Let φ be a strictly increasing convex continuous function on [0, ∞) with φ(0)=0, and let Λ{λ_k} be an increasing sequence of positive numbers such that $\sum_{k=1}^{\infty} \frac{1}{\lambda_k}=+\infty$. We say that a real valued function f on [0, 1] is of Λ_φ-bounded variation if

$\begin{gathered}V_{\varLambda_{\varphi}}(f):=\sup _{\mathcal{J}} \sum\limits_{k=1}^{\infty} \frac{\varphi\left(\left|f\left(I_k\right)\right|\right)}{\lambda_k} \\ :=\sup _{\mathcal{J}} \sum\limits_{k=1}^{\infty} \frac{\varphi\left(\left|f\left(b_k\right)-f\left(a_k\right)\right|\right)}{\lambda_k}<\infty, \end{gathered}$

where the supremum is taken over all sequences $\mathcal{J}= \left\{I_k\right\}=\left\{\left[a_k, b_k\right]\right\} $ of non-overlapping intervals in [0, 1], f(I_k)=f(b_k)-f(a_k).We say f∈Λ_φBV if there exists a constant c>0 such that V_Λφ(f/c)<∞.It is easy to see that Λ_φBV is a linear space. The linear space Λ_φBV is a Banach space, which endowed with the norm

$\|f\|_{\varLambda, \varphi}:=|f(0)|+\inf \left\{k>0: V_{\varLambda_{\varphi}}(f / k) \leqslant 1\right\}.$

We say that φ satisfies the condition Δ₂ if there exist a>0 and δ>0 such that φ(2x)≤δφ(x) for all x∈(0, a]. Musielak and Orlicz^[1] considered that Λ_φBV={f|V_Λφ(f)<∞} if and only if φ satisfies the condition Δ₂.

The class Λ_φBV was first introduced by Schramm and Waterman^[2] in 1982. In the case φ(x)=x^p(p≥1), f is said to be of bounded p-Λ-variation function. The corresponding class ΛBV^(p) was introduced in 1980 by Shiba and called by the Waterman-Shiba class. If p=1, f is said to be of Λ-bounded variation, and we denote f∈ΛBV. The corresponding class is the well-known Waterman class ΛBV. In the case Λ={1}, we get the class BV_φ of φ-bounded variation. More specifically, when φ(x)=x^p(p≥1), we get the class BV_p which is called the Wiener class. The class BV₁ is the well known class of bounded variation BV.

It is easily seen from the definition that Λ_φBV functions are bounded, and the discontinuities of a Λ_φBV function are simple and, therefore, at most denumerable. From Refs. [3-8], we know the class Λ_φBV had been studied mainly because of their applicability to the theory of Fourier series and some good approximative properties.

Kita and Yoneda^[9] introduced a new function space which is a generalization of Wiener classes (see also Refs. [10-11]). The concept was further extended by Akhobadze^[12] who studied many properties of the generalized Wiener classes BV(q, δ) thoroughly (see Ref. [13]).

Definition 1.1   Let $q=(q(n))_{n=1}^{\infty}$ be an increasing positive sequence and let $\delta=(\delta(n))_{n=1}^{\infty}$ be an increasing and unbounded positive integer sequence. We say that a function $f:[0, 1] \rightarrow \mathbb{R}$ belongs to the class BV(q, δ) if

$\begin{gathered}V(f, q, \delta):=\sup\limits_{n \geqslant 1}\sup\limits_{{I_k}}\left\{\left(\sum\limits_{k=1}^s\left|f\left(I_k\right)\right|^{q(n)}\right)^{\frac{1}{q(n)}}:\right. \\ \left.\inf _k\left|I_k\right| \geqslant \frac{1}{\delta(n)}\right\}<\infty, \end{gathered}$

where $\left\{I_k\right\}_{k=1}^s$ are non-overlapping subintervals of [0, 1].

The following statement concerning inclusion of Waterman spaces into generalize Wiener classes has been presented in Ref. [14], if $\lim\limits_{n \rightarrow \infty} q(n)=\infty$ and δ(n)=2n the inclusion ΛBV⊂BV(q, δ) holds if and only if

$\limsup\limits_{n \rightarrow \infty}\left\{\max\limits_{1 \leqslant k \leqslant \delta(n)} \frac{k^{\frac{1}{q(n)}}}{\left(\sum\limits_{i=1}^k \frac{1}{\lambda_i}\right)}\right\}<\infty.$

The following result was obtained by Hormozi et al.^[15]: for p∈[1, ∞), q and δ sequence satisfying the conditions in Definition 1.1, the inclusion ΛBV^(p)⊂BV(q, δ) holds if and only if

$\limsup\limits_{n \rightarrow \infty}\left\{\max\limits_{1 \leqslant k \leqslant \delta(n)} \frac{k^{\frac{1}{q(n)}}}{\left(\sum\limits_{i=1}^k \frac{1}{\lambda_i}\right)^{\frac{1}{p}}}\right\}<\infty .$

In this paper, we extend the results above and obtain the corresponding inclusions of Λ_φBV and BV(q, δ).Our main results can be formulated as follows.

Theorem 1.1  For q and δ satisfying the conditions in Definition 1.1, the inclusion Λ_φBV⊂BV(q, δ) holds if and only if

$\limsup\limits_{n \rightarrow \infty}\left\{\max\limits_{1 \leqslant k \leqslant \delta(n)} k^{\frac{1}{q(n)}} \varphi^{-1}\left(\left(\sum\limits_{i=1}^k \frac{1}{\lambda_i}\right)^{-1}\right)\right\}<\infty.$

(1)

Theorem 1.2   Let $\sum_{n=1}^{\infty}\left(1 / \lambda_n\right)=\infty$ and φ satisfy the condition Δ₂. Then there exists a function f∈BV(q, δ)∩C[0, 1] such that f∉Λ_φBV.It suffices to show BV(q, δ)⊂Λ_φBV does not hold if φ satisfies the condition Δ₂.

Proof of Theorem 1.1

Sufficiency  Suppose that (1) holds. Let f∈Λ_φBV. It suffice to show f∈BV(q, δ). We will prove the inequality

$\begin{aligned} & V(f, q, \delta) \\ & \leqslant 16\|f\|_{\Lambda, \varphi} \sup _{n \geqslant 1}\left\{\max _{1 \leqslant k \leqslant \delta(n)} k^{\frac{1}{q(n)}} \varphi^{-1}\left(\left(\sum\limits_{i=1}^k \frac{1}{\lambda_i}\right)^{-1}\right)\right\} \\ & =16\|f\|_{\varLambda, \varphi} \limsup _{n \rightarrow \infty}\left\{\max _{1 \leqslant k \leqslant \delta(n)} k^{\frac{1}{q(n)}} \varphi^{-1}\left(\left(\sum\limits_{i=1}^k \frac{1}{\lambda_i}\right)^{-1}\right)\right\} .\end{aligned}$

(2)

which combining with (1) implies f∈BV(q, δ). Without loss of generality, we assume that ||f||_{Λ, φ}≠0. For any $n \in \mathbb{N}$, let $\left\{I_k\right\}_{k=1}^s$ be a sequence of non-overlapping intervals in [0, 1] and $\inf\limits_{1 \leqslant k \leqslant s}\left|I_k\right| \geqslant \frac{1}{\delta(n)}$. Then s≤δ(n). Let $x_k=\frac{\left|f\left(I_k\right)\right|}{\|f\|_{\varLambda, \varphi}}$, 1≤k≤s. We rearrange {I_k} such that 0≤x_s≤…≤x₂≤x₁. It follows from the definition of ||·||_{Λ, φ} that $V_{\varLambda_{\varphi}}\left(\frac{g}{\|g\|_{\varLambda, \varphi}}\right) \leqslant 1$ holds for any g∈Λ_φBV. Hence, we have

$\sum\limits_{k=1}^s \frac{\varphi\left(x_k\right)}{\lambda_k}=\sum\limits_{k=1}^s \frac{\varphi\left(\left|f\left(I_k\right)\right| /\|f\|_{\varLambda, \varphi}\right)}{\lambda_k} \leqslant V_{\varLambda_{\varphi}}\left(\frac{f}{\|f\|_{\varLambda, \varphi}}\right) \leqslant 1.$

Let A be a positive constant, $s \in \mathbb{N}$, 1≤q<∞, 0≤x_s≤…≤x₂≤x₁, and

$I_{s, q, A}:=\sup \left\{\left(\sum\limits_{k=1}^s\left|x_k\right|^q\right)^{1 / q}: \sum\limits_{k=1}^s \frac{\varphi\left(x_k\right)}{\lambda_k} \leqslant A\right\}.$

It follows from Wang^[16] that

$I_{s, q, A} \leqslant 16 \max\limits_{1 \leqslant k \leqslant s} k^{1 / q} \varphi^{-1}\left(\left(\sum\limits_{i=1}^k 1 / \lambda_i\right)^{-1} A\right), $

where φ^-1 is the inverse function of φ.We note that

$\begin{aligned} & \left(\sum\limits_{k=1}^s\left|f\left(I_k\right)\right|^{q(n)}\right)^{1 / q(n)}=\|f\|_{\varLambda, \varphi}\left(\sum\limits_{k=1}^s\left|x_k\right|^{q(n)}\right)^{1 / q(n)} \\ & \leqslant\|f\|_{\varLambda, \varphi} I_{s, q(n), 1} \\ & \leqslant 16\|f\|_{\varLambda, \varphi} \max _{1 \leqslant k \leqslant s} k^{1 / q(n)} \varphi^{-1}\left(\left(\sum\limits_{i=1}^k 1 / \lambda_i\right)^{-1}\right) .\end{aligned}$

Taking both supremums over all sequences {I_k}_k=1^s of non-overlapping intervals in [0, 1] with $\inf\limits_{1 \leqslant k \leqslant s}\left|I_k\right| \geqslant \frac{1}{\delta(n)}$ and over all $n \in \mathbb{N}$, we obtain (2). The sufficiency is proved.

Necessity  Suppose (1) does not hold. Then there is an increasing sequence {n_k} of positive integers such that for all indices k

$\delta\left(n_k\right) \geqslant 2^{k+2}, $

(3)

and

$\max\limits_{1 \leqslant m \leqslant \delta\left(n_k\right)} m^{\frac{1}{q\left(n_k\right)}} \varphi^{-1}\left(\left(\sum\limits_{i=1}^m \frac{1}{\lambda_i}\right)^{-1}\right)>2^{2 k+1+\frac{k+1}{q(1)}}.$

(4)

We will construct a function g such that g∈Λ_φBV and g∉BV(q, δ). Let {m_k} be a sequence of positive integers such that

$1 \leqslant m_k \leqslant \delta\left(n_k\right),$

(5)

and

$\begin{gathered}\max _{1 \leqslant m \leqslant \delta\left(n_k\right)} m^{\frac{1}{q\left(n_k\right)}} \varphi^{-1}\left(\left(\sum\limits_{i=1}^m \frac{1}{\lambda_i}\right)^{-1}\right) \\ =m^{\frac{1}{q\left(n_k\right)}} \varphi^{-1}\left(\left(\sum\limits_{i=1}^{m_k} \frac{1}{\lambda_i}\right)^{-1}\right) .\end{gathered}$

(6)

Set

$\varPhi_k:=\left(\sum\limits_{i=1}^{m_k} \frac{1}{\lambda_i}\right)^{-1}.$

Consider

$\begin{array}{c} g_k(y)\\ :=\left\{\begin{array}{l}2^{-k-1} \varphi^{-1}\left(\varPhi_k\right), y \in\left[\frac{1}{2^k}+\frac{2 j-2}{\delta\left(n_k\right)}, \frac{1}{2^k}+\frac{2 j-1}{\delta\left(n_k\right)}\right) , \\ \qquad 1 \leqslant j \leqslant N_k , \\ 0, \quad \text { otherwise , }\end{array}\right.\end{array}$

where

$\begin{gathered}N_k=\min \left\{m_k, s_k\right\}, \\ s_k=\max \left\{j \in \mathbb{N}: 2 j<\frac{\delta\left(n_k\right)}{2^k}+1\right\} .\end{gathered}$

It follows from the definition of s_k that

$2\left(s_k+1\right) \geqslant \frac{\delta\left(n_k\right)}{2^k}+1.$

(7)

By (7) and (3), we have

$\frac{2 s_k-1}{\delta\left(n_k\right)} \geqslant 2^{-k-1}.$

(8)

If N_k=m_k, then

$2 N_k-1 \geqslant m_k \geqslant \frac{m_k}{2^{k+1}},$

and if N_k=s_k, then by (8) and (5) we have

$2 N_k-1 \geqslant \frac{\delta\left(n_k\right)}{2^{k+1}} \geqslant \frac{m_k}{2^{k+1}}.$

Hence, for all k,

$2 N_k-1 \geqslant \frac{m_k}{2^{k+1}}.$

(9)

Let us estimate ||g_k||_{Λ, φ}, k=1, 2, …. Since φ is convex and φ(0)=0, we get φ(t/2)≤φ(t)/2. It follows that

$\begin{aligned} V_{\varLambda_{\varphi}}\left(\frac{g_k}{2^{-k}}\right) & =\sum\limits_{j=1}^{2 N_k} \frac{\varphi\left(2^k 2^{-k-1} \varphi^{-1}\left(\varPhi_k\right)\right)}{\lambda_j} \\ \leqslant \frac{1}{2} \sum\limits_{j=1}^{2 N_k} \frac{\varPhi_k}{\lambda_j} & \leqslant \frac{1}{2}\left(\sum\limits_{j=1}^{2 N_k} \frac{1}{\lambda_j}\right)\left(\sum\limits_{i=1}^{N_k} \frac{1}{\lambda_i}\right)^{-1} \leqslant 1 .\end{aligned}$

By the definition of the norm ||·||_{Λ, φ}, we have

$\left\|g_k\right\|_{\varLambda, \varphi} \leqslant\left|g_k(0)\right|+2^{-k}=2^{-k}.$

The functions g_k have disjoint supports, and the series $\sum _{k=1}^{\infty} g_k(x)$ converges uniformly to a function g with g(0)=0 since 2^-k-1φ^-1(Φ_k)→0 as k→∞. Thus,

$\|g\|_{\varLambda, \varphi} \leqslant \sum\limits_{k=1}^{\infty}\left\|g_k\right\|_{\varLambda, \varphi} \leqslant \sum\limits_{k=1}^{\infty} 2^{-k}=1 , $

which implies g∈Λ_φBV.

Now, given any positive integer k, all intervals $\left[\frac{1}{2^k}+\frac{j-1}{\delta\left(n_k\right)}, \frac{1}{2^k}+\frac{j}{\delta\left(n_k\right)}\right]$, j=1, 2, …2N_k-1 are non-overlapping in [0, 1], and have length $\frac{1}{\delta\left(n_k\right)}$. Thus,

$\begin{aligned} & V(g, q, \delta) \geqslant\left(\sum\limits_{j=1}^{2 N_k-1} \mid g\left(\frac{1}{2^k}+\frac{j-1}{\delta\left(n_k\right)}\right)-\right. \\ & \left.\left.g\left(\frac{1}{2^k}+\frac{j}{\delta\left(n_k\right)}\right)\right|^{q\left(n_k\right)}\right)^{\frac{1}{q\left(n_k\right)}} \\ = & \left(\left(2 N_k-1\right)\left(2^{-k-1} \varphi^{-1}\left(\varPhi_k\right)\right)^{q\left(n_k\right)}\right)^{\frac{1}{q\left(n_k\right)}} \\ = & 2^{-k-1} \varphi^{-1}\left(\varPhi_k\right)\left(2 N_k-1\right)^{\frac{1}{q\left(n_k\right)}} \\ \geqslant & 2^{-k-1} \varphi^{-1}\left(\left(\sum\limits_{i=1}^{m_k} \frac{1}{\lambda_i}\right)^{-1}\right)\left(\frac{m_k}{2^{k+1}}\right)^{\frac{1}{q\left(n_k\right)}} \\ \geqslant & \frac{2^{k+(k+1) / q(1)}}{2^{(k+1) / q\left(n_k\right)}} \geqslant 2^k, \end{aligned}$

where in the second inequality, we used (9), in the third inequality we used (4) and (6). Since k was arbitrary, V(g, q, δ)=∞ which shows that g∉BV(q, δ). This completes the proof of the necessity.

Theorem 1.1 is proved.□

Proof of Theorem 1.2

We choose a strictly increasing sequence {l_k: k≥1} of positive integers such that l₁=1 and

$q\left(l_k\right) \geqslant \ln (k+1)\; {\rm{for\; all}}\; k \geqslant 2.$

Set

$\begin{aligned} & f(x) \\ & =\left\{\begin{array}{l}2 A_{k-1} \frac{\delta\left(l_k\right) \delta\left(l_k-1\right)}{\delta\left(l_k\right)-\delta\left(l_k-1\right)}\left(x-\frac{1}{\delta\left(l_k\right)}\right), \\ \text { if } \frac{1}{\delta\left(l_k\right)} \leqslant x \leqslant \frac{1}{2 \delta\left(l_k\right)}+\frac{1}{2 \delta\left(l_k-1\right)}, \\ \quad k=2, 3, \cdots, \\ 2 A_{k-1} \frac{\delta\left(l_k\right) \delta\left(l_k-1\right)}{\delta\left(l_k\right)-\delta\left(l_k-1\right)}\left(\frac{1}{\delta\left(l_k-1\right)}-x\right), \\ \text { if } \frac{1}{2 \delta\left(l_k\right)}+\frac{1}{2 \delta\left(l_k-1\right)} \leqslant x \leqslant \frac{1}{\delta\left(l_k-1\right)} , \\ \quad k=2, 3, \cdots, \\ 0, \text { otherwise, }\end{array}\right.\end{aligned}$

where

$A_k=\varphi^{-1}\left(\left(\sum\limits_{j=1}^k \frac{1}{\lambda_j}\right)^{-1 / 2}\right), k=1, 2 \cdots.$

In other words, the function f(x) is defined to be a continuous function with f(0)=0 such that f(x)=0 on the intervals $\left[\frac{1}{\delta\left(l_2-1\right)}, 1\right]$ and $\left[\frac{1}{\delta\left(l_{k+1}-1\right)}, \frac{1}{\delta\left(l_k\right)}\right]$, k=2, 3…, $f\left(\frac{1}{\delta\left(l_k\right)}\right)=f\left(\frac{1}{\delta\left(l_k-1\right)}\right)=0$, $f\left(\frac{1}{2 \delta\left(l_k\right)}+\frac{1}{2 \delta\left(l_k-1\right)}\right)=A_{k-1}$, and f is linear on the intervals $\left[\frac{1}{\delta\left(l_k\right)}, \frac{1}{2 \delta\left(l_k\right)}+\frac{1}{2 \delta\left(l_k-1\right)}\right]$, $\left[\frac{1}{2 \delta\left(l_k\right)}+\frac{1}{2 \delta\left(l_k-1\right)}, \frac{1}{\delta\left(l_k-1\right)}\right], k=2, 3 \cdots$.

First we show that f∈BV(q, δ).For any l∈$\mathbb{N}$, let $\left\{I_k\right\}_{k=1}^s=\left\{\left[a_k, b_k\right]\right\}_{k=1}^s$ be an arbitrary sequence of non-overlapping intervals in [0, 1] such that $\inf\limits_{1 \leqslant k \leqslant s}\left|I_k\right| \geqslant \frac{1}{\delta(l)}$. Without loss of generality we assume that

$0 \leqslant a_1<b_1 \leqslant a_2<b_2 \leqslant \cdots \leqslant a_s<b_s \leqslant 1.$

For this fixed l, we can choose integers l_k-1 and l_k for which l_k-1<l<l_k holds. Then it follows that q(l_k-1)≤q(l)≤q(l_k) and $\frac{1}{\delta\left(l_k\right)} \leqslant \frac{1}{\delta(l)} \leqslant \frac{1}{\delta\left(l_{k-1}\right)}$.

Then

$a_2 \geqslant b_1 \geqslant b_1-a_1 \geqslant \frac{1}{\delta(l)} \geqslant \frac{1}{\delta\left(l_k\right)}.$

Let k₀ be the integer such that $\frac{1}{\delta\left(l_{k_0}\right)} \leqslant a_2 < \frac{1}{\delta\left(l_{k_0-1}\right)}$. Then k₀≤k and

$\left|f\left(b_1\right)-f\left(a_1\right)\right| \leqslant A_{k_0}.$

We also note that {A_k} is non-increasing and

$\begin{aligned} & \sum\limits_{j=2}^s\left|f\left(b_j\right)-f\left(a_j\right)\right|^{q(l)} \\ & \leqslant \sup\limits_{\frac{1}{\delta\left(l_{k_0}\right)}=t_0<t_1<\cdots<t_{m-1}<t_m=1} \sum\limits_{j=1}^m\left|f\left(t_j\right)-f\left(t_{j-1}\right)\right|^{q(l)} \\ & =2 \sum\limits_{j=1}^{k_0-1} A_j^{q(l)}.\end{aligned}$

It then follows that

$\begin{aligned} & \left(\sum\limits_{j=1}^s\left|f\left(b_j\right)-f\left(a_j\right)\right|^{q(l)}\right)^{1 / q(l)} \\ & \leqslant\left(2 \sum\limits_{j=1}^{k_0} A_j^{q(l)}\right)^{1 / q(l)} \leqslant 2^{1 / q(l)} A_1 k_0^{1 / q(l)} \\ & \leqslant 2 A_1 k^{1 / q\left(l_{k-1}\right)} \leqslant 2 A_1 k^{\frac{1}{\ln k}}=2 \mathrm{e} A_1 .\end{aligned}$

This implies f∈BV(q, δ).

Next, we show that f∉Λ_φBV.Since φ satisfies the condition Δ₂, it suffices to prove that

$V_{\varLambda_{\varphi}}(f)=\infty.$

We have

$\begin{aligned} & V_{\varLambda_{\varphi}}(f) \\ & \geqslant \sum\limits_{j=1}^k \frac{\varphi\left(\left|f\left(\frac{1}{\delta\left(l_{j+1}\right)}\right)-f\left(\frac{1}{2 \delta\left(l_{j+1}\right)}+\frac{1}{2 \delta\left(l_{j+1}-1\right)}\right)\right|\right)}{\lambda_j} \\ & =\sum\limits_{j=1}^k \frac{\varphi\left(\left|A_j\right|\right)}{\lambda_j} \geqslant \varphi\left(A_k\right) \sum\limits_{j=1}^k \frac{1}{\lambda_j}=\left(\sum\limits_{j=1}^k \frac{1}{\lambda_j}\right)^{1 / 2} \rightarrow \infty \\ & {\rm{as}}\; k \rightarrow \infty,\end{aligned}$

which implies f∉Λ_φBV. The proof of Theorem 1.2 is complete.□