SUBDIVISION SCHEMES WITH NONNEGATIVE MASKS 1 ...
MATHEMATICS OF COMPUTATION Volume 74, Number 250, Pages 819–839 S 0025-5718(04)01712-0 Article electronically published on October 27, 2004
SUBDIVISION SCHEMES WITH NONNEGATIVE MASKS XINLONG ZHOU
Abstract. The conjecture concerning the characterization of a convergent univariate subdivision algorithm with nonnegative finite mask is confirmed.
1. Introduction Let Z denote the integer lattice. A univariate subdivision algorithm with a finitely supported mask a = {aj }j∈Z is given as follows: beginning with an initial sequence of data v 0 = {vi0 }, we set recursively new sequences of values v k , by applying the rule ai−2j vjk−1 . vik = j∈Z
This algorithm is said to converge if for each finite v 0 = {vi0 } there exists a continuous function fv such that fv ≡ 0 for at least one v 0 and i (1.1) lim sup |fv ( k ) − vik | = 0. k→∞ i∈Z 2 Obviously, the subdivision algorithm with mask a = {aj }j∈Z converges if and only if the polygon fk defined by the control points (i/2k , vik )T converges uniformly to fv . One can easily check fk (x) = vi0 akj h(2k (x − i) − j), i
j
where h is the hat function and = j ak−1 ai−2j with a1j = aj . Thus (1.1) for j all such v 0 is equivalent to the uniform convergence of akj h(2k x − j). aki
j
On the other hand, define S to be the operator given by Sf (x) = aj f (2x − j). j
One gets S k f (x) =
akj f (2k x − j), k = 1, 2, . . . .
j
Received by the editor December 13, 2002 and, in revised form, January 15, 2004. 2000 Mathematics Subject Classification. Primary 65D17, 26A15, 26A18, 39A10, 39B12. Key words and phrases. Cascade algorithm, joint spectral radius, nonnegative mask, subdivision scheme. c 2004 American Mathematical Society
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XINLONG ZHOU
The convergence of (1.1) is therefore equivalent to the uniform convergence of S k h, which leads to the following cascade algorithm: let g0 (x) = h(x); one defines recursively gk (x) by aj gk−1 (2x − j). gk (x) = j
The limit g = limk→∞ gk = limk→∞ S k h satisfies Sg = g, i.e., (1.2) g(x) = aj g(2x − j). j∈Z
This so-called two-scale delation equation and the above presented algorithms play an important role in wavelet analysis as well as in computer aided geometry design. A comprehensive discussion of this subject can be found in [2, 5, 10]. In order to characterize the convergence of the above algorithms, one uses the concept of joint spectral radius (see [11]). The joint spectral radius for two square matrices A0 and A1 is defined by 1
ρ(A0 , A1 ) = lim sup ||A1 · · · Ak || k , n→∞ k≤n
where || · || is a given matrix-norm and l ∈ {0, 1}. Thus, write a(z) = j aj z j and a(z) = (1 + z)b(z) if a(−1) = 0. The subdivision algorithm with finite mask a = {aj }j∈Z converges if and only if (see [2, 5, 10]) (i) a(1) = 2 and a(−1) = 0, (ii) ρ(B0 , B1 ) < 1 where B0 = [b2j−i ]i,j , B1 = [b2j−i+1 ]i,j and {bl } is the coefficients of b. Unfortunately, the determination of the joint spectral radius is generally NPhard by a result of Tsitsiklis and Blondel (see [13]). Thus, it seems difficult to determinate whether the considered spectral radius is less than one. Some partial results concerning the calculation of the joint spectral radius can be found in [1, 4, 7, 16] and the papers cited therein. In this paper we focus on subdivision algorithms with nonnegative finite masks, a property possessed by many applications in geometric modelling (see, e.g., [3, 12]). A remarkable fact of this class is that the convergence does not depend on the actual values of the mask but rather on the support of the mask I = {i : ai > 0} (see [10, 8]). Consequently the question was raised in [2] (see p. 55 of [2]) of identifying those I such that given any nonnegative mask supported on I, the corresponding subdivision algorithm converges. By applying a suitable translation, we may always assume in the following discussion that a nonnegative mask a = {aj } has the form a = {a0 , . . . , aN } with a0 , aN > 0. We believe that the answer to this question is (see [9]): Conjecture. The subdivision algorithm associated with the nonnegative mask a = {a0 , . . . , aN } converges if and only if the following both hold: (i) a(1) = 2, a(−1) = 0 and 0 < a0 , aN < 1, (ii) the greatest common divisor of {j : aj > 0} is 1. This conjecture is still not verified. On the other hand, it has been shown that the conditions are necessary (see [2, 9, 14]). There are various partial results which support this conjecture. Denote S(a) = {j : aj > 0}. Thus, Micchelli and Prautzsch (see [10]) prove that if (ii) is replaced by S(a) = {0, 1, . . . , N }, convergence follows.
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Gonsor (see [6]) shows that S(a) = {0, 1, . . . , N } can be weakened by {0, 1, N − 1, N } ⊆ S(a), while Melkman (see [9]), among others, proves that if instead of (ii) it holds that S(a) ⊇ {0, p, q, p + q} for gcd(p, q) = 1 or that S(a) contains two successive integers, then convergence follows. Recently, Wang proves this conjecture for a class of masks. One of the main results in [15] is the following Theorem 1.1. The subdivision algorithm with the nonnegative mask a = {aj } converges if instead of (ii) it holds that {r, p, q} ⊆ S(a) such that gcd(p−r, q−r) = 1 and q − r is even. Improving the technique introduced in [15], we can now confirm this conjecture. In next section we first collect some lemmas from [15] and prove the conjecture by using a lemma (see Lemma 2.4), which will be verified in Section 3. The proof of Lemma 2.4 is the kernel of this paper. 2. Proof of the conjecture Let us collect some results from [15]. For this goal we should denote the vector x in RN by x = (x0 , . . . , xN −1 )T . For any T ⊂ ZN , where ZN = {0, 1, . . . , N − 1}, we define IT to be the vector (x0 , x1 , . . . , xN −1 )T with xi = 1 if i ∈ T and xi = 0 otherwise. For any nonnegative N × N row stochastic matrix B, i.e., the sum for each row is one, we define a map FB such that FB (T ) = {j ∈ ZN : (BIT )j = 1}. Let A0 and A1 be N × N matrices deduced by the mask a: A0 = [a2j−i ]0≤i,j≤N −1
and
A1 = [a2j−i+1 ]0≤i,j≤N −1 .
Clearly, A0 and A1 are row stochastic matrices if a(1) = 2 and a(−1) = 0. We should write simply F0 = FA0 and F1 = FA1 . For convenience we should also use the following standard notation for algebraic sums of sets: let T be a set of integers; then for any integers α and β the set αT + β is given by αT + β := {αx + β : x ∈ T }. We have (see [15]) Lemma 2.1. Let B be a N × N nonnegative row stochastic matrix. Then FB (T1 ) and FB (T2 ) are disjoint if T1 , T2 ⊆ ZN are. Let C be an N × N nonnegative row stochastic matrix. Then FBC = FB ◦ FC . Furthermore, the subdivision algorithm with nonnegative mask a, which satisfies a(1) = 2 and a(−1) = 0, diverges if and only if there exist disjoint nonempty subsets T and T of ZN and a sequence (d1 , d2 , . . . , dm ) ∈ {0, 1}m for some m ≥ 1 such that (2.1)
T = Fdm ◦ · · · ◦ Fd1 (T ) and T = Fdm ◦ · · · ◦ Fd1 (T ).
Denote for S(a) the sets S0 = S(a) ∩ (2Z) and S1 = S(a) ∩ (2Z + 1). The map Ψ for T ⊂ Z is defined by (2T − q) ∪ (2T − q) . Ψ(T ) = q∈S0
q∈S1
The following relationship between Ψ and Fi is proved in [15].
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XINLONG ZHOU
Lemma 2.2. Let a be a nonnegative mask satisfying a(1) = 2 and a(−1) = 0. Then for any T ⊂ ZN , we have F0 (T ) = Ψ(T ) ∩ ZN , F1 (T ) = (Ψ(T ) + 1) ∩ ZN .
(2.2)
Furthermore, for any (d1 , . . . , dm ) ∈ {0, 1}m we have (2.3) where k =
m j=1
Fd1 ◦ · · · ◦ Fdm (T ) = (Ψm (T ) + k) ∩ ZN , dj 2j−1 .
We may write S(a) = {0, p1 , . . . , pk } with pk = N . Generalizing Wang’s lemma (see [15]), we have Lemma 2.3. Let T be a subset of Z and let y ∈ T . Let i,j ∈ {0, 1} such that k
i,j ≤ 1, ∀ i = 1, 2, . . . , m.
j=1
Then it holds that 2m y −
k m
2m−i i,j pj ∈ Ψm (T ).
j=1 i=1
Proof. For m = 1 the assertion is clear since 2y − xpl ∈ Ψ(T ) for all x = 0, 1 and l = 1, 2, . . . , k. On the other hand, 1,1 + · · · + 1,k ≤ 1 implies either 1,j = 0 for j = 1, . . . , k or else 1,j = 0 for j = j and 1,j = 1. Thus 2y − 1,1 p1 − 1,2 p2 − · · · − 1,k pk ∈ Ψ(T ).
The general case follows from induction on m.
Let ϕ(n) be the Euler function of the number of elements in Zn that are co-prime with n. It is known that if gcd(k, n) = 1, then k ϕ(n) ≡ 1
(mod n).
In particular, for any odd integer n one has 2ϕ(n) ≡ 1
(2.4)
(mod n).
Now we are in the position to prove the conjecture. Proof of Conjecture. The necessity of the conditions is proved in [9]. To show the sufficiency, we denote S(a) = {0, p1 , . . . , pk } with pk = N and |S(a)| for the number of S(a). The first assumption of the conjecture implies that |S(a)| ≥ 3 and there exists an odd p in S(a) such that p = pj0 < pk no matter if pk is odd or even. Hence, for |S(a)| = 3 one has S(a) = {0, p, N } with even N . Wang’s result for |S(a)| = 3 (see [15]) may be read as: the subdivision algorithm converges if gcd(p, N ) = 1. We assume in the following that |S(a)| ≥ 3. Let us observe that if the subdivision algorithms with the mask a = {aj } diverges, then by Lemma 2.1 there are a sequence (d1 , . . . , dm ) and disjoint nonempty sets T, T ⊂ ZN so that (2.5)
Fd1 ◦ · · · ◦ Fdm (T ) = T and Fd1 ◦ · · · ◦ Fdm (T ) = T .
It follows from Lemma 2.2 (see (2.3)) that
T = (Ψm (T ) + k ) ∩ ZN and T = (Ψm (T ) + k ) ∩ ZN ,
SUBDIVISION SCHEMES
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m where k = l=1 dl 2l−1 . Write F = Fd1 ◦ · · ·◦ Fdm . Then F η (T ) = T and F η (T ) = T for any η ≥ 1. Therefore, with
kη
m m m = ( dl 2l−1 ) + ( dl 2l−1 )2m + · · · + ( dl 2l−1 )2(η−1)m l=1 η−1 im
= k
l=1
l=1
2
i=0
we get for any η ≥ 1 (2.6)
T = (Ψm η (T ) + kη ) ∩ ZN and T = (Ψm η (T ) + kη ) ∩ ZN .
We should choose η in connection with S(a). For our goal we define η to be (2.7)
η=t
pk
{jϕ(j)}
j=1
with some positive integer t, which will be given later. Denote m = m η. We have 2m ≡ 1 (mod x) for odd number x from {1, . . . , pk }. Indeed, η has a factor xϕ(x), so for some integer n one has η = n xϕ(x). Since x is odd, we obtain by Euler’s formula (2.4)
2m = 2m n xϕ(x) ≡ 1
(2.8)
(mod x).
Similarly, we have kη ≡ 0
(2.9)
(mod x).
To see this, one gets by the expression of kη kη
=
k
x−1 ϕ(x)(l+1)−1 n
l=0
=
k k
2jm 2ϕ(x)lm
2jm
(mod x)
j=0
ϕ(x)−1
k n x
j=0
x−1 ϕ(x)−1 n
l=0
≡
j=ϕ(x)l
x−1 ϕ(x)−1 n
l=0
≡
2jm
2jm
(mod x)
j=0
≡
0 (mod x),
which yields (2.9). Moreover, kη = 0
⇐⇒
d1 = · · · = dm = 0
and kη = 2m − 1
⇐⇒
d1 = · · · = dm = 1.
For all other dj we can thus choose t large enough such that for some 0 < δ < 1 (2.10)
pk < kη = k
2m − 1 ≤ (1 − δ)2m 2 m − 1
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XINLONG ZHOU
and
2m η (2m − 1 − k ) + k (2.11) 2 − kη = > 2pk . 2 m − 1 Additionally, in view of (2.5) we may without loss of the generality assume that kη is even if kη = 2m − 1. In the following discussion we should always denote m and kη to be any nonnegative integers, which satisfy the above properties, i.e., 0 ≤ kη ≤ 2m ; if kη = 0, 2m − 1, then kη is even, m and kη satisfy (2.8)–(2.11). We know that the divergence of the subdivision algorithm implies that there are two disjoint nonempty sets T, T ⊂ ZN for (2.5). We may therefore assume x ∈ T and x ∈ ZN . So from Lemma 2.3 it holds that m
2m x −
k m−1
2i i,j pj ∈ Ψm (T )
j=1 i=0
for all i,j ∈ {0, 1} such that (2.12)
k
j=1 i,j
2m x + kη −
≤ 1. By (2.6) we obtain
k m−1
2i i,j pj ∈ T
j=1 i=0
for all those i,j . Denote B0 = {x} with x ∈ ZN and for l = 0, 1, . . . , Bl+1 = {2m y + kη −
k m−1
2i i,j pj ∈ ZN | y ∈ Bl , i,j ∈ {0, 1},
j=1 i=0
k
i,j ≤ 1}.
j=1
Thus, (2.12) tells us that x ∈ T implies Bl ∩ T = ∅ for all l ≥ 0. In the next section we will prove the following Lemma 2.4. Let the mask a = {aj } satisfy the conditions of the conjecture and S(a) = {0, p1, . . . , pk }. Then there is an 0 ≤ x < pk such that with B0 = {x} one has Bl = Zpk . l≥0
This lemma, together with the above consideration, implies that the set T or T must be empty. This contradiction proves the conjecture. The proof of Lemma 2.4 is much more involved and, in fact, is the kernel of this paper. We prove this lemma in the next section. 3. Proof of Lemma 2.4 The notation such as Bj , kη and m have the same mean and properties as in Sections 1 and 2. Let us begin with the observation of the first condition of the conjecture. We need the following Definition. A set of integers I = {0, q1 , . . . , qk } with 0 < qj < qj+1 has property P if either qk is even and I contains at least one odd integer or else I contains at least two odd and two even integers (including 0). Obviously, if a = {aj } satisfies the first condition of the conjecture, the set S(a) has property P. Conversely, if a set {0, q1 , . . . , qk } has property P, then one can construct a mask a = {aj } satisfying the first condition of the conjecture and
SUBDIVISION SCHEMES
825
S(a) = {0, q1 , . . . , qk }. In fact, in [9] this conjecture is formulated by using property P. We may therefore reformulate Lemma 2.4 as Lemma 2.4 . If S(a) = {0, p1 , . . . , pk } has property P and gcd(p1 , . . . , pk ) = 1, then there is an x ∈ Zpk such that with B0 = {x} (3.1) Bl = Zpk . l≥0
To prove Lemma 2.4 or Lemma 2.4 , we remember that {i,j } ∈ {0, 1} for i = 0, 1, . . . , m − 1, j = 1, . . . , k and k
i,j ≤ 1,
i = 0, 1, . . . , m − 1.
j=1
Let δi,j = i,j for i = 0, 1, . . . , m − 1, j = 1, . . . , k − 1 and δi,k = 1 −
k
i,j .
j=1
Hence, {δi,j } satisfies our conditions. We obtain (3.2)
−
2m x + kη −
k m−1
δi,j 2i pj = (pk − 1)
j=1 i=0
2m (pk − 1 − x) + (2m − 1 − kη ) −
k−1 m−1
i,j 2i (pk − pj ) −
j=1 i=0
m−1 i=0
i,k 2i pk
.
We notice that 2m − 1 − kη is between 0 and 2m − 1. Formulas (2.8) and (2.9) imply that (2.9) also holds for kη = 2m − 1 − kη instead of kη . Let Bˆj = pk − 1 − Bj . Then Bˆj is defined by {0, pk −pk−1 , . . . , pk −p1 , pk } and kη . Since gcd(p1 , . . . , pk ) = gcd(pk − pk−1 , . . . , pk − p1 , pk ), the symmetric relation (3.2) tells us that ˆl . v∈ B Bl ⇐⇒ pk − 1 − v ∈ l≥0
l≥0
Moreover, if {0, p1 , . . . , pk } has property P, then {0, pk − pk−1 , . . . , pk − p1 , pk } does also. Thus, it is enough to show Lemma 2.4 either for {0, p1 , . . . , pk } or for {0, pk − pk−1 , . . . , pk − p1 , pk }. For simplicity we define B(x) = Bl l≥0
with B0 = {x} ⊂ Zpk . The following assertions will be used frequently and are based on some special choices of {i,j }. We formulate them as four lemmas. Lemma 3.1. Let y ∈ B(x) and 0 ≤ y ≤ pl1 for some odd pl1 ∈ S(a). If 0 < (2m − 1)y + kη ≤ (2m − 1)pl1 , then y + jpl1 ∈ B(x) for j such that y + jpl1 ∈ Zpk . Proof. We choose i,j = 0 for j = l1 . By the definition of i,j the number l = m−1 i m i=0 i,l1 2 can be any number between 0 and 2 − 1. On the other hand, in view
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XINLONG ZHOU
of (2.8) and (2.9) we have (2m − 1)y + kη = αpl1 . Hence, y + (2 − 1)y + kη − m
m−1
i,l1 pl1 2i = y + (α − l)pl1 .
i=0
The choice of η (see (2.7)) implies kη > pk and 2m − kη > 2pk if kη = 0 and 2m − 1, while for kη = 0, the inequality (2m − 1)y > 0 yields y ≥ 1. Moreover, if kη = 2m − 1, then the condition implies y < pl1 . Hence α − l can be any number between −pk and pk . In particular, for l such that y + (α − l)pl1 ∈ Zpk we have y + (α − l)pl1 ∈ B(x). Lemma 3.2. Let y ∈ B(x) and pl1 ∈ S(a) be even. Assume pl1 /2 ≤ y and (2m − 1)(y − pl1 /2) + kη ≤ (2m−1 − 1)pl2 for some odd pl2 ∈ S(a). Then y − pl1 /2 ∈ B(x). Proof. We choose m−1,l1 = 1 and i,l1 = 0 if i = m−1. For other j = l2 let i,j = 0. m−2 To meet the condition on i,j , we have to choose m−1,l2 = 0. Now i=0 i,l2 2i can be any number between 0 and 2m−1 − 1. From (2.8), (2.9) and our restriction on y we conclude (2m − 1)(y − pl1 /2) + kη = αpl2 for some 0 ≤ α ≤ 2m−1 − 1. i Hence, we can choose i,l2 for i = 0, 1, . . . , m − 2 to obtain m−2 i=0 i,l2 2 = α. With this choice of i,j we get 2m y + kη −
k m−1
i,j pj 2i = y − pl1 /2.
j=1 i=0
Thus, y − pl1 /2 ∈ B(x).
Lemma 3.3. Let y ∈ B(x). If there is some odd pl1 ∈ S(a) such that (2 − 1)y + kη < (2m − 1)pl1 , then for any pj ∈ S(a) one has (i) y+(2l+1)pl1 −pj ∈ B(x) whenever y+(2l+1)pl1 −pj ∈ Zpk and (2m −1)y+kη is odd; (ii) y + 2lpl1 − pj ∈ B(x) whenever y + 2lpl1 − pj ∈ Zpk and (2m − 1)y + kη is even. m
Proof. Clearly (2m − 1)y + kη = αpl1 for some 0 ≤ α < 2m − 1. If α is odd, we m−1 choose i,l1 such that α − i=0 i,l1 2i = 2l + 1. Obviously, 0,l1 must be zero. Hence, we can choose 0,j = 1 or 0 to obtain 2m y + kη −
k m−1
i,j pj 2i = y + (2l + 1)pl1 − 0,j pj .
j=1 i=0
Since y ∈ B(x), the set B(x) contains y + (2l + 1)pl1 − 0,j pj whenever y + (2l + 1)pl1 − 0,j pj ∈ Zpk . Similarly, we have the second assertion. Lemma 3.4. Let y ∈ B(x) and pl2 −pl1 > 0 be odd for some pl1 , pl2 ∈ S(a). Assume 0 < (2m − 1)(y − pl1 ) + kη < (2m − 1)(pl2 − pl1 ). Then, y + j(pl2 − pl1 ) ∈ B(x) for j ∈ Z such that y + j(pl2 − pl1 ) ∈ Zpk . Proof. Let i,l1 = 1 − i,l2 and i,j = 0 for j = l1 , l2 . We obtain 2 y + kη − m
k m−1 j=1 i=0
i,j pj 2
i
= y + (2 − 1)(y − pl1 ) + kη − m
m−1
i,l2 (pl2 − pl1 )2i
i=0
= y + (2m − 1)(y − pl1 ) + kη − l(pl2 − pl1 ),
SUBDIVISION SCHEMES
827
where l can be any integer between 0 and 2m − 1. On the other hand, our condition, (2.8) and (2.9) imply pk ≤ (2m − 1)(y − pl1 ) + kη = α(pl2 − pl1 ) < (2m − 1)(pl2 − pl1 ). Hence, for any nonnegative integer j such that y + j(pl2 − pl1 ) ∈ Zpk we can choose l ∈ [0, 2m − 1] so that α − l = j, i.e., y + j(pl2 − pl1 ) ∈ B(x). On the other hand, the condition implies that y + 1 ≤ pl2 if kη = 2m − 1 and y + 2 ≤ pl2 if kη = 2m − 1. Thus, for any negative integer j such that y + j(pl2 − pl1 ) ∈ Zpk we can choose l ∈ [0, 2m − 1] so that α − l = j, i.e., y + j(pl2 − pl1 ) ∈ B(x). We are now ready to verify Lemma 2.4 . We prove this lemma using induction on |S(a)|. The assertion (3.1) for |S(a)| = 3 is essentially given by Wang in [15] (see the proof of Theorem 1.1 in [15]). Proof of Lemma 2.4 . The idea behind the proof is that first we prove the assertion for a small p instead of pk , i.e., {0, . . . , p − 1} ⊆ B(x). If p is odd and belongs to S(a), we can then extend this to all of {0, 1, . . . , pk −1} by using Lemma 3.1. In order to obtain the result for all of {0, 1, . . . , p − 1}, we choose carefully an x in this set and apply Lemmas 3.1–3.4 to get another number, which is greater than p. Using Lemma 3.2 or Lemma 3.4, we may reduce the number so that the new number is again contained in {0, . . . , p − 1} but is different from x. Recursively, we obtain several numbers. One of our tasks is to show that those numbers are all different. A careful choice of x is necessary. Otherwise the following calculation leads to B(x) = {0}. In fact, if kη = 0, then for x = 0 and B0 = {0} one gets Bl = {0} for all l = 1, 2, . . . . Thus, in our choice of x it may be understood that we avoid zero if kη = 0 or pk − 1 if kη = 2m − 1 during the calculation of new numbers. In other words, we should choose such an x that zero or pk − 1 occurs at the last step of our calculation. Next let us observe the set S(a) with |S(a)| ≥ 5. It is easy to check that for some nonzero p ∈ S(a) the set S(a) \ {p} also has property P. However, for sets with |S(a)| = 4 we may not have this p. The only set that does not have this p is {0, p1 , p2 , p3 } with even p1 and odd p2 , p3 . We already know that |S(a)| ≥ 3. Since the assertion for |S(a)| = 3 is essentially given in [15], we should omit the proof for this case. Our approach, therefore, has two steps; namely, the proof for the special case of |S(a)| = 4 and for |S(a)| ≥ 4. Let us begin with the special case. Step 1. S(a) = {0, p1 , p2 , p3 } with even p1 and odd p2 , p3 . We divide the proof into three cases according to the different position of p1 , p2 and p3 . Case 1. Let us prove (3.1) for 2p2 < p3 . Assume kη = 2m − 1. Denote gcd(p1 , p2 ) = d and p = p1 /2. The 0 ≤ x < p is given as follows: x + p3 ≡ 0 (mod d), d = 1, x − p2 ≡ 0 (mod p), d = 1. With this x we have (3.3)
{x + dv : 0 ≤ x + dv ≤ p} ⊆ B(x).
To see this, we apply Lemma 3.1 to get x+p2 ∈ B(x). Using Lemma 3.2 for pl1 = p1 and pl2 = p3 , we conclude x + p2 − p ∈ B(x). Repeatedly we get x + p2 − α1 p ∈ B(x) whenever 0 ≤ x + p2 − α1 p < p. We notice that x + p2 − α1 p satisfies the above restriction. Thus, recursively, i.e., replacing x by x + p2 − α1 p, we conclude
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XINLONG ZHOU
x + sp2 − αs p ∈ B(x) and 0 ≤ x + sp2 − αs p < p for s = 0, 1, . . . , p/d − 1, where α0 = 0. We note also that the choice of x ensures x + sp2 − αs p = 0 at least for s = 0, 1, . . . , p/d − 2, while for s = p/d − 1 one has x + sp2 − αs p = 0, if d = 1, x + sp2 − αs p = 0, if d = 1. Moreover, it follows from gcd(p2 , p) = d that x + sp2 − αs p = x + s p2 − αs p if 0 ≤ s < s ≤ p/d − 1. Therefore, {x + sp2 − αs p : s = 0, 1, . . . , p/d − 1} = ⊆
{x + dv : 0 ≤ x + dv < p} B(x).
Examining the way we calculate the number x + sp2 − αs p for s = p − 1 and d = 1, we obtain additionally p ∈ B(x), proving (3.3). Having (3.3), we will replace p in (3.3) by p2 . We use the same procedure as above. Let x + dv be any nonzero element from {x + dv : 0 ≤ x + dv ≤ p}. Lemma 3.1 leads to x + dv + p2 ∈ B(x) while Lemma 3.2 implies x + dv + p2 − lp ∈ B(x) for l ≥ 0 such that x + dv + p2 − lp ∈ Zp3 . Clearly, there are l and ν such that p2 /d = l p/d + ν or p2 = l p + νd. Hence, for p < x + dv ≤ p2 we get x + dv = x + d(v − ν) + p2 − l p. Now, if 0 < x + d(v − ν) ≤ p, then x + dv ∈ B(x). If x + d(v − ν) ≤ 0, there exists l such that 0 < x + d(v − ν) + l p ≤ p. Hence, x + dv = x + d(v − ν + l p/d) + p2 − (l + l )p. We still have x + dv ∈ B(x). Finally, if p < x + d(v − ν), one has l such that 0 < x + d(v − ν) − l p ≤ p. Thus, x + dv = x + d(v − ν − l p/d) + p2 − (l − l )p. We conclude in particular that (3.4)
{x + dv : 0 ≤ x + dv ≤ p2 } ⊆ B(x).
In order to replace p2 in (3.4) by p3 − 1, we again use Lemma 3.1. Examining the conditions of Lemma 3.1, we know that if d = 1, then for each 0 ≤ x + dv ≤ p2 one gets x + dv + lp2 ∈ B(x) whenever x + dv + lp2 ∈ Zp3 . Hence, (3.5)
{x + dv : 0 ≤ x + dv < p3 } ⊆ B(x).
However, if d = 1, the above calculation may not be true. We should modify this as follows: for kη = 0 let 0 ≤ x + v < p2 . Again we have (3.5). If kη = 0, let 0 < x + v ≤ p2 . We obtain x + v + lp2 ∈ B(x). Hence, (3.5) is still valid. Clearly, (3.5) implies (3.1) if d = 1. To show (3.1) for d = 1, we observe elements from (3.5) satisfying p3 − p1 ≤ x + dv < p3 −p. Obviously we can apply Lemma 3.4 for y = x+dv, pl1 = p1 and pl2 = p3 to obtain x+dv −(p3 −p1 ) ∈ B(x). Noticing 2p2 < p3 and 0 ≤ x+dv −(p3 −p1 ) < p, we get from Lemma 3.1 that x + dv − (p3 − p1 ) + p2 ∈ B(x) if kη = 0 or x − p3 ≡ 0 (mod d). Under this restriction, Lemma 3.2 tells us that x+dv −(p3 −p1 )+p2 −pl ∈ B(x) whenever x + dv − (p3 − p1 ) + p2 − pl ∈ Zp3 . Recursively, we conclude that if kη = 0 or x − p3 ≡ 0 (mod d), then x + dv − (p3 − p1 ) + jp2 − lp ∈ B(x) whenever x + dv − (p3 − p1 ) + jp2 − lp ∈ Zp3 . Let 0 ≤ x + dν − (p3 − p1 ) < p3 . Then due to gcd(p2 , p) = d, there are j ≥ 0, l ≥ 0 and 0 ≤ x + dv − (p3 − p1 ) < p such that x + dν − (p3 − p1 ) = x + dv − (p3 − p1 ) + jp2 − lp. Therefore, if kη = 0 or x − p3 ≡ 0 (mod d), {x + dv − (p3 − p1 ) : 0 ≤ x + dv − (p3 − p1 ) < p3 } ⊆ B(x).
SUBDIVISION SCHEMES
829
This observation leads to the following assertion: Let Aj = {x + dv − j(p3 − p1 ) : 0 ≤ x+dv−j(p3 −p1 ) < p3 }. Then under the assumption kη = 0 or x−(i+1)p3 ≡ 0 (mod d) we have that Ai ⊆ B(x) implies Ai+1 ⊆ B(x). We already know A0 ∪ A1 ⊆ B(x). Thus, we use the above procedure with p3 − p1 ≤ x + dv − i(p3 − p1 ) < p3 − p instead of p3 − p1 ≤ x + dv − (i − 1)(p3 − p1 ) < p3 − p to obtain Ai+1 ⊂ B(x). We remember x + p3 ≡ 0 (mod d). Hence, x − jp3 ≡ 0 (mod d) for j = 0, 1, . . . , d − 2. We conclude d−2
(3.6)
Aj ⊆ B(x).
j=0
Moreover, in the case of kη = 0 d−1
Aj ⊆ B(x).
j=0
To have (3.1) for kη = 0, we notice gcd(d, p3 ) = 1. Hence, for any v such that 0 ≤ x + v < p3 there is 0 ≤ j ≤ d − 1 with v + j(p3 − p1 ) ≡ 0 (mod d), which gives v = dv − j(p3 − p1 ) for some v. We conclude that x + v = x + dv − j(p3 − p1 ) ∈ Aj or Zp3 = {v : 0 ≤ v < p3 } ⊆ B(x). It remains to show (3.1) for kη = 0. We already have (3.6). But what is the corresponding Ad−1 for this case? We know that the first step in our procedure is to find an element in Ad−2 which satisfies p3 − p1 ≤ x + dv − (d − 2)(p3 − p1 ) < p3 − p.
(3.7)
If x + dv − (d − 2)(p3 − p1 ) = p3 − p1 , we can use our procedure to obtain x + dv − (d − 1)(p3 − p1 ) + jp2 − lp ∈ B(x). If however x + dv − (d − 2)(p3 − p1 ) = p3 − p1 , we get formally from the above procedure a set A = {jp2 − lp : 0 ≤ jp2 − lp < p3 }. But, because kη = 0, we cannot use this approach for x+ dv − (d− 1)(p3 − p1 ) (= 0). Thus, instead of Ad−1 we have only Ad−1 \ A ⊂ B(x). Hence, combining this with
(3.6), we obtain ( d−1 i=0 Ai ) \ A ⊆ B(x), which implies {v : 0 ≤ v < p3 } \ A ⊆ B(x). To show A ⊂ B(x), we have to modify (3.7). We remember that gcd(d, p) = d and x − (d − 1)p3 ≡ 0 (mod d). Hence, there is a v such that p3 − p1 < x + dv − (d − 2)(p3 − p1 ) = p3 − p, which implies that x + dv − (d − 1)(p3 − p1 ) (= p) is contained in B(x) due to Lemma 3.4. Now we can apply our procedure for p to obtain jp2 − lp ∈ B(x) or A ⊂ B(x). Consequently, Zp3 = {v : 0 ≤ v < p3 } ⊆ B(x). For kη = 2 − 1 we replace x + p3 ≡ 0 (mod d) by x + 1 + p3 ≡ 0 (mod d) and p|(x − p2 ) by p|(x + 1 − p2 ), respectively. Case 2. Now if p3 < 2p1 , we use the symmetric relation to obtain (3.1), since (0, p3 − p2 , p3 − p1 , p3 ) has property P and 2(p3 − p1 ) < p3 . m
830
XINLONG ZHOU
Case 3. Since p3 is odd, it remains to show (3.1) for 2p1 < p3 < 2p2 . Without loss of generality we may suppose p3 ≥ p2 + p1 . Now, p1 is even, so p1 = 2p for some p. Let p be odd. Denote gcd(p1 , p3 ) = d. We have gcd(p, p3 ) = d. We verify for some 0 ≤ x ≤ p that {v : 0 ≤ v < p1 } \ {p} ⊆ B(x).
(3.8)
To this end we choose 0 ≤ x ≤ p in the following way: if 0 ≤ kη < 2m − 1, then x = 0 and x − p2 ≡ 0 (mod d), if d = 1, (3.9) x + p3 ≡ 0 (mod p), if d = 1; if kη = 2m − 1, then x = p and x + 1 − p2 ≡ 0 (mod d), (3.10) x + 1 ≡ 0 (mod p),
if d = 1, if d = 1.
In the following we deal only with the case 0 ≤ kη < 2m − 1. The corresponding assertion for kη = 2m − 1 can be treated in the same way. To begin with we note p|(2m − 1) and p|kη (see (2.8) and (2.9)). Hence, for some k we get 2m x + kη = x + kp. Since kη < (1 − δ)2m for some 0 < δ < 1 (see (2.10)), the number k satisfies 2p3 < k < (2 − δ)2m . Write k = 2µ + β with β ∈ {0, 1}, so p3 < µ < (1 − δ/2)2m . Let i,j = 0 for j = 2, 3. We get 2m x + kη −
m−1 3
i,j pj 2i
=
2m x + kη − lp1
=
x + kp − lp1
=
x + µp1 − lp1 + βp
=
x + αp1 + βp.
i=0 j=1
We can choose l satisfying p3 − p1 ≤ x + αp1 + βp < p3 . Thus, x + αp1 + βp ∈ B(x) by the definition of B(x). Now the fact p1 < p3 − p1 and Lemma 3.4 imply further 0 ≤ x + αp1 + βp − p3 + p1 < p1 and x + αp1 + βp − p3 + p1 ∈ B(x). Lemma 3.2 tells us that we can substitute p if the above number is not less than p. We obtain in this way x − p3 + l1 p = x + αp1 + βp − p3 + p1 − p ∈ B(x) and 0 ≤ x− p3 + l1 p < p. As in Case 1 we conclude recursively that 0 ≤ x− sp3 + ls p < p for s = 0, . . . , p/d − 1. Clearly, by this construction x − sp3 + ls p ∈ B(x). Moreover, since gcd(p, p3 ) = d, all these numbers are different for 0 ≤ s ≤ p/d − 1. Hence, {x − sp3 + ls p : x − sp3 + ls p ∈ [0, p)} = {x + dv : 0 ≤ x + dv < p} ⊆ B(x). We notice also that the set on the left-hand side contains the zero if and only if d = 1. For d = 1 we conclude from the last relation that (3.11)
{0, 1, . . . , p − 1} ⊆ B(x).
We claim that (3.11) is still valid for d = 1. To see this, let 0 < x + dv < p. The above consideration tells us that we have α and β, where β ∈ {0, 1} depends on x + dv, such that x + dv + αp1 + βp ∈ B(x) and p3 − p1 − p2 ≤ x + dv + αp1 + βp < p3 − p2 . Now, since d = 1, we deduce p3 − p1 − p2 = 0. Hence, 0 < x + dv + αp1 + βp < p3 − p2 < p2 . We can therefore apply Lemma 3.1 for
SUBDIVISION SCHEMES
831
x + dv + αp1 + βp to obtain x + dv + αp1 + βp + p2 ∈ B(x) and by Lemma 3.4 we have x + dv + αp1 + βp + p2 − p3 + p1 ∈ B(x). This number is nonnegative and less than p1 . So by Lemma 3.2 we can again substitute p, if necessary, to obtain x + dv + (p2 − p3 + p1 ) + l1 p ∈ B(x) and 0 ≤ x + dv + (p2 − p3 + p1 ) + l1 p < p. Recursively, B(x) contains x + dv + s(p2 − p3 + p1 ) + ls,v p = x + dv + sp2 + ls,v d, ∀ s = 0, 1, . . . , d − 1.
Since gcd(p2 , d) = 1, we conclude by the second expression that they numbers are different for different s, while the first one shows that they are different if v is different. Thus, there are d×p/d = p numbers in {0, . . . , p−1}, which are contained in B(x); in other words, (3.11) also holds for d = 1. To get (3.8) from (3.11), we observe any nonzero number µ satisfying 0 < µ < p. We know that µ + αp1 + βp ∈ B(x) whenever 0 ≤ µ + αp1 + βp < p3 , where β ∈ {0, 1} relies on µ. For the µ with β = 0 let α = 1. So µ + p1 ∈ B(x) and by Lemma 3.2, µ + p1 − p ∈ B(x). For the µ with β = 1 we choose α = 0. Hence, we always have µ + p ∈ B(x), which implies (3.8). To verify our assertion from (3.8), we remember that 0 ≤ kη < 2m − 1. So kη is even. If 0 < kη < 2m − 1, then since 0 ∈ B(x) and 2m 0 + kη − lp1 = jp1 for some l, we conclude jp1 ∈ B(x) if 0 ≤ jp1 < p3 . Lemma 3.2 yields p ∈ B(x). Combining this with (3.8), we obtain {0, 1, . . . , p1 − 1, p1 , 2p1 , . . . , j0 p1 } ⊆ B(x), where j0 satisfies j0 p1 < p3 ≤ (j0 + 1)p1 . Moreover, for any 0 < 2l < p1 we have 2m 2l + kη − ip1 = 2l + αp1 with some i. Hence, B(x) contains 2l + αp1 whenever 0 < 2l + αp1 < p3 . In particular, (3.12)
{0, 1, . . . , p1 } ∪ {2j : 0 ≤ 2j < p3 } ⊆ B(x).
Let 0 < 2j < p2 . Then Lemma 3.1 shows that 2j + p2 ∈ B(x) whenever 0 < 2j + p2 < p3 . Since p2 is odd and 2p2 > p3 , we obtain from (3.12) that {l : p2 < l < p3 } ⊆ B(x). Let p1 + p2 ≤ 2j < p3 . Lemma 3.3 tells us 2j − p2 ∈ B(x). It follows from (3.8) and the last two relations that {l : 0 ≤ l ≤ p3 − p2 } ⊆ B(x). Let l be odd and let it satisfy 0 ≤ l < p3 − p2 . So using (i) of Lemma 3.3 several times, we obtain l + j(p3 − p2 ) ∈ B(x) whenever 0 ≤ l + j(p3 − p2 ) < p3 . We conclude from the last relation that {l : 0 < l < p3 , l is odd} ⊆ B(x).
(3.13)
Assertion (3.1) follows from (3.12) and (3.13). Next we prove (3.1) for kη = 0 via (3.8). The above approach tell us that we need only to have (3.12). For this goal let us observe B(p1 ). By the definition of B(p1 ) we certainly have {lp1 : 0 < l ≤ l } ⊆ B(p1 ), where l satisfies l p1 < p3 ≤ (l +1)p3 . Lemma 3.2 implies p ∈ B(p1 ). On the other hand, there is a choice of i,j such that 2 m p1 −
m−1 3 i=0 j=1
Thus,
i,j 2i pj = (2m − 1)p1 −
m−1
i,3 2i p3 = 0.
i=1
{lp1 : 0 ≤ l ≤ l } ∪ {p} ⊆ B(p1 ).
832
XINLONG ZHOU
Let p3 − p1 ≤ lp1 < p3 . It follows from Lemma 3.4 that (l + 1)p1 − p3 ∈ B(p1 ). Thus, by Lemma 3.2 there exists a j such that 0 < jp − p3 ≤ p and jp − p3 ∈ B(p1 ). Similarly, let p3 − p2 − p1 < lp1 ≤ p3 − p2 . We conclude for some j that jp − p3 + p2 ∈ B(p1 ) and 0 < jp − p3 + p2 ≤ p. Now we can choose x as follows: x = jp − p3 + p2 and 0 < jp − p3 + p2 ≤ p, if d = 1, x = jp − p3 and 0 < jp − p3 ≤ p, if d = 1. Clearly, x satisfies (3.9) and x ∈ B(p1 ). Consequently, (3.8) holds with this x and B(x) ⊆ B(p1 ). We obtain by (3.8) the desired relation, i.e., {0, 1, . . . , p1 − 1, p1 , 2p1 , . . . , l p1 } ⊆ B(p1 ). Now let p = p1 /2 be even. Denote gcd(p3 − p2 , p) = d. So d is even. We verify for some 0 ≤ x < p that (3.14)
{x + dv : 0 ≤ x + dv < p} ⊆ B(x).
To this end we choose 0 ≤ x < p as follows: x − p2 ≡ 0 (mod d), (3.15) x + 1 − p2 ≡ 0 (mod d),
if 0 ≤ kη < 2m − 1, if kη = 2m − 1.
Clearly, x is not unique. But x is odd if 0 ≤ kη < 2m −1 and even when kη = 2m −1. Moreover, any 0 ≤ x + dv < p satisfies (3.15). To verify (3.14), we remember that kη is even when 0 ≤ kη < 2m − 1. So in any case (2m − 1)x + kη is odd. We also remember p3 |(2m − 1) and p3 |kη since p3 is odd. Now by Lemma 3.3 we obtain x + p3 − p2 ∈ B(x). On the other hand, since x − p ≤ −1 and p3 − p2 ≤ (p3 − 1)/2, we have (2m − 1)(x + p3 − p2 − p) + kη < (2m−1 − 1)p3 . Thus, the condition of Lemma 3.2 is satisfied. We obtain by Lemma 3.2 with y = x + p3 − p2 , pl1 = p1 and pl2 = p3 that x + p3 − p2 − p ∈ B(x). Recursively, x + p3 − p2 − α1 p ∈ B(x) for some α1 such that 0 ≤ x + p3 − p2 − α1 p < p. Clearly, this new number also satisfies (3.15). We can therefore apply this procedure for this new number to obtain another number that satisfies (3.15) and is contained in B(x). Repeatedly, we obtain for s = 0, 1, . . . , p/d − 1 that x + s(p3 − p2 ) − αs p ∈ B(x) and 0 ≤ x + s(p3 − p2 ) − αs p < p, where α0 = 0. Since gcd(p3 − p2 , p) = d, they are all different and have the form x + dv. This shows (3.14). Next we verify (3.16)
{0, . . . , p − 1} ⊆ B(x).
To show (3.16), let 0 ≤ x + dv < p. Since (2m − 1)(x + dv) + kη is odd, it follows from (i) of Lemma 3.3 with pj = p1 and pl1 = p2 that x + dv + p2 − p1 ∈ B(x). Now (2m − 1)(x + dv + p2 − p1 ) + kη is even. We apply (ii) of Lemma 3.3 with pl1 = p2 and, if necessary, Lemma 3.3 with pl1 = p1 to obtain 0 ≤ x + dv + p2 − l1 p < p and x + dv + p2 − p1 − lp = x + dv + p2 − l1 p ∈ B(x). For this new number (2m − 1)(x + dv + p2 − l1 p) + kη is even. Clearly, we can again use (ii) of Lemma 3.3 to obtain x + dv + p2 − l1 p + 2p2 − p3 ∈ B(x). Write x + dv + p2 − l1 p + 2p2 − p3 = x + dv + 2p2 − l1 p − (p3 − p2 ).
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For this number (2m − 1)(x + dv + 2p2 − l1 p − (p3 − p2 )) + kη is odd. We get by (i) of Lemma 3.3 and, if necessary, Lemma 3.2 that 0 ≤ x+ dv + 2p2 − l2 p− l (p3 − p2 ) < p and x + dv + 2p2 − l1 p − l (p3 − p2 ) − lp = x + dv + 2p2 − l2 p − l (p3 − p2 ) ∈ B(x). Repeatedly, we conclude for s = 0, 1, . . . , d − 1 that 0 ≤ x + dv + sp2 − ls p − ls (p3 − p2 ) < p and x + dv + sp2 − ls p − ls (p3 − p2 ) ∈ B(x). By our construction it is easy to see that if (v, s) = (v , s ), then the corresponding numbers are different. There are p/d different v and d different s. Therefore, B(x) contains d × p/d numbers in {0, . . . , p − 1}. In other words, (3.16) holds. We have to show that p in (3.16) can be replaced by p3 . We first prove this for kη = 2m − 1. Let 0 < l < p be odd. So (2m − 1)l + kη is odd. As before we use Lemmas 3.3 and 3.2 to obtain l + p3 − p2 − αp ∈ B(x) for α ≥ 0 satisfying 0 ≤ l + p3 − p2 − αp < p3 . Since l + p3 − p2 − αp = l is again odd, we obtain in particular that all odd numbers in {0, . . . , p3 − p2 − 1} belong to B(x). Applying Lemma 3.3 for each odd number l from the last set, we conclude l +l(p3 −p2 ) ∈ B(x) whenever 0 ≤ x + dv + l(p3 − p2 ) < p3 and l ≥ 0. Thus, B(x) contains all odd 0 < l < p3 , i.e., (3.17)
{l : 0 ≤ l < p3 , l is odd} ⊆ B(x).
Let l be odd and 0 < l < p3 − p2 . Since p3 − p2 < p2 , we have by Lemma 3.1 with p2 that l + p2 ∈ B(x). Consequently, (3.18)
{l : p2 ≤ l < p3 } ⊆ B(x).
Furthermore, let l be even and p2 ≤ l < p3 . So (2m − 1)l + kη is even. We conclude from Lemma 3.3 that B(x) contains l − µp1 whenever µ ≥ 0 and 0 ≤ l − µp1 < p3 . It follows from the fact that p3 ≥ p2 + p1 and p1 is even that (3.19)
{l : 0 ≤ l < p2 , l is even} ⊆ B(x).
The desired assertion follows from (3.16)–(3.19). The proof for kη = 2m − 1 is the same. We omit the details here. Step 2. Let (3.1) be true for |S(a)| ≤ k with k ≥ 3. We verify (3.1) for |S(a)| = ˜ k + 1. Write S(a) = {0, p1 , . . . , pk } and S(a) = {0, pk − pk−1 , . . . , pk − p1 , pk }. ˜ Therefore, we need to prove (3.1) either for S(a) or for S(a). The special case of |S(a)| = 4, which we dealt with in Step 1, allows us to suppose that there is 0 < p ∈ S(a) such that S(a) \ {p} has property P. Denote gcd(p : p ∈ S(a) \ {p}) = d and S (a) = {q : dq ∈ S(a) \ {p}} = {0, q1 , . . . , qk−1 }. S (a) has property P. Moreover, gcd(q1 , . . . , qk−1 ) = 1 and |S (a)| = k. Let us also remark that in general m and kη depend on S(a). Hence, for different S(a) the numbers m and kη are generally different. On the other hand, it is clear that if m and kη are numbers for S(a), they are also suitable for S(a) \ {p}. Furthermore, m and kη /d are numbers for S (a). With this in mind it is now clear that the hypothesis of the induction implies for some 0 ≤ x < qk−1 that {0, . . . , qk−1 − 1} = B (x ),
834
XINLONG ZHOU
where B (x ) is the corresponding set B(x) defined by S (a), m and kη /d. By Lemma 3.1 and the definition of B(dx ) we obtain (3.20)
{ld : 0 ≤ ld < pk } ⊆ B(dx ).
Indeed, if dqk−1 = pk , (3.20) follows directly from the definition of B(dx ). If however dqk−1 = pk , then p = pk and dqk−1 = pk−1 . So {ld : 0 ≤ ld < pk−1 } ⊆ B(dx ). But there is an odd number in S(a) \ {pk , pk−1 }. Using Lemma 3.1 with this number, we can still replace pk−1 by pk . Clearly, (3.20) is (3.1) if d = 1. So in the following discussion we should always assume d = 1. In what follows we should use (3.20) to prove (3.1) for S(a). The proof will be divided into four cases according to different p and pk . Case 1. pk is even and there are at least two odd numbers in S(a). Without loss of generality let two odd numbers be p1 and p2 . Let gcd(p1 , p3 , . . . , pk ) = d and S (a) = {qi : dqi ∈ S(a) \ {p2 }} = {0, q1 , . . . , qk−1 }. We obtain (3.20) with p = p2 . To verify (3.1) for S(a) from (3.20), let 2p2 ≤ pk . We choose l ≥ 0 such that pk − p2 + ld ≤ l d + ld < pk − p2 + (l + 1)d for l = 0, . . . , p1 /d − 1. Clearly, l d + ld ∈ B(dx ). Moreover, by Lemma 3.4 the set B(dx ) contains l d + ld − pk + p2 and ld ≤ l d + ld − pk + p2 < (l + 1)d ≤ p1 . Denote y = l d−pk +p2 . We have y −p2 ≡ 0 (mod d) and y = 0. Using Lemma 3.1, we conclude that for some αl,1 > 0 and l = 0, . . . , p1 /d−1 the numbers y+ld+αl,1 p1 are in B(dx ) as well as in [pk − p2 , pk − p2 + p1 ). Again by Lemma 3.4 we obtain for l = 0, 1, . . . , p1 /d − 1 that 0 < y + ld + αl,1 p1 − (pk − p2 ) < p1 and y + ld + αl,1 p1 − (pk − p2 ) ∈ B(dx ). We note that the choice of y ensures y − s(pk − p2 ) ≡ 0 (mod d) if and only if s + 1 ≡ 0 (mod d). Thus, write αl,0 = 0. We conclude repeatedly that B(dx ) contains y + ld + αl,s p1 − s(pk − p2 ) and 0 ≤ y + ld + αl,s p1 − s(pk − p2 ) < p1 , ∀ s = 0, . . . , d − 1, l = 0, 1, . . . , p1 /d − 1. It is easy to see that all these numbers are different. Consequently, {0, . . . , p1 −1} ⊆ B(dx ). Since d|p1 , we get by (3.20) that {0, . . . , p1 } ∪ {ld : 0 ≤ ld < pk } ⊆ B(dx ). The desired assertion follows from this relation and Lemma 3.1, since any p1 < z < pk can be written as z = z + νp1 with 0 ≤ z < p1 . ˜ Let 2p1 ≥ pk . Then 2(pk − pk−2 ) ≤ pk . Hence, the above proof is valid for S(a). We again get (3.1) by the symmetric relation (3.2). Finally, let 2p1 < pk < 2p2 . Equation (3.2) allows us to suppose p1 + p2 ≤ pk . The approach is the same as for the case 2p2 ≤ pk . First we choose l ≥ 0 such that pk − p2 − p1 + ld ≤ l d + ld < pk − p2 − p1 + (l + 1)d, ∀ l = 0, . . . , p1 /d − 1. Hence, by Lemma 3.1 and (3.20) the number l d + p2 is contained in B(dx ), while Lemma 3.4 implies that y = l d + p2 + p1 − pk is in B(dx ). Moreover, for the same reason y + ld ∈ B(dx ) and ld ≤ y + ld < (l + 1)d, ∀ l = 0, . . . , p1 /d − 1.
SUBDIVISION SCHEMES
835
Clearly, y + sp2 ≡ 0 (mod d) if and only if s + 1 ≡ 0 (mod d). Therefore, using the same approach as for 2p2 ≤ pk , we conclude that B(dx ) contains y + dl + αl,s p1 − s(pk − p2 − p1 ) for s = 0, 1, . . . , d − 1 and l = 0, . . . , p1 /d − 1. These numbers are different and are in [0, p1 ). Consequently, {0, . . . , p1 } ∪ {ld : 0 ≤ ld < pk } ⊆ B(dx ) and thus we get the desired assertion. Case 2. pk is even and there is only one odd number p in S(a). Let p = ˜ and pk − p. Write p1 , pk−1 . We may assume 2p ≤ pk . Otherwise, we consider S(a) gcd(p1 , p2 , . . . , pk−1 ) = d. So with S (a) = {0, p1 /d, . . . , pk−1 /d} we obtain for some 0 ≤ x < pk−1 /d the relation (3.20). Having (3.20), we choose l such that pk − p + ld ≤ dl + ld < pk − p + (l + 1)d, ∀ l = 0, . . . , p/d − 1. Next we apply Lemma 3.4 to get y + ld = dl + ld − pk + p ∈ B(dx ). Now it is clear that 0 ≤ y + dl < p, ∀ l = 0, . . . , p/d − 1. Moreover, y −spk ≡ 0 (mod d) if and only if s+1 = 0 (mod d). The same approach as in Case 1 (with p instead of p1 ) yields {0, . . . , p} ∪ {ld : 0 ≤ ld < pk } ⊆ B(dx ). We know now how to get (3.1) from this relation. To show (3.1) for p = p1 or pk−1 , we should modify the above proof. Without loss of generality we may assume p = p1 and 2p1 > pk . So the pj , j = 2, . . . , k, are even. Denote gcd(p1 , p2 , . . . , pk−2 , pk ) = d1 . With S (a) = {0, p1 /d1 , . . . , pk−2 /d1 , pk /d1 } we obtain (3.20). To verify (3.1), let q = pk − p1 , q = pk−1 /2. So gcd(d1 , q ) = 1. We already know that ld1 +d1 ∈ B(d1 x ) for all l = 0, 1, . . . , q/d1 −1. Now, since q < p1 , it follows from Lemma 3.1 that ld1 + d1 + p1 ∈ B(d1 x ) while Lemma 3.4 with pl1 = p1 and pl2 = pk implies ld1 + d1 + p1 − αq ∈ B(d1 x ) for α such that q ≤ ld1 + d1 + p1 − αq < q + q. Lemma 3.2 with pk−1 leads to ld1 + d1 + (p1 − q ) − αq ∈ B(d1 x ) and 0 ≤ ld1 + d1 + (p1 − q ) − αq < q. Now, let y = d1 +(p1 −q )−αq. Then y ∈ B(d1 x ) and 0 < y < q. Recursively, we obtain from the above procedure that 0 ≤ y+ld1 +ls d1 −sq < q and y+ld1 +ls d1 −sq ∈ B(d1 x ) for s = 0, 1, . . . , d1 − 1 and l = 0, . . . , q/d1 − 1. Moreover, y + ld1 + ls d1 − sq = 0 if and only if s = d1 − 1. Therefore, since gcd(d1 , q ) = 1, these numbers are all different and are contained in {0, . . . , q − 1}. We conclude that {v : 0 ≤ v < q} ⊆ B(d1 x ). Again applying Lemma 3.1 with p1 , we obtain from (3.20) and the last relation that (3.21)
{0, 1, . . . , q − 1} ∪ {p1 , . . . , pk − 1} ∪ {ld1 : 0 ≤ ld1 < pk } ⊆ B(d1 x ).
Let p1 ≤ v < pk . It follows from Lemma 3.4 with pl1 = p1 and pl2 = pk that v − αq ∈ B(d1 x ) whenever v − αq ∈ Zpk . Combining this with (3.21) leads to (3.1). Case 3. pk is odd. S(a) \ {pk } does not have property P. Thus, S(a) contains an odd p = pk and an even q = 0. It is easy to see that p1 , . . . , pk−2 are even and pk−1 is odd. Suppose pk ≥ 2pk−2 , so 2pj < pk for j = 1, 2, . . . , k − 2. Clearly S(a)\ {p2} = {0, p1 , p3 , . . . , pk } has property P. Denote gcd(p1 , p3 , . . . , pk ) = d1 and S (a) = {0, p1 /d1 , p3 /d1 , . . . , pk /d1 }. We obtain (3.20) with some 0 ≤ x < pk , i.e., {ld1 : 0 ≤ ld1 < pk } ⊆ B(d1 x ).
836
XINLONG ZHOU
Let l satisfy pk − p2 + ld1 ≤ l d1 + ld1 < pk − p2 + (l + 1)d1
∀ l = 0, . . . , p1 /d1 − 1.
Clearly, for each such l it holds that (2m − 1)(l d1 + ld1 − p2 ) + kη < (2m − 1)(pk − p2 ). Hence, by Lemma 3.4 we conclude that ld1 ≤ l d1 + ld1 − (pk − p2 ) < (l + 1)d1 and l d1 + ld1 − (pk − p2 ) ∈ B(d1 x ). Write y = l d1 − (pk − p2 ). We obtain y + sp2 ≡ 0 (mod d1 ) if and only if s + 1 ≡ 0 (mod d1 ). We know that p1 = 2µ g, where g is odd. Also, d1 = 1 implies g = 1. For each y + ld1 we choose i,j = 0 for j > 1 to obtain 2m (y + ld1 ) + kη −
m−1 k
i,j 2i pj
m−1
=
2m (y + ld1 ) + kη −
=
y + ld1 + (2 − 1)(y + ld1 ) + kη − γp1
=
y + ld1 + (2m − 1)(y + ld1 ) + kη − γ2µ g,
i=0 j=1
i,1 2i p1
i=0 m
where γ can be any integer between 0 and 2m − 1. By the definition of B(d1 x ) we conclude that y + ld1 + (2m − 1)(y + ld1 ) + kη − γ2µ g ∈ B(x) whenever y + ld1 + (2m − 1)(y + ld1 ) + kη − γ2µ g ∈ Zpk . Thus, we can choose γ to obtain y + ld1 + (2m − 1)(y + ld1 ) + kη − γ2µ g = y + ld1 + rp1 + βg with some r and 0 ≤ β ≤ 2µ − 1 such that pk − p2 ≤ y + ld1 + rp1 + βg < pk . It follows from Lemma 3.4 that 0 ≤ y + ld1 + rp1 + βg − (pk − p2 ) < p2 is contained in B(d1 x ). Furthermore, with pl1 = p1 and pl2 = pk we get by Lemma 3.2 for some α ≥ 0 that 0 ≤ y + ld1 + rp1 + βg − (pk − p2 ) − αg < p1 and y + ld1 + rp1 + βg − (pk − p2 ) − αg ∈ B(d1 x ). We may write with some αl,1 y + ld1 + rp1 + βg − (pk − p2 ) − αg = y + ld1 + αl,1 d1 − (pk − p2 ). Clearly, these new numbers are different from y + ld1 . Therefore with αl,0 = 0 we conclude recursively for s = 0, 1, . . . , d1 − 1, l = 0, . . . , p1 /d1 − 1 that 0 ≤ y + ld1 + αl,s d1 − s(pk − p2 ) < p1
and
y + ld1 + αl,s d1 − s(pk − p2 ) ∈ B(d1 x ).
The choice of y ensures that these numbers are all different. Hence, together with (3.20) we get (3.22)
{0, 1, . . . , p1 } ∪ {ld1 : 0 ≤ ld1 < pk } ⊆ B(d1 x ).
To finish our proof for the case pk ≥ 2pk−2 , we have to verify {p1 + 1, . . . , pk − 1} ⊆ B(d1 x ). For this goal we assume first kη = 2m − 1. Thus, kη is even. We apply (i) of Lemma 3.3 with pl1 = pk and pj = pk−1 for each odd number from {0, . . . , p1 } to obtain {x : 0 ≤ x ≤ p1 + pk − pk−1 , x is odd} ∩ Zpk ⊆ B(d1 x ). Recursively, we conclude (3.23)
{x : 0 ≤ x < pk , x is odd} ⊆ B(d1 x ).
SUBDIVISION SCHEMES
837
Next, we use Lemma 3.1 with pk−1 for each number from {1, . . . , p1 } to get {x : pk−1 < x ≤ min{pk−1 + p1 , pk − 1}} ⊆ B(d1 x ). Now for each even number x between pk−1 and min{pk−1 + p1 , pk − 1} we apply (ii) of Lemma 3.3 with pl1 = pk and pj = p1 . So, recursively, we obtain {x : p1 ≤ x ≤ pk−1 , x is even} ⊆ B(d1 x ). Combining this with (3.22) and (3.23), we get {0, .., pk−1 } ⊆ B(d1 x ). The desired assertion follows from Lemma 3.1 with pk−1 and the last relation. Similarly, we have (3.1) for kη = 2m − 1. It remains to show (3.1) for 2pk−2 ≥ pk . Thus 2(pk −pk−2 ) ≤ pk . The symmetric relation (3.2) tells us that in this case we need only to verify (3.1) for the set S(a) with even p1 and odd p2 , . . . , pk , which satisfies 2p1 ≤ pk . Clearly {0, p1 , . . . , pk−1 } has property P. Denote gcd(p1 , . . . , pk−1 ) = d2 . We have as before {ld2 : 0 ≤ ld2 < pk } ⊆ B(d2 x ). Next, for l = 0, . . . , p1 /d2 − 1 let l be such that pk − p1 + ld2 ≤ l d2 + ld2 < pk − p1 + (l + 1)d2 . Write y = l d2 − (pk − p1 ). Lemma 3.4 tells us that y + ld2 ∈ B(d2 x ) for l = 0, . . . , p1 /d2 − 1. We now know how to use Lemmas 3.1, 3.2 and 3.4 from these numbers to obtain, for s = 0, . . . , d2 − 1 and l = 0, . . . , p1 /d2 − 1, y + ld2 + αs,l d2 − s(pk − p1 ) < p1 and y + ld2 + αs,l d2 − s(pk − p1 ) ∈ B(d2 x ), which in turn implies {0, . . . , p1 − 1} ⊆ B(d2 x ). Hence, {0, . . . , p1 } ∪ {ld2 : 0 ≤ ld2 < pk } ⊆ B(d2 x ). The assertion follows from Lemmas 3.1, 3.3 and this relation. Case 4. pk is odd. S(a) \ {pk } has property P and S(a) \ {pk } contains an odd p and an even q = 0. Denote gcd(p1 , . . . , pk−1 ) = d. Thus, we obtain (3.20), i.e., {ld : 0 ≤ ld < pk } ⊆ B(dx ). For q ≤ pk − q the proof is the same as in the last situation of Case 3. We omit the details here. Let q ≥ pk − q. We may assume that q is minimal among all nonzero even ˜ numbers in S(a) and that p satisfies 2p < pk . Otherwise we consider S(a), for which we have already the desired assertion. For the same reason we can also assume p + q ≤ pk . To this end, first let 2p ≥ q. Denote q = q/2. Furthermore, let q ≤ l d + ld < pk for some l and l = 0, . . . , q /d − 1. By Lemma 3.4 we get 2q − pk ≤ l d + ld − (pk − q) < q. We may use Lemma 3.2 to subtract q from l d + ld − (pk − q) such that 0 ≤ l d + ld − (pk − q) − αl q < q . Now write yl = l d + ld − (pk − q) − αl q . Thus, yl ∈ B(dx ) and yl = 0, ∀ l = 0, . . . , q /d − 1. Furthermore, they are all different and yl − spk ≡ 0 (mod d) if and only if s + 1 ≡ 0 (mod d). Now we obtain recursively by applying Lemma 3.1 with p, Lemma 3.4 with q and pk and finally Lemma 3.2 with q that yl + αl,s d + s(pk − q) ∈ B(dx ) and 0 ≤ yl + αl,s d + s(pk − q) < q
838
XINLONG ZHOU
for l = 0, . . . , q /d − 1 and s = 0, . . . , d − 1. Hence, together with (3.20) we conclude {0, . . . , q } ∪ {ld : 0 ≤ ld < pk } ⊆ B(dx ). We verify that q in the above relation can be replaced by p. If q ≥ p, we have nothing more to do. If q < p, we use Lemma 3.1 with p to obtain {0, . . . , q } ∪ {p, . . . , p + q } ⊆ B(dx ). For each number of {p, . . . , p + q − 1} we apply Lemma 3.2 with q to obtain {p − q , . . . , p − 1} ⊆ B(dx ). Thus, {p − q , .., p + q } ⊆ B(dx ). Again using Lemma 3.2 with q for each element from {p − q , . . . , p + q − 1}, we obtain {p − 2q , . . . , p + q } ⊆ B(dx ). Recursively, {0, . . . , p + q } ⊆ B(dx ). Thus, (3.24)
{0, . . . , p} ∪ {ld : 0 ≤ ld < pk } ⊆ B(dx ).
Since p is odd, (3.1) follows from Lemma 3.1 with p and the last relation. It remains to verify (3.1) for 2p ≤ q. We should modify the procedure for 2p ≥ q as follows: let pk − p ≤ l d + ld < pk for some l and l = 0, . . . , p/d − 1. By Lemma 3.4 with pk and q we get q − p ≤ l d + ld − (pk − q) < q. Again using Lemma 3.4 with p and q, we conclude that 0 ≤ l d + ld − (pk − p) < p. Now write yl = l d + ld − (pk − p). Thus, yl ∈ B(dx ) and yl = 0, ∀ l = 0, . . . , p/d − 1. We obtain in the same way that yl + αl,s p + s(pk − p) ∈ B(dx ) and 0 ≤ yl + αl,s p + s(pk − p) < p for l = 0. . . . , p/d − 1 and s = 0, . . . , d − 1. Hence, together with (3.20) we conclude (3.24). From (3.24) we know how to get (3.1). Acknowledgment The author is indebted to the referee for various helpful comments on his paper. References [1] M. Br¨ oker and X. Zhou, Characterization of continous, four-coefficient scaling functions via matrix spectral radius, SIAM J. Matrix Anal. Appl., 22 (2000), 242-257. MR2001h:65169 [2] A. S. Cavaretta, W. Dahmen, and C. A. Micchelli, Stationary Subdivision, Mem. Amer. Math. Soc., 453 (1991). MR92h:65017 [3] G. M. Chaikin, An algorithm for high speed curve generation, Comp. Graphics and Image. Proc., 3 (1974), 346-349. [4] D. Colella and C. Heil, Characterizations of scaling functions: continuous solutions, SIAM J. Matrix Anal. Appl., 15 (1994), 496-518. MR95f:26004 [5] I. Daubechies and J. C. Lagarias, Two-scale difference equations I. Existence and global regularity of solutions, SIAM J. Math. Anal., 22 (1991), 1388-1410. MR92d:39001 [6] D. E. Gonsor, Subdivision algorithms with nonnegative masks generally converge, Adv. Comp. Math., 1 (1993), 215-221. MR94e:65023 [7] R. -Q. Jia, Subdivision schemes in Lp spaces, Advances in Comp. Math, 3 (1995), 309-341. MR96d:65028 [8] R. -Q. Jia and D. -X. Zhou, Convergence of subdivision schemes associated with nonnegative masks, SIAM J. Matrix Anal. Appl., 21 (1999), 418-430. MR2001a:42041 [9] A. A. Melkman, Subdivision schemes with non-negative masks always converge—unless they obviously cannot?, Ann. Numer. Math., 4 (1997), 451-460. MR97i:41014 [10] C. A. Micchelli and H. Prautzsch, Uniform refinement of curves, Linear Algebra Appl., 114/115 (1989), 841-870. MR90k:65088
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[11] G. C. Rota and G. Strang, A note on the joint spectral radius, Indag. Math., 22 (1960), 379-381. MR26:5434 [12] G. de Rham, Sur une courbe plane, J. Mathem. pures et appl., 39 (1956), 25-42. MR19:842e [13] J. N. Tsitsiklis and V. D. Blondel, The Lyapunov exponent and joint spectral radius of pairs of matrices are hard, when not impossible, to compute and to approximate, Mathematics of Control, Signals and Systems, 10 (1997), 31-41. MR99h:65238a [14] Y. Wang, Two-scale dilation equations and the cascade algorithm, Random and Computational Dynamics, 3 (1995), 289-309. MR96m:42060 [15] Y. Wang, Subdivision schemes and refinement equations with nonnegative masks, J. Approx. Th., 113 (2001), 207-220. MR2002i:42054 [16] D.-Z. Zhou, The p-norm joint spectral radius for even integers, Methods and Applications of Analysis, 5 (1) (1998), 39-54. MR99e:42054 Department of Mathematics, China Jiliang University, Hangzhou, China; Institute of Mathematics, University of Duisburg-Essen, D-47057 Duisburg, Germany E-mail address: [email protected]
SUBDIVISION SCHEMES WITH NONNEGATIVE MASKS XINLONG ZHOU
Abstract. The conjecture concerning the characterization of a convergent univariate subdivision algorithm with nonnegative finite mask is confirmed.
1. Introduction Let Z denote the integer lattice. A univariate subdivision algorithm with a finitely supported mask a = {aj }j∈Z is given as follows: beginning with an initial sequence of data v 0 = {vi0 }, we set recursively new sequences of values v k , by applying the rule ai−2j vjk−1 . vik = j∈Z
This algorithm is said to converge if for each finite v 0 = {vi0 } there exists a continuous function fv such that fv ≡ 0 for at least one v 0 and i (1.1) lim sup |fv ( k ) − vik | = 0. k→∞ i∈Z 2 Obviously, the subdivision algorithm with mask a = {aj }j∈Z converges if and only if the polygon fk defined by the control points (i/2k , vik )T converges uniformly to fv . One can easily check fk (x) = vi0 akj h(2k (x − i) − j), i
j
where h is the hat function and = j ak−1 ai−2j with a1j = aj . Thus (1.1) for j all such v 0 is equivalent to the uniform convergence of akj h(2k x − j). aki
j
On the other hand, define S to be the operator given by Sf (x) = aj f (2x − j). j
One gets S k f (x) =
akj f (2k x − j), k = 1, 2, . . . .
j
Received by the editor December 13, 2002 and, in revised form, January 15, 2004. 2000 Mathematics Subject Classification. Primary 65D17, 26A15, 26A18, 39A10, 39B12. Key words and phrases. Cascade algorithm, joint spectral radius, nonnegative mask, subdivision scheme. c 2004 American Mathematical Society
819
820
XINLONG ZHOU
The convergence of (1.1) is therefore equivalent to the uniform convergence of S k h, which leads to the following cascade algorithm: let g0 (x) = h(x); one defines recursively gk (x) by aj gk−1 (2x − j). gk (x) = j
The limit g = limk→∞ gk = limk→∞ S k h satisfies Sg = g, i.e., (1.2) g(x) = aj g(2x − j). j∈Z
This so-called two-scale delation equation and the above presented algorithms play an important role in wavelet analysis as well as in computer aided geometry design. A comprehensive discussion of this subject can be found in [2, 5, 10]. In order to characterize the convergence of the above algorithms, one uses the concept of joint spectral radius (see [11]). The joint spectral radius for two square matrices A0 and A1 is defined by 1
ρ(A0 , A1 ) = lim sup ||A1 · · · Ak || k , n→∞ k≤n
where || · || is a given matrix-norm and l ∈ {0, 1}. Thus, write a(z) = j aj z j and a(z) = (1 + z)b(z) if a(−1) = 0. The subdivision algorithm with finite mask a = {aj }j∈Z converges if and only if (see [2, 5, 10]) (i) a(1) = 2 and a(−1) = 0, (ii) ρ(B0 , B1 ) < 1 where B0 = [b2j−i ]i,j , B1 = [b2j−i+1 ]i,j and {bl } is the coefficients of b. Unfortunately, the determination of the joint spectral radius is generally NPhard by a result of Tsitsiklis and Blondel (see [13]). Thus, it seems difficult to determinate whether the considered spectral radius is less than one. Some partial results concerning the calculation of the joint spectral radius can be found in [1, 4, 7, 16] and the papers cited therein. In this paper we focus on subdivision algorithms with nonnegative finite masks, a property possessed by many applications in geometric modelling (see, e.g., [3, 12]). A remarkable fact of this class is that the convergence does not depend on the actual values of the mask but rather on the support of the mask I = {i : ai > 0} (see [10, 8]). Consequently the question was raised in [2] (see p. 55 of [2]) of identifying those I such that given any nonnegative mask supported on I, the corresponding subdivision algorithm converges. By applying a suitable translation, we may always assume in the following discussion that a nonnegative mask a = {aj } has the form a = {a0 , . . . , aN } with a0 , aN > 0. We believe that the answer to this question is (see [9]): Conjecture. The subdivision algorithm associated with the nonnegative mask a = {a0 , . . . , aN } converges if and only if the following both hold: (i) a(1) = 2, a(−1) = 0 and 0 < a0 , aN < 1, (ii) the greatest common divisor of {j : aj > 0} is 1. This conjecture is still not verified. On the other hand, it has been shown that the conditions are necessary (see [2, 9, 14]). There are various partial results which support this conjecture. Denote S(a) = {j : aj > 0}. Thus, Micchelli and Prautzsch (see [10]) prove that if (ii) is replaced by S(a) = {0, 1, . . . , N }, convergence follows.
SUBDIVISION SCHEMES
821
Gonsor (see [6]) shows that S(a) = {0, 1, . . . , N } can be weakened by {0, 1, N − 1, N } ⊆ S(a), while Melkman (see [9]), among others, proves that if instead of (ii) it holds that S(a) ⊇ {0, p, q, p + q} for gcd(p, q) = 1 or that S(a) contains two successive integers, then convergence follows. Recently, Wang proves this conjecture for a class of masks. One of the main results in [15] is the following Theorem 1.1. The subdivision algorithm with the nonnegative mask a = {aj } converges if instead of (ii) it holds that {r, p, q} ⊆ S(a) such that gcd(p−r, q−r) = 1 and q − r is even. Improving the technique introduced in [15], we can now confirm this conjecture. In next section we first collect some lemmas from [15] and prove the conjecture by using a lemma (see Lemma 2.4), which will be verified in Section 3. The proof of Lemma 2.4 is the kernel of this paper. 2. Proof of the conjecture Let us collect some results from [15]. For this goal we should denote the vector x in RN by x = (x0 , . . . , xN −1 )T . For any T ⊂ ZN , where ZN = {0, 1, . . . , N − 1}, we define IT to be the vector (x0 , x1 , . . . , xN −1 )T with xi = 1 if i ∈ T and xi = 0 otherwise. For any nonnegative N × N row stochastic matrix B, i.e., the sum for each row is one, we define a map FB such that FB (T ) = {j ∈ ZN : (BIT )j = 1}. Let A0 and A1 be N × N matrices deduced by the mask a: A0 = [a2j−i ]0≤i,j≤N −1
and
A1 = [a2j−i+1 ]0≤i,j≤N −1 .
Clearly, A0 and A1 are row stochastic matrices if a(1) = 2 and a(−1) = 0. We should write simply F0 = FA0 and F1 = FA1 . For convenience we should also use the following standard notation for algebraic sums of sets: let T be a set of integers; then for any integers α and β the set αT + β is given by αT + β := {αx + β : x ∈ T }. We have (see [15]) Lemma 2.1. Let B be a N × N nonnegative row stochastic matrix. Then FB (T1 ) and FB (T2 ) are disjoint if T1 , T2 ⊆ ZN are. Let C be an N × N nonnegative row stochastic matrix. Then FBC = FB ◦ FC . Furthermore, the subdivision algorithm with nonnegative mask a, which satisfies a(1) = 2 and a(−1) = 0, diverges if and only if there exist disjoint nonempty subsets T and T of ZN and a sequence (d1 , d2 , . . . , dm ) ∈ {0, 1}m for some m ≥ 1 such that (2.1)
T = Fdm ◦ · · · ◦ Fd1 (T ) and T = Fdm ◦ · · · ◦ Fd1 (T ).
Denote for S(a) the sets S0 = S(a) ∩ (2Z) and S1 = S(a) ∩ (2Z + 1). The map Ψ for T ⊂ Z is defined by (2T − q) ∪ (2T − q) . Ψ(T ) = q∈S0
q∈S1
The following relationship between Ψ and Fi is proved in [15].
822
XINLONG ZHOU
Lemma 2.2. Let a be a nonnegative mask satisfying a(1) = 2 and a(−1) = 0. Then for any T ⊂ ZN , we have F0 (T ) = Ψ(T ) ∩ ZN , F1 (T ) = (Ψ(T ) + 1) ∩ ZN .
(2.2)
Furthermore, for any (d1 , . . . , dm ) ∈ {0, 1}m we have (2.3) where k =
m j=1
Fd1 ◦ · · · ◦ Fdm (T ) = (Ψm (T ) + k) ∩ ZN , dj 2j−1 .
We may write S(a) = {0, p1 , . . . , pk } with pk = N . Generalizing Wang’s lemma (see [15]), we have Lemma 2.3. Let T be a subset of Z and let y ∈ T . Let i,j ∈ {0, 1} such that k
i,j ≤ 1, ∀ i = 1, 2, . . . , m.
j=1
Then it holds that 2m y −
k m
2m−i i,j pj ∈ Ψm (T ).
j=1 i=1
Proof. For m = 1 the assertion is clear since 2y − xpl ∈ Ψ(T ) for all x = 0, 1 and l = 1, 2, . . . , k. On the other hand, 1,1 + · · · + 1,k ≤ 1 implies either 1,j = 0 for j = 1, . . . , k or else 1,j = 0 for j = j and 1,j = 1. Thus 2y − 1,1 p1 − 1,2 p2 − · · · − 1,k pk ∈ Ψ(T ).
The general case follows from induction on m.
Let ϕ(n) be the Euler function of the number of elements in Zn that are co-prime with n. It is known that if gcd(k, n) = 1, then k ϕ(n) ≡ 1
(mod n).
In particular, for any odd integer n one has 2ϕ(n) ≡ 1
(2.4)
(mod n).
Now we are in the position to prove the conjecture. Proof of Conjecture. The necessity of the conditions is proved in [9]. To show the sufficiency, we denote S(a) = {0, p1 , . . . , pk } with pk = N and |S(a)| for the number of S(a). The first assumption of the conjecture implies that |S(a)| ≥ 3 and there exists an odd p in S(a) such that p = pj0 < pk no matter if pk is odd or even. Hence, for |S(a)| = 3 one has S(a) = {0, p, N } with even N . Wang’s result for |S(a)| = 3 (see [15]) may be read as: the subdivision algorithm converges if gcd(p, N ) = 1. We assume in the following that |S(a)| ≥ 3. Let us observe that if the subdivision algorithms with the mask a = {aj } diverges, then by Lemma 2.1 there are a sequence (d1 , . . . , dm ) and disjoint nonempty sets T, T ⊂ ZN so that (2.5)
Fd1 ◦ · · · ◦ Fdm (T ) = T and Fd1 ◦ · · · ◦ Fdm (T ) = T .
It follows from Lemma 2.2 (see (2.3)) that
T = (Ψm (T ) + k ) ∩ ZN and T = (Ψm (T ) + k ) ∩ ZN ,
SUBDIVISION SCHEMES
823
m where k = l=1 dl 2l−1 . Write F = Fd1 ◦ · · ·◦ Fdm . Then F η (T ) = T and F η (T ) = T for any η ≥ 1. Therefore, with
kη
m m m = ( dl 2l−1 ) + ( dl 2l−1 )2m + · · · + ( dl 2l−1 )2(η−1)m l=1 η−1 im
= k
l=1
l=1
2
i=0
we get for any η ≥ 1 (2.6)
T = (Ψm η (T ) + kη ) ∩ ZN and T = (Ψm η (T ) + kη ) ∩ ZN .
We should choose η in connection with S(a). For our goal we define η to be (2.7)
η=t
pk
{jϕ(j)}
j=1
with some positive integer t, which will be given later. Denote m = m η. We have 2m ≡ 1 (mod x) for odd number x from {1, . . . , pk }. Indeed, η has a factor xϕ(x), so for some integer n one has η = n xϕ(x). Since x is odd, we obtain by Euler’s formula (2.4)
2m = 2m n xϕ(x) ≡ 1
(2.8)
(mod x).
Similarly, we have kη ≡ 0
(2.9)
(mod x).
To see this, one gets by the expression of kη kη
=
k
x−1 ϕ(x)(l+1)−1 n
l=0
=
k k
2jm 2ϕ(x)lm
2jm
(mod x)
j=0
ϕ(x)−1
k n x
j=0
x−1 ϕ(x)−1 n
l=0
≡
j=ϕ(x)l
x−1 ϕ(x)−1 n
l=0
≡
2jm
2jm
(mod x)
j=0
≡
0 (mod x),
which yields (2.9). Moreover, kη = 0
⇐⇒
d1 = · · · = dm = 0
and kη = 2m − 1
⇐⇒
d1 = · · · = dm = 1.
For all other dj we can thus choose t large enough such that for some 0 < δ < 1 (2.10)
pk < kη = k
2m − 1 ≤ (1 − δ)2m 2 m − 1
824
XINLONG ZHOU
and
2m η (2m − 1 − k ) + k (2.11) 2 − kη = > 2pk . 2 m − 1 Additionally, in view of (2.5) we may without loss of the generality assume that kη is even if kη = 2m − 1. In the following discussion we should always denote m and kη to be any nonnegative integers, which satisfy the above properties, i.e., 0 ≤ kη ≤ 2m ; if kη = 0, 2m − 1, then kη is even, m and kη satisfy (2.8)–(2.11). We know that the divergence of the subdivision algorithm implies that there are two disjoint nonempty sets T, T ⊂ ZN for (2.5). We may therefore assume x ∈ T and x ∈ ZN . So from Lemma 2.3 it holds that m
2m x −
k m−1
2i i,j pj ∈ Ψm (T )
j=1 i=0
for all i,j ∈ {0, 1} such that (2.12)
k
j=1 i,j
2m x + kη −
≤ 1. By (2.6) we obtain
k m−1
2i i,j pj ∈ T
j=1 i=0
for all those i,j . Denote B0 = {x} with x ∈ ZN and for l = 0, 1, . . . , Bl+1 = {2m y + kη −
k m−1
2i i,j pj ∈ ZN | y ∈ Bl , i,j ∈ {0, 1},
j=1 i=0
k
i,j ≤ 1}.
j=1
Thus, (2.12) tells us that x ∈ T implies Bl ∩ T = ∅ for all l ≥ 0. In the next section we will prove the following Lemma 2.4. Let the mask a = {aj } satisfy the conditions of the conjecture and S(a) = {0, p1, . . . , pk }. Then there is an 0 ≤ x < pk such that with B0 = {x} one has Bl = Zpk . l≥0
This lemma, together with the above consideration, implies that the set T or T must be empty. This contradiction proves the conjecture. The proof of Lemma 2.4 is much more involved and, in fact, is the kernel of this paper. We prove this lemma in the next section. 3. Proof of Lemma 2.4 The notation such as Bj , kη and m have the same mean and properties as in Sections 1 and 2. Let us begin with the observation of the first condition of the conjecture. We need the following Definition. A set of integers I = {0, q1 , . . . , qk } with 0 < qj < qj+1 has property P if either qk is even and I contains at least one odd integer or else I contains at least two odd and two even integers (including 0). Obviously, if a = {aj } satisfies the first condition of the conjecture, the set S(a) has property P. Conversely, if a set {0, q1 , . . . , qk } has property P, then one can construct a mask a = {aj } satisfying the first condition of the conjecture and
SUBDIVISION SCHEMES
825
S(a) = {0, q1 , . . . , qk }. In fact, in [9] this conjecture is formulated by using property P. We may therefore reformulate Lemma 2.4 as Lemma 2.4 . If S(a) = {0, p1 , . . . , pk } has property P and gcd(p1 , . . . , pk ) = 1, then there is an x ∈ Zpk such that with B0 = {x} (3.1) Bl = Zpk . l≥0
To prove Lemma 2.4 or Lemma 2.4 , we remember that {i,j } ∈ {0, 1} for i = 0, 1, . . . , m − 1, j = 1, . . . , k and k
i,j ≤ 1,
i = 0, 1, . . . , m − 1.
j=1
Let δi,j = i,j for i = 0, 1, . . . , m − 1, j = 1, . . . , k − 1 and δi,k = 1 −
k
i,j .
j=1
Hence, {δi,j } satisfies our conditions. We obtain (3.2)
−
2m x + kη −
k m−1
δi,j 2i pj = (pk − 1)
j=1 i=0
2m (pk − 1 − x) + (2m − 1 − kη ) −
k−1 m−1
i,j 2i (pk − pj ) −
j=1 i=0
m−1 i=0
i,k 2i pk
.
We notice that 2m − 1 − kη is between 0 and 2m − 1. Formulas (2.8) and (2.9) imply that (2.9) also holds for kη = 2m − 1 − kη instead of kη . Let Bˆj = pk − 1 − Bj . Then Bˆj is defined by {0, pk −pk−1 , . . . , pk −p1 , pk } and kη . Since gcd(p1 , . . . , pk ) = gcd(pk − pk−1 , . . . , pk − p1 , pk ), the symmetric relation (3.2) tells us that ˆl . v∈ B Bl ⇐⇒ pk − 1 − v ∈ l≥0
l≥0
Moreover, if {0, p1 , . . . , pk } has property P, then {0, pk − pk−1 , . . . , pk − p1 , pk } does also. Thus, it is enough to show Lemma 2.4 either for {0, p1 , . . . , pk } or for {0, pk − pk−1 , . . . , pk − p1 , pk }. For simplicity we define B(x) = Bl l≥0
with B0 = {x} ⊂ Zpk . The following assertions will be used frequently and are based on some special choices of {i,j }. We formulate them as four lemmas. Lemma 3.1. Let y ∈ B(x) and 0 ≤ y ≤ pl1 for some odd pl1 ∈ S(a). If 0 < (2m − 1)y + kη ≤ (2m − 1)pl1 , then y + jpl1 ∈ B(x) for j such that y + jpl1 ∈ Zpk . Proof. We choose i,j = 0 for j = l1 . By the definition of i,j the number l = m−1 i m i=0 i,l1 2 can be any number between 0 and 2 − 1. On the other hand, in view
826
XINLONG ZHOU
of (2.8) and (2.9) we have (2m − 1)y + kη = αpl1 . Hence, y + (2 − 1)y + kη − m
m−1
i,l1 pl1 2i = y + (α − l)pl1 .
i=0
The choice of η (see (2.7)) implies kη > pk and 2m − kη > 2pk if kη = 0 and 2m − 1, while for kη = 0, the inequality (2m − 1)y > 0 yields y ≥ 1. Moreover, if kη = 2m − 1, then the condition implies y < pl1 . Hence α − l can be any number between −pk and pk . In particular, for l such that y + (α − l)pl1 ∈ Zpk we have y + (α − l)pl1 ∈ B(x). Lemma 3.2. Let y ∈ B(x) and pl1 ∈ S(a) be even. Assume pl1 /2 ≤ y and (2m − 1)(y − pl1 /2) + kη ≤ (2m−1 − 1)pl2 for some odd pl2 ∈ S(a). Then y − pl1 /2 ∈ B(x). Proof. We choose m−1,l1 = 1 and i,l1 = 0 if i = m−1. For other j = l2 let i,j = 0. m−2 To meet the condition on i,j , we have to choose m−1,l2 = 0. Now i=0 i,l2 2i can be any number between 0 and 2m−1 − 1. From (2.8), (2.9) and our restriction on y we conclude (2m − 1)(y − pl1 /2) + kη = αpl2 for some 0 ≤ α ≤ 2m−1 − 1. i Hence, we can choose i,l2 for i = 0, 1, . . . , m − 2 to obtain m−2 i=0 i,l2 2 = α. With this choice of i,j we get 2m y + kη −
k m−1
i,j pj 2i = y − pl1 /2.
j=1 i=0
Thus, y − pl1 /2 ∈ B(x).
Lemma 3.3. Let y ∈ B(x). If there is some odd pl1 ∈ S(a) such that (2 − 1)y + kη < (2m − 1)pl1 , then for any pj ∈ S(a) one has (i) y+(2l+1)pl1 −pj ∈ B(x) whenever y+(2l+1)pl1 −pj ∈ Zpk and (2m −1)y+kη is odd; (ii) y + 2lpl1 − pj ∈ B(x) whenever y + 2lpl1 − pj ∈ Zpk and (2m − 1)y + kη is even. m
Proof. Clearly (2m − 1)y + kη = αpl1 for some 0 ≤ α < 2m − 1. If α is odd, we m−1 choose i,l1 such that α − i=0 i,l1 2i = 2l + 1. Obviously, 0,l1 must be zero. Hence, we can choose 0,j = 1 or 0 to obtain 2m y + kη −
k m−1
i,j pj 2i = y + (2l + 1)pl1 − 0,j pj .
j=1 i=0
Since y ∈ B(x), the set B(x) contains y + (2l + 1)pl1 − 0,j pj whenever y + (2l + 1)pl1 − 0,j pj ∈ Zpk . Similarly, we have the second assertion. Lemma 3.4. Let y ∈ B(x) and pl2 −pl1 > 0 be odd for some pl1 , pl2 ∈ S(a). Assume 0 < (2m − 1)(y − pl1 ) + kη < (2m − 1)(pl2 − pl1 ). Then, y + j(pl2 − pl1 ) ∈ B(x) for j ∈ Z such that y + j(pl2 − pl1 ) ∈ Zpk . Proof. Let i,l1 = 1 − i,l2 and i,j = 0 for j = l1 , l2 . We obtain 2 y + kη − m
k m−1 j=1 i=0
i,j pj 2
i
= y + (2 − 1)(y − pl1 ) + kη − m
m−1
i,l2 (pl2 − pl1 )2i
i=0
= y + (2m − 1)(y − pl1 ) + kη − l(pl2 − pl1 ),
SUBDIVISION SCHEMES
827
where l can be any integer between 0 and 2m − 1. On the other hand, our condition, (2.8) and (2.9) imply pk ≤ (2m − 1)(y − pl1 ) + kη = α(pl2 − pl1 ) < (2m − 1)(pl2 − pl1 ). Hence, for any nonnegative integer j such that y + j(pl2 − pl1 ) ∈ Zpk we can choose l ∈ [0, 2m − 1] so that α − l = j, i.e., y + j(pl2 − pl1 ) ∈ B(x). On the other hand, the condition implies that y + 1 ≤ pl2 if kη = 2m − 1 and y + 2 ≤ pl2 if kη = 2m − 1. Thus, for any negative integer j such that y + j(pl2 − pl1 ) ∈ Zpk we can choose l ∈ [0, 2m − 1] so that α − l = j, i.e., y + j(pl2 − pl1 ) ∈ B(x). We are now ready to verify Lemma 2.4 . We prove this lemma using induction on |S(a)|. The assertion (3.1) for |S(a)| = 3 is essentially given by Wang in [15] (see the proof of Theorem 1.1 in [15]). Proof of Lemma 2.4 . The idea behind the proof is that first we prove the assertion for a small p instead of pk , i.e., {0, . . . , p − 1} ⊆ B(x). If p is odd and belongs to S(a), we can then extend this to all of {0, 1, . . . , pk −1} by using Lemma 3.1. In order to obtain the result for all of {0, 1, . . . , p − 1}, we choose carefully an x in this set and apply Lemmas 3.1–3.4 to get another number, which is greater than p. Using Lemma 3.2 or Lemma 3.4, we may reduce the number so that the new number is again contained in {0, . . . , p − 1} but is different from x. Recursively, we obtain several numbers. One of our tasks is to show that those numbers are all different. A careful choice of x is necessary. Otherwise the following calculation leads to B(x) = {0}. In fact, if kη = 0, then for x = 0 and B0 = {0} one gets Bl = {0} for all l = 1, 2, . . . . Thus, in our choice of x it may be understood that we avoid zero if kη = 0 or pk − 1 if kη = 2m − 1 during the calculation of new numbers. In other words, we should choose such an x that zero or pk − 1 occurs at the last step of our calculation. Next let us observe the set S(a) with |S(a)| ≥ 5. It is easy to check that for some nonzero p ∈ S(a) the set S(a) \ {p} also has property P. However, for sets with |S(a)| = 4 we may not have this p. The only set that does not have this p is {0, p1 , p2 , p3 } with even p1 and odd p2 , p3 . We already know that |S(a)| ≥ 3. Since the assertion for |S(a)| = 3 is essentially given in [15], we should omit the proof for this case. Our approach, therefore, has two steps; namely, the proof for the special case of |S(a)| = 4 and for |S(a)| ≥ 4. Let us begin with the special case. Step 1. S(a) = {0, p1 , p2 , p3 } with even p1 and odd p2 , p3 . We divide the proof into three cases according to the different position of p1 , p2 and p3 . Case 1. Let us prove (3.1) for 2p2 < p3 . Assume kη = 2m − 1. Denote gcd(p1 , p2 ) = d and p = p1 /2. The 0 ≤ x < p is given as follows: x + p3 ≡ 0 (mod d), d = 1, x − p2 ≡ 0 (mod p), d = 1. With this x we have (3.3)
{x + dv : 0 ≤ x + dv ≤ p} ⊆ B(x).
To see this, we apply Lemma 3.1 to get x+p2 ∈ B(x). Using Lemma 3.2 for pl1 = p1 and pl2 = p3 , we conclude x + p2 − p ∈ B(x). Repeatedly we get x + p2 − α1 p ∈ B(x) whenever 0 ≤ x + p2 − α1 p < p. We notice that x + p2 − α1 p satisfies the above restriction. Thus, recursively, i.e., replacing x by x + p2 − α1 p, we conclude
828
XINLONG ZHOU
x + sp2 − αs p ∈ B(x) and 0 ≤ x + sp2 − αs p < p for s = 0, 1, . . . , p/d − 1, where α0 = 0. We note also that the choice of x ensures x + sp2 − αs p = 0 at least for s = 0, 1, . . . , p/d − 2, while for s = p/d − 1 one has x + sp2 − αs p = 0, if d = 1, x + sp2 − αs p = 0, if d = 1. Moreover, it follows from gcd(p2 , p) = d that x + sp2 − αs p = x + s p2 − αs p if 0 ≤ s < s ≤ p/d − 1. Therefore, {x + sp2 − αs p : s = 0, 1, . . . , p/d − 1} = ⊆
{x + dv : 0 ≤ x + dv < p} B(x).
Examining the way we calculate the number x + sp2 − αs p for s = p − 1 and d = 1, we obtain additionally p ∈ B(x), proving (3.3). Having (3.3), we will replace p in (3.3) by p2 . We use the same procedure as above. Let x + dv be any nonzero element from {x + dv : 0 ≤ x + dv ≤ p}. Lemma 3.1 leads to x + dv + p2 ∈ B(x) while Lemma 3.2 implies x + dv + p2 − lp ∈ B(x) for l ≥ 0 such that x + dv + p2 − lp ∈ Zp3 . Clearly, there are l and ν such that p2 /d = l p/d + ν or p2 = l p + νd. Hence, for p < x + dv ≤ p2 we get x + dv = x + d(v − ν) + p2 − l p. Now, if 0 < x + d(v − ν) ≤ p, then x + dv ∈ B(x). If x + d(v − ν) ≤ 0, there exists l such that 0 < x + d(v − ν) + l p ≤ p. Hence, x + dv = x + d(v − ν + l p/d) + p2 − (l + l )p. We still have x + dv ∈ B(x). Finally, if p < x + d(v − ν), one has l such that 0 < x + d(v − ν) − l p ≤ p. Thus, x + dv = x + d(v − ν − l p/d) + p2 − (l − l )p. We conclude in particular that (3.4)
{x + dv : 0 ≤ x + dv ≤ p2 } ⊆ B(x).
In order to replace p2 in (3.4) by p3 − 1, we again use Lemma 3.1. Examining the conditions of Lemma 3.1, we know that if d = 1, then for each 0 ≤ x + dv ≤ p2 one gets x + dv + lp2 ∈ B(x) whenever x + dv + lp2 ∈ Zp3 . Hence, (3.5)
{x + dv : 0 ≤ x + dv < p3 } ⊆ B(x).
However, if d = 1, the above calculation may not be true. We should modify this as follows: for kη = 0 let 0 ≤ x + v < p2 . Again we have (3.5). If kη = 0, let 0 < x + v ≤ p2 . We obtain x + v + lp2 ∈ B(x). Hence, (3.5) is still valid. Clearly, (3.5) implies (3.1) if d = 1. To show (3.1) for d = 1, we observe elements from (3.5) satisfying p3 − p1 ≤ x + dv < p3 −p. Obviously we can apply Lemma 3.4 for y = x+dv, pl1 = p1 and pl2 = p3 to obtain x+dv −(p3 −p1 ) ∈ B(x). Noticing 2p2 < p3 and 0 ≤ x+dv −(p3 −p1 ) < p, we get from Lemma 3.1 that x + dv − (p3 − p1 ) + p2 ∈ B(x) if kη = 0 or x − p3 ≡ 0 (mod d). Under this restriction, Lemma 3.2 tells us that x+dv −(p3 −p1 )+p2 −pl ∈ B(x) whenever x + dv − (p3 − p1 ) + p2 − pl ∈ Zp3 . Recursively, we conclude that if kη = 0 or x − p3 ≡ 0 (mod d), then x + dv − (p3 − p1 ) + jp2 − lp ∈ B(x) whenever x + dv − (p3 − p1 ) + jp2 − lp ∈ Zp3 . Let 0 ≤ x + dν − (p3 − p1 ) < p3 . Then due to gcd(p2 , p) = d, there are j ≥ 0, l ≥ 0 and 0 ≤ x + dv − (p3 − p1 ) < p such that x + dν − (p3 − p1 ) = x + dv − (p3 − p1 ) + jp2 − lp. Therefore, if kη = 0 or x − p3 ≡ 0 (mod d), {x + dv − (p3 − p1 ) : 0 ≤ x + dv − (p3 − p1 ) < p3 } ⊆ B(x).
SUBDIVISION SCHEMES
829
This observation leads to the following assertion: Let Aj = {x + dv − j(p3 − p1 ) : 0 ≤ x+dv−j(p3 −p1 ) < p3 }. Then under the assumption kη = 0 or x−(i+1)p3 ≡ 0 (mod d) we have that Ai ⊆ B(x) implies Ai+1 ⊆ B(x). We already know A0 ∪ A1 ⊆ B(x). Thus, we use the above procedure with p3 − p1 ≤ x + dv − i(p3 − p1 ) < p3 − p instead of p3 − p1 ≤ x + dv − (i − 1)(p3 − p1 ) < p3 − p to obtain Ai+1 ⊂ B(x). We remember x + p3 ≡ 0 (mod d). Hence, x − jp3 ≡ 0 (mod d) for j = 0, 1, . . . , d − 2. We conclude d−2
(3.6)
Aj ⊆ B(x).
j=0
Moreover, in the case of kη = 0 d−1
Aj ⊆ B(x).
j=0
To have (3.1) for kη = 0, we notice gcd(d, p3 ) = 1. Hence, for any v such that 0 ≤ x + v < p3 there is 0 ≤ j ≤ d − 1 with v + j(p3 − p1 ) ≡ 0 (mod d), which gives v = dv − j(p3 − p1 ) for some v. We conclude that x + v = x + dv − j(p3 − p1 ) ∈ Aj or Zp3 = {v : 0 ≤ v < p3 } ⊆ B(x). It remains to show (3.1) for kη = 0. We already have (3.6). But what is the corresponding Ad−1 for this case? We know that the first step in our procedure is to find an element in Ad−2 which satisfies p3 − p1 ≤ x + dv − (d − 2)(p3 − p1 ) < p3 − p.
(3.7)
If x + dv − (d − 2)(p3 − p1 ) = p3 − p1 , we can use our procedure to obtain x + dv − (d − 1)(p3 − p1 ) + jp2 − lp ∈ B(x). If however x + dv − (d − 2)(p3 − p1 ) = p3 − p1 , we get formally from the above procedure a set A = {jp2 − lp : 0 ≤ jp2 − lp < p3 }. But, because kη = 0, we cannot use this approach for x+ dv − (d− 1)(p3 − p1 ) (= 0). Thus, instead of Ad−1 we have only Ad−1 \ A ⊂ B(x). Hence, combining this with
(3.6), we obtain ( d−1 i=0 Ai ) \ A ⊆ B(x), which implies {v : 0 ≤ v < p3 } \ A ⊆ B(x). To show A ⊂ B(x), we have to modify (3.7). We remember that gcd(d, p) = d and x − (d − 1)p3 ≡ 0 (mod d). Hence, there is a v such that p3 − p1 < x + dv − (d − 2)(p3 − p1 ) = p3 − p, which implies that x + dv − (d − 1)(p3 − p1 ) (= p) is contained in B(x) due to Lemma 3.4. Now we can apply our procedure for p to obtain jp2 − lp ∈ B(x) or A ⊂ B(x). Consequently, Zp3 = {v : 0 ≤ v < p3 } ⊆ B(x). For kη = 2 − 1 we replace x + p3 ≡ 0 (mod d) by x + 1 + p3 ≡ 0 (mod d) and p|(x − p2 ) by p|(x + 1 − p2 ), respectively. Case 2. Now if p3 < 2p1 , we use the symmetric relation to obtain (3.1), since (0, p3 − p2 , p3 − p1 , p3 ) has property P and 2(p3 − p1 ) < p3 . m
830
XINLONG ZHOU
Case 3. Since p3 is odd, it remains to show (3.1) for 2p1 < p3 < 2p2 . Without loss of generality we may suppose p3 ≥ p2 + p1 . Now, p1 is even, so p1 = 2p for some p. Let p be odd. Denote gcd(p1 , p3 ) = d. We have gcd(p, p3 ) = d. We verify for some 0 ≤ x ≤ p that {v : 0 ≤ v < p1 } \ {p} ⊆ B(x).
(3.8)
To this end we choose 0 ≤ x ≤ p in the following way: if 0 ≤ kη < 2m − 1, then x = 0 and x − p2 ≡ 0 (mod d), if d = 1, (3.9) x + p3 ≡ 0 (mod p), if d = 1; if kη = 2m − 1, then x = p and x + 1 − p2 ≡ 0 (mod d), (3.10) x + 1 ≡ 0 (mod p),
if d = 1, if d = 1.
In the following we deal only with the case 0 ≤ kη < 2m − 1. The corresponding assertion for kη = 2m − 1 can be treated in the same way. To begin with we note p|(2m − 1) and p|kη (see (2.8) and (2.9)). Hence, for some k we get 2m x + kη = x + kp. Since kη < (1 − δ)2m for some 0 < δ < 1 (see (2.10)), the number k satisfies 2p3 < k < (2 − δ)2m . Write k = 2µ + β with β ∈ {0, 1}, so p3 < µ < (1 − δ/2)2m . Let i,j = 0 for j = 2, 3. We get 2m x + kη −
m−1 3
i,j pj 2i
=
2m x + kη − lp1
=
x + kp − lp1
=
x + µp1 − lp1 + βp
=
x + αp1 + βp.
i=0 j=1
We can choose l satisfying p3 − p1 ≤ x + αp1 + βp < p3 . Thus, x + αp1 + βp ∈ B(x) by the definition of B(x). Now the fact p1 < p3 − p1 and Lemma 3.4 imply further 0 ≤ x + αp1 + βp − p3 + p1 < p1 and x + αp1 + βp − p3 + p1 ∈ B(x). Lemma 3.2 tells us that we can substitute p if the above number is not less than p. We obtain in this way x − p3 + l1 p = x + αp1 + βp − p3 + p1 − p ∈ B(x) and 0 ≤ x− p3 + l1 p < p. As in Case 1 we conclude recursively that 0 ≤ x− sp3 + ls p < p for s = 0, . . . , p/d − 1. Clearly, by this construction x − sp3 + ls p ∈ B(x). Moreover, since gcd(p, p3 ) = d, all these numbers are different for 0 ≤ s ≤ p/d − 1. Hence, {x − sp3 + ls p : x − sp3 + ls p ∈ [0, p)} = {x + dv : 0 ≤ x + dv < p} ⊆ B(x). We notice also that the set on the left-hand side contains the zero if and only if d = 1. For d = 1 we conclude from the last relation that (3.11)
{0, 1, . . . , p − 1} ⊆ B(x).
We claim that (3.11) is still valid for d = 1. To see this, let 0 < x + dv < p. The above consideration tells us that we have α and β, where β ∈ {0, 1} depends on x + dv, such that x + dv + αp1 + βp ∈ B(x) and p3 − p1 − p2 ≤ x + dv + αp1 + βp < p3 − p2 . Now, since d = 1, we deduce p3 − p1 − p2 = 0. Hence, 0 < x + dv + αp1 + βp < p3 − p2 < p2 . We can therefore apply Lemma 3.1 for
SUBDIVISION SCHEMES
831
x + dv + αp1 + βp to obtain x + dv + αp1 + βp + p2 ∈ B(x) and by Lemma 3.4 we have x + dv + αp1 + βp + p2 − p3 + p1 ∈ B(x). This number is nonnegative and less than p1 . So by Lemma 3.2 we can again substitute p, if necessary, to obtain x + dv + (p2 − p3 + p1 ) + l1 p ∈ B(x) and 0 ≤ x + dv + (p2 − p3 + p1 ) + l1 p < p. Recursively, B(x) contains x + dv + s(p2 − p3 + p1 ) + ls,v p = x + dv + sp2 + ls,v d, ∀ s = 0, 1, . . . , d − 1.
Since gcd(p2 , d) = 1, we conclude by the second expression that they numbers are different for different s, while the first one shows that they are different if v is different. Thus, there are d×p/d = p numbers in {0, . . . , p−1}, which are contained in B(x); in other words, (3.11) also holds for d = 1. To get (3.8) from (3.11), we observe any nonzero number µ satisfying 0 < µ < p. We know that µ + αp1 + βp ∈ B(x) whenever 0 ≤ µ + αp1 + βp < p3 , where β ∈ {0, 1} relies on µ. For the µ with β = 0 let α = 1. So µ + p1 ∈ B(x) and by Lemma 3.2, µ + p1 − p ∈ B(x). For the µ with β = 1 we choose α = 0. Hence, we always have µ + p ∈ B(x), which implies (3.8). To verify our assertion from (3.8), we remember that 0 ≤ kη < 2m − 1. So kη is even. If 0 < kη < 2m − 1, then since 0 ∈ B(x) and 2m 0 + kη − lp1 = jp1 for some l, we conclude jp1 ∈ B(x) if 0 ≤ jp1 < p3 . Lemma 3.2 yields p ∈ B(x). Combining this with (3.8), we obtain {0, 1, . . . , p1 − 1, p1 , 2p1 , . . . , j0 p1 } ⊆ B(x), where j0 satisfies j0 p1 < p3 ≤ (j0 + 1)p1 . Moreover, for any 0 < 2l < p1 we have 2m 2l + kη − ip1 = 2l + αp1 with some i. Hence, B(x) contains 2l + αp1 whenever 0 < 2l + αp1 < p3 . In particular, (3.12)
{0, 1, . . . , p1 } ∪ {2j : 0 ≤ 2j < p3 } ⊆ B(x).
Let 0 < 2j < p2 . Then Lemma 3.1 shows that 2j + p2 ∈ B(x) whenever 0 < 2j + p2 < p3 . Since p2 is odd and 2p2 > p3 , we obtain from (3.12) that {l : p2 < l < p3 } ⊆ B(x). Let p1 + p2 ≤ 2j < p3 . Lemma 3.3 tells us 2j − p2 ∈ B(x). It follows from (3.8) and the last two relations that {l : 0 ≤ l ≤ p3 − p2 } ⊆ B(x). Let l be odd and let it satisfy 0 ≤ l < p3 − p2 . So using (i) of Lemma 3.3 several times, we obtain l + j(p3 − p2 ) ∈ B(x) whenever 0 ≤ l + j(p3 − p2 ) < p3 . We conclude from the last relation that {l : 0 < l < p3 , l is odd} ⊆ B(x).
(3.13)
Assertion (3.1) follows from (3.12) and (3.13). Next we prove (3.1) for kη = 0 via (3.8). The above approach tell us that we need only to have (3.12). For this goal let us observe B(p1 ). By the definition of B(p1 ) we certainly have {lp1 : 0 < l ≤ l } ⊆ B(p1 ), where l satisfies l p1 < p3 ≤ (l +1)p3 . Lemma 3.2 implies p ∈ B(p1 ). On the other hand, there is a choice of i,j such that 2 m p1 −
m−1 3 i=0 j=1
Thus,
i,j 2i pj = (2m − 1)p1 −
m−1
i,3 2i p3 = 0.
i=1
{lp1 : 0 ≤ l ≤ l } ∪ {p} ⊆ B(p1 ).
832
XINLONG ZHOU
Let p3 − p1 ≤ lp1 < p3 . It follows from Lemma 3.4 that (l + 1)p1 − p3 ∈ B(p1 ). Thus, by Lemma 3.2 there exists a j such that 0 < jp − p3 ≤ p and jp − p3 ∈ B(p1 ). Similarly, let p3 − p2 − p1 < lp1 ≤ p3 − p2 . We conclude for some j that jp − p3 + p2 ∈ B(p1 ) and 0 < jp − p3 + p2 ≤ p. Now we can choose x as follows: x = jp − p3 + p2 and 0 < jp − p3 + p2 ≤ p, if d = 1, x = jp − p3 and 0 < jp − p3 ≤ p, if d = 1. Clearly, x satisfies (3.9) and x ∈ B(p1 ). Consequently, (3.8) holds with this x and B(x) ⊆ B(p1 ). We obtain by (3.8) the desired relation, i.e., {0, 1, . . . , p1 − 1, p1 , 2p1 , . . . , l p1 } ⊆ B(p1 ). Now let p = p1 /2 be even. Denote gcd(p3 − p2 , p) = d. So d is even. We verify for some 0 ≤ x < p that (3.14)
{x + dv : 0 ≤ x + dv < p} ⊆ B(x).
To this end we choose 0 ≤ x < p as follows: x − p2 ≡ 0 (mod d), (3.15) x + 1 − p2 ≡ 0 (mod d),
if 0 ≤ kη < 2m − 1, if kη = 2m − 1.
Clearly, x is not unique. But x is odd if 0 ≤ kη < 2m −1 and even when kη = 2m −1. Moreover, any 0 ≤ x + dv < p satisfies (3.15). To verify (3.14), we remember that kη is even when 0 ≤ kη < 2m − 1. So in any case (2m − 1)x + kη is odd. We also remember p3 |(2m − 1) and p3 |kη since p3 is odd. Now by Lemma 3.3 we obtain x + p3 − p2 ∈ B(x). On the other hand, since x − p ≤ −1 and p3 − p2 ≤ (p3 − 1)/2, we have (2m − 1)(x + p3 − p2 − p) + kη < (2m−1 − 1)p3 . Thus, the condition of Lemma 3.2 is satisfied. We obtain by Lemma 3.2 with y = x + p3 − p2 , pl1 = p1 and pl2 = p3 that x + p3 − p2 − p ∈ B(x). Recursively, x + p3 − p2 − α1 p ∈ B(x) for some α1 such that 0 ≤ x + p3 − p2 − α1 p < p. Clearly, this new number also satisfies (3.15). We can therefore apply this procedure for this new number to obtain another number that satisfies (3.15) and is contained in B(x). Repeatedly, we obtain for s = 0, 1, . . . , p/d − 1 that x + s(p3 − p2 ) − αs p ∈ B(x) and 0 ≤ x + s(p3 − p2 ) − αs p < p, where α0 = 0. Since gcd(p3 − p2 , p) = d, they are all different and have the form x + dv. This shows (3.14). Next we verify (3.16)
{0, . . . , p − 1} ⊆ B(x).
To show (3.16), let 0 ≤ x + dv < p. Since (2m − 1)(x + dv) + kη is odd, it follows from (i) of Lemma 3.3 with pj = p1 and pl1 = p2 that x + dv + p2 − p1 ∈ B(x). Now (2m − 1)(x + dv + p2 − p1 ) + kη is even. We apply (ii) of Lemma 3.3 with pl1 = p2 and, if necessary, Lemma 3.3 with pl1 = p1 to obtain 0 ≤ x + dv + p2 − l1 p < p and x + dv + p2 − p1 − lp = x + dv + p2 − l1 p ∈ B(x). For this new number (2m − 1)(x + dv + p2 − l1 p) + kη is even. Clearly, we can again use (ii) of Lemma 3.3 to obtain x + dv + p2 − l1 p + 2p2 − p3 ∈ B(x). Write x + dv + p2 − l1 p + 2p2 − p3 = x + dv + 2p2 − l1 p − (p3 − p2 ).
SUBDIVISION SCHEMES
833
For this number (2m − 1)(x + dv + 2p2 − l1 p − (p3 − p2 )) + kη is odd. We get by (i) of Lemma 3.3 and, if necessary, Lemma 3.2 that 0 ≤ x+ dv + 2p2 − l2 p− l (p3 − p2 ) < p and x + dv + 2p2 − l1 p − l (p3 − p2 ) − lp = x + dv + 2p2 − l2 p − l (p3 − p2 ) ∈ B(x). Repeatedly, we conclude for s = 0, 1, . . . , d − 1 that 0 ≤ x + dv + sp2 − ls p − ls (p3 − p2 ) < p and x + dv + sp2 − ls p − ls (p3 − p2 ) ∈ B(x). By our construction it is easy to see that if (v, s) = (v , s ), then the corresponding numbers are different. There are p/d different v and d different s. Therefore, B(x) contains d × p/d numbers in {0, . . . , p − 1}. In other words, (3.16) holds. We have to show that p in (3.16) can be replaced by p3 . We first prove this for kη = 2m − 1. Let 0 < l < p be odd. So (2m − 1)l + kη is odd. As before we use Lemmas 3.3 and 3.2 to obtain l + p3 − p2 − αp ∈ B(x) for α ≥ 0 satisfying 0 ≤ l + p3 − p2 − αp < p3 . Since l + p3 − p2 − αp = l is again odd, we obtain in particular that all odd numbers in {0, . . . , p3 − p2 − 1} belong to B(x). Applying Lemma 3.3 for each odd number l from the last set, we conclude l +l(p3 −p2 ) ∈ B(x) whenever 0 ≤ x + dv + l(p3 − p2 ) < p3 and l ≥ 0. Thus, B(x) contains all odd 0 < l < p3 , i.e., (3.17)
{l : 0 ≤ l < p3 , l is odd} ⊆ B(x).
Let l be odd and 0 < l < p3 − p2 . Since p3 − p2 < p2 , we have by Lemma 3.1 with p2 that l + p2 ∈ B(x). Consequently, (3.18)
{l : p2 ≤ l < p3 } ⊆ B(x).
Furthermore, let l be even and p2 ≤ l < p3 . So (2m − 1)l + kη is even. We conclude from Lemma 3.3 that B(x) contains l − µp1 whenever µ ≥ 0 and 0 ≤ l − µp1 < p3 . It follows from the fact that p3 ≥ p2 + p1 and p1 is even that (3.19)
{l : 0 ≤ l < p2 , l is even} ⊆ B(x).
The desired assertion follows from (3.16)–(3.19). The proof for kη = 2m − 1 is the same. We omit the details here. Step 2. Let (3.1) be true for |S(a)| ≤ k with k ≥ 3. We verify (3.1) for |S(a)| = ˜ k + 1. Write S(a) = {0, p1 , . . . , pk } and S(a) = {0, pk − pk−1 , . . . , pk − p1 , pk }. ˜ Therefore, we need to prove (3.1) either for S(a) or for S(a). The special case of |S(a)| = 4, which we dealt with in Step 1, allows us to suppose that there is 0 < p ∈ S(a) such that S(a) \ {p} has property P. Denote gcd(p : p ∈ S(a) \ {p}) = d and S (a) = {q : dq ∈ S(a) \ {p}} = {0, q1 , . . . , qk−1 }. S (a) has property P. Moreover, gcd(q1 , . . . , qk−1 ) = 1 and |S (a)| = k. Let us also remark that in general m and kη depend on S(a). Hence, for different S(a) the numbers m and kη are generally different. On the other hand, it is clear that if m and kη are numbers for S(a), they are also suitable for S(a) \ {p}. Furthermore, m and kη /d are numbers for S (a). With this in mind it is now clear that the hypothesis of the induction implies for some 0 ≤ x < qk−1 that {0, . . . , qk−1 − 1} = B (x ),
834
XINLONG ZHOU
where B (x ) is the corresponding set B(x) defined by S (a), m and kη /d. By Lemma 3.1 and the definition of B(dx ) we obtain (3.20)
{ld : 0 ≤ ld < pk } ⊆ B(dx ).
Indeed, if dqk−1 = pk , (3.20) follows directly from the definition of B(dx ). If however dqk−1 = pk , then p = pk and dqk−1 = pk−1 . So {ld : 0 ≤ ld < pk−1 } ⊆ B(dx ). But there is an odd number in S(a) \ {pk , pk−1 }. Using Lemma 3.1 with this number, we can still replace pk−1 by pk . Clearly, (3.20) is (3.1) if d = 1. So in the following discussion we should always assume d = 1. In what follows we should use (3.20) to prove (3.1) for S(a). The proof will be divided into four cases according to different p and pk . Case 1. pk is even and there are at least two odd numbers in S(a). Without loss of generality let two odd numbers be p1 and p2 . Let gcd(p1 , p3 , . . . , pk ) = d and S (a) = {qi : dqi ∈ S(a) \ {p2 }} = {0, q1 , . . . , qk−1 }. We obtain (3.20) with p = p2 . To verify (3.1) for S(a) from (3.20), let 2p2 ≤ pk . We choose l ≥ 0 such that pk − p2 + ld ≤ l d + ld < pk − p2 + (l + 1)d for l = 0, . . . , p1 /d − 1. Clearly, l d + ld ∈ B(dx ). Moreover, by Lemma 3.4 the set B(dx ) contains l d + ld − pk + p2 and ld ≤ l d + ld − pk + p2 < (l + 1)d ≤ p1 . Denote y = l d−pk +p2 . We have y −p2 ≡ 0 (mod d) and y = 0. Using Lemma 3.1, we conclude that for some αl,1 > 0 and l = 0, . . . , p1 /d−1 the numbers y+ld+αl,1 p1 are in B(dx ) as well as in [pk − p2 , pk − p2 + p1 ). Again by Lemma 3.4 we obtain for l = 0, 1, . . . , p1 /d − 1 that 0 < y + ld + αl,1 p1 − (pk − p2 ) < p1 and y + ld + αl,1 p1 − (pk − p2 ) ∈ B(dx ). We note that the choice of y ensures y − s(pk − p2 ) ≡ 0 (mod d) if and only if s + 1 ≡ 0 (mod d). Thus, write αl,0 = 0. We conclude repeatedly that B(dx ) contains y + ld + αl,s p1 − s(pk − p2 ) and 0 ≤ y + ld + αl,s p1 − s(pk − p2 ) < p1 , ∀ s = 0, . . . , d − 1, l = 0, 1, . . . , p1 /d − 1. It is easy to see that all these numbers are different. Consequently, {0, . . . , p1 −1} ⊆ B(dx ). Since d|p1 , we get by (3.20) that {0, . . . , p1 } ∪ {ld : 0 ≤ ld < pk } ⊆ B(dx ). The desired assertion follows from this relation and Lemma 3.1, since any p1 < z < pk can be written as z = z + νp1 with 0 ≤ z < p1 . ˜ Let 2p1 ≥ pk . Then 2(pk − pk−2 ) ≤ pk . Hence, the above proof is valid for S(a). We again get (3.1) by the symmetric relation (3.2). Finally, let 2p1 < pk < 2p2 . Equation (3.2) allows us to suppose p1 + p2 ≤ pk . The approach is the same as for the case 2p2 ≤ pk . First we choose l ≥ 0 such that pk − p2 − p1 + ld ≤ l d + ld < pk − p2 − p1 + (l + 1)d, ∀ l = 0, . . . , p1 /d − 1. Hence, by Lemma 3.1 and (3.20) the number l d + p2 is contained in B(dx ), while Lemma 3.4 implies that y = l d + p2 + p1 − pk is in B(dx ). Moreover, for the same reason y + ld ∈ B(dx ) and ld ≤ y + ld < (l + 1)d, ∀ l = 0, . . . , p1 /d − 1.
SUBDIVISION SCHEMES
835
Clearly, y + sp2 ≡ 0 (mod d) if and only if s + 1 ≡ 0 (mod d). Therefore, using the same approach as for 2p2 ≤ pk , we conclude that B(dx ) contains y + dl + αl,s p1 − s(pk − p2 − p1 ) for s = 0, 1, . . . , d − 1 and l = 0, . . . , p1 /d − 1. These numbers are different and are in [0, p1 ). Consequently, {0, . . . , p1 } ∪ {ld : 0 ≤ ld < pk } ⊆ B(dx ) and thus we get the desired assertion. Case 2. pk is even and there is only one odd number p in S(a). Let p = ˜ and pk − p. Write p1 , pk−1 . We may assume 2p ≤ pk . Otherwise, we consider S(a) gcd(p1 , p2 , . . . , pk−1 ) = d. So with S (a) = {0, p1 /d, . . . , pk−1 /d} we obtain for some 0 ≤ x < pk−1 /d the relation (3.20). Having (3.20), we choose l such that pk − p + ld ≤ dl + ld < pk − p + (l + 1)d, ∀ l = 0, . . . , p/d − 1. Next we apply Lemma 3.4 to get y + ld = dl + ld − pk + p ∈ B(dx ). Now it is clear that 0 ≤ y + dl < p, ∀ l = 0, . . . , p/d − 1. Moreover, y −spk ≡ 0 (mod d) if and only if s+1 = 0 (mod d). The same approach as in Case 1 (with p instead of p1 ) yields {0, . . . , p} ∪ {ld : 0 ≤ ld < pk } ⊆ B(dx ). We know now how to get (3.1) from this relation. To show (3.1) for p = p1 or pk−1 , we should modify the above proof. Without loss of generality we may assume p = p1 and 2p1 > pk . So the pj , j = 2, . . . , k, are even. Denote gcd(p1 , p2 , . . . , pk−2 , pk ) = d1 . With S (a) = {0, p1 /d1 , . . . , pk−2 /d1 , pk /d1 } we obtain (3.20). To verify (3.1), let q = pk − p1 , q = pk−1 /2. So gcd(d1 , q ) = 1. We already know that ld1 +d1 ∈ B(d1 x ) for all l = 0, 1, . . . , q/d1 −1. Now, since q < p1 , it follows from Lemma 3.1 that ld1 + d1 + p1 ∈ B(d1 x ) while Lemma 3.4 with pl1 = p1 and pl2 = pk implies ld1 + d1 + p1 − αq ∈ B(d1 x ) for α such that q ≤ ld1 + d1 + p1 − αq < q + q. Lemma 3.2 with pk−1 leads to ld1 + d1 + (p1 − q ) − αq ∈ B(d1 x ) and 0 ≤ ld1 + d1 + (p1 − q ) − αq < q. Now, let y = d1 +(p1 −q )−αq. Then y ∈ B(d1 x ) and 0 < y < q. Recursively, we obtain from the above procedure that 0 ≤ y+ld1 +ls d1 −sq < q and y+ld1 +ls d1 −sq ∈ B(d1 x ) for s = 0, 1, . . . , d1 − 1 and l = 0, . . . , q/d1 − 1. Moreover, y + ld1 + ls d1 − sq = 0 if and only if s = d1 − 1. Therefore, since gcd(d1 , q ) = 1, these numbers are all different and are contained in {0, . . . , q − 1}. We conclude that {v : 0 ≤ v < q} ⊆ B(d1 x ). Again applying Lemma 3.1 with p1 , we obtain from (3.20) and the last relation that (3.21)
{0, 1, . . . , q − 1} ∪ {p1 , . . . , pk − 1} ∪ {ld1 : 0 ≤ ld1 < pk } ⊆ B(d1 x ).
Let p1 ≤ v < pk . It follows from Lemma 3.4 with pl1 = p1 and pl2 = pk that v − αq ∈ B(d1 x ) whenever v − αq ∈ Zpk . Combining this with (3.21) leads to (3.1). Case 3. pk is odd. S(a) \ {pk } does not have property P. Thus, S(a) contains an odd p = pk and an even q = 0. It is easy to see that p1 , . . . , pk−2 are even and pk−1 is odd. Suppose pk ≥ 2pk−2 , so 2pj < pk for j = 1, 2, . . . , k − 2. Clearly S(a)\ {p2} = {0, p1 , p3 , . . . , pk } has property P. Denote gcd(p1 , p3 , . . . , pk ) = d1 and S (a) = {0, p1 /d1 , p3 /d1 , . . . , pk /d1 }. We obtain (3.20) with some 0 ≤ x < pk , i.e., {ld1 : 0 ≤ ld1 < pk } ⊆ B(d1 x ).
836
XINLONG ZHOU
Let l satisfy pk − p2 + ld1 ≤ l d1 + ld1 < pk − p2 + (l + 1)d1
∀ l = 0, . . . , p1 /d1 − 1.
Clearly, for each such l it holds that (2m − 1)(l d1 + ld1 − p2 ) + kη < (2m − 1)(pk − p2 ). Hence, by Lemma 3.4 we conclude that ld1 ≤ l d1 + ld1 − (pk − p2 ) < (l + 1)d1 and l d1 + ld1 − (pk − p2 ) ∈ B(d1 x ). Write y = l d1 − (pk − p2 ). We obtain y + sp2 ≡ 0 (mod d1 ) if and only if s + 1 ≡ 0 (mod d1 ). We know that p1 = 2µ g, where g is odd. Also, d1 = 1 implies g = 1. For each y + ld1 we choose i,j = 0 for j > 1 to obtain 2m (y + ld1 ) + kη −
m−1 k
i,j 2i pj
m−1
=
2m (y + ld1 ) + kη −
=
y + ld1 + (2 − 1)(y + ld1 ) + kη − γp1
=
y + ld1 + (2m − 1)(y + ld1 ) + kη − γ2µ g,
i=0 j=1
i,1 2i p1
i=0 m
where γ can be any integer between 0 and 2m − 1. By the definition of B(d1 x ) we conclude that y + ld1 + (2m − 1)(y + ld1 ) + kη − γ2µ g ∈ B(x) whenever y + ld1 + (2m − 1)(y + ld1 ) + kη − γ2µ g ∈ Zpk . Thus, we can choose γ to obtain y + ld1 + (2m − 1)(y + ld1 ) + kη − γ2µ g = y + ld1 + rp1 + βg with some r and 0 ≤ β ≤ 2µ − 1 such that pk − p2 ≤ y + ld1 + rp1 + βg < pk . It follows from Lemma 3.4 that 0 ≤ y + ld1 + rp1 + βg − (pk − p2 ) < p2 is contained in B(d1 x ). Furthermore, with pl1 = p1 and pl2 = pk we get by Lemma 3.2 for some α ≥ 0 that 0 ≤ y + ld1 + rp1 + βg − (pk − p2 ) − αg < p1 and y + ld1 + rp1 + βg − (pk − p2 ) − αg ∈ B(d1 x ). We may write with some αl,1 y + ld1 + rp1 + βg − (pk − p2 ) − αg = y + ld1 + αl,1 d1 − (pk − p2 ). Clearly, these new numbers are different from y + ld1 . Therefore with αl,0 = 0 we conclude recursively for s = 0, 1, . . . , d1 − 1, l = 0, . . . , p1 /d1 − 1 that 0 ≤ y + ld1 + αl,s d1 − s(pk − p2 ) < p1
and
y + ld1 + αl,s d1 − s(pk − p2 ) ∈ B(d1 x ).
The choice of y ensures that these numbers are all different. Hence, together with (3.20) we get (3.22)
{0, 1, . . . , p1 } ∪ {ld1 : 0 ≤ ld1 < pk } ⊆ B(d1 x ).
To finish our proof for the case pk ≥ 2pk−2 , we have to verify {p1 + 1, . . . , pk − 1} ⊆ B(d1 x ). For this goal we assume first kη = 2m − 1. Thus, kη is even. We apply (i) of Lemma 3.3 with pl1 = pk and pj = pk−1 for each odd number from {0, . . . , p1 } to obtain {x : 0 ≤ x ≤ p1 + pk − pk−1 , x is odd} ∩ Zpk ⊆ B(d1 x ). Recursively, we conclude (3.23)
{x : 0 ≤ x < pk , x is odd} ⊆ B(d1 x ).
SUBDIVISION SCHEMES
837
Next, we use Lemma 3.1 with pk−1 for each number from {1, . . . , p1 } to get {x : pk−1 < x ≤ min{pk−1 + p1 , pk − 1}} ⊆ B(d1 x ). Now for each even number x between pk−1 and min{pk−1 + p1 , pk − 1} we apply (ii) of Lemma 3.3 with pl1 = pk and pj = p1 . So, recursively, we obtain {x : p1 ≤ x ≤ pk−1 , x is even} ⊆ B(d1 x ). Combining this with (3.22) and (3.23), we get {0, .., pk−1 } ⊆ B(d1 x ). The desired assertion follows from Lemma 3.1 with pk−1 and the last relation. Similarly, we have (3.1) for kη = 2m − 1. It remains to show (3.1) for 2pk−2 ≥ pk . Thus 2(pk −pk−2 ) ≤ pk . The symmetric relation (3.2) tells us that in this case we need only to verify (3.1) for the set S(a) with even p1 and odd p2 , . . . , pk , which satisfies 2p1 ≤ pk . Clearly {0, p1 , . . . , pk−1 } has property P. Denote gcd(p1 , . . . , pk−1 ) = d2 . We have as before {ld2 : 0 ≤ ld2 < pk } ⊆ B(d2 x ). Next, for l = 0, . . . , p1 /d2 − 1 let l be such that pk − p1 + ld2 ≤ l d2 + ld2 < pk − p1 + (l + 1)d2 . Write y = l d2 − (pk − p1 ). Lemma 3.4 tells us that y + ld2 ∈ B(d2 x ) for l = 0, . . . , p1 /d2 − 1. We now know how to use Lemmas 3.1, 3.2 and 3.4 from these numbers to obtain, for s = 0, . . . , d2 − 1 and l = 0, . . . , p1 /d2 − 1, y + ld2 + αs,l d2 − s(pk − p1 ) < p1 and y + ld2 + αs,l d2 − s(pk − p1 ) ∈ B(d2 x ), which in turn implies {0, . . . , p1 − 1} ⊆ B(d2 x ). Hence, {0, . . . , p1 } ∪ {ld2 : 0 ≤ ld2 < pk } ⊆ B(d2 x ). The assertion follows from Lemmas 3.1, 3.3 and this relation. Case 4. pk is odd. S(a) \ {pk } has property P and S(a) \ {pk } contains an odd p and an even q = 0. Denote gcd(p1 , . . . , pk−1 ) = d. Thus, we obtain (3.20), i.e., {ld : 0 ≤ ld < pk } ⊆ B(dx ). For q ≤ pk − q the proof is the same as in the last situation of Case 3. We omit the details here. Let q ≥ pk − q. We may assume that q is minimal among all nonzero even ˜ numbers in S(a) and that p satisfies 2p < pk . Otherwise we consider S(a), for which we have already the desired assertion. For the same reason we can also assume p + q ≤ pk . To this end, first let 2p ≥ q. Denote q = q/2. Furthermore, let q ≤ l d + ld < pk for some l and l = 0, . . . , q /d − 1. By Lemma 3.4 we get 2q − pk ≤ l d + ld − (pk − q) < q. We may use Lemma 3.2 to subtract q from l d + ld − (pk − q) such that 0 ≤ l d + ld − (pk − q) − αl q < q . Now write yl = l d + ld − (pk − q) − αl q . Thus, yl ∈ B(dx ) and yl = 0, ∀ l = 0, . . . , q /d − 1. Furthermore, they are all different and yl − spk ≡ 0 (mod d) if and only if s + 1 ≡ 0 (mod d). Now we obtain recursively by applying Lemma 3.1 with p, Lemma 3.4 with q and pk and finally Lemma 3.2 with q that yl + αl,s d + s(pk − q) ∈ B(dx ) and 0 ≤ yl + αl,s d + s(pk − q) < q
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for l = 0, . . . , q /d − 1 and s = 0, . . . , d − 1. Hence, together with (3.20) we conclude {0, . . . , q } ∪ {ld : 0 ≤ ld < pk } ⊆ B(dx ). We verify that q in the above relation can be replaced by p. If q ≥ p, we have nothing more to do. If q < p, we use Lemma 3.1 with p to obtain {0, . . . , q } ∪ {p, . . . , p + q } ⊆ B(dx ). For each number of {p, . . . , p + q − 1} we apply Lemma 3.2 with q to obtain {p − q , . . . , p − 1} ⊆ B(dx ). Thus, {p − q , .., p + q } ⊆ B(dx ). Again using Lemma 3.2 with q for each element from {p − q , . . . , p + q − 1}, we obtain {p − 2q , . . . , p + q } ⊆ B(dx ). Recursively, {0, . . . , p + q } ⊆ B(dx ). Thus, (3.24)
{0, . . . , p} ∪ {ld : 0 ≤ ld < pk } ⊆ B(dx ).
Since p is odd, (3.1) follows from Lemma 3.1 with p and the last relation. It remains to verify (3.1) for 2p ≤ q. We should modify the procedure for 2p ≥ q as follows: let pk − p ≤ l d + ld < pk for some l and l = 0, . . . , p/d − 1. By Lemma 3.4 with pk and q we get q − p ≤ l d + ld − (pk − q) < q. Again using Lemma 3.4 with p and q, we conclude that 0 ≤ l d + ld − (pk − p) < p. Now write yl = l d + ld − (pk − p). Thus, yl ∈ B(dx ) and yl = 0, ∀ l = 0, . . . , p/d − 1. We obtain in the same way that yl + αl,s p + s(pk − p) ∈ B(dx ) and 0 ≤ yl + αl,s p + s(pk − p) < p for l = 0. . . . , p/d − 1 and s = 0, . . . , d − 1. Hence, together with (3.20) we conclude (3.24). From (3.24) we know how to get (3.1). Acknowledgment The author is indebted to the referee for various helpful comments on his paper. References [1] M. Br¨ oker and X. Zhou, Characterization of continous, four-coefficient scaling functions via matrix spectral radius, SIAM J. Matrix Anal. Appl., 22 (2000), 242-257. MR2001h:65169 [2] A. S. Cavaretta, W. Dahmen, and C. A. Micchelli, Stationary Subdivision, Mem. Amer. Math. Soc., 453 (1991). MR92h:65017 [3] G. M. Chaikin, An algorithm for high speed curve generation, Comp. Graphics and Image. Proc., 3 (1974), 346-349. [4] D. Colella and C. Heil, Characterizations of scaling functions: continuous solutions, SIAM J. Matrix Anal. Appl., 15 (1994), 496-518. MR95f:26004 [5] I. Daubechies and J. C. Lagarias, Two-scale difference equations I. Existence and global regularity of solutions, SIAM J. Math. Anal., 22 (1991), 1388-1410. MR92d:39001 [6] D. E. Gonsor, Subdivision algorithms with nonnegative masks generally converge, Adv. Comp. Math., 1 (1993), 215-221. MR94e:65023 [7] R. -Q. Jia, Subdivision schemes in Lp spaces, Advances in Comp. Math, 3 (1995), 309-341. MR96d:65028 [8] R. -Q. Jia and D. -X. Zhou, Convergence of subdivision schemes associated with nonnegative masks, SIAM J. Matrix Anal. Appl., 21 (1999), 418-430. MR2001a:42041 [9] A. A. Melkman, Subdivision schemes with non-negative masks always converge—unless they obviously cannot?, Ann. Numer. Math., 4 (1997), 451-460. MR97i:41014 [10] C. A. Micchelli and H. Prautzsch, Uniform refinement of curves, Linear Algebra Appl., 114/115 (1989), 841-870. MR90k:65088
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[11] G. C. Rota and G. Strang, A note on the joint spectral radius, Indag. Math., 22 (1960), 379-381. MR26:5434 [12] G. de Rham, Sur une courbe plane, J. Mathem. pures et appl., 39 (1956), 25-42. MR19:842e [13] J. N. Tsitsiklis and V. D. Blondel, The Lyapunov exponent and joint spectral radius of pairs of matrices are hard, when not impossible, to compute and to approximate, Mathematics of Control, Signals and Systems, 10 (1997), 31-41. MR99h:65238a [14] Y. Wang, Two-scale dilation equations and the cascade algorithm, Random and Computational Dynamics, 3 (1995), 289-309. MR96m:42060 [15] Y. Wang, Subdivision schemes and refinement equations with nonnegative masks, J. Approx. Th., 113 (2001), 207-220. MR2002i:42054 [16] D.-Z. Zhou, The p-norm joint spectral radius for even integers, Methods and Applications of Analysis, 5 (1) (1998), 39-54. MR99e:42054 Department of Mathematics, China Jiliang University, Hangzhou, China; Institute of Mathematics, University of Duisburg-Essen, D-47057 Duisburg, Germany E-mail address: [email protected]
