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\begin {document}
\centerline{\Large\bf{Decompositions of Trigonometric Polynomials} }

\centerline{\Large\bf {Related to
Multivariate Subdivision Schemes}}
\bigskip
\centerline{Nira Dyn and M. Skopina \footnote {The paper is supported by RFBR, project N 06-01-00457}}
\bigskip






\subsection{1.Introduction}


In this  paper we study decompositions of multivariate trigonometric polynomials,
satisfying a set of constraints known as Strung-Fix conditions of order $m$,
multiplied by a factor of the form $1-exp(q(x))$, with $q(x)$ a multilinear
polynomial with integer coefficients. The decomposition is a representation of
such   a polynomial as a
sum of trigonometric polynomials, satisfying the Strung-Fix conditions
of order $m-1$, multiplied by factors of the above form.
This type of decompositions is of interest in the analysis of convergence and
smoothness of multivariate subdivision schemes. The algebraic approach to the
analysis of such schemes is investigated in~\cite{0}
relative to the expansion matrix $2I$, and in~\cite{12},  \cite{11}, relative
to general expansion matrices.
The analogous decompositions for multivariate algebraic polynomials are based
on the rich structure of ideals of multivariate polynomials. The
trigonometric decompositions are based on the rather simple idea of the
representation of a trigonometric polynomial in terms of its unique polyphase
trigonometric polynomials relative to  an expansion matrix\cite{2}.
The outline of the papers is as follows: in Section 2 we introduce our
notation, and bring basic results related to polyphase representations of
trigonometric polynomials. In Section 3 properties of trigonometric
polynomials satisfying the Strang-Fix conditions of order $m=0$ relative to a
general expansion matrix are derived, in terms of its  polyphase trigonometric
polynomials. These results are used in
Section 4 to prove the decomposition for the case $m=0$.
An  algorithm for the computation of the decomposition for $m=0$  is
presented, and the decomposition is used to derive that for the case $m=1$.
An example of this case
is also presented, with all the computational details.
The general case $m>1$, which is much more complicated is investigated in
Section 5, and an algorithm for this case is given.
In the last section it is proved that trigonometric polynomials satisfying
the Strang-Fix
conditions of order $m$  are determined by the value at the origin of the
derivatives of
their polyphase trigonometric polynomials.
This result provides a tool for the construction of such polynomials.
The application to the smoothness analysis of multivariate subdivision schemes
will be done elsewhere.










\subsection {2. Notation and  preliminary information}

Let $ \n $ be 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 $;
$ \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; $z:=(z_1\ddd z_d)$,  where $z_k:=\ex{x_k}$.


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 [p. 107] {73}).
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, $m=|\det M|$,
$D (M) =\{s_0\ddd s_{m-1}\}$, $D (M^*) =\{s^*_0\ddd s^*_{m-1}\}$, $s_0=s^*_0=\nul$, $r_k=M^{-1}s_k$,
 $k=0\ddd m-1$.
%$R (M) =\{r_0\ddd r_{m-1}\}$.
% $r^*_k=M^{-1}s^*_k$,$R^* (M) =\{r^*_0\ddd r^*_{m-1}\}$.


\noindent
{\bf Proposition  A}
{\em
The matrix $\left\{\frac1{\sqrt m}\,\ex{(r_k, s_l^*)}\right\}_{k,l=0}^{m-1}$
is unitary. In particular,}
\be
\frac1{\sqrt m}\sum\limits_{k=0}^{m-1}\ex{(r_k, s_l^*)}=\delta_{0l}.
\label{a}
\ee

A proof of this statement can be found, e.g., in~\cite{1}.

We will consider $1$-periodic trigonometric polynomials in $d$ variables
$$
t(x)=\sum\limits_{n\in\zd}\widehat t(n)\ex{(n,x)}.
$$
For any  trigonometric polynomial $t$, there exists a unique
set of   trigonometric polynomials $\tau_{\nu}$, $\nu=0\ddd m-1$,
(polyphase functions of $t$) so that
\be
t(x)=\sml \nu0{m-1}\ex{(s_\nu,x)}\tau_{\nu}(M^*x).
\label{1}
\ee
It is clear that
$$
\tau_{\nu}(x)=\sum\limits_{n\in\zd}\widehat t(Mn+s_\nu)\ex{(n,x)}.
$$

For any $n\in\z_+$, denote by $SF^n$
the set of trigonometric polynomials $t$ such that
$D^\beta t({M^*}^{-1}x)|_{x=s}=0$ for all $s\in D(M^*)$, $s\ne\nul$,
and for all $\beta\in\zd_+$, $[\beta]\le n$.
Such a notation is reasonable because the refinable function  corresponding to a polynomial $t\in SF^n$
satisfies Strang-Fix condition of order~$n$.
It will be convenient for us to define $SF^{-1}$ as the set of all trigonometric polynomials.

\subsection {3. Auxiliary results}

\begin{prop}
A trigonometric polynomial $t$
belongs to $SF^0$ if and only if
\be
\tau_\nu(\nul)=\frac{t(\nul)}{m},\ \ \ \nu=0\ddd m-1,
\label{01}
\ee
where $\tau_{0}\ddd\tau_{m-1}$ are the  polyphase functions of $t$.
\label{p1}
\end{prop}

{\bf Proof.} Let $s\in D(M^*)$, using~(\ref{1}) we have
$$
t({M^*}^{-1}s)=\sml \nu 0{m-1}\ex{(r_\nu,s)}\tau_{\nu}(s)=
\sml \nu 0{m-1}\ex{(r_\nu,s)}\tau_{\nu}(\nul).
$$
So, the relation $t\in SF^0$ can be rewritten as
$$
\sml \nu 0{m-1}\ex{(r_\nu,s_l^*)}\tau_{\nu}(\nul)=t(\nul)\delta_{\nul l}, \ \ \  l=0\ddd m-1.
$$
Consider these equalities as a linear system with unknowns
$\tau_0(\nul)\ddd\tau_{m-1}(\nul)$.
Due to Proposition A, the system has a unique solution given by~(\ref{01}).
$\Diamond$

\begin{lem}
Let $t$ be a trigonometric polynomial, $t(\nul)=0$, then
$$
t(x)=\sml k1d t_k(x)(1-\ex{(x,{\bf e}_k)}),
$$
where $t_k$, $k=1\ddd d$, are trigonometric polynomials.
\label{l1}
\end{lem}

{\bf Proof.} There exists $N\in\zd$ such that $t(x)\ex{(N,x)}=p(z)$,
where $p$ is an algebraic polynomial. It is clear that $P(1\ddd1)=0$.
By Taylor formula it follows that
$$
p(z)=\sml k1d p_k(z)(1-z_k).
$$
where $p_k$, $k=1\ddd d$, are algebraic polynomials. It remains to set
$t_k(x):=p_k(z)\exm{(N,x)}. \Diamond$

\begin{lem}
Let $t, \tilde t\in SF^0$, $t(\nul)=\tilde t(\nul)$, then
$$
t(x)-\tilde t(x)=\sml k1d t_k(x)\lll 1-\ex{(M^*x,{\bf e}_k)}\rrr,
$$
where $t_k$, $k=1\ddd d$, are trigonometric polynomials.
\label{l2}
\end{lem}

{\bf Proof.} Using~(\ref{1}) for $t, \tilde t$, we have
\be
t (x)-\tilde t(x)=\sml \nu 0{m-1}\ex{(s_\nu,x)}(\tau_{\nu}(M^*x)-\tilde \tau_{\nu}(M^*x)).
\label{2}
\ee
Due to Proposition~\ref{p1}, $\tau_{\nu}(\nul)-\tilde \tau_{\nu}(\nul)=0$, $\nu=1\ddd m-1$.
Hence, by Lemma~\ref{l1},
\be
\tau_{\nu}(y)-\tilde \tau_{\nu}(y)=\sml l1d \tau_{\nu k}(y)\lll 1-\ex{(y,{\bf e}_k)}\rrr,
\label{3}
\ee
where $\tau_{\nu k}$, $k=1\ddd d$, are trigonometric polynomials.
It remains to set $y=M^*x$ and combine~(\ref{3}) with~(\ref{2}).
$\Diamond$

\begin{lem}
Let $t\in SF^0$, $r\in\zd$, $\tilde t:=\ex{(r,\cdot)}t$, then $\tilde t\in SF^0$.
\label{l3}
\end{lem}

{\bf Proof.} By~(\ref{1}),
$$
\tilde t(x)=\ex{(r,x)}\sml \nu0{m-1}\ex{(s_\nu,x)}\tau_{\nu}(M^*x)=
\sml \nu0{m-1}\ex{(s_\nu+r,x)}\tau_{\nu}(M^*x).
$$
It is clear that $s_\nu+r=s_{l_\nu}+Mn_\nu$, $n_\nu\in\zd$, $\{l_0\ddd l_{m-1}\}=\{0\ddd m-1\}$.
So,
$$
\tilde \tau_{l_\nu}(x)=\sml \nu0{m-1}\ex{(s_{l_\nu},x)}\ex{(n_\nu,M^*x)}\tau_{\nu}(M^*x),
$$
and the trigonometric polynomial $\tilde \tau_{l_\nu}(x):=\ex{(n_\nu,x)}\tau_\nu(x)$
is the $l_\nu$-th polyphase function of $\tilde t$. If $s\in D(M^*)$, $s\ne\nul$,
due to Proposition~\ref{p1},
$$
\tilde \tau_{l_\nu}(s)=\ex{(n_\nu,s)}\tau_\nu(s)=\tau_\nu(s)=\tau_\nu(\nul)=
\frac{t(\nul)}{m}=\frac{\tilde t(\nul)}{m}.
$$
Again by Proposition~\ref{p1}, it follow that $\tilde t\in SF^0$.
$\Diamond$


\begin{lem}
Let $n\in\zd$, $ 1-\ex{(M^{-1}n,{\bf e}_j)}=0$ for each $j=1\ddd d$.
Then $n\equiv\nul\pmod M$.
\label{l4}
\end{lem}

{\bf Proof.}
Since $\ex n=1$ if and only if $n\in\z$, we have $(M^{-1}n)_j=(M^{-1}n,{\bf e}_j)\in\z$
for each $j=1\ddd d$. Hence, $M^{-1}n=l\in\zd$, i.e. $n=Ml$.~$\Diamond$

\subsection {4. Decomposition of masks}


\begin{prop}
Let $t\in SF^0$, $k=1\ddd d$, then
\be
(1-\ex{(x,{\bf e}_k)})t(x)=\sml j1d t_{jk}(x)(1-\ex{(M^*x,{\bf e}_j)}),
\label{001}
\ee
where $t_{jk}$, $j=1\ddd d$, are trigonometric polynomials. Inversely, if~(\ref{001}) holds
for a trigonometric polynomial $t$, then $t\in SF^0$.
\label{p2}
\end{prop}

{\bf Proof.} If  $t\in SF^0$, then~(\ref{001})
 follows immediately form Lemmas~\ref{l2} and~\ref{l3}. Now let $t$ be an arbitrary trigonometric
 polynomial such that~(\ref{001}) holds,  $s\in D(M^*)$, $s\ne\nul$. It follows from~(\ref{001}) that
$$
 t({M^*}^{-1}s)\lll 1-\ex{({M^*}^{-1}s,{\bf e}_k))}\rrr=
\\
\sml j1d t_{jk}({M^*}^{-1}s)(1-\ex{(s,{\bf e}_j)})=0.
$$
Taking into account Lemma~\ref{l4}, we obtain $t({M^*}^{-1}s)$, which was to be proved.~$\Diamond$

Analyzing the proofs of  Lemmas~\ref{l2} and \ref{l3} it is not difficult to
describe an algorithm for finding polynomials $t_{jk}$, $j,k=1\ddd d$, in decomposition~(\ref{001}).
We will consider that the polyphase functions $\tau_\nu$ of $t$ are given, and
the algorithm derives the polyphase functions $\tau_{jk\nu}$ of each polynomial $t_{jk}$.

\vspace{.5cm}
\noindent
{\bf ALGORITHM 1}

{\bf Input:} \ \ \ \ \  $\{\widehat\tau_\nu(l),  \ l\in\zd, \ \nu=0\ddd m-1\}$.


{\bf Output:} \ \ $\{\widehat\tau_{jk\nu}(l), \  l\in\zd,\  \nu=0\ddd m-1, \ j,k=1\ddd d\}$.

{\bf Step 0.} \ \ For each $\nu=0\ddd m-1$ find the sets $$\Omega_\nu:=\{l\in\zd:\ \widehat\tau_\nu(l)\ne0\}.$$

{\bf For each }$k=1\ddd d$ {\bf do}

\hspace{.1cm}
{\bf For each} $\nu=0\ddd m-1$ {\bf do}

\begin{quote}
{\bf Step 1.} Compute $M^{-1}({\bf e}_k-s_\nu+s_n)=:l_n$
for all $n=0\ddd m-1$ and denote by $n^*$ the unique
$n$ such that $l_n\in\zd$.
\end{quote}

\begin{quote}
{\bf Step 2.} Find the set $\tilde\Omega_{k\nu}:=\{l=r+l_{n^*}, r\in \Omega_{n^*}\}$, and for
each $l\in \tilde\Omega_{k\nu}$ put $\widehat{\tilde\tau}_\nu(l):={\widehat\tau}_{n^*}(l-l_{n^*})$,
$$
p_{k\nu}(z):=\sum\limits_{l\in\Omega_\nu\cup{\tilde\Omega}_{k\nu}}
({\widehat\tau}_\nu(l)-\widehat{\tilde\tau}_\nu(l))z^l.
$$
\end{quote}

\begin{quote}
{\bf Step 3.} Put
\ban
&&\tau_{1k\nu}(x)=\frac1{1-z_1}\lll p_{k\nu}(z)-p_{k\nu}(1,z_2\ddd z_d) \rrr,
\\
&&\tau_{2k\nu}(x)=\frac1{1-z_2}\lll p_{k\nu}(1,z_2,z_3\ddd z_d) -p_{k\nu}(1,1,z_3\ddd z_d)\rrr,
\\
&&\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots
\ldots\ldots
\\
&&\tau_{dk\nu}(x)=\frac1{1-z_d}\lll p_{k\nu}(1,1\ddd 1,z_d) -p_{k\nu}(1,1\ddd 1,1)\rrr.\Diamond
\ean
\end{quote}

\begin{theo}
If  $t\in SF^1$, then  in any decomposition~(\ref{001}) the trigonometric
polynomials $t_{jk}$, $j,k=1\ddd d$,  belong to $SF^0$.
\label{t1i}
\end{theo}

{\bf Proof.}
%First we note that decomposition~(\ref{001}) exists because $SF^{(1)}\subset SF^{(0)}$.
Let $l=1\ddd d$, $s\in D(M^*)$, $s\ne\nul$.
Since $t\in SF^{(1)}$,
\be
\frac{\partial}{\partial x_l}\lll \lll 1-\ex{({M^*}^{-1}x,{\bf e}_k))}\rrr t({M^*}^{-1}x)\rrr\Bigg|_{x=s}=0,
\ \ \ k=1\ddd d.
\label{28}
\ee
On the other hand, it follows from~(\ref{001}) that for each $k=1\ddd d$ we have
\ban
\frac{\partial}{\partial x_l}\lll\lll 1-\ex{({M^*}^{-1}x,{\bf e}_k))}\rrr t({M^*}^{-1}x)\rrr\Bigg|_{x=s}=
\hspace{3cm}
\\
\frac{\partial}{\partial x_l}\lll\sml j1d t_{jk}({M^*}^{-1}x)(1-\ex{x_j})\rrr\Bigg|_{x=s}=
\\
\sml j1d t_{jk}({M^*}^{-1}s)\frac{\partial}{\partial x_l}(1-\ex{x_j})\Bigg|_{x=s}=
-2\pi i\,t_{lk}({M^*}^{-1}s).
\ean
So, we proved that $t_{lk}({M^*}^{-1}s)=0$ for all  $s\in D(M^*)$, $s\ne\nul$, which means
$t_{lk}\in SF^{(0)}$.~$\Diamond$

 {\bf Example.\ \ \ }
Let
$
M=\left(\begin{array}{rr} 0 & 2 \\ 2 & -1
\end{array} \right),
$
%the set $D(M)$ consists of
$$
s_0=\left(\begin{array}{r} 0 \\ 0\end{array} \right),
s_1=\left(\begin{array}{r} 1 \\ 0\end{array} \right),
s_2=\left(\begin{array}{r} 0 \\ 1\end{array} \right),
s_3=\left(\begin{array}{r} 1\\ 1 \end{array} \right).
$$
Define the polyphase functions of  a polynomial $t$ by

\ban
\tau_0(x)&=&\frac1{16}(4+4z_1+4z_2+4z_1z_2),
\\
\tau_1(x)&=&\frac1{16}(5+4z_1+z_1^{-1}+2z_2+3z_2^{-1}+z_1z_2),
\\
\tau_2(x)&=&\frac1{16}(4+z_1+2z_1^{-1}+5z_2+z_2^{-1}+3z_1z_2),
\\
\tau_3(x)&=&\frac1{16}(5+z_1+4z_1^{-1}+z_2+3z_2^{-1}+z_1z_2+z_1^{-1}z_2).
\ean
It is not difficult to see that $t\in SF^0$. Using Algorithm 1, we find
$$
\begin{array}{ll}
\tau_{110}(x)=\frac1{16}(-1+2z_2+z_1z_2+2z_2^2+z_1z_2^2),
&
\tau_{210}(x)=\frac1{16}(5+3z_2),
\\
\\
\tau_{111}(x)=\frac1{16}(z_1^{-1}+3z_2),
&
\tau_{211}(x)=\frac1{16}(5+3z_2^{-1}),
\\
\\
\tau_{112}(x)=\frac1{16}(-1+2z_1^{-1}+z_2+z_2^2-z_1^{-1}z_2),
&
\tau_{212}(x)=\frac1{16}(5+3z_2+z_2^{-1}),
\\
\\
\tau_{113}(x)=\frac1{16}(2z_1^{-1}+2z_2+z_1^{-1}z_2),
&
\tau_{213}(x)=\frac1{16}(5+2z_2^{-1}),
\\
\\
\tau_{120}(x)=\frac1{16}(1+z_1+4z_2+z_2^{-1}+3z_1z_2),
&
\tau_{220}(x)=\frac1{16}z_2^{-1},
\\
\\
\tau_{121}(x)=\frac1{16}(2+z_1+z_1^{-1}+z_2+3z_2^{-1}+z_1z_2),
&
\tau_{221}(x)=0,
\\
\\
\tau_{122}(x)=\frac1{16}(3+z_2+2z_1^{-1}),
&
\tau_{222}(x)=\frac1{16}z_2^{-1},
\\
\\
\tau_{123}(x)=\frac1{16}(3+3z_1^{-1}+z_1^{-1}z_2),
&
\tau_{223}(x)=0.
\end{array}
$$
Due to Theorem 2 of \cite{10}, to prove that the subdivision scheme $S_t$ is convergent,
it sufficient to check that the norm the matrix subdivision operator $S_T$ from $\ell_\infty^2$
to $\ell_\infty^2$ is strictly less than 1, and that
 $M$ has a self-similar tile, i.e. there exists a set $E\subset\r^2$
such that $ME=\cupm\limits_{s\in D(M)}(E+s)$,
$\cupm\limits_{l\in\z^2}(E+l)=\r^2$, $\mbox{mes} E=1$. The first
requirement is fulfilled because $$
\|S_T\|_{\ell_\infty^2\to\ell_\infty^2}\le\max\limits_{(\nu,k)}
(\|\tau_{1k\nu}\|_{\ell_1}+\|\tau_{2k\nu}\|_{\ell_1})\le\frac{15}{16}
$$ (here we identify a trigonometric polynomial with its
sequence of Fourier coefficients). Set $$ E=\left\{x\in\r^2:\ \
x=\sml k1\infty M^{-k}x_k,\ x_k\in D(M)\right\}. $$ To show that
$E$ provides a self-similar tile it suffices to check (see, e.g.,
\cite{73}) that the function $$
m_0(x)=\frac14(1+\ex{x_1}+\ex{x_2}+\ex{(x_1+x_2)}) $$ is such that
$m_0({M^*}^{-k}x)$ does not vanish on $\left[-1/2,1/2\right]^2$
for all $k\in\n$. Since $$
|m_0({M^*}^{-1}x)|^2=\lll1+\cos2\pi\lll\frac{x_1}4+\frac{x_2}2\rrr\rrr
\lll1+\cos2\pi\frac{x_1}2\rrr, $$ we see that all zeros of this
function are on the lines $\frac{x_1}4+\frac{x_2}2=\pm\frac12+l$,
$\frac{x_1}2=\pm\frac12+l$, $l\in\z$, and it is clear that the
cube $\left[-1/2,1/2\right]^2$ does not intersect with these
lines. It remains to note that
$\|{M^*}^{-1}x\|_\infty\le\frac34\|x\|_\infty$, and so
${M^*}^{-1}\left[-1/2,1/2\right]^2\subset\left[-1/2,1/2\right]^2$.



Proposition \ref{p2} and Theorem~\ref{t1i} state that for $n=0,1$ condition $t\in SF^n$
implies $t_{jk}\in SF^{n-1}$. This fact
can not be extended to the case  $n>1$. If $t\in SF^2$, there exist
decompositions~(\ref{001}) whose elements are not in $SF^1$. Nevertheless, it will be shown
that decomposition of $t\in SF^n$ can be fixed up to provide  $t_{jk}\in SF^{n-1}$.

\begin{theo}
Let $n\in\z_+$. A trigonometric polynomial $t$ belongs to $SF^n$  if and only if
there exists a decomposition~(\ref{001}) with $t_{jk}\in SF^{n-1}$, $j,k=1\ddd d$.
\label{t1}
\end{theo}

{\bf Proof.} Set
\ban
a(x)=t({M^*}^{-1}x), &&a_{jk}(x)=t_{jk}({M^*}^{-1}x),
\\
b_k(x)=1-\ex{{(M^*}^{-1}x,{\bf e}_k)}, &&c_k(x)=1-\ex{(x,{\bf e}_k)}.
\ean
In these notation, (\ref{001}) can be rewritten as
\be
b_k(x)a(x)=\sml j1d a_{jk}(x)c_j(x).
\label{4}
\ee
Note  the following trivial properties of $c_j$:
\ba
&&c_j(l)=0 \ \forall l\in\zd,
\label{5}
\\
&&D^\delta c_j\equiv0\ \forall \delta\ne r{\bf e}_j, r\in\n, \delta\ne \nul,
\label{6}
\\
&&\frac{\partial c_j}{\partial x_j}(l)\ne0\ \forall l\in\zd.
\label{7}
\ea
We will prove the theorem by induction on $n$.
We have a base  for $n=0$ due to Proposition~\ref{p2}.
Let us prove the inductive step: $n\to n+1$.

Assume that each polynomial $t_{jk}$ belongs to $SF^{n}$, i.e.
$D^\beta a_{jk}(s^*_l)=0$, $l=1\ddd m-1$, for all $\beta\in\zd_+$, $[\beta]\le n$. Let
$\alpha\in\zd$, $[\alpha]=n+1$, $s\in D(M^*)$, $s\ne\nul$.
It follows from~(\ref4) and Leibniz formula that
$$
\sum\limits_{\nul\le\beta\le\alpha}
\lll\alpha\atop\beta\rrr
 D^{\alpha-\beta}b_k(s) D^\beta a(s)=D^\alpha a_{jk}(s)c_j(s).
$$
By the inductive hypotheses, $D^\beta a(s)=0$ whenever $\beta<\alpha$. Hence,
taking into account~(\ref5), we have
$$
b_k(s)D^\alpha a(s)=c_j(s)D^\alpha a_{jk}(s)=0.
$$
Due to Lemma~\ref{l4}, $b_k(s)\ne0$  for at least one $k=1\ddd d$. It follows that
$D^\alpha a(s)=0$.

Now we assume that $t\in SF^{n+1}$, i.e. $D^\alpha a(s^*_\nu)=0$, $\nu=1\ddd m-1$, for all
$\alpha\in\zd_+$, $[\alpha]\le n+1$. Due to the  inductive hypotheses, there exist
polynomials $t_{jk}\in SF^{n-1}$ satisfying~(\ref{001}), i.e. $D^\delta a_{jk}(s^*_\nu)=0$, $\nu=1\ddd m-1$,
 for all
$\delta\in\zd_+$, $[\delta]\le n-1$. We will construct new trigonometric polynomials
 satisfying~(\ref{001}) and belonging to $SF^{n}$. Note that for $n=0$ any trigonometric polynomials
$t_{jk}$  satisfying~(\ref{001})  belong to $SF^{0}$, due to Theorem~\ref{t1i}.

Let us prove the following statement. If $D^\delta a_{jk}(s^*_\nu)=0$, $\nu=1\ddd m-1$,
 for all $\delta\in\zd_+$, $[\delta]\le n$ and for all  $j=1\ddd l-1$ ($l=1\ddd d$),
then there exist polynomials $\tilde t_{jk}$, $j,k=1\ddd d$, such that
$\sml j1d \tilde a_{jk}c_j=\sml j1d a_{jk}c_j$, where $ \tilde a_{jk}(x)= \tilde t_{jk}({M^*}^{-1}x)$,
and $D^\delta \tilde a_{jk}(s^*_\nu)=0$, $\nu=1\ddd m-1$, $k=1\ddd d$,
 for all $\delta\in\zd_+$, $[\delta]\le n$, and for all  $j=1\ddd l$.


Let $s\in D(M^*), s\ne\nul$, $\beta\in \z_+^d$, $[\beta]=n$, $\sml i{l+1}d\beta_i=0$.
Set $\alpha=\beta+{\bf e}_l$ and note that $[\alpha]=n+1$.
It follows from~(\ref4) that
\be
D^\alpha\lll \sml j1d a_{jk}(x)c_j(x)\rrr\Bigg|_{x=s}=
D^\alpha(b_k(x)a(x))|_{x=s}=0.
\label8
\ee
On the other hand, due to~(\ref5), (\ref6) and the assumption of the statement,
 we have
\ban
D^\alpha\lll \sml j1d a_{jk}(x)c_j(x)\rrr\Bigg|_{x=s}=
\sml j1lD^{\alpha-{\bf e}_j} a_{jk}(s)\frac{\partial c_j}{\partial x_j}(s)=
D^{\beta} a_{lk}(s)\frac{\partial c_l}{\partial x_l}(s).
\ean
Combining this  with~(\ref7) and (\ref8), we get $D^\beta a_{lk}(s) =0$.
%whenever  $[\beta]=n$, $\beta_{l+1}=\dots=\beta_d=0$. Note that the statement
is proved already for $l=d$ (in this case $\tilde t_{jk}=t_{jk}$, $j,k=1\ddd d$).

Next let  $j=l+1\ddd d$, $\beta\in \z_+^d$, $[\beta]=n$, $\sml i{l+1}d\beta_i>0$. Define the functions
$$
q_{\beta j}(x)=\frac1{-2\pi i}g_{n-1, \beta-{\bf e}_j}(x)\sml\nu1{m-1}h_\nu(x)D^\beta a_{lk}(s^*_\nu),
$$
where $g_{N \delta}$ is a trigonometric polynomial such that
 $D^\gamma g_{N \delta}(\nul)=0$ for all $\gamma\in \z_+^d$,  $[\gamma]\le N$,
$\gamma\ne\delta$ and $D^\delta g_{N \delta}(\nul)=1$;
$$
h_\nu(x)=\frac1m\sml\mu0{m-1}\ex{(x-s_\nu^*, {M}^{-1}s_\mu)}.
$$
Since, by Proposition A, $h_\nu(s^*_\mu)=\delta_{\mu\nu}$, due  to (\ref5), (\ref6) and
Leibniz formula,
 it is not difficult to see that for each $\nu=1\ddd m-1$ we have
\ba
&&D^\beta(c_j(x)q_{\beta j}(x))\Big|_{x=s_\nu^*}=D^\beta a_{lk}(s^*_\nu);
\label{25}
\\
&&D^\delta(c_j(x)q_{\beta j}(x))\Big|_{x=s_\nu^*}=0\ \ \ \forall \delta\in \z_+^d,\ [\delta]\le n,\ \delta\ne\beta;
\label{26}
\\
&&D^\delta(c_i(x)q_{\beta j}(x))\Big|_{x=s_\nu^*}=0\ \ \ \forall \delta\in \z_+^d,\ [\delta]\le n-1, \forall i=1\ddd d.
\label{27}
\ea
Set
\ban
\tilde a_{lk}(x)&:=&a_{lk}(x)-\sml j{l+1}d
\sum\limits_{[\beta]=n, \ \beta_j>0\atop\beta_{l+1}=\dots=\beta_{j-1}=0} c_j(x)q_{\beta j}(x),
\\
\tilde a_{jk}(x)&:=&a_{jk}(x)+
\sum\limits_{[\beta]=n, \ \beta_j>0\atop\beta_{l+1}=\dots=\beta_{j-1}=0} c_l(x)q_{\beta j}(x),\ \ \ j=l+1\ddd d.
\ean
Because of construction,
$\sml jld \tilde a_{jk}c_j=\sml jld  a_{jk}c_j$, and, taking into account~(\ref{25}), (\ref{26}),
for each $\nu=1\ddd m-1$
we have $D^\beta \tilde a_{lk}(s^*_\nu)=0$  whenever $[\beta]=n$, $\sml i{l+1}d\beta_i>0$;
$D^\beta \tilde a_{lk}(s^*_\nu)=D^\beta  a_{lk}(s^*_\nu)=0$  whenever $[\beta]=n$, $\sml i{l+1}d\beta_i=0$.
At last, due to~(\ref{27}), $D^\delta \tilde a_{jk}(s^*_\nu)=0$, $j=l\ddd d$
 for all $\delta\in\zd_+$, $[\delta]\le n-1$. To complete the proof of the statement it remains
 to put $\tilde t_{jk}=\tilde a_{jk}(M^*x)$
 for $j=l\ddd d$ and $\tilde t_{jk}=t_{jk}$ for $j=1\ddd l-1$.

So, we described one step for improvement of  decomposition~(\ref{001}). Starting with $l=1$, after $(d-1)$ steps
we will obtain required polynomials.~$\Diamond$

Analyzing the proof of  Theorem~\ref{t1}  it is not difficult to
describe an algorithm for finding polynomials $t_{jk}\in SF^{n-1}$, $j,k=1\ddd d$, in decomposition~(\ref{001}).
To realize this algorithm we will need functions $g_{N\delta}$. Explicit recursive formulas for
 these functions are presented in~\cite{2}.

\vspace{.5cm}
\noindent
{\bf ALGORITHM 2}

{\bf Input:} \ \ \ \ \  $
%\mbox{A trigonometric polynomial}\
t\in SF^{N}$, $N>1$.


{\bf Output:} \ \ $
%\mbox{Trigonometric polynomials}\
t_{jk}\in SF^{N-1}, j,k=1\ddd d$.
%, satisfying~(\ref{001}).

{\bf Step 1.} \ \ Using Algorithm 1, find  $t_{jk}$, $j,k=1\ddd d$, satisfying~(\ref{001}).

{\bf Step 2.} \ \ For $n=1\ddd N-1$ do

\hspace{2.2cm} Set $t_{jk}^{(0)}:=t_{jk}$, $j,k=1\ddd d$;



\hspace{2.5cm} For each  $l=1\ddd d-1$ do

\hspace{2.8cm} For each  $k=1\ddd d$ do

$
\ \ \ \ \ a_{lk}(x)=t^{(l-1)}_{lk}({M^*}^{-1}x);
$
$$
\ \ \ \  t^{(l)}_{lk}(x):=t^{(l-1)}_{lk}(x)-\sml j{l+1}d
\lll1-\ex{(M^*x,{\bf e}_j)}\rrr\sum\limits_{[\beta]=n, \ \beta_j>0\atop\beta_{l+1}=\dots=\beta_{j-1}=0}
 q_{\beta j}(M^*x);
$$

\hspace{3.1cm} For each  $j=1\ddd l$ do
$$
\hspace{-8cm} t^{(l)}_{jk}(x):=t^{(l-1)}_{jk}(x);
$$

\hspace{3.1cm} For each  $j=l+1\ddd d$ do
$$
t^{(l)}_{jk}(x):=t^{(l-1)}_{jk}(x)+
\lll1-\ex{(M^*x,{\bf e}_l)}\rrr\sum\limits_{[\beta]=n, \ \beta_j>0\atop\beta_{l+1}=\dots=\beta_{j-1}=0}
 q_{\beta j}(M^*x).\Diamond
$$

In the case $d=2$, Step 2 of Algorithm 2 does not look so frightening, it is reduced to the following.

For $n=1\ddd N-1$ do

Set  $l=1$, $t_{jk}^{(0)}:=t_{jk}$, $j,k=1,2$;

For each  $k=1,2$ do


\ban
 a_{1k}(x)&=&t^{(0)}_{1k}({M^*}^{-1}x);
\\
 t_{1k}(x)&:=&t^{(0)}_{1k}(x)-
\lll1-\ex{(M^*x,{\bf e}_2)}\rrr\sum\limits_{[\beta]=n, \ \beta_2>0}
 q_{\beta j}(M^*x);
\\
t_{2k}(x)&:=&t^{(0)}_{2k}(x)+
\lll1-\ex{(M^*x,{\bf e}_1)}\rrr\sum\limits_{[\beta]=n, \ \beta_2>0}
 q_{\beta j}(M^*x).
\ean

In particular, for $d=2$, $N=2$, Step 2 may be realized as follows.

Set $t_{jk}^{(0)}:=t_{jk}$, $j,k=1,2$;

 For each  $k=1,2$ do
\ban
t_{1k}(x):=t^{(0)}_{lk}(x)+
\frac1{2\pi i}\lll1-\ex{(M^*x)_2}\rrr
 \sml\nu1{m-1} h_\nu(M^*x)\frac{\partial}{\partial x_2}
t^{(0)}_{1k}({M^*}^{-1}u)\Big|_{u=s^*_\nu};
\\
t_{2k}(x):=t^{(0)}_{2k}(x)-\frac1{2\pi i}
\lll1-\ex{(M^*x)_1}\rrr \sml\nu1{m-1} h_\nu(M^*x)\frac{\partial}{\partial x_2}
t^{(0)}_{1k}({M^*}^{-1}u)\Big|_{u=s^*_\nu}.
\ean

For each $n\in\n$ we introduce the set
$$
\Gamma^n:=\{k\in\r^n:\ \ k_l=1\ddd d,\  l=1\ddd n\}.
$$

\begin{theo}
Let  $n, n_0\in\n$, $n\le n_0$, $t\in SF^{n_0-1}$, $k\in\Gamma^n$, then
there exist trigonometric polynomials $t_{jk}\in SF^{n_0-n-1}$, $j\in \Gamma^n$, such that
\ba
\prod\limits_{l=1}^n\lll 1-\ex{(x,{\bf e}_{k_l})}\rrr t(x)&=&
\sml {j_1}1d \ldots \sml {j_n}1d t_{jk}(x)\prod\limits_{l=1}^n\lll 1-\ex{(M^*x,{\bf e}_{j_l})}\rrr,
\label{13}
\\
t_{jk}(\nul)&=&\prod\limits_{l=1}^n({M^*}^{-1})_{j_lk_l}\,t(\nul).
\label{14}
\ea
\label{t2}
\end{theo}




{\bf Proof.} We will proof by induction on $n$.

Base: $n=1$. Let $k\in\Gamma^1$. Due to Proposition~\ref{p2} and Theorem~\ref{t1},
there exist trigonometric polynomials $t_{jk}\in SF^{n_0-2}$, $j\in \Gamma^1$, such that~(\ref{001})
holds. Rewrite identity~(\ref{001}) as follows
$$
(1-\ex{(x,M^{-1}{\bf e}_k)})t({M^*}^{-1}x)=\sml j1d t_{jk}({M^*}^{-1}x)(1-\ex{(x,{\bf e}_j)}),
$$
Differentiating by $x_l$, we have
\ban
-2\pi i\,(M^{-1}{\bf e}_k,{\bf e}_l)\ex{(x,M^{-1}{\bf e}_k)}) t({M^*}^{-1}x)+
(1-\ex{(x,M^{-1}{\bf e}_k)})\frac{\partial}{\partial x_l}t({M^*}^{-1}x)=
\\
\sml j1d \frac{\partial}{\partial x_l}t_{jk}({M^*}^{-1}x)(1-\ex{(x,{\bf e}_j)})-
2\pi i\, t_{jk}({M^*}^{-1}x)\,\ex{(x,{\bf e}_j)}.
\ean
Substituting $x=\nul$, we obtain
$$
-2\pi i\,(M^{-1}{\bf e}_k,{\bf e}_l) t(\nul)=
-2\pi i\,t_{jk}(\nul).
$$
To prove the base it remains to note that
$$
(M^{-1}{\bf e}_k,{\bf e}_l)=({M^*}^{-1}{\bf e}_l,{\bf e}_k)=({M^*}^{-1})_{kl}.
$$

Inductive step: $n-1\to n$. Let $1<n\le n_0$, $k\in\Gamma^n$, $k^\prime:=(k_1\ddd k_{n-1})$.
By the inductive hypotheses there exist   $t_{j^\prime k^\prime}\in SF^{n_0-n}$,
$j^\prime\in \Gamma^{n-1}$, such that
\ba
\prod\limits_{l=1}^{n-1}\lll 1-\ex{(x,{\bf e}_{k_l})}\rrr t(x)=
\sml {j_1}1d \ldots \sml {j_{n-1}}1d t_{j^\prime k^\prime}(x)
\prod\limits_{l=1}^{n-1}\lll 1-\ex{(M^*x,{\bf e}_{j_l})}\rrr,
\label{9}
\\
t_{jk}(\nul)=\prod\limits_{l=1}^{n-1}({M^*}^{-1})_{j_lk_l}\,t(\nul).\hspace{4cm}
\label{10}
\ea
Since $n_0-n\ge0$ and the theorem is proved already for $n=1$,
for each $t_{j^\prime k^\prime}$, there exist
 trigonometric polynomials $t_{j_n k_n}\in SF^{n_0-n-1}$,
$j\in \Gamma^1$, such that
\ba
\lll 1-\ex{(x,{\bf e}_{k_n})}\rrr t_{j^\prime k^\prime}(x)&=&
 \sml {j_n}1d t_{j_nk_n}(x)\lll 1-\ex{(M^*x,{\bf e}_{j_n})}\rrr,
\label{11}
\\
t_{j_nk_n}(\nul)&=&({M^*}^{-1})_{j_nk_n}t_{j^\prime k^\prime}(\nul).
\label{12}
\ea
Combining~(\ref9), (\ref{10}) with (\ref{11}), (\ref{12}) we comlete the proof.~$\Diamond$


To impart a more compact form to~(\ref{13}), (\ref{14}), we introduce the following notations.
Set
\ban
&&\Delta_k(x)=\Delta_k=\lll1-\ex{({\bf e}_k,x)}\rrr,\ \ \Delta=(\Delta_1\ddd \Delta_d),
\\
&&\delta_k(x)=\delta_k=\lll1-\ex{({\bf e}_k,M^*x)}\rrr,\ \ \delta=(\delta_1\ddd \delta_d)^T.
\ean
Let $A^{[n]}$ denotes the $n$-th Kronecker degree of a matrix $A$, i.e. if $A$ is a $M\times N$ matrix with
entries $a_{jk}$, then
$$
A^{[1]}:=A,\ \ \
A^{[n+1]}:=
\lll\matrix{
a_{11}A^{[n]}&\dots&a_{1N}A^{[n]}
\cr
\vdots&\ddots&\vdots
\cr
\cr
a_{M1}A^{[n]}&\dots&a_{MN}A^{[n]}}
\rrr.
$$
Now Theorem~\ref{t2} can be rewritten as

\noindent
{\bf Theorem 8\,$^\prime$}
{\em Let  $n, n_0\in\n$, $n\le n_0$, $t\in SF^{n_0-1}$, $k\in\Gamma^n$, then
there exists a $d^n\times d^n$ matrix  $T$ whose
entries are  trigonometric polynomials $t_{jk}\in SF^{n_0-n-1}$ such that}
$$
(\Delta(x))^{[n]}t(x)=T(x)(\delta(x))^{[n]},\ \ \ T(\nul)=({M^*}^{-1})^{[n]}t(\nul).
$$

\subsection {5. Construction of desirable masks}

A simple description of the classes $SF^n$ is well known in the one-dimensional dyadic case.
A general form is given by the formula
$t(x)=(1+\ex x)^nT(x)$, where $T$ is an arbitrary trigonometric polynomial.
In the multidimensional case  $SF^n$ can not be described in a similar way.
We will give a characterization of the class $SF^n$  for arbitrary matrix dilation
which allows to construct its elements in practice.

\begin{theo}
A trigonometric polynomial $t$ belongs to $SF^n$ if and only if
the derivatives of its polyphase function $\tau_k$, $k=0\ddd m-1$, up to
order $n$ are given by
 \be
D^\alpha\tau_{k}(\nul)=\frac1{m}
\sum\limits_{\nul\le\beta\le\alpha}\lambda_\beta
\lll\alpha\atop\beta\rrr(-2\pi i r_k)^{\alpha-\beta},\ \ \alpha\in\zd_+, [\alpha]\le n,
\label{23}
\ee
where $\lambda_\alpha=D^\alpha t({M^*}^{-1}x)|_{x=0}$.
\label{t3}
\end{theo}

{\bf Remark.} For polynomials $t$ whose polyphase functions $\tau_0\ddd\tau_{m-1}$ form a
unimodular row (i.e.  there exists a dual row of trigonometric polynomials
$\tilde\tau_0\ddd\tilde\tau_{m-1}$ such that $\sum_{k=0}^{m-1}\tau_k\tilde\tau_k\equiv1$),
the statement of Theorem~\ref{t3} follows from combining the results of~\cite1 and \cite2.
It was proved in these papers
 that  both the conditions are equivalent to vanishing moments of
the corresponding wavelet system.

{\bf Proof.}  Assume that~(\ref{23}) holds with some  complex numbers $\lambda_\alpha$.
Let $s\in D(M^*)$.  By  Leibniz formula,
\ban
D^\alpha\lll \ex{(r_k,x)}\tau_k(x)\rrr\Big|_{x=s}=
\sum\limits_{\nul\le\beta\le\alpha}\lll\alpha\atop\beta\rrr
D^\beta\lll \ex{(r_k,x)}\rrr\Big|_{x=s} D^{\alpha-\beta}\tau_k(\nul)=
\\
\sum\limits_{\nul\le\beta\le\alpha}\lll\alpha\atop\beta\rrr\ex{(r_k,s)}
(2\pi ir_k)^\beta D^{\alpha-\beta}\tau_k(\nul)=
\\
\frac1{ m}\ex{(r_k,s)}\sum\limits_{\nul\le\beta\le\alpha}\lll\alpha\atop\beta\rrr
(2\pi ir_k)^\beta
\sum\limits_{\nul\le\gamma\le\alpha-\beta}\lambda_\gamma
\lll\alpha-\beta\atop\gamma\rrr(-2\pi i r_k)^{\alpha-\beta-\gamma}=
\\
\frac1{ m}\ex{(r_k,s)}\sum\limits_{\nul\le\beta\le\alpha}
\sum\limits_{\nul\le\gamma\le\alpha-\beta}\lambda_\gamma
\lll\alpha-\beta\atop\gamma\rrr\lll\alpha\atop\beta\rrr(-2\pi i r_k)^{\alpha-\gamma}
\prod\limits_{j=1}^d(-1)^{-\beta_j}=
\\
\frac1{ m}\ex{(r_k,s)}\sum\limits_{\nul\le\gamma\le\alpha}
\lambda_\gamma(-2\pi i r_k)^{\alpha-\gamma}
\lll\alpha\atop\gamma\rrr
\sum\limits_{\nul\le\beta\le\alpha-\gamma}
\lll\alpha-\gamma\atop\beta\rrr
\prod\limits_{j=1}^d(-1)^{-\beta_j}.
\ean
Since
\ban
\sum\limits_{\nul\le\beta\le\alpha-\gamma}
\lll\alpha-\gamma\atop\beta\rrr
\prod\limits_{j=1}^d(-1)^{-\beta_j}=
\prod\limits_{j=1}^d
\sum\limits_{\nul\le\beta_j\le\alpha_j-\gamma_j}
\lll\alpha_j-\gamma_j\atop\beta_j\rrr(-1)^{-\beta_j}=
\\
\prod\limits_{j=1}^d(1-1)^{\alpha_j-\gamma_j}=
\left\{
\begin{array}{ll}
0, & \alpha\ne\gamma,
\\
1, & \alpha=\gamma,
\end{array}
\right.
\ean
we have
$$
D^\alpha\lll \ex{(r_k,x)}\tau_k(x)\rrr\Big|_{x=s}=
\frac{\lambda_\alpha}{ m}\ex{(r_k,s)}, \ \ k=0\ddd m-1.
$$
It follows from~(\ref{1}) and Proposition A that
\ban
D^\alpha(t({M^*}^{-1}x)\Big|_{x=s}=
%\frac1{ m}
\sml k0{m-1}D^\alpha\lll \ex{(r_k,x)}\tau_k(x)\rrr\Big|_{x=s}=
\\
\frac{\lambda_\alpha}m\sml k0{m-1}\ex{(r_k,s)}=
\left\{
\begin{array}{ll}
\lambda_\alpha,&  \mbox{if} \ \ s=\nul,
\\
0,& \mbox{if}\ \  s\ne\nul.
\end{array}
\right.
\ean

Now let us check that~(\ref{23}) follows from the relation $t\in SF^n$.
We will prove by induction on $n$. The base for $n=0$ was established in Proposition~\ref{p1}.
To prove the inductive step $n \to n+1$, we assume that $t\in SF^{n+1}$ and~(\ref{23}) holds.
Let $\alpha\in\zd$, $[\alpha]=n+1$, $s\in D(M^*)$.
 By~(\ref{1}) and  Leibniz formula,
\ba
D^\alpha(t({M^*}^{-1}x)\Big|_{x=s}=\sml k0{m-1}
\sum\limits_{\nul\le\beta\le\alpha}\lll\alpha\atop\beta\rrr
D^{\alpha-\beta}\lll \ex{(r_k,x)}\rrr\Big|_{x=s} D^{\beta}\tau_k(\nul)=
\nonumber
\\
\sml k0{m-1}\lll D^\alpha\tau_k(0)+
\sum\limits_{\nul\le\beta<\alpha}\lll\alpha\atop\beta\rrr
D^{\alpha-\beta}(2\pi ir_k)^{\alpha-\beta} D^{\beta}\tau_k(\nul)\rrr\ex{(r_k,s)}.
\label{24}
\ea
Because of the inductive hypotheses, we have
\ban
\sum\limits_{\nul\le\beta<\alpha}\lll\alpha\atop\beta\rrr
(2\pi ir_k)^{\alpha-\beta} D^{\beta}\tau_k(\nul)=
\\
\frac1m\sum\limits_{\nul\le\beta<\alpha}\sum\limits_{\nul\le\gamma\le\beta}
\lambda_\gamma\lll\alpha\atop\beta\rrr\lll\beta\atop\gamma\rrr
(2\pi ir_k)^{\alpha-\beta}(-2\pi ir_k)^{\beta-\gamma} =
\\
\frac1m\sum\limits_{\nul\le\gamma<\alpha}
\lambda_\gamma
(2\pi ir_k)^{\alpha-\gamma}
\sum\limits_{\gamma\le\beta<\alpha}\lll\alpha\atop\beta\rrr\lll\beta\atop\gamma\rrr
\prod\limits_{j=1}^d(-1)^{\beta_j-\gamma_j} =
\\
\frac1m\sum\limits_{\nul\le\gamma<\alpha}
\lll\alpha\atop\gamma\rrr\lambda_\gamma
(2\pi ir_k)^{\alpha-\gamma}
\sum\limits_{\nul\le\delta<\alpha-\gamma}\lll\alpha-\gamma\atop\delta\rrr
\prod\limits_{j=1}^d(-1)^{\delta_j} =
\\
\frac1m\sum\limits_{\nul\le\gamma<\alpha}
\lll\alpha\atop\gamma\rrr\lambda_\gamma
(2\pi ir_k)^{\alpha-\gamma}
\lll\prod\limits_{j=1}^d(1-1)^{\alpha_j-\gamma_j}-\prod\limits_{j=1}^d(-1)^{\alpha_j}\rrr=
\\
-\frac1m\sum\limits_{\nul\le\gamma<\alpha}\lll\alpha\atop\gamma\rrr\lambda_\gamma
(-2\pi ir_k)^{\alpha-\gamma}.
\ean
Combining this with~(\ref{24}), we obtain
$$
\sml k0{m-1}\ex{(r_k,s)}\lll D^\alpha\tau_k(0)-
\frac1m\sum\limits_{\nul\le\gamma<\alpha}\lll\alpha\atop\gamma\rrr\lambda_\gamma
(-2\pi ir_k)^{\alpha-\gamma}\rrr=D^\alpha(t({M^*}^{-1}x))\Big|_{x=s}.
$$
Set
$
\lambda_\alpha=D^\alpha(t({M^*}^{-1}x))\big|_{x=0}.
$
Due to Proposition A, the linear system
$$
\sml k0{m-1}\ex{(r_k,s^*_l)}y_k=\lambda_\alpha\delta_{0l},\ \
l=0\ddd m-1,
$$
has a unique solution $y_k=\frac{\lambda_\alpha}m$, $k=0\ddd m-1$. It follows that
$$
D^\alpha\tau_k(0)-
\frac1m\sum\limits_{\nul\le\gamma<\alpha}\lll\alpha\atop\gamma\rrr\lambda_\gamma
(-2\pi ir_k)^{\alpha-\gamma}=\frac{\lambda_\alpha}m,
$$
which was to be proved. $\Diamond$

So, if we want a polynomial $t$ to belong to $SF^n$, its polyphase functions
should have derivatives at the origin given by~(\ref{23}). This can be easily realized
for an arbitrary set of parameters $\lambda_\beta$, $[\beta]\le n$.
General forms for all such polynomials $t$ are presented in~\cite{2}.

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Skopina M., {\it On Construction of Multivariate Wavelets with Vanishing Moments}
ACHA {\bf 20}  (2006), 3,  375-390

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Latour V., M\"uller  J., Nickel W.
{\it Stationry subdivision for general scaling matrices},
Math. Zeitschrift  {\bf 227} (1998),  645-661.

\bibitem {12}
 Sauer T.
{\it
Polynomial interpolation, ideals and
approximation order  of multivariate refinable functions}
Proceedings of the American Mathematical Society
{\bf 130} (2002), 11, 3335-3347.

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M\"oller H.M. and  Sauer T.
{\it Multivariate refinable functions of high approximation
order via quotient ideals of Laurent polynomials},
Adv. Comput. Math. {\bf 20 } (2004),  No.1-3, 205-228.



\end {thebibliography}
\end{document}

 Base: n=1. Differentiating~(\ref4) by $x_k$, due to~(\ref6), we have
\ban
\frac{\partial b_k}{\partial x_l}(x)a(x)+b_k(x)
\frac{\partial a}{\partial x_l}(x)=
\sml j1d \frac{\partial a_{jk}}{\partial x_l}(x)d_j(x)+
a_{lk}(x)\frac{\partial d_l}{\partial x_l}(x).
\ean
For $x=s$, $s\in D(M^*)$, $s\ne\nul$, due to~(\ref5), this yields
$$
\frac{\partial b_k}{\partial s_l}(s)a(s)+b_k(s)
\frac{\partial a}{\partial x_l}(s)=
 a_{lk}(s)\frac{\partial d_l}{\partial x_l}(s).
$$

Since $t$ belongs to $SF^0$,  $a(s)=0$. So, taking into accout~(\ref7) and Lemma~\ref{l4},
we complete the proof of the base




\bibitem {book}
Novikov I., Protassov V., Skopina M.  {\it Wavelet Theory}. Moscow: Fizmatlit (to appear).
\end {document}
 But we can not state that the wavelet systems
form Riesz bases  for $L_2(\r^2)$. It is not known if
for all biorthogonal  wavelets
$\{\psi^{(\nu)}_{jn}\}$, $\{\widetilde\psi^{(\nu)}_{jn}\}$
which are  obtained via MRA construction,  compactly supported and
have vanishing moment of order $0$,  each of the systems is
a  basis (Riesz basis)  for $L_2(\rd)$.


Let $d=2$,
$
M=\left(\begin{array}{rr} -1 & 4 \\ -1 & 1
\end{array} \right),
$
$
s_0=\left(\begin{array}{r} 0 \\ 0\end{array} \right),
s_1=\left(\begin{array}{r} 1 \\ 0\end{array} \right),
s_2=\left(\begin{array}{r} -1 \\ 0\end{array} \right).
$
The matrix dilation $M$ has a self-similar tile, i.e. there exists a set $E\subset\r^2$
such that $ME=\cupm\limits_{s\in D(M)}(E+s)$ (see fig.1).
Define the polyphase functions of  a polynomial $t$ by

\ban
\tau_0(x)&=&1,
\\
\tau_1(x)&=&1-\frac i3(\sin 2\pi x_1+\sin 2\pi x_2),
\\
\tau_1(x)&=&1+\frac i3(\sin 2\pi x_1+\sin 2\pi x_2).
\ean
It is not difficult to check  that~(\ref{23}) is fulfilled with $n=1$ and  $\lambda_\nul=3$,
$\lambda_\beta=0$ for $\beta\in\z^2$, $[\beta]=1$.
\begin {figure}%[h]
\unitlength 1mm \linethickness{1pt}
\begin{picture}(120.00,60.00)
{
\put(60.00,10.00){\vector(0,1){50.0}}%
\put(20.00,35.00){\vector(1,0){80.0}}%

\put(100.00,32.00){$x$}%
\put(62.00,62.00){$y$}%
\put(62.00,31.00){$0$}%


\put(88.00,31.00){$2$}%
\put(27.00,31.00){$-2$}%
\put(73.00,31.00){${1}$}%
\put(41.00,31.00){${-1}$}%
\put(57.00,46.00){${\frac12}$}%
\put(54.00,22.00){$-\frac12$}%

\put(75.00,27.50){\line(1,1){15.0}}%
\put(30.00,27.50){\line(1,0){45.0}}%
\put(30.00,27.50){\line(1,1){15.0}}%
\put(45.00,42.50){\line(1,0){45.0}}%
\put(45.00,27.50){\line(1,1){15.0}}%
\put(60.00,27.50){\line(1,1){15.0}}%
 }
\end{picture}
  \caption { $E$ -- small central parallelogram with the vertexes
  (0,1/2), (1,1/2), (-1,-1/2), (0,-1/2); $MP$ -- is the large parallelogram.}
\label{hf1}
\end {figure}

Using Algorithm 1, we find
\ban
t_{11}(x)&=&\frac16(1+z_1z_2+z_1^5z_2^2-z_1^{-3}+z_1+z_1^2z_2+z_1^6z_2^2-z_1^{-2}+z_1^{-4}-z_1^7z_2^2
\\
&&-z_1^8z_2^3-z_2+5z_1^3z_2-z_1^4),
\\
t_{21}(x)&=&\frac16(z_1z_2+z_1^5z_2^2+z_1^2z_2+z_1^6z_2^2-z_1^4z_2^2-z_1^{-4}+4z_2),
\\
t_{12}(x)&=&\frac16(z_1^{-2}z_2-8z_1^6z_2^5+6z_1^2z_2^2-7z_1^5z_2^5-5z_1^4z_2^3-6z_1^{3}z_2^2-5z_1^2z_2+6z_2^2+
\\
&&z_1^{-7}z_2^{-1}+z_1^7z_2^3+z_1^6z_2^2-z_1^{-2}-z_1^{-4}z_2+z_1-z_1^{-1}-7z_2-z_1^8z_2^3+z_1^{-4}),
\\
t_{22}(x)&=&\frac16(-z_1^{-2}z_2-z_1z_2+z_1^4z_2^3-4z_2^2+z_1^{-4}z_2-z_1^{-4}-z_2).
\ean
