Fixed-point theorem in classes of function with values in a dq-metric space

Abstract

We prove a fixed point result for nonlinear operators, acting on some classes of functions with values in a dq-metric space, and show some applications of it. The result has been motivated by some issues arising in Ulam stability. We use a restricted form of a contraction condition.

1 Introduction

The name of Ulam has been somehow connected with various definitions of stability (see, e.g., [1, 12, 16, 24]), but roughly speaking, the following one describes our considerations in this paper ($A^B$ denotes the family of all functions mapping a nonempty set B into a nonempty set A, ${\mathbb {R}}$ stands for the set of all real numbers and ${\mathbb {R}}_+{:}{=}[0,\infty )$).

Definition 1

Let (Y, d) be a metric space, E be a nonempty set, ${\mathcal {D}}_0\subset {\mathcal {D}}\subset Y^{E}$ and ${\mathcal {E}}\subset {{\mathbb {R}}_+}^{E}$ be nonempty, ${\mathcal {T}}:{\mathcal {D}}\rightarrow Y^E$ and ${\mathcal {S}}:{\mathcal {E}}\rightarrow {{\mathbb {R}}_+}^{E}$. We say that the equation

$$\begin{aligned} {\mathcal {T}}(\psi )(t)=\psi (t),\qquad t\in E, \end{aligned}$$

is ${\mathcal {S}}$-stable in ${\mathcal {D}}_0$ provided, for any $\psi \in {\mathcal {D}}_0$ and $\delta \in {\mathcal {E}}$ with

$$\begin{aligned} d({\mathcal {T}}(\psi )(t),\psi (t))\le \delta (t),\qquad t\in E, \end{aligned}$$

there is a solution $\phi \in {\mathcal {D}}$ of the equation, such that

$$\begin{aligned} d(\phi (t),\psi (t))\le {\mathcal {S}}\delta (t),\qquad t\in E. \end{aligned}$$

There are some close connections between Ulam stability and fixed-point theory (see, e.g., [6]). In particular, the subsequent theorem has been presented in [7, Theorem 2] and it has been shown there how to deduce some quite general Ulam stability results from it (see also [6, 9, 14]). To formulate it, we need the following hypothesis concerning operators $\Lambda :{{\mathbb {R}}_+}^E\rightarrow {{\mathbb {R}}_+}^E$ (E is a nonempty set):

$({\mathcal {C}})$ :

If $(\delta _n)_{n\in {\mathbb {N}}}$ is a sequence in ${{\mathbb {R}}_+}^E$ with $\lim _{n\rightarrow \infty } \delta _n(t)=0$ for $t\in E$, then

$$\begin{aligned} \lim _{n\rightarrow \infty } \Lambda \delta _n(t)=0,\qquad t\in E. \end{aligned}$$

Let us yet recall that $\Lambda :{{\mathbb {R}}_+}^E\rightarrow {{\mathbb {R}}_+}^E$ is non-decreasing provided

$$\begin{aligned} \Lambda \xi (t)\le \Lambda \eta (t),\qquad t\in E, \end{aligned}$$

for every $\xi ,\eta \in {{\mathbb {R}}_+}^E$ with $\xi (t)\le \eta (t)$ for every $t\in E$.

Theorem 2

Assume that (Y, d) is a complete metric space, E is a nonempty set, $\Lambda :{{\mathbb {R}}_+}^E\rightarrow {{\mathbb {R}}_+}^E$ is non-decreasing and satisfies hypothesis $({\mathcal {C}})$, and ${\mathcal {T}}:Y^E\rightarrow Y^E$ is such that

$$\begin{aligned} d\big (({\mathcal {T}}\xi )(t),({\mathcal {T}}\mu )(t)\big )\le \Lambda \big (d(\xi ,\mu )\big )(t),\qquad \xi ,\mu \in Y^E , t\in E, \end{aligned}$$

(1)

and functions $\varepsilon :E\rightarrow {\mathbb {R}}_+$ and $\varphi :E\rightarrow Y$ fulfil

$$\begin{aligned} d\big (({\mathcal {T}}\varphi )(t),\varphi (t)\big )\le \varepsilon (t),\qquad t\in E, \end{aligned}$$

and

$$\begin{aligned} \sigma (t){:}{=}\sum _{n\in {\mathbb {N}}_0} (\Lambda ^{n}\varepsilon )(t)<\infty , \qquad t\in E. \end{aligned}$$

Then, for every $t\in E$, the limit

$$\begin{aligned} \lim _{n\rightarrow \infty }({\mathcal {T}}^n\varphi )(t)=:\psi (t) \end{aligned}$$

exists and the function $\psi \in Y^E$, defined in this way, is a fixed point of ${\mathcal {T}}$ with

$$\begin{aligned} d\big (\varphi (t),\psi (t)\big )\le \sigma (t), \qquad t\in E. \end{aligned}$$

In the next section, we present a similar fixed-point theorem for dislocated quasi-metric spaces that generalizes Theorem 2 and several similar outcomes in [5, 7, 8]. In particular, we apply a restricted version of a weaker form of condition (1) (see Remark 3).

Let us recall that a dislocated quasi-metric (dq-metric, for short), in a nonempty set Y, is a function $d:Y\times Y\rightarrow [0,+\infty )$ that satisfies the following two conditions:

• (A1) if $d(x,y)=d(y,x)=0$, then $x=y$,

• (A2) $d(x,y)\leqslant d(x,z)+d(z,y)$

for all $x,y,z\in Y$. The notion of a dq-metric space is a natural generalization of the usual definitions of metric, quasi-metric, partial metric, and metric-like spaces and plays crucial roles in computer science and cryptography (see, e.g., [2, 4, 11, 13, 15, 20,21,22, 25, 26]).

Remark 1

Let $a,b\in (0,\infty )$, $n,k\in {\mathbb {N}}$ (positive integers), $\alpha :{\mathbb {R}}\rightarrow {\mathbb {R}}_+$ and $\alpha ^{-1}(\{0\})=\{0\}$. Then, it is easy to check that the function $d:{\mathbb {R}}\times {\mathbb {R}}\rightarrow [0,\infty )$, given by any of the following six formulas, is a dq-metric:

$$\begin{aligned} d(x,y)= & {} \alpha (x),\qquad x,y\in {\mathbb {R}}, \\ d(x,y)= & {} \max \,\{a|x|^k,b|y|^n\},\qquad x,y\in {\mathbb {R}}, \\ d(x,y)= & {} a|x|^k+b|y|^n,\qquad x,y\in {\mathbb {R}}, \\ d(x,y)= & {} \sqrt{a|x|^k+b|y|^n},\qquad x,y\in {\mathbb {R}}, \\ d(x,y)= & {} \root n \of {\max \,\{x-y,0\}},\qquad x,y\in {\mathbb {R}}, \\ d(x,y)= & {} \max \,\{x-[y],0\},\qquad x,y\in {\mathbb {R}}, \end{aligned}$$

where [y] denotes the integer part of y, i.e., $[y]{:}{=}\max \,\{n\in {\mathbb {Z}}: n\le y\}$ and ${\mathbb {Z}}$ stands for the set of integers. For some further examples we refer to, e.g., [2, 4, 13, 22] and the references therein.

Let d be a dq-metric in a nonempty set Y. We say that $x\in Y$ is a limit of a sequence $(x_n)_{n=1}^{\infty }$ in Y provided

$$\begin{aligned} \lim _{n\rightarrow \infty }\max \,\{d(x_n,x),d(x,x_n)\}=0; \end{aligned}$$

then we write $x_n\rightarrow x$ or $x=\lim _{n\rightarrow \infty }x_n$; in view of (A2), it is easy to note that such a limit must be unique. Next, we say that a sequence $(x_n)_{n=1}^{\infty }$ in Y is Cauchy if

$$\begin{aligned} \lim _{N\rightarrow \infty } \sup _{m,n\geqslant N}\, d(x_n,x_m)=0; \end{aligned}$$

d is complete if every Cauchy sequence in Y has a limit in Y.

Remark 2

Usually, in a dq-metric space, the Cauchy sequence is defined in a somewhat different way; e.g., in a metric-like space (Y, d), a sequence $(x_n)_{n=1}^{\infty }$ is said to be Cauchy if the limit $\lim _{N\rightarrow \infty } \sup _{m,n\geqslant N}\, d(x_n,x_m)$ exists and is finite (see [3]). However, such definitions are too weak and would exclude from our considerations the metric and quasi-metric spaces. The same concerns the notion of completeness.

Our definition of a limit of a sequence is stronger than the usual, but this seems to be necessary in the proof of the main result; moreover, it actually corresponds to our definition of the Cauchy sequence and makes such limit unique (which is not the case in general) and, therefore, more useful.

2 The main result

In what follows, we always assume that (Y, d) is a complete dq-metric space, i.e., d is a complete dq-metric in a nonempty set Y. Moreover, E denotes a nonempty set and $d:{Y}^E\times {Y}^E\rightarrow {{\mathbb {R}}_+}^{E}$ is defined by

$$\begin{aligned} d(\xi ,\mu )(t){:}{=}d(\xi (t),\mu (t)),\qquad \xi ,\mu \in {Y}^E,t\in E. \end{aligned}$$

Analogously, as in the classical metric spaces, if $(\chi _n)_{n\in {\mathbb {N}}}$ is a sequence of elements of $Y^E$, then a function $\chi \in {Y}^E$ is a pointwise limit of $(\chi _n)_{n\in {\mathbb {N}}}$ provided

$$\begin{aligned} \lim _{n\rightarrow \infty }\max \big \{d(\chi ,\chi _n)(t),d(\chi _n,\chi )(t)\big \}=0,\qquad t\in E; \end{aligned}$$

$\chi \in {Y}^E$ is a uniform limit of $(\chi _n)_{n\in {\mathbb {N}}}$ provided

$$\begin{aligned} \lim _{n\rightarrow \infty }\sup _{t\in E}\max \big \{d(\chi ,\chi _n)(t),d(\chi _n,\chi )(t)\big \}=0. \end{aligned}$$

A nonempty subset ${\mathcal {F}}$ of ${Y}^E$ is called p-closed (u-closed, respectively) if every $\chi \in {Y}^E$, which is a pointwise (uniform, resp.) limit of a sequence $(\chi _n)_{n\in {\mathbb {N}}}$ of elements of ${\mathcal {F}}$, belongs to ${\mathcal {F}}$.

Furthermore, given $f,g\in {\mathbb {R}}^E$, we write $f\le g$ if $f(t)\le g(t)$ for $t\in E$. Let $\emptyset \ne {\mathcal {C}} \subset {Y}^E$, $\Lambda :{\mathbb {R_+}}^{E} \rightarrow {\mathbb {R_+}}^{E}$, and $\omega \in {\mathbb {R_+}}^{E}$. We say that ${\mathcal {T}}:{\mathcal {C}}\rightarrow {Y}^E$ is $(\omega ,\Lambda )$—contractive provided

$$\begin{aligned} d({\mathcal {T}}\xi ,{\mathcal {T}}\mu ) \le \Lambda \delta \end{aligned}$$

for any $\xi ,\mu \in {\mathcal {C}}$ and $\delta \in {\mathbb {R_+}}^{E}$ with

$$\begin{aligned} \delta \le \omega ,\qquad d(\xi ,\mu ) \le \delta . \end{aligned}$$

Given a set $A\ne \emptyset$ and $f\in A^A$, we define $f^n\in A^A$ (for $n\in {\mathbb {N}}_0$) by

$$\begin{aligned} f^0(x)=x, \qquad f^{n+1}(x)=f(f^n(x)),\qquad x\in A, n\in {\mathbb {N}}_0. \end{aligned}$$

Finally, to simplify some formulas, we denote by $\Lambda _0$ the identity operator on ${\mathbb {R_+}}^{E}$, i.e., $\Lambda _0 \delta =\delta$ for each $\delta \in {\mathbb {R_+}}^{E}$.

Now, we are in a position to present the fixed-point theorem, which is the main result of this paper.

Theorem 3

Let ${\mathcal {C}} \subset {Y}^E$ be nonempty, $\Lambda _n:{\mathbb {R_+}}^{E}\rightarrow {\mathbb {R_+}}^{E}$ for $n\in {\mathbb {N}}$, and ${\mathcal {T}}:{\mathcal {C}}\rightarrow {\mathcal {C}}$. Assume that there exist functions $\varepsilon _1,\varepsilon _2\in {\mathbb {R_+}}^{E}$ and $\varphi \in {\mathcal {C}}$, such that

$$\begin{aligned}&\varepsilon _j^*(t){:}{=}\sum _{i=0}^{\infty }\Lambda _i\varepsilon _j(t)<\infty ,\qquad t\in E,j=1,2, \end{aligned}$$

(2)

$$\begin{aligned}&d({\mathcal {T}} \varphi ,\varphi )\le \varepsilon _1,\qquad d(\varphi ,{\mathcal {T}} \varphi )\le \varepsilon _2, \end{aligned}$$

(3)

$$\begin{aligned}&\liminf _{n\rightarrow \infty }\Lambda _1\Big (\sum _{i=n}^{\infty } \Lambda _{i}\varepsilon _j\Big )(t)=0,\qquad t\in E,j=1,2, \end{aligned}$$

(4)

and write $\varepsilon ^*(t){:}{=}\max \{\varepsilon _1(t),\varepsilon _2(t)\}$ for $t\in E$. Let ${\mathcal {T}}^n$ be $(\varepsilon ^*,\Lambda _{n})$—contractive for $n\in {\mathbb {N}}$ and one of the following two hypotheses be valid.

• (i) ${\mathcal {C}}$ is p-closed.

• (ii) ${\mathcal {C}}$ is u-closed and the sequence $\big (\sum _{i=0}^{n}\Lambda _i\varepsilon _j\big )_{n\in {\mathbb {N}}}$ tends uniformly to $\varepsilon _j^*$ on E for $j=1,2$.

Then, for each $t\in E$, there exists the limit

$$\begin{aligned} \psi (t){:}{=}\lim _{n\rightarrow \infty }{\mathcal {T}}^n\varphi (t) \end{aligned}$$

(5)

and the function $\psi \in {\mathcal {C}}$, defined in this way, is a fixed point of ${\mathcal {T}}$ with

$$\begin{aligned} d({\mathcal {T}}^n\varphi ,\psi )\le \sum _{i=n}^{\infty } \Lambda _{i}\varepsilon _1,\qquad d(\psi ,{\mathcal {T}}^n\varphi )\le \sum _{i=n}^{\infty } \Lambda _{i}\varepsilon _2,\qquad n\in {\mathbb {N}}_0. \end{aligned}$$

(6)

Moreover, the following two statements are valid:

• (a) for every sequence $(k_n)_{n\in {\mathbb {N}}}$ of positive integers with $\lim _{n\rightarrow \infty } k_n=\infty$, $\psi$ is the unique fixed point of ${\mathcal {T}}$, such that

  $$\begin{aligned} d({\mathcal {T}}^{k_n}\varphi ,\psi )\le \sum _{i=k_n}^{\infty } \Lambda _{i}\varepsilon _j,\qquad d(\psi ,{\mathcal {T}}^{k_n}\varphi )\le \sum _{i=k_n}^{\infty } \Lambda _{i}\varepsilon _l,\qquad n\in {\mathbb {N}}, \end{aligned}$$

  with some $j,l\in \{1,2\}$;

• (b) if

  $$\begin{aligned} \liminf _{n\rightarrow \infty }\Lambda _n\varepsilon _j^*(t)=0,\qquad j=1,2,t\in E, \end{aligned}$$

  (7)

  then $\psi$ is the unique fixed point of ${\mathcal {T}}$ with

  $$\begin{aligned} d(\varphi ,\psi )\le \varepsilon _1^*,\qquad d(\psi ,\varphi )\le \varepsilon _2^*, \end{aligned}$$

  and for every $j,l\in \{1,2\}$

  $$\begin{aligned} \psi (t)= \lim _{n \rightarrow \infty } {\mathcal {T}}^{k_n} \xi (t), \qquad \xi \in {\mathcal {C}},d(\xi ,\psi ) \le \varepsilon _j^*,d(\psi , \xi ) \le \varepsilon _l^*,t\in E, \end{aligned}$$

  (8)

  for every sequence $(k_n)_{n\in {\mathbb {N}}}$ of positive integers with $\lim _{n\rightarrow \infty } \Lambda _{k_n}\varepsilon _m^*(t)=0$ for $t\in E$ and $m\in \{j,l\}$.

Proof

Clearly, (3) implies that, for any $k,l\in {\mathbb {N}}$ and $n\in {\mathbb {N}}_0$

$$\begin{aligned} d({\mathcal {T}}^{n+k}\varphi ,{\mathcal {T}}^n\varphi )\le & {} \sum _{i=0}^{k-1} d({\mathcal {T}}^{n+i+1}\varphi ,{\mathcal {T}}^{n+i}\varphi )\nonumber \\\le & {} \sum _{i=n}^{n+k-1} \Lambda _{i}\varepsilon _1\le \sum _{i=n}^{\infty } \Lambda _{i}\varepsilon _1, \end{aligned}$$

(9)

$$\begin{aligned} d({\mathcal {T}}^n\varphi ,{\mathcal {T}}^{n+l}\varphi )\le & {} \sum _{i=0}^{l-1} d({\mathcal {T}}^{n+i}\varphi ,{\mathcal {T}}^{n+i+1}\varphi )\nonumber \\\le & {} \sum _{i=n}^{n+l-1} \Lambda _{i}\varepsilon _2 \le \sum _{i=n}^{\infty } \Lambda _{i}\varepsilon _2, \end{aligned}$$

(10)

whence

$$\begin{aligned} d({\mathcal {T}}^{n+k}\varphi ,{\mathcal {T}}^{n+l}\varphi )\le & {} d({\mathcal {T}}^{n+k}\varphi ,{\mathcal {T}}^n\varphi )+d({\mathcal {T}}^n\varphi ,{\mathcal {T}}^{n+l}\varphi )\\\le & {} \sum _{i=n}^{\infty } \Lambda _{i}\varepsilon _1 +\sum _{i=n}^{\infty } \Lambda _{i}\varepsilon _2. \end{aligned}$$

Therefore, by (2), $({\mathcal {T}}^n\varphi (t))_{n\in {\mathbb {N}}}$ is a Cauchy sequence in Y for each $t\in E$. Since Y is complete, this sequence is convergent. Consequently, (5) defines a function $\psi \in {\mathcal {C}}$.

Letting $k\rightarrow \infty$ in (9) and $l\rightarrow \infty$ in (10), on account of (5), we get

$$\begin{aligned} d({\mathcal {T}}^n\varphi ,\psi )\le \sum _{i=n}^{\infty } \Lambda _{i}\varepsilon _1,\qquad d(\psi ,{\mathcal {T}}^n\varphi )\le \sum _{i=n}^{\infty } \Lambda _{i}\varepsilon _2,\qquad n\in {\mathbb {N}}_0, \end{aligned}$$

(11)

which is (6). Next, using (11), we get

$$\begin{aligned} d({\mathcal {T}}^{n+1}\varphi ,{\mathcal {T}}\psi ) \le \Lambda _1\Big (\sum _{i=n}^{\infty } \Lambda _{i}\varepsilon _1\Big ),\qquad d({\mathcal {T}}\psi ,{\mathcal {T}}^{n+1}\varphi ) \le \Lambda _1\Big (\sum _{i=n}^{\infty } \Lambda _{i}\varepsilon _2\Big ) \end{aligned}$$

for $n\in {\mathbb {N}}_0$. Hence, for each $n\in {\mathbb {N}}_0$

$$\begin{aligned} d(\psi ,{\mathcal {T}}\psi )&\le d(\psi ,{\mathcal {T}}^{n+1}\varphi )+d({\mathcal {T}}^{n+1}\varphi ,{\mathcal {T}}\psi ) \le d(\psi ,{\mathcal {T}}^{n+1}\varphi )+\Lambda _1\Big (\sum _{i=n}^{\infty } \Lambda _{i}\varepsilon _1\Big ),\\ d({\mathcal {T}}\psi ,\psi )&\le \Lambda _1\Big (\sum _{i=n}^{\infty } \Lambda _{i}\varepsilon _2\Big )+d({\mathcal {T}}^{n+1}\varphi ,\psi ), \end{aligned}$$

which with $n\rightarrow \infty$ yields $d( \psi ,{\mathcal {T}}\psi )=0$ and $d({\mathcal {T}}\psi ,\psi )=0$ [in view of (4)], and consequently ${\mathcal {T}}\psi =\psi$.

Let $(k_n)_{n\in {\mathbb {N}}}$ be a sequence of positive integers with $\lim _{n\rightarrow \infty } k_n=\infty$ and $\xi \in Y^E$ be a fixed point of ${\mathcal {T}}$ with

$$\begin{aligned}&d({\mathcal {T}}^{k_n}\varphi (x),\xi (x))\le \sum _{i=k_n}^{\infty } \Lambda _{i}\varepsilon _j(t),\qquad d(\xi (x),{\mathcal {T}}^{k_n}\varphi (x))\le \sum _{i=k_n}^{\infty } \Lambda _{i}\varepsilon _l(t),\qquad \\&n\in {\mathbb {N}}, t\in E, \end{aligned}$$

with some $j,l\in \{1,2\}$. Then, by (6)

$$\begin{aligned} d(\xi (t),\psi (t))&\le \; d(\xi (t),{\mathcal {T}}^{k_n}\varphi (t))+ d({\mathcal {T}}^{k_n}\varphi (t),\psi (t))\\&\le \;\sum _{i=k_n}^{\infty } \Lambda _{i}\varepsilon _l(t)+\sum _{i=k_n}^{\infty } \Lambda _{i}\varepsilon _1(t),\qquad n\in {\mathbb {N}}_0, t\in E, \end{aligned}$$

$$\begin{aligned} d(\psi (t),\xi (t))&\le \; d(\psi (t),{\mathcal {T}}^{k_n}\varphi (t))+d({\mathcal {T}}^{k_n}\varphi (t),\xi (t))\\&\le \;\sum _{i=k_n}^{\infty } \Lambda _{i}\varepsilon _2(t)+\sum _{i=k_n}^{\infty } \Lambda _{i}\varepsilon _j(t),\qquad n\in {\mathbb {N}}_0, t\in E, \end{aligned}$$

whence letting $n\rightarrow \infty$, we get $\xi =\psi$.

It remains to prove statement (b). Therefore, assume that (7) holds and $\xi \in Y^E$ is a fixed point of ${\mathcal {T}}$ with

$$\begin{aligned} d(\varphi ,\xi )\le \varepsilon _1^*,\qquad d(\xi ,\varphi )\le \varepsilon _2^*. \end{aligned}$$

Then, for any $n\in {\mathbb {N}}_0$, we have

$$\begin{aligned} d(\psi ,\xi )&\le d(\psi ,{\mathcal {T}}^n\varphi )+d({\mathcal {T}}^n\varphi ,{\mathcal {T}}^n\xi )\nonumber \\&\le d(\psi ,{\mathcal {T}}^n\varphi )+\Lambda _{n}\varepsilon _1^*, \end{aligned}$$

$$\begin{aligned} d(\xi ,\psi )&\le d({\mathcal {T}}^n\xi ,{\mathcal {T}}^n\varphi )+d({\mathcal {T}}^n\varphi ,\psi )\nonumber \\&\le \Lambda _{n}\varepsilon _2^*+d(\psi ,{\mathcal {T}}^n\varphi ), \end{aligned}$$

whence letting $n\rightarrow \infty$, we can easily see that $\xi =\psi$.

Now, let $j,l \in \{1,2\}$ and $(k_n)_{n\in {\mathbb {N}}}$ be a sequence of positive integers with

$$\begin{aligned} \lim _{n\rightarrow \infty } \Lambda _{k_n}\varepsilon _m^*(t)=0,\qquad t\in E,m\in \{j,l\}. \end{aligned}$$

Let $\xi \in {\mathcal {C}}$ be a function such that $d(\xi ,\psi ) \le \varepsilon _j^*$ and $d(\psi , \xi ) \le \varepsilon _l^*$. Then

$$\begin{aligned} d({\mathcal {T}}^{k_n} \xi , \psi )= & {} d({\mathcal {T}}^{k_n} \xi , {\mathcal {T}}^{k_n} \psi ) \le \Lambda _{k_n} \varepsilon _j^*,\qquad n \in {\mathbb {N}}, \\ d(\psi ,{\mathcal {T}}^{k_n} \xi )= & {} d({\mathcal {T}}^{k_n} \psi ,{\mathcal {T}}^{k_n} \xi ) \le \Lambda _{k_n} \varepsilon _l^*,\qquad n \in {\mathbb {N}}. \end{aligned}$$

Letting $n \rightarrow \infty$, we get (8). $\square$

Theorem 3 implies at once the following.

Theorem 4

Let ${\mathcal {C}} \subset {Y}^E$ be nonempty, ${\mathcal {T}}:{\mathcal {C}}\rightarrow {\mathcal {C}}$ and $\Lambda :{\mathbb {R_+}}^{E}\rightarrow {\mathbb {R_+}}^{E}$. Assume that there exist functions $\varepsilon _1,\varepsilon _2\in {\mathbb {R_+}}^{E}$ and $\varphi \in {\mathcal {C}}$, such that

$$\begin{aligned}&\varepsilon _j^*(x){:}{=}\sum _{i=0}^{\infty }\Lambda ^i\varepsilon _j(t)<\infty ,\qquad t\in E,j=1,2, \nonumber \\&\quad d({\mathcal {T}} \varphi ,\varphi )\le \varepsilon _1,\qquad d( \varphi ,{\mathcal {T}}\varphi )\le \varepsilon _2, \end{aligned}$$

(12)

$$\begin{aligned}&\liminf _{n\rightarrow \infty }\Lambda \Big (\sum _{i=n}^{\infty } \Lambda ^{i}\varepsilon _j\Big )(t)=0,\qquad t\in E,j=1,2, \end{aligned}$$

(13)

and ${\mathcal {T}}$ is $(\varepsilon ^*,\Lambda )$-contractive, where $\varepsilon ^*(t){:}{=}\max \{\varepsilon _1(t),\varepsilon _2(t)\}$ for $t\in E$. Next, let one of the following two hypotheses hold.

• (i) ${\mathcal {C}}$ is p-closed.

• (ii) ${\mathcal {C}}$ is u-closed and the sequence $\big (\sum _{i=0}^{n}\Lambda ^i\varepsilon _j\big )_{n\in {\mathbb {N}}}$ tends uniformly to $\varepsilon _j^*$ on E for $j=1,2$.

Then, for each $t\in E$, there exists the limit

$$\begin{aligned} \psi (t){:}{=}\lim _{n\rightarrow \infty }{\mathcal {T}}^n\varphi (t) \end{aligned}$$

and the function $\psi \in {\mathcal {C}}$, defined in this way, is a fixed point of ${\mathcal {T}}$ with

$$\begin{aligned} d({\mathcal {T}}^n\varphi ,\psi )\le \sum _{i=n}^{\infty } \Lambda ^{i}\varepsilon _1,\qquad d(\psi ,{\mathcal {T}}^n\varphi )\le \sum _{i=n}^{\infty } \Lambda ^{i}\varepsilon _2,\qquad n\in {\mathbb {N}}_0. \end{aligned}$$

Moreover, the following two statements are valid:

• (a) For every sequence $(k_n)_{n\in {\mathbb {N}}}$ of positive integers with $\lim _{n\rightarrow \infty } k_n=\infty$, $\psi$ is the unique fixed point of ${\mathcal {T}}$ with

  $$\begin{aligned} d({\mathcal {T}}^{k_n}\varphi ,\psi )\le \sum _{i=k_n}^{\infty } \Lambda ^{i}\varepsilon _1,\qquad d(\psi ,{\mathcal {T}}^{k_n}\varphi )\le \sum _{i=k_n}^{\infty } \Lambda ^{i}\varepsilon _2,\qquad n\in {\mathbb {N}}. \end{aligned}$$

• (b) If

  $$\begin{aligned} \liminf _{n\rightarrow \infty }\Lambda ^n\varepsilon _j^*(t)=0,\qquad t\in E,j=1,2, \end{aligned}$$

  (14)

  then $\psi$ is the unique fixed point of ${\mathcal {T}}$ with

  $$\begin{aligned} d(\varphi ,\psi )\le \varepsilon _1^*,\qquad d(\psi ,\varphi )\le \varepsilon _2^*. \end{aligned}$$

  and for every $j,l\in \{1,2\}$,

  $$\begin{aligned} \psi (t)= \lim _{n \rightarrow \infty } {\mathcal {T}}^{k_n} \xi (t), \qquad \xi \in {\mathcal {C}},d(\xi ,\psi ) \le \varepsilon _j^*,d(\psi , \xi ) \le \varepsilon _l^*,t\in E, \end{aligned}$$

  for every sequence $(k_n)_{n\in {\mathbb {N}}}$ of positive integers with $\lim _{n\rightarrow \infty } \Lambda _{k_n}\varepsilon _m^*(t)=0$ for $t\in E$ and $m\in \{j,l\}$.

Proof

It is enough to notice that ${\mathcal {T}}^n$ is $(\varepsilon ^*,\Lambda ^{n})$—contractive for each $n\in {\mathbb {N}}$ and use Theorem 3. $\square$

Remark 3

There arises a natural question whether, in some situations, assumption (2) can be weaker than (12) with $\Lambda {:}{=}\Lambda _{1}$. Below, we provide a somewhat trivial example that this is the case.

Let $Y={\mathbb {R}}^3$ be endowed with the euclidean norm, $c\in {\mathbb {R}}$ and $E={\mathbb {R}}$. Define the operator ${\mathcal {T}}:Y^{{\mathbb {R}}}\rightarrow Y^{{\mathbb {R}}}$ by

$$\begin{aligned} {\mathcal {T}}\phi (x)=(0,\phi _1(x),\phi _2(x)+c),\qquad x\in {\mathbb {R}}, \end{aligned}$$

for every $\phi =(\phi _1,\phi _2,\phi _3)\in Y^{{\mathbb {R}}}$. Then

$$\begin{aligned} \Vert {\mathcal {T}}\phi (x)-{\mathcal {T}}\mu (x)\Vert&=\;\Vert (0,\phi _1(x)-\mu _1(x),\phi _2(x)-\mu _2(x))\Vert \\&\le \; \Vert \phi (x)-\mu (x)\Vert ,\qquad x\in {\mathbb {R}}, \end{aligned}$$

for every $\phi =(\phi _1,\phi _2,\phi _3),\mu =(\mu _1,\mu _2,\mu _3)\in Y^{{\mathbb {R}}}$. This shows that $\Lambda _1$ and $\Lambda _2$ exist, because it is enough to take any $\Lambda _1$ and $\Lambda _2$ with $\Lambda _i\delta \ge \delta$ for $\delta \in {{\mathbb {R}}_+}^{{\mathbb {R}}}$ and $i=1,2$. Next

$$\begin{aligned} {\mathcal {T}}^n\phi (x)=(0,0,nc),\qquad x\in {\mathbb {R}},n\in {\mathbb {N}}, n\ge 3, \end{aligned}$$

for every $\phi =(\phi _1,\phi _2,\phi _3)\in Y^{{\mathbb {R}}}$ and we can take $\Lambda _n\delta (x)=0$ for $\delta \in {{\mathbb {R}}_+}^{{\mathbb {R}}}$, $x\in {\mathbb {R}}$ and $n\in {\mathbb {N}}$, $n\ge 3$. Clearly, in such a case, (2) holds for any $\varepsilon _1,\varepsilon _2\in {\mathbb {R}}_+^{{\mathbb {R}}}$.

We show that, for any such $\Lambda _1$, we must have

$$\begin{aligned} \Lambda _1\delta \ge \delta ,\qquad \delta \in {{\mathbb {R}}_+}^{{\mathbb {R}}}. \end{aligned}$$

(15)

Therefore, take arbitrary $\delta \in {{\mathbb {R}}_+}^{{\mathbb {R}}}$ and define $\phi ,\psi \in Y^{{\mathbb {R}}}$ by

$$\begin{aligned} \phi (x)=(\delta (x),0,0),\qquad \psi (x)=(0,0,0),\qquad x\in {\mathbb {R}}. \end{aligned}$$

Then

$$\begin{aligned} {\mathcal {T}}\phi (x)= & {} (0,\delta (x),c),\qquad {\mathcal {T}}\psi (x)=(0,0,c),\qquad x\in {\mathbb {R}}, \\ \Vert \phi (x)-\psi (x)\Vert= & {} \Vert (\delta (x),0,0)\Vert =\delta (x),\qquad x\in {\mathbb {R}}, \end{aligned}$$

and

$$\begin{aligned} \Vert {\mathcal {T}}\phi (x)-{\mathcal {T}}\psi (x)\Vert =\Vert (0,\delta (x),0)\Vert =\delta (x)\le \Lambda _1 \delta (x),\qquad x\in {\mathbb {R}}. \end{aligned}$$

This shows that (15) holds, whence, by induction, we obtain that for each $n\in {\mathbb {N}}$

$$\begin{aligned} {\Lambda _1}^n\delta \ge \delta , \end{aligned}$$

and therefore

$$\begin{aligned} \sum _{i=0}^{\infty }{\Lambda _1}^i\delta (x)=\infty ,\qquad x\in {\mathbb {R}},\delta (x)\ne 0. \end{aligned}$$

3 Some comments

We need yet the following hypothesis concerning operators $\Lambda :{\mathbb {R_+}}^{E}\rightarrow {\mathbb {R_+}}^{E}$.

$({\mathcal {C}})$ :

If $(\delta _n)_{n\in {\mathbb {N}}}$ is a sequence of elements of ${{\mathbb {R}}_+}^{E}$ with

$$\begin{aligned} \lim _{n\rightarrow \infty } \delta _n(t)=0,\qquad t\in E, \end{aligned}$$

then

$$\begin{aligned} \liminf _{n\rightarrow \infty } \Lambda \delta _n(t)=0,\qquad t\in E. \end{aligned}$$

Remark 4

Note that if $\Lambda _1$ fulfils hypothesis $({\mathcal {C}})$, then (4) results at once from (2). Analogously, (12) yields (13) if $\Lambda$ fulfils $({\mathcal {C}})$.

Remark 5

Let $j\in {\mathbb {N}}$ and ${\mathbb {K}}$ be either the set of reals ${\mathbb {R}}$ or the set of complex numbers ${\mathbb {C}}$. Fix $f_{i}:E\rightarrow E$ and $L_i :E\rightarrow {\mathbb {K}}$ for $i=1,\ldots ,j$. Then, the operator ${\mathcal {T}}:{{\mathbb {K}}}^E\rightarrow {{\mathbb {K}}}^E$, given by

$$\begin{aligned} {\mathcal {T}}\phi (t){:}{=} \sum _{i=1}^{j}L_i(t)\phi (f_i(t)),\qquad \phi \in {\mathbb {K}}^E, t\in E, \end{aligned}$$

is $(\omega ,\Lambda )$ contractive, with any $\omega \in {\mathbb {R}}_+^E$ and $\Lambda :{\mathbb {R_+}}^{E}\rightarrow {\mathbb {R_+}}^{E}$ defined by the formula

$$\begin{aligned} \Lambda \delta (t){:}{=}&\;\sum _{i=1}^{j}|L_i(t)|\delta (f_i(t)),\qquad \delta \in {{\mathbb {R}}_+}^{E}, t\in E. \end{aligned}$$

Moreover, $({\mathcal {C}})$ holds.

Next, for any function $\varepsilon _0:E\rightarrow {\mathbb {R}}_+$ with $\varepsilon _0^*$ given by [see (12)]

$$\begin{aligned} \varepsilon _0^*(x){:}{=}\sum _{i=0}^{\infty }\Lambda ^i\varepsilon _0(t)<\infty ,\qquad t\in E,j=1,2, \end{aligned}$$

we have

$$\begin{aligned} \Lambda \varepsilon _0^*(t)&=\;\sum _{i=1}^{j} |L_i(t)| \sum _{k=0}^{\infty } (\Lambda ^{k}\varepsilon _0)(f_i(t))=\sum _{k=0}^{\infty } \sum _{i=1}^{j} |L_i(t)| (\Lambda ^{k}\varepsilon _0)(f_i(t))\\&=\;\sum _{k=1}^{\infty } (\Lambda ^{k}\varepsilon _0)(t),\qquad t\in E, \end{aligned}$$

and analogously, by induction, we get

$$\begin{aligned} \Lambda ^n\varepsilon _0^*(t)=\sum _{k=n}^{\infty } (\Lambda ^{k}\varepsilon _0)(t),\qquad t\in E,n\in {\mathbb {N}}_0. \end{aligned}$$

This means that (12) yields (14). Therefore, [9, Theorem 1] can be derived from Theorem 4.

Remark 6

Let $F:E \times \mathbb {R_+}\rightarrow \mathbb {R_+}$ be subadditive and non-decreasing with respect to the second variable (i.e., $F(x,a+b)\le F(x,a)+F(x,b)$ and $F(x,a)\le F(x,c)$ for $a,b,c\in \mathbb {R_+}$ with $a\le c$ and $x\in E$). Let $f:E \rightarrow E$ be given and $\Lambda :{\mathbb {R_+}}^{E}\rightarrow {\mathbb {R_+}}^{E}$ be defined by

$$\begin{aligned} \Lambda \varepsilon (x)= F(x,\varepsilon (f(x))),\qquad x\in E,n\in {\mathbb {N}}_0,\varepsilon \in {\mathbb {R_+}}^{E}. \end{aligned}$$

We show that for such $\Lambda$, condition (12) yields (13) and (14).

Therefore, assume that (12) holds for some suitable $\varepsilon _j$ with $j=1,2$. Fix $x\in E$ and define a function $F_0:\mathbb {R_+}\rightarrow \mathbb {R_+}$ by

$$\begin{aligned} F_0(a)=F(x,a),\qquad a\in \mathbb {R_+}. \end{aligned}$$

Since $F_0$ is non-decreasing and $\Lambda ^{n}\varepsilon _1(f(x))\ge 0$ for each $n\in {\mathbb {N}}_0$, we have

$$\begin{aligned} \Lambda ^{n+1}\varepsilon _1(x)= F_0\big (\Lambda ^{n}\varepsilon _1(f(x))\big )\ge F_0(0). \end{aligned}$$

Hence, by (12), we get $F_0(0)=0$.

Fix $j\in \{1,2\}$. Next, we prove that $F_0$ is continuous at 0 or there exists $l_0\in {\mathbb {N}}$ with

$$\begin{aligned} \Lambda ^{n}\varepsilon _j(f(x))=0,\qquad n\in {\mathbb {N}},n>l_0. \end{aligned}$$

To this end suppose that $F_0$ is not continuous at 0 and there is a strictly increasing sequence $\big (k_n\big )_{n\in {\mathbb {N}}}$ of positive integers, such that $\Lambda ^{k_n} \varepsilon _j(f(x))\ne 0$ for $n\in {\mathbb {N}}$. Since $F_0$ is non-decreasing and $F(0)=0$, there exists $d>0$ with $F_0(c)>d$ for every $c>0$, whence

$$\begin{aligned} \Lambda ^{k_n+1} \varepsilon _j(x)=F_0\big (\Lambda ^{k_n} \varepsilon _j(f(x))\big )\ge d,\qquad n\in {\mathbb {N}}, \end{aligned}$$

which contradicts to (12).

Thus, we have proved that

$$\begin{aligned} \lim _{j\rightarrow \infty }F_0\Big (\sum _{n=j}^{\infty } \Lambda ^{n} \varepsilon _j(f(x))\Big )=0,\qquad j=1,2. \end{aligned}$$

Furthermore, by subadditivity of $F_0$, for every $k,l\in {\mathbb {N}}_0$, $l>k$, we get

$$\begin{aligned} F_0\Big (\sum _{n=k}^{\infty } \Lambda ^{n} \varepsilon _j(f(x))\Big )\le \sum _{n=k}^{l} \Lambda ^{n+1} \varepsilon _j(x)+F_0\Big (\sum _{n=l+1}^{\infty } \Lambda ^{n} \varepsilon _j(f(x))\Big ) \end{aligned}$$

whence letting $l\rightarrow \infty$, we obtain

$$\begin{aligned} \Lambda \Big (\sum _{n=k}^{\infty } \Lambda ^{n} \varepsilon _j(x)\Big )=F_0\Big (\sum _{n=k}^{\infty } \Lambda ^{n} \varepsilon _j(f(x))\Big )\le \sum _{n=k+1}^{\infty } \Lambda ^{n} \varepsilon _j(x) \end{aligned}$$

and consequently, by induction (with $k=0$)

$$\begin{aligned} \Lambda ^l\Big (\sum _{n=0}^{\infty } \Lambda ^{n} \varepsilon _j(x)\Big )\le \sum _{n=l}^{\infty } \Lambda ^{n} \varepsilon _j(x),\qquad l\in {\mathbb {N}}. \end{aligned}$$

Clearly, using those inequalities, we can easily deduce (13) and (14) from (12).

Now, consider a very special situation when the set E has only one element, $E=\{s\}$. Then, actually, each ${\mathcal {C}}\subset Y^E$ can be considered as a subset of Y of the form $C{:}{=}\{\phi (s): \phi \in {\mathcal {C}}\}$.

Given $e\in \mathbb {R_+}$, $\lambda :\mathbb {R_+}\rightarrow \mathbb {R_+}$ and $C\subset Y$, analogously as before, we say that $T:C\rightarrow C$ is $(e,\lambda )$—contractive provided

$$\begin{aligned} d(Ty,Tz)\le \lambda ( \delta ), \end{aligned}$$

for every $y,z\in Y$ and $\delta \in {\mathbb {R}}_+$, such that $d(y,z) \le \delta \le e$.

Next, for $\lambda _1:\mathbb {R_+}\rightarrow \mathbb {R_+}$, hypothesis $({\mathcal {C}})$ takes the following form:

$({\mathcal {C}}_0)$ :

If $(\delta _n)_{n\in {\mathbb {N}}}$ is a sequence in ${\mathbb {R}}_+$ with

$$\begin{aligned} \lim _{n\rightarrow \infty } \delta _n=0, \end{aligned}$$

then

$$\begin{aligned} \liminf _{n\rightarrow \infty } \lambda _1(\delta _n)=0. \end{aligned}$$

Theorem 3, with $y_0=\varphi (s)$ and $z_0=\psi (s)$, takes the following form (we write $\lambda _{0}(\varepsilon ){:}{=}\varepsilon$ for each $\varepsilon \in \mathbb {R_+}$).

Theorem 5

Let $T:Y\rightarrow Y$, $\lambda _n:\mathbb {R_+}\rightarrow \mathbb {R_+}$ for $n\in {\mathbb {N}}$, and $\lambda _1$ satisfy hypothesis $({\mathcal {C}}_0)$. Suppose that there exist $y_0\in Y$ and $\varepsilon _1,\varepsilon _2\in \mathbb {R_+}$, such that

$$\begin{aligned}&d(T(y_0),y_0)\le \varepsilon _1,\qquad d(y_0,T(y_0))\le \varepsilon _2, \nonumber \\&\varepsilon _j^*{:}{=}\sum _{i=0}^{\infty }\lambda _i(\varepsilon _j)<\infty ,\qquad j=1,2, \end{aligned}$$

(16)

and $T^n$ is $(\varepsilon ^*,\lambda _n)$—contractive for $n\in {\mathbb {N}}$ with $\varepsilon ^*{:}{=}\max \,\{\varepsilon ^*_1,\varepsilon ^*_2\}$. Then, the limit

$$\begin{aligned} z_0{:}{=}\lim _{n\rightarrow \infty }T^n(y_0) \end{aligned}$$

exists and $z_0$ is a unique fixed point of T with

$$\begin{aligned} d(T^n(y_0),z_0)\le \sum _{i=n}^{\infty } \lambda _{i}(\varepsilon _1),\qquad d(z_0,T^n(y_0))\le \sum _{i=n}^{\infty } \lambda _{i}(\varepsilon _2),\qquad n\in {\mathbb {N}}_0. \end{aligned}$$

Moreover, the following two statements are valid:

• (a) for every sequence $(k_n)_{n\in {\mathbb {N}}}$ of positive integers with $\lim _{n\rightarrow \infty } k_n=\infty$, $z_0$ is the unique fixed point of T with

  $$\begin{aligned} d(T^{k_n}(y_0),z_0)\le \sum _{i=k_n}^{\infty } \lambda ^{i}(\varepsilon _1),\qquad d(z_0,T^{k_n}(y_0))\le \sum _{i=k_n}^{\infty } \lambda ^{i}(\varepsilon _2),\qquad n\in {\mathbb {N}}; \end{aligned}$$

• (b) if

  $$\begin{aligned} \liminf _{n\rightarrow \infty }\lambda _n(\varepsilon _j^*)=0,\qquad j=1,2, \end{aligned}$$

  then $z_0$ is the unique fixed point of T, such that

  $$\begin{aligned} d(y_0,z_0)\le \varepsilon _1^*,\qquad d(z_0,y_0)\le \varepsilon _2^*. \end{aligned}$$

Clearly, if there is $\lambda \in {\mathbb {R}}_+$, such that $\lambda _n(a)=\lambda ^n a$ for $a\in {\mathbb {R}}_+$ and $n\in {\mathbb {N}}$, then Theorem 5 becomes a natural modification of the Banach Contraction Principle (with a local contraction condition) and (16) means that $\lambda <1$.

4 Ulam stability

Now, we show how we can derive some simple Ulam stability outcomes from the results of the previous section. To this end, given $e>0$ or $e=\infty$, we need the subsequent hypothesis.

• (H1) $j\in {\mathbb {N}}$, $L_i:E \rightarrow {\mathbb {R}}_+$ for $i=1,\ldots ,j$, $\Phi :E\times Y^{j}\rightarrow Y$, and

  $$\begin{aligned} d(\Phi (t,w_1,...,w_j),\Phi (t,z_1,...,z_j))\le \sum _{k=1}^{j}L_k(t)d(w_k,z_k) \end{aligned}$$

  for any $t\in E$ and $(w_1,...,w_j),(z_1,...,z_j)\in Y^j$, such that $d(z_i,w_i)\le e$ for $i=1,\ldots ,j$.

The following corollary also can be easily deduced from Theorem  2.

Corollary 6

Assume that $\varepsilon _1,\varepsilon _2:E\rightarrow {\mathbb {R}}_+$, hypothesis (H1) is valid with $e{:}{=}\sup \,\{\varepsilon _j^*(t): t\in E, j=1,2\}$, where

$$\begin{aligned} \varepsilon _j^*(t){:}{=}\sum _{i=0}^{\infty }\Lambda ^i\varepsilon _j(t)<\infty ,\qquad t\in E,j=1,2, \end{aligned}$$

and $\Lambda :{\mathbb {R}}_+^{E}\rightarrow {\mathbb {R}}_+^{E}$ is given by

$$\begin{aligned} \Lambda \delta (t)=\sum _{k=1}^{j}L_k(t)\delta (f_k(t)),\qquad \delta \in {\mathbb {R}}_+^{E},t\in E, \end{aligned}$$

with some $f_1,\dots ,f_j:E\rightarrow E$, and $\varphi :E\rightarrow Y$ is such that

$$\begin{aligned}&d(\Phi (t,\varphi (f_1(t)),...,\varphi (f_j(t))),\varphi (t))\le \varepsilon _1(t), \qquad t\in E, \end{aligned}$$

(17)

$$\begin{aligned}&d(\varphi (t),\Phi (t,\varphi (f_1(t)),...,\varphi (f_j(t))))\le \varepsilon _2(t), \qquad t\in E. \end{aligned}$$

(18)

Then, the limit

$$\begin{aligned} \psi (t){:}{=}\lim _{n\rightarrow \infty }{\mathcal {T}}^n\varphi (t) \end{aligned}$$

(19)

exists for each $t\in E$, with ${\mathcal {T}}$ given by

$$\begin{aligned} {\mathcal {T}}\varphi (t){:}{=}\Phi (t,\varphi (f_1(t)),...,\varphi (f_j(t))), \qquad \varphi \in Y^E,\,t\in E, \end{aligned}$$

and the function $\psi :E\rightarrow Y$, defined by (19), is the unique solution of the functional equation:

$$\begin{aligned} \Phi (t,\psi (f_1(t)),...,\psi (f_j(t)))=\psi (t), \qquad t\in E, \end{aligned}$$

(20)

such that

$$\begin{aligned} d(\varphi (t),\psi (t))\le \varepsilon _1^*(t),\qquad d(\psi (t),\varphi (t))\le \varepsilon _2^*(t), \qquad t\in E. \end{aligned}$$

(21)

Proof

Let us note that inequalities (17) and (18) imply (3). Next

$$\begin{aligned}&\liminf _{n\rightarrow \infty }\Lambda ^n\varepsilon _j^*(t)=0,\qquad t\in E,j=1,2,\\&\liminf _{n\rightarrow \infty }\Lambda \Big (\sum _{i=n}^{\infty } \Lambda ^{i}\varepsilon _j\Big )(t)=0,\qquad t\in E,j=1,2, \end{aligned}$$

in view of Remarks 4 and 5. Therefore, by Theorem  4, the function $\psi$ defined by (19) is the unique fixed point of ${\mathcal {T}}$ (that is a solution of (20)) satisfying (21). $\square$

Stability of functional equations of form (20) (or related to it) has been already studied by several authors, and for further information, we refer to the survey papers [1, 8]. A particular case of (20) is the linear functional equation of the form

$$\begin{aligned} \phi (t){:}{=}\sum _{i=1}^j L_i(t)\phi (f_i(t)), \qquad \varphi \in Y^E,\,t\in E, \end{aligned}$$

under the assumptions as in Remark 5; some recent results concerning stability of less general cases of it can be found in [10, 18, 19, 23].

As an example of applications of Corollary 6 consider stability of the difference equation:

$$\begin{aligned} \psi (i)=\Phi (i,\psi (i+1)),\qquad i\in {\mathbb {N}}, \end{aligned}$$

(22)

where $\Phi :{\mathbb {N}}\times Y\rightarrow Y$ is given and $\psi :{\mathbb {N}}\rightarrow Y$ is unknown. Clearly, (22) is a very simple particular case of (20), with $E={\mathbb {N}}$, $j=1$ and $f_1(i)=i+1$ for $i\in X$.

Let $(a_n)_{n\in {\mathbb {N}}}$ be a sequence of positive reals, such that

$$\begin{aligned} \sum _{k=1}^{\infty }\prod _{l=0}^{k-1}a_{i+l}<\infty ,\qquad i\in {\mathbb {N}}. \end{aligned}$$

(23)

For instance, we can take $\rho \in (0,1)$ and write

$$\begin{aligned} a_{2n}=\frac{1}{\rho },\qquad a_{2n-1}=\rho ^2,\qquad n\in {\mathbb {N}}. \end{aligned}$$

Then

$$\begin{aligned} \prod _{l=0}^{2k}a_{i+l}=\rho \prod _{l=0}^{2k-2}a_{i+l},\qquad \prod _{l=0}^{2k+1}a_{i+l}=\rho \prod _{l=0}^{2k-1}a_{i+l},\qquad k\in {\mathbb {N}}, \end{aligned}$$

whence (23) is valid and

$$\begin{aligned} \sum _{k=1}^{\infty }\prod _{l=0}^{k-1}a_{i+l}=\frac{a_i(1+a_{i+1})}{1-\rho },\qquad i\in {\mathbb {N}}. \end{aligned}$$

Let operator $\Lambda :{\mathbb {R}}_+^{{\mathbb {N}}}\rightarrow {\mathbb {R}}_+^{{\mathbb {N}}}$ be defined by

$$\begin{aligned} \Lambda \delta (i)=a_i\delta (i+1),\qquad \delta \in {\mathbb {R}}_+^{{\mathbb {N}}},i\in {\mathbb {N}}. \end{aligned}$$

Note that

$$\begin{aligned} \Lambda ^k\delta (i)=\delta (i+k)\prod _{l=0}^{k-1}a_{i+l},\qquad k\in {\mathbb {N}},\delta \in {\mathbb {R}}_+^{{\mathbb {N}}}, \end{aligned}$$

whence

$$\begin{aligned} \sum _{k=1}^{n}\Lambda ^k\delta (i)=\sum _{k=1}^{n}\delta (i+k)\prod _{l=0}^{k-1}a_{i+l},\qquad n\in {\mathbb {N}},\delta \in {\mathbb {R}}_+^{{\mathbb {N}}}. \end{aligned}$$

(24)

Take $\gamma >0$ and let $\phi :{\mathbb {N}}\rightarrow Y$ fulfil inequalities (17) and (18) with some $\varepsilon _1,\varepsilon _2:{\mathbb {N}}\rightarrow [0,\gamma ]$, that is

$$\begin{aligned} d(\Phi (i,\phi (i+1)),\phi (i))\le \varepsilon _1(i), \qquad d(\phi (i),\Phi (i,\phi (i+1)))\le \varepsilon _2(i), \qquad i\in {\mathbb {N}}. \end{aligned}$$

Then, (24) implies that, for each $j\in \{1,2\}$

$$\begin{aligned} \varepsilon _j^*(i){:}{=}&\,\sum _{k=0}^{\infty }\Lambda ^k\varepsilon _j (i)\le \gamma \Big (1+\sum _{k=1}^{\infty }\prod _{l=0}^{k-1}a_{l+i}\Big )<\infty ,\qquad i\in {\mathbb {N}}. \end{aligned}$$

Next, if

$$\begin{aligned} d(\Phi (i,z),\Phi (i,w))\le a_id(z,w),\qquad w,z\in Y,i\in {\mathbb {N}},d(z,w)\le e, \end{aligned}$$

where $e{:}{=}\sup \,\{\varepsilon _j^*(i): i\in {\mathbb {N}}, j=1,2\}$, then $\Phi$ is as in (H1) with $j=1$, and the assumptions of Corollary 6 are satisfied with

$$\begin{aligned} L_1(i)= a_i,\qquad f_1(i)=i+1,\qquad i\in {\mathbb {N}}. \end{aligned}$$

Hence, the limit

$$\begin{aligned} \psi (i){:}{=}\lim _{n\rightarrow \infty }{\mathcal {T}}^n\phi (i) \end{aligned}$$

(25)

exists for each $i\in {\mathbb {N}}$, with

$$\begin{aligned} {\mathcal {T}}\xi (i){:}{=}\Phi (i,\xi (i+1)), \qquad \xi \in Y^E,\,i\in {\mathbb {N}}, \end{aligned}$$

and $\psi :E\rightarrow Y$, given by (25), is the unique solution of (22), such that

$$\begin{aligned} d(\phi (i),\psi (i))\le \varepsilon _1^*(i), \qquad d(\psi (i),\phi (i))\le \varepsilon _2^*(i), \qquad i\in {\mathbb {N}}. \end{aligned}$$