Optimal quantization for uniform distributions on Cantor-like sets

Transcription

1 oname manuscript o. will be inserted by the editor Optimal quantization for uniform distributions on Cantor-like sets Wolfgang Kreitmeier Received: date / Accepted: date Abstract In this paper, the problem of optimal quantization is solved for uniform distributions on some higher dimensional, not necessarily self-similar adic Cantor-like sets. The optimal codebooks are determined and the optimal quantization error is calculated. The existence of the quantization dimension is characterized and it is shown that the quantization coefficient does not exist. The special case of self-similarity is also discussed. The conditions imposed are a separation property of the distribution and strict monotonicity of the first quantization error differences. Criteria for these conditions are proved and as special examples modified versions of classical fractal distributions are discussed. Keywords optimal quantization quantization dimension quantization coefficient self-similar probabilities Mathematics Subject Classification 2000 Primary 28A80 Secondary 62H30 1 Introduction The optimal quantization of probability distributions was first discussed by electrical engineers in the 1940 s as a problem arising out of signal processing and data compression. The aim is to determine the best approximation of a d dimensional probability distribution ν by a discrete one, containing at most n supporting points. The approximation will always be generated by a set α on R d, consisting of at most n points. This set α is called codebook and with r > 0 and the Euclidean norm, the error distance is defined by Ψ α,r ν = min x a a α r dνx. 1 Subsequently, the n optimal quantization error for ν of order r is denoted by V n,r ν = inf{ψ β,r ν : β R d,cardβ n}, 2 This work contains and generalizes some parts of the authors doctoral thesis cf. [16] W. Kreitmeier Fakultät für Informatik und Mathematik, Universität Passau, Innstraße 33, Passau, Germany Tel.: / Fax: / wolfgang.kreitmeier@uni-passau.de

2 2 where card denotes cardinality. The set α is called n optimal set or n optimal codebook for ν of order r, ifψ α,r ν =V n,r ν. The problem of optimal quantization is to determine for every integer n = {1,2,..} all n optimal sets and to calculate the optimal quantization error V n,r ν. A good overview to the theory of optimal quantization and its historical development is presented by Gray and euhoff [9]. For a comprehensive mathematical treatment of the problem, the reader is referred to Graf and Luschgy [5], [7]. A more applied approach to the subject, stressing aspects of information theory and signal processing, can be found in the book of Gersho and Gray [3]. Although under weak assumptions an n optimal codebook exists for every integer n cf. [5], Theorem 4.12, the non-convex nature of the n optimal quantization problem makes it difficult to determine the n optimal codebooks and the optimal quantization error. For nonsingular distributions, exact solutions were found only in some special cases cf. [5], chapter 4.4, 5.2. In the singular case, Graf and Luschgy [4] solved the quantization problem for the classical self-similar Cantor distribution. Subsequently, for more generalized Cantor distributions, which are not necessarily self-similar, among other results the optimal codebooks were determined by Kesseböhmer and Zhu [15]. Kreitmeier [17] extended this codebook result to a class of Cantor distributions, including the classical one studied by Graf and Luschgy. Because exact solutions are difficult to find, the aim is to get high-rate asymptotics for the optimal quantization error. To this end, we consider the sequence r logn logv n,r ν, which was introduced by Zador [23]. If n tends to infinity, the strict positive limit if it exists is called quantization dimension and denoted by D r ν. If the quantization dimension exists, the question arises whether or not the sequence n r Drν V n,r ν has a strict positive and finite limit. If yes, it is called quantization coefficient and denoted by Q r ν. If the distribution ν is absolute continous with respect to the d-dimensional Lebesgue measure and x r+δ dνx < for some δ > 0, then the quantization dimension exists and equals d. The quantization coefficent exists under these conditions as well. This result goes back to Zador [22], respectively Bucklew and Wise [1]. A complete proof was given by Graf and Luschgy cf. [5], Theorem 6.2. In the case of singular distributions, the situation is different. For self-similar distributions, satisfying the so-called open set condition, Graf and Luschgy [6] have shown that the quantization dimension exist. Pötzelberger [21] proved the existence of the quantization coefficient, if the self-similar distribution satisfies the strong separation condition and a non-arithmetic condition. Using other methods, Graf and Luschgy have extended this result by replacing the strong separation condition with the open set condition cf. [8], Theorem 4.1. For singular distributions which are not self-similar, Lindsay gave an example for the non-existence of the quantization dimension cf. [19], Example 5.5. Later on, the existence of the quantization dimension for distributions on not necessarily self-similar Cantor-like sets was systematically studied and characterized by Kesseböhmer and Zhu [15], Kreitmeier [17] and Zhu [24]. In this paper, we will solve for natural 2 the quantization problem for uniform distributions on one- or higher dimensional, not necessarily self-similar adic Cantor-like sets. In section 3, for every n the n optimal codebooks are determined and a formula for the n optimal quantization error is proved cf. Theorem 1. The existence of the quantization dimension is characterized cf. Proposition 4 and under weak assumptions we show that the quantization coefficient does not exist cf. Proposition 5. Although the formula for the

3 3 quantization dimension is a special case of the results of Zhu [24], the class of distributions considered in this present paper is not completely embraced by the one studied in [24] cf. section 3.2. In the self-similar case, we get a more explicit formula for the optimal quantization errors and determine the set of accumulation points of the sequence n r Drν V n,r ν. In our proofs, the conditions imposed are a separation property of the uniform distribution on the Cantor-like set and strict monotonicity of the first differences of optimal quantization errors. Both conditions will be analyzed in detail in section 3.3. To this end, we need some results about optimal quantization for Borel probabilities, uniformly concentrated on a disjoint union of balls with equal radius. These facts are denoted and proved in section 2 and interesting on its own. To show the applicability of our results from section 3, we discuss in section 5 the optimal quantization of the uniform distribution on modified versions of several classical fractal probabilities, e.g. Cantor distribution and uniform distribution on Sierpinski gasket. For the whole paper, we denote ω as a finite subset of R d with cardω =. 2 Uniformly separated probabilities In this section, we study probability distributions which are uniformly concentrated on a disjoint union of balls with equal radius. If the radius is small enough, then for n the n optimal codebooks have a separation property cf. Definition 2, studied below in detail. Moreover, we derive an upper bound for the difference between the optimal quantization errors of the uniformly separated probability on the balls and the uniform distribution on the ball centers. Let d and α be a nonempty finite subset of R d. We call { } Wa α := x R d : x a = min x b b α the Voronoi cell of a with respect to α. Let x R d and l > 0. Let Bx,l := {z R d : x z l} be the closed ball with center x and radius l > 0. Let ν be a Borel probability measure on R d. Recall ω R d with cardω = 2. Definition 1 The distribution ν is called uniformly l, ω separated, respectively l, ω separated in short notation, if the balls Bx,l x ω are pairwise disjoint and νbx,l = 1 for every x ω. Let n and r > 0. The set of all n-optimal codebooks for ν of order r is denoted by C n,r ν. Let α C n,r ν. For x ω let and Clearly, αx αx. With we define the minimal distance in ω. αx = α Bx,l αx = {a α : Wa α Bx,l /0}. d min ω := min{ x y : x,y ω;x y}

4 4 2.1 Properties of the optimal codebooks If n, the calculation of the n optimal quantization error of ν simplifies considerably, if {αx : x ω} becomes a partition of α, i.e. if αx αy = /0 3 for every x,y ω with x y. Due to the importance of this property we make the following definition. Definition 2 Let n be an integer and n. The l,ω-separated distribution ν is called separately quantizing, if relation 3 holds for every n optimal codebook α C n,r ν. Let B,C R d be arbitrary sets and db,c := inf{ x y : x B,y C} be the distance between these two sets. For x R d we simply denote dx,b instead of d{x},b. With convb we indicate the convex hull of B, with B its topological closure. For any Borel measurable set A R d with νa > 0, let ν A = νa νa be the conditional distribution of ν w.r.t. A. The support of ν is denoted by suppν. The following Proposition states a sufficient condition for ν to be separately quantizing and a formula for the optimal quantization error. Proposition 1 Let n and n. Let r 1. Assume that ν is l,ω separated. a Let ν be separately quantizing. Then αx = αx for every x ω. The n optimal quantization error can be written as V n,r ν = x ω a αx For the special case n =, νwa α Bx,lV 1,r ν Wa α Bx,l. V,r ν = 1 V 1,r ν Bx,l. x ω b If d min ω > 10l, then ν is separately quantizing. Moreover, if r > 1, then additionally to a, V,r ν < V 1,r ν. 4 Proof Let n, r 1 and α C n,r ν. a Let x ω. First we will show that αx αx. To this end let a αx. From the definition we get Wa α Bx,l /0. Using property 3 we deduce Wa α z ω\{x} Bz,l = /0. Due to ν z ω Bz,l = 1 and [5], Theorem 4.1 we obtain 0 < νwa α = νwa α Bx,l, 5

5 5 which yields convsuppν Wa α Bx,l. 6 According to [5], Theorem 4.1., we have {a} C 1,r ν Wa α. The application of [5], Remark 4.6 a yields a convsuppν Wa α. From 6 we get a Bx,l. Hence, a αx. As already stated, the converse inclusion αx αx is obvious. Thus, we have αx = αx. ext, we will prove the two equations for the n optimal quantization error. From [5], Lemma 3.3 and [5], Theorem 4.2 we get With 3 and 5 we obtain V n,r ν = V n,r ν = νwa αv 1,r ν Wa α. a α x ω a αx = x ω a αx νwa αv 1,r ν Wa α νwa α Bx,lV 1,r ν Wa α Bx,l. From αx = αx for every x ω we deduce the first equation in Proposition 1. If n = we have cardαx = cardαx = 1 for every x ω. Hence, we get from the first equation, b Let d min ω > 10l. V,r ν = 1 V 1,r ν Bx,l. x ω b1 We will show that αx αy = /0 for every x, y ω with x y. To this end, let ω 1 = {x ω : dbx,l,α > 2l} 7 and ω 2 = ω\ω 1. We will prove that Let x ω 2. Then y Bx,l and a α exist, with Let b αx. A y Bx,l exists, with Thus, we get αx Bx,5l for every x ω 2. 8 y a = dbx,l,α 2l. y b = dy,α y a. x b x y + y b l+ y a l+ y y + y a l + 2l + 2l = 5l, which yields b Bx,5l. Thus, relation 8 is proved. Due to d min ω > 10l we obtain αx αy = /0 for every x,y ω 2,x y. 9

6 6 The set ω 2 is nonempty because otherwise we would have Ψ ω,r ν l r < 2l r < Ψ α,r ν = V n,r ν, which contradicts cardω = n. Assertion b1 is proved, if ω 1 = /0. We proceed indirectly and assume that ω 1 /0. From definition 7 of ω 1 we get dz,αx r dνz. 10 Two cases appear. V n,r ν = Ψ α,r ν > 2lr x ω cardω 1 card α\ αx. Let β = ω 1 αx. Obviously With 10 and ω 1 /0 we have Bx,l cardβ cardω 1 + card x ω2 αx 11 cardα\ x ω2 αx + card x ω2 αx = cardα. Ψ β,r ν lr x ω 1 + Bx,l Together with 11, this contradicts the optimality of α. 2. cardω 1 > card α\ αx. Let and dz,αx r dνz < Ψ α,r ν. ω 2,s = {x ω 2 : cardαx = 1} ω 2,p = {x ω 2 : cardαx > 1} = ω 2 \ω 2,s. We will derive in this second case a contradiction as well. To this end, we construct an alternative codebook β by taking points from,p αx and replacing them by ω 1. We will show that β does not contain more points than α, but generates a quantization error less than the one of α. From 9, we obtain cardα = card α\ αx + cardαx. On the other hand, cardα = cardω 1 + cardω 2 + n. Thus, we deduce 0 < cardω 1 card α\ αx = cardαx cardω 2 n = cardαx + cardω 2,s cardω 2 n,p = cardαx cardω 2,p n,p cardαx 1.,p

7 7 ow, let 2,p = cardω 2,p and {x 1,..,x 2,p } = ω 2,p an arbitrary enumeration of ω 2,p. Based on this enumeration, we recursively define the mapping by ϕ : ω 2 0 i ϕx = 0 for all x ω 2,s, ii for i {1,.., 2,p } and with 0 j=1 ϕx j := 0 it is ϕx i = min Obviously, ϕ has the following properties cardαx i 1,cardω 1 cardα\ iii ϕx {0,..,cardαx 1} for every x ω 2,p, iv x ω2 ϕx = cardω 1 card α\ αx. With this mapping ϕ, we take for every x ω 2 a and define the already mentioned codebook ow, we prove that Ψ β,r ν < V n,r ν. At first we get the upper bound From 10 and 12, we obtain βx C cardαx ϕx,r ν Bx,l β = ω 1 βx. αx i 1 j=1 ϕx j Ψ β,r ν lr x ω 1 + dz,βx r dνz 12 Bx,l = lr x ω V cardβx,r ν Bx,l. V n,r ν Ψ β,r ν > dz,αx r dνz Bx,l 1 V cardβx,r ν Bx,l + lr 2r 1 x ω 1 = 1 Ψ αx,r ν Bx,l 1 V cardβx,r ν Bx,l + lr 2r 1. x ω 1

8 8 From definition 2 of the optimal quantization error, we get V n,r ν Ψ β,r ν 13 > lr 2r 1 x ω Vcardαx,r ν Bx,l V cardβx,r ν Bx,l = lr 2r 1 x ω Vcardαx,r ν Bx,l x {z ω 2 :ϕz 0} V cardαx ϕx,r ν Bx,l. From property iii and iv of ϕ, we deduce card{z ω 2 : ϕz 0} ϕx Inequality 13 thus transforms into = cardω 1 card α\ αx cardω 1. V n,r ν Ψ β,r ν 14 > 1 cardω 1 l r 2 r cardω 1 l r 2 r 1 cardω 1 l r 2 r 1 l r 0. Property iv of the mapping ϕ and 9 implies cardβ cardω 1 + cardβx x {z ω 2 :ϕz 0} x {z ω 2 :ϕz 0} 0 V 1,r ν Bx,l Ψ {x},r ν Bx,l = cardω 1 + cardαx ϕx = cardω 1 + card αx cardω 1 cardα\ αx = cardα. Thus, 14 contradicts the optimality of α. Because each case ends in a contradiction, we have proved assertion b1. b2 We will prove inequality 4.

9 9 Our approach is to determine two distinguished points from ω for α C 1,r ν. Using the partitions of ω into the sets ω 1,ω 2,s and ω 2,p, we can deduce an upper bound for V,r ν, respectively a lower bound for V 1,r ν. This enables us to prove inequality 4. Let r > 1 and α C 1,r ν. Due to cardα = 1 there exist two different points x 1,x 2 ω and a α with dx 1,a = dx 1,α, respectively dx 2,a = dx 2,α. For every z Bx 1,l we have dz,α dx 1,α dx 1,z dx 1,α l, respectively dz,α dx 2,α l for z Bx 2,l. Hence, Bx 1,l dz,α r dνz + dz,α r dνz 15 Bx 2,l 1 max0,dx 1,α l r + max0,dx 2,α l r. Obviously, x r + y r 2 1 r x + y r 16 2 for arbitrary x,y 0 with x + y > 0. From l < 1 2 d minω and 16 we deduce for inequality 15 that Bx 1,l dz,α r dνz + Bx 2,l 2 1 r dx 1,a + dx 2,a 2l r r 2 d minω l. With statement 9, which is also valid if cardα = 1, we get The definition of ω 2,p and ω 2,s yields Thus, we have cardαx = cardα cardαx. cardαx + cardαx,s,p dz,α r dνz 17 = cardω 2,s + cardω 2,p +,p cardαx 1 cardω 2 + cardω 2,p. cardω 2 + cardω 2,p cardα = 1 = cardω 2 + cardω 1 1, which yields cardω 2,p cardω

10 10 Using 17, we conclude that V 1,r ν = dz,α r dνz 19,s \{x 1,x 2 } Bx,l + dz,α r dνz,p \{x 1,x 2 } Bx,l + x ω 1 \{x 1,x 2 } Bx,l + x {x 1,x 2 },s \{x 1,x 2 } + Bx,l,p \{x 1,x 2 } + 1 x ω 1 \{x 1,x 2 },s \{x 1,x 2 } dz,α r dνz dz,α r dνz 1 V 1,rν Bx,l 1 V cardαx,rν Bx,l 1 2lr + 2 V 1,r ν Bx,l 1 r 2 d minω l + cardω 1 22l r d minω l r. From a we get V,r ν = 1 V 1,r ν Bx,l, x ω which implies V,r ν 1 V 1,r ν Bx,l + cardω 2,p + cardω 1 l r. 20,s With 19 and 20 we deduce V 1,r ν V,rν 1 V 1,r ν Bx,l + cardω 1 22l r d minω l,s \{x 1,x 2 } V 1,r ν Bx,l + cardω 2,p + cardω 1 l r,s 1 r cardω 1 22l r d minω l cardω2,s l r cardω 2,p + cardω 1 l r 2 2l r. r

11 11 Due to d min ω > 10l, the right hand side is strict greater than cardω 1 22 r r cardω 2,s cardω 2,p + cardω r l r = cardω 1 22 r r 1 cardω 1 cardω 2,p 2 2 r l r. Due to 18 and r > 1 we obtain the lower bound which proves inequality 4. cardω 1 22 r r 2cardω r l r = cardω 1 2 r 4 2 r r 2 cardω 1 l r > 0, Remark 1 The sufficient condition d min ω > 10l in Proposition 1 b ensuring the separation property 3 of the optimal codebooks needs only the constant value 10 and does neither depend on ω nor the norm exponent r. It is also independent of the dimension d. For x R d we denote δ x as the dirac measure in x. Remark 2 One may conjecture that the condition d min ω > 10l in Proposition 1 b can be weakened, i.e. Proposition 1 b is probably still true for a smaller constant than 10. But we cannot drop the condition completely. For example, let d = 1 and ν = 2 1 δ 0 + δ 1 be the uniform distribution on the points 0 and 1. Let ω = {0, 10 7 } and l = Hence, ν is a l,ω separated measure. Clearly, C 2,r ν = {{0,1}} and W0 {0,1} =], 2 1 ]. Thus, we obtain W0 {0,1} B0,l = B0,l /0 and [ 7 4 W0 {0,1} B 10,l = 10, 5 ] /0, 10 which implies that ν is not separately quantizing. 2.2 Approximation of the optimal quantization error For d denote U ω as the uniform distribution on ω. It is well known see e.g. [5], p. 57 ff that the n optimal quantization error depends continuously from the distribution it refers to, if the distance between distributions is defined by a suitable metric. Moreover, this continuity is uniform in n. In our special case, for every ε > 0 an l > 0 exists such that for every l,ω separated distribution ν and n, we have V n,r ν V n,r U ω < ε. The following Proposition 2 refines this abstract approximation result by presenting an upper bound for the optimal quantization error difference, independent of n and tending to zero if l tends to zero. For an arbitrary set C R d and x R d let { 1, if x C 1 C x := 0, otherwise be the characteristic function on C. We denote with the diameter of C. diamc = sup{ x y : x,y C}

12 12 Proposition 2 Let n and r > 1. Let ν be an l,ω-separated probability measure. Then { lrl + diamω V n,r ν V n,r U ω r 1, if n < l r, if n. Proof Let n. Then V n,r U ω = 0 and If n <, we can identify two cases. 1. V n,r ν V n,r U ω. With an α C n,r U ω we obtain V n,r ν V n,r U ω = V n,r ν Ψ ω,r ν l r. V n,r ν V n,r U ω = x ω dy,α r dνy dy,α r du ω y 21 dy,α r dνy 1. Bx,l dx,αr Due to νbx,l = 1, for every x ω the right hand side of 21 equals x ω Bx,l x ω Bx,l dy,α r dx,α r dνy 22 dy,α r dx,α r 1 {z Bx,l:dz,α dx,α} ydνy. ow let x ω and y Bx,l such that dy,α dx,α. Choose a ξ y [dx, α, dy, α] such that From [5], Remark 4.6 a we deduce On the other hand we have which yields Equation 23 implies dy,α r dx,α r = rξ y r 1 dy,α dx,α 23 α convsuppu ω. diam convsuppu ω = diamω, maxdy,α,dx,α dx,α + dx,y diamω + l. dy,α r dx,α r rl + diamω r 1 dx,y rll + diamω r 1. Thus, we derive for 22 the upper bound rll + diamω r 1 dνy = rll + diamω r 1. x ω Bx,l

13 13 Using 21 we get V n,r ν V n,r U ω rll + diamω r V n,r ν < V n,r U ω. Let α C n,r ν. Similar to the first case, we obtain V n,r U ω V n,r ν dx,α r dy,α r dνy x ω Bx,l x ω Bx,l dx,α r dy,α r 1 {z Bx,l:dx,α dz,α} ydνy. ow we consider all x ω and y Bx,l with dy,α dx,α. As in the first case we deduce 0 < V n,r U ω V n,r ν rll + diamω r 1. The combination of both cases yields the assertion. 3 Uniform distributions on Cantor-like sets Let r > 1. Recall 2 and the finite set ω = {x 1,..,x } R d, consisting of different points x i R d. ote that we do not make any restrictions on the dimension d. Let c k ]0,1[ be a sequence of contraction ratios and S k,i contractive similitudes for k and i {1,..,} with fixpoint x i and contraction ratio c k independent of i. We assume that sup k c k = c < 1 and that the mappings S k,i satisfy the open set condition, i.e. a bounded open set V R d exists such that for every k and i {1,..,} we have and S k,i V V 24 S k,i V S k, j V = /0 for every j {1,..,}\{i}. ow we can construct the -adic Cantor-like set E and the uniform distribution µ on it. To this end, let m, recall V as the topological closure of V in R d and define for any k the set D m k V = {S m,i 1... S k+m 1,ik V : i 1,..,i k {1,..,} k }. With Ek mv = F Dk mv F, we define the Cantor-like set Em V = k Ek m V. By repeated subdivision, cf. [2], p. 9 we get a unique Borel probability measure µ m on R d, defined by µ m F = k for every F Dk mv. ote that µm is supported on E m V. We write shortly Dk 1V = D kv, respectively Ek 1V = E kv, E 1 V = E and µ 1 = µ. We denote by {1,..,} the set of all words on the alphabet 1,..,, including the empty word /0. With σ = σ 1...σ k {1,..,}, we call k the length of σ and write σ for it. The empty word /0 has length 0. We denote Sσ m = S m,σ1 S k+m 1,σk respectively shortly Sσ 1 = S σ. Moreover we write πk m = k+m 1 l=m c l respectively shortly πk 1 = π k. Lemma 1 Let m,k be integers and σ {1,..,} with length k. Then we have µ m S m σ E m V = µ m+k S m σ 1. Proof The assertion follows from the construction of µ m respectively µ m+k.

14 Solution of the optimal quantization problem To solve the quantization problem for uniform distributions on Cantor-like sets, we assume that for every m 1, the distribution µ m satisfies a separation condition which incorporates that µ m is separately quantizing cf. Definition 2. Moreover we need a monotonicity condition for the first differences of the optimal quantization errors of µ m. Let us define these two conditions. a Separate quantization SQ: we say that µ satisfies the separate quantization condition, if a set κ = {y 1,..,y } R d with cardκ = exists, such that for every m an l m > 0 exists, with a1 By i,l m S m,i E m V for every i {1,..,}, a2 the distribution µ m being an l m,κ-separated probability measure cf. Definition 1 and separately quantizing cf. Definition 2. b Monotonicity of quantization error differences MQED: we say that µ satisfies the monotonicity condition for the quantization error differences, if for every m and n {1,.., 1} the relation does hold. V n,r µ m V n+1,r µ m > V n+1,r µ m V n+2,r µ m Remark 3 If the condition SQ is satisfied, also the strong separation condition is satisfied, i.e. S m,i E m V S m, j E m V = /0 25 for every m and i, j {1,..,} with i j. Proposition 3 Let m,n with n and α C n,r µ m. If condition SQ holds, then n 1,..,n exist with i=1 n i = n and V n,r µ m = 1 i=1 c r mv ni,rµ m Moreover, for each i {1,..,} an n i optimal set β i C ni,rµ m+1 exists such that α = S m,i β i. i=1 Proof Let α C n,r µ m. Applying condition SQ, we find with κ = {y 1,..,y } that α = αy i i=1 with αy i αy j = /0 for every i, j {1,..,} with i j. Define n i = cardαy i > 0 for every i {1,..,}. From [5], Theorem 4.1 and 4.2 we deduce V n,r µ m = i=1 µ m a αy i Wa α V ni,rµ m a αy i Wa α. 27

15 15 Again by SQ and Lemma 1 we obtain µ m a αy i Wa α = µ m By i,l m 28 = µ m S m,i E m V = µ m+1 S 1 m,i respectively µ m a αy i Wa α = 1 for every i {1,..,}. The application of 28 and 29 to 27 yields 29 V n,r µ m = 1 i=1 V ni,rµ m+1 S 1 m,i. From [5], Lemma 3.2 a and the definition of S m,i we get V n,r µ m = 1 i=1 c r mv ni,rµ m+1. Define β i = S 1 m,i αy i for every i {1,..,}. From [5], Theorem 4.1 we obtain S m,i β i C ni,rµ m a αy i Wa α. Applying 28, we deduce S m,i β i C ni,rµ m+1 Sm,i 1. Finally, [5], Lemma 3.2 a implies β i C ni,rµ m+1. Remark 4 If condition SQ is satisfied, we get as a direct consequence of Proposition 3 for every m the identities V,r µ m = c r m V 1,r µ m+1 respectively V +1,r µ m = cr m 1V 1,r µ m+1 +V 2,r µ m+1. As an application, we derive the equation c r m = cr m V 1,r µ m+1 V 2,r µ m+1 1 c r V,r µ m m c r V +1,r µ m + 1V 1,r µ m+1 m 1 = 1 V,rµ m V +1,r µ m + cr m 1 = V,r µ m V +1,r µ m, c r m V,r µ m 30 which will be useful in the sequel.

16 16 Remark 5 In the self-similar case cf. section 4, Graf and Luschgy cf. [5], Lemma have shown that an n 0 exists such that identity 26 remains true for n n 0 under the strong separation condition, which is, according to Remark 3, weaker than our condition SQ. evertheless, n 0 will be greater than in many cases. For any σ {1,..,} and n < σ, we define { 0, if n = 0 σ n = σ 1...σ n, otherwise as the restriction of σ to n. For any τ {1,..,}, we define the order relation σ τ if and only if σ τ and τ σ = σ. We call σ a predecessor of τ, if σ τ. A finite set Γ {1,..,} is called a finite antichain, if any two different words σ,τ Γ are incomparable, i.e. neither σ τ nor τ σ. A finite antichain Γ is called maximal if any word τ {1,..,} with τ max{ σ : σ Γ } has a predecessor in Γ. By an iterative approach of Proposition 3 it is straightforward to see that the following statement holds. Corollary 1 Let m,n with n. Assume that condition SQ holds and let α be an n optimal set for µ m of order r > 1. Then a maximal finite antichain Γ {1,..,} and n σ {1,..,} for every σ Γ exists, with σ Γ n σ = n and V n,r µ m π σ = r σ Γ σ V nσ,rµ m+ σ. 31 Moreover, for each σ Γ an n σ -optimal set β σ for µ m+ σ of order r exists such that α = Sσ m β σ. σ Γ Remark 6 By one more application of Proposition 3, the statement of Corollary 1 does also hold with n σ {1,.., 1} for every σ Γ. For an integer k and Γ {1,..,} let Γ k be the subset of Γ consisting of all words with length k. Remark 7 If the maximal finite antichain Γ from Proposition 3 contains only words of length k, i.e. Γ = Γ k, then Γ consists of exactly k words. For σ = σ 1...σ k 1 σ k {1,..,} we denote σ = σ 1...σ k 1. We say that σ is equivalent to τ {1,..,}, if σ = τ. Remark 8 Let Γ and β σ, σ Γ as in Corollary 1. Conversely to Remark 6, we can cluster the sets β σ. We will denote exactly this. Let l := max{ σ : σ Γ } and assume that l 2. The equivalence relation described above induces the partition Γ l,1,..,γ l,w of Γ l with an integer 1 w. ow let i {1,..,w} and τ Γ l,i. Because Γ is a maximal finite antichain, we get cardγ l,i =. It is easy to see that σ Γ l,i,τ =σ S m+l 1,σl β σ is independent of the choice of τ Γ l,i and an l i optimal set for µ m+l 1 with l i max σ Γ l,i cardβ σ.

17 17 Remark 9 In the self-similar case cf. section 4, Graf and Luschgy cf. [5], Corollary 14.2 have shown that the right hand side in equation 31 is greater or equal to V n,r µ m, if condition SQ is dropped completely. If we additionally assume that the monotonicity condition MQED holds, we can sharpen identity 31 to a formula for the optimal quantization error, which depends only on the first optimal quantization errors. Theorem 1 Let r > 1 and n. Let k 1 be an integer and i {1,.., 1} such that i k n < i+1 k. Assume that condition SQ and MQED holds. Let α be an n optimal set for µ of order r. Then for every σ {1,..,} with length k a set β σ exists such that α = S σ β σ σ {1,..,}, σ =k and n i k of the sets β σ are i+1 optimal sets for µ k+1 of order r, while i+1 k n of the sets β σ are i optimal sets for µ k+1 of order r. Moreover, V n,r µ 32 = π k r k i + 1 k n V i,r µ k+1 + n i k V i+1,r µ k+1. Proof To prove Theorem 1 we will use Corollary 1. We divide the proof into two steps. 1. We will show that condition MQED ensures that the set Γ in Corollary 1 is identical with Γ k and that n σ can take only two different adjacent values for every σ Γ. Let Γ {1,..,} be as in Corollary 1. Let l be the number of all appearing lengths in Γ and denote Γ = Γ k1 Γ kl with k 1 < < k l. We study all appearing subcases. Case 1: l = 1. Case 1.1: k 1 = k. Case 1.1.1: max σ Γ n σ min σ Γ n σ 1. We will show that this case is the only one which can happen and all remaining cases end in a contradiction or can be transformed into this case. Case 1.1.2: max σ Γ n σ min σ Γ n σ > 1. Let σ 0 Γ with n σ0 = min σ Γ n σ and σ 1 Γ with n σ1 = max σ Γ n σ. According to Corollary 1 let β σ C nσ,rµ k+1 for every σ Γ such that α = σ Γ S σ β σ is an n optimal set for µ or order r. Let β σ 0 C nσ0 +1,rµ k+1 and β σ 1 C nσ1 1,rµ k+1. For σ Γ we define β σ, if σ Γ \{σ 0,σ 1 } γ σ = β σ 0, if σ = σ 0. β σ 1, if σ = σ 1 Define α = σ Γ S σ γ σ. Clearly, n = cardα = cardα. By definition 1 and the construction of µ we obtain Ψ α,rµ = min x a a α r dµx 1 σ Γ k min x a a S σ γ σ r dµ S σ Ex

18 18 Applying Lemma 1, we get Ψ α,rµ σ Γ From Corollary 1, we deduce = σ Γ = σ Γ 1 k π k r k min x a a S σ γ σ r dµ k+1 Sσ 1 x 33 min x a r dµ k+1 x a γ σ π k r k V cardγ σ µ k+1. π k V n,r µ = r σ Γ k V cardβ σ µ k The combination of 33 and 34 together with the definition of γ σ implies that V n,r µ Ψ α,rµ π k r k V nσ0,rµ k+1 +V nσ1,rµ k+1 V nσ0 +1,rµ k+1 V nσ1 1,rµ k+1. Due to the choice of σ 0 and σ 1, we have n σ0 < n σ0 + 1 n σ1 1 < n σ1. Condition MQED then yields V n,r µ Ψ α,rµ > 0, which contradicts the optimality of α. Thus Case cannot happen. Case 1.2: k 1 = k k. Corollary 1 ensures n = σ Γ n σ k +1 respectively n k. Because k n < k+1, only the case k = k 1 with n σ = for every σ Γ can happen. But in this case, we can apply Proposition 3 and will end into Case with Γ = Γ k and n σ = 1 for every σ Γ. Case 2: l > 1. Case 2.1: a σ 1 Γ k1 exists, with n σ1 <. Case 2.1.1: a σ l Γ kl exists, with n σl > 1. We proceed similar to Case Let β σ C nσ,rµ σ +1 for every σ Γ, such that α = σ Γ S σ β σ is an n optimal set for µ or order r. Let β σ 1 C nσ1 +1,rµ k1+1 and β σ l C nσl 1,rµ kl+1. For σ Γ we define γ σ = β σ, if σ Γ \{σ 1,σ l } β σ 1, if σ = σ 1 β σ l, if σ = σ l. and α = σ Γ S σ γ σ. By the same arguments as in Case we obtain π σ V n,r µ Ψ α,rµ r σ Γ π k 1 r k 1 π k l r k l σ V cardβσ µ σ +1 V cardγσ µ σ +1 V nσ1,rµ k1+1 V nσ1 +1,rµ k1+1 V nσl 1,rµ kl+1 V nσl,rµ kl+1

19 19 By iterative application of equation 30 from Remark 4 and condition MQED, we deduce V n,r µ Ψ α,rµ = π k 1 r k 1 V nσ1,rµ k 1+1 V nσ1 +1,rµ k 1+1 πk 1+1 k l r k V nσl l k 1 1,rµ kl+1 V nσl,rµ kl+1 π k 1 r k 1 V 1,r µ k1+1 V,r µ k1+1 V 1,r µ kl+1 V 2,r µ kl+1 cr k cr k l k l k 1 1 π k 1 r k V 1,r µ k1+1 V,r µ k1+1 1 V,r µ k1+1 V +1,r µ k1+1 > 0, which contradicts the optimality of α. Thus, Case does not appear. Case 2.1.2: n σ = 1 for every σ Γ kl. Case : l = 2. Case : n σ 1 for every σ Γ k1. From Remark 8 we know that the sets β σ, σ Γ k2, which are optimal for µ k 2+1, can be clustered to sets β σ, σ Γ with Γ = {σ : σ Γ k2 }, which are optimal for µ k 2. If k 2 1 = k 1 we end into case If k 2 1 > k 1 we proceed as in case and end in a contradiction. Case : a σ Γ k1 exists, with n σ < 1. First, we cluster Γ k2 to Γ as described in Case If k 2 1 = k 1, we proceed as in Case and end in a contradiction. If k 2 1 > k 1, we adopt the arguments of Case 2.1.1, yielding a contradiction as well. Case : l > 2. Due to k 2 1 > k 1, we proceed as in Case and end in a contradiction. Case 2.2: n σ = for every σ Γ k1. Case 2.2.1: l = 2. Case : n σ 2 for every σ Γ k2. Case : n σ = 2 for a σ Γ k2. We subdivide Γ k1 according to Remark 6 into Γ {1,..,} such that Γ k1 = {σ : σ Γ }. For every τ Γ k1 then the optimal set β τ is divided into 1 optimal sets β τ1,..,β τ. If k 2 1 = k 1 we end in case If k 2 1 > k 1 we proceed as in case and end in a contradiction. Case : n σ = 1 for every σ Γ k2. First we subdivide Γ k1 into Γ like in case If k 2 1 = k 1 then we end in case If k 2 1 > k 1, we cluster Γ k2 to Γ as described in Case If k 2 1 = k and = 2, then we proceed as in Case Otherwise, we adopt the arguments of Case Hence, both cases end in a contradiction.

20 20 Case : a σ Γ k2 exists, with n σ > 2. After subdivision of Γ k1 into Γ like in case , we proceed, depending on the value of k 2 k 1, like in case respectively and both ways end in a contradiction. Case 2.2.2: l > 2. Due to k 2 1 > k 1 we can proceed as in Case respectively Case and end in a contradiction. 2. We finish the proof. From Corollary 1 and 1. we deduce for n [i k,i + 1 k [ that two integers m i,m i+1 exist such that V n,r µ = σ Γ k π k r k V n σ,rµ k+1 = π k r k m i V i,r µ k+1 + m i+1 V i+1,r µ k+1 with i m i + i + 1 m i+1 = n. Applying Remark 7 we obtain m i + m i+1 = k. Thus we get m i = i + 1 k n, respectively m i+1 = n i k, which proves equation 32. The representation of the optimal sets follows from above and Corollary 1. Remark 10 Although one may conjecture that conditions SQ and MQED in Theorem 1 can be weakened, they cannot be dropped completely. The example of one-dimensional Cantor distributions cf. section 5.1 will demonstrate that SQ can not even be replaced by the strong separation condition cf. 25. Moreover, a discussion of modified versions of the Sierpinski gasket cf. section 5.2 shows that condition MQED cannot be dropped completely either. Remark 11 The special case of one-dimensional dyadic homogeneous Cantor distributions d = 1, = 2 was investigated by Kesseböhmer and Zhu [15] and Kreitmeier [17]. Although not explicitely stated, both conditions SQ and MQED were necessary in these papers as well. Lemma 3.6 in [15], respectively Proposition 3.11 in [17] indemnifies SQ. The arguments ensuring that condition MQED is satisfied are somewhat hidden, but can be found in Lemma of [15], respectively Lemma of [17]. The appropriate version of Theorem 1 is stated as Proposition 3.7 in [15], respectively Theorem 4.4 in [17]. 3.2 Quantization dimension and quantization coefficient As defined in writing on the topic see e.g. [5], Definition 11.1, we call the lower and D r µ := liminf n D r µ := limsup n r logn logv n,r µ r logn logv n,r µ the upper quantization dimension of µ of order r. If the two numbers agree, their common value D r µ = D r µ = D r µ is called quantization dimension of µ of order r. First introduced by Zador [23], the quantization dimension for singular distributions was studied by

21 21 Graf and Luschgy cf. [5], [6] in the self-similar case. They have shown that the quantization dimension always exists in this case, if the open set condition is satisfied. Lindsay has shown cf. [19], Example 5.5 that the quantization dimension needs not to exist for singular distributions that are not self-similar. Kesseböhmer and Zhu [15] studied the quantization dimension for homogeneous singular distributions on Cantor-like sets in the 1-dimensional case. To this end, they determined the structure of the optimal sets under the restriction that c = sup i c i 1 4. Kreitmeier [17] extended this codebook result and the related existence characterization of the quantization dimension to the case c 1 3. Several authors cf. [11], [12], [20] studied Hausdorff and packing dimension of even more generalized Cantor-like sets as introduced by us in section 3. They need a boundary distortion condition, which turns in our setting into inf c i > i Hua et.al. cf. [13], section 3.3 were able to determine Hausdorff and packing dimension of Cantor-like sets without using 35. But they needed in our terms the condition logc i lim i logπ i = 0. The inquiries for quantization dimension of generalized one-dimensional Cantor sets cf. [15], Theorem 1.6 3, respectively [17], Proposition 5.1 abstained from 35. Zhu [24] also characterized the quantization dimension for a large class of singular distributions on higher dimensional Cantor-like sets. Due to this more general approach, he needed the boundary condition 35 there. In our setting cf. Proposition 4, we can drop 35, but are restricted to the conditions SQ and MQED. Lemma 2 Assume that condition SQ and MQED are satisfied. Then strict positive constants 0 < M 1 M 2 < exist such that for every n with n [i k,i+1 k [, the relation does hold. M 1 π k r V n,r µ M 2 π k r Proof Immediate consequence of Theorem 1 with M 1 = min l {1,..,} V l,r µ > 0 and M 2 = max l {1,..,} V l,r µ <. Proposition 4 Assume that condition SQ and MQED holds. Then D r µ = liminf n log 1 n n i=1 logc i limsup n Proof From the definition of D r µ and n [ kn, kn+1 [, we deduce V k,r µ V n,rµ respectively D r µ = liminf n liminf k r logn logv n,r µ r log k+1 logv k,r µ kr log + r log = liminf k logv k,r µ. log 1 n n i=1 logc i = D rµ. 36

22 22 Because V n,r µ n tends to zero cf. [5], Lemma 6.1, considering Lemma 2 we obtain D r µ liminf k liminf k kr log logv k,r µ kr log logm 2 π k r = liminf n log 1 n n i=1 logc i. By the same arguments the lower inequality can be shown. The identity for D r µ follows similarly. Subsequently to the characterization of the quantization dimension we can ask whether the quantization coefficient Q r µ exists. The following Proposition shows that under weak assumptions the quantization coefficient does not exist. Proposition 5 Let r > 1 and assume that SQ and MQED does hold. Further assume that converges in R. Then D r exists and log 1 n n i=1 logc i n i limsup n n Dr r V n,r µ <, if and only if limsup k Dr k π k <, ii liminf n n Dr r V n,r µ > 0, if and only if liminf k Dr k π k > 0. iii Let sup k c k = c < 1. If 0 < liminf k Dr k π k and limsup k Dr k π k <, then 0 < liminf n r n Dr V n,r µ < limsup n n r Dr V n,r µ <. 37 Proof Proposition 4 ensures the existence of the quantization dimension D r. The assertions i and ii are an immediate consequence of Lemma 2. iii Let x ]1,2[ and consider the mapping x sx = logx log 1 2 x. 38 Clearly, sx < 1. Using de l Hospital we obtain lim x 1 sx = 1. Thus, we can choose an α ]1,2[ such that log logc < sα < 1. As a consequence, we get D r = liminf n log 1 n n i=1 logc i log < sα < 1. logc ow, let n k = k and m k = α k k with α k = [α k ]/ k for every k. Here [α k ] denotes the integer part of α k Dr. Using 32, we deduce n r k V nk,rµ = k Dr r π k r V 1,r µ k+1, respectively Dr m r k Dr = α r k V mk,rµ k r Dr π k r k 2 k α k k V 1,r µ k+1 + α k k k V 2,r µ k+1 Dr 2 α k α r k k r Dr π k r V 1,r µ k+1.

23 23 Clearly, α k k α. Together with 38, we obtain Dr m r k Dr n r k V mk,rµ Dr = α r k V nk,rµ Combining this with i and ii, we get 2 α k k α r Dr 2 α > α 1 sα 2 α = 1. 0 < liminf n Dr r V n,r µ liminf n < limsup k m Dr r k k n Dr r k V nk,rµ V mk,rµ limsupn Dr r V n,r µ <. n Remark 12 It remains an open question whether the existence characterization 36 of the quantization dimension does also hold, if only condition SQ is required. The techniques developed by Zhu [24] need the boundary condition inf i c i > 0 and therefore cannot be used here. Moreover, it remains unanswered whether it is possible to find conditions, which are equivalent to 36. Remark 13 It is also still an open question whether 37 remains valid, if we drop some or all conditions in Proposition 5 iii. 3.3 Criteria for the imposed conditions In order to make our results applicable, we need a criterion ensuring that condition SQ, respectively MQED is satisfied. Such a criterion may only depend on V, ω and c k k. ote that for every integer m and i {1,..,} the similitude S m,i can be written as S m,i = c m O m,i x i + x i 39 with an orthonormal map O m,i. cf. [14], Proposition Let us first note a simple but essential fact. Lemma 3 It is ω V. Proof Let m and i {1,..,}. Because S m,i. is a similitude, we get from relation 24 that S m,i V V. 40 ow, we proceed indirectly. Assume that x i / V. Thus, we get sup{δ > 0 : Bx i,δ V = /0} =: δ i > 0. Moreover, a w V exists such that x i w = δ i. Using 39, we obtain S m,i w S m,i Bx i,δ i = BS m,i x i,c m δ i = Bx i,c m δ i. Because c m < 1, we have Bx i,c m δ i V = /0. Hence, S m,i w / V, which contradicts relation 40. In the sequel, we will need the constant d 0 = d minω 10diamV.

24 24 Proposition 6 If c = sup i c i < d 0 and r > 1, then SQ holds with κ = ω and l m = c diamv for every m. Proof Let m and l m = c diamv. The construction of E m V and relation 24 implies E m V V. Using 39 and Lemma 3, we obtain S m,i E m V S m,i V Bx i,c diamv, 41 which proves part a1 of condition SQ with κ = ω. Due to c m < d minω 2diamV, the balls Bx,l m,x ω are pairwise disjoint. Thus, the construction of µ m and 41 yields µ m Bx,l m = 1 for every x ω. Hence, µm is an l m,ω separated probability measure. Because of l m < d 0 diamv = d minω 10, we can apply Proposition 1 b and deduce that µ m is separately quantizing. Thus, a2 of SQ is also proved. ow, we need to develop a criterion for condition MQED. To this end, we define for any Borel probability distribution ν, r 1 and > 2 the value D min r,,ν = min{v n,r ν V n+1,r ν V n+1,r ν V n+2,r ν : n {1,.., 2}} = min{v n,r ν +V n+2,r ν 2V n+1,r ν : n {1,.., 2}}. Proposition 7 Let m and > 2. Let D min r,,u ω > 0 and D min d d 1 = 0, min r,,u ω, if > 2 4r diamωr 1 diamv d 0, if = 2 If c < d 1 and r > 1, then condition MQED and SQ does hold. Proof We divide the proof into two steps. First, we will show that D min r,, µ m > 0. Due to c < d minω 2diamV, we know that µm is an l,ω separated probability measure with l = c diamv. Applying Proposition 2, we obtain D min r,, µ m > D min r,,u ω 4lrl + diamω r 1 dmin ω r 1 D min r,,u ω 4c diamv r + diamω r 1 D min r,,u ω c diamv 4r 10 diamω > 0. Secondly, we will show that V 1,r µ m V,r µ m > V,r µ m V +1,r µ m. Applying Proposition 1 b, we deduce > V 1,r µ m V,r µ m V,r µ m V +1,r µ m V,r µ m 2V,r µ m +V +1,r µ m.

25 25 With Proposition 3 and 30, the right hand side turns into 1 V,rµ m cr m V 1,r µ m+1 V 2,r µ m+1 = cr m V 2,rµ m+1 0. Combining the two steps, we recognize that MQED does hold. Remark 14 If we replace D min r,,u ω by D min r,,u ω in the definition of d 1, it is easy to see that D min r,,u ω < 0 implies D min r,, µ < 0, provided c < d 1. Remark 15 According to [5], Lemma 3.2 a, the optimal quantization error scales with s r under a similitude with scaling number s > 0. Thus, any criterion ensuring that SQ respectively MQED holds, should be invariant under such a similarity transformation of the whole model. Indeed, each of our criteria in this section satisfies this request. ote also that d 0, respectively d 1 in the criteria above do not depend on the cardinality of ω. Our criteria, ensuring that SQ does hold, is working with κ = ω and l m independent of m. evertheless SQ is not restricted to this special setting. 4 The self-similar case If c = c m is independent of m, we have µ = µ m for every m. Thus, µ becomes self-similar. In this case, we have D = D r µ = log logc for the quantization dimension cf. [6] because the similitudes satisfy the open-set condition. In this self-similar case, we can determine all accumulation points of n D r V n,r µ n. To this end, let f c : [1,] R with f c x = x r D i + 1V i,r µ iv i+1,r µ xv i,r µ V i+1,r µ, if x [i,i + 1[, i {1,.., 1} and f c x = V 1,r µ, if x =. By adapting arguments from [4] and [17] in addition to Theorem 1 we get the following result. Proposition 8 Let r > 1 and n. Let k 1 and i {1,.., 1} such that i k n < i + 1 k. Assume that condition SQ and MQED holds. Then V n,r µ = ckr k i + 1 k n V i,r µ + n i k V i+1,r µ. 42 If, furthermore, c 1, then the set of all accumulation points of the sequence n r D V n,r µ n equals the interval f c [1,], which contains more than one point. Proof Equation 42 follows immediately from Theorem 1. Due to lim x i 0 f c x = f c i for every i {2,..,}, the function f c is continuous and f c [1,] is an interval. Moreover, we see for every i {1,.., 1}, that f c is differentiable on ]i,i + 1[. Using r > 1 and c 1, we deduce D r = r logc log > 1. Thus, for all x ]i,i + 1[ we get f cx = r D x D r 1 i + 1V i iv i+1 + x D r 1 + r D V i+1 V i. Hence, f c is continuous on ]i,i + 1[ and has at most one zero in ]i,i + 1[. Thus, f c is not constant, yielding that f c [1,] contains more than one point. It remains to prove that f c [1,] equals the set of all accumulation points of the sequence n r D V n,r µ n. Let y f c[1,].

26 26 Then an i 0 {1,.., 1} and x [1,] exist, with x [i 0,i 0 +1] and y = f c x. Let k and define n k := [x k ] as the greatest integer, smaller or equal to x k. Obviously, an x k [1,] exists, with n k = x k k. The sequence x k k converges to x. Due to kn k = k and in k = i 0 we get from 42 D n r k V nk,rµ = x k k D r ckr k V i0,rµi k x k k +V i0 +1,rµx k k i 0 k. Because of c D = 1, we deduce n r D k V nk,rµ = x k r D i0 + 1V i0,rµ i 0 V i0 +1,rµ x k V i0,rµ V i0 +1,rµ = f c x k. The continuity of f c implies y = f c x = lim k f c x k. Thus, y is an accumulation point of n r D V n,r µ n. ow, let y be an accumulation point of n D r D V n,r µ n. Then a subsequence n r l V nl,rµ l exists, with D y = lim n r l l V nl,rµ. W.l.o.g. we can assume that kn l < kn l + 1 for all l. Otherwise, we take a suitable subsequence. Let x l := n l. Then x kn l l [in l,in l + 1] and 42 imply f c x l = n r D l V nl,rµ. Because x l l is bounded, a subsequence x lq q exists, converging against a x [1,]. According to the continuity of f c, we obtain yielding y f c [1,]. Remark 16 Equation 42 implies D f c x = lim f c x lq = lim n r q q lq V nlq,rµ = y, for every n. 43 turns into an equation, if V n,r µ V n+1,r µ V n+1,r µ V n+2,r µ 43 in kn n < in + 1 kn 1. Hence, the strict monotonicity for n = 1,.., 1 cf. condition MQED is not carried over to all n. Remark 17 If condition SQ and MQED are satisfied and c < 1, then the non-existence of the quantization coefficient follows from Proposition 5, but is also an immediate consequence of Proposition 8.

27 27 Remark 18 Let n. An n tuple s = s 1,..,s n with s i > 0 is called arithmetic, if a t > 0 exists such that s i /t. For such an arithmetic s, we call ϑ := max{t > 0 : s i /t for every i = 1,.,n} the period of s. Hence, the tuple logc,.., logc with c as contracting parameter is arithmetic with period logc. In this arithmetic case, Pötzelberger [21], Theorem 1 has shown that a periodic function h : R + R + with period logc exists such that lim n D r V n,r µ hlogn = 0. n In contrast to our special setting, this function h in [21] is not calculated explicitly. Remark 19 The construction of the Cantor-like set E starts with the closed set V. Hutchinson [14] has shown for the self-similar case that a replacement of V by any non-empty compact set K R d with i=1 S i K K leads always to the same Cantor-like set E. 5 Modified versions of classical fractals 5.1 One-dimensional Cantor distribution Let d = 1 and ω = {0,1}. Let r > 1 and c k be a sequence of contracting factors, satisfying c k ]0, 2 1 ] for every k. We consider the similitudes S k,1x = c k x, respectively S k,2 x = c k x + 1 c k. We choose V = ]0,1[. Obviously, the open set condition is satisfied. The uniform distribution µ on E = EV is called one-dimensional Cantor distribution. If c k = c independent of k, the distribution µ becomes self-similar. In the sub-case of c = 1 3, the measure µ is the classical Cantor distribution. If c = sup i c i < 10 1, Proposition 7 ensures that the conditions SQ and MQED are satisfied. Applying Theorem 1, we can solve the optimal quantization problem in this case. Using more refined methods, Kesseböhmer and Zhu cf. [15] proved the result for the optimal codebooks in Theorem 1 under the relaxed condition c 1 4. In another work, Kreitmeier [17] has shown that all assertions of Theorem 1 remain true, if c 1 3. ote that the solution of the optimal quantization problem in the classical case c k = 1 3 for every k goes back to Graf and Luschgy cf. [4]. Thus, our results in this paper for the one-dimensional Cantor distribution are already incorporated in other works. In this context, it is interesting to see that Theorem 1 becomes wrong, if c > 1 3. To show this, let a l l be a zero sequence with sup l a l 3 1. For k,l and c ]0, 2 1 [ let { c, if k 2 c k,l := a l, if k = 2 Denote µ l as the uniform distribution on E defined by c k,l k. Due to c 1,l = c and because c 2,l l is a zero sequence, the series µ l l converges weak against ν c = 1 4 δ 0 + δ c + δ 1 c + δ 1. Moreover, x r dµ l x x r dν c x, because the support of µ l and ν c is contained in [0,1] for every l. Using well-known stability and consistency results cf. [5], p. 57 ff, one gets V n,r µ l V n,r ν c. 44

28 28 Moreover, a sequence α l l of codebooks exists, with α l C n,r µ l for every l and ow assume that Theorem 1 is valid. Hence, d H α l,c n,r ν c l 0 45 C 3,r µ l = {{ c 2,1 c + ca l 2,1 ca l 2 },{ ca l 2,c ca l 2,1 c 2 }}. In case of c 1 3 it is easy to see that C 3,r ν c = {{ c 2,1 c,1},{0,c,1 c 2 }}. but C 3,r ν c = {{0, 1 2,1}}, if c > 1 3. In the last case, the statement about optimal codebooks in Theorem 1 gets wrong because otherwise we would derive with inf l inf{d Hα,β : α C 3,r µ l,β C 3,r ν c } 1 2 c > 0, a contradiction to 45. In the same way, the assumed validity of equation 32 in case of c > 3 1 would lead to 2 1 1V 1,r lim V 3,rµ l = lim l 2 = cr 1 2 V 1,r 2 δ δ 1 = 1 c r 1 1 r > c = V 3,rν c, l c r 1 which contradicts 44. Hence, all statements of Theorem 1 will get wrong in general, if c > 3 1. As a consequence, the condition SQ cannot be replaced by the weaker condition of strong separation cf. 25. Even in the self-similar case, it can be shown cf. [17], Example 6.3 that Theorem 1 becomes invalid, if c k = c > µ 2 l 5.2 Uniform distribution on modified Sierpinski Gasket ow, we intend to apply our results to a set ω consisting of three different points. In view of our criteria cf. section 3.3 for condition SQ and MQED it is essential to know, under which conditions D min r,3,u ω > 0 does hold. At first, we will answer this question in Proposition 9. To this end, we will use properties of the Fermat point of a triangle. After that, we introduce the uniform distribution on a modified version of the Sierpinski Gasket and discuss the optimal quantization of it. Lemma 4 Let ω R d be finite and non-empty, r = 2 and sω = 1 x ω x. Then V 1,r U ω = x sω r du ω x = 1 x sω r. 46 x ω Equation 46 also holds in case of r > 1, if x sω does not depend on x.

29 29 Proof The proof of Theorem 2.4 in [5] shows the strict convexity of R d z F r x z r du ω x R + 0. If F r has only one critical value, it is the global minimum of F r. With as gradient, we calculate F r z = r x z r 2 z x. x ω If r = 2 or if x sω is independent of x ω, we get F r sω = 0. Proposition 9 Let d 2, cardω = 3 and r > 1. Then dmin ω r D min r,3,u ω 4 dmin ω r In general D min r,3,u ω > 0 if r > 2. If and only if d min ω = diamω, 47 turns into an equation. Proof According to [5], Lemma 3.2 a, the optimal quantization error remains unchanged under isometrical mappings. Thus, we can assume w.l.o.g. that d = 2. Let ω = {A,B,C}. Let u = inf{ A P + B P + C P : P R 2 }. A point F R 2 is 1 optimal for Q ω of order 1, this means, {F} C 1,1 Q ω, if and only if u = A F + B F + C F. 48 The minimization problem defined by 48 has a unique solution cf. [10]: exactly one point F convω satisfies equation 48. a If the triangle, spanned by ω, has an angle equal or greater than 2 3 π, then F is the vertex of this angle. b Otherwise, F is the unique point, at which each of the three angles BFC, CFA and AFB - containing F as vertex - does have the value 2 3 π. The point F is called Fermat point. Due to [5], Remark 4.6 a, this minimization problem does not have another solution on R 2 \convω. Hence, the Fermat point is unique in R 2 and we have C 1,1 U ω = {{F}}. If we construct above each side of the triangle ABC an equilateral triangle cf. figure 1, we get the triangles BA C, ACB and BAC. The line segments AA, BB, and CC are all the same length u and their common point of intersection is the Fermat point cf. [10]. Obviously, V 3,r U ω = 0, respectively V 2,r U ω = 2 3 d minω 2 r. We proceed separately for each of the two cases a and b. Case a We obtain u 2d min ω. Thus, the convexity of t t r yields u r 2 r V 1,r U ω 3 3 d dmin ω r minω > Case b First we calculate u. To this end, we use the standard notation a,b and c for the sides of the