\documentclass[12pt]{article} %\documentclass{article} %\usepackage{russian} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{textcomp} %\documentstyle[12pt]{article} %\input amssym.def \setcounter{secnumdepth}{0} \font\aaa=cmsy10 at 10pt \renewcommand{\S}{{\aaa\char120\ }} \renewcommand{\phi}{\varphi} \newcommand{\be}{\begin{equation}} \newcommand{\ee}{\end{equation}} \newcommand{\ba}{\begin{eqnarray}} \newcommand{\ea}{\end{eqnarray}} \newcommand{\ban}{\begin{eqnarray*}} \newcommand{\ean}{\end{eqnarray*}} \newcommand{\too}{\mathop{\longrightarrow}} \newcommand{\nul}{{\bf0}} \newcommand{\odin}{{\bf1}} \newcommand{\cupm}{\mathop{\cup}} \newcommand{\capm}{\mathop{\cap}} \renewcommand{\t}{{\Bbb T}} \newcommand{\td}{{\Bbb T}^d} \newcommand{\rd}{{\Bbb R}^d} \newcommand{\zdd}{{\Bbb Z}^{d-1}} \newcommand{\zddp}{{\Bbb Z}^{d-1}_+} \newcommand{\rdd}{{\Bbb R}^{d-1}} \newcommand{\zd}{{\Bbb Z}^{d}} \newcommand{\tdd}{{\Bbb T}^{d-1}} \newcommand{\ld}{L(\td)} \renewcommand{\r}{{\Bbb R}} \newcommand{\z} {{\Bbb Z}} \newcommand{\n} {{\Bbb N}} \newcommand{\calt}{{\cal T}} \newcommand{\call}{{\cal L}} \newcommand{\cons}{{C\,}} \newcommand{\ddd}{,\dots,} \renewcommand{\lll}{\left(} \newcommand{\rrr}{\right)} \newcommand{\ex}[1]{e^{2\pi i{#1}}} \newcommand{\exm}[1]{e^{-2\pi i{#1}}} \newcommand{\sml}[3]{\sum\limits_{{#1}={#2}}^{#3}} \newtheorem{theo}{Theorem} \newtheorem{lem}[theo]{Lemma} \newtheorem {prop} [theo] {Proposition} \newtheorem {coro} [theo] {Corollary} \newtheorem {defi} [theo] {Definition} \begin {document} \noindent \vspace {.5cm} \begin {center} { \LARGE\bf On Construction of Multivariate Wavelet Frames \footnote {The paper is supported by RFBR, project N 03-01-00373 {\bf Keywords:} wavelet frame, matrix dilation, Unitary Extension Principle Mathematics Subject Classifications 42C40} } \end {center} \begin {center} \bf M. Skopina \end {center} \vspace {.5cm} \begin {abstract} Construction of wavelet frames with matrix dilation is studied. We found a necessary condition and a sufficient condition under which a given pair of refinable functions generates dual wavelet systems with a given number of vanishing moments. \end {abstract} %\subsection {1. Introduction} For image compression and some other applications, it is very desirable to have wavelets with vanishing moment property. %It is well known that the order of vanishing moments is one of the most %important factors for success of wavelets in various applications. In particular, vanishing moments are closely related to the approximation order of wavelet frames (see, e.g.,~\cite{35}). If wavelet system is a basis, the number of its vanishing moments depends only on the dual generating refinable function. Situation is essentially different for frames. Two pairs of dual wavelet frames may be generated by the same refinable functions and have different number of vanishing moments. The goal of this paper is to describe refinable functions generating dual wavelet systems with vanishing moments and to present an explicit method for construction compactly supported wavelet frames with arbitrary number of vanishing moments. Very close problem were investigated by Ming-Jun Lai and A. Petukhov\cite{MP} for univariate wavelet frames. Their technique is not appropriate for multi-dimensional investigations because zero properties of multivariate masks can not be described by means of factorization in contrast to the one-dimensional case . %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\subsection {2. Notations and Preliminary Information} Throughout the paper we will use the following notations. $ \n $ is the set of positive integers, $ \rd $ denotes the $d $-dimensional Euclidean space, $x = (x_1\ddd x_d) $, $y = (y_1\ddd y_d) $ are its elements (vectors), $ (x, y) = x_1y_1 +\dots+x_dy_d $, $ |x | =\sqrt {(x, x)} $, ${\bf e}_j=(0\ddd1\ddd0)$ is the $j$-th unit vector in $\rd$, \ $ \nul = (0\ddd0) \in \rd $, $ \odin = (1\ddd1) \in \rd $; $ \zd $ is the integer lattice in $ \rd $. For $x,y\in\rd$, we write $x>y$ if $x_j>y_j$, $j=1\ddd d$; $ \zd_+ = \{x\in Z^d:\ x\ge\nul \} $. If $\alpha, \beta\in\zd_+$, $a,b\in\rd$, we set $\alpha!=\prod\limits_{j=1}^d\alpha_j!$, $\lll\alpha\atop\beta\rrr= \frac{\alpha!}{\beta!(\alpha-\beta)!}$, $a^b=\prod\limits_{j=1}^d{a_j}^{b_j}$, $[\alpha]=\sml j1d \alpha_j$, $D^\alpha f=\frac{\partial^{[\alpha]}f} {\partial^{\alpha_1}x_1\dots\partial^{\alpha_d}x_d}$; $\delta_{ab}$ denotes Kronecker delta; $\Bbb T^d$ is the unit $d$-dimensional torus; $\Bbb C$ is the set of complex numbers. Let $ M $ be a non-degenerate $ d \times d $ integer matrix whose eigenvalues are bigger than 1 in module, $M^* $ is the conjugate matrix to $M $ , $I_d$ denotes the unit $ d \times d$ matrix. We say that numbers $k, n\in \zd $ are congruent modulo~ $M $ (write $k\equiv ~ n ~ \pmod {M} $) if $k-n=M\ell $, $ \ell\in\zd $. The integer lattice $ \zd $ is splitted into cosets with respect to the introduced relation of congruence. The number of cosets is equal to $ | \det M | $ (see, e.g., \cite[\S 2.7]{NPS}). Let us take an arbitrary representative from each coset, call them digits and denote the set of digits by $D (M) $. Throughout the paper we consider that such a matrix $M$ is fixed, $m=|\det M|$, $D (M) =\{s_0\ddd s_{m-1}\}$, $s_0=\nul$, $k=1\ddd m-1$, $R (M) =\{M^{-1}s_0\ddd M^{-1}s_{m-1}\}$. We will consider wavelet frames constructed in the framework of multiresolution analysis. Let a MRA in $L_2(\rd)$ be generated by a scaling function $\phi$ which satisfies the refinement equation $$ \widehat\phi(x)=m_0({M^*}^{-1}x) \widehat\phi({M^*}^{-1}x), $$ where $m_0\in L_2(\Bbb T^d)$ is its mask (refinable mask). For any $m_\nu\in L_2(\Bbb T^d)$, there exists a unique set of functions $\mu_{\nu k}\in L_2(\Bbb T^d)$, $k=0\ddd m-1$, (polyphase representatives of $m_\nu$) so that \be m_\nu(x)=\frac1{\sqrt m}\sml k0{m-1}\ex{(s_k,x)}\mu_{\nu k}(M^*x). \label{2} \ee The functions $\mu_{\nu k}$ can be expressed by $$ \mu_{\nu k}(x)=\frac1{\sqrt m}\sum\limits_{s\in D(M^*)}\exm{(M^{-1}s_k,x+s)}m_\nu({M^*}^{-1}(x+s)). $$ It is clear from these formulas that a function $m_\nu$ is differentiable ($n$ times) on $R(M^*)$ if and only if its polyphase representatives $\mu_{\nu k}$, $k=0\ddd m-1$, are differentiable ($n$ times) at the origin and $m_\nu$ is a trigonometric polynomial if and only if its polyphase representatives, are a trigonometric polynomials. Now, let another MRA be generated by a scaling function $\widetilde\phi$ with a mask $\widetilde m_0$. According to{\em Unitary Extension Principle}~\cite{RS}, to construct dual wavelet frames one finds wavelet masks $m_\nu, \widetilde m_\nu$, $\nu=1\ddd r$, $r\ge m-1$, so that the polyphase matrices \ban {\cal M}:=\left(% \begin{array}{ccc} \mu_{00} & \dots& \mu_{0, m-1} \\ \vdots & \ddots & \vdots\\ \mu_{r,0} & \dots & \mu_{r,m-1}\\ \end{array}% \right),\ \ \ \widetilde{\cal M}:=\left(% \begin{array}{ccc} \widetilde\mu_{00} & \dots& \widetilde\mu_{0, m-1} \\ \vdots & \ddots & \vdots\\ \widetilde\mu_{r,0} & \dots & \widetilde\mu_{r,m-1}\\ \end{array}% \right), \ean satisfy \be {\cal M^T}\overline{\widetilde{\cal M}}=I_m, \label{61} \ee and define wavelet functions by \ban \widehat\psi^{(\nu)}(x)&=&m_\nu({M^*}^{-1}x) \widehat\phi({M^*}^{-1}x), \\ \widehat{\widetilde\psi^{(\nu)}}(x)&=&\widetilde m_\nu({M^*}^{-1}x) \widehat{\widetilde\phi}({M^*}^{-1}x). \ean The corresponding dual wavelet systems are $\{\psi_{jk}^{(\nu)}\}$, $\{\widetilde\psi_{jk}^{(\nu)}\}$, where \newline$\psi_{jk}^{(\nu)}:=m^{j/2}\psi^{(\nu)}(M^j\cdot+k)$, $\widetilde\psi_{jk}^{(\nu)}:=m^{j/2}\widetilde\psi^{(\nu)}(M^j\cdot+k)$, $j,k\in\zd$, $\nu=0\ddd r$. It is known that if ${\cal M}=\widetilde{\cal M}$ then $\{\psi_{jk}^{(\nu)}\}$ is a tight frame in $L_2(\rd)$. If ${\cal M}, \widetilde{\cal M}$ are arbitrary matrixes satisfying~(\ref{61}), under some additional assumptions on $\phi, \widetilde\phi, m_\nu, \widetilde m_\nu$ (see \cite{98}, \cite{b16}, \cite[\S 2.7]{NPS}), we can state that $\{\psi_{jk}^{(\nu)}\}$, $\{\widetilde\psi_{jk}^{(\nu)}\}$ are dual frames in $L_2(\rd)$. Throughout the paper we will consider that wavelet systems $\{\psi_{jk}^{(\nu)}\}$, $\{\widetilde\psi_{jk}^{(\nu)}\}$ are constructed by means of Unitary Extension Principle from generating scaling functions $\phi, \widetilde\phi$ whose masks $m_0, \widetilde m_0$ are continuous at the origin and $m_0(\nul)= \widetilde m_0(\nul)=1$. %\subsection {2. Vanishing moments} \begin{defi} We say that a wavelet system $\{\psi_{jk}^{(\nu)}\}$ has vanishing moments up to order $\alpha$, $\alpha\in\zd_+$, ( has {\em $VM_\alpha$ property} in the sequel), if $D^\beta\widehat\psi^{(\nu)}(\nul)=0$, $\nu=1\ddd r$, for all $\beta\in\zd_+$, $\beta\le\alpha$. \end{defi} Assume that the functions $\widehat\phi, m_1\ddd m_r$ have derivatives up to order $\alpha$ at the origin. It easily follows from Leibniz formula that $VM_\alpha$ property holds if and only if \be D^\beta(m_\nu({M^*}^{-1}x)\Big|_{x=\nul}=0, \nu=1\ddd r,\ \ \ \forall \beta\in\zd_+, \beta\le\alpha. \label{36} \ee In the case $r=m-1$, there exist different criterions for vanishing moment. It is known~\cite{1} how to describe vanishing moment property in terms of linear identities for Fourier coefficients of the dual refinable mask (so-called {\em sum rule}). Some other descriptions of masks providing $VM_\alpha$ property are found in terms of zero-conditions~\cite{1} and in terms of containment in a quotient ideal~\cite{11}. The following polyphase criterion was given in~\cite{36}: $VM_\alpha$ property is valid for $\{\psi_{jk}^{(\nu)}\}$ if and only if there exist complex numbers $\lambda_\gamma$, $\gamma\in\zd_+$, $\gamma\le\alpha$, such that $\lambda_\nul=1$, \be D^\beta\widetilde\mu_{0k}(\nul)=\frac1{\sqrt m} \sum\limits_{\nul\le\gamma\le\beta}\lambda_\gamma \lll\beta\atop\gamma\rrr(-2\pi i M^{-1}s_k)^{\beta-\gamma} \ \ \ \forall \beta\in\zd_+, \beta\le\alpha, \label{0} \ee for each $k=0\ddd m-1$. The set of parameters $\lambda_\gamma$ in~(\ref0) is unique, and $\lambda_\gamma$ does not depend on $\alpha$ due to the following statement. \begin{prop}\cite{36} If~(\ref{0}) is valid for the polyphase representatives of $\widetilde m_0$, then \be \lambda_\beta=D^\beta\lll\widetilde m_0({M^*}^{-1}x)\rrr\Big|_{x=\nul} \label{35} \ee for all $\beta\in\zd_+$, $\beta\le\alpha$. \label{pr1} \end{prop} So, in the case $r=m-1$, $VM_\alpha$ property for $\{\psi_{jk}^{(\nu)}\}$ depends only on $\widetilde m_0$, i.e. only the first raw of the matrix $\widetilde{\cal M}$ is responsible for vanishing moments of wavelets generated by the matrix ${\cal M}$. In the case $r>m-1$, $VM_\alpha$ property for $\{\psi_{jk}^{(\nu)}\}$ depends also on the way of construction of matrixes ${\cal M}$, $\widetilde{\cal M}$. This may be illustrated by the following example. Let $d=1$, $M=m=2$, $\mu_{00}=\mu_{01}=\widetilde\mu_{00}=\widetilde\mu_{01}\equiv\frac1{\sqrt2}$, $$ {\cal M}=\widetilde{\cal M}=\left(% \begin{array}{cc} \frac1{\sqrt2} & \frac1{\sqrt2} \\ \frac12 & -\frac12 \\ \frac12 & -\frac12 \\ \end{array}% \right), \ \ \ \ \ {\cal M}^\prime=\left(% \begin{array}{cc} \frac1{\sqrt2} & \frac1{\sqrt2} \\ \frac1{\sqrt2} & 0 \\ 0 & \frac1{\sqrt2} \\ \end{array}% \right), \ \ \ \widetilde{\cal M}^\prime=\left(% \begin{array}{cc} \frac1{\sqrt2} & \frac1{\sqrt2} \\ \frac1{\sqrt2} & -\frac1{\sqrt2} \\ -\frac1{\sqrt2} & \frac1{\sqrt2} \\ \end{array}% \right). $$ Either of pairs ${\cal M},\widetilde{\cal M}$ and ${\cal M}^\prime,\widetilde{\cal M}^\prime$ satisfies~(\ref{61}). The matrixes ${\cal M},\widetilde{\cal M}$ generate wavelet masks $m_1(x)=m_2(x)=\widetilde m_1(x)=\widetilde m_2(x)=\frac1{2\sqrt2}-\frac1{2\sqrt2}\ex x$. It is clear that $m_1(0)=m_2(0)=\widetilde m_1(0)=\widetilde m_2(0)=0$, i.e. for the corresponding wavelet systems $VM_0$ property is valid. The matrixes ${\cal M}^\prime,\widetilde{\cal M}^\prime$ generate wavelet masks $ m^\prime_1(x)=\frac1{2},\ \ m^\prime_2(x)=\frac1{2}\ex x,\ \ \widetilde m^\prime_1(x)=\frac1{2}-\frac1{2}\ex x, \widetilde m^\prime_2(x)=-\frac1{2}+\frac1{22}\ex x$, and we have $m^\prime_1(0)\ne0, m^\prime_2(0)\ne0$. \begin{theo} Let $\alpha\in\zd_+$, $r\ge m-1$, the functions $\mu_{\nu,k},\widetilde\mu_{\nu k}\in L_2(\Bbb T^d)$, $\nu, k=0\ddd r$, have derivatives up to order $\alpha$ at the origin, the matrixes ${\cal N}:=\{\mu_{\nu k}\}_{\nu,k=0}^r$ and $\widetilde{\cal N}:=\{\overline\mu_{\nu k}\}_{\nu,k=0}^r$ satisfy \be {\cal N}\overline{{\widetilde{\cal N}}^T}=I_{r+1}; \label1 \ee masks $\widetilde m_0, m_1\ddd m_{m-1}$ are defined by~(\ref{2}). Then condition~(\ref{36}) is valid if and only if \item (a) there exist $\lambda_\gamma\in\Bbb C$, $\gamma\in\zd_+$, $\gamma\le\alpha$, such that $\lambda_\nul=1$ and (\ref{0})~holds for $k=0\ddd m-1$; \item (b) $D^\gamma\widetilde\mu_{0k}(\nul)=0$, $k=m\ddd r$ for all $\gamma\in\zd_+$, $\gamma\le\alpha$. \label{t1} \end{theo} {\tt Proof.} Suppose that~(\ref{36}) is valid. We will prove $(a)$ and $(b)$ by induction on $\alpha$. Check the initial step for $\alpha=0$. Let $m_\nu(\nul)=0$, $\nu=1\ddd r$. It follows form~(\ref2) that \be \sml k0{m-1}\mu_{\nu k}(\nul)=0,\ \ \ \nu=1\ddd r. \label4 \ee On the other hand, by~(\ref1), $$ \sml k0r\overline{\widetilde\mu_{0k}(\nul)}\mu_{\nu k}(\nul)=0,\ \ \ \nu=1\ddd m-1. $$ Because of linear independence of the vectors $(\mu_{\nu0}(\nul)\ddd\mu_{\nu,r}(\nul))\in\r^{r+1}$, $\nu=1\ddd r$, there exists $\lambda$ so that $$ \widetilde\mu_{00}(\nul)=\dots=\widetilde\mu_{0, m-1}(\nul)=\lambda,\ \ \widetilde\mu_{0m}(\nul)=\dots=\widetilde\mu_{0, r}(\nul)=0. $$ Taking into account the condition $\widetilde m_0(\nul)=1$ which is equivalent to $$ \frac1{\sqrt m}\lll\widetilde\mu_{\nu0}(\nul)+\dots+\widetilde\mu_{\nu,m-1}(\nul)\rrr=1, $$ we obtain $\lambda=\frac1{\sqrt m}$. For the inductive step we assume that~(\ref{36}) is valid for $\alpha>\nul$ and $(a), (b)$ holds for all $\alpha^\prime\in\zd_+$, $\alpha^\prime<\alpha$. So, due to Proposition~\ref{pr1}, there exist constants $\lambda_\gamma\in \Bbb C$, $\gamma\in\zd_+$, $\gamma<\alpha$ such that~(\ref0) holds for all $\beta<\alpha$. If $\gamma\in\zd_+$, $\gamma<\alpha$, due to~(\ref2) and Leibniz formula, we have \ba \frac1{\sqrt m}\sum\limits_{\nul\le\beta\le\alpha-\gamma} \lll\alpha-\gamma\atop\beta\rrr \sml k0{m-1}(2\pi iM^{-1}s_k)^{\alpha-\beta-\gamma}D^\beta \mu_{\nu k}(\nul)= \nonumber \\ D^{\alpha-\gamma}m_\nu({M^*}^{-1}x)\Big|_{x=\nul}=0. \label5 \ea It follows from~(\ref1) that $$ \sml k0r\overline{\widetilde\mu_{0k}}\mu_{\nu k}= 0,\ \ \nu=1\ddd m-1. $$ Differentiating this equality $\alpha$ times gives $$ \sum\limits_{\nul\le\beta\le\alpha} \lll\alpha\atop\beta\rrr \sml k0r\overline{D^{\alpha-\beta }\widetilde\mu_{0 k}(\nul)} D^\beta \mu_{\nu k}(\nul)=0. $$ Taking into account the inductive hypotheses, we have \be \sum\limits_{\nul\le\beta\le\alpha} \lll\alpha\atop\beta\rrr \sml k0{m-1}\overline{D^{\alpha-\beta }\widetilde\mu_{0 k}(\nul)} D^\beta \mu_{\nu k}(\nul)+\sml kmr\overline{D^{\alpha}\widetilde\mu_{0 k}(\nul)} \mu_{\nu k}(\nul)=0. \label3 \ee Multiply~(\ref5) by $\lll\alpha\atop\alpha-\gamma\rrr\overline{\lambda_\gamma}$ and subtract from~(\ref3). After the same manipulation with each $\gamma\in\zd_+$, $\gamma<\alpha$, we obtain \ban 0=\sml kmr\overline{D^{\alpha}\widetilde\mu_{0 k}(\nul)} \mu_{\nu k}(\nul)+\sum\limits_{\nul\le\beta\le\alpha}\lll\alpha\atop\beta\rrr \sml k0{m-1} \overline{D^{\alpha-\beta }\widetilde\mu_{0 k}(\nul)}D^\beta \mu_{\nu k}(\nul)- \\ \frac1{\sqrt m}\sum\limits_{\nul\le\gamma<\alpha}\lll\alpha\atop\alpha-\gamma\rrr\overline{\lambda_\gamma} \sum\limits_{\nul\le\beta\le\alpha-\gamma} \lll\alpha-\gamma\atop\beta\rrr \sml k0{m-1}(2\pi iM^{-1}s_k)^{\alpha-\beta-\gamma}D^\beta \mu_{\nu k}(\nul)= \\ \sml kmr\overline{D^{\alpha}\widetilde\mu_{0 k}(\nul)} \mu_{\nu k}(\nul)+ \sum\limits_{\nul\le\beta\le\alpha}\lll\alpha\atop\beta\rrr \sml k0{m-1}\lll \overline{D^{\alpha-\beta }\widetilde\mu_{0 k}(\nul)}-\right. \\ \left.\frac1{\sqrt m}\sum\limits_{\gamma\ne\alpha\atop\nul\le\gamma<\alpha-\beta} \overline{{\lll\alpha-\gamma\atop\beta\rrr \lll\alpha\atop\alpha-\gamma\rrr} {\lll\alpha\atop\beta\rrr^{-1} }\lambda_\gamma(-2\pi iM^{-1}s_k)^{\alpha-\beta-\gamma}}\right) D^\beta \mu_{\nu k}(\nul). \ean From this, taking into account that \be {\lll\alpha-\gamma\atop\beta\rrr\lll\alpha\atop\alpha-\gamma\rrr} {\lll\alpha\atop\beta\rrr ^{-1}}=\frac{(\alpha-\beta)!}{\gamma!(\alpha-\beta-\gamma)!}= \lll\alpha-\beta\atop\gamma\rrr, \label6 \ee and using the inductive hypotheses, the sum over $\beta$ is deduced to \ban \sum\limits_{\nul\le\beta\le\alpha}\lll\alpha\atop\beta\rrr \sml k0{m-1}\lll \overline{D^{\alpha-\beta }\widetilde\mu_{0 k}(\nul)}-\right.\hspace{6cm} \\ \left.\frac1{\sqrt m}\sum\limits_{\gamma\ne\alpha\atop\nul\le\gamma\le\alpha-\beta} \overline{{\lll\alpha-\beta\atop\gamma\rrr \ } \lambda_\gamma(-2\pi iM^{-1}s_k)^{\alpha-\beta-\gamma}}\right) D^\beta \mu_{\nu k}(\nul)= \\ \sml k0{m-1}\lll \overline{D^{\alpha}\widetilde\mu_{0 k}(\nul)- \frac1{\sqrt m}\sum\limits_{\nul\le\gamma<\alpha} {\lll\alpha\atop\gamma\rrr \ } \lambda_\gamma(-2\pi iM^{-1}s_k)^{\alpha-\gamma}}\right) \mu_{\nu k}(\nul). \ean So, we have \ban \sml k0{m-1}\lll \overline{D^{\alpha}\widetilde\mu_{0 k}(\nul)- \frac1{\sqrt m}\sum\limits_{\nul\le\gamma<\alpha} {\lll\alpha\atop\gamma\rrr \ } \lambda_\gamma(-2\pi iM^{-1}s_k)^{\alpha-\gamma}}\right) \mu_{\nu k}(\nul)+ \\ \sml kmr\overline{D^{\alpha}\widetilde\mu_{0 k}(\nul)} \mu_{\nu k}(\nul)=0. \ean Similarly to the arguments for the initial step, it follows from~(\ref4) that there exists $\lambda_\alpha$ such that \ban D^{\alpha }\widetilde\mu_{0 k}(\nul)- \frac1{\sqrt m}\sum\limits_{\nul\le\gamma<\alpha} {\lll\alpha\atop\gamma\rrr \lambda_\gamma(-2\pi iM^{-1}s_k)^{\alpha-\gamma}}=\frac{\lambda_\alpha}{\sqrt m}, \ \ \ k=0\ddd m-1, \\ D^{\alpha }\widetilde\mu_{0 k}(\nul)=0, \ \ \ k=m\ddd r. \ean Thus, (\ref0) is valid for $\beta=\alpha$ as was to be proved. Now we assume that $(a), (b)$ are valid. We will prove~(\ref{36}) by induction on~$\alpha$. If~(\ref0) is valid for $\alpha=\nul$, then $\widetilde\mu_{0 k}(\nul)=1/\sqrt m$, $k=0\ddd m-1$. It follows from~(\ref{1}) and $(b)$ that $$ \mu_{\nu 0}(\nul)+\dots+\mu_{\nu, m-1}(\nul)=0,\ \ \nu=1\ddd r. $$ Hence, on the basis of~(\ref{2}), $m_\nu(\nul)=0$, $\nu=1\ddd r$, what proves the initial step. For the inductive step, we assume that $(a), (b)$ is valid for $\alpha>\nul$ and~(\ref{36}) holds for all $\alpha^\prime\in\zd_+$, $\alpha^\prime<\alpha$, i.e. $$ D^{\alpha-\gamma}m_\nu({M^*}^{-1}x)\Big|_{x=\nul}=0,\ \ \gamma\in\zd_+, \ \ \gamma\ne\nul, \ \ \gamma\le\alpha. $$ This yields~(\ref5) for $\gamma\ne\nul$. Multiply~(\ref5) by $\lll\alpha\atop\alpha-\gamma\rrr\overline{\lambda_\gamma}$ and add to~(\ref2) differentiated $\alpha$ times. After the same manipulation with each $\gamma\in\zd_+$, $\gamma<\alpha$, we obtain \ban D^{\alpha}m_\nu({M^*}^{-1}x)\Big|_{x=\nul}= \frac1{\sqrt m}\sum\limits_{\nul\le\beta\le\alpha} \lll\alpha\atop\beta\rrr \sml k0{m-1}(2\pi iM^{-1}s_k)^{\alpha-\beta}D^\beta \mu_{\nu k}(\nul)+ \\ \frac1{\sqrt m}\sum\limits_{\nul<\gamma\le\alpha}\lll\alpha\atop\alpha-\gamma\rrr\overline{\lambda_\gamma} \sum\limits_{\nul\le\beta\le\alpha-\gamma} \lll\alpha-\gamma\atop\beta\rrr \sml k0{m-1}(2\pi iM^{-1}s_k)^{\alpha-\beta-\gamma}D^\beta \mu_{\nu k}(\nul)= \\ \frac1{\sqrt m}\sum\limits_{\nul\le\beta\le\alpha}\lll\alpha\atop\beta\rrr \sum\limits_{\nul\le\gamma\le\alpha-\beta}\overline{\lambda_\gamma}\lll\alpha\atop\alpha-\gamma\rrr \lll\alpha-\gamma\atop\beta\rrr\lll\alpha\atop\beta\rrr^{-1}\cdot \\ \sml k0{m-1}(2\pi iM^{-1}s_k)^{\alpha-\beta-\gamma}D^\beta \mu_{\nu k}(\nul). \ean Due to~(\ref6) and~(\ref0), this yields \ban D^{\alpha}m_\nu({M^*}^{-1}x)\Big|_{x=\nul}=\hspace{8cm} \\ \frac1{\sqrt m}\sum\limits_{\nul\le\beta\le\alpha}\lll\alpha\atop\beta\rrr\sml k0{m-1} \overline{\sum\limits_{\nul\le\gamma\le\alpha-\beta}\lambda_\gamma\lll\alpha-\beta\atop\gamma\rrr (-2\pi iM^{-1}s_k)^{\alpha-\beta-\gamma}}D^\beta \mu_{\nu k}(\nul)= \\ \sum\limits_{\nul\le\beta\le\alpha}\lll\alpha\atop\beta\rrr \sml k0{m-1}\overline{D^{\alpha-\beta}\widetilde\mu_{0k}(\nul)}D^\beta \mu_{\nu k}(\nul)= D^\alpha\lll\sml k0{m-1}\overline{\widetilde\mu_{0k}(x)} \mu_{\nu k}(x)\rrr\Bigg|_{x=\nul}= \\ D^\alpha\lll\sml k0r\overline{\widetilde\mu_{0k}(x)} \mu_{\nu k}(x)\rrr\Bigg|_{x=\nul}. \ean It follows from~(\ref1) that $ D^{\alpha}m_\nu({M^*}^{-1}x)\Big|_{x=\nul}=0 $ as was to be proved. $\Diamond$ Usually it is more useful to control univariate order of vanishing moment property (for example, to apply Taylor formula). \begin{defi} We say that a wavelet system $\{\psi_{jk}^{(\nu)}\}$ has vanishing moments up to order $n$, $n\in\z_+$, (has {\em $VM^n$ property} in the sequel) if $D^\beta\widehat\psi^{(\nu)}(\nul)=0$, $\nu=1\ddd m-1$, for all $\beta\in\zd_+$, $[\beta]\le n$. \end{defi} \begin{theo} Let $n\in\z_+$, \ \ $r\ge m-1$, the functions $\mu_{\nu,k},\ \widetilde\mu_{\nu k}\in L_2(\Bbb T^d)$, $\nu, k=0\ddd r$, have derivatives up to order $n$ at the origin, the matrixes ${\cal N}:=\{\mu_{\nu k}\}_{\nu,k=0}^r$ and $\widetilde{\cal N}:=\{{\widetilde \mu_{\nu k}}\}_{\nu,k=0}^r$ satisfy~(\ref1); masks $\widetilde m_0, m_1\ddd m_{m-1}$ are defined by~(\ref{2}). Then condition~(\ref{36}) is valid for all $\alpha\in\zd$, $[\alpha]\le n$ if and only if \item (a) there exist $\lambda_\gamma\in\Bbb C$, $\gamma\in\zd_+$, $[\gamma]\le n$, such that $\lambda_\nul=1$ and (\ref{0})~holds for $k=0\ddd m-1$; \item (b) $D^\gamma\widetilde\mu_{0k}(\nul)=0$, $k=m\ddd r$ for all $\gamma\in\zd_+$, $[\gamma]\le n$. \label{t2} \end{theo} Proof of this theorem follows immediately from Theorem~\ref{t1} and Proposition~\ref{pr1}. Let $n\in\z_+$, we will denote by $L_\infty^{(n)}$ the class of complex-valued functions which are in $L_\infty(\Bbb T^d)$ and have continuous derivatives up to order $n$ at the origin. %Functions $f,g\in L_\infty^{(n)}$ are equal if $f(\nul)=g(\nul)$ and $f=g$ a. e. \begin{lem} Let $\mu_{\nu 0},\widetilde\mu_{\nu 0}\in L_\infty^{(n)}$, $\nu=0\ddd r$, and \be \sml \nu0r\mu_{\nu 0}\overline{\widetilde\mu_{\nu 0}}= 1. \label{55} \ee Then there exist functions $\mu_{\nu k}, \widetilde\mu_{\nu k}\in L_\infty^{(n)}$, $\nu=0\ddd r$, $k=1\ddd r$, such that \be \sml \nu0r\mu_{\nu l}\overline{\widetilde\mu_{\nu k}}= \delta_{kl},\ \ \ k,l=0\ddd r. \label{54} \ee \label{l1} \end{lem} {\tt Proof.} Set $$ \mu^\prime_{\nu 0}:=\frac{\mu_{\nu 0}}{\sqrt{\sml l0r|\mu_{l0}|^2}},\ \ \ \nu=0\ddd r. $$ It is clear that the functions $\mu^\prime_{\nu 0}$ are essentially bounded and \be \sml \nu0r|\mu^\prime_{\nu 0}|^2= 1. \label{56} \ee It follows from~(\ref{55}) that $\sml l0r|\mu_{l0}(\nul)|^2\ne0$. So, $\mu^\prime_{\nu 0}\in L_\infty^{(n)}$, $\nu=0\ddd r$. Let us extend the unit vector $\mu^\prime_{0 0}\ddd \mu^\prime_{r 0}$ to a unitary matrix. Due to~(\ref{56}), there exist $\nu_0$ so that $\mu^\prime_{\nu_0 0}(\nul)\ne1$. We may consider that $\nu_0=0$ (else we will interchange $\mu^\prime_{\nu_0 0}$ and $\mu^\prime_{0 0}$, extend this new vector to a unitary matrix and interchange its $0$-th and $\nu_0$-th rows). Due to Householder transform, an extension to a unitary matrix may be realized by: $$ \mu^\prime_{0 k}=\overline{\mu^\prime_{k 0}}\frac{1-\mu^\prime_{00}}{1-\overline{\mu^\prime_{00}}},\ \ \mu^\prime_{\nu k}=\delta_{lk}- \frac{\mu^\prime_{\nu 0}\overline{\mu^\prime_{k 0}}}{1-\overline{\mu^\prime_{00}}}, \ \ \ \nu, k=1\ddd r. $$ Because of~(\ref{56}), we have $|\mu^\prime_{\nu 0}|\le\sqrt{1-|\mu^\prime_{00}|^2}$, $\nu=1\ddd r$. This yields essential boundedness of the the functions $\mu^\prime_{\nu k}$. Since $1-\mu^\prime_{00}(\nul)\ne0$, it follows that $\mu^\prime_{\nu k}\in L_\infty^{(n)}$, $\nu,k=0\ddd r$. Set \ban &&\widetilde\mu_{\nu k}:=\mu^\prime_{\nu k}, \ \ \ \nu=0\ddd r, \ k=1\ddd, r, \\ &&\widetilde Q_k:=(\widetilde \mu_{0k}\ddd\widetilde \mu_{rk}), k=0\ddd r, \\ &&Q_0:=(\mu_{00}\ddd\mu_{r0}), Q_k:=\widetilde Q_k-\widetilde Q_k\widetilde Q^T_0Q_0,\ \ \ k=1\ddd r. \ean It is not difficult to see that the entries of $Q_k$ are in $ L_\infty^{(n)}$ and $Q_k\overline{\widetilde Q_l^T}=\delta_{kl}$, \ \ \ $k,l=0\ddd r$. It remains to denote by $\mu_{\nu k}$ the $\nu$-th component of $Q_k$.$\Diamond$ \begin{lem} Let $A$ be a class of complex-valued functions such that \item (i) if $f,g\in A$, $a,b\in{\Bbb C}$ then $ af+bg\in A$, \item (ii) if $f,g\in A$, then $fg\in A$,\newline and let $\cal A$ be a class of matrixes whose entries are in $A$. If any two $n\times 1$ matrixes $Q$, $\widetilde Q\in{\cal A}$ satisfying $Q^T\overline{\widetilde{Q}}=1$ can be extended to $n\times n$ matrixes ${\cal N}, \widetilde{\cal N}\in{\cal A}$ satisfying ${\cal N}^T\overline{\widetilde{\cal N}}=I_n$, then any two $n\times j$ matrixes $\cal M$, $\widetilde {\cal M}\in{\cal A}$, $1