RECURRENT ITERATED FUNCTION SYSTEMS

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1 RECURRENT ITERATED FUNCTION SYSTEMS ALEXANDRU MIHAIL The theory of fractal sets is an old one, but it also is a modern domain of research. One of the main source of the development of the theory of fractal sets is the discovery of a new types of fractal sets. One of the well-known examples are the attractors of iterated function systems. There are several generalization of iterated functions systems: directed-graph iterated function systems, random iterated function systems and so on. The aim of the paper is to present a generalization of iterated function systems (IFS) and of their attractors, the recurrent iterated function systems. AMS 2000 Subject Classification: 28A80. Key words: recurrent iterated function system, fractal set, Lipschitz function, contraction, autosimilar set. 1. INTRODUCTION In this section we review concepts, notation and basic results concerning iterated function systems (IFS). Details can be found in [1], [2] or [3]. Let (X, d X ) and (Y, d Y ) be two metric spaces and C(X, Y ) the set of continuous functions from X in to Y. On C(X, Y ) we will consider the generalized metric d : C(X, Y ) C(X, Y ) R + = R + {+ } defined by d(f, g) = sup d Y (f(x), g(x)). x X s For a sequence (f n ) n of elements from C(X, Y ) and f C(X, Y ), f n f u.c denotes pointwise convergence, f n f uniform convergence on compact sets, u and f n f uniform convergence (that is, convergence in the generalized metric d). For a function f : X X we denote by Lip(f) R + the Lipschitz constant associated with f, that is, Lip(f) = sup x,y X; x y d(f(x), f(y)). d(x, y) A function f is said to be a Lipschitz function if Lip(f) < + and a contraction if Lip(f) < 1. REV. ROUMAINE MATH. PURES APPL., 53 (2008), 1, 43 53

2 44 Alexandru Mihail 2 Let Con(X) = {f : X X Lip(f) < 1}. Let (X, d) be a metric space and K(X) the set of nonvoid compact subsets of X. Then K(X) with the Hausdorff-Pompeiu distance h : K(X) K(X) [0, ) defined by h(e, F ) = max(d(e, F ), d(f, E)) = min{r E B(F, r) and F B(E, r)}, where d(e, F ) = sup d(x, F ) = sup ( inf d(x, y)), is a metric space. x E x E y F (K(X), h) is a complete metric space if (X, d) is a complete metric space. (K(X), h) is compact if (X, d) is compact (see [4]). The Hausdorff-Pompeiu distance can be extended to a semidistance on P (X) with values in R + with the same definition, where P (X) is the set of nonvoid subsets of X. Concerning the Hausdorff-Pompeiu distance we have the following important properties. Proposition 1.1. Let (X, d) be a metric space, f : X X a Lipschitz function and K(X) the set of compact subsets of X. If (H i ) i I and (K i ) i I are two families of subsets of X, then 1) h( H i, K i ) = h( H i, K i ) sup h(h i, K i ); i I i I i I i I i I 2) h(f(k), f(h)) Lip(f)h(K, H) for any K, H K(X). An iterated function system, (IFS for short) consists of a metric space (X, d) and a finite family of contractions f k : X X, k {1, 2,..., n}, n N.We denote such an IFS by S = (X, (f k ) ). Its contraction constant is c S = max Lip(f k ). There are three important mathematical results concerning contractions used in the IFS theory. Theorem 1.1 (Banach Contraction Principle). Let (X, d X ) be a complete metric space and f Con(X). Then there exists a unique α X such that f(α) = α. Moreover, lim n d X(f [n] (x), α) = 0 for any x X. Theorem 1.2 (Continuity Theorem). Let (X, d X ) be a complete metric space, f, g Con(X), α the fixed point of f and β the fixed point of g. Then 1 d X (α, β) d(f, g) 1 min{lip(f), Lip(g)}. Theorem 1.3 (Collage Theorem). Let (X, d X ) be a complete metric space, f Con(X) and x X. Let α X be the unique fixed point of f.

3 3 Recurrent iterated function systems 45 Then max d X (x, α) d X(f(x), x) 1 Lip(f). With an IFS S = (X, (f k ) ) with contraction constant c S = Lip(f k ) one can associate the function F S : K(X) K(X) defined by F S (B) = n f k (B), B K(X), which is a c S -contraction. Therefore, by the Banach contraction principle, we have Theorem 1.4. Let (X, d X ) be a complete metric space and S = (X, (f k ) ) an IFS with c S = max Lip(f k ) < 1. Then there exists a unique set A(S), such that F S (A(S)) = A(S), and for every H 0 K(X) the sequence (H n ) n, where H n+1 = F S (H n ), n N, is convergent to A(S) in K(X). As for the speed of convergence we have for all n N. In particular, h(h n, A(S)) cn S 1 c S h(h 0, H 1 ), h(h 0, A(S)) 1 1 c S h(h 0, H 1 ). We also have the following result concerning the continuity of the attractor of an IFS. Theorem 1.5. Let (X, d X ) be a complete metric space and x X. Let S = (X, (f k ) ) and S = (X, (g k ) ) be two iterated function systems with contraction constants c S and c S. Then h(a(s), A(S )) max 1 d(f k, g k ) 1 min(c S, c S ). We present now the idea of our generalization. Instead of considering contractions f from X in to X in the definition of an IFS, we shall consider contractions from X 2 = X X in to X. We shall call recurrent iterated function systems (RIFS for short) this kind of iterated function systems. On X 2 we consider the distance d 2 : X 2 X 2 [0, + ) given by d 2 ((x, x 1 ), (y, y 1 )) = max{d(x, y), d(x 1, y 1 )}. Definition 1.1. Let (X, d) be a complete metric space. For a function f : X X X the number Lip(f) = d(f(x, x 1 ), f(y, y 1 )) sup x,y,x 1,y 1 X; x y or x 1 y 1 max{d(x, y), d(x 1, y 1 )} is called the Lipschitz constant of f. We have Lip(f) [0, + ].

4 46 Alexandru Mihail 4 A function f : X X X is said to be a Lipschitz function if Lip(f) < and a contraction if Lip(f) < 1. Let Con 2 (X) = {f : X X X Lip(f) < 1}. Definition 1.2. Let (X, d) be a complete metric space. A recurrent iterated function systems (RIFS for short) on X consists of a finite family of contractions (f k ), n N, with f k : X X X, and is denoted by S = (X, (f k ) ). The contraction constant of such an RIFS is c S = max Lip(f k ). 2. MAIN RESULT In this section we prove the existence of the attractor of an RIFS and also present its main properties. Definition 2.1. Let (X, d) be a metric space and f : X X X a continuous function. The function F f : K(X) K(X) K(X) defined by F f (K, H) = f(k, H) = f(k H) = {f(x, y) x K and y H} is called the set function associated with f. Let S = (X, (f k ) ) be an RIFS. The function F S : K(X) K(X) K(X) defined by F S (K, H) = n f k (K, H) is called the set function associated with S. The next result is obvious. Lemma 2.1. If (X, d X ) is a complete metric space f n, f : X X u X are continuous functions, n N, such that f n f and H, K K(X), then lim f n(h, K) = f(h, K) in K(X). n The next result is a consequence of Lemma 2.1 and Proposition 1.1. Lemma 2.2. Let (X, d X ) be a complete metric space. Let S m = (X, (fk m) ) and S = (X, (f k) ), m N, be recurrent iterated func- u f k on a dense set in X as m +, k {1, 2,..., n}, tion systems. If fk m then lim F S n n (K, H) = F S (K, H) for every H, K K(X). Lemma 2.3. Let f : X X X be a Lipschitz function. Then Lip(F f ) Lip(f). Proof. Let K, H, K 1, H 1 K(X) and r = max(h(k, K 1 ), h(h, H 1 )). Then K B(K 1, r) and H B(H 1, r). Let x K and y H. There are

5 5 Recurrent iterated function systems 47 x 1 K 1 and y 1 H 1 such that d(x, x 1 ) < r and d(y, y 1 ) < r. Then d(f(x 1, y 1 ), f(x, y)) Lip(f) max{d(x, x 1 ), d(y, y 1 )} < Lip(f)r. This means that f(x, y) B(f(x 1, y 1 ),Lip(f)r) B(f(K 1, H 1 ),Lip(f)r) which implies that f(k, H) B(f(K 1, H 1 ), Lip(f)r). In a similar way, we obtain the inclusion f(k 1, H 1 ) B(f(K, H), Lip(f)r). This means that h(f(k, H), f(k 1, H 1 )) Lip(f)r. The next result is a key one. Lemma 2.4. Let S =(X, (f k ) ) be an RIFS with c S = max Lip(f k ) < < 1. Then the set function F S : K(X) K(X) K(X) associated with S, is a c S contraction. Proof. We have ( n h(f S (K, H), F S (K 1, H 1 )) = h f k (K, H), ) f k (K 1, H 1 ) maxh(f k (K, H), f k (K 1, H 1 )) c S max(h(k, K 1 ), h(h, H 1 )). For a contraction of type f : X X X we have results similar to those of type f : X X. We shall sketch the proof and present some applications. Theorem 2.1 (Banach Contraction Principle). Let (X, d) be a complete metric space and f Con 2 (X) with contractivity factor c [0, 1). Then there exists a unique α X such that f(α, α) = α. Moreover, for any x 0, x 1 X the sequence (x n ) n 0 defined by x n+1 = f(x n, x n 1 ), n N, is convergent to α. As for the speed of convergence of (x n ) n 0 we have In particular, d(x n, α) 2c[n/2] 1 c max{d(x 0, x 1 ), d(x 1, x 2 )}. d(x 0, α) 2 1 c max{d(x 0, x 1 ), d(x 1, x 2 )}. Proof. Let z n = max{d(x n+1, x n ), d(x n, x n 1 )}. We have d(x n+1, x n ) = d(f(x n, x n 1 ), f(x n 1, x n 2 )) c max{d(x n 1, x n 2 ), d(x n, x n 1 )} = cz n 1 and d(x n+2, x n+1 ) = d(f(x n+1, x n ), f(x n, x n 1 )) c max{d(x n+1, x n ), d(x n, x n 1 )} c max{cz n 1, z n 1 } = cz n 1. Then z n+1 cz n 1. It follows that d(x n+1, x n ) c [n/2] z 1

6 48 Alexandru Mihail 6 and d(x n+p, x n ) d(x n+p, x n+p 1 ) + d(x n+p 1, x n+p 2 ) + + d(x n+1, x n ) hence c [(n+p 1)/2] z 1 + c [(n+p 2)/2] z c [n/2] z 1, d(x n+p, x n ) 2c[n/2] z 1 1 c for every p N. The rest of the proof goes in the same manner as that for the Banach contraction principle. Using Lemma 2.4 in the case of an RIFS, Theorem 2.1 becomes Theorem 2.2. Let (X, d X ) be a complete metric space and S = (X, (f k ) ) an RIFS with contraction constant c S = max Lip(f k ) < 1. Then there exists a unique A(S) K(X) such that F S (A(S), A(S)) = A(S). Moreover, for any H 0, H 1 K(X), the sequence (H n ) n 1 defined by H n+1 = F S (H n, H n 1 ), n N, is convergent to A(S). As for the speed of convergence we have h(h n, A(S)) 2c[n/2] S max{h(h 0, H 1 ), h(h 2, H 1 )}. 1 c S In particular (Collage Theorem), h(h 0, A(S)) 2 1 c S max{h(h 0, H 1 ), h(h 2, H 1 )}. We remark that A(S) = f k (A(S), A(S)) = = x A(S) f k (A(S), {x}) = x A(S) x,y A(S) f k ({x}, A(S)) = f k ({x}, {y}). Theorem 2.3 (Continuity Theorem). Let (X, d X ) be a complete metric space and f, g Con 2 (X) with fixed points α and β and contraction constants c f and c g. Then d X (α, β) min { 1 1 c f d X (f(β, β), g(β, β)), 1 d(f, g). 1 min(c f, c g ) 1 } d X (f(α, α), g(α, α)) 1 c g

7 7 Recurrent iterated function systems 49 Proof. We have d X (α, β) = d X (f(α, α), g(β, β)) d X (f(α, α), f(β, β)) + d X (f(β, β), g(β, β)) whence Similarly, we have c f d X (α, β) + d X (f(β, β), g(β, β)) c f d X (α, β) + d(f, g), d X (α, β) 1 1 c f d X (f(β, β), g(β, β)) 1 1 c f d(f, g). d X (α, β) 1 1 c g d X (f(α, α), g(α, α)) 1 1 c g d(f, g) and the conclusion follows. Using Lemma 2.4 in the case of an RIFS, Theorem 2.3 becomes Theorem 2.4 (Continuity Theorem). Let (X, d X ) be a complete metric space, and S = (X, (f k ) ) and S = (X, (g k ) ) be two RIFS with c = max Lip(f k ) < 1 and c = max Lip(g k ) < 1. Then h(a(s), A(S )) 1 1 min(c, c ) max d(f k, g k ). Another version of the continuity theorem is Theorem 2.5. Let (X, d X ) be a complete metric space and a function f : X X X. Let us consider f n Con 2 (X), where n N, with fixed point s α n. Assume c = sup c fn < 1, where c fn = Lip(f n ), and f n f on a dense n 1 set in X. Then f Con 2 (X), Lip(f) c and if α is the fixed point of f, then lim α n = α. n Proof. It follows from the hypotheses that f n u.c f on X X and c f = Lip(f) c < 1. Hence f is a contraction. By Theorem 2.3 we have d X (α, α n ) 1 1 c d X(f n (α, α), f(α, α)). Using Lemma 2.4 in the case of an RIFS, Theorem 2.5 becomes Theorem 2.6. Let (X, d X ) be a complete metric space, n N, and S m = (X, (fk m) ), m N, and S = (X, (f k ) ) recurrent iterated function systems with contraction constants c n and c 0. If sup c n = c < 1 and f m k s f k on a dense set in X X, k {1, 2,..., n}, then lim A(S n) = A(S). n n 1

8 50 Alexandru Mihail 8 3. EXAMPLES AND REMARKS Let (X, d X ) be a complete metric space, S = (X, (f k ) ) an RIFS, A(S) K(X) the attractor of S and (H n ) n N a sequence defined by H n+1 = F S (H n, H n 1 ), n N, which is convergent to A(S). We make some remarks on the monotony of the sequence (H n ) n N. If H 0 H 1 H 2, then H n 1 H n, for all n N, and A(S) = H n. In particular, if H F S (H, H) then H A(S). Let A = {x 0 there exists k such that f k (x 0, x 0 ) = x 0 } be the set of fixed points of the functions (f k ). Then A f(a, A), and so A A(S). In other words, every fixed point of a function from S is in A(S). If H 0 H 1 H 2, then H n 1 H n, for all n N, and A(S) = H n. Proposition 3.1. Let (X, d X ) be a complete metric space, S = (X, (f k ) ) a recurrent iterated function system with contraction constant c, and H K(X). If H B(F S (H, H), r) then H B ( A(S), r 1 c). Proof. Let (H n ) n N be the sequence defined by H n+1 = F S (H n, H n ), n N, and H 0 = H. As in Lemmas 2.3 and 2.4, we have by induction that H n B(H n+1, rc n ), so H B (H n, r n 1 c ). k Since H n A(S) and lim r n 1 c k = n k=0 r r 1 c, we get H B(A(S), k=0 1 c ). Proposition 3.2. Let (X, d X ) be a complete metric space, S = (X, (f k ) ) an RIFS, x 1, x 2,..., x l, y 1, y 2,..., y m A(S) and t : {1, 2,..., l + m} {1, 2,..., n} a fixed function. Let S 1 = (X, (g k ),l+m ) be the IFS defined by { ft(k) (x k, x) if k n g k (x) = f t(k) (x, y k n ) if k n + 1. Then A(S 1 ) A(S). Proof. We have g k (A(S)) = f t(k) (x k, A(S)) A(S) if k n and g k (A(S)) = f t(k) (A(S), y k n ) A(S) if k n + 1. Hence n 1 n 1 It follows that F S1 (A(S)) = g k (A(S)) A(S). l+m A(S 1 ) A(S).

9 9 Recurrent iterated function systems 51 Let (X, d X ) be a metric space and a function. f : X X X. Let us consider the function (f) : X X defined by (f)(x) = f(x, x), x X. For an RIFS S = (X, (f k ) ), let us define (S) = (X, ( (f k )) ). If S has contraction constant c, then (S) has a contraction constant less than c. Proposition 3.3. Let (X, d X ) be a complete metric space and S = (X, (f k ) ) an RIFS. Then A( (S)) A(S). Proof. It is easy to see that F (S) (H) F S (H, H) for every H K(X). Indeed F (S) (H) = (f k )(H) = {f(x, x) x H} = = {f(x, y) x H and y H} = f k (H, H) = F S (H, H). Also, for H, K K(X) such that H K we have F (S) (H) F S (K, K) since F (S) (H) F (S) (K) F S (K, K). Let H K(X) be fixed. Consider the sequences (H n ) n N and (K n ) n N defined by H 0 = H, K 0 = H, H n+1 = F S (H n, H n ), and F (S) (K n ) = K n+1, where n N. It is clear that K 0 H 0. If K n H n then K n+1 = F (S) (K n ) F S (H n, H n ) = H n+1, for all n N. Since lim n K n = A( (S)) and lim n H n = A(S), we have A( (S)) A(S). Example 1. Every IFS can be seen as an RIFS. Indeed, let S = (X, (f k ) ) be an IFS. Let S = (X, (f k ) ) be the RIFS defined by f k (x, y) = f k (x). Then A(S) = A(S). Example 2. Let X = R and let f, g : R R R be defined by f(x, y) = x 3 + y 3 and g(x, y) = x Let the RIFS S = (R, (f, g)), and let F S : K(R) K(R) K(R) be defined by F S (K, H) = f(k, H) g(k, H). The attractor of S is [0, 1]. Indeed f([0, 1], [0, 1]) = [0, 2/3] [0, 1] and g([0, 1], [0, 1]) = [2/3, 1] [0, 1], so that It follows that A(S) = [0, 1]. F S ([0, 1], [0, 1]) = [0, 1]. Example 3. Let X = R and let f, g : R R R be defined by f(x, y) = x 5 + y 5 and g(x, y) = x 5 + y Let the RIFS S = (R, (f, g)), and let F : K(R) K(R) K(R) be defined by F S (K, H) = f(k, H) g(k, H). The attractor of S is [0, 2/5] [3/5, 1].

10 52 Alexandru Mihail 10 Indeed f([0, 1], [0, 1]) = [0, 2/5] [0, 1] and g([0, 1], [0, 1]) = [3/5, 1] [0, 1], so that F S ([0, 1], [0, 1]) = [0, 2/5] [3/5, 1] [0, 1]. It follows that A(S) [0, 1] and A(S) = H n, where (H n ) n 1 is the sequence defined by H n+1 = f(h n, H n ), n N, and H 0 = [0, 1]. We remark that f(x, y) = f(y, x), g(x, y) = g(y, x), F (K, H) = F (H, K), g(x, y) = f(x, y) and g(k, H) = f(k, H) Let also t : R R be defined by t(x) = x Then (t t)(x) = x, f(t(x), t(y)) = t(g(x, y)), g(t(x), t(y)) = t(f(x, y)) and t(a(s)) = A(S). In other words the attractor of the RIFS S = (R, (f, g)) is symmetric with respect to 1/2. Then n 1 H 1 = F S ([0, 1], [0, 1]) = [0, 2/5] [3/5, 1], f([0, 2/5] [3/5, 1], [0, 2/5] [3/5, 1]) = = f([0, 2/5], [0, 2/5]) f([0, 2/5], [3/5, 1]) f([3/5, 1], [3/5, 1]) = [0, 4/25] [3/25, 7/25] [6/25, 2/5] = [0, 2/5], g([0, 2/5] [3/5, 1]) = [3/5, 1], H 2 = F (H 1, H 1 ) = f(h 1, H 1 ) g(h 1, H 1 ) = [0, 2/5] [3/5, 1] = H 1. It follows that A(S) = H 1 = [0, 2/5] [3/5, 1]. We remark that A(S) cannot be the attractor of a classical IFS with two injective functions. Such an attractor, if disconnected, must be totally disconnected. However, A(S) is the attractor of a classical IFS with four functions, namely f 0 : R R defined by f 1 : R R defined by f 2 : R R defined by f 1 (x) = f 2 (x) = and f 3 : R R defined by f 3 (x) = f 0 (x) = min{ x, 2/5}, 2 min{2/5 x 2/5, 2/5} 2 min{3/5 + x 3/5, 1} 2 min{1 x 1, 2/5} , Example 4. Let X = R and f 0, f 1 : R 2 R 2 be defined by f 0 (x, y) = ( x 3, 0) and f 1(x, y) = ( x , 0).

11 11 Recurrent iterated function systems 53 Let the IFS S = (R, (f 0, f 1 )) and A(S) = C the Cantor set. Let C 0 = f 0 (C) and C 1 = f 1 (C). Let us note that C 0 C, C 1 C, f 0 (C 0 ) C 0 and f 1 (C 1 ) C 1. Consider the RIFS S = (R, (f 0, f 1 )), where f 0 : R 2 R 2 R 2 and f 1 : R 2 R 2 R 2 are given by f 0 ((x 1, x 2 ), (y 1, y 2 )) = ( x 1 3, y 1 3 ) and f 1 ((x 1, x 2 ), (y 1, y 2 )) = ( x , 0). Then A(S) = C 0 C 0 C 1 {0}. Indeed, since f 0 (C 0 C 0 C 1 {0}) = f 0 (C 0 C 0 ) f 0 (C 1 {0}) = = f 0 (C 0 ) f 0 (C 0 ) f 0 (C 1 ) {0} C 0 C 0 C 1 {0} and f 1 (C 0 C 0 C 1 {0}) = f 1 (C 0 C 0 ) f 1 (C 1 {0}) = = f 1 (C 0 ) {0} f 1 (C 1 ) {0} = C 1 {0}, we have A(S) C 0 C 0 C 1 {0}. Since f 0 ((x 1, x 2 ), (0, 0)) = f 0 (x 1, x 2 ) and f 1 ((x 1, x 2 ), (0, 0)) = f 1 (x 1, x 2 ), we get C A(S). Also, f 0 (C C) = C 0 C 0 and f 1 (C C) = C 1 {0}. It follows that C 0 C 0 C 1 {0} A(S), which completes the proof. REFERENCES [1] M.F. Barnsley, Fractals Everywhere. Academic Press, New York, [2] K.J. Falconer, The Geometry of Fractal Sets. Cambridge Univ. Press, Cambridge, [3] K.J. Falconer, Fractal Geometry-Foundations and Applications. Wiley, [4] N Secelean, Măsură şi fractali. Ed. Univ. Sibiu, Received 1st March 2006 University of Bucharest Faculty of Mathematics and Computer Science Str. Academiei Bucharest, Romania mihail alex@yahoo.com