a尺度二维四向双正交多小波的构造和Mallat算法 The Construction and the Mallat Algorithm of Biorthogonal Two Dimensional Four-Direction Multi-Wavelet with Dilation Factor a

doi:10.12677/PM.2019.99128

Pure Mathematics
Vol. 09 No. 09 ( 2019 ), Article ID: 32935 , 21 pages
10.12677/PM.2019.99128

The Construction and the Mallat Algorithm of Biorthogonal Two Dimensional Four-Direction Multi-Wavelet with Dilation Factor a

Xu Zhang, Wanshe Li^*

●How to Cite this Article

School of Mathematics and Information Science, Shaanxi Normal University, Xi’an Shaanxi

Received: Oct. 18^th, 2019; accepted: Nov. 6^th, 2019; published: Nov. 13^th, 2019

ABSTRACT

Through the concept of two dimensional four-direction multi-scaling function and two dimensional four-direction multi-wavelet, the two dimensional four-direction orthogonal multi-wavelet with dilation factor two is generalized to two dimensional biorthogonal multi-wavelet with dilation factor a. Furthermore, the construction algorithm of two dimensional biorthogonal multi-scaling function and multi-wavelet with dilation factor a is given. Finally, the Mallat algorithm of two dimensional four-direction multi-wavelet with dilation factor a is studied.

Keywords:Two Dimensional Four-Direction Multi-Scaling Function, Two Dimensional Four-Direction Multi-Wavelet, Biorthogonality, Mallat Algorithm

a尺度二维四向双正交多小波的构造和Mallat算法

张旭，李万社^*

陕西师范大学，数学与信息科学学院，陕西西安

收稿日期：2019年10月18日；录用日期：2019年11月6日；发布日期：2019年11月13日

摘要

通过二维四向多加细函数以及二维四向多小波的概念，推广二尺度二维四向正交多小波为a尺度二维四向双正交多小波，更进一步对于a尺度二维四向双正交多加细函数和多小波的构造算法做出了相应给出，最后研究了a尺度二维四向多小波的Mallat算法。

关键词 :二维四向多加细函数，二维四向多小波，双正交性，Mallat算法

This work is licensed under the Creative Commons Attribution International License (CC BY).

http://creativecommons.org/licenses/by/4.0/

1. 引言

小波分析是这些年来发展起来的一门新兴数学理论以及方法，在信号处理，语音处理，图像处理，数据压缩，微分方程求解，地震勘探等各个领域有着广泛的应用。Haar小波是同时具有正交性，对称性和紧支撑性的单小波，但是其它单小波并不能具有这样好的性质，所以人们引入了多小波。从带宽来看，二尺度小波高频端的带宽比较窄，那么从小波分析的效果来看二尺度小波效果相对比较差，所以人们提出了a尺度小波。双向小波的概念是杨守志等人首先提出的 [1]，后来进一步得出了一系列好的理论和结果 [2] [3]。本文在引入二维四向多小波基础上，建立了a尺度二维四向多加细函数和a尺度二维四向多小波，给出了a尺度二维四向多加细函数和a尺度二维四向多小波的正交和双正交准则，以及它们的构造算法，最后讨论了a尺度二维四向多小波的分解与重构的Mallat算法。

2. 预备知识

先给出文章要提到的记号： $C^{N}$ 表示N维复欧几里德空间。 $I$ 表示 $r \times r$ 阶单位矩阵， $O$ 表示 $r \times r$ 阶零矩阵， $T$ 表示向量或矩阵的转置，向量值函数信号空间 $L^{2} (R, C^{N}) ≜ L^{2}$ 可以表示为 $L^{2} (R, C^{N}) ≜ {F (t) = {(f_{1} (t), f_{2} (t), \dots, f_{N} (t))}^{T} : t \in R, f_{k} (t) \in L^{2} (R), k = 1,2, \dots, N}$ ，(根据文献 [4] )对于 $F (t) \in L^{2} (R, C^{N})$ 它的积分和Fourier变换分别定义为

$\int_{R} F (t) d t ≜ {[\int_{R} f_{1} (t) d t, \int_{R} f_{2} (t) d t, \dots, \int_{R} f_{N} (t) d t]}^{T}$

和

$F (ω) ≜ {[\int_{R} f_{1} (t) e^{- i ω t} d t, \int_{R} f_{2} (t) e^{- i ω t} d t, \dots, \int_{R} f_{N} (t) e^{- i ω t} d t]}^{T}$

若F和G都是一元函数空间，它们两个的基底分别为 ${f_{k} (x)}_{k \in Z}$ 和 ${g_{k} (x)}_{k \in Z}$ ，二元函数空间H表示为 $H = F \otimes G$ 是F和G的张量积空间，H的基底可以表示为 ${f_{k} (x) g_{k} (x)}_{k \in Z}$ 。

定义 ${V_{j}}$ 的张量积空间为 $V_{j}^{2} = V_{j} \otimes V_{j} = {f_{k} (x) g_{k} (x)}_{f (x), g (x) \in V_{j}}$ ，而 ${V_{j}}$ 表示为由一元尺度函数 $φ$ 生成的一个正交多分辨分析。那么关于二元函数 $f (x, y) \in L^{2} (R^{2})$ ，引入记号 $φ (x, y) = φ (x) φ (y)$ 。

二维四向加细尺度函数

基于双向尺度函数的概念，现在假设有 $r + 1$ 个双向尺度函数 $φ_{1} (x), φ_{2} (x), \dots, φ_{r} (x), φ (y) \in L^{2} (R)$ ，记 $Φ (x) = {[φ_{1} (x), φ_{2} (x), \dots, φ_{r} (x)]}^{T}$ ，那么通过 $Φ (x)$ 和 $φ (y)$ 的张量积来构造二维四向多尺度函数。假设双向加细函数 $φ_{i} (x)$ 和 $φ (y)$ 都符合细分方程

$φ_{i} (x) = \sum_{k} P_{1, k}^{+} φ_{i} (a x - k) + \sum_{k} P_{1, k}^{-} φ_{i} (k - a x), i = 1, \dots, r$

$φ (y) = \sum_{k} P_{2, k}^{+} φ (a y - k) + \sum_{k} P_{2, k}^{-} φ (k - a y)$

设 $Φ (x, y) = φ_{i} (x) φ (y)$ ，就能得出

$\begin{matrix} Φ (x, y) = {\sum_{k} P_{1, k}^{+} φ_{i} (a x - k) + \sum_{k} P_{1, k}^{-} φ_{i} (k - a x)} \otimes {\sum_{l} P_{2, l}^{+} φ (a y - l) + \sum_{l} P_{2, l}^{-} φ (l - a y)} \\ = \sum_{k, l} P_{1, k}^{+} P_{2, l}^{+} φ_{i} (a x - k) φ (a y - l) + \sum_{k, l} P_{1, k}^{+} P_{2, l}^{-} φ_{i} (a x - k) φ (l - a y) \\ + \sum_{k, l} P_{1, k}^{-} P_{2, l}^{+} φ_{i} (k - a x) φ (a y - l) + \sum_{k, l} P_{1, k}^{-} P_{2, l}^{-} φ_{i} (k - a x) φ (l - a y) \\ = \sum_{k, l} P_{1, k}^{+} P_{2, l}^{+} φ (a x - k, a y - l) + \sum_{k, l} P_{1, k}^{+} P_{2, l}^{-} φ (a x - k, l - a y) \\ + \sum_{k, l} P_{1, k}^{-} P_{2, l}^{+} φ (k - a x, a y - l) + \sum_{k, l} P_{1, k}^{-} P_{2, l}^{-} φ (k - a x, l - a y) . \end{matrix}$

从而根据适合的 ${P_{l, k}^{+, +}}, {P_{l, k}^{+, -}}, {P_{l, k}^{-, +}}, {P_{l, k}^{-, -}}$ ，就有

$\begin{matrix} Φ (x, y) = \sum_{k, l} P_{l, k}^{+, +} φ (a x - k, a y - l) + \sum_{k, l} P_{l, k}^{+, -} φ (a x - k, l - a y) \\ + \sum_{k, l} P_{l, k}^{-, +} φ (k - a x, a y - l) + \sum_{k, l} P_{l, k}^{-, -} φ (k - a x, l - a y) . \end{matrix}$

令 $Φ (x, y) = Φ (x) \cdot φ (y)$ ，则

$Φ (x, y) = {[Φ_{1} (x, y), Φ_{2} (x, y), \dots, Φ_{r} (x, y)]}^{T}$

那么就可以有a尺度多小波细分方程

$\begin{matrix} Φ (x, y) = \sum_{k, l} P_{l, k}^{+, +} Φ (a x - k, a y - l) + \sum_{k, l} P_{l, k}^{+, -} Φ (a x - k, l - a y) \\ + \sum_{k, l} P_{l, k}^{-, +} Φ (k - a x, a y - l) + \sum_{k, l} P_{l, k}^{-, -} Φ (k - a x, l - a y) . \end{matrix}$ (1)

接下来对(1)式进行Fourier变换就可以有

$\begin{matrix} \hat{Φ} (ω_{1}, ω_{2}) = P^{+, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) + P^{+, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \\ + P^{-, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) + P^{-, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}), \end{matrix}$ (2)

其中

$P^{+, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) = \frac{1}{a^{2}} \sum_{k, l} P_{k, l}^{+, +} e^{- i (\frac{ω_{1} k}{a} + \frac{ω_{2} l}{a})},$

$P^{+, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) = \frac{1}{a^{2}} \sum_{k, l} P_{k, l}^{+, -} e^{- i (\frac{ω_{1} k}{a} + \frac{ω_{2} l}{a})},$

$P^{-, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) = \frac{1}{a^{2}} \sum_{k, l} P_{k, l}^{-, +} e^{- i (\frac{ω_{1} k}{a} + \frac{ω_{2} l}{a})},$

$P^{-, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) = \frac{1}{a^{2}} \sum_{k, l} P_{k, l}^{-, -} e^{- i (\frac{ω_{1} k}{a} + \frac{ω_{2} l}{a})},$

令

$P_{k, l}^{-, +} = [\begin{matrix} P_{1, k, l}^{-, +} & P_{r, k, l}^{-, +} \\ 0 & P_{r^{2}, k, l}^{-, +} \end{matrix}], P_{k, l}^{-, -} = [\begin{matrix} P_{1, k, l}^{-, -} & P_{r, k, l}^{-, -} \\ 0 & P_{r^{2}, k, l}^{-, -} \end{matrix}],$

为双正，正负和双负矩阵符号。下面对(1)进行变形有

$\begin{matrix} Φ (x, - y) = \sum_{k, l} P_{l, k}^{+, +} Φ (x) (a x - k, - a y - l) + \sum_{k, l} P_{l, k}^{+, -} Φ (x) (a x - k, l + a y) \\ + \sum_{k, l} P_{l, k}^{-, +} Φ (x) (k - a x, - a y - l) + \sum_{k, l} P_{l, k}^{-, -} Φ (x) (k - a x, l + a y) . \end{matrix}$ (3)

$\begin{matrix} Φ (- x, y) = \sum_{k, l} P_{l, k}^{+, +} Φ (x) (- a x - k, a y - l) + \sum_{k, l} P_{l, k}^{+, -} Φ (x) (- a x - k, l - a y) \\ + \sum_{k, l} P_{l, k}^{-, +} Φ (x) (a x + k, a y - l) + \sum_{k, l} P_{l, k}^{-, -} Φ (x) (a x + k, l - a y) . \end{matrix}$ (4)

$\begin{matrix} Φ (- x, - y) = \sum_{k, l} P_{l, k}^{+, +} Φ (x) (- a x - k, - a y - l) + \sum_{k, l} P_{l, k}^{+, -} Φ (x) (- a x - k, l + a y) \\ + \sum_{k, l} P_{l, k}^{-, +} Φ (x) (a x + k, - a y - l) + \sum_{k, l} P_{l, k}^{-, -} Φ (x) (a x + k, l + a y) . \end{matrix}$ (5)

下面对(3)~(5)式都进行Fourier变换就有

$\begin{matrix} \hat{Φ} (ω_{1}, - ω_{2}) = P^{+, +} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \hat{Φ} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) + P^{+, -} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \hat{Φ} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \\ + P^{-, +} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \hat{Φ} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) + P^{-, -} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \hat{Φ} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}), \end{matrix}$ (6)

$\begin{matrix} \hat{Φ} (- ω_{1}, ω_{2}) = P^{+, +} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) + P^{+, -} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \\ + P^{-, +} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) + P^{-, -} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}), \end{matrix}$ (7)

$\begin{matrix} \hat{Φ} (- ω_{1}, - ω_{2}) = P^{+, +} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \hat{Φ} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) + P^{+, -} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \hat{Φ} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \\ + P^{-, +} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \hat{Φ} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) + P^{-, -} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \hat{Φ} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}), \end{matrix}$ (8)

根据(2)(6)~(8)式可以有，令

$Γ (x, y) = {[Φ (x, y), Φ (x, - y), Φ (- x, y), Φ (- x, - y)]}^{T},$

$\hat{Γ} (ω_{1}, ω_{2}) = [\begin{matrix} P^{+, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) & P^{+, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) & P^{-, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) & P^{-, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \\ P^{+, -} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) & P^{+, +} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) & P^{-, -} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) & P^{-, +} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \\ P^{-, +} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) & P^{-, -} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) & P^{+, +} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) & P^{+, -} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \\ P^{-, -} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) & P^{-, +} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) & P^{+, -} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) & P^{+, +} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \end{matrix}] \hat{Γ} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}),$ (9)

方程(2)有解，当且仅当式(9)有解。

设

$Γ (x, y) = \sum_{k, l} [\begin{matrix} P_{k, l}^{+, +} & P_{k, l}^{+, -} & P_{k, l}^{-, +} & P_{k, l}^{-, -} \\ P_{k, - l}^{+, -} & P_{k, - l}^{+, +} & P_{k, - l}^{-, -} & P_{k, - l}^{-, +} \\ P_{- k, l}^{-, +} & P_{- k, l}^{-, -} & P_{- k, l}^{+, +} & P_{- k, l}^{+, -} \\ P_{- k, - l}^{-, -} & P_{- k, - l}^{-, +} & P_{- k, - l}^{+, -} & P_{- k, - l}^{+, +} \end{matrix}] Γ (a x - k, a y - l),$ (10)

则(9)式为 $Γ (x, y)$ 在频域里的a尺度加密方程，它的加细面具符号为

$P (ω_{1}, ω_{2}) = [\begin{matrix} P^{+, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) & P^{+, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) & P^{-, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) & P^{-, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \\ P^{+, -} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) & P^{+, +} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) & P^{-, -} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) & P^{-, +} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \\ P^{-, +} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) & P^{-, -} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) & P^{+, +} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) & P^{+, -} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \\ P^{-, -} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) & P^{-, +} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) & P^{+, -} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) & P^{+, +} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \end{matrix}]$ (11)

定义方程(1)的自相关矩阵符号

$Ω (x, y) = \sum_{k, l} [\begin{matrix} Ω_{11} & Ω_{12} & Ω_{13} & Ω_{14} \\ Ω_{21} & Ω_{22} & Ω_{23} & Ω_{24} \\ Ω_{31} & Ω_{32} & Ω_{33} & Ω_{34} \\ Ω_{41} & Ω_{42} & Ω_{43} & Ω_{44} \end{matrix}] e^{- i (\frac{ω_{1} k}{a} + \frac{ω_{2} l}{a})},$

其中

$Ω_{11} = 〈 Φ (x, y), Φ (x - k, y - l) 〉; Ω_{12} = 〈 Φ (x, y), Φ (x - k, l - y) 〉;$

$Ω_{13} = 〈 Φ (x, y), Φ (k - x, y - l) 〉; Ω_{14} = 〈 Φ (x, y), Φ (k - x, l - y) 〉;$

$Ω_{21} = 〈 Φ (x, - y), Φ (x - k, y - l) 〉; Ω_{22} = 〈 Φ (x, - y), Φ (x - k, l - y) 〉;$

$Ω_{23} = 〈 Φ (x, - y), Φ (k - x, y - l) 〉; Ω_{24} = 〈 Φ (x, - y), Φ (k - x, l - y) 〉;$

$Ω_{31} = 〈 Φ (- x, y), Φ (x - k, y - l) 〉; Ω_{32} = 〈 Φ (- x, y), Φ (x - k, l - y) 〉;$

$Ω_{33} = 〈 Φ (- x, y), Φ (k - x, y - l) 〉; Ω_{34} = 〈 Φ (- x, y), Φ (k - x, l - y) 〉;$

$Ω_{41} = 〈 Φ (- x, - y), Φ (x - k, y - l) 〉; Ω_{42} = 〈 Φ (- x, - y), Φ (x - k, l - y) 〉;$

$Ω_{43} = 〈 Φ (- x, - y), Φ (k - x, y - l) 〉; Ω_{44} = 〈 Φ (- x, - y), Φ (k - x, l - y) 〉 .$

下面引入变换算子 $τ$ ：

$τ A ({(ω_{1})}^{a}, {(ω_{2})}^{a}) = \sum_{k, l = 0}^{a - 1} P (ω_{1} + \frac{2 k π}{a}, ω_{2} + \frac{2 l π}{a}) A (ω_{1} + \frac{2 k π}{a}, ω_{2} + \frac{2 l π}{a}) P^{*} (ω_{1} + \frac{2 k π}{a}, ω_{2} + \frac{2 l π}{a}),$ (12)

其中 $A ({(ω_{1})}^{a}, {(ω_{2})}^{a})$ 是 $P (ω_{1}, ω_{2})$ 的Laurent多项式， $P (ω_{1}, ω_{2})$ 由(12)式可知。

关于 $Ω (ω_{1}, ω_{2})$ 和 $τ$ 可以有下面的引理。

引理1：矩阵符号 $Ω (ω_{1}, ω_{2})$ 和变换算子 $τ$ 的定义如上，那么由Poisson求和公式可得

$Ω (ω_{1}, ω_{2}) = \sum_{k, l} [\begin{matrix} Φ (ω_{1} + \frac{2 k π}{a}, ω_{2} + \frac{2 l π}{a}) \\ Φ (ω_{1} + \frac{2 k π}{a}, - ω_{2} - \frac{2 l π}{a}) \\ Φ (- ω_{1} - \frac{2 k π}{a}, ω_{2} + \frac{2 l π}{a}) \\ Φ (- ω_{1} - \frac{2 k π}{a}, - ω_{2} - \frac{2 l π}{a}) \end{matrix}] {[\begin{matrix} Φ (ω_{1} + \frac{2 k π}{a}, ω_{2} + \frac{2 l π}{a}) \\ Φ (ω_{1} + \frac{2 k π}{a}, - ω_{2} - \frac{2 l π}{a}) \\ Φ (- ω_{1} - \frac{2 k π}{a}, ω_{2} + \frac{2 l π}{a}) \\ Φ (- ω_{1} - \frac{2 k π}{a}, - ω_{2} - \frac{2 l π}{a}) \end{matrix}]}^{T}$

进一步，我们可以得出 $Ω (ω_{1}, ω_{2})$ 是 $τ$ 相应于特征值为1的特征矩阵。

定理1：加细方程(1)有紧支撑解当且仅当它的面具符号满足下面情况之一

1) $(\begin{array}{l} \sum_{k, l} (P_{k, l}^{+, +} + P_{k, l}^{+, -} + P_{k, l}^{-, +} + P_{k, l}^{-, -}) = a^{2} I, \\ | \sum_{k, l} (P_{k, l}^{+, +} - P_{k, l}^{+, -} - P_{k, l}^{-, +} + P_{k, l}^{-, -}) | \leq a^{2}, \\ | \sum_{k, l} (P_{k, l}^{+, +} + P_{k, l}^{+, -} - P_{k, l}^{-, +} - P_{k, l}^{-, -}) | \leq a^{2}, \\ | \sum_{k, l} (P_{k, l}^{+, +} - P_{k, l}^{+, -} + P_{k, l}^{-, +} - P_{k, l}^{-, -}) | \leq a^{2}; \end{array}$

2) $(\begin{array}{l} | \sum_{k, l} (P_{k, l}^{+, +} + P_{k, l}^{+, -} + P_{k, l}^{-, +} + P_{k, l}^{-, -}) | \leq a^{2}, \\ \sum_{k, l} (P_{k, l}^{+, +} - P_{k, l}^{+, -} - P_{k, l}^{-, +} + P_{k, l}^{-, -}) = a^{2} I, \\ | \sum_{k, l} (P_{k, l}^{+, +} + P_{k, l}^{+, -} - P_{k, l}^{-, +} - P_{k, l}^{-, -}) | \leq a^{2}, \\ | \sum_{k, l} (P_{k, l}^{+, +} - P_{k, l}^{+, -} + P_{k, l}^{-, +} - P_{k, l}^{-, -}) | \leq a^{2}; \end{array}$

3) $(\begin{array}{l} | \sum_{k, l} (P_{k, l}^{+, +} + P_{k, l}^{+, -} + P_{k, l}^{-, +} + P_{k, l}^{-, -}) | \leq a^{2}, \\ | \sum_{k, l} (P_{k, l}^{+, +} - P_{k, l}^{+, -} - P_{k, l}^{-, +} + P_{k, l}^{-, -}) | \leq a^{2}, \\ \sum_{k, l} (P_{k, l}^{+, +} + P_{k, l}^{+, -} - P_{k, l}^{-, +} - P_{k, l}^{-, -}) = a^{2} I, \\ | \sum_{k, l} (P_{k, l}^{+, +} - P_{k, l}^{+, -} + P_{k, l}^{-, +} - P_{k, l}^{-, -}) | \leq a^{2}; \end{array}$

4) $(\begin{array}{l} | \sum_{k, l} (P_{k, l}^{+, +} + P_{k, l}^{+, -} + P_{k, l}^{-, +} + P_{k, l}^{-, -}) | \leq a^{2}, \\ | \sum_{k, l} (P_{k, l}^{+, +} - P_{k, l}^{+, -} - P_{k, l}^{-, +} + P_{k, l}^{-, -}) | \leq a^{2}, \\ | \sum_{k, l} (P_{k, l}^{+, +} + P_{k, l}^{+, -} - P_{k, l}^{-, +} - P_{k, l}^{-, -}) | \leq a^{2}, \\ \sum_{k, l} (P_{k, l}^{+, +} - P_{k, l}^{+, -} + P_{k, l}^{-, +} - P_{k, l}^{-, -}) = a^{2} I; \end{array}$

证明：根据文献 [5] [6] [7] [8]，方程(10)存在紧支撑分布解当且仅当1是(11)式定义矩阵 $P (1,1)$ 的一个特征值， $P (1,1)$ 的其他特征值的模都不大于1。另外， $P (1,1)$ 的4个特征值分别是， $\frac{1}{a^{2}} \sum_{k, l} (P_{k, l}^{+, +} - P_{k, l}^{+, -} - P_{k, l}^{-, +} + P_{k, l}^{-, -})$ ， $\frac{1}{a^{2}} \sum_{k, l} (P_{k, l}^{+, +} + P_{k, l}^{+, -} - P_{k, l}^{-, +} - P_{k, l}^{-, -})$ 以及 $\frac{1}{a^{2}} \sum_{k, l} (P_{k, l}^{+, +} - P_{k, l}^{+, -} + P_{k, l}^{-, +} - P_{k, l}^{-, -})$ 。定理易证。

定理2：若要 $Φ (x, y)$ 有紧支撑性，则要证明每个分量是紧支撑的。设 $Φ (x) = {[φ_{1} (x), φ_{2} (x), \dots, φ_{r} (x)]}^{T}$ ，其中 $φ_{i} (x)$ 和 $φ (y)$ 是双向细分函数满足

$φ_{i} (x) = \sum_{k = 0}^{N_{1}} P_{k}^{+} φ_{i} (a x - k) + \sum_{k = - N_{1}}^{0} P_{k}^{-} φ_{i} (k - a x)$

$φ (y) = \sum_{k = 0}^{N_{2}} p_{k}^{+} φ (a y - k) + \sum_{k = - N_{2}}^{0} p_{k}^{-} φ (k - a y)$

如果 $Φ (x)$ 和 $φ (y)$ 是紧支撑的，那么可根据 $Φ (x)$ 和 $φ (y)$ 的张量积生成二维四向加细函数 $S u p p Φ (x) \subseteq [- \frac{N_{1}}{a - 1}, \frac{N_{1}}{a - 1}]$ ， $S u p p φ (y) \subseteq [- \frac{N_{2}}{a - 1}, \frac{N_{2}}{a - 1}]$ ，则有

$S u p p Φ (x, y) = Φ (x) \cdot φ (y) \subseteq [- \frac{N_{1}}{a - 1}, \frac{N_{1}}{a - 1}] \times [- \frac{N_{2}}{a - 1}, \frac{N_{2}}{a - 1}]$

证明：根据文献 [9] 中定理4可知。

3. a尺度二维四向多分辨分析

定义子空间序列 ${V_{j}}_{j \in Z} \subset L^{2} (R^{2})$ ，

$\begin{array}{l} V_{j} = C l o s e_{L^{2} (R^{2})} 〈 a^{j} Φ (a^{j} x - k, a^{j} y - l), a^{j} Φ (a^{j} x - k, l - a^{j} y), \\ a^{j} Φ (k - a^{j} x, a^{j} y - l), a^{j} Φ (k - a^{j} x, l - a^{j} y), k, l \in Z 〉, \end{array}$ (13)

那么要产生在 $L^{2} (R^{2})$ 中的一个多分辨分析 ${V_{j}}_{j \in Z}$ 当且仅当(13)式里的 $V_{j}$ 应当满足以下条件：

1) $\dots \subset V_{- 1} \subset V_{0} \subset V_{1} \subset \dots$ ；

2) $C l o s e_{L^{2} (R^{2})} \cup_{j \in Z} V_{j} = L^{2} (R^{2})$ ；

3) $\cap_{j \in Z} V_{j} = {0}$ ；

4) $f (x, y) \in V_{j} \Leftrightarrow f (a x, a y) \in V_{j + 1}$ ；

5) 存在 $L^{2} (R^{2})$ 里的一个函数 $Φ (x, y)$ ，使得集合 ${Φ (x - k, y - l), Φ (x - k, l - y), Φ (k - x, y - l), Φ (k - x, l - y) : k, l \in Z}$ 是 $V_{0}$ 的Riesz基，那么就有两个常数 $0 < A \leq B < \infty$ ，则对于系数向量序列

${c^{k, l}}_{k, l \in Z} = {[c_{1}^{k, l}, c_{2}^{k, l}, c_{3}^{k, l}, c_{4}^{k, l}]}_{k, l \in Z} \subset l^{2} (Z^{4})$ 有

$\begin{matrix} A {‖ \sum_{k, l \in Z} c_{k, l} c_{k, l}^{*} ‖}_{2}^{2} \leq ‖ \sum_{k, l \in Z} c_{1}^{k, l} Φ (x - k, y - l) + \sum_{k, l \in Z} c_{2}^{k, l} Φ (x - k, l - y) + \sum_{k, l \in Z} c_{3}^{k, l} Φ (k - x, y - l) + {\sum_{k, l \in Z} c_{4}^{k, l} Φ (k - x, l - y) ‖}_{2}^{2} \\ \leq B {‖ \sum_{k, l \in Z} c_{k, l} c_{k, l}^{*} ‖}_{2}^{2}, \end{matrix}$ (14)

称式(14)为稳定性条件。

根据多重多分辨分析的性质 $f (x, y) \in V_{j} \Leftrightarrow f (x + \frac{k}{a^{j}}, y + \frac{l}{a^{j}}) \in V_{j}$ ，可以定义：

$Φ_{j, k, l}^{+, +} = a^{j} Φ (a^{j} x - k, a^{j} y - l), Φ_{j, k, l}^{+, -} = a^{j} Φ (a^{j} x - k, l - a^{j} y),$

$Φ_{j, k, l}^{-, +} = a^{j} Φ (k - a^{j} x, a^{j} y - l), Φ_{j, k, l}^{-, -} = a^{j} Φ (k - a^{j} x, l - a^{j} y) .$

则 ${Φ_{j, k, l}^{+, +}, Φ_{j, k, l}^{+, -}, Φ_{j, k, l}^{-, +}, Φ_{j, k, l}^{-, -} : k, l \in Z}$ 同样可以构成 $V_{j}$ 的Riesz基，有

$A {‖ \sum_{k, l \in Z} c_{k, l} c_{k, l}^{*} ‖}_{2}^{2} \leq {‖ \sum_{k, l \in Z} c_{1}^{k, l} Φ_{j, k, l}^{+, +} + \sum_{k, l \in Z} c_{2}^{k, l} Φ_{j, k, l}^{+, -} + \sum_{k, l \in Z} c_{3}^{k, l} Φ_{j, k, l}^{-, +} + \sum_{k, l \in Z} c_{4}^{k, l} Φ_{j, k, l}^{-, -} ‖}_{2}^{2} \leq B {‖ \sum_{k, l \in Z} c_{k, l} c_{k, l}^{*} ‖}_{2}^{2},$

因为 $Φ (x, y) \in V_{0} \subset V_{1}$ ，并且 ${Φ_{j, k, l}^{+, +}, Φ_{j, k, l}^{+, -}, Φ_{j, k, l}^{-, +}, Φ_{j, k, l}^{-, -} : k, l \in Z}$ 同样也可以构成 $V_{1}$ 的Riesz基，故有 $c_{1}^{k, l}, c_{2}^{k, l}, c_{3}^{k, l}, c_{4}^{k, l} \in l^{2} (Z^{4})$ ，从而使 $Φ (x, y)$ 满足(1)式。

定理3：如果尺度函数 $Φ (x, y) \in L^{2} (R^{2})$ 符合多分辨分析，现定义 $V_{j} = {Φ_{j, k, l}^{+, +}, Φ_{j, k, l}^{+, -}, Φ_{j, k, l}^{-, +}, Φ_{j, k, l}^{-, -} : k, l \in Z}$ 构成 $V_{0}$ 的Riesz基。若存在函数集 ${Φ (x - k, y - l), Φ (x - k, l - y), Φ (k - x, y - l), Φ (k - x, l - y) : k, l \in Z}$ 构成 $V_{0}$ 的Riesz基，那么 $\cap_{j \in Z} V_{j} = {0}$ 。

证明：由于 ${Φ_{j, k, l}^{+, +}, Φ_{j, k, l}^{+, -}, Φ_{j, k, l}^{-, +}, Φ_{j, k, l}^{-, -} : k, l \in Z}$ 构成 $V_{0}$ 的Riesz基，故存在两个常数 $0 < A \leq B < \infty$ ，对任意的 $f (x, y) \in V_{0}$ ，有

$A {‖ f ‖}_{2}^{2} \leq \sum_{k, l \in Z} {| 〈 f, Φ_{0, k, l}^{+, +} 〉 |}^{2} + \sum_{k, l \in Z} {| 〈 f, Φ_{0, k, l}^{+, -} 〉 |}^{2} + \sum_{k, l \in Z} {| 〈 f, Φ_{0, k, l}^{-, +} 〉 |}^{2} + \sum_{k, l \in Z} {| 〈 f, Φ_{0, k, l}^{-, -} 〉 |}^{2} \leq B {‖ f ‖}_{2}^{2},$

所以对于所有的 $f (x, y) \in V_{j}$ ，就有

$A {‖ f ‖}_{2}^{2} \leq \sum_{k, l \in Z} {| 〈 f, Φ_{j, k, l}^{+, +} 〉 |}^{2} + \sum_{k, l \in Z} {| 〈 f, Φ_{j, k, l}^{+, -} 〉 |}^{2} + \sum_{k, l \in Z} {| 〈 f, Φ_{j, k, l}^{-, +} 〉 |}^{2} + \sum_{k, l \in Z} {| 〈 f, Φ_{j, k, l}^{-, -} 〉 |}^{2} \leq B {‖ f ‖}_{2}^{2},$

对于 $\forall ε > 0$ ， $f (x, y) \in \cap_{j \in Z} V_{j}$ ，存在一个紧支撑连续函数 $\tilde{f} (x, y)$ ，使 ${‖ f - \tilde{f} ‖}_{L^{2}} \leq ε$ 。

若 $P_{j}$ 是 $V_{j}$ 的正交投影算子，则 $‖ f ‖ - ‖ P_{j} \tilde{f} ‖ \leq ‖ f - P_{j} \tilde{f} ‖ = ‖ P_{j} f - P_{j} \tilde{f} ‖ \leq ‖ f - \tilde{f} ‖ \leq ε$ ，故 $‖ f ‖ \leq ε + ‖ P_{j} \tilde{f} ‖, \forall j \in Z$ 。

设 $S u p p \tilde{f} (x, y) = [- r, r] \times [- r, r] = S$ ，则

$\begin{array}{l} \sum_{k, l \in Z} {| 〈 \tilde{f}, Φ_{j, k, l}^{+, +} 〉 |}^{2} \leq a^{2 j} \sum_{k, l \in Z} {[\int_{(x, y) \in S} | \tilde{f} (x, y) | | Φ (a^{j} x - k, a^{j} y - l) | d x d y]}^{2} \\ \leq a^{2 j} {‖ \tilde{f} (x, y) ‖}_{L^{\infty}}^{2} \sum_{k, l \in Z} {[\int_{(x, y) \in S} | Φ (a^{j} x - k, a^{j} y - l) | d x d y]}^{2} \\ \leq a^{2 j} {‖ \tilde{f} (x, y) ‖}_{L^{\infty}}^{2} \sum_{k, l \in Z} \int_{(x, y) \in S} {| Φ (a^{j} x - k, a^{j} y - l) |}^{2} d x d y \\ = {‖ \tilde{f} (x, y) ‖}_{L^{\infty}}^{2} \int_{D^{j}} {| Φ (x, y) |}^{2} d x d y \end{array}$

其中 $D^{j} = \underset{k, l}{\cup} [k - a^{j} r, k + a^{j} r] \times [l - a^{j} r, l + a^{j} r]$ 。所以，当 $j \to - \infty$ 时 $\sum_{k, l \in Z} {| 〈 \tilde{f}, Φ_{j, k, l}^{+, +} 〉 |}^{2} \to 0$ 。同理，可以证明： $\sum_{k, l \in Z} {| 〈 \tilde{f}, Φ_{j, k, l}^{-, +} 〉 |}^{2} \to 0$ ， $\sum_{k, l \in Z} {| 〈 \tilde{f}, Φ_{j, k, l}^{+, -} 〉 |}^{2} \to 0$ ， $\sum_{k, l \in Z} {| 〈 \tilde{f}, Φ_{j, k, l}^{-, -} 〉 |}^{2} \to 0$ 。又由于

${‖ P_{j} \tilde{f} ‖}^{2} \leq \frac{1}{A} \sum_{k, l \in Z} [{| 〈 f, Φ_{0, k, l}^{+, +} 〉 |}^{2} + {| 〈 f, Φ_{0, k, l}^{+, -} 〉 |}^{2} + {| 〈 f, Φ_{0, k, l}^{-, +} 〉 |}^{2} + {| 〈 f, Φ_{0, k, l}^{-, -} 〉 |}^{2}]$

所以 $\lim_{j \to - \infty} ‖ P_{j} \tilde{f} ‖ = 0$ 。根据 $‖ f ‖ \leq ε + ‖ P_{j} \tilde{f} ‖, \forall j \in Z$ ，可以知道 $f (x, y) = 0$ 。

定理4：若 $Φ (x, y) \in L^{2} (R^{2})$ 符合定义式(1)，根据(13)式定义的 $V_{j}$ ，若 $Φ (x, y)$ 满足1) 集合 ${Φ (x - k, y - l), Φ (x - k, l - y), Φ (k - x, y - l), Φ (k - x, l - y) : k, l \in Z}$ 是 $V_{0}$ 的Riesz基；2) 对所有的 $(ω_{1}, ω_{2}) \in R^{2}$ ， $\hat{Φ} (ω_{1}, ω_{2})$ 有界；3) $\hat{Φ} (ω_{1}, ω_{2})$ 在 $(ω_{1}, ω_{2}) = (0,0)$ 附近连续， $\hat{Φ} (ω_{1}, ω_{2}) \neq 0$ ，那么 $C l o s e_{L^{2} (R^{2})} \cup_{j \in Z} V_{j} = L^{2} (R^{2})$ 。

所以对于所有的 $f (x, y) \in V_{j}$ ，就有

对于， $f (x, y) \in {(\cup_{j \in Z} V_{j})}^{⊥}$ ，那么就有紧支撑连续函数 $\tilde{f} (x, y)$ ，使 ${‖ f - \tilde{f} ‖}_{L^{2}} \leq ε$ 。

若 $P_{j}$ 是 $V_{j}$ 的正交投影算子，则 $‖ P_{j} \tilde{f} ‖ = {‖ P_{j} (\tilde{f} - f) ‖}_{L^{2}} \leq ε$ 。又因为 $P_{j} \tilde{f} \in V_{j}$ ，则

$B {‖ P_{j} \tilde{f} ‖}^{2} \geq \sum_{k, l \in Z} {| 〈 P_{j} \tilde{f}, Φ_{j, k, l}^{+, +} 〉 |}^{2} + \sum_{k, l \in Z} {| 〈 P_{j} \tilde{f}, Φ_{j, k, l}^{+, -} 〉 |}^{2} + \sum_{k, l \in Z} {| 〈 P_{j} \tilde{f}, Φ_{j, k, l}^{-, +} 〉 |}^{2} + \sum_{k, l \in Z} {| 〈 P_{j} \tilde{f}, Φ_{j, k, l}^{-, -} 〉 |}^{2},$

进一步可以得到

$\begin{array}{l} \sum_{k, l \in Z} {| 〈 P_{j} \tilde{f}, Φ_{j, k, l}^{+, +} 〉 |}^{2} + \sum_{k, l \in Z} {| 〈 P_{j} \tilde{f}, Φ_{j, k, l}^{+, -} 〉 |}^{2} + \sum_{k, l \in Z} {| 〈 P_{j} \tilde{f}, Φ_{j, k, l}^{-, +} 〉 |}^{2} + \sum_{k, l \in Z} {| 〈 P_{j} \tilde{f}, Φ_{j, k, l}^{-, -} 〉 |}^{2} \\ = \frac{1}{{(2 π)}^{4}} \int_{- \infty}^{+ \infty} \int_{- \infty}^{+ \infty} ({| \hat{Φ} (a^{- j} ω_{1}, a^{- j} ω_{2}) |}^{2} + {| \hat{Φ} (a^{- j} ω_{1}, - a^{- j} ω_{2}) |}^{2} + {| \hat{Φ} (- a^{- j} ω_{1}, a^{- j} ω_{2}) |}^{2} \\ + {| \hat{Φ} (- a^{- j} ω_{1}, - a^{- j} ω_{2}) |}^{2}) \hat{P_{j} \tilde{f}} (ω_{1}, ω_{2}) d ω_{1} d ω_{2} + D \end{array}$

其中

$D \leq ({‖ \hat{Φ} (ω_{1}, ω_{2}) ‖}_{L^{\infty}}^{2} + {‖ \hat{Φ} (ω_{1}, - ω_{2}) ‖}_{L^{\infty}}^{2} + {‖ \hat{Φ} (- ω_{1}, ω_{2}) ‖}_{L^{\infty}}^{2} + {‖ \hat{Φ} (- ω_{1}, - ω_{2}) ‖}_{L^{\infty}}^{2}) \cdot \sum_{k \neq 0, l \neq 0} {| P_{j} \tilde{f} (a^{j} k, a^{j} l) |}^{2} .$

因为 $\tilde{f} (x, y) \in C^{\infty}$ ，所以存在一个常数M，使 $| \tilde{f} (x, y) | \leq M {(1 + | x + y |)}^{- 2}$ 故

$\begin{matrix} D \leq M^{2} ({‖ \hat{Φ} (ω_{1}, ω_{2}) ‖}_{L^{\infty}}^{2} + {‖ \hat{Φ} (ω_{1}, - ω_{2}) ‖}_{L^{\infty}}^{2} + {‖ \hat{Φ} (- ω_{1}, ω_{2}) ‖}_{L^{\infty}}^{2} + {‖ \hat{Φ} (- ω_{1}, - ω_{2}) ‖}_{L^{\infty}}^{2}) \cdot \sum_{k \neq 0, l \neq 0} {| P_{j} \tilde{f} (a^{j} k, a^{j} l) |}^{2} \\ \leq M^{2} ({‖ \hat{Φ} (ω_{1}, ω_{2}) ‖}_{L^{\infty}}^{2} + {‖ \hat{Φ} (ω_{1}, - ω_{2}) ‖}_{L^{\infty}}^{2} + {‖ \hat{Φ} (- ω_{1}, ω_{2}) ‖}_{L^{\infty}}^{2} + {‖ \hat{Φ} (- ω_{1}, - ω_{2}) ‖}_{L^{\infty}}^{2}) \cdot \sum_{k \neq 0, l \neq 0} {(1 + | a^{j} k + a^{j} l |)}^{- 2} \\ \leq {M^{'}}^{2} a^{- 2 j} \end{matrix}$

由以上的推导，可以得到

$\begin{array}{l} \frac{1}{{(2 π)}^{4}} \int_{- \infty}^{+ \infty} \int_{- \infty}^{+ \infty} ({| \hat{Φ} (a^{- j} ω_{1}, a^{- j} ω_{2}) |}^{2} + {| \hat{Φ} (a^{- j} ω_{1}, - a^{- j} ω_{2}) |}^{2} + {| \hat{Φ} (- a^{- j} ω_{1}, a^{- j} ω_{2}) |}^{2} \\ + {| \hat{Φ} (- a^{- j} ω_{1}, - a^{- j} ω_{2}) |}^{2}) \hat{P_{j} \tilde{f}} (ω_{1}, ω_{2}) d ω_{1} d ω_{2} \leq \sum_{k, l \in Z} {| 〈 P_{j} \tilde{f}, Φ_{j, k, l}^{+, +} 〉 |}^{2} \\ + \sum_{k, l \in Z} {| 〈 P_{j} \tilde{f}, Φ_{j, k, l}^{+, -} 〉 |}^{2} + \sum_{k, l \in Z} {| 〈 P_{j} \tilde{f}, Φ_{j, k, l}^{-, +} 〉 |}^{2} + \sum_{k, l \in Z} {| 〈 P_{j} \tilde{f}, Φ_{j, k, l}^{-, -} 〉 |}^{2} + | D | \\ \leq B ε^{2} + {M^{'}}^{2} a^{- 2 j} . \end{array}$

由于 $Φ (ω_{1}, ω_{2})$ 有界且在 $(ω_{1}, ω_{2}) = (0,0)$ 附近处连续且 $Φ (0,0) \neq 0$ ，故当 $j \to + \infty$ 时，以上不等式收敛于

$\frac{1}{{(2 π)}^{4}} ({| \hat{Φ} (0,0) |}^{2} + {| \hat{Φ} (0,0) |}^{2} + {| \hat{Φ} (0,0) |}^{2} + {| \hat{Φ} (0,0) |}^{2}) \cdot {‖ \hat{P_{j} \tilde{f}} ‖}_{L^{2}}^{2} = \frac{1}{4 π^{4}} {| \hat{Φ} (0,0) |}^{2} \cdot {‖ \hat{P_{j} \tilde{f}} ‖}_{L^{2}}^{2}$

从而 ${‖ \hat{P_{j} \tilde{f}} ‖}_{L^{2}}^{2} \leq \frac{{(2 π)}^{4} B ε^{2}}{4 {| \hat{Φ} (0,0) |}^{2}} = \frac{4 π^{4} B ε^{2}}{{| \hat{Φ} (0,0) |}^{2}}$ ，所以 ${‖ f ‖}_{L^{2}} \leq ε + {‖ \hat{P_{j} \tilde{f}} ‖}_{L^{2}} \leq ε + \frac{2 π^{2} \sqrt{B} ε}{\sqrt{{| \hat{Φ} (0,0) |}^{2}}}$ ，因为 $ε$ 是任意小的，所以 $f = 0$ ，则说明 $C l o s e_{L^{2} (R^{2})} \cup_{j \in Z} V_{j} = L^{2} (R^{2})$ 。

4. 双正交二维四向加细函数和小波

定理5：如果二维四向加细函数是正交的，那么应该满足下列等式

$\begin{array}{l} 〈 Φ (x, y), Φ (x - k, y - l) 〉 = δ_{0, k} δ_{0, l} I_{r}, 〈 Φ (x, y), Φ (k - x, y - l) 〉 = O_{r}, \\ 〈 Φ (x, y), Φ (x - k, l - y) 〉 = O_{r}, 〈 Φ (x, y), Φ (k - x, l - y) 〉 = O_{r} . \end{array}$

定理6：如果二维四向加细函数 $Φ (x, y)$ 和 $\tilde{Φ} (x, y)$ 是双正交的，那么应该满足下列等式

$\begin{array}{l} 〈 Φ (x, y), \tilde{Φ} (x - k, y - l) 〉 = δ_{0, k} δ_{0, l} I_{r}, 〈 Φ (x, y), \tilde{Φ} (k - x, y - l) 〉 = O_{r}, \\ 〈 Φ (x, y), \tilde{Φ} (x - k, l - y) 〉 = O_{r}, 〈 Φ (x, y), \tilde{Φ} (k - x, l - y) 〉 = O_{r} . \end{array}$

定理7：如果二维四向加细函数 $Φ (x, y)$ 和 $\tilde{Φ} (x, y)$ 是双正交的尺度函数，那么它的双正面具，正负面具和双负面具符号都应该满足

$(\begin{array}{l} \sum_{k, l = 0}^{a - 1} {P^{+, +} (p_{1}, p_{2}) {\tilde{P}}^{+, +} (- p_{1}, - p_{2}) + P^{-, +} (p_{1}, p_{2}) {\tilde{P}}^{-, +} (- p_{1}, - p_{2}) \\ + P^{+, -} (p_{1}, p_{2}) {\tilde{P}}^{+, -} (- p_{1}, - p_{2}) + P^{-, -} (p_{1}, p_{2}) {\tilde{P}}^{-, -} (- p_{1}, - p_{2})} = I_{r} \\ \sum_{k, l = 0}^{a - 1} {P^{+, +} (p_{1}, p_{2}) {\tilde{P}}^{+, -} (- p_{1}, p_{2}) + P^{+, -} (p_{1}, p_{2}) {\tilde{P}}^{+, +} (- p_{1}, p_{2}) \\ + P^{-, +} (p_{1}, p_{2}) {\tilde{P}}^{-, -} (- p_{1}, p_{2}) + P^{-, -} (p_{1}, p_{2}) {\tilde{P}}^{-, +} (- p_{1}, p_{2})} = O_{r} \\ \sum_{k, l = 0}^{a - 1} {P^{+, +} (p_{1}, p_{2}) {\tilde{P}}^{-, +} (p_{1}, - p_{2}) + P^{+, -} (p_{1}, p_{2}) {\tilde{P}}^{-, -} (p_{1}, - p_{2}) \\ + P^{-, +} (p_{1}, p_{2}) {\tilde{P}}^{+, +} (p_{1}, - p_{2}) + P^{-, -} (p_{1}, p_{2}) {\tilde{P}}^{+, -} (p_{1}, - p_{2})} = O_{r} \\ \sum_{k, l = 0}^{a - 1} {P^{+, +} (p_{1}, p_{2}) {\tilde{P}}^{-, -} (p_{1}, p_{2}) + P^{+, -} (p_{1}, p_{2}) {\tilde{P}}^{-, +} (p_{1}, p_{2}) \\ + P^{-, +} (p_{1}, p_{2}) {\tilde{P}}^{+, -} (p_{1}, p_{2}) + P^{-, -} (p_{1}, p_{2}) {\tilde{P}}^{+, +} (p_{1}, p_{2})} = O_{r} \end{array}$ (15)

其中： $p_{1} = ω_{1} + \frac{2 k π}{a}$ ； $p_{2} = ω_{2} + \frac{2 l π}{a}$ 。

证明：由文献 [7] 的定理1以及本文定理5定理6正交双正交定义易证。

假设 $Φ (x, y)$ 和 $\tilde{Φ} (x, y)$ 是双正交二维四向加细函数，对任意的 $j \in Z$ ，定义 $V_{j + 1} = V_{j} \oplus W_{j}$ ，其中 $W_{j}$ 是 $V_{j}$ 在 $V_{j + 1}$ 中的正交补。那么当 $j \neq k$ 时，就有 $W_{j} ⊥ W_{k}$ 并且 $L^{2} (R)= \oplus_{j \in Z} W_{j}$ ，其中 $W_{j} = \oplus_{h = 1}^{a - 1} {W_{j}^{h,1} \oplus W_{j}^{h,2} \oplus W_{j}^{h,3}}$ ， $W_{j}^{h,1} ⊥ W_{j}^{h,2}$ ， $W_{j}^{h,1} ⊥ W_{j}^{h,1}$ ， $W_{j}^{h,2} ⊥ W_{j}^{h,3}$ 。

如果有r个小波函数 $ψ_{1} (x), ψ_{2} (x), \dots, ψ_{r} (x)$ ，则记 $Ψ^{h, i} (x) = {[ψ_{1} (x), ψ_{2} (x), \dots, ψ_{r} (x)]}^{T}$ ，则通过张量积的构造就有

$Ψ^{h, i} (x, y) = Ψ^{h, i} (x) ψ (y) = {[Ψ_{1} (x, y), Ψ_{2} (x, y), \dots, Ψ_{r} (x, y)]}^{T}$

和

${\tilde{Ψ}}^{h, i} (x, y) = {[{\tilde{Ψ}}_{1} (x, y), {\tilde{Ψ}}_{2} (x, y), \dots, {\tilde{Ψ}}_{r} (x, y)]}^{T}$

$(h = 1, 2, \dots, a - 1; i = 1, 2, 3)$ ，使得集合 ${Ψ^{h, i} (x - k, y - l), Ψ^{h, i} (x - k, l - y), Ψ^{h, i} (k - x, y - l), Ψ^{h, i} (k - x, l - y) : k, l \in Z, h = 1,2, \dots, a - 1; i = 1,2,3}$ 和集合 ${{\tilde{Ψ}}^{h, i} (x - k, y - l), {\tilde{Ψ}}^{h, i} (x - k, l - y), {\tilde{Ψ}}^{h, i} (k - x, y - l), {\tilde{Ψ}}^{h, i} (k - x, l - y) : k, l \in Z, h = 1,2, \dots, a - 1; i = 1,2,3}$ 构成 $W_{0}$ 的一组双正交基，则 $Ψ^{h, i} (x, y)$ 和 ${\tilde{Ψ}}^{h, i} (x, y)$ 是与 $Φ (x, y)$ 和 $\tilde{Φ} (x, y)$ 对应的双正交二维四向多小波函数，应该满足

$\begin{array}{l} \tilde{Φ} (x, y), Ψ^{h, i} (x - k, y - l) = 〈 \tilde{Φ} (x, y), Ψ^{h, i} (x - k, l - y) 〉 = O_{r}, \\ Φ (x, y), {\tilde{Ψ}}^{h, i} (k - x, y - l) = 〈 Φ (x, y), {\tilde{Ψ}}^{h, i} (k - x, l - y) 〉 = O_{r}, \\ \tilde{Φ} (x, y), Ψ^{h, i} (k - x, y - l) = 〈 \tilde{Φ} (x, y), Ψ^{h, i} (k - x, l - y) 〉 = O_{r}, \\ 〈 Ψ^{n, m} (x, y), {\tilde{Ψ}}^{s, t} (x - k, y - l) 〉 = δ_{n, s} δ_{m, t} δ_{0, k} δ_{0, l} I_{r}, \end{array}$

$\begin{array}{l} 〈 Ψ^{n, m} (x, y), {\tilde{Ψ}}^{s, t} (x - k, l - y) 〉 = O_{r}, \\ 〈 {\tilde{Ψ}}^{n, m} (x, y), {\tilde{Ψ}}^{s, t} (x - k, y - l) 〉 = δ_{n, s} δ_{m, t} δ_{0, k} δ_{0, l} I_{r}, \\ 〈 {\tilde{Ψ}}^{n, m} (x, y), {\tilde{Ψ}}^{s, t} (x - k, l - y) 〉 = O_{r}, \\ 〈 Ψ^{n, m} (x, y), Ψ^{s, t} (k - x, y - l) 〉 = O_{r}, \end{array}$

$\begin{array}{l} 〈 Ψ^{n, m} (x, y), Ψ^{s, t} (k - x, l - y) 〉 = O_{r}, \\ 〈 {\tilde{Ψ}}^{n, m} (x, y), {\tilde{Ψ}}^{s, t} (k - x, y - l) 〉 = O_{r}, \\ 〈 {\tilde{Ψ}}^{n, m} (x, y), {\tilde{Ψ}}^{s, t} (k - x, l - y) 〉 = O_{r}, \end{array}$ (16)

其中： $h, n, s = 1, 2, \dots, a - 1; i, m, t = 1, 2, 3$ 。

假如 $Ψ^{h, i} (x, y)$ 是 $Φ (x, y)$ 对应的多小波函数，相应的 ${\tilde{Ψ}}^{h, i} (x, y)$ 是 $\tilde{Φ} (x, y)$ 对应的多小波函数，那么就存在 ${Q_{k, l}^{h, i, +, +}}_{k, l \in Z}$ ， ${Q_{k, l}^{h, i, +, -}}_{k, l \in Z}$ ， ${Q_{k, l}^{h, i, -, +}}_{k, l \in Z}$ ， ${Q_{k, l}^{h, i, -, -}}_{k, l \in Z}$ 和 ${{\tilde{Q}}_{k, l}^{h, i, +, +}}_{k, l \in Z}$ ， ${{\tilde{Q}}_{k, l}^{h, i, +, -}}_{k, l \in Z}$ ， ${{\tilde{Q}}_{k, l}^{h, r, -, +}}_{k, l \in Z}$ ， ${{\tilde{Q}}_{k, l}^{h, i, -, -}}_{k, l \in Z}$ ，满足

$\begin{matrix} Ψ^{h, i} (x, y) = \sum_{k, l} Q_{k, l}^{h, i, +, +} Φ (a x - k, a y - l) + \sum_{k, l} Q_{k, l}^{h, i, +, -} Φ (a x - k, l - a y) \\ + \sum_{k, l} Q_{k, l}^{h, i, -, +} Φ (k - a x, a y - l) + \sum_{k, l} Q_{k, l}^{h, i, -, -} Φ (k - a x, l - a y); \end{matrix}$ (17)

$\begin{matrix} {\tilde{Ψ}}^{h, i} (x, y) = \sum_{k, l} {\tilde{Q}}_{k, l}^{h, i, +, +} \tilde{Φ} (a x - k, a y - l) + \sum_{k, l} {\tilde{Q}}_{k, l}^{h, i, +, -} \tilde{Φ} (a x - k, l - a y) \\ + \sum_{k, l} {\tilde{Q}}_{k, l}^{h, i, -, +} \tilde{Φ} (k - a x, a y - l) + \sum_{k, l} {\tilde{Q}}_{k, l}^{h, i, -, -} \tilde{Φ} (k - a x, l - a y) . \end{matrix}$ (18)

对(17)式和(18)式两边做Fourier变换

$\begin{matrix} {\hat{Ψ}}^{h, i} (ω_{1}, ω_{2}) = Q^{h, i, +, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) + Q^{h, i, +, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \\ + Q^{h, i, -, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) + Q^{h, i, -, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}); \end{matrix}$

$\begin{matrix} {\hat{\tilde{Ψ}}}^{h, i} (ω_{1}, ω_{2}) = {\tilde{Q}}^{h, i, +, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) + {\tilde{Q}}^{h, i, +, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (\frac{ω_{1}}{a}, - \frac{ω_{2}}{a}) \\ + {\tilde{Q}}^{h, i, -, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (- \frac{ω_{1}}{a}, \frac{ω_{2}}{a}) + {\tilde{Q}}^{h, i, -, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) \hat{Φ} (- \frac{ω_{1}}{a}, - \frac{ω_{2}}{a}); \end{matrix}$

其中

$\begin{array}{l} Q^{h, i, +, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) = \frac{1}{a^{2}} \sum_{k, l} Q_{k, l}^{h, i, +, +} e^{- i (\frac{ω_{1} k}{a} + \frac{ω_{2} l}{a})}, Q^{h, i, +, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) = \frac{1}{a^{2}} \sum_{k, l} Q_{k, l}^{h, i, +, -} e^{- i (\frac{ω_{1} k}{a} + \frac{ω_{2} l}{a})}, \\ Q^{h, i, -, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) = \frac{1}{a^{2}} \sum_{k, l} Q_{k, l}^{h, i, -, +} e^{- i (\frac{ω_{1} k}{a} + \frac{ω_{2} l}{a})}, Q^{h, i, -, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) = \frac{1}{a^{2}} \sum_{k, l} Q_{k, l}^{h, i, -, -} e^{- i (\frac{ω_{1} k}{a} + \frac{ω_{2} l}{a})}, \end{array}$

$(h = 1, 2, \dots, a - 1; i = 1, 2, 3)$ 为 $Ψ^{h, i} (x, y)$ 的双正，正负，双负面具符号。

$\begin{array}{l} {\tilde{Q}}^{h, i, +, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) = \frac{1}{a^{2}} \sum_{k, l} {\tilde{Q}}_{k, l}^{h, i, +, +} e^{- i (\frac{ω_{1} k}{a} + \frac{ω_{2} l}{a})}, {\tilde{Q}}^{h, i, +, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) = \frac{1}{a^{2}} \sum_{k, l} {\tilde{Q}}_{k, l}^{h, i, +, -} e^{- i (\frac{ω_{1} k}{a} + \frac{ω_{2} l}{a})}, \\ {\tilde{Q}}^{h, i, -, +} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) = \frac{1}{a^{2}} \sum_{k, l} {\tilde{Q}}_{k, l}^{h, i, -, +} e^{- i (\frac{ω_{1} k}{a} + \frac{ω_{2} l}{a})}, {\tilde{Q}}^{h, i, -, -} (\frac{ω_{1}}{a}, \frac{ω_{2}}{a}) = \frac{1}{a^{2}} \sum_{k, l} {\tilde{Q}}_{k, l}^{h, i, -, -} e^{- i (\frac{ω_{1} k}{a} + \frac{ω_{2} l}{a})}, \end{array}$

$(h = 1, 2, \dots, a - 1; i = 1, 2, 3)$ 为 ${\tilde{Ψ}}^{h, i} (x, y)$ 的双正，正负，双负面具符号。

定理8：若 $Φ (x, y)$ 和 $\tilde{Φ} (x, y)$ 是双正交二维四向多加细函数， $Ψ^{h, i} (x, y)$ 和 ${\tilde{Ψ}}^{h, i} (x, y)$ 是相应的双正交二维四向多小波函数，则它们的面具符号满足

(19)

(20)

其中： $h, m, n = 1, 2, \dots, a - 1; i, s, t = 1, 2, 3; p_{1} = ω_{1} + \frac{2 k π}{a}; p_{2} = ω_{2} + \frac{2 l π}{a}$ 。

证明：根据(16)式的正交性易得。

5. 构造算法

定理9：如果 $Φ (x, y)$ 和 $\tilde{Φ} (x, y)$ 是双正交二维四向多加细函数， $Ψ^{h, i} (x, y)$ 和 ${\tilde{Ψ}}^{h, i} (x, y)$ 是相应的双正交二维四向多小波函数， $P (ω_{1}, ω_{2}), \tilde{P} (ω_{1}, ω_{2})$ 和 $Q^{h, i} (ω_{1}, ω_{2}), {\tilde{Q}}^{h, i} (ω_{1}, ω_{2})$ 是矩阵符号，构造

$(\begin{array}{l} P^{+, +} (ω_{1}, ω_{2}) = λ_{0} (ω_{1}, ω_{2}) P (ω_{1}, ω_{2}) + \sum_{j = 1}^{a - 1} λ_{j} (ω_{1}, ω_{2}) Q^{h, i} (ω_{1}, ω_{2}), \\ P^{+, -} (ω_{1}, ω_{2}) = λ_{a} (ω_{1}, ω_{2}) P (ω_{1}, - ω_{2}) + \sum_{j = a + 1}^{2 a - 1} λ_{j} (ω_{1}, ω_{2}) Q^{h, i} (ω_{1}, - ω_{2}), \\ P^{-, +} (ω_{1}, ω_{2}) = λ_{2 a} (ω_{1}, ω_{2}) P (- ω_{1}, ω_{2}) + \sum_{j = 2 a + 1}^{3 a - 1} λ_{j} (ω_{1}, ω_{2}) Q^{h, i} (- ω_{1}, ω_{2}), \\ P^{-, -} (ω_{1}, ω_{2}) = λ_{3 a} (ω_{1}, ω_{2}) P (- ω_{1}, - ω_{2}) + \sum_{j = 3 a + 1}^{4 a - 1} λ_{j} (ω_{1}, ω_{2}) Q^{h, i} (- ω_{1}, - ω_{2}); \end{array}$

$(\begin{array}{l} {\tilde{P}}^{+, +} (ω_{1}, ω_{2}) = {\tilde{λ}}_{0} (ω_{1}, ω_{2}) \tilde{P} (ω_{1}, ω_{2}) + \sum_{j = 1}^{a - 1} {\tilde{λ}}_{j} (ω_{1}, ω_{2}) {\tilde{Q}}^{h, i} (ω_{1}, ω_{2}), \\ {\tilde{P}}^{+, -} (ω_{1}, ω_{2}) = {\tilde{λ}}_{a} (ω_{1}, ω_{2}) \tilde{P} (ω_{1}, - ω_{2}) + \sum_{j = a + 1}^{2 a - 1} {\tilde{λ}}_{j} (ω_{1}, ω_{2}) {\tilde{Q}}^{h, i} (ω_{1}, - ω_{2}), \\ {\tilde{P}}^{-, +} (ω_{1}, ω_{2}) = {\tilde{λ}}_{2 a} (ω_{1}, ω_{2}) \tilde{P} (- ω_{1}, ω_{2}) + \sum_{j = 2 a + 1}^{3 a - 1} {\tilde{λ}}_{j} (ω_{1}, ω_{2}) {\tilde{Q}}^{h, i} (- ω_{1}, ω_{2}), \\ {\tilde{P}}^{-, -} (ω_{1}, ω_{2}) = {\tilde{λ}}_{3 a} (ω_{1}, ω_{2}) \tilde{P} (- ω_{1}, - ω_{2}) + \sum_{j = 3 a + 1}^{4 a - 1} {\tilde{λ}}_{j} (ω_{1}, ω_{2}) {\tilde{Q}}^{h, i} (- ω_{1}, - ω_{2}) . \end{array}$

$(h :0 \leq h \leq a - 10 \leq h \leq 4 a - 1; i :1 \leq i \leq 3; i, h \in Z^{+})$ 。函数 $λ_{j} (ω_{1}, ω_{2})$ 和 ${\tilde{λ}}_{j} (ω_{1}, ω_{2})$ 以 $\frac{2 π}{a}$ 为周期，且满足

1) $(\begin{array}{l} \sum_{j = 0}^{4 a - 1} λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{j} (- ω_{1}, - ω_{2}) = 1, \\ \sum_{j = 0}^{a - 1} {λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{a + j} (- ω_{1}, - ω_{2}) + λ_{2 a + j} (ω_{1}, ω_{2}) {\tilde{λ}}_{3 a + j} (- ω_{1}, - ω_{2})} = 0, \\ \sum_{j = 0}^{a - 1} {λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{2 a + j} (- ω_{1}, - ω_{2}) + λ_{a + j} (ω_{1}, ω_{2}) {\tilde{λ}}_{3 a + j} (- ω_{1}, - ω_{2})} = 0, \\ \sum_{j = 0}^{a - 1} {λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{3 a + j} (- ω_{1}, - ω_{2}) + λ_{a + j} (ω_{1}, ω_{2}) {\tilde{λ}}_{2 a + j} (- ω_{1}, - ω_{2})} = 0, \\ λ_{0} (0, 0) + λ_{a} (0, 0) + λ_{2 a} (0, 0) + λ_{3 a} (0, 0) = 1, \\ {\tilde{λ}}_{0} (0, 0) + {\tilde{λ}}_{a} (0, 0) + {\tilde{λ}}_{2 a} (0, 0) + {\tilde{λ}}_{3 a} (0, 0) = 1, \\ | λ_{0} (0, 0) - λ_{a} (0, 0) - λ_{2 a} (0, 0) + λ_{3 a} (0, 0) | < 1, \\ | {\tilde{λ}}_{0} (0, 0) - {\tilde{λ}}_{a} (0, 0) - {\tilde{λ}}_{2 a} (0, 0) + {\tilde{λ}}_{3 a} (0, 0) | < 1, \\ | λ_{0} (0, 0) + λ_{a} (0, 0) - λ_{2 a} (0, 0) - λ_{3 a} (0, 0) | < 1, \\ | {\tilde{λ}}_{0} (0, 0) + {\tilde{λ}}_{a} (0, 0) - {\tilde{λ}}_{2 a} (0, 0) - {\tilde{λ}}_{3 a} (0, 0) | < 1, \\ | λ_{0} (0, 0) - λ_{a} (0, 0) + λ_{2 a} (0, 0) - λ_{3 a} (0, 0) | < 1, \\ | {\tilde{λ}}_{0} (0, 0) - {\tilde{λ}}_{a} (0, 0) + {\tilde{λ}}_{2 a} (0, 0) - {\tilde{λ}}_{3 a} (0, 0) | < 1; \end{array}$

2) $(\begin{array}{l} \sum_{j = 0}^{4 a - 1} λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{j} (- ω_{1}, - ω_{2}) = 1, \\ \sum_{j = 0}^{a - 1} {λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{a + j} (- ω_{1}, - ω_{2}) + λ_{2 a + j} (ω_{1}, ω_{2}) {\tilde{λ}}_{3 a + j} (- ω_{1}, - ω_{2})} = 0, \\ \sum_{j = 0}^{a - 1} {λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{2 a + j} (- ω_{1}, - ω_{2}) + λ_{a + j} (ω_{1}, ω_{2}) {\tilde{λ}}_{3 a + j} (- ω_{1}, - ω_{2})} = 0, \\ \sum_{j = 0}^{a - 1} {λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{3 a + j} (- ω_{1}, - ω_{2}) + λ_{a + j} (ω_{1}, ω_{2}) {\tilde{λ}}_{2 a + j} (- ω_{1}, - ω_{2})} = 0, \\ | λ_{0} (0, 0) + λ_{a} (0, 0) + λ_{2 a} (0, 0) + λ_{3 a} (0, 0) | < 1, \\ | {\tilde{λ}}_{0} (0, 0) + {\tilde{λ}}_{a} (0, 0) + {\tilde{λ}}_{2 a} (0, 0) + {\tilde{λ}}_{3 a} (0, 0) | < 1, \\ λ_{0} (0, 0) - λ_{a} (0, 0) - λ_{2 a} (0, 0) + λ_{3 a} (0, 0) = 1, \\ {\tilde{λ}}_{0} (0, 0) - {\tilde{λ}}_{a} (0, 0) - {\tilde{λ}}_{2 a} (0, 0) + {\tilde{λ}}_{3 a} (0, 0) = 1, \\ | λ_{0} (0, 0) + λ_{a} (0, 0) - λ_{2 a} (0, 0) - λ_{3 a} (0, 0) | < 1, \\ | {\tilde{λ}}_{0} (0, 0) + {\tilde{λ}}_{a} (0, 0) - {\tilde{λ}}_{2 a} (0, 0) - {\tilde{λ}}_{3 a} (0, 0) | < 1, \\ | λ_{0} (0, 0) - λ_{a} (0, 0) + λ_{2 a} (0, 0) - λ_{3 a} (0, 0) | < 1, \\ | {\tilde{λ}}_{0} (0, 0) - {\tilde{λ}}_{a} (0, 0) + {\tilde{λ}}_{2 a} (0, 0) - {\tilde{λ}}_{3 a} (0, 0) | < 1; \end{array}$

3) $(\begin{array}{l} \sum_{j = 0}^{4 a - 1} λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{j} (- ω_{1}, - ω_{2}) = 1, \\ \sum_{j = 0}^{a - 1} {λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{a + j} (- ω_{1}, - ω_{2}) + λ_{2 a + j} (ω_{1}, ω_{2}) {\tilde{λ}}_{3 a + j} (- ω_{1}, - ω_{2})} = 0, \\ \sum_{j = 0}^{a - 1} {λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{2 a + j} (- ω_{1}, - ω_{2}) + λ_{a + j} (ω_{1}, ω_{2}) {\tilde{λ}}_{3 a + j} (- ω_{1}, - ω_{2})} = 0, \\ \sum_{j = 0}^{a - 1} {λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{3 a + j} (- ω_{1}, - ω_{2}) + λ_{a + j} (ω_{1}, ω_{2}) {\tilde{λ}}_{2 a + j} (- ω_{1}, - ω_{2})} = 0, \\ | λ_{0} (0, 0) + λ_{a} (0, 0) + λ_{2 a} (0, 0) + λ_{3 a} (0, 0) | < 1, \\ | {\tilde{λ}}_{0} (0, 0) + {\tilde{λ}}_{a} (0, 0) + {\tilde{λ}}_{2 a} (0, 0) + {\tilde{λ}}_{3 a} (0, 0) | < 1, \\ | λ_{0} (0, 0) - λ_{a} (0, 0) - λ_{2 a} (0, 0) + λ_{3 a} (0, 0) | < 1, \\ | {\tilde{λ}}_{0} (0, 0) - {\tilde{λ}}_{a} (0, 0) - {\tilde{λ}}_{2 a} (0, 0) + {\tilde{λ}}_{3 a} (0, 0) | < 1, \\ λ_{0} (0, 0) + λ_{a} (0, 0) - λ_{2 a} (0, 0) - λ_{3 a} (0, 0) = 1, \\ {\tilde{λ}}_{0} (0, 0) + {\tilde{λ}}_{a} (0, 0) - {\tilde{λ}}_{2 a} (0, 0) - {\tilde{λ}}_{3 a} (0, 0) = 1, \\ | λ_{0} (0, 0) - λ_{a} (0, 0) + λ_{2 a} (0, 0) - λ_{3 a} (0, 0) | < 1, \\ | {\tilde{λ}}_{0} (0, 0) - {\tilde{λ}}_{a} (0, 0) + {\tilde{λ}}_{2 a} (0, 0) - {\tilde{λ}}_{3 a} (0, 0) | < 1; \end{array}$

4) $(\begin{array}{l} \sum_{j = 0}^{4 a - 1} λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{j} (- ω_{1}, - ω_{2}) = 1, \\ \sum_{j = 0}^{a - 1} {λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{a + j} (- ω_{1}, - ω_{2}) + λ_{2 a + j} (ω_{1}, ω_{2}) {\tilde{λ}}_{3 a + j} (- ω_{1}, - ω_{2})} = 0, \\ \sum_{j = 0}^{a - 1} {λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{2 a + j} (- ω_{1}, - ω_{2}) + λ_{a + j} (ω_{1}, ω_{2}) {\tilde{λ}}_{3 a + j} (- ω_{1}, - ω_{2})} = 0, \\ \sum_{j = 0}^{a - 1} {λ_{j} (ω_{1}, ω_{2}) {\tilde{λ}}_{3 a + j} (- ω_{1}, - ω_{2}) + λ_{a + j} (ω_{1}, ω_{2}) {\tilde{λ}}_{2 a + j} (- ω_{1}, - ω_{2})} = 0, \\ | λ_{0} (0, 0) + λ_{a} (0, 0) + λ_{2 a} (0, 0) + λ_{3 a} (0, 0) | < 1, \\ | {\tilde{λ}}_{0} (0, 0) + {\tilde{λ}}_{a} (0, 0) + {\tilde{λ}}_{2 a} (0, 0) + {\tilde{λ}}_{3 a} (0, 0) | < 1, \\ | λ_{0} (0, 0) - λ_{a} (0, 0) - λ_{2 a} (0, 0) + λ_{3 a} (0, 0) | < 1, \\ | {\tilde{λ}}_{0} (0, 0) - {\tilde{λ}}_{a} (0, 0) - {\tilde{λ}}_{2 a} (0, 0) + {\tilde{λ}}_{3 a} (0, 0) | < 1, \\ | λ_{0} (0, 0) + λ_{a} (0, 0) - λ_{2 a} (0, 0) - λ_{3 a} (0, 0) | < 1, \\ | {\tilde{λ}}_{0} (0, 0) + {\tilde{λ}}_{a} (0, 0) - {\tilde{λ}}_{2 a} (0, 0) - {\tilde{λ}}_{3 a} (0, 0) | < 1, \\ λ_{0} (0, 0) - λ_{a} (0, 0) + λ_{2 a} (0, 0) - λ_{3 a} (0, 0) = 1, \\ {\tilde{λ}}_{0} (0, 0) - {\tilde{λ}}_{a} (0, 0) + {\tilde{λ}}_{2 a} (0, 0) - {\tilde{λ}}_{3 a} (0, 0) = 1; \end{array}$

则 $P^{+, +} (ω_{1}, ω_{2}), P^{+, -} (ω_{1}, ω_{2}), P^{-, +} (ω_{1}, ω_{2}), P^{-, -} (ω_{1}, ω_{2})$ 和 ${\tilde{P}}^{+, +} (ω_{1}, ω_{2}), {\tilde{P}}^{+, -} (ω_{1}, ω_{2}), {\tilde{P}}^{-, +} (ω_{1}, ω_{2}), {\tilde{P}}^{-, -} (ω_{1}, ω_{2})$ ，产生一个双正交二维四向多细分函数 $Φ (x, y)$ 并满足

$Φ (x, y) = \sum_{k, l} P_{l, k}^{+, +} Φ (a x - k, a y - l) + \sum_{k, l} P_{l, k}^{+, -} Φ (a x - k, l - a y) + \sum_{k, l} P_{l, k}^{-, +} Φ (k - a x, a y - l) + \sum_{k, l} P_{l, k}^{-, -} Φ (k - a x, l - a y) .$

$\tilde{Φ} (x, y) = \sum_{k, l} {\tilde{P}}_{l, k}^{+, +} \tilde{Φ} (a x - k, a y - l) + \sum_{k, l} {\tilde{P}}_{l, k}^{+, -} \tilde{Φ} (a x - k, l - a y) + \sum_{k, l} {\tilde{P}}_{l, k}^{-, +} \tilde{Φ} (k - a x, a y - l) + \sum_{k, l} {\tilde{P}}_{l, k}^{-, -} \tilde{Φ} (k - a x, l - a y) .$

证明：将上面已经构造的 $P^{+, +} (ω_{1}, ω_{2}), P^{+, -} (ω_{1}, ω_{2}), P^{-, +} (ω_{1}, ω_{2}), P^{-, -} (ω_{1}, ω_{2})$ 和 ${\tilde{P}}^{+, +} (ω_{1}, ω_{2}), {\tilde{P}}^{+, -} (ω_{1}, ω_{2}), {\tilde{P}}^{-, +} (ω_{1}, ω_{2}), {\tilde{P}}^{-, -} (ω_{1}, ω_{2})$ ，代入(15)式化简可得。再令(11)式 $P (ω_{1}, ω_{2}) = P (0,0)$ ，则化简 $P (0,0)$ 并求特征值可证得定理。

定理10：如果 $Φ (x, y)$ 和 $\tilde{Φ} (x, y)$ 是双正交二维四向多加细函数， $Ψ^{h, i} (x, y)$ 和 ${\tilde{Ψ}}^{h, i} (x, y)$ 是相应的双正交二维四向多小波函数， $P (ω_{1}, ω_{2}), \tilde{P} (ω_{1}, ω_{2})$ 和 $Q^{j, s} (ω_{1}, ω_{2}), {\tilde{Q}}^{j, s} (ω_{1}, ω_{2})$ 是面具符号，构造

$(\begin{array}{l} Q^{h, i, +, +} (ω_{1}, ω_{2}) = μ_{0}^{h, i} (ω_{1}, ω_{2}) P (ω_{1}, ω_{2}) + \sum_{j = 1}^{a - 1} μ_{j}^{h, i} (ω_{1}, ω_{2}) Q^{j, s} (ω_{1}, ω_{2}), \\ Q^{h, i, +, -} (ω_{1}, ω_{2}) = μ_{a}^{h, i} (ω_{1}, ω_{2}) P (ω_{1}, - ω_{2}) + \sum_{j = a + 1}^{2 a - 1} μ_{j}^{h, i} (ω_{1}, ω_{2}) Q^{j, s} (ω_{1}, - ω_{2}), \\ Q^{h, i, -, +} (ω_{1}, ω_{2}) = μ_{2 a}^{h, i} (ω_{1}, ω_{2}) P (- ω_{1}, ω_{2}) + \sum_{j = 2 a + 1}^{3 a - 1} μ_{j}^{h, i} (ω_{1}, ω_{2}) Q^{j, s} (- ω_{1}, ω_{2}), \\ Q^{h, i, -, -} (ω_{1}, ω_{2}) = μ_{3 a}^{h, i} (ω_{1}, ω_{2}) P (- ω_{1}, - ω_{2}) + \sum_{j = 3 a + 1}^{4 a - 1} μ_{j}^{h, i} (ω_{1}, ω_{2}) Q^{j, s} (- ω_{1}, - ω_{2}); \end{array}$

$(\begin{array}{l} {\tilde{Q}}^{h, i, +, +} (ω_{1}, ω_{2}) = {\tilde{μ}}_{0}^{h, i} (ω_{1}, ω_{2}) \tilde{P} (ω_{1}, ω_{2}) + \sum_{j = 1}^{a - 1} {\tilde{μ}}_{j}^{h, i} (ω_{1}, ω_{2}) {\tilde{Q}}^{j, s} (ω_{1}, ω_{2}), \\ {\tilde{Q}}^{h, i, +, -} (ω_{1}, ω_{2}) = {\tilde{μ}}_{a}^{h, i} (ω_{1}, ω_{2}) \tilde{P} (ω_{1}, - ω_{2}) + \sum_{j = a + 1}^{2 a - 1} {\tilde{μ}}_{j}^{h, i} (ω_{1}, ω_{2}) {\tilde{Q}}^{j, s} (ω_{1}, - ω_{2}), \\ {\tilde{Q}}^{h, i, -, +} (ω_{1}, ω_{2}) = {\tilde{μ}}_{2 a}^{h, i} (ω_{1}, ω_{2}) \tilde{P} (- ω_{1}, ω_{2}) + \sum_{j = 2 a + 1}^{3 a - 1} {\tilde{μ}}_{j}^{h, i} (ω_{1}, ω_{2}) {\tilde{Q}}^{j, s} (- ω_{1}, ω_{2}), \\ {\tilde{Q}}^{h, i, -, -} (ω_{1}, ω_{2}) = {\tilde{μ}}_{3 a}^{h, i} (ω_{1}, ω_{2}) \tilde{P} (- ω_{1}, - ω_{2}) + \sum_{j = 3 a + 1}^{4 a - 1} {\tilde{μ}}_{j}^{h, i} (ω_{1}, ω_{2}) {\tilde{Q}}^{j, s} (- ω_{1}, - ω_{2}) . \end{array}$

$(h :0 \leq h \leq a - 10 \leq h \leq 4 a - 1; i, s :1 \leq i, s \leq 3; i, h, s \in Z^{+})$ 。函数 $λ_{j} (ω_{1}, ω_{2})$ ， $μ_{j}^{h, i} (ω_{1}, ω_{2})$ 和 ${\tilde{μ}}_{j}^{h, i} (ω_{1}, ω_{2})$ 以 $\frac{2 π}{a}$ 为周期，且满足

2) $(\begin{array}{l} \sum_{j = 0}^{4 a - 1} μ_{j}^{h, k} (ω_{1}, ω_{2}) {\tilde{μ}}_{j}^{s, t} (- ω_{1}, - ω_{2}) = δ_{h, s} δ_{k, t}, \\ \sum_{j = 0}^{a - 1} {μ_{j}^{h, i} (ω_{1}, ω_{2}) {\tilde{μ}}_{a + j}^{h, i} (- ω_{1}, - ω_{2}) + μ_{2 a + j}^{h, i} (ω_{1}, ω_{2}) {\tilde{μ}}_{3 a + j}^{h, i} (- ω_{1}, - ω_{2})} = 0, \\ \sum_{j = 0}^{a - 1} {μ_{j}^{h, i} (ω_{1}, ω_{2}) {\tilde{μ}}_{2 a + j}^{h, i} (- ω_{1}, - ω_{2}) + μ_{a + j}^{h, i} (ω_{1}, ω_{2}) {\tilde{μ}}_{3 a + j}^{h, i} (- ω_{1}, - ω_{2})} = 0, \\ \sum_{j = 0}^{a - 1} {μ_{j}^{h, i} (ω_{1}, ω_{2}) {\tilde{μ}}_{3 a + j}^{h, i} (- ω_{1}, - ω_{2}) + μ_{a + j}^{h, i} (ω_{1}, ω_{2}) {\tilde{μ}}_{2 a + j}^{h, i} (- ω_{1}, - ω_{2})} = 0; \end{array}$

那么就可以得出 $Q^{h, i, +, +} (ω_{1}, ω_{2}), Q^{h, i, +, -} (ω_{1}, ω_{2}), Q^{h, i, -, +} (ω_{1}, ω_{2}), Q^{h, i, -, -} (ω_{1}, ω_{2})$ 和 ${\tilde{Q}}^{h, i, +, +} (ω_{1}, ω_{2}), {\tilde{Q}}^{h, i, +, -} (ω_{1}, ω_{2}), {\tilde{Q}}^{h, i, -, +} (ω_{1}, ω_{2}), {\tilde{Q}}^{h, i, -, -} (ω_{1}, ω_{2})$ 产生一个双正交二维四向多小波函数 $Φ (x, y)$ 并满足

证明：由于 $Φ (x, y)$ ， $\tilde{Φ} (x, y)$ 和 $Ψ^{h, i} (x, y)$ ， ${\tilde{Ψ}}^{h, i} (x, y)$ 的双正交性，可以将已构造的 $Q^{h, i, +, +} (ω_{1}, ω_{2}), Q^{h, i, +, -} (ω_{1}, ω_{2}), Q^{h, i, -, +} (ω_{1}, ω_{2}), Q^{h, i, -, -} (ω_{1}, ω_{2})$ 和 ${\tilde{Q}}^{h, i, +, +} (ω_{1}, ω_{2}), {\tilde{Q}}^{h, i, +, -} (ω_{1}, ω_{2}), {\tilde{Q}}^{h, i, -, +} (ω_{1}, ω_{2}), {\tilde{Q}}^{h, i, -, -} (ω_{1}, ω_{2})$ ，代入(16)式和(19)式化简可证得定理。

6. Mallat算法

如果 $Φ (x, y)$ 和 $\tilde{Φ} (x, y)$ 是双正交二维四向多加细函数， $Ψ^{h, i} (x, y)$ 和 ${\tilde{Ψ}}^{h, i} (x, y)$ 是相应的双正交二维四向多小波函数，对于能量有限信号 $f (x, y) \in L (R^{2})$ 在分辨率 $a_{j}$ 下的近似函数 $f_{j} (x, y) \in V_{j}$ ，则 $V_{j + 1}$ 可分解为 $V_{j + 1} = V_{j} \oplus W_{j} = V_{j} \oplus_{h = 1}^{a - 1} {W_{j}^{h, 1} \oplus W_{j}^{h, 2} \oplus W_{j}^{h, 3}}$ ，则 $\forall f_{j + 1} (x, y) \in V_{j + 1}$ ，记 $c_{j, k_{1}, k_{2}} = {[c_{j, k_{1}, k_{2}}^{1}, c_{j, k_{1}, k_{2}}^{2}, \dots, c_{j, k_{1}, k_{2}}^{r}]}^{T}$ ， $d_{j, k_{1}, k_{2}} = {[d_{j, k_{1}, k_{2}}^{1}, d_{j, k_{1}, k_{2}}^{2}, \dots, d_{j, k_{1}, k_{2}}^{r}]}^{T}$ 有

$\begin{matrix} f_{j + 1} (x, y) = \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, +} Φ_{j + 1, k_{1}, k_{2}}^{+, +} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, -} Φ_{j + 1, k_{1}, k_{2}}^{+, -} \\ + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, +} Φ_{j + 1, k_{1}, k_{2}}^{-, +} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, -} Φ_{j + 1, k_{1}, k_{2}}^{-, -} \\ = \sum_{k_{1}, k_{2}} c_{j, k_{1}, k_{2}}^{+, +} Φ_{j, k_{1}, k_{2}}^{+, +} + \sum_{k_{1}, k_{2}} c_{j, k_{1}, k_{2}}^{+, -} Φ_{j, k_{1}, k_{2}}^{+, -} + \sum_{k_{1}, k_{2}} c_{j, k_{1}, k_{2}}^{-, +} Φ_{j, k_{1}, k_{2}}^{-, +} \\ + \sum_{k_{1}, k_{2}} c_{j, k_{1}, k_{2}}^{-, -} Φ_{j, k_{1}, k_{2}}^{-, -} + \sum_{h = 1}^{a - 1} \sum_{r = 1}^{3} [\sum_{k_{1}, k_{2}} d_{j, k_{1}, k_{2}}^{h, i, +, +} Ψ_{j, k_{1}, k_{2}}^{h, i, +, +} + \sum_{k_{1}, k_{2}} d_{j, k_{1}, k_{2}}^{h, i, +, -} Ψ_{j, k_{1}, k_{2}}^{h, i, +, -} \\ + \sum_{k_{1}, k_{2}} d_{j, k_{1}, k_{2}}^{h, i, -, +} Ψ_{j, k_{1}, k_{2}}^{h, i, -, +} + \sum_{k_{1}, k_{2}} d_{j, k_{1}, k_{2}}^{h, i, -, -} Ψ_{j, k_{1}, k_{2}}^{h, i, -, -}] \end{matrix}$

6.1. 分解算法

如果 $f (x, y)$ 在分辨率 $a_{j + 1}$ 下的近似函数为 $f_{j + 1} (x, y)$ ，那么它可以进一步分解为 $f (x, y)$ 在 $a_{j}$ 下的主要部分(通过低通滤波器得到)和细节部分(通过高通滤波器得到)。即分解算法要实现的目标就是：已知 $c_{j + 1, k_{1}, k_{2}}^{+, +}, c_{j + 1, k_{1}, k_{2}}^{+, -}, c_{j + 1, k_{1}, k_{2}}^{-, +}, c_{j + 1, k_{1}, k_{2}}^{-, -}$ ，求 $c_{j, k_{1}, k_{2}}^{+, +}, c_{j, k_{1}, k_{2}}^{+, -}, c_{j, k_{1}, k_{2}}^{-, +}, c_{j, k_{1}, k_{2}}^{-, -}$ 和 $d_{j, k_{1}, k_{2}}^{+, +}, d_{j, k_{1}, k_{2}}^{+, -}, d_{j, k_{1}, k_{2}}^{-, +}, d_{j, k_{1}, k_{2}}^{-, -}$ 。

定理11：若 $f_{j + 1} (x, y) = \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, +} Φ_{j + 1, k_{1}, k_{2}}^{+, +} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, -} Φ_{j + 1, k_{1}, k_{2}}^{+, -} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, +} Φ_{j + 1, k_{1}, k_{2}}^{-, +} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, -} Φ_{j + 1, k_{1}, k_{2}}^{-, -}$ 并且由(1)式和(17)式，我们可以得到下列分解公式

$\begin{matrix} c_{j, n_{1}, n_{2}}^{+, +} = \frac{1}{a^{2}} [\sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, +} \bar{P_{k_{1} - a n_{1}, k_{2} - a n_{2}}^{+, +}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, -} \bar{P_{k_{1} - a n_{1}, k_{2} - a n_{2}}^{+, -}} \\ + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, +} \bar{P_{k_{1} - a n_{1}, k_{2} - a n_{2}}^{-, +}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, -} \bar{P_{k_{1} - a n_{1}, k_{2} - a n_{2}}^{-, -}}]; \end{matrix}$

$\begin{matrix} c_{j, n_{1}, n_{2}}^{+, -} = \frac{1}{a^{2}} [\sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, +} \bar{P_{k_{1} - a n_{1}, a n_{2} - k_{2}}^{+, -}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, -} \bar{P_{k_{1} - a n_{1}, a n_{2} - k_{2}}^{+, +}} \\ + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, +} \bar{P_{k_{1} - a n_{1}, a n_{2} - k_{2}}^{-, -}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, -} \bar{P_{k_{1} - a n_{1}, a n_{2} - k_{2}}^{-, +}}]; \end{matrix}$

$\begin{matrix} c_{j, n_{1}, n_{2}}^{-, +} = \frac{1}{a^{2}} [\sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, +} \bar{P_{a n_{1} - k_{1}, k_{2} - a n_{2}}^{-, +}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, -} \bar{P_{a n_{1} - k_{1}, k_{2} - a n_{2}}^{-, -}} \\ + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, +} \bar{P_{a n_{1} - k_{1}, k_{2} - a n_{2}}^{+, +}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, -} \bar{P_{a n_{1} - k_{1}, k_{2} - a n_{2}}^{+, -}}]; \end{matrix}$

$\begin{matrix} c_{j, n_{1}, n_{2}}^{-, -} = \frac{1}{a^{2}} [\sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, +} \bar{P_{a n_{1} - k_{1}, a n_{2} - k_{2}}^{-, -}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, -} \bar{P_{a n_{1} - k_{1}, a n_{2} - k_{2}}^{-, +}} \\ + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, +} \bar{P_{a n_{1} - k_{1}, a n_{2} - k_{2}}^{-, +}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, -} \bar{P_{a n_{1} - k_{1}, a n_{2} - k_{2}}^{+, +}}]; \end{matrix}$

$\begin{matrix} d_{j, n_{1}, n_{2}}^{h, i, +, +} = \frac{1}{a^{2}} [\sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, +} \bar{Q_{k_{1} - a n_{1}, k_{2} - a n_{2}}^{h, i, +, +}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, -} \bar{Q_{k_{1} - a n_{1}, k_{2} - a n_{2}}^{h, i, +, -}} \\ + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, +} \bar{Q_{k_{1} - a n_{1}, k_{2} - a n_{2}}^{h, i, -, +}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, -} \bar{Q_{k_{1} - a n_{1}, k_{2} - a n_{2}}^{h, i, -, -}}]; \end{matrix}$

$\begin{matrix} d_{j, n_{1}, n_{2}}^{h, i, +, -} = \frac{1}{a^{2}} [\sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, +} \bar{Q_{k_{1} - a n_{1}, a n_{2} - k_{2}}^{h, i, +, -}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, -} \bar{Q_{k_{1} - a n_{1}, a n_{2} - k_{2}}^{h, i, +, +}} \\ + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, +} \bar{Q_{k_{1} - a n_{1}, a n_{2} - k_{2}}^{h, i, -, -}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, -} \bar{Q_{k_{1} - a n_{1}, a n_{2} - k_{2}}^{h, i, -, +}}]; \end{matrix}$

$\begin{matrix} d_{j, n_{1}, n_{2}}^{h, i, -, +} = \frac{1}{a^{2}} [\sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, +} \bar{Q_{a n_{1} - k_{1}, k_{2} - a n_{2}}^{h, i, -, +}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, -} \bar{Q_{a n_{1} - k_{1}, k_{2} - a n_{2}}^{h, i, -, -}} \\ + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, +} \bar{Q_{a n_{1} - k_{1}, k_{2} - a n_{2}}^{h, i, +, +}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, -} \bar{Q_{a n_{1} - k_{1}, k_{2} - a n_{2}}^{h, i, +, -}}]; \end{matrix}$

$\begin{matrix} d_{j, n_{1}, n_{2}}^{h, i, -, -} = \frac{1}{a^{2}} [\sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, +} \bar{Q_{a n_{1} - k_{1}, a n_{2} - k_{2}}^{h, i, -, -}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, -} \bar{Q_{a n_{1} - k_{1}, a n_{2} - k_{2}}^{h, i, -, +}} \\ + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, +} \bar{Q_{a n_{1} - k_{1}, a n_{2} - k_{2}}^{h, i, +, -}} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, -} \bar{Q_{a n_{1} - k_{1}, a n_{2} - k_{2}}^{h, i, +, +}}]; \end{matrix}$

其中： $h = 1, 2, \dots, a - 1; i = 1, 2, 3$ 。

证明：因为 ${Φ (x - k, y - l), Φ (x - k, l - y), Φ (k - x, y - l), Φ (k - x, l - y) : k, l \in Z}$ 是 $V_{0}$ 的标准正交基，所以就有

$\begin{matrix} c_{j, n_{1}, n_{2}}^{+, +} = 〈 f_{j + 1}, Φ_{j, n_{1}, n_{2}}^{+, +} 〉 \\ = \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, +} 〈 Φ_{j + 1, n_{1}, n_{2}}^{+, +}, Φ_{j, n_{1}, n_{2}}^{+, +} 〉 \\ + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, -} 〈 Φ_{j + 1, n_{1}, n_{2}}^{+, -}, Φ_{j, n_{1}, n_{2}}^{+, +} 〉 \\ + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, +} 〈 Φ_{j + 1, n_{1}, n_{2}}^{-, +}, Φ_{j, n_{1}, n_{2}}^{+, +} 〉 \\ + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, -} 〈 Φ_{j + 1, n_{1}, n_{2}}^{-, -}, Φ_{j, n_{1}, n_{2}}^{+, +} 〉, \end{matrix}$

则根据(1)式和(20)式可以知道， $〈 Φ_{j + 1, n_{1}, n_{2}}^{+, +}, Φ_{j, n_{1}, n_{2}}^{+, +} 〉 = \frac{1}{a^{2}} \bar{P_{k_{1} - a n_{1}, k_{2} - a n_{2}}^{+, +}}$ 。同理可以证得：

$\begin{array}{l} 〈 Φ_{j + 1, n_{1}, n_{2}}^{+, -}, Φ_{j, n_{1}, n_{2}}^{+, +} 〉 = \frac{1}{a^{2}} \bar{P_{k_{1} - a n_{1}, k_{2} - a n_{2}}^{+, -}}; \\ 〈 Φ_{j + 1, n_{1}, n_{2}}^{-, +}, Φ_{j, n_{1}, n_{2}}^{+, +} 〉 = \frac{1}{a^{2}} \bar{P_{k_{1} - a n_{1}, k_{2} - a n_{2}}^{-, +}}; \\ 〈 Φ_{j + 1, n_{1}, n_{2}}^{-, -}, Φ_{j, n_{1}, n_{2}}^{+, +} 〉 = \frac{1}{a^{2}} \bar{P_{k_{1} - a n_{1}, k_{2} - a n_{2}}^{-, -}}, \end{array}$

故我们就可以得到

其它等式同理可证。

6.2. 重构算法

通过 $f (x, y)$ 在分辨率 $a_{j}$ 下的主要部分和细节部分，来重构 $f (x, y)$ 在分辨率 $a_{j + 1}$ 下的主要部分，也即重构算法是分解算法的逆过程，那么要实现的目标就是：已知 $c_{j, k_{1}, k_{2}}^{+, +}, c_{j, k_{1}, k_{2}}^{+, -}, c_{j, k_{1}, k_{2}}^{-, +}, c_{j, k_{1}, k_{2}}^{-, -}$ 和 $d_{j, k_{1}, k_{2}}^{+, +}, d_{j, k_{1}, k_{2}}^{+, -}, d_{j, k_{1}, k_{2}}^{-, +}, d_{j, k_{1}, k_{2}}^{-, -}$ 求 $c_{j + 1, k_{1}, k_{2}}^{+, +}, c_{j + 1, k_{1}, k_{2}}^{+, -}, c_{j + 1, k_{1}, k_{2}}^{-, +}, c_{j + 1, k_{1}, k_{2}}^{-, -}$ 。

定理12：若 $f_{j + 1} (x, y) = \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, +} Φ_{j + 1, k_{1}, k_{2}}^{+, +} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{+, -} Φ_{j + 1, k_{1}, k_{2}}^{+, -} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, +} Φ_{j + 1, k_{1}, k_{2}}^{-, +} + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, -} Φ_{j + 1, k_{1}, k_{2}}^{-, -}$ 并且由(1)式和(17)式，我们可以得到下列重构公式

$\begin{matrix} c_{j + 1, n_{1}, n_{2}}^{+, +} = \frac{1}{a^{2}} [\sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{+, +} P_{n_{1} - a k_{1}, n_{2} - a k_{2}}^{+, +} + \sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{+, -} P_{n_{1} - a k_{1}, n_{2} - a k_{2}}^{+, -} \\ + \sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{-, +} P_{n_{1} - a k_{1}, n_{2} - a k_{2}}^{-, +} + \sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{-, -} P_{n_{1} - a k_{1}, n_{2} - a k_{2}}^{-, -} \\ + \sum_{h = 1}^{a - 1} \sum_{r = 1}^{3} (\sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, +, +} Q_{n_{1} - a k_{1}, n_{2} - a k_{2}}^{h, i, +, +} + \sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, +, -} Q_{n_{1} - a k_{1}, n_{2} - a k_{2}}^{h, i, +, -} \\ + \sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, -, +} Q_{n_{1} - a k_{1}, n_{2} - a k_{2}}^{h, i, + -, +} + \sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, -, -} Q_{n_{1} - a k_{1}, n_{2} - a k_{2}}^{h, i, -, -})]; \end{matrix}$

$\begin{matrix} c_{j + 1, n_{1}, n_{2}}^{+, -} = \frac{1}{a^{2}} [\sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{+, +} P_{n_{1} - a k_{1}, a k_{2} - n_{2}}^{+, -} + \sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{+, -} P_{n_{1} - a k_{1}, a k_{2} - n_{2}}^{+, +} \\ + \sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{-, +} P_{n_{1} - a k_{1}, a k_{2} - n_{2}}^{-, -} + \sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{-, -} P_{n_{1} - a k_{1}, a k_{2} - n_{2}}^{-, +} \\ + \sum_{h = 1}^{a - 1} \sum_{r = 1}^{3} (\sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, +, +} Q_{n_{1} - a k_{1}, a k_{2} - n_{2}}^{h, i, +, -} + \sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, +, -} Q_{n_{1} - a k_{1}, a k_{2} - n_{2}}^{h, i, +, +} \\ + \sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, -, +} Q_{n_{1} - a k_{1}, a k_{2} - n_{2}}^{h, i, -, -} + \sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, -, -} Q_{n_{1} - a k_{1}, a k_{2} - n_{2}}^{h, i, -, +})]; \end{matrix}$

$\begin{matrix} c_{j + 1, n_{1}, n_{2}}^{-, +} = \frac{1}{a^{2}} [\sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{+, +} P_{a k_{1} - n_{1}, n_{2} - a k_{2}}^{-, +} + \sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{+, -} P_{a k_{1} - n_{1}, n_{2} - a k_{2}}^{-, -} \\ + \sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{-, +} P_{a k_{1} - n_{1}, n_{2} - a k_{2}}^{+, +} + \sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{-, -} P_{a k_{1} - n_{1}, n_{2} - a k_{2}}^{+, -} \\ + \sum_{h = 1}^{a - 1} \sum_{r = 1}^{3} (\sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, +, +} Q_{a k_{1} - n_{1}, n_{2} - a k_{2}}^{h, i, -, +} + \sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, +, -} Q_{a k_{1} - n_{1}, n_{2} - a k_{2}}^{h, i, -, -} \\ + \sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, -, +} Q_{a k_{1} - n_{1}, n_{2} - a k_{2}}^{h, i, +, +} + \sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, -, -} Q_{a k_{1} - n_{1}, n_{2} - a k_{2}}^{h, i, +, -})]; \end{matrix}$

$\begin{matrix} c_{j + 1, n_{1}, n_{2}}^{-, -} = \frac{1}{a^{2}} [\sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{+, +} P_{a k_{1} - n_{1}, a k_{2} - n_{2}}^{-, -} + \sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{+, -} P_{a k_{1} - n_{1}, a k_{2} - n_{2}}^{-, +} \\ + \sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{-, +} P_{a k_{1} - n_{1}, a k_{2} - n_{2}}^{+, -} + \sum_{k_{1}, k_{2}} c_{j, n_{1}, n_{2}}^{-, -} P_{a k_{1} - n_{1}, a k_{2} - n_{2}}^{+, +} \\ + \sum_{h = 1}^{a - 1} \sum_{r = 1}^{3} (\sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, +, +} Q_{a k_{1} - n_{1}, a k_{2} - n_{2}}^{h, i, -, -} + \sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, +, -} Q_{a k_{1} - n_{1}, a k_{2} - n_{2}}^{h, i, -, +} \\ + \sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, -, +} Q_{a k_{1} - n_{1}, a k_{2} - n_{2}}^{h, i, +, -} + \sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, -, -} Q_{a k_{1} - n_{1}, a k_{2} - n_{2}}^{h, i, +, +})] . \end{matrix}$

证明：由于

$\begin{matrix} c_{j + 1, n_{1}, n_{2}}^{+, +} = 〈 f_{j + 1}, Φ_{j + 1, n_{1}, n_{2}}^{+, +} 〉 \\ = \sum_{k_{1}, k_{2}} c_{j, k_{1}, k_{2}}^{+, +} 〈 Φ_{j, n_{1}, n_{2}}^{+, +}, Φ_{j + 1, n_{1}, n_{2}}^{+, +} 〉 + \sum_{k_{1}, k_{2}} c_{j, k_{1}, k_{2}}^{+, -} 〈 Φ_{j, n_{1}, n_{2}}^{+, -}, Φ_{j + 1, n_{1}, n_{2}}^{+, +} 〉 \\ + \sum_{k_{1}, k_{2}} c_{j, k_{1}, k_{2}}^{-, +} 〈 Φ_{j, n_{1}, n_{2}}^{-, +}, Φ_{j + 1, n_{1}, n_{2}}^{+, +} 〉 + \sum_{k_{1}, k_{2}} c_{j + 1, k_{1}, k_{2}}^{-, -} 〈 Φ_{j, n_{1}, n_{2}}^{-, -}, Φ_{j + 1, n_{1}, n_{2}}^{+, +} 〉 \\ + \sum_{h = 1}^{a - 1} \sum_{r = 1}^{3} [\sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, +, +} 〈 Ψ_{j, k_{1}, k_{2}}^{h, i, +, +}, Φ_{j + 1, n_{1}, n_{2}}^{+, +} 〉 + \sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, +, -} 〈 Ψ_{j, k_{1}, k_{2}}^{h, i, +, -}, Φ_{j + 1, n_{1}, n_{2}}^{+, +} 〉 \\ + \sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, -, +} 〈 Ψ_{j, k_{1}, k_{2}}^{h, i, -, +}, Φ_{j + 1, n_{1}, n_{2}}^{+, +} 〉 + \sum_{k_{1}, k_{2}} d_{j, n_{1}, n_{2}}^{h, i, -, -} 〈 Ψ_{j, k_{1}, k_{2}}^{h, i, -, -}, Φ_{j + 1, n_{1}, n_{2}}^{+, +} 〉], \end{matrix}$

而且 $〈 Φ_{j, n_{1}, n_{2}}^{+, +}, Φ_{j + 1, n_{1}, n_{2}}^{+, +} 〉 = \frac{1}{a^{2}} P_{n_{1} - a k_{1}, n_{2} - a k_{2}}^{+, +}$ ，类似可得

$\begin{array}{l} 〈 Φ_{j, n_{1}, n_{2}}^{+, -}, Φ_{j + 1, n_{1}, n_{2}}^{+, +} 〉 = \frac{1}{a^{2}} P_{n_{1} - a k_{1}, n_{2} - a k_{2}}^{+, -}; \\ 〈 Φ_{j, n_{1}, n_{2}}^{-, +}, Φ_{j + 1, n_{1}, n_{2}}^{+, +} 〉 = \frac{1}{a^{2}} P_{n_{1} - a k_{1}, n_{2} - a k_{2}}^{-, +}; \\ 〈 Φ_{j, n_{1}, n_{2}}^{-, -}, Φ_{j + 1, n_{1}, n_{2}}^{+, +} 〉 = \frac{1}{a^{2}} P_{n_{1} - a k_{1}, n_{2} - a k_{2}}^{-, -} . \end{array}$

故我们可以证得

其它等式同理可证。

文章引用

张旭,李万社. a尺度二维四向双正交多小波的构造和Mallat算法
The Construction and the Mallat Algorithm of Biorthogonal Two Dimensional Four-Direction Multi-Wavelet with Dilation Factor a[J]. 理论数学, 2019, 09(09): 1015-1035. https://doi.org/10.12677/PM.2019.99128

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