Higher order multifractal analysis

Texte intégral

(1)INSTITUT NATIONAL DE RECHERCHE EN INFORMATIQUE ET EN AUTOMATIQUE. Higher order multifractal analysis Robert VOJAK, Jacques LE´VY VE´HEL. N˚ 2796 Fe´vrier 1996. PROGRAMME 5. ISSN 0249-6399. apport de recherche.

(2) Higher order multifractal analysis Robert VOJAK, Jacques LE VY VE HEL Programme 5 | Traitement du signal, automatique et productique Projet Fractales Rapport de recherche n2796 | Fevrier 1996 | 34 pages. Abstract: We extend the multifractal analysis to the description of the distribution. of the n-th order correlation of singularities. For this purpose, new multifractal spectra are de

(3) ned, and some general relations between them are given. We

(4) nally perform explicit computations in the case of deterministic multinomial measures, and show that the multifractal formalism does not hold for correlations on this class of measures. Key-words: Hausdor dimension, multifractal spectrum, multifractal correlation, multinomial measures.. (Resume : tsvp). Jacques Levy Vehel : e-mail: jlv@bora.inria.fr Robert Vojak : e-mail: Robert.Vojak@inria.fr Unite´ de recherche INRIA Rocquencourt Domaine de Voluceau, Rocquencourt, BP 105, 78153 LE CHESNAY Cedex (France) Te´le´phone : (33 1) 39 63 55 11 – Te´le´copie : (33 1) 39 63 53 30.

(5) Analyse multifractale d'ordre superieur Resume : Nous etendons l'analyse multifractale a la description de la distri-. bution des correlations d'ordre n des singularites. Dans cette optique, de nouveaux spectres multifractals sont de

(6) nis, et nous donnons quelques resultats generaux liant ces spectres. Nous donnons egalement des expressions explicites de ces spectres dans le cas des mesures multinomiales deterministes, et montrons que le formalisme multifractal du deuxieme ordre n'a pas lieu pour cette classe de mesures. Mots-cle : dimension de Hausdor , spectre multifractal, correlations multifractales, mesures multinomiales..

(7) 2. VEHEL Robert VOJAK, Jacques LEVY. 1. Introduction Multifractal analysis has recently drawn much attention as a tool for studying the structure of singular measures, both in theory and in applications. E orts have mainly been focused on special cases such as self-similar and self-ane measures, both in the deterministic case [1, 2, 3, 4, 5, 6, 7] and in the random case [8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]. Multifractal analysis has also been extended to a larger class of set functions, namely the class of Choquet capacities [21]. Recent e orts have also been made to extend the analysis to point functions [22, 23, 24]. In the multifractal scheme, the pointwise structure of a singular measure is analyzed through the so-called \multifractal spectrum", which gives either geometrical or probabilistic information about the distribution of points having the same singularity. Several de

(8) nitions of a multifractal spectrum (of a measure) appear in the litterature. Some are related to measure theory (Hausdor and packing measures), other to large deviation theory and to probability (Renyi exponents and Legendre spectrum). The so-called \multifractal formalism" assesses that, in some situations, all the spectra coincide. One motivation for studying this multifractal formalism stems from the fact that both Hausdor and packing dimensions are known to be very awkward to compute. The other de

(9) nitions are more suited to applications, as they are easier to evaluate. The multifractal formalism is known to hold for the class of multiplicative measures [7] but it fails in general [7, 21]. Our concern in this work is di erent. The full characterization of the singularity structure of a sequence of Choquet capacities (and of measures) goes beyond the information contained in the multifractal spectra [21]. Other tools are needed if one wants to describe more precisely this singularity structure. A natural extension is to recognize that the usual spectra correspond to one point statistics, and to generalize the analysis to higher order statistics. In this view, we introduce here the notion of multifractal correlation. We extend the de

(10) nitions of the aforementionned spectra (section 2), and give general results relating them (sections 4 and 3). In section 5, we perform a two point statistics on a class of deterministic multiplicative measures (namely the multinomial measures) and show that the multifractal formalism does not hold in higher order statistics. The random case will be treated in a forthcoming paper. 2. Definitions In what follows, we give de

(11) nitions of quantities measuring the multifractal correlations between singularities of a given probability measure. The parameters we. INRIA.

(12) Higher order multifractal analysis. 3. consider are just extensions of the Renyi exponents and the various multifractal spectra.. 2.1. Notations. Let be a probability measure with support in [0; 1). For n 2 N? , S let Pn := fIkn ; 0 k < n g be a partition of [0; 1), P := n2N Pn, and xnk := inf Ikn ?. for all k. Throughout the paper, we assume that the conditions (C1) through (C4) are met : (C1) limn!1 max0j<n jIjnj = 0, (C2) 8n; k; Ikn is an interval, semi-open to the right. (C3) 8 n; 8 j; 0 j < n 9 k such that Ijn $ Ikn?1, where I00 := E . (C4) 8 > 0, lim sup jI j k(I ) 1. (. I 2P ;jI j!0. ) nj j I j where k(Ijn ) := sup n+1 ; Ikn+1 Ijn . jIk j. We stress the fact that, in our case, a multifractal analysis is relative to a

(13) xed sequence of partitions P . In particular, if P changes, all the quantities de

(14) ned below (i.e. , fh, fg , and fl) may vary. For x 2 [0; 1) and n 2 N , let I n(x) be the interval Ikn containing x. Let p 2 N, " > 0, r := (rn)n2N a sequence of (0; 1)p , with rn := (rn1 ; rn2 ; : : : ; rnp ) for all n. The sequence r is such that either lim ri = 0 for all i, or the sequence is n n constant. Let D := limninf Dn, where Dn := [0; 1 ? max1ip rni ). For all x 2 Dn, set (I n(x)) n(x) := log log jI n (x)j which is de

(15) ned when (I n(x)) 6= 0, and. 8 n 2 N n (x; r) := ( n(x); n (x + rn1 ); : : : ; n (x + rnp )) 1 ); : : : ; lim (x + rp ) 8 x 2 D (x; r) := lim. ( x ) ; lim. ( x + r n n n n n n n n when the considered limits exist. Let r1 := (rn)n2N be a sequence of (0; 1)1 , with rn := (rn1 ; rn2 ; : : : ; rnp ; : : : ), every sequence (rni )n converging towards 0 for all i. We de

(16) ne. n (x; r1) := ( n (x); n(x + rn1 ); : : : ; n(x + rnp ); : : :). . for all n. (x; r1 ) := lim (x); lim (x + rn1 ); : : : ; lim (x + rnp ); : : : n n n n n n RR n2796. .

(17) VEHEL Robert VOJAK, Jacques LEVY. 4. where x belongs to limninf [0; 1 ? supi rni ). Throughout the rest of the paper, we shall always refer to the case p = 0 by replacing the sequence r with the symbol ;. Thus,. n(x; ;) := n (x) (x; ;) := lim (x) n n. 2.2. Hausdor spectrum of order p (or Hausdor p-spectrum). We will use. the following de

(18) nition of the dimension of a set E , dim E , which is similar to that of the classical Hausdor dimension, except the fact that the coverings are restricted to the elements of P . Let. Ms (E ). dim E :=. [. =E Ei; i=1 i s (E ) lim M !0 inf fs = Ms(E ) = 0g = supfs =. := inf. Ms (E ) :=. (X +1. diam(Ei)s. diam(Ei ) ; Ei 2 P 8i. ). Ms(E ) = +1g. Since the elements of P satisfy conditions (C1) through (C4), the de

(19) nition of dim coincides with that of the classical Hausdor dimension [25]. Let := ( 0 ; 1 ; : : : ; p ) 2 (R+ )p+1 . Set E ;r := fx 2 [0; 1) = (x; r) = g for all 2 (R+ )p+1 . The Hausdor p-spectrum of is de

(20) ned as. fh( ; r) := dim E ;r. 2.3. Large deviation p-spectrum. Let n 2 N and ! p ? \ \ Dkn ;::: ;k := Ikn Ikn ? rni 0. p. 0. i=1. i. Notice that the Dkn ;:::;kp form a partition of Dn := [0; 1 ? max1ip rni ), and that Dn = [0; 1) if p = 0. Hence, for all x 2 Dn, we note Dn (x) the (unique) interval Dkn ;:::;kp containing x. Q Let U n := fx 2 Dn ; (I n (x)) pi=1 (I n (x + rni )) 6= 0g. 0. 0. INRIA.

(21) Higher order multifractal analysis. Set. Gn"( ; r). :=. ( . x 2 Un. 5. = n(x; r) 2. Yp i=1. [ i ? "; i + "]. ). K"n( ; r) := (kn(x); kn (x + rn1 ); : : : ; kn (x + rnp )) ; x 2 Gn"( ; r). where kn(x) is the (unique) integer k such that x 2 Ikn. De

(22) ne, for all ";

(23) > 0,. S"n( ;

(24) ; r) :=. X. (k0 ;:::;kp )2K ;r) n "(. jDkn ;::: ;kp j

(25) 0. S"( ;

(26) ; r) := lim sup S"n ( ;

(27) ; r) n. P. (with the convention ; = 0). Using (C1), it is easy to show, by analogy with the Hausdor dimension, that there exists a real number fg"( ; r) such that.

(28) < fg"( ; r) =) S"( ;

(29) ; r) = +1

(30) > fg"( ; r) =) S"( ;

(31) ; r) = 0 fg"( ; r) is non decreasing in ", and we note fg ( ; r) := lim f "( ; r) = "> inf0 fg"( ; r) "#0 g the large deviation p-spectrum of . Remark : when p = 0 (no correlation), and when all the intervals have the same size n?1, it is straightforward (lemma 4) that, for all 2 R+ , log card fIkn ; jIknj +" (Ikn) jIkn j ?"g fg ( ; ;) = lim lim sup "#0 log n n which is the usual quantity considered in the litterature [26][14][3].. 2.4. Legendre p-spectrum. Here we extend the de

(32) nitions given in [7].. Let (n)n1 be a sequence of positive integers such that. X. n>0 Recall that U n := fx 2 Dn ;. exp(?n) < 1 for all > 0. (I n (x)) Qpi=1 (I n(x + rni )) 6= 0g. For all x 2 Dn,. X n ?1 Yp p Yp n n i n (Ik0 ) (Iki )1lDkn0;::: ;kp (x) (I (x)) (I (x + rn)) = i=1 i=1 k0 ;k1 ;:::;kp =0. which shows that the left-hand side quantity is a step function of x. This proves that U n is a

(33) nite union of semi-open intervals, and hence a Borel set. For q := (q0 ; : : : ; qp ) 2 Rp+1 and 2 R, we can now de

(34) ne. Cn(q; ; r) := RR n2796. Z. log jD (x)j (I n(x))q0 log jIn(x)j. n. Un. Yp. i=1. qi loglogjIjnD(xn+(xr)ij )j n i n (I (x + rn)). jDn (x)j?( +1)dx.

(35) VEHEL Robert VOJAK, Jacques LEVY. 6 and. C (q; ; r) = lim sup ?n 1 log Cn(q; ; r) n. We suppose that C (q; ; r) is not constantly equal to 0 or 1 (this imposes the growth of the sequence (n)n ). Set. := f(q; ) 2 Rp+2 = C (q; ; r) < 0g One veri

(36) es that C is convex, non decreasing in , and non increasing in qi for all i. A similar argument as the one found in [7] allows to show that there exists a concave map : q 7! (q; r) such that . . = f(q; ) 2 Rp+2 = < (q; r)g. ( is the interior of ). We suppose that is

(37) nite on an open set I of Rp+1 containing 0. We de

(38) ne the Legendre p-spectrum fl of as being the following Legendre transform of : fl( ; r) := infq [<q ; > ? (q; r)]. where <: ; : > denotes the inner product in Rp+1 .. Remark 1 : the de

(39) nition of Cn can be also stated as follows : since the Dkn ;:::;kp form a partition of Dn, we can write 0. Z. P n (x)jq n (x)+ pi j D Un X n ?1 Pp 0 jDkn ;:::;kp j i =. Cn(q; ; r) =. k0 ;k1 ;::: ;kp =0. 0. 0. =1. =0. qi n (x+rni )?( +1) dx qi n (xnki )?. where 0 means that the summation runs through those indices k0; k1 ; : : : ; kp such Q that pi=0 (Ikni ) 6= 0. Remark 2 : when p = 0, we have Dkn = Ikn, and we obtain the de

(40) nitions considered in [7]. In particular, if all the intervals have the same size exp(?n) = n?1 , one can easily show (lemma 5) that we come up with the usual formulae X n(q; ;) = ? log1 log (Ikn)q n k = (Ikn )6=0 (q; ;) = limninf n(q; ;) 3. Particular cases The de

(41) nitions given above may be simpli

(42) ed when, for large n, all the intervals Ikn have the same size exp(?n) = n?1 for all k, and when rni n 2 N. INRIA.

(43) Higher order multifractal analysis. Lemma 1. Dkn ;:::;kp 0. 7. 8 > k :; otherwise 6 ;, and let x 2 Dkn ;::: ;k . Then kn(x) = k0 and kn(x + = 0. Proof. Assume Dkn ;:::;k rni ) = ki for all 1 i p. Since rni n 2 N, we have ki = kn(x + rni ) = kn(x) + rni n = k0 + rni n (recall that kn(x) = [n x]). If ki = k0 + rni n , then Dkn ;:::;k = Ikn . Otherwise, Dkn ;:::;k = ;. Corollary 1. 8 x 2 Dn Dkn (x);k (x+r );:::;k (x+r ) = I n(x) Proof. Since kn(x + rni ) = kn(x) + rni n for all 1 i p, we deduce, using lemma 0. p. p. 0. n. 1,. p. 0. n. 1. n. 0. 0. n. p. p n. Dknn(x);kn(x+rn );:::;kn(x+rnp ) = Iknn(x) = I n(x) 1. Set . G~ n"( ; r) := 0 k < n ; n(xnk) 2 [ 0 ? "; 0 + "]; n (xnk + rni ) 2 [ i ? "; i + "]; 1 i p. Lemma 2.. G~ n"( ; r) = fkn (x); x 2 Gn"( ; r)g. Proof. Let k 2 fkn(x); x 2 Gn"( ; r)g. There exists x such that n (x) 2 [ 0 ? "; 0 + "] n(x + rni ) 2 [ i ? "; i + "]; 1 i p. with k = kn (x). Since rni n 2 N , we have kn(x + rni ) = kn(x) + rni n and n(x) = n (xnk) n (x + rni ) = n (xnk + rni ) 1 i p Thus, k 2 G~ n"( ; r).. Conversely, let k 2 G~ n"( ; r), and let x be any element of Ikn (thus, kn(x) = k). Then n (x) = n (xnk) and n (x + rni ) = n (xnk + rni ) 1 i p yielding x 2 Gn"( ; r) We conclude k 2 fkn(x); x 2 Gn"( ; r)g.. RR n2796.

(44) VEHEL Robert VOJAK, Jacques LEVY. 8. Lemma 3. Proof.. n. K"n( ; r) = (k; k + rn1 n ; : : : ; k + rnp n); k 2 G~ n"( ; r). o. . K"n( ; r) := (kn(x); kn (x + rn1 ); : : : ; kn (x + rnp )) ; x 2 Gn"( ; r) . = (kn(x); kn (x) + rn1 n; : : : ; kn(x) + rnp n) ; x 2 Gn"( ; r) n o = (k; k + rn1 n; : : : ; k + rnp n) ; k 2 G~ n"( ; r) (lemma 2). Lemma 4. Proof.. ~n. G" ( ; r) fg ( ; r) = lim lim sup log card "#0 log n. X. S"( ;

(45) ; r) = lim sup n. =. n. jIkn j

(46) (lemmas 1 and 3). k2G~ n" ( ;r) lim sup n?

(47) card G~ n"( ; r) n. G~ n"( ; r) = lim sup exp ? log n

(48) ? log card log n n. !!. It is then straightforward that, for a given " > 0, G~ n"( ; r) =) S ( ;

(49) ; r) = 0

(50) > lim sup log card " log n n G~ n"( ; r) =) S ( ;

(51) ; r) = +1

(52) < lim sup log card " log yielding. n. n. ~n. G" ( ; r) fg"( ; r) = lim sup log card log n. n. Lemma 5. fl( ; r) = inf q f< ; q > ? (q; r)g where P Q log 0 (I n )q p (I n + ri )q k. k. (q; r) = limninf. 0. i=1. k. n. i. log n?1 P0 meaning that the summation runs through those indices k such that (I n ) Qp (I n + k i=1 k rni ) 6= 0.. Proof. We have, using lemma 1, X0 Cn(q; ; r) =. Pp jIkn jq n(xnk )+ i k p X0 n q Y. = n. k. 0. (Ik ). =1. 0. i=1. qi n (xnk +rni )?. (Ikn + rni )qi. INRIA.

(53) Higher order multifractal analysis. 9. and. C (q; ; r) = ? limninf. log. P 0 (I n)q Qp (I n + ri )q i=1 k n k k 0. i. log n?1. 4. General properties Let us recall some properties on the \classical" spectra fh (:; ;), fg (:; ;) and fl(:; ;).. Theorem 1. [7, 21] We have fh(:; ;) fg (:; ;) fl(:; ;). Theorem 2. [21] Let If. (I n) k ;0 k < An := log n log jIkn j log card An = 0 lim n n. then. fl(:; ;) = fg??(:; ;). where ? denotes the Legendre transform. Therefore, fl(:; ;) is the concave hull of fg (:; ;).. Our aim in this section is to extend these results to higher order statistics.. 4.1. Comparison of fh and fg . The following result holds : Theorem 3. fh(:; r) fg (:; r) for all sequence r of (0; 1)p, p 2 N . Proof. Consider p 1, 2 (R+ )p+1 , and assume that E ;r 6= ; (the case E ;r = ; is trivial). Let x 2 E ;r and " > 0. Then there exists n0 := n0 (x; ; r) such that, for all n n0 ,. n (x) 2 [ 0 ? "; 0 + "] n (x + rni ) 2 [ i ? "; i + "] 1 i p Hence, x 2 Gn"( ; r) and (kn(x); kn (x + rn1 ); : : : ; kn (x + rnp )) 2 K"n( ; r). Since in addition x 2 Dknn(x);kn(x+rn);::: ;kn(x+rnp ) , we obtain 1. 8">0 x2 RR n2796. [\. [. l nl (k0 ;:::;kp )2K ;r) n "(. Dkn ;:::;kp 0.

(54) VEHEL Robert VOJAK, Jacques LEVY. 10. T S. " " " := n Set E ;l;r nl (k ;:::;kp )2K"n ( ;r) Dk ;::: ;kp and s ;l;r := dim E ;l;r . S " and Clearly, for all " > 0, E ;r l E ;l;r 0. 0. 8 " > 0 dim E ;r sup s" ;l;r l. Let s; 2 R+ . For large n, we have max(k ;::: ;kp) jDkn ;:::;kp j (condition (C1)), and thus 0. " Hs E ;l;r . which yields. X. (k0 ;::: ;kp )2K"n ( ;r). 0. jDkn ;:::;kp js =: S"n ( ; s; r) 0. " Hs E ;l;r S"( ; s; r). and. 8 " > 0 8 l s" ;l;r fg"( ; r) We conclude. fh( ; r) fg ( ; r). 4.2. Comparison of fg and fl. Theorem 4. Under conditions (C1) and (C2), we have fg (:; r) fl(:; r) for all sequence r of (0; 1)p, p 2 N .. Theorem 5. Assume that conditions (C1) and (C2) are met. Let p 2 N? , and let r. be a sequence of (0; 1)p. Set If. (I n) k ;0 k < An := log n log jIkn j log card An = 0 lim n n. then. fl(:; r) = fg??(:; r) where ? denotes the Legendre transform. Therefore, fl is the concave hull of fg .. INRIA.

(55) Higher order multifractal analysis. 11. Proof of theorem 4. Let 2 R; q 2 Rp+1 ; " > 0 and such that fg ( ; r) 6= ?1. For all n 2 N , X ?1 P jDkn ;:::;k jq (x )+ q (x )? Cn (q; ; r) = k ;k ;:::;k =0 P X jDkn ;::: ;k jq (x )+ q (x )? (k ;:::;k )2K ( ;r) P X jDkn ;::: ;k j q ( ")? n. p. 0. 0. 0. p i=1 i n. n k0. n. p. 1. n ". p. 0. (k0 ;:::;kp )2K"n ( ;r). 0. p. 0. p. 0. p i=1 i n. n k0. n. p i=0 i. n ki. n ki. i. (choose + if qi 0, ? if qi < 0). Cn (q; ; r) S"n ( ;. p X i=0. qi ( i ") ? ; r). Let < (q; r). Then there exists c > 0 such that, for large n, Cn(q; ; r) 0. Letting " tend to 0 gives. < ; q > ? (q; r) fg ( ; r) 8q We conclude. fg ( ; r) fl( ; r). Proof of theorem 5. From theorem 4, we already have fg?? fl. Let us prove the. opposite inequality. Let q := (q0 ; q1 ; : : : ; qp ) 2 Rp+1 . By assumption, there exists an integer N (n) such that. n. o. ( n (xnk ); : : : ; n (xnkp )) ; k0 ; : : : ; kp = 0; : : : ; n ? 1 = 0. with. n. log N (n) = 0 lim n . n n ; kp ) ; n;i = ( n (xk0 ); : : :. Let Kn;i := (k0; : : : Then NX (n) Cn (q; ; r) = Set. X. i=1 (k0 ;::: ;kp )2Kn;i. i(n) := argmaxi=1;:::;N (n). N[ (n) i=1. f n;ig. o. ; (xnkp )) .. jDkn ;::: ;kp j<q ; n;i> ? 0. X (k0 ;:::;kp )2Kn;i. jDkn ;:::;kp j<q ; n;i> ? 0. (i(n) may be not unique. In that case, take the smallest one, for instance). This yields (set n for n;i(n) and Kn for Kn;i(n)). Cn (q; ; r) N (n) RR n2796. X. (k0 ;:::;kp )2Kn. jDkn ;::: ;kp j<q ; n> ? 0. (1).

(56) VEHEL Robert VOJAK, Jacques LEVY. 12. This inequality holds in particular for those indices n0 such that C (q; ; r) = limn0 ?n01 log Cn0 (q; ; r). Furthermore, we can extract from ( n0 ) a sequence ( j ) converging towards a limit denoted := ( 0 ; 1 ; : : : ; p ). More precisely,. Yp 8 " > 0 9 j0 8 j j0 j 2 [ i ? "; i + "] i=0. Since j 2 Kj for all j , we deduce. 8 " > 0 9 j0 8 j j0 Kj K"j ( ; r) Inequality (1) reads, for large j ,. 8q. Cj (q; ; r) N (j ) Cj (q; ; r) N (j ). X (k0 ;:::;kp )2K"j. Pp jDkj ;::: ;kp j i. =0. 0. qi ( i")?. p X j S" ( ; qi ( i ") ? ; r) i=0. This inequality, along with the hypothesis of the theorem, yields. 8">0 and. (q; r) . p X i=0. qi( i ") ? fg"( i ; r). (q; r) < ; q > ?fg ( ; r) fg?(q; r). We conclude. fg?? fl 5. Two point statistics of multinomial measures. In this section, our aim is to compute the spectra de

(57) ned in the section 2 on a speci

(58) c class of measures, namely the multinomial measures for the case p = 1. We also show that the multifractal formalism does not hold in higher order statistics. We

(59) rst recall the de

(60) nitions of such measures and some of their properties. We close this section by giving the values of the spectra.. 5.1. Multinomial measures : de

(61) nitions and properties. P Let B be an integer, B 2. Let (m0; m1; : : : ; mB?1 ) 2 (0; 1)B such that iB=0?1 mi =. 1. We construct a sequence (n )n of probability measures on [0; 1) as follows: start with 0 = Lebesgue measure. Split the interval [0; 1) in B intervals of equal size, say fIi ; 0 i1 < B g. De

(62) ne a set function 1 on [0; 1) by 1 (Ii ) = mi . A well known theorem of Caratheodory's asserts that this set function de

(63) nes a probability measure on [0; 1). Iterate the process on each interval Ii , so as to obtain B 2 intervals fIi ;i ; i1; i2 = 0; : : : ; B ? 1g of size B ?2 . Then de

(64) ne a probability measure 2 such that 2 (Ii ;i ) = mi mi . 1. 1. 1. 1. 1. 2. 1. 2. 1. 2. INRIA.

(65) Higher order multifractal analysis. 13. This iterative construction yields a sequence (n )n of probability measures de

(66) ned on [0; 1), and a sequence fIi ;i ;::: ;in ; i1; i2; : : : ; in = 0; : : : ; B ? 1g of partitions of [0; 1) in B -adic intervals such that jIi ;i ;:::;in j = B ?n, and n(Ii ;i ;::: ;in ) = mi mi : : : min . This sequence possesses a weak? limit called multinomial measure of base B , with weights (m0; m1; : : : ; mB?1 ). 1. 2. 1. 1. 2. 2. 1. 2. The multinomial measures possess the following well-known properties [27, 26, 7] :. Proposition 1. P Consider x 2 [0; 1) and its B -adic expansion x = k1 xk B ?k with xk 2 f0; : : : ; B ? 1g for all k (non-terminating expansion are not considered here). For all i < B , let 'i (x) := nlim !1 card fk n ; xk = ig=n be the proportion of i's in the expansion of x. (when this limit exists). The following results hold : P 1. (x) exists almost everywhere, and equals ? iB=0?1 'i (x) logB mi. P 2. For all q 2 R, (q; ;) = ? logB ( iB=0?1 mqi ). 3. fl ( ; ;) = ?1 if and only if 62 D := [ min ; max ], with min := ? logB maxi (mi), max := ? logB mini (mi). o 4. For all 2D, 9 ! q 2 R such that. = 0(q ; ;) and fl( ; ;) = q ? (q ; ;). 5. q!lim 0(q; ;) = min , q!?1 lim 0(q; ;) = max . +1 P 6. fl ( ; ;) = ? iB=0?1 'i( ) logB 'i ( ) where ('0 ( ); : : : ; 'B?1 ( )) is a solution of the constrained problem. 8 >< min(PiB=0?1 'i logB 'i ) PB?1 >: PB?1i=0 'i = 1 ? i=0 'i logB mi = . where is known (with the convention 0 log 0 = 0). 7. fh (:; ;) = fg (:; ;) = fl(:; ;).. Proof. See [27, 26, 7] for instance. Let be a multinomial measure with weights (m0; m1; : : : ; mB?1 ), where B 2. In what follows, we consider a sequence r := (rn)n whose general term is rn := B ?p(n), (p(n))n being a non decreasing sequence of integers chosen such that

(67) := p(n) exists. In the following, we consider only the case p(n) n, the case lim n!+1 n p(n) > n leading to a trivial analysis.. RR n2796.

(68) VEHEL Robert VOJAK, Jacques LEVY. 14. For ( ; 0 ) 2 (R+ )2 , we shall write fh( ; 0 ; r), fg ( ; 0 ; r) and fl ( ; 0 ; r) for fh (( ; 0 ); r), fg (( ; 0 ); r) and fl (( ; 0 ); r) respectively. In addition, when p = 0, the symbol ; will be omitted. Since the computations of these spectra require several lemmas, we state the results in this section and postpone all the proofs until section 6. We have the following results :. Proposition 2. (Hausdor spectrum) 8 ><fh( ; ;) if = 0 0 fh( ; ; r) = > :?1 if 6= 0 The computation of fh is rather simple, and leads to a trivial result which does not give much new information. For this reason, we shall compute another function, also de

(69) ned in terms of a Hausdor dimension, which allows a

(70) ner analysis of the measure in terms of two point statistics. Let 0 ~ 0 E ; ;r := t 2 [0; 1) ; lim (t) = and lim sup n (t + rn) = n n This aforementionned function is de

(71) ned as follows:. n. Fh( ; 0 ; r) := dim E~ ; 0;r Set 0 = ? logB m0 and B?1 = ? logB mB?1 . If m0 6= mB?1 , set. := ?? 0. 0. B?1. 0 and c := c ( ; 0 ) := ( ? B ?1) ?+ 0 ( ? ) B?1 0. 0. Proposition 3. (Fh spectrum) If m0 mB?1 , If m0 < mB?1 ,. 8 ><fh( ) 0 Fh( ; ; r) = > :?1. 8 > fh( ) > < Fh( ; 0 ; r) = >0 > :?1. if 0 = if 0 6= . if ( 0 = ) if = B?1 and 0 < (1 ?

(72) ) +

(73) 0 otherwise. INRIA.

(74) Higher order multifractal analysis. 15. Thus, if for instance

(75) = 1, all correlations between ( ; 0 ) occur when = B?1 and 0 2 [ B?1 ; 0 ]. REMARK 1 : In the de

(76) nition of E~ ; 0;r , we could as well consider a lim inf instead of a lim sup. The relation between n(t + rn) and n(t) (see page 28) shows that this is equivalent to interchanging the role of the weitghts m0 and mB?1 in the computation of all the three spectra.. Proposition 4. (Legendre spectrum) If m0 6= mB?1 , 8 > (1 ? ) fl( c ) > > > <f ( ) fl( ; 0 ; r) = > l 0 > > > :?1 If m0 = mB?1 ,. if c 2 ( min; max ) and 6= 1; 0 if = 0 2 f min ; max g if = B?1 and 0 = 0 otherwise. 8 ><fl ( ) if = 0 0 fl( ; ; r) = > :?1 if 6= 0. Proposition 5. (Large deviation spectrum) fg f l REMARK 2 : In particular, when m0 > mB?1, we have. fh( B?1 ; 0 ; r) = Fh( B?1 ; 0 ; r) = ?1 < 0 = fl( B?1 ; 0 ; r) i.e. a second order analog of the \multifractal formalism" does not hold for multinomial measures. REMARK 3 : Let B be an integer greater than 1, be a multinomial measure with weights (m0; : : : ; mB?1 ), and be a multinomial measure with weights (p0; : : : ; pB?1 ). and have the same

(77) rst order spectra if and only if fm0; : : : ; mB?1 g = fp0; : : : ; pB?1 g. Thus, for instance if B = 2, the

(78) rst order spectra do not di erentiate between 2 with weights M = (m0; m1), and 2 with weights M~ = (m1; m0). It is straightforward to check that the second order spectra of 2 and 2 do di er. It is also easy to prove that the second order spectra allow to di erentiate two trinomial measures as long as all the weights are di erent. More generally, we conjecture that for 2 p 1, the p-th order spectra are di erent for all p-nomial measures, even when the sets of weights are the same.. RR n2796.

(79) VEHEL Robert VOJAK, Jacques LEVY. 16. 6. Proofs In the computations of the spectra, we widely use B -adic expansions of real numbers of [0; 1). Throughout the paper, only terminating expansions are considered.. 6.1. Computation of (q; q0; r). Note xnk := 0:a1a2 : : : an the B-adic expansion of xnk, with ai 2 f0; 1; : : : ; B ? 1g. For sake of simplicity, let us write p for p(n).. n can be written X n(q; q0 ; r) =. xnk =ap <B?1. 0 (Ikn)q (Ikn + B ?p )q +. := S0 + S1. X xnk =ap =B?1. 0 (Ikn )q (Ikn + B ?p )q. 6.1.1. Computation of S0 . In this case, only the p-th digit of xnk and xnk + B ?p are di erent (one can notice that xnk + B ?p remains in [0; 1)). Set i := ap and jN j := n0 + n1 + : : : + nB?1 with N := (n0 ; n1 ; : : : ; nB?1 ), each ni being the number of i's in the B -adic expansion of xnk . Clearly,. (Ikn + B ?p ) = mmi+1 (Ikn) i. and thus. S0 = =. S0 =. BX ?2. 1. 0 X @. n?1 A (mn0 0 mn1 1 : : : mnBB??11 mi)q+q0 i=0 jN j=n?1 n0 ;n1 ;:::;nB?1. "BX ?2 i=0 BX ?2 i=0. mi+1 q0 mi. 0 mqi mqi+1. !. #2 X 0 @ miq+q0 4. n?1 jN j=n?1 n0 ;n1 ;:::;nB?1 !n?1 BX ?1 0 miq+q i=0. mi+1 q0 mi. 1 3 0 0 0 A (mq0+q )n (mq1+q )n : : : (mqB+?q1 )n ? 5 0. B. 1. 6.1.2. Computation of S1 . Suppose that within the

(80) rst p ? 1

(81) rst digits, the last l digits have value B ? 1, and that ap?l?1 B ? 2 :. xnk = 0: a| 1 : : :{zap?l?2} ap?l?1 (|B ? 1) :{z: : (B ? 1)} (B ? 1) |ap+1{z: : : an} ap?l through ap?1 When adding 1 to the p-th digit, ap changes to 0, and so do the l preceding digits, whereas the (p ? l ? 1)-th digit changes its value to ap?l?1 + 1 B ? 1 : xnk + B ?p = 0: a| 1 : : :{zap?l?2} (ap?l?1 + 1) 0| :{z: : 0} 0 a| p+1{z: : : an} We are led to the following conditions on l : since we assume that there are l +1 digits having value B ? 1, the

(82) rst condition we get is l +1 nB?1 . The second condition is also straightforward: l +1 p ? 1. Thus, we must have ?1 l min(nB?1 ? 1; p ? 2) INRIA. 1.

(83) Higher order multifractal analysis. 17. (by convention, the case l = ?1 refers to the case ap B ? 2, which leads to the computation of S0). We then come up with the following result (set i := ap?l?1) : l+1 (Ikn + B ?p ) = m0l+1mi+1 (Ikn ) mB?1 mi. 0 For a

(84) xed l, the number of xnk's having such an expansion is @ jN j = n ? l ? 2. We deduce that 0 ?2 X BX ?2 pX @ n?l?2 S1 = i=0 l=0 jN j=n?l?2 n ;n ;:::;nB? 0 ?2 X BX ?2 pX @ n?l?2 = 0. 1. 1. i=0 l=0 jN j=n?l?2 n0 ;n1 ;:::;nB?1. 0 mq0 mqB?1. "BX ?2. miq (mi+1)q0. 1. n?l?2 A n0 ;n1 ;::: ;nB?1. with. 1 0 l+1 m !q 0 ? m q + q i+1 0 A mn0 mn1 : : : mnB??1 mlB+1?1 mi mlB+1?1mi 1 A (mq0+q0 )n (mq1+q0 )n : : : (mqB+?q10 )n ? 1. 0. B. 0. 1. B. 1. 1. 0. (mq0 )l+1(mqB?1 )l+1 miq (mi+1)q0. #. 8p?2 39= <X q0 q l 2 X 0 n?l?2 1 q+q0 n q+q0 n q+q0 nB? : l=0 (m0 mB?1 ) 4jN j=n?l?2 @ n ;n ;:::;nB? A (m0 ) (m1 ) : : : (mB?1 ) 5; !n?l?23 "BX # 2pX ?2 0 BX ?1 ?2 0 0 q 0 q + q q q q q 5 mk = m0 mB?1 miq mi+1 4 (m0 mB?1 )l i=0 k=0 l=0 #l "BX # " # n ? 2 p ? 2" ?2 BX ?1 q 0 mq X 0 q 0 0 m 0 q q q + q B ? 1 = m0 mB?1 miq mi+1 mk PB?1 mq+q0 k=0 k i=0 k=0 l=0 "BX # "BX #n?2 ? 2 ? 1 0 0 0 = mq0 mqB?1 miq mqi+1 (p ? 1; M (q; q0 )) mqk+q =. i=0. 0. 0. i=0. 1. 1. k=0. with. M (q; q0 ) := and. 1. 0. mq0 mqB?1 PB?1 mq+q0 k=0 k. 8 Mp ? 1 >< (p; M ) = > M ? 1 if M 6= 1 : p if M = 1. Combining this result with S0 gives. n (q; q0 ; B ?p(n)) = RR n2796. "BX ?2 i=0. 0 miq mqi+1. # "BX ?1 k=0. 0 mqk+q. #n?1. (p(n); M (q; q0)). 1.

(85) VEHEL Robert VOJAK, Jacques LEVY. 18 and. 1 log (q; q0 ; B ?p(n)) (q; q0 ; r) = ? n!lim B n +1 n. 1 log (p(n); M (q; q0)) = (q + q0 ) ? n!lim B +1 n 0 0 (q; q ; r) = (q + q ) ?

(86) max(0; L(q; q0 ; r)). with L(q; q0 ; r) := (q + q0 ) + q logB mB?1 + q0 logB m0.. 6.2. Computation of fl ( ; 0 ; r).. For sake of simplicity, we shall omit the sequence r in L(q; q0 ). Set. J (q; q0 ) := q + q0 ? (q + q0 ) . Q := (q; q0 ) 2 R2 ; L(q; q0 ) 0 Recall that (q; q0; r) = (q+q0)?

(87) max(0; L(q; q0 )). Therefore, fl( ; 0 ; r) = inf (q;q0)2Q J (q; q0 ). The computation of fl is quite tedious, and its value depends upon whether m0 and mB?1 are equal. Since the in

(88) mum is seeked over Q, we need to derive some properties of this set depending on the values of m0 and mB?1 . This is done in the following lemmas.. Lemma 6. 1. 8i 2 f0; : : : ; B ? 1g, 8q 2 R (q) < ?q logB mi. 2. m0 = mB?1 =) Q = R2 . Proof. BP ?1. 1. Clearly, for all i and q, mqk > mqi and thus (q) < ?q logB mi. k=0 2. m0 = mB?1 =) L(q; q0 ) = (q + q0 ) + (q + q0 ) logB m0 . From the property 1, we deduce L(q; q0 ) < 0 for all q; q0 2 R2 .. Lemma 7. 8u 2 R 9q 2 R such that (q; u ? q) 2 Q. Proof. By de

(89) nition of L and using lemma 6, L(q; u ? q) = (u) + q logB mB?1 + (u ? q) logB m0 < (u ? q)(logB m0 ? logB mB?1 ) Then choose q very

(90) ng (u ? q)(logB m0 ? logB mB?1 ) 0 (if m0 = mB?1 , the result is straightforward).. INRIA.

(91) Higher order multifractal analysis. 19. We will also need the following lemma. Lemma 8. If m0 6= mB?1 , 8 u 2 R 9 !q 2 R such that L(q; u ? q) = 0 Proof. Trivial. Let a; b be two

(92) xed elements of [ min ; max ], a = 6 b, and de

(93) ne, for all x; y 2 + [ min ; max ] such that (x ? y)=(b ? a) 2 R n f1g, ? by (E ) A(x; y) := x ?axy + a?b The following results hold :. Lemma 9.. ? Assume a > b, and note (x0 ; y0) a solution of A(x; y) = min . If b 6= min , (x0; y0) = ( min ; min ). If b = min , (x0; y0) takes the form ( min ; y0 ), where y0 spans [ min ; max] n fag. The same results hold when replacing min with max . ? Assume a < b, and note (x0 ; y0) a solution of A(x; y) = min . If a 6= min , (x0 ; y0) = ( min; min ). If a = min , (x0 ; y0) takes the form (x0; min ), where x0 spans [ min ; max ] n fbg. The same results hold when replacing min with max .. Proof.. Let us notice some properties of the mapping A : A is strictly increasing in x if y < a. A is strictly decreasing in x if y > a. A is strictly increasing in y if x > b. A is strictly decreasing in y if x < b. A(x; a) = a for all x 6= b. A(b; y) = b for all y 6= a. A(x; x) = x for all x. Let us seek for solutions to (E) other than ( min; min ), and under the assumption a > b. This implies that A is de

(94) ned only when y x. (1) If b 6= min (i.e. b > min ). CLAIM 1. If x0 6= min (i.e. x0 > min ), then y0 < a. Indeed, if y0 = a, then b > a = A(x0 ; y0) = min which is contradictory. If y0 > a, we have min = A(x0; y0 ) A(y0; y0) = y0 > a, which is also contradictory. CLAIM 2. x0 = min. Assume that x0 > min . RR n2796.

(95) VEHEL Robert VOJAK, Jacques LEVY. 20. ? If x0 < b, then min = A(x0; y0) > A(x0; a) = a, which is contradictory. ? If x0 > b, and since y0 x0, we have min = A(x0; y0 ) A(x0 ; x0) = x0 , which is contradictory. ? If x0 = b, min = A(x0 ; y0) = b, which is contradictory. Thus we proved x0 = min. CLAIM 3. y0 = min . Assume y0 > min . Since x0 = min , we have x0 x0, which implies min = A(x0; y0 ) < A(x0 ; x0) = x0 which is contradictory. Hence, if b 6= min , then x0 = y0 = min . (2) If b = min , one only needs to solve the equation A(x; y) = min which yields x0 = min and y0 is any value in [ min ; max ] n fag. The reasoning is similar when searching for solutions to A(x; y) = max : prove that if a 6= max, then y0 6= max implies x0 > b, and then deduce that y0 = max. The result will then follow after proving that x0 6= max cannot hold. When a < b, the result is obtained from the previous one by interchanging the role of a and b, and of x and y. If (q; q0 ) in Q minimizes J , then either one of the Kuhn and Tucker conditions is met : (H1) rJ (q; q0 ) = 0 if L(q; q0 ) < 0 (H2) 9 0 rJ (q; q0 ) + rL(q; q0 ) = 0 if L(q; q0 ) = 0. ? Assume that (H1) is satis

(96) ed. This implies. 8 < ? (q + q0) = 0 : 0 ? (q + q0 ) = 0. If 6= 0 , there are no elements in Q minimizing J and satisfying (H1), and we are then led to treating the case (H2). If = 0 , then. fl ( ; 0 ; r) =. inf f (q + q0 ) ? (q + q0 )g = qinf f q ? (q)g (lemma 7) 2R = fl( ) (q;q0 )2Q. ? Assume that (H2) is satis

(97) ed. The corresponding equality reads. 8 > < 0? 0(0 q + q0 )0 + ( 0(0 q + q0 )0 + logB mB?1 ) = 0 (E ) 9 0 > ? (q + q ) + ( (q + q ) + logB m0) = 0 : L(q; q0 ) = 0 INRIA.

(98) Higher order multifractal analysis. 21. which gives. + logB mB?1 = 0 + logB m0. (2). If m0 = mB?1 , (2) yields = 0 and, by lemma 6, Q = R2 . { If =6 0 , then, as in the case (H1), the in

(99) mum is not reached, and fl( ; 0 ; r) = ?1. { If = 0 , then 0 0 fl( ; ; r) = (q;qinf 0)2R f (q + q ) ? (q + q )g = fl ( ) 2. If m0 6= mB?1 , (2) reads. := ( ; 0 ) := ?? 0. 0. B?1. { If ( ; 0 ) < 0, then (E) does not hold, and fl ( ; 0 ; r) = ?1. { If ( ; 0 ) = 1, (E) implies = ? logB mB?1 and 0 = ? logB m0. If these two equalities are met, then for all q; q0 , J (q; q0 ) = q + q0 ? (q + q0 ) = ?L(q; q0 ) = 0, and hence fl( ; 0 ; r) = 0. If ( ; 0 ) = 6 (? logB mB?1 ; ? logB m0), then fl ( ; 0 ; r) = ?1. { If ( ; 0 ) 0 and ( ; 0 ) =6 1, then (E) implies 0(q + q0 ) = c ( ; 0 ). (3). If c ( ; 0 ) 2 ( min ; max ), then there exists q 2 R such that 0(q) = c, and thus, using lemma 8,. 9 (q; q0 ) 2 @Q such that 0 (q + q0 ) = c ; q = q + q0 Furthermore, we know that (proposition 1). fl ( c) = c q ? (q) = c (q + q0 ) ? (q + q0 ) = c (q + q0 ) + q logB mB?1 + q0 logB m0 0 B mB?1 + q 0 + logB m0 = q + log 1? 1?. = 1 ?1 ( q + 0 q0 + q logB mB?1 + q0 logB m0) = 1 ?1 ( q + 0 q0 ? (q + q0 )) = 1 ?1 fl( ; 0 ; r) which can be rewritten. fl( ; 0 ; r) = (1 ? )fl( c ( ; 0 )). RR n2796.

(100) VEHEL Robert VOJAK, Jacques LEVY. 22. If c = min , then, by lemma 9, either = 0 = min, which. gives fl ( ; ; r) = fl ( ), or = min and 0 > . In this case, fl( ; 0 ; r) = ?1. Indeed, let u 2 R+ and qu such that L(u ? qu ; qu ) = 0 (as in lemma 8). Notice than qu 0. Then J (u ? qu ; qu ) = min u ? (u)+( 0 ? min )qu , and limu!+1 J (u ? qu ; qu ) = limu!+1 fl( min )+( 0 ? min)qu = ?1, and we conclude fl( ; 0 ; r) = ?1. In the same manner, if c = max , then either = 0 = max , which implies fl( ; 0 ; r) = fl( ), or = max and 0 > , which in turn implies fl( ; 0 ; r) = ?1.. 6.3. Computation of Fh( ; 0 ; r). Let t 2 [0; 1), t := P1i=1 ti B ?i with ti 2 f0; : : : ; B ? 1g, and n 2 N . From now on, we shall use the following de

(101) nitions :. ni (t; n) := #fk n ; tk = ig for i 2 f0; : : : ; B ? 1g. n (t; n)=n when the limit exists. 'i (t) := lim n i l(t; n) := n ? 1 ? maxfi < n ; ti < B ? 1g = number of succesive digits equal to B ? 1, starting backwards from the (n ? 1)-th digit. Notice that l(t; n) = 0 if tn?1 < B ? 1. l(t) := lim sup l(t; p(n))=n. n!+1 It will be also useful to consider packets of B ? 1's in the expansion of t. In that view, we de

(102) ne two sequences (an)n and (bn)n as follows:. a1 := minfi ; ti = B ? 1g. b1 := minfi > a1 ; ti 1,. an := minfi > bn?1 ; ti = B ? 1g Set :=. [ n1. bn := minfi > an ; ti < B ? 1g. fan; an+1; : : : ; bn ? 1g. is the set of indices of all packets of B ? 1's in the B -adic expansion of t, i.e.. i 2 , ti = B ? 1. Lemma 10. Let t 2 [0; 1). Then lim inf l(t; pn(n)) = 0 n!+1. INRIA.

(103) Higher order multifractal analysis. 23. Proof. If

(104) = 0, the result follows from the property l(t; p(n)) < p(n).. Assume

(105) > 0, and de

(106) ne the sequence (n)n as follows: 1 := 1, n+1 := minfi ; p(i) > p(n)g.. CLAIM 1. lim n+1 = 1. n Clearly, n+1 > n since p(n+1) > p(n). Thus, limn n = +1 and lim infn n+1 =n 1. In addition, we have p(n) p(n+1 ? 1) ) lim sup n+1 = lim sup p(pn(+1 ?) 1) 1 n n n n. l(t; p(n)) = 0. CLAIM 2. lim inf n!+1 n l ( t; p ( n Assume lim inf n )) > 0. There exists a constant A such that, for large n, n!+1 l(t; p(n)) A n > 0. In particular, this implies tp(n)?1 = B ? 1 when n is large. Moreover, since non terminating expansions are not considered here, we have. 8 n 9 n0 := n0(n) n such that l(t; p(n )) < p(n ) ? p(n ?1) 0. 0. 0. (4). Indeed, if (4) does not hold, then. 9 n0 8 n n0 l(t; p(n)) p(n) ? p(n?1) Since tp(n)?1 = B ? 1, this implies. l(t; p(n)) = p(n) ? p(n?1) and tn = B ? 1 for large n, i.e. the expansion is non terminating, which is contradictory. Set un := n (n). Inequality (4) reads 0. We conclude. l(t; p(un)) < p(un) 1 ? p(un?1) un un?1 un un un?1 p(un) un. l ( t; p ( n )) u n ? 1 lim inf n

(107) 1 ? lim =0 n!+1 n un. which is in contradiction with the hypothesis of claim 2.. Lemma 11. Let t 2 [0; 1). If 'B?1 (t) exists, then l(t) > 0 ) 'B?1 (t) = 1 RR n2796.

(108) VEHEL Robert VOJAK, Jacques LEVY. 24. Proof.. an < 1. CLAIM 1. l(t) > 0 =) lim inf n!+1 bn Let (n)n be a sequence of N such that. l(t; p(n)) l(t) = lim n n. and set I := fn ; tp(n)?1 = B ? 1g. Clearly, I is not empty, nor

(109) nite. Otherwise, l(t; p(n)) = 0 for large n, and l(t) = 0. We can assume, without loss of generality, that tp(n)?1 = B ? 1 for all n, i.e. p(n) ? 1 2 . Denote by i(n) the (unique) integer such that ai(n) < p(n) bi(n) We have l(t; p(n)) = p(n) ? ai(n) bi(n) ? ai(n) a n. implying. n. i(n). 0 < l(t) lim sup bn a? an n!+1. n. an < 1 ) ' (t) = 1. CLAIM 2. lim inf B?1 n!+1 bn Clearly, limn an = +1. Otherwise, for large n, (an)n would be constant and l(t; p(n)) = 0 for large n, yielding l(t) = 0. an < 1. By de

(110) nition of (a ) and (b ) , Set L := lim inf nn nn n!+1 b n. which gives. 8 n nB?1 (t; bn) = nB?1 (t; an) + bn ? an ? 1. 8 n nB?1b(t; bn) = ab n nB?1a(t; an) + 1 ? ab n ? b1 n n n n n. Since 'B?1 (t) exists, we obtain. 'B?1 (t) = L 'B?1 (t) + 1 ? L which reads. 'B?1 (t) = 1. Lemma 12. Let l 2 [0;

(111) ]. There exists t 2 [0; 1) such that 'B?1 (t) = 1 and l(t) = l In order to prove lemma 12, we shall

(112) rst prove the following result :. Lemma 13. Assume limn p(n) = +1. Let (un)n2N be a sequence of N such that un+1 un. limn un = +1. lim inf n pp(u(un) ) = r 2 [0; 1] n+1. INRIA.

(113) Higher order multifractal analysis If the real number t :=. 25. P t B ?i is such that i1 i ti < B ? 1 () 9 n 2 N i = p(un). then. l(t) =

(114) (1 ? r) Proof. The real t has the following B -adic expansion t = 0: (|B ? 1) :{z: : (B ? 1)} (|B ? 1) :{z: : (B ? 1)} (|B ? 1) :{z: : (B ? 1)} (|B ?{z1) : : }:. ". ". ". digit #p(u1) digit #p(u2) digit #p(u3 ) where every can represent any integer other that B ? 1. Clearly, p(un) ? p(un?1) ? 1 l ( t; p ( u )) n = lim sup =

(115) (1 ? r) l(t) lim sup. un. n. n. N?. un. Let us prove the opposite inequality. Let k 2 and set n(k) := maxfi ; p(ui) < p(k)g, i.e. n(k) veri

(116) es p(un(k)) < p(k) p(un(k)+1). Notice that limk n(k) = +1. Indeed, we have i n(k) , p(ui) < p(k). Thus, the inequality p(k) p(k + 1) p(un(k+1)) gives n(k) < n(k + 1). Hence, limk n(k) = +1. Furthermore, l(t; p(k)) = p(k) ? p(un(k)) ? 1 and p(k) p(un(k)) 1 p ( u n (k) ) l(t) = lim sup k ? k ? k =

(117) 1 ? limkinf p(k) k. Let us show that lim inf k p(pu(nk()k)) r. The de

(118) nition of n(k) yields and since limk n(k) = +1, We obtain. p(un(k)) p(un(k)) p(k) p(un(k)+1) limkinf p(pu(nk()k)) r. l(t)

(119) (1 ? r). which

(120) nishes the proof.. Proof of lemma 12. We shall treat the cases l = 0, l 2 (0;

(121) ) and l =

(122) separately. CASE l = 0. A possible choice of t is. t := 0:(B ?1)0(B ?1)(B ?1)0(B ?1)(B ?1)(B ?1)0(B ?1)(B ?1)(B ?1)(B ?1)0 : : : It satis

(123) es l(t) = 0. To see this, denote by (zn)n1 the sequence of indices of P the null digits, i.e. zn := ni=1 i + n = n(n + 3)=2. Let k > 1 and n(k) be the RR n2796.

(124) VEHEL Robert VOJAK, Jacques LEVY. 26. (unique) integer such that zn(k) < k zn(k)+1. It is straightforward to verify that limk n(k)=k = 0. Then l(t; k) = k ? zn(k) ? 1 zn(k)+1 ? zn(k) ? 1 = n(k)2 ? 1 yields l(t; k) = 0 lim k k This in turn shows that l(t) = lim sup l(t; pn(n)) = lim sup l(t;p(pn(n) )) p(nn) 0

(125) = 0 n!+1 n!+1 The real t also satis

(126) es 'B?1 (t) = 1. Indeed,. nB?1 (t; k) = k ? n(k) and we conclude. nB?1 (t; k) = 1 'B?1 (t) := lim k k. CASE l 2 (0;

(127) ). Set r := 1 ?

(128) l 2 (0; 1). Let (un)n2N be a sequence of N de

(129) ned as un := minfi ; p(i) . n X. k=1. [r?k ] + ng. (5). Such a sequence is well-de

(130) ned, since the hypothesis

(131) > 0 implies limn p(n) = +1. Thus,. p(un ? 1) <. n X. k=1. [r?k ] + n p(un). (6). The sequence (un)n2N posseses the following properties :. Inequalities (6) give. p(un+1) . un+1 un lim u = +1 n n p(un) = r lim n p(un+1 ). (7) (8) (9). X [r?i ] + n + 1 > [r?k ] + n > p(un ? 1). nX +1. n. k=1. k=1. Since the mapping n 7! p(n) is not decreasing, we deduce un+1 > un ? 1 which proves (7). Equality (8) follows from the de

(132) nition (5) : un p(un) n. To prove (9), notice that p(n ? 1) = lim p(n ? 1) n n ? 1 = 1 lim n p(n) n n ? 1 p(n) n INRIA.

(133) Higher order multifractal analysis. 27. Inequalities (6) yield p(un ? 1) < Pni=1 [r?i ] + n < p(un) p(un+1) Pni=1+1[r?i ] + n + 1 p(un+1 ? 1) which gives p(un) lim sup pp(u(un ? 1) r lim inf n p(u ) ? 1) n. n+1. n+1. Using (8), we deduce (9). We can now use lemma 13 to deduce. l(t) =

(134) (1 ? r) = l Let us now show that 'B?1 (t) exists and equals 1. Let k 2 N and set n(k) := maxfi ; p(ui) kg, i.e. n(k) is such that p(un(k)) k < p(un(k)+1). log( k + 1) Let n ? log r . We have r?(n+1) > k + 1 and. p(un+1) which yields. nX +1 i=1. [r?i ] + n + 1 > r?(n+1) > k. k + 1) n(k) log( ? log r It is straightforward that nB?1 (t; k) = k ? n(k), which gives 'B?1 (t) = 1. CASE l =

(135) . For n 2 N , set n X un := minfi ; p(i) kk + ng k=1. As for the l 2 (0;

(136) ) case, one can show that the hypotheses of lemma 13 are satis

(137) ed, with r = 0. Thus, l(t) =

(138) = l In addition, we have 'B?1 (t) = 1. To see this, let k 2 N and n(k) := maxfi ; p(ui )p kg. Choose n [ k]. We have. p(un+1 ) (n + 1)(n+1) (n + 1)2 > k. p. yielding n(k) n. Thus, for all k 2 N , n(k) [ k]. As in the case treated above, we deduce 'B?1 (t) = 1.. Lemma 14. Let B > 2, I 2 f0; : : : ; B ?1g and P := f('0; : : : ; 'B?1 ) 2 [0; 1]B ; P 'j = j mk for all k 6= I , or mI < mk for all k 6= I , then. X. k<B. 'k logB mk = logB mI () 'I = 1; 'k = 0 for all k 6= I. P. 2. Otherwise, there exists ('0 ; : : : ; 'B?1 ) 2 P such that 'k logB mk = logB mI k mk for all k 6= I (the other case can be. easily deduced from this one). Suppose that 'I < 1. Then there exists j 6= I such that 'j 6= 0. We have. X. k<B. 'k logB mk =. X. k<B ;'k 6=0. 'k logB mk < logB mI. X. k<B ;'k 6=0. 'k = logB mI. Proof of 2. If there exists k 6= I such that mk = mI , then the proof is straightforward : choose 'k = 1 and 'j = 0 for all j = 6 k. Now consider the case where mk = 6 mI for all k =6 I , mI =6 maxk mk and mI =6 mink mk . Let k; j be such that mk < mI < mj . Then choose 'k ; 'j in [0; 1] such that. 8 > <'k log mk + 'j log mj = 0 mI mI > :'k + 'j 2 (0; 1]. This guarantees that the vector ('0; : : : ; 'B?1 ), de

(140) ned by 'i = 0 for all i 62 fk; j; I g P and 'I = 1 ? 'j ? 'k , belongs to P, and that i<B 'i logB mi = logB mI .. T ft ; ' (t) < 1g 6= ;, then B?1 \ dim E? log m ? ft ; 'B?1 (t) < 1g = dim E? log m ? Proof. Clearly, f?t ; 'B?1 (t) = 1g E? log m ?, and thus, using a theorem of T Billingsley's, dim E? log m ? ft ; 'B?1 (t) = 1g = 0. ? T ft ; ' (t) < 1g, T ft ; ' (t) = 1g S ?E = E Since E. Lemma 15. If E? log . B. mB?1 B. B. B. 1. B. B. ? logB mB?1. we deduce. B. B. B. 1. 1. 1. ? logB mB?1. . dim E? logB mB?. 1. \. B?1. B?1. ? logB mB?1. ft ; 'B?1 (t) < 1g = dim E? logB mB?. 1. We can now turn to the proof of propositions 2 and 3. Let t 2 [0; 1) be such that (t) exists. Set l := l(t; p(n)) and i := ap(n)?l?1. We have (see page 17) n (t + B ?p(n) )) n(t + B ?p(n) ) = log(log( B ?n) = ? l(t; p(nn)) + 1 logB mm0 ? n1 logB mmi+1 + n(t) (10) B?1. i. INRIA.

(141) Higher order multifractal analysis. 29. Proof of proposition 2.. Recall that fh( ; 0 ; r) := dim E ; 0;r where. E ; 0;r = ft ; (t) = ; lim (t + B ?p(n) ) = 0 when both limits existg n n. Equality (10) and lemma 10 yield n!lim (t + B ?p(n)) = n!lim (t) whenever +1 n +1 n lim l(t; p(n)) exists. Hence, if 6= 0 , it is clear that f ( ; 0 ; r) = ?1. If = 0 , n. h n fh( ; ; r) = fh( ; ;). Therefore, the fh spectrum is very trivial, and a need for. another function capable of a

(142) ner analysis arise. For this reason, we have de

(143) ned the Fh spectrum (see page 14).. Proof of proposition 3.. As mentionned in the remark at page 14, we can consider a lim as well as a lim in the de

(144) nition of E~ ; 0;r without loosing in generality. Switching from one de

(145) nition to another is equivalent to switching the values of m0 and mB?1 . Indeed, let us examine the two following cases : (1) m0 mB?1 : Equation (10) gives lim sup n(t + B ?p(n) ) = l(t) logB mmB?1 + (t) n!+1. lim inf n!+1 n. (t + B ?p(n) ). 0. = (t) (see lemma 10). (2) m0 mB?1 : lim sup n(t + B ?p(n)) = (t) n!+1. lim inf (t + B ?p(n)) = l(t) logB mmB?1 + (t) (see lemma 10) n!+1 n 0. Let us consider the two following cases : m0 mB?1 and m0 < mB?1 . CASE m0 mB?1 Thus, by lemma 10, lim sup n (t + B ?p(n) ) = (t) n!+1. which leads to. E~ ; 0;r. 8 < ft ; (t) = g =: ;. We deduce the Fh spectrum :. 8 < Fh( ; 0 ; r) := : fh( ) ?1. if = 0 if 6= 0 if = 0 if 6= 0. CASE m0 < mB?1 In this case, lim sup n (t + B ?n ) = l(t) c0 + (t) n!+1. RR n2796.

(146) VEHEL Robert VOJAK, Jacques LEVY. 30 where c0 := logB mmB?1 . 0 Set. L ; 0;r. Clearly,. . 0 ? . := t 2 [0; 1); ; l(t) = c 0. \. E~ ; 0;r = E L ; 0;r. 0 Case c? 62 [0;

(147) ] : then E~ ; 0;r = ; and Fh( ; 0 ; r) = ?1. 0 Case 0 = . { If = ? logB mB?1 : by lemma 12, there exists t0 2 [0; 1) such that 'B?1 (t0 ) = 1 and l(t0 ) = 0. Thus E T L ; 0 ;r = 6 ;, and dim E~ ; 0;r 0. Either one of the following cases holds : If mB?1 > mi for all i < B ? 1, or mB?1 < mi for all i < B ? 1, we have (lemma 14). E = ft ; 'B?1 (t) = 1g Hence,. Fh( ; ; r) = fh( ) = 0 Otherwise, according to lemma 14, E T ft ; 'B?1 (t) < 1g 6= ;. From lemma 11, we have. E. \. ft ; 'B?1 (t) < 1g E~ ; ;r E. and lemma 15 yields. Fh( ; ; r) = fh( ). { If 6= ? logB mB?1 . We have t 2 E . Then 'B?1 (t) < 1 and, by lemma 11, l(t) = 0, and hence t 2 L ; ;r . This gives E L ; ;r , and E~ ; ;r = E . We conclude. Fh( ; ; r) = fh( ) In both cases, Fh( ; ; r) = fh( ). 0 Case c? 2 (0;

(148) ] 0 We have. t 2 E. \. 8 < L ; 0;r ) : l(t) > 0 'B?1 (t) exists. lemma 11. ). 'B?1 (t) = 1. { If = ? logB mB?1 Then E~ ; 0;r = 6 ; by lemma 12, and since E~ ; 0;r ft ; 'B?1 (t) = 1g, we conclude Fh( ; 0 ; r) = 0.. INRIA.

(149) Higher order multifractal analysis. 31. { If 6= ? logB mB?1 Let t 2 E . Then 'B?1 (t) < 1, which gives l(t) = 0 (lemma 12). In T other words, E L 0 = ; and. ; ;r. Fh( ; 0 ; r) = ?1. 6.4. Computation of fg ( ; 0 ; r). In the case of one point statistics, it is well known that fh(:; ;) fl(:; ;), and hence fh(:; ;) fg (:; ;) fl(:; ;) (theorem 1, page 9. See also [21]). The fg spectrum can be computed independently of this result, using the Gartner-Ellis theorem, and the di erentiability of (see for instance [3]). In the case of two point statistics, we shall also use the large deviation technique, despite the fact that the function is neither di erentiable nor steep (the vectors (q; q0 ) where has no derivatives are exactly the elements of @Q). Nevertheless, a close look at the proof of Ellis' theorem [28] shows that these conditions on are not necessarely required in our case. We will discuss this point in more detail later in the computation of fg . Let X be a random variable uniformly distributed on [0; 1), and de

(150) ne. Yn(X) := (? logB (I n (X)); ? logB (I n (X) + rn)) and, for all " > 0, set F" := [ ? "; + "] [ 0 ? "; 0 + "]. Then Yn(X) n 0 1 2 F = lim 1 log card K" ( ; ; r) lim log Pr n!+1 n. ". n. n!+1 n. = log B. Bn " 0 (fg ( ; ; r) ? 1). Let (1 ; 2) 2 R2 . The moment generating function Mn(1; 2 ) of Yn(X) is. Mn(1; 2 ) := E(e<( ; );Yn(X)>) = E (I n(X))? B (I n(X + rn))? 1. 2. 1 log. =. B ?n. n ?1 BX. k=0. (Ikn )?. 1 log B. (Ikn + rn)?. 2 log B. . 2 log B. Set 'n(1 ; 2 ) := n1 log Mn(1 ; 2 ). Since exists, we obtain. 1 2 '(1 ; 2 ) := lim ' ( ; ) = ? log B 1 + ? ; ? n n 1 2 log B log B Clearly, ' is a convex and continuous on R2 . Now consider the rate function I (s; t) := sup ; (s1 + t2 ? '(1 ; 2 )). Notice that I (s; t) = ? log B (fl(s; t; r) ? 1). Ellis' theorem asserts that, if the supremum in I (s; t) is reached on F" for " su1. 2. ciently small, then. Yn(X) 1 lim inf log Pr n 2 F" ? inf I (s; t) n!+1 n (s;t)2F " o. RR n2796.

(151) VEHEL Robert VOJAK, Jacques LEVY. 32 which yields. fg ( ; 0 ; r) "lim sup fl(s; t; r) !0 (s;t)2F " o. which, combined with theorem 4, gives. lim sup fl (s; t; r) fg ( ; 0 ; r) fl( ; 0 ; r). "!0. (s;t)2F " o. (11). 0 In what follows, we will prove that "lim !0 sup fl (s; t; r) = fl ( ; ; r) by examining (s;t)2F " o. all the possible values of and 0 as in proposition 4.. First of all, let us treat the trivial case fl ( ; 0 ; r) = ?1. Theorem 4 gives fg ( ; 0 ; r) = ?1, and hence fg ( ; 0 ; r) = fl ( ; 0 ; r).. CASE = 0 .. Then fl( ; 0 ; r) = fl ( ) = fh( ) = fh( ; 0 ; r) and, by theorems 3 and 4, fg ( ; 0 ; r) = fl ( ; 0 ; r).. CASE 0 6= , m0 6= mB?1 .. { Assume that c 2 ( min ; max); 6= 1; 0. Let " > 0 be small enough such that, for all (s; t) 2 F", we have c (s; t) 2 ( min ; max). In addition, choose " < 21 j 0 ? B?1 + ? 0j. This condition ensures that for all (s; t) 2 F", we have (s; t) = 6 1 and (s; t) > 0. Thus, for all (s; t) 2 F", the in

(152) mum in fl(s; t; r) := inf q;q0 fsq + tq0 ? (q; q0; r)g is reached (see the computation of fl(:; :; r) in this case, page. 21). Since fl(s; t; r) = (1 ? )fl( c (s; t)) (see page 15), fl(:; :; r) is continuous on F". Using inequalities (11), we conclude fg ( ; 0 ; r) = fl( ; 0 ; r).. { Assume that = ? logB mB?1 and 0 = ? logB m0 (then = 1). Take 0 < " j 0 ? B?1 j=2. This ensures that, for all (s; t) 2 F", (s; t) 0,. and that the in

(153) mum in fl (s; t; r) is reached (here again, see the computation of fl for the case (s; t) 0, page 21). Since fl(:; :; r) is continuous on F", we conclude fg ( ; 0 ; r) = fl( ; 0 ; r).. This

(154) nishes the proof of proposition 5. References. 1. B.B. Mandelbrot. Intermittent turbulence in self similar cascades : divergence of high moments and dimension of the carrier. J. Fluid Mech., 62:331{358, 1974.. INRIA.

(155) Higher order multifractal analysis. 33. 2. R. Riedi and B. B. Mandelbrot. Multifractal formalism for in

(156) nite multinomial measures. Advances in Applied Mathematics, (16):132{150, 1995. 3. R. Riedi. An improved multifractal formalism and self-similar measures. Journal Math. Anal. Appl., 189, 1995. 4. R. Cawley and R.D. Mauldin. Multifractal decomposition of Moran fractals. Advances in Mathematics, (92):196{236, 1992. 5. G.A. Edgar and R.D. Mauldin. Multifractal decompositions of digraph recursive fractals. Proc. London Math. Soc., (65):604{628, 1992. 6. L Olsen. A Multifractal Formalism. Advances in Mathematics. to appear. 7. G. Brown, G. Michon, and J. Peyriere. On the Multifractal Analysis of Measures. Journal of Stat. Phys., t. 66:775{790, 1992. 8. B.B. Mandelbrot. A class of multinomial multifractal measures with negative (latent) values for the dimension f ( ). In Fractals' Physical Origin and Properties (Erice, 1988), pages 3{29, New York, Plenum, 1989. Edited by L. Pietronero. 9. B.B. Mandelbrot. Multiplications aleatoires iterees et distributions invariantes par moyenne ponderee aleatoire. C.R. Acad. Sci. Paris, A(278):289{292, 1974. 10. B.B. Mandelbrot. Multiplications aleatoires iterees et distributions invariantes par moyenne ponderee aleatoire : quelques extensions. C.R. Acad. Sci. Paris, 278:355{358, 1974. 11. B.B. Mandelbrot. Random multifractals : negative dimensions and the resulting limitations of the thermodynamical formalism. Proc. Royal Soc. London, A(434):79{88, 1991. 12. J. P. Kahane. Sur le modele de turbulence de Benoit Mandelbrot. C.R. Acad. Sci. Paris, 278:621{623, 1974. 13. J.-P. Kahane and J. Peyriere. Sur Certaines Martingales de B. Mandelbrot. Advances in Mathematics, 22:131{145, 1979. 14. R. Holley and E. Waymire. Multifractal dimensions and scaling exponents for strongly bounded random cascades. Ann of App. Prob., 2(4):819{845, 1992. 15. K.J. Falconer. The Multifractal Spectrum of Statistically Self-Similar Measures. Journal of Theoretical Probability, 7(3):681{702, July 1994. 16. S. Graf. Statistically self-similar fractals. Probability Theory and Related Fields, 74:357{ 392, 1987. 17. R.D. Mauldin and S.C. Williams. Random recursive constructions: asymptotic geometric and topological properties. Trans. Amer. Math. Soc., (295):325{346, 1986. 18. M. Arbeiter and N. Patzschke. Random Self-Similar Multifractals. Advances in Mathematics, to appear. 19. L. Olsen. Random Geometrically Graph Directed Self-similar Multifractals. Pitman Research Notes in Mathematics, 1994. Series 307. 20. L. Olsen. Self-ane multifractal Sierpinski sponges in Rd . 1994. preprint. 21. Jacques Levy Vehel and Robert Vojak. Multifractal Analysis of Choquet Capacities : Preliminary Results. A para^tre dans Advances in Applied Mathematics, 1995. 22. S. Ja ard. Multifractal Formalism for Functions, Part 1 and Part 2. SIAM Journal of Mathematical Analysis. to appear. 23. S. Ja ard. Construction de Fonctions Multifractales ayant un Spectre de Singularites Prescrit. C.R. Acad. Sci. Paris, pages 19{24, 1992. T. 315, Serie I.. RR n2796.

(157) 34. VEHEL Robert VOJAK, Jacques LEVY. 24. A. Arneodo, J.-F. Muzy, and E. Bacry. Multifractal Formalism for Fractal Signals. Technical report, Paul-Pascal Research Center, France, 1992. 25. J. Peyriere. Calculs de Dimensions de Hausdor . Duke Mathematical Journal, 44(3):591{ 601, September 1977. 26. K. J. Falconer. Fractal Geometry : Mathematical Foundation and Applications. John Wiley & Sons, 1990. 27. P. Billingsley. Ergodic Theory and Information. Robert E. Krieger Publishing Company, 1978. 28. R. Ellis. Entropy, Large Deviations, and Statistical Mechanics. Springer Verlag, 1985.. INRIA.

(158) Unite´ de recherche INRIA Lorraine, Technopoˆle de Nancy-Brabois, Campus scientifique, 615 rue du Jardin Botanique, BP 101, 54600 VILLERS LE`S NANCY Unite´ de recherche INRIA Rennes, Irisa, Campus universitaire de Beaulieu, 35042 RENNES Cedex Unite´ de recherche INRIA Rhoˆne-Alpes, 46 avenue Fe´lix Viallet, 38031 GRENOBLE Cedex 1 Unite´ de recherche INRIA Rocquencourt, Domaine de Voluceau, Rocquencourt, BP 105, 78153 LE CHESNAY Cedex Unite´ de recherche INRIA Sophia-Antipolis, 2004 route des Lucioles, BP 93, 06902 SOPHIA-ANTIPOLIS Cedex. E´diteur INRIA, Domaine de Voluceau, Rocquencourt, BP 105, 78153 LE CHESNAY Cedex (France) ISSN 0249-6399.

(159)