Research Article | Open Access

Yanmei Xue, Ning Bi, "Construction of Compactly Supported Refinable Componentwise Polynomial Functions in ", *Abstract and Applied Analysis*, vol. 2011, Article ID 321428, 14 pages, 2011. https://doi.org/10.1155/2011/321428

# Construction of Compactly Supported Refinable Componentwise Polynomial Functions in

**Academic Editor:**Natig Atakishiyev

#### Abstract

We provide a sufficient condition for constructing a class of compactly supported refinable functions with componentwise polynomial property in . 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.

#### 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 as -splines in and the box splines in 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 -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 ? And what is a componentwise polynomial in ?

*Definition 1.1 (see [7]). *A compactly supported function in with supp is called *a spline* if there exists a partition into simplices in such that is a polynomial on each .

*Definition 1.2 (see [16]). *A compactly supported function is *a componentwise polynomial *if there exists an open set such that the Lebesgue measure of is zero and the restriction of on any connected open component of 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 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 -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 ?

In this paper, we present a sufficient condition of the compactly supported refinable functions being componentwise polynomials in . 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

Let be a compactly supported function in , and let be an expanding matrix; that is, the modules of all eigenvalues are greater than 1. For all with only finitely many and , we say that a measurable function is a *measurable function solution *of the refinement equation
if satisfies the refinement equation (2.1) in the sense of almost everywhere, where is called the *mask symbol* of the refinable function .

Theorem 2.1. *Let be a compactly supported function satisfying the refinement equation
**
where . If the mask symbol of is given by
**
where , , , , and , , , . Then is a refinable componentwise polynomial function supported on .*

In what follows we need Lemma 2.2 formulated below.

Lemma 2.2. *Let be defined as Theorem 2.1 and . Denote . Then is a polynomial at most degree on .*

*Proof. *Denote . It is easy to check that
Since , it follows from (2.3) that holds for all , , . Then by the following Poisson summation formula in :
we get that , , . Denote and . Then according to (2.4), we get that
when . By calculating, it is not difficult to get that
Then according to (2.6) and (2.7), is a polynomial at most degree on .

To accomplish Theorem 2.1, an iteration method is carried out to the proof. For convenience, we use to denote that is a (componentwise) polynomial. We first form a reasonable partition of the region which can be divided into pairwise disjoint components, and then we prove that in all these components by Steps 1–4.

*Proof of Theorem 2.1. *Let us divide the interval into parts with points of division , where , , . Let us also divide the interval into parts with points of division , where , , . This results in the partition of the region into components. For convenience, we give for these components some notations as follows:
Denote
where , , , , and , ,
Applying Lemma 2.2, it follows that in . In the following, we mainly study the function in , and for the case , . In such case, we denote the three regions as , , and , respectively.

Divide into .

Divide into + .

Divide into − .

From [16, 17], the above division can ensure that all the components of , , and are pairwise disjoint, respectively, and the areas of , , and satisfy the equality , and . Therefore, we prove that in , , and , respectively.*Step 1. *Let be the th row block of . Let , , be the th column block of ; let , , be the th column block of . And is the th row and the th column block of according to the defined rules of , and . According to Lemma 2.2, is a polynomial in . Then it follows from the refinement equation (2.2) that
where ; .

As in [16], by (2.2), it can be shown that the components where in are from those in . And it is easy to check that the distribution of the components where in is the same as that in . For simplicity, we call the components where are unknown as the blanks.

In the following, we first consider the case since the components where in can be gotten from the translation of those in . Then we get the components where in :
where and , is defined as above, and all the components of are pairwise disjoint, respectively. The above formula (2.12) is different from formula (15) in [16] since we cannot get that in , . In fact, the coefficients of mask symbol in (2.3) do not satisfy condition (7) in [16]; then, we just get in . Therefore, the blanks in which is , lead to the blanks in . Then . Step by step, the blanks in lead to the blanks in and . For simplicity, we denote the area of the components where in by . Denoting the above total area in (2.12) by , we obtain

Let be the th column block of . Let , , be the th row block of ; let , , be the th row block of . And is the th row and the th column block of according to the defined rules of , , and . Symmetrically, we can get the total area of the components where in :
where are the components where in , respectively.*Step 2. *Let be the th row block of . Let be the th column block of . And is the th row and the th column block of according to the defined rules of and . Applying the same method in [16], we get the first incremental area of the components where in : , where , . And it is easy to obtain that the first incremental area of the components where in is .*Step 3. *Applying the refinement equation (2.2), it can be shown that the components where in can be got from those in , and . Then the first incremental area of the components where in is . Similarly, , . Furthermore, the components where in can be got from those in , respectively. Since , we get the first incremental area of the components where in : . Similarly, , . Therefore, we obtain the first incremental area of the components where in :
Symmetrically, we deduce that

Since the components where in and the blanks in are one-to-one correspondence with those in , respectively, then the parts of blanks in can be filled with , , correspondingly, and the incremental components where in from are disjoint with . Therefore, by (2.12), it follows that
Similarly,
Then we get the first incremental area of the components where in :
Symmetrically, we deduce that
It follows from Step 2 that the second incremental areas of the components where in and , respectively, are
Symmetrically, we deduce that
*Step 4. *Repeating Step 3, we get the second incremental areas of the components where in and , respectively:
Then
Therefore, we get the third incremental areas of the components where in and , respectively:
Furthermore, by calculating, we can obtain that the fourth incremental area of the components where in is
Step by step, applying mathematical induction, it can be shown that
Since we always use the form of filling in the blank to calculate the incremental areas, there is no repeated area in . Then . So we let . According to the above recursive formula, it follows that
Substituting the values of , , and into the above formula and simplifying, we get
Again substituting the values of , , , and into the above formula and simplifying, we conclude that
where
Finally, we obtain all the incremental areas of the components where in :
Substituting the values of , , , , and into the above formula, it follows that . This implies that all the blanks in can be filled by the above iteration method. And there is no repeated area in .

Hence, in , and from the above steps, it can be shown that in and , respectively. It is obvious that , , and can be obtained by units of right translation and units of up translation of , , and , respectively, where , . Then applying the refinement equation (2.2), in , , and , respectively. Similarly, in , , , , and , respectively. Therefore, is a componentwise polynomial function supported on .

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 in and : and .

*Step 2. *From and , we can get the first incremental areas of the components where in and : and .

*Step 3. *From and , we can get the first incremental area of the components where in : . Then again applying (2.2), we obtain the first incremental areas of the components where in and : and . Repeating Step 2, we get the second incremental areas of the components where in and : and .

*Step 4. *Repeating Step 3, we get the third, the fourth,…, the th incremental areas of the components where in : , and there is no repeated area in . Finally, we can obtain all the incremental areas of the components where in and prove that the total area of the components is equal to the area of the region .

Applying Lemma 1 in [16] and Theorem 2.1, we conclude the following.

Corollary 2.3. *Let be a compactly supported function satisfying the refinement equation (2.1). If there exists a matrix such that , and , where and are defined as Theorem 2.1. Then is a compactly supported refinable componentwise polynomial function.*

#### 3. Examples

*Example 3.1. *Let be the refinable function with dilation matrix , and its mask symbol is given by
Applying Lemma 2.2, it is easy to check that in . Then, according to Theorem 2.1, is a componentwise polynomial function supported on . The refinable function is given in Figure 1.

*Example 3.2. *Let be the refinable function with dilation matrix , where is the identity matrix of order 2. And its mask symbol is given by
Applying Lemma 2.2, it is easy to check that in . Then, according to Theorem 2.1, is a componentwise polynomial function supported on . The refinable function is given in Figure 2.

#### Acknowledgments

The authors express their gratitude to the referees for providing valuable comments and suggestions leading to several improvements in the manuscript. Research of the authors is supported in part by NSFC 10631080 and Guangdong 2007B090400091.

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#### Copyright

Copyright © 2011 Yanmei Xue and Ning Bi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.