Construction of Compactly Supported Refinable Componentwise Polynomial Functions in R 2

and Applied Analysis 3 if f x satisfies the refinement equation 2.1 in the sense of almost everywhere, whereH ξ 1/|A| ∑k∈Z2h k e−ik T ξ is called the mask symbol of the refinable function f . Theorem 2.1. Let g x ∈ L1 R2 be a compactly supported function satisfying the refinement equation


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 as B-splines in R and the box splines in R n 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 R n ?And what is a componentwise polynomial in R n ?Definition 1.1 see 7 .A compactly supported function f x in R n with supp f Ω is called a spline if there exists a partition Ω N j 1 R j into simplices {R j } in R n such that f is a polynomial on each {R j }.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 R 2 ?
In this paper, we present a sufficient condition of the compactly supported refinable functions being componentwise polynomials in R 2 .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.

Preliminaries and Main Results
Let f x be a compactly supported function in R 2 , and let A ∈ M 2 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∈Z 2 h k | det A|, we say that a measurable function f x : R 2 → C is a measurable function solution of the refinement equation where Q r 1 ,r 2 ξ : In what follows we need Lemma 2.2 formulated below.
Lemma 2.2.Let g x be defined as Theorem 2.1 and x 2.4 Since Then by the following Poisson summation formula in R 2 : Then according to 2.4 , we get that Then according to 2.6 and 2.7 , g x is a polynomial at most N 1 N 2 − 2 degree on To accomplish Theorem 2.1, 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 2N 1 1 2N 2 1 pairwise disjoint components, and then we prove that g ∈ P in all these components by Steps 1-4.
This results in the partition of the region Λ into 2N 1 1 2N 2 1 components.For convenience, we give for these components some notations as follows: A 0 :

2.10
Applying Lemma 2.2, it follows that g x ∈ P in D 0 .In the following, we mainly study the function g x in A 0 , B 0 , and C 0 for the case j 1 0, j 2 0. In such case, we denote the three regions as A 0 , B 0 , and C 0 , respectively.Divide A 0 into A : 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 S A S A 0 , S B S B 0 and S C S C 0 .Therefore, we prove that g ∈ P in A, B, and C, respectively.
Step 1.Let b r s : Then it follows from the refinement equation 2.2 that As in 16 , by 2.2 , it can be shown that the components where And it is easy to check that the distribution of the components where g x ∈ P in b k,2i 1 is the same as that in b r k−1 .For simplicity, we call the components where g x are unknown as the blanks.
In the following, we first consider the case i 0 since the components where g x 1 /λ 1 , x 2 /λ 2 ∈ P in b k,2i 1 can be gotten from the translation of those in b k1 .Then we get the components where g x 1 /λ 1 , x 2 /λ 2 ∈ P in b c 1 : where r 1 ≤ s 2 , . . ., s k−1 ≤ λ 1 − 1 and s i ∈ Z, j is defined as above, and all the components of b 11 , b 21 , . . ., b k1 are pairwise disjoint, respectively.The above formula 2.12 is different from formula 15 in 16 since we cannot get that g In fact, the coefficients of mask symbol in 2.3 do not satisfy condition 7 in 16 ; then, we just get For simplicity, we denote the area of the components where g x ∈ P in X by S X .Denoting the above total area in 2.12 by S b c 1 , we obtain

2.13
Let 0, 1, . . ., λ 2 − r 2 − 2, be the 2j 2 th row block of C.And c ij is the ith row and the jth column block of C according to the defined rules of c c k , c r 2j 1 , and c r 2j 2 .Symmetrically, we can get the total area of the components where g x ∈ P in c r 1 :
Step 2. Let a r s : Ω ∞ × Ω 1 , . . ., s be the sth row block of A. Let a c k : Ω 1 , . . ., k × Ω ∞ be the kth column block of A. And a ij is the ith row and the jth column block of A according to the defined rules of a r s and a c k .Applying the same method in 16 , we get the first incremental area of the components where g x ∈ P in A: And it is easy to obtain that the first incremental area of the components where Step 3. Applying the refinement equation 2.2 , it can be shown that the components where g x ∈ P in b 12 can be got from those in C, and S b 12 /S C r 2 /λ 1 λ 2 .Then the first incremental area of the components where Furthermore, the components where g x ∈ P in b j2 can be got from those in a r j−1 , j 2, . . ., k, respectively.Since S b j2 /S a r j−1 r 2 1 /λ 1 λ 2 , we get the first incremental area of the components where g x ∈ P in b j2 : Therefore, we obtain the first incremental area of the components where g x ∈ P in b c 2 :

2.15
Symmetrically, we deduce that Since the components where g x ∈ P in b j1 and the blanks in b j1 are one-to-one correspondence with those in b r j−1 , j 2, . . ., k, respectively, then the parts of blanks in b j1 can be filled with Δ 1 b j−1,2i 2 , i 0, 1, . . ., λ 1 −r 1 −2, correspondingly, and the incremental components Δ 1 b j1 where g x ∈ P in b j1 from Δ 1 b j−1,2i 2 are disjoint with b j1 .Therefore, by 2.12 , it follows that 2.17 Similarly, Then we get the first incremental area of the components where g x ∈ P in b c 1 :

2.19
Symmetrically, we deduce that 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
Symmetrically, we deduce that

2.22
Step 4. Repeating Step 3, we get the second incremental areas of the components where g x ∈ P in b c 2 and c r 2 , respectively:

2.24
Therefore, we get the third incremental areas of the components where g x ∈ P in A and C, respectively:

2.25
Furthermore, by calculating, we can obtain that the fourth incremental area of the components where g x ∈ P in A is

2.26
Step by step, applying mathematical induction, it can be shown that

2.27
Since we always use the form of filling in the blank to calculate the incremental areas, there is no repeated area in According to the above recursive formula, it follows that

2.28
Substituting the values of q, Δ 2 S A , and Δ 3 S A into the above formula and simplifying, we get

2.29
Again substituting the values of Δ 1 S c r 1 , Δ 1 S b c 1 , Δ 1 S c r 2 , and Δ 1 S b c 2 into the above formula and simplifying, we conclude that where

2.31
Finally, we obtain all the incremental areas of the components where g x ∈ P in A:

2.32
Substituting the values of q, S r c 1 , S c b 1 , S a 11 , and S into the above formula, it follows that This implies that all the blanks in A can be filled by the above iteration method.And there is no repeated area in Δ 1 S A , Δ 2 S A , . . ., Δ n S A .
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 A 0 , B 0 , and C 0 can be obtained by j 1 units of right translation and j 2 units of up translation of A, B, and C, respectively, where j 1 0, . . ., N 1 − 1, j 2 0, . . ., N 2 − 1.Then applying the refinement equation 2.2 , g x ∈ P in A 0 , B 0 , and C 0 , respectively.Similarly, g x ∈ P in E 0 , F 0 , G 0 , H 0 , and I 0 , respectively.Therefore, g x is a componentwise polynomial function supported on 0, 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 b c 1 and c r 1 : S b c 1 and S c r 1 .
Step 2. From S b c 1 and S c r 1 , we can get the first incremental areas of the components where g ∈ P in A and C: Δ 1 S A and Δ 1 S C .
Step 3. From Δ 1 S A and Δ 1 S C , we can get the first incremental area of the components where g ∈ P in b c 2 : Δ 1 S b c 2 .Then again applying 2.2 , we obtain the first incremental areas of the components where g ∈ P in b c 1 and c r 1 : Δ 1 S b c 1 and Δ 1 S c r 1 .Repeating Step 2, we get the second incremental areas of the components where g ∈ P in A and C: Δ 2 S A and Δ 2 S C .
Step 4. Repeating Step 3, we get the third, the fourth,. .., the nth incremental areas of the components where g ∈ P in A: Δ 3 S A , Δ 4 S A , . . ., Δ n S A , and there is no repeated area in Δ 1 S A , Δ 2 S A , . . ., Δ n S A .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.Let f x ∈ L 1 R 2 be a compactly supported function satisfying the refinement equation 2.1 .If there exists a matrix P such that P −1 AP B, and f x : g P −1 x , where B and g x are defined as Theorem 2.1.Then f x is a compactly supported refinable componentwise polynomial function.

Definition 1 . 2
see 16 .A compactly supported function f : R n → C is a componentwise polynomial if there exists an open set G such that the Lebesgue measure of R n \ G is zero and the restriction of f on any connected open component of G coincides with some polynomial.
be the 2i 2 th column block of B. And b ij is the ith row and the jth column block of B according to the defined rules of b r s , b c 2i 1 , and b lead to the blanks in b 21 .Then b 21 ⊂ b 21 .Step by step, the blanks in b r k−1 lead to the blanks in b k1 and b k1 ⊂ b k1 .
R 2 , 2.1 if f x satisfies the refinement equation 2.1 in the sense of almost everywhere, where H ξ 1/|A| k∈Z 2 h k e −ik T ξ is called the mask symbol of the refinable function f.