Continuity Dependence of Attractors of Iterated Function Systems with Orbital Condition

Abstract

In the article Secelean and Wardowski (Qual Theory Dyn Syst 21:162, 2022), there were introduced iterated function systems involving certain condition of orbital type. The attractors of such IFSs substantially extend the family of fractal objects investigated so far. In the present work we provide some further properties and give the answer regarding the continuity dependence of this type of attractors with respect to the continuous change of the parameters that describe the maps of the given orbital IFS and the initially chosen compact subset.

1 Introduction

In the classical Hutchinson-Barnsley theory, which is based on works by Hutchinson [2] and Barnsley [1], there is considered a finite family of contractions $f_i :X \rightarrow X$ on a complete metric space X. Based on such the so called hyperbolic iterated function system (IFS for short) it can be defined on the family $\mathscr {K}(X)$ of all non-empty compact subsets of X the fractal operatorF given by $K\mapsto \bigcup _i f_i(K)$. This is a contraction with respect to Hausdorff–Pompeiu metric and so has a unique fixed point $A_F$ referred to as the attractor of the system. Barnsley [1] and Jachymski [4] have shown that the attractor varies continuously with respect to continuous perturbation of the family of contractions.

A different approach to the concept of IFS was proposed by Miculescu et al. [5] and this was extended by Secelean and Wardowski [7]. Here we consider the effect of perturbation on the (non-unique) attractors for these IFSs of orbital type.

We use d for the metric of all considered base metric spaces X, P, etc. On finite products we use the max metric so that, for example, $d((p,x),(p',x')):=\max (d(p,p'),d(x,x'))$. For a subset B of X and $\epsilon > 0$, we let $V_{\epsilon }(B):= \{x\in X:d(x,y)<\epsilon \text { for some } y\in B\}$. For subsets A, B of X we define

$$\begin{aligned} \delta (A,B):= & {} \sup _{x\in A} \inf _{y\in B} d(x,y) = \inf \{\epsilon >0:A\subseteq V_{\epsilon }(B)\},\\ H(A,B):= & {} \max \{\delta (A,B),\delta (B,A)\}. \end{aligned}$$

If $\{A_i\}$, $\{B_i\}$ are two families of subsets of X with the same index set and $\epsilon > 0$ such that $A_i \subseteq V_{\epsilon }(B_i)$ for all i, then $\bigcup _i A_i \subseteq V_{\epsilon }(\bigcup _i (B_i))$. Consequently, we have

$$\begin{aligned} \delta \left( \bigcup _i A_i, \bigcup _i B_i \right) \le \sup _i \delta (A_i,B_i), \end{aligned}$$

(1)

(see, e.g. [6]).

It is easy to see that $\delta (A,B) = \delta (\overline{A},\overline{B})$ and from now on we will restrict to closed subsets.

It may happen that $\delta (A,B)$ is infinite, e.g. if B is bounded and A is not. The set of pairs of closed subsets (A, B) such that $H(A,B) < \infty$ is an equivalence relation and on each equivalence class, e.g. the class of bounded subsets, H is a metric called the Hausdorff–Pompeiu metric. We shall be primarily interested in $\mathscr {K}(X)$ the set of non-empty compact subsets of X equipped with the metric H. The metric space $\mathscr {K}(X)$ is complete or compact when X satisfies the corresponding property (see Chapter 1 in [3]).

We will use the following properties of the metric space $\mathscr {K}(X)$.

• H1 If U is an open subset of X, then $\langle U \rangle := \{K\in \mathscr {K}(X):K\subset U \}$ is open in $\mathscr {K}(X)$.

Proof

If $K \in \mathscr {K}(X)$, $K\subset U$, then, by compactness, there exists $\epsilon > 0$ such that $V_{\epsilon }(K) \subset U$ and one can see that $\{L \in \mathscr {K}(X):H(K,L) < \epsilon \} \subset \langle U \rangle$. $\square$

• H2 If $\mathscr {A}$ is a compact subset of $\mathscr {K}(X)$, then $\bigcup \mathscr {A}$ is compact and so is a member of $\mathscr {K}(X)$.

Proof

Let $\mathscr {U}$ be an open cover of $\bigcup \mathscr {A}$. For each $A \in \mathscr {A}$ choose a finite subcover $\mathscr {U}_A$ of A with union $U_A$. Thus, $U_A$ is open with $A\subset U_A$. It follows that $\{\langle U_A \rangle :A\in \mathscr {A} \}$ is an open cover of $\mathscr {A}$. If $\{\langle U_{A_1} \rangle ,\ldots , \langle U_{A_k}\rangle \}$ is a finite subcover of $\mathscr {A}$, then $\bigcup _{i=1}^k \mathscr {U}_{A_i}$ is a finite subcover of $\bigcup \mathscr {A}$. $\square$

• H3 The map $\mathscr {K}(X)^n \rightarrow \mathscr {K}(X)$ given by $(K_1,\ldots , K_n) \mapsto \bigcup _{i=1}^n K_i$ is a Lipschitz map with Lipschitz constant 1.

Proof

This is clear from (1). $\square$

• H4 If $f:X\rightarrow Y$ is a continuous map, then the induced map $f:\mathscr {K}(X) \rightarrow \mathscr {K}(Y)$, given by $K \mapsto f(K)$, is continuous. Furthermore, if f is a Lipschitz map with Lipschitz constant $\alpha$, then the induced map is Lipschitz with the same constant.

Proof

If $\{K_n\}$ is a sequence in $\mathscr {K}(X)$ converging to K, then since a sequence together with its limit is compact, it follows from [H2] that $K\cup \bigcup _n K_n$ is compact and so on it f is uniformly continuous. Given $\epsilon > 0$, let $\delta > 0$ be an $\epsilon$ modulus of uniform continuity for the restriction of f to this set. There exists positive integer N such that for $n\ge N$, $K\subset V_\delta (K_n)$ and $K_n \subset V_\delta (K)$. It follows that $f(K) \subset V_{\epsilon }(f(K_n))$ and $f(K_n) \subset V_\epsilon (f(K))$. Thus, $\{f(K_n)\}$ converges to f(K).

If f has Lipschitz constant $\alpha$, then for all $A, B, \epsilon > 0$, $A\subseteq V_\epsilon (B)$ implies $f(A) \subseteq V_{\alpha \epsilon } (f(B))$. Thus, the induced map has Lipschitz constant $\alpha$. $\square$

Recall that if $f:X\rightarrow X$ is a contraction, i.e. a Lipschitz map with constant $\alpha < 1$, then the Banach Fixed Point Theorem says that f has at most one fixed point $x^*$. Furthermore, for any $x\in X$, and positive integer n, $d(f^n(x),f^{n+1}(x))\le \alpha ^n d(x,f(x))$. Hence, the sequence $\{f^n(x)\}$ is Cauchy and if X is complete the limit point $x^*$ exists, is fixed by f and satisfies

$$\begin{aligned} d(x,x^*) \le \frac{d(x,f(x))}{1-\alpha }. \end{aligned}$$

(2)

If $\{f_i:X\rightarrow X\}$ indexed by $\{i\in I\}$ is a finite family of continuous maps, then the associated fractal operator $F:\mathscr {K}(X)\rightarrow \mathscr {K}(X)$ is given by

$$\begin{aligned} F(K) := \bigcup _{i\in I} f_i(K), \ \ \text {for} \ \ K \in \mathscr {K}(X). \end{aligned}$$

(3)

From [H3] and [H4] it follows that F is continuous. Furthermore, if each $f_i$ is Lipschitz with constant at most $\alpha$, then F has Lipschitz constant at most $\alpha$. If X is complete and $\alpha < 1$, then $\mathscr {K}(X)$ is complete and so F has a unique fixed point $A_F \in \mathscr {K}(X)$, which we call the attractor of F (see, Hutchinson [2]).

1.1 Blowing in the Wind

We consider functions of the form $f:P\times X\rightarrow X$ which are continuous in each variable separately. We write $f_p(x):= f(p,x)$ so that each $f_p$ is a continuous function on X. We define $\hat{f} :P\times X \rightarrow P\times X$ by $\hat{f}(p,x):=(p,f(p,x))=(p,f_p(x))$. We say that f is uniformly Lipschitz in x if there exists $\alpha$ such that each $f_p$ has Lipschitz constant at most $\alpha$.

Lemma 1.1

If f(p, x) is continuous in the variable p and uniformly Lipschitz in x, then $f:P\times X \rightarrow X$ is continuous, i.e. jointly continuous.

Proof

For $(p,x) \in P\times X$ we have

$$\begin{aligned} d(f(p,x),f(p',x'))&\le d(f(p,x),f(p',x)) + d(f(p',x),f(p',x'))\\&\le d(f(p,x),f(p',x)) + \alpha d(x,x'). \end{aligned}$$

Given $\epsilon > 0$ there exists $\delta$ so that $d(p,p')<\delta$ implies $d(f(p,x),f(p',x)) < \epsilon /2$. So if $d(p,p') < \delta$ and $d(x,x')<\epsilon /2 \alpha$, then $d(f(p,x),f(p',x'))<\epsilon$. $\square$

If $f,g:P\times X\rightarrow X$, we define $f\cdot g:P\times X \rightarrow X$ by $(f\cdot g)(p,x):= f(p,g(p,x))$ so that $(f\cdot g)_p = f_p \circ g_p$ for all $p\in P$. For a positive integer k we thus define $f^k :P\times X\rightarrow X$ so that $f^k(p,x) = (f^k)_p(x):= (f_p)^k(x)$. Observe that if f and g are jointly continuous, then $f\cdot g$ and the iterates $f^k$ are jointly continuous. Furthermore, if f and g are uniformly Lipschitz in x with constants at most $\alpha$, then $f\cdot g$ and $f^k$ are uniformly Lipschitz with constants at most $\alpha ^2$ and at most $\alpha ^k$ respectively.

If f(p, x) is continuous in the x variable we define for $f:P\times X\rightarrow X$ the induced map $f:P\times \mathscr {K}(X) \rightarrow \mathscr {K}(X)$ by $(p,K)\mapsto f_p(K)$.

Lemma 1.2

If $f:P\times X\rightarrow X$ is jointly continuous, then the induced map $f:P\times \mathscr {K}(X) \rightarrow \mathscr {K}(X)$ is jointly continuous. Furthermore, if f is uniformly Lipschitz in the variable x with constants at most $\alpha$, then the induced map is uniformly Lipschitz in the second variable with constants at most $\alpha$.

Proof

The map $P\times \mathscr {K}(X)\rightarrow \mathscr {K}(P\times X)$ given by $(p,K)\mapsto \{p\}\times K$ is an isometric inclusion. The induced map $\mathscr {K}(P\times X)\rightarrow \mathscr {K}(X)$ is continuous by [H4]. The composition is the induced map f. If $f_p$ has Lipschitz constant at most $\alpha$, then the induced map $f_p$ on $\mathscr {K}(X)$ has Lipschitz constant at most $\alpha$ by [H4] again. $\square$

If $\{f_i:P\times X\rightarrow X\}$ indexed by $\{i\in I\}$ is a finite family of maps each continuous in the second variable, then the associated fractal operator $F:P\times \mathscr {K}(X)\rightarrow \mathscr {K}(X)$ is given by

$$\begin{aligned} F_p(K) := \bigcup _i (f_i)_p(K), \ \ \text {for} \ \ K\in \mathscr {K}(X). \end{aligned}$$

(4)

That is, $F_p$ is the fractal operator associated for the family $\{(f_i)_p:X\rightarrow X\}$ for each p. From Lemmas 1.1 and 1.2 it follows that if each $f_i(p,x)$ is continuous in the variable p and uniformly Lipschitz in the variable x with constants at most $\alpha$, then F is jointly continuous and is uniformly Lipschitz in the variable K with constant at most $\alpha$.

Theorem 1.1

(Barnsley [1]) Let $f:P\times X\rightarrow X$ be a map such that f(p, x) is continuous in the p variable and is uniformly Lipschitz in the x variable with constants at most $\alpha <1$. If X is complete, then for each $p\in P$, there is a unique fixed point $x^*(p)$ for $f_p$. Furthermore, the map $x^*:P\rightarrow X$ defined by $p\mapsto x^*(p)$ is continuous. That is, the fixed point of $f_p$ depends continuously on p.

Proof

Assume the sequence $\{p_n\}$ converges to $p_{\infty }$ in P. For $\epsilon > 0$ the set

$$\begin{aligned} U_{\epsilon }:= \{p:d(f(p,x^*(p_\infty )),x^*(p_\infty )) < \epsilon \} \end{aligned}$$

is open in P and contains $p_\infty$. Hence, there exists positive integer N so that $n\ge N$ implies $p_n\in U_{\epsilon }$. That is, for such n, $d(f_{p_n}(x^*(p_\infty )),x^*(p_\infty )) < \epsilon$. From equation (2) it follows that $d(x^*(p_n),x^*(p_\infty ))<\epsilon /(1-\alpha )$. Thus $\{x^*(p_n)\}$ converges to $x^*(p_\infty )$. $\square$

Corollary 1.1

(Jachymski [4]) Let $\{f_i :P\times X \rightarrow X\}$ be a finite family of maps such that each $f_i(p,x)$ is continuous in the p variable and is uniformly Lipschitz in the x variable with constants at most $\alpha <1$. Let F be the fractal operator defined by (4). If X is complete, then for each $p\in P$, there is a unique attractor $A_p$ for $F_p$. Furthermore, the map from P to $\mathscr {K}(X)$ defined by $p\mapsto A_p$ is continuous. That is, the attractor of $F_p$ depends continuously on p.

Proof

This is immediate from Theorem 1.1 and the preceding remarks. $\square$

1.2 IFSs of Orbital Type

For a finite non-empty set I and positive integer n, let $\Lambda _n(I)$ denote the family of all sequences of the form $\omega =(\omega _1,\omega _2,\ldots ,\omega _n)$, where $\omega _i\in I$ for all i, written as words $\omega _1 \omega _2 \ldots \omega _n$. The number n, denoted by $|\omega |$, is called the length of $\omega$. By $\emptyset$ we mean the empty word, which is the sequence of zero elements from I, i.e. $\Lambda _0(I)=\{ \emptyset \}.$ $\Lambda ^*(I)$ denotes the set of all finite sequences of elements from I. For $\alpha = \alpha _1 \alpha _2 \ldots \alpha _m \in \Lambda _m(I)$ and $\beta = \beta _1 \beta _2 \ldots \beta _n \in \Lambda _n(I)$, by $\alpha \beta$ we denote the concatenation of $\alpha$ and $\beta$ of the form $\alpha _1 \alpha _2 \ldots \alpha _m \beta _1 \beta _2 \ldots \beta _n$ being an element of $\Lambda _{m+n}(I)$. For a family of maps $\{f_i:X\rightarrow X\}$ indexed by $\{i\in I\}$ and $\omega =\omega _1 \omega _2 \ldots \omega _n \in \Lambda _n(I)$ we will use the notations $f_\omega := f_{\omega _1} \circ f_{\omega _2}\circ \ldots \circ f_{\omega _n}, \ f_{\emptyset }:=id_X.$

In [5] the authors considered the finite family $\{f_i:X\rightarrow X\}$ of continuous maps defined on a complete metric space X with the following condition of orbital type

$$\begin{aligned} d(f_\omega (x),f_{\omega i}(x))\le C^{|\omega |}d(x,f_i(x)) \end{aligned}$$

(5)

for all $x\in X$, $i \in I$, and $\omega \in \Lambda ^*(I)$, with a positive constant $C < 1$. They proved that the fractal operator associated for the IFS consisting of maps satisfying (5) is weakly Picard, i.e. the obtained attractor depends on an initially chosen compact set.

Secelean and Wardowski [7] assumed the existence of $\gamma > 0$ such that for all $x\in X$, $i\in I$, and $\omega \in \Lambda ^*(I)$

$$\begin{aligned} d(f_{\omega }(x),f_{\omega i}(x))\le \Phi (\gamma |\omega |,d(x,f_i(x))), \end{aligned}$$

(6)

where $\Phi :\mathbb {R}_+\times \mathbb {R}_+\rightarrow \mathbb {R}_+$ satisfies

• (i) $\Phi (r,\cdot )$ is non-decreasing, $\Phi (\cdot ,s)$ is non-increasing,

• (ii) $\sum _{i=0}^{\infty } \Phi (i,t) < \infty$,

for all $r, s, t \ge 0$. With such assumptions the authors obtained the following theorem.

Theorem 1.2

(Secelean and Wardowski [7]) Let $\{f_i:X\rightarrow X\}$ be a finite family of continuous maps satisfying (6). If X is complete, then the fractal operator (3) has a fixed point. More precisely, for each $K\in \mathscr {K}(X)$ there is the attractor $A_K \in \mathscr {K}(X)$ and the sequence $\{F^{n}(K)\}$ converges to $A_K$.

Example 1.1

Assume that $\{f_i:P\times X\rightarrow X\}$ is a finite family of maps such that each $f_i(p,x)$ is continuous in the p variable and is uniformly Lipschitz in the x variable with constants at most $\alpha < 1$. Assume that P and X are both complete. The family of maps $\{\hat{f}_i:P\times X\rightarrow P\times X\}$ is defined by $\hat{f}_i(p,x)=(p,f_i(p,x))$.

Following (3), we define $\hat{F}:\mathscr {K}(P\times X)\rightarrow \mathscr {K}(P\times X)$ by

$$\begin{aligned} \hat{F}(K) := \bigcup _i \hat{f}_i (K), \ \ \text {for} \ \ K\in \mathscr {K}(P\times X). \end{aligned}$$

From [H3] and [H4] it follows that $\hat{F}$ is continuous. However, it is certainly not a contraction (unless P is a singleton). On the other hand, it is clearly true that for all (p, x), $d(\hat{f}_{\omega }(p,x),\hat{f}_{\omega i}(p,x)) \le \alpha ^{|\omega |}d((p,x),\hat{f}_i(p,x))$ because each $(f_i)_p$ is an $\alpha$ contraction. Thus, Theorem 1.2 applies.

Let $\pi _1 :\mathscr {K}(P\times X)\rightarrow \mathscr {K}(P)$ be the induced map associated with the first coordinate projection $\pi _1 :P\times X\rightarrow P$, $(p,x)\mapsto p$. For each $B\in \mathscr {K}(P)$ the fiber $(\pi _1)^{-1}(\{B\})$ is invariant with respect to $\hat{F}$, i.e. $\pi _1(\hat{F}(K))=\pi _1(K)$ for all $K\in \mathscr {K}(P\times X)$. With $i_p:X\rightarrow P\times X$ the injection given by $x\mapsto (p,x)$ we define for all $K\in \mathscr {K}(P\times X)$ and $p\in \pi _1(K)$, the set $K_p:= i_p^{-1}(K)\in \mathscr {K}(X)$. Note that $i_p^{-1}(K)$ is non-empty if and only if $p\in \pi _1(K)$. We define $F_p :\mathscr {K}(X)\rightarrow \mathscr {K}(X)$ to be the fractal operator given in (4) associated with the family of $\alpha$ contractions $\{(f_i)_p:X\rightarrow X\}$. In consequence, each $F_p$ has an attractor $A_p$ and by Corollary 1.1 the $A_p$’s vary continuously with p. It follows that as p varies over $B\in \mathscr {K}(P)$, the subset $\{\{p\}\times A_p:p\in B\}$ is compact in $\mathscr {K}(P\times X)$ and so $A_B$ defined to be $\bigcup _{p\in B} \{p\}\times A_p$ is compact and so is an element of $\mathscr {K}(P\times X)$ (see [H2]).

Proposition 1.1

The restriction of $\hat{F}$ to the invariant set $(\pi _1)^{-1}(\{B\})$ has a unique attractive fixed point. That is, for any $K\in \mathscr {K}(P\times X)$ with $\pi _1(K) = B$, the sequence $\{\hat{F^n}(K)\}$ converges to $A_B$. To be precise, there exists a real number M such that

$$\begin{aligned} H(\hat{F}^n(K),A_B) \le \alpha ^n M. \end{aligned}$$

(7)

Proof

Let M be the diameter of the compact subset $A_B\cup K \subset P\times X$. Clearly, for each $p\in B$, $H(A_p,K_p) \le M$. Observe that for each $p\in B$ and each positive integer n

$$\begin{aligned} (\hat{F}^n(K))_p = (F_p)^n(K_p). \end{aligned}$$

Thus, we have

$$\begin{aligned} K = \bigcup _{p\in B} \{p\} \times K_p, \ \ \text {and} \ \ \hat{F}^n(K) = \bigcup _{p\in B} \{p\}\times (F_p)^n (K_p). \end{aligned}$$

Because each $F_p$ is an $\alpha$ contraction, we have for all $p\in B$, that $H(A_p, (F_p)^n(K_p)) \le \alpha ^n H(A_p,K_p) \le \alpha ^n M$. The inequality (7) then follows from (1). $\square$

Remark 1.1

If B is not a singleton, the restriction of $\hat{F}$ to $\pi _1^{-1}(B)$ need not be a contraction.

Based on Theorem 1.2 the authors defined the operator $\mathscr {F}:\mathscr {K}(X)\rightarrow \mathscr {K}(X)$ of the form

$$\begin{aligned} \mathscr {F}(K):=A_K \end{aligned}$$

and provided some of its properties.

Theorem 1.3

(Secelean and Wardowski [7]) The operator $\mathscr {F}$ is continuous and monotone.

Theorem 1.4

(Secelean and Wardowski [7]) Let $\{K_i\}$ be a sequence of subsets in $\mathscr {K}(X)$. Then

$$\begin{aligned} {\mathscr {F}}\left( \bigcup _{i=1}^nK_i\right) =\bigcup _{i=1}^n{\mathscr {F}}(K_i)\ \ \text { for every } n\ge 1. \end{aligned}$$

If, further, $K:=\overline{\bigcup _{i}K_i}\in \mathscr {K}(X)$, then

$$\begin{aligned} \overline{\bigcup _{i}{\mathscr {F}}(K_i)}\in \mathscr {K}(X) \text { and } {\mathscr {F}}(K)=\overline{\bigcup _{i}{\mathscr {F}}(K_i)}. \end{aligned}$$

Observe that we can extend Theorem 1.4 as follows.

Theorem 1.5

If $\mathscr {A} \subset \mathscr {K}(X)$ is compact, then ${\mathscr {F}}(\bigcup \mathscr {A}) = \bigcup {\mathscr {F}}(\mathscr {A})$.

Proof

Let $\{A_i\}$ be a dense sequence in $\mathscr {A}$, then $\overline{\bigcup _i A_i} = \bigcup \mathscr {A}$ and so $\{\bigcup _{i=1}^n A_i\}$ converges to $\bigcup \mathscr {A}$. By monotonicity, ${\mathscr {F}}(\bigcup \mathscr {A}) \supseteq \bigcup {\mathscr {F}}(\mathscr {A}) \supseteq \bigcup _i {\mathscr {F}}(A_i)$. By Theorem 1.4, the terms on the two ends are equal and so ${\mathscr {F}}(\bigcup \mathscr {A}) = \bigcup {\mathscr {F}}(\mathscr {A})$. $\square$

An immediate consequence of Theorem 1.5 is the following result.

Corollary 1.2

For every $K\in \mathscr {K}(X)$, ${\mathscr {F}}(K) = \bigcup _{x\in K} {\mathscr {F}}(\{x\}).$

When we consider in (6) the function

$$\begin{aligned} \Phi (r,s)=sC^r, \ r,s\ge 0, \end{aligned}$$

for some positive $C<1$, then we obtain condition (5). In [7] the authors obtained new classes of IFSs and their attractors which were based on condition (6) with two additional conditions posed on the function $\Phi$, i.e.

• (iii) $\Phi \big (r,\Phi (s,t)\big ) \le \Phi (r+s,t)$,

• (iv) $\Phi (0,t)=t$

for all $r,s,t \ge 0$. For more details the reader is referred to Theorem 3.1 and Theorem 3.2 in [7].

In the present article first we provide a fixed point result concerning some class of Lipschitz maps induced by the family of functions $\Phi$, next we continue the study about the IFSs consisting of functions satisfying orbital condition (6) and give the answer to the substantial question posed in [7] concerning the continuity dependence property.

2 $\Phi$-Contractions

Assume $\Phi :\mathbb {R}_+ \times \mathbb {R}_+ \rightarrow \mathbb {R}_+$ satisfies (i)–(iv). Observe that (i) and (ii) imply $\sum _{n=0}^{\infty } \Phi (n\delta , t) < \infty$, for all $\delta > 0$, $t\ge 0$. Also, from (i) and (ii) the following condition holds

• (v) $\lim _{u\rightarrow \infty } \Phi (u,t) = 0$ for all $t\ge 0$.

Call $f:X\rightarrow X$ a $\Phi$-contraction if there exists $\delta > 0$ such that $d(f(x),f(y))\le \Phi (\delta ,d(x,y))$ for all $x,y \in X$. From (i) and (iv) it follows that a $\Phi$-contraction is Lipschitz with constant at most 1. From (i) and (iii) we have $d(f^n(x),f^n(y)) \le \Phi (n\delta ,d(x,y))$ for all $x,y \in X$ and any positive integer n.

Theorem 2.1

If $f:X\rightarrow X$ is a $\Phi$-contraction, then it has at most one fixed point and for any x the sequence $\{f^n(x)\}$ is Cauchy. If X is complete, then the limit $x^*$ exists, is fixed by f, and

$$\begin{aligned} d(x^*,x) \le \sum _{n=0}^{\infty } \Phi (n\delta , d(f(x),x)) \ \ \text {for all} \ \ x\in X. \end{aligned}$$

(8)

In addition, the induced map $f:\mathscr {K}(X)\rightarrow \mathscr {K}(X)$ is a $\Phi$-contraction, i.e. for all $A,B\in \mathscr {K}(X)$

$$\begin{aligned} H(f(A),f(B)) \le \Phi (\delta , H(A,B)). \end{aligned}$$

(9)

Proof

If x, y are fixed points, then $d(x,y)\le \Phi (n\delta ,d(x,y))$ for all n and so from (v) $d(x,y)=0$. Since $d(f^{n+1}(x),f^n(x)) \le \Phi (n\delta ,d(f(x),x))$ for all $x\in X$ and n, the series $\sum _{n=0}^{\infty }d(f^{n+1}(x),f^n(x))$ converges and so $\{f^n(x)\}$ is Cauchy. If the limit $x^*$ exists, which it does when X is complete, then the limit is a fixed point. Since $d(f^{n+1}(x),x)\le \sum _{k=0}^n \Phi (k\delta ,d(f(x),x))$, the estimate (8) follows.

If for some $\epsilon > 0$, $A\subseteq V_{\epsilon }(B)$, then $f(A)\subseteq V_{\Phi (\delta ,\epsilon )}(f(B))$ and so the estimate (9) follows. $\square$

Corollary 2.1

If $\{f_i:X\rightarrow X\}$ is a finite family of $\Phi$-contractions, then the associated fractal operator $F:\mathscr {K}(X)\rightarrow \mathscr {K}(X)$ is a $\Phi$-contraction.

Proof

If $d(f_i(x),f_i(y)) \le \Phi (\delta _i,d(x,y))$, then with $\delta = \text {min}_i \delta _i$, (9) implies that for $A,B \in \mathscr {K}(X)$, $H(f_i(A),f_i(B))\le \Phi (\delta , H(A,B))$ for all i and so from [H3] and definition (3) it follows that

$$\begin{aligned} H(F(A),F(B))\le \Phi (\delta ,H(A,B)). \end{aligned}$$

$\square$

It is clear that if $f:X\rightarrow X$ and $g:Y\rightarrow Y$ are $\Phi$-contractions, then the product $f\times g:X\times Y\rightarrow X\times Y$ is a $\Phi$-contraction.

3 Continuity of Attractors

Consider a finite family of maps $\{f_i:P\times X\rightarrow X\}$ indexed by $\{i\in I\}$ with X being complete. For any $p\in P$, $x\in X$, and $\alpha , \beta \in \Lambda ^*(I)$, by $f_{\alpha \beta }(p,x)$ we mean $f_\alpha (p,f_\beta (p,x))$, $f_{\emptyset }(p,x):= x$. We impose the following conditions

• (C1) $f_i(p,\cdot )$ is continuous for every $p\in P$ and $i\in I$;

• (C2) $f_\omega (\cdot ,x)$ is continuous for every $\omega \in \Lambda _n(I)$, $n\in \mathbb {N}$, and $x\in X$;

• (C3) For each $p\in P$ there exists $\gamma (p) > 0$ satisfying

  $$\begin{aligned} d(f_ \omega (p,x),f_{\omega i}(p,x))\le \Phi (\gamma (p){| w|},d(x,f_i(p,x))), \end{aligned}$$

  for all $x\in X$, $i\in I$, and $\omega \in \Lambda ^*(I)$, with $\Phi$ satisfying (i)–(ii).

Observe that if each $f_i$ is jointly continuous then both (C1) and (C2) hold.

By Theorem 1.2, for every $p\in P$ and $K\in \mathscr {K}(X)$, there exists the attractor $A_K^p \in \mathscr {K}(X)$ i.e. a fixed point of the map

$$\begin{aligned} F(p,K):= \bigcup _{i} f_i(p,K) \end{aligned}$$

and $\{F^n(p,K)\}$, defined by

$$\begin{aligned} F^0(p,K)&:=K,\\ F^n(p,K)&:=F(p,F^{n-1}(p, K)),\ n=1, 2,\ldots , \end{aligned}$$

converges to $A^p_K$ with respect to the metric H.

Before we enunciate our main result, let us consider the following example, where there is considered an IFS consisting of one parametrized map.

Example 3.1

Let $f(p,x):= px$, $p\in [0,1]$, $x\in [0,\infty )$ and consider

$$\begin{aligned} \Phi (r,s):=s\lambda ^r,\ \text {for some } \lambda \in (0,1) \end{aligned}$$

with

$$\begin{aligned} \gamma (p)= \left\{ \begin{array}{ll} \log p/\log \lambda , \ {} &{}p \in (0,1), \\ 1, \ {} &{}p \in \{0,1\}. \end{array} \right. \end{aligned}$$

For $p\in \{0,1\}$, a positive integer n, and any $x\ge 0$, we have

$$\begin{aligned} |f^{n}(p,x)-f^{n+1}(p,x)|=0 \le \Phi (n,|x-f(p,x)|). \end{aligned}$$

When $p\in (0,1)$, then we get the following equalities

$$\begin{aligned} |f^{n}(p,x)-f^{n+1}(p,x)|&=p^nx-p^{n+1}x =(\lambda ^{\log p/\log \lambda })^n |x-f(p,x)|\\&= \Phi (n\gamma (p),|x-f(p,x)|). \end{aligned}$$

Now, observe that taking a sequence $\{p_k\}$ of the form

$$\begin{aligned} p_k=\frac{k}{k+1}, \ \ k = 0, 1, \ldots \end{aligned}$$

then, for any $x\ge 0$ we have

$$\begin{aligned}{} & {} f^{n}(p_k,x)=(p_k)^nx \rightarrow 0, \ \ n\rightarrow \infty \\{} & {} f^{n}(1,x) \rightarrow x, \ \ n\rightarrow \infty . \end{aligned}$$

In the above example the map f satisfies conditions (C1)–(C3) and the continuity dependence does not hold, even if f is jointly continuous with respect to both variables. Indeed, for every $x > 0$, we have

$$\begin{aligned} A^{p_k}_{\{x\}}&= \{0\} \text { for } k=0,1,\ldots ,\\ A^1_{\{x\}}&= \{x\}. \end{aligned}$$

Proposition 3.1

For every $K \in \mathscr {K}(X)$ and $n\in \mathbb {N}$ the map $F^{n}(\cdot , K)$ is continuous.

Proof

Let n be a positive integer, $\{p_k\} \subset P$, and $p_k\rightarrow p$ in P. By (1) we have

$$\begin{aligned}&H(F^{n}(p_k,K), F^{n}(p,K)) = H \left( \bigcup _{\alpha \in \Lambda _n(I)}f_\alpha (p_k,K), \bigcup _{\beta \in \Lambda _n(I)}f_\beta (p,K)\right) \\&\quad = H \left( \bigcup _{x\in K}\bigcup _{\alpha \in \Lambda _n(I)}f_\alpha (p_k,\{x\}),\bigcup _{y\in K} \bigcup _{\beta \in \Lambda _n(I)}f_\beta (p,\{y\})\right) \\&\quad \le \max _{\alpha \in \Lambda _n(I)} H \left( \bigcup _{x\in K}f_\alpha (p_k,\{x\}),\bigcup _{y\in K}f_\alpha (p,\{y\})\right) . \end{aligned}$$

Therefore, there exists $\alpha _0 \in \Lambda _n(I)$ such that

$$\begin{aligned} H(F^{n}(p_k,K), F^{n}(p,K)) \le H \left( \bigcup _{x\in K}f_{\alpha _0} (p_k,\{x\}),\bigcup _{y\in K}f_{\alpha _0} (p,\{y\})\right) . \end{aligned}$$

K is compact and $f_{\alpha _0}(p_k,\cdot )$, $f_{\alpha _0}(p,\cdot )$ are continuous (from (C1)), hence

$$\begin{aligned} H(F^{n}(p_k,K), F^{n}(p,K))\le & {} \sup _{x\in K} d(f_{\alpha _0} (p_k,x),f_{\alpha _0} (p,x)) \\= & {} d(f_{\alpha _0} (p_k,x_0),f_{\alpha _0} (p,x_0)) \end{aligned}$$

for some $x_0 \in K$. Now, using continuity of $f_{\alpha _0}(\cdot , x_0)$ (from (C2)), going with $k\rightarrow \infty$, we get

$$\begin{aligned} H(F^{n}(p_k,K), F^{n}(p,K)) \rightarrow 0. \end{aligned}$$

$\square$

Lemma 3.1

(Secelean and Wardowski [7]) Let $\Phi :\mathbb {R}_+\times \mathbb {R}_+\rightarrow \mathbb {R}_+$ be a function satisfying (i). For every fractal operator F related to the IFS $\{f_i:X\rightarrow X\}$ satisfying (6), the following condition holds

$$\begin{aligned} H(F^{n}(\{x\}),F^{n+1}(\{x\})) \le \Phi (\gamma n,H(\{x\},F(\{x\}))) \end{aligned}$$

for all $x\in X$ and any positive integer n.

Lemma 3.2

For every positive integer n, $p\in P$, and $K\in \mathscr {K}(X)$ the following inequality holds

$$\begin{aligned} H(F^{n}(p,K), F^{n+1}(p,K))\le \Phi (\gamma (p)n,C_p), \end{aligned}$$

where $C_p:= \sup _{x\in K}H(\{x\},F(p,\{x\})).$

Proof

We have

$$\begin{aligned} H(F^{n}(p,K), F^{n+1}(p,K))&= H \left( \bigcup _{x\in K}F^{n}(p,\{x\}), \bigcup _{x\in K} F^{n+1}(p,\{x\})\right) \\&\le \sup _{x\in K} H(F^{n}(p,\{x\}), F^{n+1}(p,\{x\})). \end{aligned}$$

By Lemma 3.1 and (i), we get

$$\begin{aligned} H(&F^{n}(p,K), F^{n+1}(p,K)) \le \sup _{x\in K} \Phi (\gamma (p)n, H(\{x\}, F(p,\{x\}))) \le \Phi (\gamma (p)n, C_p). \end{aligned}$$

$\square$

Now, we can formulate and prove our main result, which is an answer to the open question posed in [7].

Theorem 3.1

Assume that (C1)–(C3) hold. If

$$\begin{aligned} \gamma := \inf \{\gamma (p):p\in P\} > 0, \end{aligned}$$

(10)

then the map $P\rightarrow \mathscr {K}(X)$, $p \mapsto A^p_K$ is continuous i.e. for every $\{p_k\}\subset P$, $p\in P$, and $K \in \mathscr {K}(X)$,

$$\begin{aligned} p_k \rightarrow p \text { implies } H(A^{p_k}_K,A^{p}_K) \rightarrow 0. \end{aligned}$$

Proof

Let $K \in \mathscr {K}(X)$, $\{p_k\} \subset P$, and $p_k\rightarrow p$ in P. For every positive integers k, m, n, by Lemma 3.2, we have

$$\begin{aligned}&H(F^{m+n}(p_k,K),F^{m+n}(p,K))\le H(F^{m+n}(p_k,K),F^{m+n-1}(p_k,K))\\&+H(F^{m+n-1}(p_k,K),F^{m+n-1}(p,K))+ H(F^{m+n-1}(p,K),F^{m+n}(p,K)) \\&\le \Phi ((m+n-1)\gamma (p_k), C_{p_k}) + \Phi ((m+n-1)\gamma (p), C_p)\\&\quad +H(F^{m+n-1}(p_k,K),F^{m+n-1}(p,K))\\&\le \\&\ldots \\&\le \sum _{i=m}^{m+n-1}\Phi (i\gamma (p_k), C_{p_k}) + \sum _{i=m}^{m+n-1}\Phi (i\gamma (p), C_p) + H(F^{m}(p_k,K),F^{m}(p,K)). \end{aligned}$$

Going with $n\rightarrow \infty$ and using (10), we receive

$$\begin{aligned} H(A^{p_k}_K,A^p_K)&\le \sum _{i=m}^{\infty }\Phi (i\gamma , C_{p_k}) + \sum _{i=m}^{\infty }\Phi (i\gamma , C_p) + H(F^{m}(p_k,K),F^{m}(p,K)). \end{aligned}$$

Due to the continuity of $F(\cdot ,\{x\})$ for every $x\in X$ (Proposition 3.1), we get $C_{p_k}\rightarrow C_p$. In consequence, we can choose $C>0$ such that $C_{p_k} < C$ for every positive integer k. Therefore, we get

$$\begin{aligned} H(A^{p_k}_K,A^p_K)&\le \sum _{i=m}^{\infty }\Phi (i\gamma , C) + \sum _{i=m}^{\infty }\Phi (i\gamma , C_p) + H(F^{m}(p_k,K),F^{m}(p,K)). \end{aligned}$$

Using in the above inequality the continuity of $F^{m}(\cdot ,K)$ and (ii), we finally receive

$$\begin{aligned} H(A^{p_k}_K,A^p_K) \rightarrow 0. \end{aligned}$$

$\square$

Remark 3.1

Example 3.1 shows that condition (10) is essential.

Remark 3.2

Corollary 1.1 is a consequence of Theorem 3.1. Indeed, observe that (C1) and (C2) are satisfied since $f_i$’s are jointly continuous (see Lemma 1.1).

Now, take $x\in X$, $i\in I$, and $\omega \in \Lambda ^*(I)$. Since each $f_i(p,\cdot )$ is an $\alpha (p)$ contraction with

$$\begin{aligned} \alpha (p)< \alpha < 1, \end{aligned}$$

(11)

we get

$$\begin{aligned} d(f_ \omega (p,x),f_{\omega i}(p,x))\le \alpha (p)^{| w|}d(x,f_i(p,x)), \end{aligned}$$

which is (C3) with $\Phi (r,s) = s\lambda ^r$ for any fixed $\lambda \in (0,1)$ and $\gamma (p):= \log _{\lambda }\alpha (p) > 0$.

Finally, observe that (10) follows from (11).

Note that with stronger assumptions we can obtain the following result.

Fig. 1

The attractors $A^p_K$ from Example 3.2 in the unit square for $K=\{(1,1)\}$ and some values of parameter p

Theorem 3.2

Assume that $\{f_i:P\times X\rightarrow X\}$ is a finite family of jointly continuous maps with P and X complete and (C3) holds with $\gamma := \inf _{p\in P} \gamma (p) > 0$. Then the map $(p,K)\mapsto A_K^p$ is a continuous function from $P\times \mathscr {K}(X) \rightarrow \mathscr {K}(X)$.

Proof

The family $\{\hat{f}_i:P\times X\rightarrow P\times X\}$ satisfies

$$\begin{aligned} d(\hat{f}_w(p,x),\hat{f}_{\omega i}(p,x))\le & {} \Phi (\gamma |\omega |,d((p,x),\hat{f}_i(p,x))) \ \ \text {for all} \ \ x\in X,\ i\in I,\\{} & {} \text { and } \omega \in \Lambda ^*(I). \end{aligned}$$

With $\hat{{\mathscr {F}}}:\mathscr {K}(P\times X)\rightarrow \mathscr {K}(P\times X)$ the attractor function for this family, it follows from Theorem 1.3 that $\hat{{\mathscr {F}}}$ is continuous. The map given by $p\times K \mapsto \{p\}\times K$ is an isometric injection from $P\times \mathscr {K}(X)$ to $\mathscr {K}(P\times X)$ and the projection $\pi _2:P\times X\rightarrow X$ induces a continuous map from $\mathscr {K}(P\times X)$ to $\mathscr {K}(X)$. Since $\hat{{\mathscr {F}}}(\{p\}\times K) = \{p\}\times A_K^p$, it follows that the function defined by $(p,K)\mapsto A^p_K$ is continuous. $\square$

The following example illustrates the results obtained within this article. A slight change of parameters does not affect the given attractor too much, i.e. changes it in a continuous way.

Example 3.2

Consider the maps $f_1, f_2 :[0,1]\times [0,1]^2 \rightarrow [0,1]^2$ given by

$$\begin{aligned} f_1(p,(x,y)) = ( p(1-y^2)/(1+x^2),\ 0.7y), \end{aligned}$$

$$\begin{aligned} f_2(p,(x,y))= (x^2/(1+x^2),\ 0.8y+0.2). \end{aligned}$$

Conditions (C1) and (C2) are satisfied since both maps are jointly continuous. (C3) can be verified with $\Phi (r,s) = s/(1+rs^{1/2})^{2}$ and $\gamma = 1/\sqrt{0.8}-1$. The assumptions of Theorems 3.1 and 3.2 are fulfilled. Figure 1 illustrates the continuous change of the attractors $A^p_K$ the subsets of $\mathscr {K}([0,1]^2)$ with fixed K and p varying in [0, 1].