Introduction · SUBDIVISION SCHEMES WITH NONNEGATIVE MASKS XINLONG ZHOU Abstract. The conjecture concerning the characterization of a convergent univariate subdivision algorithm with

MATHEMATICS OF COMPUTATIONVolume 74, Number 250, Pages 819–839S 0025-5718(04)01712-0Article electronically published on October 27, 2004

SUBDIVISION SCHEMES WITH NONNEGATIVE MASKS

XINLONG ZHOU

Abstract. The conjecture concerning the characterization of a convergentunivariate subdivision algorithm with nonnegative finite mask is confirmed.

1. Introduction

Let Z denote the integer lattice. A univariate subdivision algorithm with afinitely supported mask a = {aj}j∈Z is given as follows: beginning with an initialsequence of data v0 = {v0

i }, we set recursively new sequences of values vk, byapplying the rule

vki =

∑j∈Z

ai−2jvk−1j .

This algorithm is said to converge if for each finite v0 = {v0i } there exists a contin-

uous function fv such that fv �≡ 0 for at least one v0 and

(1.1) limk→∞

supi∈Z

|fv(i

2k) − vk

i | = 0.

Obviously, the subdivision algorithm with mask a = {aj}j∈Z converges if andonly if the polygon fk defined by the control points (i/2k, vk

i )T converges uniformlyto fv. One can easily check

fk(x) =∑

i

v0i

∑j

akj h(2k(x − i) − j),

where h is the hat function and aki =

∑j ak−1

j ai−2j with a1j = aj . Thus (1.1) for

all such v0 is equivalent to the uniform convergence of∑j

akj h(2kx − j).

On the other hand, define S to be the operator given by

Sf(x) =∑

j

ajf(2x − j).

One getsSkf(x) =

∑j

akj f(2kx − j), k = 1, 2, . . . .

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, subdivi-

sion scheme.

c©2004 American Mathematical Society

819

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use

820 XINLONG ZHOU

The convergence of (1.1) is therefore equivalent to the uniform convergence of Skh,which leads to the following cascade algorithm: let g0(x) = h(x); one defines recur-sively gk(x) by

gk(x) =∑

j

ajgk−1(2x − j).

The limit g = limk→∞ gk = limk→∞ Skh satisfies Sg = g, i.e.,

(1.2) g(x) =∑j∈Z

ajg(2x − j).

This so-called two-scale delation equation and the above presented algorithms playan 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 theconcept of joint spectral radius (see [11]). The joint spectral radius for two squarematrices A0 and A1 is defined by

ρ(A0, A1) = limn→∞ sup

k≤n||Aε1 · · ·Aεk

|| 1k ,

where || · || is a given matrix-norm and εl ∈ {0, 1}. Thus, write a(z) =∑

j ajzj

and a(z) = (1 + z)b(z) if a(−1) = 0. The subdivision algorithm with finite maska = {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 coeffi-

cients of b.Unfortunately, the determination of the joint spectral radius is generally NP-

hard by a result of Tsitsiklis and Blondel (see [13]). Thus, it seems difficult todeterminate whether the considered spectral radius is less than one. Some partialresults 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, aproperty 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 actualvalues 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 identifyingthose I such that given any nonnegative mask supported on I, the correspondingsubdivision algorithm converges. By applying a suitable translation, we may alwaysassume in the following discussion that a nonnegative mask a = {aj} has the forma = {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 thatthe conditions are necessary (see [2, 9, 14]). There are various partial results whichsupport 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 twosuccessive integers, then convergence follows. Recently, Wang proves this conjecturefor 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) = 1and 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 byusing a lemma (see Lemma 2.4), which will be verified in Section 3. The proof ofLemma 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 vectorx in R

N 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 = 0otherwise. For any nonnegative N × N row stochastic matrix B, i.e., the sum foreach 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. Weshould write simply F0 = FA0 and F1 = FA1 . For convenience we should also usethe 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 rowstochastic matrix. Then FBC = FB ◦ FC . Furthermore, the subdivision algorithmwith nonnegative mask a, which satisfies a(1) = 2 and a(−1) = 0, diverges ifand 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 mapΨ for T ⊂ Z is defined by

Ψ(T ) =

⋂q∈S0

(2T − q)

∪

⋂q∈S1

(2T − q)

.

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

(2.2) F0(T ) = Ψ(T ) ∩ ZN , F1(T ) = (Ψ(T ) + 1) ∩ ZN .

Furthermore, for any (d1, . . . , dm) ∈ {0, 1}m we have

(2.3) Fd1 ◦ · · · ◦ Fdm(T ) = (Ψm(T ) + k) ∩ ZN ,

where k =∑m

j=1 dj2j−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 εi,j ∈ {0, 1} such that

k∑j=1

εi,j ≤ 1, ∀ i = 1, 2, . . . , m.

Then it holds that

2my −k∑

j=1

m∑i=1

2m−iεi,jpj �∈ Ψm(T ).

Proof. For m = 1 the assertion is clear since 2y − xpl �∈ Ψ(T ) for all x = 0, 1 andl = 1, 2, . . . , k. On the other hand, ε1,1 + · · · + ε1,k ≤ 1 implies either ε1,j = 0 forj = 1, . . . , k or else ε1,j = 0 for j �= j′ and ε1,j′ = 1. Thus

2y − ε1,1p1 − ε1,2p2 − · · · − ε1,kpk �∈ 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-primewith n. It is known that if gcd(k, n) = 1, then

kϕ(n) ≡ 1 (mod n).

In particular, for any odd integer n one has

(2.4) 2ϕ(n) ≡ 1 (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 showthe sufficiency, we denote S(a) = {0, p1, . . . , pk} with pk = N and |S(a)| for thenumber of S(a). The first assumption of the conjecture implies that |S(a)| ≥ 3 andthere exists an odd p in S(a) such that p = pj0 < pk no matter if pk is odd oreven. Hence, for |S(a)| = 3 one has S(a) = {0, p, N} with even N . Wang’s resultfor |S(a)| = 3 (see [15]) may be read as: the subdivision algorithm converges ifgcd(p, N) = 1. We assume in the following that |S(a)| ≥ 3. Let us observe that ifthe subdivision algorithms with the mask a = {aj} diverges, then by Lemma 2.1there 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 ,



where k′ =∑m′

l=1 dl2l−1. Write F = Fd1 ◦· · ·◦Fdm′ . Then F η(T ) = T and F η(T ′) =T ′ for any η ≥ 1. Therefore, with

kη = (m′∑l=1

dl2l−1) + (m′∑l=1

dl2l−1)2m′+ · · · + (

m′∑l=1

dl2l−1)2(η−1)m′

= k′η−1∑i=0

2im′

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∏j=1

{jϕ(j)}

with some positive integer t, which will be given later. Denote m = m′η. We have2m ≡ 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’sformula (2.4)

(2.8) 2m = 2m′n′xϕ(x) ≡ 1 (mod x).

Similarly, we have

(2.9) kη ≡ 0 (mod x).

To see this, one gets by the expression of kη

kη = k′n′x−1∑

l=0

ϕ(x)(l+1)−1∑j=ϕ(x)l

2jm′

= k′n′x−1∑

l=0

ϕ(x)−1∑j=0

2jm′2ϕ(x)lm′

≡ k′n′x−1∑

l=0

ϕ(x)−1∑j=0

2jm′(mod x)

≡ k′n′xϕ(x)−1∑

j=0

2jm′(mod x)

≡ 0 (mod x),

which yields (2.9). Moreover,

kη = 0 ⇐⇒ d1 = · · · = dm′ = 0

andkη = 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 − 12m′ − 1

≤ (1 − δ)2m


824 XINLONG ZHOU

and

(2.11) 2m − kη =2m′η(2m′ − 1 − k′) + k′

2m′ − 1> 2pk.

Additionally, in view of (2.5) we may without loss of the generality assume thatkη is even if kη �= 2m − 1. In the following discussion we should always denotem 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). Weknow that the divergence of the subdivision algorithm implies that there are twodisjoint nonempty sets T, T ′ ⊂ ZN for (2.5). We may therefore assume x �∈ T andx ∈ ZN . So from Lemma 2.3 it holds that

2mx −k∑

j=1

m−1∑i=0

2iεi,jpj �∈ Ψm(T )

for all εi,j ∈ {0, 1} such that∑k

j=1 εi,j ≤ 1. By (2.6) we obtain

(2.12) 2mx + kη −k∑

j=1

m−1∑i=0

2iεi,jpj �∈ T

for all those εi,j . Denote B0 = {x} with x ∈ ZN and for l = 0, 1, . . . ,

Bl+1 = {2my + kη −k∑

j=1

m−1∑i=0

2iεi,jpj ∈ ZN | y ∈ Bl, εi,j ∈ {0, 1},k∑

j=1

εi,j ≤ 1}.

Thus, (2.12) tells us that x �∈ T implies Bl ∩T = ∅ for all l ≥ 0. In the next sectionwe will prove the following

Lemma 2.4. Let the mask a = {aj} satisfy the conditions of the conjecture andS(a) = {0, p1, . . . , pk}. Then there is an 0 ≤ x < pk such that with B0 = {x} onehas ⋃

l≥0

Bl = Zpk.

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 thispaper. 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 inSections 1 and 2. Let us begin with the observation of the first condition of theconjecture. We need the following

Definition. A set of integers I = {0, q1, . . . , qk} with 0 < qj < qj+1 has propertyP if either qk is even and I contains at least one odd integer or else I contains atleast two odd and two even integers (including 0).

Obviously, if a = {aj} satisfies the first condition of the conjecture, the setS(a) has property P. Conversely, if a set {0, q1, . . . , qk} has property P, then onecan construct a mask a = {aj} satisfying the first condition of the conjecture and



S(a) = {0, q1, . . . , qk}. In fact, in [9] this conjecture is formulated by using propertyP. 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)

⋃l≥0

Bl = Zpk.

To prove Lemma 2.4 or Lemma 2.4′, we remember that {εi,j} ∈ {0, 1} for i =0, 1, . . . , m − 1, j = 1, . . . , k and

k∑j=1

εi,j ≤ 1, i = 0, 1, . . . , m − 1.

Let δi,j = εi,j for i = 0, 1, . . . , m − 1, j = 1, . . . , k − 1 and

δi,k = 1 −k∑

j=1

εi,j .

Hence, {δi,j} satisfies our conditions. We obtain

2mx + kη −k∑

j=1

m−1∑i=0

δi,j2ipj = (pk − 1)(3.2)

−2m(pk − 1 − x) + (2m − 1 − kη) −

k−1∑j=1

m−1∑i=0

εi,j2i(pk − pj) −m−1∑i=0

εi,k2ipk

.

We notice that 2m−1−kη is between 0 and 2m−1. Formulas (2.8) and (2.9) implythat (2.9) also holds for k′

η = 2m − 1 − kη instead of kη. Let Bj = pk − 1 − Bj .Then Bj is defined by {0, pk−pk−1, . . . , pk−p1, pk} and k′

η. Since gcd(p1, . . . , pk) =gcd(pk − pk−1, . . . , pk − p1, pk), the symmetric relation (3.2) tells us that

v ∈⋃l≥0

Bl ⇐⇒ pk − 1 − v ∈⋃l≥0

Bl.

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) =⋃l≥0

Bl

with B0 = {x} ⊂ Zpk.

The following assertions will be used frequently and are based on some specialchoices of {εi,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 εi,j = 0 for j �= l1. By the definition of εi,j the number l =∑m−1i=0 εi,l12

i can be any number between 0 and 2m−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 + (2m − 1)y + kη −m−1∑i=0

εi,l1pl12i = y + (α − l)pl1 .

The choice of η (see (2.7)) implies kη > pk and 2m − kη > 2pk if kη �= 0 and2m − 1, while for kη = 0, the inequality (2m − 1)y > 0 yields y ≥ 1. Moreover, ifkη = 2m − 1, then the condition implies y < pl1 . Hence α − l can be any numberbetween −pk and pk. In particular, for l such that y + (α − l)pl1 ∈ Zpk

we havey + (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 εm−1,l1 = 1 and εi,l1 = 0 if i �= m−1. For other j �= l2 let εi,j = 0.To meet the condition on εi,j , we have to choose εm−1,l2 = 0. Now

∑m−2i=0 εi,l22

i

can be any number between 0 and 2m−1 − 1. From (2.8), (2.9) and our restrictionon y we conclude (2m − 1)(y − pl1/2) + kη = αpl2 for some 0 ≤ α ≤ 2m−1 − 1.Hence, we can choose εi,l2 for i = 0, 1, . . . , m− 2 to obtain

∑m−2i=0 εi,l22i = α. With

this choice of εi,j we get

2my + kη −k∑

j=1

m−1∑i=0

εi,jpj2i = y − pl1/2.

Thus, y − pl1/2 ∈ B(x). �Lemma 3.3. Let y ∈ B(x). If there is some odd pl1 ∈ S(a) such that (2m − 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′ ∈ Zpkand (2m−1)y+kη

is odd;(ii) y + 2lpl1 − pj′ ∈ B(x) whenever y + 2lpl1 − pj′ ∈ Zpk

and (2m − 1)y + kη iseven.

Proof. Clearly (2m − 1)y + kη = αpl1 for some 0 ≤ α < 2m − 1. If α is odd, wechoose εi,l1 such that α − ∑m−1

i=0 εi,l12i = 2l + 1. Obviously, ε0,l1 must be zero.Hence, we can choose ε0,j′ = 1 or 0 to obtain

2my + kη −k∑

j=1

m−1∑i=0

εi,jpj2i = y + (2l + 1)pl1 − ε0,j′pj′ .

Since y ∈ B(x), the set B(x) contains y + (2l + 1)pl1 − ε0,j′pj whenever y + (2l +1)pl1 − ε0,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). Assume0 < (2m − 1)(y − pl1) + kη < (2m − 1)(pl2 − pl1). Then, y + j(pl2 − pl1) ∈ B(x) forj ∈ Z such that y + j(pl2 − pl1) ∈ Zpk

.

Proof. Let εi,l1 = 1 − εi,l2 and εi,j = 0 for j �= l1, l2. We obtain

2my + kη −k∑

j=1

m−1∑i=0

εi,jpj2i = y + (2m − 1)(y − pl1) + kη −m−1∑i=0

εi,l2(pl2 − pl1)2i

= y + (2m − 1)(y − pl1) + kη − l(pl2 − pl1),



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) ∈ Zpkwe 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 choosel ∈ [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 inductionon |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 assertionfor 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} byusing Lemma 3.1. In order to obtain the result for all of {0, 1, . . . , p−1}, we choosecarefully an x in this set and apply Lemmas 3.1–3.4 to get another number, whichis greater than p. Using Lemma 3.2 or Lemma 3.4, we may reduce the numberso that the new number is again contained in {0, . . . , p − 1} but is different fromx. Recursively, we obtain several numbers. One of our tasks is to show that thosenumbers are all different. A careful choice of x is necessary. Otherwise the followingcalculation leads to B(x) = {0}. In fact, if kη = 0, then for x = 0 and B0 = {0} onegets Bl = {0} for all l = 1, 2, . . . . Thus, in our choice of x it may be understoodthat we avoid zero if kη = 0 or pk − 1 if kη = 2m − 1 during the calculation of newnumbers. In other words, we should choose such an x that zero or pk − 1 occurs atthe last step of our calculation.

Next let us observe the set S(a) with |S(a)| ≥ 5. It is easy to check that forsome nonzero p ∈ S(a) the set S(a) \ {p} also has property P. However, for setswith |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. Sincethe assertion for |S(a)| = 3 is essentially given in [15], we should omit the proof forthis case. Our approach, therefore, has two steps; namely, the proof for the specialcase 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 proofinto 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. Denotegcd(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 − α1p ∈B(x) whenever 0 ≤ x + p2 − α1p < p. We notice that x + p2 − α1p satisfies theabove restriction. Thus, recursively, i.e., replacing x by x + p2 − α1p, we conclude


828 XINLONG ZHOU

x + sp2 − αsp ∈ B(x) and 0 ≤ x + sp2 − αsp < p for s = 0, 1, . . . , p/d − 1, whereα0 = 0. We note also that the choice of x ensures x + sp2 − αsp �= 0 at least fors = 0, 1, . . . , p/d − 2, while for s = p/d − 1 one has{

x + sp2 − αsp = 0, if d = 1,x + sp2 − αsp �= 0, if d �= 1.

Moreover, it follows from gcd(p2, p) = d that x + sp2 − αsp �= x + s′p2 − αs′p if0 ≤ s < s′ ≤ p/d − 1. Therefore,

{x + sp2 − αsp : s = 0, 1, . . . , p/d − 1} = {x + dv : 0 ≤ x + dv < p}⊆ B(x).

Examining the way we calculate the number x+ sp2 −αsp 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 asabove. Let x+ dv be any nonzero element from {x+ dv : 0 ≤ x+ dv ≤ p}. Lemma3.1 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 suchthat p2/d = l′p/d + ν or p2 = l′p + νd. Hence, for p < x + dv′ ≤ p2 we getx+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 particularthat

(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 theconditions of Lemma 3.1, we know that if d �= 1, then for each 0 ≤ x + dv ≤ p2 onegets 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 thisas follows: for kη �= 0 let 0 ≤ x + v < p2. Again we have (3.5). If kη = 0, let0 < 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 ifkη �= 0 or x − p3 �≡ 0 (mod d), then x + dv − (p3 − p1) + jp2 − lp ∈ B(x) wheneverx + dv − (p3 − p1) + jp2 − lp ∈ Zp3 . Let 0 ≤ x + dν − (p3 − p1) < p3. Then due togcd(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).



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−pinstead of p3 − p1 ≤ x + dv − (i − 1)(p3 − p1) < p3 − p to obtain Ai+1 ⊂ B(x). Weremember x + p3 ≡ 0 (mod d). Hence, x − jp3 �≡ 0 (mod d) for j = 0, 1, . . . , d− 2.We conclude

(3.6)d−2⋃j=0

Aj ⊆ B(x).

Moreover, in the case of kη �= 0

d−1⋃j=0

Aj ⊆ B(x).

To have (3.1) for kη �= 0, we notice gcd(d, p3) = 1. Hence, for any v′ such that0 ≤ x+ v′ < p3 there is 0 ≤ j ≤ d−1 with v′ + j(p3−p1) ≡ 0 (mod d), which givesv′ = dv− j(p3 −p1) for some v. We conclude that x+ v′ = x+dv− j(p3 −p1) ∈ Aj

orZp3 = {v : 0 ≤ v < p3} ⊆ B(x).

It remains to show (3.1) for kη = 0. We already have (3.6). But what is thecorresponding Ad−1 for this case? We know that the first step in our procedure isto find an element in Ad−2 which satisfies

(3.7) p3 − p1 ≤ x + dv − (d − 2)(p3 − p1) < p3 − p.

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−1i=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 andx − (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 toLemma 3.4. Now we can apply our procedure for p to obtain jp2 − lp ∈ B(x) orA′ ⊂ B(x). Consequently,

Zp3 = {v : 0 ≤ v < p3} ⊆ B(x).

For kη = 2m − 1 we replace x + p3 ≡ 0 (mod d) by x + 1 + p3 ≡ 0 (mod d) andp|(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.


830 XINLONG ZHOU

Case 3. Since p3 is odd, it remains to show (3.1) for 2p1 < p3 < 2p2. Withoutloss of generality we may suppose p3 ≥ p2 + p1. Now, p1 is even, so p1 = 2p forsome p.

Let p be odd. Denote gcd(p1, p3) = d. We have gcd(p, p3) = d. We verify forsome 0 ≤ x ≤ p that

(3.8) {v : 0 ≤ v < p1} \ {p} ⊆ B(x).

To this end we choose 0 ≤ x ≤ p in the following way: if 0 ≤ kη < 2m − 1, thenx �= 0 and

(3.9){

x − p2 ≡ 0 (mod d), if d �= 1,x + p3 ≡ 0 (mod p), if d = 1;

if kη = 2m − 1, then x �= p and

(3.10){

x + 1 − p2 ≡ 0 (mod d), if d �= 1,x + 1 ≡ 0 (mod p), if d = 1.

In the following we deal only with the case 0 ≤ kη < 2m − 1. The correspondingassertion 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 somek we get 2mx + 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}, sop3 < µ < (1 − δ/2)2m. Let εi,j = 0 for j = 2, 3. We get

2mx + kη −m−1∑i=0

3∑j=1

εi,jpj2i = 2mx + kη − lp1

= x + kp − lp1

= x + µp1 − lp1 + βp

= x + αp1 + βp.

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 thanp. We obtain in this way x − p3 + l1p = x + αp1 + βp − p3 + p1 − p ∈ B(x) and0 ≤ x−p3+ l1p < p. As in Case 1 we conclude recursively that 0 ≤ x−sp3+ lsp < pfor s = 0, . . . , p/d−1. Clearly, by this construction x−sp3 + lsp ∈ B(x). Moreover,since gcd(p, p3) = d, all these numbers are different for 0 ≤ s ≤ p/d − 1. Hence,

{x − sp3 + lsp : x − sp3 + lsp ∈ [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 ifd = 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} dependson 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



x + dv + αp1 + βp to obtain x + dv + αp1 + βp + p2 ∈ B(x) and by Lemma 3.4 wehave x + dv + αp1 + βp + p2 − p3 + p1 ∈ B(x). This number is nonnegative andless than p1. So by Lemma 3.2 we can again substitute p, if necessary, to obtainx + dv + (p2 − p3 + p1) + l1p ∈ B(x) and 0 ≤ x + dv + (p2 − p3 + p1) + l1p < p.Recursively, B(x) contains

x + dv + s(p2 − p3 + p1) + ls,vp = x + dv + sp2 + l′s,vd, ∀ s = 0, 1, . . . , d − 1.

Since gcd(p2, d) = 1, we conclude by the second expression that they numbersare different for different s, while the first one shows that they are different if v isdifferent. Thus, there are d×p/d = p numbers in {0, . . . , p−1}, which are containedin 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, whereβ ∈ {0, 1} relies on µ. For the µ with β = 0 let α = 1. So µ + p1 ∈ B(x) and byLemma 3.2, µ + p1 − p ∈ B(x). For the µ with β = 1 we choose α = 0. Hence, wealways 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 iseven. If 0 < kη < 2m − 1, then since 0 ∈ B(x) and 2m0 + kη − lp1 = jp1 for somel, we conclude jp1 ∈ B(x) if 0 ≤ jp1 < p3. Lemma 3.2 yields p ∈ B(x). Combiningthis with (3.8), we obtain

{0, 1, . . . , p1 − 1, p1, 2p1, . . . , j0p1} ⊆ B(x),

where j0 satisfies j0p1 < p3 ≤ (j0 + 1)p1. Moreover, for any 0 < 2l < p1 we have2m2l + kη − ip1 = 2l + αp1 with some i. Hence, B(x) contains 2l + αp1 whenever0 < 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 severaltimes, we obtain l + j(p3 − p2) ∈ B(x) whenever 0 ≤ l + j(p3 − p2) < p3. Weconclude from the last relation that

(3.13) {l : 0 < l < p3, l is odd} ⊆ B(x).

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 εi,j such that

2mp1 −m−1∑i=0

3∑j=1

εi,j2ipj = (2m − 1)p1 −m−1∑i=1

εi,32ip3 = 0.

Thus,{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 thatjp − 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 andB(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 verifyfor 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:

(3.15){

x − p2 ≡ 0 (mod d), if 0 ≤ kη < 2m − 1,x + 1 − p2 ≡ 0 (mod d), 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 thatkη is even when 0 ≤ kη < 2m − 1. So in any case (2m − 1)x + kη is odd. We alsoremember p3|(2m − 1) and p3|kη since p3 is odd. Now by Lemma 3.3 we obtainx + 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 withy = x + p3 − p2, pl1 = p1 and pl2 = p3 that x + p3 − p2 − p ∈ B(x). Recursively,x + p3 − p2 −α1p ∈ B(x) for some α1 such that 0 ≤ x + p3 − p2 −α1p < p. Clearly,this new number also satisfies (3.15). We can therefore apply this procedure forthis new number to obtain another number that satisfies (3.15) and is contained inB(x). Repeatedly, we obtain for s = 0, 1, . . . , p/d − 1 that

x + s(p3 − p2) − αsp ∈ B(x) and 0 ≤ x + s(p3 − p2) − αsp < p,

where α0 = 0. Since gcd(p3 − p2, p) = d, they are all different and have the formx + 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 followsfrom (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 − l1p < p and

x + dv + p2 − p1 − lp = x + dv + p2 − l1p ∈ B(x).

For this new number (2m − 1)(x+ dv + p2− l1p)+ kη is even. Clearly, we can againuse (ii) of Lemma 3.3 to obtain x + dv + p2 − l1p + 2p2 − p3 ∈ B(x). Write

x + dv + p2 − l1p + 2p2 − p3 = x + dv + 2p2 − l1p − (p3 − p2).



For this number (2m−1)(x+dv+2p2− l1p− (p3−p2))+kη is odd. We get by (i) ofLemma 3.3 and, if necessary, Lemma 3.2 that 0 ≤ x+dv+2p2− l2p− l′(p3−p2) < pand

x + dv + 2p2 − l1p − l′(p3 − p2) − lp = x + dv + 2p2 − l2p − l′(p3 − p2) ∈ B(x).

Repeatedly, we conclude for s = 0, 1, . . . , d − 1 that

0 ≤ x+dv +sp2− lsp− l′s(p3−p2) < p and x+dv +sp2− lsp− l′s(p3−p2) ∈ B(x).

By our construction it is easy to see that if (v, s) �= (v′, s′), then the correspondingnumbers 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 thisfor kη �= 2m − 1. Let 0 < l < p be odd. So (2m − 1)l + kη is odd. As before weuse Lemmas 3.3 and 3.2 to obtain l + p3 − p2 − αp ∈ B(x) for α ≥ 0 satisfying0 ≤ l + p3 − p2 − αp < p3. Since l + p3 − p2 − αp = l′ is again odd, we obtain inparticular that all odd numbers in {0, . . . , p3 − p2 − 1} belong to B(x). ApplyingLemma 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 odd0 < 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 withp2 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 concludefrom 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}) = dand

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 alsoremark that in general m and kη depend on S(a). Hence, for different S(a) thenumbers m and kη are generally different. On the other hand, it is clear that if mand 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 thehypothesis 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 Lemma3.1 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′). Ifhowever 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 thisnumber, we can still replace pk−1 by pk. Clearly, (3.20) is (3.1) if d = 1. So in thefollowing 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 accordingto different p and pk.

Case 1. pk is even and there are at least two odd numbers in S(a). Without lossof 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 forl = 0, . . . , p1/d − 1. Clearly, l′d + ld ∈ B(dx′). Moreover, by Lemma 3.4 the setB(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,1p1

are in B(dx′) as well as in [pk − p2, pk − p2 + p1). Again by Lemma 3.4 we obtainfor l = 0, 1, . . . , p1/d − 1 that

0 < y + ld + αl,1p1 − (pk − p2) < p1 and y + ld + αl,1p1 − (pk − p2) ∈ B(dx′).

We note that the choice of y ensures y − s(pk − p2) ≡ 0 (mod d) if and only ifs + 1 ≡ 0 (mod d). Thus, write αl,0 = 0. We conclude repeatedly that B(dx′)contains y + ld + αl,sp1 − s(pk − p2) and

0 ≤ y + ld + αl,sp1 − 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′), whileLemma 3.4 implies that y = l′d + p2 + p1 − pk is in B(dx′). Moreover, for the samereason

y + ld ∈ B(dx′) and ld ≤ y + ld < (l + 1)d, ∀ l = 0, . . . , p1/d − 1.



Clearly, y + sp2 ≡ 0 (mod d) if and only if s+1 ≡ 0 (mod d). Therefore, using thesame approach as for 2p2 ≤ pk, we conclude that B(dx′) contains y + dl + αl,sp1 −s(pk − p2 − p1) for s = 0, 1, . . . , d − 1 and l = 0, . . . , p1/d − 1. These numbers aredifferent 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 �=p1, pk−1. We may assume 2p ≤ pk. Otherwise, we consider S(a) and pk − p. Writegcd(p1, p2, . . . , pk−1) = d. So with S′(a) = {0, p1/d, . . . , pk−1/d} we obtain for some0 ≤ 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 clearthat

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 approachas 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, areeven. 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. Sogcd(d1, q

′) = 1. We already know that ld1+d1 ∈ B(d1x′) for all l = 0, 1, . . . , q/d1−1.

Now, since q < p1, it follows from Lemma 3.1 that ld1 + d1 + p1 ∈ B(d1x′) while

Lemma 3.4 with pl1 = p1 and pl2 = pk implies ld1 + d1 + p1 − αq ∈ B(d1x′) for

α such that q′ ≤ ld1 + d1 + p1 − αq < q′ + q. Lemma 3.2 with pk−1 leads told1 + d1 + (p1 − q′) − αq ∈ B(d1x

′) and 0 ≤ ld1 + d1 + (p1 − q′)− αq < q. Now, lety = d1+(p1−q′)−αq. Then y ∈ B(d1x

′) and 0 < y < q. Recursively, we obtain fromthe above procedure that 0 ≤ y+ld1+lsd1−sq′ < q and y+ld1+lsd1−sq′ ∈ B(d1x

′)for s = 0, 1, . . . , d1 − 1 and l = 0, . . . , q/d1 − 1. Moreover, y + ld1 + lsd1 − sq′ = 0if and only if s = d1 − 1. Therefore, since gcd(d1, q

′) = 1, these numbers are alldifferent and are contained in {0, . . . , q − 1}. We conclude that

{v : 0 ≤ v < q} ⊆ B(d1x′).

Again applying Lemma 3.1 with p1, we obtain from (3.20) and the last relationthat

(3.21) {0, 1, . . . , q − 1} ∪ {p1, . . . , pk − 1} ∪ {ld1 : 0 ≤ ld1 < pk} ⊆ B(d1x′).

Let p1 ≤ v < pk. It follows from Lemma 3.4 with pl1 = p1 and pl2 = pk thatv−αq ∈ B(d1x

′) 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) containsan odd p �= pk and an even q �= 0. It is easy to see that p1, . . . , pk−2 are evenand pk−1 is odd. Suppose pk ≥ 2pk−2, so 2pj < pk for j = 1, 2, . . . , k − 2. ClearlyS(a)\{p2} = {0, p1, p3, . . . , pk} has property P. Denote gcd(p1, p3, . . . , pk) = d1 andS′(a) = {0, p1/d1, p3/d1, . . . , pk/d1}. We obtain (3.20) with some 0 ≤ x′ < pk, i.e.,

{ld1 : 0 ≤ ld1 < pk} ⊆ B(d1x′).


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(d1x′).

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 = 2µg, where g is odd. Also, d1 �= 1 implies g �= 1. Foreach y + ld1 we choose εi,j = 0 for j > 1 to obtain

2m(y + ld1) + kη −m−1∑i=0

k∑j=1

εi,j2ipj = 2m(y + ld1) + kη −m−1∑i=0

εi,12ip1

= y + ld1 + (2m − 1)(y + ld1) + kη − γp1

= y + ld1 + (2m − 1)(y + ld1) + kη − γ2µg,

where γ can be any integer between 0 and 2m − 1. By the definition of B(d1x′)

we conclude that y + ld1 + (2m − 1)(y + ld1) + kη − γ2µg ∈ B(x) whenever y + ld1

+ (2m − 1)(y + ld1) + kη − γ2µg ∈ Zpk. Thus, we can choose γ to obtain

y + ld1 + (2m − 1)(y + ld1) + kη − γ2µg = y + ld1 + rp1 + βg

with some r and 0 ≤ β ≤ 2µ − 1 such that pk − p2 ≤ y + ld1 + rp1 + βg < pk. Itfollows from Lemma 3.4 that 0 ≤ y + ld1 + rp1 +βg− (pk −p2) < p2 is contained inB(d1x

′). Furthermore, with pl1 = p1 and pl2 = pk we get by Lemma 3.2 for someα ≥ 0 that

0 ≤ y + ld1 + rp1 + βg − (pk − p2) − αg < p1

andy + ld1 + rp1 + βg − (pk − p2) − αg ∈ B(d1x

′).We may write with some αl,1

y + ld1 + rp1 + βg − (pk − p2) − αg = y + ld1 + αl,1d1 − (pk − p2).

Clearly, these new numbers are different from y + ld1. Therefore with αl,0 = 0 weconclude recursively for s = 0, 1, . . . , d1 − 1, l = 0, . . . , p1/d1 − 1 that

0 ≤ y+ ld1 +αl,sd1−s(pk−p2) < p1 and y+ ld1 +αl,sd1−s(pk −p2) ∈ B(d1x′).

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(d1x′).

To finish our proof for the case pk ≥ 2pk−2, we have to verify {p1 +1, . . . , pk −1} ⊆B(d1x

′). 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(d1x

′).

Recursively, we conclude

(3.23) {x : 0 ≤ x < pk, x is odd} ⊆ B(d1x′).



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(d1x′).

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(d1x′).

Combining this with (3.22) and (3.23), we get

{0, .., pk−1} ⊆ B(d1x′).

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 symmetricrelation (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(d2x′).

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(d2x′) for l = 0, . . . ,

p1/d2 − 1. We now know how to use Lemmas 3.1, 3.2 and 3.4 from these numbersto obtain, for s = 0, . . . , d2 − 1 and l = 0, . . . , p1/d2 − 1,

y + ld2 + αs,ld2 − s(pk − p1) < p1 and y + ld2 + αs,ld2 − s(pk − p1) ∈ B(d2x′),

which in turn implies {0, . . . , p1 − 1} ⊆ B(d2x′). Hence,

{0, . . . , p1} ∪ {ld2 : 0 ≤ ld2 < pk} ⊆ B(d2x′).

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 thedetails here.

Let q ≥ pk − q. We may assume that q is minimal among all nonzero evennumbers in S(a) and that p satisfies 2p < pk. Otherwise we consider S(a), forwhich we have already the desired assertion. For the same reason we can alsoassume 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 weget 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) − αlq′ < q′. Now write

yl = l′d + ld − (pk − q) − αlq′. 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 3.4with q and pk and finally Lemma 3.2 with q that

yl + αl,sd + s(pk − q) ∈ B(dx′) and 0 ≤ yl + αl,sd + 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 havenothing 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}, weobtain {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. ByLemma 3.4 with pk and q we get q − p ≤ l′d + ld − (pk − q) < q. Again usingLemma 3.4 with p and q, we conclude that 0 ≤ l′d + ld − (pk − p) < p. Now writeyl = 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,sp + s(pk − p) ∈ B(dx′) and 0 ≤ yl + αl,sp + 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.

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Department of Mathematics, China Jiliang University, Hangzhou, China; Instituteof Mathematics, University of Duisburg-Essen, D-47057 Duisburg, Germany

E-mail address: [email protected]


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Documents

Introduction · SUBDIVISION SCHEMES WITH NONNEGATIVE MASKS XINLONG ZHOU Abstract. The conjecture concerning the characterization of a convergent univariate subdivision algorithm with