PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 125, Number 7, July 1997, Pages 2119{2125 S (97) THE STRONG OPEN SET CONDITION

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PROCDINGS OF TH AMRICAN MATHMATICAL SOCITY Volume 125, Number 7, July 1997, Pages 2119{2125 S 0002-9939(97)03816-1 TH STRONG OPN ST CONDITION IN TH RANDOM CAS NORBRT PATZSCHK (Communcated by Palle. T. Jorgensen) Abstract. To descrbe some fractal propertes of a self{smlar set or measure, such as the Hausdor dmenson and the multfractal spectrum, t s useful that t satses the strong open set condton, whch means, there s an open set satsfyng the open set condton and, addtonally, a part of the self{smlar set must meet the open set. It s known that n the non{random case the strong open set condton and the open set condton are equvalent. Ths paper treats the random case. If the open set condton s assumed, we show that there s a random open set satsfyng the strong open set condton. Further, we gve an applcaton to multfractal analyss of the random self{smlar fractal. 1. Notatons In ths secton we recall the denton of self{smlar sets and self{smlar measures, and gve some propertes. A more detaled ntroducton and the proofs of the propertes may be found n [PZ] and n [AP]. Let K R d be a xed compact set wth K = nt K. We are gven a postve nteger N 2 and a probablty measure on Sm N [0; 1] N, where Sm s the space of all smlartes of R d equpped wth the usual topology of unform convergence on compact sets. In ths paper we assume the followng. Assumpton 1. (I) Z N p (d(s 1 ; : : : ; S N ; p 1 ; : : : ; p N )) = 1, (II) S (nt K) nt K for all = 1; : : : ; N and S (nt K) \ S j (nt K) = ; for all 6= j for {a. a. (S 1 ; : : : ; S N ; p 1 ; : : : ; p N ), (III) there exst p mn > 0 and r mn > 0 such that p p mn and Lp S r mn for all = 1; : : : ; N and {a. a. (S 1 ; : : : ; S N ; p 1 ; : : : ; p N ). Condton (II) s known as the open set condton. In the sequel we often make use of symbolc dynamcs. Let = f1; : : : ; Ng N be the code space over the ndces S 1; : : : ; N, n = f1; : : : ; Ng n the space of all 1 sequences of length n, and = n=0 n. For 2 n denote by jj = n the length of, and by jk the truncaton of to the rst k entres, k n. For 2 and 2 [ we wrte f there s a 0 2 [ wth = 0. Further, let [] = f 2 : g be the cylnder sets n, 2. Receved by the edtors January 16, 1996 and, n revsed form, February 7, 1996. 1991 Mathematcs Subject Classcaton. Prmary 28A80, Secondary 60D05, 60G57. Key words and phrases. Random fractals, (strong) open set condton, multfractals. 2119 c1997 Amercan Mathematcal Socety

2120 NORBRT PATZSCHK Dene the space =? Sm N [0; 1] N. Let F be the product {algebra on. Takng P the product measure wth on each component we get our prmary probablty space (; F; P). For! 2 and 2 wrte!() = (S 1 (!); : : : ; S N (!); p 1 (!); : : : ; p N (!)) and S ; (!) = d; p ; (!) = 1. By F k we denote the {algebra generated by all S and p wth jj k. For brevty wrte S = S j1 S jjj r = Lp S r = Lp S = r j1 r jjj p = p j1 p jjj K = S K for 2. In [AP] t s shown that the random varables = lm n!1 2 n p (j1) p (jn) exst for all 2 and!([]) = p (!) (!) extends to a random measure! on for P{a. a.! 2, wth () = ; = 1 and ; 2 < 1. Let us dene random mappngs! :! K by! () = lm n!1 S jn(!)(x 0 ): Ths lmt exsts for P{almost all! and does not depend on the choce of x 0. The random measure wth! =!!?1 s called the random self{smlar measure, and the random set wth! =! () the random self{smlar set. We call a random subset? a Markov stoppng, f (I) for each 2 and each! 2 there s a unque 2?(!) wth, and (II) f! 2 : 2?(!)g 2 F jj for all 2. A smple example s the random set? r = f 2 : r < r r jjj?1 g for r 2 (0; 1). If? s a Markov stoppng then, by the open set condton, fnt K (!) : 2?(!)g s a famly of mutually dsjont sets for almost all! 2. Let F? be the sub{{ algebra of F generated by fs ; p : there s a 2? wth g. Further, let us ntroduce shft operators : [! [ by () = and wrte T (!) =!. Then S (T!) = S (!) and p (T!) = p (!). We wll denote the objects generated by T wth a superscrpt, e. g. S (!) = S (T!);! = T!, and so on. The measures and and the set fulll the followng nvarances (cf. [AP, PZ]).

TH STRONG OPN ST CONDITION IN TH RANDOM CAS 2121 Theorem 2. Let? be a Markov stoppng. Then () = 2? of F?. () = 2? F?. [ () = 2? p?1, where the are.. d. copes of and ndependent p S?1, where the are.. d. copes of and ndependent of S where the are.. d. copes of and ndependent of F?. ; and are characterzed by the propertes of the theorem above and by the requrement (R d ) = () = 1. 2. The Man Theorem Now we are able to state the man theorem of ths paper. The proof s analogous to one of Schef [S]. Theorem 3. Let the assumptons 1 be satsed. Then there s a random open set U wth U = nt U, such that the followng hold. () S (U ) U for all = 1; : : : ; N wth probablty one (where U (!) = U(T!)). () S (U ) \ S j (U j ) = ; for all 6= j wth probablty one. () \ U 6= ; wth probablty one. Proof. 1. Fx " > 0 and let K (") = S x2k B(x; "). For 2 wrte G (!) = S (!)(K (") ) and dene I()(!) = f 2? r(!)(!) : G (!) \ K (!) 6= ;g: Fx x 0 2 nt K. Snce nt K s open and bounded, there are 0 < r < R < 1 such that B(x 0 ; r) nt K K B(x 0 ; R). Let 2 I()(!). By denton of K and of? r (!)(!), L d (K (!)) r (!) d L d (B(x 0 ; r)) r (!) d r d mn Ld (B(x 0 ; r)): On the other hand, snce G (!) B(S (!)x 0 ; r (!)(R + ")), we have K (!) B(S (!)x 0 ; r (!)(3R + ")). By volume estmatng therefore #I()(!) r (!) d r d mn L d B(x 0 ; r) L d (K (!)) 2I()(!) L d? B(S (!)x 0 ; (3R + ")r (!)) = r (!) d L d (B(x 0 ; 3R + ")): That means the cardnalty of I()(!) s bounded above, ndependent of! 2 and 2. Let M = ess sup sup 2 #I(). Snce the cardnalty s a dscrete value, there s an 0 2 such that P(#I( 0 ) = M) > 0. 2. Take! 2 and 2 such that #I()(!) = M. Let 2 and ~! 2 T?1f!g. Then r (~!) = r (~!)r (~!) = r (~!)r (!) and K (~!) = S (~!)K (!). Hence, 2 I()(~!) for all 2 I()(!). By maxmalty of I()(!) ths mples I()(~!) = I()(!) and #I()(~!) = M.

2122 NORBRT PATZSCHK 3. Denote 0 = f! 2 : there s an 2 wth #I()(!) = Mg: Then q = P( 0 ) > 0. Furthermore, T?1 0 0 for all and, by denton of P, ft 1?1 0 ; : : : ; T?1 0g s an ndependent famly. Therefore, N q = P( 0 ) P = 1? P N[ N\ T?1 0 = 1? P( c 0) N = 1? (1? q) N T?1 c 0 whch mples (snce q > 0), that q = 1. Hence, for almost all! there s an (!) 2 such that #I()(!) = M. The mappng! 7! (!) may be assumed to be measurable. Wth step 2 we get I( (!))(!) = I ( (!))(!) for all 2. Denote = f! 2 : T! 2 0 for all 2 g: Snce s countable and P a product measure we nfer that P() = 1. 4. Dene U(!) = [ 2 S (!) S (!)(!)(K ("=2) ) for! 2. It remans to show that the assertons hold. () If! 2 then [ S (!)(U (!)) = S (!) S (!) S (!)(!)(K ("=2) ) [ 2 = S (!) S (!)(!)(K ("=2) ) 2 U(!) for all = 1; : : : ; N. () Assume there s an! 2 and a par 6= j wth S (U (!)) \ S j (U j (!)) 6= ;. By denton of U there are ; 2 and = (!) and = j (!) and y 2 S (!) S (K ("=2) ) \ S j (!) S j (!)(K ("=2) ). Wthout loss of generalty we assume r j (!) r (!). Choose 0 such that j 0 2? r (!)(!). By denton there are y 1 2 K (!) wth d(y 1 ; y) < r (!) "=2 and y 2 2 K j 0(!) wth d(y 2 ; y) < r j (!) "=2 < r (!) "=2. Hence, d(y 1 ; y 2 ) < r (!) "=2, whch mples K j 0(!) \ G (!) 6= ;; hence j 0 2 I( (!))(!), beng a contradcton to step 3. () Clearly, K (!) (!) U(!) for all! 2 and, hence, (!) \ U(!) 6= ;. 5. Replacng U by U 0 = nt U, we nfer that U 0 = nt U 0. Moreover, (), () and () reman vald for U 0, concludng the proof.

TH STRONG OPN ST CONDITION IN TH RANDOM CAS 2123 Remark 1. In [AP] we sad that the strong open set condton s satsed, f the assertons of the theorem above hold wth a non{random set U. But many proofs reman vald replacng ths non{random set by a random set. There are some derences between the sets U and nt K. Whle K s gven a pror, the random set U depends (n general) on the whole {algebra F. A dsadvantage s also the fact that there s no constant r 0 such that for almost each! there s a ball wth radus r 0 nsde U(!). 3. An Applcaton An applcaton s the determnaton of the multfractal spectrum. Let us consder the sets a = fx 2 : lm r#0 log (B(x; r)) log r exsts and equals ag of ponts wth local dmenson a. The multfractal spectrum s dened by the Hausdor dmensons of these sets. In [AP] we dened a functon : R! R by N p q r(q) = 1: In [F] some propertes of ths functon are lsted. Let (q) =? 0 (q). Then s ether constant or strctly decreasng. The multfractal spectrum s related to the Legendre transform f = of satsfyng f((q)) = q(q) + (q). The key of the proof was the fact that log d(x; @K) s ntegrable f the strong open set condton s satsed (cf. [AP, 2.8]). An analogue s vald n our case. Lemma 4. Let U be the random open set from Theorem 3. Then Z j log d(x; @U)j (dx) < 1: Proof. By Theorem 3 there are an 2 and a > 0 such that P(d(K ; @U) > =r mn ) > 0. Let r = ess nf r r mn > 0 and 0 < z = 1fd(K ; @U) > r=r mn g ([]) < 1: Wrte?(n) =? r n and G n = f 2?(n) : d(k ; @U) r n g: Let 2 G n. Then r r n+1. If d(k ; @U ) > r=r mn, then, snce 2?(n), d(k ; @U) d(k ; @S U ) = r d(k ; @U ) > r n+1 :

2124 NORBRT PATZSCHK Hence " 2G n+1; = p " p 1? [] F jj # # []=p F jj " #! (K )=p F jj 2G n+1; 2?(n+1)nG n+1; p? 1? 1fd(K ; @U ) > r=r mn g = p (1? z) = (1? z) [ [] j F jj ]: [] Fjj Let 2 G n+1. Then there exsts a unque 2?(n) wth. Furthermore, d(k ; @U) d(k ; @U) r n+1 r n : Hence, 2 G n. Ths mples [] (1? z) [] 2G n+1 2G n and, by nducton, [] (1? z) n 2G n for all n. Snce?1 fx 2 : d(x; @U) r n g S 2G n [], fx 2 : d(x; @U) r n g whch mples the asserton. 2G n [] (1? z) n ; The proof of the next theorem s, wth lttle changes (replacng K by U f necessary), the same as n [AP, 3.10]. Let mn = nff : f((q)) > 0g and max = supf(q) : f((q)) > 0g. Theorem 5. Let the assumptons 1 be satsed. Then ether case I or case II holds. Case I: mn = max = D. Then dm = dm D = D wth probablty one and a = ; for all a 6= D wth probablty one. Case II: mn < max. Let a 2 R, then () f a < mn or a > max then a = ; wth probablty one, () f mn < a < max then dm a = f(a) wth probablty one, () dm = f((0)) wth probablty one. Analogous results as n [AP] hold for generalzed dmensons (cf. [AP, 4.11]) and tangental dstrbutons (cf. [AP, 5.2]).

TH STRONG OPN ST CONDITION IN TH RANDOM CAS 2125 References [AP] M. Arbeter and N. Patzschke, Random self{smlar multfractals, Math. Nachr. 181 (1996), 5 { 42. [F] K. J. Falconer, The multfractal spectrum of statstcally self{smlar measures, J. Theoret. Probab. 7 (1994), 681{702. MR 95m:60076 [PZ] N. Patzschke and U. Zahle, Self{smlar random measures IV. The recursve constructon model of Falconer, Graf, and Mauldn and Wllams. Math. Nachr. 149 (1990), 285 { 302. MR 92j:28007 [S] A. Schef, Separaton propertes of self{smlar sets, Proc. Amer. Math. Soc. 122 (1994), 111 { 115. MR 94k:28012 Fakultat fur Mathematk und Informatk, Fredrch{Schller{Unverstat Jena, D{ 07740 Jena, Germany -mal address: patzschke@mnet.un-jena.de