized) and equivalent to a block basic sequence of $(\mathbf{d}_{n})_{n=1}^{\infty}$. Again by [ACL73, Corollary 2] we may assume, passing to a further subsequence, that $(T(\mathbf{g}_{n_{k}}))_{k=1}^{\infty}\gtrsim(\mathbf{d}_{n})_{n=1}^{\infty}$. Hence,

[TABLE]

∎

Remark 5.2*.*

Notice that, by Theorem 3.1 (viii), the inclussion map $I_{d,g}\colon d(\mathbf{w},p)\to g(\mathbf{w},p)$ is never strictly singular.

Remark 5.3*.*

Observe that $g(\mathbf{w},p)$ is never isometric in the natural way to $d(\mathbf{w},p)$. Just consider the minimum $k\in\mathbb{N}$ such that $w_{k+1}<1$, so that, letting

[TABLE]

and $f=\mathbf{e}_{k+1}+\alpha\sum_{n=1}^{k}\mathbf{e}_{n}$ we have $\|f\|_{g}=1<\|f\|_{d}$.

Before constructing weights for which $(\mathbf{g}_{n})_{n=1}^{\infty}$ is not symmetric we need a few definitions.

Definition 5.4**.**

Let $\mathbf{w}=(w_{n})_{n=1}^{\infty}$ be a weight.

1. (a)

$\mathbf{w}$ is said to be essentially decreasing if $\sup_{k\leq n}w_{n}/w_{k}<\infty$. 2. (b)

$\mathbf{w}$ is said to be regular if

[TABLE] 3. (c)

The conjugate weight $\mathbf{w}^{*}=(w_{n}^{*})_{n=1}^{\infty}$ is defined by

[TABLE] 4. (d)

$\mathbf{w}$ is said to be bi-regular if both $\mathbf{w}$ and $\mathbf{w}^{*}$ are regular weights.

Lemma 5.5**.**

A weight $\mathbf{w}$ is essentially decreasing if and only if there exists a normalized non-increasing weight $\mathbf{v}$ such that $\mathbf{w}\approx\mathbf{v}$.

Proof.

If $(w_{n})_{n=1}^{\infty}$ is an essentially decreasing weight then the weight $\mathbf{v}=(v_{n})_{n=1}^{\infty}$ given by $v_{n}=(\inf_{k\leq n}w_{k})/w_{1}$ does the trick. The converse is (also) trivial. ∎

Lemma 5.6**.**

If a weight $\mathbf{w}$ is regular then $\mathbf{w}\notin\ell_{1}$ and $\mathbf{w}^{*}\in c_{0}$.

Proof.

Assume, by contradiction, that $\mathbf{w}=(w_{n})_{n=1}^{\infty}\in\ell_{1}$, that is, $\sum_{n=1}^{\infty}w_{n}\approx 1$. Then, $1/m\lesssim w_{m}$ for $m\in\mathbb{N}$, and so $\mathbf{w}\notin\ell_{1}$. Now assume, again by contradiction, that $\mathbf{w}^{*}=(w_{n})_{n=1}^{\infty}\notin c_{0}$. Then there is an infinite subset $A\subseteq\mathbb{N}$ with $nw_{n}\approx 1$ for $n\in A$. Consequently, $\sum_{n=1}^{m}w_{n}\lesssim 1$ for $m\in A$ and so $\mathbf{w}\in\ell_{1}$. ∎

Lemma 5.7**.**

Let $\mathbf{w}=(w_{n})_{n=1}^{\infty}$ be an essentially decreasing regular weight. Then:

1. (i)

$mw_{m}\approx\sum_{n=1}^{m}w_{n}$* for $m\in\mathbb{N}$.* 2. (ii)

$\mathbf{w}^{*}$* also is essentially decreasing.*

Proof.

(i) Let $C=\sup_{k\leq n}w_{n}/w_{k}$. For any $m\in\mathbb{N}$ we have

[TABLE]

(ii) By Part (i), $\mathbf{w}^{*}\approx(1/(\sum_{n=1}^{m}w_{n}))_{m=1}^{\infty}$. We infer from Lemma 5.6 that $\mathbf{w}^{*}$ is essentially decreasing. ∎

Lemma 5.8**.**

A bi-regular weight is essentially decreasing if and only if $mw_{m}\lesssim\sum_{n=1}^{m}w_{n}$ for $m\in\mathbb{N}$.

Proof.

The “only if” part is a consequence of Lemma 5.7. Assume that $mw_{m}\approx\sum_{n=1}^{m}w_{n}$ for $m\in\mathbb{N}$. By Lemma 5.6 $\mathbf{w}^{*}$ is essentially decreasing. Then, by Lemma 5.7, $\mathbf{w}=\mathbf{w}^{**}$ is essentially decreasing. ∎

Proposition 5.9**.**

Let $\mathbf{w}=(w_{n})_{n=1}^{\infty}$ be a weight. The following are equivalent:

• (i)

$\mathbf{w}$* is essentially decreasing and bi-regular;*

• (ii)

$\mathbf{w}$* is bi-regular and $\mathbf{w}^{*}$ is essentially decreasing;*

• (iii)

$\mathbf{w}$* is essentially decreasing and*

[TABLE]

• (iv)

$\mathbf{w}^{*}$* is essentially decreasing and (5.1) holds.*

Proof.

(i) $\Rightarrow$ (ii) is straightforward from Lemma 5.7 (ii), while (ii) $\Rightarrow$ (i) is a consequence of (i) $\Rightarrow$ (ii) and the fact that $\mathbf{w}^{**}=\mathbf{w}$.

(i) $\Rightarrow$ (iii) Let $\mathbf{w}^{*}=(w_{n}^{*})_{n=1}^{\infty}$ and put

[TABLE]

Let $m\in\mathbb{N}$ and choose $r\in\mathbb{N}$ such that $2r-1\leq m\leq 2r$. We have

[TABLE]

as desired.

(iii) $\Rightarrow$ (i) Let $\mathbf{w}^{*}=(w_{n}^{*})_{n=1}^{\infty}$ and put

[TABLE]

Let us prove that $\mathbf{w}^{*}$ is essentially decreasing. Let $k$ and $n$ be integers with $1\leq n\leq k$. Since the function $\eta$ given by $\eta(x):=\log(1+x)/x$ is decreasing on $(0,\infty)$,

[TABLE]

Hence

[TABLE]

Now, for any $m\in\mathbb{N}$,

[TABLE]

and

[TABLE]

The equivalence (ii) $\Leftrightarrow$ (iv) is a consequence of (i) $\Leftrightarrow$ (iii) and the simple fact that

[TABLE]

where $\mathbf{w}^{*}=(w_{n}^{*})_{n=1}^{\infty}$. ∎

Theorem 5.10**.**

Let $\mathbf{w}$ be a normalized non-increasing bi-regular weight and $1\leq p<\infty$. Then $\mathbf{w}\in\mathcal{W}$ and the canonical basis is not a symmetric basis for $g(\mathbf{w},p)$.

Proof.

The fact that $\mathbf{w}\in\mathcal{W}$ is straightforward from Lemma 5.6. For each $r\in\mathbb{N}$, consider sequences $f^{(r)}=(a_{r,n})_{n=1}^{\infty}$ and $g^{(r)}=(b_{r,n})_{n=1}^{\infty}$ given by

[TABLE]

Notice that $g^{(r)}$ is a rearrangement of $f^{(r)}$. We have

[TABLE]

By Lemma 5.7 (ii), $\mathbf{w}^{*}=(w_{n}^{*})_{n=1}^{\infty}$ is an essentially decreasing sequence. Let $r\in\mathbb{N}$ and $\phi\in\mathcal{O}$. Denote by $m(r,\phi)$ the largest integer $m$ such that $\phi(m)\leq r$. We have $\phi(n)\leq n+r-m$ for $1\leq n\leq m$. Therefore,

[TABLE]

Proposition 5.9 yields that the canonical basis is not symmetric. ∎

To give relief to Theorem 5.10 we need to exhibit examples of normalized non-increasing bi-regular weights. To that end, in light of Lemma 5.6, if suffices to give examples of essentially decreasing bi-regular weights. Next, we give a very general procedure for constructing this kind of weights. For instance, we will prove that, for $0<a<1$ and $b\in\mathbb{R}$, $((\log(1+n))^{b}n^{-a})_{n=1}^{\infty}$ is a essentially decreasing bi-regular weight.

Definition 5.11**.**

A weight $(u_{n})_{n=1}^{\infty}$ is said to be asymptotically constant if

[TABLE]

Proposition 5.12**.**

Let $0<a<1$ and $(u_{n})_{n=1}^{\infty}$ be an asymptotically constant weight. Then $\mathbf{w}=(n^{-a}u_{n})_{n=1}^{\infty}$ is an essentially decreasing bi-regular weight.

Proof.

Let $b=1-a$ and consider $\mathbf{v}=(v_{n})_{n=1}^{\infty}$ given by $v_{n}=1/u_{n}$. We have $\mathbf{w}^{*}=(n^{-b}v_{n})_{n=1}^{\infty}$, with $0<b<1$ and $\mathbf{v}$ asymptotically constant. Then, taking into account Lemma 5.8 it suffices to prove

[TABLE]

To that end, put

[TABLE]

and let $(a_{n})_{n=1}^{\infty}$ be the unique weight such that $a_{2^{j-1}}=c_{j}$ for $j\in\mathbb{N}$ and $(a_{n})_{n=2^{j-1}}^{2^{j}}$ is in arithmetic progression. Define $(s_{n})_{n=0}^{\infty}$ by $s_{0}=0$ and $s_{n}=n^{1-a}a_{n}$ for $n\in\mathbb{N}$. Since

[TABLE]

we have $t_{n}\approx s_{n}$ for $n\in\mathbb{N}$. Thus we need only show that

[TABLE]

Given $n\geq 2$, pick $j=j(n)\in\mathbb{N}$ such that $2^{j-1}<n\leq 2^{j}$. We have

[TABLE]

∎

We close with a most natural question that our work leaves open:

Question 5.13*.*

Do there exist $\mathbf{w}\in\mathcal{W}$ and $1\leq p<\infty$ so that $g(\mathbf{w},p)$ admits a symmetric basis?

Acknowledgment

Ben Wallis thanks Bünyamin Sarı for his helpful comments.

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