# Polynomial Reproduction of Vector Subdivision Schemes

Authors: Y. F. Shen; D. H. Yuan; S. Z. Yang
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We discuss the polynomial reproduction of vector subdivision schemes with general integer dilationm≥2. We first present a simple algebraic condition for polynomial reproduction of such schemes with standard subdivision symbol. We then extend it to general subdivision symbol satisfying certain order of sum rules. We also illustrate our results with several examples. Our results show that such kind of scheme can produce exactly the same scalar polynomial from which the data is sampled by convolving with a finite nonzero sequence of vectors.


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## Body

## 1. Introduction

Subdivision schemes are efficient algorithmic methods in approximation theory, computer aid geometric design, and wavelet construction; see [1, 2] and references therein. The subdivision scheme is calledvector subdivision scheme if the corresponding subdivision symbol is a matrix of trigonometric polynomial. Note that the vector subdivision scheme plays an important role in multiwavelet and multichannel wavelets analysis; see [3–5] and references therein. In this paper we would like to derive simple algebraic conditions on the subdivision symbol that allow us to determine the degree of its polynomial reproduction, which is different from polynomial generalization.The polynomial generalization is the capability of vector subdivision scheme to generate the full space of scalar polynomials with desired degree. This property is equivalent to approximation order (defined inL2-norm), accuracy, and sum rulers; see [6, 7]. The so-called polynomial reproduction is the capability of vector subdivision scheme to produce exactly the same scalar polynomial p(x) from which the data is sampled by convolving with a special sequence of vectors h0={h0,j∈Rs,j∈S},
(1)d0,kT:=(p*h0)(k)=∑j∈Z‍p(k-j)h0,j,k∈Z,
where S is some finite index set containing 0, * denotes convolution operator, and T denotes transpose operator.There are many valuable works that were done when scalar subdivision scheme was considered [8–11]. Hormann and Sabin [8] presented some algebraic conditions which can be used to determine the degree of polynomial reproduction for a family of schemes. Polynomial reproduction was studied by Dyn et al. [9] for arbitrary prime and dual binary schemes and extended to univariate subdivision scheme of any arity by Conti and Hormann [10]. Further, Charina and Conti [11] yielded simple algebraic conditions on the subdivision symbol for the multivariate setting of scalar subdivision with dilation matrix mI, |m|>1; Charina and Romani [12] extended to the general expanding dilation matrix M∈Zs×s case. Motivated by all these works, we discuss the property of polynomial reproduction for vector subdivision schemes.The remainder of this paper is organized as follows. Section2 sets the notations concerning vector subdivision scheme with general integer dilation m≥2 and stresses the connection of convergence between the vector subdivision scheme and traditional cascade algorithm. In Section 3, we provide algebraic tools for determining the degree of polynomial reproduction of the vector subdivision scheme with standard subdivision symbol; basing on two scale similarity transform (TST), we also establish the relationship of vector subdivision schemes between general subdivision symbol and the standard subdivision symbol. In Section 4, several examples are provided to illustrate our results.


## 2. Background and Notation

A vector subdivision schemeSP,h0,Φ0 with integer dilation m≥2 is given by a finitely supported matrix sequence P={Pk∈Rs×s,k∈Z} (so-called subdivision mask), a constant column vector h0={h0,j∈Rs,j∈S}, and an initial function Φ0(x). The constant vector h0 and the initial function Φ0 are determined by P and will be given by (39) or (55) in the next section. The symbols of h0 and P (denoted by h0(z) and P(z), resp.) are given by the Laurent polynomials
(2)h0(z)=∑j∈S‍h0,jzj,P(z)=∑k∈Z‍Pkzk,
respectively.In this paper, we use two vector spaces of Laurent polynomial,L(Rs) and L'(Rs). We say h0(z)∈L(Rs), if
(3)h0T(z)=hTA(z)=hT[1∑j∈S‍rjzj0Is-1],
with some h∈Rs and h≠0, where {rj,j∈S} is some finite 1×(s-1)-vector sequence and Is-1 is the (s-1)×(s-1) identity matrix. Similarly, we say h0(z) belongs to L'(Rs) if h0(z)=A(z)h.For a data sequence{dk∈Rs,k∈Z}, thevector subdivision operatorSP acting on it is defined by
(4)(SPd)kT=∑j∈Z‍djTPk-mj,k∈Z.
Actually the vector subdivision scheme SP,h0,Φ0 is a recursive algorithm based on the vector subdivision operator
(5)dl+1,kT=SPdl,kT=∑j∈Z‍dl,jTPk-mj,k∈Z,l∈N
and a convolution operator
(6)fl+1,k=∑j∈Z‍dl+1,jTΦ0(k-j),k∈Z,l∈N,
with the initial data d0={d0,k∈Rs:k∈Z} and the function
(7)Φ0(x)=∑j∈S‍yjX(x-j),
where X(x) is the characteristic function of interval [0,1) and y={yj∈Rs,j∈S} is selected such that y(z)∈L'(Rs) and h0(z)Ty(z)=1.Lemma 1.
Ifh0(z)∈L(Rs), then there exist linearly independent vectors y1(z),…,ys(z)∈L'(Rs) such that
(8)h0T(z)yj(z)=1,j=1,…,s.Proof.
Suppose thath0T=hT[1∑j∈Srjzj0Is-1]. It is not difficult to find that there exist linearly independent constant vectors y1,…,ys satisfying
(9)hTyj=1,j=1,…,s.
Let
(10)yj(z)=[1-∑j∈S‍rjzj0Is-1]yj,j=1,…,s.
Then, by calculation the conclusion follows directly.In order to define a sequence of continuous functions, we also need parameterization. As in [10] for the scalar case, we chose the associated parameter values:
(11)tkl:=t0l+kml,t0l=t0l-1-τml,t00=0,k∈Z,l∈N.
We call {tkl,k∈Z,l∈N} the sequence ofparameter values associated with the vector subdivision scheme. Let us define continuous functions fl by linearly interpolating the data fl,k to the parameter values tkl,
(12)fl(tkl)=fl,k,k∈Z,l∈N.Definition 2.
If the sequence of continuous functions{fl,l∈N} converges (in uniform norm), then we define the limit function as
(13)gd0=lim⁡l→∞fl.We say thatgd0 is the limit function of the vector subdivision scheme SP,h0,Φ0 acting on the initial data d0. If s=1 and h0=1, one may find that the vector subdivision scheme is the same as the scalar one in [10]. The limit functions ϕ0:ϕδ0(p):=gδ0(p), p=1,…,s, are calledbasic limit functions if the initial data is selected by
(14)δ0,k(p)={ep,k=0,0,otherwise,k∈Z,p=1,…,s,
where ep denotes pth column of the s×s identity matrix. It is easy to check that Φ=(ϕ1,…,ϕs)T is compactly supported and satisfies the refinement equation
(15)Φ(t)=∑k∈Z‍PkΦ(mt-k),t∈R.In this paper, we consider vector subdivision scheme that is convergent andnonsingular, so that gd0=0 if and only if d0=0. Define the linear operator QP (which is called cascade operator) as follows:
(16)QPΦ=∑k∈Z‍PkΦ(m·-k),Φ∈(L2(R))s.
By induction on n, it is easy to verify that
(17)QPnΦ0(x)=∑k∈Z‍Pn,kΦ0(mnx-k),n∈N,
where Φ0 is defined as (7),
(18)Pn,k=∑j∈Z‍Pn-1,jPk-Mj,P1,k=Pk,k∈Z.Under the convergent condition of cascade algorithm (see [13]), we can prove that QPnΦ0 converges to the normalized solution Φ of (15) in (Lp(R))s (1≤p≤∞).Proposition 3.
A vector subdivision schemeSP,h0,Φ0 is convergent and nonsingular if and only if (15) has compactly supported solution Φ and the integer translations of Φ are linearly independent.Proof.
Firstly, we claim that the convergence of vector subdivision scheme is the same as that of the cascade algorithm. Actually, by induction, for the initial dataδ0(p),p=1,…,s, we have
(19)f1,j(p)=∑k∈Z‍δ1,k(p)TΦ0(j-k)=∑k∈Z‍∑n∈Z‍δ0,n(p)TPk-mnΦ0(j-k)=∑n∈Z‍δ0,n(p)T∑k∈Z‍PkΦ0(m(m-1j-n)-k)=∑n∈Z‍δ0,n(p)TQPΦ0(m-1j-n)=epTQPΦ0(m-1j),
and
(20)fn,j(p)=∑n∈Z‍δ0,n(p)TQPnΦ0(j-n)=epTQPnΦ0(m-nj).
Then, our declaration follows by a standard method in analysis. Note that the convergence of cascade algorithm with finite supported mask is equivalent to the existence of compactly supported solution of (15) (see [14]). We can obtain the desired result according to the proof of [11, Proposition 1.3].Following [10, 11], we give the definitions of polynomial generation, polynomial reproduction, and stepwise polynomial reproduction with respect to vector subdivision scheme. We denote by Πk the space of polynomials up to degree k. Suppose pl={pl,k,k∈Z}:={p(tkl),k∈Z}, for l∈N, and let (l(Z))s be the space of s-vector sequence indexed by Z.Definition 4 (polynomial generation).
A convergent stationary vector subdivision schemeSP,h0,Φ0 generates polynomials up to degree nG (ΠnG-generating), if, for any polynomial p(x)∈ΠnG, there exists some initial data d0∈(l(Z))s such that gd0=p(x).Definition 5 (polynomial reproduction).
A convergent stationary vector subdivision schemeSP,h0,Φ0 with parameter values t(l),l∈N is reproducing polynomials up to degree nR (ΠnR-reproducing), if, for the initial data d0={(p*h0)(tk(0)),k∈Z}, with any polynomial p(x)∈ΠnR, the limit of the vector subdivision scheme satisfies gd0=p(x).Definition 6 (stepwise polynomial reproduction).
A  convergent stationary vector subdivision schemeSP,h0,Φ0 with parameter values t(l),l∈N is stepwise reproducing polynomials up to degree nR (stepwise ΠnR-reproducing), if, for the data dl={(pl*h0)(k),k∈Z}, with any polynomial p(x)∈ΠnR,
(21)dl+1T=(SPdT)lor,equivalently,(pl+1*h0T)(k)=∑j∈Z‍dlT(tj(l))Pk-mj,k∈Z.Noting basic limit functionΦ defined by (14), it is easy to show that
(22)gd0=∑k∈Z‍d0,kTΦ(·-k).
Thus the equivalence of polynomial generation and accuracy (approximation order, sum rulers) is clear; see [6, 7]. The following proposition shows that for a nonsingular convergent vector subdivision scheme the concepts of polynomial reproduction and stepwise polynomial reproduction are equivalent.Proposition 7.
A nonsingular convergent vector subdivision scheme is stepwiseΠnR reproducing if and only if it is ΠnR reproducing.Proof.
Since thath0T(z)y(z)=1, then by discrete Fourier transform we have ∑u∈S‍h0,n-uTyu=δn. Thus under the stepwise polynomial reproduction, from (6) we can easily get, for p(x)∈ΠnR,
(23)fl,k=∑j∈S‍dl,k-jTΦ0(j)=∑j∈S‍(pl*h0T)(k-j)yj=∑j∈S‍p(tk-jl)∑u∈S‍h0,uTyj-u=p(tkl),k∈Z,l∈N.
As a result, fl,k=p(tkl) implies fl+1,k=p(tkl+1). The rest of necessary part one may refer to [11, Proposition 1.7]. For the sufficient part, noting (6) and (7), we have
(24)fl,k=∑j∈S‍dl,k-jTyj,k∈Z,l∈N.
Consider p(x)∈ΠnR. Taking fl,k=p(tkl) and using the same methods of [11, Proposition 2.1], one obtains from (6) that
(25)fl+1,k=p(tkl+1).
Noting that h0(z)Ty(z)=1, we deduce from fl,k=p(tkl) that
(26)0=∑j∈S‍dl,k-jTyj-∑n‍p(tk-nl)δn=∑j∈S‍dl,k-jTyj-∑n‍p(tk-nl)∑j∈S‍h0,n-jTyj=∑j∈S‍(dl,k-jT-(pl*h0T)(k-j))yj≜∑j∈S‍xl,k-jyj,
where xl,k=[xk,1l,xk,2l,…,xk,sl]. Similarly, the equation fl+1,k=p(tkl+1) implies
(27)∑j∈S‍(dl+1,k-jT-(pl+1*h0T)(k-j))yj=0,k∈Z.
Following Lemma 1, we select linearly independent vectors  yt(z) as (10) satisfying h0(z)Tyt(z)=1,t=1,…,s. Then,
(28)[0,…,0]=∑j∈S‍xl,k-j[yj1,…,yjs]=∑j∈S∖{0}‍xl,k-j[0rj0Os-1]Y+[xk,1l,…,xk,sl][1r00Is-1]Y=[xk,1l,xk,2l,…,xk,sl]Y+[0,∑j∈S‍xk-j,1lrj]Y,k∈Z,
where Y=[y1,…,ys]. Since Y is an invertible matrix, then fl,k=p(tkl) is equivalent to xk,1l=xk,2l=⋯=xk,sl=0. The conclusion follows directly.


## 3. Algebraic Condition for Polynomial Reproduction

In this section we firstly consider the nonsingular subdivision scheme with standard subdivision symbolP(z) in (33) and provide algebraic conditions on P(z) for checking the polynomial reproduction. With the two-scale similarity transform (TST) in the hand, we further discuss the properties of polynomial reproduction with respect to the general subdivision symbol satisfying certain order of sum rules. For TST, see [6, 13, 15, 16] and references therein.We denote thesubsymbols of a vector subdivision symbol P(z) by
(29)Pα(z)=∑k∈Z‍Pmk+αzmk+α,α=0,…,m-1,z∈C∖{0},
and remark that the jth derivative of a subsymbol is
(30)D(j)Pα(z)=∑k∈Z‍qj(mk+α)zmk+α-j,
where qj∈πj is the polynomial
(31)qj(x)=∏u=0j-1‍(x-u).
We further assume that
(32)P(z)=[P11(z)P12(z)P21(z)P22(z)],
where P11(z), P12(z), P21(z), and P22(z) are 1×1, 1×(s-1), (s-1)×1, and (s-1)×(s-1) matrices of Laurent polynomials, respectively. The subsymbols of P11(z), (P12(z)) are defined by P11α(z), (P12α(z)),  α=0,…,m-1. We definestandard subvision symbol with sum rules of order n by
(33)P(z)=[(1+z+⋯+zm-1)nQ11(z)(1-zm)nQ12(z)P21(z)P22(z)],
where Q11(1)=1/mn-1 and Q11(z), Q12(z) are 1×1, 1×(s-1) matrices, respectively. Similar to [10, Lemma 2.1] or [11, Proposition 2.1], we have the following proposition.Proposition 8.
The following are equivalent.(i)
The vector subdivision symbolP(z) takes the form of (33).(ii)
The entries ofP(z) satisfy
(34)P11(1)=m,D(j)P11(ϵ)=0,j=0,…,n-1,llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllϵ∈E′;D(j)P12(ϵ)=0,j=0,…,n-1,ϵ∈E,with0∈Rs-1, where E={e-i(2πk/m),k=0,…,m-1} and E′=E∖{1}.(iii)
The subsymbols of the entries ofP(z) satisfy
(35)mD(j)P11α(1)=D(j)P11(1),j=0,…,n-1,lllllllllllllllllllllllα=0,…,m-1;D(j)P12α(1)=0,j=0,…,n-1,α=0,…,m-1.(iv)
The coefficients of the entries ofP(z) satisfy
(36)m∑k∈Z‍qj(l-mk)P11,α-mk=D(j)P11(1),∑k∈Z‍qj(l-mk)P12,α-mk=0,l∈Z,llllj=0,…,n-1,α=0,…,m-1.Similar to [10, Lemma 4.2] or [11, Proposition 2.5], we also have the following proposition. It allows us to express the polynomial reproduction of  SP,h0,Φ0 in terms of the properties of its symbol, where P(z) takes the form of (33).Proposition 9.
Letn∈N, τ∈R, and qj as in (31). A vector subdivision symbol P(z) satisfies
(37)D(j)P11(1)=mqj(τ),D(j)P11(ϵ)=0,j=0,…,n,llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllϵ∈E′,D(j)P12(ϵ)=0,j=0,…,n,ϵ∈E,
if and only if
(38)∑k∈Z‍P11,u-mkkj=(u-τm)j,u∈Z,j=0,…,n,∑k∈Z‍P12,u-mkkj=0,u∈Z,j=0,…,n.


### 3.1. Standard Subdivision Symbol

Suppose that subdivision symbol has the standard form (33) in this subsection. Then the constant vector h0 and initial function Φ0(x) in SP,h0,Φ0 can be chosen by
(39)h0=e1,Φ0(x)=yX(x),
respectively, where y satisfies h0Ty=1. We provide a simple algebraic condition for determining τ∈R, which appears in (11) and guarantees the reproduction of linear polynomial. In Theorem 11, we then provide algebraic conditions on P(z) for checking the reproduction of polynomials of higher degree.Proposition 10.
Leth0, Φ0 be defined by (39) and let P take the form of (33) with n=2. Let SP,h0,Φ0 be a nonsingular vector subdivision scheme. Then, SP,h0,Φ0 reproduces linear polynomials if and only if its parameter values are given by (11) with
(40)τ=DP11(1)m.Proof.
We follow the lines of Theorem 3.1 of [10] or Proposition 2.3 of  [11]. According to Proposition 7, polynomial reproduction is equivalent to stepwise polynomial reproduction. Hence it is sufficient to show that dl,k=h0p(tkl) implies dl+1,k=h0p(tkl+1),k∈Z, for any l∈N. Note that any convergent vector subdivision scheme reproduces constants; we only consider the monomial p(x)=x. Let l∈N and dl,k=h0p(tkl)=h0·(t0l+(j/ml)),j∈Z. For any k∈Z and α=0,…,m-1, we have
(41)dl+1,mk+αT=∑j‍dl,jTPmk+α-mj=∑j‍dl,k-jTPmj+α=∑j‍(h0T(t0l+k-jml))Pmj+α=∑j‍(h0T(t0l+mk+αml+1))Pmj+α-∑j‍(h0T(t0l+mj+αml+1))Pmj+α=(h0Tt0l+mk+αml+1)-h0TDP(1)ml+1,dl+1,mk+αT=h0Tt0l-h0Tτml+1+h0Tmk+αml+1.
As a result, the vector subdivision scheme reproduces linear polynomials stepwisely if and only if(42)h0Tτ=h0TDP(1).
Then, (40) follows directly.Theorem 11.
LetnR∈N0. A convergent subdivision scheme with symbol P(z) in (33) and associated parameterization in (11), for some τ∈R, reproduces polynomials of degree up to nR if and only if
(43)D(k)P11(1)=mqk(τ),D(k)P11(ϵ)=0,k=0,…,nR,llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllϵ∈E′,D(k)P12(ϵ)=0,k=0,…,nR,ϵ∈E.Proof.
The proof is induction. In the casek=0 the conclusion follows by Proposition 8. By Proposition 7, it suffices to prove the results for the step polynomial reproduction.
“⇐:” Let k≤nR and l>0. Suppose that p(x)=ckxk+p~(x), with p~∈Πk-1. We show that
(44)dl={h0p(tjl)=h0ck(tjl)k+h0p~(tjl),j∈Z}
satisfies dl+1T=SPdlT={h0Tp(tjl+1),j∈Z}. Actually by induction and stepwise polynomial reproducing, we have
(45)dl+1,jT=∑u∈Z‍dl,uTPj-mu=∑u∈Z‍h0Tck(t0l+uml)kPj-mu+p~(tjl+1)=∑u∈Z‍h0Tck∑p=0k‍(kp)(uml)p(t0l)k-pPj-mu+p~(tjl+1)=h0Tck∑p=0k‍(t0l)k-p∑u∈Z‍(kp)(uml)pPj-mu+p~(tjl+1).
By Proposition 9, (43) is equivalent to
(46)h0T∑u∈Z‍upPj-mu=∑u∈Z‍up[P11,j-mu,P12,j-mu]=h0T(j-τm)p.
Then,
(47)dl+1,jT=h0Tck∑p=0k‍(kp)(t0l)k-p(j-τml+1)p+h0Tp~(tjl+1)=h0Tck(t0l+j-τml+1)k+h0Tp~(tjl+1)=h0Tck(tjl+1)k+h0Tp~(tjl+1).
“⇒:” Assume that the vector subdivision scheme is ΠnR reproducing. Let p(x)=xk(k≤nR) and dl={h0p(tjl),j∈Z}, with h0=e1. Using the same method as above, we have
(48)dl+1,jT=h0T∑p=0k‍(kp)(t0l)k-p∑u∈Z‍(uml)pPj-mu.
On the other hand,
(49)p(tjl+1)=∑p=0k‍(kp)(t0l)k-p(j-τml+1)p.
Combining the above two equations, we deduce
(50)∑p=0k‍(kp)(t0l)k-p(1ml)p[∑u∈Z‍uph0TPj-mu-h0T(j-τm)p]=0.
By induction, we obtain
(51)∑u∈Z‍ukh0TPj-mu-h0T(j-τm)k=0.
As a result, by Proposition 9 the conclusion (43) follows.


### 3.2.  General Subdivision Symbol

We devote this section to discussing the polynomial reproduction of vector subdivision scheme with general subdivision symbol. Suppose that subdivision symbolP(z) has sum rules of order n (see [6, 7]), that is, if there exists a row vector y(ξ) of trigonometric polynomials such that y(0)≠0 and
(52)[y(d·)P(·)](j)(2πkm)=δky(j)(0),j=0,…,n-1;lllllllllllllllllllllllllllllllllllllllllllllllllllllllk=0,…,m-1.In [16, Theorem 2.2], they showed that subdivision symbol P~(ξ) satisfies the sum rules of order n, if and only if there exists an invertible s×s matrix
(53)U(z)=[1∑j∈Srjzj0Is-1]U0
such that P~(z)=U(zm)P(z)U-1(z) with P(z) taking the form (33), where U0 is a constant invertible matrix. It is evident that
(54)U-1(z)=V(z):=V0[1-∑j∈S‍rjzj0Is-1]
with V0U0=Is.Let the constant vectorh0 and initial function Φ0(x) in SP,h0,Φ0 be defined by (39). Then we choose h~0 and Φ~0(x) in SP~,h~0,Φ~0 by
(55)h~0T(z)=h0TV(z),Φ~0(x)=∑j∈S‍y~jX(x-j),
with y~(z)=U(z)y. It is easy to see that h~0(z)∈L(Rs),  y~(z)∈L'(Rs), and h~0T(z)y~(z)=1.Theorem 12.
The vector subdivision schemeSP~,h~0,Φ~0 is ΠnR reproducing, if and only if  SP,h0,Φ0 is ΠnR reproducing.Proof.
By Proposition7, we only need to consider the equivalence under the condition of stepwise ΠnR reproducing. We just show the sufficient part (the necessary part can be proved similarly).
Supposep(x)∈ΠnR and dl={dl,k:=h0p(tk(l)),k∈Z}, where l∈N and parameter values tk(l) are defined by (11). By the definition of stepwise polynomial reproduction, we have
(56)dl+1,k=∑j∈Z‍dlT(tj(l))Pk-mj,k∈Z.
Since V(z)U(z)=1, then we get ∑u∈S‍VuUj-u=δjIs, where Uj,j∈Z are the coefficients of  U(z) with Uj=Os,j∉S and Vj,j∈Z are defined similarly. For the new scheme SP~,h~0,Φ~, we have
(57)d~l,jT=(pl*h~0T)(j)=∑q∈S‍h0Tp(tj-ql)Vq.
Then
(58)(SP~d~lT)k=∑j∈Z‍d~lT(tj(l))P~k-mj=∑j∈Z‍d~lT(tj(l))∑v∈Z‍∑u∈S‍UuPv-muVk-mj-v=∑v∈Z‍∑u∈S‍∑j∈Z‍∑q∈S‍h0Tp(tj-ql)VqUuPv-muVk-mj-v=∑v∈Z‍∑u∈Z‍∑j∈Z‍h0Tp(tjl)∑q∈S‍VqUu-j-qPv-muVk-v=∑v∈Z‍∑u∈Z‍h0Tp(tul)Pv-muVk-v.
Here, necessary variable substitution is used. Following the stepwise polynomial reproduction of SPh0,Φ0, we know that
(59)(SP~d~l)k=∑v∈Z‍h0Tp(tvl+1)Vk-v=d~l+1(tkl+1).
Thus the proof is completed.


## 3.1. Standard Subdivision Symbol

Suppose that subdivision symbol has the standard form (33) in this subsection. Then the constant vector h0 and initial function Φ0(x) in SP,h0,Φ0 can be chosen by
(39)h0=e1,Φ0(x)=yX(x),
respectively, where y satisfies h0Ty=1. We provide a simple algebraic condition for determining τ∈R, which appears in (11) and guarantees the reproduction of linear polynomial. In Theorem 11, we then provide algebraic conditions on P(z) for checking the reproduction of polynomials of higher degree.Proposition 10.
Leth0, Φ0 be defined by (39) and let P take the form of (33) with n=2. Let SP,h0,Φ0 be a nonsingular vector subdivision scheme. Then, SP,h0,Φ0 reproduces linear polynomials if and only if its parameter values are given by (11) with
(40)τ=DP11(1)m.Proof.
We follow the lines of Theorem 3.1 of [10] or Proposition 2.3 of  [11]. According to Proposition 7, polynomial reproduction is equivalent to stepwise polynomial reproduction. Hence it is sufficient to show that dl,k=h0p(tkl) implies dl+1,k=h0p(tkl+1),k∈Z, for any l∈N. Note that any convergent vector subdivision scheme reproduces constants; we only consider the monomial p(x)=x. Let l∈N and dl,k=h0p(tkl)=h0·(t0l+(j/ml)),j∈Z. For any k∈Z and α=0,…,m-1, we have
(41)dl+1,mk+αT=∑j‍dl,jTPmk+α-mj=∑j‍dl,k-jTPmj+α=∑j‍(h0T(t0l+k-jml))Pmj+α=∑j‍(h0T(t0l+mk+αml+1))Pmj+α-∑j‍(h0T(t0l+mj+αml+1))Pmj+α=(h0Tt0l+mk+αml+1)-h0TDP(1)ml+1,dl+1,mk+αT=h0Tt0l-h0Tτml+1+h0Tmk+αml+1.
As a result, the vector subdivision scheme reproduces linear polynomials stepwisely if and only if(42)h0Tτ=h0TDP(1).
Then, (40) follows directly.Theorem 11.
LetnR∈N0. A convergent subdivision scheme with symbol P(z) in (33) and associated parameterization in (11), for some τ∈R, reproduces polynomials of degree up to nR if and only if
(43)D(k)P11(1)=mqk(τ),D(k)P11(ϵ)=0,k=0,…,nR,llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllϵ∈E′,D(k)P12(ϵ)=0,k=0,…,nR,ϵ∈E.Proof.
The proof is induction. In the casek=0 the conclusion follows by Proposition 8. By Proposition 7, it suffices to prove the results for the step polynomial reproduction.
“⇐:” Let k≤nR and l>0. Suppose that p(x)=ckxk+p~(x), with p~∈Πk-1. We show that
(44)dl={h0p(tjl)=h0ck(tjl)k+h0p~(tjl),j∈Z}
satisfies dl+1T=SPdlT={h0Tp(tjl+1),j∈Z}. Actually by induction and stepwise polynomial reproducing, we have
(45)dl+1,jT=∑u∈Z‍dl,uTPj-mu=∑u∈Z‍h0Tck(t0l+uml)kPj-mu+p~(tjl+1)=∑u∈Z‍h0Tck∑p=0k‍(kp)(uml)p(t0l)k-pPj-mu+p~(tjl+1)=h0Tck∑p=0k‍(t0l)k-p∑u∈Z‍(kp)(uml)pPj-mu+p~(tjl+1).
By Proposition 9, (43) is equivalent to
(46)h0T∑u∈Z‍upPj-mu=∑u∈Z‍up[P11,j-mu,P12,j-mu]=h0T(j-τm)p.
Then,
(47)dl+1,jT=h0Tck∑p=0k‍(kp)(t0l)k-p(j-τml+1)p+h0Tp~(tjl+1)=h0Tck(t0l+j-τml+1)k+h0Tp~(tjl+1)=h0Tck(tjl+1)k+h0Tp~(tjl+1).
“⇒:” Assume that the vector subdivision scheme is ΠnR reproducing. Let p(x)=xk(k≤nR) and dl={h0p(tjl),j∈Z}, with h0=e1. Using the same method as above, we have
(48)dl+1,jT=h0T∑p=0k‍(kp)(t0l)k-p∑u∈Z‍(uml)pPj-mu.
On the other hand,
(49)p(tjl+1)=∑p=0k‍(kp)(t0l)k-p(j-τml+1)p.
Combining the above two equations, we deduce
(50)∑p=0k‍(kp)(t0l)k-p(1ml)p[∑u∈Z‍uph0TPj-mu-h0T(j-τm)p]=0.
By induction, we obtain
(51)∑u∈Z‍ukh0TPj-mu-h0T(j-τm)k=0.
As a result, by Proposition 9 the conclusion (43) follows.


## 3.2.  General Subdivision Symbol

We devote this section to discussing the polynomial reproduction of vector subdivision scheme with general subdivision symbol. Suppose that subdivision symbolP(z) has sum rules of order n (see [6, 7]), that is, if there exists a row vector y(ξ) of trigonometric polynomials such that y(0)≠0 and
(52)[y(d·)P(·)](j)(2πkm)=δky(j)(0),j=0,…,n-1;lllllllllllllllllllllllllllllllllllllllllllllllllllllllk=0,…,m-1.In [16, Theorem 2.2], they showed that subdivision symbol P~(ξ) satisfies the sum rules of order n, if and only if there exists an invertible s×s matrix
(53)U(z)=[1∑j∈Srjzj0Is-1]U0
such that P~(z)=U(zm)P(z)U-1(z) with P(z) taking the form (33), where U0 is a constant invertible matrix. It is evident that
(54)U-1(z)=V(z):=V0[1-∑j∈S‍rjzj0Is-1]
with V0U0=Is.Let the constant vectorh0 and initial function Φ0(x) in SP,h0,Φ0 be defined by (39). Then we choose h~0 and Φ~0(x) in SP~,h~0,Φ~0 by
(55)h~0T(z)=h0TV(z),Φ~0(x)=∑j∈S‍y~jX(x-j),
with y~(z)=U(z)y. It is easy to see that h~0(z)∈L(Rs),  y~(z)∈L'(Rs), and h~0T(z)y~(z)=1.Theorem 12.
The vector subdivision schemeSP~,h~0,Φ~0 is ΠnR reproducing, if and only if  SP,h0,Φ0 is ΠnR reproducing.Proof.
By Proposition7, we only need to consider the equivalence under the condition of stepwise ΠnR reproducing. We just show the sufficient part (the necessary part can be proved similarly).
Supposep(x)∈ΠnR and dl={dl,k:=h0p(tk(l)),k∈Z}, where l∈N and parameter values tk(l) are defined by (11). By the definition of stepwise polynomial reproduction, we have
(56)dl+1,k=∑j∈Z‍dlT(tj(l))Pk-mj,k∈Z.
Since V(z)U(z)=1, then we get ∑u∈S‍VuUj-u=δjIs, where Uj,j∈Z are the coefficients of  U(z) with Uj=Os,j∉S and Vj,j∈Z are defined similarly. For the new scheme SP~,h~0,Φ~, we have
(57)d~l,jT=(pl*h~0T)(j)=∑q∈S‍h0Tp(tj-ql)Vq.
Then
(58)(SP~d~lT)k=∑j∈Z‍d~lT(tj(l))P~k-mj=∑j∈Z‍d~lT(tj(l))∑v∈Z‍∑u∈S‍UuPv-muVk-mj-v=∑v∈Z‍∑u∈S‍∑j∈Z‍∑q∈S‍h0Tp(tj-ql)VqUuPv-muVk-mj-v=∑v∈Z‍∑u∈Z‍∑j∈Z‍h0Tp(tjl)∑q∈S‍VqUu-j-qPv-muVk-v=∑v∈Z‍∑u∈Z‍h0Tp(tul)Pv-muVk-v.
Here, necessary variable substitution is used. Following the stepwise polynomial reproduction of SPh0,Φ0, we know that
(59)(SP~d~l)k=∑v∈Z‍h0Tp(tvl+1)Vk-v=d~l+1(tkl+1).
Thus the proof is completed.


## 4. Examples

Example 1 (see [17, Example 4.2], GHM refinable mask).
The vector subdivision schemeSP~,h~0,Φ~0 is based on the following mask symbol,
(60)P~(z)=12∑k=03‍Pkzk=140[12(1+z)162-2(z+1)(z2-10z+1)-6+20z-6z2].It is not difficult to find that the subdivision maskP~ satisfies the sum rules of order 2. The corresponding vector y(z) in (52) is selected by
(61)y(z)=[22(1+z-1)z-1].
We construct a trigonometric matrix U(z) by
(62)U(z)=[22(3-z)2-z12].
Then, P(z) takes the form of (33),
(63)[(1+z)2Q11(z)(1-z2)2Q12(z)Q21(z)Q22(z)]
with
(64)Q11(z)=110(z3-14z2+20z-2);Q12(z)=220(z2-14z+22);Q21(z)=-210(z3-12z2-5z);Q22(z)=-110(z4-14z3+22z2-5).
Using Proposition 10 we get that, for τ=0, the scheme reproduces linear polynomials. Since D(2)P11(1)=-59/5≠2q2(τ)=0, the scheme cannot reproduce polynomials of degree 2. For their graphs, see Figure 1.Polynomial reproduction of schemeSP~,h~0,Φ~0 with scheme mask symbol (60); solid line denotes the true polynomial and asterisk point denotes the discrete data using 5 times iteration of subdivision operator; (a) p(t)=t; (b) p(t)=t2.
(a)
(b)Example 2 (see [17, Example 5]).
The vector subdivision schemeSP~,h~0,Φ~0 is based on the mask symbol P~(z) with
(65)P~11(z)=(10-310)80(1+z)(1+(38+1210)z+z2),P~12(z)=(56-215)80(1-z)(1+z)2,P~21(z)=(56-315)80(1-z)(1-10(3+3)z+z2),P~22(z)=(5-310)1040(1+z)(13-(10+610)z+13z2).The subdivision maskP~ satisfies the sum rules of order 3; see [17]. The corresponding vector y(z) in (52) is selected by
(66)y(z)=[6(-2+10)12-2+1012(1+10z2+4+10z-1)lllll×6(-2+10)12(z-2-z-1)].
If U(z) is chosen by
(67)U(z)=[115-612(z-1)(z-3)01],
then P(z)=U(zm)P~(z)U-1(z) takes the form of (33), with  n=3,m=2. Using Proposition 10 we get that, for τ=3/2, the scheme reproduces linear polynomials. Since D(2)P11(1)=10-2≠2q2(τ)=3/2, the scheme cannot reproduce polynomials of degree 2. For their graphs, see Figure 2.Polynomial reproduction of schemeSP~,h~0,Φ~0 with scheme mask symbol (65); solid line denotes the true polynomial and asterisk point denotes the discrete data using 5 times iteration of subdivision operator; (a) p(t)=t; (b) p(t)=t2.
(a)
(b)Example 3 (see [16, Example 5.2]).
The vector subdivision schemeSP~,h~0,Φ~0 is based on the following mask symbol:
(68)P~(z)=[12+z4+14z38z-3z8z16-116z14-116z-z16].The subdivision maskP~ satisfies the sum rules of order 4; see [16] or [18]. If U(z) is chosen by
(69)U(z)=[1112(z2-z-2)-23(z-z-1)01],
then P(z) takes the form of (33),
(70)[(1+z)4Q11(z)(1-z2)4Q12(z)Q21(z)Q22(z)],
with
(71)Q11(z)=196z5(z6-4z5+z4+16z3+z2-4z+1);Q12(z)=11152z7(1-z2)(z4-8z3+8z2-8z+1);Q21(z)=-18z(z-1)(z+1);Q22(z)=-196z3(z6-8z5+11z4-32z3+11z2-8z+1).
Using Proposition 10 we get that, for τ=0, the scheme reproduces linear polynomials. Since D(3)P11(1)=D(2)P11(1)=0, the scheme also can reproduce polynomials up to degree 3. For their graphs, see Figure 3.Polynomial reproduction of schemeSP~,h~0,Φ~0 with scheme mask symbol (68); solid line denotes the true polynomial and asterisk point denotes the discrete data using 5 times iteration of subdivision operator; (a) p(t)=t; (b) p(t)=t2; (c) p(t)=t3; (d) p(t)=t4.
(a)
(b)
(c)
(d)



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