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\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 $\cal M$,
$\widetilde {\cal M}\in{\cal A}$, $1<j<n$,  satisfying
$ {\cal M}^T\overline{\widetilde{ {\cal M}}}=I_j$.
 can be extended to $n\times n$ matrixes ${\cal N}, \widetilde{\cal N}\in{\cal A}$ satisfying
${\cal N}^T\overline{\widetilde{\cal N}}=I_n$,
  \label{l2}
\end{lem}


{\tt Proof.} We will prove by induction on $j$. The base for $j=1$
is given. Let us check the inductive step $j-1\to j$. Let $j\times n$ matrixes
$\cal M, \widetilde {\cal M}\in \cal A$ satisfy $ {\cal M}^T\overline{\widetilde{ {\cal M}}}=I_j$. Denote by
$Q_k$, $\widetilde Q_k$ the $k$-th columns respectively  of
$\cal M$, $\widetilde {\cal M}$. Due to the statement of the
base, the matrixes $Q$, $\widetilde Q\in{\cal A}$  can be extended to
$n\times n$ matrixes ${{\cal N^\prime}}, \widetilde{\cal N^\prime}\in{\cal A}$
satisfying ${\cal N^\prime}^T\overline{\widetilde{\cal N^\prime}}=I_n$,
 Let $Q^\prime_k$, $\widetilde Q^\prime_k$, $k=2\ddd n$, denote
  the $k$-th columns respectively  of
  ${{\cal N^\prime}}, \widetilde{\cal N^\prime}\in{\cal  A}$.
Fix a point $x$ for which ${Q^\prime_l(x)}^T\overline{\widetilde Q^\prime}_k(x)=\delta_{kl}$.
Since the vectors $Q^\prime_2(x)\ddd Q^\prime_n(x)$
form a basis for the orthogonal complement to $\widetilde Q^\prime_1(x)$
in $\r^n$, we have
$$
Q_k(x)=\sml l2n\alpha_{lk}(x)Q^\prime_l(x),\ \ \ k=2\ddd j.
$$
Similarly,
$$
\widetilde Q_k(x)=\sml l2n\widetilde\alpha_{lk}(x)\widetilde Q^\prime_l(x),\ \ \ k=2\ddd j.
$$
It is clear that $\alpha_{lk}, \widetilde\alpha_{lk}\in A$ and
$\sml l2n \alpha_{lk}\overline{\widetilde\alpha}_{lk^\prime}=\delta_{k k^\prime}$, $k, k^\prime=2\ddd j$.
Due to the inductive hypotheses, there exist functions $\alpha_{lk}, \widetilde\alpha_{lk}\in A$,
$l=2\ddd n$, $k=j+1\ddd n$, such that
\be
\sml l2n \alpha_{lk}\overline{\widetilde\alpha}_{lk^\prime}=\delta_{k k^\prime},\ \ \ k, k^\prime=2\ddd n.
\label{53}
\ee
Set
$$
Q_k:=\sml l2n\alpha_{lk}Q^\prime_l,\ \ \
\widetilde Q_k:=\sml l2n\widetilde\alpha_{lk}\widetilde Q^\prime_l,\ \ \
 k=j+1\ddd n.
$$

Because of~(\ref{53}) and  biorthogonality of the systems $Q^T_1, {Q^\prime_2}^T\ddd {Q^\prime_n}^T$ and
$\overline{\widetilde Q_1},\overline{\widetilde Q^\prime_2}\ddd \overline{\widetilde Q^\prime_n}$,
we obtain
\ban
{Q_l(x)}^T\overline{\widetilde Q}_k(x)=\delta_{kl}, \ \ k,l=1\ddd n.
\ean
To complete the proof it remains to introduce matrixes ${\cal N}$ and  $\widetilde{\cal N}$
whose columns are respectively $Q_1\ddd Q_n$ and  $\widetilde Q_1\ddd \widetilde Q_n$. $\Diamond$

Now we are ready to give a necessary condition for $VM^n$ property.

\begin{theo}
Let dual wavelet systems $\{\psi_{jk}^{(\nu)}\}$, $\{\widetilde\psi_{jk}^{(\nu)}\}$ with  $VM^n$ property
be generated by refinable functions $\phi,  \widetilde\phi$ whose  Fourier transforms have  derivatives up
to order $n$ at the origin, and let the entries  $\mu_{\nu k},\widetilde \mu_{\nu k}$ of the corresponding
 polyphase matrixes ${\cal M}, \widetilde{\cal M}$ be in $L^{(n)}_\infty$. Then
 there exist complex numbers $\lambda_\gamma, \widetilde\lambda_\gamma$,
$\gamma\in\zd_+$,  $[\gamma]\le n$, such that
\ba
\widetilde\lambda_\nul=\lambda_\nul=1,\ \ \
\sum\limits_{\nul\le\gamma\le \alpha}\lll\alpha\atop\gamma\rrr
\lambda_\gamma \overline{\widetilde\lambda_{\alpha-\gamma}}=0 \ \ \
\forall \alpha\in\zd_+, 0<[\alpha]\le n;
\label{57}
\\
D^\beta\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 n,
\label{58}
\\
D^\beta\widetilde\mu_{0k}(\nul)=\frac1{\sqrt m}
\sum\limits_{\nul\le\gamma\le\beta}\widetilde\lambda_\gamma
\lll\beta\atop\gamma\rrr(-2\pi i M^{-1}s_k)^{\beta-\gamma} \ \ \ \forall    \beta\in\zd_+,  [\beta]\le n,
\label{59}
\ea
 for $k=0\ddd m-1$.
\label{t3}
\end{theo}


{\tt Proof.} Due to Lemmas~\ref{l1}, \ref{l2}, the $(r+1)\times m$ matrixes ${\cal M}, \widetilde{\cal M}$
can be extended to $(r+1)\times (r+1)$ matrixes ${\cal N}, \widetilde{\cal N}$ such that their entries
are in $L^{(n)}_\infty$ and
${\cal N}^T\overline{\widetilde{\cal N}}=I_{r+1}$.
It follows from conditions (a) of Theorem~\ref{t2} that there exist complex numbers
$\lambda_\gamma, \widetilde\lambda_\gamma$, $\gamma\in\zd_+$,  $[\gamma]\le n$, such that
$\widetilde\lambda_\nul=\lambda_\nul=1$ and~(\ref{58}), (\ref{59})
 is valid  for $k=0\ddd m-1$. It remains to check~(\ref{57}).

Let $\alpha\in\zd_+$, $0<[\alpha]\le n$, $k=0\ddd m-1$, $\rho:=2\pi M^{-1}s_k$.
Due to~(\ref{58}), (\ref{59}), we have
\ba
m D^\alpha\lll \mu_{0 k}(x)\overline{\widetilde\mu_{0 k}(x)} \rrr\Big|_{x=\nul}=
\sum\limits_{\nul\le\beta\le \alpha}\lll\alpha\atop\beta\rrr
D^\beta\mu_{0k}(\nul)\overline{D^{\alpha-\beta}\widetilde\mu_{0k}(\nul)}=\nonumber
\\
\sum\limits_{\nul\le\beta\le \alpha}\lll\alpha\atop\beta\rrr
\sum\limits_{\nul\le\gamma\le\beta}
\lll\beta\atop\gamma\rrr\lambda_{\beta-\gamma}(-i\rho)^\gamma
\sum\limits_{\nul\le\delta\le\alpha-\beta}
\lll\alpha-\beta\atop\delta\rrr\overline{\widetilde\lambda_{\alpha-\beta-\delta}}(i\rho)^\delta=\nonumber
\\
\sum\limits_{\nul\le\gamma\le\alpha}\sum\limits_{\gamma\le\beta\le \alpha}\sum\limits_{\nul\le\delta\le\alpha-\beta}
\lll\alpha\atop\beta\rrr\lll\beta\atop\gamma\rrr\lll\alpha-\beta\atop\delta\rrr
(-\odin)^\gamma(i\rho)^{\gamma+\delta}\lambda_{\beta-\gamma}
\overline{\widetilde\lambda_{\alpha-\beta-\delta}}=\nonumber
\\
\sum\limits_{\nul\le\gamma\le\alpha}\sum\limits_{\nul\le\delta\le\alpha-\gamma}
\sum\limits_{\gamma\le\beta\le \alpha-\delta}
\lll\alpha\atop\beta\rrr\lll\beta\atop\gamma\rrr\lll\alpha-\beta\atop\delta\rrr
(-\odin)^\gamma(i\rho)^{\gamma+\delta}\lambda_{\beta-\gamma}
\overline{\widetilde\lambda_{\alpha-\beta-\delta}}=\nonumber
\\
\sum\limits_{\nul\le\gamma\le\alpha}\sum\limits_{\gamma\le\epsilon\le\alpha}
\sum\limits_{\gamma\le\beta\le \alpha-\epsilon+\gamma}
\lll\alpha\atop\beta\rrr\lll\beta\atop\gamma\rrr\lll\alpha-\beta\atop\epsilon-\gamma\rrr
(-\odin)^\gamma(i\rho)^\epsilon\lambda_{\beta-\gamma}
\overline{\widetilde\lambda_{\alpha-\beta-\epsilon+\gamma}}=\nonumber
\\
\sum\limits_{\nul\le\epsilon\le\alpha}(i\rho)^\epsilon
\sum\limits_{\nul\le\gamma\le\epsilon}
\sum\limits_{\gamma\le\beta\le \alpha-\epsilon+\gamma}
\lll\alpha\atop\beta\rrr\lll\beta\atop\gamma\rrr\lll\alpha-\beta\atop\epsilon-\gamma\rrr
(-\odin)^\gamma\lambda_{\beta-\gamma}
\overline{\widetilde\lambda_{\alpha-\beta-\epsilon+\gamma}}=\nonumber
\\
\sum\limits_{\nul\le\epsilon\le\alpha}(i\rho)^\epsilon
\sum\limits_{\nul\le\gamma\le\epsilon}
\sum\limits_{\nul\le\kappa\le \alpha-\epsilon}
\lll\alpha\atop\kappa+\gamma\rrr\lll\kappa+\gamma\atop\gamma\rrr\lll\alpha-\kappa-\gamma\atop\epsilon-\gamma\rrr
(-\odin)^\gamma\lambda_{\kappa}
\overline{\widetilde\lambda_{\alpha-\kappa-\epsilon}}=\nonumber
\\
\sum\limits_{\nul\le\epsilon\le\alpha}(i\rho)^\epsilon
\sum\limits_{\nul\le\kappa\le \alpha-\epsilon}
\lambda_{\kappa}\overline{\widetilde\lambda_{\alpha-\kappa-\epsilon}}
\sum\limits_{\nul\le\gamma\le\epsilon}
\frac{\alpha!}{\kappa!\gamma!(\epsilon-\gamma)!(\alpha-\kappa-\epsilon)!}
(-\odin)^\gamma=\nonumber
\\
\sum\limits_{\nul\le\epsilon\le\alpha}(i\rho)^\epsilon
\sum\limits_{\nul\le\kappa\le \alpha-\epsilon}
\lambda_{\kappa}\overline{\widetilde\lambda_{\alpha-\kappa-\epsilon}}
\lll\alpha\atop\kappa+\epsilon\rrr\sum\limits_{\nul\le\gamma\le\epsilon}
\lll\epsilon\atop\gamma\rrr(-\odin)^\gamma=\nonumber
\\
\sum\limits_{\nul\le\epsilon\le\alpha}(i\rho)^\epsilon
\sum\limits_{\nul\le\kappa\le \alpha-\epsilon}
\lambda_{\kappa}\overline{\widetilde\lambda_{\alpha-\kappa-\epsilon}}
\lll\alpha\atop\kappa+\epsilon\rrr(\odin-\odin)^\epsilon=
\sum\limits_{\nul\le\kappa\le \alpha}
\lambda_{\kappa}\overline{\widetilde\lambda_{\alpha-\kappa}}.\
\label{63}
\ea
So, $D^\alpha\lll \mu_{0 k}(x)\overline{\widetilde\mu_{0 k}(x)} \rrr\Big|_{x=\nul}$
does not depend on $k$ and
\ban
\sum\limits_{\nul\le\kappa\le \alpha}\lambda_{\kappa}\overline{\widetilde\lambda_{\alpha-\kappa}}=
\sml k0{m-1}D^\alpha\lll \mu_{0 k}(x)\overline{\widetilde\mu_{0 k}(x)} \rrr\Bigg|_{x=\nul}=
D^\alpha\lll \sml k0{m-1}\mu_{0 k}(x)\overline{\widetilde\mu_{0 k}(x)} \rrr\Bigg|_{x=\nul}.
\ean
Due to  conditions (b) of Theorem~\ref{t2},
$$
D^\gamma\mu_{0k}(\nul)=0, D^\gamma\widetilde\mu_{0k}(\nul)=0, \ k=m\ddd r,
\ \ \ \forall \gamma\in\zd_+, [\gamma]\le n.$$
From this, taking into account that $\sml k0{r}\mu_{0 k}(x)\overline{\widetilde\mu_{0 k}(x)}=1$,
we obtain
\ban
D^\alpha\lll \sml k0{m-1}\mu_{0 k}(x)\overline{\widetilde\mu_{0 k}(x)} \rrr\Bigg|_{x=\nul}=
D^\alpha\lll \sml k0{r}\mu_{0 k}(x)\overline{\widetilde\mu_{0 k}(x)} \rrr\Bigg|_{x=\nul}=0. \Diamond
\ean


\begin{coro}
Let  $\mu_{0k},\widetilde\mu_{0k}\in L_\infty^{(n)}$, $\nu=0\ddd m-1$.
If  there exist complex numbers $\lambda_\gamma, \widetilde\lambda_\gamma$,
$\gamma\in\zd_+$,  $[\gamma]\le n$, such that~(\ref{57}), (\ref{58}), (\ref{59}) are fulfilled,
 then
$$D^\alpha\lll\overline{\mu_{0k}(x)}\widetilde\mu_{0k}(x)\rrr\Big|_{x=\nul}=0,\ \ \  k=0\ddd m-1, \ \
%\forall \alpha\in \zd_+, \ \|\alpha\|_1\le n
$$  for all
  $\alpha\in \zd_+$, $0<[\alpha]\le n$.
  \label{c1}
\end{coro}

The proof of Corollary~\ref{c1}  follows from~(\ref{63}).

It is not difficult to see that conditions~(\ref{57})-(\ref{59}) are not sufficient for refinable functions
$\phi,  \widetilde\phi$ to generate dual wavelet systems $\{\psi_{jk}^{(\nu)}\}$, $\{\widetilde\psi_{jk}^{(\nu)}\}$
with  $VM^n$ property. Really, let $d=1$, $M=m=2$, $s_0=0, s_1=1$,
$\lambda_0=\widetilde\lambda_0=1$, $\lambda_1=\widetilde\lambda_1=0$,
 $\mu_{00}=\widetilde\mu_{00}\equiv\frac1{\sqrt2},\ \ \
  \mu_{01}=\widetilde\mu_{01}=\frac1{\sqrt2}\lll1-\frac i2\sin2\pi x\rrr$. It is clear that~(\ref{58}), (\ref{59})
  hold for $n=1$, and~(\ref{57}) is fulfilled. Assume that dual wavelet systems  with  $VM^1$ property
are generated by these functions. This means that polyphase matrixes ${\cal M}, \widetilde{\cal M}$ whose first
rows are $(\mu_{00}, \mu_{01})$, $(\widetilde\mu_{00}, \widetilde\mu_{01})$ respectively satisfy~(\ref{61}).
and such that~(\ref{36}) holds for the corresponding wavelet masks.
Due to Lemmas~\ref{l1}, \ref{l2}, the  matrixes ${\cal M}, \widetilde{\cal M}$
can be extended to  matrixes ${\cal N}, \widetilde{\cal N}$ such that their entries
$\mu_{\nu k}$, $\widetilde\mu_{\nu k}$, $\nu, k=0\ddd r$, are in $L^{(1)}_\infty$ and
${\cal N}^T, \overline{\widetilde{\cal N}}$ are mutually inverse. It follows from Theorem~\ref{t2} that
\be
\frac{d^2}{dx^2}\lll\sml k2r\mu_{0 k}(x)\overline{\widetilde\mu_{0 k}(x)}\rrr\Bigg|_{x=0}=0.
\label{62}
\ee
But
\ban
\frac{d^2}{dx^2}\lll\sml k2r\mu_{0 k}(x)\overline{\widetilde\mu_{0 k}(x)}\rrr=
\frac{d^2}{dx^2}\lll1-\sml k01\mu_{0 k}(x)\overline{\widetilde\mu_{0 k}(x)}\rrr=
\\
\frac{d^2}{dx^2}\lll1-\frac12-\frac12\lll1+\frac14\sin^22\pi x\rrr\rrr=
\frac{d^2}{dx^2}\lll-\frac18\sin^22\pi x\rrr=-\pi^2\cos4\pi x.
\ean
So, we see that~(\ref{62}) is not true.

A sufficient condition is given in the following statement.

\begin{theo}
Let $\phi,  \widetilde\phi$ be refinable functions, their Fourier transforms
$\widehat\phi,  \widehat{\widetilde\phi}$  have  derivatives up to order $n$
at the origin, $\widehat\phi_0(\nul)= \widehat{\widetilde \phi}_0(\nul)=1$,
and let $\mu_{00}\ddd\mu_{0, m-1}$, $\widetilde \mu_{00}\ddd\widetilde \mu_{0, m-1}\in L^{(n)}_\infty$ be
the  polyphase representatives of their masks. If  there exist complex numbers
$\lambda_\gamma, \widetilde\lambda_\gamma$, $\gamma\in\zd_+$,  $[\gamma]\le n$, such that
$\lambda_\nul=\widetilde\lambda_\nul=1$, (\ref{58}), (\ref{59})
 holds  for $k=0\ddd m-1$  and there exist functions
 $\mu_{0k}$, $\widetilde \mu_{0k}\in L^{(n)}_\infty$, $k=m\ddd r$, such that
 \ban
 &&\sml k0r\mu_{0 k}\overline{\widetilde\mu_{0k}}=1,
 \\
&&D^\beta\mu_{0k}(\nul)=D^\beta\widetilde\mu_{0k}(\nul)=0,\ k=m\ddd r,\ \ \ \forall \beta\in\zd_+, [\beta]\le n.
 \ean
then the  functions  $\phi,  \widetilde\phi$
generate dual wavelet systems $\{\psi_{jk}^{(\nu)}\}$, $\{\widetilde\psi_{jk}^{(\nu)}\}$
with  $VM^n$ property.
\label{t4}
\end{theo}

{\tt Proof.} Set $Q=(\mu_{00}\ddd\mu_{0r})$, $\widetilde Q=(\widetilde\mu_{00}\ddd\widetilde\mu_{0r})$.
Due to Lemma~\ref{l1}, the $1\times (r+1)$ matrixes $Q, \widetilde Q$
can be extended to $(r+1)\times (r+1)$ matrixes
${\cal N}=\{\mu_{\nu k}\}_{\nu,k=0}^r, \widetilde{\cal N}=\{\overline\mu_{\nu k}\}_{\nu,k=0}^r$ such that their entries
are in $L^{(n)}_\infty$ and
${\cal N}\overline{{\widetilde{\cal N}}^T}=I_{r+1}$.
So, the matrixes
\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~(\ref{61}). It follows from Theorem~\ref{t2} that the corresponding wavelet
masks $m_1\ddd m_{m-1}, \widetilde m_1\ddd \widetilde m_{m-1}$
satisfy~(\ref{36}) for all $\alpha\in\zd$, $[\alpha]\le n$, what was to be proved.
$\Diamond$

Applied mathematicians and engineers are especially
interested in construction of compactly supported wavelet systems.
To provide this property  generating refinable functions should be  compactly supported
and wavelet masks should be trigonometric polynomials.

\begin{theo}
Let $\phi,  \widetilde\phi$ be  compactly supported refinable functions with polynomial masks,
 $\widehat\phi_0(\nul)= \widehat{\widetilde \phi}_0(\nul)=1$,
and let $\mu_{00}\ddd\mu_{0, m-1}$, $\widetilde \mu_{00}\ddd\widetilde \mu_{0, m-1}$ be
the  polyphase representatives of their masks. If  there exist complex numbers
$\lambda_\gamma, \widetilde\lambda_\gamma$, $\gamma\in\zd_+$,  $[\gamma]\le n$, such that~(\ref{58}), (\ref{59})
 holds  for $k=0\ddd m-1$  and there exist trigonometric polynomials
 $\mu_{0k}$, $\widetilde \mu_{0k}$, $k=m\ddd r$, such that
 \ban
 &&\sml k0r\mu_{0 k}\overline{\widetilde\mu_{0k}}=1,
 \\
&&D^\beta\mu_{0k}(\nul)=D^\beta\widetilde\mu_{0k}(\nul)=0,\ k=m\ddd r,\ \ \ \forall \beta\in\zd_+, [\beta]\le n.
 \ean
then the  functions  $\phi,  \widetilde\phi$
generate dual  compactly supported  wavelet systems $\{\psi_{jk}^{(\nu)}\}$, $\{\widetilde\psi_{jk}^{(\nu)}\}$
with  $VM^n$ property.
\label{t5}
\end{theo}



{\tt Proof.} Set $Q=(\mu_{00}\ddd\mu_{0r})$, $\widetilde Q=(\widetilde\mu_{00}\ddd\widetilde\mu_{0r})$.
Due to Suslin's solution of a generalized  Serre conjecture\cite{3}, the row  $Q$
can be extended to a unimodular matrix with polynomial entries. After this it is not
difficult to find $(r+1)\times (r+1)$ matrices ${\cal N}, \widetilde{\cal N}$ extending
 $Q, \widetilde Q$. such that their entries are trigonometric polynomials and
${\cal N}\overline{{\widetilde{\cal N}}^T}=I_{r+1}$  (see~\cite{40},~\cite{32}, \cite[\S 2.6]{NPS}).
Next we repeat the arguments of the previous proof.
$\Diamond$


Finally, we will describe a method for construction  compactly supported
wavelet frames with $VM^n$ property.

{\bf Step 1.} Given $n\in\zd$ and given  set of parameters $\lambda_\beta\in{\Bbb C}$,
$\beta\in \zd_+$, $[\beta]\le n$, $\lambda_\nul=1$,
 find a dual set of parameters $\widetilde\lambda_\beta\in{\Bbb C}$ satisfying~(\ref{57})
 by the following recursive  formulas
$$
\widetilde\lambda_\nul=1,\ \ \ \ {\widetilde\lambda_\alpha}=-\overline{\lambda_\alpha}
-\sum\limits_{\nul<\beta\le\alpha}
\lll\alpha\atop\beta\rrr\overline{\lambda_\beta}{\widetilde\lambda_{\alpha-\beta}}.
$$



{\bf Step 2.} Chose functions $\mu^\prime_{00}\ddd \mu^\prime_{0\,m-1}$ and
$\widetilde\mu_{00}\ddd \widetilde\mu_{0\,m-1}$ defined by
\ban
\mu^\prime_{0k}(x)=
\frac1{\sqrt m}\sum\limits_{[\alpha]\le n}g_\alpha(x)
\sum\limits_{\nul\le\beta\le\alpha}
\lll\alpha\atop\beta\rrr\lll-2\pi i M^{-1}s_k)\rrr^{\beta}\lambda_{\alpha-\beta}+
\\
\sum\limits_{[\alpha]= n+1}T_\alpha(x)\prod\limits_{j=1}^{\alpha_j}\lll 1-\ex{x_j}\rrr^{\alpha_j},
\ean
\ban
\widetilde\mu_{0k}(x)=
\frac1{\sqrt m}\sum\limits_{[\alpha]\le n}g_\alpha(x)
\sum\limits_{\nul\le\beta\le\alpha}
\lll\alpha\atop\beta\rrr\lll-2\pi i M^{-1}s_k)\rrr^{\beta}\widetilde\lambda_{\alpha-\beta}+
\\
\sum\limits_{[\alpha]= n+1}\widetilde T_\alpha(x)\prod\limits_{j=1}^{\alpha_j}\lll 1-\ex{x_j}\rrr^{\alpha_j},
\ean
where $T_\alpha, \widetilde T_\alpha$ are arbitrary trigonometric polynomials,
$g_\alpha$ are trigonometric polynomials such that
$D^\alpha g_\alpha(\nul)=1$, $D^\beta g_\alpha(\nul)=0$
for all $\beta\in\zd_+$, $\beta\ne\alpha$, $[\beta]\le n$ (recursive formulas for computing $g_\alpha$
are given in~\cite{36}). It is clear that~(\ref{59}) are fulfilled



{\bf Step 3.}
Set $\sigma:=\sml l0{m-1}\overline{\mu^\prime_{0l}}\,\widetilde\mu_{0l}$,
$\mu_{0k}:=(2-\sigma)\mu^\prime_{0k}, k=0\ddd m-1$.



Due to Corollary~\ref{c1}, we have $D^\beta\sigma(\nul)=0$
for all   $\beta\in\zd_+$,  $0<[\beta]\le n$. It follows that
$D^\beta\mu_{0k}(\nul)=D^\beta\mu^\prime_{0k}(\nul)$
for all  $\beta\in\zd_+$,  $[\beta]\le n$.
It is not difficult to see that~(\ref{58}) holds and
$$
1-\sml k0{m-1}\overline{\mu_{0k}(x)}\widetilde\mu_{0k}=(1-\sigma)^2.
$$

Set $\mu_{0m}:=1-\sigma, \widetilde\mu_{0m}:=1-\overline{1-\sigma}$.




{\bf Step 4.}\
Find  matrixes
$$
\cal M=\lll\matrix{
\mu_{00}&\dots&\mu_{0, m-1}&\mu_{0, m}
\cr
\mu_{10}&\dots&\mu_{1, m-1}&*
\cr
\dots&\dots&\dots&\dots
\cr
\mu_{m,0}&\dots&\mu_{m, m-1}&*
}\rrr,\ \ \
\widetilde {\cal M}=\lll\matrix{
\widetilde \mu_{00}&\dots&\widetilde \mu_{0, m-1}&\widetilde \mu_{0, m}
\cr
\widetilde\mu_{10}&\dots&\widetilde\mu_{1, m-1}&*
\cr
\dots&\dots&\dots&\dots
\cr
\widetilde \mu_{m,0}&\dots&\widetilde\mu_{m, m-1}&*
}\rrr
$$
such that their entries are trigonometric polynomials and  ${\cal M}\overline{\widetilde {\cal M}^T}=I_{m+1}$.

Though matrixes $\cal M, \widetilde {\cal M}$ can be constructed theoretically (see the proof of
Theorem~\ref{t5}), it is very complicate to implement the algorithm in practice.
Instead, we suggest the following explicit way (the payment of  simplicity of this way is
increasing of the redundancy of the frames).

Set $\mu_{0,m+1}:=0, \widetilde\mu_{0, m+1}:=0$. For each $\nu=1\ddd m+1$, define
\ban
\widetilde\mu_{\nu, m+1}:=\overline{\mu_{0,  m+1-\nu}},\ \
\mu_{\nu, m+1}:=\overline{\widetilde\mu_{0,  m+1-\nu}},\ \ \hspace{6.7cm}
\\
\mu_{\nu k}:=\delta_{m+1-\nu, k}-
{\mu_{0 k }\overline{\widetilde\mu_{0, m+1-\nu}}}, \ \ \
\widetilde\mu_{\nu k}:=\delta_{m+1-\nu, k}-
{\widetilde\mu_{0 k }\overline{\mu_{0, m+1-\nu}}}, \ \ \    k=0\ddd m.
\ean
It is not difficult to check that the matrixes ${\cal M}:=\{\mu_{\nu k}\}_{\nu,k=0}^{m+1}$,
$\widetilde{\cal M}:=\{{\widetilde \mu_{\nu k}}\}_{\nu,k=0}^{m+1}$
satisfy ${\cal M}\overline{\widetilde {\cal M}^T}=I_{m+2}$.



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 \vspace{.5cm}
\noindent
Maria Skopina, Department of Applied Mathematics and Control
Processes, Saint Petersburg State University, 

\noindent
e-mail:\ skopina@MS1167.spb.edu
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