THE HUTCHINSON BARNSLEY THEORY FOR INFINITE ITERATED FUNCTION SYSTEMS

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Bull. Austral. Math. Soc. Vol. 72 2005) [441 454] 39b12, 47a35, 47h09, 54h25 THE HUTCHINSON BARNSLEY THEORY FOR INFINITE ITERATED FUNCTION SYSTEMS Gertruda Gwóźdź- Lukawska and Jacek Jachymski We show that some results of the Hutchinson Barnsley theory for finite iterated function systems can be carried over to the infinite case. Namely, if {F i : i N} is a family of Matkowski s contractions on a complete metric space X, d) such that F i x 0 ) is bounded for some x 0 X, then there exists a non-empty bounded and separable set K which is invariant with respect to this family, that is, K = F i K). Moreover, given σ N N and x X, the limit lim F σ 1 F σn x) exists and does not depend on x. We also study separately the case in which X, d) is Menger convex or compact. Finally, we answer a question posed by Máté concerning a finite iterated function system {F 1,..., F N } with the property that each of F i has a contractive fixed point. 1. Introduction Let {F i : i I} be a countable family of selfmaps of a complete metric space X, d), where either I = {1,..., N} for some N N, or I = N. In case in which I is finite, Hutchinson [11] proved that if all F i are Banach contractions, then the mapping FA) := i I F i A) for A X has a unique fixed point K in the hyperspace of all nonempty compact subsets of X. Subsequently, this result was popularised by Barnsley [4], and therefore, in the literature, F is usually said to be the Hutchinson Barnsley operator, whereas K is called a fractal in the sense of Barnsley associated with the iterated function system iterated function system {F i : i I}. It is worth emphasising that two different proofs of the above result were given in [11]. The first one which is the most familiar is based on an application of the Banach Contraction Principle to operator F with a use of the Hausdorff metric in the above hyperspace. Recently, this approach was extended to iterated multifunction systems by Andres and Górniewicz [3] see also [1] and [2] for further generalisations Received 19th July, 2005 We are grateful to Andrzej Komisarski for some useful discussion. Copyright Clearance Centre, Inc. Serial-fee code: 0004-9729/05 441 $A2.00+0.00.

442 G. Gwóźdź- Lukawska and J. Jachymski [2] involving weakly contractive mappings). The second Hutchinson s proof is more elementary and it uses neither the Hausdorff metric, nor the hyperspace see [11, Section 3.1]). Yet another proof based on ordering techniques is given in [14].) Also the advantage of this proof is that it gives another characterisation of a fractal. To see that, let us define first Γσ, n, x) := F σ1 F σn x) for σ I N and x X. Following Máté [15], we say that a family {F i : i I} has property P) if the limit 1) Γσ) := lim Γσ, n, x) exists for all σ I N and x X, and does not depend on x. In fact, Máté considered only the case in which I is finite.) Then Hutchinson s argument shows that a finite family of Banach contractions has property P), and the fractal K associated with this family coincides with the set ΓI N ), that is, { } K = lim F σ 1 F σn x) : σ {1,..., N} N, where x X is fixed or may vary with σ. This result was partially extended by Máté [15] in the following way. Let ϕ be an upper semi-continuous and non-decreasing function from R +, the set of all non-negative reals, into R + such that ϕt) < t for all t > 0. Let F 1,..., F N be ϕ-contractions, that is, 2) df i x, F i y) ϕ dx, y) ) for x, y X and i {1,..., N}. The class of such ϕ-contractions was introduced by Browder [6].) If, given σ {1,..., N} N and x X, the sequence Γσ, n, x) ) is bounded, then the family {F 1,..., F N } has property n N P). However, in practice the latter assumption is inconvenient for verifying unless X, d) is bounded. Our first purpose here is to show that, in fact, it can be dropped if ϕ is such that ) 3) lim sup t ϕt) =. t Moreover, the result is still true if we consider an infinite iterated function system as stated in Theorem 1, in which we also give a list of equivalent conditions for the existence of a non-empty bounded set K X such that 4) K = F i K). Following Hutchinson [11], every non-empty set K not necessarily bounded) satisfying 4) is said to be invariant with respect to {F i : i N}. Simple examples show that

[3] The Hutchinson Barnsley theory 443 a family {F i : i N} of ϕ-contractions need not possess a compact invariant set see Example 2). Also note that the limit condition 3) introduced in the metric fixed point theory by Matkowski [18] is unnecessary if X, d) is metrically convex or compact see Theorems 3 and 4, respectively). In this paper we also study the following problem see Theorem 3) posed by Máté [15]: Let each F i i = 1,..., N) have a contractive fixed point x i, that is, x i = F i x i and given x X, Fi n x) n N converges to x i. Is it true that the family {F 1,..., F N } has property P)? We answer this question in the negative; moreover, our Example 4 inspired us to find another sufficient condition for property P) see Theorem 5). That extends the following well-known result from the theory of contractive mappings: If F : X X is such that, for some p N, F p has a contractive fixed point x, then x is also a contractive fixed point of F. 2. Invariant sets of infinite families of ϕ-contractions Throughout this section we assume that ϕ: R + R + is non-decreasing and such that lim ϕ n t) = 0 for t > 0. Then it is easy to show that ϕt) < t for t > 0. Matkowski [17, Theorem 1.2] also see [9, page 15]) proved that each ϕ-contraction has a contractive fixed point. For a detailed discussion of several classes of mappings satisfying nonlinear contractive conditions, see [12]; in particular, Matkowski s theorem extends an earlier result of Browder [6]. The main result of this section is the following theorem dealing with infinite families of Matkowski s contractions. Theorem 1. Let X, d) be a complete metric space and F i : X X i N) be ϕ-contractions with ϕ independent of i N) satisfying the limit condition 3). The following statements are equivalent: i) ii) there exists an x 0 X such that F i x 0 ) is bounded; given x X, F i x) is bounded; iii) x i ) is bounded, where x i is a fixed point of F i ; iv) there exists a non-empty bounded set K X such that K = F i K). Moreover, if i) holds, then the family {F i : i N} has property P) and given x X, sequences Γσ, n, x) ) converge to Γσ) uniformly with respect to σ n N NN. Furthermore, the set K := ΓN N ) satisfies the following conditions: a) K is bounded and K = F i K ); b) if K X is bounded and K = F i K), then K K, that is, K is the greatest with respect to the inclusion ) fixed point of the Hutchinson Barnsley operator F in the hyperspace of all bounded subsets of X.

444 G. Gwóźdź- Lukawska and J. Jachymski [4] The proof of Theorem 1 will be preceded by the series of auxiliary results. Lemma 1. Let X, d) be a metric space and F i : X X i N) be Lipschitzian L i <. Assume that each F i has a with the Lipschitz constant L i such that L := sup fixed point x i not necessarily unique). Consider i), ii) and iii) of Theorem 1. Then i) ii) iii). Proof: i) ii): Given x X, we have df i x, F i x 0 ) L dx, x 0 ) for all i N. Set A := {F i x 0 : i N} and r := L dx, x 0 ). Then {F i x : i N} A r := a A Ba, r). Since A is bounded, so is A r and hence ii) holds. ii) implies i) a fortiori. iii) ii): Given x X, df i x, x i ) = df i x, F i x i ) L dx, x i ). By iii), r := L sup { dx, x i ) : i N } <, so {F i x : i N} {x i : i N} r which yields ii). Lemma 2. Let X, d) be a metric space. Assume that F i : X X i N) are continuous and each of them has a contractive fixed point x i. Consider iii) and iv) of Theorem 1. Then iv) iii). Proof: iv) implies that given i N, F i K) K. Hence and by continuity, we get F i K) F i K) K. Let x K. By hypothesis, F n i x x i as n. Since F n i x K, we infer x i K. Since, by iv), also K is bounded, so is {x i : i N}. Let s: N N N N be the shift operator, that is, sσ 1,..., σ n,...) := σ 2,..., σ n+1,...) for σ N N. Proposition 1. Let X, d) be a metric space. Assume that F i : X X i N) are continuous and such that {F i : i N} has property P). Let Σ N N be such that Σ, sσ) Σ and s 1 Σ) Σ. Then the set ΓΣ) is invariant with respect to {F i : i N}. Proof: Set K := ΓΣ). Clearly, K is non-empty. If x K, then x = Γσ) = lim F σ1 F σn x) for some σ Σ.

[5] The Hutchinson Barnsley theory 445 By property P), there exists the limit y := lim F σ2 F σn x) = Γ sσ) ). Then, by continuity of F σ1, we have x = F σ1 y). Since sσ) Σ, we infer y K and thus x F σ1 K). This yields the inclusion: K F i K). Conversely, if x F i K) for some i N, then there is a y K such that x = F i y) and y = Γσ) for some σ Σ. Then, by continuity of F i, we get x = lim F i F σ1 F σn y) = Γσ ), where σ := i, σ 1,..., σ n,...). Since σ s 1 Σ) Σ, we infer x K. Example 1. Fix a σ N N and set Σ σ := {σ N N : i,j N n N σ i+n = σ j+n}. Then Σ σ is the least element of the family { Σ N N : σ Σ, sσ) Σ and s 1 Σ) Σ }. x X, Lemma 3. sup Under the assumptions of Theorem 1, if i) holds, then, given { d x, F i1 F in x) ) } : n N, i 1,..., i n N <, that is, the semigroup generated by mappings F i is pointwise bounded on X. Proof: Let x X. Given n N, set { a n := sup d x, F i1 F in x) ) } : i 1,..., i n N. Then a 1 = sup { dx, F i x) : i N }. Since all F i are, in particular, non-expansive, Lemma 1 yields a 1 is finite. By 3), there is an M > 0 such that M ϕm) a 1. Using induction we show that a n M for all n N. Clearly, a 1 M. Let n N be such that a n M. Then, given i 1,..., i n+1 N, ) d x, F i1 F in+1 x) dx, F i1 x) + ϕ dx, F i2 F in+1 x) ) a 1 + ϕa n ) a 1 + ϕm) M, since F i1 is a ϕ-contraction, ϕ is non-decreasing and d x, F i2 F in+1 x) ) a n. Since i 1,..., i n+1 were arbitrary, we infer a n+1 M. Thus, by induction, a n ) n N is bounded which completes the proof. Proof of Theorem 1. Assume that i) holds. We show {F i : i N} has property P). Let σ N N and x X. By Lemma 3, the constant { 5) M := sup d x, F i1 F in x) ) } : n N, i 1,..., i n N

446 G. Gwóźdź- Lukawska and J. Jachymski [6] is finite. Hence, given j, k N, we have 6) d F σ1 F σj x), F σ1 F σj+k x) ) ϕ j d x, F σj+1 F σj+k x) )) ϕ j M), since F σ1,..., F σj are ϕ-contractions and ϕ is non-decreasing. Since ϕ j M) 0 as j, given ε > 0, there is an l > 0 such that ϕ j M) < ε for all j l. Hence and by 6), we infer 7) d Γσ, j, x), Γσ, j + k, x) ) < ε for all j l and k N which means Γσ, n, x) ) is a Cauchy sequence, so it converges to some y X. We n N show the limit does not depend on x. Given x X, d Γσ, n, x), Γσ, n, x ) ) ϕ n dx, x ) ) 0 as n. Thus we may set Γσ) := lim Γσ, n, x). Letting k in 7), we obtain d Γσ, j, x), Γσ) ) ε for all j l. Note that l does not depend on σ which means Γσ, n, x) ) converges to Γσ) uniformly n N with respect to σ N N. In particular, {F i : i N} has property P). Now, since all F i are continuous, Proposition 1 yields K is invariant with respect to {F i : i N}. Moreover, if x X and M is defined by 5), then given σ N N, d x, F σ1 F σn x) ) M for all n N. Letting n, we get d x, Γσ) ) M which yields the boundedness of K since M does not depend on σ. Thus K satisfies a); in particular, the above argument shows that i) iv) holds. Hence and by Lemmas 1 and 2, we infer i), ii), iii) and iv) are equivalent. Finally, we prove b). Assume that K is a non-empty bounded subset of X and K = F i K). Let x K. Then there are σ 1 N and x 1 K such that x = F σ1 x 1. Similarly, x 1 = F σ2 x 2 for some σ 2 N and x 2 K. Continuing in this fashion, we get sequences σ N N and x n ) n N such that x n K and x = F σ1 F σn x n ) for all n N. Set y n := Γσ, n, x). Then dx, y n ) = d F σ1 F σn x n ), F σ1 F σn x) ) ϕ n dx, x n ) ) ϕ n diam K) 0 as n. Hence y n x, that is, x = Γσ) which means x K. This yields K K. Theorem 1 implies the following somewhat surprising property of the Hutchinson Barnsley operator F.

[7] The Hutchinson Barnsley theory 447 Corollary 1. Let H b X) denote the hyperspace of all non-empty bounded subsets of X. Under the assumptions of Theorem 1, consider the operator FA) := F i A) for A X. The following statements are equivalent: i) ii) F has a fixed point in H b X); F maps H b X) into H b X). Proof: i) ii): By Theorem 1 iv) iii)), x i ) is bounded. Set B := {x i : i N}. Let A H b X) and x A. Given i N, we have df i x, x i ) = df i x, F i x i ) ϕ dx, x i ) ) ϕr), where r := diam A B). Clearly, r is finite and F i x B ϕr) which yields FA) B ϕr). Thus FA) is bounded. ii) i) follows immediately from Theorem 1 since ii) implies that F i x) is bounded whenever x X. The following example shows that an infinite iterated function system need not have a compact invariant set even if X, d) is compact. Moreover, a bounded invariant set need not be unique. Example 2. Let X := [0, 1] and F i x := x/2 + 1/2 i for i N and x X. Suppose, on the contrary, there exists a non-empty compact set K X such that K = F i K). Then F i K) K, so by the Contraction Principle, F i has a fixed point in K which means K {1/2 i 1 : i N}. Hence 0 K since K is closed. However, [1/2 i, 1/2 i + 1/2] = 0, 1], F i K) F i X) = so 0 F i K) which yields a contradiction. On the other hand, by Theorem 1, the above family has bounded invariant sets as, for example, 0, 1] = ΓN N )) or 0, 1] Q. Though ΓN N ) need not be compact, it is always separable as shown below. Lemma 4. Endow N with the discrete topology, and the product N N with the Tychonoff topology τ T. Under the assumptions of Theorem 1, if i) holds, then Γ is continuous from N N, τ T ) into X, d). Proof: Let σ N N and ε > 0. By Theorem 1, K = ΓN N ) ) is bounded and invariant. Let x K. There is a p N such that ϕ p diam K ) < ε. Set U := {σ 1 } {σ p } N N.

448 G. Gwóźdź- Lukawska and J. Jachymski [8] Then U is a neighbourhood of σ. If σ U and n N, then d Γσ, p + n, x), Γσ, p + n, x) ) = d F σ1 F σp F σp+n x), F σ1 F σp F σ p+1 F σ p+n x) ) ϕ p d F σp+1 F σp+n x), F σ p+1 F σ p+n x) )) ϕ p diam K ) < ε, because σ i = σ i for i = 1,..., p; moreover, since x is in K, so are F σp+1 F σp+n x) and F σ p+1 F σ p+n x). Letting n, we obtain which yields the continuity of Γ. Corollary 2. d Γσ), Γσ ) ) ε Under the assumptions of Theorem 1, if i) holds, then the set ΓN N ) is bounded, separable and invariant with respect to {F i : i N}. Proof: In virtue of Theorem 1, we only need to show ΓN N ) is separable. This follows immediately from Lemma 4 and [8, 1.4.11 and 2.3.16] since Γ is continuous and N N, τ T ) is separable. As a consequence of Theorem 1, we get the following extension of [15, Theorem 2] by omitting the assumption of boundedness of Γσ, n, x) ) n N. Theorem 2. Let X, d) be a complete metric space and N N. Let F 1,..., F N be ϕ-contractions on X with ϕ satisfying 3). Then {F 1,..., F N } has property P), and the set Γ{1,..., N} N ) is compact and invariant with respect to this family. Proof: Set F i := F 1 for i > N. Then F i ) satisfies the assumptions of Theorem 1 including i), so it has property P). In particular, Γσ) is well defined for all σ {1,..., N} N which means {F 1,..., F N } has property P). Moreover, if σ N N and we define σ i := σ i if σ i {1,..., N}; σ i := 1 otherwise, then it follows from the definition of F i ) that Γσ) = Γσ ). Consequently, K = ΓN N ) = Γ {1,..., N} N). Since, by Tychonoff s theorem, {1,..., N} N is a compact subset of N N and, by Lemma 4, Γ is continuous, we infer K is compact. Also, by Theorem 1, K = F i K ) which yields K = N F i K ). The following theorem illuminates connections between invariant sets of an infinite iterated function system {F i : i N} and invariant sets of its finite subfamilies {F 1,..., F n } n N). Proposition 2. Let the assumptions of Theorem 1 be satisfied including condition i). For n N, let K n = Γ {1,..., n} N)) be a compact invariant set with respect to {F 1,..., F n }. Then the set K := K n is bounded and invariant with respect to n N {F i : i N}. i=1

[9] The Hutchinson Barnsley theory 449 Proof: Set and Σ := Σ n. Then n N Σ n := {1,..., n} N for all n N, ΓΣ) = n N ΓΣ n ) = n N K n = K, so K ΓN N ). Since, by Theorem 1, ΓN N ) is bounded, so is K. In virtue of Proposition 1 it suffices to show that sσ) Σ and s 1 Σ) Σ. Clearly, sσ n ) Σ n for n N which yields sσ) Σ. Now assume σ s 1 Σ), that is, sσ) Σ n for some n N. If σ 1 n, then σ Σ n ; otherwise, σ Σ σ1. Thus in both cases we have σ Σ. That means s 1 Σ) Σ. Now we discuss a question whether condition i) of Theorem 1 is necessary for property P). Also one may ask if the assumption that all F i are ϕ-contractions with ϕ independent of i is essential. It turns out that for some families of mappings both conditions are also necessary for property P) as shown in the following Example 3. Let X := R be endowed with the Euclidean metric. Let α i R \ {0} and β i R i N). Set F i x := α i x + β i for x X and i N. We show that {F i : i N} has property P) if and only if α := sup { α i : i N } < 1 and β := sup { β i : i N } <, that is, F i are Banach α-contractions satisfying i) of Theorem 1. The sufficiency part follows of course from Theorem 1. So assume {F i : i N} has property P). Suppose, on the contrary, α 1. Then, given n N, 2 n /2 n + 1) < α, so there is a σ n N such that It is easy to verify that α σn 2 n /2 n + 1). 8) F σ1 F σn x) = α σ1... α σn x + β σ1 + n α σ1... α σj 1 β σj. Denote a n := α σ1... α σn, b n := β σ1 + n α σ1... α σj 1 β σj. By property P), F σ1 j=2 F σn 0) Γσ), that is, b n Γσ), and F σ1 F σn 1) Γσ), that is, a n +b n Γσ). Hence a n 0. On the other hand, a n since n k=1 n k=1 j=2 2 k /1 + 2 k ) which yields a contradiction 2 k /1 + 2 k ) 0. Thus α < 1. Now suppose, on the contrary, β =. By induction we define a sequence σ n ) n N. Set σ 1 := 1. For n 2, take σ n such that β σn 1/ α σ1... α σn 1.

450 G. Gwóźdź- Lukawska and J. Jachymski [10] Then the series α σ1... α σn 1 β σn diverges. On the other hand, F σ1 F σn 0) n=2 Γσ) which, by 8), implies the convergence of the above series; a contradiction. Thus β is finite. Remark 1. Note that the assumption that F i of Example 3 are not constant functions is essential. Indeed, any family of constant functions does have property P) though it need not satisfy i) of Theorem 1. So, in general, condition i) is not necessary for P). 3. Invariant sets in Menger convex spaces and compact spaces Recall that X, d) is said to be metrically convex or Menger convex see, for example, [5]) if given x, y X, x y, there is a z X such that x z y and dx, y) = dx, z)+dz, y). In this case, with the help of Matkowski s [19] result, the assumptions of Theorem 1 may be weakened in the following way. Theorem 3. Let X, d) be a complete metrically convex space and F i : X X i N) be ϕ-contractions, where ϕ: R + R + is such that ϕt) < t for t > 0. Then the assertion of Theorem 1 holds. Proof: By hypothesis, we have df i x, F i y) ϕ dx, y) ) for x, y X and i N. Without loss of generality, we may assume ϕ0) = 0. Then ϕ is continuous at 0, so by [19, Proposition 3], there is an increasing concave function γ : R + R + such that γt) < t for t > 0 and df i x, F i y) γ dx, y) ) for x, y X and i N. Then γ is continuous and lim γ n t) = 0 see, for example, [6]). Moreover, by [10, Theorem 7.2.5], γ is subadditive. Using [10, Theorem 7.6.1] we infer γt) { γt) } lim = inf : t > 0 γ1) < 1 t t t which yields lim t t γt) ) =. Thus we may apply Theorem 1 substituting there γ for ϕ. In the same way we could restate Theorem 2 for finite iterated function system. Moreover, in case in which X, d) is compact, we need not use a function ϕ at all letting F i be Edelstein s [7] contractions, according to the following Theorem 4. Let X, d) be a compact metric space, N N and F i : X X ) i {1,..., N} be such that df i x, F i y) < dx, y) for all x, y X, x y. Then {F 1,..., F N } has property P), and there exists a compact and invariant set with respect to this family.

[11] The Hutchinson Barnsley theory 451 Proof: By [13, Proposition 1], given i {1,..., N}, there is a subadditive and non-decreasing function ϕ i : R + R + such that ϕ i t) < t for all t > 0 and df i x, F i y) ϕ i dx, y) ) for x, y X. Set ϕt) := max { ϕ 1 t),..., ϕ N t) } for t R +. Clearly, the above properties of ϕ i carry ) over to ϕ. Thus repeating the argument of the preceding proof, we infer lim t ϕt) t =. Now the assertion follows directly from Theorem 2. Remark 2. Though the assumptions of Theorem 3 on function ϕ are essentially weaker than those of Theorem 1, both theorems deal, in fact, with the same class of mappings as can be deduced from the proof of Theorem 3. The same comment concerns relations between Theorems 2 and 4. 4. Around Máté s question Example 4 given below settles in the negative a question posed by Máté [15]. Following [15], we say that a mapping F : X X has property Q) if, for every x X, there exists the limit lim F n x and does not depend on x. Example 4. Let X := [0, 1] be endowed with the Euclidean metric. Set F x := x + 1/2 if x [0, 1/2); F x := 1 if x [1/2, 1], Gx := 0 if x [0, 1/2); Gx := x 1/2 if x [1/2, 1]. Clearly, F and G are continuous selfmaps of X. Since F 2 and G 2 are constant F 2 1, G 2 0), we infer F and G have property Q). We show, however, {F, G} has no property P). Moreover, sequences Γσ, n, x) ) need not converge. To see it, set n N σ 2j 1 := 1 and σ 2j := 2 for j N. Then, given x X, Γσ, 2n, x) = F G) n x) = F Gx), because F Gx) = 1/2 if x [0, 1/2); F Gx) = x if x [1/2, 1], so F G is an involution. On the other hand, Γσ, 2n + 1, x) = F G F ) n x) = F G F x) since G F x) = x if x [0, 1/2); G F x) = 1/2 if x [1/2, 1]. Hence we get Γσ, 2n + 1, x) = x + 1/2 if x [0, 1/2); Γσ, 2n + 1, x) = 1 if x [1/2, 1]. Thus we may conclude that Γσ, n, x) ) is convergent if and only if x = 0 or x = 1. n N So P) is not valid for the family {F, G} though Q) holds for each of these mappings.

452 G. Gwóźdź- Lukawska and J. Jachymski [12] Observe that the essence of Example 4 lies in the fact that F 2 and G 2 are Banach contractions whereas F G and G F are not contractive. Indeed, if both these compositions had been Banach contractions, then P) would have been valid for {F, G}. That can be deduced from the following more general result mentioned by Máté [16] only for families of affine selfmaps of R n ): If F 1,..., F N are selfmaps of a complete metric space X, d) such that for some p N all mappings from the family 9) F p := { F i1 F ip : i 1,..., i p {1,..., N} } are Banach contractions, then {F 1,..., F N } has property P). Note that, by the Hutchinson [11] theorem, the above assumption implies P) is valid for F p. It turns out that the latter condition is also sufficient for property P) of {F 1,..., F N }, according to the following Theorem 5. Let X, d) be a metric space not necessarily complete), N N and F 1,..., F N be selfmaps of X. If, for some p N, F p defined by 9) has property P), then so does {F 1,..., F N }. Moreover, Γ{1,..., N} N ) = Γ p {1,..., N p } N), where Γ p is defined below by 10). Proof: Clearly, F p is finite. Denote its elements by G i, i {1,..., N p }. Given σ {1,..., N p } N, n N and x X, set 10) Γ p σ, n, x) := G σ1 G σn x). Now let σ {1,..., N} N and x X. Then, for each n N, Γσ, pn, x) = F σ1 F σp ) F σpn 1)+1 F σpn )x) = G σ 1 G σ n x), where σ {1,..., N p } N is such that G σ i = F σpi 1)+1 F σpi Γσ, pn, x) = Γ p σ, n, x). Since P) is valid for F p, we infer for i N. Hence we get lim Γσ, pn, x) = Γ pσ ). Since this limit does not depend on x, we may conclude substituting successively F 1 x,..., F N x for x that 11) ε>0 k N n k j {1,...,N} d Γσ, pn, F j x), Γ p σ ) ) < ε. Fix an ε > 0. By 11), there is a k N such that if n k, then the inequality of 11) holds for all j {1,..., N}; in particular, for j := σ pn+1, we have d Γσ, pn, F σpn+1 x), Γ p σ ) ) < ε. Since Γσ, pn, F σpn+1 x) = Γσ, pn + 1, x), the above argument shows lim Γσ, pn + 1, x) = Γ pσ ).

[13] The Hutchinson Barnsley theory 453 Continuing in this fashion, we obtain lim Γσ, pn + k, x) = Γ pσ ) for all k {0, 1,..., p 1} which yields lim Γσ, n, x) = Γ p σ ). That means {F 1,..., F N } has property P). Moreover, the above argument shows Γ{1,..., N} N ) Γ p {1,..., N p } N). Since the opposite inclusion is obvious, the proof is completed. Remark 3. Since property P) of singleton {F } is equivalent to property Q) of F, Theorem 5 yields the result stated in the last sentence of the introduction. As an immediate consequence of Theorems 2 and 5, we obtain the following Corollary 3. Let X, d) be a complete metric space and N N. Let F 1,..., F N be selfmaps of X not necessarily continuous) such that, for some p N, all compositions F i1 F ip i1,..., i p {1,..., N} ) are ϕ-contractions with ϕ as in Theorem 2. Then {F 1,..., F N } has property P), and there exists a compact and invariant set with respect to this family. To illuminate the assumptions of Corollary 3, consider the following Example 5. Let X := R be endowed with the Euclidean metric. Let a be any irrational number, and D denote the Dirichlet function. Set F 1 := D and F 2 := D + a. It is easily seen that the assumptions of Corollary 3 are satisfied with p = 2 since all compositions F i F j i, j {1, 2} ) are constant. On the other hand, Theorem 2 is not applicable here since F 1 and F 2 are discontinuous. So Corollary 3 does extend Theorem 2. References [1] J. Andres and J. Fišer, Metric and topological multivalued fractals, Internat. J. Bifur. Chaos Appl. Sci. Engrg. 14 2004), 1277 1289. [2] J. Andres, J. Fišer, G. Gabor and K. Leśniak, Multivalued fractals, Chaos Solitons Fractals 24 2005), 665 700. [3] J. Andres and L. Górniewicz, On the Banach contraction principle for multivalued mappings, in Approximation, optimization and mathematical economics Pointe-à Pitre) Physica, Heidelberg, 2001), pp. 1 23. [4] M.F. Barnsley, Fractals everywhere Academic Press, New York, 1988). [5] L.M. Blumenthal, Theory and applications of distance geometry Clarendon Press, Oxford, 1953). [6] F.E. Browder, On the convergence of successive approximations for nonlinear functional equation, Indag. Math. 30 1968), 27 35.

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