Accumulation constants of iterated function systems with ... - CUNY

Accumulation constants of iterated functionsystems with Bloch target domainsSeptember 29, 20051 IntroductionLinda Keen and Nikola Lakic 1Suppose that we are given a random sequence of holomorphic maps f 1 , f 2 , f 3 , . . .of the unit disk ∆ onto a subdomain X ⊂ ∆. We consider the compositionsF n = f 1 ◦ f 2 ◦ . . . f n−1 ◦ f n .The sequence {F n } is called the iterated function system coming from the sequencef 1 , f 2 , f 3 , . . .; we abbreviate this to IFS. By Montel’s theorem (see forexample [3]), the sequence F n is a normal family, and every convergent subsequenceconverges uniformly on compact subsets of ∆ to a holomorphic functionF . The limit functions F are called accumulation points. Therefore every accumulationpoint is either an open self map of ∆ or a constant map. The constantaccumulation points may be located either inside X or on its boundary.Note that for the iterated systems we consider here, the compositions aretaken in the reverse of the usual order; that is, backwards. There is a theoryfor forward iterated function systems that is somewhat simpler and is dealt within [5]. For example, for forward iterated function systems, by using constantfunctions, it is easy to construct systems with non-unique limits.The first results for (backward) iterated function systems were found byLorentzen and Gill ([8], [4]) who, independently proved that if X is relativelycompact in ∆, the limit functions are always constant and each IFS has a uniquelimit.In [2] the authors considered iterated function systems for which the targetdomain is non-relatively compact. Using techniques from hyperbolic geometry,they defined a hyperbolic Bloch condition for the target domain and provedthat any X satisfying this condition has only constant limit functions. In [6] weproved that this Bloch condition is also necessary.In [7] we turned to non-Bloch target domains. Using Blaschke products,we proved that any holomorphic map from ∆ to X can be realized as the limit1 The first author partially supported by a PSC-CUNY grant. The second author partiallysupported by the grant DMS 0200733 from the National Science Foundation1

function of some IFS. We also proved that many sets of open maps and constantsin ¯X can be realized as limit functions of an IFS.In this paper we turn our attention to the possible limit constants for Blochtarget domains. We ask what points in X or ∂X can occur as limits. Our mainresult is that for a non-relatively compact Bloch domain X, any finite set ofdistinct points in X can be realized as the full set of limits of an IFS.We also give a new proof of the Lorentzen Gill theorem based on reasoning in[2] and [6]. This theorem shows that if the target domain is relatively compact,the limit function must always lie inside X and not on its boundary. For non-Bloch domains, we saw in [7] that boundary points may be limit points. It is anopen question whether boundary points can be limit points for arbitrary nonrelativelycompact Bloch domains. Here we give two examples of special classesof non-relatively compact Bloch domains for which any boundary point may bea limit point.The paper is organized as follows. In section 2 we provide a proof of theLorentz-Gill theorem and show that the limit point must be in X. In section 3we prove the main result that for a non-relatively compact Bloch domain any ndistinct points can be the limit set of an IFS. Finally, in section 4 we find twoclasses of Bloch domains for which any boundary point can be a limit point.2 Relatively Compact SubdomainsIn this section we consider iterated function systems where the target domainis relatively compact. We remark that if the function f 1 of any IFS is a constantmap, then f 1 ◦ . . . f n is the same constant map and this constant is theunique accumulation point. Similarly, if f k (z) ≡ c, c constant, then the uniqueaccumulation point of the IFS is the constant f 1 ◦ . . . f k−1 (c).We now make the tacit assumption that the functions in our IFS are nonconstant.We begin by recalling a theorem of Lorentz and Gill.Theorem 1 (Lorentz-Gill) If X is a relatively compact subset of the unit disk,then every IFS has a unique constant limit inside X. Moreover, every constantin X is the limit of some IFS.ρProof. Set µ(X) = sup ∆ (z)z∈X ρ X (z) . Because X ⊂ ∆, for all z ∈ X, ρ ∆(z)

Integrating the left and right sides of this inequality along a path γ ⊂ Xjoining points z, w ∈ X that closely approximates the infimum and taking alimit gives us the strong contraction property,∫∫ρ ∆ (f(z), f(w)) = infγγρ ∆ (f(z))|f ′ (z)||dz| ≤ k infγγρ ∆ (z)|dz| = kρ ∆ (z, w)(2)Now suppose that some subsequence F nj of the IFS converges to a holomorphicmap g. The derivatives F n ′ j(z) converge to g ′ (z). Showing that g is aconstant map is equivalent to showing that F n ′ j(z) converges to zero. By theinfinitesimal strong contraction property, (1), we have|F ′ n j(z)| = |f ′ 1(f 2 f 3 . . . f nj z)f ′ 2(f 3 f 4 . . . f nj z) . . . f ′ n j−1(f nj z)f ′ n j(z)| ≤On the one hand, k n jρ ∆ (z)ρ ∆(g(z))kρ ∆ (f 2 f 3 . . . f nj z) kρ ∆ (f 3 f 4 . . . f nj z)ρ ∆ (f 1 f 2 f 3 . . . f nj z) ρ ∆ (f 2 f 3 f 4 . . . f nj z) . . . kρ ∆ (z)ρ ∆ (f nj z) =k n jρ ∆ (z)ρ ∆ (F nj z) .tends to zero, and on the other,ρ ∆ (z)ρ ∆(F nj (z))converges to< ∞. Therefore g′ (z) = 0 for all z in ∆, and every limit function isconstant.To show the constant is unique note that we only need to look at F n (0). SinceX is relatively compact, there is some positive B such that sup z,w∈X ρ ∆ (z, w) n, we apply the strong contraction property (2) m − n times toget,ρ ∆ (F m (0), F n (0)) = ρ ∆ (f 1 ◦ . . . f m (0), f 1 ◦ . . . f n (0))≤ k m−n ρ ∆ (f n+1 ◦ . . . f m (0), 0) < k m−n BThus, F n (0) is a Cauchy sequence and so has a unique limit.To see that the limit cannot be in ∂X, look at the IFS G n = f 2 ◦ f 3 ◦ . . . f n ;then, by the above, G n converges to a point a in ¯X. Since F n = f 1 ◦G n , however,F n converges to f(a) and f(a is in X.Finally, to show that any point a in X is the limit of some IFS, set f 1 ≡ a.Note that these arguments use the compactness and don’t work for nonrelativelycompact domains. They do still work, however, if we replace the unitdisk ∆ with an arbitrary source domain. □3 Non-relatively compact subdomainsWe turn now to the question of subdomains X that are not relatively compactin ∆. Although most of the arguments work for non-Bloch domains, we areinterested in the case when they are Bloch.3

In [5], in addition to our discussion of forward iterated systems, we showedthat if X is any non relatively compact subset of ∆, we could find a (backward)IFS that had two limit functions. Here we generalize this construction to showthat for every integer n, we can find iterated function systems with any givenset of n distinct points as the full set of accumulation points. A key to theconstruction isLemma 3.1 Let X be any non relatively compact subset of ∆, and for anyfixed n, let a 1 , . . . , a n be any distinct points in ∆ \ {0}. Then there exists afunction f : ∆ → ∆ and points x 1 , . . . , x n ∈ X such that for all i = 1, . . . , n,f(x i ) = a i /x i .Proof. We use the notation:and note that A(a, A(−a, z)) = z.A(a, z) = z − a1 − āzStep 1: Since X is not relatively compact we choose an x 1 ∈ X such that|x 1 | > |a 1 |. Let g 1 (z) be a self map of the unit disk to be determined. Definef(z) = A(x 1, z)g 1 (A(x 1 , z)) + a1x 11 + ā1¯x 1A(x 1 , z)g 1 (A(x 1 , z))It follows that f(x 1 ) = a 1 /x 1 as required. Because we want to work inductivelywe rewrite this definition implicitly as followsA(x 1 , z)g 1 (A(x 1 , z)) = A( a 1x 1, f(z)) (3)If n = 1 we set g 1 (z) ≡ 0 and we are done. From now on we assume that n > 1and that we have chosen x 1 .Step 2: Before we proceed, we set up some further notation:For 1 ≤ j ≤ k ≤ n set a jk = A(x j , x k ). Next, for k = 2, . . . n setFor j = 2, . . . n − 1 and k = j, j + 1, . . . n setb 1k = A( a 1x 1, a kx k) (4)b jk = A( b (j−1)ja (j−1)j, b (j−1)ka (j−1)k) (5)In order that our construction work we need to choose the x i so that thefollowing inequalities hold:In step 1 we chose x 1 so this holds for i = 1.| a ix i| < 1, i = 1, . . . n (6)4

We repeat this process for the n new points x 1 , b 2 , . . . b 2...(n−1) and obtain asecond set of n−1 distinct points x 2 , x 21 and b 3 , b 3...(n−1) and a function f 2 suchthat f 2 (x 2 ) = 0, f 2 (x 21 ) = x 1 andf 2 (0) = b 2 , . . . f 2 (b 3 ) = b 23 , . . . , f 2 (b 3...(n−1) ) = b 2...(n−1)We continue in this way. For i = 3, . . . , n − 1 start with the n − 1 pointsx i−1 , x (i−1)(i−2) , . . . x (i−1)...1 , b i , b i(i+1) , b i...(n−1)and obtain a function f i and n − 1 new pointssuch thatandx i , x i(i−1) , . . . , x i(i−1)...1 , b i+1 , b (i+1)(i+2) , . . . b (i+1)...(n−1)f i (x i ) = 0, f i (x i(i−1) ) = x i−1 , . . . , f i (x i(i−1)...1 ) = x (i−1)...1f i (0) = b i , f i (b i+1 ) = b i(i+1) , . . . , f i (b (i+1)...(n−1) ) = b i...(n−1)We thus obtain the first n − 1 maps and check that they satisfy f 1 (0) = c 1 ,f 1 ◦ f 2 (0) = c 2 , . . . f 1 ◦ . . . ◦ f n−1 (0) = c n−1 . Moreover, we have the points of thecycles such that• x 1 , . . . , x n−1 ∈ X satisfying f i (x i ) = 0, i = 1, . . . , n − 1.• x 21 , x 32 , . . . x (n−1)(n−2) ∈ X satisfying f i (x i(i−1) ) = x i−1 , i = 2, . . . , n − 1.• x 321 , x 432 , . . . x n(n−1)(n−2) ∈ X satisfying f i (x i(i−1)(i−2) ) = x (i−1)(i−2) , i =2, . . . , n − 1.• . . .• x (n−1)...21 ∈ X satisfying f n−1 (x (n−1)...21 ) = x (n−2)...21We now have n − 1 points of the first cycle, n − 2 points of the second andso forth. The next step is the general step; we need to complete the cycles.We construct a holomorphic map f n from ∆ to X to complete the first cycle;that is, so that f n (0) = x (n−1)...21 and f 1 ◦ . . . ◦ f n (0) = 0. We also obtain newpoints x n(n−1) , . . . , x n(n−1)...32 in X \ {0} in each of the second through n − 1-stcycles and start a new cycle with a new point x n . That is,. . .andf n (0) = x (n−1)...21 (11)f n (x n(n−1)...32 ) = x (n−1)(n−2)...32 (12)f n (x n(n−1)...43 ) = x (n−1)(n−2)...43 (13)f n (x n(n−1) ) = x (n−1) (14)f n (x n ) = 0 (15)7

For the construction of f n , we again begin with a covering map. Let π 1 bea holomorphic covering map from ∆ onto X such that π 1 (0) = x (n−1)...21 . Wenow choose any n − 1 points in ∆ that are preimages under π 1 of the danglingpoints of the cycles we are constructing as follows:. . .y (n−1)...2 such that π 1 (y (n−1)...2 ) = x (n−1)...2y (n−1)...3 such that π 1 (y (n−1)...3 ) = x (n−1)...3y n−1 such that π 1 (y (n−1) ) = x (n−1)y n such that π 1 (y n ) = 0These n−1 points together with 0 form a set of n distinct points in ∆. Usinglemma 3.1 we can find n − 1 points x n , x n(n−1) , . . . , x n(n−1)...32 in X \ {0} and afunction g such that. . .g(x n(n−1)...2 ) = y (n−1)...2x n(n−1)...2g(x n(n−1)...3 ) = y (n−1)...3x n(n−1)...3g(x n(n−1) ) =y n−1x n(n−1)andg(x n ) = y nx nFinally, let f n (z) = π 1 (zg(z)). We have completed the first cycle so thatthe composition f 1 ◦ f 2 ◦ . . . ◦ f n fixes zero. We now repeat this construction adinfinitum to obtain f n+1 , f n+2 , . . .. At each stage we complete one cycle and addpoints to the next n − 1 cycles. Thus, the cycle relation, (9), holds for each i.Let F k = f 1 ◦ . . . f k . Then, F k (0) = c r where r = k mod n. The accumulationpoints are limits of subsequences {F nk }. For any such limit F , F (0) = c r forsome r = 0, . . . , n − 1. Because the c r are distinct, we have at least n distinctaccumulation points.If X is Bloch, all the limit functions of this IFS are constant. Since F (0) = c kfor some k for every limit function there are exactly n possible constant functions.□4 Boundary points as limiting valuesAs we saw in section 2, if X is relatively compact, all limit functions lie insideX. As we mentioned in the introduction, in [7] we proved that if X is non-Bloch,we can find an IFS whose limit functions take on any or all boundary points. IfX is Bloch and not relatively compact, however, it is not known whether we canobtain limits that are boundary points of X.In this section we exhibit two special classes of subdomains that do admit anIFS whose limit point does lie on the boundary. This gives an affirmative answerto our question for those non-relatively compact Bloch domains in these classes.8

Theorem 3 Let X be a subdomain of ∆ formed by removing an infinite collectionof isolated points from ∆. For any boundary point b ∈ ∂X, there is an IFSwith a limit function that takes the value b.Proof. Choose some b ∈ ∂X; either b is one of the isolated boundary pointsof X or b ∈ ∂∆. Let c 1 , c 2 , . . . be a sequence of points in X that tend to b.Assume, without loss of generality that the origin belongs to X. The idea of theproof is similar to the one above, and works because, although the arguments inthe proof of lemma 3.1 do not extend to an infinte number of points, we can usethe special nature of X to obtain an infinite point version of lemma 3.1.Let g 1 : ∆ → X be a covering map such that g 1 (0) = c 1 . It is uniquelydetermined up to pre-composition by a rotation about the origin. Since X is notsimply connected, we may pick points a 2 , a 3 , . . . in ∆ such that g 1 (a j ) = c j and|a j | < |a j+1 |. The setsA θ = {e −iθ a j : j = 2, 3, . . .}are disjoint for 0 < θ < 2π. Since ∆ \ X is countable, there exists θ such thatA θ ⊂ X. Let c 1j = e −iθ a j and let f 1 (z) = g 1 (e iθ z). Then f 1 (0) = c 1 , andf 1 (c 1j ) = c j for j > 1.We next construct f 2 in the same way. We choose a covering map f 2 so thatf 2 (0) = c 12 ; then f 1 ◦ f 2 (0) = c 2 . We choose preimages c 2j , j = 3, 4, . . . suchthat f 2 (c 2j ) = c 1j . We use the same argument as above to adjust f 2 so that allthese preimages lie in X.We repeat the construction for each n. We take f n as a covering map suchthat f n (0) = c (n−1)n and adjust so that we can find points c nj , j = n + 1, n +2, . . . , ∈ X with f n (c nj ) = c (n−1)j . Then f 1 ◦ . . . f n (0) = c n .Set F n (z) = f 1 ◦ . . . f n (z). Since c n → b, if G is a limit function of F n , thenG(0) = b.Note that if X is Bloch, then G must be constant, G(z) ≡ b. □Theorem 4 Suppose Y is non relatively compact subdomain of ∆ with locallyconnected boundary. Then, for any boundary point c ∈ ∂Y , there is an IFS witha limit function that takes the value c.Proof. Let c ∈ ∂∆ ⋂ ∂Y. We will construct an IFS whose accumulation pointis c. All our maps f i will map the unit disk conformally onto Y. Let f be aRiemann map from the unit disk onto Y. By Caratheodory’s theorem f extendscontinuously to the boundary of the unit disk (see Theorem 2.1 in [3]). Thepreimage of c under this extension is a point on the unit circle, and precomposingby a Mobius map if necessary, we may assume that the continuous extension off, which we will still call f, fixes c. Take a sequence z n of points in Y such thatz n converges to c. Then f(z n ) converges to c. Therefore there exists a point z n1such that |f(z n1 ) − c| < 1 2 . Let A 1 be a hyperbolic isometry of the unit disk suchthat A 1 (c) = c and A 1 (0) = z n1 . Let f 1 = f ◦ A 1 . Then|f 1 (0) − c| < 1 29

andlim f 1(z) = c.z→cTherefore f 1 (f(z n )) converges to c, and we may choose z n2 such that |f 1 f(z n2 )−c| < 1 4 . Now we take a hyperbolic isometry A 2 of the unit disk such that A 2 (c) = cand A 2 (0) = z n2 . Let f 2 = f ◦ A 2 . Then|f 1 f 2 (0) − c| < 1 4andlim f 2(z) = c.z→cIn this way, we obtain a sequence of maps f n from ∆ onto Y ⊂ X such that|f 1 f 2 . . . f n (0) − c| ≤ 12. Therefore c is the accumulation point of the IFSnf 1 f 2 . . . f n . Suppose now that c is any point on the boundary of Y and let fbe a Riemann map from the unit disk onto Y. Then there exists a point c 0 onthe unit circle such that f(c 0 ) = c. Precomposing f by a rotation if necessary,we may assume that c 0 ∈ ∂∆ ⋂ ∂Y. By the above, there exists an IFS F n whoseaccumulation functions all map 0 to c 0 . Then every accumulation function of theIFS G n = f ◦ F n maps 0 to c. □Examples of domains satisfying conditions in Theorem 4 are a those that meetthe boundary in a Stolz angle and polygons with ideal boundary.References[1] L. V. Ahlfors, Complex Analysis, McGrawHill, 1953[2] A. F. Beardon, T. K. Carne, D. Minda and T. W. Ng, Random iteration ofanalytic maps, J.Ergod. Th. and Dyn. Systems[3] L. Carleson and T. W. Gamelin, Complex Dynamics, Springer-Verlag(1993).[4] J. Gill, Compositions of analytic functions of the form F n (z) =F n−1 (f n (z)), f n (z) → f(z), J. Comput. Appl. Math., 23 (2), 1988, 179–184[5] L. Keen and N. Lakic Forward Iterated Function Systems,In Complex Dynamics and Related Topics, Lectures atthe Morningside Center of Mathematics, New Studies in Advanced Mathematics,IP Vol 5 2003.[6] L. Keen and N. Lakic Random holomorphic iterations and degenerate subdomainsof the unit disk To appear, Proc. Amer.Math. Soc.[7] L. Keen and N. Lakic Limit functions for Iterated Functions Systems onNon-Bloch Domains In preparation[8] L. Lorentzen, Compositions of contractions, J. Comput. Appl. Math., 321990, 169–17810