1 Introduction

The aim of this paper is to study the generalized notion of bounded variation. The paper can be subdivided into two parts. In the first part, we focus on power oriented bounded variation. While in the second half we study distance dependent variation. At the beginning of each section, a brief discussion about structural characterization and various inclusion properties are also provided.

Let \(r\in ]0,\infty [\) be fixed. A function \(f:[a,b]\rightarrow \mathbb {R}\) is said to be satisfying r-bounded variation if for any partition \(P=\{a=x_0,\ldots ,x_n=b\}\) with \(x_0<\cdots <x_n,\) the expression \(\sum \nolimits ^{n}_{i=1}|f(x_i)-f(x_{i-1})|^{r}\) is bounded. It can be easily observed that for the case \(r=1\), this notion implies the well-known definition of bounded variation.

The term function of bounded variation was first introduced by Jordan in his paper [1]. Later the concept was generalized by Wiener in the paper [2] where he investigated several topics related to Fourier Analysis. His generalized form is widely known as p-variation and its definition is the same as the above mentioned r-variation with \(r\ge 1\). Since then many more generalized forms in this direction have been introduced including Young’s well known generalized bounded variation in [3]. For more details, one can look into the articles and books [7,8,9,10,11,12].

One of our main objectives in this paper is to find a possible decomposition for functions satisfying r-bounded variation. We establish that for \(r\in ]0,1[\), if a function is equipped with r-bounded variation then f is nothing but a function of ordinary bounded variation. In other words, in such cases f can be expressed as the difference of two monotone functions. We generalize Jordan’s decomposition theorem and show that a function \(f:[0,b]\rightarrow \mathbb {R}\) satisfying Wiener variation/p-variation (r-variation with \(r\ge 1\)) can be treated as approximately monotone. Moreover under some minimal assumptions, such a function f can be precisely approximated by a nondecreasing function g that satisfies the following inequality

$$\begin{aligned} \Phi (x)\le g(x)-f(x)\le 2\Phi \Big (\dfrac{x}{2}\Big )\quad \text{ where } \end{aligned}$$
$$\begin{aligned} {\Phi (u):=\sup _{x,x+u\in [0,b]}\Big \{ {^{r}}v_{f}(x+u)-{^{r}}v_{f}(x)\Big \}^{\dfrac{1}{r}}, \quad \quad \text{( }{^{r}v}_{f} \text{ is } \text{ r-variation } \text{ function } \text{ of } \text{ f) }}. \end{aligned}$$

We also prove that for \(0<r_1<r_2\), a function satisfying \(r_1\)-bounded variation is also a function of \(r_2\)-bounded variation. We show that if \(\Phi ^{r}\) is superadditive, then a \(\Phi \)-Hölder function satisfies r-bounded variation. The definition and other essentials for \(\Phi \)-Hölder functions are discussed below.

Let \(p\ge 0\) be fixed and I be an arbitrary interval in \(\mathbb {R}\) such that \(f:I\rightarrow \mathbb {R}\) satisfies the following functional inequality

$$f(x)\le f(y)+(y-x)^{p}\quad \text{ for } \text{ all } x,y\in I \quad \text{ with } x<y.$$

For \(p=1\), it is clearly visible that \(f(t)+t\) is nondecreasing in I, while for \(p>1\), it can be shown that the function f possesses the usual monotonicity(increasingness). In the case \(p=0\), f can be treated as a \(\epsilon \)-monotone function. Using the result on the decomposition of such functions mentioned by Páles in his paper [4]; f is expressible as \(f:=g+h\), where g is nondecreasing and \(||h||\le \dfrac{1}{2}\). The decomposition of f when \(p\in ]0,1[\) is still unknown. In this paper we show that if \(I\subseteq \mathbb {R}_+\) with \(0\in I\), then for any \(p\in ]0,1[\), the function f can be tightly approximated with a nondecreasing majorant g that satisfies

$$f(x)+x^p\le g(x)\le f(x)+2^{(1-p)}x^p.$$

Motivated by this concept, we studied \(\Phi \)-monotonicity in depth in our papers [5, 6]. A function \(f:I\rightarrow \mathbb {R}\) is said to be \(\Phi \)-monotone if for any \(x,y\in I\) with \(x<y\), the following inequality holds

$$\begin{aligned} f(x)\le f(y)+\Phi (y-x)\quad \text{ where }\quad \Phi :[0,\ell (I)]\rightarrow \mathbb {R}_+ \end{aligned}$$

On the other hand a function \(f:I\rightarrow \mathbb {R}\) is termed as \(\Phi \)-Hölder if both f and \(-f\) are \(\Phi \)-monotone, that is for any \(x,y\in I\), the following inequality must be satisfied

$$\begin{aligned} |f(x)-f(y)|\le \Phi (|y-x|). \end{aligned}$$

Let \(d>0\) be fixed and \(\ell (I)\ge d\) such that for any \(x,y\in I\) with \(y-x \ge d\), \(\Phi (y-x)=0\) holds. Then the notion of \(\Phi \)-monotonicity for f can be treated as distance dependent monotonicity(increasingness).

Inspired by this, we introduce the following concept. For a fixed \(d>0\), a function \(f: I\rightarrow \mathbb {R}\) with \(\ell (I)\ge d\) is said to be increasing by a period d (or d- periodically increasing) if the following holds

$$f(x)\le f(y) \qquad \text{ for } \text{ all } \quad x,y\in I\quad \text{ with } \quad y-x\ge d.$$

In our recently submitted paper, we studied the fundamental structural properties of such functions. For any given function f we give a precise formula to obtain the largest d-periodically increasing minorant. We also show an approximation of a d-periodically increasing function with an ordinary monotone function. Furthermore, a basic characterization and a conditional decomposition of d-monotone functions was also provided.

The main objective in the last section of the paper is to establish a relationship between functions satisfying d-bounded variation and d-periodically increasing functions. We show that a function that possesses d-bounded variation can be decomposed as the difference of a monotone and a d-periodically increasing function. For this, we introduce a distance oriented concept of variation, named d-variation.

Let \(\ell ([a,b])\ge d\) and the interval [a, b] be partitioned as follows

$$P=\{a=x_0,\ldots ,x_n=b\} \text{ such } \text{ that } x_0<\cdots <x_n$$
$$ \text{ with } x_i-x_{i-1}\ge d\text{, } (i\in \{1,\ldots ,n\}\text{) }.$$

Then for any function \(f:[a,b]\rightarrow \mathbb {R}\) we define d-variation on the partition P as

$$\begin{aligned} {\overset{b}{\underset{a}{V_{d}}}(f,P):=\sum _{i=1}^{n}|f(x_i-f(x_{i-1})|\quad \text{ with }\quad x_i-x_{i-1}\ge d \quad (i\in \{1,\ldots ,n\}).} \end{aligned}$$

If for such possible partitions of [a, b], the above expression remains bounded, we say that f is a function of d-bounded variation. We will later see that a function of d-bounded variation can be unbounded as well. We also show that the variation function \(_{d}v_{f}\) is nondecreasing. Moreover, the superadditive characteristic of d-variation functions is also established.

2 On approximately monotone and Hölder functions

Before proceeding, we need to recall some notions and terminology. We consider I to be a non-empty and non singleton interval. \(\ell (I)(\in \overline{\mathbb {R}_+})\) denotes the length of the interval in the extended real line. Then the function \(\Phi :[0,\ell (I)]\rightarrow \mathbb {R}_+\) is said to be an error function.

Based upon \(\Phi \), we can formulate the concept of \(\Phi \)-monotonicity. A function \(f: I\rightarrow \mathbb {R}\) is said to be \(\Phi \)-monotone(increasing) if for any \(x,y\in I\) with \(x\le y\) the following functional inequality holds

$$f(x)\le f(y)+\Phi (y-x).$$

On the other hand if both f and \(-f\) are \(\Phi \)-monotone, we will say that f is \(\Phi \)-Hölder. In other words a function \(f:I\rightarrow \mathbb {R}\) is said to be \(\Phi \) Hölder if for any \(x,y\in I\), it satisfies the inequality

$$|f(x)- f(y)|\le \Phi (|y-x|).$$

Now we will go through some inequalities associated with the error function \(\Phi \) provided it has some structural properties.

Proposition 2.1

Suppose \(\Phi :[0,\ell (I)[\rightarrow \mathbb {R}_+\) is subadditive and concave. Then for any \(x,y\in [0,\ell (I)]\) with \(x\le y\), the following inequality holds

$$\begin{aligned} {\Phi (y)-\Phi (x)\le \Phi (y-x)\le 2\Phi \Big (\dfrac{y}{2}\Big )-\Phi (x).} \end{aligned}$$
(1)

Additionally, under the before-mentioned conditions \((-\Phi )\) is \(\Phi \)-monotone.

Proof

By using the subadditivity of \(\Phi \), we get

$$\begin{aligned} \Phi (y)=\Phi (y-x+x)\le \Phi (y-x)+\Phi (x). \end{aligned}$$

This is the first inequality in (1). Now applying the concavity property of \(\Phi \), we can compute the following inequality

$$\begin{aligned} \dfrac{\Phi (y-x)}{2}+\dfrac{\Phi (x)}{2}\le \Phi \Big (\dfrac{(y-x)+x}{2}\Big )=\Phi \Big (\dfrac{y}{2}\Big ). \end{aligned}$$

Multiplying the above inequality by 2 and rewriting it, we will obtain the remaining part of the inequality in (1).

The second assertion can be easily seen from the first part of the inequality in (1). By rearranging its terms we get

$$(-\Phi )(x)\le (-\Phi )(y)+\Phi (y-x).$$

This yields the \(\Phi \)-monotonicity of \((-\Phi )\) and completes the proof of the first assertion.\(\square \)

Based upon the above theorem, we can propose the following corollary related to power functions.

Corollary 2.2

For any \(0\le x\le y\), the following inequality will hold true for any \(p\in [0,1]\)

$$\begin{aligned} {y^p-x^p\le (y-x)^{p}\le 2^{1-p}y-x^p.} \end{aligned}$$
(2)

Proof

For \(u\ge 0\) and for any fixed \(p\in [0,1]\), the function \(u\rightarrow u^{p}\) is subadditive and concave. Hence, the result directly follows from 2.1.\(\square \)

We are now ready to present a possible decomposition for \(\Phi \)-monotone functions. The following theorem shows the tight approximation of a \(\Phi \)-monotone function by an ordinary monotone function.

Theorem 2.3

Let \(I\subseteq \mathbb {R}_+\) be a non-empty and non-singleton interval such that \(0\in I\).

\(\Phi :[0,\ell (I)]\rightarrow \mathbb {R}_+\) is a subadditive error function. If \(g:I\rightarrow \mathbb {R}\) is non-decreasing then \(f:=g-\Phi \) is a \(\Phi \)-monotone function in I. Conversely if, \(f:I\rightarrow \mathbb {R}\) is \(\Phi \)-monotone where \(\Phi \) is subadditive and concave, then there exists a nondecreasing majorant \(g:I\rightarrow \mathbb {R}\) of f that satisfies the following inequality

$$\begin{aligned} {\Phi (x)\le g(x)-f(x)\le 2\Phi \Big (\dfrac{x}{2}\Big ).} \end{aligned}$$
(3)

Proof

To prove the first part of the theorem, we assume \(x,y\in I\) with \(x\le y\). By using the first inequality of (1), we get the following for f

$$\begin{aligned} f(x){} & {} =g(x)-\Phi (x)\\{} & {} \le g(y)-\Phi (y)+\big (\Phi (y)-\Phi (x)\big )\\{} & {} =f(y)+\Phi (y-x). \end{aligned}$$

This establishes that f is \(\Phi \)-monotone. To prove the converse assertion, we use the inequality associated with the \(\Phi \)-monotonicity of f. Applying the last inequality in (1), we observe that

$$\begin{aligned} {f(x)+\Phi (x)\le f(y)+2\Phi \bigg (\dfrac{y}{2}\bigg )\qquad ( \text{ for } \text{ all } x,y\in I \,\, \text{ with } \, x\le y).} \end{aligned}$$
(4)

We can define \(g:I\rightarrow \mathbb {R}\) as follows

$$g(u):=\inf _{z\in I, u\le z}\bigg (f(z)+2\Phi \Big (\dfrac{z}{2}\Big )\bigg ).$$

We assume \(x,y\in I\) with \(x\le y\). Then we can formulate the following inequality

$$\begin{aligned} g(x)=\inf _{z\in I, x\le z}\bigg (f(z)+2\Phi \Big (\dfrac{z}{2}\Big )\bigg )\le \inf _{z\in I, y\le z}\bigg (f(z)+2\Phi \Big (\dfrac{z}{2}\Big )\bigg )=g(y). \end{aligned}$$

This shows that g is nondecreasing. Now from (4), we obtain the following

$$\begin{aligned} f(x)+\Phi (x)\le g(x)\le f(x)+2\Phi \bigg (\dfrac{x}{2}\bigg ). \end{aligned}$$

Subtracting f from all parts of the above inequality, we get (3) and it validates the result.\(\square \)

From the above theorem, we can establish a conditional decomposition of the function \(f: I\rightarrow \mathbb {R}\) satisfying the following inequality

$$\begin{aligned} {f(x)\le f(y)+c(y-x)^{p} \quad x,y\in I\quad \text{ with } \quad x\le y. \quad (p\in ]0,1[).} \end{aligned}$$
(5)

We have already discussed such functions in the Introduction of this paper.

Corollary 2.4

Let \(I\subseteq \mathbb {R}_+\) be a nonempty and nonsingleton interval with \(0\in I\). Then for \(f:I\rightarrow \mathbb {R}\) satisfying (5), there exists a nondecreasing majorant g of f such that the following inequality holds in I.

$$cx^p\le g(x)-f(x)\le \big (2^{1-p}c\big )x^p.$$

Proof

One can easily show that for a fixed \(c>0\), the function \(u\rightarrow c u^{p}\) (\(u\in I\)) is subadditive and concave. Hence, the result is a direct consequence of Theorem 2.3 and corollary 2.2.\(\square \)

In the next section, we will discuss functions of r-bounded variation.

3 On functions of power dependent variation

Throughout this section I will denote the non-empty and non singleton interval [a, b]. Let P be a partition of I as follows

$$\begin{aligned} {P=\{x_0,\ldots ,x_n\} \quad \text{ with } \quad a=x_0<\cdots <x_n=b.} \end{aligned}$$
(6)

Let \(r\in ]0,\infty [\) be fixed. For a real valued function \(f: I\rightarrow \mathbb {R}\), we define r-bounded variation \(\overset{b}{\underset{a}{V}^{r}}(f,P)\) for the partition P of I as follows

$$\overset{b}{\underset{a}{V}^{r}}(f,P):=\sum _{i=1}^{n}|f(x_i)-f(x_{i-1})|^{r}.$$

If the above mentioned expression is bounded with respect to all such possible partitions of I, we say that f possesses r-bounded variation. The supremum of all r-variations of f is called total r-variation which is denoted by

$$\overset{b}{\underset{a}{V^{r}}}(f):=\sup \overset{b}{\underset{a}{V}^{r}}(f,P).$$

We term \({^{r}v}_{f}:I\rightarrow \mathbb {R}\), defined by \({^{r}v}_{f}(x):=\overset{x}{\underset{a}{V^{r}}}(f)\) as the r-variation function for f. It can be clearly observed that if a function is unbounded then it can’t be a function of r-bounded variation. Also, there are bounded functions which cannot be categorized as functions satisfying r-bounded variation for any \(r\in ]0,\infty [\). For example \(f:I\rightarrow \mathbb {R}\) defined by

$$f(x)={\left\{ \begin{array}{ll} 1, \quad \text{ if } \text{ x } \text{ is } \text{ rational }\\ 0 \quad \text{ if } \text{ x } \text{ is } \text{ irrational } \end{array}\right. } $$

is bounded but not a function of r-bounded variation. We will utilize the following subadditive (superadditive) property in the upcoming result extensively.

For any \(r\le 1\) and \(a_1,\ldots ,a_n\in \mathbb {R}_+\), the following inequality is satisfied

$$\begin{aligned} \bigg (\sum _{i=1}^{n} a_i\bigg )^r\le \sum _{i=1}^{n} a_i^{r}. \end{aligned}$$

For \(r\ge 1\) under the same assumptions on \({a_{i}}'s\), the reverse inequality holds.

Now we are going to show that the r- variation function possesses monotonicity.

Proposition 3.1

\({^{r}v}_{f}\) is a nondecreasing function.

Proof

Let \(x,y\in I\) with \(x\le y\). For any \(\epsilon >0\), we will have a partition P of I as defined in (6) satisfying the following inequality

$$\begin{aligned} {{^{r}v}_{f}(x)< \sum _{i=1}^{n}|f(x_i)-f(x_{i-1})|^{r}+\epsilon \le {^{r}v}_{f}(y)+\epsilon .} \end{aligned}$$

Upon taking \(\epsilon \rightarrow 0\), we get the monotonicity of \({^{r}v}_{f}\).\(\square \)

The next proposition will show that the total r-bounded variation of a function defined in a compact interval of \(\mathbb {R}\) possesses characteristics similar to superadditivity.

Proposition 3.2

Let \(r>0\) be fixed and \(f:I\rightarrow \mathbb {R}\) is a function of r- bounded variation. Then for any \(c\in I^{\circ }\), the following inequality holds

$$\begin{aligned} {\overset{c}{\underset{a}{V}^{r}}(f)+\overset{b}{\underset{c}{V}^{r}}(f)\le \overset{b}{\underset{a}{V}^{r}}(f).} \end{aligned}$$
(7)

Additionally, in the case of \(r\in ]0,1],\) the above inequality turns into equality.

Proof

To establish the inequality, we assume that \(\epsilon >0\) is arbitrary. Then there must exist two partitions \(P_1=\{a=x_0,\ldots ,x_k=c\}\) and \(P_2=\{c=x_k,\ldots ,x_n=b\}\) respectively for [a, c] and [c, b] with \(x_0<\cdots <x_n\) satisfying the following two inequalities

$$\begin{aligned} \overset{c}{\underset{a}{V}^{r}}(f)<\overset{c}{\underset{a}{V}^{r}}(f,P_1)+\dfrac{\epsilon }{2}\quad \text{ and }\quad \overset{c}{\underset{a}{V}^{r}}(f)<\overset{b}{\underset{c}{V}}(f,P_2)+\dfrac{\epsilon }{2}. \end{aligned}$$

Upon adding up the two inequalities side by side, we obtain

$$\begin{aligned} \overset{c}{\underset{a}{V}^{r}}(f)+\overset{b}{\underset{c}{V}^{r}}(f)<\overset{c}{\underset{a}{V}^{r}}(f,P_1)+\overset{c}{\underset{a}{V}^{r}}(f,P_2)+\epsilon =\overset{b}{\underset{a}{V}^{r}}(f,P_1\cup P_2)+\epsilon \le \overset{b}{\underset{a}{V}}(f)+\epsilon . \end{aligned}$$

Since \(\epsilon \) is arbitrary, taking \(\epsilon \rightarrow 0\), we obtain (7). This verifies the first statement.

To prove the second assertion it will be enough to show that for any \(r\in ]0,1]\), the reverse inequality of (7) also holds. For any arbitrary \(\epsilon >0\), there must exist a partition like P of I, as defined in (6) satisfying the following inequality

$$\begin{aligned} \overset{b}{\underset{a}{V}^{r}}(f)<\sum _{i=1}^{n}|f(x_i)-f(x_{i-1})|^r+\epsilon . \end{aligned}$$

Then for any \(c\in I^{\circ }\) there must exist \(k\in \{1,\ldots , n\}\) such that \(c\in [x_{k-1},x_k]\) where \(x_{k-1},x_k\in P\). Then by using the subadditive property for \(r\in ]0,1]\), we can expand the above inequality as follows

$$\begin{aligned}&\overset{b}{\underset{a}{V}^{r}}(f)<\sum _{i=1}^{k-1}|f(x_i)-f(x_{i-1})|^r+|f(x_{k-1}-f(c)+f(c)-f(x_k)|^{r}\\&\qquad +\sum _{i=k-1}^{n}|f(x_i)-f(x_{i-1})|^r+\epsilon \\&\quad \le \sum _{i=1}^{k-1}|f(x_i)-f(x_{i-1})|^r+|f(x_{k-1}-f(c)|^{r}+|(f(c)-f(x_{k+1})|^{r}\\&\qquad +\sum _{i=k}^{n}|f(x_i)-f(x_{i-1})|^r+\epsilon \\&\quad \le \overset{c}{\underset{a}{V}^{r}}(f)+\overset{b}{\underset{c}{V}^{r}}(f)+\epsilon . \end{aligned}$$

Since \(\epsilon \) is arbitrary, taking it negligibly small we obtain

$$\begin{aligned} {\overset{b}{\underset{a}{V}^{r}}(f)\le \overset{c}{\underset{a}{V}^{r}}(f)+\overset{b}{\underset{c}{V}^{r}}(f).} \end{aligned}$$

This proves the assertion and completes the proof.\(\square \)

The next proposition will show that the algebraic multiplication of two r-bounded functions results in a function with the same variation characteristic provided \(r\in ]0,1].\)

Proposition 3.3

Let \(r_1,r_2\in ]0,1]\). If \(f,g:I\rightarrow \mathbb {R}\) are two functions of \(r_1\) and \(r_2\)-bounded variations, respectively, then \(fg:I\rightarrow \mathbb {R}\) will satisfy \(\max \{r_1,r_2\}\)-bounded variation.

Proof

We assume \(r=\max \{r_1,r_2\}\). Since f and g are functions of \(r_1\) and \(r_2\)-bounded variations, respectively, by Theorem 3.9, f and g will satisfy r-bounded variation. Therefore, there exists \(k\in \mathbb {R}_+\) such that both the inequalities \(|f(x)|\le k\) and \(|g(x)|\le k\) hold for all \(x\in I\). Now, we consider a partition P as in (6). Since \(r\in ]0,1]\), utilizing the subadditive property of \(V^r\big (P,(fg)\big )\), we compute the upper bound as follows

$$\begin{aligned} V^r\big (P,(fg)\big ){} & {} =\sum _{i=1}^n|(fg)(x_i)-(fg)(x_{i-1})|^r\\{} & {} \le \sum _{i=1}^n\Big (|f(x_i)||g(x_i)-g(x_{i-1})|+|f(x_i)-f(x_{i-1})||g(x_{i-1})|\Big )^r\\{} & {} \le k^r\bigg (\sum _{i=1}^{n}|f(x_i)-f(x_{i-1})|^r+\sum _{i=1}^{n}|g(x_i)-g(x_{i-1})|^r\bigg )\\{} & {} \le k^r\Big [V^r\big (f\big )+V^r\big (g\big )\Big ]. \end{aligned}$$

We will end up with the same bound as above for any arbitrary partition P. And hence it is evident that fg satisfies r-bounded variation. This completes the proof of the statement.\(\square \)

Next we will show the relationship between a function satisfying r-bounded variation and an approximately Hölder function. But before that we need the below proposition.

Proposition 3.4

For \(r\in [1,\infty [\), the function \(\Phi :[0,\ell (I)]\rightarrow \mathbb {R}_+\) is defined as follows

$$\begin{aligned} { \Phi (u):=\sup _{x,x+u\in I}\Big \{ {^{r}}v_{f}(x+u)-{^{r}}v_{f}(x)\Big \}^{\dfrac{1}{r}}.} \end{aligned}$$
(8)

Then \(\Phi \) is a subadditive function.

Proof

To prove the statement we assume \(u,v\in \mathbb {R}_+\) such that \(u+v\in \ell (I).\) Then from (8), for any given \(\epsilon > 0\) there must exist \(x\in I\) with \(x+u+v\in I\) satisfying the following functional inequality which is extended using the subadditive property due to 1/r.

$$\begin{aligned}{} & {} \Phi (u+v)< \Big \{{^{r}}v_{f}(x+u+v)-{^{r}}v_{f}(x)\Big \}^{\dfrac{1}{r}}+\epsilon \\{} & {} \quad =\Big \{{^{r}}v_{f}(x+u+v)-{^{r}}v_{f}(x+u)\Big \}^{\dfrac{1}{r}}+\Big \{{^{r}}v_{f}(x+u)+{^{r}}v_{f}(x)\Big \}^{\dfrac{1}{r}}+\epsilon \\{} & {} \quad \le \sup _{x,x+v\in I}\Big \{ {^{r}}v_{f}(x+v)-{^{r}}v_{f}(x)\Big \}^{\dfrac{1}{r}}+\sup _{x,x+u\in I}\Big \{ {^{r}}v_{f}(x+u)-{^{r}}v_{f}(x)\Big \}^{\dfrac{1}{r}}+\epsilon \\{} & {} \quad =\Phi (u)+\Phi (v)+\epsilon . \end{aligned}$$

Since \(\epsilon \) is arbitrary, upon taking \(\epsilon \rightarrow 0\), it is evident from the above inequality that \(\Phi \) possesses subadditivity.\(\square \)

Proposition 3.5

Let \(r\in [1,\infty [\) and the function \(f:I\rightarrow \mathbb {R}\) satisfies r-bounded variation. Then f is a \(\Phi \)-Hölder function where \(\Phi \) is defined as in (8).

Proof

To prove the theorem we assume \(x,y\in I\) with \(x<y\). Assuming \(u=y-x\) and utilizing Proposition 3.2 and Proposition 3.4 we compute the following inequality

$$\begin{aligned} {|f(x)-f(y)|\le \Bigg (\overset{y}{\underset{x}{V}^{r}}(f)\Bigg )^{\dfrac{1}{r}}\le \Big \{{^{r}}v_{f}(y)-{^{r}}v_{f}(x)\Big \}^{\dfrac{1}{r}}\le \Phi (y-x).} \end{aligned}$$

This validates our statement. \(\square \)

The next result is a generalization of Jordan’s decomposition theorem. We will see that depending upon the value of r, an r-bounded function can be expressed as the difference of two monotone functions or under some minimal assumptions can be approximated by a nondecreasing function.

Theorem 3.6

Let \(r\in ]0,\infty [\) be fixed, \(f: I\rightarrow \mathbb {R}\) be a function of r-bounded variation. If \(r\in ]0,1]\), f can be written as the difference of two monotone functions. Additionally, for \(r=1\), the inverse implication also holds true.

On the other hand, for \(r\in ]1,\infty [\), if \(\inf (I)=0\) and \(\Phi \) is concave then f can be approximated by a monotone majorant g satisfying the same functional inequality as in (3). That is

$$\Phi (x)\le g(x)-f(x)\le 2\Phi \bigg (\dfrac{x}{2}\bigg ),$$

where the function \(\Phi \) is as defined in (8).

Proof

First we will consider the case for \(r\in ]0,1]\).

We can express f as \(f:=\big ({^{r}v}_{f}\big )^{\dfrac{1}{r}}-\Big (\big ({^{r}v}_{f}\big )^{\dfrac{1}{r}}-f\Big )\). Let, \(x,y\in I\) with \(x<y\). Due to the non-negativity of \({^{r}v}_{f}\) and by Proposition 3.1, it is evident that the function \(\big ({^{r}v}_{f}\big )^{\dfrac{1}{r}}\) is monotone. Now to prove the first assertion, we only need to show that \(\big ({^{r}v}_{f}\big )^{\dfrac{1}{r}}-f\) is also nondecreasing. Let \(x,y\in I\) with \(x<y\). Then from Proposition 3.2, we obtain the following inequality

$$\begin{aligned} |f(y)-f(x)|^r\le \overset{y}{\underset{x}{V}^{r}}(f)={^{r}v}_{f}(y)-{^{r}v}_{f}(x). \end{aligned}$$

Since \(r\in ]0,1]\), by using the superadditivity and non-negativity of \({^{r}v}_{f}\), the above inequality can be expanded as follows

$$\begin{aligned} f(y)-f(x)\le |f(y)-f(x)|\le \Big ({^{r}v}_{f}(y)-{^{r}v}_{f}(x)\Big )^{\dfrac{1}{r}}\le \big ({^{r}v}_{f}(y)\big )^{\dfrac{1}{r}}-\big ({^{r}v}_{f}(x)\big )^{\dfrac{1}{r}}. \end{aligned}$$

Rearranging the terms in the above expression we obtain

$$\begin{aligned} {\big ({^{r}v}_{f}(x)\big )^{\dfrac{1}{r}}-f(x)\le \big ({^{r}v}_{f}(y)\big )^{\dfrac{1}{r}}-f(y).} \end{aligned}$$

This yields the monotonicity of \(\big ({^{r}v}_{f}\big )^{\dfrac{1}{r}}-f\) and completes the proof of the first statement.

For r=1, the function \(f\in \mathbb {B}\mathbb {V}(I)\) and hence by Jordan’s Decomposition Theorem the inverse implication is obvious.

Finally, we arbitrarily select \(r\in ]1,\infty ]\). From Proposition 3.5, we have that f is a \(\Phi \)-Hölder function. Hence f satisfies \(\Phi \)-monotonicity. Again from Proposition 8 and our assumption, we have that \(\Phi \) is a concave and subadditive error function. Thus by the converse part of Theorem 2.3 the inequality holds. And this completes the proof. \(\square \)

However, the reverse implication of the above theorem does not hold. Based on the above results, we are now going to study some inclusion properties of different variation classes. The class of functions satisfying r-bounded variation where \(r\in ]0,1[\) will be denoted by \(\underset{]0,1[}{\mathbb {B}\mathbb {V}}(I)\). The notation \({\mathbb {B}\mathbb {V}}(I)\) is used for the class of functions of ordinary bounded variation. On the other hand, for \(r>1\), such a class is represented by \(\underset{]1,\infty [}{\mathbb {B}\mathbb {V}}(I)\). Based upon the definition of these classes we can establish the following inclusion theorem. One can also obtain the below result as a direct consequence of Theorem 3.9. From Proposition 3.3and the theorem below, one can easily see that the class of functions \(\underset{]0,1[}{\mathbb {B}\mathbb {V}}(I)\) is a ring under the usual functional addition and multiplication, while \(\underset{]1,\infty [}{\mathbb {B}\mathbb {V}}(I)\) is a commutative group.

Theorem 3.7

Let \(r\in ]0,1[\), then the following proper inclusion holds

$$\begin{aligned} {\underset{]0,1[}{\mathbb {B}\mathbb {V}}(I)\subsetneq {\mathbb {B}\mathbb {V}}(I)\subsetneq \underset{]1,\infty [}{\mathbb {B}\mathbb {V}}(I).} \end{aligned}$$
(9)

Proof

To establish the first inclusion property, we assume that \(f \in \underset{]0,1[}{\mathbb {B}\mathbb {V}}(I).\) Then from Theorem 3.6, we know that f can be decomposed as the algebraic difference of two monotone functions. Utilizing Jordan’s decomposition theorem, we can conclude \(f \in {\mathbb {B}\mathbb {V}}(I)\). This shows the first part of the inclusion in (9).

Now to show \({\mathbb {B}\mathbb {V}}(I)\subseteq \underset{]1,\infty [}{\mathbb {B}\mathbb {V}}(I)\), we assume \(f\in {\mathbb {B}\mathbb {V}}(I)\) and \(r\in ]1,\infty [\) arbitrary. Then at the partition P defined as in (6), utilizing superadditivity, we calculate \(V^r(f,P)\) as follows

$$\begin{aligned} {V^r(f,P)=\sum _{i=1}^{n}|f(x_i)-f(x_{i-1}|^r\le \bigg (\sum _{i=1}^{n}|f(x_i)-f(x_{i-1})|\bigg )^r.} \end{aligned}$$

Since the expression \(\sum \nolimits ^{n}_{i=1}|f(x_i)-f(x_{i-1}|\) is bounded, this implies the boundedness of \(V^r(f,P)\) as well. Likewise, the same holds true for any other partition of I. This implies \(f\in \underset{]1,\infty [}{\mathbb {B}\mathbb {V}}(I)\) and completes the proof of the inclusion property.

Now, we are going to show that the inclusions are proper through examples. First we show that there exists a function \(f\in {\mathbb {B}\mathbb {V}}\) such that \(f\notin \underset{]0,1[}{\mathbb {B}\mathbb {V}}\). Similarly, we also construct an example such that \(f\in \underset{]1,\infty [}{\mathbb {B}\mathbb {V}}(I)\setminus {\mathbb {B}\mathbb {V}}(I)\).

The identity function (\(t\rightarrow t\)) is a function of bounded variation in I. We assume \(r\in ]0,1[\) is fixed and consider a partition of I as mentioned in (6) with the condition \(x_i-x_{i-1}=\frac{b-a}{n}\) for all \(i\in \{1,\ldots ,n\}.\) We calculate \(V^{r}\big (f,P\big )\) as follows

$$\begin{aligned}{} & {} {V^{r}\big (f,{P}\big )=\sum _{i=1}^{n}|x_i-x_{i-1}|^{r} =n^{(1-r)}(b-a)^{r}.} \end{aligned}$$

Upon taking \(n\rightarrow \infty \), we can see that \(V^{r}\big (f,{P}\big )\) is not bounded. Through the arbitrariness of r, we conclude that the identity function does not satisfy r-bounded variation for any \(r<1\). It shows the validity of our first assertion.

To justify the second assertion, we take I to be [0, 1] and define the following function \(f:[0,1]\rightarrow \mathbb {R}\) as follows

$$\begin{aligned} { f(x)={\left\{ \begin{array}{ll} x\cos \Big (\dfrac{\pi }{x}\Big ) \text{ if } x\in ]0,1]\\ 0 \quad \text{ if } \quad x=0. \end{array}\right. }} \end{aligned}$$

We choose \(m\in \mathbb {N}\) such that \(\frac{1}{m}<1.\) Then for any \(n\in N\), we partition [0, 1] as

\(\hat{P}=\Big \{0,\dfrac{1}{m+n},\dfrac{1}{m+n-1},\ldots \dfrac{1}{m},1\Big \}.\) Based on the partition \(\hat{P}\), we calculate the ordinary variation of f as follows

$$\begin{aligned}{ V^{1}(f,\hat{P})\ge \Big (\dfrac{1}{m+n}+\dfrac{1}{m+n-1}\Big )+\cdots +\Big (\dfrac{1}{m}+\dfrac{1}{m+1}\Big ).} \end{aligned}$$

Upon taking \(n\rightarrow \infty \), it is evident that \(V^{1}(f,\hat{P})\) is not bounded. So, we can conclude \(f\notin {\mathbb {B}\mathbb {V}}(I).\)

Now we assume \(r>1\) to be arbitrary. Then we compute \(V^{r}(f,\hat{P})\) to obtain the following inequality

$$\begin{aligned} V^{r}(f,\hat{P})\le & {} \Big (1+\dfrac{1}{m+n}\Big )^{r}+\Big (\dfrac{1}{m+n}+\dfrac{1}{m+n-1}\Big )^{r} \\{} & {} +\cdots +\Big (\dfrac{1}{m}+\dfrac{1}{m+1}\Big )^{r}+\Big (1+\dfrac{1}{m}\Big )^{r}\\\le & {} \Big (1+\dfrac{1}{m+n}\Big )^{r}+2^{r}\Big (\dfrac{1}{m+n-1}\Big )^{r}\cdots +2^{r}\Big (\dfrac{1}{m}\Big )^{r}+\Big (1+\dfrac{1}{m}\Big )^{r}. \end{aligned}$$

Since for any \(r>1\), the series \(\underset{n\in \mathbb {N}}{\sum }\bigg (\dfrac{1}{n}\bigg )^{r}\) is convergent, by the comparison theorem, we have that \(V^{r}(f,\hat{P})\) is bounded. The same assertion can be shown for any other partition of [0, 1] just by approximating that partition with \(\hat{P}\). This implies f is a function of r-bounded variation for any \(r>1\). In other words \(f\in \underset{]1,\infty [}{\mathbb {B}\mathbb {V}}.\) This validates that the second inclusion is proper and completes the proof.\(\square \)

The next proposition shows the linkage between \(\Phi \)-Hölder function and r-bounded variation.

Proposition 3.8

Let \(\Phi : [0,\ell (I)]\rightarrow \mathbb {R}_+\) be an error function such that \({\Phi }^{r}\) is superadditive. Then any \(\Phi \)-Hölder function will also be a function of r-bounded variation.

Proof

Let \(f:I\rightarrow \mathbb {R}\) be a \(\Phi \) -Hölder function. We partition P as described in (6). Now by using the superadditivity of \({\Phi }^{r}\), we obtain the r-variation of f as follows

$$\begin{aligned} \underset{a}{\overset{b}{V^{r}}}(f,P){} & {} =\sum _{i=1}^{r}|f(x_i)-f(x_{i-1})|^{r}\\{} & {} \le \sum _{i=1}^{n}\Phi ^{r}(x_i-x_{i-1})\\{} & {} \le \Phi ^{r}(b-a). \end{aligned}$$

This holds true with respect to any possible partition of I and hence f satisfies r- bounded variation.\(\square \)

For the next proposition we assume that \(\mathbb {B}\mathbb {V}_{r}\) represents classes of functions satisfying r- bounded variation. The inclusion statement of Theorem 3.7 is a direct consequence of the following theorem.

Theorem 3.9

Let \(0<r_1<r_2\). Then \(\mathbb {B}\mathbb {V}_{r_1}\subseteq \mathbb {B}\mathbb {V}_{r_2}\) holds.

Proof

To prove the statement we assume that \(f\in \mathbb {B}\mathbb {V}_{r_1}.\) We consider a partition P as defined in (6). We compute \(V^{r_2}\big (f,P\big )\) as follows

$$\begin{aligned} {V^{r_2}\big (f,P\big ):=\sum _{i=1}^{n}|f(x_i)-f(x_{i-1})|^{r_2}.} \end{aligned}$$

Since \(\dfrac{r_2}{r_1}>1\), using the superadditivity of non-negative numbers we compute the following inequality

$$\begin{aligned}{} & {} {V^{r_2}\big (f,P\big )=\sum _{i=1}^{n}\bigg (|f(x_i)-f(x_{i-1})|^{r_1}\bigg )^{\frac{r_2}{r_1}}\le \bigg (\sum _{i=1}^{n}|f(x_i)-f(x_{i-1})|^{r_1}\bigg )^{\frac{r_2}{r_1}}}\\{} & {} \qquad \qquad {\le \bigg (V^{r_1}\big (f)\bigg )^{\frac{r_2}{r_1}}.} \end{aligned}$$

This shows that the above inequality also implies the boundedness of \(V^{r_2}\big (f,P\big )\). Since P is any arbitrary partition, we can conclude \(f\in \mathbb {B}\mathbb {V}_{r_2}\). This completes the proof.\(\square \)

Through an example we can easily observe that the above inclusion is proper. For the justification, here we only consider the case where \(0<r_1<r_2<1\). One can easily get the same validation for the remaining cases. For simplicity we consider \(I=[0,1]\) and select any \(r>1\). Now we define the function \(f:\mathbb {R}\rightarrow \mathbb {R}\) as follows

$$\begin{aligned} { f(x)={\left\{ \begin{array}{ll} (-1)^n\dfrac{1}{n^{r}}, \text{ if } x=\dfrac{1}{n} \text{ for } \text{ all } n\in \mathbb {N}\\ 0 \quad \text{ otherwise } \end{array}\right. }.} \end{aligned}$$
(10)

Since \(r>1\) the series \(\underset{n\in \mathbb {N}}{\sum }\bigg (\dfrac{1}{n^r}\bigg )\) is convergent. Using it, one can easily show that \(f\in {\mathbb {B}\mathbb {V}}\). Now we select \(r_1,r_2\in ]0,1[\) such that \(\dfrac{1}{r_2}<r<\dfrac{1}{r_1}\). Now we choose a fixed \(m\in \mathbb {N}\) and consider the partition \(\hat{P}\) for any arbitrary \(n\in \mathbb {N}\) as mentioned in Theorem 3.7

$$\hat{P}=\bigg \{0,\dfrac{1}{m+n},\dfrac{1}{m+n-1}\cdots \dfrac{1}{m},1\bigg \}.$$

By calculating the \(r_1\)-variation of \(f:I\rightarrow \mathbb {R}\) at \(\tilde{P}\), we obtain the following inequality

$$\begin{aligned} V^{r_1}(f,\hat{P})&\ge 2^{rr_1}\Big (\dfrac{1}{m+n-1}\Big )^{rr_1}+\cdots +2^{rr_1}\Big (\dfrac{1}{m+1}\Big )^{rr_1}. \end{aligned}$$

As \(rr_1<1\), the series \(\underset{n\in \mathbb {N}}{\sum }\bigg (\dfrac{1}{n^{rr_1}}\bigg )\) is divergent. Thus, upon taking \(n\rightarrow \infty \), we can clearly notice that \(V^{r_1}(f,P)\) is unbounded. Therefore, f is not an \(r_1\)-bounded function. In an analogous way we can easily establish that \(f\in \mathbb {B}\mathbb {V}_{r_2}\). This justifies the statement.

4 On functions of d-bounded variation

Throughout this section d will be a fixed positive number and \(I=[a,b]\) will represent a non-empty and non singleton closed interval with \(\ell (I)\ge 2d\). However for some of the results this condition can be relaxed to \(\ell (I)\ge d\) and it is very much self explanatory. A function \(f: I\rightarrow \mathbb {R}\) is said to be d-periodically increasing if for any \(x,y\in I\), with \(y-x\ge d\), \(f(x)\le f(y)\) holds. It is evident that the class of d-periodically increasing functions is a convex cone as it is closed under addition and multiplication by non-negative scalars. Moreover, if f is d-periodically increasing and non-negative then for any \(r(\in \mathbb {R}_+)\) the function \(f^{r}\) is also periodically increasing with the same period d. It can also be shown that the class of d-periodically increasing functions is also closed under pointwise limit operations.

Let \(f: I\rightarrow \mathbb {R}\) be a real valued function. We consider an arbitrary partition of I as \(P=\{x_0,\ldots ,x_n\}\) with \( a=x_0<\cdots <x_n=b\) satisfying \( x_i-x_{i-1}\ge d\) where \(i\in \{1,\ldots ,n\}\). Based on the partition P, we derive the d-variation of f in I as follows

$$\begin{aligned} {\overset{b}{\underset{a}{V_{d}}}(f,P):=\sum _{i=1}^{n}|f(x_i)-f(x_{i-1})|.} \end{aligned}$$
(11)

The supremum of (11) with respect to all such d-variations of f, is termed the total d-variation of f. In other words, we can define the total d-variation of f as follows

$$\begin{aligned} {\overset{b}{\underset{a}{V_{d}}}(f):=\sup \overset{b}{\underset{a}{V_{d}}}(f,P).} \end{aligned}$$

By using the notion of total d-variation of f we can formulate the d-variation function \(_{d}v_{f}\) as follows

$$\begin{aligned} {_{d}v_{f}(x):= {\left\{ \begin{array}{ll} 0\quad \qquad \text{ if }\quad x\in [a,a+d[\\ \overset{x}{\underset{a}{V_{d}}}(f)\quad \text{ if }\quad x\in [a+d,b] \end{array}\right. }} \end{aligned}$$

It is evident that any bounded function f will satisfy d- bounded variation since

$${\overset{b}{\underset{a}{V_{d}}}}(f,P)\le \bigg \lceil {\dfrac{b-a}{d}}\bigg \rceil \sup _{u,v\in [a,b]} |f(u)-f(v)|$$

holds. However the converse is not true. For example \(f:[0,1]\rightarrow \mathbb {R}\)

$$\begin{aligned} {f(x):= {\left\{ \begin{array}{ll} \dfrac{1}{x} \quad \text{ if }\quad x\in ]0,1]\\ 0\quad \text{ if } \quad x=0 \end{array}\right. }} \end{aligned}$$

is an unbounded function but possesses d-bounded variation. This implies that the class of functions satisfying d-bounded variation is bigger than the class of bounded functions.

The following results give a structural overview of this newly defined variation. The proposition below shows the d-periodic monotonicity of a d-variation function.

Proposition 4.1

Let \(f: I\rightarrow \mathbb {R}\) possesses d-bounded variation. Then the d-variation function \(_{d}v_{f}\) is d-periodically increasing.

Proof

To validate the statement, we consider two cases. First let \(x\in [a,a+d[\) and \(y\in [a+d,b]\) such that \(y-x\ge d\). Then the d-periodical monotonicity of \({_{d}v}_{f}\) is clearly evident from the following inequality

$${_{d}v}_{f}(x)=0\le \overset{y}{\underset{a}{V_{d}}}(f)={_{d}v}_{f}(y).$$

Next, we take \(x,y\in [a+d,b]\subseteq I\) such that \(y-x\ge d\). Let \(\epsilon > 0\) be arbitrary. Then there must exist a partition P of [a, x] as \(P=\{a=x_0,\ldots ,x_n=x\}\) such that \(a=x_0<x_1<\cdots <x_n=x\) with \(x_i-x_{i-1}\ge d\) for all \(i\in \{1,\ldots n\}\) satisfying

$$ {_{d}v}_{f}(x)=\overset{x}{\underset{a}{V_{d}}}(f)< \overset{x}{\underset{a}{V_{d}}}(f,P)+\epsilon .$$

Using the above inequality we obtain the following

$$\begin{aligned} \overset{y}{\underset{a}{V_{d}}}(f)+\epsilon\ge & {} \bigg (\sum _{i=1}^{n}|f(x_i)-f(x_{i-1})|+|f(y)-f(x)|\bigg )+\epsilon \\= & {} \overset{x}{\underset{a}{V_{d}}}(f,P)+\epsilon +|f(y)-f(x)|\\> & {} \overset{x}{\underset{a}{V_{d}}}(f). \end{aligned}$$

Taking \(\epsilon \rightarrow 0\); we see that \(\overset{x}{\underset{a}{V_{d}}}(f)\le \overset{y}{\underset{a}{V_{d}}}(f)\) holds. It proves that \(_{d}v_{f}\) is d-periodically increasing in the interval I.\(\square \)

The next proposition shows that the total d-variation of a function possesses the superadditive property in a restricted interval.

Proposition 4.2

Let f be a function of d-bounded variation in I. Then, for any \(c\in [a+d, b-d]\subseteq I\), the following inequality holds

$$\overset{c}{\underset{a}{V_{d}}}(f)+\overset{b}{\underset{c}{V_{d}}}(f)\le \overset{b}{\underset{a}{V_{d}}}(f).$$

Proof

Since at the beginning of the section we assumed that \(\ell (I)\ge 2d\), \([a+d,b-d]\) is non-empty. Let \(c\in [a+d,b-d]\subseteq I\), this implies that \(\ell ([a,c])\) and \(\ell ([c,b])\ge d\). We assume \(\epsilon >0\) to be arbitrary. Then there must exist two partitions for [a, c] and [c, b], respectively as \(P_1=\{x_0,\ldots ,x_{r-1},c=x_r\}\) and \(P_2=\{c=x_r, x_{r+1},\ldots ,x_n\}\) such that \(a=x_0<\cdots<x_r(=c)<\cdots <x_n=b\) with \(x_i-x_{i-1}\ge d\) for all \(i\in \{1,\ldots ,n\}\) satisfying the following two inequalities

$$\begin{aligned} { \overset{c}{\underset{a}{V_{d}}}(f)<\overset{c}{\underset{a}{V_{d}}}(f,P_1)+\dfrac{\epsilon }{2}\qquad \text{ and }\qquad \overset{b}{\underset{c}{V_{d}}}(f)<\overset{b}{\underset{c}{V_{d}}}(f,P_2)+\dfrac{\epsilon }{2}} \end{aligned}$$

Summing up these two inequalities side by side,we obtain

$$\begin{aligned} \overset{c}{\underset{a}{V_{d}}}(f)+\overset{b}{\underset{c}{V_{d}}}(f)< & {} \overset{c}{\underset{a}{V_{d}}}(f,P_1)+\overset{b}{\underset{c}{V_{d}}}(f,P_2)+\epsilon \\= & {} \overset{b}{\underset{a}{V_{d}}}(f,P_1\cup P_2)+\epsilon \\\le & {} \overset{b}{\underset{a}{V_{d}}}(f)+\epsilon . \end{aligned}$$

Due to the arbitrariness of \(\epsilon \), upon taking \(\epsilon \rightarrow 0\), we obtain

$$\begin{aligned} {\overset{c}{\underset{a}{V_{d}}}(f)+\overset{b}{\underset{c}{V_{d}}}(f)\le \overset{b}{\underset{a}{V_{d}}}(f).} \end{aligned}$$
(12)

This yields the inequality to be shown and validates the statement.\(\square \)

We will use the following proposition in the result related to decomposition.

Proposition 4.3

Every d-periodically increasing function satisfies d-bounded variation.

Proof

We assume \(f:I\rightarrow \mathbb {R}\) is a d-periodically increasing function. Then for the partition P as defined in (11), the following equation can be computed

$$\begin{aligned} { \overset{b}{\underset{a}{V_{d}}}(f,P)=\sum _{i=1}^{n}|f(x_i)-f(x_{i-1})|=\sum _{i=1}^{n}\Big (f(x_i)-f(x_{i-1}\Big )=f(b)-f(a).} \end{aligned}$$

Upon taking the supremum of all such possible partitions P, we can conclude that

$$\overset{b}{\underset{a}{V_{d}}}(f)=f(b)-f(a).$$

It shows that f is a function of d-bounded variation and completes the proof.\(\square \)

Now we are ready to state a possible decomposition for any function that possesses d-bounded variation in terms of a monotone and a d-periodically increasing function. The result somewhat resembles the classical Jordan decomposition theorem.

Theorem 4.4

A function of d-bounded variation can be expressed as the difference of a mononotone and a d-periodically increasing function.

Conversely the difference of two d-periodically increasing functions is a function of d-bounded variation.

Proof

First we assume that \(f:I\rightarrow \mathbb {R}\) is a function of d-bounded variation. Now we construct a function \({_{d}w}_f(x):I\rightarrow \mathbb {R}_+\) as

$${_{d}w}_f:= {\left\{ \begin{array}{ll} 0\qquad \text{ if }\quad x\in [a,a+d[\\ \underset{z\in [a, x-d]}{\sup }\overset{x}{\underset{z}{V_{d}}}(f)\,\, \text{ if }\quad x\in [a+d, b] \end{array}\right. } $$

and rewrite f in the following way

$$f:={_{d}}w_{f}-({_{d}}w_{f}-f).$$

To validate the theorem it will be enough to show that \({_{d}}w_{f}\) is increasing while \({_{d}}w_{f}-f\) is d periodically increasing.

Let \(x,y\in I\) with \(x<y\). If \(x,y\in [a,a+d[\) then there is nothing to show. For \(x\in [a,a+d[\) and \(y\in [a+d,b]\), the monotonicity follows from the definition of \({_{d}}w_{f}\). On the other hand if both \(x,y\in [a+d,b]\), then the following inequality yields that \({_{d}}w_{f}\) is nondecreasing

$${_{d}}w_{f}(x)=\underset{z\in [a, x-d]}{\sup }\overset{x}{\underset{z}{V_{d}}}(f)\le \underset{z\in [a, y-d]}{\sup }\overset{y}{\underset{z}{V_{d}}}(f)={_{d}w}_f(y).$$

To show \({_{d}}w_{f}-f\) is d-periodically increasing, we consider \(x,y\in I\) such that \(y-x\ge d.\) This implies \(y\in [a+d, b]\) and we have to consider two cases. For \(x\in [a, a+d[\), by definition \({_{d}}w_{f}(x)=0\). On the other hand if \(x\in [a+d, b]\) then \({_{d}}w_{f}(x)=\underset{z\in [a, x-d]}{\sup }\overset{x}{\underset{z}{V_{d}}}(f)\). For both of these cases we can compute the following inequality

$${_{d}}w_{f}(y)\ge {_{d}}w_{f}(x)+|f(y)-f(x)|\ge {_{d}}w_{f}(x)+f(y)-f(x).$$

Rearranging the terms of the above inequality we see that \({_{d}}w_{f}-f\) is d-periodically increasing. This proves the first part of the theorem.

To show the converse assertion, let g and h be two d-periodically increasing functions. Now as in Proposition 4.3, we can compute the following inequality

$$\begin{aligned} {\overset{b}{\underset{a}{V_{d}}}(g-h)\le g(b)+h(b)-g(a)-h(a)} \end{aligned}$$

This implies \(g-h\) is a function of d-bounded variation and completes the proof of the theorem.\(\square \)

d-periodically increasing functions often occur in the stock market prices of a reputed company’s share. In most of the cases it varies in a short time frame but eventually it gets an upward trend after a distinct time interval. A similar type of periodic monotonicity is observable in the population growth graph of most East European nations. Therefore, an interdisciplinary study of such delayed monotonicity is also very much possible.