Local uniform non-squareness of Orlicz–Lorentz function spaces

Abstract

We give criteria for local uniform non-squareness of Orlicz–Lorentz function spaces $\varLambda _{\varphi ,\omega }$ equipped with the Luxemburg norm, where the widest possible class of convex Orlicz functions is admitted. As immediate consequences, criteria for local uniform non-squareness of Orlicz function spaces $L^{\varphi }$, which complete the already known results, are deduced.

1 Preliminaries

We say that a Banach space X is non-square if for any x and y from S(X) (the unit sphere of X), we have $\min (\Vert \frac{x-y}{2}\Vert ,\Vert \frac{x+y}{2}\Vert )<1$. A Banach space X is said to be locally uniformly non-square if for any $x\in S(X)$ there exists $\delta =\delta (x)\in (0,1)$ such that $\min (\Vert \frac{x-y}{2}\Vert ,\Vert \frac{x+y}{2}\Vert )\le 1-\delta$ for any $y\in B(X)$ (the unit ball of X). Finally, we say that a Banach space X is uniformly non-square if there exists $\delta \in (0,1)$ such that $\min (\Vert \frac{x-y}{2}\Vert ,\Vert \frac{x+y}{2}\Vert )\le 1-\delta$ for any $x, y\in B(X)$.

Recall that uniform non-squareness of Banach spaces was defined by James as a geometric property which implies super-reflexivity (see [9, 10]). Hence, knowing that a Banach space possesses this property, even without any characterization of its dual space, we can infer that it is super-reflexive, and so reflexive. Additionally, as García-Falset, Llorens-Fuster and Mazcu$\tilde{\mathrm {n}}$an-Navarro have shown, uniformly non-square Banach spaces possess the fixed-point property (see [5]). Therefore, it is natural and interesting to ask about criteria of non-squareness of various well-known classes of Banach spaces. The purpose of this paper is to complete the study started in [3] (see also [2]).

Let $L^{0}=L^{0}([0,\gamma ))$ be the space of all (equivalence classes of) Lebesgue measurable real-valued functions defined on the interval $[0,\gamma )$, where $\gamma \le \infty$. For any $x,y\in L^{0}$, we write $x\le y$ if $x(t)\le y(t)$ almost everywhere with respect to the Lebesgue measure m on the interval $[0,\gamma )$.

Given any $x\in L^{0}$, we define its distribution function $\mu _{x}:[0,+\infty )\rightarrow [0,\gamma ]$ by

$$\begin{aligned} \mu _{x}(\lambda )=m\{t\in [0,\gamma ):|x(t)|>\lambda \} \end{aligned}$$

(see [1, 14, 15]) and the non-increasing rearrangement $x^{*}:[0,\gamma )\rightarrow [0,\infty ]$ of x as

$$\begin{aligned} x^{*}(t)=\inf \{\lambda \ge 0:\mu _{x}(\lambda )\le t\} \end{aligned}$$

(under the convention $\inf \emptyset =\infty$). We say that two functions $x,y\in L^{0}$ are equimeasurable if $\mu _{x}(\lambda )=\mu _{y}(\lambda )$ for all $\lambda \ge 0$. Then we obviously have $x^{*}=y^{*}$.

A Banach space $E=(E,\le ,\Vert \cdot \Vert )$, where $E\subset L^{0}$, is said to be a Köthe space if the following conditions are satisfied:

• $\mathrm{(i)}$ if $x\in E$, $y\in L^{0}$ and $|y|\le |x|$, then $y\in E$ and $\Vert y\Vert \le \Vert x\Vert$,

• $\mathrm{(ii)}$ there exists a function x in E that is strictly positive on the whole $[0,\gamma )$.

Recall that the Köthe space E is called a symmetric space if E is rearrangement invariant in the sense that if $x\in E$, $y\in L^{0}$ and $x^{*}=y^{*}$, then $y\in E$ and $\Vert x\Vert =\Vert y\Vert$. For basic properties of symmetric spaces we refer to [1, 14, 15].

In the whole paper $\varphi$ will denote an Orlicz function, that is, $\varphi :[-\infty ,\infty ]\rightarrow [0,\infty ]$ (our definition is extended from R into $R^{e}$ by assuming $\varphi (-\infty )=\varphi (\infty )=\infty$) and $\varphi$ is convex, even, vanishing and continuous at zero, left continuous on $(0,\infty )$ and not identically equal to zero on $(-\infty ,\infty )$. Let us denote

$$\begin{aligned} a_{\varphi }=\sup \{u\ge 0:\varphi (u)=0\} \quad \mathrm{and}\quad b_{\varphi }=\sup \{u\ge 0:\varphi (u)<\infty \} \end{aligned}$$

and

$$\begin{aligned} \beta =\sup \left\{ u\ge 0:\varphi \left( \frac{u}{2}\right) =\frac{1}{2}\varphi \left( u\right) \right\} . \end{aligned}$$

(1)

Let us note that if $a_{\varphi }>0$, then $\beta =a_{\varphi }$, while the left-continuity of $\varphi$ on $(0,\infty )$ is equivalent to the fact that $\lim _{u\rightarrow (b_{\varphi })^{-}}\varphi (u)=\varphi (b_{\varphi })$.

Recall that an Orlicz function $\varphi$ satisfies the condition $\varDelta _{2}$ for all $u\in {\mathbb {R}}$ ($\varphi \in \varDelta _{2}({\mathbb {R}})$, for short) if there exists a constant $K>0$ such that the inequality

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

(2)

holds for any $u\in {\mathbb {R}}$ (then we have $a_{\varphi }=0$ and $b_{\varphi }=\infty$). Analogously, we say that an Orlicz function $\varphi$ satisfies the condition $\varDelta _{2}$ at infinity ($\varphi \in \varDelta _{2}(\mathbb {\infty })$, for short) if there exist a constant $K>0$ and a constant $u_{0}\ge 0$ such that $\varphi (u_{0})<\infty$ and inequality (2) holds for any $u\ge u_{0}$ (then we obtain $b_{\varphi }=\infty$).

For any Orlicz function $\varphi$ we define its complementary function in the sense of Young by the formula

$$\begin{aligned} \psi (u)=\sup _{v>0}\{|u|v-\varphi (v)\} \end{aligned}$$

(3)

for all $u\in {\mathbb {R}}$. It is easy to show that $\psi$ is also an Orlicz function.

Let $\omega :[0,\gamma )\rightarrow {\mathbb {R}}_{+}$ be a non-increasing and locally integrable function, which will be called a weight function. Define

$$\begin{aligned} \alpha&=\sup \left\{ t\ge 0:\omega \left( t\right) >0\right\} , \\ \gamma _{0}&=\sup \left\{ t\ge 0:\omega \text { is constant on }(0,t)\right\} . \end{aligned}$$

Recall that Orlicz–Lorentz spaces—a natural generalization of Orlicz and Lorentz spaces—were introduced at the beginning of the 1990s [11, 12, 16,17,18]. These investigations spurred further investigations of Orlicz–Lorentz spaces.

Given an Orlicz function $\varphi$ and a weight function $\omega$, we define a convex modular $I_{\varphi ,\omega }:L^{0}\rightarrow {\mathbb {R}}_{+}^{e}=[0,\infty ]$ by the formula

$$\begin{aligned} {I_{\varphi ,\omega }\left( x\right) }\mathrel {\mathop :}=\int _{0}^{\gamma } \varphi (x^{*}(t))\omega (t)dt. \end{aligned}$$

(4)

The modular space

$$\begin{aligned} {\varLambda _{\varphi ,\omega }}={\varLambda _{\varphi ,\omega }([0,\gamma ))}=\{x\in L^{0}:{I_{\varphi ,\omega }\left( \lambda x\right) }<\infty \,\, \mathrm{for}\,\, \mathrm{some}\,\, \lambda >0\}, \end{aligned}$$

generated by the modular $I_{\varphi ,\omega }$, is called the Orlicz–Lorentz function space. Since the modular unit ball $\{x\in L^{0}:I_{\varphi ,\omega }(x)\le 1\}$ is an absolutely convex and absorbing subset of $\varLambda _{\varphi ,\omega }([0,\gamma ))$, its Minkowski functional $\Vert \cdot \Vert _{\varphi ,\omega }$, defined by

$$\begin{aligned} {\Vert x\Vert _{\varphi ,\omega }}=\inf \{\lambda >0:{I_{\varphi ,\omega } \left( x/\lambda \right) }\le 1\}, \end{aligned}$$

is a seminorm in $\varLambda _{\varphi ,\omega }([0,\gamma ))$. It is easy to check that, thanks to the condition $\varphi (u)\rightarrow \infty$ as $u\rightarrow \infty$, it is a norm in $\varLambda _{\varphi ,\omega }([0,\gamma ))$, called the Luxemburg norm.

In our investigations we apply the results concerning the monotonicity properties of Lorentz function spaces first presented in [8] and [4]. Let us recall that a Lorentz function space $\varLambda _{\omega }$ is defined by the formula

$$\begin{aligned} \varLambda _{\omega }=\varLambda _{\omega }([0,\gamma ))=\left\{ x\in L^{0}:\Vert x\Vert _{\omega }=\int _{0}^{\gamma }x^{*}(t)\omega (t)dt<\infty \right\} . \end{aligned}$$

A Banach lattice $E=(E,\le ,\Vert \cdot \Vert )$ is said to be locally uniformly monotone, whenever for any $x\in (E)_{+}$ (the positive cone of E) with $\Vert x\Vert =1$ and any $\varepsilon \in (0,1)$ there is $\delta (x, \varepsilon )\in (0,1)$ such that the conditions $0\le y\le x$ and $\Vert y\Vert \ge \varepsilon$ imply that $\Vert x-y\Vert \le 1-\delta (x,\varepsilon )$.

Theorem 1

([4], Proposition 4.1) The Lorentz function space $\varLambda _{\omega }$ is locally uniformly monotone if and only if $\omega$ is positive on $[0,\gamma )$ and $\int _{0}^{\gamma }\omega (t)dt=\infty$ whenever $\gamma =\infty$.

In our investigations we will also apply Lemma 1. It is proved analogously as in the case of Orlicz spaces using the convexity of the modular $I_{\varphi ,\omega }$ (cf. also [13] for a more general case).

Lemma 1

Suppose that an Orlicz function $\varphi$ satisfies a suitable condition $\varDelta _{2}$, that is, $\varphi \in \varDelta _{2}({\mathbb {R}})$ if $\gamma =\infty$ and $\int _{0}^{\infty }\omega (t)dt=\infty$, and $\varphi \in \varDelta _{2}(\mathbb \infty )$ otherwise.

Then for any $\varepsilon \in (0,1)$ there exists $\delta =\delta (\varepsilon )\in (0,1)$ such that $\Vert x\Vert \le 1-\delta$ for any $x\in \varLambda _{\varphi ,\omega }$ whenever $I_{\varphi ,\omega }(x)\le 1-\varepsilon$.

In particular, for any $x\in \varLambda _{\varphi ,\omega }$, we then have $\Vert x\Vert =1$ if and only if $I_{\varphi ,\omega }\left( x\right) =1$.

2 Results

We start with the following

Theorem 2

Let $\gamma =\infty$. Then the Orlicz–Lorentz function space $\varLambda _{\varphi ,\omega }$ is locally uniformly non-square if and only if $\int _{0}^{\infty }\omega (t)dt=\infty$, $\varphi \in \varDelta _{2}({\mathbb {R}})$ and $\varphi (\frac{u}{2})<\frac{1}{2}\varphi (u)$ for every $u>0$.

Proof. Sufficiency. Let us fix $x\in S(\varLambda _{\varphi ,\omega })$. Since $\varphi \in \varDelta _2({\mathbb {R}})$, by Lemma 1 it is enough to show that there exists $\delta \in (0, 1)$ such that $\min \left( I_{\varphi ,\omega }\left( \frac{x-y}{2}\right) , I_{\varphi ,\omega }\left( \frac{x+y}{2}\right) \right) \le 1-\delta$ for every $y\in B\left( \varLambda _{\varphi ,\omega }\right)$. There exist $t_1, t_2$ such that $0\le t_1<t_2<\infty$, $0<x^{*}(t_2)\le x^{*}(t_1)<\infty$ and

$$\begin{aligned} \int _{t_1}^{t_2}\varphi (x^{*}(t))\omega (t)dt\ge \xi \end{aligned}$$

(5)

for some $\xi \in (0, 1]$. Without loss of generality we may assume that $x^{*}(s)>x^{*}(t)>x^{*}(w)$ for all $s\in (0, t_1), t\in (t_1, t_2)$ and $w>t_2$ if $t_1>0$, and $x^{*}(t)>x^{*}(w)$ for all $t\in (t_1, t_2)$ and $w>t_2$, otherwise. By [14], property $7^{\circ }$, page 64 there exist sets $e_{t_1}$ and $e_{t_2}$ with $m(e_{t_1})=t_1$, $m(e_{t_2})=t_2$,

$$\begin{aligned} \int _{0}^{t_1} x^{*}(t)dt=\int _{e_{t_1}} \left| x(t)\right| dt \quad \mathrm{and}\quad \int _{0}^{t_2} x^{*}(t)dt=\int _{e_{t_2}} \left| x(t)\right| dt \end{aligned}$$

(in the case $t_1=0$ we have $e_{t_1}=\emptyset$), whence it follows that $e_{t_1}\subsetneqq e_{t_2}$. Moreover, by the properties of the function $\varphi$, we get

$$\begin{aligned} \varphi \left( \frac{1}{2}u\right) \le \frac{1-\eta _{x}}{2}\varphi \left( u\right) \end{aligned}$$

for $u\in \left[ x^{*}\left( t_1\right) , x^{*}\left( t_2\right) \right]$, where $\eta _x \in (0, 1)$. For any $y\in B\left( \varLambda _{\varphi , \omega }\right)$ we define

$$\begin{aligned} B_1=\Big \{ t \in e_{t_2}{\setminus } e_{t_1}: x(t) y(t) \ge 0 \Big \} {,} \\ B_2=\Big \{ t \in e_{t_2}{\setminus } e_{t_1}: x(t) y(t) < 0 \Big \}. \end{aligned}$$

By the inequality (5), we obtain $I_{\varphi , \omega } \left( x \chi _{e_{t_2}{\setminus } e_{t_1}}\right) \ge \xi$, whence $I_{\varphi , \omega } \left( x \chi _{B_1}\right) \ge \frac{\xi }{2}$ or $I_{\varphi , \omega } \left( x \chi _{B_2}\right) \ge \frac{\xi }{2}$. Let us assume that the first inequality holds. Then we have

$$\begin{aligned} \varphi \left( \frac{x(t)-y(t)}{2}\right)&\le \varphi \left( \frac{\max \Big \{\vert x(t)\vert ,\vert y(t) \vert \Big \}}{2}\right) \le \varphi \left( \frac{x(t)}{2}\right) +\varphi \left( \frac{y(t)}{2}\right) \\&\le \frac{1}{2}\varphi \left( x(t)\right) -\frac{\eta _{x}}{2}\varphi \left( x(t)\right) +\frac{1}{2}\varphi \left( y(t)\right) \end{aligned}$$

for m-a.e. $t \in B_1$. Consequently,

$$\begin{aligned} \varphi \circ \left( \frac{x-y}{2}\right) \le \frac{1}{2}\varphi \circ y + \frac{1}{2}\varphi \circ x - \frac{\eta _x}{2}\varphi \circ x \chi _{B_1}. \end{aligned}$$

By the local uniform monotonicity of the Lorentz space $\varLambda _{\omega }$ we get

$$\begin{aligned} I_{\varphi , \omega }\left( \frac{x-y}{2}\right)&=\left\| \varphi \circ \left( \frac{x-y}{2}\right) \right\| _{\omega } \le \left\| \frac{1}{2}\varphi \circ y+\frac{1}{2}\varphi \circ x-\frac{\eta _x}{2}\varphi \circ x\chi _{B_1} \right\| _{\omega } \\&\le \frac{1}{2}\Vert \varphi \circ y\Vert _{\omega }+\frac{1}{2}\Vert \varphi \circ x-\eta _x\varphi \circ x\chi _{B_1}\Vert _{\omega } \le 1-\frac{1}{2}\delta \left( \varphi \circ x,\frac{\xi \cdot \eta _{x}}{2}\right) , \end{aligned}$$

where $\delta \left( \varphi \circ x,\frac{\xi \cdot \eta _{x}}{2}\right)$ is a constant from the definition of local uniform monotonicity of the Lorentz space $\varLambda _{\omega }$ corresponding to $\varphi \circ x$ and $\varepsilon =\frac{\xi \cdot \eta _{x}}{2}$. In the second case, that is if $I_{\varphi ,\omega }\left( x\chi _{B_2}\right) \ge \frac{\xi }{2}$, proceeding analogously we can prove that $I_{\varphi ,\omega }\left( \frac{x+y}{2}\right) \le 1-\delta \left( \varphi \circ x,\frac{\xi \cdot \eta _x}{2}\right)$.

Necessity. If $\int _{0}^{\infty }\omega \left( t\right) dt<\infty$ or $\varphi \notin \varDelta _{2}({\mathbb {R}})$, then $\varLambda _{\varphi ,\omega }$ contains an order isometric copy of $l^{\infty }$ (see [11, Theorem 2.4]). Suppose now that $\int _{0}^{\infty }\omega \left( t\right) dt=\infty$, $\varphi \in \varDelta _{2}({\mathbb {R}})$ and there exists $u_0>0$ such that $\varphi \left( \frac{1}{2}u_0\right) =\frac{1}{2}\varphi \left( u_0\right)$. Then $\varphi \left( \frac{1}{2}u\right) =\frac{1}{2}\varphi \left( u\right)$ for any $u\in \left[ 0, u_0\right)$. Moreover, we can find $t_0>0$ such that $\int _{0}^{t_0}\varphi \left( u_0\right) \omega \left( t\right) dt=1$ and $t_n$ such that $\int _{0}^{t_n}\varphi \left( u_n\right) \omega \left( t\right) dt=1$ for all $n\in {\mathbb {N}}$, where $u_n=\frac{1}{2^n}u_0$. Taking

$$\begin{aligned} x=u_0 \chi _{\left[ 0,t_{0}\right) }\quad \mathrm{and}\quad y_n=u_n \chi _{\left[ t_{0},t _{0}+t_n\right) }, \end{aligned}$$

we get $I_{\varphi , \omega }\left( x\right) =I_{\varphi , \omega }\left( y_n\right) =1$ for all $n\in {\mathbb {N}}$. Since

$$\begin{aligned} \int _{0}^{t_0}\varphi \left( u_n\right) \omega \left( t\right) dt=\int _{0}^{t_0} \varphi \left( \frac{u_0}{2^n}\right) \omega \left( t\right) dt=\frac{1}{2^n}\int _{0}^{t_0}\varphi \left( u_0\right) \omega \left( t\right) dt=\frac{1}{2^n}, \end{aligned}$$

we get

$$\begin{aligned} \int _{t_0}^{t_0+t_n}\varphi \left( u_n\right) \omega \left( t\right) dt\ge \int _{t_0}^{t_n}\varphi \left( u_n\right) \omega \left( t\right) dt=1-\frac{1}{2^{n}}. \end{aligned}$$

Therefore

$$\begin{aligned} I_{\varphi , \omega }\left( \frac{x\pm y_n}{2}\right)&=\int _{0}^{t_0}\varphi \left( \frac{u_0}{2}\right) \omega \left( t\right) dt+\int _{t_0}^{t_0+t_n}\varphi \left( \frac{u_n}{2}\right) \omega \left( t\right) dt \\&=\frac{1}{2}\int _{0}^{t_0}\varphi \left( u_0\right) \omega \left( t\right) dt+\frac{1}{2}\int _{t_0}^{t_0+t_n}\varphi \left( u_n\right) \omega \left( t\right) dt \\&\quad \ge \frac{1}{2}+\frac{1}{2}\left( 1-\frac{1}{2^n}\right) =1-\frac{1}{2^{n+1}}. \end{aligned}$$

Consequently

$$\begin{aligned} \lim _{n\rightarrow \infty }\min \left( \left\| \frac{x-y_n}{2}\right\| _{\varphi , \omega }, \left\| \frac{x+y_n}{2}\right\| _{\varphi , \omega }\right) =1 , \end{aligned}$$

which means that the space $\varLambda _{\varphi ,\omega }$ is not locally uniformly non-square.

Theorem 3

Let $\alpha =\gamma <\infty$ and $\int _{0}^{\gamma }\varphi (\beta )\omega (t)dt<1$. Then the following conditions are equivalent:

• (i) $\varphi \in \varDelta _2(\infty )$,

• (ii) the Orlicz–Lorentz space $\varLambda _{\varphi ,\omega }$ is non-square,

• (iii) the Orlicz–Lorentz space $\varLambda _{\varphi ,\omega }$ is locally uniformly non-square.

Proof

The implication (iii)$\Rightarrow$(ii) is obvious. The implication (ii)$\Rightarrow$(i) has been proved in Theorem 2.2 in [3]. We shall show (i)$\Rightarrow$(iii). Let us take an arbitrary $x\in S\left( \varLambda _{\varphi ,\omega }\right)$. By the assumption that $\int _{0}^{\gamma }\varphi (\beta )\omega (t)dt<1$ and $\varphi \in \varDelta _2(\infty )$, we get

$$\begin{aligned} m\left( \left\{ t\in \left[ 0, \gamma \right) : \vert x(t)\vert>\beta \right\} \right) =t_0>0. \end{aligned}$$

Hence, there exist $t_1, t_2$ such that $0\le t_1 < t_2 \le t_0$, $\beta<x^{*}(t_2)\le x^{*}(t_1)<\infty$ and

$$\begin{aligned} \int _{t_1}^{t_2}\varphi \left( x^{*}(t)\right) \omega (t)dt>0. \end{aligned}$$

Proceeding analogously as in the proof of sufficiency in Theorem 2, we can find $\delta >0$ depending only on x such that $\min \left( \left\| \frac{x-y}{2}\right\| _{\varphi , \omega }, \left\| \frac{x+y}{2}\right\| _{\varphi , \omega }\right) \le 1-\delta$ for every $y\in B\left( \varLambda _{\varphi , \omega }\right)$.

Theorem 4

Let $\alpha =\gamma <\infty$ and $\int _{0}^{\gamma }\varphi (\beta )\omega (t)dt\ge 1$. Then the Orlicz–Lorentz function space $\varLambda _{\varphi ,\omega }$ is locally uniformly non-square if and only if $\varphi \in \varDelta _{2}(\infty )$, $\psi \in \varDelta _{2}(\infty )$, where $\psi$ denotes the complementary function of $\varphi$ in the sense of Young (see formula (3)), and $\int _{0}^{\gamma _{0}/2}\varphi (\beta )\omega (t)dt<1$.

Proof

Sufficiency. Since $\varphi \in \varDelta _{2}(\infty )$, by Lemma 1, it is sufficient to prove the corresponding inequalities for the modular.

Case 1. Suppose that $\gamma _{0}>0$. Let us take an arbitrary $x\in S\left( \varLambda _{\varphi ,\omega }\right)$. If

$$\begin{aligned} m\left( \Big \{ t\in \left[ 0, \gamma \right) : \vert x(t)\vert>\beta \Big \}\right) =t_0>0, \end{aligned}$$

(6)

then we proceed analogously as in the proof of the implication (i)$\Rightarrow$(iii) in Theorem 3.

Let us now assume that inequality (6) does not hold or, equivalently, $x^{*}(0)\le \beta$. Define

$$\begin{aligned} c:=\frac{1-\int _{0}^{\gamma _{0}/2}\varphi (\beta )\omega (t)dt}{8}. \end{aligned}$$

By the assumptions we get $c\in \left( 0, \frac{1}{8}\right)$. Let $u_0$ be such that

$$\begin{aligned} \int _{0}^{\gamma _{0}/2}\varphi (u_0)\omega (t)dt=1-4c. \end{aligned}$$

(7)

We have $u_0>\beta$ and, by the condition $\psi \in \varDelta _2(\infty )$, there exists $\eta =\eta (u_0)\in (0, 1)$ such that

$$\begin{aligned} \varphi \left( \frac{u}{2}\right) \le \frac{1-\eta }{2}\varphi (u) \end{aligned}$$

(8)

for $u\ge u_0$. Moreover, in virtue of $\varphi \in \varDelta _2(\infty )$, for $K=\frac{2-\eta }{2(1-\eta )}$ we can find $\tau \in (0, 1)$ such that

$$\begin{aligned} \varphi ((1+\tau )u)\le K\varphi (u) \end{aligned}$$

for $u \ge u_0$. Thus, for $u\ge u_1$, where $u_1=\max \left( u_0, \frac{\beta }{\tau }\right)$, we get

$$\begin{aligned} \varphi (u+\beta )\le \varphi (u+u_1\tau )\le \varphi (u+u\tau )\le K\varphi (u). \end{aligned}$$

(9)

In virtue of (8) and (9), we obtain

$$\begin{aligned} \varphi \left( \frac{u+\beta }{2}\right) \le \frac{1-\eta }{2} \varphi (u+\beta )\le \frac{1-\eta }{2}K\varphi (u)=\frac{2-\eta }{4}\varphi (u) \end{aligned}$$

(10)

for $u\ge u_1$. Let us take an arbitrary $y\in S\left( \varLambda _{\varphi ,\omega }\right)$ and define

$$\begin{aligned} e\left( \frac{u_0+\beta }{2}\right)&=\left\{ t\in [0, \gamma ): \vert y(t)\vert \ge \frac{u_0+\beta }{2}\right\} ,\nonumber \\ e(u_0)&=\Big \{ t\in [0, \gamma ): \vert y(t)\vert \ge u_0\Big \}, \\ e(u_1+2\beta )&=\Big \{ t\in [0, \gamma ): \vert y(t)\vert \ge u_1+2\beta \Big \}.\nonumber \end{aligned}$$

(11)

Case 1.1. Suppose

$$\begin{aligned} \int _{0}^{\gamma }\varphi \left( \left( y\chi _{e(u_1+2\beta )}\right) ^{*}(t) \right) \omega (t)dt=\int _{0}^{m(e(u_1+2\beta ))}\varphi \left( y^{*}(t)\right) \omega (t)dt\ge \frac{c}{2}. \end{aligned}$$

(12)

Define

$$\begin{aligned} f(u_1+\beta )=\Big \{ t\in [0, \gamma ): \left| x(t)+y(t)\right| \ge u_1+\beta \Big \}. \end{aligned}$$

We have $\left| y(t) \right| \ge u_1$ for m-a.e. $t\in f(u_1+\beta )$ and $e(u_1+2\beta )\subset f(u_1+\beta )$. Moreover, in virtue of (10), we obtain

$$\begin{aligned} \varphi \left( \frac{x(t)+y(t)}{2}\right) \le \varphi \left( \frac{\vert y(t)\vert +\beta }{2}\right) \le \frac{2-\eta }{4}\varphi (y(t)) \end{aligned}$$

for m-a.e. $t\in f(u_1+\beta )$, whence

$$\begin{aligned} \varphi \left( \left( \frac{x+y}{2}\chi _{f(u_1+\beta )} \right) ^{*}(t)\right) \le \frac{2-\eta }{4} \varphi \left( \left( y\chi _{f(u_1+\beta )}\right) ^{*}(t)\right) \end{aligned}$$

and in consequence

$$\begin{aligned}&\int _{0}^{m(f(u_1+\beta ))} \varphi \left( \left( \frac{x+y}{2}\right) ^{*}(t)\right) \omega (t)dt \nonumber \\&\quad =\int _{0}^{m(f(u_1+\beta ))} \varphi \left( \left( \frac{x+y}{2}\chi _{f(u_1+\beta )}\right) ^{*}(t)\right) \omega (t)dt \nonumber \\&\quad \le \frac{2-\eta }{4}\int _{0}^{m(f(u_1+\beta ))} \varphi \left( \left( y\chi _{f(u_1+\beta )}\right) ^{*}(t)\right) \omega (t)dt \nonumber \\&\quad \le \frac{2-\eta }{4}\int _{0}^{m(e(u_1+2\beta ))} \varphi \left( \left( y\chi _{e(u_1+2\beta )}\right) ^{*}(t)\right) \omega (t)dt \nonumber \\&\qquad +\,\, \frac{2-\eta }{4}\int _{m(e(u_1+2\beta ))}^{m(f(u_1+\beta ))} \varphi \left( \left( y\chi _{f(u_1+\beta )}\right) ^{*}(t)\right) \omega (t)dt. \end{aligned}$$

(13)

Define

$$\begin{aligned} t(x)&=m\left( \left( \left[ 0, \gamma \right) {\setminus } f\left( u_1+\beta \right) \right) \cap {{\,\mathrm{supp}\,}}x\right) ,\\ t(y)&=m\left( \left( \left[ 0, \gamma \right) {\setminus } f\left( u_1+\beta \right) \right) \cap {{\,\mathrm{supp}\,}}y\right) . \end{aligned}$$

By [3, Remark 1.1], we get

$$\begin{aligned}&\int _{m(f(u_1+\beta ))}^{\gamma } \varphi \left( \left( \frac{x+y}{2}\right) ^{*}(t)\right) \omega (t)dt \nonumber \\&\quad =\int _{m(f(u_1+\beta ))}^{\gamma }\varphi \left( \left( \frac{x+y}{2}\chi _{([0, \gamma ){\setminus } f(u_1+\beta ))}\right) ^{*} \left( t-m\left( f\left( u_1+\beta \right) \right) \right) \right) \omega (t)dt \nonumber \\&\quad \le \frac{1}{2}\int _{m(f(u_1+\beta ))}^{m(f(u_1+\beta ))+t(y)}\varphi \left( \left( y\chi _{([0, \gamma ){\setminus } f(u_1+\beta ))}\right) ^{*}(t-m(f(u_1+\beta )))\right) \omega (t)dt \nonumber \\&\qquad +\,\,\frac{1}{2}\int _{m(f(u_1+\beta ))}^{m(f(u_1+\beta ))+t(x)}\varphi \left( \left( x\chi _{([0, \gamma ){\setminus } f(u_1+\beta ))}\right) ^{*}(t-m(f(u_1+\beta )))\right) \omega (t)dt. \end{aligned}$$

(14)

By (12), (13) and (14) we obtain

$$\begin{aligned}&I_{\varphi , \omega }\left( \frac{x+y}{2}\right) =\int _{0}^{\gamma }\varphi \left( \left( \frac{x+y}{2}\right) ^{*}(t)\right) \omega (t)dt\nonumber \\&\quad \le \frac{2-\eta }{4}\int _{0}^{m(e(u_1+2\beta ))} \varphi \left( \left( y\chi _{e(u_1+2\beta )}\right) ^{*}(t)\right) \omega (t)dt \nonumber \\&\qquad +\,\frac{2-\eta }{4}\int _{m(e(u_1+2\beta ))}^{(m(f(u_1+\beta ))} \varphi \left( \left( y\chi _{f(u_1+\beta )}\right) ^{*}(t)\right) \omega (t)dt \nonumber \\&\qquad +\,\,\frac{1}{2}\int _{m(f(u_1+\beta ))}^{m(f(u_1+\beta ))+t(y)}\varphi \left( \left( y\chi _{([0, \gamma ){\setminus } f(u_1+\beta ))}\right) ^{*}(t-m(f(u_1+\beta )))\right) \omega (t)dt \nonumber \\&\qquad +\,\,\frac{1}{2}\int _{m(f(u_1+\beta ))}^{m(f(u_1+\beta ))+t(x)}\varphi \left( \left( x\chi _{([0, \gamma ){\setminus } f(u_1+\beta ))}\right) ^{*}(t-m(f(u_1+\beta )))\right) \omega (t)dt \nonumber \\&\quad \le \frac{2-\eta }{4}\int _{0}^{m(e(u_1+2\beta ))} \varphi \left( y^{*}(t)\right) \omega (t)dt \nonumber \\&\qquad +\,\,\frac{1}{2}\int _{m(e(u_1+2\beta ))}^{(m(f(u_1+\beta ))}\varphi \left( \left( y\chi _{(f(u_1+\beta ){\setminus } e(u_1+2\beta ))}\right) ^{*}(t-m(e(u_1+2\beta )))\right) \omega (t)dt \nonumber \\&\qquad +\,\,\frac{1}{2}\int _{m(f(u_1+\beta ))}^{m(f(u_1+\beta ))+t(y)}\varphi \left( \left( y\chi _{([0, \gamma ){\setminus } f(u_1+\beta ))}\right) ^{*}(t-m(f(u_1+\beta )))\right) \omega (t)dt \nonumber \\&\qquad +\,\,\frac{1}{2}\int _{m(f(u_1+\beta ))}^{m(f(u_1+\beta ))+t(x)}\varphi \left( \left( x\chi _{([0, \gamma ){\setminus } f(u_1+\beta ))}\right) ^{*}(t-m(f(u_1+\beta )))\right) \omega (t)dt \end{aligned}$$

(15)

Since the function

$$\begin{aligned} {\bar{y}}(t)\mathrel {\mathop :}=\left\{ \begin{array}{l}{\left( y\chi _{(f(u_1+\beta ){\setminus } e(u_1+2\beta ))}\right) ^{*}(t-m(e(u_1+2\beta )))} \\ \qquad \quad \text{ for } t\in [m(e(u_1+2\beta )),m(f(u_1+\beta ))), \\ \\ \left( y\chi _{([0, \gamma ){\setminus } f(u_1+\beta ))}\right) ^{*}(t-m(f(u_1+\beta ))) \\ \qquad \quad \text{ for } t\in [m(f(u_1+\beta )),m(f(u_1+\beta ))+t(y)), \\ \\ 0\qquad \text{ otherwise, } \\ \end{array} \right. \end{aligned}$$

is equimeasurable with $y\chi _{([0,\gamma ){\setminus } e(u_1+2\beta )}$, by Hardy-Littlewood inequality, we have

$$\begin{aligned}&\int _{m(e(u_1+2\beta ))}^{\gamma }\varphi \left( {\bar{y}}(t)\right) \omega (t)dt\nonumber \\&\quad =\int _{m(e(u_1+2\beta ))}^{(m(f(u_1+\beta ))}\varphi \left( \left( y\chi _{(f(u_1+\beta ){\setminus } e(u_1+2\beta ))}\right) ^{*}(t-m(e(u_1+2\beta )))\right) \omega (t)dt \nonumber \\&\qquad +\int _{m(f(u_1+\beta ))}^{m(f(u_1+\beta ))+t(y)}\varphi \left( \left( y\chi _{([0, \gamma ){\setminus } f(u_1+\beta ))}\right) ^{*}(t-m(f(u_1+\beta )))\right) \omega (t)dt \nonumber \\&\quad \le \int _{m(e(u_1+2\beta ))}^{(m(f(u_1+\beta ))+t(y)}\varphi \left( \left( y\chi _{[0,\gamma ){\setminus } e(u_1+2\beta ))}\right) ^{*}(t-m(e(u_1+2\beta )))\right) \omega (t)dt \nonumber \\&\quad =\int _{m(e(u_1+2\beta ))}^{\gamma }\varphi \left( y^{*}(t)\right) \omega (t)dt. \end{aligned}$$

(16)

Simultaneously, by

$$\begin{aligned} \varphi \circ \left( x\chi _{([0,\gamma ){\setminus } f(u_1+\beta ))}\right) \le \varphi \circ x, \end{aligned}$$

we get

$$\begin{aligned} \varphi \circ \left( \left( x\chi _{([0,\gamma ){\setminus } f(u_1+\beta ))}\right) ^{*}\right) \le \varphi \circ x^{*}, \end{aligned}$$

whence

$$\begin{aligned}&\int _{m(f(u_1+\beta ))}^{m(f(u_1+\beta ))+t(x)}\varphi \left( \left( x\chi _{([0, \gamma ){\setminus } f(u_1+\beta ))}\right) ^{*}(t-m(f(u_1+\beta )))\right) \omega (t)dt\nonumber \\&\quad \le \int _{0}^{t(x)}\varphi \left( \left( x\chi _{([0, \gamma ){\setminus } f(u_1+\beta ))}\right) ^{*}(t)\right) \omega (t)dt\le \int _{0}^{\gamma }\varphi \left( x^{*}(t)\right) \omega (t)dt. \end{aligned}$$

(17)

Eventually, by (12), (15), (16) and (17), we have

$$\begin{aligned}&I_{\varphi , \omega }\left( \frac{x+y}{2}\right) =\int _{0}^{\gamma }\varphi \left( \left( \frac{x+y}{2}\right) ^{*}(t)\right) \omega (t)dt\nonumber \\&\quad \le \frac{2-\eta }{4}\int _{0}^{m(e(u_1+2\beta ))} \varphi \left( y^{*}(t)\right) \omega (t)dt \nonumber \\&\qquad +\,\frac{1}{2}\int _{m(e(u_1+2\beta ))}^{\gamma }\varphi \left( (y^{*}(t)\right) \omega (t)dt \nonumber \\&\qquad +\,\frac{1}{2}\int _{0}^{\gamma }\varphi \left( x^{*}(t)\right) \omega (t)dt \nonumber \\&\quad = \frac{1}{2}\int _{0}^{\gamma }\varphi \left( y^{*}(t)\right) \omega (t)dt+\frac{1}{2}\int _{0}^{\gamma }\varphi \left( x^{*}(t)\right) \omega (t)dt\nonumber \\&\qquad -\frac{\eta }{4}\int _{0}^{m(e(u_1+2\beta ))} \varphi \left( y^{*}(t)\right) \omega (t)dt \le 1-\frac{\eta c}{8}. \end{aligned}$$

(18)

Case 1.2. Let now

$$\begin{aligned} \int _{0}^{\gamma }\varphi \left( \left( y\chi _{e(u_1+2\beta )}\right) ^{*}(t) \right) \omega (t)dt=\int _{0}^{m(e(u_1+2\beta ))}\varphi \left( y^{*}(t)\right) \omega (t)dt<\frac{c}{2} \end{aligned}$$

(19)

and

$$\begin{aligned} \int _{0}^{\gamma }\varphi \left( \left( y\chi _{e(u_0)}\right) ^{*}(t) \right) \omega (t)dt=\int _{0}^{m(e(u_{0}))}\varphi \left( y^{*}(t)\right) \omega (t)dt\ge c. \end{aligned}$$

(20)

Let $t_1$ and $t_2$ be such that the following equalities hold:

$$\begin{aligned} \int _{0}^{t_1}\varphi (u_1+2\beta )\omega (t)dt=\frac{c}{2},\nonumber \\ \int _{0}^{t_2}\varphi \left( \frac{u_0+\beta }{2}\right) \omega (t)dt=1. \end{aligned}$$

(21)

We have $0<t_1<t_2<\gamma$. Moreover, by (19) and (20), we get

$$\begin{aligned} \int _{m(e(u_{1}+2\beta ))}^{m(e(u_{0}))}\varphi (u_{1}+2\beta ) \omega (t)dt\ge \int _{m(e(u_{1}+2\beta ))}^{m(e(u_{0}))} \varphi (y^{*}(t))\omega (t)dt>\frac{c}{2}, \end{aligned}$$

whence, by the definition of $t_{1}$, we obtain $m(e(u_0))\ge t_1$. Define

$$\begin{aligned} A_{y}^{1}=\Big \{ t\in e(u_0): y(t)x(t)\ge 0\Big \},\\ A_{y}^{2}=\Big \{ t\in e(u_0): y(t)x(t)<0\Big \}. \end{aligned}$$

We have $m\left( A_{y}^{1}\right) \ge \frac{t_1}{2}$ or $m\left( A_{y}^{2}\right) \ge \frac{t_1}{2}$. Let us assume that the first inequality holds. Then, by inequality (8),

$$\begin{aligned} \varphi \left( \frac{x(t)-y(t)}{2}\right) \le \varphi \left( \frac{y(t)}{2}\right) \le \frac{1-\eta }{2}\varphi (y(t))\le \frac{1}{2}\varphi (x(t))+\frac{1}{2}\varphi (y(t))-\frac{\eta }{2}\varphi (y(t)) \end{aligned}$$

for m-a.e. $t\in A_{y}^{1}$. Moreover,

$$\begin{aligned} \varphi \left( \frac{x(t)-y(t)}{2}\right) \le \frac{1}{2} \varphi \left( x(t)\right) +\frac{1}{2}\varphi \left( y(t)\right) \end{aligned}$$

for m-a.e. $t\in [0, \gamma ){\setminus } A_{y}^{1}$. Hence

$$\begin{aligned} \varphi \circ \left( \frac{x-y}{2}\right) \le \frac{1}{2}\varphi \circ x+\frac{1}{2}\varphi \circ y-\frac{\eta }{2}\varphi \circ y\chi _{A_{y}^{1}} \le \frac{1}{2}\varphi \circ x+\frac{1}{2}\varphi \circ y-\frac{\nu }{2}\varphi \circ y\chi _{A_{y}^{1}}, \end{aligned}$$

where $\nu =\min \left( \eta , 1-\frac{\varphi \left( \frac{u_0+\beta }{2}\right) }{\varphi (u_0)}\right)$. Since $y(t)\ge u_0$ for m-a.e. $t\in A_{y}^{1}$, we obtain

$$\begin{aligned} \varphi (y(t))-\nu \varphi (y(t))&\ge \varphi (u_0)(1-\nu )\ge \varphi (u_0)\left( 1-\left( 1-\frac{ \varphi \left( \frac{u_0+\beta }{2}\right) }{\varphi (u_0)}\right) \right) \nonumber \\&=\varphi \left( \frac{u_0+\beta }{2}\right) \end{aligned}$$

(22)

for the same t. It follows from the inequality $0\le \varphi \circ y-\nu \varphi \circ y\chi _{A_{y}^{1}}\le \varphi \circ y$ that

$$\begin{aligned} \left( \varphi \circ y - \nu \varphi \circ y\chi _{A_{y}^{1}}\right) ^{*}\le \left( \varphi \circ y\right) ^{*}=\varphi \circ y^{*}. \end{aligned}$$

(23)

Furthermore, it follows from (22) that for any $t\in [0,\gamma )$ such that $y^{*}(t)<\frac{u_0+\beta }{2}$ we have equality in (23). Simultaneously

$$\begin{aligned}&\int _{0}^{\gamma }\left( \varphi \circ y - \nu \varphi \circ y\chi _{A_{y}^{1}}\right) (t)dt \le \int _{0}^{\gamma }\left( \varphi \circ y - \nu \varphi (u_0)\chi _{A_{y}^{1}}\right) (t)dt \\ =&\int _{0}^{\gamma }\varphi (y(t))dt-\int _{0}^{\gamma }\nu \varphi (u_0)\chi _{A_{y}^{1}}(t)dt \le \int _{0}^{\gamma }\varphi (y(t))dt-\nu \varphi (u_0)\cdot \frac{t_1}{2}. \end{aligned}$$

Hence, by [14, property $1^\circ$, p. 63] we obtain

$$\begin{aligned} \int _{0}^{\gamma }\left( \varphi \circ y - \nu \varphi \circ y\chi _{A_{y}^{1}}\right) ^{*}(t)dt \le \int _{0}^{\gamma }\varphi (y^{*}(t))dt-\frac{\nu t_1\varphi (u_0)}{2}. \end{aligned}$$

Since $I_{\varphi ,\omega }(y)=1$, by the definitions of $e(\frac{u_{0}+\beta }{2})$ and $t_2$ (see formulas (11) and (21), respectively), we have $m(e(\frac{u_{0}+\beta }{2}))\le t_{2}$. Hence, by the remark concerning inequality (23), we obtain

$$\begin{aligned} \int _{t_2}^{\gamma }\left( \varphi \circ y - \nu \varphi \circ y\chi _{A_{y}^{1}}\right) ^{*}(t)dt =\int _{t_2}^{\gamma }\varphi (y^{*}(t))dt \end{aligned}$$

and in consequence

$$\begin{aligned}&\int _{0}^{\gamma }\left( \varphi \left( y^{*}(t)\right) -\left( \varphi \circ y-\nu \varphi \circ y\chi _{A_{y}^{1}}\right) ^{*}(t)\right) \omega (t)dt\\&\quad = \int _{0}^{t_{2}}\left( \varphi \left( y^{*}(t)\right) -\left( \varphi \circ y-\nu \varphi \circ y\chi _{A_{y}^{1}}\right) ^{*}(t)\right) \omega (t)dt\\&\quad \ge \int _{0}^{t_{2}}\left( \varphi \left( y^{*}(t)\right) -\left( \varphi \circ y-\nu \varphi \circ y\chi _{A_{y}^{1}}\right) ^{*}(t)\right) \omega (t_{2})dt\\&\quad \ge \frac{\nu t_1\varphi (u_0)\omega (t_2)}{2}. \end{aligned}$$

Therefore,

$$\begin{aligned} I_{\varphi , \omega }\left( \frac{x-y}{2}\right)&=\left\| \varphi \circ \left( \frac{x-y}{2}\right) \right\| _{\omega }\le \left\| \frac{1}{2}\varphi \circ x+\frac{1}{2}\varphi \circ y-\frac{\nu }{2}\varphi \circ y\chi _{A_{y}^{1}}\right\| _{\omega } \nonumber \\&\le \frac{1}{2}\Vert \varphi \circ x\Vert _{\omega }+\frac{1}{2}\Vert \varphi \circ y-\nu \varphi \circ y\chi _{A_{y}^{1}}\Vert _{\omega } \nonumber \\&\le \frac{1}{2}\Vert \varphi \circ x\Vert _{\omega }+\frac{1}{2}\Vert \varphi \circ y\Vert _{\omega }-\frac{\nu t_1\varphi (u_0)\omega (t_2)}{4} \nonumber \\&=1-\frac{\nu t_1\varphi (u_0)\omega (t_2)}{4}. \end{aligned}$$

(24)

If $m(A_{y}^{1})<\frac{t_1}{2}$, then $m(A_{y}^{2})\ge \frac{t_1}{2}$, and proceeding analogously we obtain

$$\begin{aligned} I_{\varphi , \omega }\left( \frac{x+y}{2}\right) \le 1-\frac{\nu t_1 \varphi (u_0)\omega (t_2)}{4}. \end{aligned}$$

(25)

Case 1.3. Suppose that

$$\begin{aligned} \int _{0}^{\gamma }\varphi \left( \left( y\chi _{e(u_0)}\right) ^{ *}(t)\right) \omega (t)dt=\int _{0}^{m(e(u_{0}))} \varphi \left( y^{*}(t)\right) \omega (t)dt<c. \end{aligned}$$

(26)

Let $t_3$ and $u_2$ be such that

$$\begin{aligned} \int _{0}^{t_3}\varphi (u_0)\omega (t)dt=1-2c \end{aligned}$$

and

$$\begin{aligned} \int _{0}^{\gamma }\varphi (u_2)\omega (t)dt=c. \end{aligned}$$

By the definition of $u_0$ (see (7)) we get $\frac{1}{2}\gamma _{0}<t_3<\frac{3}{4}\gamma _{0}$. Simultaneously, since $\int _{0}^{\gamma }\varphi (\beta )\omega (t)dt\ge 1$, we have $0<u_2<\beta <u_0$. Define

$$\begin{aligned} B_{x}&=\Big \{ t\in [0, \gamma ): \vert x(t) \vert \ge u_2\Big \}, \end{aligned}$$

(27)

$$\begin{aligned} B_{y}&=\Big \{ t\in [0, \gamma ): \vert y(t) \vert \ge u_2\Big \}. \end{aligned}$$

(28)

By the inequalities $x^{*}(0)\le \beta$ and (26) we obtain $m(B_x)\ge t_3$ and $m(B_y)\ge t_3$. Define

$$\begin{aligned} t_4:=\frac{t_3-\frac{\gamma _{0}}{2}}{2}<\frac{1}{8}\gamma _{0} \end{aligned}$$

and

$$\begin{aligned} C_{x,y}^{+}&=\Big \{ t\in [0,\gamma ): \min (\vert x(t) \vert , \vert y(t) \vert )\ge \frac{u_2}{4} \wedge x(t)y(t)>0\Big \},\end{aligned}$$

(29)

$$\begin{aligned} C_{x,y}^{-}&=\Big \{ t\in [0,\gamma ): \min (\vert x(t) \vert , \vert y(t) \vert )\ge \frac{u_2}{4} \wedge x(t)y(t)<0\Big \}. \end{aligned}$$

(30)

Case 1.3.1. Assume that $m(C_{x,y}^{+})\ge \frac{t_4}{2}$. Then we have

$$\begin{aligned} \varphi \left( \frac{x(t)-y(t)}{2}\right) \le \frac{1}{2}\varphi (y(t))+\frac{1}{2}\varphi (x(t))-\frac{1}{2} \cdot \frac{u_2}{4\beta }\varphi (x(t)) \end{aligned}$$

for m-a.e. $t\in C_{x,y}^{+}$. Indeed, if $\vert x(t) \vert \ge \vert y(t)\vert$, then

$$\begin{aligned} \varphi \left( \frac{x(t)-y(t)}{2}\right)&\le \varphi \left( \frac{x(t)-\frac{u_2}{4}{{\,\mathrm{sgn}\,}}(x(t))}{2}\right) \le \varphi \left( \frac{x(t)-\frac{u_2}{4}\cdot \frac{x(t)}{\beta }}{2}\right) \\&=\varphi \left( \frac{1}{2}\left( 1-\frac{u_2}{4\beta }\right) x(t) \right) \le \frac{1}{2}\left( 1-\frac{u_2}{4\beta }\right) \varphi (x(t))\\&=\frac{1}{2}\varphi (x(t))-\frac{1}{2}\cdot \frac{u_2}{4\beta }\varphi (x(t)). \end{aligned}$$

On the other hand, if $\vert x(t) \vert <\vert y(t)\vert$, we get

$$\begin{aligned} \varphi \left( \frac{x(t)-y(t)}{2}\right) \le \varphi \left( \frac{y(t)}{2}\right) \le \frac{1}{2}\varphi (y(t))+ \frac{1}{2}\varphi (x(t))-\frac{1}{2}\cdot \frac{u_2}{4\beta }\varphi (x(t)). \end{aligned}$$

It is obvious that for $t\in [0,\gamma ){\setminus } C_{x,y}^{+}$ we have

$$\begin{aligned} \varphi \left( \frac{x(t)-y(t)}{2}\right) \le \frac{1}{2}\varphi (x(t))+\frac{1}{2}\varphi (y(t)). \end{aligned}$$

Thus,

$$\begin{aligned} \varphi \circ \left( \frac{x(t)-y(t)}{2}\right) \le \frac{1}{2}\varphi \circ y+\frac{1}{2}\varphi \circ x -\frac{1}{2}\cdot \frac{u_2}{4\beta }\varphi \circ x\chi _{C_{x,y}^{+}}. \end{aligned}$$

Since

$$\begin{aligned} I_{\varphi , \omega }\left( x\chi _{C_{x,y}^{+}}\right) =\left\| \varphi \circ x\chi _{C_{x,y}^{+}}\right\| _{\omega } \ge \int _{0}^{t_4/2}\varphi \left( \frac{u_2}{4}\right) \omega (t)dt=:\zeta , \end{aligned}$$

by the local uniform monotonicity of the Lorentz space $\varLambda _\omega$, we get

$$\begin{aligned} I_{\varphi , \omega }\left( \frac{x-y}{2}\right)&=\left\| \varphi \circ \left( \frac{x(t)-y(t)}{2}\right) \right\| _{\omega }\le \frac{1}{2}\left\| \varphi \circ y \right\| _{\omega }+\frac{1}{2} \left\| \varphi \circ x -\frac{u_2}{4\beta }\varphi \circ x\chi _{C_{x,y}^{+}}\right\| _{\omega }\nonumber \\&\le 1-\frac{1}{2}\delta \left( \varphi \circ x,\frac{u_2\zeta }{4\beta }\right) , \end{aligned}$$

(31)

where $\delta \left( \varphi \circ x,\frac{u_2\zeta }{4\beta }\right)$ is a constant from the definition of the local uniform monotonicity of the Lorentz space $\varLambda _\omega$ corresponding to $\varphi \circ x$ and $\varepsilon =\frac{u_2 \zeta }{4\beta }$.

Analogously, if $m(C_{x,y}^{-})>\frac{t_4}{2}$, we have

$$\begin{aligned} I_{\varphi , \omega }\left( \frac{x+y}{2}\right) \le 1-\frac{1}{2}\delta \left( \varphi \circ x,\frac{u_2\zeta }{4\beta }\right) . \end{aligned}$$

(32)

Case 1.3.2. Finally assume that $m(C_{x,y}^{+})<\frac{t_4}{2}$ and $m(C_{x,y}^{-})<\frac{t_4}{2}$. The proof of this case is analogous to the proof of Case 2.2.2 in Theorem 2.4 in [3]. We shall present it for completeness’ sake. Let us define

$$\begin{aligned} C_x&=B_x{\setminus }\left( C_{x,y}^{+}\cup C_{x,y}^{-}\right) , \end{aligned}$$

(33)

$$\begin{aligned} C_y&=B_y{\setminus }\left( C_{x,y}^{+}\cup C_{x,y}^{-}\right) , \end{aligned}$$

(34)

where $B_{x}$ and $B_{y}$ are defined by formulas (27) and (28). We have $C_x\cap C_y=\emptyset$ and

$$\begin{aligned} \min (m(C_x),m(C_y))\ge t_3-t_4=\frac{1}{2}t_3+\frac{1}{4}\gamma _{0}>\frac{1}{2}\gamma _{0}, \end{aligned}$$

(35)

whence

$$\begin{aligned} m(C_x \cup C_y)\ge t_3+\frac{1}{2}\gamma _{0}>\gamma _{0}. \end{aligned}$$

(36)

Define

$$\begin{aligned} a:=\frac{t_3-\frac{\gamma _{0}}{2}}{8}<\frac{1}{32}\gamma _{0}\quad \mathrm{and}\quad t_5:=\gamma _{0}+a. \end{aligned}$$

By [14, property $7^{\circ }$, p. 64] there exist sets $e_{\gamma _{0}}=e_{\gamma _{0}}\left( \frac{x+y}{2}\right)$ and $e_{t_{5}}=e_{t_{5}}\left( \frac{x+y}{2}\right)$ such that $m\left( e_{\gamma _{0}}\right) =\gamma _{0}$, $m\left( e_{t_{5}}\right) =t_{5}$,

$$\begin{aligned} \int _{0}^{\gamma _{0}}\left( \frac{x+y}{2}\right) ^{*}(t)dt=\int _{e_{\gamma _{0}}} \left| \frac{x+y}{2}\right| (t)dt \end{aligned}$$

and

$$\begin{aligned} \int _{0}^{t_{5}}\left( \frac{x+y}{2}\right) ^{*}(t)dt=\int _{e_{t_{5}}}\left| \frac{x+y}{2}\right| (t)dt. \end{aligned}$$

Without loss of generality we may assume that $e_{\gamma _{0}}\subset e_{t_{5}}$. Denoting $D_{\gamma _{0}}=e_{t_{5}}{\setminus } e_{\gamma _{0}}$ and $D_{t_{5}}=\left[ 0, \gamma \right) {\setminus } e_{t_{5}}$, by [3, Remark 1.1], we obtain

$$\begin{aligned} I_{\varphi ,\omega }\left( \frac{x+y}{2}\right)&=\int _{0}^{\gamma _{0}}\varphi \left( \left( \frac{x+y}{2}\right) ^{*}(t)\right) \omega (t)dt +\int _{\gamma _{0}}^{t_{5}}\varphi \left( \left( \frac{x+y}{2}\right) ^{*}(t)\right) \omega (t)dt \\&\quad +\,\int _{t_{5}}^{\gamma }\varphi \left( \left( \frac{x+y}{2}\right) ^{*}(t)\right) \omega (t)dt \\&=\int _{0}^{\gamma _{0}}\varphi \left( \left( \frac{x+y}{2}\chi _{e_{\gamma _{0}}}\right) ^{*}(t)\right) \omega (t)dt\\&\quad +\int _{\gamma _{0}}^{t_{5}}\varphi \left( \left( \frac{x+y}{2}\chi _{D_{\gamma _{0}}}\right) ^{*}(t-\gamma _{0})\right) \omega (t)dt \\&\quad +\,\int _{t_{5}}^{\gamma }\varphi \left( \left( \frac{x+y}{2}\chi _{D_{t_{5}}}\right) ^{*}(t-t_{5})\right) \omega (t)dt \\&\le \frac{1}{2}\int _{0}^{\gamma _{0}}\varphi \left( \left( x\chi _{e_{\gamma _{0}}}\right) ^{*}(t)\right) \omega (t)dt +\frac{1}{2}\int _{0}^{\gamma _{0}}\varphi \left( \left( y\chi _{e_{\gamma _{0}}}\right) ^{*}(t)\right) \omega (t)dt \\&\quad +\,\frac{1}{2}\int _{\gamma _{0}}^{t_{5}}\varphi \left( \left( x\chi _{D_{\gamma _{0}}}\right) ^{*}(t-\gamma _{0})\right) \omega (t)dt\\&\quad +\frac{1}{2}\int _{\gamma _{0}}^{t_{5}}\varphi \left( \left( y\chi _{D_{\gamma _{0}}}\right) ^{*}(t-\gamma _{0})\right) \omega (t)dt\\&\quad +\,\frac{1}{2}\int _{t_{5}}^{\gamma }\varphi \left( \left( x\chi _{D_{t_{5}}}\right) ^{*}(t-t_{5})\right) \omega (t)dt\\&\quad +\frac{1}{2}\int _{t_{5}}^{\gamma }\varphi \left( \left( y\chi _{D_{t_{5}}}\right) ^{*}(t-t_{5})\right) \omega (t)dt. \end{aligned}$$

By inequality (36) we have

$$\begin{aligned} m((C_x \cup C_y)\cap D_{t_{5}})&=m((C_x \cup C_y){\setminus } e_{t_{5}})\ge \left( t_3+\frac{\gamma _{0}}{2}\right) -\left( \gamma _{0}+a\right) \\&=t_3-\frac{\gamma _{0}}{2}-a=7a. \end{aligned}$$

Hence, without loss of generality, it may be assumed that $m(C_{x}\cap D_{t_{5}})>a$. Then, since $C_{x}\subset B_{x}$ we have $m(\{t:|x|\chi _{D_{t_{5}}}(t)\ge u_{2}\})>a$ and thus $\left( x\chi _{D_{t_5}}\right) ^{*}(a)\ge u_2$. If $\left( x\chi _{e_{\gamma _{0}}}\right) ^{*}(\gamma _{0}-a)\le \frac{u_2}{4}$, then

$$\begin{aligned}&\int _{\gamma _{0}-a}^{\gamma _{0}}\left[ \varphi \left( \left( x\chi _{D_{t_{5}}}\right) ^{*}(t-\gamma _{0}+a)\right) -\varphi \left( \left( x\chi _{e_{\gamma _{0}}}\right) ^{*}(t)\right) \right] \omega (t)dt\nonumber \\&\qquad -\int _{t_{5}}^{t_{5}+a}\left[ \varphi \left( \left( x\chi _{D_{t_{5}}}\right) ^{*}(t-t_{5})\right) -\varphi \left( \left( x\chi _{e_{\gamma _{0}}}\right) ^{*}(t-(t_{5}-\gamma _{0}+a))\right) \right] \omega (t)dt\nonumber \\&\quad =\int _{\gamma _{0}-a}^{\gamma _{0}}\left[ \varphi \left( \left( x\chi _{D_{t_{5}}}\right) ^{*}(t-\gamma _{0}+a)\right) -\varphi \left( \left( x\chi _{e_{\gamma _{0}}}\right) ^{*}(t)\right) \right] (\omega (t)-\omega (t+t_{5}-\gamma _{0}+a))dt\nonumber \\&\quad \ge (\omega _{0}-\omega (t_{5}))\int _{\gamma _{0}-a}^{\gamma _{0}}\left( \varphi \left( \left( x\chi _{D_{t_{5}}}\right) ^{*}(t-\gamma _{0}+a)\right) -\varphi \left( \left( x\chi _{e_{\gamma _{0}}}\right) ^{*}(t)\right) \right) dt\nonumber \\&\quad \ge a\left( \varphi (u_{2})-\varphi \left( \frac{u_{2}}{4}\right) \right) (\omega _{0}-\omega (t_{5})) \ge \frac{3\,a\,p\,u_{2}(\omega _{0}-\omega (t_{5}))}{4}, \end{aligned}$$

(37)

where p is in fact the constant value of the derivative of $\varphi$ on the interval $(0,\beta )$ (the interval of linearity of $\varphi$) and $\omega _0=\omega (t)$ for any $t\in (0, \gamma _{0})$. By the definition of $\gamma _{0}$ we get $\omega _0-\omega (t_5)>0$. Hence and by Hardy-Littlewood inequality (proceeding analogously as in formula (16)), we obtain

$$\begin{aligned}&\int _{0}^{\gamma _{0}}\varphi \left( \left( x\chi _{e_{\gamma _{0}}}\right) ^{*}(t)\right) \omega (t)dt+ \int _{\gamma _{0}}^{t_{5}}\varphi \left( \left( x\chi _{D_{\gamma _{0}}}\right) ^{*}(t-\gamma _{0})\right) \omega (t)dt\nonumber \\&\qquad +\,\int _{t_{5}}^{\gamma }\varphi \left( \left( x\chi _{D_{t_{5}}}\right) ^{*}(t-t_{5})\right) \omega (t)dt \nonumber \\&\quad \le \int _{0}^{\gamma _{0}-a}\varphi \left( \left( x\chi _{e_{\gamma _{0}}}\right) ^{*}(t)\right) \omega (t)dt+ \int _{\gamma _{0}-a}^{\gamma _{0}}\varphi \left( \left( x\chi _{D_{t_{5}}}\right) ^{*}(t-(\gamma _{0}-a))\right) \omega (t)dt\nonumber \\&\qquad +\,\int _{\gamma _{0}}^{t_{5}}\varphi \left( \left( x\chi _{D_{\gamma _{0}}}\right) ^{*}(t-\gamma _{0})\right) \omega (t)dt\nonumber \\&\qquad +\,\int _{t_{5}}^{t_{5}+a}\varphi \left( \left( x\chi _{e_{\gamma _{0}}}\right) ^{*} (t-(t_{5}-\gamma _{0}+a))\right) \omega (t)dt \nonumber \\&\qquad +\,\int _{t_{5}+a}^{\gamma }\varphi \left( \left( x\chi _{D_{t_{5}}}\right) ^{*}(t-t_{5})\right) \omega (t)dt -\frac{3\,a\,p\,u_{2}(\omega _{0}-\omega (t_{5}))}{4}\nonumber \\&\quad \le \int _{0}^{\gamma }\varphi (x^{*}(t))\omega (t)dt-\frac{3\,a\,p\,u_{2}(\omega _{0}-\omega (t_{5}))}{4}=1-\frac{3\,a\,p\,u_{2}(\omega _{0}-\omega (t_{5}))}{4}. \end{aligned}$$

(38)

Suppose now that $\left( x\chi _{e_{\gamma _{0}}}\right) ^{*}(\gamma _{0}-a)>\frac{u_2}{4}$. Since $m(C_{x,y}^{+}\cup C_{x,y}^{-})<t_4<\frac{1}{8}\gamma _{0}$, we have $m \left( \Big \{ t\in e_{\gamma _{0}}: \vert y(t) \vert >\frac{u_2}{4}\Big \}\right) \le t_4+a\le \frac{5}{32}\gamma _{0}$. Thus $\left( y\chi _{e_{\gamma _{0}}}\right) ^{*}(\gamma _{0}-a)\le \frac{u_2}{4}$. Simultaneously,

$$\begin{aligned} m(C_y\cap e_{\gamma _{0}})\le m(B_y\cap e_{\gamma _{0}})\le t_4+a \le \frac{5}{32}\gamma _{0}. \end{aligned}$$

Hence, by (35) we get

$$\begin{aligned} m(C_y\cap D_{t_{5}})\ge \frac{1}{2}t_3+\frac{1}{4}\gamma _{0}-(t_4+a)-a=\frac{\gamma _{0}}{2}-2a \end{aligned}$$

and we conclude that $(y\chi _{D_{t_{5}}})^{*}(a)\ge u_{2}$. Repeating the procedure from (37) and (38) for y we obtain

$$\begin{aligned} I_{\varphi , \omega }\left( \frac{x+y}{2}\right) \le 1-\frac{3\,a\,p\,u_{2}(\omega _{0}-\omega (t_{5}))}{8}. \end{aligned}$$

(39)

Summarizing Case 1, for any $x\in S(\varLambda _{\varphi , \omega })$ such that $x^{*}(0)\le \beta$, by (18), (24), (25), (31), (32) and (39) we have

$$\begin{aligned}&\min \left( I_{\varphi , \omega }\left( \frac{x+y}{2}\right) , I_{\varphi , \omega }\left( \frac{x-y}{2}\right) \right) \\&\quad \le 1-\min \left( \frac{\eta c}{8}, \frac{\nu t_1 \varphi (u_0)\omega (t_2)}{4}, \frac{1}{2}\delta \left( \varphi \circ x,\frac{u_2 \zeta }{4\beta }\right) , \frac{3\,a\,p\,u_{2}(\omega _{0}-\omega (t_{5}))}{8}\right) . \end{aligned}$$

Case 2. Let $\gamma _{0}=0$. If $x^{*}(0)>\beta$, we can proceed analogously as in the proof of Case 1. Assume now that $x^{*}(0)\le \beta$. Let $c=\frac{1}{8}$ and $u_0>\beta$. Since $\psi \in \varDelta _2(\infty )$, there exists $\eta \in (0,1)$ such that inequality (8) holds for $u\ge u_0$. We define $u_1$, $e(u_0)$ and $e(u_1+2\beta )$ as in Case 1. If

$$\begin{aligned} \int _{0}^{\gamma }\varphi \left( \left( y\chi _{e(u_1+2\beta )}\right) ^{*}(t)\right) \omega (t)dt\ge \frac{1}{16}, \end{aligned}$$

then proceeding analogously as in Case 1.1 we obtain

$$\begin{aligned} I_{\varphi , \omega }\left( \frac{x+y}{2}\right) \le 1-\frac{\eta }{64}. \end{aligned}$$

(40)

Suppose now that

$$\begin{aligned} \int _{0}^{\gamma }\varphi \left( \left( y\chi _{e(u_1+2\beta )}\right) ^{ *}(t)\right) \omega (t)dt<\frac{1}{16} \end{aligned}$$

and

$$\begin{aligned} \int _{0}^{\gamma }\varphi \left( \left( y\chi _{e(u_0)}\right) ^{*}(t)\right) \omega (t)dt\ge \frac{1}{8}. \end{aligned}$$

We define $t_1$ and $t_2$ by the equalities

$$\begin{aligned} \int _{0}^{t_1}\varphi \left( u_1+2\beta \right) \omega (t)dt=\frac{1}{16} \end{aligned}$$

and

$$\begin{aligned} \int _{0}^{t_2}\varphi \left( \frac{u_0+\beta }{2}\right) \omega (t)dt=1. \end{aligned}$$

We have $0<t_1<t_2<\gamma$. Proceeding analogously as in Case 1.2, we get

$$\begin{aligned} \min \left( I_{\varphi , \omega }\left( \frac{x-y}{2}\right) , I_{\varphi , \omega }\left( \frac{x+y}{2}\right) \right) =1-\frac{\nu t_1\varphi (u_0)\omega (t_2)}{4}, \end{aligned}$$

(41)

where $\nu =\min \left( \eta , 1-\frac{\varphi \left( \frac{u_0+\beta }{2}\right) }{\varphi (u_0)}\right)$.

Finally let us assume that

$$\begin{aligned} \int _{0}^{\gamma }\varphi \left( \left( y\chi _{e(u_0)}\right) ^{*}(t) \right) \omega (t)dt<\frac{1}{8} \end{aligned}$$

and define $t_3$ and $u_2$ by the equalities

$$\begin{aligned} \int _{0}^{t_3}\varphi \left( u_0\right) \omega (t)dt=\frac{3}{4} \end{aligned}$$

and

$$\begin{aligned} \int _{0}^{\gamma }\varphi \left( u_2\right) \omega (t)dt=\frac{1}{8}. \end{aligned}$$

It is obvious that $0<u_2<\beta <u_0$. Defining the sets $B_x$ and $B_y$ as in Case 1.3 (see (27) and (28)), we obtain $m(B_x)\ge t_3$ and $m(B_y)\ge t_3$. By the assumption that $\gamma _{0}=0$, we can find $t_{\gamma }$ such that $0<t_{\gamma }<t_3$ and $\omega (t_{\gamma })>\omega (t_{3})$. Let now $t_4:=\frac{t_{\gamma }}{4}<\frac{t_3}{4}$ and let $C_{x,y}^{+}$ and $C_{x,y}^{-}$ be defined by formulas (29) and (30). If $m(C_{x,y}^{+})\ge \frac{t_4}{2}$ or $m(C_{x,y}^{-})\ge \frac{t_4}{2}$, then, analogously as in Case 1.3.1, we obtain

$$\begin{aligned} \min \left( I_{\varphi , \omega }\left( \frac{x-y}{2}\right) , I_{\varphi , \omega }\left( \frac{x+y}{2}\right) \right) \le 1-\frac{1}{2}\delta \left( \varphi \circ x,\frac{u_2\zeta }{4\beta }\right) , \end{aligned}$$

(42)

where $\zeta =\int _{0}^{t_4/2}\varphi \left( \frac{u_2}{4}\right) \omega (t)dt$ and $\delta \left( \varphi \circ x,\frac{u_2\zeta }{4\beta }\right)$ is the constant from the definition of local uniform monotonicity of the Lorentz space $\varLambda _{\omega }$ corresponding to $\varphi \circ x$ and $\varepsilon =\frac{u_2\zeta }{4\beta }$.

In the case when $\max \left( m(C_{x,y}^{+}), m(C_{x,y}^{-})\right) <\frac{t_4}{2}$, we define the sets $C_x$ and $C_y$ as in Case 1.3.2 (see (33) and (34)). We have $C_x\cap C_y=\emptyset$ and, moreover,

$$\begin{aligned} \min \left( m(C_x),m(C_y)\right) \ge t_3-t_4\ge t_3-\frac{t_3}{4}=\frac{3}{4}t_3, \end{aligned}$$

whence

$$\begin{aligned} m(C_x\cup C_y)\ge \frac{3}{2}t_3. \end{aligned}$$

Defining $a=\frac{1}{32}t_{\gamma }<\frac{1}{32}t_3$ and putting $t_{\gamma }$ in place of $\gamma _{0}$ and $t_3$ in place of $t_5$, we can repeat the procedure from Case 1.3.2. We get

$$\begin{aligned} I_{\varphi , \omega }\left( \frac{x+y}{2}\right) \le 1-\frac{3apu_2\left( \omega (t_{\gamma })-\omega (t_3)\right) }{8}, \end{aligned}$$

(43)

where p is the derivative of $\varphi$ on the interval $(0,\beta )$.

Recapitulating Case 2, for any x such that $x^{*}(0)\le \beta$, by (40), (41), (42) and (43) we get

$$\begin{aligned}&\min \left( I_{\varphi , \omega }\left( \frac{x-y}{2}\right) , I_{\varphi , \omega }\left( \frac{x+y}{2}\right) \right) \\&\quad \le 1-\min \left( \frac{\eta }{64}, \frac{\nu t_1\varphi (u_0)\omega (t_2)}{4}, \frac{1}{2}\delta \left( \varphi \circ x,\frac{u_2\zeta }{4\beta }\right) , \frac{3apu_2\left( \omega (t_{\gamma })-\omega (t_3)\right) }{8}\right) . \end{aligned}$$

Necessity. The necessity of conditions $\varphi \in \varDelta _{2}(\infty )$ and $\int _{0}^{\gamma _{0}/2}\varphi (\beta )\omega (t)dt<1$ has been shown in Theorem 2.2 of [3]. Suppose that $\psi \notin \varDelta _{2}(\infty )$. Then there exists a sequence $(u_n)_{n=1}^{\infty }$ such that $u_n \rightarrow \infty$ as $n\rightarrow \infty$ and

$$\begin{aligned} \varphi \left( \frac{u_n}{2}\right) >\frac{1-\frac{1}{n}}{2}\varphi \left( u_n\right) \end{aligned}$$

for any $n\in {\mathbb {N}}$. We can assume that $u_1>2\beta$.

There exist $t_0>0$ such that $\int _{0}^{t_0}\varphi (\beta )\omega (t)dt=1$ and $t_n>0$ such that $\int _{0}^{t_n}\varphi (u_n)\omega (t)dt=1$ for all $n\in {\mathbb {N}}$. Then we obtain $t_n<t_0$ for all $n\in {\mathbb {N}}$ and $t_n\rightarrow 0$ as $n\rightarrow \infty$. Define

$$\begin{aligned} x&=\beta \chi _{[0,t_0)} \\ y_n&=u_n\chi _{[0,t_n)}, \quad \text {for any } n\in {\mathbb {N}}. \end{aligned}$$

For any $n\in {\mathbb {N}}$ consider a straight line passing through the points $\left( \frac{1}{2}u_n, \frac{n-1}{2n}\varphi (u_n)\right)$ and $\left( u_n, \varphi (u_n)\right)$, given by the equation

$$\begin{aligned} y(t)=\frac{n+1}{n}\frac{\varphi (u_n)}{u_n}t-\frac{1}{n}\varphi (u_n) \quad \text {for } t\in {\mathbb {R}}. \end{aligned}$$

It is easy to observe that for any $u\in (0, u_n)$ the graph of the function $\varphi$ lies above the graph of this line. Hence, we obtain

$$\begin{aligned} \varphi \left( \frac{u_n-\beta }{2}\right)&\ge \frac{n+1}{n}\frac{\varphi (u_n)}{u_n}\left( \frac{u_n-\beta }{2}\right) -\frac{1}{n}\varphi (u_n)\\&=\frac{n-1}{2n}\varphi (u_n)-\frac{n+1}{2n}\varphi (u_n)\frac{\beta }{u_n} \end{aligned}$$

and

$$\begin{aligned}&\lim _{n\rightarrow \infty } I_{\varphi , \omega }\left( \frac{y_n-x}{2}\right) =\lim _{n\rightarrow \infty }\left( \int _{0}^{t_n}\varphi \left( \frac{u_n-\beta }{2}\right) \omega (t)dt +\int _{t_n}^{t_0}\varphi \left( \frac{\beta }{2}\right) \omega (t)dt\right) \\&\quad \ge \lim _{n\rightarrow \infty }\left( \frac{n-1}{2n}\int _{0}^{t_n}\varphi \left( u_n\right) \omega (t)dt -\frac{n+1}{2n}\frac{\beta }{u_n}\int _{0}^{t_n}\varphi \left( u_n\right) \omega (t)dt +\frac{1}{2}\int _{t_n}^{t_0}\varphi \left( \beta \right) \omega (t)dt\right) \\&\quad =\frac{1}{2}+\frac{1}{2}=1. \end{aligned}$$

In consequence,

$$\begin{aligned} \lim _{n\rightarrow \infty }\min \left( \left\| \frac{x-y_n}{2}\right\| _{\varphi , \omega }, \left\| \frac{x+y_n}{2}\right\| _{\varphi , \omega }\right) =1. \end{aligned}$$

3 Applications to Orlicz function spaces

Note that in the case when $\omega (t)=1$ for any $t\in (0,\gamma )$ the corresponding Orlicz–Lorentz function spaces become the well-known Orlicz function spaces $L^{\varphi }=L^{\varphi }([0,\gamma ))$. On the basis of the results presented in [3] and the foregoing part of the present paper we can easily deduce criteria for non-squareness properties of Orlicz function spaces $L^{\varphi }$.

In the case when $\gamma =\infty$, by Theorem 2.1 from [3] and Theorem 2, we get the following

Corollary 1

([7], Theorem 2) Let $\gamma =\infty$. Then the following conditions are equivalent:

• 1. $\varphi \in \varDelta _2({\mathbb {R}})$ and $\varphi (\frac{u}{2})<\frac{1}{2}\varphi (u)$ for any $u>0$,

• 2. the Orlicz space $L^{\varphi }$ is non-square,

• 3. the Orlicz space $L^{\varphi }$ is locally uniformly non-square.

In turn, by Theorem 2.3 from [3], we have

Corollary 2

([6], Theorem 1.2.(i)) Let $\gamma =\infty$. Then the Orlicz space $L^{\varphi }$ is uniformly non-square if and only if $\varphi \in \varDelta _2({\mathbb {R}})$ and $\psi \in \varDelta _2({\mathbb {R}})$, where $\psi$ denotes the complementary function of $\varphi$ in the sense of Young (see formula (3)).

Let now $\gamma <\infty$ and let $\beta$ be defined by formula (1) (in the paper [3] the same constant was denoted by $\delta$). Then by Theorems 2.2 and 2.4 from [3] and Theorems 3 and 4 , we obtain

Corollary 3

Let $\gamma <\infty$. Then the following assertions are true:

• 1. ([7], Theorem 3) The Orlicz space $L^{\varphi }$ is non-square if and only if $\varphi \in \varDelta _2(\infty )$ and $\varphi (\beta )\cdot \gamma <2$.

• 2. The Orlicz space $L^{\varphi }$ is locally uniformly non-square if and only if $\varphi \in \varDelta _2(\infty )$ and $(\varphi (\beta )\cdot \gamma <1$ or $\varphi (\beta )\cdot \gamma \in [1,2)$ and $\psi \in \varDelta _2(\infty ))$.

• 3. [6], Theorem 1.2.(ii)) The Orlicz space $L^{\varphi }$ is uniformly non-square if and only if $\varphi \in \varDelta _2(\infty )$, $\psi \in \varDelta _2(\infty )$ and $\varphi (\beta )\cdot \gamma <2$.

Therefore, if $\varphi (\beta )\cdot \gamma <1$, then the Orlicz space $L^{\varphi }$ is locally uniformly non-square if and only if it is non-square, while if $\varphi (\beta )\cdot \gamma \in [1,2)$, then the Orlicz space $L^{\varphi }$ is locally uniformly non-square if and only if it is uniformly non-square.