On some nonlinear operators in ${\varvec{\Lambda }} {\varvec{BV}}$-spaces

Abstract

In this paper we investigate autonomous as well as nonautonomous superposition operators acting between spaces of functions of bounded $\Lambda$-variation. A particular emphasis is put on acting conditions as well as on continuity problems for such operators. In particular, we give necessary and sufficient conditions for nonautonomous superposition operators to map a space of functions of bounded $\Lambda$-variation into itself. Moreover, we prove the continuity of certain autonomous superposition operators acting between various spaces of functions of bounded $\Lambda$-variation. We also examine the existence of $\Lambda BV$-solutions as well as the topological structure of such solution sets to classical nonlinear integral equations.

1 Introduction

One of the most interesting generalizations of the classical variation in the sense of Jordan seems to be the notion of $\Lambda$-variation. It was introduced by Waterman in 1972 [18], in connection with his investigation of Fourier series. More precisely, it appears that Fourier series of functions of bounded harmonic variation which is a particular $\Lambda$-variation converge pointwise to the arithmetic mean of the left- and the right-hand side limits and converge uniformly to the functions under consideration on closed intervals of their continuity (see [18] for more details).

It appears that functions of bounded $\Lambda$-variation possess some properties which are entitled to functions of bounded variation in the sense of Jordan. For example, functions of bounded $\Lambda$-variation are bounded and sets of points of their discontinuities are at most denumerable (see [14], Theorem 3 and Theorem 4). Moreover, for such functions a Helly-type theorem holds (see [11]). Let us add also that the space of functions of bounded $\Lambda$-variation from a functional point of view was investigated, for example, in the paper [16]. It has appeared that the so-called Shao–Sablin index plays an essential role in that approach.

The autonomous superposition operators acting in the spaces of functions of bounded $\Lambda$-variation have been investigated for example in the paper [15]. The authors proved in it an analogue of the well-known Josephy theorem, which describes the behavior of such operators in the space of functions of bounded variation in the sense of Jordan. On the other hand, the sufficient conditions for nonautonomous superposition operators to act in the spaces of functions of bounded $\Lambda$-variation were given in the paper [2].

The existence and the existence and uniqueness of global and local $\Lambda BV$-solutions, that is, solutions being functions of bounded $\Lambda$-variation, to nonlinear Hammerstein as well as to nonlinear Volterra–Hammerstein integral equations were investigated, for example, in the paper [4]. Let us also add that linear and semilinear differential equations in the spaces of continuous functions of bounded $\Lambda$-variation were investigated in the paper [2].

In this paper we would like to achieve a few goals. First, in Sect. 3 we focus on nonautonomous superposition operators acting in the spaces of functions of bounded $\Lambda$-variation. The main result in this section is Theorem 7 which gives necessary and sufficient conditions for nonautonomous superposition operators to map a space of functions of bounded $\Lambda$-variation into itself. The results included in this section are mainly inspired by the paper [3]. However, let us emphasize the fact that nonautonomous superposition operators in the spaces of functions of bounded $\Lambda$-variation behave in a slightly different way than in spaces of functions of bounded variation in the sense of Jordan (see Example 1).

In Sect. 4 we deal with the problem of continuity of autonomous superposition operators acting between spaces of functions of bounded $\Lambda$-variation. It appears that in some cases generators of such operators reduce to constant mappings (see Theorem 10). Let us recall that a necessary and sufficient condition for an autonomous superposition operator to map a space of functions of bounded $\Lambda$-variation into itself is the requirement of its generator to be a locally Lipschitz function (see [15] or [1], Theorem 5.14). We show that in case of generators being continuously differentiable functions the corresponding superposition operator is continuous. However, if one considers generators that do not have to be locally Lipschitz functions, then it appears that for a given domain it is possible to indicate a space of functions of bounded $\Lambda$-variation that the superposition operator acts to (see Theorem 11). According to Theorem 12 it appears that such operators are continuous without additional assumptions. Let us add that the problem of continuity of the superposition operators still remains open (even in the autonomous case) because we still cannot say anything about the continuity of operators with a general Lipschitz generator and acting from a space of functions of bounded $\Lambda$-variation into itself.

In Sect. 5 we focus on $\Lambda BV$-solutions to some classical nonlinear integral equations. We prove therein an existence result to the nonlinear Hammerstein integral equation, extending Theorem 1 from the paper [4]. Moreover, we prove an Aronszajn-type theorem concerning the topological structure of continuous $\Lambda BV$-solution sets to the nonlinear Volterra–Hammerstein integral equations.

Finally, in Sect. 6 we examine embeddings of compact subsets of a given space of functions of bounded $\Lambda$-variation into another space of this type. It appears, roughly speaking, that sets of functions which are continuous in $\Lambda$-variation possess some nice properties, as far as compactness is concerned (see Theorems 18, 19).

2 Preliminaries

In this section we collect some basic definitions and facts, which will be needed in the sequel.

Notation

Throughout the paper, we will denote the unit interval [0, 1] by I. By $S_n$ we denote the symmetric group of the set $\{1,2,\ldots , n\}$.

Definition 1

Let us consider a nondecreasing sequence of positive real numbers $\Lambda = (\lambda _n)_{n\in \mathbb {N}}$. We call such sequence a Waterman sequence if

$$\begin{aligned} \sum _{n=1}^\infty \frac{1}{\lambda _n} = +\infty . \end{aligned}$$

Definition 2

Let $(\lambda _n)_{n\in \mathbb {N}}$ be a Waterman sequence, $J = [a,b]\subset I$ and let $x:I\rightarrow \mathbb {R}$. We say that x is of bounded $\Lambda$-variation over J if there exists a positive constant M such that for any finite sequence of nonoverlapping¹ subintervals $\{[a_1, b_1], [a_2, b_2], \ldots , [a_n, b_n]\}$ of J, the following inequality holds

$$\begin{aligned} \sum _{i=1}^n \frac{|x(b_i) - x(a_i)|}{\lambda _i} \le M. \end{aligned}$$

The supremum of the above sums taken over the family of all the finite collections of nonoverlapping subintervals of J is called the $\Lambda$-variation of x over J and it is denoted by $\mathrm{{var}}_\Lambda (x;J)$. If $J=I$, then we simply write $\mathrm{{var}}_\Lambda (x)$.

Remark 1

In the above definition the finite collections of intervals and the finite sums may be replaced by countable collections and series, respectively (see [19], Theorem 1, p. 34).

The vector space of all the functions defined on the interval I and of bounded $\Lambda$-variation, endowed with the norm $\left\| x\right\| _{\Lambda BV} {\mathrel {\mathop :}=}|x(0)| + \mathrm{{var}}_\Lambda (x)$ forms a Banach space ${\Lambda BV}(I)$ (see [19], Section 3).

Below the norm in the normed space E will be denoted by $\left\| \cdot \right\| _E$. The exception is the supremum norm in the space B(I) of all bounded functions on the interval I (and its subspaces) which will be denoted by $\left\| \cdot \right\| _\infty$. To shorten slightly the notation, we will also write $\left\| \cdot \right\| _{\Lambda BV}$ instead of $\Vert \cdot \Vert _{\Lambda BV(I)}$. The open ball, in the space E, centered at $x_0$ with radius $r>0$ will be denoted as $B_E(x_0,r)$, while the closed ball as $\overline{B}_E(x_0,r)$.

Let $a = t_0< t_1<\cdots<t_{n-1}<t_n=b$ be a partition of the interval J. We will denote such partition by $\pi = \{t_0,t_1,\ldots ,t_{n-1},t_n\}$.

Now we are going to rephrase the definition of $\Lambda$-variation (in a slightly different way than it was done before by Prus-Wiśniowski—see [16]). This, very simple, observation will play an essential role in our considerations.

Proposition 1

Let $x:J\rightarrow \mathbb {R}$ and let $\Lambda$ be a Waterman sequence. The following conditions are equivalent:

• (1) there exists a positive constant M such that for any finite partition $\pi = \{c_0, c_1, \ldots , c_n\}$ of J and any permutation $\sigma \in S_{n}$, we have

  $$\begin{aligned} \sum _{i=1}^{n}\frac{|x(c_{i}) - x(c_{i-1})|}{\lambda _{\sigma (i)}}\le M; \end{aligned}$$

• (2) x is of bounded $\Lambda$-variation.

Proof

Let x be as in (1). Let us choose an arbitrary finite sequence $\{[a_1, b_1], [a_2, b_2], \ldots , [a_n, b_n]\}$ of nonoverlapping subintervals of J. We complete it to be also a finite partition $\pi = \{c_0, c_1, \ldots , c_m\}$ of J. We have

$$\begin{aligned} \sum _{i=1}^{n}\frac{|x(b_i) - x(a_i)|}{\lambda _i} \le \sup _{\sigma \in S_m}\sum _{i=1}^{m}\frac{|x(c_{i}) - x(c_{i-1})|}{\lambda _{\sigma (i)}} \le \sup _{\sigma \in S_m}M = M. \end{aligned}$$

The verification of the inverse direction is trivial. $\square$

In the next lemma, to simplify the notation a little bit, instead of taking any interval $J\subset I$ we will focus on the entire interval I.

Lemma 1

There exists a positive constant $\tilde{c}_\Lambda$ such that for any function $x\in {\Lambda BV}(I)$ we have

$$\begin{aligned} \left\| x\right\| _\infty \le \tilde{c}_\Lambda \left\| x\right\| _{\Lambda BV}. \end{aligned}$$

Proof

Taking any $t\in I$, we get

$$\begin{aligned} |x(t)|\le & {} |x(0)|+|x(t)-x(0)| = |x(0)| + \lambda _1 \frac{|x(t)-x(0)|}{\lambda _1} \le |x(0)|\\&+\, \lambda _1\mathrm{{var}}_\Lambda (x) \le \max \{1,\lambda _1\}\left\| x\right\| _{\Lambda BV}. \end{aligned}$$

Finally, for completeness, let us recall the definition of nonlinear superposition operators.

Definition 3

Let J be an arbitrary interval.

• (a) Let $f:J \times \mathbb R \rightarrow \mathbb R$. The operator

  $$\begin{aligned} F(g)(u) {\mathrel {\mathop :}=}f(u, g(u)), \end{aligned}$$

  where $g:J \rightarrow \mathbb R$ is an arbitrary function, is called the nonautonomous superposition operator generated by the function f.

• (b) Let $f:\mathbb R \rightarrow \mathbb R$. The operator

  $$\begin{aligned} F(g)(u) {\mathrel {\mathop :}=}f( g(u)), \end{aligned}$$

  where $g:J \rightarrow \mathbb R$ is an arbitrary function, is called the autonomous superposition operator generated by the function f.

3 Nonautonomous superposition operators in the space of functions of bounded $\Lambda$-variation.

In this section, for the sake of simplicity, instead of an arbitrary interval J, we will consider the unit interval I. In the paper [3] Bugajewska et al. have stated the necessary and sufficient conditions which guarantee that the nonautonomous superposition operator generated by a function $f:I \times \mathbb R\rightarrow \mathbb R$ maps the space $\textit{BV}(I)$ of all functions of bounded variation in the sense of Jordan into itself and is locally bounded.

Theorem 1

([3, Theorem 3.8]) Suppose that $f:I \times \mathbb R\rightarrow \mathbb R$ is a given function. The following conditions are equivalent:

• (i) the nonautonomous superposition operator F, generated by f, maps the space BV(I) into itself and is locally bounded;

• (ii) for every $r>0$ there exists a constant $M_r>0$ such that for every $k \in \mathbb N$, every finite partition $0=t_0<\cdots <t_k=1$ of the interval I and every finite sequence $u_0, u_1,\ldots , u_k \in [-r,r]$ with $\sum _{i=1}^k |u_i - u_{i-1}|\le r$, the following inequalities hold

  $$\begin{aligned} \sum _{i=1}^k\bigl |f(t_i,u_i)-f(t_{i-1},u_i)\bigr | \le M_r \quad \text {and} \quad \sum _{i=1}^k\bigl |f(t_{i-1},u_i)-f(t_{i-1},u_{i-1})\bigr | \le M_r. \end{aligned}$$

Now we are going to present some results that show that the similar conditions may be stated for the more general case of nonlinear superposition operators between spaces $\Lambda BV(I)$ and $\Gamma BV(I)$ (to avoid confusion, let us explain that $\Lambda$ and $\Gamma$ in the above symbols mean that those spaces may be constructed with the help of two various Waterman sequences). But before we proceed to that more general case we are going to begin with the generators f satisfying the local Lipschitz condition with respect to the second variable.

First, let us explicitly write the definition which will be used in the sequel.

Definition 4

We say that a function $f:I\times \mathbb R\rightarrow \mathbb R$ is locally bounded if the image $f(I\times [-M,M])$ is bounded for any $M>0$. Similarly we say that the superposition operator $F:\Lambda BV(I)\rightarrow \Gamma BV(I)$ is locally bounded if the image of each ball $F(B_{{\Lambda BV}}(0,M))$ is bounded in $\Gamma BV(I)$.

Theorem 2

Let $f:I\times \mathbb {R}\rightarrow \mathbb {R}$ satisfy the following conditions:

• (i) f satisfies the local Lipschitz condition on $\mathbb {R}$ uniformly in $t\in I$;

• (ii) for any $r>0$ there exists $M_r>0$ such that for any positive number n, any partition $\pi$ of the interval I, any $u_0, u_1, \ldots , u_{n}\in \mathbb R$ and any permutation $\sigma \in S_n$, the following implication holds

  $$\begin{aligned} \sup _{\sigma \in S_n}\sum _{i=1}^{n}\frac{|u_i - u_{i-1}|}{\lambda _{\sigma (i)}} \le r \Longrightarrow \sup _{\sigma \in S_n}\sum _{i=1}^n\frac{|f(t_i, u_{i-1}) - f(t_{i-1},u_{i-1})|}{\lambda _{\sigma (i)}}\le M_r. \end{aligned}$$

Then the nonautonomous superposition operator F, generated by f, maps the space ${\Lambda BV}(I)$ into itself and is locally bounded.

Proof

Let $x\in {\Lambda BV}(I)$ be such that $\left\| x\right\| _{\Lambda BV}\le r$. Let $L_r$ be the uniform Lipschitz constant corresponding to the function $u\mapsto f(t,u)$ restricted to the interval $[-\tilde{r},\tilde{r}]$, where $\tilde{r} = r\tilde{c}_\Lambda$. For any partition $\pi = \{t_0, t_1, \ldots , t_n\}$ we have

$$\begin{aligned}&\sum _{i=1}^n\frac{|f(t_i, x(t_i)) - f(t_{i-1}, x(t_{i-1}))|}{\lambda _{\sigma (i)}}\\&\quad \le \sum _{i=1}^n\frac{|f(t_i, x(t_i)) - f(t_{i}, x(t_{i-1}))|}{\lambda _{\sigma (i)}} + \sum _{i=1}^n\frac{|f(t_i, x(t_{i-1})) - f(t_{i-1}, x(t_{i-1}))|}{\lambda _{\sigma (i)}}\\&\quad \le L_r\sum _{i=1}^n\frac{|x(t_i) - x(t_{i-1}) |}{\lambda _{\sigma (i)}} + M_r \le L_r\cdot r + M_r, \end{aligned}$$

which proves that F maps the space ${\Lambda BV}(I)$ into itself. Moreover, we have:

$$\begin{aligned} \left\| F(x)\right\| _{\Lambda BV}&= |F(x(0))| + \mathrm{{var}}_\Lambda (F(x)) \le |f(0, x(0))| + L_r\cdot \mathrm{{var}}_\Lambda (x) + M_r\\&\le |f(0, x(0)) - f(0, 0)| + |f(0,0)| + L_r\mathrm{{var}}_\Lambda (x) + M_r \\&\le L_r\left\| x\right\| _{\Lambda BV} + M_r + |f(0,0)|\le L_r\cdot r + M_r + |f(0,0)|, \end{aligned}$$

which means that F is locally bounded.

It appears that in the case of functions satisfying the local Lipschitz condition in view of the first variable, the condition (ii) of Theorem 2 plays an important role, if one considers the space BV(I). More precisely, the following result holds.

Theorem 3

([3], Proposition 3.3) Let us assume that $f:I\times \mathbb {R}\rightarrow \mathbb {R}$ satisfies the condition (i) of Theorem 2. If the nonautonomous superposition operator F, generated by f, maps the space BV(I) into itself and is locally bounded, then f satisfies (ii).

It is quite surprising that in the above result one cannot replace the space BV(I) by an arbitrary space ${\Lambda BV}(I)$. To establish that, let us consider the following.

Example 1

Let us consider the function $f:I\times \mathbb {R}\rightarrow \mathbb {R}$, defined by the formula

$$\begin{aligned} f(t,x) = {\left\{ \begin{array}{ll} 0, &{} \mathrm{if }\; t<1; \\ x, &{} \mathrm{if } \; t = 1. \end{array}\right. } \end{aligned}$$

The nonautonomous superposition operator, generated by the function f, maps any space $\Lambda BV(I)$ into any space $\Gamma BV(I)$ and it is locally bounded. Let $\lambda _n=n$ and $a_n=\frac{1}{\sqrt{n}}$, for $n \in \mathbb {N}$. Obviously $\sum _{n=1}^{\infty }{a_n} = +\infty$ and there exists $r>0$ such that $\sum _{n=1}^{\infty } \frac{a_n}{\lambda _n} \le r$. Therefore, one can take an arbitrarily large interval $[0, \sum _{n=1}^N a_n]$ and its partition consisting of points $u_i = \sum _{n=1}^i a_n$ $(i=1,\ldots ,N)$ and $u_0=0$. For that partition, the predecessor of the implication which appears in the condition (ii) of Theorem 2 is satisfied. On the other hand, we have

$$\begin{aligned} \sum _{i=1}^N\frac{|f(t_i, u_{i-1}) - f(t_{i-1},u_{i-1})|}{\lambda _{\sigma (i)}} = \frac{|f(1,u_{N-1})|}{\lambda _{\sigma (N)}}, \end{aligned}$$

so for some permutation $\sigma \in S_{N}$ the above sum is equal to $\frac{|u_{N-1}|}{\lambda _1}$, which means that it can be arbitrarily large.

Remark 2

Let us notice that the local Lipschitz condition imposed on a function $f:\mathbb R\rightarrow \mathbb R$ is a necessary and sufficient condition for the autonomous superposition operator to act from ${\Lambda BV}(I)$ to ${\Lambda BV}(I)$ (cf. [15]), so Theorem 2 presented above is the direct generalization of that result to the nonautonomous case.

The following three results describe some properties of nonautonomous superposition operators acting between spaces $\Lambda BV(I)$ and $\Gamma BV(I)$.

Theorem 4

Assume that $f:I\times \mathbb {R}\rightarrow \mathbb {R}$ generates a nonautonomous superposition operator F which maps a space $\Lambda BV(I)$ into $\Gamma BV(I)$. If f is not locally bounded, then neither is F.

Proof

Since f is not locally bounded, there exists $r>0$ and sequences $(t_n )_{n\in \mathbb {N}}$, $t_n\in I$ for $n\in \mathbb N$ and $(u_n)_{n\in \mathbb {N}}$, $u_n\in [-r, r]$ for $n\in \mathbb N$, such that

$$\begin{aligned} \lim _{n\rightarrow + \infty }| f(t_n, u_n) | = + \infty . \end{aligned}$$

For every $n\in \mathbb {N}$ we define

$$\begin{aligned} x_n(t) = {\left\{ \begin{array}{ll} u_n, &{}\qquad \text {if} \quad t=t_n,\\ 0, &{}\qquad \text {otherwise}. \end{array}\right. } \end{aligned}$$

We have $\left\| x_n\right\| _{\Lambda BV} = |x_n(0)| + \mathrm{{var}}_{\Lambda }(x_n) \le r + \frac{1}{\lambda _1} r + \frac{1}{\lambda _2} r$ for each n. By the assumptions, we infer that $F(x_n)\in \Gamma BV(I)$ and for fixed $s\ne t_n$ we have

$$\begin{aligned}&|f(t_n, u_n)| - |f(s,0)| \le |f(t_n, u_n) - f(s,0)|\\&\quad = \gamma _1 \frac{|f(t_n, x_n(t_n)) - f(s, x_n(s))|}{\gamma _1} \le \gamma _1 \mathrm{{var}}_{\Gamma }(F(x_n)), \end{aligned}$$

where s is an arbitrary point of the interval I. Hence, $\lim \limits _{n\rightarrow \infty } \mathrm{{var}}_{\Gamma }(F(x_n)) = + \infty$.

Theorem 5

If the nonautonomous superposition operator F, generated by f, maps a space $\Lambda BV(I)$ into a space $\Gamma BV(I)$, then for every $r>0$ the set $T_r = \{ t\in I: \sup \limits _{u\in [-r,r]} |f(t,u)| = + \infty \}$ is finite.

Proof

Let us assume that $F:\Lambda BV(I)\rightarrow \Gamma BV(I)$ and there exists such $r>0$ that the set $T_r$ is at least denumerable (countable and infinite). There exists a sequence $((t_n, u_n))_{n\in \mathbb {N}}$, $(t_n, u_n)\in I\times [-r,r]$ for $n\in \mathbb N$ such that:

• $| f(t_{n+1}, u_{n+1}) | \ge | f(t_n, u_n) | + 1$;

• $t_n$, $u_n$ are monotone;

• $t_n \rightarrow t_0$, $u_n \rightarrow u_0$ as $n \rightarrow +\infty$;

• $t_m \ne t_n$ if $m \ne n$.

Let us define the function

$$\begin{aligned} x(t) = {\left\{ \begin{array}{ll} u_n, \qquad &{}\text {if}\quad t=t_n,\\ u_0, \qquad &{}\text {if}\quad t=t_0,\\ \text {linear}, \qquad &{}\text {if}\quad t\in (\min \{ t_n, t_{n+1} \}, \max \{ t_n, t_{n+1} \}),\\ x(\sup _{n\in \mathbb {N}}t_n), \qquad &{}\text {if} \quad t\in ( \sup _{n\in \mathbb {N}} t_n, 1 ],\\ x(\inf _{n\in \mathbb {N}}t_n), \qquad &{}\text {if} \quad t\in [0, \inf _{n\in \mathbb {N}} t_n ). \end{array}\right. } \end{aligned}$$

This function is monotone, so it is of bounded variation in the sense of Jordan and, therefore, $x\in \Lambda BV(I)$. For any n we have

$$\begin{aligned} \mathrm{{var}}_{\Gamma }(F(x))\ge & {} \sum _{i=1}^{n}\frac{| F(x)(t_{i+1}) - F(x)(t_i) |}{\gamma _i}\\\ge & {} \sum _{i=1}^n \frac{| f(t_{i+1}, x(t_{i+1})) - f(t_i, x(t_i)) |}{\gamma _i} \ge \sum _{i=1}^n\frac{1}{\gamma _i}, \end{aligned}$$

which tends to infinity as $n \rightarrow +\infty$. This gives a contradiction.

Theorem 6

Let F be a nonautonomous superposition operator, generated by $f:I\times \mathbb {R}\rightarrow \mathbb {R}$, which maps a space $\Lambda BV(I)$ into $\Gamma BV(I)$. Then for every $u\in \mathbb R$ the function $t\mapsto f(t,u)$ is of bounded $\Gamma$-variation. Furthermore, in general, nothing can be said about the function $u\mapsto f(t,u)$, where $t\in I$ is fixed.

Proof

For every $u\in \mathbb {R}$ let us set $x_u = u$. In view of the assumption, $F(x_u)\in \Gamma BV(I)$, that is, the function $t\mapsto f(t,u)$ is of bounded $\Gamma$-variation.

To show the second claim, let us consider the function $f:I\times \mathbb {R}\rightarrow \mathbb {R}$ given by the formula $f(t,u) = h(t)g(u)$, where $g:\mathbb {R}\rightarrow \mathbb {R}$ is an arbitrary function and $h:I\rightarrow \mathbb {R}$ is defined as follows:

$$\begin{aligned} h(t) = {\left\{ \begin{array}{ll} 0, \qquad &{}\text {if}\quad t\in (0,1],\\ 1, \qquad &{}\text {if}\quad t = 0. \end{array}\right. } \end{aligned}$$

Then the nonautonomous superposition operator F, generated by the function f, is defined as follows:

$$\begin{aligned} F(x)(t) = {\left\{ \begin{array}{ll} 0, \qquad &{}\text {if}\quad t\in (0,1],\\ g(x(0)), \qquad &{}\text {if}\quad t=0, \end{array}\right. } \end{aligned}$$

where $x\in \Lambda BV(I)$. Therefore, F maps the space $\Lambda BV(I)$ into $\Gamma BV(I)$ and nothing can be said about the function $u\mapsto f(0,u) = g(u)$.

Now, let us introduce the following.

Definition 5

Let $u = \{u_0, u_1, \ldots , u_k\}$ be a finite sequence of real numbers. The number

$$\begin{aligned} \mathrm{{var}}_{\Lambda }(u) = \sup _{\tilde{u}}\left\{ \sup _{\sigma \in S_n}\sum _{i=1}^n \frac{|u_{k_{i}} - u_{k_{i-1}}|}{\lambda _{\sigma (i)}} \right\} , \end{aligned}$$

where $\tilde{u} = \{ u_{k_0}, u_{k_1}, \ldots , u_{k_n}\}$ is any subsequence of u, is called the $\Lambda$-variation of the sequence u.

We are ready to prove the main result of this section.

Theorem 7

Let $f:I\times \mathbb {R}\rightarrow \mathbb {R}$. The following conditions are equivalent:

• (i) the nonautonomous superposition operator F, generated by f, maps the space $\Lambda BV(I)$ into $\Gamma BV(I)$ and is locally bounded,

• (ii) for every $r>0$ there exists $M_r>0$ such that for every $k\in \mathbb {N}$, every partition $\pi =\{t_0, t_1,\ldots ,t_k\}$ of the interval I and every sequence $u=\{u_0, u_1, \ldots , u_k\}$ of elements of the interval $[-r\tilde{c}_{\Lambda },r\tilde{c}_{\Lambda }]$ such that $\mathrm{{var}}_{\Lambda }(u)\le r$, the following inequalities hold

  $$\begin{aligned}&\sup _{\sigma \in S_k}\sum _{i=1}^k \frac{| f(t_i, u_i) - f(t_{i-1}, u_i) |}{\gamma _{\sigma (i)}}\le M_r \quad \text {and}\\&\sup _{\sigma \in S_k}\sum _{i=1}^k \frac{| f(t_{i-1}, u_i) - f(t_{i-1}, u_{i-1}) |}{\gamma _{\sigma (i)}}\le M_r. \end{aligned}$$

Proof

(ii) $\Rightarrow$ (i) Let $x\in \Lambda BV(I)$ be such that $\left\| x\right\| _{\Lambda BV}\le r$. Let $M_r$ and $\pi$ be as in (ii). We have

$$\begin{aligned}&\sum _{i=1}^k \frac{| f(t_i, x(t_i)) - f(t_{i-1}, x(t_{i-1})) |}{\gamma _{\sigma (i)}}\\&\quad \le \sum _{i=1}^k \frac{| f(t_i, u_i) - f(t_{i-1}, u_i) |}{\gamma _{\sigma (i)}} + \sum _{i=1}^k \frac{| f(t_{i-1}, u_i) - f(t_{i-1}, u_{i-1}) |}{\gamma _{\sigma (i)}}\\&\quad \le 2M_r, \end{aligned}$$

where $u_i = x(t_i)$ and $\sigma \in S_k$. Furthermore, we have

$$\begin{aligned} \left\| F(x)\right\| _{\Gamma BV}&= |F(x(0))| + \mathrm{{var}}_{\Gamma }(F(x)) = |f(0, x(0))| + \mathrm{{var}}_{\Gamma }(F(x))\\&\le | f(0, x(0)) - f(0,0) | + | f(0,0) | + \mathrm{{var}}_{\Gamma }(F(x))\\&= \gamma _1\frac{| f(0,x(0)) - f(0,0) |}{\gamma _1} + | f(0,0) | + \mathrm{{var}}_{\Gamma }(F(x))\\&\le \gamma _1\cdot M_r + |f(0,0)| + 2M_r = (\gamma _1 + 2)M_r + |f(0,0)|, \end{aligned}$$

which means that F is locally bounded.

(i) $\Rightarrow$ (ii) Assume that F satisfies the condition (i) and there exists $r>0$ such that for every $n\in \mathbb {N}$ there exists a partition $\pi _n = \{ t_0^{(n)}, t_1^{(n)}, \ldots , t_{k_n}^{(n)} \}$ of the interval I and a sequence $u_0^{(n)}, u_1^{(n)}, \ldots , u_{k_n}^{(n)}$ of elements of the interval $[-r\tilde{c}_{\Lambda },r\tilde{c}_{\Lambda }]$ such that $\mathrm{{var}}_{\Lambda }(u)\le r$ and

$$\begin{aligned} \sup _{\sigma \in S_{k_n}} \sum _{i=1}^{k_n} \frac{| f( t_{i}^{(n)}, u_i^{(n)} ) - f( t_{i-1}^{(n)}, u_i^{(n)} ) |}{\gamma _{\sigma (i)}} > n \end{aligned}$$

(1)

or

$$\begin{aligned} \sup _{\sigma \in S_{k_n}} \sum _{i=1}^{k_n} \frac{| f( t_{i-1}^{(n)}, u_i^{(n)} ) - f( t_{i-1}^{(n)}, u_{i-1}^{(n)} ) |}{\gamma _{\sigma (i)}} > n. \end{aligned}$$

(2)

For each $n\in \mathbb {N}$ we define the function $\xi ^n :[0,1]\rightarrow \mathbb {R}$ by the formula

$$\begin{aligned} \xi ^n(t) = {\left\{ \begin{array}{ll} u_i^{(n)}, &{}\qquad \text {if}\quad t=t_i^{(n)},\\ \text {linear}, &{}\qquad \text {if} \quad t\in (t_{i-1}^{(n)}, t_i^{(n)}). \end{array}\right. } \end{aligned}$$

Obviously, we have

$$\begin{aligned} \mathrm{{var}}_{\Lambda }{(\xi ^{(n)})} = \mathrm{{var}}_{\Lambda }(u^{(n)}) \le r. \end{aligned}$$

Since F is locally bounded, there exists R (corresponding to r) such that $\sup \limits _{n\in N}\left\| F(\xi ^{n})\right\| _{\Gamma BV}\le ~R$. For any $n\in \mathbb {N}$ we have

$$\begin{aligned}&\sum _{i=1}^{k_n}\frac{| f(t_i^{(n)}, u_i^{(n)}) - f(t_{i-1}^{(n)}, u_i^{(n)}) |}{\gamma _{\sigma (i)}}\\&\quad \le \sum _{i=1}^{k_n}\frac{| f(t_i^{(n)}, u_i^{(n)}) - f(t_{i-1}^{(n)}, u_{i-1}^{(n)}) |}{\gamma _{\sigma (i)}} + \sum _{i=1}^{k_n}\frac{| f(t_{i-1}^{(n)}, u_{i-1}^{(n)}) - f(t_{i-1}^{(n)}, u_i^{(n)}) |}{\gamma _{\sigma (i)}}\\&\quad \le \sup _n \mathrm{{var}}_{\Gamma }({F(\xi ^{n})}) + \sum _{i=1}^{k_n}\frac{| f(t_{i-1}^{(n)}, u_{i-1}^{(n)}) - f(t_{i-1}^{(n)}, u_i^{(n)}) |}{\gamma _{\sigma (i)}}. \end{aligned}$$

Taking the supremum over $\sigma \in S_{k_n}$ we get

$$\begin{aligned}&\sup _{\sigma \in S_{k_n}}\sum _{i=1}^{k_n}\frac{| f(t_i^{(n)}, u_i^{(n)}) - f(t_{i-1}^{(n)}, u_i^{(n)}) |}{\gamma _{\sigma (i)}} \le \sup _n \mathrm{{var}}_{\Gamma }{(F(\xi ^{n}))} \\&\quad + \sup _{\sigma \in S_{k_n}}\sum _{i=1}^{k_n}\frac{| f(t_{i-1}^{(n)}, u_{i-1}^{(n)}) - f(t_{i-1}^{(n)}, u_i^{(n)}) |}{\gamma _{\sigma (i)}}. \end{aligned}$$

Similarly, we obtain

$$\begin{aligned}&\sum _{i=1}^{k_n}\frac{| f(t_{i-1}^{(n)}, u_i^{(n)}) - f(t_{i-1}^{(n)}, u_{i-1}^{(n)}) |}{\gamma _{\sigma (i)}}\\&\quad \le \sum _{i=1}^{k_n}\frac{| f(t_{i-1}^{(n)}, u_i^{(n)}) - f(t_{i}^{(n)}, u_{i}^{(n)}) |}{\gamma _{\sigma (i)}} + \sum _{i=1}^{k_n}\frac{| f(t_{i}^{(n)}, u_{i}^{(n)}) - f(t_{i-1}^{(n)}, u_{i-1}^{(n)}) |}{\gamma _{\sigma (i)}}\\&\quad \le \sup _n \mathrm{{var}}_{\Gamma }{(F(\xi ^{n}))} + \sum _{i=1}^{k_n}\frac{| f(t_{i-1}^{(n)}, u_{i}^{(n)}) - f(t_{i}^{(n)}, u_i^{(n)}) |}{\gamma _{\sigma (i)}}. \end{aligned}$$

Taking the supremum over $\sigma \in S_{k_n}$ we get

$$\begin{aligned}&\sup _{\sigma \in S_{k_n}}\sum _{i=1}^{k_n}\frac{| f(t_{i-1}^{(n)}, u_i^{(n)}) - f(t_{i-1}^{(n)}, u_{i-1}^{(n)}) |}{\gamma _{\sigma (i)}} \le \sup _n \mathrm{{var}}_{\Gamma }{(F(\xi ^{n}))}\\&\quad + \sup _{\sigma \in S_{k_n}}\sum _{i=1}^{k_n}\frac{| f(t_{i-1}^{(n)}, u_{i}^{(n)}) - f(t_{i}^{(n)}, u_i^{(n)}) |}{\gamma _{\sigma (i)}}. \end{aligned}$$

Hence, $\sup _{\sigma \in S_{k_n}}\sum _{i=1}^{k_n}\frac{| f(t_{i-1}^{(n)}, u_i^{(n)}) - f(t_{i-1}^{(n)}, u_{i-1}^{(n)}) |}{\gamma _{\sigma (i)}}$ tends to infinity (as $n\rightarrow \infty$) if and only if $\sup _{\sigma \in S_{k_n}}\sum _{i=1}^{k_n}\frac{| f(t_{i-1}^{(n)}, u_{i}^{(n)}) - f(t_{i}^{(n)}, u_i^{(n)}) |}{\gamma _{\sigma (i)}}$ tends to infinity. By this observation, we get

$$\begin{aligned} \lim _{n\rightarrow +\infty } \sup _{\sigma \in S_{k_n}}\sum _{i=1}^{k_n}\frac{| f(t_{i-1}^{(n)}, u_i^{(n)}) - f(t_{i-1}^{(n)}, u_{i-1}^{(n)}) |}{\gamma _{\sigma (i)}} = + \infty \end{aligned}$$

Let us fix $n\in \mathbb {N}$ and let us consider the finer partition $\pi$ with a sequence of additional points $\tau _i^{(n)}\in (t_{i-1}^{(n)}, t_{i}^{(n)})$, $i=1,2,\ldots ,k_n$. Let us define a sequence $(s_i^n)$ in the following manner:

$$\begin{aligned} s_0^n {\mathrel {\mathop :}=}t_0^{(n)},\quad s_1^n {\mathrel {\mathop :}=}\tau _1^{(n)},\quad s_2^n {\mathrel {\mathop :}=}t_1^{(n)},\quad s_3^n {\mathrel {\mathop :}=}\tau _2^{(n)}, \ldots , s_{2k_n-1}^n {\mathrel {\mathop :}=}\tau _{k_n}^{(n)},\quad s_{2k_n}^n{\mathrel {\mathop :}=}t_{k_n}^{(n)}. \end{aligned}$$

Let us define two functions:

$$\begin{aligned}&x^n(t) \\&\quad = {\left\{ \begin{array}{ll} u_i^{(n)}, \quad &{}\text {if}\quad t\in [s_{2i-2}^n, s_{2i-1}^n]\quad \text {for some}\quad i\in \{1,\ldots ,k_n\},\\ \frac{s_{2i}^n - t}{s_{2i}^n - s_{2i-1}^n}u_i^{(n)} + \frac{t-s_{2i-1}^n}{s_{2i}^n - s_{2i-1}^n}u_{i+1}^{(n)} \quad &{}\text {if}\quad t\in [s_{2i-1}^n, s_{2i}^n]\quad \text {for some}\quad i\in \{1,\ldots ,k_n-1\},\\ u_{k_n}^{(n)}, \quad &{}\text {if}\quad t\ge s_{2k_n-1}^n, \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned}&y^n(t)\\&\quad = {\left\{ \begin{array}{ll} \frac{s_{2i-1}^n - t}{s_{2i-1}^n - s_{2i-2}^n}u_{i-1}^{(n)} + \frac{t-s_{2i-2}^n}{s_{2i-1}^n - s_{2i-2}^n}u_{i}^{(n)} \quad &{}\text {if}\quad t\in [s_{2i-2}^n, s_{2i-1}^n]\quad \text {for some}\quad i\in \{1,\ldots ,k_n\},\\ u_{i}^{(n)}, \quad &{}\text {if}\quad t\in [s_{2i-1}^n, s_{2i}^n]\quad \text {for some}\quad i\in \{1,\ldots ,k_n\}. \end{array}\right. } \end{aligned}$$

Actually, by defining the above functions we were going to build two functions that are continuous, piecewise linear and constant in the left (the case of $x^{(n)}$) or in the right (the case of $y^{(n)}$) part of each interval $[t_{i-1}^{(n)}, t_i^{(n)}]$, and achieving values $u_i^{(n)}$ in points $t_i^{(n)}$ of the partition $\pi$.

We have $x^n(0) = u_1^{(n)}$, $y^n(0)=u_0^{(n)}$ and

$$\begin{aligned} \mathrm{{var}}_{\Lambda }{(x^n)} \le \mathrm{{var}}_{\Lambda }{(\xi ^{(n)})}\le r, \end{aligned}$$

and

$$\begin{aligned} \mathrm{{var}}_{\Lambda }{(y^n)} \le \mathrm{{var}}_{\Lambda }{(\xi ^{(n)})}\le r. \end{aligned}$$

Hence, $\left\| x^n\right\| _{\Lambda BV}\le 2r$ and $\left\| y^n\right\| _{\Lambda BV}\le 2r$ for each $n\in \mathbb {N}$. By the local boundedness of F we get $\left\| F(z^n)\right\| _{\Gamma BV}\le R$ for $z^n\in \{x^n, y^n\}$. On the other hand, we have

$$\begin{aligned}&\sum _{i=1}^{k_n}\frac{| f( t_{i-1}^{(n)}, u_i^{(n)} ) - f( t_{i-1}^{(n)}, u_{i-1}^{(n)} ) |}{\gamma _{\sigma (i)}}\\&\quad \le \sum _{i=1}^{k_n}\frac{| f( t_{i-1}^{(n)}, u_i^{(n)} ) - f( \tau _{i}^{(n)}, u_{i}^{(n)} ) |}{\gamma _{\sigma (i)}} + \sum _{i=1}^{k_n}\frac{| f( \tau _{i}^{(n)}, u_i^{(n)} ) - f( t_{i-1}^{(n)}, u_{i-1}^{(n)} ) |}{\gamma _{\sigma (i)}}\\&\quad = \sum _{i=1}^{k_n}\frac{| f( t_{i-1}^{(n)}, x^n(t_{i-1}^{(n)}) ) - f( \tau _{i}^{(n)}, x^{n}(\tau _i^{(n)}) ) |}{\gamma _{\sigma (i)}}\\&\qquad + \sum _{i=1}^{k_n}\frac{| f( \tau _{i}^{(n)}, y^n(\tau _{i}^{(n)}) ) - f( t_{i-1}^{(n)}, y^{n}(t_{i-1}^{(n)}) ) |}{\gamma _{\sigma (i)}}\\&\quad \le \mathrm{{var}}_{\Gamma }{(F(x^n))} + \mathrm{{var}}_{\Gamma }{(F(y^n))} \le 2R, \end{aligned}$$

which contradicts (2).

4 Continuity of the autonomous superposition operator

Now we are going to concentrate on the continuity of an autonomous superposition operator mapping the space ${\Lambda BV}(I)$ into itself.

First, we will show that the space ${\Lambda BV}(I)$ is a Banach algebra under a certain norm, equivalent to the norm $\left\| \cdot \right\| _{\Lambda BV}$.

Let us consider an arbitrary Waterman sequence $(\lambda _n)_{n\in \mathbb N}$. If that sequence is bounded by M, then for any $\lambda _i$, $i\in \mathbb N$, we have

$$\begin{aligned} \frac{1}{M} \le \frac{1}{\lambda _i} \le \frac{1}{\lambda _1} \end{aligned}$$

and, therefore, the norm $\left\| \cdot \right\| _{\Lambda BV}$ is equivalent to the norm $\left\| \cdot \right\| _{BV}$, what means that ${\Lambda BV}(I) = BV(I)$ and this space is a Banach algebra under certain equivalent norm.

If the sequence $(\lambda _n)_{n\in \mathbb N}$ is proper, that is, unbounded, then by Lemma 1 there exists such constant $\tilde{c}_\Lambda \ge 1$ that for any function $x\in {\Lambda BV}(I)$ we have

$$\begin{aligned} \left\| x\right\| _\infty \le \tilde{c}_\Lambda \left\| x\right\| _{\Lambda BV}. \end{aligned}$$

Hence, for any two functions $x,y\in {\Lambda BV}(I)$, any partition $\pi =\{ c_0,c_1, c_2,\ldots , c_n\}$ of I and any permutation $\sigma \in S_n$ we have

$$\begin{aligned}&|x(0) y(0)| + \sum _{k=1}^n\frac{| x(c_{i})y(c_{i}) - x(c_{i-1})y(c_{i-1}) |}{\lambda _{\sigma (i)}}\\&\quad \le \left\| y\right\| _\infty \left( |x(0)| + \sum _{i=1}^n\frac{| x(c_{i}) - x(c_{i-1})| }{\lambda _{\sigma (i)}} \right) \\&\qquad + \left\| x\right\| _\infty \left( |y(0)| + \sum _{i=1}^n\frac{| y(c_{i}) - y(c_{i-1})| }{\lambda _{\sigma (i)}} \right) \\&\quad \le \left\| y\right\| _\infty \left\| x\right\| _{\Lambda BV} + \left\| x\right\| _\infty \left\| y\right\| _{\Lambda BV}\\&\quad \le 2\tilde{c}_\Lambda \left\| x\right\| _{\Lambda BV}\left\| y\right\| _{\Lambda BV}. \end{aligned}$$

Therefore, we get

$$\begin{aligned} \left\| xy\right\| _{\Lambda BV}\le 2\tilde{c}_\Lambda \left\| x\right\| _{\Lambda BV}\left\| y\right\| _{\Lambda BV}. \end{aligned}$$

Let us multiply both sides of the above inequality by $2\tilde{c}_\Lambda$ and let us denote $\left\| x\right\| _{\Lambda BV}' = 2\tilde{c}_\Lambda \left\| x\right\| _{\Lambda BV}$. Then we get

$$\begin{aligned} \left\| xy\right\| _{\Lambda BV}' \le \left\| x\right\| _{\Lambda BV}'\left\| y\right\| _{\Lambda BV}' \end{aligned}$$

and, therefore, ${\Lambda BV}(I)$ is a Banach algebra under a norm, $\left\| \cdot \right\| _{\Lambda BV}'$ which is equivalent to the norm $\left\| \cdot \right\| _{\Lambda BV}$.

We begin our considerations concerning the problem of continuity of autonomous superposition operators with the case of analytic generators of such operators.

Theorem 8

Let $f:\mathbb {R}\rightarrow \mathbb {R}$ be a sum of a power series centered at 0 with the radius of convergence $\rho = +\infty$, that is, there exist real numbers $a_0, a_1, \ldots$ such that

$$\begin{aligned} f(u) = \sum _{i=0}^\infty a_i u^i \quad \text {for}\quad u\in \mathbb {R}. \end{aligned}$$

Then the autonomous superposition operator F, generated by f, which maps the Banach space ${\Lambda BV}(I)$ into itself is continuous.

Proof

Let us note that the operator F is well defined, since the function f satisfies a local Lipschitz condition (see Remark 2).

Write

$$\begin{aligned} f_n(u) = \sum _{i=1}^n a_i u^i, \qquad g_n(u) = f_n'(u) = \sum _{i=1}^n ia_i u^{i-1}, \end{aligned}$$

where $u\in \mathbb {R}$ and $u^0 = 1$. Furthermore, let $x^0(t)\equiv 1$ and

$$\begin{aligned} F_n(x) = \sum _{i=1}^n a_i x^i \quad \text {for every }n\in \mathbb N \text {and }x\in {\Lambda BV}(I) . \end{aligned}$$

Since ${\Lambda BV}(I)$ is a Banach algebra under a certain norm equivalent to $\left\| \cdot \right\| _{\Lambda BV}$, the mapping $F_n:{\Lambda BV}(I)\rightarrow {\Lambda BV}(I)$ is continuous. Therefore, to show the continuity of F it suffices to show that the sequence of mappings $(F_n)_{n\in \mathbb N}$ converges to F uniformly on bounded sets. Since the function $u\mapsto (f_n - f)(u)$ satisfies the Lipschitz condition on every interval $[-a, a]$ with the constant $L_n(a) = \sup _{u\in [-a, a]}|f'(u) - g_n(u)|$, we obtain

$$\begin{aligned} \mathrm{{var}}_{\Lambda }{(F_n(x) - F(x))} \le L_n(b)\mathrm{{var}}_{\Lambda }{(x)}, \end{aligned}$$

where $b = \tilde{c}_\Lambda \left\| x\right\| _{\Lambda BV}$, hence

$$\begin{aligned} \left\| F_n(x) - F(x)\right\| _{\Lambda BV}&= |f_n(x(0)) - f(x(0))| + \mathrm{{var}}_{\Lambda }{(F_n(x) - F(x))}\\&\le |f_n(x(0)) - f(x(0))| + L_n(a\tilde{c}_\Lambda )\mathrm{{var}}_{\Lambda }{(x)}, \end{aligned}$$

which shows that $F_n(x) \rightarrow F(x)$ as $n\rightarrow +\infty$ uniformly for every $x\in \overline{B}_{{\Lambda BV}}(0, a)$, where a is an arbitrary, fixed real number.

Now, we are going to consider generators of $C^1$-class. For that we will use the well-known concept of Bernstein polynomials. For the definition and properties (especially those required in the proof below) of Bernstein polynomials we would like to refer the reader to [5] and references therein.

Theorem 9

Let $f:\mathbb R\rightarrow \mathbb R$ be a continuously differentiable function. Then the autonomous superposition operator $F:{\Lambda BV}(I) \rightarrow {\Lambda BV}(I)$, generated by the function f, is continuous.

Proof

First observe that the operator F is well defined due to the continuity of $f'$ and the mean-value theorem.

To show its continuity we will approximate F by an almost uniformly convergent sequence of continuous mappings on the space ${\Lambda BV}(I)$.

For a given $a>0$ let $\phi _a$ denote the restriction of f to the interval $[-a, a]$. Moreover, let $F_n:\overline{B}_{{\Lambda BV}}(0,r)\rightarrow {\Lambda BV}(I)$ be the autonomous superposition operator generated by the nth Bernstein polynomial $B^a_n(\phi _a)$ of the function $\phi _a$, where $a = \tilde{c}_\Lambda r$. Since ${\Lambda BV}(I)$ is a Banach algebra, the operators $F_n$ are continuous.

Now we are going to show that the sequence $(F_n)_{n\in \mathbb {N}}$ converges uniformly to F on $\overline{B}_{{\Lambda BV}}(0,r)$. Note that the function $u\mapsto [f-B_n^a(\phi _a)](u)$ satisfies the Lipschitz condition on the interval $[-a, a]$ with the constant $L_n(a)=\sup _{u\in [-a, a]}|f'(u) - \frac{\text {d}}{\text {d} u}B_n^a(\phi _a)(u)|$, and thus we have

$$\begin{aligned} \mathrm{{var}}_{\Lambda }{(F(x) - F_n(x))} \le L_n(a)\mathrm{{var}}_{\Lambda }{(x)}\quad \text {for }x\in \overline{B}_{{\Lambda BV}}(0,r). \end{aligned}$$

Therefore, by [5] Proposition 1 and Proposition 2, for $x\in \overline{B}_{{\Lambda BV}}(0, r)$, we get

$$\begin{aligned} \left\| F(x) - F_n(x)\right\| _{\Lambda BV}&\le |f(x(0)) - B_n^a(\phi _a)( x(0) )| + L_n(a)\mathrm{{var}}_{\Lambda }{(x)}\\&\le |\phi _a(x(0)) - B_n^a(\phi _a)(x(0))| + r L_n(a)\rightarrow 0, \end{aligned}$$

as $n\rightarrow + \infty$, which ends the proof.

We should note here that we cannot expect a similar result to hold for the codomain being the proper subspace of the domain. To show it, first we introduce some useful notation. Given two Waterman sequences $\Lambda =(\lambda _n)_{n\in \mathbb {N}}$ and $\Gamma =(\gamma _n)_{n\in \mathbb {N}}$ we will write $\Gamma <\Lambda$, whenever

$$\begin{aligned} \lim _{n\rightarrow +\infty } \dfrac{\gamma _n}{\lambda _n} = 0. \end{aligned}$$

It is quite easy to show that if $\Gamma < \Lambda$, then

$$\begin{aligned} \lim _{n\rightarrow +\infty } \dfrac{\sum _j^n\frac{1}{\lambda _j}}{\sum _j^n \frac{1}{\gamma _j}}=0, \end{aligned}$$

what implies that $\Gamma BV(I)\subsetneq {\Lambda BV}(I)$ (see [13], Theorem 3).

We may also observe that since there exists such $M>0$ that $\frac{1}{\lambda _n}\le M\frac{1}{\gamma _n}$ we have

$$\begin{aligned} \left\| x\right\| _{\Lambda BV} \le M\left\| x\right\| _{\Gamma BV}, \end{aligned}$$

for all $x\in \Gamma BV(I)$, which makes the inclusion $\Gamma BV(I)\subsetneq {\Lambda BV}(I)$ a continuous map.

Theorem 10

If $\Gamma <\Lambda$ is a Waterman sequence and $F:\Lambda BV(I)\rightarrow \Gamma BV(I)$ is an autonomous superposition operator, generated by $f:\mathbb {R}\rightarrow \mathbb {R}$, then f is a constant function.

Proof

We may assume that $\gamma _n\ge 1$ for every $n\in \mathbb {N}$. We will prove that f is differentiable at every point of its domain and that its derivative is equal to zero.

Assume, contrary to our claim, that there exists $x_0\in \mathbb {R}$, $\varepsilon >0$ and a sequence $(x_n)_{n\in \mathbb {N}}$ of real numbers such that $x_n\rightarrow 0$ and

$$\begin{aligned} \left| \frac{f(x_0+x_n)-f(x_0)}{x_n}\right| \ge \varepsilon . \end{aligned}$$

(3)

We may assume that $(x_n)_{n\in \mathbb {N}}$ is monotone and, moreover, that all terms are of the same sign. To abbreviate a notation, let us assume that they are positive. Since $(x_n)_{n\in \mathbb {N}}$ tends to zero, there exists an infinite set $P\subseteq \mathbb {N}$ such that for every $p\in P$ there is m such that $p\le x^{-1}_m<{p+1}$.

Now, we are going to define some sequences $(N_k)_{k\in \mathbb {N}}, (m_k)_{k\in \mathbb {N}}, (p_k)_{k\in \mathbb {N}}$ of positive integers. Put $N_1=1$. Choose a positive integer $N_2>N_1$ such that $\lambda _n\ge 2\gamma _n$ for every $n\ge N_2$ and

$$\begin{aligned} p_1\le \sum _{n=N_1}^{N_2-1} \frac{1}{\gamma _n}<p_1+1 \end{aligned}$$

(4)

for some positive integer $p_1\in P, p_1\ge 2$. Such $N_2$ must exist since $\sum _{n=N_1}^\infty \frac{1}{\gamma _n} = +\infty$ and $\gamma _n\ge 1$ for every positive integer n. Let $m_1$ be such a positive integer that $p_1\le x^{-1}_{m_1}<p_1+1$.

Assume that $k>1$ and that $N_k, m_{k-1}, p_{k-1}$ are already defined. Choose a positive integer $N_{k+1}>N_k$ such that $\lambda _n\ge 2^k\gamma _n$ for every $n\ge N_{k+1}$ and

$$\begin{aligned} p_{k}\le \sum _{N_k}^{N_{k+1}-1} \frac{1}{\gamma _n}<p_{k}+1 \end{aligned}$$

(5)

for some positive integer $p_{k}\in P$ such that $p_k>\max \{p_{k-1}, 2^k\}$. As above, such $N_{k+1}$ must exist since $\sum _{n=N_k}^\infty \frac{1}{\gamma _n} = +\infty$ and $\gamma _n\ge 1$ for every positive integer n. Let $m_{k}$ be such a positive integer that $p_{k}\le x^{-1}_{m_{k}}<p_{k}+1$, that is

$$\begin{aligned} \frac{1}{p_k+1}<x_{m_k}\le \frac{1}{p_k}. \end{aligned}$$

(6)

Let us notice that

$$\begin{aligned} \frac{1}{p_k}<\frac{1}{2^k} \end{aligned}$$

(7)

for every positive integer k. Put $y_n=x_{m_k}$ for n such that $N_k\le n<N_{k+1}$. By (5) and (6), we get

$$\begin{aligned} \sum _{n=1}^\infty \frac{y_n}{\gamma _n}=\sum _{k=1}^\infty x_{m_k}\sum _{n=N_k}^{N_{k+1}-1}\frac{1}{\gamma _n}>\sum _{k=1}^\infty \frac{1}{p_k+1}\cdot p_k=+\infty . \end{aligned}$$

(8)

Using additionally (7) and the fact that $\lambda _n\ge 2^{k-1}\gamma _n$ for every $n\ge N_{k}$, we get

$$\begin{aligned} \sum _{n=1}^\infty \frac{y_n}{\lambda _n}\le \sum _{k=1}^\infty \frac{x_{m_k}}{2^{k-1}}\sum _{n=N_k}^{N_{k+1}-1}\frac{1}{\gamma _n}< \sum _{k=1}^\infty \frac{1}{2^{k-1}} \frac{p_k+1}{p_k}<+\infty . \end{aligned}$$

(9)

Now we are ready to define a function $g:[0,1]\rightarrow \mathbb {R}$ such that $g\in \Lambda BV(I)$ and $F(g)\not \in \Gamma BV(I)$ what will contradict our assumption. Let

$$\begin{aligned} g(t) = \left\{ \begin{array}{ll} y_n+x_0 &{} \text {if }t=\frac{1}{2n} \text {for} n\in \mathbb {N},\\ x_0 &{} \text {if } t\not =\frac{1}{2n}. \end{array} \right. \end{aligned}$$

Let $t_n=\frac{1}{n}$ for $n\in \mathbb {N}$. The function g is of bounded $\Lambda$-variation if and only if $\sum _{j=1}^\infty \frac{|g(t_{j+1})-g(t_j)|}{\lambda _j}<\infty$; however, by (9), we get

$$\begin{aligned} \sum _{n=1}^\infty \frac{|g(t_{n+1})-g(t_n)|}{\lambda _n}\le \sum _{n=1}^\infty \frac{|g(t_{2n})-g(t_{2n-1})|+|g(t_{2n+1})-g(t_{2n})|}{\lambda _n}<+\infty \end{aligned}$$

since $|g(t_{2n})-g(t_{2n-1})|=|g(t_{2n+1})-g(t_{2n})|=y_n$. Furthermore, by (3) and (8), we have

$$\begin{aligned} \sum _{n=1}^\infty \frac{|F(g)(t_{2n})-F(g)(t_{2n-1})|}{\gamma _n}=\sum _{n=1}^\infty \frac{|f(g(t_{2n}))-f(g(t_{2n-1}))|}{\gamma _n}= \end{aligned}$$

$$\begin{aligned} =\sum _{n=1}^\infty \frac{|f(x_0+y_{n})-f(x_0)|}{\gamma _n}\ge \varepsilon \sum _{n=1}^\infty \frac{y_{n}}{\gamma _n}=\infty . \end{aligned}$$

It means that F(g) is not of bounded $\Gamma$-variation, which gives a contradiction.

We will end this section considering the general case of continuous functions $f:\mathbb R\rightarrow \mathbb R$. Since f is not necessarily Lipschitz continuous, then the superposition operator generated by f and defined in $\Lambda BV(I)$ does not necessarily act in $\Lambda BV(I)$. However, we can show that it acts in some space $\Gamma BV(I)$. For that, we will need the following three lemmas.

Let $\Gamma _k=(\gamma ^k_n)_{n\in \mathbb N}$ for $k\in \mathbb {N}$.

Lemma 2

Assume that $\Gamma _k$ is a Waterman sequence for every $k\in \mathbb {N}$ and that

$$\begin{aligned} \gamma ^{k+1}_n\ge \gamma ^k_n \end{aligned}$$

(10)

for $n, k\in \mathbb {N}$. Then there exists a Waterman sequence $\Gamma$ such that $\Gamma _k BV(I)\subseteq \Gamma BV(I)$ for all $k\in \mathbb {N}$.

Proof

There exists a sequence $(N_k)_{k \in \mathbb {N}}$ of positive integers such that $N_1=1$ and

$$\begin{aligned} \sum _{n=N_k}^{N_{k+1}-1}\frac{1}{\gamma _n^k}\ge 1. \end{aligned}$$

Let us define $\gamma _n=\gamma ^k_n$ for $N_k\le n<N_{k+1}$. The sequence $\Gamma$ is monotone by (10) and the monotonicity of each $\Gamma _k$. Furthermore, $\sum _{n=1}^{\infty } \frac{1}{\gamma _n}=+\infty$, by the definition of the sequence $(N_k)_{k \in \mathbb {N}}$ and therefore $\Gamma$ is a Waterman sequence.

Inequality (10) implies that for every $k \in \mathbb {N}$ there exists $N \in \mathbb {N}$ such that $\gamma ^k_n\le \gamma _n$ for $n\ge N$ and hence $\Gamma _k BV(I)\subseteq \Gamma BV(I)$ for every positive integer k, which ends the proof.

Lemma 3

If $(w_n)_{n\in \mathbb N}$ is a monotone sequence of positive real numbers tending to zero and $\Gamma$ is a Waterman sequence, then there exists a Waterman sequence $\Gamma '=(\gamma '_n)_{n\in \mathbb N}$ such that $\gamma '_n\ge \gamma _n$ for every positive integer n and $\sum _{n=1}^{\infty }\frac{w_n}{\gamma '_n}<+\infty$.

Proof

Let $(N_k)_{k \in \mathbb {N}}$ be such a strongly increasing sequence of positive integers that

• (i) $N_1=1$;

• (ii) $\forall _{k>1}\forall _{n\ge N_k} w_n\le \frac{1}{2^k}$;

• (iii) $\forall _{k\ge 1} \sum _{n=N_{k+1}}^{N_{k+2}-1} \frac{1}{\gamma _n}\ge \sum _{n=N_{k}}^{N_{k+1}-1} \frac{1}{\gamma _n}\ge 1$.

Let us define

$$\begin{aligned}\gamma '_n=\left( \sum _{j=N_{k}}^{N_{k+1}-1} \frac{1}{\gamma _j}\right) \gamma _n\end{aligned}$$

for $N_k\le n<N_{k+1}$ and $k \in \mathbb {N}$. It is readily seen from (iii) that $\Gamma '$ is monotone and $\gamma '_n\ge \gamma _n$ for every positive integer n. It is a Waterman sequence, since

$$\begin{aligned}\sum _{n=1}^\infty \frac{1}{\gamma '_n}=\sum _{k=1}^\infty \sum _{n=N_k}^{N_{k+1}-1} \frac{1}{\gamma '_n}=\sum _{k=1}^\infty \sum _{n=N_k}^{N_{k+1}-1}\left( \sum _{j=N_{k}}^{N_{k+1}-1} \frac{1}{\gamma _j}\right) ^{-1}\frac{1}{\gamma _n}=+\infty .\end{aligned}$$

Finally, we get

$$\begin{aligned}\sum _{n=2}^\infty \frac{w_n}{\gamma '_n}=\sum _{k=2}^\infty \sum _{n=N_k}^{N_{k+1}-1} \frac{w_n}{\gamma '_n}\le \sum _{k=2}^\infty \frac{1}{2^k}\sum _{n=N_k}^{N_{k+1}-1} \left( \sum _{N_{k}}^{N_{k+1}-1} \frac{1}{\gamma _j}\right) ^{-1}\frac{1}{\gamma _n}=\sum _{k=2}^\infty \frac{1}{2^k}<+\infty .\end{aligned}$$

Lemma 4

If $\Lambda$ is a Waterman sequence, then there exists a Waterman sequence $\Lambda '=(\lambda '{_n})_{n \in \mathbb {N}}$ such that $\Lambda <\Lambda '$. Moreover, the sequence $(\frac{\lambda _n}{\lambda _n'})_{n \in \mathbb {N}}$ is monotone.

Proof

Let $(N_k)_{k \in \mathbb {N}}$ be such a sequence that $N_1=1$ and $\sum _{n=N_k}^{N_{k+1}-1}\frac{1}{\lambda _n}\ge 2^k$ for $k \in \mathbb {N}$. Let us define $\lambda '_n=2^k\lambda _n$ for $N_k\le n<N_{k+1}$ and $k \in \mathbb {N}$. Obviously, $\Lambda '$ is monotone, $\frac{\lambda _n}{\lambda '_n}\rightarrow 0$, as $n \rightarrow \infty$ and

$$\begin{aligned}\sum _{n=1}^\infty \frac{1}{\lambda '_n}=\sum _{k=1}^\infty \sum _{n=N_k}^{N_{k+1}-1}\frac{1}{\lambda '_n}= \sum _{k=1}^\infty \frac{1}{2^k}\sum _{n=N_k}^{N_{k+1}-1}\frac{1}{\lambda _n}=+\infty \end{aligned}$$

Hence, $\Lambda '$ is a Waterman sequence having all the desired properties.

Theorem 11

If $f:\mathbb {R}\rightarrow \mathbb {R}$ is continuous, F is the autonomous superposition operator, generated by f and $\Lambda$ is a Waterman sequence, then there exists a Waterman sequence $\Gamma$ such that $F(\Lambda BV(I))\subseteq \Gamma BV(I)$.

Proof

First, let $\Lambda '=(\lambda _{n}^{'})_{n \in \mathbb {N}}$ be the Waterman sequence given in Lemma 4. For every positive integer k let $\omega _k$ denote a modulus of continuity of the function $f|_{[-k,k]}$. We have $\omega _k(\delta )\rightarrow 0$, as $\delta \rightarrow 0$, because obviously a continuous function on compact interval is uniformly continuous.

We shall define a sequence $(\Gamma _k)_{k \in \mathbb {N}}$ of Waterman sequences such that $\gamma ^k_n\le \gamma ^{k+1}_n$ for all positive integers n, k. Let $\Gamma _1$ be a Waterman sequence such that

• (i) $\gamma ^1_n\ge \lambda '_n$;

• (ii) $\sum _{n=1}^{\infty }\frac{\omega _1(\frac{\lambda _n}{\lambda '_n})}{\gamma ^1_n}<+\infty$.

Assume that $\Gamma _k$ is already defined. Let $\Gamma _{k+1}$ be a Waterman sequence such that

• (i) $\gamma ^{k+1}_n\ge \gamma _n^k$;

• (ii) $\sum _{n=1}^{\infty }\frac{\omega _{k+1}(\frac{\lambda _n}{\lambda '_n})}{\gamma ^{k+1}_n}<+\infty$.

The above construction is possible by Lemma 3. Let $\Gamma$ be such Waterman sequence that $\Gamma _k BV(I)\subseteq \Gamma BV(I)$ for all $k\in \mathbb {N}$. Lemma 2 implies the existence of such a sequence.

Now it suffices to prove that if $x:I\rightarrow \mathbb {R}$ is of bounded $\Lambda$-variation, then F(x) is of bounded $\Gamma$-variation. Let $x\in \Lambda BV(I)$. Obviously, there exists a positive integer k such that $\left| x(t)\right| \le k$ for $t\in [0,1]$. Let $(I_n)_{n\in \mathbb N}\subseteq [0,1]$ be a sequence of nonoverlapping intervals. Let us define $A=\{n\in \mathbb {N}: |x(I_n)|\le \frac{\lambda _n}{\lambda '_n}\}$. Let us notice that $\sum _{n\not \in A}\frac{1}{\lambda '_n}\le \sum _{n\not \in A}\frac{|x(I_n)|}{\lambda _n}<+\infty$, since $x\in \Lambda BV(I)$. Using this fact, properties (i), (ii), and the monotonicity of $\omega _k$ for every $k\in \mathbb {N}$, we get

$$\begin{aligned}\sum _{n\in \mathbb {N}}\frac{|(f\circ x)(I_n)|}{\gamma ^k_n}\le \sum _{n\in \mathbb {N}}\frac{\omega _k(|x(I_n)|)}{\gamma ^k_n}=\sum _{n\in A}\frac{\omega _k(|x(I_n)|)}{\gamma ^k_n}+\sum _{n\not \in A}\frac{\omega _k(|x(I_n)|)}{\gamma ^k_n}\end{aligned}$$

$$\begin{aligned}\le \sum _{n\in A}\frac{\omega _k(\frac{\lambda _n}{\lambda '_n})}{\gamma ^k_n}+\sum _{n\not \in A}\frac{\omega _k(2k)}{\gamma ^k_n}\le \sum _{n\in A}\frac{\omega _k(\frac{\lambda _n}{\lambda '_n})}{\gamma ^k_n}+\omega _k(2k)\sum _{n\not \in A}\frac{1}{\lambda '_n}<+\infty .\end{aligned}$$

Hence, F(x) is of bounded $\Gamma _k$-variation and, therefore, it is also of bounded $\Gamma$-variation.

Now we are going to prove that the superposition operator considered above is actually continuous. We will use again Bernstein polynomials and will keep the notation used in Theorem 9. More precisely, we will need one more property of Bernstein polynomials which we will state below. But first let us remind the important definition.

Definition 6

(cf. [7], p. 40) The optimal modulus of continuity of a function $f:[-a,a]\rightarrow \mathbb R$ is a function $\omega _f:[0,2a]\rightarrow \mathbb R$ given by

$$\begin{aligned} \omega _f(\delta ) = \sup \bigl \{|f(t)-f(s)|:t,s \in [-a,a] \text { and } |t-s|\le \delta \bigr \}, \qquad \delta \ge 0. \end{aligned}$$

Proposition 2

(cf. [10]) If $\phi _a :[-a,a] \rightarrow \mathbb R$ is a continuous function with the optimal modulus of continuity $\omega _{\phi _a}$, then $\omega _{B_n^a(\phi _a)}(t) \le 4\omega _{\phi _a}(t)$ for $t \ge 0$.

As an obvious consequence of the above proposition, we get

$$\begin{aligned} \omega _{\phi _a-B_n^a(\phi _a)}(t) \le 5\omega _{\phi _a}(t), \quad \text {for}\quad t \in [-a,a]. \end{aligned}$$

An easy observation (the proof will be omitted) is the following.

Proposition 3

$\omega _{\phi _a - B_n^a(\phi _a)}(2a)\rightarrow 0$    as    $n\rightarrow +\infty$.

The following theorem refers to the superposition operator acting from the space $\Lambda BV(I)$ to certain space $\Gamma BV(I)$, which actually depends on the generator f (as it is constructed in Theorem 11).

Theorem 12

If $f:\mathbb {R}\rightarrow \mathbb {R}$ is continuous, F is the autonomous superposition operator, generated by f and $\Lambda$ is a Waterman sequence, then there exists such a Waterman sequence $\Gamma$ that the autonomous superposition operator $F:\Lambda BV(I)\rightarrow \Gamma BV(I)$ is continuous.

Proof

Let $\Gamma$ and $\Gamma _k$ $(k\in \mathbb N)$ be the Waterman sequences constructed in the proof of Theorem 11. Hence, $F(\Lambda BV(I))\subseteq \Gamma BV(I)$ and we may focus on the proof of the continuity of the superposition operator F.

Let us fix $r>0$ and put $a = \tilde{c}_\Lambda r$. Let us define $F_n:\overline{B}_{\Lambda BV}(0,r)\rightarrow \Gamma BV(I)$ as the sequence of autonomous superposition operators, generated by Bernstein polynomials $B_n^a(\phi _a)$, where $\phi _a = f|_{[-a,a]}$. Let us fix $k> a$. Since $\Gamma _k BV(I)$ is a Banach algebra, we conclude that the superposition operator $F_n$ is continuous, for every $n\in \mathbb {N}$. Now we are going to show that $F_n(x)$ converges to F(x), uniformly in view of $x\in \overline{B}_{\Lambda BV}(0,r)$, in the norm of the space $\Gamma _k BV(I)$ (and hence in the norm of the space $\Gamma BV(I)$).

Let us take any $x\in \overline{B}_{\Lambda BV}(0,r)$ and estimate the norm $\Vert F(x) - F_n(x)\Vert _{\Gamma _k BV}$. Let us take any collection $(I_m)_{m\in \mathbb N}$ of nonoverlapping intervals, contained in I. Let $\Lambda '$ be the Waterman sequence given as in the Lemma 4; without loss of generality we may assume that $\frac{\lambda _m}{\lambda _{m}^{'}}\le 2a$ for every $m\in \mathbb {N}$. Let $A=\{m\in \mathbb {N}: |x(I_m)|\le \frac{\lambda _m}{\lambda '_m}\}$. As above one can check that $\sum _{m\not \in A} \frac{1}{\lambda _m'} \le \sum _{m\not \in A} \frac{|x(I_m)|}{\lambda _m}\le \Vert x\Vert _{\Lambda BV}\le r$. Hence, we get

$$\begin{aligned} \sum _{m\in \mathbb {N}}\frac{|((\phi _a - B_n^a(\phi _a))\circ x)(I_m)|}{\gamma ^k_m}\le \sum _{m\in \mathbb {N}}\frac{\omega _{\phi _a - B_n^a(\phi _a)}(|x(I_m)|)}{\gamma ^k_m} \end{aligned}$$

$$\begin{aligned} = \sum _{m\in A}\frac{\omega _{\phi _a - B_n^a(\phi _a)}(|x(I_m)|)}{\gamma ^k_m}+\sum _{m\not \in A}\frac{\omega _{\phi _a - B_n^a(\phi _a)}(|x(I_m)|)}{\gamma ^k_m} \end{aligned}$$

$$\begin{aligned} \le \sum _{m\in \mathbb N}\frac{\omega _{\phi _a - B_n^a(\phi _a)}( \frac{\lambda _m}{\lambda '_m})}{\gamma ^k_m}+\omega _{\phi _a - B_n^a(\phi _a)}(2a)\sum _{m\not \in A}\frac{1}{\lambda '_m} \end{aligned}$$

$$\begin{aligned} \le \sum _{m\in \mathbb N}\frac{\omega _{\phi _a - B_n^a(\phi _a)}( \frac{\lambda _m}{\lambda '_m})}{\gamma ^k_m} + r\omega _{\phi _a - B_n^a(\phi _a)}(2a). \end{aligned}$$

Now, let us concentrate on the sum

$$\begin{aligned} \sum _{m\in \mathbb N}\frac{\omega _{\phi _a - B_n^a(\phi _a)}( \frac{\lambda _m}{\lambda '_m})}{\gamma ^k_m}. \end{aligned}$$

Fix $\varepsilon >0$. Since the series $\sum _{m\in \mathbb N}\frac{\omega _{k}( \frac{\lambda _m}{\lambda '_m})}{\gamma ^k_m}$ converges, we can find such $m_0\in \mathbb N$ that

$$\begin{aligned} \sum _{m\ge m_0} \frac{5\omega _{k}( \frac{\lambda _m}{\lambda '_m})}{\gamma ^k_m} < \frac{\varepsilon }{3} \end{aligned}$$

Then also

$$\begin{aligned} \sum _{m\ge m_0}\frac{\omega _{\phi _a - B_n^a(\phi _a)}( \frac{\lambda _m}{\lambda '_m})}{\gamma ^k_m} \le \sum _{m\ge m_0} \frac{5\omega _{k}( \frac{\lambda _m}{\lambda '_m})}{\gamma ^k_m} < \frac{\varepsilon }{3}. \end{aligned}$$

Now let us take such large $n\in \mathbb N$ that

$$\begin{aligned} \omega _{\phi _a - B_n^a(\phi _a)}(2a)\sum _{m\not \in A}\frac{1}{\lambda '_m}< \frac{\varepsilon }{3}, \end{aligned}$$

and

$$\begin{aligned} \sum _{m\le m_0}\frac{\omega _{\phi _a - B_n^a(\phi _a)}( \frac{\lambda _m}{\lambda '_m} )}{\gamma ^k_m} \le \omega _{\phi _a - B_n^a(\phi _a)}( 2a)\sum _{m\le m_0}\frac{1}{\gamma ^k_m} < \frac{\varepsilon }{3}. \end{aligned}$$

Now for n big enough we have

$$\begin{aligned} \sum _{m\in \mathbb {N}}\frac{|((\phi _a - B_n^a(\phi _a))\circ x)(I_m)|}{\gamma ^k_m} \end{aligned}$$

$$\begin{aligned} \le r\omega _{\phi _a - B_n^a(\phi _a)}(2a) + \sum _{m\ge m_0}\frac{\omega _{\phi _a - B_n^a(\phi _a)}( \frac{\lambda _m}{\lambda '_m})}{\gamma ^k_m} + \sum _{m\le m_0}\frac{\omega _{\phi _a - B_n^a(\phi _a)}( \frac{\lambda _m}{\lambda '_m} )}{\gamma ^k_m} < \varepsilon . \end{aligned}$$

which proves that $F_n$ converges to F, as $n\rightarrow \infty$, uniformly on the ball $\overline{B}_{\Lambda BV}(0,r)$. It implies the continuity of the operator F.

5 Applications

As applications of our results from previous sections we are going to give some theorems concerning the so-called $\Lambda BV$-solutions to some nonlinear integral equations.

5.1 Hammerstein integral equation

In this subsection we will be interested in the problem of the existence of solutions of bounded $\Lambda$-variation to the following nonlinear Hammerstein integral equation

$$\begin{aligned} x(t) := \lambda \int _0^1 k(t,s)f(x(s))d s,\qquad t\in I, \end{aligned}$$

(11)

where $\lambda \in \mathbb R$. Let us make the following assumptions:

• 1 $^{\circ }$ $f:\mathbb R\rightarrow \mathbb R$ is a continuously differentiable function which is sub-linear, that is, $\lim _{u\rightarrow \infty }|f(u)|/|u| = 0$;

• 2 $^{\circ }$ the kernel $k:I \times I\rightarrow \mathbb R$ is such that:

  • (a) for every $t\in I$ the function $s\mapsto k(t,s)$ is Lebesgue measurable;

  • (b) the function $s\mapsto k(0,s)$ is Lebesgue integrable;

  • (c) there exists such a Waterman sequence $\Lambda$ that $\mathrm{{var}}_{\Lambda }{(k(\cdot , s))}\le m(s)$ for a. e. $s\in I$, where $m:I \rightarrow [0,+\infty ]$ is a Lebesgue integrable function.

Remark 3

Let us note that if the function k satisfies assumptions 2 $^{\circ }$ (a)–(c), then for every $t\in I$, the function $s\mapsto k(t,s)$ is Lebesgue integrable on I. Indeed

$$\begin{aligned} |k(t,s)|\le & {} |k(0,s)| + |k(0,s) - k(t,s)| \le |k(0,s)| + \lambda _1 \mathrm{{var}}_{\Lambda }{(k(\cdot ,s))}\\\le & {} \lambda _1 m(s) + |k(0,s)|\text { for a.e. }s\in I, \end{aligned}$$

which confirms our claim.

Proposition 4

Let $k:I\times I\rightarrow \mathbb R$ satisfy 2 $^{\circ }$(a) – (c). Then the operator K, defined by the formula

$$\begin{aligned} Kx(t) = \int _0^1 k(t,s)x(s)d s\quad t\in I,\quad x\in \Lambda BV(I), \end{aligned}$$

(12)

maps the space $\Lambda BV(I)$ into $\Lambda BV(I)$ and is compact.

Proof

Let $x\in \Lambda BV(I)$. Then, by Remark 3, we have

$$\begin{aligned} \biggl |\int _0^1 k(t,s)x(s)d s\biggr | \le \int _0^1 |k(t,s)x(s)|d s \le \left\| x\right\| _\infty m_1 \le \tilde{c}_{\Lambda } m_1 \left\| x\right\| _{\Lambda BV}, \end{aligned}$$

where $m_1 = \int _0^1(\lambda _1 m(s) + |k(0,s)|)d s$. Hence, Kx(t) exists and is finite for every $t\in I$. Moreover, for any partition $\pi =\{t_0, t_1,\ldots ,t_n\}$ of the interval I and any permutation $\sigma \in S_n$ we have

$$\begin{aligned} \sum _{i=1}^n \frac{| Kx(t_i) - Kx(t_{i-1}) |}{\lambda _{\sigma (i)}}&\le \int _0^1\sum _{i=1}^n\frac{| (k(t_i, s) - k(t_{i-1},s))x(s) |}{\lambda _{\sigma (i)}}d s\\&\le \int _0^1 \mathrm{{var}}_{\Lambda }{(k(\cdot ,s))}|x(s)|d s\\&\le \left\| x\right\| _\infty \int _0^1 m(s)d s \le \tilde{c}_{\Lambda }m_2\left\| x\right\| _{\Lambda BV}, \end{aligned}$$

where $m_2 = \int _0^1 m(s)d s$. Since

$$\begin{aligned} \left\| Kx\right\| _{\Lambda BV}&\le \int _0^1 |k(0,s)x(s)|d s + \tilde{c}_{\Lambda }m_2\left\| x\right\| _{\Lambda BV} \\&\le \tilde{c}_{\Lambda }\left( \int _0^1 |k(0,s)|d s + m_2 \right) \left\| x\right\| _{\Lambda BV} = c\left\| x\right\| _{\Lambda BV}, \end{aligned}$$

where

$$\begin{aligned} c=\tilde{c}_{\Lambda }\left( \int _0^1 |k(0,s)|d s + m_2 \right) , \end{aligned}$$

we conclude that K is well defined and continuous.

Now we will show that K is compact. Let $(x_n)_{n\in \mathbb N}$ be a sequence of elements of $\overline{B}_{\Lambda BV}(0,1)$. In view of Helly’s extraction theorem (cf. [11], Theorem 3.2), there exists a subsequence $(x_{n_k})_{k\in \mathbb {N}}$ of $(x_n)_{n\in \mathbb {N}}$, pointwise convergent to some $x\in \overline{B}_{\Lambda BV}(0,1)$. Let us set $y_k = x_{n_k} - x$ for $k\in \mathbb N$. We will show that $Ky_k\rightarrow 0$ with respect do the $\Lambda BV$-norm. Given $\varepsilon > 0$ let $k_0$ be such that

$$\begin{aligned} \int _0^1|k(0,s)y_k(s)|d s\le \frac{\varepsilon }{2} \quad \text {and}\quad \int _0^1 m(s)|y_k(s)|d s \le \frac{\varepsilon }{2}\quad \text {for }k\ge k_0. \end{aligned}$$

Let $\pi$ and $\sigma$ be as above. Then

$$\begin{aligned} \sum _{i=1}^n\frac{| Ky(t_i) - Ky(t_{i-1}) |}{\lambda _{\sigma (i)}}&\le \int _0^1\sum _{i=1}^n\frac{| k(t_i, s) - k(t_{i-1},s) |}{\lambda _{\sigma (i)}}|y_k(s)|d s\\&\le \int _0^1 m(s)|y_k(s)|d s \le \frac{\varepsilon }{2}, \end{aligned}$$

and therefore $\mathrm{{var}}_{\Lambda }{(Ky_k)}\le \frac{\varepsilon }{2}$ for $k\ge k_0$. Finally, we obtain

$$\begin{aligned} \left\| Ky_k\right\| _{\Lambda BV} \le \int _0^1 |k(0,s)y_k(s)|d s + \int _0^1 m(s)|y_k(s)|d s \le \varepsilon \end{aligned}$$

which ends the proof.

Since we have the completely continuous linear map we can apply the Schauder fixed point theorem to get the following existence result.

Theorem 13

Let $k:I\times I\rightarrow \mathbb R$ and $f:\mathbb R\rightarrow \mathbb R$ satisfy 2 $^{\circ }$ and 1 $^{\circ }$, respectively. Then for every $\lambda \in \mathbb R$ there exists a $\Lambda BV$-solution to equation (11).

Proof

If $\lambda = 0$ the claim is obvious. Without loss of generality we can assume that $\lambda = 1$. Let us consider the operator $G = K \circ F :\Lambda BV(I) \rightarrow \Lambda BV(I)$, where the integral operator $K:\Lambda BV(I) \rightarrow \Lambda BV(I)$ is defined in (12) and $F:\Lambda BV \rightarrow \Lambda BV$ is the autonomous superposition operator generated by the function f. Due to Proposition 4, the operator K is continuous and due to Theorem 9, the operator F is also continuous, so the G is continuous, too. Moreover, it is completely continuous, because F maps bounded sets into bounded ones as a consequence of f being a locally Lipschitz function.

Therefore, it is enough to find a closed ball $\overline{B}_{\Lambda BV}(0,a)\subset {\Lambda BV}(I)$ which is invariant under the completely continuous map G. First, let us observe that for any ball $\overline{B}_{{\Lambda BV}}(0,a)$ and $x\in \overline{B}_{{\Lambda BV}}(0,a)$ there is $\left\| x\right\| _\infty \le \tilde{c}_\Lambda a$ and

$$\begin{aligned} \left\| F(x)\right\| _\infty \le \sup _{u\in [-\tilde{c}_\Lambda a, \tilde{c}_\Lambda a]}|f(u)|. \end{aligned}$$

By assumption 1 $^{\circ }$ (a) there exists $R>0$ such that

$$\begin{aligned} \sup _{u\in [-\tilde{c}_\Lambda R,\tilde{c}_\Lambda R]}|f(u)|\left( \int _0^1 |k(0,s)|d s + \int _0^1 m(s) d s \right) \le R. \end{aligned}$$

Otherwise, there would exist a sequence $(u_n)_{n\in \mathbb N}$ of real numbers for which we would have

$$\begin{aligned} |f(u_n)|\cdot \left( \int _0^1 |k(0,s)|d s + \int _0^1 m(s) d s \right) > n\quad \text {and} \quad |u_n| \le \tilde{c}_\Lambda n. \end{aligned}$$

The sequence $(u_n)_{n\in \mathbb N}$ is not bounded, so there must exist an appropriate subsequence $(w_n)_{n\in \mathbb N}$, such that $|w_n|\rightarrow + \infty$ and

$$\begin{aligned} \frac{|f(w_n)|}{|w_n|}\ge \left( \int _0^1 |k(0,s)|d s + \int _0^1 m(s) d s \right) ^{-1} \tilde{c}_\Lambda ^{-1}, \end{aligned}$$

which contradicts 1 $^{\circ }$ (a). Therefore,

$$\begin{aligned} \left\| G(x)\right\| _{\Lambda BV}&= |G(x)(0)| + \mathrm{{var}}_{\Lambda }{(G(x))} \le \int _0^1 |k(0,s)||f(x(s))|d s\\&\quad + \int _0^1 m(s)|f(x(s))|d s \\&\le \sup _{u\in [ -\tilde{c}_\Lambda R, \tilde{c}_\Lambda R ]} |f(u)| \left( \int _0^1 |k(0,s)|d s + \int _0^1 m(s) d s \right) \le R \end{aligned}$$

for $x\in \overline{B}_{{\Lambda BV}}(0,R)$, which implies that the ball $\overline{B}_{{\Lambda BV}}(0,R)$ is invariant under the mapping G. By the Schauder fixed point theorem, this implies that there exists a fixed point of G, which completes the proof.

Let us focus on the continuity of the operator K which appears in Proposition 4. In the paper [6] the authors gave the sufficient and necessary conditions for the integral operator $K:BV(I)\rightarrow BV(I)$ of the form (12) to be continuous (see [6], Theorem 4). It appears that similar conditions may be stated for such integral operators $K:BV(I)\rightarrow {\Lambda BV}(I)$.

Notation

For a given function $f\in BV(I)$, by $v_f$ we will denote the function $t\rightarrow \mathrm{{var}}(f,[0,t]), t\in I$, where $\mathrm{{var}}(f,[0,t])$ denotes the variation in the sense Jordan of the function f over the interval [0, t]. Moreover, to avoid misunderstandings in the next theorem as well as in its proof, the Lebesgue integral will be denoted by “$(L)\int$” while the Riemann–Stieltjes integral is denoted by “$(RS)\int$”.

Theorem 14

The linear integral operator K, defined by (12), maps continuously the space BV(I) into $\Lambda BV(I)$ if and only if the following conditions are satisfied:

• (i) for every $t \in I$ the function $s \mapsto k(t,s)$ is Lebesgue integrable on I,

• (ii) there exists a constant $M>0$ such that $\displaystyle \sup _{\xi \in I} \mathrm{{var}}_\Lambda \biggl ((L)\int _0^{\xi } k(\cdot ,s)d s\biggr ) \le M$.

Proof

Suppose that the conditions (i) and (ii) hold. We will show that the operator K maps BV(I) into $\Lambda BV(I)$ and is continuous.

Let us note that by [6], Remark 5, the function Kx is well defined for every $x \in BV(I)$. Suppose now that $I_1,\ldots , I_n$ is an arbitrary finite family of nonoverlapping subintervals of I of the form $I_i=[a_i,b_i]$. Then, by [6], Proposition 2 and the condition (ii), for every $x \in BV(I)$, we get

$$\begin{aligned} \sum _{i=1}^n \frac{\bigl |Kx(I_i)\bigr |}{\lambda _i}&= \sum _{i=1}^n \frac{1}{\lambda _i}\biggl |(L)\int _0^1 [k(b_i,s) - k(a_{i},s)]x(s) d s\biggr | \\&\le |x(1)| \cdot \sum _{i=1}^n \frac{1}{\lambda _i} \biggl |(L)\int _0^1 [k(b_i,s) - k(a_{i},s)]d s\biggr |\\&\quad + (RS) \int _0^1 \sum _{i=1}^n \frac{1}{\lambda _i}\biggl |(L)\int _0^\xi [k(b_i,s) - k(a_{i},s)]d s\biggr | dv_x(\xi )\\&\le M\left\| x\right\| _{\infty } + Mv_x(1) \le M\left\| x\right\| _{BV}\!\!, \end{aligned}$$

which implies that $Kx \in \Lambda BV(I)$ and $\mathrm{{var}}_{\Lambda }(Kx) \le 2M\left\| x\right\| _{BV}$. Therefore, $\left\| Kx\right\| _{\Lambda BV} \le \bigl (\left\| k(0,\cdot )\right\| _{L^1}+2M\bigr )\cdot \left\| x\right\| _{BV}$ for $x \in BV(I)$. This shows that the linear operator K is continuous.

Now, we shall show that the conditions (i) and (ii) are necessary. The function $x \equiv 1$ is clearly of bounded Jordan variation, and hence the function

$$\begin{aligned} t \mapsto (L)\int _0^1 k(t,s) d s \end{aligned}$$

is of bounded $\Lambda$-variation. Thus, for every $t \in I$, the function $s \mapsto k(t,s)$ is Lebesgue integrable on I. It remains to show the condition (ii). Since the linear operator K is continuous, there exists a positive number $M>0$ such that $\left\| Kx\right\| _{\Lambda BV} \le M\left\| x\right\| _{BV}$ for every $x \in BV(I)$. Hence,

$$\begin{aligned} \mathrm{{var}}_{\Lambda }(K\chi _{[0,\xi ]}) \le \left\| K\chi _{[0,\xi ]}\right\| _{\Lambda BV} \le M\left\| \chi _{[0,\xi ]}\right\| _{BV} \le 2M. \end{aligned}$$

Therefore,

$$\begin{aligned} \sup _{\xi \in I} \mathrm{{var}}_{\Lambda } \biggl ( (L)\int _0^{\xi } k(t,s) d s \biggr ) = \sup _{\xi \in I} \mathrm{{var}}_{\Lambda }(K\chi _{[0,\xi ]}) \le 2M, \end{aligned}$$

which ends the proof. $\square$

5.2 Volterra–Hammerstein integral equation.

In this subsection we are going to focus not only on the existence of $\Lambda BV$-solutions to the nonlinear Volterra–Hammerstein integral equation but we would like to examine also the topological structure of such solution sets. More precisely, let us consider the following nonlinear Volterra–Hammerstein integral equation

$$\begin{aligned} x(t)= g(t) + \int _0^t k(t,s)f(s,x(s))ds, \quad t\in I, \end{aligned}$$

(13)

where

• 3$^\circ$ $g\in C(I)\cap {\Lambda BV}(I)$;

• 4$^\circ$ $f:I\times \mathbb R\rightarrow \mathbb R$ satisfies the Carathéodory conditions, that is:

  • (a) for every $u\in \mathbb R$ the function $t\mapsto f(t,u)$ is Lebesgue measurable;

  • (b) for a.e. $t\in I$ the function $u\mapsto f(t,u)$ is continuous;

  • (c) $|f(t,u)|\le m_1(t)$ for $(t,u)\in I\times \mathbb R$ with $m_1\in L^p(I)$, where $p\in (1,+\infty ]$;

• 5$^\circ$ the kernel $k:\Delta \rightarrow \mathbb R$, where $\Delta := \{ (t,s)\in I\times I : 0\le s\le t\le 1 \}$, is such that:

  • (a) for every $t\in I$ the function $s\mapsto k(t,s)$ is Lebesgue measurable on [0, t];

  • (b) $|k(s,s)| + \mathrm{{var}}_{\Lambda }({k(\cdot , s)}, [s,1]) \le m_2(s)$ for a.e $s\in I$ with $m_2\in L^q(I), \text {where } q^{-1} + p^{-1} = 1$;

  • (c) for every $\varepsilon > 0$ there exists $\delta > 0$ such that

    $$\begin{aligned} \int _0^1 |k(\tau , s) - k(t,s)| m_1(s)ds \le \varepsilon , \end{aligned}$$

    for all $(\tau , t)\in \Delta$ such that $0\le \tau - t\le \delta$.

Remark 4

Let us note that if the kernel k satisfies the assumptions 5$^\circ$ (a) and (b), then for every $t\in I$ the function $s\mapsto k(t,s)$ belongs to $L^q[0,t]$. Indeed, given any $t\in I$, we have

$$\begin{aligned} |k(t,s)|\le & {} \frac{\lambda _1 |k(s,s) - k(t,s)|}{\lambda _1} + |k(s,s)| \le \lambda _1 \mathrm{{var}}_{\Lambda }({k(\cdot , s)}, [s,1]) + |k(s,s)|\\\le & {} \max ( \lambda _1,1) m_2(s), \end{aligned}$$

for almost every $s\in [0,t]$, which confirms our claim.

First, let us focus on the continuity of a certain linear Volterra integral operator.

Lemma 5

Let $p\in (1,+\infty ]$. If the kernel k satisfies assumptions 5$^{\circ }$ (a) and (b), then the linear Volterra integral operator K defined by

$$\begin{aligned} Kx(t) = \int _0^t k(t,s)x(s)ds,\quad t\in I, \end{aligned}$$

(14)

maps the space $L^p(I)$ into ${\Lambda BV}(I)$ and is continuous.

Proof

Let $x\in L^p(I)$. First let us observe that the integral

$$\begin{aligned} \int _0^t k(t,s)x(s)ds \end{aligned}$$

exists and is finite for every $t\in I$ (cf. Remark 4), hence the operator K is well defined. Given an arbitrary partition $0 = t_1 \le t_2\le \ldots \le t_n = 1$ and $\sigma \in S_n$ we have

$$\begin{aligned} \sum _{i=1}^n \frac{|Kx(t_i) - Kx(t_{i-1})|}{\lambda _{\sigma (i)}}= & {} \sum _{i=1}^n \frac{|\int _0^{t_i} k(t_i, s)x(s)ds - \int _0^{t_{i-1}} k(t_{i-1}, s)x(s)ds| }{\lambda _{\sigma (i)}}\\\le & {} \int _0^1 \sum _{i=1}^n \frac{| \tilde{k}(t_i,s) - \tilde{k}(t_{i-1}, s) |}{\lambda _{\sigma (i)}}|x(s)|ds, \end{aligned}$$

where

$$\begin{aligned} \tilde{k}(t,s) = {\left\{ \begin{array}{ll} k(t,s), \qquad &{}\text {if}\quad (t,s)\in \Delta ,\\ 0, \qquad &{}\text {if}\quad (t,s) \notin \Delta . \end{array}\right. } \end{aligned}$$

(15)

Since

$$\begin{aligned} \sum _{i=1}^n \frac{| \tilde{k}(t_i, s) - \tilde{k}(t_{i-1},s) |}{\lambda _{\sigma (i)}}\le & {} \frac{|k(s,s)|}{\lambda _{1}} + \mathrm{{var}}_{\Lambda }(k(\cdot , s),[s,1])\\\le & {} \max (\frac{1}{\lambda _1}, 1) m_2(s) \quad \text {for a.e. }s\in I, \end{aligned}$$

we infer that

$$\begin{aligned} \sum _{i=1}^n \frac{|Kx(t_i) - Kx(t_{i-1})|}{\lambda _{\sigma (i)}} \le \max (\frac{1}{\lambda _1},1) \int _0^1 m_2(s)|x(s)|ds, \end{aligned}$$

which proves that $\left\| Kx\right\| _{\Lambda BV} \le \max (\frac{1}{\lambda _1},1) \left\| m_2\right\| _{L^q}\cdot \left\| x\right\| _{L^p}$, which means that K maps the space $L^p(I)$ into ${\Lambda BV}(I)$ and is continuous. $\square$

We will also need the following.

Lemma 6

Let $p\in (1, +\infty ]$. Suppose the assumptions 5$^{\circ }$ (a) and (b) hold. If a bounded sequence $(x_n)_{n\in \mathbb N}$, $x_n\in L^p(I)$ for $n\in \mathbb N$ converges almost everywhere (or in measure) to a function $x\in L^p(I)$, then the sequence $(Kx_n)_{n\in \mathbb N}$, where K is given by (14), converges to Kx with respect to the ${\Lambda BV}$-norm.

Proof

First let us note that Kx, $Kx_n \in {\Lambda BV}(I)$, by Lemma 5. Let us take an arbitrary partition $0 = t_0 \le t_1 \le \ldots \le t_n = 1$ of the interval I and a permutation $\sigma \in S_n$. Then for the function $\tilde{k}:I\times I \rightarrow \mathbb R$, defined by (15), we have

$$\begin{aligned}&\sum _{i=1}^n \frac{|Kx(t_i) - Kx_n(t_i) - Kx(t_{i-1}) + Kx_n(t_{i-1}))|}{\lambda _{\sigma (i)}} \\&\le \int _0^1\sum _{i=1}^n\frac{| \tilde{k}(t_i, s) - \tilde{k}(t_{i-1},s) |}{\lambda _{\sigma (i)}}|x_n(s) - x(s)|ds. \end{aligned}$$

Hence,

$$\begin{aligned} \mathrm{{var}}_{\Lambda }(Kx_n - Kx) \le \max (\frac{1}{\lambda _1}, 1) \int _0^1 m_2(s)|x_n(s) - x(s)| ds. \end{aligned}$$

Therefore, in view of the assumptions and Vitali’s Convergence Theorem we get $\left\| Kx_n - Kx\right\| _{\Lambda BV}\rightarrow 0$, as $n\rightarrow + \infty$. $\square$

Now we are going to present the Aronszajn-type result for $\Lambda BV$-solutions to the Volterra–Hammerstein equation in ${\Lambda BV}(I)$. We are going to apply the certain Vidossich-type result. In what follows, assume that $\mathcal {K}$ is a bounded and convex subset of a normed space, and E is a Banach space. Denote by $C(\mathcal {K},E)$ the space of all bounded and continuous functions $x :\mathcal {K} \rightarrow E$ endowed with the supremum norm.

Theorem 15

([17], Theorem 2) Let $F:C(\mathcal {K},E)\rightarrow C(\mathcal {K},E)$ be a continuous mapping satisfying the following conditions:

• (i) the set $F(C(\mathcal {K},E))$ is equiuniformly continuous;

• (ii) there exists $t_0\in \mathcal {K}$ and $x_0\in E$ such that $F(x)(t_0) = x_0$ for every $x\in C(\mathcal {K}, E)$;

• (iii) for every $\varepsilon > 0$ and every $x,y\in C(\mathcal {K},E)$ the following implication holds

  $$\begin{aligned} x|_{\mathcal {K}_\varepsilon } = y|_{\mathcal {K}_\varepsilon } \Rightarrow F(x)|_{\mathcal {K}_\varepsilon } = F(y)|_{\mathcal {K}_\varepsilon }, \end{aligned}$$

  where $\mathcal {K}_\varepsilon = \{ t\in \mathcal {K} : \left\| t- t_0\right\| \le \varepsilon \}$;

• (iv) every sequence $(x_n)_{n\in \mathbb N}$ in $C(\mathcal {K},E)$ such that $\lim _{n\rightarrow + \infty }(x_n - F(x_n)) = 0$ has a limit point.

Then the set of fixed points of the mapping F is a compact $R_\delta$, that is, it is homeomorphic to the intersection of a decreasing sequence of compact absolute retracts.

Theorem 16

If the assumptions 3$^\circ$ - 5$^\circ$ hold, then the set T of all continuous solutions of bounded $\Lambda$-variation to the nonlinear Volterra–Hammerstein integral equation (13) is a compact $R_\delta$ in the Banach space $C(I)\cap {\Lambda BV}(I)$, endowed with the ${\Lambda BV}$-norm.

Proof

The proof falls into two parts. First, we shall show that the mapping

$$\begin{aligned} F(x)(t) = g(t) + \int _0^t k(t,s)f(s,x(s))ds,\quad t\in I, \end{aligned}$$

defined for $x\in C(I)$ satisfies the assumptions of Theorem 15. Let $x\in C(I)$ and $\varepsilon > 0$. In view of the assumptions, there exists $\delta > 0$ such that:

• $|g(t) - g(\tau )| \le \frac{\varepsilon }{3}$ for $t,\tau \in I, |t-\tau |< \delta$;

• $\int _0^t | k(\tau , s) - k(t,s) |m_1(s)ds \le \frac{\varepsilon }{3}$ for $(\tau , t) \in \Delta , |t-\tau | \le \delta$;

• $\max \{1,\lambda _1\}\int _A m_1(s)m_2(s)ds\le \frac{\varepsilon }{3}$ for any Lebesgue measurable set $A\subset I$ such that $\mu (A)\le \delta$.

Therefore, for $t,\tau \in I$ such that $0\le \tau - t \le \delta$, we have

$$\begin{aligned}&| F(x)(t) - F(x)(\tau ) | \le | g(t) - g(\tau ) | + \biggl |\int _0^t k(t,s)f(s,x(s))ds - \int _0^\tau k(\tau , s)f(s,x(s))ds\biggr |\\&\quad \le | g(t) - g(\tau ) | + \int _0^t |k(t,s) - k(\tau , s)||f(s,x(s))|ds + \int _t^\tau |k(\tau ,s)||f(s,x(s))|ds\\&\quad \le | g(t) - g(\tau ) | + \int _0^t |k(t,s) - k(\tau , s)|m_1(s)ds + \max \{1,\lambda _1\}\int _t^\tau m_1(s)m_2(s)ds\\&\quad \le \frac{\varepsilon }{3} + \frac{\varepsilon }{3} + \frac{\varepsilon }{3} = \varepsilon . \end{aligned}$$

This shows that $F(x)\in C(I)$. Furthermore, let us observe that the number $\delta$ is independent of x, which implies, that the set F(C(I)) is equiuniformly continuous.

The continuity of the mapping F is a consequence of Lemma 6 and the fact that if a sequence $(x_n)_{n\in \mathbb N}$ is uniformly convergent to $x\in C(I)$, then the sequence $( f(\cdot , x_n(\cdot )) )_{n\in \mathbb N}$, bounded in $L^p(I)$, converges almost everywhere to the function $t\mapsto f(t,x(t)), t\in I$.

The assumptions (ii) and (iii) of Theorem 15 are obviously satisfied for $t_0 = 0$ and $x_0 = g(0)$.

Hence, it suffices to prove that the mapping F satisfies the Palais–Smale condition (iv). Let $(x_n)_{n\in \mathbb N}$ be a sequence in C(I) such that $\lim _{n\rightarrow + \infty }(x_n - F(x_n)) = 0$ with respect to the supremum norm. In view of the Assumption 5.2 and Lemma 5 we get

$$\begin{aligned} \mathrm{{var}}_{\Lambda }(F(x)) \le \left\| g\right\| _{\Lambda BV} + \max (\frac{1}{\lambda _1}, 1) \left\| m_1\right\| _{L^p}\left\| m_2\right\| _{L^q}, \quad \text {for } x\in C(I). \end{aligned}$$

(16)

Therefore, by Helly’s Extraction Theorem, there exists a subsequence $(F(x_{n_k}))_{k\in \mathbb N}$ of $(F(x_n))_{n\in \mathbb N}$ pointwise convergent to a function $y\in {\Lambda BV}(I)$. Thus, $(x_{n_k})_{k\in \mathbb N}$ is also pointwise convergent to y. Hence, for almost all $t\in I$, we have $f(t, x_{n_k}(t))\rightarrow f(t, y(t))$ as $k\rightarrow + \infty$, and the sequence $(f(\cdot , x_{n_k}(\cdot )))_{k\in \mathbb N}$ is bounded in $L^p(I)$. This, by Lemma 6, implies that $(F(x_{n_k}))_{k\in \mathbb N}$ converges to $F(y) = y$ with respect to the ${\Lambda BV}$-norm, and so, since the supremum norm is weaker than the ${\Lambda BV}$-norm, the sequence $(x_n)_{n\in \mathbb N}$ has a limit point in C(I).

All the assumptions of Theorem 15 are satisfied and, therefore, the set S of all continuous solutions of the equation (13) is a compact $R_\delta$ in C(I). To end the proof, it suffices to show that S endowed with the metric $d_\infty$ induced by the supremum norm is homeomorphic to the set T of all continuous solutions of (13) of bounded $\Lambda$-variation, endowed with the metric $d_{\Lambda BV}$ induced by the ${\Lambda BV}$-norm. Note that $S = T$ as sets, and since the ${\Lambda BV}$-norm is stronger than the supremum norm, we get that the identity map $id:S\rightarrow T$ is continuous. Now, we shall show that $id:T\rightarrow S$ is also continuous. Let us take a sequence $(x_n)_{n\in \mathbb N}$ in S convergent to $x_0 \in S$. Reasoning as above, we infer that the sequence $(F(x_n)_{n\in \mathbb {N}})$ converges to $F(x_0)$ with respect to the metric $d_{\Lambda BV}$ induced by the norm $\left\| \cdot \right\| _{\Lambda BV}$. But $F(x_n) = x_n$ for all $n\in \mathbb N \cup \{ 0 \}$, and hence $\left\| x_n - x_0\right\| _{\Lambda BV}\rightarrow 0$ as $n\rightarrow + \infty$. This shows that the identity map constitutes a homeomorphism between the metric spaces T and S and, in consequence, proves that the set of all continuous solutions of (13) of bounded $\Lambda$-variation is a compact $R_\delta$ set in $C(I)\cap {\Lambda BV}(I)$ with respect to the ${\Lambda BV}$-norm.

6 Compactness

Let us start with some notation. Let $\omega :[0,+\infty ) \rightarrow [0,+\infty )$ be a continuous strictly increasing function such that $\omega (0)=0$. By $H^{\omega }(I)$ we will denote the vector space of real-valued functions defined on I which satisfy the following condition:

$$\begin{aligned} |x(t)-x(s)| \le k\omega (|t-s|) for t,s \in I, \text {where}\, k \ge 0. \end{aligned}$$

In the paper [6] the authors gave a sufficient condition for a subset of ${\Lambda BV}(I)$ space to be relatively compact. More precisely, let $\Lambda BV^{\omega }(I)$ denote the Banach space $H^{\omega }(I) \cap \Lambda BV(I)$, endowed with the norm $\left. \!\!\left\bracevert \!\! x\!\! \right\bracevert \!\!\right. _{\Lambda }=\max \{\left\| x\right\| _{\omega },\left\| x\right\| _{\Lambda }\}$, where

$$\begin{aligned} \left\| x\right\| _{\omega } = \inf \bigl \{k \ge 0:{|x(t)-x(s)| \le k\omega (|t-s|) \text { for } t,s \in I}\bigr \}. \end{aligned}$$

Proposition 5

(cf. [6], Proposition 5) Let $\Lambda =(\lambda _n)_{n \in \mathbb N}$ and $\Gamma =(\gamma _n)_{n \in \mathbb N}$ be two Waterman sequences such that $\Lambda <\Gamma$. Then the space $\Lambda BV^{\omega }(I)$ can be compactly embedded into the space $\Gamma BV(I)$.

It is interesting to know if it is possible, roughly speaking, to reverse Proposition 5, that is, if it is possible to prove that for each compact subset K of $\Lambda BV(I)$ there exists such Waterman sequence $\Gamma =(\gamma _n)_{n\in \mathbb N}$, that $\Gamma < \Lambda$, $K\subset \Gamma BV(I)$ and K is bounded in the norm $\Vert \cdot \Vert _{\Gamma BV}$. Below we are going to show that in general the answer is negative, but for the wide range of Waterman sequences the answer is positive. First, let us recall two definitions. For given Waterman sequence $\Lambda$, let us denote by $\Lambda _{(m)}$ the Waterman sequence constructed by deleting the first $m-1$ terms of $\Lambda$.

Definition 7

(cf. [16]) A function $x\in \Lambda BV(I)$ is said to be continuous in $\Lambda$-variation, if $\lim _{m\rightarrow \infty }\mathrm{{var}}_{\Lambda _{(m)}}(x)=0$.

Definition 8

(cf. [16]) If $\Lambda$ is a Waterman sequence, then:

$$\begin{aligned} S_\Lambda =\limsup _{n\rightarrow \infty }\frac{\sum _{i=1}^{2n}\frac{1}{\lambda _i}}{\sum _{i=1}^{n}\frac{1}{\lambda _i}} \end{aligned}$$

is called the Shao–Sablin index of $\Lambda$.

The set of functions which are continuous in $\Lambda$-variation will be denoted by $\Lambda BV_c(I)$ while its subset consisting of all the functions which are also continuous will be denoted by $C\Lambda BV_c(I)$. Let us recall that this concept was introduced by Waterman in [20] to provide a sufficient condition for the so-called $(C,\beta )$-summability of the Fourier series of a function. Waterman conjectured in [21] that not every function of bounded $\Lambda$-variation is continuous in $\Lambda$-variation. An example of such function was given in the paper [8]. Moreover, Prus-Wiśniowski in [16] described the relation between the Shao–Sablin index of a Waterman sequence $\Lambda$ and the existence of functions that are not continuous in $\Lambda$-variation.

Theorem 17

(cf. [16]) For every proper Waterman sequence $\Lambda$, the following statements are equivalent:

• (i) the space $C\Lambda BV(I)$ is separable,

• (ii) $C\Lambda BV_c(I)=C\Lambda BV(I)$,

• (iii) $\Lambda BV_c(I)=\Lambda BV(I)$,

• (iv) $S_\Lambda <2.$

Surprisingly, the concept of a function continuous in $\Lambda$-variation is closely related to the compactness in the space $\Lambda BV(I)$. We are going to begin with the following.

Lemma 7

For every proper Waterman sequence $\Lambda$ and $K\subseteq \Lambda BV(I)$, the following conditions are equivalent:

• (i) there exists such $\Gamma <\Lambda$ that $K\subseteq \Gamma BV(I)$ and K is bounded in $\Gamma BV$-norm,

• (ii) K is bounded and the following condition holds:

  $$\begin{aligned} \forall _{\varepsilon >0}\exists _{N\in \mathbb {N} }\forall _{n\ge N}\forall _{x\in K}\mathrm{{var}}_{\Lambda _{(n)}}(x)<\varepsilon . \end{aligned}$$

  (17)

Proof

Let us start with the implication $(ii)\Rightarrow (i)$ There exists such $M>0$ that $\left\| x\right\| _{\Lambda BV}<2M$ for any $x\in K$. By (17), there exists a strongly increasing sequence $(N_k)_{k\in \mathbb {N}}$ of positive integers such that $N_1=1$ and $\mathrm{{var}}_{\Lambda _{(N_k)}}(x)<\frac{M}{2^k}$ for any $k\in \mathbb {N}$ and $x\in K$. Let us define the sequence $\Gamma$ in the following way:

$$\begin{aligned} \frac{1}{\gamma _1}=\frac{1}{\lambda _1}, \end{aligned}$$

$$\begin{aligned} \frac{1}{\gamma _{k+1}}=\min \left\{ \frac{1}{\gamma _k}, m\cdot \frac{1}{\lambda _{k+1}}\right\} , \end{aligned}$$

where m is a positive integer such that $N_m\le k<N_{m+1}$.

The sequence $\Gamma <\Lambda$ is monotone directly by its definition, and $\frac{1}{\gamma _k} \ge \frac{1}{\lambda _k}$ and, therefore, it is a Waterman sequence. Let $(I_n)_{n\in \mathbb {N}}$ be any sequence of nonoverlapping intervals. Then, for any $x\in K$ we have

$$\begin{aligned} \sum _{n=1}^\infty \frac{\left| x(I_n)\right| }{\gamma _n}= & {} \sum _{k=1}^\infty \sum _{n=N_k}^{N_{k+1}-1}\frac{\left| x(I_n)\right| }{\gamma _n}\le \sum _{k=1}^\infty \sum _{n=N_k}^{N_{k+1}-1}\frac{k\cdot \left| x(I_n)\right| }{\lambda _n}\\\le & {} \sum _{k=1}^\infty k\cdot \mathrm{{var}}_{\Lambda _{N_k}}(x)<\sum _{k=1}^\infty k\cdot \frac{M}{2^k}<+\infty . \end{aligned}$$

This implies that x is of bounded $\Gamma$-variation and, moreover, that

$$\begin{aligned}\left\| x\right\| _{\Gamma BV}\le |x(0)|+\sum _{k=1}^\infty k\cdot \frac{M}{2^k}\le 2M+\sum _{k=1}^\infty k\cdot \frac{M}{2^k}<+\infty \end{aligned}$$

for any $x\in K$, that is, that K is bounded in $\Gamma BV$-norm which ends the first part of the proof.

Now let us proceed to the proof of the implication $(i)\Rightarrow (ii)$ and let us assume that there exists $\Gamma <\Lambda$ such that $K\subseteq \Gamma BV(I)$ is bounded in $\Gamma BV$-norm and that (17) does not hold. Then there exists such $\delta >0$ and a sequence $(N_n)_{n\in \mathbb N}$, $N_n\rightarrow +\infty$ and functions $x_n\in K$ such that $\mathrm{{var}}_{\Lambda _{(N_n)}}(x_n)>\delta$. Take any $M>0$. There exists $N\in \mathbb {N}$ such that

$$\begin{aligned}\frac{1}{\gamma _n}\ge M\frac{1}{\lambda _n},\end{aligned}$$

for $n\ge N$. We may also assume that $N_n\ge N$. Take $x_n\in K$ and a sequence of nonoverlapping intervals $(I_k)_{k\in \mathbb {N}}$, $(I_k)\subset I$ for every $k\in \mathbb {N}$, such that $\sum _{k=N_n}^\infty \frac{|x_n(I_k)|}{\lambda _k}\ge \delta$. Then, we get

$$\begin{aligned}\left\| x_n\right\| _{\Gamma BV} \ge \sum _{k=1}^\infty \frac{|x(I_k)|}{\gamma _k}\ge \sum _{k=N_n}^\infty \frac{|x(I_k)|}{\gamma _k}\ge \sum _{k=N_n}^\infty M\cdot \frac{|x(I_k)|}{\lambda _k}\ge M\delta \end{aligned}$$

and, therefore, K is not bounded in $\Gamma BV$-norm, which gives a contradiction. $\square$

Corollary 1

If $x\in \Lambda BV$, then the following conditions are equivalent

• (i) there exists a Waterman sequence $\Gamma <\Lambda$ such that $x\in \Gamma BV(I)$,

• (ii) x is continuous in $\Lambda$-variation.

Remark 5

As we mentioned above, if $S_\Lambda =2$, then there exists such a continuous function $x_0\in {\Lambda BV}(I)$ that is not continuous in $\Lambda$-variation and thus it does not belong to any $\Gamma BV(I)\subsetneq {\Lambda BV}(I)$. The set $\{x_0\}$ is compact but, at the same time, there does not exist a space $\Gamma BV(I)$ where $\Gamma < \Lambda$ and $x_0\in \Gamma BV(I)$. This shows that if $S_\Lambda = 2$, then not every compact subset of $\Lambda BV(I)$ is a bounded subset of some proper subspace $\Gamma BV(I)\subsetneq \Lambda BV(I)$. As one can check, there exist proper Waterman sequences having the Shao–Sablin index equal to 2. As an example of such a sequence one can consider the sequence $(\ln (n+1))_{n\in \mathbb {N}}$.

Theorem 18

If $K\subset \Lambda BV(I)$ is compact and $K\subseteq \Lambda BV_c(I)$, then there exists a Waterman sequence $\Gamma <\Lambda$ such that $K\subseteq \Gamma BV(I)$ and K is bounded in $\Gamma BV(I)$.

Proof

Fix $\varepsilon >0$. Since K is compact, there exist $x_1, x_2,\ldots , x_k\in K$ such that

$$\begin{aligned}K\subseteq \bigcup _{n=1}^k B_{\Lambda BV}\left( x_n,\frac{\varepsilon }{2}\right) .\end{aligned}$$

Since $x_1,\ldots , x_k\in \Lambda BV_c(I)$, there exists $N\in \mathbb {N}$ such that $\mathrm{{var}}_{\Lambda _{(n)}}(x_m)<\frac{\varepsilon }{2}$ for every $n\ge N$ and $m=1,2,\ldots k$. Take any $x\in K$. There is such m that $x\in B(x_m, \frac{\varepsilon }{2})$. Then for $n\ge N$ we have

$$\begin{aligned} \mathrm{{var}}_{\Lambda _{(n)}}(x)= & {} \mathrm{{var}}_{\Lambda _{(n)}}(x-x_m+x_m)\le \mathrm{{var}}_{\Lambda _{(n)}}(x-x_m)+\mathrm{{var}}_{\Lambda _{(n)}}(x_m)\\\le & {} \mathrm{{var}}_{\Lambda _{(n)}}(x-x_m)+\mathrm{{var}}_{\Lambda _{(n)}}(x_m)<\varepsilon . \end{aligned}$$

We have just proved (17) so by Lemma 7 there exists such $\Gamma <\Lambda$ that $K\subset \Gamma BV(I)$ and K is bounded in $\left\| \cdot \right\| _{\Gamma BV}$. $\square$

Corollary 2

If $S_\Lambda <2$ and $K\subseteq \Lambda BV(I)$ is compact, then there exists a Waterman sequence $\Gamma <\Lambda$ such that $K\subseteq \Gamma BV(I)$.

Proof

By Theorem 17, if $S_\Lambda <2$, then $\Lambda BV(I)=\Lambda BV_c(I)$. $\square$

Now we are going to consider $\sigma$-compact subsets of the space $\Lambda BV(I)$.

Denote $\Gamma _k=(\gamma ^k_n)_{n\in \mathbb N}$ for $k\in \mathbb {N}$, $\Gamma =(\gamma _n)_{n\in \mathbb N}$ and $\Lambda =(\lambda _n)_{n\in \mathbb N}$. We will start with the slight modification of the Lemma 2.

Lemma 8

If $\Lambda$ is a Waterman sequence, $\Gamma _k<\Lambda$ are Waterman sequences for every $k\in \mathbb {N}$ and

$$\begin{aligned} \lambda _n\ge \gamma ^{k+1}_n\ge \gamma ^k_n \end{aligned}$$

(18)

for $n, k\in \mathbb {N}$, then there exists a Waterman sequence $\Gamma <\Lambda$ such that $\Gamma _k BV(I)\subseteq \Gamma BV(I)$ for every $k\in \mathbb {N}$.

Proof

There exists a strongly increasing sequence $(N_k)_{k\in \mathbb N}$ of positive integers such that $N_1=1$ and

$$\begin{aligned} \lambda _n\ge k\cdot \gamma ^k_n \end{aligned}$$

(19)

for every $n\ge N_k$, $k\in \mathbb {N}$.

Put $\gamma _n=\gamma ^k_n$ for $N_k\le n<N_{k+1}$ and $k\in \mathbb {N}$. Then

• (i) $\Gamma =(\gamma _n)_{n\in \mathbb N}$ is monotone, by (18) and the monotonicity of $\Gamma _k$ for $k\in \mathbb {N}$,

• (ii) $\sum _{n=1}^{\infty }\frac{1}{\gamma _n}\ge \sum _{n=1}^{\infty }\frac{1}{\lambda _n}=+\infty$, by (18),

which implies that $\Gamma$ is a Waterman sequence. Moreover, (19) implies that for an arbitrary $k\in \mathbb {N}$ and $n\in \mathbb {N}$ such that $N_k\le n<N_{k+1}$:

$$\begin{aligned}\frac{\lambda _n}{\gamma _n}=\frac{\lambda _n}{\gamma ^k_n}\ge k,\end{aligned}$$

so $\Gamma <\Lambda$. Finally, by the definition of $\Gamma$ and (18), we get $\gamma _n\ge \gamma ^k_n$ for every $n\ge N_k$, $k\in \mathbb {N}$, so $\Gamma _k BV(I)\subseteq \Gamma BV(I)$ for all $k\in \mathbb {N}$.

Lemma 9

If $\Gamma _1, \Gamma _2<\Lambda$ are Waterman sequences, then there exists a Waterman sequence $\Gamma <\Lambda$ such that $\gamma ^i_n \le \gamma _n$ for every $n\in \mathbb {N}$ $(i=1,2)$.

Proof

We may assume that $\gamma ^1_n, \gamma ^2_n\le \lambda _n$ for every $n\in \mathbb {N}$, since $\Gamma _1<\Lambda$, $\Gamma _2<\Lambda$. Let $\gamma _n=\max \{\gamma ^1_n, \gamma ^2_n\}$ for $n\in \mathbb {N}$. Then $\Gamma$ is monotone and $\sum _{n=1}^{\infty }\frac{1}{\gamma _n}\ge \sum _{n=1}^{\infty }\frac{1}{\lambda _n}=+\infty$, so $\Gamma$ is a Waterman sequence. Obviously, $\Gamma <\Lambda$. $\square$

Theorem 19

If K is $\sigma$-compact and $K\subseteq \Lambda BV_c(I)$, then there exists $\Gamma <\Lambda$ such that $K\subseteq \Gamma BV(I)$.

Proof

Let $K=\bigcup _{n\in \mathbb N} K_n$, where $K_n$ is compact for every positive integer n. By Theorem 18, for every $n\in \mathbb {N}$ there exists a Waterman sequence $\Lambda _n<\Lambda$ such that $K_n\subseteq \Lambda _n BV(I)$. We shall define a sequence $(\Gamma _k)_k$ of Waterman sequences, fulfilling the assumptions of the Lemma 8 and such that $\Lambda _kBV(I)\subseteq \Gamma _kBV(I)$ for $k\in \mathbb {N}$. Let $\Gamma _1=\Lambda _1$. Assume that $\Gamma _k$ is a Waterman sequence such that $\Gamma _k<\Lambda$. Applying Lemma 9 to the sequences $\Lambda _{k+1}, \Gamma _k<\Lambda$ we obtain a Waterman sequence $\Gamma _{k+1}$, such that $\gamma _n^k\le \gamma _n^{k+1}$ and $\lambda _n^{k+1}\le \gamma _n^{k+1}$ for every $n\in \mathbb {N}$. Moreover, $\Lambda _{k+1} BV(I)\subseteq \Gamma _{k+1} BV(I)$ for every $k \in \mathbb {N} \cup \{0\}$. This completes the construction.

Finally, applying Lemma 8 to the sequence $(\Gamma _k)_{k\in \mathbb N}$ we get the desired Waterman sequence $\Gamma$. $\square$