Isomorphic copies of $l^\infty$ in Cesàro–Orlicz function spaces

Abstract

We characterize Cesàro–Orlicz function spaces $Ces_\varphi$ containing isomorphic copy of $l^\infty$. We also describe the subspaces $(Ces_\varphi )_a$ of all order continuous elements of $Ces_\varphi$. Finally, we study the monotonicity structure of the spaces $Ces_\varphi$ and $(Ces_\varphi )_a$.

1 Introduction

The structure of different types of Cesàro spaces has been widely investigated during the last decades from the isomorphic as well as isometric point of view. Firstly, the Cesàro sequence spaces $ces_p$ has been studied by many authors starting from 1970. Then the attention has been moved to the Cesàro function spaces $Ces_p$. It was interesting among others that some properties are fulfilled in the sequence case and do not in the function case. Moreover, it happens that the cases $Ces_p[0,1]$ and $Ces_p[0,\infty )$ are essentially different (see the description of the Köthe dual of Cesàro spaces in [1] and [19]. Note also that isometric description of the dual space to $Ces_p$ when $1< p < \infty$ can be found in [15]). Probably the most important papers concerning the structure of the spaces $Ces_p$ are two papers by Astashkin and Maligranda (see [1, 2]). Recall that the space $Ces_p[0,1]$, for $1\le p \le \infty$, consists of those Lebesgue measurable real functions f on [0, 1] for which the Cesàro-means $C|f|(x)=\frac{1}{x} \int _0^x|f(t)|dt$ belong to $L_p[0,1]$ (similarly for the spaces $ces_p$ and $Ces_p[0,\infty )$). It is natural to investigate the space

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

for any Banach ideal space X on I, where $I=[0,1]$ or $I=[0,\infty )$. These spaces have been defined in [27] for X being a Banach ideal space on $[0,\infty )$ and, for example, in [12, 19, 20] for X being a symmetric space on I. If we take instead of X the Orlicz sequence space $l^\varphi$ (the generalization of $l^p$) then the space CX becomes the Cesàro–Orlicz sequence space denoted by $ces_\varphi$. Cleary, $ces_\varphi = ces_p$ if $\varphi (u) = u^p$. The spaces $ces_\varphi$ has been studied by many authors (e.g. [9–11, 14, 16]). In particular, the criteria for existence of an isomorphic as well as isometric copies of $l^\infty$ in the spaces $ces_\varphi$ have been established in [14, 16]. If we put $X=L^\varphi (I)$ the Orlicz function space then CX becomes the Cesàro–Orlicz function space $Ces_\varphi (I)$. As far as we know the structure of these spaces has not been investigated until now. We want to characterize these spaces $Ces_\varphi$ which contain an order isomorphic copy of $l^\infty$ (equivalently, are not order continuous). Clearly, such results are very useful in further studying isomorphic structure of these spaces. We characterize the subspaces $(Ces_\varphi )_a$ of all order continuous elements in $Ces_\varphi$. Finally, we prove criteria for strict monotonicity of the spaces $(Ces_\varphi )_a$. We also give a characterization of uniform monotonicity in Cesàro–Orlicz function spaces. We consider the largest possible class of Orlicz functions giving the maximal generality of spaces under consideration.

2 Preliminaries

The symbols $\mathbb {R}$, $\mathbb {R_+}$ and $\mathbb {N}$ denote the sets of reals, nonnegative reals and natural numbers, respectively. Let $L^0=L^0 (I)$ be the space of all classes of real-valued Lebesgue measurable functions defined on I, where $I=[0,1]$ or $I=[0,\infty )$. A Banach space $E=(E,\Vert \cdot \Vert )$ is said to be a Banach ideal space on I if E is a linear subspace of $L^0(I)$ and satisfies two conditions:

• (i) if $g\in E$, $f\in L^0$ and $|f|\le |g|$ a.e. on I then $f \in E$ and $\Vert f\Vert \le \Vert g\Vert$,

• (ii) there is an element $f \in E$ that is positive on whole I .

Sometimes we write $\left||\cdot \right||_{E}$ to be sure in which space the norm has been taken.

For two Banach ideal spaces E and F on I the symbol $E \hookrightarrow F$ means that the embedding $E\subset F$ is continuous, i.e., there exists constant a $C>0$ such that $\left||x\right||_{F} \le C\left||x\right||_{E}$ for all $x\in E$. Moreover, $E=F$ means that the spaces are the same as the sets and the norms are equivalent.

Recall that for $f\in E$ the distribution function $d_f$ is defined by

$$\begin{aligned} d_f(\lambda )=m(\{t\in I : |f(t)|>\lambda \}) \end{aligned}$$

for all $\lambda >0$, where m is the Lebesgue measure. The non-increasing rearrangement of f is denoted by $f^*$ and is defined as

$$\begin{aligned} f^*(t)=\inf \{\lambda >0 : d_f(\lambda )<t\} \end{aligned}$$

for $t \ge 0$. We say that two functions $f,g \in L^0(I)$ are equimeasurable if they have the same distribution functions $d_f \equiv d_g$. Then we write $f\sim g$. By a symmetric function space (symmetric Banach function space or rearrangement invariant Banach function space) on I we mean a Banach ideal space $(E,\left||\cdot \right||_{E})$ with an additional property that for any two functions $f\in E$, $g\in L^0(I)$ with $f\sim g$ we have $g\in E$ and $\left||f\right||_{E} = \left||g\right||_{E}$. In particular, $\left||f\right||_{E} = \left||f^*\right||_{E}$.

A point $f\in E$ is said to have an order continuous norm (or to be an order continuous element) if for each sequence $(f_n)\subset E$ satisfying $0\le f_n\le |f|$ and $f_n\rightarrow 0$ a.e. on I, one has $\left||f_n\right||\rightarrow 0$. By $E_a$ we denote the subspace of all order continuous elements of E. It is worth to notice that in case of Banach ideal spaces on I, $x\in E_a$ if and only if $\left||x\chi _{A_n}\right||\rightarrow 0$ for any decreasing sequence of Lebesgue measurable sets $A_n\subset I$ with empty intersection. A Banach ideal space is called order continuous ($E\in \text {(OC)}$ shortly) if every element of E is order continuous, i.e., $E=E_a$.

We say a Banach ideal space X is strictly monotone $(X\in \text {(SM)})$, if $\left||x\right|| < \left||y\right||$ for all $x,y\in X$ such that $0\le x\le y$ and $x\ne y$. A Banach ideal space X is uniformly monotone $(X\in \text {(UM)})$ if for each $\epsilon \in (0,1)$ there exists $\delta (\epsilon )\in (0,1)$ such that for all $0\le y\le x$, $\left||x\right||=1$ and $\left||y\right||\ge \epsilon$, we have $\left||x-y\right||\le 1-\delta (\epsilon )$. X is called lower locally uniformly monotone $(X\in \text {(LLUM)})$ if for any $x\in X$, $\left||x\right||=1$ and $\epsilon \in (0,1)$ there exists $\delta (x,\epsilon ) \in (0,1)$) such that $\left||x-y\right||\le 1-\delta (x,\epsilon )$ whenever $0\le y\le x$ and $\left||y\right||\ge \epsilon$. Note that monotonicity properties of Banach ideal spaces are useful in dominated best approximation problems, see [8] for further references.

The next well known theorem shows the relation between the order continuity of E and the existing of an isomorphic copy of $l^\infty$.

Theorem A

(G. Ya. Lozanovskiĭ, see [23]) A Banach ideal space E is order continuous if and only if E contains no isomorphic copy of $l^\infty$.

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

• (i) $\varphi$ is convex,

• (ii) $\varphi (0)=0$,

• (iii) $\varphi$ is neither identically equal to zero nor infinity on $(0,\infty )$,

• (iv) $\varphi$ is left continuous on $(0,\infty )$, i.e., $\lim _{u\rightarrow b^{-}_{\varphi }} \varphi (u)=\varphi (b_\varphi )$ if $b_\varphi <\infty$, where

  $$\begin{aligned} b_\varphi = \sup \{u>0 : \varphi (u)<\infty \}. \end{aligned}$$

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

If we denote

$$\begin{aligned} a_\varphi =\sup \{u\ge 0 : \varphi (u)=0\}, \end{aligned}$$

then $0\le a_\varphi \le b_\varphi \le \infty$, $a_\varphi <\infty$, $b_\varphi >0$, since an Orlicz function is neither identically equal to zero nor infinity on $(0,\infty )$. The function $\varphi$ is continuous and non-decreasing on $[0,b_\varphi )$ and is strictly increasing on $[a_\varphi ,b_\varphi )$. We use notations $\varphi >0$, $\varphi <\infty$ when $a_\varphi =0$, $b_\varphi <\infty$, respectively.

We say an Orlicz function $\varphi$ satisfies the condition $\Delta _2$ for large arguments $(\varphi \in \Delta _2(\infty ) \text { for short})$ if there exists $K>0$ and $u_0>0$ such that $\varphi (u_0 )<\infty$ and

$$\begin{aligned} \varphi (2u)\le K\varphi (u) \end{aligned}$$

for all $u\in [u_0,\infty )$. Similarly, we can define the condition $\Delta _2$ for small, with $\varphi (u_0)>0$ $(\varphi \in \Delta _2 (0))$ or for all arguments $(\varphi \in \Delta _2 (\mathbb {R_+}))$. These conditions play a crucial role in the theory of Orlicz spaces, see [7, 18, 25, 26]. We will write $\varphi \in \Delta _2$ in two cases: $\varphi \in \Delta _2(\infty )$ if $I=[0,1]$ and $\varphi \in \Delta _2 (\mathbb {R_+})$ if $I=[0,\infty )$.

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

$$\begin{aligned} L^\varphi = \{f\in L^0(I) : I_\varphi (f/\lambda )<\infty \,\, \text {for some}\,\, \lambda =\lambda (f)>0\}, \end{aligned}$$

where $I_\varphi (f)=\int _I \varphi (|f(t)|)dt$ is a convex modular (for the theory of Orlicz spaces and modular spaces see [25, 26]). The space $L^\varphi$ is a Banach ideal space with the Luxemburg–Nakano norm

$$\begin{aligned} \left||f\right||_{\varphi }=\inf \{\lambda > 0 : I_\varphi (f/\lambda )\le 1\}. \end{aligned}$$

It is well known that $\left||f\right||_{\varphi }\le 1$ if and only if $I_\varphi (f)\le 1$.

The continuous Cesàro operator is defined for $0<x\in I$ as

$$\begin{aligned} Cf(x)=\frac{1}{x}\int _0^x f(t)dt. \end{aligned}$$

For a Banach ideal space X on I we define an abstract Cesàro space $CX=CX(I)$ as

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

with a norm $\left||f\right||_{CX}=\left||C|f|\right||_{X}$.

The Cesàro–Orlicz function space $Ces_\varphi =Ces_\varphi (I)$ is defined as $Ces_\varphi (I)=CL^\varphi (I)$. Consequently, the norm in the space $Ces_\varphi$ is given by the formula

$$\begin{aligned} \left||f\right||_{Ces(\varphi )}=\inf \{\lambda > 0 : \rho _\varphi (f/\lambda )\le 1\} \end{aligned}$$

where $\rho _\varphi (f)=I_\varphi (C|f|)$ is a convex modular.

The dilation operator $D_s$, $s>0$ defined on $L^0 (I)$ by

$$\begin{aligned} D_s x(t)=x(t/s) \chi _I (t/s)=x(t/s) \chi _{[0,\min \{1,s\} )} (t), \end{aligned}$$

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

$$\begin{aligned} p(E)=\lim \limits _{s\rightarrow 0^{+}} \frac{\mathrm{{ln}} \left||D_s\right||_{E\rightarrow E}}{\mathrm{ln}s}, \end{aligned}$$

$$\begin{aligned} q(E)=\lim \limits _{s\rightarrow \infty } \frac{\mathrm{ln} \left||D_s\right||_{E\rightarrow E}}{\mathrm{{ln}}s}. \end{aligned}$$

In particular, they satisfy the inequalities

$$\begin{aligned} 1\le p(E)\le q(E) \le \infty . \end{aligned}$$

In the case when E is the Orlicz space $L^\varphi$, these indices correspond to the so-called Orlicz-Matuszewska indices of Orlicz functions generating the Orlicz spaces, i.e., $\alpha _\varphi =p(L^\varphi )$ and $\beta _\varphi =q(L^\varphi )$, where $\alpha _\varphi$ and $\beta _\varphi$ are the lower and upper Orlicz-Matuszewska indices of the Orlicz space $L^\varphi$ (see [22, Proposition 2.b.5 and Remark 2 on page 140]). For more details see [5, 6, 24–26].

In this paper we accept the convention that $\sum _{n=m}^{k} x_n=0$ if $k<m$.

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

Theorem B

[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 $p(E)>1$.

The immediate consequence of Theorem B and the above discussion is a next corollary.

Corollary 1

The embedding $L^\varphi \hookrightarrow Ces_\varphi$ holds if and only if $\alpha _\varphi >1$.

Remark

Let X,Y be Banach ideal spaces on I. If $X\hookrightarrow Y$ then $CX\hookrightarrow CY$. Indeed, suppose $X\hookrightarrow Y$. Then for all $x\in CX$

$$\begin{aligned} \left||x\right||_{CY}=\left||C|x|\right||_{Y}\le A\left||C|x|\right||_{X}=A\left||x\right||_{CX}, \end{aligned}$$

for some constant $A>0$. This means that $CX\hookrightarrow CY$.

Let $L^\varphi$ and $L^\psi$ be the Orlicz spaces with $\varphi ,\psi <\infty$. The criteria for the embeddings $L^\varphi \hookrightarrow L^\psi$ can be found in [25, Theorem 3.4 p. 18]. Consequently, we conclude that:

• (i) if there exists $k>0$ with $\psi (u)\le \varphi (ku)$ for all $u\in [0,\infty )$ then $Ces_\varphi [0,\infty )\hookrightarrow Ces_\psi [0,\infty )$,

• (ii) if there exists $k,u_0>0$ such that $\psi (u)\le \varphi (ku)$ for all $u\in [u_0,\infty )$ then $Ces_\varphi [0,1]\hookrightarrow Ces_\psi [0,1]$,

• (iii) if there exists $k,u_0>0$ such that $\psi (u)\le \varphi (ku)$ for all $u\in [0,u_0 ]$ then $ces_\varphi \hookrightarrow ces_\psi$.

Note that case (iii) is exactly Proposition 1 in [16] but now it is an immediate consequence of our above remark.

3 Results

Fact

Let X be a Banach ideal space on I. By the definition, the order continuity of X gives the same for CX (for proof see [21, Lemma 1]). The case of strict monotonicity and uniform monotonicity is similar. The converse is not true in general (see [12, Proposition 2.1] and [21, Example 1]).

Proof

We proof only the implication for uniform monotonicity. We apply Theorem 6 (ii) in [13]. Suppose that $X\in \text {(UM)}$. Take $\epsilon >0$, $x,y\in CX$, $x,y\ge 0$, $\left||x\right||_{CX}=1$ and $\left||y\right||_{CX}\ge \epsilon$. We have

$$\begin{aligned} \left||x+y\right||_{CX}=\left||C|x+y|\right||_{X}=\left||C|x|+C|y|\right||_{X}. \end{aligned}$$

Since X is uniformly monotone we have that there is a $\sigma (\epsilon )>0$ such that

$$\begin{aligned} \left||C|x|+C|y|\right||_{X}\ge 1+\sigma (\epsilon ). \end{aligned}$$

But this means that $CX\in \text {(UM)}$. $\square$

Proposition 2

Suppose X is a symmetric space on $I=[0,1]$ and $C : X\rightarrow X$. Then $X\in \mathrm {(OC)}$ if and only if $CX\in \mathrm {(OC)}$.

Proof

Necessity It follows from Fact above.

Sufficiency We thank Professor Karol Leśnik for giving the proof. Suppose that $X\notin \text {(OC)}$, i.e. there exists $0< f\in X\backslash X_a$. Because X is symmetric we can assume that $f=f^*$ (see [8, Lemma 2.6]). Then $Cf\in X$ and $Cf\ge f$. Applying the symmetry of X and passing to subsequence, if necessary, we can assume that there is a $\delta >0$ such that

$$\begin{aligned} \left||f\chi _{[0,1/n)}\right||_{X}\ge \delta , \end{aligned}$$

for all $n\in \mathbb {N}$. Since

$$\begin{aligned} C(f\chi _{[0,1/n)})(t)\ge f\chi _{[0,1/n)}(t), \end{aligned}$$

for all $t\in I$ and $n\in \mathbb {N}$, we have

$$\begin{aligned} \left||f\chi _{[0,1/n)}\right||_{CX}=\left||C(f\chi _{[0,1/n)})\right||_{X}\ge \left||f\chi _{[0,1/n)}\right||_{X}\ge \delta >0, \end{aligned}$$

which means that $X\notin \text {(OC)}$. $\square$

Proposition 3

Let $\varphi$ be an Orlicz function. The following conditions are equivalent:

• (i) the space $Ces_\varphi [0,\infty )\ne \{0\}$,

• (ii) there exists $\lambda _0>0$ and $x_0\in [0,\infty )$ such that $\int _{x_0}^\infty \varphi (\lambda _0/t)dt<\infty$,

• (iii) for each $\lambda >0$ there exist $y_0\in [0,\infty )$ with $\int _{y_0}^\infty \varphi (\lambda /t)dt<\infty$.

Proof

The equivalence of conditions (i) and (ii) follows from [19, Theorem 1 (a)]. Clearly, $\text {(iii)}\Rightarrow \text {(ii)}$.

$\text {(ii)}\Rightarrow \text {(iii)}$. Let $\lambda > 0$. We have to consider two cases. If $\lambda \le \lambda _0$, there is nothing to proof. Now suppose $\lambda > \lambda _0$. Set $y_0 = \lambda x_0/\lambda _0$. Then

$$\begin{aligned}&\int _{y_0}^\infty \varphi (\lambda /t)dt=\int _{y_0}^\infty \varphi \left( \frac{\lambda _0}{\lambda _0t/\lambda }\right) dt=\frac{\lambda }{\lambda _0}\int _{\lambda _0 y_0/\lambda }^{\infty }\varphi \left( \frac{\lambda _0}{u}\right) du \\&\quad =\frac{\lambda }{\lambda _0}\int _{x_0}^{\infty }\varphi \left( \frac{\lambda _0}{u}\right) du < \infty . \end{aligned}$$

$\square$

Remark 3.1 Let $\varphi$ be an Orlicz function satisfying condition (S):

$$\begin{aligned} \liminf \limits _{t\rightarrow 0} t\varphi ^{\prime }(t)/\varphi (t)>1, \end{aligned}$$

where the symbol $\varphi ^{\prime }$ means the right derivative of the function $\varphi$. Then $Ces_\varphi [0,\infty )\ne \{0\}$.

Proof

Firstly, condition (S) implies (ii) in Proposition 3. Indeed, we can use the same argument as in the proof of implication $\text {(a)}\Rightarrow \text {(c)}$ in [9, Theorem 2.2]. But condition (ii) is equivalent to $Ces_\varphi [0,\infty )\ne \{0\}$. $\square$

Remark 4

Let $\varphi$ be an Orlicz function. Then $Ces_\varphi [0,1]\ne \{0\}$.

Proof

It follows from [19, Theorem 1 (b)] that $Ces_\varphi [0,1]\ne {0}$ if and only if there exists $0<a<1$ such that $\chi _{[a,1]} \in L^\varphi [0,1]$. But $L^\varphi [0,1]$ is a symmetric space and

$$\begin{aligned} L^\infty [0,1]=L^1 [0,1]\cap L^\infty [0,1]\hookrightarrow L^\varphi [0,1]\hookrightarrow L^1 [0,1]+L^\infty [0,1]=L^1 [0,1] \end{aligned}$$

(see [4]). Therefore $\chi _{[a,1]} \in L^\varphi [0,1]$ for all $a\in (0,1)$. $\square$

The following theorem gives a similar information as in the sequence space $ces_\varphi$ one can get from [9, Theorem 2.3]. Before we formulate the theorem define a set

$$\begin{aligned} C_\varphi =C_\varphi (I)=\{x\in Ces_\varphi (I) : \rho _\varphi (kx)<\infty \ \text {for all}\ k>0\}. \end{aligned}$$

Note that $C_\varphi = \{0\}$ whenever $b_\varphi <\infty$.

Theorem 5

Suppose $\varphi <\infty$. Then the following conditions are true:

• (i) $C_\varphi$ is the subspace of all order continuous elements of $Ces_\varphi$,

• (ii) $C_\varphi$ is a closed separable subspace of $Ces_\varphi$.

Proof

(i). We will show that $C_\varphi \hookrightarrow (Ces_\varphi )_a$. Take any $x\in C_\varphi$ and a sequence $(A_n )$ with $A_n\searrow \emptyset$. It is enough to show, that $\left||x\chi _{A_n}\right||_{Ces(\varphi )}\rightarrow 0$ or equivalently $\rho _\varphi (kx \chi _{A_n} )\rightarrow 0$ for any $k>0$. Set $k>0$. Clearly,

$$\begin{aligned} \lim \limits _{n\rightarrow \infty } m((0,t]\cap A_n )=0, \end{aligned}$$

for each $t>0$. Then for almost each $t>0$

$$\begin{aligned} 0\le C(|x\chi _{A_n} ) |)(t)=\frac{1}{t} \int _0^t|x(s) \chi _{A_n}(s) |ds \rightarrow 0. \end{aligned}$$

Moreover,

$$\begin{aligned} \rho _\varphi (kx\chi _{A_n})\le \rho _\varphi (kx)<\infty , \end{aligned}$$

for each k. Thus, by the Lebesgue dominated convergence theorem, we have $\rho _\varphi (kx\chi _{A_n})\rightarrow 0$, so $x\in (Ces_\varphi (I))_a$.

Next we prove the reverse inclusion. Define a set

$$\begin{aligned} B_\varphi =B_\varphi (I)=\text {cl}\{f\in Ces_\varphi (I) : f \ \text {is simple function and} \ m(\text {supp}(f))<\infty \}. \end{aligned}$$

We claim that:

Consider the inclusion $B_\varphi \subset C_\varphi$. Of course, if $B_\varphi =\emptyset$ then $B_\varphi \subset C_\varphi$ so we can assume that $B_\varphi \ne \emptyset$. We divide the proof in three cases.

• (1) Suppose $x=\chi _{[a,b]}$ , where $[a,b]\subset I$, $0\le a<b<\infty$. Since $\varphi < \infty$ and $Ces_\varphi \ne \{0\}$, by Proposition 3 we conclude that $\rho _\varphi (kx)<\infty$ for all $k>0$.

• (2) Let $x=\sum _{k=1}^{n} c_k \chi _{A_k}$, $A_k=[a_k,b_k ]\subset I$ for $n,k\in \mathbb {N}$ and $0\le a_k<b_k<\infty$. Using the convexity of modular $\rho _\varphi$ and argument from case (1), we obtain $\rho _\varphi (\lambda x)<\infty$ for all $\lambda >0$.

• (3) Assume that there exists a sequence $(x_n )$ such that $x_n\rightarrow x$ in $Ces_\varphi$ and $x_n$ are of the form (2). Let $k>0$. For sufficiency large $n\in \mathbb {N}$ we have

  $$\begin{aligned} \rho _\varphi (kx)= & {} \rho _\varphi (k(x-x_n+x_n))=\rho _\varphi \left( \frac{1}{2}2k(x-x_n )+\frac{1}{2}2kx_n\right) \\\le & {} \frac{1}{2}\rho _\varphi (2k(x-x_n ))+\frac{1}{2}\rho _\varphi (2kx_n)\le 1+\rho _\varphi (2kx_n)<\infty . \end{aligned}$$

  Therefore $B_\varphi \subset C_\varphi$.

Now, we will prove the reverse inclusion $C_\varphi \subset B_\varphi$. Let $x\in C_\varphi$, $k>0$ and $\epsilon >0$. Since $\int _I \varphi (C(k|x|))dt<\infty$ there exists $\beta =\beta (k)\in \mathbb {R_+}$ with

$$\begin{aligned} \int _{\beta }^\infty \varphi (C(2k|x(t)|))dt<\epsilon /2. \end{aligned}$$

Put $z_n=\varphi (C(2k|x|\chi _{[0,1/n)}))$ and $z=\varphi (C(2k|x|))$. Then $z\in L^1$, $0\le z_n \le z$ and $z_n \rightarrow 0$ a.e. on I. Note that $L^1\in \text {(OC)}$ whence $\left||z_n\right||_{L^1}\rightarrow 0$. Consequently, there exists $\alpha =\alpha (k)\in \mathbb {R_+}$ such that

$$\begin{aligned} \int _{0}^\infty \varphi (C(2k|x|\chi _{[0,\alpha )}))dt<\epsilon /2. \end{aligned}$$

Define an element $x_n=x\chi _{[1/n,n]}$ for $n\in \mathbb {N}$. For sufficiently large $n\in \mathbb {N}$ satisfying $1/n<\alpha$ and $n>\beta$ we have

$$\begin{aligned}&\rho _\varphi (k(x_n-x))=\rho _\varphi (kx\chi _{[0,1/n)\cup (n,\infty )})=\int _0^\infty \varphi (C(k|x(t)|\chi _{[0,1/n)\cup (n,\infty )}(t)))dt\\&\quad =\int _{0}^{\infty } \varphi (C(k|x(t)|\chi _{[0,1/n)}(t))+C(k|x(t)|\chi _{(n,\infty )}(t)))dt\\&\quad \le \frac{1}{2}\int _{0}^{\infty } \varphi (C(2k|x(t)|\chi _{[0,1/n)}(t)))dt+\frac{1}{2}\int _{0}^{\infty } \varphi (C(2k|x(t)|\chi _{(n,\infty )}(t)))dt\\&\quad \le \frac{1}{2}\int _{0}^{\infty } \varphi (C(2k|x(t)|\chi _{[0,\alpha )}(t)))dt+\frac{1}{2}\int _{0}^{\infty } \varphi (C(2k|x(t)|\chi _{(\beta ,\infty )}(t)))dt<\epsilon \end{aligned}$$

i.e. $\left||x_n-x\right||_{Ces(\varphi )}\rightarrow 0$. Now, it is enough to prove that for each interval [a, b], $0\le a<b<\infty$, there is a sequence of functions $y_n$ of the form (2) with

$$\begin{aligned} \left||x\chi _{[a,b]}-y_n\right||_{Ces(\varphi )}\rightarrow 0, \end{aligned}$$

as $n\rightarrow 0$. Without loss of generality we may assume that $x\chi _{[a,b]}\ge 0$. Since $x\in L^0$ we can find a sequence $(y_n )$ of simple functions such that $y_n\nearrow x \chi _{[a,b]}$. Thus

$$\begin{aligned} 0\le x\chi _{[a,b]}-y_n\le x\chi _{[a,b]}, \end{aligned}$$

and $x\chi _{[a,b]}-y_n\rightarrow 0$ a.e. on I. But $x\in C_\varphi \subset (Ces_\varphi (I))_a$, whence

$$\begin{aligned} \left||x\chi _{[a,b]}-y_n\right||_{Ces(\varphi )}\rightarrow 0. \end{aligned}$$

This proves the claim (A).

Take $x\in (Ces_\varphi (I))_a$. Define an element $x_n=x\chi _{[1/n,n]}$. Then $\left||x_n-x\right||_{Ces(\varphi )}\rightarrow 0$ by the definition. Moreover, $x_n\in B_\varphi$ for all $n\in \mathbb {N}$ by the above proof. Since $B_\varphi$ is closed, so $x\in B_\varphi =C_\varphi$.

(ii) Clearly, since $\rho _\varphi$ is a convex modular, $C_\varphi$ is a linear subspace of $Ces_\varphi$. $C_\varphi$ is closed, since $B_\varphi =C_\varphi$ and $B_\varphi$ is closed by the definition (note that $X_a$ is closed for each Banach ideal space X, see [4, proof of Theorem 3.8, p. 16]). Separability of the space $C_\varphi$ follows from (i) and the fact, that Lebesgue measure is separable, see [4, Theorem 5.5, p. 27]. $\square$

Remark 6

Theorem 5 (i) can be shortened. First note that $(Ces_\varphi )_a \subset B_\varphi$ (see [4, Theorem 3.11, p. 18], the set $B_\varphi$ is denoted by $(Ces_\varphi )_b$ in [4], cf. [4, Definition 3.9, p. 17]). Moreover, $B_\varphi \subset C_\varphi \subset (Ces_\varphi )_a$ as we showed above. Thus $(Ces_\varphi )_a = C_\varphi$.

Theorem 7

Let $\varphi$ be an Orlicz function and suppose $C : L^\varphi \rightarrow L^\varphi$. Then the following conditions are equivalent:

• (i) $\varphi \in \Delta _2$,

• (ii) $Ces_\varphi \in \text {(OC)}$,

• (iii) $Ces_\varphi$ contains no isomorphic copy of $l^\infty$.

Proof

Equivalence $\text {(ii)}\Leftrightarrow \text {(iii)}$ follows from Theorem A. Now we will proof the equivalence of conditions (i) and (ii).

$\text {(i)}\Rightarrow \text {(ii)}$. If $\varphi \in \Delta _2$, then $L^\varphi$ is order continuous, see e.g. [25, pp. 21–22]. Therefore also $Ces_\varphi$ is order continuous by the Fact.

$\text {(ii)}\Rightarrow \text {(i)}$. We have to consider two cases.

I. Suppose $I=[0,\infty )$.

• (1) Let $b_\varphi <\infty$ and $\varphi (b_\varphi )=\infty$. Let $(u_n)\subset \mathbb {R_+}$ be the sequence with $u_n\nearrow b_\varphi ^{-}$. Of course, $\varphi (u_n)\rightarrow \infty$. Passing to a subsequence if necessary, we can assume that $\varphi (u_n)\ge 1$ for all $n\in \mathbb {N}$. Let $a_n=1/2^n\varphi (u_n)$ for $n\in \mathbb {N}$ and denote $a=\sum _{n=1}^\infty a_n$. Define a sequence of pairwise disjoint open intervals $(A_n )_{n\in \mathbb {N}}\subset [0,1]$ as follows

  $$\begin{aligned} A_n=\left( a-\sum _{k=1}^n \frac{1}{2^k\varphi (u_k)}, a-\sum _{k=1}^{n-1} \frac{1}{2^k\varphi (u_k)}\right) , \end{aligned}$$

  for all $n\in \mathbb {N}$. Then $m(A_n )=a_n$ for all $n\in \mathbb {N}$. Let

  $$\begin{aligned} x=\sum _{n=1}^{\infty } u_n\chi _{A_n}. \end{aligned}$$

  We claim that $x\in Ces_\varphi$. In fact,

  $$\begin{aligned} I_\varphi (x)= & {} \int _0^\infty \varphi (x) dm=\int _0^\infty \sum _{n=1}^\infty \varphi (u_n)\chi _{A_n}dm\\= & {} \sum _{n=1}^\infty \int _{A_n} \varphi (u_n)dm=\sum _{n=1}^\infty \varphi (u_n)m(A_n)=\sum _{n=1}^\infty \frac{1}{2^n}=1. \end{aligned}$$

  The assumption $C : L^\varphi \rightarrow L^\varphi$ implies that $L^\varphi \hookrightarrow Ces_\varphi$, so $x\in Ces_\varphi$. Note that there exists $N\in \mathbb {N}$ such that for all $k\ge N$, $2u_k>b_\varphi$. Consequently, since x is non-increasing

  $$\begin{aligned} \rho _\varphi (2x)=I_\varphi (C(2x))=I_\varphi (2Cx)\ge I_\varphi (2x)=\sum _{n=1}^\infty \varphi (2u_n)m(A_n)=\infty , \end{aligned}$$

  and, by Theorem 5 (i), we conclude that $Ces_\varphi \notin \text {(OC)}$.

• (2) Assume that $b_\varphi <\infty$ and $\varphi (b_\varphi )<\infty$. Define an element $x=b_\varphi \chi _{[0,1]}$. Since $I_\varphi (x)<\infty$ so $x\in L^\varphi$ and $x\in Ces_\varphi$ because $C : L^\varphi \rightarrow L^\varphi$. Moreover,

  $$\begin{aligned} \rho _\varphi (2x)=I_\varphi (C(2x))\ge I_\varphi (2x)=\varphi (2b_\varphi )=\infty , \end{aligned}$$

  thus, $Ces_\varphi [0,\infty )\notin \text {(OC)}$, by Theorem 5 (i).

• (3) Let $a_\varphi >0$. Put $x=a_\varphi \chi _{[0,\infty )}$. Then $x\in Ces_\varphi [0,\infty )$ and

  $$\begin{aligned} \rho _\varphi (2x)=I_\varphi (C(2x))=I_\varphi (2x)=\varphi (2a_\varphi )m([0,\infty ))=\infty . \end{aligned}$$

  Therefore, $Ces_\varphi [0,\infty )\notin \text {(OC)}$ by Theorem 5 again.

• (4) Now we assume that $\varphi >0$, $\varphi <\infty$ and $\varphi \notin \Delta _2 (\mathbb {R_+})$. This means, that $\varphi \notin \Delta _2(0)$ or $\varphi \notin \Delta _2 (\infty )$. If $\varphi \notin \Delta _2 (\infty )$ then

  $$\begin{aligned} \varphi (2u_n )\ge 2^n \varphi (u_n ), \end{aligned}$$

  for some sequence $(u_n )\nearrow \infty$. Therefore in this case we can use the arguments from I (1) above. Now assume that $\varphi \notin \Delta _2(0)$. That means, that there exist a decreasing sequence $(u_n )_{n\in \mathbb {N}}\subset \mathbb {R_+}$, $u_n\searrow 0$ and $\varphi (2u_n )\ge 2^n \varphi (u_n )$ for all $n\in \mathbb {N}$. Define an element

  $$\begin{aligned} x=\sum _{n=1}^\infty u_n \chi _{B_n}, \end{aligned}$$

  where $B_n=(\sum _{k=1}^{n-1} \frac{1}{2^k \varphi (u_k )}, \sum _{k=1}^{n} \frac{1}{2^k \varphi (u_k )})$. Then

  $$\begin{aligned} I_\varphi (x)= & {} \int _{0}^\infty \varphi (x)dm=\int _{0}^\infty \sum _{n=1}^\infty \varphi (u_n)\chi _{B_n} dm\\= & {} \sum _{n=1}^\infty \int _{B_n} \varphi (u_n)dm = \sum _{n=1}^\infty \varphi (u_n)m(B_n) = \sum _{n=1}^\infty \frac{1}{2^n} = 1, \end{aligned}$$

  so $x\in Ces_\varphi$ by the assumption $C : L^\varphi \rightarrow L^\varphi$. Furthermore, since x is non-increasing,

  $$\begin{aligned} \rho _\varphi (2x)= & {} I_\varphi (C(2x))\ge I_\varphi (2x)\\= & {} \sum _{n=1}^\infty \varphi (2u_n)m(B_n) \ge \sum _{n=1}^\infty 2^n \varphi (u_n)m(B_n) = \sum _{n=1}^\infty 1 = \infty , \end{aligned}$$

  which means that the space $Ces_\varphi [0,\infty )$ is not order continuous by Theorem 5 above.

II. Suppose $I=[0,1]$.

• (1) Let $b_\varphi <\infty$ and $\varphi (b_\varphi )=\infty$. In this case we can use construction from case I (1) to show that $Ces_\varphi \notin \text {(OC)}$.

• (2) Assume that $b_\varphi <\infty$ and $\varphi (b_\varphi )<\infty$. Now we follow as in the proof of case I (2) but in this case we don’t need the assumption that $C : L^\varphi \rightarrow L^\varphi$. In fact, put $x=b_\varphi \chi _{[0,1]}$ . Then

  $$\begin{aligned} \rho _\varphi (x)=I_\varphi (Cx)=I_\varphi (x)=\int _{0}^{1} \varphi (b_\varphi )dm=\varphi (b_\varphi )<\infty , \end{aligned}$$

  so $x\in Ces_\varphi$. Furthermore,

  $$\begin{aligned} \rho _\varphi (2x)=I_\varphi (C(2x))=I_\varphi (2C(x))=I_\varphi (2x)=\varphi (2b_\varphi )=\infty , \end{aligned}$$

  which means that $Ces_\varphi \notin \text {(OC)}$.

• (3) Now suppose $\varphi <\infty$ and $\varphi \notin \Delta _2 (\infty )$. It is well known that $L^\varphi [0,1]\in \text {(OC)}$ if and only if $\varphi \in \Delta _2 (\infty )$ (see, e.g. [25, pp. 21–22]). Therefore, from Proposition 2, $Ces_\varphi [0,1]\notin \text {(OC)}$. Additionally, the same direct proof of case I (4) above works also for $I=[0,1]$.$\square$

An immediate consequence of Fact, Theorems 5 and 7 is the next corollary.

Corollary 8

Let $\varphi$ be an Orlicz function.

• (i) If $\varphi \in \Delta _2$, then $Ces_\varphi =(Ces_\varphi )_a=C_\varphi$.

• (ii) If $C : L^\varphi \rightarrow L^\varphi$ then $C_\varphi =Ces_\varphi$ if and only if $\varphi \in \Delta _2$.

Lemma 9

If $x\in C_\varphi$, then $\left||x\right||_{Ces(\varphi )}=1$ if and only if $\rho _\varphi (x)=1$.

Proof

See the proof of Lemma 2.1 in [9]. Note that the implication: $\rho (x) = 1$, then $\left||x\right||_{\rho } = 1$ for $x \in X_\rho$, holds in any modular space $X_\rho$ generated by a convex modular $\rho$, see [26]. $\square$

Theorem 10

Let $\varphi$ be an Orlicz function satisfying $\varphi <\infty$.

• (i) If $I=[0,\infty )$, then the space $C_\varphi$ is strictly monotone if and only if $\varphi >0$.

• (ii) If $I=[0,1]$ and $\lim _{u\rightarrow \infty } \varphi (u)/u= \infty$, then the space $C_\varphi$ is strictly monotone if and only if $\varphi >0$.

Proof

Necessity Assume that $a_\varphi >0$. We have to consider two cases:

• (i) If $I=[0,\infty )$ then take $y_t=t\chi _{[0,1)}$ for $t\in \mathbb {R_+}$. We have

  $$\begin{aligned} (Cy_t)(u)=t\chi _{[0,1)}(u)+\frac{t}{u} \chi _{[1,\infty )}(u). \end{aligned}$$

  We define the function

  $$\begin{aligned} f(t)=\int _{0}^{\infty } \varphi \left( t\chi _{[0,1)}(u)+\frac{t}{u} \chi _{[1,\infty )}(u)\right) du=\rho _\varphi (y_t), \end{aligned}$$

  for $t>0$. Since $t/x \rightarrow 0$ as $x\rightarrow \infty$ and $a_\varphi >0$, so there exists $x_0=x_0(t)\in \mathbb {R_+}$ such that $t/x_0 \le a_\varphi$. Consequently, $f(t)\le \varphi (t) x_0$, which means that f(t) takes finite values for $t>0$. Moreover, f is convex function, so f is continuous function on $\mathbb {R_+}$. This means that $f[\mathbb {R_+}]=\mathbb {R_+}$. In that case, from the Darboux property we find $\lambda >0$ satisfying $f(\lambda )=1$. Take $x_1>1$ with $\lambda /x_1 \le a_\varphi /2$. Let $z=a_\varphi \chi _{[x_1,x_1+1)}/2$. Then for any $x\ge x_1$ we have

  $$\begin{aligned} C(y_\lambda +z)(x)=Cy_\lambda (x)+Cz(x)\le a_\varphi , \end{aligned}$$

  whence $\rho _\varphi (y_\lambda +z)=1$ and, from Lemma 9, we get $\left||y_\lambda +z\right||_{Ces(\varphi )}=1$. Summing up, we built elements $u=y_\lambda$ i $v=y_\lambda +z$ such that $u\ne v$, $u\le v$ i $\left||u\right||_{Ces(\varphi )}=\left||v\right||_{Ces(\varphi }=1$. Moreover, $u,v\in C_\varphi$. This follows easily from the equality $C_\varphi =B_\varphi$ (see Proof of Theorem 5). This means, that the space $C_\varphi$ isn’t strictly monotone.

• (ii) Suppose $I=[0,1]$. For $a,b\in \mathbb {R}$ set $x=b\chi _{[0,a)}$. Then

  $$\begin{aligned} (Cx)(t)=b\chi _{[0,a)}(t)+\frac{ab}{t}\chi _{[a,1)}(t). \end{aligned}$$

  Since $\lim _{u\rightarrow \infty } \varphi (u)/u = \infty$, there is a number $b_0\in \mathbb {R}$ with $b_0>a_\varphi$ and $a_\varphi \varphi (b_0)/b_0 >1$. We define the function

  $$\begin{aligned} f(a)=a\varphi (b_0)+\int _{a}^{\frac{ab_0}{a_\varphi }}\varphi \left( \frac{ab_0}{t}\right) dt, \end{aligned}$$

  for $a\in [0,a_\varphi /b_0 ]$. Clearly, $f(0)=0$ and $f(a_\varphi /b_0 )>1$. Moreover, we claim that f is continuous on $[0,a_\varphi /b_0]$. Take $a_n\rightarrow a$ with $a\in (0,a_\varphi /b_0]$ and put

  $$\begin{aligned} A= & {} \left[ a,\frac{ab_0}{a_\varphi }\right] , \\ A_n= & {} \left[ a_n,\frac{ab_0}{a_\varphi }\right] , \end{aligned}$$

  for $n\in \mathbb {N}$. Then

  $$\begin{aligned} 0\le & {} |f(a_n)-f(a)|\le |a_n-a|\varphi (b_0) +\left| \int _{a}^{\frac{ab_0}{a_\varphi }} \varphi \left( \frac{ab_0}{t}\right) dt-\int _{a_n}^{\frac{a_nb_0}{a_\varphi }} \varphi \left( \frac{a_nb_0}{t}\right) dt\right| \\= & {} |a_n-a|\varphi (b_0)+\bigg | \int _{A\backslash A_n} \varphi \left( \frac{ab_0}{t}\right) dt\\&+\int _{A\cap A_n} \varphi \left( \frac{ab_0}{t}\right) -\varphi \left( \frac{a_nb_0}{t}\right) dt -\int _{A_n\backslash A} \varphi \left( \frac{a_nb_0}{t}\right) dt\bigg |\\\le & {} |a_n-a|\varphi (b_0)+\bigg |\int _{A\backslash A_n} \varphi \left( \frac{ab_0}{t}\right) dt\bigg | +\bigg |\int _{A\cap A_n} \varphi \left( \frac{ab_0}{t}\right) -\varphi \left( \frac{a_nb_0}{t}\right) dt\bigg |\\&+\bigg |\int _{A_n\backslash A} \varphi \left( \frac{a_nb_0}{t}\right) dt\bigg |. \end{aligned}$$

Now we have

$$\begin{aligned} 0\le \left| \int _{A\backslash A_n} \varphi \left( \frac{ab_0}{t}\right) dt\right| \le m(A\backslash A_n)\sup \limits _{t\in A}\varphi \left( \frac{ab_0}{t}\right) =m(A\backslash A_n)\varphi (b_0)\rightarrow 0, \end{aligned}$$

as $n\rightarrow \infty$. Similarly, we can show that $|\int _{A_n\backslash A} \varphi (\frac{a_nb_0}{t})dt |\rightarrow 0$ as $n\rightarrow \infty$. Moreover, for n large enough $a/2<a_n<2a$. Therefore, for $t\in A$

$$\begin{aligned} \left| \varphi \left( \frac{ab_0}{t}\right) -\varphi \left( \frac{a_nb_0}{t}\right) \right| \le \text {max}_{t_1,t_2\in [a_\varphi ,2b_0]} |\varphi (t_1)-\varphi (t_2)| =\eta (a,b_0)=\eta <\infty , \end{aligned}$$

because $\varphi$ is continuous on the compact set $[a_\varphi ,2b_0]$. Since $|\varphi (\frac{ab_0}{t})-\varphi (\frac{a_nb_0}{t})|\rightarrow 0$ pointwise on A, $\eta \chi _A$ is integrable majorant and $L^1 [0,1]$ is order continuous, so from Dominated Convergence Theorem we get

$$\begin{aligned} 0\le \left| \int _{A\cap A_n} \varphi \left( \frac{ab_0}{t}\right) -\varphi \left( \frac{a_nb_0}{t}\right) dt\right| \le \left| \int _{A} \varphi \left( \frac{ab_0}{t}\right) -\varphi \left( \frac{a_nb_0}{t}\right) dt\right| \rightarrow 0, \end{aligned}$$

as $n\rightarrow \infty$. This means that $|f(a_n)-f(a)|\rightarrow 0$, so f is continuous. This proves the claim.

By the Darboux property of f, there exists $a_1\in (0,a_\varphi /b_0 )$ such that $f(a_1 )=1$. Consider two elements

$$\begin{aligned} x_1=b_0\chi _{[0,a_1)}, \end{aligned}$$

and

$$\begin{aligned} x_2=b_0\chi _{[0,a_1)}+\left( a_\varphi -\frac{a_1b_0}{\delta }\right) \chi _{[\frac{\delta +1}{2},1)}, \end{aligned}$$

where $\delta >0$ is such that $a_1b_0/a_\varphi<\delta <1$. Obviously, $x_1\le x_2$ and $x_1\ne x_2$ and $x_1,x_2\in C_\varphi$. Furthermore, $\rho _\varphi (x_1)=\rho _\varphi (x_2)=1$. Indeed, we have the following inequalities

$$\begin{aligned} a_1<\frac{b_0}{a_\varphi }a_1<\delta<\frac{\delta +1}{2}<1, \end{aligned}$$

so for $t\in [(\delta +1)/2,1)$

$$\begin{aligned} (Cx_2)(t)=\frac{a_1b_0}{t}+\left( a_\varphi -\frac{a_1b_0}{\delta }\right) \frac{1}{t}\int _{\frac{\delta +1}{2}}^{t}ds <\frac{a_1b_0}{t}+a_\varphi -\frac{a_1b_0}{\delta }\le a_\varphi . \end{aligned}$$

Thus, by the Lemma 9, $\left||x_1\right||_{Ces(\varphi )}=\left||x_2\right||_{Ces(\varphi )}=1$, whence the space $C_\varphi$ is not strictly monotone.

Sufficiency Assume, that $a_\varphi =0$, $0\le x\le y$, $x\ne y$ and $x,y\in C_\varphi$. Without loss of generality, we can assume, that $\left||x\right||_{Ces(\varphi )}=1$. From Lemma 9, $\rho _\varphi (x)=1$. We need only to show that $\rho _\varphi (y)>1$. Since $\varphi$ is superadditive, so for any $u,v\in (C_\varphi )_{+}$ we have

$$\begin{aligned} \rho _\varphi (u+v)\ge \rho _\varphi (u)+\rho _\varphi (v). \end{aligned}$$

Since $y-x\ge 0$, $y-x\ne 0$ and $\varphi >0$, we have $\rho _\varphi (y-x)>0$, and consequently

$$\begin{aligned} \rho _\varphi (y)=\rho _\varphi (x+(y-x))\ge \rho _\varphi (x)+\rho _\varphi (y-x)=1+\rho _\varphi (y-x)>1. \end{aligned}$$

$\square$

Theorem 11

Let $\varphi$ be an Orlicz function with $\varphi <\infty$ and $C : L^\varphi \rightarrow L^\varphi$. Suppose additionally $\lim _{u\rightarrow \infty } \varphi (u)/u= \infty$ if $I=[0,1]$. The following conditions are equivalent:

• (i) $Ces_\varphi \in \text {(UM)}$,

• (ii) $Ces_\varphi \in \text {(LLUM)}$,

• (iii) $\varphi >0$ and $\varphi \in \Delta _2$.

Proof

$\text {(i)}\Rightarrow \text {(ii)}$. This implication follows by the definition.

$\text {(ii)}\Rightarrow \text {(iii)}$. If $a_\varphi >0$ then $C_\varphi \notin \text {(SM)}$ by Theorem 10. Moreover, see [3, Proposition 2.1], the following implication is true: if $X\in \text {(LLUM)}$ then $X\in \text {(OC)}$. Therefore, if $\varphi \notin \Delta _2$ then $Ces_\varphi \notin \text {(OC)}$ by Theorem 7, whence $Ces_\varphi \notin \text {(LLUM)}$.

$\text {(iii)}\Rightarrow \text {(i)}$. If $\varphi >0$ and $\varphi \in \Delta _2$ then $L_\varphi \in \text {(UM)}$, see e.g. [13, Theorem 7]. Consequently $Ces_\varphi \in \text {(UM)}$ by Fact. $\square$