Local approach to order continuity in Ces\`aro function spaces

TL;DR

This paper characterizes points of order continuity in Cesàro function spaces built over symmetric spaces, providing conditions for when these spaces inherit order continuity and applying results to specific Cesàro spaces.

Contribution

It offers a complete characterization of order continuity points in Cesàro function spaces over symmetric spaces, including new criteria and equivalences.

Findings

$(CX)_a = C(X_a)$ under certain conditions

Order continuity of $X$ implies order continuity of $CX$ when the Cesàro operator is bounded

Criteria for order continuity points in Cesàro-Orlicz, Lorentz, and Marcinkiewicz spaces

Abstract

The goal of this paper is to present a complete characterisation of points of order continuity in abstract Ces\`aro function spaces $CX$ for $X$ being a symmetric function space. Under some additional assumptions mentioned result takes the form $(CX)_a = C(X_a)$. We also find simple equivalent condition for this equality which in the case of $I=[0,1]$ comes to $X\neq L^\infty$. Furthermore, we prove that $X$ is order continuous if and only if $CX$ is, under assumption that the Ces\`aro operator is bounded on $X$. This result is applied to particular spaces, namely: Ces\`aro-Orlicz function spaces, Ces\`aro-Lorentz function spaces and Ces\`aro-Marcinkiewicz function spaces to get criteria for OC-points.

Equations

$$C:f\mapsto Cf(x):=\frac{1}{x}{\int }_0^{x}f(t)\text{d}t.$$

$$\mathrm{infess}(A):=\inf \{ϵ\ge 0:m([0,ϵ)\cap A)=0\}.$$

$${\phi }_E(t):={\Vert {\chi }_{[0,t]}\Vert }_E,$$

$${\Vert f^{\ast }{\chi }_{[0,\frac{1}{n})}\Vert }_E\to 0\text{ and }{\Vert f^{\ast }{\chi }_{[n,\infty )}\Vert }_E\to 0\text{ for }n\to \infty .$$

$${\phi }_E(0^{+})$$

$${\Vert {\chi }_{[0,t_n]}\Vert }_E\le {\Vert \frac{1}{n}{f}^{\ast }{\chi }_{[0,t_n]}\Vert }_E\le \frac{1}{n}{\Vert f^{\ast }\Vert }_E\to 0,$$

$$D_{s}f(t):=f(\frac{t}{s}){\chi }_I(\frac{t}{s})=f(\frac{t}{s}){\chi }_{[0,\min \{1,s\})}(t),$$

$$p(E):=\underset{s\to \infty }{\lim }\frac{\text{ln}s}{\text{ln}{\Vert D_s\Vert }_{E\to E}},$$

$$q(E):=\underset{s\to 0^{+}}{\lim }\frac{\text{ln}s}{\text{ln}{\Vert D_s\Vert }_{E\to E}}.$$

$$L^p\cap L^q\hookrightarrow E\hookrightarrow L^p+L^q.$$

$$L^p\cap L^{\infty }\hookrightarrow E\hookrightarrow L^p+L^{\infty }.$$

$$Cf(x):=\frac{1}{x}{\int }_0^{x}f(t)\text{d}t,$$

$$CX:=\{f\in L^0(I):C|f|\in X\},$$

$$Cf\le \left|Cf\right|\le C\left|f\right|\le C(f^{\ast }):=f^{\ast \ast }.$$

$$\{0\}\ne C{L}^{\infty }[0,1]:=Ce{s}_{\infty }[0,1]\hookrightarrow CX[0,1],$$

$$\frac{1}{x}{\chi }_{(t,\infty )}=\frac{1}{x}{\chi }_{(t,t_0]}+\frac{1}{x}{\chi }_{(t_0,\infty )}\in X,$$

$$C(L_a^{\Phi })$$

$$=\{f:I_{\Phi }(\lambda C\left|f\right|)<\infty \text{for all }\lambda (f):=\lambda >0\}$$

$$=\{f:{\rho }_{\Phi }(\lambda f)<\infty \text{for all }\lambda (f):=\lambda >0\}=C_{\Phi },$$

$$(C{L}^{\Phi }{)}_a=(Ce{s}_{\Phi }{)}_a=C_{\Phi }=C(L_a^{\Phi }).$$

$$C(L_a^{\infty })=C\{0\}=\{0\}(Ce{s}_{\infty }{)}_a:=(C{L}^{\infty }{)}_a\ne \{0\},$$

$$C{\chi }_{B_n}(x)$$

$$=\frac{{\chi }_{(a,b)}(x)}{x}{\int }_{(a,\min \{x,b\})\cap A_n}\text{dt}+\frac{m(B_n)}{x}{\chi }_{(b,\infty )}(x)$$

$$\le \frac{m(B_n)}{b}{\chi }_{(a,b)}(x)+\frac{m(B_n)}{x}{\chi }_{(b,\infty )}(x).$$

$${\Vert {\chi }_{(a,b)}{\chi }_{A_n}\Vert }_{CX}$$

$$={\Vert C{\chi }_{B_n}\Vert }_X\le {\Vert \frac{m(B_n)}{b}{\chi }_{(a,b)}\Vert }_X+{\Vert \frac{m(B_n)}{x}{\chi }_{(b,\infty )}\Vert }_X$$

$$=m(B_n)(\frac{1}{b}{\Vert {\chi }_{(a,b)}\Vert }_X+{\Vert \frac{1}{x}{\chi }_{(b,\infty )}\Vert }_X)\to 0\text{as }n\to \infty .$$

$$C({\chi }_{B_n})\le C({\chi }_{[a,m(B_n)+a]}).$$

$${\Vert {\chi }_A{\chi }_{A_n}\Vert }_{CX}$$

$$\le {\Vert C{\chi }_{[a,m(B_n)+a]}\Vert }_X$$

$$={\Vert {\chi }_{[a,m(B_n)+a]}\Vert }_{CX}\to 0\text{as }n\to \infty .$$

$${\Vert f{\chi }_{[\delta ,\infty )}{\chi }_{A_n}\Vert }_{CX}\le ϵ,$$

$$C(f{\chi }_{A_n\cap [0,\delta )})\le \sup \{Cf(s):s\in [0,\delta )\}({\chi }_{[0,1)}+\frac{1}{t}{\chi }_{[1,\infty )})\le ϵ({\chi }_{[0,1)}+\frac{1}{t}{\chi }_{[1,\infty )}),$$

$${\Vert f{\chi }_{A_n}\Vert }_{CX}$$

Full text

00footnotetext: 2010 Mathematics Subject Classification: 46B20, 46B42, 46E30.

Key words and phrases: Cesàro function spaces; Cesàro-Orlicz function spaces; Cesàro-Lorentz function spaces; Cesàro-Marcinkiewicz function spaces; order continuity; local structure of a separated point.

1. Introduction

Much has been said about the Cesàro spaces both from isomorphic and isometric point of view, see for example the papers [1], [2] by Astashkin and Maligranda, [9] by Curbera and Ricker, [10] by Delgado and Soria and [22], [23], [24] by Leśnik and Maligranda and references given there.

For a Banach ideal space $X$ the abstract Cesàro space $CX$ is the space of all functions $f$ such that $C|f|\in X$, equipped with the norm $\left\lVert f\right\rVert_{CX}=\left\lVert C|f|\right\rVert_{X}$, where $C$ denotes the continuous Cesàro operator

[TABLE]

General considerations for this construction has been initiated in [22]. The abstract Cesàro spaces $CX$ are neither symmetric nor reflexive. Surprisingly, the descriptions of Kothe duals of $CX$ spaces is different for the case of $I=[0,1]$ and $I=[0,\infty)$, see [22].

Here we will focus on considering the abstract Cesàro spaces $CX$ for those function spaces $X$ which are symmetric. We study the local structure of this spaces in the terms of order continuity property.

The paper is organized as follows.

In Section 2 we collect some necessary preliminaries on Banach ideal spaces, symmetric function spaces, Cesàro function spaces and order continuity property. Here, we specify Theorems A, B, C, D and E, Fact 1 and Lemma 2 because we will used them often in this article.

Section 3 contains the main results of this paper. Curbera and Ricker in [9] proved that $(CX)_{a}=C(X_{a})$ for symmetric spaces $X\neq L^{\infty}[0,1]$ on $I=[0,1]$. Moreover, Kiwerski and Kolwicz in [15] have shown analogous equality in the case of the Cesàro-Orlicz function spaces $Ces_{\Phi}:=CL^{\Phi}$, see also Remark 6 for a more accurate discussion. We extend these results to the class of symmetric spaces and we get a full characterization of order continuous points in abstract Cesàro function spaces $CX$.

In the last Section 4, we show applications of our characterization for particular cases of symmetric spaces. Some results from this section were proved earlier directly for $X=L^{p}$ by Hassard and Hussein in [12], Shiue in [31], for $X=L^{\infty}$ by Zaanen in [32] and for $X=L^{\Phi}$ by Kiwerski and Kolwicz in [15].

2. Notation and preliminaries

Denote by $m$ the Lebesgue measure on $I$ and by $L^{0}=L^{0}(I)$ the space of all classes of real-valued Lebesgue measurable functions defined on $I$, where $I=[0,1]$ or $I=[0,\infty)$. Through all the paper when we pick a subset $A\subset I$ we assume that $A$ is a Lebesgue measurable set. For a subset $A\subset I$ we define the essential infimum $\operatorname*{infess}(A)$ of $A$ as follows

[TABLE]

A Banach space $E=(E,\|\cdot\|)$ is said to be a Banach ideal space on $I$ (we write $E[0,1]$ or $E[0,\infty)$) if $E$ is a linear subspace of $L^{0}(I)$ and satisfies the condition that if $g\in E$, $f\in L^{0}$ and $|f|\leq|g|$ a.e. on $I$ then $f\in E$ and $\|f\|\leq\|g\|$. Sometimes we write $\left\|\cdot\right\|_{E}$ to be sure in which space the norm has been taken. We say that $E$ is non-trivial if $E\neq\{0\}$.

For $f\in L^{0}$ we define support as $\operatorname{supp}(f):=\{t\in I:f(t)\neq 0\}.$ Recall that the support $\operatorname{supp}(X)$ of Banach ideal space $X$ is defined as measurable subset of $I$ such that $m(\operatorname{supp}(f)\setminus\operatorname{supp}(X))=0$ for every $f\in X$, and for every $U\subset\operatorname{supp}(X)$ of finite positive measure we have $\chi_{U}\in X$.

For two Banach spaces $E$ and $F$ on $I$ the symbol $E\hookrightarrow F$ means that the embedding $E\subset F$ is continuous, i.e. there exists a constant $A>0$ (we call it the embedding constant) such that $\left\|f\right\|_{F}\leq A\left\|f\right\|_{E}$ for all $f\in E$. Recall that for two Banach ideal spaces $E$ and $F$ the embedding $E\subset F$ is always continuous. Moreover, $E=F$ (resp. $E\equiv F$) means that the spaces are the same as the sets and the norms are equivalent (resp. equal). By $E\simeq F$ we denote the fact that the Banach spaces $E$ and $F$ are isomorphic.

For function $f\in L^{0}$ we define distribution functions as $d_{f}(\lambda):=m(\{t\in I:|f(t)|>\lambda\})$ for $\lambda\geq 0$. We say that two functions $f,g\in L^{0}(I)$ are equimeasurable when they have the same distribution functions, i.e. $d_{f}\equiv d_{y}$. By a symmetric function space (symmetric Banach function space or rearrangement invariant Banach function space) on $I$ we mean a Banach ideal space $E=(E,\left\lVert\cdot\right\rVert_{E})$ with the additional property that for any two equimeasurable functions $f,g\in L^{0}(I)$ if $f\in E$ then $g\in E$ and $\left\lVert f\right\rVert_{E}=\left\lVert g\right\rVert_{E}$ (we also accept the convention to write ”symmetric space” within the meaning of ”symmetric function space” because in this paper we focus only on the consideration of this case). In particular, $\left\lVert f\right\rVert_{E}=\left\lVert f^{*}\right\rVert_{E}$, where $f^{*}(t):=\inf\{\lambda>0:d_{f}(\lambda)\leq t\}$ for $t\geq 0$. Note that, if a symmetric function space $E$ on $I$ is non-trivial then $\operatorname{supp}(E)=I$. For the theory of symmetric spaces the reader is referred to [4] and [19].

For a symmetric function space $E$ on $I$ its fundamental function $\varphi_{E}$ is defined by the formula

[TABLE]

for $0<t\in I$. Writing $\varphi_{E}(0^{+})$ or $\varphi_{E}(\infty)$ we understand $\lim\limits_{t\rightarrow 0^{+}}\varphi_{E}(t)$ or $\lim\limits_{t\rightarrow\infty}\varphi_{E}(t)$, respectively.

A point $f\in X$ is said to have an order continuous norm ($f$ is an OC-point) if for any sequence $(f_{n})\subset X$ with $0\leq f_{n}\leq|f|$ and $f_{n}\rightarrow 0$ a.e. on $I$, we have $\left\lVert f_{n}\right\rVert_{X}\rightarrow 0$. By $X_{a}$ we denote the subspace of all order continuous elements of $X$. A Banach ideal space $X$ is called order continuous (we write $X\in(OC)$ for short) if every element of $X$ has an order continuous norm, i.e. if $X_{a}=X$. It is worth to notice that in case of Banach ideal spaces on $I$, $f\in X_{a}$ if and only if $\left\lVert f\chi_{A_{n}}\right\rVert_{X}\rightarrow 0$ for any sequence $(A_{n})$ satisfying $A_{n}\downarrow\emptyset$ (that is, decreasing sequence of Lebesgue measurable sets $A_{n}\subset I$ with intersection of measure zero, see [4, Proposition 3.5, p. 15]). The subspace $X_{a}$ is always closed, see [4, Th. 3.8, p. 16]. Characterization of order continuity given in Theorem A (iii) and (iv) is well known.

Theorem A.

1. (i)

([8, Lemma 2.6]) Let $E$ be symmetric space. Then $f\in E$ is a point of order continuity if and only if $f^{*}$ is also. 2. (ii)

([3, Lemma 2.5], [8, Lemma 2.5], cf. [19]) Let $E$ be symmetric space. Then $f\in E_{a}$ if and only if

[TABLE]

In particular if $f$ is a point of order continuity then $f^{*}(\infty)=0$. 3. (iii)

([27]) A Banach ideal space $E$ is order continuous if and only if $E$ contains no isomorphic copy of $l^{\infty}$. 4. (iv)

([4, Th. 5.5, p. 27]) A Banach ideal space $E$ is order continuous if and only if is separable.

$\hfill\square$

Let $X$ be a Banach ideal space. The closure in $X$ of the set of simple functions is denoted by $X_{b}$. It is well known, that the subspace $X_{b}$ is the closure in $X$ of the set of bounded functions supported in sets of finite measure, cf. [4, Prop. 3.10. p. 17]. Of course, $X_{b}$ is always non-trivial for non-trivial space $X$, moreover the subspace $X_{b}$ is an order ideal of $X$. We always have $X_{a}\subset X_{b}$, see [4, Th. 3.11, p. 18], and the inclusion $\{0\}\neq X_{a}\subset X_{b}$ may be proper, cf. [4, Ex. 3, p. 30]. The fact, that this example is based on non-symmetric construction is essential. In the symmetric case we have only two possibilities, more precisely $X_{a}=\{0\}$ or $X_{a}=X_{b}$. The subspaces $X_{a}$ and $X_{b}$ coincide if and only if the characteriztic functions of the sets of finite measure all have absolutely continuous norms, cf. [4, Th. 3.13, p. 19], [20, Prop. 2.2] and Theorem B. Next theorem describes the opposite extreme, when $X_{a}$ is trivial.

Theorem B. Let $E$ be a symmetric space. The following conditions are equivalent:

1. (i)

$E_{a}=\{0\}$, 2. (ii)

$E\hookrightarrow L^{\infty}$, 3. (iii)

$E_{a}\neq E_{b}$, 4. (iv)

$\varphi_{E}(0^{+})>0$.

In particular, if $I=[0,1]$ then condition (ii) is equivalent to the statement $E=L^{\infty}$.

Proof.

Obviously, $E\hookrightarrow L^{\infty}$ is equivalent to $E=L^{\infty}$ if $I=[0,1]$ because the embedding $L^{\infty}\hookrightarrow E$ holds for each symmetric space $E$ on $I=[0,1]$, see [4, Corollary 6.7, p. 78]. Equivalence of conditions (i), (iii) and (iv) follows immediately from [4, Th. 5.5, p. 67] and our discussion preceding this theorem. For the implication (ii) $\Rightarrow$ (iv) it is enough to observe, that

[TABLE]

where $A$ is the embedding constant. Finally, suppose $\varphi_{E}(0^{+})>0$ and $E\nsubseteq L^{\infty}$. This means, that there is an unbounded element $f\in E$. Then the $f^{*}$ is unbounded in zero and $f^{*}\in E$ by the symmetry of $E$. Therefore we can find a sequence $(t_{n})\subset I$, $t_{n}\rightarrow 0^{+}$ as $n\rightarrow\infty$, with $f^{*}(t_{n})>n$. We have

[TABLE]

as $n\rightarrow\infty$ contrary to the assumption that $\varphi_{E}(0^{+})>0$. ∎

For $s>0$ the dilation operator $D_{s}$ is defined, on $L^{0}(I)$, by

[TABLE]

for $t\in I$, is bounded in any symmetric space $E$ on $I$ and $\left\|D_{s}\right\|_{E\rightarrow E}\leq\text{max}\{1,s\}$ (see [4, p. 148]). The lower and upper Boyd indices of $E$ are defined by

[TABLE]

[TABLE]

They satisfy the inequalities $1\leq p(E)\leq q(E)\leq\infty$. For more details see [6], [5] and [28].

Next result will be used in Section 3.

Theorem C. ([26, Prop. 2.b.3, p. 132]) Let $E$ be a symmetric function space. Then for every $1\leq p<p(E)$ and $q(E)<q\leq\infty$ we have

[TABLE]

In particular, if we take $q=\infty$ then for every $1\leq p<p(E)$ we get

[TABLE]

$\hfill\square$

The continuous Cesàro operator $C:L^{0}(I)\rightarrow L^{0}(I)$ is defined by

[TABLE]

for $0<x\in I$. For a Banach ideal space $X$ on $I$ we define an abstract Cesàro space $CX=CX(I)$ by

[TABLE]

with the norm $\left\|f\right\|_{CX}=\left\|C|f|\right\|_{X}$ (see [9], [10], [22], [23], [24]). Let us note that for non-symmetric space $X$ the space $CX$ need not have a weak unit even if $X$ has it (see [22, Example 2]), so in general $\operatorname{supp}(CX)\subset\operatorname{supp}(X)$. Of course, if we assume that $C:X\rightarrow X$ then $\operatorname{supp}(CX)=\operatorname{supp}(X)$, cf. [22, Remark 1]. Moreover, if $CX$ is non-trivial and $X$ is symmetric space on $I$ then $\operatorname{supp}(CX)=I$, see Lemma 2.

Let us mention the important result about boundedness of the Cesàro operator.

Theorem D. [17, p. 127] For any symmetric space $E$ on $I$ the operator $C:E\rightarrow E$ is bounded if and only if the lower Boyd index satisfies $p(E)>1$. $\hfill\square$

Note that if $(X,\left\|\cdot\right\|_{X})$ is a Banach ideal space then the assumption $C:X\rightarrow X$ is equivalent to the statement $C$ is bounded. In fact, if $C:X\rightarrow X$, then $X\hookrightarrow CX$. This means that there is $M>0$ with $\left\|f\right\|_{CX}=\left\|C|f|\right\|_{X}\leq M\left\|f\right\|_{X}$ for all $f\in X$, i.e. $C$ is bounded. On the other hand, if $C$ is bounded then for all $f\in X$ we have $\left\lVert f\right\rVert_{CX}=\left\lVert C|f|\right\rVert_{X}\leq\|C\|\left\lVert|f|\right\rVert_{X}=\|C\|\left\lVert f\right\rVert_{X},$ which means that $X\subset CX$ and $C:X\rightarrow X$. However, it may happen that $X\not\hookrightarrow CX$ (see [10, Proposition 2.1], [16, Example 14]). Moreover, $C:CX\rightarrow X$ is always bounded (from the definition of $CX$) and $CX$ is so-called optimal domain of $C$ for $X$ (cf. [10] and [23]).

The immediate consequence of Theorem D and the above discussion about boundedness of the Cesàro operator is a next

Theorem E. For any symmetric space $E$ the embedding $E\hookrightarrow CE$ holds if and only if $p(E)>1$. In particular, if $p(E)>1$ then the space $CE$ is non-trivial.$\hfill\square$

It is worth to notice the following useful observation.

Fact 1**.**

(cf. [4, Prop. 3.2, p. 52]) For every function $f:I\rightarrow\mathbb{R}$ we have the following inequalities

[TABLE]

Moreover, $(Cf)^{*}\leq C(f^{*})$. $\hfill\square$

Let us note some basic fact about the non-triviality of the space $CX$ when $X$ is symmetric function space.

Lemma 2**.**

Let $X$ be symmetric space on $I$. If $I=[0,1]$ then $CX\neq\{0\}$. If $I=[0,\infty)$ then the following conditions are equivalent

1. (i)

$CX\neq\{0\}$, 2. (ii)

$\frac{1}{x}\chi_{(t_{0},\infty)}\in X$ for some $t_{0}>0$, 3. (iii)

$\frac{1}{x}\chi_{(t,\infty)}\in X$ for all $t>0$.

Proof.

We give an easy proof for the convenience of the reader. Assume that $X$ is symmetric space on $I$. Because $L^{\infty}[0,1]\hookrightarrow X[0,1]$, so we have that

[TABLE]

see [4, Corollary 6.7, p. 78] and [15, Remark]. This means that $CX[0,1]$ is always non-trivial. Equivalence of conditions (i) and (ii) follows from [22, Theorem 1 (a)]. Of course, (iii) implies (ii). Therefore, we will only show that (ii) implies (iii). In fact, take $0<t<t_{0}$ (if $t_{0}<t$ there is nothing to prove). Then

[TABLE]

because $\frac{1}{x}\chi_{(t_{0},\infty)}\in X$ from the assumption and $\frac{1}{x}\chi_{(t,t_{0}]}\in X$ since $X$ is symmetric. ∎

3. On the OC-points in Cesàro function spaces CX

Theorem 3**.**

Let $X$ be symmetric space such that Cesàro operator is bounded on $X$. Then $X\in(OC)$ if and only if $CX\in(OC)$.

The proof of the above theorem will be given at the end of this section as a consequence of characterization of subspace $(CX)_{a}$. We will begin with some observations and examples.

Remark 4**.**

The implication: $X$ is order continuous then $CX$ is also, follows easily (in fact from the definition) without any assumption, see [24, Lemma 1 (a)]. Moreover, in the case of $I=[0,1]$ Theorem 3 has been proved already, see [15, Proposition 2].

Leaving the assumption of symmetry Theorem 3 ceases to be true, as the following example shows.

Example 5**.**

(see [24, Example 1]) If $X=L^{2}[0,\frac{1}{4}]\oplus L^{\infty}[\frac{1}{4},\frac{1}{2}]\oplus L^{2}[\frac{1}{2},1]$, then of course $X\notin(OC)$ but $CX=Ces_{2}[0,1]\in(OC)$, see Proposition 20.

Remark 6**.**

Curbera and Ricker showed in [9, Prop. 3.1 (c)], using the methods of vector measures and integral representation, that (adapting into our notation) $C(X_{a})=CX_{a}$ whenever $X$ is a symmetric space on $[0,1]$ and $X\neq L^{\infty}$. The last condition can be expressed equivalently in several ways, cf. Theorem B. Similar results are also known in the case of $I=[0,\infty)$ but only for a certain class of the Orlicz spaces. Namely, suppose that $\Phi$ is an Orlicz function with $\Phi<\infty$, i.e. $b_{\Phi}=\infty$ and $L^{\Phi}$ is the Orlicz space generated by function $\Phi$ (cf. the definitions in the subsection 4.2 and references therein). From [15, Theorem 5 (i)] we have the following equalities

[TABLE]

where $C_{\Phi}$ is the space of all order continuous elements of space $CL^{\Phi}:=Ces_{\Phi}$, so

[TABLE]

Example 7**.**

The above observation leads naturally to the question - is it true, that $C(X_{a})=(CX)_{a}$ for any symmetric space $X$? Moment of thought show us that in general that is not the case. Taking $L^{\infty}$ we observe that

[TABLE]

since elementary computation shows that, for example $\chi_{[\frac{1}{2},1)}\in(CL^{\infty})_{a}$.

However, in Theorem 16 we will give an answer when this representation is possible and describe the space $(CX)_{a}$ in the other cases. We start with the following lemma.

Lemma 8**.**

Let $X$ be symmetric space such that $CX\neq\{0\}$ and let $A\subset I$. If $\operatorname*{infess}(A)>0$ and $m(A)<\infty$ then $\chi_{A}\in(CX)_{a}$.

Proof.

At the beginning we will show that $\chi_{(a,b)}\in(CX)_{a}$, when $0<a<b<\infty$. Let $(A_{n})\subset I$ be sequence of measurable sets with $A_{n}\downarrow\emptyset$. Put $B_{n}=A_{n}\cap(a,b)$. Simply $m(B_{n})\rightarrow 0$ when $n\rightarrow\infty$. We have

[TABLE]

From Lemma 2, assumption $CX\neq\{0\}$ implies $\left\lVert\frac{1}{x}\chi_{(b,\infty)}\right\rVert_{X}<\infty$. Therefore

[TABLE]

Since $A_{n}$ was chosen arbitrary, we get $\chi_{(a,b)}\in(CX)_{a}$.

Let $A\subset I$ be a subset with $\operatorname*{infess}(A)=a>0$ and $m(A)<\infty$. Take a sequence of measurable sets $(A_{n})\subset I$ with $A_{n}\downarrow\emptyset$. Denote $B_{n}=A_{n}\cap A$. Then, like above, $m(B_{n})\rightarrow 0$ when $n\rightarrow\infty$. Note that

[TABLE]

Since $\chi_{[a,m(B_{n})+a]}\leq\chi_{[a,m(A)+a]}$ and $\chi_{[a,m(B_{n})+a]}(t)\downarrow 0$ as $n\rightarrow\infty$ for a.e. $t\in I$ thus, from a previous case, we obtain

[TABLE]

Thus $\chi_{A}\in(CX)_{a}$. ∎

Corollary 9**.**

Let $X$ be symmetric space such that $CX\neq\{0\}$. If $f\in(CX)_{b}$ and $\operatorname*{infess}(A)>0$, where $A\subset I$ then $f\chi_{A}\in(CX)_{a}$.

Proof.

Let $f\in(CX)_{b}$, $(f_{n})$ be sequence of simple functions with $\lim\limits_{n\rightarrow\infty}f_{n}=f$ and $A\subset I$ be like in the assumptions above. Of course, $f_{n}\chi_{A}\rightarrow f\chi_{A}$. Moreover, $m(\operatorname{supp}(f_{n})\cap A)<\infty$ thus $f_{n}\chi_{A}\in(CX)_{a}$ from Lemma 8. Since $C:CX\rightarrow X$ is always bounded and $(CX)_{a}$ is closed hence $f\chi_{A}\in(CX)_{a}$. ∎

We will use the notation $\Delta_{0}=\Delta_{0}(E):=\{f\in E:\lim\limits_{t\rightarrow 0}Cf(t)=0\},$ where $E$ is a Banach ideal space.

Lemma 10**.**

Let $X$ be symmetric space. Then $(CX)_{b}\cap\Delta_{0}\subset(CX)_{a}.$

Proof.

If $CX=\{0\}$ there is nothing to prove. Assume that $CX\neq\{0\}$ and take $0\leq f\in(CX)_{b}\cap\Delta_{0}$. Let $(A_{n})\subset I$ be arbitrary sequence of measurable subsets with $A_{n}\downarrow\emptyset$. Take $\epsilon>0$. Since $Cf(0^{+})=0$ there exist $\delta>0$ such that $\sup_{t\in[0,\delta)}Cf(t)<\epsilon$. From Corollary 9 we get $f\chi_{[\delta,\infty)}\in(CX)_{a}$. Therefore there exist $n_{\delta}\in\mathbb{N}$ with

[TABLE]

for every $n\geq n_{\delta}$. Without loss of generality suppose that $\delta\leq 1$. Then

[TABLE]

for $n\in\mathbb{N}$. Moreover, $\left\lVert\frac{1}{t}\chi_{[1,\infty)}\right\rVert_{X}<\infty$ in view of our assumption $CX\neq\{0\}$ and Lemma 2. We have

[TABLE]

for $n\geq n_{\delta}$. Since $\epsilon>0$ was arbitrary, we conclude that

[TABLE]

which ends the proof. ∎

Lemma 11**.**

Let $X$ be Banach ideal function space, then

[TABLE]

Proof.

Suppose $0\leq f\in C(X_{a})$. Let $(A_{n})$ be arbitrary sequence of measurable subsets of $I$ with $A_{n}\downarrow\emptyset$. First, note that $f(t)\chi_{A_{n}}(t)\downarrow 0$ as $n\rightarrow\infty$ for a.e. $t\in I$ thus, from Lebesgue dominated convergence theorem, we obtain that $C(f\chi_{A_{n}})(t)\downarrow 0$ as $n\rightarrow\infty$ for a.e. $t\in I$. Since $Cf\in X_{a}$ and $C(f\chi_{A_{n}})\leq Cf$ for every $n\in\mathbb{N}$ we have

[TABLE]

i.e. $f\in(CX)_{a}$. ∎

Lemma 12**.**

Let $X$ be symmetric function space such that Cesàro operator is bounded on $X$ and $X_{a}=X_{b}$. Then $(CX)_{b}\subset C(X_{a})$.

Proof.

We divide the proof into three steps. First, we will show that $\chi_{[0,a]}\in C(X_{a})$ for every $0<a<\infty$. Fix $a>0$ and note that

[TABLE]

for $0<t\in I$. Since $X_{a}=X_{b}$ thus $\chi_{[0,a]}\in X_{a}$ and we only need to show that $f(t):=\frac{1}{t}\chi_{[a,\infty]}(t)\in X_{a}$. Let $(A_{n})$ be sequence of measurable subsets of $I$ with $A_{n}\downarrow\emptyset$. Without loss of generality we can assume that $A_{n}\subset\operatorname{supp}(f)=[a,\infty)$ for every $n\in\mathbb{N}$. Let $\delta>0$ be arbitrary. From Lemma 8 we have $\chi_{[a,a+\delta]}\in X_{a}$ therefore there exist $n_{\delta}\in\mathbb{N}$ such that $\left\lVert\frac{1}{a}\chi_{[a,a+\delta]}\chi_{A_{n}}\right\rVert_{X}\leq\frac{1}{\delta}$ for all $n\geq n_{\delta}$. From Theorem C and our assumption that Cesàro operator is bounded there exist $p>1$ with $L^{p}\cap L^{\infty}\hookrightarrow X$. Since $L^{p}\in(OC)$, we can find $n_{\delta}^{\prime}\in\mathbb{N}$ satisfying

[TABLE]

for every $n\geq n_{\delta}^{\prime}$. Let $n\geq\max\{n_{\delta},n_{\delta}^{\prime}\}$ we have

[TABLE]

where $D>0$ is inclusion constant. Since $\delta>0$ was arbitrary, we obtain

[TABLE]

Consequently, $f\in X_{a}$ and hence $\chi_{[0,a]}\in C(X_{a})$.

Now let $A\subset I$ be arbitrary set of finite measure. From Fact 1 we have

[TABLE]

From a previous step of proof we also know that $C(\chi_{[0,m(A)]})\in X_{a}$ thus from ideal property of $X_{a}$ we get $C(\chi_{A})\in X_{a}$, i.e. $\chi_{A}\in C(X_{a}).$

Finally, let $f\in(CX)_{b}$ and $(f_{n})$ be a sequence of simple functions with $\lim\limits_{n\rightarrow\infty}f_{n}=f$ pointwise. We already know that $Cf_{n}\in X_{a}$ for every $n\in\mathbb{N}$. From continuity of Cesàro operator $C:CX\rightarrow X$ we have that $Cf_{n}\rightarrow Cf$. Therefore $Cf\in X_{a}$ because $X_{a}$ is closed and the proof is complete. ∎

Remark 13**.**

We can prove more, namely: if $X$ is a symmetric function space with $X_{a}=X_{b}$ then $(CX)_{b}\subset C(X_{a})$ if and only if $\frac{1}{t}\chi_{(\lambda_{0},\infty)}(t)\in X_{a}$ for some $\lambda_{0}>0$. In above proof we in fact proved the sufficiency. The necessity is even simpler, if $\frac{1}{t}\chi_{(1,\infty)}(t)\notin X_{a}$ then $C(\chi_{(0,1)})(t)\notin X_{a}$, i.e. $\chi_{(0,1)}(t)\notin C(X_{a})$, but of course $\chi_{(0,1)}(t)\in(CX)_{b}$. It follows from the proof of remaining lemma that if $C:X\rightarrow X$ and $X_{a}=X_{b}$ then $\frac{1}{t}\chi_{[1,\infty)}\in X_{a}$. Is also worth noting that in the case of $I=[0,1]$ the assumption $X\neq L^{\infty}$ is equivalent to $\frac{1}{t}\chi_{(\lambda,\infty)}(t)\in X_{a}$ for all $0<\lambda<1$, see Theorem B.

Lemma 14**.**

Let $X$ be symmetric space with $X_{a}=\{0\}$. Then $(CX)_{a}\subset\Delta_{0}.$

Proof.

If $CX=\{0\}$ then inclusion is trivial, so we can assume that $CX\neq\{0\}$. Take $0\leq f\notin\Delta_{0}$. Then

[TABLE]

We can find a sequence $(a_{n})\subset(0,1)$ such that $a_{n}\rightarrow 0$ and $Cf(a_{n})>\frac{\delta}{2}$ for every $n\in\mathbb{N}$. Since $Cf$ is continuous function, therefore there exist open neighborhood $U_{n}$ of $a_{n}$ with $\inf\limits_{t\in U_{n}}\{Cf(t)\}\geq\frac{\delta}{4}$ for $n\in\mathbb{N}$. We can chose a subsequence $n_{k}\in\mathbb{N}$ such that $U_{n_{k}}\cap[0,\frac{1}{k})\neq\emptyset$ for $k\in\mathbb{N}$. Now we have

[TABLE]

which shows that $f\notin(CX)_{a}$. ∎

Lemma 15**.**

Let $X$ be symmetric space with $X_{a}\neq\{0\}$. Then

[TABLE]

Proof.

Take $\delta>0$. Let $(f_{n})\subset(CX)_{b}$ be a sequence chosen so that $f_{n}\rightarrow f$ almost everywhere. Using Corollary 9 we conclude that $f\chi_{[\delta,\infty)}\in(CX)_{a}$. Since $C(X_{a})\subset(CX)_{a}$ and $X_{a}\neq\{0\}$ we have

[TABLE]

Observe that

[TABLE]

thus $f_{n}\chi_{[0,\delta)}\in(CX)_{a}$ using the ideal property of $(CX)_{a}$. From continuity of $C:CX\rightarrow X$ we have $f_{n}\chi_{[0,\delta)}\rightarrow f\chi_{[0,\delta)}$ in $CX$. This means that $f\chi_{[0,\delta)}\in(CX)_{a}$ because $(CX)_{a}$ is closed. ∎

Theorem 16**.**

Let $X$ be symmetric function space. Then one of following holds:

1. (i)

$(CX)_{a}=(CX)_{b}$ if $X_{a}\neq\{0\}$, 2. (ii)

$(CX)_{a}=(CX)_{b}\cap\Delta_{0}$ if $X_{a}=\{0\}$.

In particular, if Cesàro operator is bounded on $X$ and $X_{a}\neq\{0\}$ then

[TABLE]

Proof.

Suppose $X_{a}\neq\{0\}$. Then $X_{a}=X_{b}$ from Theorem B. This is the case of Lemma 15 so we have $(CX)_{a}=(CX)_{b}$. If we additionally assume that the Cesàro operator $C$ is bounded on $X$, using Lemma 11 and Lemma 12 we have

[TABLE]

Therefore $(CX)_{a}=C(X_{a})$.

Assume now that $X_{a}=\{0\}.$ From Lemma 14 we have $(CX)_{a}\subset\Delta_{0}$. Since we always have that $(CX)_{a}\subset(CX)_{b}$ hence $(CX)_{a}\subset(CX)_{b}\cap\Delta_{0}$. Combining this with Lemma 10 we obtain

[TABLE]

thus $(CX)_{a}=(CX)_{b}\cap\Delta_{0}$. ∎

Remark 17**.**

The previous theorem can be formulated in a more concise form if we assume that the Cesàro operator is bounded on symmetric space $X$. In this case

[TABLE]

Moreover, it follows from Remark 13 that if $X$ is symmetric space and $X_{a}$ is non-trivial then $(CX)_{a}=C(X_{a})$ if and only if $\frac{1}{t}\chi_{(\lambda_{0},\infty)}(t)\in X_{a}$ for some $\lambda_{0}>0$. Additionally, as consequence of Remark 13, we have $(CX)_{a}=C(X_{a})$ if and only if $X\neq L^{\infty}$ in the case of $I=[0,1]$.

Lemma 18**.**

Let $X$ be a symmetric space such that the Cesàro operator is bounded on $X$. If $f\in X\setminus X_{a}$ then $f^{*}\in CX\setminus(CX)_{a}$.

Proof.

Take an element $f\in X\setminus X_{a}$. From a symmetry of $X$ we get $f^{*}\in X$ and $C(f^{*})\in X$ since the operator $C$ is bounded on $X$. Now we have to consider two cases:

1. (i)

$X_{a}\neq\{0\}$. Then, from Theorem 16, $(CX)_{a}=C(X_{a})$. From Theorem A (i) $f^{*}\notin X_{a}$. From ideal property of $X_{a}$ and since $C(f^{*})\geq f^{*}$, we have that $C(f^{*})\notin X_{a}$, i.e.

[TABLE] 2. (ii)

$X_{a}=\{0\}$. Observe that $0\neq f$, since $0\in X_{a}$ thus

[TABLE]

From Theorem 16 we obtain $f^{*}\notin\Delta_{0}\cap(CX)_{b}=(CX)_{a}$.

∎

It is time to give proof of the Theorem 3 announced at the beginning of this section.

Proof of Theorem 3. Necessity. Let $X$ be a symmetric space such that Cesàro operator is bounded on $X$ and suppose $X\in(OC)$. Then $X=X_{a}$ and from Theorem 16 we get

[TABLE]

which means that $CX\in(OC)$.

Sufficiency. If $X\notin(OC)$ then there exist an element $f\notin X_{a}$. From Lemma 18 $f^{*}\notin(CX)_{a}$ and consequently $CX\notin(OC)$ which completes the proof. $\hfill\square$

4. Applications

Although each of the following spaces belongs to the class of symmetric spaces, in this special cases, the criteria for OC-points become more specified.

4.1. The Cesàro function spaces $Ces_{p}$

Remark 19**.**

We have the following characterization of the closure of the set of simple functions in the space $L^{\infty}$

[TABLE]

and

[TABLE]

Of course, $(L^{\infty})_{a}=\{0\}$ and $p(L^{\infty})>1$. Let us define a set $\Delta_{\infty}=\Delta_{\infty}(E):=\{f\in E:\lim\limits_{t\rightarrow\infty}Cf(t)=0\}$ for $E$ being a Banach ideal space on $[0,\infty)$. With this notation, we get the following reformulation of Theorem 16 (ii):

[TABLE]

The space $Ces_{\infty}[0,1]$ is known as Korenblyum-Krein-Levin space $K$ and it is also known that

[TABLE]

see [32, pp. 469-471]. Thus, the above characterization is a generalization of this classical result.

It is worth to mention that $Ces_{1}[0,1]\equiv L^{1}(\ln(1/t))$. From Lebesgue dominated convergence theorem we obtain that $L^{1}(\ln(1/t)$ is order continuous. Moreover, $Ces_{1}[0,\infty)=\{0\}$, see [2, Th. 3.1 (a)]. Since the spaces $L^{p}$ are order continuous for $1\leq p<\infty$ and $L^{\infty}_{a}=\{0\}$, so as a consequence of Theorem 16 and Remark 19 we obtain well known result (cf. [2, Th. 3.1 (b)]).

Proposition 20**.**

We have the following characterization

[TABLE]

In particular, the Cesàro function space $Ces_{p}$ is order continuous if and only if $1\leq p<\infty$.

4.2. The Cesàro-Orlicz function spaces $Ces_{\Phi}$

A function $\Phi:[0,\infty)\rightarrow[0,\infty]$ is called an Orlicz function if:

1. (i)

$\Phi$ is convex, 2. (ii)

$\Phi(0)=0$, $\Phi(\infty)=\infty$, 3. (iii)

$\Phi$ is neither identically equal to zero nor infinity on $(0,\infty)$.

For more information about Orlicz functions see [7] and [18].

Let us denote by

[TABLE]

and

[TABLE]

We have, $a_{\Phi}<\infty$ and $b_{\Phi}>0$, since an Orlicz function is neither identically equal to zero nor infinity on $[0,\infty)$. The function $\varphi$ is continuous and non-decreasing on $[0,b_{\Phi})$ and is strictly increasing on $[a_{\Phi},b_{\Phi})$.

We say an Orlicz function $\Phi$ satisfies the condition $\Delta_{2}$ for large arguments ($\Phi\in\Delta_{2}(\infty)$ for short) if there exists $K>0$ and $u_{0}>0$ such that $\Phi(u_{0})<\infty$ and

[TABLE]

for all $u\in[u_{0},\infty)$. Similarly, we can define the condition $\Delta_{2}$ for small, with $\Phi(u_{0})>0$ $(\Phi\in\Delta_{2}(0))$ or for all arguments $(\Phi\in\Delta_{2}(\mathbb{R_{+}}))$. These conditions play a crucial role in the theory of Orlicz spaces, see [7], [18], [29] and [30].

The Orlicz function space $L^{\Phi}=L^{\Phi}(I)$ generated by an Orlicz function $\Phi$ is defined by

[TABLE]

where $I_{\Phi}(f)=\int_{I}\Phi(|f(t)|)\text{d}t$ is a convex modular (for the theory of Orlicz spaces and modular spaces see [29] and [30]). The space $L^{\Phi}$ is a Banach ideal function space with the Luxemburg-Nakano norm

[TABLE]

Let us recall that

[TABLE]

where $S_{0}$ is the ideal of bounded functions with support of finite measure. It is also easy to see that

[TABLE]

when $b_{\Phi}=\infty$, cf. [11], [29, Th. 3.3 (a), p. 17] and Theorem B.

The Cesàro-Orlicz function space $Ces_{\Phi}=Ces_{\Phi}(I)$ is defined by $Ces_{\Phi}(I):=CL^{\Phi}(I)$. Consequently, the norm in the space $Ces_{\Phi}$ is given by the formula

[TABLE]

where $\rho_{\Phi}(f)=I_{\Phi}(C|f|)$ is a convex modular. Note that $Ces_{\Phi}[0,1]\neq\{0\}$ for any Orlicz function $\Phi$, see Lemma 2, [15, Remark 4]. Moreover, for the equivalent conditions to the non-triviality of $Ces_{\Phi}[0,\infty)$ see [15, Proposition 3].

Lemma 21**.**

Let $\Phi$ be an Orlicz function. Then

[TABLE]

Proof.

The idea of this proof comes from [11, Th. 3.1]. Without loss of generality, we can consider only the case when the Cesàro-Orlicz function space is non-trivial. Like in mentioned article we use the notation

[TABLE]

We claim that:

1. (i)

$H(S_{0})$ is closed, 2. (ii)

$H(S_{0})\subset\text{cl}{S_{0}}$, where by $\text{cl}{S_{0}}$ we mean the closure of $S_{0}$ in $Ces_{\Phi}$.

Take $f\in\text{cl}H(S_{0})$ and fix $\lambda>0$. There exist a sequence $(f_{n})\subset H(S_{0})$ with $f_{n}\rightarrow f$. From [29, Th. 1.3, p. 8] there exist $n_{0}\in\mathbb{N}$ such that

[TABLE]

Let $s\in S_{0}$ be such that

[TABLE]

Since modular $\rho_{\Phi}$ is convex, we have

[TABLE]

and (i) follows.

Pick $f\in H(S_{0})_{+}$ and $\lambda>0$. From the definition of $H(S_{0})$ there exist $0\leq s_{\lambda}\leq f$ such that

[TABLE]

Take a sequence $(y_{n})\subset S_{0}$ with $y_{n}\uparrow f-s_{\lambda}$ a.e. on $I$. Since $0\downarrow f-s_{\lambda}-y_{n}\leq f-s_{\lambda}$ a.e. on $I$ thus using Lebesgue dominated convergence theorem we get $C(f-s_{\lambda}-y_{n})\downarrow 0$ a.e. on $I$. Because $L^{1}\in(OC)$ and

[TABLE]

for all $n\in\mathbb{N}$, so

[TABLE]

Therefore there exist $n_{0}\in\mathbb{N}$ with

[TABLE]

i.e. $\text{dist}(f,S_{0})\leq\lambda$, see [11]. Since $\lambda>0$ was arbitrary so $f\in\text{cl}{S_{0}}$. This completes the proof of part (ii).

It is clear that $\text{cl}{S_{0}}=(Ces_{\Phi})_{b}$ and $S_{0}\subset H(S_{0})$. Combining this observation with (i) and (ii) we conclude that

[TABLE]

and

[TABLE]

i.e. $(Ces_{\Phi})_{b}=H(S_{0})$. ∎

Next proposition is a generalisation of Theorem 5, Theorem 7 and Corollary 8 from [15].

Proposition 22**.**

Let $\Phi$ be an Orlicz function. Then

[TABLE]

Moreover, if $p(L^{\Phi})>1$ then the Cesàro-Orlicz function space $Ces_{\Phi}$ is order continuous if and only if $\varphi\in\Delta_{2}$ (i.e. if $p(L^{\Phi})>1$ then $Ces_{\Phi}\in(OC)$ if and only if $L^{\Phi}\in(OC)$).

Proof.

Suppose $Ces_{\Phi}$ is non-trivial and

1. (i)

$b_{\Phi}=\infty$. Then there exist $\lambda_{0}>0$ with $I_{\Phi}(\frac{\eta_{0}}{x}\chi_{(\lambda_{0},\infty)})<\infty$ for some $\eta_{0}>0$. We will show that this inequality holds for all $\eta>0$. There exist $\gamma>0$ such that $I_{\Phi}(\frac{1}{x}\chi_{(\gamma,\infty)})<\infty$. Indeed, taking $\gamma=\frac{\lambda_{0}}{\eta_{0}}$ we have

[TABLE]

Take $\eta>0$. Then

[TABLE]

which proves the claim. Therefore, $\frac{1}{x}\chi_{(\lambda_{0},\infty)}(x)\in L^{\Phi}_{a}$. Using Remark 13 we get

[TABLE] 2. (ii)

$a_{\Phi}=0$ and $b_{\Phi}<\infty$. In this case $L^{\Phi}_{a}=\{0\}$, thus from Theorem 16 (ii) and Lemma 21 we get

[TABLE] 3. (iii)

$a_{\Phi}>0$ and $b_{\Phi}<\infty$. Then $L^{\Phi}=L^{\infty}$, cf. [29, Ex. 1, p. 98] thus $Ces_{\Phi}=Ces_{\infty}$ and it is enough to apply Remark 19.

Moreover, the Orlicz space $L^{\Phi}$ is order continuous if and only if $\Phi\in\Delta_{2}$, see [29, Remark 1, p. 22]. Therefore, if $p(L^{\Phi})>1$ then the Cesàro-Orlicz function space $Ces_{\Phi}$ is order continuous if and only if $\varphi\in\Delta_{2}$ from Theorem 3. ∎

Remark 23**.**

Note that $p(L^{\Phi})=\alpha_{\Phi}$ and $q(L^{\Phi})=\beta_{\Phi}$, where $\alpha_{\Phi}$ and $\beta_{\Phi}$ are the so-called lower and upper Orlicz-Matuszewska indices, see [26, Prop. 2.b.5 and Remark 2, p. 140]) and [29], [30] for more information.

4.3. The Cesàro-Lorentz function spaces $C\Lambda_{\varphi}$

The fundamental function $\varphi_{E}$ of every symmetric function space $E$ on $I$ is quasi-concave on $I$, that is, $\varphi_{E}(t)=0$ if and only if $t=0$, $\varphi_{E}(t)$ is increasing on $I$ and $\varphi_{E}(t)/t$ is non-increasing for $t\in(0,m(I))$. Moreover, for any quasi-concave function $\varphi$ there is a symmetric function space whose fundamental function is $\varphi$. The smallest symmetric function space with fundamental function $\varphi$ is called the Lorentz function space $\Lambda_{\varphi}=\Lambda_{\varphi}(I)$ and is defined as

[TABLE]

with a norm given as $\left\lVert f\right\rVert_{\Lambda_{\varphi}(I)}=\int_{0}^{m(I)}f^{*}\text{d}\varphi$. The fundamental function of a symmetric function space $E$ is not necessary concave but $E$ can be equivalently renormed in such a way that the resulting fundamental function is concave. In this case the Riemann-Stieltjes integral in the definition of the Lorentz function space $\Lambda_{\varphi_{E}}$ my be rewritten in the form

[TABLE]

where $\varphi_{E}(t)=\int_{0}^{t}\phi_{E}(s)\text{d}s$. Then we have embedding $\Lambda_{\varphi_{E}}\hookrightarrow E$ with embedding constant equal 1.

Proposition 24**.**

Let $\varphi$ be quasi-concave function. Then

[TABLE]

In particular, the Cesàro-Lorentz function space $C\Lambda_{\varphi}$ is order continuous if and only if $\varphi(0^{+})=0$ and $\varphi(\infty)=\infty$ (i.e. $C\Lambda_{\varphi}\in(OC)$ if and only if $\Lambda_{\varphi}\in(OC)$).

Proof.

We can assume that $C\Lambda_{\varphi}$ is non-trivial because if $C\Lambda_{\varphi}=\{0\}$ there is noting to prove. To prove the first part we have to consider the following cases. Suppose

1. (i)

$\varphi(0^{+})=0$ and $\varphi(\infty)=\infty$. It follows from [19, Lemma 5.1] and Theorem A (iv) that $\Lambda_{\varphi}\in(OC)$. Moreover, from [21, Corollary 4.13] we have $\frac{1}{x}\chi_{(1,\infty)}(x)\in(\Lambda_{\varphi})_{a}$ so using Remark 17 we obtain $(C\Lambda_{\varphi})_{a}=C(\Lambda_{\varphi})_{a}=C\Lambda_{\varphi}$. 2. (ii)

$\varphi(0^{+})=0$ and $\varphi(\infty)<\infty$. In this situation $\Lambda_{\varphi}\notin(OC)$ but $(\Lambda_{\varphi})_{a}$ is non-trivial. More precisely,

[TABLE]

see [21, Corollary 4.13]. Since $C$ is bounded, using Theorem 16 we get

[TABLE] 3. (iii)

$\varphi(0^{+})>0$ and $\varphi(\infty)=\infty$. From Theorem B we get $(\Lambda_{\varphi})_{a}=\{0\}$ and it is enough to use Theorem 16 (ii). 4. (iv)

$\varphi(0^{+})>0$ and $\varphi(\infty)<\infty$. In this case $\Lambda_{\varphi}=L^{\infty}$, see [19, p. 108] and we can use Remark 19.

The second part is a direct consequence of the foregoing considerations. ∎

4.4. The Cesàro-Marcinkiewicz function spaces $CM_{\varphi}$

For any quasi-concave function $\varphi$ on $I$ the Marcinkiewicz function space (called also weak Lorentz space) $M_{\varphi}=M_{\varphi}(I)$ is defined as

[TABLE]

with a norm $\left\lVert f\right\rVert_{M_{\varphi}(I)}=\sup_{t\in I}\varphi(t)f^{**}(t)$, where $f^{**}(t):=\frac{1}{t}\int_{0}^{t}f^{*}(s)\text{d}s$ is a maximal function. The Marcinkiewicz function space $M_{\varphi}$ is the largest symmetric function space with fundamental function $\varphi$, i.e. for any symmetric function space $E$ we have $E\hookrightarrow M_{\varphi_{E}}$ with embedding constant equal 1.

Proposition 25**.**

Let $\varphi$ be quasi-concave function. Then

[TABLE]

In particular, if we assume that $p(M_{\varphi})>1$ then the non-trivial Cesàro-Marcinkiewicz function space $CM_{\varphi}$ is never order continuous.

Proof.

Let us consider the following cases:

1. (i)

$\varphi(0^{+})=0$ and $p(M_{\varphi})>1$. Operator $C$ is bounded from Theorem D and

[TABLE]

see [13, Def. 1.3 and Th. 1.3] so applying Theorem 16 we get

[TABLE] 2. (ii)

$\varphi(0^{+})=0$ and $p(\Lambda_{\varphi})=1$. Because $(M_{\varphi})_{a}\neq\{0\}$ and the Cesàro operator $C$ is not bounded on $M_{\varphi}$ we can use Theorem 16 (i). 3. (iii)

$\varphi(0^{+})>0$ and $\varphi(\infty)=\infty$. In this situation $(M_{\varphi})_{a}$ is trivial in view of Theorem B, so we simply use Theorem 16 (ii). 4. (iv)

$\varphi(0^{+})>0$ and $\varphi(\infty)<\infty$. It is easy to see that $M_{\varphi}=L^{\infty}$. Indeed, inclusion $M_{\varphi}\hookrightarrow L^{\infty}$ is a consequence of Theorem B. Now, if $f\in L^{\infty}$ then $f^{**}$ also is a bounded function and

[TABLE]

since $\varphi(\infty)<\infty$. Therefore also $L^{\infty}\hookrightarrow M_{\varphi}$. From Remark 19 we get

[TABLE]

and the proof is finished.

∎

Question 26**.**

It seems that non-trivial Cesàro-Marcinkiewicz function space $CM_{\varphi}$ is never order continuous but this is only our conjecture.

4.5. The spaces $C(L^{1}\cap L^{\infty})$ and $C(L^{1}+L^{\infty})$

The spaces $L^{1}\cap L^{\infty}$ and $L^{1}+L^{\infty}$ occupy a special place in the theory of symmetric spaces because they are respectively the largest and the smallest of all symmetric function spaces, i.e. $L^{1}\cap L^{\infty}\hookrightarrow E[0,\infty)\hookrightarrow L^{1}+L^{\infty}$ and $L^{\infty}\hookrightarrow E[0,1]\hookrightarrow L^{1}$, cf. [4, Th. 6.6 and Th. 6.7, pp. 77-78], $\left\lVert f\right\rVert_{L^{1}\cap L^{\infty}}=\max\{\left\lVert f\right\rVert_{L^{1}},\left\lVert f\right\rVert_{L^{\infty}}\}$, and

[TABLE]

cf. [4, Th. 6.2 and Th. 6.4, p. 74 and 76].

The case of $L^{1}\cap L^{\infty}$ is not very interesting because if $I=[0,\infty)$ then $C(L^{1}\cap L^{\infty})=\{0\}$ and if $I=[0,1]$ then $L^{\infty}\hookrightarrow L^{1}$ so $L^{1}\cap L^{\infty}=L^{\infty}$ and $L^{1}+L^{\infty}=L^{1}$. Therefore for $I=[0,1]$ we have $C(L^{1}\cap L^{\infty})=C(L^{\infty})=Ces_{\infty}$ and $C(L^{1}+L^{\infty})=C(L^{1})=Ces_{1}$ - these cases were already discussed in the Section 4.1.

Proposition 27**.**

$(C(L^{1}+L^{\infty}))_{a}=\{f\in C(L^{1}+L^{\infty}):(Cf)^{*}(\infty)=0\}$.

Proof.

We claim that

[TABLE]

This characterization is well known (see e.g. note in Section 2 in [13] and references therein) but we will give a short proof for the sake of completeness. Inclusion $\subset$ is a direct consequence of Theorem A (ii). For the reverse inclusion $\supset$ take $f\in L^{1}+L^{\infty}$ and observe that

[TABLE]

Since $f^{*}\chi_{[0,\frac{1}{n}]}\rightarrow 0$ almost everywhere and is dominated by integrable function thus, from Lebesgue dominated convergence theorem

[TABLE]

as $n\rightarrow\infty$. Therefore, if $f\notin(L^{1}+L^{\infty})_{a}$ then according to Theorem A (ii) we get

[TABLE]

Passing to subsequence if necessary we can assume that $\left\lVert f^{*}\chi_{[n,\infty)}\right\rVert_{L^{1}+L^{\infty}}\geq\delta$ for some $\delta>0$. We have

[TABLE]

so $f^{*}(\infty)>0$ and the claim follows. Note that from (4.1)

[TABLE]

Using Remark 13 we obtain $(C(L^{1}+L^{\infty}))_{a}=C((L^{1}+L^{\infty})_{a})$ and the proof is complete. ∎