Ring-like structures of frequency domains of wavelets

Transcription

1 Ring-like structures of frequency domains of wavelets Zhihua Zhang and Naoki aito Dept. of Math., Univ. of California, Davis, California, 95616, UA. Abstract. It is well known that the global frequency domain Ω of any orthonormal wavelet has a hole which contains the origin, viz. the frequency domain Ω possesses a ring-like structure Ω = \ (0 ). We show that under some weak conditions, the set and the hole are determined uniquely by Ω, where the size of the hole satisfies 0 4 and the union of 4πν translations (ν Zd ) of is the whole space R d. Meanwhile, we give the corresponding converse theorem. We also show an interesting result: there is no orthonormal wavelet whose global frequency domain is the difference set of two balls. Finally, in order to explain our theory, we construct various global frequency domains and explain a general method of the construction of a wavelet with a given frequency domain. Key words: global frequency domain, orthonormal wavelet, regular set 1. Introduction A set Ψ = {ψ µ } d 1 1 of functions is called an orthonormal wavelet if their integral translations and dyadic dilations form an orthonormal basis of L (R d ). The orthonormal wavelets have widely applications in signal and image processing. The support of the Fourier transform of a function is called the frequency domain of this function. For an orthonormal wavelet Ψ = {ψ µ } d 1 1, the union of frequency domains of all ψ µ is called the global frequency domain of the orthonormal wavelet Ψ. The orthonormal wavelets with compact global frequency domains have an important property: their frequency domains have a hole which contains the origin. Precisely say, if the global frequency domain of an orthonormal wavelet {ψ µ } d 1 1 is bounded and for each µ, the Fourier transform of ψ µ satisfies some weak This research is partially supported by NF grant DM and the ONR grant N

2 conditions, then there exists a neighborhood D of the origin such that for each µ, ψµ (ω) = 0, ω D [1,,6]. From this, we see that the global frequency domain of an orthonormal wavelet possesses a ring-like structure. For example, for the tensor product hannon s wavelet, its global frequency domain is Ω = [ π, π] d \( π, π) d ; for the tensor product Meyer s wavelet, the global frequency domain is Ω = [ 8π 3, 8π 3 ]d \( π 3, π 3 )d [5,6]. Further researches on holes of frequency domains show [,4] that for an orthonormal wavelet ψ L (R), if its frequency domain Ω [ 8π 3, 8π 3 ], then ψ µ (ω) = 0, ω [ π 3, π 3 ]; if ψ(ω) = 0, ω ( π, π), then ψ is a hannon wavelet. For an orthonormal wavelet, the size of the global frequency domain Ω and the size of hole D which contains the origin are described as follows [6]. (i) If [A, B] d is the smallest cube containing Ω, then B A 4π. (ii) If (a, b) d is the largest cube which lies in D, then b a π. From the tensor product hannon wavelet, we know that the above conclusions (i) and (ii) cannot be improved. In general, there is no estimate of the lower bound of b a. For any ε > 0, there exists an orthonormal wavelet ψ L (R) such that the intersection of its frequency domain Ω and [ ǫ, ǫ] is a set with positive measure []. In this paper, we discuss ring-like structures of global frequency domains of orthonormal wavelets and determine the size and the position of the holes in global frequency domains. Precisely say, we show that if the global frequency domains Ω of orthonormal wavelets possess the following ring-like structures Ω = \ (0 ) and the origin is an interior point of the hole, then under some weak conditions, we have (i) the set can be determined by Ω: = m Ω; (ii) the hole can be determined by : \ ( + νπ); ν Z d \{0} (iii) the size of the hole satisfies 4 ; (iv) the union of the 4νπ translations of is just the whole space R d (ν Z d ).

3 As an example, for Meyer s wavelet, its frequency domain is Ω = \, where = [ 8π 3, 8π 3 ] and = [ π 3, π 3 ] satisfy the conditions (i)-(iv). Conversely, if and satisfy (ii)-(iv), then there exists an orthonormal wavelet Ψ with the global frequency domain Ω = \. Furthermore we research the case that is a convex region in the frequency domain Ω = \. When is a cuboid, the hole must be a cuboid. For ball, we show an interesting result: there is no orthonormal wavelet whose global frequency domain is the difference set of two balls. We present several examples at the end of this paper. From these examples, we see that according to our theory, one can construct orthonormal wavelets with various global frequency domains. In Example 5.5, we explain a general method of the construction of a wavelet with a given frequency domain. This paper is organized as follows. In ection, we recall some known concepts and results. In ection 3, we introduce the concepts of regular sets and the kernels of sets, and then we characterize ring-like structures of global frequency domains of orthonormal wavelets. In ection 4, we give many corollaries and converse theorems. In ection 5, we present several examples to explain our theory.. Preliminaries Throughout this paper, the closure, the boundary, and the interior of a set E R d are denoted by E, E, and E o, respectively. The diameter of E is defined as Diam E = sup{ x y ; x, y E}. Let E be a closed region. If a point ω 0 E is such that ω 0 + λ(ω ω 0 ) E (0 λ 1) for any ω E, then E is called a star region with the center ω 0. If, for arbitrary two points ω 1, ω E, ω 1 + λ(ω ω 1 ) E (0 λ 1), then E is called a convex region. For a set E R d, a R d, and λ R, denote E + a := {ω + a : ω E}, λe := {λω, ω E}, E + πz d = (E + πν), ν Z d and ν 0 means ν Z d \ {0}. For two sets E, F R d, E F means F \ E = 0; E F means the measure E \ F = F \ E = 0, where is the Lebesgue measure. 3

4 The support of a function f on R d is defined as suppf = {t R d f(t) 0}. We always assume f(t) 0 for a.e. t suppf. In this paper, for convenience, the notation a.e. is always omitted. Denote the Fourier transform of f L (R d ) by f: f(ω) = domain of f. R d f(t)e i(t ω) dt, ω R d. The support of f is called the frequency Below, we will recall some known concepts and results in the wavelet analysis. The set of functions Ψ = {ψ µ } d 1 1 in the space L (R d ) is called an orthonormal wavelet if the system { md ψµ ( m n)} m Z, n Zd, µ=1,..., d 1 is an orthonormal basis for L (R d ). The global frequency domain Ω of an orthonormal wavelet Ψ = {ψ µ } d 1 1 is defined as the union of frequency domains of all ψ µ, i.e., Ω = d 1 µ=1 supp ψ µ. Most of orthonormal wavelets are generated by multiresolution analyses. A multiresolution analysis (MRA) consists of a sequence of closed subspaces {V m } m Z of L (R d ) satisfying (i) V m V m+1 (m Z), (ii) V m = L (R d ), m Z (iv) f V m f( ) V m+1 (m Z), (iii) m Z V m = {0}, (v) there exists a ϕ V 0 such that {ϕ( n)} n Z d is an orthonormal basis of V 0. The function ϕ in (v) is called a scaling function [5,6]. For an orthonormal wavelet Ψ = {ψ µ } d 1 1, if there exists a MRA such that {ψ µ ( n)} n Zd, µ=1,..., d 1 is an orthonormal basis of the space W 0 (V 0 W0 = V 1 ), then Ψ is called a MRA wavelet, where is the orthogonal sum. It is well-known that there exist many non-mra orthonormal wavelets, for example, Journe wavelet [6]. The relation between MRA and orthonormal wavelet is as follows. Lemma.1 [8]. Let Ψ = {ψ µ } d 1 1 be an orthonormal wavelet with the bounded global frequency domain Ω. Denote F = m Ω. If, for each µ, ψ µ is continuous at every point on F Ω, then Ψ is a MRA wavelet. In the paper [8], we proved Lemma.1 in the case d = 1. For d > 1, the argument of Lemma.1 is similar. 4

5 Lemma. [6]. Let Ψ = {ψ µ } d 1 1 be a MRA wavelet derived by a scaling function ϕ. Then ϕ(ω) = d 1 ψ µ ( m ω), ω R d. µ=1 m 1 In [9], we gave a characterization of the frequency domains of scaling functions. Lemma.3 [9]. Let G be a bounded closed set. Then there is a scaling function ϕ with supp ϕ = G if and only if (i) G G, (ii) m Z m G R d, (iii) G + πz d R d, and (iv) (G \ G ) ( G + πν) (ν Zd ). We also need the following lemma. It gives a necessary condition of orthonormal wavelets. Lemma.4 [6]. Let Ψ = {ψ µ } d 1 1 be an orthonormal wavelet. Then d 1 ψ µ ( m ω) = 1, ω R d \ {0}. µ=1 m Z Based on these four lemmas, we will characterize the ring-like structures of global frequency domains of orthonormal wavelets in ection Ring-like structure of global frequency domains of orthonormal wavelets We first introduce the following concepts of regular sets and kernels of sets. Definition 3.1. Let a bounded closed set E be a union of finitely many disjoint closed regions of R d. If E satisfies the following two conditions (i) 0 E Eo and (ii) ω 0 + 8πν E for any ω 0 E and ν 0, then E is called a regular set of R d, where ν 0 means ν Z d \ {0}. For example, the set E 1 = [ 4π, π] [ 10π 4, 15π 4 ] [6π, 7π] is a regular set of R; the set E = [ π, π] ( [ 5π, 3π ] [ π, π] ) is a regular set of R. uppose that a closed star region F of R d with the center ω = 0 satisfies the conditions 0 F o and Diam F < 8π. Then F is a regular set of R d. 5

6 Definition 3.. Let E be a closed set of R d. The set E ker : = E o \ ν 0(E + πν) is called the kernel of E, i.e., E ker = {ω E o : ω + πν E, ν 0}. uppose that B is a closed ball of R d with the radius r. We can check that if r π, then B ker = ; if r π, then B ker = B o ; if π < r < π and d > 1, then B ker is a simply connected region but it is not a ball. uppose that E is a cube and E = [a, b] d. We can check that if b a 4π, then E ker = ; if b a π, then E ker = E o ; if π < b a < 4π, then E ker = (b π, π + a) d. Now we characterize the ring-like structures of global frequency domains of orthonormal wavelets. Theorem 3.3. Let Ψ = {ψ µ } d 1 1 be an orthonormal wavelet with the global frequency domain Ω \ o (0 o ), where is a regular set and 0 o. Again, let for each µ, ψ µ be continuous at every point on. Then (i) = m Ω, (ii) ( ), (iii) ker 4, (iv) + 4πZd R d. (3.1) Remark. In (3.1), (i) shows that is determined uniquely by Ω. (ii) shows that the hole is determined uniquely by. (iii) shows that the relative size of the hole in. (iv) shows that the union of the 4πν translations of covers the space R d (ν Z d ). By + 4πZ d R d, we conclude a known result: if a cube [A, B] d Ω, then B A 4π [6]. By the hole ( ) ker and the definition of the kernel, we conclude other known result: if a cube (a, b)d, then b a π [6]. Proof of Theorem 3.3. The proof is divided into six steps as follows. tep 1. We prove m Ω =. ince Ψ = {ψ µ } d 1 1 is an orthonormal wavelet with global frequency domain Ω, by Lemma.4, we have d 1 supp ψ µ ( m ω) = R d. (3.) µ=1 m Z 6

7 Again, by supp ψ µ ( m ω) = m supp ψ µ and the global frequency domain Ω = d 1 supp ψ µ, we conclude First, we claim. µ=1 m Ω = R d. (3.3) m Z Let E = \. ince E, by the assumption:, we obtain m E m (m 0). From this and Ω = \, we have m E Ω = (m 0). Furthermore, E m Ω = (m 0). (3.4) ince is a regular set, for m 1, we have m 1. ince E =, we have m E m 1 =. From this, we obtain m E =, so m E Ω =. Furthermore, E m Ω = (m 1). Combining this with (3.4), we have E ( ) m Ω =. Again, by (3.3), we have E =. o we obtain the claim m Z. (3.5) From (3.5) and Ω = \, we conclude that \ Ω. (3.6) Furthermore, we have ( m \ m 1 ) m Ω (m 0). Again, since is bounded and 0 o, we obtain m = {0} and \ {0} = \ m = ( m \ m 1 ) m Ω. ince m Ω m (m 0), we have \ {0} m Ω. o m Ω =. (3.7) tep. We prove = Ω. ince is a regular set, by (3.5), we have o, so =. From this, we know that the distance between and is greater than zero, i.e., δ = dist(, ) > 0. (3.8) 7

8 ince is a closed set and Ω = \, o we know that Ω is a closed set. Now we claim Ω =. (i) First we prove that Ω ( ). Let ω 0 Ω. By (3.8), we know that dist(ω 0, ) δ or dist(ω 0, ) δ. In the case dist(ω 0, ) δ. For any ball B(ω 0, δ) (δ < δ ), by ω 0 Ω, we have B(ω 0, δ). ince ω 0 Ω, there exists two points ω 1 and ω in B(ω 0, δ) such that ω 1 Ω o and ω Ω. From ω 1 Ω o and Ω = \, o we have ω 1. o From ω Ω, we have ω or ω. o Again by B(ω 0, δ), we get ω, o so ω 0. In the case dist(ω 0, ) δ. By ω 0 Ω Ω and Ω = \, o we have ω 0. o o any ball B(ω 0, δ) (δ < δ ) satisfies B(ω 0, δ) =. ince ω 0 Ω, there exist two points ω 1 and ω in B(ω 0, δ) such that ω 1 Ω o and ω Ω. From ω 1 Ω o and Ω, we have ω 1 o. From ω Ω and Ω = \, o we have ω or ω. o Again by B(ω 0, δ) =, we have ω, so ω 0. Now we have concluded that if ω 0 Ω, then ω 0 or ω 0, i.e., Ω ( ). (ii) Conversely, we prove that Ω ( ). Let ω 0. For any ball B(ω 0, δ) (δ < δ ), by (3.8), we have B(ω 0, δ) =. ince ω 0, there exist two points ω 1 and ω in B(ω 0, δ) such that ω 1 o, ω, so ω Ω. From ω 1 o and B(ω 0, δ) =, we have ω 1 Ω o, so ω 0 Ω. Let ω 0. For any ball B(ω 0, δ) (δ < δ ), by (3.8) and ω 0, we have B(ω 0, δ). ince ω 0, there exists two points ω 1 and ω in B(ω 0, δ) such that ω 1 o and ω. ince ω and B(ω 0, δ), we have ω \ o = Ω. By ω 1, o we have ω 1 Ω, so ω 0 Ω. Therefore, when ω 0 or ω 0, we have ω 0 Ω, i.e., Ω ( ). By (i) and (ii), we get the claim Ω =. By this claim and =, we get Ω =. (3.9) tep 3. Let G =. We prove that (Go \ G ) ( Go + πν) =, ν Zd. 8

9 By (3.7) and (3.9), we have = ( m Ω ) Ω. From this and the assumption that for each µ, ψ µ is continuous on, by Lemma.1, we conclude that Ψ is a MRA wavelet. However, By Lemma., the corresponding scaling function ϕ satisfies ϕ(ω) = m 1 d 1 µ=1 supp ϕ = m 1 d 1 µ=1 supp ψ µ ( m ) d 1 = m supp ψ µ ( m ). o supp ϕ = m Ω = 1 m Ω. Therefore, by (3.7), we obtain m 1 µ=1 supp ψ µ = m Ω. d 1 µ=1 ψ µ ( m ω). This implies supp ϕ = = G. (3.10) ince is a regular set and G =, we see that G Go, ω 0 + 4πν G (ω 0 G, ν 0). (3.11) These properties are used in the following argument. ince ϕ is a scaling function with the frequency domain G, by Lemma.3, we have (G \ G ) ( G + πν), ν Zd. Noticing that G = and is a union of finitely many disjoint closed regions, we know that G = 0. o (G o \ G ) ( Go + πν), ν Zd. ince the set of the left-hand side of this formula is an open set, the following precisely equality holds (G o \ G ) ( Go + πν) =, ν Zd. (3.1) tep 4. We further prove that G o G o ( + πν) =, ν 0. (3.13) 9

10 If (3.13) is not valid, then there exist two points ξ 1, ξ Go and ξ 1 ξ = πν 0 for some ν 0 0. Let L be a straight line passing though the points ξ 1 and ξ. Let points ω 1 and ω be the nearest points of intersection of line L and the boundary ( G ) away from ξ 1 and ξ, respectively. Again by (3.11), we have ω 1, ω ( G o ( G ) L ). Denote the distance between two points x and y by ρ(x, y). Without loss of generality, we assume ρ(ω 1, ξ 1 ) ρ(ω, ξ ). (a) uppose that ρ(ω 1, ξ 1 ) < ρ(ω, ξ ). ince ω 1 (G o ( G ) L), we can take ξ 1 near ω 1 such that ξ 1 G o, ξ 1 G, ξ 1 L, and ρ(ω 1, ξ 1) < ρ(ω, ξ ) ρ(ω 1, ξ 1 ). Let ξ = ξ 1 πν 0. By ξ = ξ 1 πν 0, we have ρ(ξ, ξ ) = ρ(ξ 1, ξ 1 ) ρ(ξ 1, ω 1 ) + ρ(ω 1, ξ 1) < ρ(ω, ξ ). Again, since ξ Go and ω is a nearest point of intersection of L and ( G ) away from ξ, we have ξ Go. From this and ξ 1 G o \ G, ξ + πν 0 = ξ 1, we have This is contrary to (3.1). ξ 1 ((G o \ G ) ( Go + πν 0)). (b) uppose that ρ(ω 1, ξ 1 ) = ρ(ω, ξ ). Let ξ 0 = ω 1 πν 0. Now we prove ξ 0 ( G ). In fact, if ξ 0 ( G ), then ξ0 G. However, ξ0 + 4πν 0 = ω 1 G, this is contrary to (3.11). By ξ = ξ 1 πν 0, we have ρ(ξ 0, ξ ) = ρ(ω 1, ξ 1 ) = ρ(ω, ξ ). Again, by ξ Go and the definition of ω, we have ξ 0 G. From this and ξ0 ( G ), we have ξ0 Go. ince ω 1 (G o ( G ) L) and ω 1 πν 0 = ξ o Go, we know that there exists a point ξ0 1 G o \ G near ω 1 such that ξ 0 1 πν 0 Go, i.e., ξ 0 1 ( (G o \ G ) ) ( Go + πν 0). 10

11 This is also contrary to (3.1). o (3.13) holds. tep 5. We will prove Ω \ ( ) ker. From (3.1) and (3.13), we obtain G o ( G o + πν ) =, ν 0. By G = 0, we obtain G ( G + πν), ν 0. Again by (3.10), we have ϕ(ω + πν) = 0 (ν 0) for ω G. Combining this with a necessary condition of the scaling functions [3]: we obtain ν Z d ϕ(ω + πν) = 1, ω R d, (3.14) ϕ(ω) = 1, ω G. (3.15) ince G ker : = {ω G o : ω + πν G, ν 0}, by supp ϕ = G and (3.14), we obtain G ker {ω R d : ϕ(ω) = 1}. (3.16) From this and (3.15), we have G G ker G. (3.17) Below we prove Ω = \ G ker. ince R d = (R d \ G) (G \ G) (G \ G ker ) G ker, we have Ω = R d Ω = ( (R d \ G) ) ( Ω (G \ G) ) ( Ω (G \ G ker ) ) ) Ω (G ker Ω. (3.18) By Lemma., we see that ϕ( ω ) ϕ(ω) = d 1 supp µ=1 ψ µ (ω), ω R d. o we have ( ϕ( ω ) d 1 ) ϕ(ω) = supp ψ µ = Ω. This implies that ω Ω if and only if ϕ( ω ) ϕ(ω). Based on this claim, we have (α) For ω (R d \ G), by G G and supp ϕ = G, we see that ϕ(ω) = ϕ( ω ) = 0, so, ω Ω, i.e., µ=1 (R d \ G) Ω. 11

12 (β) For ω (G \ G), by supp ϕ = G, we obtain ϕ(ω) = 0 and ϕ( ω ) 0, so ω Ω, i.e., (G \ G) Ω G \ G. (γ) For ω G \ G ker, by (3.15), we have ϕ( ω ) = 1. By (3.16), we have ϕ(ω) 1, so ω Ω, i.e., (G \ G ker ) Ω G \ G ker. (δ) For ω G ker, by (3.16), we have ϕ(ω) = 1. From (3.17), it follows that G ker G ker, further, ϕ( ω ) = 1, so ω Ω, i.e., G ker Ω. Combining (3.18) with (α)-(δ), noticing that G G and G ker G, we obtain Ω (G \ G) (G \ G ker ) G \ G ker. Again by G =, we have Ω \ ( ) ker. tep 6. We prove (3.1). By (3.7), we get (i). By the assumption, we have Ω \ and. Again, by Ω \ ( ) ker and ( ) ker, we get (ii). By (3.17) and G =, we get (iii). By (3.10), we know that is the frequency domain of the scaling function ϕ. Using Lemma.3, we have + πzd R d. o (iv) holds. Theorem 3.3 is proved. 4. ome corollaries and converse theorems In this section, as the corollaries of Theorem 3.3, we discuss a special case that the set is a convex region in the ring-like structure Ω = \ o of the frequency domain of an orthonormal wavelet. On the other hand, we give two converse theorems of Theorem ome corollaries 1

13 If is a bounded convex region and 0 o, then o, but may not be a regular set. However, if, in Theorem 3.3, we replace the condition is a regular set by the condition is a bounded convex region, then Theorem 3.3 is still valid. This is because the condition ω 0 + 8πν (ω 0, ν 0) is only used in argument of Formula (3.13). However when is a bounded convex region, G = is also a bounded convex region. In this case, (3.1) implies clearly (3.13) and this argument does not need the condition ω 0 + 8πν (ω 0, ν 0). o we have Corollary 4.1. In Theorem 3.3, if the condition is a regular set is replaced by the condition is a bounded convex region, then the conclusion (3.1) is still valid. In particular, we discuss the case that is a cuboid. Corollary 4.. Let = d [a i, b i ], and let Ψ = {ψ µ } d 1 1 be an orthonormal wavelet with global i=1 frequency domain Ω \ o (0 o ) and for each µ, ψ µ is continuous on. If the set satisfies 0 o, then (i) is also a cuboid and = d ( b i i=1 π, a i + π). (ii) For each i, we have 4π + b i a i < 0, 0 < b i 4π + a i, and 4π b i a i 16π 3. Proof. ince is a bounded convex region, by Corollary 4.1, we know that (3.1) holds. By +4πZ d R d and = d [a i, b i ], it follows that b i a i 4π for each i. By the definition of the kernel and = ( ) ker, we i=1 get (i). By 4 and 0 o, we have We get (ii). Corollary 4. is proved. b i 4 a i + π b i, a i b i π a i 4, a i < 0 < b i (i = 1,..., d). It is natural to ask whether there exists an orthonormal wavelet Ψ with the global frequency domain Ω = \ (0 o ), where and are both balls. Corollary 4.3. For d > 1, there is no orthonormal wavelet Ψ = {ψ µ } d 1 1 with the global frequency domain Ω \ o (0 o ), where and are both balls and ψ µ is continuous on for each µ. Proof. uppose that there exists an orthonormal wavelet Ψ such that the above conditions hold. Using 13

14 Corollary 4.1, we can obtain (3.1). From + 4πZ d R d, we know that Diam > 4π. o the radius of the ball 1 is greater than π. Furthermore, by the definition of the kernel, we know that = ( ) ker is not a ball in R d. This is a contradiction. Corollary 4.3 is proved. 4.. Converse theorems Now we discuss the converse theorems of Theorem 3.3. We first prove the following: Theorem 4.4. uppose that is a bounded closed set and 0 o,. Again if the set satisfies the conditions 4 ( ) ker and + 4πZd R d, then there exists an orthonormal wavelet Ψ with the global frequency domain Ω \ ( ) ker. Proof. Let G =. We will check that G satisfies four conditions of Lemma.3. By the definition of the kernel, we know that ( ) ker = o \ ( ) ker ( ν 0 + πν). This implies that ( + πν) =, ν 0. (4.1) By the known condition 4 ( ) and G = ker, we have G ( ) ker. Again, by (4.1), G (G + πν) = (ν 0). Furthermore, we have G ( G + πν) = 0, ν 0. This implies that (G \ G ) ( G + πν) = 0, ν Zd. (4.) ince 0 o, we know that ω = 0 is an interior point of G. o there exists a ball B with the center ω = 0 such that B G. From m Z m B = R d, we have m Z m G = R d. (4.3) By and + 4πZd R d, we have G G and G + πz d R d. From this and (4.)-(4.3), by using Lemma.3, we know that there exists a scaling function ϕ with supp ϕ = G =. Let Ψ be an orthonormal wavelet generated by the scaling function ϕ. Noticing that G ( G +πν) = 0 (ν 0), similar to the arguments of tep 5 in Theorem 3.3, we conclude that the global frequency domain of Ψ is Ω \ ( ) ker. Theorem 4.4 is proved. 14

15 From Theorem 4.4, we get the converse theorem of Theorem 3.3 immediately. Theorem 4.5. Let be a regular set. If the sets and satisfy the conditions (ii)-(iv) in (3.1), then there exists an orthonormal wavelet Ψ with the global frequency domain Ω = \. o Observing the argument of Theorem 3.3, we see that in tep 1 and tep, we have proved that = m Ω and = Ω. Based on these two conclusions, in tep 3, tep 4, and tep 5, we derived the conclusions (ii)-(iv) in (3.1) by using the conditions: is a regular set and each ψ µ is continuous on. o we get another converse theorem. Theorem 4.6. Let Ψ = {ψ µ } d 1 1 be an orthonormal wavelet with the global frequency domain Ω. Denote = m Ω. If is a regular set, = Ω, and for each µ, ψ µ is continuous on, then the global frequency domain has the form Ω \ ( ) ker, where 4 ( ) ker, + 4πZd R d. As an example, it is easy to check that the tensor product Meyer s wavelet satisfies the condition of Theorem 4.6. In fact, its Fourier transform is infinitely differentiable and its global frequency domain is Ω = [ 8π 3, 8π 3 ]d \ ( π 3, π 3 )d. By m Ω = [ 8π 3, 8π 3 ]d \ {0}, we have = m Ω = [ 8π 3, 8π 3 ]d. o is a regular set. Again by Ω = ( [ 8π 3, 8π 3 ]d ) ( [ π 3, π 3 ]d ), we have = Ω. 5. Examples In this section, we will construct various global frequency domains and explain a method of the construction of an orthonormal wavelet with a given frequency domain. From Theorem 4.4, we see that if a bounded closed set satisfies the conditions 0 4 ( ) ker o and + 4πZ d R d, 15

16 then Ω = \ ( ) ker is a global frequency domain of some orthonormal wavelet. In the following examples, we start from a union of closed regions satisfying the above conditions. Removing a hole ( ) ker in the set, we obtain \ ( ) ker which must be the global frequency domain of some orthonormal wavelet. Example 5.1. Let be a union of two rectangles of R : = [ π, π] ( [ 5π, 3π ] [ π, π] ). Then we have 0 o and + 4πZ = R. A direct calculation shows that the kernel ( ) ker = ( π, π) \ ( [ 3π 4, π ] [ π, π] ) = ( ( π, 3π 4 ) ( π, π) ) ( ( π, π) ( π, π) ) is a union of two rectangles where the rectangle ( π, π) ( π, π) contains the origin. From this we get 4 ( ) ker. Therefore satisfies the conditions of Theorem 4.4. By using Theorem 4.4, there exists an orthonormal wavelet Ψ = {ψ µ } 3 1 with the global frequency domain Ω 1 = \ ( ) ker. Furthermore, we have Ω 1 = ( [ π, π] \ ( ) ) ker ([ 5π ), 3π] [ π, π] = : Ω 1 Ω 1, where Ω 1 is a triply connected region and Ω 1 is a rectangle. Example 5.. Denote a disk with the center ω 0 = (3π, 3π ) and the radius π 4 by B(ω π 0, 4 ). Let be a union of the square [ π, π] and the closed disk B(ω 0, π 4 ). Then 0 o,, and + 4πZ = R. A direct calculation shows that the kernel ( ( ) ker = ( π, π) ω0 \ B π(1, 0), π ) 8 is a doubly connected region and 4 ( ) ker. From this, we see that satisfies the conditions of Theorem 4.4. o there exists an orthonormal wavelet Ψ with the global frequency domain Ω = \ ( ) ker = ( [ π, π] \ ( π, π) ) ( B( ω 0 π(1, 0), π 8 ) B(ω 0, where Ω is the difference set of two squares and Ω is a union of two disks. π ) 4 ) =: Ω Ω, 16

17 Example 5.3. Let R be a trapezoid with two curved sides: = { (x, y) R : π x π, f(x) y f(x) + 14π 3 }, where f is a continuous function and satisfies 7π 3 f(x) π ( π x π). Then { = (x, y) R : π x π, f(x) y f(x) + 7π 3 }. From 7π 3 f(x) π ( π x π), we have [ π, π] [ 7π 6, 4π 3 ] and [ π, π] [ π, 7π 3 ]. o 0 o and. A calculation shows that the kernel of 1 is ( ) ker = { (x, y) R : π < x < π, f(x) + π 3 < y < f(x) } + π. Again, by the definition, we have ( 4 [ π ) (, π ] [ 7π 1, 4π 6 ] [ π ), π ] [ 8π 1, 5π 6 ] ( ) ker. ince 7π 3 f(x) π ( π x π), we have 7π 3 f(x) + 14π 3 8π 3 ( π x π). This implies [ π, π] [ π, 7π 3 ] [ π, π]. o + 4πZ = R. By Theorem 4.4, we see that there exists an orthonormal wavelet Ψ = {ψ µ } 3 1 such that its global frequency domain is in the doubly connected domain Ω 3 = \ ( ) ker. Example 5.4. Let be a union of three closed intervals [ 5π, π], [ 7π 4, 7π 4 [ = 5π4 ] [, π 7π ] 8, 7π [ π, 8 ] 5π. 4 ], and [π, 5π ]. Then o 0 o,, and +4πZ = R. It is easy to check that the kernel ( ) ker = ( 9π 8, π) ( 3π 4, 3π 4 ) (π, 9π 8 ) satisfies 1 4 ( 3π 4, 3π 4 ) ( ) ker. By Theorem 4.4, there exists an orthonormal wavelet ψ with the frequency domain Ω 4 = \ ( ) ker = [ 5π, π] [ 7π 4, 9π 8 ] [ π, 3π 4 ] [ 3π 4, π] [ 9π 8, 7π 4 ] [π, 5π ]. 17

18 Now we present a typical example to explain a method of the construction of an orthonormal wavelet with a given frequency domain. Example 5.5. Define a simply connected closed region R by = [ π, π] ( 4 1 : = { (x, y) R π x π, π } 0π x y π, ν=1 ν ), where : = { (x, y) R π } 0π y x π, π y π, 3 : = 1 + 4π(0, 1), 4 : = + 4π(1, 0). (5.1) We see that each i is a region whose boundary is a side of the square [ π, π] and a circular arc through two ends of this side. (i) Check that satisfies the conditions of Theorem 4.4. ince 0 o and is a closed star region with the center ω = 0, we know that. A direct calculation shows that ( ) ker = o \ 4 ν=1 E ν and 4 ν=1 E o ν o, (5.) where E 1 : = { (x, y) R π x π, π 5π x y 3π + 5π x } E : = { (x, y) R π 5π y x 3π + 5π y, π y π } E 3 : = E 1 + π(0, 1), E 4 : = E + π(1, 0). We see easily that each E i is a region whose boundary is two circular arcs through two vertexes of a side of the square [ π, π]. By (5.1) and (5.), we know that ( ) ker is a simply connected region and + 4πZ = R. By the definition of, we know that 4 [(1 5)π, ( 5 1)π ] [( 5 3)π, (3 5)π] ( ) ker. (5.3) By Theorem 4.4, we know that there exists an orthonormal wavelet Ψ with the global frequency domain Ω = \ ( ) ker. Now we discuss how to construct such a wavelet Ψ. 18

19 (ii) Construct a scaling function ϕ such that supp ϕ =. Define ν(x, y) as a harmonic function satisfying ν x + ν y = 0 ((x, y) E o 1) with boundary conditions ν(x, y) = 0, π x π, y = 3π + 5π x, ν(x, y) = π, π x π, y = π 5π x. (5.4) By (5.), we have R = ( ) ( ) 4 ( ) ker E ν R \. ν=1 Define ϕ such that ϕ(ω) = 1 (ω ( ) ker), ϕ(ω) = 0 (ω R \ ), (5.5) and cos ν(x, y), ω E 1, cos ν(y, x), ω E, ϕ(ω) = where ω = (x, y). (5.6) sin ν(x, y π), ω E 3, sin ν(y, x π), ω E 4, Here ν(x, y) has been defined on E 1. By the definition, we know that for (x, y) E, we have (y, x) E 1 ; for (x, y) E 3, we have (x, y π) E 1 ; for (x, y) E 4, we have (x π, y) E, so (y, x π) E 1. Hence ϕ(x, y) is well-defined on 4 E ν. From this and (5.4)-(5.6), we know that ϕ(ω + πk) = 1, ω R and ν=1 k Z ϕ(ω) is a continuous function in R except four points ( π, π), ( π, π), (π, π), and (π, π). Define H 0 (ω) as 0, ω \ 4, H 0 (ω) = ϕ(ω), ω 4, H 0 (ω + πν) = H 0 (ω) (ω R, ν Z ). (5.7) ince 4 [ π, π], H 0 is well-defined. By (5.4), we have 4 ( ) ker. Again by (5.5), we have ϕ(ω) = 1 (ω ). From this and (5.7), we obtain that ϕ(ω) = H 0 (ω) ϕ(ω), ω R and H 0 is a bounded, πz periodic function. Again, since ϕ is continuous at ω = 0 and ϕ(0) = 1, by a known result [3], we know that ϕ is a scaling function. ince = ( ) 4 ker ( E ν ), ν=1 by (5.5) and (5.6), we have supp ϕ =. 19

20 (iii) Construct an orthonormal wavelet Ψ with the global frequency domain Ω = \ ( ) ker. Let the scaling function ϕ and the filter H 0 be stated in (5.5)-(5.7). By a known result [7], we obtain a corresponding orthonormal wavelet Ψ = {ψ µ } 3 1 satisfying ψ µ (ω) = H µ ( ω ) ( ω ) ϕ (µ = 1,, 3), (5.8) where H 1 (ω) = e iω H 0 (ω + π(0, 1)), H (ω) = e i(ω 1+ω ) H 0 (ω + π(1, 0)), H 3 (ω) = e iω 1 H 0 (ω + π(1, 1)) (ω = (ω 1, ω ) R ). ince supp ϕ = and supph 0 = 4 + πz, by (5.1) and (5.8), we obtain supph 1 = 4 π(0, 1) + πz supp ψ ) ( ) 1 = (supp H 1 supp ϕ = π(0, 1) + 4πZ = A, where A = ( ( + π(0, 1)) ( + π(0, 1)) ). imilarly, we have supp ψ = B and supp ψ 3 = C, where B = ( + π( 1, 0)) ( + π(1, 0)), C = ( + π( 1, 1)) ( + π(1, 1)) ( + π( 1, 1)) ( + π(1, 1)). Hence the global frequency domain of Ψ is Ω = 3 supp ψ µ = (A B C). (5.9) µ=1 ince [ π, π], we have (A B C) ( [ 3π, 3π] \ ( π, π) ). Again, by (A B) ( 4 [ 3π, 3π], we obtain ((A B C) ) ( ( \ ( π, π) ) ( 4 ( π, π) ( ) ker, we have Ω = ν=1 ν=1 E ν ) and E ν ) ). Again, by (5.) and o ( (A B C) ) ( ( \ ( π, π) ) ) ( o \ ( ) ker) \ ( ) ker. (5.10) 0

21 On the other hand, since Ω and (A B C) ( ) ker, we have Ω \ ( ) ker. From this, by (5.9) and (5.10), we get Ω \ ( ) ker. References [1] Bonami, A., oria, F. and Weiss, G., Band-limited wavelets, J. Geom. Anal. 3 (1993) [] Brandolini, L., Garrigos, G., Rzeszotnik, Z. and Weiss, G., The behaviour at the origin of a class of band-limited wavelets, Contemp. Math. 47 (1999) [3] Daubechies, I., Ten lectures on wavelets, CBM- Conference Lecture Notes, IAM, Philadelphia 6 (199) [4] Hernandez, E., Wang, X. and Weiss, G., moothing minimally supported frequency wavelets I, J. Fourier Anal. Appl. (4) (1996) [5] Hernandez, E. and Weiss, G., A first course on wavelet CRC Press (1996). [6] Long, R. L., Multidimensional wavelet analysis, International Publishing Co., Beijing (1995). [7] Riemenschneider,. D., and hen, Z. W., Wavelets and prewavelets in low dimensions, J. Approx. Theory 71 (199) [8] Zhang, Z., Continuity of Fourier transforms of band-limited wavelets, J. Comp. Anal. Appl. 9(4) (007) [9] Zhang, Z., upports of Fourier transforms of scaling functions, Appl. Comput. Harmonic Anal. () (007) [10] Zhou, D. X., Construction of real-valued wavelets by symmetry, J. Approx. Theory, 81(3) (1995)