Kadec–Klee properties of Orlicz–Lorentz sequence spaces equipped with the Orlicz norm

Abstract

The necessary and sufficient conditions for both the Kadec–Klee property as well as the Kadec–Klee property with respect to the coordinatewise convergence in Orlicz–Lorentz sequence spaces equipped with the Orlicz norm and generated by arbitrary Orlicz functions as well as any non-increasing weight sequences are given. Moreover, for their subspaces of elements with an order continuous norm the full characterization of the Kadec–Klee property with respect to the coordinatewise convergence is presented. Some tools useful in the proofs of the main results are also provided.

1 Introduction

Throughout this paper $\mathbb {R}$, $\mathbb {R}_+$ and $\mathbb {N}$ denote the sets of reals, nonnegative reals and natural numbers, respectively. By S(X) and B(X) denote the unit sphere and the unit ball of a Banach space $X=(X,\Vert \cdot \Vert )$, respectively.

The triple $(\mathbb {N}, 2^\mathbb {N},m)$ stands for the counting measure space while $\ell ^0=\ell ^0(m)$ denotes the space of all real sequences $x:\mathbb {N}\rightarrow \mathbb {R}$. For every $x=(x(i))_{i=1}^\infty \in \ell ^0$, let ${\text {supp}}(x):=\{i\in \mathbb {N}:x(i)\ne 0\}$ and $|x|(i):=|x(i)|$ for all $i\in \mathbb {N}$, that is, |x| denotes the absolute value of x.

Given any $x\in \ell ^0$ we define its distribution function $\mu _{x}:[0,+\infty )\rightarrow \{0,\infty \}\cup \mathbb {N}$ by

$$\begin{aligned} \mu _{x}(\lambda )=m(\{i\in \mathbb {N}:|x(i)|>\lambda \}), \end{aligned}$$

where $\lambda \ge 0$, m is the counting measure on $2^\mathbb {N}$ (see [2, 21, 23]) and its non-increasing rearrangement $x^{*}=(x^*(i))_{i=1}^\infty$ as

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

for all $i\in \mathbb {N}$ (under the convention $\inf \emptyset =\infty$). We say that two sequences $x,y\in \ell ^{0}$ are equimeasurable if $\mu _{x}(\lambda )=\mu _{y}(\lambda )$ for all $\lambda \ge 0$. It is obvious that equimeasurability of x and y gives the equality $x^{*}=y^{*}$.

In the whole paper $\Phi$ denotes an Orlicz function (see [4, 25, 28]), that is, $\Phi :[-\infty ,\infty ]\rightarrow [0,\infty ]$ (our definition is extended from R into $R^{e}:=[-\infty ,\infty ]$ by assuming $\Phi (-\infty )=\Phi (\infty )=\infty$) and $\Phi$ 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

$$\begin{aligned} a_{\Phi }= & {} \sup \{u\ge 0:\Phi (u)=0\},\\ b_{\Phi }= & {} \sup \{u\ge 0:\Phi (u)<\infty \}. \end{aligned}$$

Note that the left continuity of $\Phi$ on $(0,\infty )$ is equivalent to the fact that $\lim _{u\rightarrow (b_{\Phi })^{-}}\Phi (u)=\Phi (b_{\Phi })$.

We say that an Orlicz function $\Phi$ satisfies condition $\delta _2$ ($\Phi \in \delta _2$ for short) if there exist constants $u_0>0$ with $\Phi (u_0)>0$ and $K>0$ such that $\Phi (2u)\le K\Phi (u)$ for any $u\in [0,u_0]$.

For any Orlicz function $\Phi$ we define its complementary (in the sense of Young) function $\Psi$ by the formula

$$\begin{aligned} \Psi (u)=\sup \nolimits _{v>0}\{|u|v-\Phi (v)\} \end{aligned}$$

for all $u\in \mathbb {R}$.

Let $\omega :\mathbb {N}\rightarrow \mathbb {R}_{+}$ be a non-increasing sequence, called a weight sequence. In the whole paper we will assume that $\omega \left( 1\right) >0$.

Recall that Orlicz–Lorentz spaces, which are a natural generalization of Orlicz and Lorentz spaces, were introduced in the early nineties of the twentieth century [18, 24, 27]. These investigations gave an impulse to further study of the spaces, results of which have been published among others in papers [1, 3, 7, 10,11,12,13,14,15, 19, 22, 33].

Let us define a modular $I_{\Phi ,\omega }:\ell ^0\rightarrow [0,\infty ]$ by formula

$$\begin{aligned} I_{\Phi ,\omega }(x)=\overset{\infty }{\underset{i=1}{\sum }}\Phi (x^*(i))\omega (i) \end{aligned}$$

and the Orlicz–Lorentz sequence space

$$\begin{aligned} \lambda _{\Phi ,\omega }=\{x\in \ell ^0:I_{\Phi ,\omega }(\beta x)<\infty \text { for some }\beta >0\}. \end{aligned}$$

In most of the previous papers one studied Orlicz–Lorentz spaces equipped with the Luxemburg norm

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

It is easy to show that if $\sum \nolimits _{i=1}^{\infty }\omega \left( i\right) <\infty$ or $a_{\varphi }>0$, then $\lambda _{\Phi ,\omega }=l^{\infty }$ with equivalent norms. Otherwise, we have $\lambda _{\Phi ,\omega }\hookrightarrow c_{0}$.

Several years ago the first papers considering different properties in Orlicz–Lorentz spaces endowed with the Orlicz norm defined as

$$\begin{aligned} \Vert x\Vert _{\Phi ,\omega }^O=\sup \left\{ \underset{n=1}{\overset{\infty }{\sum }}x^*(i)y^*(i)\omega (i):I_{\Psi ,\omega }(y)\le 1\right\} \end{aligned}$$

appeared (see [15, 33] and also [7]). In [8] it was shown that for arbitrary Orlicz function $\Phi$ and any non-increasing weight function $\omega$, equality

$$\begin{aligned} \Vert x\Vert _{\Phi ,\omega }^O=\Vert x\Vert _{\Phi ,\omega }^A \end{aligned}$$

holds for any $x\in \lambda _{\Phi ,\omega }$, where the Amemiya norm is defined by the formula

$$\begin{aligned} \Vert x\Vert _{\Phi ,\omega }^A=\inf \limits _{k>0}\frac{1}{k}\left\{ 1+I_{\Phi ,\omega }(kx)\right\} . \end{aligned}$$

(1)

Since $\Vert x\Vert _{\Phi ,\omega }=\inf \nolimits _{k>0}\{\frac{1}{k}\max (1,I_{\Phi ,\omega }(kx))\}$ for any $x\in \lambda _{\Phi ,\omega }$ (see [8]), the both norms, that is, the Luxemburg and the Orlicz norms are equivalent. More precisely, for any $x\in \lambda _{\Phi ,\omega }$ we have $\Vert x\Vert _{\Phi ,\omega }\le \Vert x\Vert _{\Phi ,\omega }^O\le 2\Vert x\Vert _{\Phi ,\omega }$. It is well known that both $(\lambda _{\Phi ,\omega },\Vert \cdot \Vert _{\Phi ,\omega })$ and $(\lambda _{\Phi ,\omega },\Vert \cdot \Vert _{\Phi ,\omega }^O)$ are symmetric Banach spaces (see [8, 18, 24]).

A natural question about the attainability by any $x\in \lambda _{\Phi ,\omega }\backslash \{0\}$ of the infimum in formula (1) arises, that is, whether there is $k=k(x)>0$ for which equality

$$\begin{aligned} \Vert x\Vert _{\Phi ,\omega }^O=\Vert x\Vert _{\Phi ,\omega }^A=\frac{1}{k}\left\{ 1+I_{\Phi ,\omega }(kx)\right\} \end{aligned}$$

holds. In order to answer this question, let us define for any $x\in \lambda _{\Phi ,\omega }\backslash \{0\}$ the following constants

$$\begin{aligned}&\lambda _{\infty }=\lambda _{\infty }(x)=\sup \{\lambda>0:I_{\Phi ,\omega }(\lambda x)<\infty \},\\&k^*=k^*(x)=\inf \{k>0:I_{\Psi ,\omega }(p(kx^{*}))\ge 1\},\\&k^{**}=k^{**}(x)=\sup \{k>0:I_{\Psi ,\omega }(p(kx^{*}))\le 1\}, \end{aligned}$$

where p is the right-hand side derivative of $\Phi$. Define the function f on $(0,\infty )$,

$$\begin{aligned} f(k)=f_{x}(k)=\frac{1}{k}\left( 1+I_{\Phi ,\omega }(kx)\right) . \end{aligned}$$

(2)

The function f is continuous on the interval $(0,\lambda _\infty )$ and left-continuous at $\lambda _\infty$ provided that $\lambda _\infty <\infty$. We also have $0<k^*\le k^{**}\le \lambda _\infty \le \infty$. Moreover, as it was shown in [8], the function f is strictly decreasing on the interval $(0,k^*)$, strictly increasing on the interval $(k^{**},\lambda _\infty )$ if $k^{**}<\lambda _\infty$ as well as constant and equal to $\Vert x\Vert _{\Phi ,\omega }^O$ on the interval $[k^*,k^{**}]$ if $k^*<k^{**}$. Therefore, assuming for any $x\in \lambda _{\Phi ,\omega }\backslash \{0\}$ that

$$\begin{aligned} K(x)=\left\{ \begin{array}{ll} \left[ k^*,k^{**}\right] &{} \text { if } k^{**}<\infty \\ \left[ k^*,\infty \right) &{} \text { if } k^*<k^{**}=\infty \\ \emptyset &{} \text { if }k^*=\infty , \end{array} \right. \end{aligned}$$

(3)

we get that if $k^*<\infty$, then $\Vert x\Vert _{\Phi ,\omega }^O=\frac{1}{k}\left\{ 1+I_{\Phi ,\omega }(kx)\right\}$ for any $k\in K(x)$ and $\Vert x\Vert _{\Phi ,\omega }^O=\lim \nolimits _{k\rightarrow \infty }\frac{1}{k}\left\{ 1+I_{\Phi ,\omega }(kx)\right\}$ if $k^*=\infty$ (see [8]).

In the further part of this paper we will need to use the following

Remark 1

Note that if $b_\Phi <\infty$, then for any $x\in \lambda _{\Phi ,\omega }\backslash \{0\}$, we have that $k^{**}\le \lambda _\infty \le \frac{b_\Phi }{x^*(1)}$. In the case of $b_\Phi =\infty$ and $\lim \nolimits _{u\rightarrow \infty }\frac{\Phi (u)}{u}=\lim \nolimits _{u\rightarrow \infty }\frac{1}{u}\int \nolimits _{0}^{u}p(s)ds=\lim \nolimits _{u\rightarrow \infty }p(u)=\infty$, we have also that $k^{**}<\infty$ for any $x\in \lambda _{\Phi ,\omega }\backslash \{0\}$. Indeed, if $\lambda _\infty =\infty$, then there exists $k_{x}>0$ such that $\Psi (p(k_{x}x^*(1)))\omega (1)>1$ and $k^{**}\le k_{x}$.

Now, assume that $\lim \nolimits _{u\rightarrow \infty }\frac{\Phi (u)}{u}=B <\infty$ and let $x\in \lambda _{\Phi ,\omega }\backslash \{0\}$. First, we will show that if $\Psi (B){\sum _{i=1}^{{m({\text {supp }}x)}}}\omega (i)>1$, then $k^{**}<\infty$. In order to do it, let $n_x\in \mathbb {N}$ be such that

$$\begin{aligned} \Psi (B)\underset{i=1}{\overset{n_x}{\sum }}\omega (i)>1. \end{aligned}$$

(4)

In the case when $m({\text {supp }}x)<\infty$, we take $n_x=m({\text {supp }}x)$ and in the case when $m({\text {supp }}x)=\infty$, we can find $n_x \in \mathbb {N}$, for which inequality (4) holds. Since $\lim \nolimits _{u\rightarrow \infty }p(u)=B$, there exists $u_x>0$ for which $\Psi (p(u_x))\sum _{i=1}^{n_x}\omega (i)>1$. Defining $k_x=\frac{u_x}{x^*(n_x)}$, we obtain that $I_{\Psi ,\omega }(p(k_{x}x^{*}))>1$, so $k^{**}\le k_x$.

In the case of $\Psi (B)\sum \nolimits _{i=1}^{m({\text {supp }}x)}\omega (i)<1$ or $\Psi (B)\sum \nolimits _{i=1}^{m({\text {supp }}x)}\omega (i)=1$ and $p(u)<B$ for any $u>0$, we get that $k^*=\infty$, which means $K(x)=\emptyset$.

Finally, assume that $\Psi (B)\sum _{i=1}^{m({\text {supp }}x)}\omega (i)=1$ and $p(u)=B$ starting from some $u_0>0$. If $m({\text {supp }}x)<\infty$, then $k^*\le \frac{u_0}{x^*(n_x)}<k^{**}=\infty$, where as above $n_{x}=m({\text {supp }}x)$. Let now $m({\text {supp }}x)=\infty$. If $i_{0}=\sup \{i\in \mathbb {N}:\omega (i)>0\}<\infty$, then $k^*\le \frac{u_0}{x^*(i_{0})}<k^{**}=\infty$. However, if $\omega (i)>0$ for any $i\in \mathbb {N}$, then $k^*\le \frac{u_0}{x^*(\infty )}<k^{**}=\infty$, whenever $x^*(\infty )=\mathop {\lim }\nolimits _{i\rightarrow \infty } x^*(i)>0$, and $k^*=\infty$ otherwise.

Let $(E,\le ,\Vert \cdot \Vert _E)$ be a Banach sequence lattice. An element $x\in E$ is said to be order continuous if for any sequence $(x_n)$ in $E_+$ (the positive cone of E) such that $x_n\le |x|$ for any $n\in \mathbb {N}$ and $x_n\rightarrow 0$ coordinatewisely, there holds $\Vert x_n\Vert _E\rightarrow 0$ as $n\rightarrow \infty$. The subspace $E_a$ of all order continuous elements in E is an order ideal in E. Space E is called order continuous if $E_a=E$.

Since the Luxemburg and the Orlicz norms are equivalent, the subspace of order continuous elements for both these norms is the same (as a subset of the elements of the space $\lambda _{\Phi ,\omega }$). We will denote it as $(\lambda _{\Phi ,\omega })_a$. The following theorem holds true:

Theorem 1

The space $(\lambda _{\Phi ,\omega },\Vert \cdot \Vert _{\Phi ,\omega }^{O})$ is order continuous if and only if $\sum \limits _{i=1}^{\infty }\omega (i)$ $=\infty$ and $\Phi \in \delta _2$.

This theorem is the consequence of the fact that the space $(\lambda _{\Phi ,\omega },\Vert \cdot \Vert _{\Phi ,\omega })$ (with the Luxemburg norm) is order continuous if and only if $\sum _{i=1}^{\infty }\omega (i)=\infty$ and $\Phi \in \delta _2$ (cf. [18, Theorem 2.4]) as well as the equivalence of the Luxemburg and the Orlicz norms.

A Banach space $X=(X,\Vert \cdot \Vert )$ is said to have the Kadec–Klee property or property H (resp. the Kadec–Klee property with respect to the coordinatewise convergence or property $H_c$) if for any sequence $(x_n)_{n=1}^\infty$ in X and $x\in X$ such that $\lim _{n\rightarrow \infty }\Vert x_n\Vert =\Vert x\Vert$, we have $\Vert x_n-x\Vert \rightarrow 0$ provided $x_n\rightarrow x$ weakly as $n\rightarrow \infty$ (resp. provided $x_n(i)\rightarrow x(i)$ coordinatewisely). Note that this property was originally considered by Radon [29] and next by Riesz [30, 31], where it was proved that $L^p$ spaces $(1<p<\infty )$ have property H, while $L^1 [0,1]$ has not. The newer results concerning the Kadec–Klee properties a reader can find for example in [3, 5, 6, 9, 10, 17, 32].

2 Auxiliary results

Remembering Remark 1, we can formulate the following

Lemma 1

Let $\lim \nolimits _{u\rightarrow \infty }\frac{\Phi (u)}{u}=B<\infty$ and $x\in \lambda _{\Phi ,\omega }$. If $\Psi (B)\sum _{i=1}^{m({\text {supp }}x)}\omega (i)\le 1$ (equivalently $k^{**}=\infty )$, then

$$\begin{aligned} \Vert x\Vert _{\Phi ,\omega }^{O}=B\mathop {\sum }\limits _{i=1}^{\infty } x^*(i)\omega (i). \end{aligned}$$

Moreover, if $\Psi (B)\sum _{i=1}^{\infty }\omega (i)> 1$, then $m({\text {supp }}x)<\infty$.

Proof

Note that $\Psi (B)<\infty$ and $\Psi (u)=\infty$ for any $u>B$. Hence, for any $y\in \lambda _{\Psi ,\omega }$ such that $I_{\Psi ,\omega }(y)\le 1$, we get that $y(i)\le B$ for any $i\in \mathbb {N}$. Therefore, for the same y, we obtain

$$\begin{aligned} \mathop {\sum }\limits _{i=1}^{\infty }x^*(i)y^*(i)\omega (i)\le B \mathop {\sum }\limits _{i=1}^{\infty }x^*(i)\omega (i). \end{aligned}$$

Defining

$$\begin{aligned} z=(\underset{m({\text {supp }}x)\text { times}}{\underbrace{B,B,\ldots ,B}},0,0,\ldots ), \end{aligned}$$

we get that $I_{\Psi ,\omega }(z)=\sum _{i=1}^{m({\text {supp }}x)}\Psi (B)\omega (i)\le 1$ and

$$\begin{aligned} \underset{i=1}{\overset{m({\text {supp }}x)}{\sum }}x^*(i)z^*(i)\omega (i)=B\underset{i=1}{\overset{m({\text {supp }}x)}{\sum }}x^*(i)\omega (i)=B\mathop {\sum }\limits _{i=1}^{\infty }x^*(i)\omega (i). \end{aligned}$$

Hence it follows that

$$\begin{aligned} \Vert x\Vert _{\Phi ,\omega }^{O}=B\mathop {\sum }\limits _{i=1}^{\infty } x^*(i)\omega (i). \end{aligned}$$

Assume now additionally that $\Psi (B)\sum _{i=1}^{\infty }\omega (i)>1$. Then, $a_\Psi <b_\Psi =B$ and $0<\Psi (B)<\infty$. Hence, there exists $n_0\in \mathbb {N}$ such that $\Psi (B)\sum _{i=1}^{n_0}\omega (i)>1$, whence we get that $m({\text {supp }}x)<n_0$. $\square$

Let us note the following observation. It is known that if a sequence $(x_n)_{n=1}^\infty$ from $c_0$ is uniformly convergent to x, then $x^*_n(i)\rightarrow x^*(i)$ for all $i\in \mathbb {N}$. Consequently, the sequence $(x_n^*)_{n=1}^\infty$ is uniformly convergent to $x^*$. However, the above implication is not true in general, if the assumption about uniform convergence of $(x_n)_{n=1}^\infty$ to x is replaced by the assumption that $(x_n)_{n=1}^\infty$ is convergent to x coordinatewisely. This fact can be easily visualized by the following example. Let us take the sequence $(x_n)_{n=1}^\infty$ such that $x_n(n)=1$ and $x_n(i)=0$ for $i\ne n$, where $n\in \mathbb {N}$. Then $(x_n)_{n=1}^\infty$ converges to zero coordinatewisely but $x_n^*(1)$ does not converge to zero. However, Lemma 2 holds. This Lemma is a kind of generalization of Lemma 3.1 from [10]. We will present its proof below just for the completeness of this presentation as well as for the convenience of a reader.

Lemma 2

Let $\Phi$ vanishes only at zero and $\omega (i)>0$ for any $i\in \mathbb {N}$. Then, for any $x\in c_0$ and any sequence $(x_n)_{n=1}^\infty$ from $c_0$ such that $x_n\rightarrow x$ coordinatewisely and $I_{\Phi ,\omega }(x_n)\rightarrow I_{\Phi ,\omega }(x)=\alpha <\infty$ as $n\rightarrow \infty$, we obtain that $x^*_n(i)\rightarrow x^*(i)$ for any $i\in \mathbb {N}$. Consequently, $x_n^*\rightarrow x^*$ uniformly.

Proof

Assume that x and a sequence $(x_n)_{n=1}^\infty$ satisfy the assumptions of the lemma and that, passing to a subsequence if necessary, there exists $i_0\in \mathbb {N}$ such that $|x_n^*(i_0)-x^*(i_0)|>\varepsilon$ for some $\varepsilon >0$ and all $n\in \mathbb {N}$. We will consider now three cases (passing to a subsequence if necessary).

Case 1. Assume that $0\le x_n^*(i_0)<x^*(i_0)$ for each $n\in \mathbb {N}$. Then

$$\begin{aligned} x^*(i_0)>x_n^*(i_0)+\varepsilon \end{aligned}$$

(5)

for the same n. Note that the set $A=\{i\in \mathbb {N}:|x(i)|>\frac{\varepsilon }{2}\}$ is finite because $x\in c_0$. Let us denote $m_0=m(A)$, $y=x\chi _{A}$ and $y_n=x_n\chi _A$ for each $n\in \mathbb {N}$. Since $y_n\rightarrow y$ uniformly, we get that $y^*_n$ tends to $y^*$ coordinatewisely (so uniformly as well). Simultaneously, $x^*(i)=y^*(i)$ for $i=1,2,\ldots ,m_0$ and $i_0\le m_0$. Then $x^*(i_0)=y^*(i_0)$. Consequently, we can find $n_0\in \mathbb {N}$ such that $|y_n^*(i_0)-x^*(i_0)|=|y_n^*(i_0)-y^*(i_0)|\le \frac{\varepsilon }{2}$ for $n\ge n_0$, whence, by (5), there holds that $y_n^*(i_0)\ge x_n^*(i_0)+\frac{\varepsilon }{2}$ for $n\ge n_0$. On the other hand, $y_n\le x_n$ for any $n\in \mathbb {N}$, whence $y_n^*\le x_n^*$ for any $n\in \mathbb {N}$, which arrive us to a contradiction.

Case 2. Assume that $0<x^*(i_0)<x^*_n(i_0)$ for all $n\in \mathbb {N}$ and let us define $u=x^*(i_0)+\varepsilon$, $v=x^*(i_0)+\frac{\varepsilon }{2}$ and

$$\begin{aligned} \delta =\Phi (u)\omega (i_0)-\Phi (v)\omega (i_0). \end{aligned}$$

(6)

Then, there exists $m_0\in \mathbb {N}$, $m_0\ge i_0$ such that $\sum _{i=1}^{m_0}\Phi (x^*(i))\omega (i)\ge \alpha -\frac{\delta }{4}$ and $x^*(m_0+1)<x^*(m_0)$. Define the set $B=\{i\in \mathbb {N}:|x(i)|\ge x^*(m_0)\}$, $z=x\chi _B$ and $z_n=x_n\chi _B$ for any $n\in \mathbb {N}$. We get that $m(B)=m_0<\infty$, so $z^*(i)=x^*(i)$ for $i=1,2,\ldots ,m_0$ and $z_n\rightarrow z$ uniformly, whence $z_n^*\rightarrow z^*$ coordinatewisely and, consequently, $\Phi \circ z_n^*\rightarrow \Phi \circ z^*$ coordinatewisely. Therefore, there exists $n_0\in \mathbb {N}$ such that

$$\begin{aligned} \underset{i=1}{\overset{m_0}{\sum }}\Phi (z_n^*(i))\omega (i)\ge \alpha -\frac{\delta }{2} \end{aligned}$$

(7)

for any $n\ge n_0$. Without loss of generality, we can also assume that

$$\begin{aligned} z_n^*(i_0)<z^*(i_0)+\frac{\varepsilon }{2}= x^*(i_0)+\frac{\varepsilon }{2} \end{aligned}$$

(8)

for $n\ge n_0$. Since $z_n\le x_n$, we get that $z_n^*\le x_n^*$ and, consequently, applying (6) and (8), we get

$$\begin{aligned} I_{\Phi ,\omega }(x_{n})\ge & {} \underset{i=1}{\overset{i_0-1}{\sum }}\Phi (x_n^*(i))\omega (i)+\Phi (u)\omega (i_0)+\underset{i=i_0+1}{\overset{m_0}{\sum }}\Phi (x_n^*(i))\omega (i)\\= & {} \underset{i=1}{\overset{i_0-1}{\sum }}\Phi (x_n^*(i))\omega (i)+\Phi (v)\omega (i_0)+\delta +\underset{i=i_0+1}{\overset{m_0}{\sum }}\Phi (x_n^*(i))\omega (i)\\\ge & {} \underset{i=1}{\overset{m_0}{\sum }}\Phi (z_n^*(i))\omega (i)+\delta \ge \alpha +\frac{\delta }{2} \end{aligned}$$

for $n\ge n_{0}$, where the last inequality comes from (7), a contradiction.

Case 3. Assume that $0=x^*(i_0)<x_n^*(i_0)$ for any $n\in \mathbb {N}$. Let $C=\{i\in \mathbb {N}:|x(i)|>0\}$. Then $m(C)=m_0<i_0$. Taking $u=\varepsilon$, $v=0$ and $\delta =\sum _{i=m_0+1}^{i_0} \Phi (u)\omega (i)$, similarly as before, we will get a contradiction. $\square$

Remark 2

In Lemma 2 (also in Lemma 3) the assumptions that $\Phi$ vanishes only at zero and $\omega (i)>0$ for any $i\in \mathbb {N}$ can not be omitted. Indeed, assume firstly that $a_\Phi >0$ and define $x=a_\Phi e_1$ and $x_n=a_\Phi e_1+a_\Phi e_n$ for $n\ge 2$. Then $I_{\Phi ,\omega }(x)=I_{\Phi ,\omega }(x_n)=0$ for $n\ge 2$, $x_n\rightarrow x$ coordinatewisely and $x_n^*(2)=a_\Phi >0=x^*(2)$ for $n\ge 2$.

Now, assume that $a_\Phi =0$ and there exists $j\in \mathbb {N}$ such that $w(j)>w(j+1)=0$. Let $u>0$ be such that $\Phi (u)<\infty$ and let us define $x=\sum _{i=1}^{j} u e_i$ and $x_n=\sum _{i=1}^{j} u e_i + u e_n$ for $n\ge j+1$. Then $I_{\Phi ,\omega }(x)=I_{\Phi ,\omega }(x_n)=\Phi (u)\sum _{i=1}^{j}\omega (i)$ for $n\ge j+1$, $x_n\rightarrow x$ coordinatewisely and $x_n^*(j+1)=u>0=x^*(j+1)$ for $n\ge j+1$.

Next lemma can be proved analogously as Lemma 3.2. from [10].

Lemma 3

Let $\Phi$ vanishes only at zero and $\omega (i)>0$ for any $i\in \mathbb {N}$. Then $(x_n-x)^*\rightarrow 0$ uniformly as $n\rightarrow \infty$ for any $x\in c_0$ and any sequence $(x_n)_{n=1}^\infty$ from $c_0$ such that $x_n\rightarrow x$ coordinatewisely and $I_{\Phi ,\omega }(x_n)\rightarrow I_{\Phi ,\omega }(x)=\alpha <\infty$ as $n\rightarrow \infty$.

Remark 3

Note that in Lemma 3 assumptions $x\in c_0$ and $x_n\in c_0$ for any $\mathbb {N}$ are essential. Indeed, for any fixed $u>0$ let us define $x(i)=u$ for $i\in \mathbb {N}$, $x_n(n)=0$ and $x_n(i)=u$ for $n\in \mathbb {N}$ and $i\ne n$. Obviously, $x\in \ell ^\infty \backslash c_0$, $x_n\in \ell ^\infty \backslash c_0$ for $n\in \mathbb {N}$ and $x_n\rightarrow x$ coordinatewisely. We also have that $x^*(i)=u=x_n^*(i)$ for $i,n\in \mathbb {N}$. Therefore, $x^*_n\rightarrow x^*$ coordinatewisely. What is more, if $\Phi (u)<\infty$ and $\sum \nolimits _{i=1}^{\infty }\omega (i)<\infty$, we get that $I_{\Phi ,\omega }(x)=I_{\Phi ,\omega }(x_n)<\infty$ for $n\in \mathbb {N}$. At the same time $(x_n-x)^*(1)=u$ for $n\in \mathbb {N}$.

The proof of the lemma below can be found in [8].

Lemma 4

Let $(x_n)$ be a sequence of elements of the space $\lambda _{\Phi ,\omega }$. Then the following properties hold.

• (1) If $\lim \nolimits _{n\rightarrow \infty }\Vert x_n\Vert _{\Phi ,\omega }^O=0$, then $\lim \nolimits _{n\rightarrow \infty }I_{\Phi ,\omega }(x_n)=0$.

• (2) Assume that $\sum _{i=1}^{\infty }\omega (i)=\infty$. Then the implication that if $\lim \nolimits _{n\rightarrow \infty }I_{\Phi ,\omega }(x_n)=0$, then $\lim \nolimits _{n\rightarrow \infty }\Vert x_n\Vert _{\Phi ,\omega }^O=0$ holds true if and only if $\Phi \in \delta _2$.

• (3) If $\sum _{i=1}^{\infty }\omega (i)<\infty$, then the implication that if $\lim \nolimits _{n\rightarrow \infty }I_{\Phi ,\omega }(x_n)=0$, then $\lim \nolimits _{n\rightarrow \infty }\Vert x_n\Vert _{\Phi ,\omega }^O=0$ holds true if and only if $a_\Phi =0$.

3 Main results

Before we present the main results of our paper, we will recall what so far has been known about the property of the Kadec–Klee type in Orlicz, Lorentz and Orlicz–Lorentz spaces. In the case of Orlicz spaces the suitable criteria for both the Orlicz as well as the Luxemburg norms was given in [4, 9, 26]. Note that in the case of non-atomic measure in place of the Kadec–Klee property with respect to the coordinatewise convergence we consider the Kadec–Klee property for the local convergence in measure. Analogous results for Lorentz spaces can be found in [5]. Whereas, in the case of Orlicz–Lorentz spaces the research connected to the properties of the Kadec–Klee type were led only for the Luxemburg norm (see [3, 16, 17] and also [10]). In particular, we get that the Orlicz–Lorentz sequence space $\lambda _{\Phi ,\omega }$ equipped with the Luxemburg norm $\Vert \cdot \Vert _{\Phi ,\omega }$ has the Kadec–Klee property with respect to the coordinatewise convergence if and only if $\Phi \in \delta _2$, $\sum _{i=1}^{\infty }\omega (i)=\infty$ and $\Phi (b_\Phi )\omega (1)\ge 1$.

Theorem 2

The following conditions are equivalent:

• (1) The function $\Phi$ satisfies condition $\delta _2$ and $\sum _{i=1}^{\infty }\omega (i)=\infty$.

• (2) The Orlicz–Lorentz sequence space $(\lambda _{\Phi ,\omega },\Vert \cdot \Vert _{\Phi ,\omega }^O)$ has the Kadec–Klee property with respect to the coordinatewise convergence.

• (3) The Orlicz–Lorentz sequence space $(\lambda _{\Phi ,\omega },\Vert \cdot \Vert _{\Phi ,\omega }^O)$ has the Kadec–Klee property.

Proof

It is obvious that (2) implies (3). Since the Kadec–Klee property implies an order continuity of the space $(\lambda _{\Phi ,\omega },\Vert .\Vert _{\Phi ,\omega }^O)$ (see [9, Proposition 2.1]) and the last means that $\Phi \in \delta _2$ and $\sum _{i=1}^{\infty }\omega (i)=\infty$ (see Theorem 1), we obtain that (3) implies (1). Now, we will show (1) implies (2).

Case 1. At the beginning let us assume that $b_{\Phi }<\infty$ or $b_{\Phi }=\infty$ and $\lim _{u\rightarrow \infty }\frac{\Phi (u)}{u}=\infty$. Let $x\in S(\lambda _{\Phi ,\omega },\Vert .\Vert _{\Phi ,\omega }^O)$, $x_n\in S(\lambda _{\Phi ,\omega },\Vert .\Vert _{\Phi ,\omega }^O)$ for $n\in \mathbb {N}$ and $x_n\rightarrow x$ coordinatewisely. By the assumptions of $\Phi$ (see Remark 1), for any $n\in \mathbb {N}$ we can find $k_n>1$ such that $1=\Vert x_n\Vert _{\Phi ,\omega }^{O}=\frac{1}{k_n}\left( 1+I_{\Phi ,\omega }(k_nx_n)\right)$. First, we will show the boundedness of the sequence $(k_n)_{n=1}^\infty$. Define

$$\begin{aligned} m_1=\max \{i\in \mathbb {N}:x^*(i)=x^*(1)\}. \end{aligned}$$

(9)

By $\mathop {\sum }\nolimits _{i=1}^{\infty }\omega (i)=\infty$, we have $m_1<\infty$ and $x^*(m_1+1)<x^*(1)$. By virtue of the rearrangement definition, there exist natural numbers $i_1,i_2,\ldots ,i_{m_1}$ such that

$$\begin{aligned} |x(i_1)|=|x(i_2)|=\cdots =|x(i_{m_1})|=x^*(1) \end{aligned}$$

and $|x(i)|\le x^*(m_1+1)$ for $i\in \mathbb {N}\backslash \{i_1,i_2,\ldots ,i_{m_1}\}$. Since $x_n\rightarrow x$ coordinatewisely, there exists $n_0\in \mathbb {N}$ such that

$$\begin{aligned} x_n(i_j)\ge b:=\frac{x^*(1)+x^*(m_1+1)}{2} \end{aligned}$$

for $j=1,2,\ldots ,m_1$ and $n\ge n_0$. Hence $x_n^*(i)\ge b$ for $i=1,2,\ldots ,m_1$ and $n\ge n_0$. Therefore, if $b_{\Phi }<\infty$, then we have $k_{n}\le b_{\Phi }/b$ for any $n\ge n_{0}$. Let now $b_{\Phi }=\infty$ and $\lim _{u\rightarrow \infty }\frac{\Phi (u)}{u}=\infty$. If $\lim \nolimits _{n\rightarrow \infty }k_n=\infty$ (passing to a subsequence if necessary), then by the assumption that $\lim \nolimits _{u\rightarrow \infty }\frac{\Phi (u)}{u}=\infty$, we obtain

$$\begin{aligned} 1=\frac{1}{k_n}\left( 1+I_{\Phi ,\omega }(k_nx_n)\right) \ge \frac{1}{k_n}\underset{i=1}{\overset{m_1}{\sum }}\Phi (k_nx_n^*(i))\omega (i)\ge b\frac{\Phi (b k_n )}{bk_n}\underset{i=1}{\overset{m_1}{\sum }}\omega (i)>1 \end{aligned}$$

beginning from some natural $n_{1}\ge n_{0}$, a contradiction. So $(k_n)_{n=1}^{\infty }$ is bounded.

Now, we will show that any cluster point $k_{0}$ of the sequence $(k_n)_{n=1}^{\infty }$ belongs to K(x), that is,

$$\begin{aligned} \frac{1}{k_0}\left( 1+I_{\Phi ,\omega }(k_0x)\right) =\Vert x\Vert _{\Phi ,\omega }^{O}=1. \end{aligned}$$

(10)

From the properties of the Orlicz norm

$$\begin{aligned} \frac{1}{k_0}\left( 1+I_{\Phi ,\omega }(k_0x)\right) \ge 1. \end{aligned}$$

(11)

In order to get the opposite inequality, let us define the sequence $(m_l)$ in the following way: $m_1$ is given by formula (9) and

$$\begin{aligned} m_l=\max \{i\in \mathbb {N}:x^*(i)=x^*(m_{l-1}+1)\} \end{aligned}$$

for $l=2,3,4,\ldots$. Note that sequence $(m_l)$ is increasing. If $m({\text {supp}} x)=\infty$, then $\lim \nolimits _{l\rightarrow \infty } m_l=\infty$. In the case of $m({\text {supp}}x)=p<\infty$, we can find $l_x\le p$ such that $m_{l_x}=p$. For any l the set $N_l=\{i\in \mathbb {N}:|x(i)|\ge x^*(m_l)\}$ is finite, more precisely, $m(N_l)=m_l$. Let $\lim _{s\rightarrow \infty }k_{n(s)}=k_{0}$. Since $k_{n(s)}x_{n(s)}\rightarrow k_{0}x$ coordinatewisely, then $k_{n(s)}x_{n(s)}\chi _{N_l}\rightarrow k_{0}x\chi _{N_l}$ uniformly for any fixed l, whence $k_{n(s)}x_{n(s)}^*\chi _{N_l}\rightarrow k_{0}x^*\chi _{N_l}$ coordinatewisely for any fixed l and, consequently,

$$\begin{aligned} \underset{i=1}{\overset{m_{l}}{{\sum }}}\Phi (k_{n(s)} x_{n(s)}^*(i))\omega (i)\rightarrow \underset{i=1}{\overset{m_{l}}{{\sum }}}\Phi (k_0 x^*(i))\omega (i) \end{aligned}$$

as $s\rightarrow \infty$. Hence, for any fixed l, there holds

$$\begin{aligned} 1\ge \lim \nolimits _{s\rightarrow \infty }\frac{1}{k_{n(s)}}\left( 1+\underset{i=1}{\overset{m_{l}}{{\sum }}}\Phi (k_{n(s)} x_{n(s)}^*(i))\omega (i)\right) =\frac{1}{k_0}\left( 1+\underset{i=1}{\overset{m_{l}}{{\sum }}}\Phi (k_0 x^*(i))\omega (i)\right) . \end{aligned}$$

From the arbitrariness of $l\in \mathbb {N}$, we get

$$\begin{aligned} \frac{1}{k_0}\left( 1+\underset{i=1}{\overset{\infty }{{\sum }}}\Phi (k_0 x^*(i))\omega (i)\right) \le 1. \end{aligned}$$

This inequality together with inequality (11) give us equality (10).

Let now $(x_{n(s)})$ be an arbitrary subsequence of the sequence $(x_{n})$. Without loss of generality, passing to a subsequence if necessary, we can assume that sequence $(k_{n(s)})$ is convergent to some $k_{0}$. Thus $k_{n(s)}x_{n(s)}\rightarrow k_{0}x$ coordinatewisely and $k_{0}\in K(x)$, whence

$$\begin{aligned} \frac{1}{k_0}\left( 1+I_{\Phi ,\omega }(k_0x)\right) =1=\frac{1}{k_{n(s)}}\left( 1+I_{\Phi ,\omega }(k_{n(s)}x_{n(s)})\right) \quad \mathrm{for}\quad s\in \mathbb {N}\end{aligned}$$

and, consequently,

$$\begin{aligned} \lim \limits _{s\rightarrow \infty }I_{\Phi ,\omega }(k_{n(s)}x_{n(s)})=I_{\Phi ,\omega }(k_0x). \end{aligned}$$

By virtue of Lemmas 2 and 3, we get

$$\begin{aligned} k_{n(s)}x_{n(s)}^*(i)\rightarrow k_0x^*(i) \end{aligned}$$

(12)

for any $i\in \mathbb {N}$ and

$$\begin{aligned} (k_{n(s)}x_{n(s)}-k_0x)^*\rightarrow 0 \end{aligned}$$

(13)

uniformly as $s\rightarrow \infty$.

Since $\ell ^1$ has the Kadec–Klee property with respect to the coordinatewise convergence, we obtain

$$\begin{aligned} \mathop {\sum }\limits _{i=1}^{\infty }\left| \Phi (k_{n(s)} x^*_{n(s)}(i))-\Phi (k_0x^*(i))\right| \omega (i)=\Vert (\Phi (k_{n(s)}x^*_{n(s)})-\Phi (k_0x^*))\omega \Vert _{\ell ^1}\rightarrow 0 \end{aligned}$$

as $s\rightarrow \infty$. By virtue of Lemma 2, p. 97 from [20], there exist $y\in \ell ^1_+$ (the positive cone of $\ell ^1$) and a subsequence $(k_{n(s(t))}x^*_{n(s(t))})$ of the sequence $(k_{n(s)}x^*_{n(s)})$ such that

$$\begin{aligned} |\Phi (k_{n(s(t))}x^*_{n(s(t))})-\Phi (k_0x^*)|\omega \le y. \end{aligned}$$

(14)

If $k_{0}x^{*}(1)<u_{0}$ (the constant from condition $\delta _{2}$), then $k_{n(s(t))}x^{*}_{n(s(t))}(1)\le u_{0}$, starting from some $t_{0}\in \mathbb {N}$. Consequently, applying the rearrangement properties and the condition $\delta _{2}$, we get for $i=2j$ and $t\ge t_{0}$ that

$$\begin{aligned}&\Phi ((k_{n(s(t))}x_{n(s(t))}-k_0x)^*(2j))\omega (2j)\\&\quad \le \Phi (k_{n(s(t))}x_{n(s(t))}^*(j)+k_0x^*(j))\omega (2j)\\&\quad \le \Phi \left( 2\cdot \frac{k_{n(s(t))} x^*_{n(s(t))}(j)+k_0 x^*(j)}{2}\right) \omega (j)\\&\quad \le \frac{K}{2}\left( \Phi (k_{n(s(t))}x^*_{n(s(t))}(j))+\Phi (k_0x^*(j))\right) \omega (j)\\&\quad \le \frac{K}{2}\left( y(j)+2\Phi (k_0x^*(j))\omega (j)\right) . \end{aligned}$$

Similarly, for $i=2j+1$ and $t\ge t_{0}$, we have

$$\begin{aligned}&\Phi ((k_{n(s(t))}x_{n(s(t))}-k_0x)^*(2j+1))\omega (2j+1)\\&\quad \le \Phi ((k_{n(s(t))}x_{n(s(t))}-k_0x)^*(2j))\omega (2j)\\&\quad \le \frac{K}{2}\left( y(j)+2\Phi (k_0x^*(j))\omega (j)\right) . \end{aligned}$$

Moreover, for $i=1$ and $t\ge t_{0}$, we get

$$\begin{aligned} \Phi ((k_{n(s(t))}x_{n(s(t))}-k_0x)^*(1))\omega (1)\le & {} \Phi (k_{n(s(t))}x_{n(s(t))}^*(1)+k_0x^*(j))\omega (1)\\\le & {} \frac{K}{2}\left( y(1)+2\Phi (k_0x^*(1))\omega (1)\right) . \end{aligned}$$

Let us define the sequence $z=(z(i))_{i=1}^\infty$ by the formula

$$\begin{aligned} z(i)=\left\{ \begin{array}{ll}{\frac{K}{2}\left( y(1)+2\Phi (k_0x^*(1))\omega (1)\right) }&{}\hbox {for}\, i=1,\\ {\frac{K}{2}\left( y(j)+2\Phi (k_0x^*(j))\omega (j)\right) }&{}\hbox {if}\, i=2j \,\hbox {or}\, i=2j+1, j\in \mathbb {N}.\end{array}\right. \end{aligned}$$

Then $z\in \ell ^1$ and $\Phi \circ (k_{n(s(t))} x_{n(s(t))}-k_0x)^*\cdot \omega \le z$ for $t\ge t_{0}$. Moreover, by (13) we have $\Phi \circ (k_{n(s(t))} x_{n(s(t))}-k_0x)^*\cdot \omega \rightarrow 0$ uniformly as $t\rightarrow \infty$. Hence, by the Lebesgue dominated convergence theorem, $I_{\Phi ,\omega }(k_{n(s(t))}x_{n(s(t))}-k_0x)\rightarrow 0$, whence, by Lemma 4, $\Vert k_{n(s(t))}x_{n(s(t))}-k_0x\Vert _{\Phi ,\omega }^{O}\rightarrow 0$. Since $k_{n(s(t))}\rightarrow k_0$, we obtain $\Vert x_{n(s(t))}-x\Vert _{\Phi ,\omega }^{O}\rightarrow 0$.

Let now $k_{0}x^{*}(1)\ge u_{0}$. Since $x\in c_{0}$, there exists $j_{0}\in \mathbb {N}$ such that $k_{0}x^{*}(j_{0})<u_{0}$. Hence, by (12) and (13), we can find $t_{0}\in N$ such that for any $t\ge t_{0}$, we have $(k_{n(s(t))}x_{n(s(t))}-k_0x)^*(1)\le u_{0}$ and $k_{n(s(t))}x^{*}_{n(s(t))}(j_{0})\le u_{0}$. Defining $z=(z(i))_{i=1}^\infty$ by the formula

$$\begin{aligned} z(i)=\left\{ \begin{array}{l@{\quad }l}{\Phi (u_{0})\omega (i)}&{}\hbox {for}\, i=1,\ldots ,2j_{0}-1,\\ {\frac{K}{2}\left( y(j)+2\Phi (k_0x^*(j))\omega (j)\right) }&{}\hbox {if}\, i=2j \,\hbox {or}\, i=2j+1, j\ge j_{0},\end{array}\right. \end{aligned}$$

we get again that $z\in \ell ^1$ and $\Phi \circ (k_{n(s(t))} x_{n(s(t))}-k_0x)^*\cdot \omega \le z$ for $t\ge t_{0}$. Proceeding analogously as above we obtain $\Vert x_{n(s(t))}-x\Vert _{\Phi ,\omega }^{O}\rightarrow 0$.

Finally, applying the double extract convergence theorem, we get $\Vert x_n-x\Vert _{\Phi ,\omega }^{O}\rightarrow 0$.

Case 2. Assume that $\lim \nolimits _{u\rightarrow \infty }\frac{\Phi (u)}{u}=B<\infty$ and let $x\in S(\lambda _{\Phi ,\omega },\Vert .\Vert _{\Phi ,\omega }^O)$, $x_n\in S(\lambda _{\Phi ,\omega },\Vert .\Vert _{\Phi ,\omega }^O)$ for $n\in \mathbb {N}$ and $x_n\rightarrow x$ coordinatewisely.

First, assume additionally that $\Psi (B)\sum _{i=1}^{m({\text {supp }}x)}\omega (i)>1$ or $\Psi (B)\sum _{i=1}^{m({\text {supp }}x)}\omega (i)$ $=1$ and $p(u)=B$ starting with some $u_0$. If $\Psi (B)\sum _{i=1}^{m({\text {supp }}x)}\omega (i)>1$, then there exist $j_x\in \mathbb {N}$, $j_x\le m({\text {supp }}x)$ and $u_x>0$ such that $\Psi (p(u_x))\sum _{i=1}^{j_x}\omega (i)>1$. Without loss of generality we can assume that $x^*(j_x)>x^*(j_x+1)$, whence $m(N_x)=j_x$, where $N_x=\{i\in \mathbb {N}:|x(i)|\ge x^*(j_x)\}$. Since $(x_n)$ is coordinatewise convergent to x, there exists $n_0\in \mathbb {N}$ such that $|x_n(i)|\ge \frac{x^*(j_x)+x^*(j_x+1)}{2}$ for $i\in N_x$ and $n\ge n_0$. Hence, $I_{\Psi ,\omega } (p(k_{1}x^{*}_{n}))>1$ for $k_1=\frac{2u_x}{x^*(j_x)+x^*(j_x+1)}$ and $n\ge n_0$ and, consequently, $k_n^{**}:=k^{**}(x_n)\le k_1$ for the same n. Next, proceeding analogously as in Case 1, we obtain that $x_n\rightarrow x$ in norm.

We proceed analogously in the case when $\Psi (B)\sum _{i=1}^{m({\text {supp }}x)}\omega (i)=1$ and there exists $u_0>0$ such that $p(u)=B$ for all $u\ge u_0$. Indeed, note that in this case $\Psi (B)>0$ and $m({\text {supp }}x)<\infty$ by virtue of the assumption that $\sum _{i=1}^{\infty }\omega (i)=\infty$. Taking this time $j_x=m({\text {supp }}x)$ and using again the fact that $x_n\rightarrow x$ coordinatewisely, we can find $n_0\in \mathbb {N}$ such that $|x_n(i)|\ge \frac{1}{2}x^*(j_x)$ for any $i\in {\text {supp }}x$ and $n\ge n_0$. Hence, taking $k_1=\frac{2 u_0}{x^*(j_x)}$, we get that $I_{\Psi ,\omega }(p(k_{1}x^{*}_{n}))\ge \Psi (p(u_{0}))\sum _{i=1}^{m({\text {supp }}x)}\omega (i)=1$ for all $n\ge n_0$, which means that for any $n\ge n_0$ there is $k_n\le k_1$ such that $\Vert x_n\Vert _{\Phi ,\omega }^O=\frac{1}{k_n}(1+I_{\Phi ,\omega }(k_nx_n))$. Next, proceeding again in the same way as in Case 1, we can finish this part of the proof.

Finally, assume that $\Psi (B)\sum _{i=1}^{m({\text {supp }}x)}\omega (i)<1$ or $\Psi (B)\sum _{i=1}^{m({\text {supp }}x)}\omega (i)=1$ and $p(u)<B$ for any $u>0$, which means that $K(x)=\emptyset$, or equivalently,

$$\begin{aligned} \frac{1}{k}(1+I_{\Phi ,\omega }(kx))>1 \end{aligned}$$

(15)

for any $k>0$ (see Remark 1). Let $(x_{n(s)})$ be an arbitrary subsequence of the sequence $(x_n)$. We will consider two subcases.

Subcase 1. Assume, passing to a subsequence if necessary, that there exists $s_0\in \mathbb {N}$ such that $\Psi (B)\sum _{i=1}^{m({\text {supp }}x_{n(s)})}\omega (i)\le 1$ for any $s\ge s_0$. By virtue of Lemma 1, we obtain

$$\begin{aligned} B \mathop {\sum }\limits _{i=1}^{\infty }x^*(i)\omega (i)=1=B \mathop {\sum }\limits _{i=1}^{\infty }x_{n(s)}^*(i)\omega (i) \end{aligned}$$

for all $s\ge s_0$. Hence, by Lemma 2 (for $\Phi _1(u)=|u|$, we have $\lim \nolimits _{s\rightarrow \infty }I_{\Phi _{1},\omega }(x_{n(s)})=I_{\Phi _{1},\omega }(x)=\frac{1}{B}$), we get $x_{n(s)}^*(i)\rightarrow x^*(i)$ for any $i\in \mathbb {N}$. Applying again the fact that $\ell ^1$ has the Kadec–Klee property with respect to coordinatewise convergence, we obtain

$$\begin{aligned} \mathop {\sum }\limits _{i=1}^{\infty } |x^*_{n(s)}(i)-x^*(i)|\omega (i)=\Vert (x^*_{n(s)}-x^*)\omega \Vert _{\ell ^1}\rightarrow 0 \end{aligned}$$

as $s\rightarrow \infty$. Hence, by virtue of Lemma 4, p. 97 from [20], there exist $y\in \ell ^1_+$ and a subsequence $\left( x_{n(s(t))}^*\right)$ of the sequence $\left( x_{n(s)}^*\right)$ such that

$$\begin{aligned} |x_{n(s(t))}^*-x^*|\omega \le y. \end{aligned}$$

Hence, similarly to Case 1, we get

$$\begin{aligned} (x_{n(s(t))}-x)^*(2j)\omega (2j)\le & {} (x_{n(s(t))}^*(j)+x^*(j))\omega (j) \\\le & {} y(j)+2x^*(j)\omega (j). \end{aligned}$$

Consequently, defining

$$\begin{aligned} z(i)=\left\{ \begin{array}{ll} y(1)+2x^*(1)\omega (1) &{} \text {for }i=1 \\ y(j)+2x^*(j)\omega (j) &{} \text {for }i=2j\text { or }i=2j+1,\,j\in \mathbb {N}, \end{array}\right. \end{aligned}$$

we have that $z=(z(i))_{i=1}^{\infty }\in \ell ^1$ and $\left( x_{n(s(t))} -x\right) ^*\omega \le z$. Since, by virtue of Lemma 3, $(x_{n(s(t))}-x)^*\rightarrow 0$ uniformly as $t\rightarrow \infty$, applying again the Lebesgue dominated convergence theorem, we get that $\lim \nolimits _{t\rightarrow \infty } \sum _{i=1}^{\infty } (x_{n(s(t))}-x)^*(i)\omega (i) =0$. Hence, it is enough to show that, starting with some $t_0$, there holds

$$\begin{aligned} \Vert x_{n(s(t))}-x\Vert _{\Phi ,\omega }^O=B \mathop {\sum }\limits _{i=1}^{\infty } (x_{n(s(t))}-x)^*(i)\omega (i), \end{aligned}$$

(16)

whence we finally get

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }\Vert x_{n(s(t))}-x\Vert _{\Phi ,\omega }^O=0. \end{aligned}$$

Obviously, formula (16) is satisfied if $\Psi (B)=0$. Now, assume that $\Psi (B)>0$. Since $\sum _{i=1}^{\infty } \omega (i)=\infty$, we get that $m({\text {supp }}x)<\infty$. Hence, and by the assumption that $x_n$ is coordinatewise convergent to x, we obtain that ${\text {supp }}x \subset {\text {supp }}x_{n(s(t))}$ starting with some $t_0$ and, without loss of generality, we can assume that $s(t_0)\ge s_0$. Consequently, $\Psi (B) \sum _{i=1}^{m({\text {supp }}(x_{n(s(t))}-x ))}\omega (i)\le 1$ for all $t\ge t_0$, whence from Lemma 1 formula (16) arrives for the same t.

Subcase 2. Assume, starting with some $s_0\in \mathbb {N}$, that $\Psi (B)\sum _{i=1}^{m({\text {supp }}(x_{n(s)}))}\omega (i)$ $>1$ holds. Then $\Psi (B)>0$. Since $\sum _{i=1}^{\infty } \omega (i)=\infty$, we get that $i_x:=m({\text {supp }}x)<\infty$. Let

$$\begin{aligned} b_{n(s)}=\sup \{|x_{n(s)}(i)| :i\in \mathbb {N}\backslash {\text {supp }}x \}. \end{aligned}$$

Since $x_{n(s)}\in c_0$, then for any $s\in \mathbb {N}$ there exists $i_s$ such that $b_{n(s)}=|x_{n(s)}(i_s)|$. We will show indirectly that $\lim \nolimits _{s\rightarrow \infty } b_{n(s)}=0$. In order to do it suppose, passing to a subsequence if necessary, that there is $\delta >0$ such that $b_{n(s)}\ge \delta$ starting with some $s_1\ge s_0$. Without loss of generality we can assume that $\delta \le \frac{1}{2} x^*(i_x)$. Moreover, since $x_{n(s)}\rightarrow x$ coordinatewisely, we can assume that $|x_{n(s)}(i)|\ge \frac{1}{2} x^*(i_x)$ for any $i\in {\text {supp }}x$ and $s\ge s_1$.

If $\Psi (B)\sum _{i=1}^{i_x +1}\omega (i)\le 1$, then by virtue of Lemma 1, for any $s\ge s_1$ there holds

$$\begin{aligned} 1= & {} \Vert x_{n(s)}\Vert _{\Phi ,\omega }^O\ge \Vert x_{n(s)}\chi _{{\text {supp }}x}+\delta e_{i_s}\Vert _{\Phi ,\omega }^O\nonumber \\= & {} B\underset{i=1}{\overset{i_x}{\sum }}(x_{n(s)}\chi _{{\text {supp }}x})^*(i)\omega (i)+B\delta \omega (i_x +1) . \end{aligned}$$

(17)

Since $\lim \nolimits _{s\rightarrow \infty }(x_{n(s)}\chi _{{\text {supp }}x})^*(i)=x^*(i)$ for any $i\in {\text {supp }}x$, we have

$$\begin{aligned} \lim \limits _{s\rightarrow \infty } B\underset{i=1}{\overset{i_x }{\sum }}(x_{n(s)}\chi _{{\text {supp }}x})^*(i)\omega (i)=B\underset{i=1}{\overset{i_x}{\sum }} x^*(i)\omega (i)=1, \end{aligned}$$

whence we conclude that there is $s_2\ge s_1$ such that

$$\begin{aligned} B\underset{i=1}{\overset{i_x}{\sum }}(x_{n(s)}\chi _{{\text {supp }}x})^*(i)\omega (i)\ge 1-\frac{1}{2}B\delta \omega (i_x+1) \end{aligned}$$

for any $s\ge s_2$. Hence,

$$\begin{aligned} B \underset{i=1}{\overset{i_x}{\sum }}(x_{n(s)}\chi _{{\text {supp }}x})^*(i)\omega (i) +B\delta \omega (i_x+1)>1 \end{aligned}$$

for any $s\ge s_2$, a contradiction to (17).

Now, assume that $\Psi (B)\sum _{i=1}^{i_x +1}\omega (i)>1$. Then, there exists $u_x>0$ such that $\Psi (p(u_x))\sum _{i=1}^{i_x +1}\omega (i)>1$. Defining $k_1=\frac{u_x}{\delta }$, we have that $I_\Psi (p(k_1(x_{n(s)}\chi _{{\text {supp }}x}+\delta e_{i_s})^{*}))>1$ for all $s\ge s_1$. Therefore, for each $s\ge s_{1}$ there is $k_s\in [1,k_1]$ such that

$$\begin{aligned} 1= & {} \Vert x_{n(s)}\Vert _{\Phi ,\omega }^O \ge \Vert x_{n(s)}\chi _{{\text {supp }}x} + \delta e_{i_s}\Vert _{\Phi ,\omega }^O \nonumber \\= & {} \frac{1}{k_s}\left\{ 1+\underset{i=1}{\overset{i_x}{\sum }}\Phi (k_s(x_{n(s)}\chi _{{\text {supp }}x})^*(i))\omega (i) + \Phi (k_s\delta ) \omega (i_x +1)\right\} \nonumber \\\ge & {} \frac{1}{k_s} \left\{ 1+\underset{i=1}{\overset{i_x}{\sum }}\Phi (k_s(x_{n(s)}\chi _{{\text {supp }}x})^*(i))\omega (i)\right\} +\Phi (\delta )\omega (i_x +1) , \end{aligned}$$

(18)

where the inequality $\Phi (\delta )\le \frac{\Phi (k_s\delta )}{k_s}$ is a consequence of the fact that function $\frac{\Phi (u)}{u}$ is non-decreasing.

On the other hand, by (15),

$$\begin{aligned}&\lim \limits _{s\rightarrow \infty } \frac{1}{k_1} \left\{ 1+\underset{i=1}{\overset{i_x}{\sum }}\Phi (k_1(x_{n(s)}\chi _{{\text {supp }}x})^*(i))\omega (i)\right\} \\&\quad =\frac{1}{k_1} \left\{ 1+\underset{i=1}{\overset{i_x}{\sum }}\Phi (k_1(x^*(i)))\omega (i)\right\} >1. \end{aligned}$$

Hence, by the fact that function $f_{x_{n(s)}\chi _{{\text {supp }}x}}$ (see formula (2)) is strictly decreasing, there is $s_2\ge s_1$ such that for all $s\ge s_2$, we have

$$\begin{aligned}&\frac{1}{k_s} \left\{ 1+\underset{i=1}{\overset{i_x}{\sum }}\Phi (k_s(x_{n(s)}\chi _{{\text {supp }}x})^*(i))\omega (i)\right\} \\&\quad \ge \frac{1}{k_1} \left\{ 1+\underset{i=1}{\overset{i_x}{\sum }}\Phi (k_1(x_{n(s)}\chi _{{\text {supp }}x})^*(i))\omega (i)\right\} > 1, \end{aligned}$$

a contradiction to (18).

Therefore, $\lim \nolimits _{s\rightarrow \infty } b_{n(s)}=0$ holds. Let $s_1\ge s_0$ be such that $b_{n(s)}<\frac{1}{2} x^*(i_x)$ for $s\ge s_1$ and $|x_{n(s)}(i)|>\frac{1}{2} x^*(i_x)$ for any $i\in {\text {supp }}x$ and $s\ge s_1$. Hence, $x_{n(s)}^*(i)=(x_{n(s)}\chi _{{\text {supp }}x})^*(i)$ for $i=1,2,\ldots ,i_x$ and $x_{n(s)}^*(i)=(x_{n(s)}\chi _{\mathbb {N}\backslash {\text {supp }}x})^*(i-i_x)$ for $i>i_x$. At the same time $x_{n(s)}^*(i)\rightarrow x^*(i)$ for $i=1,2,\ldots ,i_x$. Now, we will show that $\lim \nolimits _{s\rightarrow \infty }k_{n(s)}=\infty$ where

$$\begin{aligned} 1= \Vert x_{n(s)}\Vert _{\Phi ,\omega }^O=\frac{1}{k_{n(s)}}\left\{ 1+I_{\Phi ,\omega }(k_{n(s)}x_{n(s)})\right\} . \end{aligned}$$

Indeed, in the opposite case, there exists a subsequence $(k_{n(s(t))})$ such that $\lim \nolimits _{t\rightarrow \infty }k_{n(s(t))}=k<\infty$. Then, by inequality (15), we get

$$\begin{aligned} 1= & {} \lim \limits _{t\rightarrow \infty }\Vert x_{n(s(t))}\Vert _{\Phi ,\omega }^O=\lim \limits _{t\rightarrow \infty } \frac{1}{k_{n(s(t))}}\left\{ 1+ \mathop {\sum }\limits _{i=1}^{\infty }\Phi (k_{n(s(t))}x_{n(s(t))}^*(i))\omega (i)\right\} \\\ge & {} \lim \limits _{t\rightarrow \infty } \frac{1}{k_{n(s(t))}}\left\{ 1+\underset{i=1}{\overset{i_x}{\sum }}\Phi (k_{n(s(t))}(x_{n(s(t))}\chi _{{\text {supp }}x})^*(i))\omega (i)\right\} \\= & {} \frac{1}{k}\left\{ 1+\underset{i=1}{\overset{i_x}{\sum }}\Phi (kx^*(i))\omega (i)\right\} >\Vert x\Vert _{\Phi ,\omega }^O=1, \end{aligned}$$

a contradiction. We also have for any $s\ge s_1$ that

$$\begin{aligned} 1= & {} \Vert x_{n(s)}\Vert _{\Phi ,\omega }^O = \frac{1}{k_{n(s)}}\left\{ 1+ \mathop {\sum }\limits _{i=1}^{\infty }\Phi (k_{n(s)}x_{n(s)}^*(i))\omega (i)\right\} \\= & {} \frac{1}{k_{n(s)}}\Bigg \{ 1+ \underset{i=1}{\overset{i_x}{\sum }}\Phi (k_{n(s)}(x_{n(s)}\chi _{{\text {supp }}x})^*(i))\omega (i)\\&+ \underset{i=i_x+1}{\overset{\infty }{\sum }}\Phi (k_{n(s)}(x_{n(s)}\chi _{\mathbb {N}\backslash {\text {supp }}x})^*(i-i_x))\omega (i)\Bigg \}. \end{aligned}$$

Since

$$\begin{aligned}&\lim \limits _{s\rightarrow \infty }\frac{1}{k_{n(s)}} \underset{i=1}{\overset{i_x}{\sum }}\Phi (k_{n(s)}(x_{n(s)}\chi _{{\text {supp }}x})^*(i))\omega (i)\\&\quad =\lim \limits _{s\rightarrow \infty } \underset{i=1}{\overset{i_x}{\sum }}\frac{\Phi (k_{n(s)}(x_{n(s)}\chi _{{\text {supp }}x})^*(i))}{k_{n(s)}(x_{n(s)}\chi _{{\text {supp }}x})^*(i)}(x_{n(s)}\chi _{{\text {supp }}x})^*(i)\omega (i)\\&\quad =B \underset{i=1}{\overset{i_x}{\sum }}x^*(i)\omega (i)=1, \end{aligned}$$

we obtain

$$\begin{aligned} \lim \limits _{s\rightarrow \infty }\frac{1}{k_{n(s)}}\left\{ 1+ \underset{i=i_x+1}{\overset{\infty }{\sum }}\Phi (k_{n(s)}(x_{n(s)}\chi _{\mathbb {N}\backslash {\text {supp }}x})^*(i-i_x))\omega (i)\}\right\} =0. \end{aligned}$$

(19)

Defining

$$\begin{aligned} \overline{x}_{n(s)}= b_{n(s)}\chi _{{\text {supp }}x}+x_{n(s)}\chi _{\mathbb {N}\backslash {\text {supp }}x} \end{aligned}$$

for $s\ge s_1$, we have

$$\begin{aligned} |x-x_{n(s)}|\le |(x-x_{n(s)})\chi _{{\text {supp }}x}|+ |\overline{x}_{n(s)}|. \end{aligned}$$

Consequently,

$$\begin{aligned} \Vert x-x_{n(s)}\Vert _{\Phi ,\omega }^O\le & {} \Vert |(x-x_{n(s)})\chi _{{\text {supp }}x}|+ |\overline{x}_{n(s)}| \Vert _{\Phi ,\omega }^O \\\le & {} \Vert (x-x_{n(s)})\chi _{{\text {supp }}x} \Vert _{\Phi ,\omega }^O+ \Vert \overline{x}_{n(s)} \Vert _{\Phi ,\omega }^O \end{aligned}$$

for the same s. Note that for any $s\ge s_1$ there holds

$$\begin{aligned}&\Vert \overline{x}_{n(s)} \Vert _{\Phi ,\omega }^O \le \frac{1}{k_{n(s)}}\left\{ 1+ I_{\Phi ,\omega }(k_{n(s)}\overline{x}_{n(s)}) \right\} \\&\quad = \frac{1}{k_{n(s)}}\left\{ 1+ \underset{i=1}{\overset{i_x}{\sum }}\Phi (k_{n(s)}b_{n(s)})\omega (i)+ \underset{i=i_x+1}{\overset{\infty }{\sum }}\Phi (k_{n(s)}(x_{n(s)}\chi _{\mathbb {N}\backslash {\text {supp }}x})^*(i-i_x))\omega (i)\right\} \\&\quad = \frac{1}{k_{n(s)}}\underset{i=1}{\overset{i_x}{\sum }}\Phi (k_{n(s)}b_{n(s)})\omega (i)+\frac{1}{k_{n(s)}}\left\{ 1+\underset{i=i_x+1}{\overset{\infty }{\sum }}\Phi (k_{n(s)}(x_{n(s)}\chi _{\mathbb {N}\backslash {\text {supp }}x})^*(i-i_x))\omega (i)\right\} . \end{aligned}$$

Hence, by virtue of

$$\begin{aligned} \lim \limits _{s\rightarrow \infty }\frac{1}{k_{n(s)}}\underset{i=1}{\overset{i_x}{\sum }}\Phi (k_{n(s)}b_{n(s)})\omega (i)= & {} \lim \limits _{s\rightarrow \infty }\frac{\Phi (k_{n(s)}b_{n(s)})}{k_{n(s)}}\underset{i=1}{\overset{i_x}{\sum }}\omega (i)\\= & {} \lim \nolimits _{s\rightarrow \infty } B b_{n(s)}\underset{i=1}{\overset{i_x}{\sum }}\omega (i)=0 \end{aligned}$$

and equality (19), we get that $\lim \nolimits _{s\rightarrow \infty }\Vert \overline{x}_{n(s)} \Vert _{\Phi ,\omega }^O=0$. Meanwhile, $\lim \nolimits _{s\rightarrow \infty }\Vert (x-x_{n(s)})\chi _{{\text {supp }}x} \Vert _{\Phi ,\omega }^O=0$, whence we finally obtain that $\lim \nolimits _{s\rightarrow \infty }\Vert x-x_{n(s)} \Vert _{\Phi ,\omega }^O=0$.

Summarizing both subcases and applying the double extract convergence theorem, we conclude that $\lim \nolimits _{n\rightarrow \infty }\Vert x-x_{n} \Vert _{\Phi ,\omega }^O=0$, which finishes the proof. $\square$

Finally, let us present a criterion for the subspace of order continuous elements $(\lambda _{\Phi ,\omega })_a$ of the space $(\lambda _{\Phi ,\omega },\Vert \cdot \Vert _{\Phi ,\omega }^O)$ to possess the Kadec–Klee property with respect to the coordinatewise convergence.

Theorem 3

(i) If $\mathop {\sum }\nolimits _{i=1}^{\infty }\omega (i)=\infty$, then $((\lambda _{\Phi ,\omega })_a,\Vert \cdot \Vert _{\Phi ,\omega }^O)$ has the Kadec–Klee property with respect to the coordinatewise convergence if and only if $\Phi$ satisfies the $\delta _2$-condition.

(ii) If $\mathop {\sum }\nolimits _{i=1}^{\infty }\omega (i)<\infty$, then $((\lambda _{\Phi ,\omega })_a,\Vert \cdot \Vert _{\Phi ,\omega }^O)$ has the Kadec–Klee property with respect to the coordinatewise convergence if and only if $\omega (i)>0$ for any $i\in \mathbb {N}$ and, $a_\Phi =0$ if there exists $j\in \mathbb {N}$ such that $K(x_j)\ne \emptyset$ where $x_j=(\underset{j\;\text {times}}{\underbrace{1,1,\ldots ,1}},0,0,\ldots )$.

Proof

(i) If $\mathop {\sum }\nolimits _{i=1}^{\infty }\omega (i)=\infty$ and $\Phi \in \delta _2$, then by virtue of Theorem 1, we obtain that $(\lambda _{\Phi ,\omega })_a=\lambda _{\Phi ,\omega }$, whence by Theorem 2, we get that $((\lambda _{\Phi ,\omega })_a,\Vert \cdot \Vert _{\Phi ,\omega }^O)$ has the Kadec–Klee property with respect to the coordinatewise convergence.

Now, we will show the necessity of the $\delta _2$-condition. Assume that $\Phi \notin \delta _2$. Let $j\in \mathbb {N}$ be such that $K(x_j)\ne \emptyset$ where $x_j=(\underset{j\;\text {times}}{\underbrace{1,1,\ldots ,1}},0,0,\ldots )$. If $b_{\Phi }<\infty$ or $b_{\Phi }=\infty$ and $\lim _{u\rightarrow \infty }\frac{\Phi (u)}{u}=\infty$, then we have $j=1$ (see Remark 1). While, in the case when $B=\lim \nolimits _{u\rightarrow \infty }\frac{\Phi (u)}{u}<\infty$, we get $\Psi (B)>0$ (note here that the case when $\Psi (B)=0$ is equivalent to the equality $\Phi (u)=Bu$ for any $u\ge 0$ and this function satisfies the $\delta _2$-condition). Since $\sum _{i=1}^{\infty }\omega (i)=\infty$, we can find $j\in \mathbb {N}$ such that $\Psi (B)\underset{i=1}{\overset{j}{\sum }}\omega (i)>1$ and, in consequence, $K(x_j)\ne \emptyset$ (see Remark 1). Let $h\in K(x_j)$.

First, assume that $a_\Phi >0$. For $n\ge j+1$ let us define

$$\begin{aligned} x_{j,n}=\underset{i=1}{\overset{j}{\sum }}e_i + \min \left( 1,\frac{a_\Phi }{h}\right) e_n. \end{aligned}$$

Note that $x_j$ and $x_{j,n}$ for $n\ge j+1$ belong to $(\lambda _{\Phi ,\omega })_a$. Since $0\le x_j\le x_{j,n}$ for $n\ge j+1$, we have $\Vert x_j\Vert _{\Phi ,\omega }^O\le \Vert x_{j,n}\Vert _{\Phi ,\omega }^O$ for the same n. On the other hand

$$\begin{aligned} \Vert x_j\Vert _{\Phi ,\omega }^O=\frac{1}{h}\left( 1+I_{\Phi ,\omega }(h x_j)\right) =\frac{1}{h}\left( 1+I_{\Phi ,\omega }(h x_{j,n})\right) \ge \Vert x_{j,n}\Vert _{\Phi ,\omega }^O \end{aligned}$$

for $n\ge j+1$ and, finally, $\Vert x_j\Vert _{\Phi ,\omega }^O=\Vert x_{j,n}\Vert _{\Phi ,\omega }^O$ for the same n. Obviously, $x_{j,n}\rightarrow x_j$ coordinatewisely. Simultaneously, $\Vert x_{j,n}-x_j\Vert _{\Phi ,\omega }^O=\Vert \min \left\{ 1,\frac{a_\Phi }{h}\right\} e_n \Vert _{\Phi ,\omega }^O>0$, so we conclude that $((\lambda _{\Phi ,\omega })_a,\Vert \cdot \Vert _{\Phi ,\omega }^O)$ does not possess the Kadec–Klee property with respect to the coordinatewise convergence.

Assume right now that $a_\Phi =0$. Since $\Phi \notin \delta _2$, we find a decreasing to zero sequence $(u_n)_{n=1}^\infty$ such that $\frac{u_1}{h}<1$, $\Phi (2u_n)> 2^{n+1}\Phi (u_n)$ and $\Phi (u_n)\omega (1)\le \frac{1}{2^{n+1}}$. Let $j_n$, where $n\in \mathbb {N}$, be such that

$$\begin{aligned} \frac{1}{2^{n+1}}<\Phi (u_n)\underset{i=1}{\overset{j_n}{\sum }}\omega (i)\le \frac{1}{2^n}. \end{aligned}$$

Define

$$\begin{aligned} x_{j,n}=\underset{i=1}{\overset{j}{\sum }}e_i + \underset{i=j+1}{\overset{j+j_n}{\sum }}\frac{u_n}{h}e_i \end{aligned}$$

for $n\in \mathbb {N}$. Obviously, $x_j$ and $x_{j,n}$ for $n\in \mathbb {N}$ belong to $(\lambda _{\Phi ,\omega })_a$ and $x_{j,n}\rightarrow x_j$ coordinatewisely. Applying the orthogonal subadditivity of the modular, we get

$$\begin{aligned} \Vert x_{j,n}\Vert _{\Phi ,\omega }^O\le & {} \frac{1}{h}\left( 1+I_{\Phi ,\omega }(h x_{j,n})\right) \le \frac{1}{h}\left( 1+I_{\Phi ,\omega }(h x_j)\right) + \frac{1}{h}\underset{i=1}{\overset{j_n}{\sum }}\Phi (u_n)\omega (i)\\\le & {} \Vert x_j\Vert _{\Phi ,\omega }^O+\frac{1}{h2^n} \end{aligned}$$

for $n\in \mathbb {N}$, whence and by the fact that $\Vert x_j\Vert _{\Phi ,\omega }^O\le \Vert x_{j,n}\Vert _{\Phi ,\omega }^O$ for $n\in \mathbb {N}$, we consequently obtain that $\Vert x_{j,n}\Vert _{\Phi ,\omega }^O\rightarrow \Vert x_j\Vert _{\Phi ,\omega }^O$. Simultaneously,

$$\begin{aligned} I_{\Phi ,\omega }(2h(x_{j,n}-x_j))=\underset{i=1}{\overset{j_n}{\sum }}\Phi (2u_n)\omega (i)>\underset{i=1}{\overset{j_n}{\sum }}2^{n+1}\Phi (u_n)\omega (i)>1 \end{aligned}$$

for $n\in \mathbb {N}$, whence $\Vert h(x_{j,n}-x_j)\Vert _{\Phi ,\omega }\ge \frac{1}{2}$ for the same n, where $\Vert \cdot \Vert _{\Phi ,\omega }$ is the Luxemburg norm. Therefore, $\Vert x_{j,n}-x_j\Vert _{\Phi ,\omega }^O\ge \Vert x_{j,n}-x_j\Vert _{\Phi ,\omega }\ge \frac{1}{2h}$ for $n\in \mathbb {N}$ and we conclude again that $((\lambda _{\Phi ,\omega })_a,\Vert \cdot \Vert _{\Phi ,\omega }^O)$ does not possess the Kadec–Klee property with respect to the coordinatewise convergence.

(ii) Sufficiency. Assume that, if $\sum _{i=1}^{\infty }\omega (i)<\infty$, then, as it was shown in [12, Theorem 4.2], we have that $(\lambda _{\Phi ,\omega })_a=c_0$. Generally, the proof of sufficiency is similar to the proof of Theorem 2 (implication $(1)\Rightarrow (2)$). Therefore, we will focus only on the essential differences.

First, let us note that if $\omega (i)>0$ for any $i\in \mathbb {N}$, then condition $K(x_j)=\emptyset$ for any $j\in \mathbb {N}$, where $x_j=(\underset{j\;\text {times}}{\underbrace{1,1,\ldots ,1}},0,0,\ldots )$, is equivalent to conditions $\lim \nolimits _{u\rightarrow \infty }\frac{\Phi (u)}{u}=B<\infty$ and $\Psi (B)\sum _{i=1}^{\infty }\omega (i)\le 1$ (see Remark 1). According to the conditions of the theorem, inequality $a_\Phi >0$ can be true in that case only. But then, for all elements $x\in (\lambda _{\Phi ,\omega })_a=c_0$, by virtue of Lemma 1, we get that $\Vert x\Vert _{\Phi ,\omega }^{O}=B\sum _{i=1}^{\infty } x^*(i)\omega (i)$ and the proof can be led similarly as in Case 2, Subcase 1 (see below).

Case 1. Using the assumption $(\lambda _{\Phi ,\omega })_a=c_0$ in place of assumption $\sum _{i=1}^{\infty }\omega (i)=\infty$, we get that $k_{n(s)}x^*_{n(s)}(i)\rightarrow k_0x^*(i)$ for any $i\in \mathbb {N}$. Hence, by virtue of Lemma 3, we obtain that $(k_{n(s)}x_{n(s)}- k_0x)^*$ tends to zero uniformly as $s\rightarrow \infty$. Therefore, there exists $s_0$ such that $(k_{n(s)}x_{n(s)}- k_0x)^*(1)<\frac{1}{2}\min \{1,b_\Phi \}$ for $s\ge s_0$. Thus, for any $i\in \mathbb {N}$ and $s\ge s_0$, we have

$$\begin{aligned} \Phi ((k_{n(s)}x_{n(s)}- k_0x)^*(i))\le \Phi ((k_{n(s)}x_{n(s)}- k_0x)^*(1))\le \Phi \left( \frac{1}{2}\min \{1,b_\Phi \}\right) . \end{aligned}$$

Defining $z=(z(i))_{i=1}^\infty$ by formula $z(i)=\Phi \left( \frac{1}{2}\min \{1,b_\Phi \}\right) \omega (i)$, we obtain that $\Phi \circ (k_{n(s)}x_{n(s)}- k_0x)^*\omega \le z$ and $z\in \ell ^1$, whence by virtue of the Lebesgue dominated convergence theorem we get again that $I_{\Phi ,\omega }(k_{n(s)}x_{n(s)}- k_0x)\rightarrow 0$ as $s\rightarrow \infty$. Hence, by virtue of Lemma 4, we finally get that $\Vert k_{n(s)}x_{n(s)}- k_0x\Vert _{\Phi ,\omega }^O\rightarrow 0$ as $s\rightarrow \infty$.

Case 2. Having regard to the above considerations, the proof when $\Psi (B)\cdot \sum _{i=1}^{m({\text {supp }}(x))}\omega (i)>1$ is similar to that of Theorem 2. Also in the case when $\Psi (B)\sum _{i=1}^{m({\text {supp }}(x))}\omega (i)= 1$, $p(u)=B$ starting with some $u_0$ and $m({\text {supp }}x)<\infty$, we can repeat reasoning from the proof of Theorem 2.

The case when $(\Psi (B)\sum _{i=1}^{m({\text {supp }}(x))}\omega (i)=1$, $p(u)=B$ starting with some $u_0$ and $m({\text {supp }}x)=\infty )$ we consider jointly with cases $\Psi (B)\sum _{i=1}^{m({\text {supp }}(x))}\omega (i)<1$ and $(\Psi (B)\sum _{i=1}^{m({\text {supp }}(x))}\omega (i)=1$ and $p(u)<B$ for all $u>0)$. Similarly as in Theorem 2, we need to consider two subcases.

In Subcase 1, defining the majorant z, we apply inequality $(x_{n(s)}-x)^*(i)\le (x_{n(s)}-x)^*(1)$ for $i\in \mathbb {N}$ and the fact that $(x_{n(s)}-x)^*(1)\le 1$ starting with some $s_0$ (by virtue of Lemma 3, $(x_{n(s)}-x)^*\rightarrow 0$ uniformly as $s\rightarrow \infty$). Therefore, defining $z=(z(i))_{i=1}^\infty$ by formula $z(i)=\omega (i)$, we get that $z\in \ell ^1$ and $(x_{n(s)}-x)^*\omega \le z$ for any $s\ge s_0$. Next, formula (16) is proved in the same way as in the proof of Theorem 2. Let us only note that if $\Psi (B)>0$ and $m({\text {supp }}x)=\infty$, then $\Psi (B)\sum _{i=1}^{\infty }\omega (i)\le 1$, which allow us to apply Lemma 1.

Coming to the proof of Subcase 2, note that from condition $\Psi (B) \sum _{i=1}^{m({\text {supp }}(x_{n(s)}))} \omega (i)$ $>1$, we conclude that $\Psi (B)>0$. Hence and from condition $\Psi (B)\sum _{i=1}^{m({\text {supp }}(x))} \omega (i)$ $\le 1$, we get that $m({\text {supp }}(x))<\infty$. Further, the proof can be led and finished similarly as this for Theorem 2.

Necessity. First, suppose that there is $j\in \mathbb {N}$ such that $\omega (j)>\omega (j+1)=0$. Define

$$\begin{aligned} x_j =\underset{i=1}{\overset{j}{\sum }}e_i \quad \text { and }\quad x_{j,n}=\underset{i=1}{\overset{j}{\sum }}e_i + e_n \end{aligned}$$

for $n\ge j+1$. Then, $x_j$ and $x_{j,n}$ for $n\ge j+1$ belong to $(\lambda _{\Phi ,\omega })_a$, $x_{j,n}\rightarrow x_j$ coordinatewisely and

$$\begin{aligned} \Vert x_j\Vert _{\Phi ,\omega }^O=\inf \limits _{k>0}\frac{1}{k}\left( 1+I_{\Phi ,\omega }(k x_j)\right) =\inf \limits _{k>0}\frac{1}{k}\left( 1+I_{\Phi ,\omega }(k x_{j,n})\right) = \Vert x_{j,n}\Vert _{\Phi ,\omega }^O, \end{aligned}$$

but $\Vert x_{j,n}-x_j\Vert _{\Phi ,\omega }^O=\Vert e_n \Vert _{\Phi ,\omega }^O>0$, which means that $((\lambda _{\Phi ,\omega })_a,\Vert \cdot \Vert _{\Phi ,\omega }^O)$ does not have the Kadec–Klee property with respect to the coordinatewise convergence.

Finally, suppose that $\omega (i)>0$ for all $i\in \mathbb {N}$, there exists $j\in \mathbb {N}$ such that $K(x_j)\ne \emptyset$, where $x_j=\sum _{i=1}^{j}e_i$, and $a_\Phi >0$ . Let $h\in K(x_j)$. Then, defining

$$\begin{aligned} x_{j,n}=\underset{i=1}{\overset{j}{\sum }}e_i +\min \left( 1,\frac{a_\Phi }{h}\right) e_n \end{aligned}$$

for $n\ge j+1$, in the same way as in the proof of the necessity of Case (i), we obtain again that $((\lambda _{\Phi ,\omega })_a,\Vert \cdot \Vert _{\Phi ,\omega }^O)$ does not have the Kadec–Klee property with respect to the coordinatewise convergence.