# Construction of Compactly Supported Refinable Componentwise Polynomial Functions inℝ2

Authors: Yanmei Xue; Ning Bi
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We provide a sufficient condition for constructing a class of
compactly supported refinable functions with componentwise polynomial property inℝ2. An iteration algorithm is developed to compute the polynomial on each component of the functions' support. Finally, two examples for constructing the symmetric refinable componentwise polynomial functions are given.


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## Body

## 1. Introduction

Refinable functions are among the most important functions; they form the foundation of wavelet theory and subdivision scheme theory. The details can be found in [1–3]. Other areas in which refinable functions play important roles are fractal geometry and self-affine tilings (see [4–6]).Splines, as well as refinable functions, have been widely used in numerical solutions of differential and integral equations, digital signal processing, image compression, and many others. Particularly, the refinable splines such asB-splines in ℝ and the box splines in ℝn play a key role in approximation theory and in computer-aided geometric design. In recent years, some researches on refinable splines have made great progress (see [7–15]).As we know, a refinable function with an analytic expression is of great importance, especially in those requiring high-precision domains. However, refinable functions are solutions of refinement equations and are usually defined by some cascade algorithms. Due to the iterative process, most of refinable functions do not have explicit analytic expression. In [9, 11], the authors have proved that there are no other piecewise smooth compactly supported refinable functions with explicit analytic expression than B-splines. Then much attention has been given in the literature to the componentwise polynomial since it is a polynomial on each connected component of the function’s support, and there are some preliminary researches on this field (see [16–20]). What is a spline in ℝn? And what is a componentwise polynomial in ℝn?Definition 1.1 (see [7]).
A compactly supported functionf(x) in ℝn with supp(f)=Ω is called a spline if there exists a partition Ω=⋃j=1NRj into simplices {Rj} in ℝn such that f is a polynomial on each {Rj}.Definition 1.2 (see [16]).
A compactly supported functionf:ℝn↦ℂ is a componentwise polynomialif there exists an open set G such that the Lebesgue measure of ℝn∖G is zero and the restriction of f on any connected open component of G coincides with some polynomial.Generally, the componentwise polynomial is not a spline; it is much broader than a spline. The difference between the componentwise polynomial and the spline is that the former’s support consists of infinitely many simplices, and the latter’s is just a finitely union of some simplices. In one dimension, there have been some good results on the componentwise polynomials (see [17–20]). The concept of componentwise polynomials in ℝ is first introduced in [17, 18] under the name of local polynomials, and the componentwise polynomial property of a one-dimensional refinable function has been well studied and demonstrated to be useful there. In particular, an iteration algorithm is given to compute the polynomial on each component (see [18]). In [19], a few examples of compactly supported refinable componentwise constant functions, which satisfy either orthogonality or interpolation property and which are continuous and symmetric, are given. Additional examples of refinable componentwise linear functions that are differentiable and symmetric are also given in [19]. In [20], the authors present another example of a componentwise constant function that is not locally integrable, and they further assert that any refinable componentwise polynomial function with the dilation factor 2 must be a finite linear combination of the integer shifts of some B-spline. In two dimensions, the authors study a class of compactly supported refinable componentwise constant functions and give a sufficient condition of the refinable function being a componentwise constant function, which is much harder than that in one dimension (see [16]). How about the refinable componentwise polynomial functions in ℝ2?In this paper, we present a sufficient condition of the compactly supported refinable functions being componentwise polynomials inℝ2. By using a reasonable partition of the functions’ support, we provide an iteration algorithm to compute the polynomial on each connected component of the functions’ support. Compared with [16], the algorithm has been improved broader than that in [16] since the technical condition  (7) in [16] can be removed.


## 2. Preliminaries and Main Results

Letf(x) be a compactly supported function in ℝ2, and let A∈M2(ℤ) be an expanding matrix; that is, the modules of all eigenvalues are greater than 1. For all h(k)∈ℝ with only finitely many h(k)≠0 and ∑k∈ℤ2h(k)=|detA|, we say that a measurable function f(x):ℝ2→ℂ is a measurable function solutionof the refinement equation
(2.1)f(x)=∑k∈Z2h(k)f(Ax-k),a.e.x∈R2,
if f(x) satisfies the refinement equation (2.1) in the sense of almost everywhere, where H(ξ)=(1/|A|)∑k∈ℤ2h(k)e-ikTξ is called the mask symbol of the refinable function f.Theorem 2.1.
Letg(x)∈L1(ℝ2) be a compactly supported function satisfying the refinement equation
(2.2)g(x)=∑k∈Z2c(k)g(Bx-k),a.e.x∈R2,
where B=diag(λ1,λ2). If the mask symbol of g(x) is given by
(2.3)H1(ξ)=[1-e-iλ1ξ1λ1(1-e-iξ1)]N1[1-e-iλ2ξ2λ2(1-e-iξ2)]N2Q(r1,r2)(ξ),
where Q(r1,r2)(ξ):=∑s=(0,0)(r1,r2)q(s)e-isTξ, Q(r1,r2)(0)=1, s=(s1,s2)T, ξ=(ξ1,ξ2)T, and 0<r1<λ1-1, 0<r2<λ2-1, λ1,λ2≥3, r1,r2,s1,s2,λ1,λ2,N1,N2∈ℤ. Then g(x) is a refinable componentwise polynomial function supported on (0,N1+r1/(λ1-1))×(0,N2+r2/(λ2-1)).In what follows we need Lemma2.2 formulated below.Lemma 2.2.
Letg(x) be defined as Theorem 2.1 and x=(x1,x2)T. Denote G(x1,x2):=(g(x1,x2),…,g(x1,x2+N2-1),…,g(x1+N1-1,x2),…,g(x1+N1-1,x2+N2-1))T. Then g(x) is a polynomial at most N1+N2-2 degree on ⋃j1=0N1-1⋃j2=0N2-1(j1+(r1/(λ1-1),1))×(j2+(r2/(λ2-1),1)).Proof.
DenoteΛ:=(0,N1+r1/(λ1-1))×(0,N2+r2/(λ2-1)). It is easy to check that
(2.4)supp⁡(g)⊆Λ.
Since g(x)∈L1(ℝ2), it follows from (2.3) that (∂j+sĝ/∂ξ1j∂ξ2s)(2n1π,2n2π)=0 holds for all (n1,n2)∈ℤ2-(0,0), 0≤j≤N1-1, 0≤s≤N2-1. Then by the following Poisson summation formula in ℝ2:
(2.5)∑k1,k2(x1+k1)j(x2+k2)sg(x1+k1,x2+k2)=∑n1,n2ij+s∂j+sĝ∂ξ1j∂ξ2s(2n1π,2n2π)ei2π(n1x1+n2x2),
we get that ∑k1,k2(x1+k1)j(x2+k2)sg(x1+k1,x2+k2)=mjs, 0≤j≤N1-1, 0≤s≤N2-1. Denote Ai(t):=ti((x2+j)k)0≤j,k≤N2-1 and M(x1,x2):=(Ai(x1+j))0≤i,j≤N1-1. Then according to (2.4), we get that
(2.6)M(x1,x2)G(x1,x2)=(m00,…,m0,N2-1,…,mN1-1,0,…,mN1-1,N2-1)T
when (x1,x2)∈(r1/(λ1-1),1)×(r2/(λ2-1),1). By calculating, it is not difficult to get that
(2.7)det⁡M(x1,x2)=|12⋯N2-1122⋯(N2-1)2⋮⋮⋮⋮12N2-1⋯(N2-1)N2-1|N1⋅(2!⋅3!⋅⋯⋅(N1-1)!)N2≠0.
Then according to (2.6) and (2.7), g(x) is a polynomial at most N1+N2-2 degree on ⋃j1=0N1-1⋃j2=0N2-1(j1+(r1/(λ1-1),1))×(j2+(r2/(λ2-1),1)).To accomplish Theorem2.1, an iteration method is carried out to the proof. For convenience, we use g∈P to denote that g is a (componentwise) polynomial. We first form a reasonable partition of the region Λ which can be divided into (2N1+1)(2N2+1) pairwise disjoint components, and then we prove that g∈P in all these components by Steps 1–4.Proof of Theorem2.1.
Let us divide the interval(0,N1+r1/(λ1-1)) into 2N1+1 parts with points of division 0=u1<u2<⋯<u2N1+1<u2N1+2=N1+r1/(λ1-1), where u2k+1=k, u2k+2=k+r1/(λ1-1), k=0,1,…,N1. Let us also divide the interval (0,N2+r2/(λ2-1)) into 2N2+1 parts with points of division 0=v1<v2<⋯<v2N2+1<v2N2+2=N2+r2/(λ2-1), where v2k+1=k, v2k+2=k+r2/(λ2-1), k=0,1,…,N2. This results in the partition of the region Λ into (2N1+1)(2N2+1) components. For convenience, we give for these components some notations as follows:
(2.8)A0′:=⋃j1=0N1-1⋃j2=0N2-1(j1+(0,r1λ1-1))×(j2+(0,r2λ2-1)),B0′:=⋃j1=0N1-1⋃j2=0N2-1(j1+(r1λ1-1,1))×(j2+(0,r2λ2-1)),C0′:=⋃j1=0N1-1⋃j2=0N2-1(j1+(0,r1λ1-1))×(j2+(r2λ2-1,1)),D0′:=⋃j1=0N1-1⋃j2=0N2-1(j1+(r1λ1-1,1))×(j2+(r2λ2-1,1)),E0′:=⋃j2=0N2-1(N1+(0,r1λ1-1))×(j2+(0,r2λ2-1)),F0′:=⋃j2=0N2-1(N1+(0,r1λ1-1))×(j2+(r2λ2-1,1)),G0′:=⋃j1=0N1-1(j1+(0,r1λ1-1))×(N2+(0,r2λ2-1)),H0′:=⋃j1=0N1-1(j1+(r1λ1-1,1))×(N2+(0,r2λ2-1)),I0′:=(N1+(0,r1λ1-1))×(N2+(0,r2λ2-1)).
Denote
(2.9)Ω(ϵ1,…,ϵk):=(∑i=1kϵiλ1i+r1(λ1-1)λ1k,∑i=1kϵiλ1i+1λ1k),Ω′(ϵ1′,…,ϵs′):=(∑j=1sϵj′λ2j+r2(λ2-1)λ2s,∑j=1sϵj′λ2j+1λ2s),
where ϵi∈{0,1,…,r1}, ϵj′∈{0,1,…,r2}, ϵk≠r1, ϵs′≠r2, and 1≤i≤k, 1≤j≤s,
(2.10)Ω∞:=⋃k=1∞⋃(ϵ1,…,ϵk-1)∈{0,1,…,r1}k-1⋃ϵk∈{0,1,…,r1-1}Ω(ϵ1,…,ϵk),Ω∞′:=⋃s=1∞⋃(ϵ1′,…,ϵs-1′)∈{0,1,…,r2}s-1⋃ϵs′∈{0,1,…,r2-1}Ω′(ϵ1′,…,ϵs′).
Applying Lemma 2.2, it follows that g(x)∈P in 𝒟0′. In the following, we mainly study the function g(x) in 𝒜0′,ℬ0′, and 𝒞0′ for the case j1=0, j2=0. In such case, we denote the three regions as 𝒜0, ℬ0, and 𝒞0, respectively.
Divide𝒜0 into 𝒜:=⋃k=1∞⋃s=1∞Ω(ϵ1,…,ϵk)×Ω′(ϵ1′,…,ϵs′).
Divideℬ0 into ℬ:=⋃i=r1λ1-2⋃s=1∞[(r1/λ1(λ1-1)+i/λ1,(i+1)/λ1)⋃((i+1)/λ1,(i+1)/λ1+r1/λ1(λ1-1))⋃(r1/λ1(λ1-1) + (λ1-1)/λ1,1)]×Ω′(ϵ1′,…,ϵs′).
Divide𝒞0 into 𝒞:=⋃k=1∞⋃j=r2λ2-2Ω(ϵ1,…,ϵk)×[(r2/λ2(λ2-1)+j/λ2,(j+1)/λ2)⋃((j+1)/λ2,(j+1)/λ2+r2/λ2(λ2 − 1))⋃(r2/λ2(λ2-1)+(λ2-1)/λ2,1)].
From [16, 17], the above division can ensure that all the components of 𝒜, ℬ, and 𝒞 are pairwise disjoint, respectively, and the areas of 𝒜, ℬ, and 𝒞 satisfy the equality 𝕊𝒜=𝕊𝒜0, 𝕊ℬ=𝕊ℬ0 and 𝕊𝒞=𝕊𝒞0. Therefore, we prove that g∈P in 𝒜, ℬ, and 𝒞, respectively.
Step 1.
Letbsr:=(r1/λ1(λ1-1)+r1/λ1,1)×Ω′(ϵ1′,…,ϵs′) be the sth row block of ℬ. Let b2i+1c:=(r1/λ1(λ1-1)+(r1+i)/λ1,(r1+i+1)/λ1)×Ω∞′, i=0,1,…,λ1-r1-1, be the (2i+1)th column block of ℬ; let b2i+2c:=((r1+i+1)/λ1,r1/λ1(λ1-1)+(r1+i+1)/λ1)×Ω∞′, i=0,1,…,λ1-r1-2, be the (2i+2)th column block of ℬ. And bij is the ith row and the jth column block of ℬ according to the defined rules of bsr,b2i+1c, and b2i+2c. According to Lemma 2.2, g(x) is a polynomial in (r1/(λ1-1),1)×(r2/(λ2-1),1). Then it follows from the refinement equation (2.2) that
(2.11)∀(x1λ1,x2λ2)∈b1,2i+1=(r1λ1(λ1-1)+r1+iλ1,r1+i+1λ1)×Ω′(ϵ1′),g(x1λ1,x2λ2)∈P,
where i=0,1,…,λ1-r1-1; ϵ1′∈{0,1,…,r2-1}.
As in [16], by (2.2), it can be shown that the components where g(x1/λ1,x2/λ2)∈P in bk,2i+1 are from those in bk-1r. And it is easy to check that the distribution of the components where g(x)∈P in bk,2i+1 is the same as that in bk-1r. For simplicity, we call the components where g(x) are unknown as the blanks.
In the following, we first consider the casei=0 since the components where g(x1/λ1,x2/λ2)∈P in bk,2i+1 can be gotten from the translation of those in bk1. Then we get the components where g(x1/λ1,x2/λ2)∈P in b1c:
(2.12)∀(x1λ1,x2λ2)∈{b11:[(r1λ1(λ1-1),1λ1)+r1λ1]×Ω′(ϵ1′);b21′:[(r1λ12(λ1-1),1λ12)+r1λ1+s2λ12]×Ω′(ϵ1′,ϵ2′)⊂b21;⋮bk1′:[(r1λ1k(λ1-1),1λ1k)+r1λ1+s2λ12+⋯+skλ1k]×Ω′(ϵ1′,…,ϵk′)⊂bk1,
where r1≤s2,…,sk-1≤λ1-1 and si∈ℤ, ϵj′ is defined as above, and all the components of b11,b21′,…,bk1′ are pairwise disjoint, respectively. The above formula (2.12) is different from formula (15) in [16] since we cannot get that g(x1/λ1,x2/λ2)∈P in b1,2i+2, i=0,1,…,λ1-r1-2. In fact, the coefficients of mask symbol in (2.3) do not satisfy condition  (7) in [16]; then, we just get g(x1/λ1,x2/λ2)∈P in b1,2i+1,i=0,1,…,λ1-r1-1. Therefore, the blanks in b1r which is b1,2i+2,i=0,1,…,λ1-r1-2, lead to the blanks in b21. Then b21′⊂b21. Step by step, the blanks in bk-1r lead to the blanks in bk1 and bk1′⊂bk1. For simplicity, we denote the area of the components where g(x)∈P in 𝒳 by S𝒳. Denoting the above total area in (2.12) by Sb1c, we obtain
(2.13)Sb1c=lim⁡k→∞(Sb11+Sb21′+⋯+Sbk1′)=λ1-r1-1λ1-1⋅λ2-r2-1λ2-1⋅r2λ1λ2-(r2+1)(λ1-r1).
Letckc:=Ω(ϵ1,…,ϵk)×(r2/λ2(λ2-1)+r2/λ2,1) be the kth column block of 𝒞. Let c2j+1r:=Ω∞×(r2/λ2(λ2-1)+(r2+j)/λ2,(r2+j+1)/λ2), j=0,1,…,λ2-r2-1, be the (2j+1)th row block of 𝒞; let c2j+2r:=Ω∞×((r2+j+1)/λ2,r2/λ2(λ2-1)+(r2+j+1)/λ2), j=0,1,…,λ2-r2-2, be the (2j+2)th row block of 𝒞. And cij is the ith row and the jth column block of 𝒞 according to the defined rules of ckc, c2j+1r, and c2j+2r. Symmetrically, we can get the total area of the components where g(x)∈P in c1r:
(2.14)Sc1r=lim⁡k→∞(Sc11+Sc12′+⋯+Sc1k′)=λ1-r1-1λ1-1⋅λ2-r2-1λ2-1⋅r1λ1λ2-(r1+1)(λ2-r2),
where c12′,…,c1k′ are the components where g(x)∈P in c12,…,c1k, respectively.Step 2.
Letasr:=Ω∞×Ω′(ϵ1′,…,ϵs′) be the sth row block of 𝒜. Let akc:=Ω(ϵ1,…,ϵk)×Ω∞′ be the kth column block of 𝒜. And aij is the ith row and the jth column block of 𝒜 according to the defined rules of asr and akc. Applying the same method in [16], we get the first incremental area of the components where g(x)∈P in 𝒜: Δ1S𝒜=(r2Sc1r+r1Sb1c-Sa11)/(1-q), where q=(r1+1)(r2+1)/λ1λ2, Sa11=((λ1-r1-1)/λ1(λ1-1))·((λ2-r2-1)/λ2(λ2-1))·r1r2. And it is easy to obtain that the first incremental area of the components where g(x)∈P in 𝒞 is Δ1S𝒞=(λ2-r2)Sc1r.Step 3.
Applying the refinement equation (2.2), it can be shown that the components where g(x)∈P in b12 can be got from those in 𝒞, and Sb12/S𝒞=r2/λ1λ2. Then the first incremental area of the components where g(x)∈P in b12 is Δ1Sb12=(r2/λ1λ2)Δ1S𝒞=(r2/λ1λ2)(λ2-r2)Sc1r. Similarly, Δ1Sb1,2i+2=Δ1Sb12, i=0,1,…,λ1-r1-2. Furthermore, the components where g(x)∈P in bj2 can be got from those in aj-1r,j=2,…,k, respectively. Since Sbj2/Saj-1r=(r2+1)/λ1λ2, we get the first incremental area of the components where g(x)∈P in bj2: Δ1Sbj2=((r2+1)/λ1λ2)Saj-1r. Similarly, Δ1Sbj,2i+2=Δ1Sbj2, i=0,1,…,λ1-r1-2. Therefore, we obtain the first incremental area of the components where g(x)∈P in b2c:
(2.15)Δ1Sb2c=lim⁡k→∞(Δ1Sb12+Δ1Sb22+⋯+Δ1Sbk2)=r2λ1λ2(λ2-r2)Sc1r+r2+1λ1λ2Δ1SA.
Symmetrically, we deduce that
(2.16)Δ1Sc2r=r1λ1λ2(λ1-r1)Sb1c+r1+1λ1λ2Δ1SA.
Since the components whereg(x)∈P in bj1 and the blanks in bj1 are one-to-one correspondence with those in bj-1r,j=2,…,k, respectively, then the parts of blanks in bj1 can be filled with Δbj-1,2i+21, i=0,1,…,λ1-r1-2, correspondingly, and the incremental components Δ1bj1 where g(x)∈P in bj1 from Δbj-1,2i+21 are disjoint with bj1′. Therefore, by (2.12), it follows that
(2.17)Δ1Sb21Δ1Sb12=Δ1Sb31Δ1Sb22=⋯=Δ1Sbk1Δ1Sbk-1,2=(r2+1)(λ1-r1-1)λ1λ2.
Similarly,
(2.18)Δ1Sb31Δ1Sb21=Δ1Sb41Δ1Sb31=⋯=Δ1Sbk1Δ1Sbk-1,1=(r2+1)(λ1-r1)λ1λ2.
Then we get the first incremental area of the components where g(x)∈P in b1c:
(2.19)Δ1Sb1c=lim⁡k→∞(Δ1Sb21+Δ1Sb31+⋯+Δ1Sbk1)=lim⁡k→∞Δ1Sb2c⋅(r2+1)(λ1-r1-1)λ1λ2[1+(r2+1)(λ1-r1)λ1λ2+⋯+((r2+1)(λ1-r1)λ1λ2)k-3]=[r2λ1λ2(λ2-r2)Sc1r+r2+1λ1λ2Δ1SA]⋅(r2+1)(λ1-r1-1)λ1λ2-(r2+1)(λ1-r1).
Symmetrically, we deduce that
(2.20)Δ1Sc1r=[r1λ1λ2(λ1-r1)Sb1c+r1+1λ1λ2Δ1SA]⋅(r1+1)(λ2-r2-1)λ1λ2-(r1+1)(λ2-r2).
It follows from Step 2 that the second incremental areas of the components where g(x)∈P in 𝒜 and 𝒞, respectively, are
(2.21)Δ2SA=r2Δ1Sc1r+r1Δ1Sb1c1-q,Δ2SC=(λ2-r2)Δ1Sc1r+(λ2-r2-1)Δ1Sc2r=(λ2-r2)⋅(r1+1)(λ2-r2-1)λ1λ2-(r1+1)(λ2-r2)Δ1Sc2r+(λ2-r2-1)Δ1Sc2r=Δ1Sc2r⋅λ1λ2(λ2-r2-1)λ1λ2-(r1+1)(λ2-r2).
Symmetrically, we deduce that
(2.22)Δ2SB=Δ1Sb2c⋅λ1λ2(λ1-r1-1)λ1λ2-(r2+1)(λ1-r1).Step 4.
Repeating Step3, we get the second incremental areas of the components where g(x)∈P in b2c and c2r, respectively:
(2.23)Δ2Sb2c=r2λ1λ2Δ2SC+r2+1λ1λ2Δ2SA,Δ2Sc2r=r1λ1λ2Δ2SB+r1+1λ1λ2Δ2SA.
Then
(2.24)Δ2Sb1c=Δ2Sb2c⋅(r2+1)(λ1-r1-1)λ1λ2-(r2+1)(λ1-r1),Δ2Sc1r=Δ2Sc2r⋅(r1+1)(λ2-r2-1)λ1λ2-(r1+1)(λ2-r2).
Therefore, we get the third incremental areas of the components where g(x)∈P in 𝒜 and 𝒞, respectively:
(2.25)Δ3SA=r2Δ2Sc1r+r1Δ2Sb1c1-q=r1r2(λ1-r1-1)(λ2-r2-1)[(r2+1)Δ1Sc2r+(r1+1)Δ1Sb2c](1-q)[λ1λ2-(r2+1)(λ1-r1)][λ1λ2-(r1+1)(λ2-r2)]+r2Δ1Sc1r+r1Δ1Sb1cλ1λ2(1-q)2[r1(r2+1)2(λ1-r1-1)λ1λ2-(r2+1)(λ1-r1)+r2(r1+1)2(λ2-r2-1)λ1λ2-(r1+1)(λ2-r2)],Δ3SC=(λ2-r2)Δ2Sc1r+(λ2-r2-1)Δ2Sc2r=[r1λ1λ2⋅Δ1Sb2c⋅λ1λ2(λ1-r1-1)λ1λ2-(r2+1)(λ1-r1)+r1+1λ1λ2⋅r2Δ1Sc1r+r1Δ1Sb1c1-q]×λ1λ2(λ2-r2-1)λ1λ2-(r1+1)(λ2-r2).
Furthermore, by calculating, we can obtain that the fourth incremental area of the components where g(x)∈P in 𝒜 is
(2.26)Δ4SA=r1r2(λ1-r1-1)(λ2-r2-1)(r2Δ1Sc1r+r1Δ1Sb1c)(1-q)[λ1λ2-(r2+1)(λ1-r1)][λ1λ2-(r1+1)(λ2-r2)]+r2Δ1Sc1r+r1Δ1Sb1cλ1λ2(1-q)2⋅2r1r2(r1+1)(r2+1)(λ1-r1-1)(λ2-r2-1)[λ1λ2-(r2+1)(λ1-r1)][λ1λ2-(r1+1)(λ2-r2)]+r1r2(λ1-r1-1)(λ2-r2-1)[(r2+1)Δ1Sc2r+(r1+1)Δ1Sb2c]λ1λ2(1-q)2[λ1λ2-(r2+1)(λ1-r1)][λ1λ2-(r1+1)(λ2-r2)]×[r1(r2+1)2(λ1-r1-1)λ1λ2-(r2+1)(λ1-r1)+r2(r1+1)2(λ2-r2-1)λ1λ2-(r1+1)(λ2-r2)]+r2Δ1Sc1r+r1Δ1Sb1c(λ1λ2)2(1-q)3⋅[r1(r2+1)2(λ1-r1-1)λ1λ2-(r2+1)(λ1-r1)+r2(r1+1)2(λ2-r2-1)λ1λ2-(r1+1)(λ2-r2)]2.
Step by step, applying mathematical induction, it can be shown that
(2.27)ΔnSA=r1r2(λ1-r1-1)(λ2-r2-1)[λ1λ2-(r2+1)(λ1-r1)][λ1λ2-(r1+1)(λ2-r2)][1+2(r1+1)(r2+1)λ1λ2(1-q)]Δn-2SA+1λ1λ2(1-q)[r1(r2+1)2(λ1-r1-1)λ1λ2-(r2+1)(λ1-r1)+r2(r1+1)2(λ2-r2-1)λ1λ2-(r1+1)(λ2-r2)]Δn-1SA(n≥4).
Since we always use the form of filling in the blank to calculate the incremental areas, there is no repeated area in Δ1S𝒜,Δ2S𝒜,…ΔnS𝒜. Then limn→∞∑k=1nΔkS𝒜≤𝕊𝒜. So we let S=limn→∞∑k=2nΔkS𝒜. According to the above recursive formula, it follows that
(2.28)S-(Δ2SA+Δ3SA)=r1r2(λ1-r1-1)(λ2-r2-1)[λ1λ2-(r2+1)(λ1-r1)][λ1λ2-(r1+1)(λ2-r2)]×[1+2(r1+1)(r2+1)λ1λ2(1-q)]S+1λ1λ2(1-q)×[r1(r2+1)2(λ1-r1-1)λ1λ2-(r2+1)(λ1-r1)+r2(r1+1)2(λ2-r2-1)λ1λ2-(r1+1)(λ2-r2)](S-Δ2SA).
Substituting the values of q, Δ2S𝒜, and Δ3S𝒜 into the above formula and simplifying, we get
(2.29){[r1(r2+1)2(λ1-r1-1)λ1λ2-(r2+1)(λ1-r1)+r2(r1+1)2(λ2-r2-1)λ1λ2-(r1+1)(λ2-r2)]1-λ1λ2+(r1+1)(r2+1)λ1λ2-(r1+1)(r2+1)⋅r1r2(λ1-r1-1)(λ2-r2-1)[λ1λ2-(r2+1)(λ1-r1)][λ1λ2-(r1+1)(λ2-r2)]-1λ1λ2-(r1+1)(r2+1)⋅[r1(r2+1)2(λ1-r1-1)λ1λ2-(r2+1)(λ1-r1)+r2(r1+1)2(λ2-r2-1)λ1λ2-(r1+1)(λ2-r2)]}S=λ1λ2λ1λ2-(r1+1)(r2+1)×{r2Δ1Sc1r+r1Δ1Sb1c+r1r2(λ1-r1-1)(λ2-r2-1)[(r2+1)Δ1Sc2r+(r1+1)Δ1Sb2c][λ1λ2-(r2+1)(λ1-r1)][λ1λ2-(r1+1)(λ2-r2)]}.
Again substituting the values of Δ1Sc1r, Δ1Sb1c, Δ1Sc2r, and Δ1Sb2c into the above formula and simplifying, we conclude that
(2.30)S=r1r2(λ1-r1-1)(λ2-r2-1)(λ1-1)(λ2-1)⋅m1m2,
where
(2.31)m1=r1(r2+1)(λ2-r2)(λ1-r1-1)+r2(r1+1)(λ1-r1)(λ2-r2-1)+r1r2(λ1-r1-1)(λ2-r2-1)[(r1+1)(λ2-r2)λ1λ2-(r1+1)(λ2-r2)+(r2+1)(λ1-r1)λ1λ2-(r2+1)(λ1-r1)]+1λ1λ2-(r1+1)(r2+1){λ1λ2[2λ1λ2-(r1+1)(λ2-r2)-(r2+1)(λ1-r1)]-[λ1λ2-(r1+1)(λ2-r2)][λ1λ2-(r2+1)(λ1-r1)]}×{r1(r2+1)2(λ1-r1-1)λ1λ2-(r2+1)(λ1-r1)+r2(r1+1)2(λ2-r2-1)λ1λ2-(r1+1)(λ2-r2)+2r1r2(r1+1)(r2+1)(λ1-r1-1)(λ2-r2-1)[λ1λ2-(r2+1)(λ1-r1)][λ1λ2-(r1+1)(λ2-r2)]r1(r2+1)2(λ1-r1-1)λ1λ2-(r2+1)(λ1-r1)},m2=[λ1λ2-(r1+1)(r2+1)][λ1λ2-(r2+1)(λ1-r1)][λ1λ2-(r1+1)(λ2-r2)]-r1r2(λ1-r1-1)(λ2-r2-1)[λ1λ2+(r1+1)(r2+1)]-r1(r2+1)2(λ1-r1-1)[λ1λ2-(r1+1)(λ2-r2)]-r2(r1+1)2(λ2-r2-1)[λ1λ2-(r2+1)(λ1-r1)].
Finally, we obtain all the incremental areas of the components where g(x)∈P in 𝒜:
(2.32)ΘSA=lim⁡n→∞(Δ1SA+Δ2SA+⋯+ΔnSA)=Δ1SA+S=r2Sc1r+r1Sb1c-Sa111-q+S.
Substituting the values of q, Sc1r, Sb1c, Sa11, and S into the above formula, it follows that ΘS𝒜=(r1/(λ1-1))·(r2/(λ2-1))=𝕊𝒜. This implies that all the blanks in 𝒜 can be filled by the above iteration method. And there is no repeated area in Δ1S𝒜,Δ2S𝒜,…,ΔnS𝒜.
Hence,g(x)∈P in 𝒜, and from the above steps, it can be shown that g(x)∈P in ℬ and 𝒞, respectively. It is obvious that 𝒜0′, ℬ0′, and 𝒞0′ can be obtained by j1 units of right translation and j2 units of up translation of 𝒜, ℬ, and 𝒞, respectively, where j1=0,…,N1-1, j2=0,…,N2-1. Then applying the refinement equation (2.2), g(x)∈P in 𝒜0′, ℬ0′, and 𝒞0′, respectively. Similarly, g(x)∈P in ℰ0′, ℱ0′, 𝒢0′, ℋ0′, and ℐ0′, respectively. Therefore, g(x) is a componentwise polynomial function supported on (0,N1+r1/(λ1-1))×(0,N2+r2/(λ2-1)).To make it easier to understand, the above algorithm is briefly summarized as follows.Step 1.
Applying the refinement equation (2.2), we get the areas of the components where g∈P in b1c and c1r: Sb1c and Sc1r.Step 2.
FromSb1c and Sc1r, we can get the first incremental areas of the components where g∈P in 𝒜 and 𝒞: Δ1S𝒜 and Δ1S𝒞.Step 3.
FromΔ1S𝒜 and Δ1S𝒞, we can get the first incremental area of the components where g∈P in b2c: Δ1Sb2c. Then again applying (2.2), we obtain the first incremental areas of the components where g∈P in b1c and c1r: Δ1Sb1c and Δ1Sc1r. Repeating Step 2, we get the second incremental areas of the components where g∈P in 𝒜 and 𝒞: Δ2S𝒜 and Δ2S𝒞.Step 4.
Repeating Step3, we get the third, the fourth,…, the nth incremental areas of the components where g∈P in 𝒜: Δ3S𝒜,Δ4S𝒜,…,ΔnS𝒜, and there is no repeated area in Δ1S𝒜,Δ2S𝒜,…,ΔnS𝒜. Finally, we can obtain all the incremental areas of the components where g∈P in 𝒜 and prove that the total area of the components is equal to the area of the region Λ.Applying Lemma  1 in [16] and Theorem 2.1, we conclude the following.Corollary 2.3.
Letf(x)∈L1(ℝ2) be a compactly supported function satisfying the refinement equation (2.1). If there exists a matrix P such that P-1AP=B, and f(x):=g(P-1x), where B and g(x) are defined as Theorem 2.1. Then f(x) is a compactly supported refinable componentwise polynomial function.


## 3. Examples

Example 3.1.
Letf(x)∈L1(ℝ2) be the refinable function with dilation matrix (3004), and its mask symbol is given by
(3.1)H(ξ1,ξ2)=[1-e-3iξ13(1-e-iξ1)]2[1-e-4iξ24(1-e-iξ2)]×16(1+e-iξ1+e-iξ2+e-iξ1e-iξ2+e-2iξ2+e-iξ1e-2iξ2).
Applying Lemma 2.2, it is easy to check that f(x)∈P in ⋃j1=01(j1+(1/2,1))×(2/3,1). Then, according to Theorem 2.1, f(x) is a componentwise polynomial function supported on (0,5/2)×(0,5/3). The refinable function f(x) is given in Figure 1.Figure 1
The symmetric refinable componentwise polynomial functionf.Example 3.2.
Letf(x)∈L1(ℝ2) be the refinable function with dilation matrix 3I2, where I2 is the identity matrix of order 2. And its mask symbol is given by
(3.2)H(ξ1,ξ2)=[1-e-3iξ13(1-e-iξ1)]2[1-e-3iξ23(1-e-iξ2)]2(14+14e-iξ1+14e-iξ2+14e-iξ1e-iξ2).
Applying Lemma 2.2, it is easy to check that f(x)∈P in ⋃j1=01(j1+(1/2,1))×⋃j2=01(j2+(1/2,1)). Then, according to Theorem 2.1, f(x) is a componentwise polynomial function supported on (0,5/2)×(0,5/2). The refinable function f(x) is given in Figure 2.Figure 2
The symmetric refinable componentwise polynomial functionf.



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*Source: 321428-2011-12-07.xml*