Quotients, $\ell _\infty$ and abstract Cesàro spaces

Abstract

Investigating some re-arrangement properties of the norm in the quotient spaces $X/X_a$ we determine the properties of the codomain space X guaranteeing the existence of a lattice isometric copy of $\ell _\infty$ in the abstract Cesàro space CX.

1 Introduction and preliminaries

Let ${{\mathscr {P}}}$ be a property for which it makes sense to ask whether a Banach space has this property or not. For a Banach ideal space X we can now pose the following problem: what condition, say $\varepsilon$, should we impose on the space X so that the implication:

holds true? Here CX stands for the Cesàro space which can be viewed, for example, as an optimal domain for the Hardy operator $C :f \mapsto C(f)(x) := \frac{1}{x}\int _0^x f(t) dt$, that is, the biggest in the sense of inclusion Banach ideal space such that the operator C with fixed codomain space X is still bounded. The Cesàro spaces themselves constitute only a special, but important, subclass of a very broad family of spaces generated via sublinear operators (see [3, 18, 19, 21, 34, 35] and [58]); this class of spaces includes, for example:

• Köthe–Bochner spaces,

• real interpolation spaces,

• extrapolation spaces,

• Besov and Triebel–Sobolev spaces,

• some other optimal domains associated with the kernel operators (like multiplication operator, Volterra operator, Cesàro operator, Copson operator, Poisson operator or Riemann–Liouville operator), differential operators, convolutions, the Fourier transform, the (finite) Hilbert transform and the Sobolev embedding,

cf. [62] and references given there. Many, more or less classical results, can be seen through the prism ($\Delta$) of the above scheme, where the Cesàro operator C is replaced by the appropriate sublinear operator, say, S; let us give few concrete examples of this type of results, which, however, remain close enough to the Cesàro spaces: (a) Astashkin [3], solving the problem posed in [58], proved that every non-trivial subspace of a Banach space $D_{p}(S)$ generated by some positive sublinear operator S and an $L_p$-space with $1 \leqslant p < \infty$ contains, for any $\varepsilon > 0$, an $(1+\varepsilon )$-copy of $\ell _p$ which is $(1 + \varepsilon )$-complemented in $D_p(S)$ (note that this result generalizes the well-known Levy result for the Lions–Petree interpolation spaces $(X_0,X_1)_{\theta ,p}$ and, on the other hand, the earlier Astashkin and Maligranda result [5] for the classical Cesàro function spaces $Ces_p$); (b) Hudzik and Wlaźlak in [34] and [35] studied various convexity and monotonicity properties, say ${{\mathscr {G}}}$, of the space $D_X(S)$, where X is a Banach ideal space. Essentially, they were able to show that if $X \in ({{\mathscr {G}}})$ and the injective positive sublinear operator S has some special geometric properties corresponding to the property ${{\mathscr {G}}}$, then also $D_X(S) \in ({{\mathscr {G}}})$. This result implies immediate applications for the Köthe–Bochner spaces and the Cesàro–Orlicz spaces (cf. [43] for a more direct approach in the case of different types of the Cesàro function spaces); (c) Mastyło in [58] and [59] investigated some structural properties of the space $D_X(S)$.

In this paper we give a general and rather natural condition to ensure that the Cesàro space CX contains a lattice isometric copy of $\ell _\infty$, as long as a codomain space X is rearrangement invariant and contains such a copy as well.

A Banach space $(X, \Vert \cdot \Vert _X)$ is said to be a Banach lattice if there is a lattice order on X, say $\leqslant$, such that for $x,y \in X$ with $\left|x\right| \leqslant \left|y\right|$ we have $\Vert x\Vert _X \leqslant \Vert y\Vert _X$. By $X_+$ we denote the positive cone of a Banach lattice X, that is, $X_+ = \{x \in X :x \geqslant 0 \}$.

A mapping T between two Banach lattices X and Y is said to be a lattice (or an order) isomorphism if it is a linear topological isomorphism which preserves the order $\leqslant$. If, additionally, the operator norm of a lattice isomorphism $T :X \rightarrow Y$ is equal one, then we will emphasize this fact by calling T a lattice (or an order) isometry.

Let $(\Omega ,\Sigma ,\mu )$ be a complete $\sigma$-finite measure space. Denote by $L_{0}(\Omega ) = L_{0}(\Omega ,\Sigma ,\mu )$ the set of all (equivalence classes of) real-valued $\mu$-measurable functions defined on a measure space $(\Omega ,\Sigma ,\mu )$. A Banach ideal space $X = (X,\Vert \cdot \Vert _X)$ on $(\Omega ,\Sigma ,\mu )$ is understood to be a Banach space X such that X is a linear subspace of $L_{0}(\Omega ,\Sigma ,\mu )$ satisfying the so-called ideal property, which means that if $f,g\in L_{0}(\Omega ,\Sigma ,\mu )$, $\left|f(t)\right| \leqslant \left|g(t)\right|$ $\mu$-almost everywhere on $\Omega$ and $g\in X$, then $f\in X$ and $\Vert f\Vert _{X}\leqslant \Vert g\Vert _{X}$. If it is not stated otherwise we assume that a Banach ideal space X on $(\Omega ,\Sigma ,\mu )$ contains a function which is positive $\mu$-almost everywhere on $\Omega$ (such a function is called the weak unit in X). For a function $f\in L_{0}(\Omega ,\Sigma ,\mu )$ we define a support of f as a set ${{\,\mathrm{supp}\,}}(f):= \{x\in \Omega :f(x) \ne 0\}$. Moreover, by a support ${{\,\mathrm{supp}\,}}(X)$ of the Banach ideal space X on $(\Omega ,\Sigma ,\mu )$ we mean the smallest (in the sense of inclusion) $\mu$-measurable subset, say A, of $\Omega$ such that $f\chi _{\Omega {\setminus } A} = 0$ for all $f\in X$. Recall also that the Banach ideal space X has a weak unit if and only if ${{\,\mathrm{supp}\,}}(X)=\Omega$. We say that a Banach ideal space X is non-trivial if $X \ne \left\{ 0 \right\}$.

For two Banach ideal spaces X and Y on $\Omega$ the symbol $X \overset{E}{\hookrightarrow } Y$ means that the embedding $X \subset Y$ is continuous with the norm not bigger than $E > 0$, that is to say, $\Vert f\Vert _Y \leqslant E\Vert f\Vert _X$ for all $f \in X$. If the embedding $X \overset{E}{\hookrightarrow } Y$ holds with some (maybe unknown) constant $E > 0$ we simply write $X \hookrightarrow Y$. Moreover, the symbol $X = Y$ ($X \equiv Y$) means that the spaces are the same as a sets and the norms are equivalent (equal, respectively).

We say that an element x in a Banach lattice X is order continuous (or has an order continuous norm) if for any sequence $\{x_n\}_{n=1}^\infty$ in X satisfying $0 \leqslant x_n \leqslant \left|x\right|$ and $x_n \downarrow 0$ (that is, $x_{n+1} \leqslant x_n$ and $\inf _{n \in {\mathbb {N}}}\{x_n\} = 0$) we have $\Vert x_n\Vert \rightarrow 0$ as $n \rightarrow \infty$. By $X_a$ we denote the subspace of all order continuous elements of X. If $X = X_a$, that is, every element of X is order continuous, then the space X is said to be order continuous (we write $X \in (OC)$ shortly).

By $X^*$ we denote the topological dual space of a Banach lattice X. If $X^{dd} = X$ then we have the Yosida–Hewitt decomposition of the space $X^*$, namely

$$\begin{aligned} X^* = X_a^* \oplus X^s, \end{aligned}$$

where $X_a^*$ is the space of all order continuous functionals, that is to say, $x^* \in X_a^*$ if and only if $x^*(x_n) \rightarrow 0$ as $n \rightarrow \infty$ for any sequence $\{x_n\}_{n=1}^\infty \subset X_+$ and $x_n \downarrow 0$, and $X^s$ is the space of all singular functionals, that is to say, $x^* \in X^s$ if and only if $x^*(x) = 0$ for any $x \in X_a$. For an order continuous Banach lattice X the spaces $X_a^*$ and $X^*$ coincide. If X is a Banach ideal space on $(\Omega ,\Sigma ,\mu )$ such that ${{\,\mathrm{supp}\,}}(X_a) = {{\,\mathrm{supp}\,}}(X)$ then the space $X_a^*$ coincide with the Köthe dual space (or associated space) $X'$

$$\begin{aligned} X' := \left\{ f \in L_0(\Omega ) :\sup \limits _{\Vert g\Vert _X \leqslant 1} \int _\Omega \left|f(t)g(t)\right|d\mu < \infty \right\} . \end{aligned}$$

Recall that $X \overset{1}{\hookrightarrow } X''$ and we have the equality $X \equiv X''$ if and only if the norm in X has the Fatou property, that is, the conditions $0 \leqslant f_{n}\uparrow f \in L_{0}$ with $\{f_{n}\}_{n=1}^\infty \subset X$ and $\sup _{n\in {\mathbb {N}}}\Vert f_{n}\Vert _{X}<\infty$ imply that $f\in X$ and $\Vert f_{n}\Vert _{X}\uparrow \Vert f\Vert _{X}$.

A closed linear subspace J of a Banach lattice X is called an order ideal (or simply ideal) if it has the so-called ideal property which means that if $f \in J$, $g \in X$ and $\left|g\right| \leqslant \left|f\right|$ then also $g \in J$ (subspaces with this property are also called solid). It is clear that $\{ 0 \}$ and the space X itself are order ideals of X. Non-trivial example of an order ideal is the closure (in norm topology of the Banach ideal space X) of the set of simple functions $X_b$ (see [7,  Theorem 3.11, p. 18]) and $X_a$, that is, the subspace of all functions with order continuous norm in X (see [7,  Theorem 3.8, p. 16]). An intersection and a sum of any two ideals is again an ideal (cf. [60,  Proposition 1.2.2, p. 12]). If J is a closed ideal in a Banach lattice X, then X/J is a Banach lattice with respect to the quotient norm, that is, $\Vert x\Vert _{X / J} = \inf \{\Vert y\Vert _X :Q(y) = Q(x) \}$, where $Q :J \rightarrow X / J$ is the canonical quotient map (cf. [60,  Corollary 1.3.14, p. 32]).

In this paper we will deal with Banach lattices but the main focus will be on rearrangement invariant spaces (or symmetric spaces), that is to say, the Banach ideal spaces with the additional property that for any two equimeasurable functions $f, g \in L_{0}$ (which means that they have the same distribution functions $d_{f} \equiv d_{g}$, where $d_{f}(\lambda ) := \mu (\{t\in \Omega :|f(t)| > \lambda \})$ for $\lambda \geqslant 0$) and if $f\in X$ then $g \in X$ and $\Vert f\Vert _{X}=\Vert g\Vert _{X}$. Due to the Luxemburg representation theorem, it is enough to consider rearrangement invariant spaces only on three separable measure spaces, namely, the unit interval $\left( 0,1\right)$ or the semi-axis $\left( 0,\infty \right)$ with the Lebesgue measure m (in this case we will talk about function spaces) and the set of positive integers ${\mathbb {N}}$ with the counting measure $\#$ (and then we will refer to sequence spaces). Moreover, by a non-increasing rearrangement of a function $f :\Omega \rightarrow {\mathbb {R}}$ we mean the function $f^* :[0,\infty ) \rightarrow [0,\infty ]$ defined as

$$\begin{aligned} f^{*}(t) = \mathrm {inf}\{\lambda > 0 :d_{f}(\lambda ) \leqslant t\} \quad \text {for } t \geqslant 0, \end{aligned}$$

under the convention $\mathrm {inf}\{\emptyset \} = \infty$. In the sequence case, however, we need a little modification of the above definition, namely $x_n^* = \mathrm {inf}\{\lambda > 0 :d_x(\lambda ) < n\}$, where $x = \{x_n\}_{n=1}^\infty$.

The fundamental function $\varphi _{X}$ of a rearrangement invariant space X on $\Omega$ is defined by the following formula

$$\begin{aligned} \varphi _{X}(t)=\Vert \chi _{(0,t)}\Vert _{X}\quad \text {for } t > 0, \end{aligned}$$

(with a fairly obvious modification when $\Omega = {\mathbb {N}}$, that is, $\varphi _X(n) = \Vert \sum _{k=1}^n e_k\Vert _X$ for $n \in {\mathbb {N}}$, where $\{e_n\}_{n=1}^\infty$ is the canonical basic sequence of X), the symbol $\chi _A$, throughout, will denote the characteristic function of a set A. It is well-known that the fundamental function is quasi-concave on $\Omega$, that is: $\varphi _{X}(0) = 0$; $\varphi _{X}$ is positive and non-decreasing; $t \mapsto \varphi _{X}(t)/t$ is non-increasing for $t > 0$ or, equivalently, $\varphi _{X}(t) \leqslant \max \{1,t/s\}\varphi _{X}(s)$ for all $s,t \in \Omega$.

For some general properties of Banach ideal and rearrangement invariant spaces we refer, for example, to [7, 48, 52, 53] and [57]. More information about Banach lattices can be found, for example, in [2, 41] and [60]

Let us recall some examples of rearrangement invariant spaces. Each increasing concave function $\varphi$ on $\Omega$ generates the Lorentz function space $\Lambda _\varphi$ on $\Omega$ endowed with the norm

$$\begin{aligned} \Vert f\Vert _{\Lambda _\varphi } = \int _\Omega f^*(t)d\varphi (t) = \varphi (0^+)\Vert f\Vert _{\infty } + \int _0^{m(\Omega )} f^*(t)\varphi '(t)dt < \infty . \end{aligned}$$

Recall also that for a quasi-concave function $\varphi$ the Marcinkiewicz function space $M_{\varphi }$ on $\Omega$ is defined in the following way

$$\begin{aligned} M_{\varphi } = \left\{ f\in L_{0}(\Omega ) :\Vert f\Vert _{M_{\varphi }} = \sup _{t \in \Omega }\frac{\varphi (t)}{t}\int _0^t f^*(s)ds < \infty \right\} . \end{aligned}$$

Lorentz and Marcinkiewicz spaces play quite a special role among rearrangement invariant spaces, namely they are the smallest and, respectively, the largest (in a sense of inclusion) rearrangement invariant spaces with a given fundamental function. More precisely, for a given rearrangement invariant space X with the fundamental function $\varphi$ (note that every such a function is equivalent to a concave function) we have the following embeddings

$$\begin{aligned} \Lambda _\varphi \overset{2}{\hookrightarrow } X \overset{1}{\hookrightarrow } M_\varphi . \end{aligned}$$

Let $F :[0,\infty ) \rightarrow [0,\infty ]$ be an non-decreasing, convex function on $[0,\infty )$ with $F(0) = 0$ (such a function is called an Orlicz function) and let X be a Banach ideal space on $\Omega$. The Calderón–Lozanovskiĭ space $X_F$ on $\Omega$ is defined as the space of all measurable functions $f :\Omega \rightarrow {\mathbb {R}}$ for which the following norm, the so-called Luxemburg–Nakano norm, is finite

$$\begin{aligned} \Vert f\Vert _{X_F} = \inf \left\{ \lambda > 0 :\Vert F(\left|f\right|/\lambda )\Vert _X \leqslant 1 \right\} < \infty . \end{aligned}$$

The spaces $X_F$ were introduced by Calderón [13,  p. 122] and Lozanovskiĭ [54]. This is a Banach ideal space. Moreover, if X is a rearrangement invariant space with the Fatou property, then the space $X_F$ is like that as well. Note also that in the case when $X = L_1$ the space $X_F$ is just the Orlicz space $L_F$ equipped with the Luxemburg–Nakano norm. On the other hand, if X is a Lorentz space $\Lambda _\varphi$, then $X_F$ is the corresponding Orlicz–Lorentz space $\Lambda _{F,\varphi }$. It is also clear that in the special situation, when $F(t) = t$, the Orlicz–Lorentz space $\Lambda _{F,\varphi }$ coincide, up to equality of norms, with the Lorentz space $\Lambda _\varphi$. Moreover, if $F(t) = t^{p}$, where $1 \leqslant p < \infty$, then the space $X_F$ is the p -convexification $X^{(p)}$ of the space X equipped with the norm $\Vert f\Vert _{X^{(p)}} = \Vert \left|f\right|^p\Vert _X^{1/p}$. Finally, if $F(t) = 0$ for $0 \leqslant t \leqslant 1$ and $F(t) = \infty$ for $t > 1$, then $X_F \equiv L_\infty$.

In the case of sequence spaces we prefer a slight modification of the introduced notation, that is, if this make sense, we will use small letters to denote the space, for example, we will just denote the Lorentz space $\Lambda _\varphi$ on ${\mathbb {N}}$ as $\lambda _\varphi$ and the Orlicz space $L_F$ on ${\mathbb {N}}$ as $\ell _F$, etc.

We will use the notation $A \preccurlyeq B$ or $B \succcurlyeq A$ to denote an estimate of the form $A \leqslant CB$ for some constant $C > 0$ depending on the involved parameters only. We also write $A \asymp B$ for $A \preccurlyeq B \preccurlyeq A$. Other notations and definitions will be introduced as needed.

Now we give a brief overview of the paper.

Section 2 is slightly different in nature from the rest of this article. Here we revisit some Hudzik’s [30] characterizations of Banach lattices containing a lattice isometric copies of $\ell _\infty$ and complete one of them (Theorem 2.1). This connection between, on the one hand, the existence of a lattice isometric copy of $\ell _\infty$ in a Banach lattice X and, on the other, with the fact that the unit sphere in the space X must contain an element, say x, with the property that the distance from x to the order ideal of all order continuous elements in X is equal exactly one, will be our starting point and the leitmotif to which we will come back in later parts of our work.

Section 3 is rather technical in character. In short, we show there that the quotient spaces $X/X_a$, where $X_a$ is the ideal of all order continuous elements in a rearrangement invariant space X, behave somewhat similar to a rearrangement invariant space, that is to say, they have the monotone quotient norm (Corollary 3.2) and they are invariant with respect to a non-increasing rearrangement (Theorem 3.5). We will also explain how our formulas are related to some of de Jonge’s earlier results from [20].

In Sect. 4 we will focus on Cesàro spaces. Using the results from the previous sections we prove some general theorems about a lattice isometric copies of $\ell _\infty$ in these spaces (Theorem 4.1 and Theorem 4.6) and justify (which is not so straightforward) that they really generalize the so far known results obtained in the class of Cesàro–Orlicz spaces in [39] and [44].

Finally, at the end of this paper, we include Appendix, which was created for the purposes of Sect. 4 (and, at the same time, it complements the results obtained by the first and last author in [42]). It contains a fairly satisfactory description of the ideal of all order continuous elements in the Cesàro sequence spaces CX expressed in the language of the ideal $X_a$.

2 About isometric copies of $\ell _{\infty }$ one more time

Lozanowskiĭ’s well-known result from [55] gives some geometric characterization of the order continuity property. More precisely, it states that a Banach lattice X contains a lattice isomorphic copy of $\ell _\infty$ if and only if the space X is not order continuous. Of course, every isomorphic copy of $\ell _\infty$ is automatically complemented in the ambient space due to the fact that the space $\ell _\infty$ is isometrically injective. Note also that a Banach lattice X contains a (lattice) almost isometric copy of $\ell _\infty$ whenever it contains a (lattice) isomorphic copy of $\ell _\infty$ (see [32,  Theorem 1] and [64,  Theorem 3]). It seems to be worth comparing this statement with the classical James result [38] (often called simply the James’s distortion theorem), which describes a similar phenomenon but for spaces $c_0$ and $\ell _1$. Moreover, if a Banach space X contains an asymptotically isometric copy of $\ell _\infty$, then it contains even an isometric copy of $\ell _\infty$ (see [24,  Theorem 6] and [16,  Theorem 2.5] for a “lattice version”). However, in general, the lack of order continuity does not guarantee the existence of any isometric copy of $\ell _\infty$. Indeed, let X and Y be two Banach ideal spaces on $\Omega$ with non-trivial intersection $X \cap Y$ such that the first one is not order continuous and the second one is strictly convex (for example, we can take $X = L_\infty$ and $Y = L_2$). Then the space

$$\begin{aligned} {\mathcal {X}} = X \cap Y \text { with the norm } \Vert f\Vert _{{\mathcal {X}}} = \Vert f\Vert _X + \Vert f\Vert _Y, \end{aligned}$$

is strictly convex and is not order continuous. Consequently, in view of the Lozanowskiĭ result [55], it contains a lattice isomorphic copy of $\ell _\infty$ but clearly cannot contain an isometric copy of $\ell _\infty$. Note also that the same trick will work if instead of strict convexity we will require that the second space is strictly monotone. In this case, however, the space Z will be strictly monotone and not order continuous, so it will contains a lattice isomorphic but not lattice isometric copy of $\ell _\infty$. Let us now recall the following classical result.

Theorem A

(Riesz’s lemma) If Y is a proper closed linear subspace of a normed space $X = (X, \Vert \cdot \Vert )$, then for any $0< \varepsilon < 1$ there exists $x \in S(X)$ such that $d_X({x, Y}) \geqslant 1 - \varepsilon$.

It can be demonstrated that the above result is in general not true for $\varepsilon = 0$. Actually, for a Banach space X to have the property that given a proper closed linear subspace Y of X there exists $x \in S(X)$ with $d _X(x, Y) = 1$ it is necessary and sufficient that the space X is reflexive (cf. [23,  Notes and Remarks on p. 6]). In fact, James [37] proved that a Banach space X is reflexive if and only if every continuous linear functional on X is norm attaining. Therefore, if X is nonreflexive, then there is a linear functional, say $\varphi$, in $S(X^*)$ which does not achieve its norm. Taking $Y = \text {ker}(\varphi )$ we see immediately that Y is a proper closed linear subspace of X and there is no such element $x \in S(X)$ with $d _X(x, Y) = 1$. On the other hand, if X is reflexive and Y is a closed linear subspace of X, then applying the Hahn–Banach theorem and bearing in mind the result of James we see that there exists $\psi \in S(X^*)$ such that $Y \subset \text {ker}(\psi )$ and $x \in S(X)$ with $\psi (x) = 1$. Consequently, we have the following inequalities:

$$\begin{aligned} \Vert x - y\Vert \geqslant \left|\psi (x) - \psi (y)\right| = \left|\psi (x)\right| = 1 \text { for all } y \in Y, \end{aligned}$$

that is, $d _X(x, Y) = 1$. On the other hand, the existence of an element realizing the distance from the ideal $X_a$, where now X is a Banach ideal space, is closely related to the fact that the space X contains a lattice isometric copy of $\ell _\infty$. Anyhow, for $X = \ell _\infty$ and $f = \chi _{{\mathbb {N}}}$ we have $\Vert f\Vert _{X} = d _X(\chi _{{\mathbb {N}}}, X_a = c_0) = 1$ and, in a sense, this is the generic case.

Theorem B

(Hudzik [30]) Let X be a super $\sigma$-Dedekind complete Banach lattice with the semi-Fatou property such that $(X_a^d)^d = X$.¹ If we can find an element $f \in X$ such that $\Vert f\Vert _X = d_{X}({f, X_a}) = 1$, then X contains a lattice isometric copy of $\ell _\infty$.

Let’s us consider the space ${\mathfrak {X}} = (L_\infty , \Vert \cdot \Vert _{{\mathfrak {X}}})$, where $\Vert \cdot \Vert _{{\mathfrak {X}}}$ is an equivalent norm on $L_\infty$ given by the formula: $\Vert f\Vert _{{\mathfrak {X}}} = \Vert f\Vert _{L_\infty } + \Vert f\Vert _{Y}$ for $f \in L_\infty$, and Y is strictly convex space such that $L_\infty \hookrightarrow Y$. Then the subspace ${\mathfrak {X}}_a$ is trivial and

$$\begin{aligned} d_{\mathfrak {X}}({f, {\mathfrak {X}}_a}) = d_{\mathfrak {X}}({f, \{ 0 \}}) = \Vert f\Vert _{{\mathfrak {X}}} = 1 \quad \text {for all } f \in S({\mathfrak {X}}). \end{aligned}$$

However, the space ${\mathfrak {X}}$ cannot contain a lattice isometric copy of $\ell _\infty$ as we have already explained above. Therefore the assumption $(X_a^d)^d = X$ in Theorem B about the support of the ideal $X_a$ cannot be omitted in general.

Needless to say, the question, whether the condition from Theorem B is also sufficient, imposes itself. We believe that, this fact belongs to folklore but, as far as we know, it was never explicitly mentioned, even in [30]. To fill this gap and avoid the impression which the authors themselves had that this condition may only be necessary, we will give a short proof which completes Hudzik’s result.

Theorem 2.1

Let X be a Banach ideal space over a $\sigma$-finite measure space with the semi-Fatou property such that ${{\,\mathrm{supp}\,}}(X_a) = {{\,\mathrm{supp}\,}}(X)$. The space X contains a lattice isometric copy of $\ell _\infty$ if and only if there exists an element $f \in X$ with $\Vert f\Vert _X = d_{X}({f, X_a}) = 1$.

Proof

Using Theorem 1 from [30] we can find a sequence $\{ f_n \}_{n=1}^\infty \subset S(X)$ with ${{\,\mathrm{supp}\,}}(f_n) \cap {{\,\mathrm{supp}\,}}(f_m) = \emptyset$ for $n \ne m$ such that $\Vert \sum _{n=1}^{\infty } f_n\Vert _X = 1$. Take $g \in X_a$ and observe that

$$\begin{aligned} \left\| {\sum _{n=1}^{\infty } f_n - g}\right\| _X&\geqslant \sup \limits _{n \in {\mathbb {N}}} \Vert f_n - g\chi _{{{\,\mathrm{supp}\,}}(f_n)}\Vert _X \\&\geqslant \left|\Vert f_n\Vert _X - \Vert g\chi _{{{\,\mathrm{supp}\,}}(f_n)}\Vert _X\right| \rightarrow 1, \end{aligned}$$

because $\Vert g\chi _{{{\,\mathrm{supp}\,}}(f_n)}\Vert _X \rightarrow 0$ as $n \rightarrow \infty$. $\square$

Referring to Theorem 2.1, if a Banach ideal space X contains a lattice isometric copy of $\ell _\infty$, then we can find $f \in X$ with $\Vert f\Vert _X = d _X(f, X_a) = 1$. So, looking from a slightly different perspective, this means that $\Vert f\Vert _{X / X_a} = 1$ and there exist a non-zero singular functional $S \in S(X^*)$ with $S(f) = 1$ (note only that $(X / X_a)^* \approx X_a^{\perp } = X^s$, where $X^s$ is the space of all singular functionals, that is, those $S \in X^*$ for which $S(x) = 0$ for all $x \in X_a$). The converse of this statement is also true but before we give a short proof let us note the following result.

Lemma 2.2

Let X be a Banach lattice. If S is a singular functional on X, then

$$\begin{aligned} \Vert S\Vert _{X^{*}} = \sup \left\{ \frac{S(f)}{d_{X}({f, X_a})} :f \in X {\setminus } X_a \text { with } \Vert f\Vert _X = 1 \right\} . \end{aligned}$$

Proof

First, observe that $S(f) = 0$ for all $f \in X_a$, so

$$\begin{aligned} \Vert S\Vert _{X^{*}}&= \sup \left\{ S(f) :f \in X \text { with } \Vert f\Vert _X = 1 \right\} \\&= \sup \left\{ S(f) :f \in X {\setminus } X_a \text { with } \Vert f\Vert _X = 1 \right\} \\&\leqslant \sup \left\{ \frac{S(f)}{d _X(f,X_a)} :f \in X {\setminus } X_a \text { with } \Vert f\Vert _X = 1 \right\} , \end{aligned}$$

where the last inequality follows from the fact that if $\Vert f\Vert _X = 1$, then $d _X(f, X_a) \leqslant 1$. Therefore, we only need to prove the reverse inequality. To do this, take $f \in X {\setminus } X_a$ and let $\{ f_n \}_{n=1}^\infty \subset X_a$ be a sequence that realizes the distance of the function f from the ideal $X_a$, that is, $\lim _{n\rightarrow \infty } \Vert f - f_n\Vert _X = d _X(f, X_a)$. Now, since $S(f) = S(f - f_n)$, so

$$\begin{aligned} S(f) \leqslant \Vert S\Vert _{X^{*}} \Vert f - f_n\Vert _X \rightarrow \Vert S\Vert _{X^{*}} d _X(f, X_a) \text { as } n \rightarrow \infty . \end{aligned}$$

Thus, we obtain that

$$\begin{aligned} \Vert S\Vert _{X^{*}} \geqslant \frac{S(f)}{d _X(f,X_a)}, \end{aligned}$$

which ends the proof. $\square$

Note that the above lemma for Orlicz–Lorentz spaces was proved in [40] basing on the modular space structure.

Consequently, if there exists a non-zero norm attaining singular functional on X, say S, then $S(f) = \Vert S\Vert _{X^*}$ for some $f \in B(X)$, that is, $d _X(f, X_a) = 1$ (cf. Lemma 2.2) and the space X contains a lattice isometric copy of $\ell _\infty$ in view of Theorem 2.1. What we have said so far may be summarized as follows.

Proposition 2.3

Let X be a Banach ideal space over a $\sigma$-finite measure space with the semi-Fatou property such that ${{\,\mathrm{supp}\,}}(X_a) = {{\,\mathrm{supp}\,}}(X)$. Then X contains a lattice isometric copy of $\ell _\infty$ if and only if there exists a non-zero singular functional on X which attains its norm.

We will now (seemingly) deviate for a moment from the course chosen so far. In [28] (see also [68]; cf. [67]) Granero and Hudzik extended and completed the results obtained in [50] for the space $\ell _\infty /c_0$² considering a more general construction $\ell _{F}/(\ell _{F})_a$, where $\ell _{F}$ is the Orlicz sequence space (in fact, they even allowed for $\ell _{F}$ to be a Fréchet space). Of course, we can go a step further and consider a neo-classical Banach space $X/X_a$, where X is a Banach lattice. Despite this somewhat baroque name, it is clear that $X/X_a$ is sometimes “classical”, for example, if X is a rearrangement invariant Banach sequence space with the Fatou property such that $\varphi _X(\infty ) < \infty$, then $X / X_a = \ell _\infty / c_0 \approx C(\beta {\mathbb {N}} {\setminus } {\mathbb {N}})$.

Proposition 2.4

Let X be a Banach sequence space such that $X \hookrightarrow \ell _\infty$, the ideal $X_a$ is non-trivial and $X_a \ne X$. Then $X/X_a$ is not a dual space.

Proof

It is well-known that the space $\ell _\infty /c_0$ contains an isomorphic copy of $c_0(\Gamma )$, where $\text {card}(\Gamma ) = {\mathfrak {c}}$ and ${\mathfrak {c}}$ denotes the cardinal of the continuum, that is, ${\mathfrak {c}} = 2^{\aleph _0}$. Let us briefly remind that we can proceed as follows: (1) for each irrational number r, take a sequence $\{ q_n^r \}_{n=1}^\infty$ of distinct rational numbers converging to r and put $F_r = \{ q_n^r :n \in {\mathbb {N}}\}$; (2) the sets $\{ F_r \}_{r \in {\mathbb {R}}}$ form an uncountable family of infinite sets such that $F_r \cap F_{r'}$ is finite whenever $r \ne r'$; (3) observe that $\ell _\infty \equiv \ell _\infty ({\mathbb {Q}})$; (4) the closed linear span of $\{ \pi (\chi _{F_r}) :r \in {\mathbb {R}} \}$, where $\pi :\ell _\infty \rightarrow \ell _\infty /c_0$ is the canonical quotient map, is isometric to $c_0({\mathbb {R}})$ (cf. footnote on page 19 in [66]). Now, if $X/X_a$ were a dual space, then thanks to Rosenthal’s generalization of the classical Bessaga and Pełczyński result [66,  Corollary 1.5], the space $X/X_a$ should have a copy of $\ell _\infty (\Gamma )$. But this is impossible, because

$$\begin{aligned} \text {card}(\ell _\infty (\Gamma )) = 2^{\mathfrak {c}} > {\mathfrak {c}} = \text {card}(\ell _\infty ) \geqslant \text {card}(X) \geqslant \text {card}(X/X_a), \end{aligned}$$

where the second inequality follows from the fact that $X \hookrightarrow \ell _\infty$. $\square$

The above result is a direct generalization of [28,  Proposition 6.2] which was proved in the case when $X = \ell _F$.

Let us now recall the well-known Phillips–Sobczyk theorem, which states that $c_0$ is not complemented in $\ell _\infty$ (to be precise, what Phillips proved was that c is not complemented in $\ell _\infty$; look at [12], cf. [1,  pp. 44–48]). We will close this paragraph by re-proving a “lattice” variant of this theorem and exploring the Bourgain’s approach [10] to this phenomenon.

Proposition 2.5

(Lozanovskiĭ, 1973) Let X be a Banach ideal space on a complete $\sigma$-finite measure space $(\Omega , \Sigma , \mu )$ such that $X_a \ne X$. Then the subspace $X_a$ is not complemented in X except for the case when $X_a$ is trivial.

Proof

Suppose that X is as above and consider the following diagram

$$\begin{aligned} c_0 \hookrightarrow \ell _\infty \overset{T}{\longrightarrow } X \overset{Q}{\longrightarrow } X / X_a, \end{aligned}$$

where T is a lattice isomorphism and Q is the canonical quotient map. It follows that $T(c_0) \subset X_a$ (see Proposition 1 and Theorem 1 in [70]; cf. [65]). Thus, if $q :\ell _\infty \rightarrow \ell _\infty / c_0$ is a quotient mapping, then there exists a unique mapping $U :\ell _\infty / c_0 \rightarrow X / X_a$ defined in the following way $U :x + c_0 \mapsto T(x) + X_a$. It is not hard to see that $QT = Uq$. In short, the following diagram commutes

and the space $X / X_a$ contains a lattice isomorphic copy of $\ell _\infty / c_0$. Now, observe that X can be renormed to be a strictly convex space. In fact, let us consider the multiplication operator $M_y$ given by $M_y :X \ni x \mapsto xy \in L_1(\mu )$, where $y \in X'$ is chosen so that $\Vert y\Vert _{X'} = 1$ and $y(t) > 0$ for $\mu$-a.e. $t \in \Omega$ (the Köthe dual space always has a weak unit; cf. [71,  p. 472]). Then we can define the functional $\sharp \cdot \sharp$ on X as follows

$$\begin{aligned} \sharp x \sharp = \Vert x\Vert _X + \Vert M_y(x)\Vert _{L_1(\mu )}, \end{aligned}$$

Now, since the space $\ell _\infty$ is universal for all separable Banach spaces (see [1,  Theorem 2.5.7, p. 46]) and admit strictly convex renorming (see [22,  p. 94]), so without losing generality we can assume that $L_1(\mu )$ is already strictly convex; in other words, we can always equip the space $L_1(\mu )$ with a new norm, say $\left|\cdot \right|$, such that the space $(L_1(\mu ), \left|\cdot \right|)$ is strictly convex. But this gives that $\sharp \cdot \sharp$ is an equivalent norm on X and that the space $(X, \sharp \cdot \sharp )$ is strictly convex as well (this follows from the well-known fact that if $T :X \rightarrow Y$ is an injective operator and Y is strictly convex Banach space then X can be renormed to be also strictly convex; see [22,  p. 94] for more details). On the other hand, Bourgain [10] proved that the space $\ell _\infty /c_0$ does not admit an equivalent strictly convex norm while, as we showed above, X does. Consequently, the space $X/X_a$, which contains an isomorphic copy of $\ell _\infty / c_0$, is not isomorphic to a subspace of X and this implies that $X_a$ is not complemented in X. $\square$

Another proof of this fact can be found, for example, in [69,  Theorem 1.21, pp. 62–63].

Note also that Partington [64,  Theorem 1] proved that the space $X / X_a$ not only cannot be renormed to be strictly convex, but even contains a lattice isometric copy of $\ell _\infty$ (cf. [28,  Proposition 5.1] and [70,  Theorem 2]).

3 Some properties of the norm in the quotient space

It is clear that the quotient space X/M, where X is a normed space and M is a closed subspace of X, is also a normed space with respect to the norm $\Vert \cdot \Vert _{X/M}$ defined as

$$\begin{aligned} \Vert x + M\Vert _{X/M} := \inf \{ \Vert y\Vert _X :y \in x + M \}. \end{aligned}$$

In simple words, this norm measures the distance from a coset to the origin of the space X/M, that is,

$$\begin{aligned} \Vert x + M\Vert _{X/M} = d _X(x + M, 0 + M) = d _X(x, M) := \inf \{ \Vert x - m\Vert _X :m \in M \}. \end{aligned}$$

Therefore, the study of the function $x \mapsto d _X(x, M)$ is parallel to the study of certain properties of the quotient space X/M and vice versa. For example, de Jonge [20] proved that a Banach lattice X is a semi-M space (without going into detail, this class includes Orlicz spaces $L_F$ with both the Orlicz and the Luxemburg–Nakano norm and Lorentz spaces $\Lambda _\varphi$; see [20,  Examples 2.4 and 9.6]) if and only if $X/X_a$ is an AM-space (or, which is one thing, $(X/X_a)^* = X^s$ is an AL-space), that is to say,

$$\begin{aligned} d _X( \sup \{ x, y \}, X_a) = \max \{ d _X(x, X_a), d _X(y, X_a) \}\quad \text {for } x, y \in X_+. \end{aligned}$$

From this perspective, we will show that if X is a rearrangement invariant Banach space, then the quotient space $X/X_a$ is, in a sense, also “rearrangement invariant” (Theorem 3.5). This result will turn out to be crucial in the next section, where we will deal with abstract Cesàro spaces looking for isometric copies of $\ell _\infty$. But let’s start with some auxiliary lemmas.

Lemma 3.1

Let X be a Banach ideal space on $(\Omega , \Sigma , \mu )$ and let J be an order ideal of X. If $f \in X$ then we have the following formula

$$\begin{aligned} d_X({f, J}) = \inf \{ \Vert f-g\Vert _X :g\in J \text { with } \left|g\right| \leqslant \left|f\right| \}. \end{aligned}$$

Proof

Without loss of generality we can assume that $J \ne \{0\}$ and $f > 0$, that is to say, $f \geqslant 0$ and $f \ne 0$. Of course, to show the above equality we need only to prove that

$$\begin{aligned} \inf \{ \Vert f-g\Vert _X :g\in J \text { such that } 0 \leqslant g \leqslant f \} \leqslant d _X(f, J). \end{aligned}$$

Let $\{ f_n \}_{n=1}^\infty \subset J_+$ be a sequence realizing the distance $d _X(f, J)$, that is, $\lim _{n\rightarrow \infty }\Vert f - f_n\Vert _X = d _X(f, J)$. Set

$$\begin{aligned} \Omega _n = \{ x \in \Omega :f_n(x) \leqslant f(x) \} \text { for } n \in {\mathbb {N}}, \end{aligned}$$

and put

$$\begin{aligned} {\widetilde{f}}_n = f_n\chi _{\Omega _n} + f\chi _{\Omega {\setminus } \Omega _n} \quad \text {for } n \in {\mathbb {N}}. \end{aligned}$$

Observe that ${\widetilde{f}}_n \leqslant f$ for every $n\in {\mathbb {N}}$. Consequently,

$$\begin{aligned} \Vert f - {\widetilde{f}}_n\Vert _X = \Vert (f - f_n)\chi _{\Omega _n}\Vert _X \leqslant \Vert \left|f - f_n\right|\Vert _X = \Vert f - f_n\Vert _X. \end{aligned}$$

Taking limits on both sides of the above inequality, we see immediately that

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\Vert f - {\widetilde{f}}_n\Vert _X \leqslant \lim \limits _{n\rightarrow \infty }\Vert f - f_n\Vert _X = d _X(f, J). \end{aligned}$$

Moreover, since J is an order ideal of X, it follows that $\{ {\widetilde{f}}_n \}_{n=1}^\infty \subset J$ and the proof is finished. $\square$

The immediate conclusion from the above lemma is the following result showing that the quotient norm is order-preserving.

Corollary 3.2

Let X be a Banach ideal space on $(\Omega , \Sigma , \mu )$ and let J be an order ideal of X. If $0 \leqslant g \leqslant f \in X$ then

$$\begin{aligned} d_X({g, J}) \leqslant d_X({f, J}). \end{aligned}$$

Proof

Take $0 < g \leqslant f \in X$. Let $\{ f_n \}_{n=1}^\infty \subset J$ be a sequence that realizes the distance $d _X(f, J)$, that is to say, $\lim _{n\rightarrow \infty } \Vert f - f_n\Vert = d _X(f,J)$. It follows from Lemma 3.1 that we can additionally assume that $f_n \leqslant f$ for every $n\in {\mathbb {N}}$. Put

$$\begin{aligned} g_n = \min \{ g, f_n \} \quad \text {for } n \in {\mathbb {N}}. \end{aligned}$$

Since J is an order ideal of X, it is clear that $\{ g_n \}_{n=1}^\infty \subset J$. Observe also that $0 \leqslant g - g_n \leqslant f - f_n$ for all $n \in {\mathbb {N}}$, so due to the ideal property of X, we get that

$$\begin{aligned} \Vert g - g_n\Vert _X \leqslant \Vert f - f_n\Vert _X. \end{aligned}$$

But this simply means that $d _X(g, J) \leqslant d _X(f, J)$. $\square$

Moreover, the quotient norm is invariant with respect to the absolute value.

Lemma 3.3

Let X be a Banach ideal space on $(\Omega , \Sigma , \mu )$ and let J be an order ideal of X. If $f \in X$ then

$$\begin{aligned} d_X({f, J}) = d_X({\left|f\right|, J}). \end{aligned}$$

Proof

To show one of the inequalities let’s take $f\in X$ and notice that

$$\begin{aligned} d _X(f, J) \geqslant \inf \{\Vert \left|f\right| - \left|g\right|\Vert _X :g \in J \} \geqslant \inf \{\Vert \left|f\right| - h\Vert _X :h \in J \} = d _X(\left|f\right|, J). \end{aligned}$$

On the other hand, for each $g \in J_{+}$ there is $h \in J$ such that $\Vert f - h\Vert _X \leqslant \Vert \left|f\right| - g\Vert _X$. Indeed, for $g \in J_{+}$ set

$$\begin{aligned} h \left( x \right) =\left\{ \begin{array}{ccc} g\left( x \right) &{}\quad \text {if } &{} f\left( x\right) \geqslant 0, \\ -g\left( x \right) &{}\quad \text {if } &{} f\left( x\right) < 0. \end{array} \right. \end{aligned}$$

Then $h \in J$ and $\Vert f - h\Vert _X = \Vert \left|f\right| - g\Vert _X$, whence $d _X(\left|f\right|, J_{+}) \geqslant d _X(f, J)$. Since $d _X(\left|f\right|, J) = d _X(\left|f\right|, J_{+})$ we conclude that $d _X(f, J) \leqslant d _X(\left|f\right|, J)$ and this finishes the proof. $\square$

It will turn out right away that the formula expressing the distance of the non-increasing rearrangement of f from the ideal $X_a$ can be given in a fairly computable form.

Theorem 3.4

Let X be a rearrangement invariant space on $\Omega$, where $\Omega = (0,1)$, $\Omega = (0,\infty )$ or $\Omega = {\mathbb {N}}$, such that the ideal $X_a$ is non-trivial. Then for $f \in X$ we have the following equality

$$\begin{aligned} d_X({f^*,X_a}) = \lim \limits _{n \rightarrow \infty } \Vert f^* \chi _{\Omega _n}\Vert _X, \end{aligned}$$

where $\Omega _n = \Omega \cap \left( (0, \frac{1}{n}) \cup (n,\infty )\right)$.

Proof

We will only give a proof when $\Omega = (0, \infty )$, because the other cases are completely analogous.

To begin with, due to the fact that the sequence $\{ \Vert f^* \chi _{(0,1/n) \cup (n,\infty )}\Vert _X \}_{n=1}^{\infty }$ is non-increasing and bounded from below by 0, the limit of the above sequence must exists. Moreover, we can assume that $X \ne X_a$, because otherwise there is nothing to prove.

Denote

$$\begin{aligned} L = \lim \limits _{n \rightarrow \infty } \Vert f^* \chi _{\left( 0,\frac{1}{n} \right) \cup (n,\infty )}\Vert _X. \end{aligned}$$

First of all, it is not hard to see that

$$\begin{aligned} d _X(f^*,X_a) \geqslant L, \end{aligned}$$

or, equivalently, $\Vert f^* - g\Vert _X \geqslant L$ for each $g \in X_a$. Indeed,

$$\begin{aligned} \Vert f^* - g\Vert _X&\geqslant \lim \limits _{n \rightarrow \infty } \Vert (f^* - g) \chi _{\left( 0,\frac{1}{n} \right) \cup (n,\infty )}\Vert _X \\&\geqslant \lim \limits _{n \rightarrow \infty } \left|\Vert f^* \chi _{\left( 0,\frac{1}{n} \right) \cup (n,\infty )}\Vert _X - \Vert g \chi _{\left( 0,\frac{1}{n} \right) \cup (n,\infty )}\Vert _X\right| = L, \end{aligned}$$

where the equality follows from the fact that $g \in X_a$.

On the other hand, our assumption that $X_a \ne \{ 0 \}$ implies that $\{ f^* \chi _{(1/n,n)} \}_{n=1}^{\infty } \subset X_a$ (cf. [20,  Lemma 7.3 (ii)] or [42,  Theorem B]), so

$$\begin{aligned} d _X(f^*,X_a) = \inf \{\Vert f^* - g\Vert :g \in X_a \} \leqslant \lim \limits _{n \rightarrow \infty } \Vert f^* - f^* \chi _{\left( \frac{1}{n},n \right) }\Vert _X = L, \end{aligned}$$

and the proof is done. $\square$

Now we prove the main result of this section which shows that the quotient norm in the space $X/X_a$ is invariant with respect to the non-increasing rearrangement.

Theorem 3.5

Let X be a rearrangement invariant space on $\Omega$, where $\Omega = (0,1)$ or $\Omega = (0,\infty )$, with the Fatou property. Then for $f \in X$ we have the following equality

$$\begin{aligned} d_X({f^*, X_a}) = d_X({f, X_a}). \end{aligned}$$

Proof

We can assume that $X_{a} \ne \left\{ 0 \right\}$, $X_{a} \ne X$ and $f \notin X_{a}$, because otherwise there is nothing to prove. Moreover, in view of Lemma 3.3, we can also assume that $f > 0$.

Let’s start by noticing that the inequality

is true even without additional assumptions on the function $f \in X$. In fact, we have

$$\begin{aligned} d _X(f, X_a)&= \inf \{\Vert f - g\Vert _X :g \in X_a \} \geqslant \inf \{\Vert f^* - g^*\Vert _X :g \in X_a \} \\&= \inf \{\Vert f^* - g\Vert _X :g \in X_a \quad \text {and}\quad g = g^* \} \geqslant d _X(f^*, X_a), \end{aligned}$$

where the first inequality follows essentially from the Calderón–Ryff theorem (see [7,  Theorem 2.10, p. 114]; cf. also Lemma 4.6 in [48,  p. 95]) and the second equality follows from Lemma 2.6 in [15]. Consequently, it remains to prove the reverse inequality. To do this, we divide the proof into several parts.

$1^{\circ }$. Suppose that $f^{*}\left( \infty \right) = 0$. Then, as a consequence of Ryff’s theorem (see [7,  Corollary 7.6, p. 83]), there is a measure preserving transformation $\sigma$ from the support of f onto the support of $f^*$ such that $f^{*}\circ \sigma = f$ m-almost everywhere on the support of f. If we were able to show that for each $g \in X_{a}$ there exists $h_g \in X_{a}$ such that $\Vert f - h_g\Vert _X \leqslant \Vert f^* - g\Vert _X$, the proof of this part would be complete. For this, let’s take $g \in X_{a}$ and put $h_g = g \circ \sigma$. Due to [15,  Lemma 2.6] and the fact that $h_g^* = (g \circ \sigma )^*$, we conclude that $h_g \in X_a$. Now, since we have the equalities

$$\begin{aligned} \Vert f^* - g\Vert _X = \Vert (f^* - g) \circ \sigma \Vert _X = \Vert f - h_g\Vert _X, \end{aligned}$$

it follows that $d _X(f^*,X_a) \geqslant d _X(f,X_a)$. Combining this inequality with the inequality ($\heartsuit$), we see that we have just finished the proof in the first case.

$2^{\circ }$. Moving on to the second case, let’s assume that $f^*(\infty ) > 0$. Denote

$$\begin{aligned} \Omega _{\Box } = \{x \in \Omega :f(x) \Box f^*(\infty ) \}, \end{aligned}$$

where $\Box$ is one of the following relations: $\geqslant$, >, $=$, < or $\leqslant$ (for example, $\Omega _> = \{x \in \Omega :f(x) > f^*(\infty )\}$). Let us consider three subcases.

Case 1: Assume that $m(\Omega _>) = \infty$. Then there exists a measure preserving transformation $\omega :\Omega _> \rightarrow [0,\infty )$ such that $f^{*} \circ \omega = f$ m-almost everywhere on $\Omega _>$. Take $g \in X_a$ and put

$$\begin{aligned} h_g = (g \circ \omega )\chi _{\Omega _>}. \end{aligned}$$

We claim that $d_{f\chi _{\Omega _>} - h_g} = d_{f - h_g}$. Indeed, for all $0 \leqslant x < f^*(\infty )$ we have

$$\begin{aligned} d_{f\chi _{\Omega _>} - h_g}(x) = \infty = d_{f - h_g}(x), \end{aligned}$$

because $h_g \in X_a$, whence $h_g^*(\infty ) = 0$. Furthermore, for $t \in \Omega _{\leqslant }$

$$\begin{aligned} h_g(t) = 0 \quad \text {and}\quad f(t) \leqslant f^*(\infty ), \end{aligned}$$

so for each $x \geqslant f^*(\infty )$ we obtain that

$$\begin{aligned} d_{f - h_g}(x)&= m\{t \in \Omega :\left|f(t) - h_g(t)\right|> x \} = m\{t \in \Omega _> :\left|f(t) - h_g(t)\right|> x \} \\&= m\{t \in \Omega _> :\left|f(t)\chi _{\Omega _>} - h_g(t)\right|> x \} = d_{f\chi _{\Omega _>} - h_g}(x), \end{aligned}$$

and the claim follows. In consequence, we have

$$\begin{aligned} \Vert f^* - g\Vert _X&= \Vert (f^* - g) \circ \omega \Vert _X = \Vert f\chi _{\Omega _>} - h_g\Vert _X \\&= \Vert (f\chi _{\Omega _>} - h_g)^*\Vert _X = \Vert (f - h_g)^*\Vert _X = \Vert f - h_g\Vert _X. \end{aligned}$$

Case 2: Suppose now that $m(\Omega _>) < \infty$ and $f\chi _{\Omega _{>}} \in X_a$. Put

$$\begin{aligned} \Gamma _f = \{ g \in X_a :g(x) = f(x) \text { for } x \in \Omega _{>} \text { and } 0 \leqslant g(x) \leqslant f^*(\infty ) \text { otherwise} \}. \end{aligned}$$

It follows directly from the above definition that $f\chi _{\Omega _{>}} \in \Gamma _f$ and $\Gamma _f \subset X_a$. We claim that

To see this, for each $g \in X_a$ let us define a function $g_f$ in the following way

$$\begin{aligned} g_f(x) =\left\{ \begin{array}{ccc} f\left( x\right) &{}\quad \text {if} &{} x\in \Omega _{>} \\ f\left( x\right) &{}\quad \text {if} &{} x\in \Omega {\setminus } \Omega _{>} \text { and } g(x) > f^*(\infty ) \\ g\left( x\right) &{}\quad \text {otherwise}&{} \end{array} \right. . \end{aligned}$$

Now, it is clear that $\Vert f - g_f\Vert _X \leqslant \Vert f - g\Vert _X$ and, due to the fact that $m\{x \in \Omega {\setminus } \Omega _{>} :g(x) > f^*(\infty ) \} < \infty$, also $g_f \in \Gamma _f$. But this means that $d _X(f,X_a) \geqslant d _X(f,\Gamma _f)$. Since $\Gamma _f \subset X_a$, the opposite inequality is evident and the equality $(\spadesuit )$ is proved. Consequently, since $g \in X_a$, so

$$\begin{aligned} d _X(f,X_a) = \inf \{\Vert f - g\Vert :g \in \Gamma _f \} = \Vert f^*\chi _{(m(\Omega _{>}), \infty )}\Vert _X = f^*(\infty )\Vert \chi _{(0,\infty )}\Vert _X. \end{aligned}$$

On the other hand, since $m(\Omega _>) < \infty$ and $f\chi _{\Omega _{>}} \in X_a$, applying Theorem 3.4, we conclude that

$$\begin{aligned} d _X(f^*, X_a) = \lim \limits _{n \rightarrow \infty }\Vert f^*\chi _{\left( 0, \frac{1}{n} \right) \cup (n,\infty )}\Vert = f^*(\infty )\Vert \chi _{(0,\infty )}\Vert _X, \end{aligned}$$

that is to say, $d _X(f,X_a) = d _X(f^*,X_a)$.

Case 3: Let $m(\Omega _>) < \infty$ and $f\chi _{\Omega _{>}} \in X{\setminus } X_{a}$. Observe that $f^*(0^+) = \infty$, because otherwise, keeping in mind that $m(\Omega _{>}) < \infty$, we would get $f\chi _{\Omega _{>}} \in X_a$ which is undoubtedly a contradiction. We will show that

Take $n_0 \in {\mathbb {N}}$ satisfying $f^*(1/n_0) > f^*(\infty )$ and note that this limit exists because the sequence $\left\{ \Vert f^*\chi _{(0,\frac{1}{n})} + f^*(\infty )\chi _{(\frac{1}{n}, \infty )}\Vert _X \right\} _{n=1}^\infty$ is non-increasing for $n \geqslant n_0$ and bounded from below. Now, there is a measure preserving transformation $\tau :\Omega _{>} \rightarrow (0,m(\Omega _{>}))$ such that $f^* \circ \tau = f$. Put $T_n = \tau ^{-1}((0,1/n))$ for $n = n_0, n_0+1, \ldots$ and denote

$$\begin{aligned} f_n = f^*\left( \frac{1}{n_0} \right) \chi _{T_n} \end{aligned}$$

It is clear that $f_n \in X_a$ for $n \geqslant n_0$. Moreover, since $f^*(0^+) = \infty$, so

$$\begin{aligned} \left( f^* - f^*\left( \frac{1}{n_0} \right) \right) \chi _{(0,\frac{1}{n})} \geqslant f^*(\infty )\chi _{(0,\frac{1}{n})} \text { for } n \in {\mathbb {N}} \text { big enough}, \end{aligned}$$

and, consequently, we have

$$\begin{aligned} d _X(f,X_a) \leqslant \lim \limits _{n \rightarrow \infty } \Vert f - f_n\Vert _X = \lim \limits _{n \rightarrow \infty } \Vert \left( f^* - f^*\left( \frac{1}{n_0} \right) \right) \chi _{(0,\frac{1}{n})} + f^*(\infty )\chi _{(\frac{1}{n}, \infty )}\Vert _X \leqslant L. \end{aligned}$$

On the other hand, we get

$$\begin{aligned} L \geqslant d _X(f,X_a)&= \inf \{\Vert f - g\Vert _X :g \in X_a \text { and } \left|g\right| \leqslant \left|f\right| \}\\&\geqslant \inf \{\Vert f^* - g^*\Vert _X :g \in X_a \text { and } \left|g\right| \leqslant f \} \\&\geqslant \inf \{\Vert f^* - g^*\Vert _X :g \in X_a \text { and } 0 \leqslant g^* \leqslant f^* \}, \end{aligned}$$

where the first equality follows from Lemma 3.1 and second inequality follows from the Calderón–Ryff theorem. Thus, to finish the proof of $(\diamondsuit )$, it is enough to show that

To see this, take $\varepsilon > 0$ and $h \in X_a$ as above. We can find a number, say $m \in {\mathbb {N}}$, such that

$$\begin{aligned} \Vert h^*\chi _{(0,\frac{1}{m}) \cup (m,\infty )}\Vert < \epsilon , \end{aligned}$$

because $h^* \in X_a$ (see [15,  Lemma 2.6]). Hence,

$$\begin{aligned} \Vert f^* - h^*\Vert _X&= \Vert f^* - h^*\chi _{(\frac{1}{m}, m)} - h^*\chi _{(0,\frac{1}{m}) \cup (m,\infty )}\Vert _X \\&\geqslant \Vert f^* - h^*\chi _{(\frac{1}{m}, m)}\Vert _X - \Vert h^*\chi _{(0,\frac{1}{m}) \cup (m,\infty )}\Vert _X \\&> \Vert f^* - h^*\chi _{(\frac{1}{m}, m)}\Vert _X - \varepsilon \\&= \Vert f^*\chi _{(0,\frac{1}{m}) \cup (m,\infty )} + (f^* - h^*)\chi _{(\frac{1}{m}, m)}\Vert _X - \varepsilon \geqslant \Vert f^*\chi _{(0,\frac{1}{m}) \cup (m,\infty )}\Vert _X - \varepsilon , \end{aligned}$$

and $(\clubsuit )$ follows. But this means that the proof of $(\diamondsuit )$ is finished as well. However, using Theorem 3.4 again, we see that

$$\begin{aligned} d _X(f^*,X_a) = \lim \limits _{n \rightarrow \infty } \Vert f^*\chi _{\left( 0,\frac{1}{n} \right) \cup (n,\infty )}\Vert _X, \end{aligned}$$

that is, $d _X(f,X_a) = d _X(f^*,X_a)$. But this, finally, completes the proof of the last case and, at the same time, also of the whole theorem. $\square$

Theorem 3.6

Let X be a rearrangement invariant space on ${\mathbb {N}}$ with the Fatou property. Then for $f \in X$ we have the following equality

$$\begin{aligned} d_X({f^*, X_a}) = d_X({f, X_a}). \end{aligned}$$

Proof

The argument in the sequence case (which is probably not a surprise) is analogous to the proof of Theorem 3.5 and, in fact, it is even easier. But we want to give a few clues that (we hope) will allow to easily modify the mentioned proof.

$1^{\circ }$.:

For a rearrangement invariant sequence space X the ideal $X_a$ is always non-trivial.

$2^{\circ }$.:

Since $X \hookrightarrow \ell _\infty$ (cf. [7,  Corollary 6.8, p. 78]), so $X_a \hookrightarrow c_0$ and this means that $f^*(\infty ) = 0$, whenever $f \in X_a$.

$3^{\circ }$.:

Every function supported on the set of finite measure belongs to $X_a$.

$4^{\circ }$.:

The inequality (coming from the Calderón–Ryff theorem):

$$\begin{aligned} \Vert f^* - g^*\Vert _X \leqslant \Vert f - g\Vert _X, \end{aligned}$$

works also for every rearrangement invariant sequence space X (cf. [7,  Corollary 7.6, p. 83]). $\square$

Corollary 3.7

Let X be a rearrangement invariant space on $\Omega$, where $\Omega = (0,1)$, $\Omega = (0,\infty )$ or $\Omega = {\mathbb {N}}$, such that the ideal $X_a$ is non-trivial. Then for $f \in X$ we have the following equality

$$\begin{aligned} d_X({f,X_a}) = \lim \limits _{n \rightarrow \infty } \Vert f^* \chi _{\Omega _n}\Vert _X, \end{aligned}$$

where $\Omega _n = \Omega \cap \left( (0, \frac{1}{n}) \cup (n,\infty )\right)$.

The above corollary can be interpreted in the following way: the function $f \in X$ can be approximated by a bounded function supported in a set of finite measure to within $d _X(f,X_a) + \varepsilon$, where $\varepsilon > 0$ can be arbitrarily small, and this function can be selected in an explicit way. A similar result, but for Orlicz spaces equipped with an Orlicz norm, can be found in [47,  Lemma 10.1, p. 84].

Corollary 3.8

(de Jonge [20]) Let X be a rearrangement invariant space on $(0,\infty )$ with the Fatou property such that $L_\infty \hookrightarrow X$. Suppose that

Then for all $f \in X$ we have the following formula

$$\begin{aligned} d_X({f,X_a}) = f^*(\infty )\varphi _X(\infty ). \end{aligned}$$

Before we provide the proof, let us note that the above condition (${{\mathscr {D}}}_\infty$) itself plays in rearrangement invariant spaces an analogous role to the $\Delta _2(\infty )$-condition in the case of Orlicz spaces. In fact, if X is an Orlicz space $L_F$ on $(0,\infty )$ generated by the Orlicz function F which vanishes outside zero and $F \in \Delta _2(\infty )$, then $L_\infty \hookrightarrow X$ and $X \in ({{\mathscr {D}}}_\infty )$. However, the assumptions of the above corollary are true also in the case of Lorentz spaces $\Lambda _\varphi$ provided $\varphi (0^+) = 0$ and $\varphi (\infty ) < \infty$.

Proof

Suppose that X is as above and observe that the sequence $\left\{ (f^*\chi _{(n,\infty )})^*\right\} _{n=1}^\infty$ is bounded and converges uniformly to a bounded function $f^*(\infty )\chi _{(0,\infty )}$, so in view of the embedding $L_\infty \hookrightarrow X$ we get

$$\begin{aligned} d _X(f,X_a)= & {} \lim \limits _{n\rightarrow \infty } \Vert f^*\chi _{(n,\infty )}\Vert _X = \lim \limits _{n\rightarrow \infty } \Vert (f^*\chi _{(n,\infty )})^*\Vert _X = \Vert f^*(\infty )\chi _{(0,\infty )}\Vert _X \\= & {} f^*(\infty )\varphi _X(\infty ), \end{aligned}$$

where the first equality follows from Corollary 3.7. $\square$

Although we believe that the below results (Corollaries 3.9 and 3.11 ) are expected, we will present proofs of these facts as illustrations of the techniques we developed in this paragraph.

Corollary 3.9

Let $\varphi$ be a quasi-concave function such that $\lim _{t \rightarrow 0^{+}} t/\varphi (t) = 0$. Then the Marcinkiewicz space $M_\varphi$ contains a lattice isometric copy of $\ell _\infty$.

Proof

Let us consider the function $\psi$ defined in the following way

$$\begin{aligned} \psi (0) = 0 \quad \text {and} \quad \psi (t) = \frac{t}{\varphi (t)} \quad \text {for } t > 0. \end{aligned}$$

It follows from [7,  Theorem 5.2, p. 66] that $\psi$ is also a quasi-concave function as the fundamental function of the space $(M_{\varphi })'$. In addition, due to the condition $\lim _{t \rightarrow 0^{+}} t/\varphi (t) = 0$, the function $\psi$ is continuous (cf. [7,  Corollary 5.3, p. 67]). Therefore, we can find $0< a < 1$ such that the function $\psi$ is strictly increasing on (0, a). Consequently, the function $\psi '$ is non-increasing and positive m-almost everywhere on (0, a). Put $g = \psi '\chi _{(0,a)}$. Then

$$\begin{aligned} \frac{\varphi (t)}{t}\int _0^t g^*(s)ds = \frac{\varphi (t)}{t}\int _0^t g(s)ds = \frac{\varphi (t)}{t}\psi (t) = 1 \quad \text {for } 0< t < a, \end{aligned}$$

and, because the function $t \mapsto \varphi (t)/t$ is non-increasing, we also have that

$$\begin{aligned} \frac{\varphi (t)}{t}\int _0^t g^*(s)ds = \frac{\varphi (t)}{t} \frac{a}{\varphi (a)} \leqslant 1 \quad \text {for } t \geqslant a. \end{aligned}$$

This means that $\Vert g\Vert _{M_\varphi } = 1$. Moreover, we can repeat the above argument with the same result for the function $\psi _\varepsilon = \psi \chi _{(0,\varepsilon )}$, where $0< \varepsilon < a$, in place of the function $\psi$. Therefore, using Corollary 3.7, we get

$$\begin{aligned} d_{M_\varphi }(g, (M_\varphi )_a) \geqslant \lim _{n\rightarrow \infty }\Vert g^*\chi _{(0,\frac{1}{n})}\Vert _{M_\varphi } = 1, \end{aligned}$$

and we are done. $\square$

Lemma 3.10

Let X be a rearrangement invariant space on $(0,\infty )$. If $L_\infty \hookrightarrow X$, then X contains a lattice isometric copy of $\ell _\infty$.

Let us note in addition that the converse is not true, for example, the Zygmund space $e^L$ (recall that $e^L = L_F$, where $F(t) = e^t -1$; cf. [7,  p. 243]) contains a lattice isometric copy of $\ell _\infty$ (cf. [31]), but it is clear that $L_\infty \not \hookrightarrow e^L$.

Proof

Take $f = \chi _{(0,\infty )}/\Vert \chi _{(0,\infty )}\Vert _X$ and let $\{N_i\}_{i=1}^\infty$ be a pairwise disjoint family of an infinite subsets of ${\mathbb {N}}$ such that $\bigcup _{i=1}^\infty N_i = {\mathbb {N}}$. Moreover, let $\{\omega _j\}_{j=1}^\infty$ be a sequence of real numbers such that

• $0 = \omega _1< \omega _2 < \cdots$,

• $\lim _{j \rightarrow \infty } \omega _j = \infty$,

• $\inf \nolimits _{\begin{array}{c} i,j \in {\mathbb {N}} \\ i \ne j \end{array}} \left|\omega _j - \omega _i\right| > \delta$ for some $\delta > 0$.

Put $\Omega _j^i = (\omega _j,\omega _{j+1})$ for $i \in {\mathbb {N}}$ and $j \in N_i$. Consider the following sequence $\left\{ f\chi _{\bigcup _{j \in N_i} \Omega _j^i}\right\} _{i=1}^\infty$. It is clear that this sequence is pairwise disjoint, $\Vert \sum _{i=1}^\infty f\chi _{\bigcup _{j \in N_i} \Omega _j^i}\Vert _X = 1$ and $\Vert f\chi _{\bigcup _{j=1}^\infty \Omega _j^i}\Vert _X = 1$, because $\left( f\chi _{\bigcup _{j \in N_i} \Omega _j^i}\right) ^* = f$ for every $i \in {\mathbb {N}}$ and $\Vert f\Vert _X = 1$. Consequently, in order to complete the proof we can refer to [30,  Theorem 1]. $\square$

Corollary 3.11

Let $\varphi$ be a concave function on $(0,\infty )$ such that $\varphi (0^{+}) = 0$. Then the Lorentz space $\Lambda _\varphi$ contains a lattice isometric copy of $\ell _\infty$ if and only if $\varphi (\infty ) < \infty$.

Proof

It follows from [48,  Lemma 5.1, p. 110] that the space $\Lambda _{\varphi }$ is order continuous if $\varphi (\infty ) = \infty$. Therefore, due to the well-known Lozanovskiĭ result, it does not contain even an lattice isomorphic copy of $\ell _{\infty }$ and proves the necessity.

Now, suppose that $\varphi (\infty ) < \infty$. But this means that $L_\infty \hookrightarrow X$ and it is enough to use Lemma 3.10. $\square$

Let us add that it may happen that the Lorentz space $\Lambda _\varphi$ contains an lattice isomorphic copy of $\ell _\infty$ but not an lattice isometric copy of $\ell _\infty$. Indeed, if $\varphi (0^+) > 0$ and $\varphi (\infty ) = \infty$, then the space $\Lambda _\varphi$ is not order continuous (see [48,  Lemma 5.1, p. 110]) but is strictly monotone (see [46,  Lemma 3.1(i)]; however, a slight modification of the proof is required).

4 On isometric copies of $\ell _\infty$ in abstract Cesàro spaces

For a Banach ideal space X on (0, 1) or $(0,\infty )$ we define the abstract Cesàro function space CX as a set

$$\begin{aligned} CX = \{f \in L_0 :C(\left|f\right|) \in X \} \text { with the natural norm } \Vert f\Vert _{CX} = \Vert C(\left|f\right|)\Vert _X, \end{aligned}$$

where C denotes the continuous Cesàro (Hardy) operator defined as

$$\begin{aligned} C :f \mapsto C(f)(x) = \frac{1}{x} \int _0^x f(t) dt \quad \text {for } x > 0. \end{aligned}$$

On the other hand, in the sequence case, by the discrete Cesàro (Hardy) operator we will understand the operator $C_d$ defined as

$$\begin{aligned} C_d :\{x_n\}_{n=1}^\infty \mapsto \left\{ \frac{1}{n} \sum _{k=1}^n x_k \right\} _{n=1}^\infty , \end{aligned}$$

and then the corresponding abstract Cesàro sequence space CX is defined in an analogous way as above, that is to say, as a set of all sequences $x = \{x_n\}_{n=1}^\infty$ such that $C_d(\left|x\right|) \in X$ for which the following norm

$$\begin{aligned} \Vert x\Vert _{CX} = \Vert C_d(\left|x\right|)\Vert _X \end{aligned}$$

is finite.

Because it always should be clear from the context which Cesàro construction we use, we will abuse the notation a little by writing simply C for the continuous as well as the discrete Cesáro operator. Moreover, to avoid some trivialities and unnecessary complications, let’s assume that we are only interested in non-trivial Cesàro spaces. It just so happens that, if the Cesàro operator C is bounded on X, then $X \hookrightarrow CX$ and, consequently, the space CX is non-trivial.

The classical Cesàro spaces $ces_p$ and $Ces_p$ are only a special case of this general construction $X \leadsto CX$, when $X = \ell _p$ or $X = L_p$, respectively. We will not say more about the history and the current development of the Cesaro space theory, referring to [6] and the references given there (see also [4, 5, 8, 17, 19, 21, 36, 39, 42, 45, 51] and [61]; cf. [49]).

Let us now formulate and prove the main result of this section.

Theorem 4.1

Let X be a rearrangement invariant Banach space on (0, 1), $(0,\infty )$ or on ${\mathbb {N}}$ with the Fatou property such that the Cesàro operator C is bounded on X and the ideal $X_a$ is non-trivial. Suppose there exists a function $f \in X$ and $0 \leqslant a < b \leqslant \infty$ with:

• $\Vert f\Vert _X = d_X({f,X_a}) = 1$,

• $\Vert f^*\chi _{(0,a)\cup (b,\infty )}\Vert _{CX} = 1$.

Then the space CX contains a lattice isometric copy of $\ell _\infty$.

The proof of the above theorem makes strong use of Theorem 3.5 from §3 and Theorem 16 from [42]. Note that Theorem 16 from [42] has been proved for rearrangement invariant function, but not sequence, spaces. For the missing proof, we refer to Appendix.

Proof

Let $f \in X$ be as above. First of all, let’s simplify the notation a little by putting $A := (0,a) \cup (b,\infty )$ and $A^c := (a,b)$. In view of Theorem 3.5, we have the equality

$$\begin{aligned} d _X(f^*, X_a) = d _X(f, X_a) = 1. \end{aligned}$$

It follows from [42,  Lemma 8] that $f^*\chi _{A^c} \in (CX)_a$ (again, Lemma 8 in [42] is about rearrangement invariant function spaces, but simple modification of this argument works in the sequence case as well). Consequently, taking $h = f^*\chi _{A}$, we have

$$\begin{aligned} d _{CX}(h, (CX)_a)&= \inf \{ \Vert h - g\Vert _{CX} :g \in (CX)_a \text { and } 0< g \leqslant h \} \\&\quad \quad \quad (\text {by Lemma~3.1}) \\&= \inf \{ \Vert C(h) - C(g)\Vert _{X} :g \in C(X_a) \text { and } 0< g \leqslant h \} \\&\quad \quad \quad (\text {because } h - g \geqslant 0 \text { and } CX_a = C(X_a) \text {, see [KT17, Theorem 16]}) \\&= \inf \{ \Vert C(h+f^*\chi _{A^c}) - C(g+f^*\chi _{A^c})\Vert _{X} :C(g) \in X_a \text { and } 0 < g \leqslant h \} \\&\geqslant \inf \{ \Vert C(f^*) - C({\widetilde{g}})\Vert _{X} :{\widetilde{g}} \in CX \text { and } C({\widetilde{g}}) \in X_a \} \\&\quad \quad \quad (\text {since } g+f^*\chi _{A^c} \in C(X_a) ) \\&\geqslant \inf \{ \Vert C(f^*) - {{\widehat{g}}}\Vert _{X} :{\widehat{g}} \in X_a \} \\&\quad \quad \quad (\text {we take the infimum on a larger set}) \\&= d _X(C(f^*), X_a) \geqslant d _X(f^*, X_a) = 1, \end{aligned}$$

where the last inequality follows from Corollary 3.2 and the fact that $f^* \leqslant C(f^*)$ (cf. [7,  Proposition 3.2, p. 52] - just note that $C(f^*) \equiv f^{**}$). Now, since ${{\,\mathrm{supp}\,}}((CX)_a) = {{\,\mathrm{supp}\,}}(CX)$ (see, once again, [42,  Lemma 8]) we can apply Theorem B and finish the proof. $\square$

It should be mentioned here that Theorem 4.1 covers all previous results concerning the problem of existence of a lattice isometric copy of $\ell _\infty$ in Cesàro spaces. To be more precise, Kamińska and Kubiak in [39] considered this problem for Orlicz sequence spaces $\ell _F$ and Kiwerski and Kolwicz in [44] for Orlicz function spaces $L_F$.

A common feature of all these results is a certain assumption about the Orlicz class, namely, about the closedness of the Orlicz class under the action of the Cesàro operator, which comes down to the inclusion

$$\begin{aligned} \left\{ \{x_n\}_{n=1}^\infty \in \ell _F :\sum _{n=1}^\infty F(x_n)< \infty \right\} \subset \left\{ \{x_n\}_{n=1}^\infty \in \ell _F :\sum _{n=1}^\infty F\left( \frac{1}{n} \sum _{k=1}^n x_k \right) < \infty \right\} \end{aligned}$$

in the sequence case, or the modular-type inequality

$$\begin{aligned} \int _0^\infty F\left( \frac{1}{x}\int _0^x\left|f(t)\right|dt\right) dx \leqslant M\int _0^\infty F(\left|f(x)\right|)dx \end{aligned}$$

in the function case (which looks a little like the boundedness³ of the Cesàro operator on the Orlicz class or like the Hardy inequality with the Orlicz function F instead of the power function; cf. [49,  §3]). Note that both assumptions imply the boundedness of the Cesàro operator on the space $\ell _F$ and $L_F$, respectively (cf. [39,  Proposition 2] and [44]).

Our Theorem 4.1, at least formally, improves both of these results. To show this, we will take an even more general point of view, because we will consider the Calderón–Lozanovskiĭ spaces $X_F$ (recall, that the spaces $(\ell _1)_F$ and $(L_1)_F$ coincide, up to the equality of norms, with the Orlicz spaces $\ell _F$ and $L_F$, respectively).

Before we proceed to the formulation of this result, let us remind that by the Calderón–Lozanovskiĭ class we understand the set $\{f \in L_0 :\Vert F(\left|f\right|)\Vert _X < \infty \}$ (cf. [57,  §3] for some comments and key properties in the particular case when $X = L_1$).

Corollary 4.2

Let X be an order continuous rearrangement invariant function space with the Fatou property and let F be a finitely valued Orlicz function. Assume that there exists $M > 0$ such that for all functions, say f, from the Calderón–Lozanovskiĭ class we have the following inequality:

If F does not satisfy the $\Delta _2^X$-condition,⁴ then the space $C(X_F)$ contains a lattice isometric copy of $\ell _\infty$.

Proof

Since $F \notin \Delta _2^X$, it follows from Theorem 1 and Theorem 2 in [31] that the Calderón–Lozanovskiĭ space $X_F$ contains a lattice isometric copy of $\ell _\infty$. Therefore, in view of Theorem 2.1 (let us note the fact that ${{\,\mathrm{supp}\,}}((X_F)_a) = {{\,\mathrm{supp}\,}}(X_F)$) and Theorem 3.5, there exists a function $f \in X_F$ such that

$$\begin{aligned} \Vert f\Vert _{X_F} = d_{X_F}(f, (X_F)_a) = d_{X_F}(f^*, (X_F)_a) = 1. \end{aligned}$$

Because the space $X_F$ is rearrangement invariant, so $\Vert f^*\Vert _{X_F} =1$ and, consequently, $\Vert F(f^*)\Vert _X \leqslant 1$. Now, we divide the proof into two parts.⁵

$1^\circ$. Suppose that $f^*\chi _{(c,\infty )} \in (X_F)_a$ for some (equivalently, for all) $c > 0$. Consequently, we have that

$$\begin{aligned} 1 = d_{X_F}(f^*, (X_F)_a) = \lim \limits _{n \rightarrow \infty }\Vert f^*\chi _{(0,\frac{1}{n})\cup (n,\infty )}\Vert _{X_F} = \lim \limits _{n \rightarrow \infty }\Vert f^*\chi _{(0,\frac{1}{n})}\Vert _{X_F}, \end{aligned}$$

where the second equality follows from Theorem 3.4. Since $X \in (OC)$, so we can find $b > 0$ with

$$\begin{aligned} \Vert (F(f^*))\chi _{(0,b)}\Vert _X \leqslant \frac{1}{M}, \end{aligned}$$

where the constant $M > 0$ is from the condition (${{\mathscr {M}}}$). Therefore, we have the following inequalities

$$\begin{aligned} \Vert F(C(f^*\chi _{(0,b)}))\Vert _X \leqslant M \Vert F(f^*\chi _{(0,b)})\Vert _X = M\Vert (F(f^*))\chi _{(0,b)}\Vert _X \leqslant 1. \end{aligned}$$

But this means that $\Vert f^*\chi _{(0,b)}\Vert _{C(X_F)} \leqslant 1$. What’s more, thanks to our assumptions, $f^*\chi _{(b,\infty )} \in (X_F)_a$, whence

$$\begin{aligned} 1 = d_{X_F}(f^*, (X_F)_a) \leqslant \Vert f^* - f^*\chi _{(b,\infty )}\Vert _{X_F} = \Vert f^*\chi _{(0,b)}\Vert _{X_F} \leqslant \Vert C(f^*\chi _{(0,b)})\Vert _{X_F} \leqslant 1, \end{aligned}$$

that is, $\Vert C(f^*\chi _{(0,b)})\Vert _{X_F} = 1$. Now we can apply Theorem 4.1 and finish the proof of this part.

$2^\circ$. Assume that $f^*\chi _{(c,\infty )} \in X_F {\setminus } (X_F)_a$ and $f^*\chi _{(0,c)} \in (X_F)_a$ for some (equivalently, for all) $c > 0$. Referring once again to Theorem 3.4, we have the following equalities

$$\begin{aligned} 1 = d_{X_F}(f^*, (X_F)_a) = \lim \limits _{n \rightarrow \infty }\Vert f^*\chi _{(0,\frac{1}{n})\cup (n,\infty )}\Vert _{X_F} = \lim \limits _{n \rightarrow \infty }\Vert f^*\chi _{(n,\infty )}\Vert _{X_F}. \end{aligned}$$

Just as before, since $X \in (OC)$, so

$$\begin{aligned} \Vert (F(f^*))\chi _{(b,\infty )}\Vert _X \leqslant \frac{1}{M}\quad \text {for some } b > 0 \text { big enough}, \end{aligned}$$

where $M > 0$ is from the condition (${{\mathscr {M}}}$). Moreover, using the condition (${{\mathscr {M}}}$), we get

$$\begin{aligned} \Vert F(C(f^*\chi _{(b,\infty )}))\Vert _X \leqslant M \Vert F(f^*\chi _{(b,\infty )})\Vert _X = M\Vert (F(f^*))\chi _{(b,\infty )}\Vert _X \leqslant 1. \end{aligned}$$

that is, $\Vert f^*\chi _{(b,\infty )}\Vert _{C(X_F)} \leqslant 1$. Now, we claim that

To prove ($=$) it is enough to show that

$$\begin{aligned} \Vert F(C((1+\varepsilon )f^*\chi _{(b,\infty )}))\Vert _X> 1 \quad \text {for all } \varepsilon > 0. \end{aligned}$$

Take $\varepsilon > 0$. Note that

$$\begin{aligned} C\left( (1+\varepsilon )f^*\chi _{(b,\infty )}\right) (x) = (1+\varepsilon ) \frac{1}{x}\int _b^x f^*(t)dt \geqslant (1+\varepsilon )\left( 1 - \frac{b}{x}\right) f^*(x), \end{aligned}$$

for $x \geqslant b$. Moreover, there exists $B > b$ such that

$$\begin{aligned} (1+\varepsilon )\left( 1 - \frac{b}{x}\right) f^*(x) \geqslant \left( 1 + \frac{\varepsilon }{2}\right) f^*(x) \quad \text {for } x \geqslant B. \end{aligned}$$

In consequence, we have

$$\begin{aligned} \Vert F(C((1+\varepsilon )f^*\chi _{(b,\infty )}))\Vert _X\geqslant & {} \Vert F(C((1+\varepsilon )f^*\chi _{(B,\infty )}))\Vert _X \\\geqslant & {} \Vert F\left( \left( 1+\frac{\varepsilon }{2}\right) f^*\chi _{(B,\infty )}\right) \Vert _X > 1, \end{aligned}$$

where the last inequality follows from the fact that $\Vert f^*\chi _{(B,\infty )}\Vert _{X_F} = 1$ (to see this just note that due to our assumptions $f^*\chi _{(0,B)} \in (X_F)_a$ and then $1 = d_{X_F}(f^*, (X_F)_a) \leqslant \Vert f^* - f^*\chi _{(0,B)}\Vert _{X_F} = \Vert f^*\chi _{(B,\infty )}\Vert _{X_F} \leqslant 1$) and the definition of the Luxemburg–Nakano norm. But this proves the claim ($=$). Again, we are left to use Theorem 4.1. We have therefore completed the proof. $\square$

Corollary 4.3

Let X be an order continuous rearrangement invariant sequence space with the Fatou property and let F be an Orlicz function. Assume that the Calderón–Lozanovskiĭ class is closed under the Cesàro operator C. If F vanishes only at zero and does not satisfy the $\Delta _2(0)$-condition, then the space $C(X_F)$ contains a lattice isometric copy of $\ell _\infty$.

Proof

To begin with, since $X = X_a \hookrightarrow c_0 \hookrightarrow \ell _\infty$ and the Orlicz function F vanishes only at 0, it follows that the condition $\Delta _2(0)$ is equivalent to the condition $\delta _2^X$ (see [33,  Lemma 4]).⁶ Due to the lack of the $\delta _2^X$-condition and Lemma 2.4 from [27], the Calderón–Lozanovskiĭ space $X_F$ contains a lattice isometric copy of $\ell _\infty$ (note only that the results from [27] have been proven for a wider than $X_F$ class of spaces, the so-called generalized Calderón–Lozanovskiĭ spaces $X_M$, in which the Orlicz function F is replaced by the much more general Musielak–Orlicz function M). Exactly as in the function’ case (cf. Corollary 4.2), citing Theorem 2.1 along with Theorem 3.5, we can find a sequence $f \in X_F$ with

$$\begin{aligned} \Vert f\Vert _{X_F} = d_{X_F}(f, (X_F)_a) = d_{X_F}(f^*, (X_F)_a) = 1. \end{aligned}$$

What’s more $\Vert f^*\Vert _{X_F} =1$ and, consequently, $\Vert F(f^*)\Vert _X \leqslant 1$. Now, because the Calderón–Lozanovskiĭ class is closed under the Cesàro operator C we conclude that $\Vert F(C(f^*))\Vert _{X} < \infty$. Since $X \in (OC)$, there exists $n \in {\mathbb {N}}$ such that

$$\begin{aligned} \Vert (F(C(f^*)))\chi _{\{n, n+1, \ldots \}}\Vert _{X} \leqslant 1. \end{aligned}$$

Moreover, it is not hard to see that

$$\begin{aligned} C(f^*\chi _{\{n, n+1, \ldots \}}) \leqslant (C(f^*))\chi _{\{n, n+1, \ldots \}}, \end{aligned}$$

whence also

$$\begin{aligned} \Vert F\left( C(f^*\chi _{\{n, n+1, \ldots \}})\right) \Vert _{X} \leqslant \Vert F\left( (C(f^*))\chi _{\{n, n+1, \ldots \}}\right) \Vert _{X} \leqslant 1. \end{aligned}$$

But this means that $\Vert f^*\chi _{\{n, n+1, \ldots \}}\Vert _{C(X_F)} \leqslant 1$. Note also that $f^*\chi _{\{1, 2, \ldots , n-1\}} \in (X_F)_a$, so

$$\begin{aligned} 1 = d_{X_F}(f^*, (X_F)_a) \leqslant \Vert f - f^*\chi _{\{1, 2, \ldots , n-1\}}\Vert _{X_F} = \Vert f^*\chi _{\{n, n+1, \ldots \}}\Vert _{X_F} \leqslant 1. \end{aligned}$$

Now, after some obvious modifications (such as converting integrals with sums or using the discrete Cesàro operator in place of his continuous counterpart) in the proof of equality ($=$) from Corollary 4.2, we can show that

$$\begin{aligned} \Vert f^*\chi _{\{n, n+1, \ldots \}}\Vert _{C(X_F)} = 1. \end{aligned}$$

Keeping in mind that the order continuous part of a rearrangement invariant sequence space is always non-trivial we can complete the proof by referring once again to Theorem 4.1. $\square$

The assumption that appeared in Theorem 4.1 may at first glance look quite wishful, but it follows from the proof of the next result that rather natural subclass of the class of rearrangement invariant function spaces satisfies this condition.

Proposition 4.4

Let X be a rearrangement invariant space on $(0,\infty )$ with the Fatou property such that the Cesàro operator C is bounded on X. Assume that $L_\infty \hookrightarrow X$ and that the space X satisfies the condition $({{\mathscr {D}}}_\infty )$ from Corollary 3.8. Then the space CX contains a lattice isometric copy of $\ell _\infty$.

Proof

To begin with, take $f = \chi _{(0,\infty )}/\Vert \chi _{(0,\infty )}\Vert _X$. The assumption $L_\infty \hookrightarrow X$ implies that $\Vert f\Vert _X = 1$. Moreover,

$$\begin{aligned} d _X(f,X_a) = \lim _{n\rightarrow \infty }\Vert f^*\chi _{(n,\infty )}\Vert _X = \lim _{n\rightarrow \infty }\Vert (f^*\chi _{(n,\infty )})^*\Vert _X = \Vert f\Vert _X = 1, \end{aligned}$$

where the first equality follows from Corollary 3.7 and the fact due to the condition $({\mathscr {D}}_\infty )$ that every function supported on the set of finite measure is order continuous. Consequently, we have the equalities $\Vert f\Vert _X = d _X(f,X_a) = 1$. Again, due to the condition $({{\mathscr {D}}}_\infty )$, the ideal $X_a$ is non-trivial, so ${{\,\mathrm{supp}\,}}(X_a) = {{\,\mathrm{supp}\,}}(X)$ and we can use Theorem B to conclude that the space X contains a lattice isometric copy of $\ell _\infty$. On the other hand, it is not hard to see that

$$\begin{aligned} \Vert f^*\chi _{(b,\infty )}\Vert _{CX} = 1\quad \text {for every } b > 0. \end{aligned}$$

Indeed, we have

$$\begin{aligned} \Vert f^*\chi _{(b,\infty )}\Vert _{CX} = \frac{\Vert C(\chi _{(b,\infty )})\Vert _X}{\Vert \chi _{(0,\infty )}\Vert _X} = \frac{\Vert (C(\chi _{(b,\infty )}))^*\Vert _X}{\Vert \chi _{(0,\infty )}\Vert _X} = \frac{\Vert \chi _{(0,\infty )}\Vert _X}{\Vert \chi _{(0,\infty )}\Vert _X} = 1. \end{aligned}$$

But this means that the condition in Theorem 4.1 is satisfied with $a = 0$ and $b > 0$. We therefore conclude that the space CX contains a lattice isometric copy of $\ell _\infty$ and the proof is done. $\square$

Some spaces, such as Orlicz spaces (and certain variants of Orlicz spaces like Orlicz–Lorentz spaces or Musielak–Orlicz spaces; cf. [30,  Remarks on p. 526]), have this self-improving property that if they contain a lattice isomorphic copy of $\ell _\infty$, then they actually contain a lattice isometric copy of $\ell _\infty$. This simple observation along with the results from [42] leads us to the following

Theorem 4.5

Let X be an order continuous rearrangement invariant space with the Fatou property and let F be an Orlicz function. Suppose that the Cesàro operator C is bounded on the Calderón–Lozanovskiĭ space $X_F$. If the space $C(X_F)$ contains a lattice isometric copy of $\ell _\infty$, then the space $X_F$ contains it also.

Proof

Assume that the space $X_F$ does not contain a lattice isometric copy of $\ell _\infty$. Then $X_F$ does not contain even an lattice isomorphic copy of $\ell _\infty$. However, in view of the Lozanovskiĭ theorem [55], this means that $X_F$ is order continuous. At this stage, applying [42,  Theorem 3], we obtain that the space $C(X_F)$ is order continuous as well. Consequently, $C(X_F)$ cannot contain a lattice isometric copy of $\ell _\infty$ and we are done. $\square$

As we already mentioned, if X is a Banach ideal space with a trivial ideal $X_a$, then the characterization given in Theorem B is no longer valid in general. What seems to be quite interesting, the case of Cesàro spaces CX (at least, when X is a rearrangement invariant space) is different, because a support of the ideal $(CX)_a$ is equal to a support of the space X even if the ideal $X_a$ is trivial (see [42,  Lemma 8]). Therefore, we can still use Theorem B and obtain the next

Theorem 4.6

Let X be a rearrangement invariant space with the Fatou property such that the ideal $X_a$ is trivial. Suppose that there exists a function $f > 0$ with

• $C(f)(0^+) = 1$,

• $\Vert f\Vert _{CX} = \Vert id :X \rightarrow L_\infty \Vert$.

Then the Cesàro space CX contains a lattice isometric copy of $\ell _\infty$.

Proof

At the beginning observe that $X \hookrightarrow L_\infty$, because $X_a = \{ 0 \}$ (cf. [42,  Theorem B]). Moreover, since we can always replace the norm $\Vert \cdot \Vert _X$ by a constant multiple of itself, it follows that without loss of generality we can assume that $\Vert \text {id} :X \rightarrow L_\infty \Vert = 1$.

Let $f > 0$ be a function whose existence is guaranteed by our assumptions, that is to say,

$$\begin{aligned} C(f)(0^+) = 1 \quad \text {and} \quad \Vert f\Vert _{CX} = \Vert \text {id} :X \rightarrow L_\infty \Vert = 1. \end{aligned}$$

We will show that

Indeed, we have the following inequalities

where the first equality is due to Lemma 3.1 and the last equality is the consequence of the fact that $C(g)(0^+) = 0$, whenever $g \in (CX)_a$ and the ideal $X_a$ is trivial (see [42,  Lemma 14]). In summary, we proved the claim $(*)$, because we have shown that

$$\begin{aligned} 1 \leqslant d _{CX}(f, (CX)_a) \leqslant \Vert f\Vert _{CX} \leqslant 1. \end{aligned}$$

Now, to complete the proof, it is enough to note that ${{\,\mathrm{supp}\,}}((CX)_a) = {{\,\mathrm{supp}\,}}(CX)$ (see [42,  Lemma 8]) and use once again Hudzik’s result from [30] (cf. Theorem B). $\square$

Remark 4.7

• (a) If the space X is not rearrangement invariant, then there is no chance for an analogue of Theorem 4.6. In fact, for example, the space $L_\infty (w)$ with the weight w defined as $w(x) = x$ (for $x > 0$) contains a lattice isometric copy of $\ell _\infty$, but $C(L_\infty (w)) \equiv L_1$ and the space $L_1$ is even separable.

• (b) It may happen that the space X contains a lattice isometric copy of $\ell _\infty$, but the Cesàro space CX is trivial. It is enough to consider, for example, the space $X = L_1 \cap L_\infty$ on $(0,\infty )$.

If X is a rearrangement invariant space on (0, 1) with the Fatou property such that the ideal $X_a$ is trivial, then the space X is just $L_\infty (0,1)$ up to equivalence of norms (for the proof see, e.g., [42,  Theorem B]). Therefore, in such a case, Theorem 4.6 is about the space $Ces_\infty$. However, the structure of rearrangement invariant spaces on a semi-axis with $X_a = \{0\}$ is not so transparent, because we only know that they are subspaces of $L_\infty (0,\infty )$. On the other hand, given a rearrangement invariant space Z, it is not difficult to produce such a subspace - just take $Z \cap L_\infty (0,\infty )$.

Corollary 4.8

Let X be a rearrangement invariant space on $(0,\infty )$ with the Fatou property such that the Cesàro space CX is non-trivial. Suppose that either the ideal $X_a$ is non-trivial and the Cesàro operator C is bounded on X or $X_a = X$. Then the space $C(X \cap L_\infty )$ contains a lattice isometric copy of $\ell _\infty$.

Proof

This fact follows (more or less) directly from Theorem 4.6. We just need to check a few details. Moreover, we will give the proof only in the case when the ideal $X_a$ is non-trivial and the operator C is bounded on X, because the remaining case is much easier.

First of all, let us note that

$$\begin{aligned} X \cap L_\infty \overset{1}{\hookrightarrow } L_\infty \text { (so, also } (X \cap L_\infty )_a = \{0\}) \text { and } C(X \cap L_\infty ) \ne \{0\}. \end{aligned}$$

Put $f_\varepsilon = \chi _{(0,\varepsilon )}$, where $\varepsilon > 0$. We claim that

To show $(\star )$, it is enough to observe that $\varphi _X(0^+) = 0$ (this is actually equivalent to the fact that the ideal $X_a$ is non-trivial, see [7,  Theorem 5.5 (a), pp. 67–68]; cf. [42,  Theorem B]) and the Cesàro operator C is bounded on X, hence

$$\begin{aligned} \Vert f_{\varepsilon }\Vert _{CX} = \Vert C(\chi _{(0,\varepsilon )})\Vert _X \leqslant \varphi _X(\varepsilon ) \Vert C\Vert _{X \rightarrow X} \rightarrow 0 \text { as } \varepsilon \rightarrow 0. \end{aligned}$$

Consequently, we have the following equalities

$$\begin{aligned} \Vert f_{\varepsilon _0}\Vert _{C(X \cap L_\infty )} = \max \{\Vert f_{\varepsilon _0}\Vert _{CX}, \Vert f_{\varepsilon _0}\Vert _{Ces_\infty }\} = \max \{\Vert f_{\varepsilon _0}\Vert _{CX}, 1\} = 1. \end{aligned}$$

This means that we have defined the function $f_{\varepsilon _0}$ with

$$\begin{aligned} C(f_{\varepsilon _0})(0^+) = 1 \text { and } \Vert f_{\varepsilon _0}\Vert _{C(X \cap L_\infty )} = 1 = \Vert \text {id} :X \cap L_\infty \rightarrow L_\infty \Vert . \end{aligned}$$

Now, because the ideal $(X \cap L_\infty )_a$ is trivial, we can use Theorem 4.6 and end the proof. $\square$

Remark 4.9

• (a) Let $X = L_p + L_\infty$ on $(0,\infty )$, where $1< p < \infty$. It is clear that in this case the ideal $X_a$ is non-trivial⁷ and the Cesàro operator is bounded on X. However, it is also clear⁸ that $X \cap L_\infty \equiv L_\infty$. Therefore, using Theorem 4.6 (via Corollary 4.8) we re-prove Proposition 4.9 from [45] which states that the space $Ces_\infty$ contains, as one would expect, a lattice isometric copy of $\ell _\infty$.

• (b) Let $X = L_F$ be an Orlicz space generated by an Orlicz function F such that $b_F := \sup \{x > 0 :F(x) < \infty \} = 1$, $F(1) \leqslant 1$ and the left derivative of F at $x = 1$ is finite. Then there exists an Orlicz function, say G, with $b_G = \infty$ such that $L_F \equiv L_G \cap L_\infty$. The proof of this fact is analogous to the proof of [57,  Theorem 12.1 (a), p. 99]. However, to get an isometry instead of isomorphism, we need to notice two things: (1) due to the assumption regarding the derivative of F, we can always extend F (but not necessarily in a unique way!) to an Orlicz function, say ${\widetilde{F}}$, with $b_{{\widetilde{F}}} = \infty$; (2) the function $({\widetilde{F}} \vee F_\infty )(x) := \max \{{\widetilde{F}}(x), F_\infty (x)\}$, where

  $$\begin{aligned} F_\infty (x) = {\left\{ \begin{array}{ll} 0 &{}\quad \text {for } 0 \leqslant x \leqslant 1\\ \infty &{}\quad \text {for } x > 1 \end{array}\right. }, \end{aligned}$$

  is equal to F. Now, the embedding $L_F \overset{1}{\hookrightarrow } L_{{\widetilde{F}}} \cap L_\infty$ follows directly from [57,  Theorem 12.1]. Therefore, it remains to show that also

  

  Take $f \in L_{{\widetilde{F}}} \cap L_\infty$ with $\max \{\Vert f\Vert _{L_{{\widetilde{F}}}}, \Vert f\Vert _{L_\infty }\} = 1$. Let us consider two cases:

  $1^{\circ }$.:

  $\Vert f\Vert _{L_\infty } = 1$. Then $\left|f(x)\right| \leqslant 1 = b_F$, so $\int _0^\infty F(\left|f(x)\right|)dx = \int _0^\infty {\widetilde{F}}(\left|f(x)\right|)dx \leqslant 1$. Moreover, since $\Vert f\Vert _{L_\infty } = 1 = b_F$, it follows that $\int _0^\infty F(\left|f(x)\right|/\lambda )dx = \infty$ for every $\lambda < 1$. Therefore, $\Vert f\Vert _{L_F} = 1$ as well.

  $2^{\circ }$.:

  $\Vert f\Vert _{L_{{\widetilde{F}}}} = 1$ and $\Vert f\Vert _{L_\infty } < 1$. Then $\int _0^\infty F(\left|f(x)\right|)dx = \int _0^\infty {\widetilde{F}}(\left|f(x)\right|)dx \leqslant 1$, so $\Vert f\Vert _{L_F} \leqslant 1$. In order to obtain a contradiction, suppose that $\Vert f\Vert _{L_F} < 1$. Then $\int _0^\infty F(\left|f(x)\right|/\lambda _0)dx \leqslant 1$ for some $\lambda _0 < 1$. Hence, $\left|f\right|/\lambda _0 \leqslant 1 = b_F$ and, consequently, also $\int _0^\infty {\widetilde{F}}(\left|f(x)\right|/\lambda _0)dx \leqslant 1$. However, this implies that $\Vert f\Vert _{L_{{\widetilde{F}}}} < 1$, which is impossible. The proof of $(\sharp )$ has been completed.

In particular, if $1 \leqslant p < \infty$ and the Orlicz function $F_{p,\infty }$ is given in the following way

$$\begin{aligned} F_{p,\infty }(x) = {\left\{ \begin{array}{ll} x^p &{}\quad \text {for } 0 \leqslant x \leqslant 1\\ \infty &{}\quad \text {for } x > 1 \end{array}\right. }, \end{aligned}$$

then $L_{F_{p,\infty }} \equiv L_p \cap L_\infty$.

The above spaces provided some natural examples of spaces which satisfy the assumptions of Corollary 4.8 (of course, in order to use Corollary 4.8, the functions F and G cannot be completely arbitrary; however, translating the assumptions of Corollary 4.8 into the language of the Orlicz spaces - the $\Delta _2$-condition and the Matuszewska–Orlicz indices (cf. [53,  Proposition 2.b.5, p. 139]) will surely appear - does not present much difficulty).

• (c) Let $\varphi$ be an increasing concave function with $\varphi (0^+) = 1$ (this normalization is basically inessential). Since the norm in the Lorentz space $\Lambda _\varphi$ is given by the formula

  $$\begin{aligned} \Vert f\Vert _{\Lambda _\varphi } = \Vert f\Vert _{L_\infty } + \int _0^\infty f^*(t)\varphi '(t)dt = \left\| {\left( \Vert f\Vert _{\Lambda _{\int _0^t \varphi '(s)ds}}, \Vert f\Vert _{L_\infty }\right) }\right\| _{\ell _1^2}, \end{aligned}$$

  it follows that

  $$\begin{aligned} \Lambda _\varphi \equiv \Lambda _{\int _0^t \varphi '(s)ds} \cap L_\infty \overset{1}{\hookrightarrow } L_\infty \quad \text {and}\quad \left( \Lambda _{\int _0^t \varphi '(s)ds}\right) _a \ne \{0\}, \end{aligned}$$

  where, of course, the intersection space is considered with the equivalent two-dimensional $\ell _1$-norm given above (after all, all norms on ${\mathbb {R}}^2$ are equivalent). However, in this case the proof of Corollary 4.8 does not work. On the other hand, if we consider the general family of equivalent norms on the space $X \cap L_\infty$ of the form

  $$\begin{aligned} \Vert f\Vert _{X \oplus _F L_\infty } = \Vert (\Vert f\Vert _X, \Vert f\Vert _{L_\infty })\Vert _{F}\quad \text {for } f \in X \cap L_\infty , \end{aligned}$$

  where $\Vert \cdot \Vert _F$ is a norm on ${\mathbb {R}}^2$ with the ideal property, then for our proof to work, it is enough to require that

  $$\begin{aligned} \Vert (x_0,1)\Vert _{F} = \Vert (0,1)\Vert _F = 1\quad \text {for some } x_0 > 0, \end{aligned}$$

  which means that the unit sphere in $({\mathbb {R}}^2, \Vert \cdot \Vert _F)$ contains an order interval beginning at point (0, 1) (in particular, $({\mathbb {R}}^2, \Vert \cdot \Vert _F)$ is not strictly monotone). For example, the “hexagonal” norm

  $$\begin{aligned} \Vert (x,y)\Vert _{\text {hex}} = \max \left\{ \left|x + \frac{\sqrt{3}}{3}y\right|, \left|x - \frac{\sqrt{3}}{3}y\right|,\frac{2\sqrt{3}}{3}\left|y\right|\right\} \quad \text {for } x,y \in {\mathbb {R}}, \end{aligned}$$

  seems to be a good candidate for such a norm. But we will stop here.

• (d) Boundedness of the Cesàro operator C on the space X and the fact that X is order continuous are incomparable. Indeed, the Cesàro operator C is not bounded on the Zygmund space $L\log L$ ($L\log L = \Lambda _\varphi$, where $\varphi (t) = t(1+\log (1/t))$; cf. [7,  p. 246]), but the Cesàro space $C(L\log L)$ is non-trivial (cf. [45]) and the space $L\log L$ is order continuous. On the other hand, the Cesàro operator is bounded on $L_\infty$, but the ideal $(L_\infty )_a$ is trivial.

Appendix: Local approach to order continuity in abstract Cesàro sequence spaces

We are going to present here a result characterizing the ideal of order continuous elements in the Cesàro sequence spaces. This result, as well as it’s proof, is analogous to Theorem 7 from [42] (to be fair, due to the fact that $X_a$ is always non-trivial subspace of X, whenever X is a Banach sequence space, it is much simpler than its function counterpart). However, in order to relieve the reader from the tedious obligation to check all the details of this proof on his own, we rather prefer to provide a brief sketch of the argument.

Proposition 5.1

Let X be a rearrangement invariant sequence space such that the Cesàro operator C is bounded on X. Then $(CX)_a = C(X_a)$.

Proof

To prove the equality $(CX)_a = C(X_a)$ we will show the following inclusions

$$\begin{aligned} C(X_a) \subset (CX)_a \subset (CX)_b \subset C(X_a). \end{aligned}$$

The first inclusion $C(X_a) \subset (CX)_a$ holds true even for Banach sequence spaces and its proof is mutatis mutandis the same as Lemma 11 in [42], while the second inclusion $(CX)_a \subset (CX)_b$ follows directly from [7,  Theorem 3.11, p. 18]. Consequently, it remains only to show that $(CX)_b \subset C(X_a)$. We will do this in two steps.

$1^{\circ }$. Take $\chi _{A}$, where $A \subset {\mathbb {N}}$ and $\max (A) := n_0 < \infty$. Let $n>n_0$. Since $p_X > 1$, so⁹$\ell _p \hookrightarrow X$ for all $1< p < p_X$ and we have the following inequalities

$$\begin{aligned} \Vert C(\chi _A)\chi _{\{n,n+1,\dots \}}\Vert _{X} \leqslant \left\| {\frac{1}{k}\chi _{\{n,n+1,\dots \}}(k)}\right\| _{X} \preccurlyeq \left\| {\frac{1}{k}\chi _{\{n,n+1,\dots \}}(k)}\right\| _{\ell _p} \rightarrow 0, \end{aligned}$$

as $n \rightarrow \infty$.

$2^{\circ }$. Take $x = \{x_n\}_{n=1}^\infty \in (CX)_b$. Let $\{s_n\}_{n=1}^\infty$ be a sequence of sequences such that $s_n$ converges to x in the norm of the space CX and $\#({{\,\mathrm{supp}\,}}(s_n)) < \infty$ for all $n \in {\mathbb {N}}$. It follows from the previous step that $\{s_n\}_{n=1}^\infty \subset C(X_a)$. Now, using the reverse triangle inequality, we get

$$\begin{aligned} \Vert C(\left|x\right|) - C(\left|s_n\right|)\Vert _X \leqslant \Vert C(\left|x - s_n\right|)\Vert _X = \Vert x - s_n\Vert _{CX} \rightarrow 0, \end{aligned}$$

as $n \rightarrow \infty$. Since $X_a$ is a closed ideal of X, so $C(\left|x\right|) \in X_a$ and the proof is completed. $\square$

The immediate consequence of the above result is the following fact (which, in the case of the abstract Cesàro function spaces, was the main result of [42]).

Corollary 5.2

Let X be a rearrangement invariant sequence space such that the Cesàro operator C is bounded on X. Then CX is order continuous if and only if X is order continuous as well.