Nemytskii operator on $(\phi ,2,\alpha )$-bounded variation space in the sense of Riesz

Abstract

In this paper we show that if the Nemytskii operator maps the $(\phi ,2,\alpha )$-bounded variation space into itself and satisfies some Lipschitz condition, then there are two functions g and h belonging to the $(\phi ,2,\alpha )$-bounded variation space such that $f(t,y)=g(t)y+h(t)$ for all $t\in [a,b]$, $y\in {\mathbb {R}}$.

1 Introduction

According to Lakoto [16], functions of bounded variation were discovered by Camille Jordan around 1880 through a “critical” re-examination of Dirichlet’s famous flawed proof that arbitrary functions can be represented by Fourier series (see, [14]). It was Jordan who gave the characterization of such functions as differences of increasing functions, but, as point out by Hawkins [15], the key observation that Dirichlet’s proof was valid for differences of increasing functions had already been made by Dubois-Raymond [13]. In the same vein, in 1905 G. Vitali [32] introduce the absolutely continuous functions of one variable. Since then, the concept has been generalized in many ways.

Some of those generalizations were motivated by problems in areas such that geometric measures theory and mathematical physics. (For applications of function variation in mathematical physics, see the monograph [33]). One of those generalizations appeared in 1908, when de la Vallée Poussin [12] defined the second bounded variation of a function f in the interval [a, b] by

$$\begin{aligned} V^2(f)=V^2(f,[a,b])=\sup _\Pi \sum _{j=1}^{n-1}\left| \frac{f(x_{j+1})-f(x_j)}{x_{j+1}-x_j}-\frac{f(x_j)-f(x_{j-1})}{x_j-x_{j-1}}\right| , \end{aligned}$$

where the supremun is taken over all partition

$$\begin{aligned} \Pi =\{a=x_0<x_1<\cdots <x_n=b\} \end{aligned}$$

of the interval [a, b].

Another generalization is due to F. Riesz in his 1910 paper (see, [28]), he defined the p -variation of a function on an interval [a, b] as

$$\begin{aligned} V_p^R(f)=V_p^R(f,[a,b])=\sup _\Pi \sum _{j=1}^{n}\frac{|f(x_j)-f(x_{j-1})|^p}{|x_j-x_{j-1}|^{p-1}}. \end{aligned}$$

(1.1)

Again, the supremun is taken over all partition $\Pi =\{a=x_0<x_1<\cdots <x_n=b\}$ of the interval [a, b]. Riesz proved that, for $1<p<+\infty$, the class of functions of bound p-variation (i.e., the class of functions for which $V_p^R(f)<+\infty$) coincides with the class of absolutely continuous functions with derivative belonging to $L_p([a,b])$. Moreover, the p-variation of a function f on [a, b] is given by

$$\begin{aligned} V_p^R(f)=V_p^R(f,[a,b])=\int \limits _a^b|f'(x)|^pdx. \end{aligned}$$

(1.2)

One may replace the p-th power in (1.1) by a function $\phi$ behaves similar to $x^p$ for $p\ge 1$ as follows: A function $\phi :[0,+\infty )\longrightarrow [0,+\infty )$ such that

• (a) $\phi (x)=0$ if and only if $x=0$.

• (b) $\lim _{x\rightarrow +\infty } \phi (x)=+\infty$,

is known as Young function. In 1953 Y.T. Medvedev [26] introduced the concept of $\phi$-bounded variation in the following way: given a Young function $\phi$, a partition $\Pi =\{a=x_0<x_1<\cdots <x_n=b\}$ of [a, b] and a function $f:[a,b]\longrightarrow {\mathbb {R}}$, the $\phi$-variation of f is defined as

$$\begin{aligned} V_\phi ^R(f)=V_\phi ^R(f,[a,b])=\sup _\Pi \sum _{j=1}^{n}\phi \left( \frac{|f(x_j)-f(x_{j-1})|}{|x_j-x_{j-1}|^{p-1}}\right) |x_{j-1}-x_j|, \end{aligned}$$

(1.3)

where the supremun is taken over all partition $\Pi$ of [a, b]. We might observe that, when $\phi (x)=x^p$, $p\ge 1$, $x\ge 0$ we get back the p-variation concept. In other words, the Medvedev characterization generalizes the one made by Riesz. In such a sense that (1.3) is called the Riesz-Medvedev variation of f on [a, b]. Again, in case $V_\phi ^R(f)<\infty$, we say that f has bounded Riesz-Medvedev variation (or bounded $\phi$-variation in Riesz’s sense) on [a, b], and we write $f\in BV_\phi ^R([a,b])$.

In the same paper [26], for a convex Young function $\phi$ which satisfies the $\infty _1$-condition (that is $\lim _{x\rightarrow +\infty }\frac{\phi (x)}{x}=+\infty$), the following remarkable result was proven: $f\in BV_\phi ^R([a,b])$ if and only if f is absolutely continuous on [a, b] and $\int \limits _a^b\phi (|f'(x)|)dx<+\infty$.

Moreover, the $\phi$-variation of f on [a, b] is given by

$$\begin{aligned} V_\phi ^R(f)=V_\phi ^R(f,[a,b])=\int \limits _a^b\phi (|f'(x)|)dx. \end{aligned}$$

(1.4)

Also note that (1.4) generalizes (1.2). In [6] the first and third named authors, together with H. Rafeiro, introduced the $(2,\alpha )$-variation in the sense of de la Vallée Poussin, combining the second bounded variation with the $(p,\alpha )$-variation (see [3] and [8]).

Definition 1.1

Let $\phi$ be a $\phi$-function (Young function), f a real function defined on [a, b] and let $\alpha$ be any strictly increasing continuous function defined on [a, b]. Let $\Pi$ be a block partition of the interval [a, b], that is,

$$\begin{aligned} \Pi&:a=x_{1,1}<x_{1,2}\le x_{1,3}<x_{1,4}=x_{2,1}<x_{2,2}\le x_{2,3}<x_{2,4}=x_{3,1}<\cdots \\&<x_{n-1,4}=x_{n,2}\le x_{n,3}<x_{n,4}=b. \end{aligned}$$

Let

$$\begin{aligned} \sigma ^R_{(\phi ,2,\alpha )}(f,\Pi )=&\sum _{j=1}^n\phi \left( \left| \frac{f(x_{j,4})-f(x_{j,3})}{(\alpha (x_{j,4})-\alpha (x_{j,3}))(\alpha (x_{j,4})-\alpha (x_{j,1}))}\right. \right. \\&\left. \left. -\frac{f(x_{j,2})-f(x_{j,1})}{(\alpha (x_{j,2})-\alpha (x_{j,1}))(\alpha (x_{j,4})-\alpha (x_{j,1}))}\right| \right) |\alpha (x_{j,4})-\alpha (x_{j,1})|\\ =&\sum _{j=1}^n\phi \left( \frac{\left| \frac{f(x_{j,4})-f(x_{j,3})}{\alpha (x_{j,4})-\alpha (x_{j,3})}-\frac{f(x_{j,2})-f(x_{j,1})}{\alpha (x_{j,2})-\alpha (x_{j,1})}\right| }{|\alpha (x_{j,4})-\alpha (x_{j,1})|}\right) |\alpha (x_{j,4})-\alpha (x_{j,1})|\\ =&\sum _{j=1}^n\phi \left( \frac{|f_\alpha [x_{j,4},x_{j,3}]-f_\alpha [x_{j,2},x_{j,1}]|}{|\alpha (x_{j,4})-\alpha (x_{j,1})|}\right) |\alpha (x_{j,4})-\alpha (x_{j,1})| \end{aligned}$$

where

$$\begin{aligned} f_\alpha [p,q]=\frac{f(q)-f(p)}{\alpha (q)-\alpha (p)} \end{aligned}$$

and

$$\begin{aligned} V_{(\phi ,2,\alpha )}^R(f,[a,b])=V_{(\phi ,2,\alpha )}^R(f)=\sup _{\Pi }\sigma ^R_{(\phi ,2,\alpha )}(f,\Pi ), \end{aligned}$$

where the supremum is taken over all possible block partition of [a, b].

$V_{(\phi ,2,\alpha )}^R(f)$ is called $(\phi ,2,\alpha )$-variation in the sense of Riesz on the interval [a, b]. If $V_{(\phi ,2,\alpha )}^R(f)<\infty$, the function f is said to be of $(\phi ,2,\alpha )$-variation in the sense of Riesz. The set of all this functions is denoted by $V_{(\phi ,2,\alpha )}^R([a,b])$. $RV_{(\phi ,2,\alpha )}([a,b])$ is the space generated by $V_{(\phi ,2,\alpha )}^R([a,b])$.

2 Definitions and some needed results

In this section, we gather definitions and notations that will be used throughout the paper. Let $\alpha$ be any strictly increasing continuous function defined on [a, b].

Definition 2.1

Let $\phi$ be a convex $\phi$-function, then

$$\begin{aligned} RV_{(\phi ,2,\alpha )}^0([a,b])=\{f:[a,b]\longrightarrow {\mathbb {R}}\;:\; f\in RV_{(\phi ,2,\alpha )}([a,b])\text { and }f(a)=0\} \end{aligned}$$

is the linear space of $(\phi ,2,\alpha )$-bounded variation in the sense of Riesz which vanish at a.

Definition 2.2

Let $\phi$ be a convex $\phi$-function,

$$\begin{aligned} |\cdot |_{(\phi ,2,\alpha )}^R:RV_{(\phi ,2,\alpha )}^0([a,b])\longrightarrow {\mathbb {R}}^+ \end{aligned}$$

given by

$$\begin{aligned} |f|_{(\phi ,2,\alpha )}^R=|f'_\alpha (a)|+\inf \{\varepsilon >0\;:\; V_{(\phi ,2,\alpha )}^R(f/\varepsilon )\le 1\}. \end{aligned}$$

Definition 2.3

Suppose f and $\alpha$ are real-valued functions on the same open interval (bounded or unbounded). Suppose $x_0$ is a point in this interval. We say f is $\alpha$-derivable at $x_0$ if

$$\begin{aligned} \lim _{\alpha \rightarrow x_0}\frac{f(x)-f(x_0)}{\alpha (x)-\alpha (x_0)}\quad \text {exists.} \end{aligned}$$

We denote its value by $f_\alpha '(x_0)$, which we call the $\alpha$-derivative of f at $x_0$.

Definition 2.4

A function $\upsilon :[a,b]\longrightarrow {\mathbb {R}}$ is $\alpha$-convex in [a, b], if for all $a\le \lambda \le \xi \le \mu \le b$ the following holds

$$\begin{aligned} \upsilon (\xi )\le \frac{\alpha (\xi )-\alpha (\lambda )}{\alpha (\mu )-\alpha (\lambda )}\upsilon (\mu )+\frac{\alpha (\mu )-\alpha (\xi )}{\alpha (\mu )-\alpha (\lambda )}\upsilon (\lambda ). \end{aligned}$$

As a consequence, we have, from the properties of $\alpha$-convex functions, the existence of the lateral derivatives $f_{\alpha ^+}'(x_0)$ and $f_{\alpha ^-}'(x_0)$ in each point $x_0\in (a,b)$ and the existence of $f_{\alpha ^+}'(a)$ and $f_{\alpha ^-}'(b)$.

Theorem 2.1

Let $\phi$ be a convex $\phi$-function. If $f\in V_{(\phi ,2,\alpha )}^R([a,b])$, then $f\in BV^{(2,\alpha )}([a,b])$. Moreover

$$\begin{aligned} V^{(2,\alpha )}(f)\le \frac{1}{\phi (1)}V_{(\phi ,2,\alpha )}^R(f)+\alpha (b)-\alpha (a). \end{aligned}$$

For the proof see [2].

Theorem 2.2

If $\phi$ is a convex $\phi$-function such that satisfy the $(\infty _1)$-condition, then we have the following embedding results

$$\begin{aligned}&RV_{(\phi ,2,\alpha )}([a,b])\subset BV^{(2,\alpha )}([a,b])\subset \alpha -Lip([a,b])\subset RV_{(\phi ,\alpha )}([a,b])\subset \\&\quad \cdots \subset \alpha -AC([a,b])\subset BV([a,b])\subset B([a,b]). \end{aligned}$$

Theorem 2.3

$f\in BV_{(2,\alpha )}([a,b])$ if and only if $f=f_1-f_2$ where $f_1$ and $f_2$ are $\alpha$-convex functions.

Theorem 2.4

Let $\phi$ be a convex function, which satisfy the $(\infty _1)$-condition. If $f\in V_{(\phi ,2,\alpha )}^R([a,b])$, then there exists $f_\alpha '(x_0)$ on each point $x_0\in [a,b]$.

Proof

From Theorem 2.2 we know the $V_{(\phi ,2,\alpha )}^R([a,b])\subset BV^{(2,\alpha )}([a,b])$ and by Theorem 2.3 we have that there are lateral $\alpha$-derivatives at each point of the interval [a, b]. Suppose there is $x_0\in (a,b)$ such that $f_{\alpha ^+}(x_0)\ne f_{\alpha ^-}(x_0)$. From the definition of $V_{(\phi ,2,\alpha )}([a,b])$, let us consider in the partition the points $\cdots \le x_0+h\le x_0<x_0+h<\cdots$ in order to obtain

$$\begin{aligned} V_{(\phi ,2,\alpha )}^R([a,b])\ge&\phi \left( \frac{\left| \frac{f(x_0+h)-f(x_0)}{\alpha (x_0+h)-\alpha (x_0)}-\frac{f(x_0)-f(x_0-h)}{\alpha (x_0)-\alpha (x_0-h)}\right| }{|\alpha (x_0+h)-\alpha (x_0-h)|}\right) |\alpha (x_0+h)-\alpha (x_0-h)|\\ =&\frac{ \phi \left( \frac{\left| \frac{f(x_0+h)-f(x_0)}{\alpha (x_0+h)-\alpha (x_0)}-\frac{f(x_0)-f(x_0-h)}{\alpha (x_0)-\alpha (x_0-h)}\right| }{|\alpha (x_0+h)-\alpha (x_0-h)|}\right) }{\frac{|f_{\alpha ^+}'(x_0)-f_{\alpha ^-}(x_0)|}{\alpha (x_0+h)-\alpha (x_0+h)}}|f_{\alpha ^+}(x_0)-f_{\alpha ^-}(x_0)| \end{aligned}$$

letting $h\rightarrow 0$ and using the fact that $\phi$ and $\alpha$ are continuous, then we have

$$\begin{aligned} V_{(\phi ,2,\alpha )}^R([a,b])\ge&\frac{\phi \left( \frac{|f_{\alpha ^+}'(x_0)-f_{\alpha ^-}'(x_0)|}{\displaystyle \lim _{h\rightarrow 0}|\alpha (x_0+h)-\alpha (x_0-h)|}\right) }{\frac{|f'_{\alpha ^+}(x_0)-f_{\alpha ^-}'(x_0)|}{\displaystyle \lim _{h\rightarrow 0}|\alpha (x_0+h)-\alpha (x_0-h)|}}\cdot |f_{\alpha ^+}'(x_0)-f_{\alpha ^-}(x_0)|\\ =&\lim _{h\rightarrow 0}\frac{\phi \left( \frac{|f_{\alpha ^+}'(x_0)-f_{\alpha ^-}'(x_0)|}{|\alpha (x_0+h)-\alpha (x_0-h)|}\right) }{\frac{|f'_{\alpha ^+}(x_0)-f_{\alpha ^-}'(x_0)|}{|\alpha (x_0+h)-\alpha (x_0-h)|}}\cdot |f_{\alpha ^+}'(x_0)-f_{\alpha ^-}(x_0)|=+\infty . \end{aligned}$$

That is,

$$\begin{aligned} \lim _{h\rightarrow 0}\frac{|f'_{\alpha ^+}(x_0)-f_{\alpha ^-}(x_0)|}{|\alpha (x_0+h)-\alpha (x_0-h)|}=+\infty \end{aligned}$$

and

$$\begin{aligned} |\alpha (x_0+h)-\alpha (x_0-h)|\ne 0. \end{aligned}$$

This contradicts the fact that $f\in V_{(\phi ,2,\alpha )}^R([a,b])$, so f is $\alpha$-derivable at each point of (a, b) and there exist $f_{\alpha ^+}'(a)$ and $f_{\alpha ^-}(b)$. $\square$

Theorem 2.5

Let $\phi$ be a convex $\phi$-function which satisfies the $(\infty _1)$-condition and $f:[a,b]\longrightarrow {\mathbb {R}}$. If $f\in V_{(\phi ,2,\alpha )}^R([a,b])$, that is $V_{(\phi ,2,\alpha )}^R(f)<+\infty$, then $f_\alpha '\in V_{(\phi ,2,\alpha )}^R([a,b])$, that is $V_{(\phi ,2,\alpha )}^R(f_\alpha ')<+\infty$. Moreover,

$$\begin{aligned} \int \limits _a^b\phi (|f_\alpha ''(t)|)d\alpha (t)\le V_{(\phi ,2,\alpha )}(f). \end{aligned}$$

Proof

Let $\Pi : a<x_0<x_1\cdots <x_n=b$ be a partition of [a, b] and $0<h\le \min \left\{ \frac{x_j-x_{j-1}}{2},\; j=1,2,\dots ,n\right\}$. We have that

$$\begin{aligned}&a=x_0<x_0+h<x_1-h<x_1<x_1+h\le x_2-h<\\&\quad \cdots<x_{n-1}<x_{n-1}+h\le x_n-h<x_n=b \end{aligned}$$

is a block partition of [a, b]. By definition we have

$$\begin{aligned} \sum _{j=1}^n\phi \left( \frac{\left| \frac{f(x_j)-f(x_j-h)}{\alpha (x_j)-\alpha (x_j-h)}-\frac{f(x_{j-1}+h)-f(x_{j-1})}{\alpha (x_{j-1}+h)-\alpha (x_{j-1})}\right| }{|\alpha (x_j)-\alpha (x_{j-1})|}\right) |\alpha (x_j)-\alpha (x_{j-1})|\le V_{(\phi ,2,\alpha )}^R(f). \end{aligned}$$

Allowing that h goes to 0 in the previous expression, we deduce

$$\begin{aligned} \lim _{h\rightarrow 0}\sum _{j=1}^n\phi \left( \frac{\left| \frac{f(x_j)-f(x_j-h)}{\alpha (x_j)-\alpha (x_j-h)}-\frac{f(x_{j-1}+h)-f(x_{j-1})}{\alpha (x_{j-1}+h)-\alpha (x_{j-1})}\right| }{|\alpha (x_j)-\alpha (x_{j-1})|}\right) |\alpha (x_j)-\alpha (x_{j-1})|\le V_{(\phi ,2,\alpha )}^R(f). \end{aligned}$$

By continuity of $\phi$ and the definition of $f_{\alpha ^+}'$ and $f_{\alpha ^-}'$

$$\begin{aligned} \sum _{j=1}^n\phi \left( \frac{|f'_{\alpha ^-}(x_j)-f'_{\alpha ^+}(x_{j-1})|}{|\alpha (x_j)-\alpha (x_{j-1})|}\right) |\alpha (x_j)-\alpha (x_{j-1})|\le V_{(\phi ,2,\alpha )}^R(f). \end{aligned}$$

This result holds for all partition $\Pi$ of [a, b]. In consequence

$$\begin{aligned} V_{(\phi ,\alpha )}^R(f'_\alpha )\le V_{(\phi ,2,\alpha )}^R(f). \end{aligned}$$

Hence

$$\begin{aligned} f_\alpha '\in V_{(\phi ,\alpha )}^R([a,b]). \end{aligned}$$

By virtue of the Theorem of Medved’ev (see [26]) we conclude that $f'_\alpha \in \alpha -AC([a,b])$ and therefore there exists $f''_\alpha$ a.e. in [a, b]. Moreover

$$\begin{aligned} \int \limits _a^b\phi \left( |f_\alpha ''(t)|\right) d\alpha (t)\le V_{(\phi ,\alpha )}^R(f_\alpha ')\le V_{(\phi ,2,\alpha )}^R(f). \end{aligned}$$

$\square$

The next result is the reciprocal.

Theorem 2.6

Let $\phi$ be a convex function which satisfy the $(\infty _1)$-condition and $f:[a,b]\longrightarrow {\mathbb {R}}$. If $f_\alpha '$ is $\alpha$-absolutely continuous and

$$\begin{aligned} \int \limits _a^b\phi (|f''_\alpha (t)|)d\alpha (t)<+\infty , \end{aligned}$$

then

$$\begin{aligned} V_{(\phi ,2,\alpha )}^R(f)\le \int \limits _a^b\phi (|f''_\alpha (t)|)d\alpha (t). \end{aligned}$$

Proof

Let $\Pi : a=x_{1,1}<x_{1,2}\le x_{1,3}<x_{1,4}=x_{2,1}<\cdots<x_{n-1,4}=x_{n,1}<x_{n,2}\le x_{n,3}<x_{n,4}=b$ be a block partition of [a, b].

Since $f_\alpha '$ exists, then f is continuous in (a, b) and using the mean value Theorem, we deduce that there exists $x_j^+\in (x_{j,3},x_{j,4})$ and $x_j^-\in (x_{j,1},x_{j,4})$ such that

$$\begin{aligned} f'_\alpha (x_j^-)=\frac{f(x_{j,2})-f(x_{j,3})}{\alpha (x_{j,2})-\alpha (x_{j,1})} \end{aligned}$$

and

$$\begin{aligned} f'_\alpha (x_j^+)=\frac{f(x_{j,4})-f(x_{j,3})}{\alpha (x_{j,4})-\alpha (x_{j,3})},\quad j=1,2,\dots ,n. \end{aligned}$$

In this way we obtain the following estimation

$$\begin{aligned}&\phi \left( \frac{\left| \frac{f(x_{j,4})-f(x_{j-3})}{\alpha (x_{j,4})-\alpha (x_{j,3})}-\frac{f(x_{j,2})-f(x_{j-1})}{\alpha (x_{j,2})-\alpha (x_{j,1})}\right| }{|\alpha (x_{j,4})-\alpha (x_{j,1})|}\right) |\alpha (x_{j,4})-\alpha (x_{j,1})| \\&\quad =\phi \left( \frac{|f'\alpha (x_j^+)-f'\alpha (x_j^-)|}{|\alpha (x_{j,4})-\alpha (x_{j,1}|}\right) |\alpha (x_{j,4})-\alpha (x_{j,1})|\\&\quad =\phi \left( \frac{1}{|\alpha (x_{j,4})-\alpha (x_{j,1})|}\int \limits _{x_j^-}^{x_j^+}|f''_\alpha (\xi )|d\alpha (\xi )\right) |\alpha (x_{j,4})-\alpha (x_{j,1})| \\&\quad \le \phi \left( \frac{1}{|\alpha (x_{j,4})-\alpha (x_{j,1})|}\int \limits _{x_{j,1}}^{x_{j,4}}|f''_\alpha (\xi )|d\alpha (\xi )\right) |\alpha (x_{j,4})-\alpha (x_{j,1})|\\&\quad =\phi \left( \frac{\displaystyle \int \limits _{x_{j,1}}^{x_{j,4}} |f_\alpha ''(\xi )|d\alpha (\xi ) }{\displaystyle \int \limits _{x_{j,1}}^{x_{j,4}}d\alpha (\xi )}\right) |\alpha (x_{j,4})-\alpha (x_{j,1})|. \end{aligned}$$

By the Jensen inequality

$$\begin{aligned} \le&\frac{\displaystyle \int \limits _{x_{j,1}}^{x_{j,4}} \phi (|f_\alpha ''(\xi )|)d\alpha (\xi ) }{\displaystyle \int \limits _{x_{j,1}}^{x_{j,4}}d\alpha (\xi )}|\alpha (x_{j,4})-\alpha (x_{j,1})|.\\ =&\frac{1}{|\alpha (x_{j,4})-\alpha (x_{j,1})|}\left[ \displaystyle \int \limits _{x_{j,1}}^{x_{j,4}}\phi (|f''_\alpha (\xi )|)d\alpha (\xi )\right] |\alpha (x_{j,4})-\alpha (x_{j,1})|\\ =&\displaystyle \int \limits _{x_{j,1}}^{x_{j,4}}\phi (|f''_\alpha (\xi )|)d\alpha (\xi ). \end{aligned}$$

Next, adding for $j=1,2,\dots ,n$

$$\begin{aligned} \sigma _{(\phi ,2,\alpha )}^R(f,\Pi )=&\sum _{j=1}^n \phi \left( \frac{\left| \frac{f(x_{j,4})-f(x_{j-3})}{\alpha (x_{j,4})-\alpha (x_{j,3})}-\frac{f(x_{j,2})-f(x_{j-1})}{\alpha (x_{j,2})-\alpha (x_{j,1})}\right| }{|\alpha (x_{j,4})-\alpha (x_{j,1})|}\right) |\alpha (x_{j,4})-\alpha (x_{j,1})|\\ \le&\sum _{j=1}^n\int \limits _{x_{j,1}}^{x_{j,4}}\phi (|f''_\alpha (\xi )|)d\alpha (\xi )\\ =&\int \limits _a^b\phi (|f''_\alpha (\xi )|)d\alpha (\xi ). \end{aligned}$$

From this expression we deduce that

$$\begin{aligned} V_{(\phi ,2,\alpha )}^R(f)\le \int \limits _a^b\phi (|f_\alpha ''(\xi )|)d\alpha (\xi ), \end{aligned}$$

therefore $f\in V_{(\phi ,2,\alpha )}([a,b])$. $\square$

From the previous theorems, we deduce the following.

Corollary 2.7

Let $\phi$ be a convex $\phi$-function which satisfy the $(\infty _1)$-condition and $f:[a,b]\longrightarrow {\mathbb {R}}$. Then the following propositions are equivalent

• 1. $f\in V_{(\phi ,2,\alpha )}^R([a,b])$ if and only if $f'_\alpha \in V_{(\phi ,\alpha )}^R([a,b])$.

• 2. $f\in V_{(\phi ,2,\alpha )}^R([a,b])$ if and only if $f_\alpha '\in \alpha -AC([a,b])$ and $\int \limits _a^b\phi (|f_\alpha '(t)|)d\alpha (t)<+\infty$.

Moreover

$$\begin{aligned} V_{(\phi ,2,\alpha )}^R(f)=V_{(\phi ,\alpha )}^R(f_\alpha ')=\int \limits _a^b\phi (|f''_\alpha (\xi )|)d\alpha (\xi ). \end{aligned}$$

Corollary 2.8

Let $\phi$ be a convex $\phi$-function which satisfy the $(\infty _1)$-condition and $f\in RV_{(\phi ,2,\alpha )}^0([a,b])$. Then

$$\begin{aligned} |f|_{(\phi ,2,\alpha )}^R=|f_\alpha '(a)|+\inf \{\varepsilon >0\;:\; \int \limits _a^b\phi \left( \frac{|f_\alpha ''(t)|}{\varepsilon }\right) d\alpha (t)\le 1\}. \end{aligned}$$

Proof

By Corollary 2.7 and Definition 2.2 we obtain

$$\begin{aligned} |f|^R_{(\phi ,2,\alpha )}=&|f_\alpha '(a)|+\inf \{\varepsilon>0\;:\;V_{(\phi ,2,\alpha )}(f/\varepsilon )\le 1\}\\ =&|f_\alpha '(a)|+\inf \{\varepsilon >0\;:\;\int \limits _a^b\phi \left( \frac{|f_\alpha ''(t)|}{\varepsilon }\right) d\alpha (t)\le 1\}. \end{aligned}$$

$\square$

Remark 2.1

From Corollary 2.8 we might derive the following result:

$$\begin{aligned} \Vert f\Vert _{(\phi ,2,\alpha )}^R=|f(a)|+|f_\alpha '(a)|+\inf \{\varepsilon >0\;:\; \int \limits _a^b\phi \left( \frac{|f_\alpha ''(t)|}{\varepsilon }\right) d\alpha (t)\le 1\}. \end{aligned}$$

Corollary 2.9

Let $\phi$ be a convex $\phi$-function which satisfy the $(\infty _1)$-condition and $f\in RV_{(\phi ,2,\alpha )}^0([a,b])$. Then

$$\begin{aligned} |f|_{(\phi ,2,\alpha )}^R=\Vert f_\alpha '\Vert _{(\phi ,\alpha )}^R. \end{aligned}$$

Proof

Applying Corollary 2.7 we have

$$\begin{aligned} |f|_{(\phi ,2,\alpha )}^R=&|f_\alpha '(a)|+\inf \{\varepsilon>0\;:\; V_{(\phi ,2,\alpha )}^R(f/\varepsilon )\le 1\}\\ =&|f_\alpha '(a)|+\inf \{\varepsilon >0\;:\; V_{(\phi ,\alpha )}^R(f'_\alpha /\varepsilon )\le 1\}\\ =&\Vert f_\alpha '\Vert _{(\phi ,2,\alpha )}^R. \end{aligned}$$

$\square$

Definition 2.5

Let $\phi$ be a convex $\phi$-function, then

$$\begin{aligned}&\{f:[a,b]\longrightarrow {\mathbb {R}}\;:\; \exists \lambda>0\text { such that } \lambda f\in V_{(\phi ,2,\alpha )}^R([a,b])\}\\&\quad = \{f:[a,b]\longrightarrow {\mathbb {R}}\;:\; \exists \lambda >0\text { such that } V_{(\phi ,2,\alpha )}^R(\lambda f)<+\infty \} \end{aligned}$$

it is called the linear space of the $(\phi ,2,\alpha )$-bounded variation functions in the sense of Riesz, and it is denoted by $RV_{(\phi ,2,\alpha )}([a,b])$.

Definition 2.6

Let $\phi$ be a convex $\phi$-function, then

$$\begin{aligned}&\{f:[a,b]\longrightarrow {\mathbb {R}}\;:\; \exists \lambda>0\text { such that } \lambda f\in V_{(\phi ,\alpha )}^R([a,b])\}\\ =&\{f:[a,b]\longrightarrow {\mathbb {R}}\;:\; \exists \lambda >0\text { such that } V_{(\phi ,\alpha )}^R(\lambda f)<+\infty \} \end{aligned}$$

it is called the linear space of the $(\phi ,\alpha )$-bounded variation functions in the sense of Riesz, and it is denoted by $RV_{(\phi ,\alpha )}([a,b])$.

Corollary 2.10

Let $\phi$ be a convex $\phi$-function which satisfy the $(\infty _1)$-condition and let $f:[a,b]\longrightarrow {\mathbb {R}}$, then $f\in RV_{(\phi ,2,\alpha )}([a,b])$ if and only if $f'_\alpha \in RV_{(\phi ,\alpha )}([a,b])$.

Proof

$f\in RV_{(\phi ,2,\alpha )}([a,b])$ iff there exits $\lambda >0$ such that $V_{(\phi ,2,\alpha )}^R(\lambda f)<+\infty$. By Definition 2.5

$$\begin{aligned} \Longleftrightarrow \exists \lambda >0\text { such that }V_{(\phi ,\alpha )}^R((\lambda f)'_\alpha )<+\infty \end{aligned}$$

by Corollary 2.7

$$\begin{aligned}&\Longleftrightarrow \exists \lambda >0\text { such that } V_{(\phi ,\alpha )}^R(\lambda f'_\alpha )<+\infty \\&\Longleftrightarrow f'_\alpha \in RV_{(\phi ,\alpha )}([a,b]), \text { by Definition}~2.6. \end{aligned}$$

$\square$

Definition 2.7

If $\phi$ is a convex $\phi$-function $\Vert \cdot \Vert _{(\phi ,2,\alpha )}^R:RV_{(\phi ,2,\alpha )}([a,b])\longrightarrow {\mathbb {R}}^+$ given by

$$\begin{aligned} \Vert f\Vert ^R_{(\phi ,2,\alpha )}=|f(a)|+|f-f(a)|_{(\phi ,2,\alpha )}^R. \end{aligned}$$

Lemma 2.11

Let $\phi$ be a $\phi$-function, then $f\in RV_{(\phi ,2,\alpha )}([a,b])$ if and only if $f-f(a)\in RV_{(\phi ,2,\alpha )}^0([a,b])$.

Proof

By Definition 1.1, we observe that $\sigma _{(\phi ,2,\alpha )}(f-f(a),\Pi )=\sigma _{(\phi ,2,\alpha )}^R(f,\Pi )$ for any partition $\Pi$ of [a, b] where

$$\begin{aligned} V_{(\phi ,2,\alpha )}(f-f(a))=V_{(\phi ,2,\alpha )}^R(f). \end{aligned}$$

$\square$

Observe that

$$\begin{aligned} \Vert f\Vert ^R_{(\phi ,2,\alpha )}= & {} |f(a)|+|f-f(a)|_{(\phi ,2,\alpha )}^R\\= & {} |f(a)|+|(f-f(a))_\alpha '(a)|+|f-f(a)|_{(\phi ,2,\alpha )}^0, \text {by Definition}~2.2\\= & {} |f(a)|+|f_\alpha '(a)|+\inf \{\varepsilon>0\;:\; V_{(\phi ,2,\alpha )}^R\left( \frac{f-f(a)}{\varepsilon }\right) \le 1\}, \text {by Definition}~2.1\\= & {} |f(a)|+|f_\alpha '(a)|+\inf \{\varepsilon >0\;:\; V_{(\phi ,2,\alpha )}^R\left( f/\varepsilon \right) \le 1\}, \text {by Lemma}~2.11.\\ \end{aligned}$$

3 $RV_{(\phi ,2,\alpha )}([a,b])$ as a Banach algebra

In this section we will show that $RV_{(\phi ,2,\alpha )}([a,b])$ is closed under the product of functions. To attain such a goal, we will use a criterion given in 1987 by L. Maligranda and W. Orlicz [17], which supplies a test to check if some function space is a Banach algebra, namely.

Lemma 3.1

(Maligranda Orlicz criterion) Let $(X,\Vert \cdot \Vert )$ be a Banach space whose elements are bounded functions and the space is closed under multiplication of functions. Let us asssume that

$$\begin{aligned} f\cdot g\in X\text { and } \Vert fg\Vert \le \Vert f\Vert _\infty \cdot \Vert g\Vert +\Vert f\Vert \cdot \Vert g\Vert _\infty \end{aligned}$$

for any $f,g\in X$. Then the space X equipped with the norm

$$\begin{aligned} \Vert f\Vert _1=\Vert f\Vert _\infty +\Vert f\Vert \end{aligned}$$

is a normed Banach algebra. Also if $X\hookrightarrow B([a,b])$, then the norms $\Vert \cdot \Vert _1$ and $\Vert \cdot \Vert$ are equivalent. Moreover, if $\Vert f\Vert _\infty \le M\Vert f\Vert$ for $f\in X$, then $(X,\Vert \cdot \Vert _2)$ is a normed Banach algebra with $\Vert f\Vert _2=2M\Vert f\Vert$, $f\in X$ and the norms $\Vert \cdot \Vert _2$ and $\Vert \cdot \Vert$ are equivalent.

In [7] the first and third named authors generalized the Maligranda Orlicz Lemma, in the following way.

Theorem 3.2

(Generalized Maligranda-Orlicz’s Lemma) Let $(X,\Vert \cdot \Vert )$ be a Banach space whose elements are bounded functions, which is closed under pointwise multiplication of functions. Let us assume that $f\cdot g\in X$ such that

$$\begin{aligned} \Vert fg\Vert \le \Vert g\Vert _\infty \cdot \Vert g\Vert +\Vert f\Vert \cdot \Vert g\Vert _\infty +K\Vert f\Vert \cdot \Vert g\Vert ,\quad K>0. \end{aligned}$$

Then $(X,\Vert \cdot \Vert )$ equipped with the norm

$$\begin{aligned} \Vert f\Vert _1=\Vert f\Vert _\infty +K\Vert f\Vert ,\quad f\in X \end{aligned}$$

is a Banach algebra, if $X\hookrightarrow B([a,b])$, then $\Vert \cdot \Vert _1$ and $\Vert \cdot \Vert$ are equivalent.

Theorem 3.3

Let $\phi$ be a convex $\phi$-convex function which satisfy the $(\infty _1)$-condition. Let $f,g\in RV_{(\phi ,2,\alpha )}([a,b])$, then $f\cdot g\in RV_{(\phi ,2,\alpha )}([a,b])$.

Proof

Let $f,g\in RV_{(\phi ,2,\alpha )}([a,b])$, then by Corollary 2.10, we have $f_\alpha ',g_\alpha '\in RV_{(\phi ,\alpha )}([a,b])$ since $RV_{(\phi ,2,\alpha )}([a,b])\subset RV_{\phi ,\alpha }([a,b])$ by Theorem 2.2. Also, $f,g\in RV_{(\phi ,\alpha )}([a,b])$ since $RV_{(\phi ,\alpha )}$ is an algebra, (see [4]), we obtain that

$$\begin{aligned} f_\alpha 'g+g'_\alpha f=(fg)_\alpha '\in RV_{(\phi ,\alpha )}([a,b]). \end{aligned}$$

One more time from Theorem 2.2 we conclude that $f\cdot g\in RV_{(\phi ,2,\alpha )}([a,b])$. $\square$

Lemma 3.4

Let $\phi$ be a convex $\phi$-function which satisfy the $(\infty _1)$-condition. Let $f,g\in RV_{(\phi ,2,\alpha )}^0([a,b])$, then there exists $K>0$ such that

$$\begin{aligned} |f\cdot g|_{(\phi ,2,\alpha )}^R\le \Vert f\Vert _\infty \cdot |g|_{(\phi ,2,\alpha )}^R+\Vert g\Vert _\infty \cdot |f|_{(\phi ,2,\alpha )}^R+K|f|_{(\phi ,2,\alpha )}^R\cdot |g|_{(\phi ,2,\alpha )}^R. \end{aligned}$$

Proof

Let $f,g\in RV_{(\phi ,2,\alpha )}^0([a,b])$, then by Corollary 2.9 we have

$$\begin{aligned} |f\cdot g|_{(\phi ,2,\alpha )}^R=&\Vert (f g)_\alpha '\Vert _{(\phi ,\alpha )}^R\\ \le&\Vert f g_\alpha '\Vert _{(\phi ,\alpha )}^R+\Vert f_\alpha 'g\Vert _{(\phi ,\alpha )}^R\\ \le&|f\Vert _\infty \cdot \Vert g'_\alpha \Vert _{(\phi ,\alpha )}^R+\Vert g'_\alpha \Vert _\infty \cdot \Vert f\Vert _{(\phi ,\alpha )}^R+\Vert g\Vert _\infty \Vert f_\alpha '\Vert _{(\phi ,\alpha )}^R\\&+\Vert f'\alpha \Vert _\infty \cdot \Vert g\Vert _{(\phi ,\alpha )}^R\\ =&\Vert f\Vert _\infty \cdot |g|_{(\phi ,2,\alpha )}^R+\Vert g\Vert _\infty \cdot |f|_{(\phi ,2,\alpha )}^R+\Vert g'_\alpha \Vert _\infty \Vert f\Vert _{(\phi ,\alpha )}^R\\&+\Vert f'_\alpha \Vert _\infty \cdot \Vert g\Vert _{(\phi ,\alpha )}^R. \end{aligned}$$

$f(a)=0$ implies $\Vert f\Vert _{(\phi ,\alpha )}^R=|f|_{(\phi ,\alpha )}^R$. Then

$$\begin{aligned} |f\cdot g|_{(\phi ,2,\alpha )}^R\le & {} \Vert f\Vert _\infty \cdot |g|_{(\phi ,2,\alpha )}^R+\Vert g\Vert _\infty \cdot |f|_{(\phi ,2,\alpha )}^R\\&\quad +\Vert g'_\alpha \Vert _\infty |f|_{(\phi ,\alpha )}^R+\Vert f'_\alpha \Vert _\infty \cdot \Vert g\Vert _{(\phi ,\alpha )}^R. \end{aligned}$$

Since $RV_{(\phi ,\alpha )}([a,b])\hookrightarrow B([a,b])$, there exists $M_2>0$ such that

$$\begin{aligned} \Vert f_\alpha '\Vert _\infty \le M_2\Vert f_\alpha '\Vert _{(\phi ,\alpha )}^R=M_2|f|_{(\phi ,2,\alpha )}^R,\text { since } f_\alpha '\in RV_{(\phi ,\alpha )}([a,b]). \end{aligned}$$

Hence,

$$\begin{aligned} |f\cdot g|_{(\phi ,2,\alpha )}^R\le&\Vert f\Vert _\infty |g|_{(\phi ,2,\alpha )}^R+\Vert g\Vert _\infty |f|_{(\phi ,2,\alpha )}^R+M_1M_2|g|_{(\phi ,2,\alpha )}^R|f|_{(\phi ,\alpha )}^R\\&+M_1M_2|f|_{(\phi ,2,\alpha )} |g|_{(\phi ,2,\alpha )}^R\\ =&\Vert f\Vert _\infty |g|_{(\phi ,2,\alpha )}^R+\Vert g\Vert _\infty |f|_{(\phi ,2,\alpha )}^R+K|g|_{(\phi ,2,\alpha )}^R|f|_{(\phi ,\alpha )}^R \end{aligned}$$

with $K=2M_1M_2$. $\square$

Lemma 3.5

Let $\phi$ be a convex $\phi$-function which satisfy the $(\infty _1)$-condition, let $f,g\in RV_{(\phi ,2,\alpha )}([a,b])$, then there exists $K>0$ such that

$$\begin{aligned} \Vert f g\Vert _{(\phi ,2,\alpha )}^R\le \Vert f\Vert _\infty \Vert g\Vert _{(\phi ,2,\alpha )}^R+\Vert g\Vert _\infty \Vert f\Vert _{(\phi ,2,\alpha )}^R+K\Vert f\Vert _{(\phi ,2,\alpha )}^R\Vert g\Vert _{(\phi ,2,\alpha )}^R. \end{aligned}$$

Proof

Let $f,g\in RV_{(\phi ,2,\alpha )}^R([a,b])$, then

$$\begin{aligned} \Vert f g\Vert _{(\phi ,2,\alpha )}^R= & {} |(fg)(a)|+|fg-(fg)(a)|_{(\phi ,2,\alpha )}^R \text {By Definition}~2.7 \\= & {} |(fg)(a)|+\Vert [fg-(fg)(a)]_\alpha '\Vert _{(\phi ,2,\alpha )} \\&\qquad \text { since }fg-(fg)(a)\in RV_{(\phi ,2,\alpha )}^0([a,b]) \text { and by Corollary}~2.9, \text { we have }\\= & {} |(fg)(a)|+\Vert (fg)_\alpha '\Vert _{(\phi ,\alpha )}^R\\\le & {} |f(a)|\cdot |g(a)|+\Vert fg_\alpha '\Vert _{(\phi ,\alpha )}^R+\Vert gf_\alpha '\Vert _{(\phi ,\alpha )}^R\\\le & {} 2|f(a)|\cdot |g(a)|+\Vert f\Vert _\infty \Vert g_\alpha '\Vert _{(\phi ,\alpha )}^R+\Vert g_\alpha '\Vert _\infty \Vert f\Vert _{(\phi ,\alpha )}^R+\Vert g\Vert _\infty \Vert f_\alpha '\Vert _{(\phi ,\alpha )}^R\\&\qquad +\Vert f_\alpha '\Vert _\infty \Vert g\Vert _{(\phi ,\alpha )}^R\\\le & {} \Vert f\Vert _\infty |g(a)|+|f(a)|\cdot \Vert g\Vert _\infty +\Vert f\Vert _\infty \Vert (g-g(a))'_\alpha \Vert _{(\phi ,\alpha )}^R\\&\qquad +\Vert g\Vert _\infty \Vert (f-f(a))_\alpha '\Vert _{(\phi ,\alpha )}^R+\Vert g_\alpha '\Vert _\infty \Vert f\Vert _{(\phi ,\alpha )}^R+\Vert f'_\alpha \Vert _\infty \Vert g\Vert _{(\phi ,\alpha )}. \end{aligned}$$

Since $RV_{(\phi ,2,\alpha )}([a,b]) \hookrightarrow RV_{(\phi ,\alpha )}([a,b])$ (Theorem 2.2) there exists $M_1>0$ such that $\Vert \cdot \Vert _{(\phi ,\alpha )}^R\le M_1\Vert \cdot \Vert _{(\phi ,2,\alpha )}^R$ and by Remark 2.1 we have

$$\begin{aligned} \Vert f g\Vert _{(\phi ,2,\alpha )}^R\le & {} \Vert f\Vert _\infty |g(a)|+\Vert g\Vert _\infty |f(a)|+\Vert f\Vert _\infty |g-g(a)|_{(\phi ,2,\alpha )}^R\\&\quad +\Vert g\Vert _\infty |f-f(a)|_{(\phi ,2,\alpha )}^R+M_1\Vert g_\alpha '\Vert _\infty \Vert f\Vert _{(\phi ,2,\alpha )}^R+M_1\Vert f_\alpha '\Vert _\infty \Vert g\Vert _{(\phi ,2,\alpha )}^R. \end{aligned}$$

Since $RV_{(\phi ,\alpha )}([a,b])\hookrightarrow B([a,b])$ (Theorem 2.2) there exists $M_2>0$ such that

$$\begin{aligned} \Vert f_\alpha '\Vert _\infty =\Vert (f-f(a))_\alpha '\Vert _\infty \le M_2\Vert (f-f(a))_\alpha '\Vert _{(\phi ,\alpha )}R=M_2|f-f(a)|_{(\phi ,2,\alpha )}^R \end{aligned}$$

(by Corollary 2.9). Since $f-f(a)\in RV_{(\phi ,\alpha )}([a,b])$ (by Corollary 2.10)

$$\begin{aligned} \Vert f_\alpha '\Vert _\infty \le&\Vert f\Vert _\infty \left( |g(a)|+|g-g(a)|_{(\phi ,2,\alpha )}^R\right) +\Vert g\Vert _\infty \left( |f(a)|+|f-f(a)|_{(\phi ,2,\alpha )}^R\right) \\&+M_1M_2|f-f(a)|_{(\phi ,2,\alpha )}^R\Vert g\Vert _{(\phi ,2,\alpha )}^R+M_1M_2|g-g(a)|_{(\phi ,2,\alpha )}^R\Vert f\Vert _{(\phi ,2,\alpha )}^R\\ \le&\Vert f\Vert _\infty \Vert g\Vert _{(\phi ,2,\alpha )}^R+\Vert g\Vert _\infty \Vert f\Vert _{(\phi ,2,\alpha )}^R\\&+M_1M_2\left( |f(a)|+|f-f(a)|_{(\phi ,2,\alpha )}^R\right) \Vert f\Vert _{(\phi ,2,\alpha )}\\ =&\Vert f\Vert _\infty \Vert g\Vert _{(\phi ,2,\alpha )}^R+\Vert g\Vert _\infty \Vert f\Vert _{(\phi ,2,\alpha )}+K\Vert f\Vert _{(\phi ,2,\alpha )}^R\Vert g\Vert _{(\phi ,2,\alpha )}^R \end{aligned}$$

with $K=2M_1M_2$. $\square$

Theorem 3.6

Let $\phi$ be a convex $\phi$-function which satisfy the $(\infty _1)$-condition. Then $RV_{(\phi ,2,\alpha )}([a,b])$ with the norm

$$\begin{aligned} \Vert f\Vert _{(\phi ,2,\alpha )}^1=\Vert f\Vert _\infty +K\Vert f\Vert _{(\phi ,2,\alpha )}^R,\quad f\in RV_{(\phi ,2,\alpha )}([a,b]) \end{aligned}$$

is a Banach algebra. The norms $\Vert \cdot \Vert _{(\phi ,2,\alpha )}^R$ and $\Vert \cdot \Vert _{(\phi ,2,\alpha )}^1$ are equivalents, that is there exists $\gamma ,\delta >0$ such that

$$\begin{aligned} \gamma \Vert \cdot \Vert _{(\phi ,2,\alpha )}^1\le \Vert \cdot \Vert _{(\phi ,2,\alpha )}^R\le \delta \Vert \cdot \Vert _{(\phi ,2,\alpha )}^1. \end{aligned}$$

Proof

We just need to check the hypotheses of Theorem 3.2. $\square$

4 Nemytskii operator on $RV_{(\phi ,2,\alpha )}([a,b])$

The superposition operator, or Nemytskii operator, defined by $F(u(s))=f(s,u(s))$, is the simplest among the nonlinear operators. It appeared for the first time in 1934 in the paper of V.V. Nemytskii [27], in connection with the study of solutions of some nonlinear integral equations. Due to its simplicity, this operator have largely studied, since it is very useful in diverse modeling applications in differential and integral equations, variational calculus, probability theory and statistics, optimization theory, among others. In the monograph [1] can be found the fundamental properties of this operators like boundedness, compactness etc, in the general setting of ideal spaces of measurable functions. In this section we study under what conditions the Nemytskii operator acts in the $(\phi ,s,\alpha )$-bounded variation space.

In his 1982 paper, J. Matkowski [18] has show that the operator F generated by $f:[a,b]\times {\mathbb {R}}\longrightarrow {\mathbb {R}}$ maps Lip([a, b]) into itself and it is globally Lipschitz, that is, there exists a positive constant K such that

$$\begin{aligned} \Vert F(u)-F(v)\Vert _{Lip([a,b])}\le K\Vert u-v\Vert _{Lip([a,b])} \end{aligned}$$

where $u,v\in Lip([a,b])$ if and only if there exist $g,h\in Lip([a,b])$ such that

$$\begin{aligned} f(t,x)=g(t)x+h(t) \quad \text {for } t\in [a,b],\; x\in {\mathbb {R}}. \end{aligned}$$

(4.1)

Remark 4.1

Note that there are function spaces where the Matkowski result does no remain valid. For example, on the spaces C([a, b]) and $L_p([a,b])$ with $p\ge 1$ take $g:{\mathbb {R}}\longrightarrow {\mathbb {R}}$ given by $g(x)=\sin (x)$ and define $f(t,x)=g(x)$, $t\in [a,b]$, $x\in {\mathbb {R}}$.

The function g is Lipschitz on ${\mathbb {R}}$, but does not satisfy the relation (4.1), however, the operator F generated by f maps each the above spaces into itself and

$$\begin{aligned} \Vert F(u)-F(v)\Vert =\Vert \sin (u(\cdot ))-\sin (v(\cdot ))\Vert _\infty \le K\Vert u-v\Vert _\infty \end{aligned}$$

with $u,v\in C([a,b])$, and

$$\begin{aligned}&\Vert F(u)-F(v)\Vert _{Lip([a,b])}\le \left( \int \limits _a^b |\sin (u(t))-\sin (v(t))|^p dt\right) ^{1/p}\\&\quad \le K\Vert u-v\Vert _{Lip([a,b])} \end{aligned}$$

with $u,v\in L_p([a,b])$, where K is Lipschitz result has been extended in the framework of various function spaces for single-valued as well as multivalued Lipschitzian Nemytskii operators c.f. [9,10,11, 17, 19,20,21,22,23,24,25, 29,30,31, 34]. In this section we extend the Matkowski result in the framework of the function space $RV_{(\phi ,2,\alpha )}([a,b])$.

Theorem 4.1

Let $\phi$ be a convex $\phi$-function which satisfies the $(\infty _1)$ condition. Let $f:[a,b]\times {\mathbb {R}}\longrightarrow {\mathbb {R}}$. Then Nemyskii operator associated to f defined by

$$\begin{aligned} F:RV_{(\phi ,2,\alpha )}([a,b])&\longrightarrow {\mathbb {R}}\\ u&\longmapsto F(u) \end{aligned}$$

with $F(u)=f(t,u(t))$, $t\in [a,b]$ act on $RV_{(\phi ,2,\alpha )}([a,b])$ and is globally Lipschitz, that is there exists $K>0$ such that

$$\begin{aligned} \Vert F(u_1)-F(u_2)\Vert _{(\phi ,2,\alpha )}^R\le K\Vert u_1-u_2\Vert _{(\phi ,2,\alpha )}^R,\quad t\in [a,b] \end{aligned}$$

if and only if there exist $g,h\in RV_{(\phi ,2,\alpha )}([a,b])$ such that

$$\begin{aligned} f(t,y)=g(t)y+h(t),\quad t\in [a,b],\; y\in {\mathbb {R}}. \end{aligned}$$

Proof

From Theorem 2.2$RV_{(\phi ,2,\alpha )}([a,b]) \hookrightarrow \alpha -Lip([a,b])$, then there exists $N>0$ such that

$$\begin{aligned} \Vert \cdot \Vert _{\alpha -Lip([a,b])} \le N\Vert \cdot \Vert _{(\phi ,2,\alpha )}^R. \end{aligned}$$

By hypothesis, there exists $K>0$ such that

$$\begin{aligned} \Vert Fu_1-Fu_2\Vert _{\alpha -Lip([a,b])}\le&N\Vert Fu_1-Fu_2\Vert _{(\phi ,2,\alpha )}^R\\ \le&KN\Vert u_1-u_2\Vert _{(\phi ,2,\alpha )}^R,\quad u_1,u_2\in RV_{(\phi ,2,\alpha )}([a,b]). \end{aligned}$$

Let us define two particular polynomials $u_1,u_2$ such a way that $u_1,u_2\in RV_{(\phi ,2,\alpha )}([a,b])$.

To define that fix $t,t'\in [a,b]$, $t<t'$, $y_1,y_2,y_1',y_2'\in {\mathbb {R}}$. Let us define $u_i:[a,b]\longrightarrow {\mathbb {R}}$, $i=1,2$ by

$$\begin{aligned} u_i(s)=&\frac{y_i'-y_i}{2(\alpha (t')-\alpha (t))}(\alpha (s)-\alpha (a))^2\\&+\frac{y_i'-y_i}{\alpha (t')-\alpha (t)}\left[ 1-\frac{(\alpha (t')-\alpha (t))^2-(\alpha (t)-\alpha (a))^2}{2(\alpha (t')-\alpha (t))}\right] (\alpha (s)-\alpha (a))\\&+y_i-\frac{y_i'-y_i}{2(\alpha (t')-\alpha (t))}(\alpha (t)-\alpha (a))^2-\frac{y_i'-y_i}{\alpha (t')-\alpha (t)}\\&\times \left[ 1-\frac{(\alpha (t')-\alpha (t))^2-(\alpha (t)-\alpha (a))^2}{2(\alpha (t')-\alpha (t))}\right] (\alpha (t)-\alpha (a)),\quad s\in [a,b]. \end{aligned}$$

The functions $u_1$ and $u_2$ satisfies the following conditions:

$$\begin{aligned} u_i(t)=&y_i,\qquad i=1,2\\ u_i(t')=&y_i',\qquad i=1,2. \end{aligned}$$

Moreover,

$$\begin{aligned} (u_i)'(s)= & {} \frac{y_i'-y_i}{\alpha (t')-\alpha (t)}(\alpha (s)-\alpha (a))\\&\quad +\frac{y_i'-y_i}{\alpha (t')-\alpha (t)}\left[ 1-\frac{(\alpha (t')-\alpha (t))^2-(\alpha (t)-\alpha (a))^2}{2(\alpha (t')-\alpha (t))}\right] ,\quad s\in [a,b],\; i=1,2 \end{aligned}$$

and

$$\begin{aligned} (u_i)_\alpha ''(s)=\frac{y_i'-y_i}{\alpha (t')-\alpha (t)},\quad s\in [a,b],\; i=1,2. \end{aligned}$$

Let us calculate $\Vert u_1-u_2\Vert _{(\phi ,2,\alpha )}^R$. We observe that $(u_i)'_\alpha$, $i=1,2$ is absolutely continuous with respect to $\alpha$ in [a, b], and

$$\begin{aligned} \int \limits _a^b\phi (|(u_i)''_\alpha (s)|)d\alpha (s)=&\int \limits _a^b\phi \left( \left| \frac{y_i'-y_i}{\alpha (t')-\alpha (t)}\right| \right) d\alpha (s)\\ =&\phi \left( \left| \frac{y_i'-y_i}{\alpha (t')-\alpha (t)}\right| \right) (\alpha (b)-\alpha (a))<+\infty ,\quad i=1,2. \end{aligned}$$

By Corollary 2.7 we conclude that $u_i\in RV_{(\phi ,2,\alpha )}([a,b])$, $i=1,2$.

To calculate $\Vert u_1-u_2\Vert _{(\phi ,2,\alpha )}^R$:

$$\begin{aligned} u_1(a)-u_2(a)=y_1-y_2-\frac{y_1'-y_1-y_2'+y_2}{\alpha (t')-\alpha (t)}(\alpha (t)-\alpha (a))\left( \frac{\alpha (t)-\alpha (t')}{2}+1\right) , \\ (u_1)_\alpha '-(u_2)_\alpha '=\frac{y_1'-y_1-y_2'+y_2}{\alpha (t')-\alpha (t)}[2-\alpha (t')-\alpha (t)+2\alpha (a)]. \end{aligned}$$

One more time by Corollary 2.7

$$\begin{aligned} V_{(\phi ,2,\alpha )}^R\left( \frac{u_1-u_2}{\varepsilon }\right) =&\int \limits _a^b\phi \left( \left| \frac{(u_1-u_2)''_\alpha (s)}{\varepsilon }\right| \right) d\alpha (s)\\ =&\int \limits _a^b\phi \left( \left| \frac{y_1'-y_1-y_2'+y_2}{\varepsilon (\alpha (t')-\alpha (t))}\right| \right) d\alpha (s)\\ =&\phi \left( \left| \frac{y_1'-y_1-y_2'+y_2}{\varepsilon (\alpha (t')-\alpha (t))}\right| \right) (\alpha (b)-\alpha (a)). \end{aligned}$$

Then

$$\begin{aligned}&V_{(\phi ,2,\alpha )}^R\left( \frac{u_1-u_2}{\varepsilon }\right) \le 1\\&\quad \Longleftrightarrow \; \phi \left( \left| \frac{y_1'-y_1-y_2'+y_2}{\varepsilon (\alpha (t')-\alpha (t))}\right| \right) (\alpha (b)-\alpha (a))\le 1\\&\quad \Longleftrightarrow \; \left| \frac{y_1'-y_1-y_2'+y_2}{\varepsilon (\alpha (t')-\alpha (t))}\right| \le \phi ^{-1}\left( \frac{1}{\alpha (b)-\alpha (a)}\right) \\&\quad \Longleftrightarrow \; \frac{|y_1'-y_1-y_2'+y_2|}{\phi ^{-1}\left( \frac{1}{\alpha (b)-\alpha (a)}\right) |\alpha (t')-\alpha (t)|}\le \varepsilon \end{aligned}$$

and then

$$\begin{aligned} \inf \left\{ \varepsilon >0\;:\; V_{(\phi ,2,\alpha )}^R\left( \frac{u_1-u_2}{\varepsilon }\right) \le 1\right\} = \frac{|y_1'-y_1-y_2'+y_2|}{\phi ^{-1}\left( \frac{1}{\alpha (b)-\alpha (a)}\right) |\alpha (t')-\alpha (t)|}. \end{aligned}$$

Since $Fu_1$ and $Fu_2$ are in $RV_{(\phi ,2,\alpha )}([a,b]) \hookrightarrow \alpha -Lip([a,b])$ also $Fu_1-Fu_2\in \alpha -Lip([a,b])$ with $Fu_i:[a,b]\longrightarrow {\mathbb {R}}$ given by

$$\begin{aligned} (Fu_i)(s)=f(s,u_i(s)),\quad i=1,2. \end{aligned}$$

In particular,

$$\begin{aligned} (Fu_i)(t)=&f(t,u_i(t))=f(s,u_i(s))\quad i=1,2\\ (Fu_i)(t')=&f(t',u_i(t'))=f(t',u_i(t'))\quad i=1,2. \end{aligned}$$

Then

$$\begin{aligned} \frac{|(Fu_1-Fu_2)(t')-(Fu_1-Fu_2)(t)|}{|\alpha (t')-\alpha (t)|}\le&\Vert Fu_1-Fu_2\Vert _{\alpha -Lip([a,b])}\\ \le&KN\Vert u_1-u_2\Vert _{(\phi ,2,\alpha )}^R. \end{aligned}$$

Replacing

$$\begin{aligned}&\frac{|f(t',y_1')-f(t',y_2')-(f(t,y_1)-f(t,y_2))|}{|\alpha (t')-\alpha (t)|}\\&\quad \le KN\left\{ \left| y_1-y_2-\frac{y_1'-y_1-y_2'+y_2}{\alpha (t')-\alpha (t)}(\alpha (t)-\alpha (a))\left( \frac{\alpha (t)-\alpha (t')}{2}+1\right) \right| \right. \\&\qquad +\left| \frac{y_1'-y_2'-(y_1-y_2)}{2(\alpha (t')-\alpha (t))}[2-\alpha (t')-\alpha (t)+2\alpha (a)]\right| \\&\qquad \left. +\frac{|y_1'-y_1-y_2'+y_2|}{\phi ^{-1}\left( \frac{1}{\alpha (b)-\alpha (a)}\right) |\alpha (t')-\alpha (t)|}\right\} . \end{aligned}$$

Multiplying the inequality by $|\alpha (t')-\alpha (t)|$ and applying the triangular inequality, we obtain

$$\begin{aligned}&|f(t',y_1')-f(t',y_2')-(f(t,y_1)-f(t,y_2))| \\&\quad \le KN\left\{ |y_1-y_2||\alpha (t')-\alpha (t)|+|y_1'-y_2'-(y_1-y_2)||\alpha (t)\right. \\&\qquad \left. -\alpha (a)|\left| \frac{\alpha (t)-\alpha (t')}{2}+1\right| \right. \\&\qquad \left. +|y_1'-y_2'-(y_1-y_2)|\left| 1-\frac{\alpha (t')-\alpha (t)}{2}+\alpha (a)\right| +\frac{|y_1'-y_1-y_2'+y_2|}{\phi ^{-1}\left( \frac{1}{|\alpha (b)-\alpha (a)|}\right) }\right\} . \end{aligned}$$

For $y\in {\mathbb {R}}$ the constant function $u_0(t)=y$, $t\in [a,b]$ belong to $RV_{(\phi ,2,\alpha )}([a,b])$ by hypothesis the function $(Fu_0)(t)=f(t,u_0(t))=f(t,y)$ belong to $RV_{(\phi ,2,\alpha )}([a,b])$ and therefore the function $f(\cdot ,y)$ is continuous in [a, b]. Since $\alpha$ is continuous $\alpha (t')\rightarrow \alpha (t)$ whenever $t'\rightarrow t$, then

$$\begin{aligned}&|f(t',y_1')-f(t',y_2')-(f(t,y_1)-f(t,y_2))| \\&\quad \le KN|y_1'-y_2'\\&\qquad -(y_1-y_2)|\left\{ |\alpha (t)-\alpha (a)|+|\alpha (t)-\alpha (a)-1|+\frac{1}{\phi ^{-1}\left( \frac{1}{|\alpha (b)-\alpha (a)|}\right) }\right\} \\&\quad \le KN|y_1'-y_2'\\&\qquad -(y_1-y_2)|\left\{ |\alpha (b)-\alpha (a)|+|\alpha (b)-\alpha (a)-1|+\frac{1}{\phi ^{-1}\left( \frac{1}{|\alpha (b)-\alpha (a)|}\right) }\right\} . \end{aligned}$$

Arguing as in Theorem 3.1 in [5] we get the result.

Reciprocally, let $g,h\in RV_{(\phi ,2,\alpha )}([a,b])$ such that $f(t,y)=g(t)y+h(t)$. The Nemytskii operator generated by f is given by

$$\begin{aligned} (Fu)(t)=f(t,u(t))=g(t)u(t)+h(t),\quad t\in [a,b]. \end{aligned}$$

Since $RV_{(\phi ,2,\alpha )}([a,b])$ is an algebra (Theorem 3.3) we conclude the F acts in the space $RV_{(\phi ,2,\alpha )}([a,b])$. We will show that F satisfies a globally Lipschitz condition, let $u_1,u_2\in RV_{(\phi ,2,\alpha )}([a,b])$ and so

$$\begin{aligned} \Vert Fu_1-Fu_2\Vert _{(\phi ,2,\alpha )}^R=&\Vert f(\cdot ,u_1(\cdot ))-f(\cdot ,u_2(\cdot ))\Vert _{(\phi ,2,\alpha )}^R\\ =&\Vert g(\cdot )u_1(\cdot )-h(\cdot )-(g(\cdot )u_2(\cdot )-h(\cdot ))\Vert _{(\phi ,2,\alpha )}^R\\ =&\Vert g(\cdot )[u_1(\cdot )-u_2(\cdot )]\Vert _{(\phi ,2,\alpha )}^R=\Vert g(u_1-u_2)\Vert _{(\phi ,2,\alpha )}^R. \end{aligned}$$

By Theorem 3.6 the norms $\Vert \cdot \Vert _{(\phi ,2,\alpha )}^R$ and $\Vert \cdot \Vert _{(\phi ,2,\alpha )}^1$ are equivalents, thus

$$\begin{aligned} \Vert g(u_1-u_2)\Vert _{(\phi ,2,\alpha )}^R\le&\delta \Vert g(u_1-u_2)\Vert _{(\phi ,2,\alpha )}^1. \end{aligned}$$

Since $( RV_{(\phi ,2,\alpha )}([a,b]), \Vert \cdot \Vert _{(\phi ,2,\alpha )}^1)$ is a Banach algebra, we have

$$\begin{aligned} \delta \Vert g(u_1-u_2)\Vert _{(\phi ,2,\alpha )}^1\le&\delta \Vert g\Vert _{(\phi ,2,\alpha )}^1\Vert u_1-u_2\Vert _{(\phi ,2,\alpha )}^1\\ \le&\delta \left( \frac{1}{\gamma }\right) \Vert g\Vert _{(\phi ,2,\alpha )}^R\frac{1}{\gamma }\Vert u_1-u_2\Vert _{(\phi ,2,\alpha )}^R\\ =&\frac{\delta }{\gamma ^2}\Vert g\Vert _{(\phi ,2,\alpha )}^R\Vert u_1-u_2\Vert _{(\phi ,2,\alpha )}^R. \end{aligned}$$

Considering $\frac{\delta }{\gamma ^2}\Vert g\Vert _{(\phi ,2,\alpha )}^R$ as Lipschitz constant, it is concluded.$\square$