Sandwich results for periodicity and conjugacy

Abstract

Let P be a nonconstant selfmap of a set $\mathcal {M}$. A sandwich-type theorem for generalized sub-P-periodic functions defined on $\mathcal {M}$ with values in a reflexive Banach space is proved. In particular, given functions $f,g:\mathcal {M}\rightarrow \mathbb {R}$, we obtain necessary and sufficient conditions for the existence of a generalized P-periodic function $F:\mathcal {M}\rightarrow \mathbb {R}$ such that $f\le F\le g$. The formula for F is given and its Lipschitz constant is discussed. Moreover the solvability of the functional equation $f\circ p= r\circ f$ with the help of a new sandwich method, is considered.

1 Introduction

In a recent paper Izumi et al. [5] proved the following sandwich (separation) type result: functions$f,g:\mathbb {R}\rightarrow \mathbb {R}$satisfy the inequalities

$$\begin{aligned} \ f(x+np)\le g(x), \quad n\in \mathbb {Z}, x\in \mathbb {R} , \end{aligned}$$

(1)

for some $p\in \mathbb {R}{\setminus } \{0\}$if, and only if, there is a p-periodic function $F:\mathbb {R} \rightarrow \mathbb {R}$ separating f and g,  that is such that $F\left( x+p\right) =F\left( x\right)$ for all $x\in \mathbb {R}$ and $f\le F\le g$.

In Sect. 2 of this paper, applying Gajda’s theorem on the existence of a translative invariant mean in the Banach space of bounded vector valued functions [4], we prove far reaching generalizations of this result. Let $\mathcal {M}$ be a nonempty set, $P:\mathcal { M}\rightarrow \mathcal {M}$ a nonconstant mapping, and $\mathfrak {X}$ a reflexive Banach space. Assuming that a function $f:\mathcal {M}\rightarrow \mathfrak {X}$ and a set-valued function $\Gamma :\mathcal {M}\rightarrow 2^{ \mathfrak {X}}$ with bounded closed and convex values are such that

$$\begin{aligned} f(P^{n}(x))\in \Gamma (x),\quad n\in \mathbb {N}_{0}, x\in \mathbb {R} \end{aligned}$$

(where $\mathbb {N}_{0}:=\left\{ 0,1,\ldots ,\right\}$), we prove that there is a P-periodic selection F of $\Gamma$ (i.e. such that $F\circ P=F)$ (Theorem 2). Assume that $h:\mathcal {M}\rightarrow \mathbb {R}$ is P-super-periodic (i.e. $h\circ P\ge h)$ and $g:\mathcal { M}\rightarrow \mathbb {R}$ is P-sub-periodic (i.e. $h\circ P\le h)$. Applying Theorem 2 with $\mathfrak {X}\mathbb {=R}$ we prove the inequality $h\le g$ holds if and only if there is a P-periodic function separating h and g (Theorem 3). A strengthening of Theorem 3 is given. At the end of this section, assuming that $\mathcal {M}$ is a metric space, we give conditions under which the Lipschitz continuity of a $\left( P\text {-}\Gamma \right)$-periodic function f (see Sect. 2), implies the existence of the Lipschitz continuous P-periodic function F.

In the context of the periodic equation $f\circ P=f,$ with the notation $p=P$, it becomes quite natural to consider a more general functional equation $f\circ p=g\circ f$ linked both to conjugacy and commutativity. A description of solutions of this equations, as well as some related topics, can be found in the papers [2, 3, 7, 8, 10]. They focus on the existence and uniqueness of continuous and homeomorphic solutions in the case when p and g are given self-maps of a real interval or the unit circle $S_{1}$. In [9], Zdun investigated bounded and continuous solutions $f:\left[ 0,1 \right] \rightarrow Y$, where Y is a complete metric space, of the system of equations $f\circ p_{k}=g_{k}\circ f$, $k=0,1,\ldots ,n-1$. Ciepliński and Leśniak dealt with these problems in [1]. Moreover, Kopeček [6] investigated the equation $f\circ p=g\circ f$ for given functions p and g defined on subsets of all real numbers. He obtained a nice characterization of solvability of the above equation in the case when p, g are strictly increasing and continuous. Motivated by results of Kopeček, we investigate similar topics in Sect. 3. Namely, we prove sandwich type results for the functional equation $f\left( p\left( x\right) \right) = r\left( f\left( x\right) \right)$. We conclude this paper with results concerning the functional equation

$$\begin{aligned} \alpha f\left( p\left( x\right) \right) =g\left( f\left( x\right) \right) , \quad x\in \mathcal {M}, \end{aligned}$$

and the functional inequality

$$\begin{aligned} \alpha f\left( p^{n}\left( x\right) \right) \ge g^{n}\left( f\left( x\right) \right) , \quad x\in \mathcal {M}, \end{aligned}$$

where the constant $\alpha$ and the functions $p:\mathcal {M}\rightarrow \mathcal {M}$ and $g:\mathfrak {X\rightarrow X}$ are given.

2 Generalized periodicity

Let S be a semigroup and $\mathfrak {X}$ be a Banach space. By $\mathcal {B}(S,\mathfrak {X})$ we denote the space of bounded $\mathfrak {X}$-valued functions with the sup-norm $\Vert \cdot \Vert _{\infty }$. In our consideration the following result due to Gajda [4, Th.2.1, p. 20] is important.

Theorem 1

[4] Let $(S,+)$ be a commutative semigroup. Suppose that $\mathfrak {X}$ is reflexive. There is a continuous linear operator $M:\mathcal {B}(S,\mathfrak {X})\rightarrow \mathfrak {X}$ such that

• (i) $M(\varphi _a)=M(\varphi )$ for all $a\in S$ and $\varphi \in \mathcal {B}(S,\mathfrak {X})$, where $\varphi _a$ denotes the translation of $\varphi$ defined by $\varphi _a(x):=\varphi (x+a)$, $x\in S$;

• (ii) for every $\varphi \in \mathcal {B}(S,\mathfrak {X})$ and closed convex bounded subset V of $\mathfrak {X}$, if $\varphi (S)\subseteq V$ , then $M(\varphi )\in V$.

The linear operator M is called an invariant mean.

In the sequel $\mathcal {M}$ is a nonempty set, $P:\mathcal {M} \rightarrow \mathcal {M}$ is a nonconstant function, $\mathfrak {X}$ is a reflexive Banach space, and $\Gamma :\mathcal {M}\rightarrow 2^{ \mathfrak {X}}$ a set-valued function such that, for every x, the value $\Gamma (x)\varsubsetneq \mathfrak {X}$ is nonempty and convex.

We say that the function $F:\mathcal {M}\rightarrow \mathfrak {X}$ is P-periodic [resp. P-$\Gamma$-periodic] if

$$\begin{aligned} \forall _{x\in \mathcal {M}}\ \ F(P(x))=F(x)\quad \left[ \mathrm {resp.}\ F(P(x))\in \Gamma (x)\right] . \end{aligned}$$

(2)

To summarize, the paper [5] and conditions (1), (2) motivate this paper.

Put $\mathbb {N}_0:=\mathbb {N}\cup \{0\}$. Now we prove the first main result of this paper.

Theorem 2

Let $\mathcal {M}$ be a nonempty set and let $\mathfrak {X}$ be a reflexive Banach space. Assume that $P:\mathcal {M} \rightarrow \mathcal {M}$ is not a constant mapping. Suppose that $\Gamma :\mathcal {M}\rightarrow 2^{\mathfrak {X}}$ is a set-valued function such that for all $x\in \mathcal {M}$ the set $\Gamma (x)$ is nonempty, bounded, closed and convex. Suppose that a function $f:\mathcal {M} \rightarrow \mathfrak {X}$ satisfies the condition

$$\begin{aligned} \ f(P^{n}(x))\in \Gamma (x), \quad x\in \mathcal {M},\ n\in \mathbb {N}_{0}. \end{aligned}$$

(3)

Then there is a P-periodic function $F:\mathcal {M}\rightarrow \mathfrak {X}$ such that

$$\begin{aligned} F(x)\in \bigcap \limits _{m\in \mathbb {N}}\overline{\mathrm {conv}} \{f(P^{n}(x)):n>m\}, \quad x\in \mathcal {M}; \end{aligned}$$

(4)

in particular, $F(x)\in \Gamma (x)$ for all $x\in \mathcal {M}$.

Proof

Fix $a\in \mathfrak {X}$, $\varepsilon \in (0,1)$, $x\in \mathcal {M}$. By the definition of $\mathrm{dist}(a,A)$, there is a $d(x)\in \mathfrak {X}$ such that

$$\begin{aligned}&d(x)\in \overline{\mathrm {conv}}\{f(P^{n}(x)):n\in \mathbb {N}_{0}\}\ \ \mathrm {and} \\&\Vert d(x)-a\Vert -\varepsilon < \mathrm {dist}\left( a,\mathrm {conv} \{f(P^{n}(x)):n\in \mathbb {N}_{0}\}\right) . \end{aligned}$$

Given $y\in \mathcal {M}$ we define a function $\varphi ^{y}:\mathbb {N}_{0}\rightarrow \mathfrak {X}$ by $\varphi ^{y}(k):=d\left( P^{k}(y)\right)$. Put $\varphi _{1}^{y}(k):=\varphi ^{y}(k+1)$; see (i). It follows that $\varphi ^{P(y)}(k)=d(P^{k}(P(y)))=d(P^{k+1}(y))=\varphi ^{y}(k+1)=\varphi _{1}^{y}(k)$, which means

$$\begin{aligned} \varphi ^{P(y)}=\varphi _{1}^{y}\quad \hbox {for all}\ y\in \mathcal {M}. \end{aligned}$$

(5)

By Theorem 1 (with $S=\mathbb {N}_{0}$) there is an invariant mean $M:\mathcal {B}(\mathbb {N}_{0},\mathfrak {X} )\rightarrow \mathfrak {X}$. Now we define $F:\mathcal {M} \rightarrow \mathfrak {X}$ by

$$\begin{aligned} F(x):=M\!\left( \varphi ^{x}\right) \quad \mathrm {for}\ \ x\in \mathcal {M}. \end{aligned}$$

(6)

Applying in turn (5) and (i), we have $F(P(x))=M\!\left( \varphi ^{P(x)}\right) =M\!\left( \varphi _{1}^{x}\right) =M\!\left( \varphi ^{x}\right) =F(x)$, so the function F is P-periodic. It is easy to see that

$$\begin{aligned} d(P^k(x))\in \mathrm {conv}\{f(P^{n+k}(x)):n\in \mathbb {N}_0\}\subseteq \mathrm {conv}\{f(P^n(x)):n\in \mathbb {N}_0\}. \end{aligned}$$

Thus we get $d(P^k(x))\in \overline{\mathrm {conv}}\{f(P^n(x)) :n\in \mathbb {N}_0\}\subseteq \Gamma (x)$ for all k in $\mathbb {N}_0$. Therefore, $\varphi ^x(\mathbb {N}_0)\subseteq \overline{\mathrm {conv}}\{f(P^n(x)):n\in \mathbb {N}_0\}\subseteq \Gamma (x)$, and so by (ii) we obtain $F(x)=M(\varphi ^x)\in \overline{\mathrm {conv}}\{f(P^n(x)):n\in \mathbb {N}_0\}$.

Putting $P^{k}(x)$ in place of x in the above equality we get $F(P^{k}(x))\in \overline{\mathrm {conv}}\{f(P^{n}(P^{k}(x))):n\in \mathbb {N}_{0}\}$. On the other hand, we also have $F(P^{k}(x))=F(x)$, hence $F(x)\in \overline{\mathrm {conv}}\{f(P^{n+k}(x))):n\in \mathbb {N} _{0}\}$. Since $k\in \mathbb {N}_{0}$ is arbitrarily chosen, we conclude that $F(x)\in \bigcap _{m\in \mathbb {N}}\overline{\mathrm {conv}} \{f(P^{n}(x))):n>m\}$. $\square$

Theorem 3

Let $\mathcal {M}$ be a nonempty set and assume that $P:\mathcal {M}\rightarrow \mathcal {M}$ is not a constant mapping. Suppose that $h:\mathcal {M}\rightarrow \mathbb {R}$ is a P-super-periodic function, i.e. $h(P(\cdot ))\ge h(\cdot )$, and $g:\mathcal {M}\rightarrow \mathbb {R}$ is a P-sub-periodic, i.e. $g(P(\cdot ))\le g(\cdot )$. Then the functions $h,g:\mathcal {M}\rightarrow \mathbb {R}$ satisfy the condition

$$\begin{aligned} \forall _{x\in \mathcal {M}} \ \ h(x)\le g(x), \end{aligned}$$

(7)

if and only if there is a P-periodic function $F:\mathcal {M}\rightarrow \mathbb {R}$ such that $h\le F\le g$.

Proof

Assume (7). Define $f:\mathcal {M} \rightarrow \mathbb {R}$ by $f(x):=\frac{1}{2}h(x)+\frac{1}{2}g(x)$. By the assumptions we have $h(P^n(x))\ge h(x)$ and $g(P^n(x))\le g(x)$. It follows that $h(x)\le f(P^n(x)) \le g(x)$ for all x in M. For the set-valued function $\Gamma :\mathcal {M}\rightarrow 2^{ \mathbb {R}}$ given by $\Gamma (x):=[h(x),g(x)]$ we have $f(P^n(x))\in \Gamma (x)$. Now applying Theorem 2 we arrive at the desired assertion. The converse implication has a trivial verification.$\square$

The above result can be strengthened (in some circumstances) as follows.

Theorem 4

Let $\mathcal {M}$ be a nonempty set assume that $P:\mathcal {M}\rightarrow \mathcal {M}$ is not a constant mapping. Suppose that $h:\mathcal {M}\rightarrow \mathbb {R}$ is a P-super-periodic function. Then the functions $h,g:\mathcal {M}\rightarrow \mathbb {R}$ satisfy the condition

$$\begin{aligned} \forall _{x\in \mathcal {M}}\ \forall _{n\in \mathbb {N}_0} \ \ h(P^n(x))\le g(x), \end{aligned}$$

(8)

if and only if there is a P-periodic function $F:M\rightarrow \mathbb {R}$ such that $h\le F\le g$.

Proof

Assume (8). It is clear that $h(x)\le h(P^{n}(x))\le g(x)$. Define $\widehat{g}:\mathcal {M}\rightarrow \mathbb {R}$ by $\widehat{g}(x):=\sup \{h(P^{n}(x)):n\in \mathbb {N} _{0}\}$. It is easy to check that $h(x)\le \widehat{g}(x)\le g(x)$ and $\{h(P^{n+1}(x)):n\in \mathbb {N}_{0}\}\subseteq \{h(P^{n}(x)):n\in \mathbb {N}_{0}\}$. Therefore we obtain $\widehat{g}(P(x))=\sup \{h(P^{n+1}(x)):n\in \mathbb {N}_{0}\}\le \sup \{h(P^{n}(x)):n\in \mathbb {N}_{0}\}=\widehat{g}(x)$, which means that $\widehat{g}$ is P -sub-periodic. Moreover $h(x)\le \widehat{g}(x)$. Applying Theorem 3, we complete the proof. The converse result is immediate.$\square$

The next result shows that the Lipschitz property of a P-$\Gamma$-periodic function can be inherited by some P-periodic function.

Theorem 5

Let $(\mathcal {M},\varrho )$ be a metric space and $\mathfrak {X}$ a reflexive Banach space. Suppose that $P:\!\mathcal {M}\rightarrow \mathcal {M}$ is a uniformly Lipschitz function, i.e. there is $\gamma >0$ such that

$$\begin{aligned} \varrho \left( P^{n}(x),P^{n}(y)\right) \leqslant \gamma \cdot \varrho (x,y) \quad \mathrm{for }~ x,y\in \mathcal {M}; \end{aligned}$$

$\Gamma :\mathcal {M} \rightarrow 2^{\mathfrak {X}}$ is a set-valued function such that for all $x\in \mathcal {M}$ the set $\Gamma (x)$ is bounded, closed, convex; and $f:\mathcal {M}\rightarrow \mathfrak {X}$ is a P-$\Gamma$-periodic function. If f is a Lipschitz function with Lipschitz constant $\beta$, then there is a P-periodic function $F:\mathcal {M}\rightarrow \mathfrak {X}$ such that F is a function with Lipschitz constant $\beta \gamma$.

Proof

Given $y\in \mathcal {M}$ we define a function $\varphi ^y:\mathbb {N}_0\rightarrow \mathfrak {X}$ by $\varphi ^y(k):=f \left( P^k(y)\right)$. Put $\varphi _{1}^{y}(k):=\varphi ^{y}(k+1)$. It follows that $\varphi ^{P(y)}(k)=f(P^k(P(y)))= f(P^{k+1}(y))=\varphi ^y(k+1)=\varphi ^y_1(k)$, which means

$$\begin{aligned} \varphi ^{P(y)}=\varphi ^y_1 \quad \hbox {for all}\ y\in \mathcal {M}. \end{aligned}$$

(9)

Now Theorem 1 implies that there is an invariant mean $M:\mathcal {B}(\mathbb {N}_0, \mathfrak {X})\rightarrow \mathfrak {X}$. Define a function $F:\mathcal {M}\rightarrow \mathfrak {X}$ by the formula $F(x):=M\!\left( \varphi ^x\right)$ for $x\in \mathcal {M}$. Hence, by (9) and (i) we get $F(P(x))=M\!\left( \varphi ^{P(x)}\right) =M\! \left( \varphi ^{x}_1\right) =M\!\left( \varphi ^{x}\right) =F(x)$, so F is P-periodic.

Finally, we show that f is a Lipschitz function. Recall that the function M is a continuous operator with $\Vert M\Vert =1$. From the definition of F, we have

$$\begin{aligned} \Vert F(x)-F(y)\Vert _{\mathfrak {X}}&=\left\| M\!\left( \varphi ^x\right) -M\!\left( \varphi ^y\right) \right\| _{\mathfrak {X}}\leqslant \Vert M\Vert \cdot \Vert \varphi ^x- \varphi ^y\Vert _\infty \\&=\Vert M\Vert \cdot \sup \left\{ \left\| \varphi ^x(k)-\varphi ^y(k)\right\| _{ \mathfrak {X}}:k\in \mathbb {N}_0 \right\} \\&=\sup \left\{ \left\| f(P^k(x))-f(P^k(y))\right\| _{\mathfrak {X}}:k\in \mathbb {N}_0 \right\} \\&\leqslant \sup \left\{ \beta \varrho \left( P^k(x),P^k(y)\right) :k\in \mathbb {N }_0 \right\} \le \beta \gamma \varrho (x,y). \end{aligned}$$

The Proof of Theorem 5 is complete. $\square$

3 Sandwich with functional equation $f\circ p=r\circ f$

In the previous section we saw that the problem of determining periodic (resp. p-super-periodic) functions leads to the functional equation $f(p(x))=f(x)$ (resp. the functional inequality $f(p(x))\ge f(x)$). So, it becomes quite natural to consider the functional equation $f(p(x))=r(f(x))$ (resp. the functional inequality $f(p(x))\ge r(f(x))$).

The results presented in this section are of the type as in the previous section, but focused on the functional equation $f(p(x))=r(f(x))$ instead of periodic functions. First we prove a sandwich type result for the functional equation $f(p(x))=g(f(x))$.

Theorem 6

Let $I\subseteq \mathbb {R}$ be a compact interval. Assume that functions $p:I\rightarrow I$ and $r:I\rightarrow I$ satisfy the following conditions: p is invertible and

$$\begin{aligned} \left| p^{-1}\left( x\right) -p^{-1}\left( y\right) \right| \le \alpha \left( \left| x-y\right| \right) , \quad x,y\in I, \end{aligned}$$

(10)

where $\alpha :\left[ 0,\infty \right) \rightarrow \left[ 0,\infty \right)$ is right-continuous at 0 and $\alpha \left( 0\right) =0$; r is increasing, continuous and

$$\begin{aligned} \left| r\left( x\right) -r\left( y\right) \right| \le \beta \left( \left| x-y\right| \right) , \quad x,y\in I; \end{aligned}$$

(11)

where $\beta :\left[ 0,\infty \right) \rightarrow \left[ 0,\infty \right)$ is increasing, right continuous at 0 and $\beta \left( 0\right) =0$. Assume that there is a right-continuous at 0 function $\gamma :\left[ 0,\infty \right) \rightarrow \left[ 0,\infty \right) ,$ with $\gamma \left( 0\right) =0$ such that

$$\begin{aligned} \beta \circ \gamma \circ \alpha \le \gamma . \end{aligned}$$

(12)

If there exist continuous functions $f,g:I\rightarrow I$ such that

$$\begin{aligned} f\le & {} g, \end{aligned}$$

(13)

$$\begin{aligned} f\circ p\le & {} r\circ f, \end{aligned}$$

(14)

$$\begin{aligned} r\circ g\le & {} g\circ p, \end{aligned}$$

(15)

and for some $h_0:I\rightarrow I$, we have $f\le h_0\le g$ and $|h_0(x)-h_0(y)|\le \gamma (|x - y|)$ for all $x,y\in I$, then there is a continuous function $h:I\rightarrow I$ such that

$$\begin{aligned} f\le h\le g \end{aligned}$$

(16)

and satisfying the equality

$$\begin{aligned} h\circ p=r\circ h. \end{aligned}$$

Proof

Let $\mathcal {C}\left( I\right)$ be the Banach space of all continuous functions $\varphi :I\rightarrow \mathbb {R}$ with the supremum norm $\left\| \varphi \right\| _{\infty }:=\sup \left\{ \left| \varphi \left( x\right) \right| :x\in I\right\}$, and denote by $\mathcal {F}$ the family of all continuous functions $\varphi :I\rightarrow I$ such that

$$\begin{aligned} f\le \varphi \le g, \end{aligned}$$

(17)

and

$$\begin{aligned} \left| \varphi \left( x\right) -\varphi \left( y\right) \right| \le \gamma \left( \left| x-y\right| \right) , \quad x,y\in I. \end{aligned}$$

(18)

By the assumption, $\mathcal {F}$ is nonempty. Define the mapping $T:\mathcal {F}\rightarrow \mathcal {C}\left( I\right)$ by

$$\begin{aligned} T\left( \varphi \right) :=r\circ \varphi \circ p^{-1}. \end{aligned}$$

Then, for every $\varphi \,$ and $x\in I,$ applying in turn: inequality (14) together with the invertibility of p; the monotonicity of r and the inequality in (17); again the monotonicity of r and the second inequality in (17); the inequality in (15), we obtain

$$\begin{aligned} f\left( x\right) \le r\left( f\left( p^{-1}\left( x\right) \right) \right) \le r\left( \varphi \left( p^{-1}\left( x\right) \right) \right) \le r\left( g\left( p^{-1}\left( x\right) \right) \right) \le g\left( x\right) , \end{aligned}$$

which shows that

$$\begin{aligned} f\le T\left( \varphi \right) \le g. \end{aligned}$$

Moreover, using in turn: the definition of T; inequality (11); inequality (10); and (12), we get for arbitrary $\varphi \in \mathcal {F}$ and for all $x,y\in I,$

$$\begin{aligned} \left| T\left( \varphi \right) \left( x\right) -T\left( \varphi \right) \left( y\right) \right|= & {} \left| r\left( \varphi \left( p^{-1}\left( x\right) \right) \right) -r\left( \varphi \left( p^{-1}\left( y\right) \right) \right) \right| \\\le & {} \beta \left( \left| \varphi \left( p^{-1}\left( x\right) \right) -\varphi \left( p^{-1}\left( y\right) \right) \right| \right) \\\le & {} \beta \left( \gamma \left( \left| p^{-1}\left( x\right) -p^{-1}\left( y\right) \right| \right) \right) \\\le & {} \beta \left( \gamma \left( \alpha \left( \left| x-y\right| \right) \right) \right) \\\le & {} \gamma \left( \left| x-y\right| \right) , \end{aligned}$$

which shows that, $T\left( \varphi \right)$ satisfies inequality (18) for every $\varphi \in \mathcal {F}$. This proves that T maps $\mathcal { F}$ into itself. Clearly, $\mathcal {F}$ is a closed subset of $C\left( I\right)$. It follows from the continuity of $\gamma$ at 0 and $\gamma \left( 0\right) =0,$ and the Arzela Theorem that $\mathcal {F}$ is a compact subset of $C\left( I\right)$. Moreover, one should be able to verify with little effort that $\mathcal {F}$ is convex.

Now we show that T is continuous. Take $\varphi _{n}\in \mathcal {F},n\in \mathbb {N}$; $\varphi \in \mathcal {F}$ such $\lim _{n\rightarrow \infty }\varphi _{n}=\varphi$. In turns making use the definition of the norm $\Vert \cdot \Vert _{\infty }$; the bijectivity of p; inequality (11); the monotonicity of $\beta$ and the continuity of the involved functions on the compact interval, we have

$$\begin{aligned} \left\| T\left( \varphi _{n}\right) -T\left( \varphi \right) \right\| _{\infty }= & {} \left\| r\circ \varphi _{n}\circ p^{-1}-r\circ \varphi \circ p^{-1}\right\| _{\infty } \\= & {} \sup \left\{ \left| r\left( \varphi _{n}\left( p^{-1}\!\left( x\right) \right) \right) - r\left( \varphi \left( p^{-1}\!\left( x\right) \right) \right) \right| :x\in I\right\} \\= & {} \sup \left\{ \left| r\left( \varphi _{n}\left( x\right) \right) -r\left( \varphi \left( x\right) \right) \right| :x\in I\right\} \\\le & {} \sup \left\{ \beta \left( \left| \varphi _{n}\left( x\right) -\varphi \left( x\right) \right| \right) :x\in I\right\} \\\le & {} \beta \left( \max \left\{ \left| \varphi _{n}\left( x\right) -\varphi \left( x\right) \right| :x\in I\right\} \right) \\= & {} \beta \left( \Vert \varphi _n-\varphi \Vert _{\infty } \right) , \end{aligned}$$

whence $\lim _{n\rightarrow \infty }T(\varphi _{n})=T(\varphi )$, so T is continuous. The Schauder Fixed-Point Theorem implies that there is a functional $h\in \mathcal {F}$ such that $T(h)=h$, i.e. $r\circ h \circ p^{-1}=h$. Thus $h\circ p=r\circ h$.$\square$

Next we would like to present an application of the above theorem to Hölder functions. For a number $q\in \left( 0,1\right]$, by $\mathcal {H}_{q}\left( I\right)$ denote the space of Hölder continuous functions $\varphi :I\rightarrow \mathbb {R}$, that is such that

$$\begin{aligned} \left| \varphi \left( x\right) -\varphi \left( y\right) \right| \le C\left| x-y\right| ^{q}, \quad x,y\in I, \end{aligned}$$

for some $C\ge 0.$ As an immediate consequence of the above result we obtain:

Corollary 7

Let $I\subseteq \mathbb {R}$, $p,r, f, g, h_{0}$ satisfy the conditions of Theorem 6 with $\alpha \left( t\right) =\alpha \left( 1\right) t,\beta \left( t\right) =\beta \left( 1\right) t$ and $\gamma \left( t\right) =\gamma \left( 1\right) t^{q}$, for some $q\in \left( 0,1\right]$. If

$$\begin{aligned} \beta \left( 1\right) \le \left[ \alpha \left( 1\right) \right] ^{-q}, \end{aligned}$$

then there exists a function $h\in \mathcal {H}_{q}\left( I\right)$ such that

$$\begin{aligned} f\le h\le g, \end{aligned}$$

and satisfying the equality

$$\begin{aligned} h\circ p=r\circ h. \end{aligned}$$

We conclude this work by proving two results concerning the functional equation $\alpha f(p(x))=g(f(x))$ and the functional inequality $\alpha f(p^n(x))\ge g^n(f(x))$.

Theorem 8

Let $\mathcal {M}\subseteq (\mathbb {R}^{n},\Vert \cdot \Vert )$ be a compact subset with $\mathrm {int}\mathcal {M}\ne \emptyset$, and suppose that $\mathfrak {X},\mathcal {W},\mathcal {Z}$ are nontrivial Banach spaces such that $\mathfrak {X}=\mathcal {W}+ \mathcal {Z}$ and $\mathcal {W}\cap \mathcal {Z}=\{0\}$, i.e. $\mathfrak {X}$ is the direct sum of $\mathcal {W}$ and $\mathcal {Z}$. Assume that $p:\mathcal {M}\rightarrow \mathcal {M}$ is continuous. Suppose that $g:\mathfrak {X}\rightarrow \mathfrak {X}$ is a continuous surjection such that the restriction $g|_{\mathcal {W}}:\mathcal {W}\rightarrow \mathfrak {X}$ is a surjective isometry (not necessarily linear), and $\mathcal {Z}\subseteq \{x\in \mathfrak {X}:g(x)=0\}$. Then for every $\alpha \in (0,1)$ there exists a continuous function $f:\mathcal {M} \rightarrow \mathfrak {X}$ such that

$$\begin{aligned} \forall _{x\in \mathcal {M}}\ \ \alpha f(p(x))=g(f(x)) \end{aligned}$$

and f is not a constant function. Moreover, if $\dim \mathcal {Z}\ge n$, then the solution can be chosen in such a way that f is injective.

Proof

Let $N:\mathbb {R}^{n}\rightarrow \mathcal {Z}$ be a continuous linear mapping with $\Vert N\Vert \le 1$. The operator N can be chosen in such a way that $N|_{\mathcal {M}}$ is not constant. Since $N( \mathcal {M})$ is compact and $N(\mathcal {M})\subseteq \mathcal {Z}\subseteq \mathfrak {X}$, without loss of generality, it may be assumed that the set $\mathcal {M}$ is contained in the closed unit ball $B_{\mathfrak {X}}:=\{y\in \mathfrak {X}:\Vert y\Vert \le 1\}$. By $\mathcal {C}(\mathcal {M},\mathfrak {X})$ we denote the space of continuous $\mathfrak {X}$-valued functions with the supremum norm $\Vert \cdot \Vert _{\infty }$, and set $B_{\mathcal {C}}:=\{f\in \mathcal {C}(\mathcal {M}, \mathfrak {X}):\Vert f\Vert _{\infty } \le 1\}$.

Put $g_{1}:=\left( g|_{\mathcal {W}}\right) ^{-1}:\mathfrak {X} \rightarrow \mathcal {W}$ and define a mapping $\varphi :B_{ \mathcal {C}}\rightarrow B_{\mathcal {C}}$ by the formula $\varphi (f):=(1-\alpha )N|_{\mathcal {M}}+\alpha g_{1}\circ f\circ p$. It is easy to verify that $\varphi$ is a contraction. By the Banach Fixed-Point Theorem, there is a function $f_{\alpha }:\mathcal {M}\rightarrow \mathfrak {X }$ such that $(1-\alpha )N|_{\mathcal {M}}+\alpha g_{1}\circ f_{\alpha }\circ p=f_{\alpha }$.

Since $g|_{\mathcal {W}}:\mathcal {W}\rightarrow \mathfrak {X}$ is a surjective isometry, obviously, so is $g_{1}:\mathfrak {X}\rightarrow \mathcal {W}$. The Mazur-Ulam Theorem implies that $g_{1}$ is an affine mapping. Since $g\circ g_{1}=\mathrm {Id}_{ \mathfrak {X}}$, we have

$$\begin{aligned} g(f_{\alpha }(x))&=g\left( (1-\alpha )N|_{\mathcal {M}}(x)+\alpha (g_{1}\circ f_{\alpha }\circ p)(x)\right) \\&=(1-\alpha )g\left( N|_{\mathcal {M}}(x)\right) +\alpha g(g_{1}(f_{\alpha }\circ p(x))) \\&=(1-\alpha )\cdot 0+\alpha f_{\alpha }(p(x))=\alpha f_{\alpha }(p(x)), \end{aligned}$$

whence $g\circ f_{\alpha }=\alpha f_{\alpha }\circ p$. Now, we prove that $f_{\alpha }$ is not constant. Assume, contrary to our claim, that $f_{\alpha }(\cdot )=c$ for some $c\in \mathfrak {X}$. It follows that $(g_{1}\circ f_{\alpha }\circ p)(\cdot )=g_{1}(c)$, whence $(1-\alpha )N|_{\mathcal {M} }(\cdot )=c-\alpha g_{1}(c)$. Thus the function $N|_{\mathcal {M}}:\mathcal {M}\rightarrow \mathcal {Z}$ would be constant, which is a contradiction.

Now assume that $\dim \mathcal {Z}\ge n$. Then there is a continuous injective linear operator $N:\mathbb {R}^n\rightarrow \mathcal {Z}$. Repeating the above reasoning we obtain the equality $(1-\alpha )N|_{\mathcal {M} }+\alpha g_1\circ f_{\alpha }\circ p=f_{\alpha }$. We show that $f_\alpha$ is an injection. Suppose that $f_\alpha (x)=f_\alpha (y)$. Then

$$\begin{aligned} (1-\alpha )N(x)+\alpha g_{1}(f_{\alpha }p(x))=f_{\alpha }(x)=f_{\alpha }(y)=(1-\alpha )N(y)+\alpha g_{1}(f_{\alpha }p(y)). \end{aligned}$$

Note that $\mathfrak {X}=\mathcal {W}\oplus \mathcal {Z}$, $g_{1}(\mathfrak {X} )\subseteq \mathcal {W}$ and $N(x),N(y)\in \mathcal {Z}$. So the uniqueness of the decomposition implies that $N(x)=N(y)$. Since N is injective, $x=y$.$\square$

Bearing the Proof of Theorem 8 in mind, we can formulate and prove another result. More precisely, now we want to study the existence of solutions f of the functional inequality $f(p(x))\ge g(f(x))$ .

Theorem 9

Let $\mathcal {M}\subseteq (\mathbb {R} ^n,\Vert \cdot \Vert )$ be a compact convex subset. Assume that $p:\mathcal { M}\rightarrow \mathcal {M}$ is a continuous mapping. Let $0\le d< b< c<a$. Suppose that $g:[0,a]\rightarrow [0,a]$ is a continuous function such that $[d,b]=\{x\in [0,a]:g(x)=0\}$ and assume that $g|_{[0,c]}:[0,c]\rightarrow [0,a]$ is convex. Moreover, let $g|_{[b,c]}:[b,c]\rightarrow [0,a]$ be an increasing bijection and assume that $g_1:=(g|_{[b,c]})^{-1}$ is a Lipschitz function with Lipschitz constant $\mathrm {Lip}(g_1)\le 1$. Then there exists a continuous function $f:\mathcal {M}\rightarrow \mathbb {R}$ such that

$$\begin{aligned} \forall _{x\in \mathcal {M}}\ \forall _{n\in \mathbb {N}}\ \ g^{n}(f(x))\le \alpha f(p^{n}(x)); \end{aligned}$$

in particular $g\circ f\le f\circ p$. Moreover, the solution f can be chosen to be not constant. If $\mathcal {M}\subseteq \mathbb {R}$ is a compact interval and $p:\mathcal {M}\rightarrow \mathcal {M}$ is increasing, then f can be chosen to be nondecreasing and not constant.

Proof

Fix a number $\alpha \in \left( \frac{b}{g_{1}(b)} ,1\right)$ and define a set

$$\begin{aligned} W:=\{f:\mathcal {M}\rightarrow [b,a]\,|\ f\ \mathrm {is\ continuous}\}. \end{aligned}$$

Consider a continuous mapping $\mu :\mathcal {M}\rightarrow [d,b]$. Suppose that $\mu$ is not constant. Define a mapping $\varphi :W\rightarrow W$ by $\varphi (f):=(1-\alpha )\mu +\alpha g_{1}\circ f\circ p$ for all $f\in W$.

The mapping $\varphi$ is well-defined. Indeed, since the range of of $f\circ p$ is bounded from below by b,  the monotonicity of $g_{1}$ implies that $\varphi (f)(t)=(1-\alpha )\mu (t)+\alpha g_{1}(f(p(t)))\ge \alpha g_{1}(b)\ge b$, so $\varphi \left( f\right) \left( t\right) \in \left[ b,a\right]$ for every $t\in \mathcal {M}$. By the Banach Fixed-Point Theorem we get

$$\begin{aligned} f=(1-\alpha ) \mu +\alpha g_{1}\circ f\circ p \end{aligned}$$

(19)

for some $f\in W$. Since g is convex and $g\circ \mu =0$, it follows from (19) that $g\circ f\le \alpha f\circ p$. Combining this inequality and again (19), we get

$$\begin{aligned} g\circ f\le \alpha f\circ p\le f\circ p=(1-\alpha ) \mu \circ p+g_{1}\circ f\circ p^{2}. \end{aligned}$$

Since g is increasing in $\left[ b,c\right]$ and convex, we obtain $g\circ g\circ f\le \alpha \circ f\circ p^{2}$, that $g^{2}\circ f\le \alpha \circ f\circ p^{2}$. Now, taking into account that $g^{n}$ is increasing for $n\ge 2,$ we can prove, by induction, that $g^{n}\circ f\le \alpha \circ f\circ p^{n}$.

Moreover, if $\mathcal {M}\subseteq \mathbb {R}$ is a compact interval, then we may chose an increasing function $\mu :\mathcal {M}\rightarrow [d,b]$. Next we define

$$\begin{aligned} W:=\{f:\mathcal {M}\rightarrow [b,a]\,|\ f\ \hbox {is continuous and nondecreasing}\}. \end{aligned}$$

and consider a similar mapping $\varphi :W\rightarrow W$. The rest run similarly. Finally, in this case it is easy to notice that the fixed point f is a nondecreasing function.$\square$