On Relating Interpolatory Wavelets to Interpolatory Scaling Functions in Multiresolution Analyses

Abstract

Construction of interpolatory wavelets is an important topic in discrete signal processing. In classical wavelet sampling theories, interpolatory wavelets are constructed from Riesz bases in wavelet spaces. Since analytical expressions for such Riesz bases are generally complex or unavailable, it has been difficult to obtain suitable interpolatory wavelets in practice. In this paper, interpolatory scaling functions are used to determine and construct interpolatory wavelets. We first show that there may not exist interpolatory wavelets even when interpolatory scaling functions exist. Then, an inequality in terms of interpolatory scaling functions, denoted as the two-scale condition, is given as a necessary and sufficient condition for existence of interpolatory wavelets. Finally, based on the two-scale condition, a filter bank is constructed for obtaining interpolatory wavelets directly from interpolatory scaling functions. In examples, our theorems are applied to some typical wavelet spaces, demonstrating our construction algorithm for interpolatory wavelets.

Appendices

Appendix 1: Proof of Lemma 1

Proof

In (9), the change of variable $y=2^jx$ gives

$$\begin{aligned} \sum \limits _{k\in \mathbb {Z}} {b_k^0 \tilde{S}^\phi (y-k)} =\sum \limits _{k\in \mathbb {Z}} {c_k^0 \tilde{S}^\phi (2y-2k)} +\sum \limits _{k\in \mathbb {Z}} {d_k^0 \tilde{S}^\phi (2y-2k-1)}. \end{aligned}$$

(37)

Obviously Eq. (9) holds if and only if (37) does. Let $P_{\tilde{S}^\phi } (w)=\frac{1}{2}\sum \limits _{k\in \mathbb {Z}} {\tilde{p}_k^s \mathrm{e}^{-iwk/2}}$ be a two-scale symbol corresponding to $\tilde{S}^\phi (x)$ such that

$$\begin{aligned} \left\{ {\begin{array}{l} \tilde{S}^\phi (x)=\sum \limits _{k\in \mathbb {Z}} {\tilde{p}_k^s \tilde{S}^\phi (2x-k)} \\ \hat{\tilde{S}}^\phi (w)=\frac{1}{2}\sum \limits _{k\in \mathbb {Z}} {\tilde{p}_k^s \mathrm{e}^{-iwk/2}} \hat{\tilde{S}}^\phi (\frac{w}{2})=P_{\tilde{S}^\phi } (w)\hat{\tilde{S}}^\phi (\frac{w}{2}) \\ \end{array}} \right. \end{aligned}$$

(38)

where $\hat{\tilde{S}}^\phi (w)$ is the Fourier transform of $\tilde{S}^\phi (x)$.

First, we show that

$$\begin{aligned} 0<\left| {\sum \nolimits _k {\tilde{p}_{2k}^s \mathrm{e}^{-iwk}} } \right| ^2<+\infty ,\text{ with } w\in [-\pi ,\pi ] \end{aligned}$$

for the coefficients $\{\tilde{p}_k^s \}_{k\in \mathbb {Z}}$ in (38).

In [5, 6, 18], it is shown that the Fourier transform $\hat{S}^\phi (w)$ can be expressed in terms of the Fourier transform $\hat{\phi }(w)$ as

$$\begin{aligned} \hat{S}^\phi (w)=\frac{\hat{\phi }(w)}{\sum \limits _{k\in \mathbb {Z}} {\phi (k)\mathrm{e}^{-iwk}} }=\frac{\hat{\phi }(w)}{\sum \limits _{k\in \mathbb {Z}} {\hat{\phi }(w+2k\pi )} } \end{aligned}$$

(39)

with $1/\sum \nolimits _{k\in \mathbb {Z}} {\hat{\phi }(w+2k\pi )} \in L^2[0,2\pi ]$ .

Applying (39) to (3) gives

$$\begin{aligned} P_s (z)=\frac{\sum \nolimits _{k\in \mathbb {Z}} {\hat{\phi }(w+4k\pi )} }{\sum \nolimits _{n\in \mathbb {Z}} {\hat{\phi }(w+2n\pi )} }. \end{aligned}$$

(40)

Let

$$\begin{aligned} P_\phi (w)=\frac{1}{2}\sum \limits _{k\in \mathbb {Z}} {p_k \mathrm{e}^{-iwk/2}} , \end{aligned}$$

(41)

denote a two-scale symbol related to a scaling function $\phi (x)$. We have

$$\begin{aligned} \hat{\phi }(w+4k\pi )=P_\phi (w+4k\pi )\hat{\phi }(w/2+2k\pi )=P_\phi (w)\hat{\phi }(w/2+2k\pi ),k\in \mathbb {Z}.\nonumber \\ \end{aligned}$$

(42)

Inserting (42) into (40) yields

$$\begin{aligned} P_s (w)=\frac{P_\phi (w)\sum \nolimits _{k\in \mathbb {Z}} {\hat{\phi }(w/2+2k\pi )} }{\sum \nolimits _{n\in \mathbb {Z}} {\hat{\phi }(w+2n\pi )} }. \end{aligned}$$

(43)

Since $\tilde{S}^\phi (x)$ is also a scaling function, it follows from (38) and (43) that

$$\begin{aligned} P_s (w)=\frac{P_{\tilde{S}^\phi } (w)\sum \nolimits _{k\in \mathbb {Z}} {\hat{\tilde{S}}^\phi (w/2+2k\pi )} }{\sum \nolimits _{n\in \mathbb {Z}} {\hat{\tilde{S}}^\phi (w+2n\pi )} }. \end{aligned}$$

(44)

Since $\{\tilde{S}^\phi (x-k)\}_{k\in \mathbb {Z}}$ forms the dual basis of $\{S^\phi (x-k)\}_{k\in \mathbb {Z}}$, we have $\hat{\tilde{S}}^\phi (w)=\hat{S}^\phi (w)/\sum \nolimits _{k=-\infty }^{+\infty } {\left| {\hat{S}^\phi (w+2k\pi )} \right| ^2}$ (see Eq. 5.6.8 in [7]). Hence, it follows from (44) that

$$\begin{aligned} P_s (w)=\frac{P_{\tilde{S}^\phi } (w)\sum \nolimits _{k\in \mathbb {Z}} {\hat{S}^\phi (w/2+2k\pi )} }{\sum \nolimits _{k\in \mathbb {Z}} {\hat{S}^\phi (w+2k\pi )} }\times \frac{\sum \nolimits _{k\in \mathbb {Z}} {\left| {\hat{S}^\phi (w+2k\pi )} \right| ^2} }{\sum \nolimits _{k\in \mathbb {Z}} {\left| {\hat{S}^\phi (w/2+2k\pi )} \right| ^2} }. \end{aligned}$$

(45)

Since $\sum \nolimits _{k\in \mathbb {Z}} {\hat{S}^\phi (w+2k\pi )} =1$ by (39), Eqs. (45) and (7) imply

$$\begin{aligned} P_s (w)=P_{\tilde{S}^\phi } (w)\frac{\sum \nolimits _{k\in \mathbb {Z}} {\left| {\hat{S}^\phi (w+2k\pi )} \right| ^2} }{\sum \nolimits _{k\in \mathbb {Z}} {\left| {\hat{S}^\phi (w/2+2k\pi )} \right| ^2} }=P_{\tilde{S}^\phi } (w)\frac{E_s (2w)}{E_s (w)}. \end{aligned}$$

(46)

Equations (38) and (46) give

$$\begin{aligned} \left| {\sum \nolimits _k {\tilde{p}_{2k}^s \mathrm{e}^{-iwk}} } \right| ^2&=\left| {\overline{\sum \nolimits _k {\tilde{p}_{2k}^s \mathrm{e}^{-iwk}} } } \right| ^2=\left| {\overline{P_{\tilde{S}^\phi } (w+2\pi )+P_{\tilde{S}^\phi } (w)} } \right| ^2 \nonumber \\&=\left| {\overline{\frac{P_s (w+2\pi )E_s (w+2\pi )+P_s (w)E_s (w)}{E_s (2w)}} } \right| ^2\nonumber \\&=\frac{\left| {\overline{P_s (w+2\pi )} E_s (w+2\pi )+\overline{P_s (w)} E_s (w)} \right| ^2}{\left| {E_s (2w)} \right| ^2}. \end{aligned}$$

(47)

Since $\{S^\phi (x-k)\}_{k\in \mathbb {Z}}$ forms a Riesz basis of $V_0$, there are two constants $A_{Es}$ and $B_{Es}$ with $0<A_{Es} \le B_{Es} <+\infty$ such that

$$\begin{aligned} 0<A_{Es} \le E_s (2w)\le B_{Es} <+\infty ( \mathrm{{Theorem }}\, 9\, \mathrm{{in}}\, [19]). \end{aligned}$$

(48)

It follows from (6), (47) and (48) that

$$\begin{aligned} 0<\frac{A_s }{B_{Es}^2 }\le \left| {\sum \nolimits _k {\tilde{p}_{2k}^s \mathrm{e}^{-iwk}} } \right| ^2\le \frac{B_s }{A_{Es}^2 }<+\infty . \end{aligned}$$

(49)

Equation (49) implies $0<\left| {\sum \nolimits _k {\tilde{p}_{2k}^s \mathrm{e}^{-iwk}} } \right| ^2<+\infty$ for every $w\in [-\pi ,\pi ]$.

Second, we show there exist unique coefficients $\{b_k^0 \}_{k\in \mathbb {Z}}$ and $\{d_k^0 \}_{k\in \mathbb {Z}}$ for which (37) holds.

Assume $\{c_k^0 \}_{k\in \mathbb {Z}}$ is any element of $l^2(\mathbb {R})$. Since $0<\left| {\sum \nolimits _k {\tilde{p}_{2k}^s \mathrm{e}^{-iwk}} } \right| ^2<+\infty$ by (49), the series $2\sum \nolimits _{k\in \mathbb {Z}} {c_k^0 \mathrm{e}^{-iwk}} /\sum \nolimits _{k\in \mathbb {Z}} {\tilde{p}_{2k}^s \mathrm{e}^{-iwk}}$ is an element of $L^2[-\pi ,\pi ]$ for any given $\{c_k^0 \}_{k\in \mathbb {Z}} \in l^2(\mathbb {R})$. Hence there are unique coefficients $\{b_k^0 \}_k \in L^2(\mathbb {R})$ such that

$$\begin{aligned} \sum \nolimits _{k\in \mathbb {Z}} {b_k^0 \mathrm{e}^{-iwk}} =2\sum \nolimits _{k\in \mathbb {Z}} {c_k^0 \mathrm{e}^{-iwk}} /\sum \nolimits _{k\in \mathbb {Z}} {\tilde{p}_{2k}^s \mathrm{e}^{-iwk}} . \end{aligned}$$

(50)

On the other hand, since $\{b_k^0 \}_k \in l^2(\mathbb {R})$ in (50), we have that $\frac{1}{2}\sum \nolimits _{k\in \mathbb {Z}} {b_k^0 \mathrm{e}^{-iwk}} \sum \nolimits _{k\in \mathbb {Z}} {\tilde{p}_{2k+1}^s \mathrm{e}^{-iwk}}$ is also an element of $L^2[-\pi ,\pi ]$. Hence, there are unique coefficients $\{d_k^0 \}_k \in l^2(\mathbb {R})$ such that

$$\begin{aligned} \frac{1}{2}\sum \nolimits _{k\in \mathbb {Z}} {b_k^0 \mathrm{e}^{-iwk}} \sum \nolimits _{k\in \mathbb {Z}} {\tilde{p}_{2k+1}^s \mathrm{e}^{-iwk}} =\sum \nolimits _{k\in \mathbb {Z}} {d_k^0 \mathrm{e}^{-iwk}} . \end{aligned}$$

(51)

Equations (50) and (51) imply there always exist two unique sets of coefficients $\{b_k^0 \}_{k\in \mathbb {Z}}$ and $\{d_k^0 \}_{k\in \mathbb {Z}}$ in $l^2(\mathbb {R})$ such that

$$\begin{aligned} \sum \nolimits _{k\in \mathbb {Z}} {c_k^0 \mathrm{e}^{-iwk}} +\mathrm{e}^{-iw/2}\sum \nolimits _{k\in \mathbb {Z}} {d_k^0 \mathrm{e}^{-iwk}} =\frac{1}{2}\sum \nolimits _{k\in \mathbb {Z}} {b_k^0 \mathrm{e}^{-iwk}} \sum \nolimits _{n\in \mathbb {Z}} {\tilde{p}_n^s \mathrm{e}^{-iwn/2}}\nonumber \\ \end{aligned}$$

(52)

for any given $\{c_k^0 \}_{k\in \mathbb {Z}} \in l^2(\mathbb {R})$.

Multiplying (52) by $\hat{\tilde{S}}^\phi (w/2)$ and inserting (38) yield

$$\begin{aligned}&\left( {\sum \nolimits _{k\in \mathbb {Z}} {c_k^0 \mathrm{e}^{-iwk}} +\mathrm{e}^{-iw/2}\sum \nolimits _{k\in \mathbb {Z}} {d_k^0 \mathrm{e}^{-iwk}} }\right) \hat{\tilde{S}}^\phi (w/2)\nonumber \\&\quad =\frac{1}{2}\sum \nolimits _{k\in \mathbb {Z}} {b_k^0 \mathrm{e}^{-iwk}} \sum \nolimits _{n\in \mathbb {Z}} {\tilde{p}_n^s \mathrm{e}^{-iwn/2}} \hat{\tilde{S}}^\phi (w/2)=\sum \nolimits _{k\in \mathbb {Z}} {b_k^0 \mathrm{e}^{-iwk}} \hat{\tilde{S}}^\phi (w).\nonumber \\ \end{aligned}$$

(53)

for any $\{c_k^0 \}_k \in l^2(\mathbb {R})$. Taking the inverse Fourier transform on both sides in (53), we obtain

$$\begin{aligned} \sum \nolimits _k {c_k^0 \tilde{S}^\phi (2x-2k)} +\sum \nolimits _k {d_k^0 \tilde{S}^\phi (2x-2k-1)} =\sum \nolimits _k {b_k^0 \tilde{S}^\phi (x-k)} . \end{aligned}$$

(54)

Since the coefficients $\{b_k^0 \}_k$ and $\{d_k^0 \}_k$ for which (52) holds are unique, it follows from (54) that, for any given $\{c_k^0 \}_k \in l^2(\mathbb {R})$, there exist unique $\{b_k^0 \}_{k\in \mathbb {Z}}$, $\{d_k^0 \}_{k\in \mathbb {Z}} \in l^2(\mathbb {R})$ for which (37) holds. Hence for a given set of coefficients $\{c_k^0 \}_k \in l^2(\mathbb {R})$, there exist unique $\{b_k^0 \}_{k\in \mathbb {Z}}$, $\{d_k^0 \}_{k\in \mathbb {R}} \in l^2(\mathbb {R})$ for which (9) holds. $\square$

Appendix 2: Proof of Theorem 1

Proof

Since $S^\phi (x)$ is a scaling function, $f_s (x)\in V_j$ can be written as

$$\begin{aligned} f_s (x)=\sum \limits _{k\in \mathbb {Z}} {f_s \left( \frac{k}{2^j}\right) S_{j,k}^\phi (x)} =\sum \limits _{k\in \mathbb {Z}} \left( {F_k S_{j,2k+1}^\phi (x)+f_s (\frac{2k}{2^j})S_{j,2k}^\phi (x)}\right) . \end{aligned}$$

(55)

where $S_{j,k}^\phi (x)=S^\phi (2^jx-k)$ and $\{F_k \}_{k\in \mathbb {Z}} =\{f_s ((2k+1)/2^j)\}_{k\in \mathbb {Z}}$.

It follows from (55) that

$$\begin{aligned} \sum \limits _{n\in \mathbb {Z}} {\left\langle {f_s ,S_{j-1,n}^\phi } \right\rangle \tilde{S}_{j-1,n}^\phi }&=\sum \limits _{n\in \mathbb {Z}} {\sum \limits _{k\in \mathbb {Z}} {F_k \left\langle {S^\phi _{j,2k+1} (x),S_{j-1,n}^\phi } \right\rangle \tilde{S}_{j-1,n}^\phi }}\nonumber \\&\quad +\sum \limits _{n\in \mathbb {Z}} {\sum \limits _{k\in \mathbb {Z}} {f_s \left( \frac{2k}{2^j}\right) \left\langle {S^\phi _{j,2k} (x),S_{j-1,n}^\phi } \right\rangle \tilde{S}_{j-1,n}^\phi }} \end{aligned}$$

(56)

where $\{\tilde{S}_{j-1,k}^\phi (x)=\tilde{S}^\phi (2^{j-1}x-k)\}_k$ is the dual basis of $\{S_{j-1,k}^\phi (x)=S^\phi (2^{j-1}x-k)\}_k$.

Obviously $\sum \nolimits _{n\in \mathbb {Z}} {\left\langle {f_s ,S_{j-1,n}^\phi } \right\rangle \tilde{S}_{j-1,n}^\phi }$ in (56) is the projection of $f_s (x)$ on $V_{j-1}$. Hence $f_s (x)$ is an element of $W_{j-1}$ if and only if

$$\begin{aligned} \sum \limits _{n\in \mathbb {Z}} {\left\langle {f_s ,S_{j-1,n}^\phi } \right\rangle \tilde{S}_{j-1,n}^\phi } =0. \end{aligned}$$

(57)

Now we show that for given $\left\{ {F_k } \right\} _k$, there exists unique $\{f_s (2k/2^j)\}_{k\in \mathbb {Z}}$ in (56) for (57) to hold, i.e., the sequence $\left\{ {F_k } \right\} _k$ in (55) determines a unique function $f_s (x)$ whose projection $\sum \nolimits _{n\in \mathbb {Z}} {\left\langle {f_s ,S_{j-1,n}^\phi } \right\rangle \tilde{S}_{j-1,n}^\phi }$ on $V_{j-1}$ vanishes.

Letting

$$\begin{aligned} a_l =\left\langle {S^\phi (2^jx),S^\phi (2^jx-l)} \right\rangle =\frac{1}{2^j}\left\langle {S^\phi (x),S^\phi (x-l)} \right\rangle , \end{aligned}$$

(58)

then

$$\begin{aligned} S^\phi (2^jx-k)\!=\!\sum \limits _l {\left\langle {S^\phi (2^jx),S^\phi (2^jx-l)} \right\rangle \tilde{S}^\phi (2^jx\!-\!l\!-\!k)} \!=\!\sum \limits _l {a_l \tilde{S}^\phi (2^jx\!-\!l\!-\!k)} .\nonumber \\ \end{aligned}$$

(59)

Substituting (59) into (55) yields that

$$\begin{aligned} f_s (x)&= \sum \limits _{k\in \mathbb {Z}} {F_k \sum \limits _{n\in \mathbb {Z}} {a_n \tilde{S}^\phi (2^jx-n-2k-1)} }\nonumber \\&\quad +\sum \limits _{k\in \mathbb {Z}} {f_s \left( \frac{2k}{2^j}\right) \sum \limits _{n\in \mathbb {Z}} {a_n \tilde{S}^\phi (2^jx-n-2k)} } . \end{aligned}$$

(60)

It follows from (57) and (60) that

$$\begin{aligned} \left\langle {f_s ,S_{j-1,l}^\phi } \right\rangle&=\sum \limits _{k\in \mathbb {Z}} {F_k \sum \limits _{n\in \mathbb {Z}} {a_n \left\langle {\tilde{S}_{j,n+2k+1}^\phi ,S_{j-1,l}^\phi } \right\rangle } }\nonumber \\&\quad +\sum \limits _{k\in \mathbb {Z}} {f_s \left( \frac{2k}{2^j}\right) \sum \limits _{n\in \mathbb {Z}} {a_n \left\langle {\tilde{S}_{j,n+2k}^\phi ,S_{j-1,l}^\phi } \right\rangle } } =0. \end{aligned}$$

(61)

Since $\{\tilde{S}_{j,k}^\phi (x)\}_k$ is the dual basis of the interpolatory basis $\{S_{j,k}^\phi (x)\}_k$ in $V_j$, we have

$$\begin{aligned} \left\{ {\begin{array}{c} \left\langle {\tilde{S}_{j,n+2k+1}^\phi ,S_{j-1,l}^\phi } \right\rangle =\int _{-\infty }^{+\infty } {\tilde{S}_{j,n+2k+1}^\phi (x)\bar{S}_{j-1,l}^\phi (x)dx} =\bar{S}^\phi \left( \frac{n+2k-2l+1}{2}\right) \\ \left\langle {\tilde{S}_{j,n+2k}^\phi ,S_{j-1,l}^\phi } \right\rangle =\int _{-\infty }^{+\infty } {\tilde{S}_{j,n+2k}^\phi (x)\bar{S}_{j-1,l}^\phi (x)dx} =\bar{S}^\phi \left( \frac{n+2k-2l}{2}\right) \\ \end{array}} \right. \!,\nonumber \\ \end{aligned}$$

(62)

where $\bar{S}^\phi (x)$ is the complex conjugate of $S^\phi (x)$.

Inserting (62) into (61) yields

$$\begin{aligned} \left\langle {f_s ,S_{j-1,l}^\phi } \right\rangle&=\sum \limits _{k\in \mathbb {Z}} {F_k \sum \limits _{n\in \mathbb {Z}} {a_n \bar{S}^\phi \left( \frac{n+2k-2l+1}{2}\right) } }\nonumber \\&\quad +\sum \limits _{k\in \mathbb {Z}} {f_s \left( \frac{2k}{2^j}\right) \sum \limits _{n\in \mathbb {Z}} {a_n \bar{S}^\phi \left( \frac{n+2k-2l}{2}\right) } } =0. \end{aligned}$$

(63)

The Fourier transform of (63) yields

$$\begin{aligned} \sum \limits _{l\in \mathbb {Z}} {\left\langle {f_s ,S_{j-1,l}^\phi } \right\rangle \mathrm{e}^{-iwl}}&= A_f +B_f =\sum \limits _{k\in \mathbb {Z}} {F_k \sum \limits _{n\in \mathbb {Z}} {a_n \sum \limits _{l\in \mathbb {Z}} {\bar{S}^\phi \left( \frac{n+2k-2l+1}{2}\right) \mathrm{e}^{-iwl}} } } \nonumber \\&+\sum \limits _{k\in \mathbb {Z}} {f_s \left( \frac{2k}{2^j}\right) \sum \limits _{n\in \mathbb {Z}} {a_n \sum \limits _{l\in \mathbb {Z}} {\bar{S}^\phi \left( \frac{n+2k-2l}{2}\right) \mathrm{e}^{-iwl}} } } =0,\nonumber \\ \end{aligned}$$

(64)

where $A_f =\sum \limits _{k\in \mathbb {Z}} {F_k \sum \limits _{n\in \mathbb {Z}} {a_n \sum \limits _{l\in \mathbb {Z}} {\bar{S}^\phi (\frac{n+2k-2l+1}{2})\mathrm{e}^{-iwl}} } }$ and $B_f =\sum \limits _{k\in \mathbb {Z}} f_s (\frac{2k}{2^j})\sum \limits _{n\in \mathbb {Z}} a_n \sum \limits _{l\in \mathbb {Z}} {\bar{S}^\phi (\frac{n+2k-2l}{2})\mathrm{e}^{-iwl}}$.

By the Poisson summation formula

$$\begin{aligned} \sum \limits _{l\in \mathbb {Z}} {\bar{S}^\phi \left( \frac{n+2k-2l+1}{2}\right) \mathrm{e}^{-iwl}}&= \overline{\sum \limits _{l\in \mathbb {Z}} {S^\phi \left( \frac{n+1}{2}+k+l\right) \mathrm{e}^{-iwl}} }\nonumber \\&= \overline{\sum \limits _{l\in \mathbb {Z}} {\hat{S}^\phi \left( w+2l\pi \right) \mathrm{e}^{i(w+2l\pi )(n+1+2k)/2}} }\nonumber \\ \end{aligned}$$

(65)

where $\hat{S}^\phi (w)$ is the Fourier transform of $S^\phi (x)$.

Inserting (65) into $A_f$ in (64) yields

$$\begin{aligned} A_f =\sum \limits _{k\in \mathbb {Z}} {F_k \mathrm{e}^{-ikw}\sum \limits _{n\in \mathbb {Z}} {a_n \mathrm{e}^{-\frac{n+1}{2}wi}\overline{\sum \limits _{l\in \mathbb {Z}} {\hat{S}^\phi (w+2l\pi )\mathrm{e}^{i(n+1)l\pi }} } } } . \end{aligned}$$

(66)

Cutting the summation (66) into two parts, one for $n=2m$ and the other for $n=2m-1$, yields

$$\begin{aligned} A_f&= \sum \limits _{k\in \mathbb {Z}} {F_k \mathrm{e}^{-ikw}} \left( {\sum \limits _{m\in \mathbb {Z}} {a_{2m} \mathrm{e}^{-mwi}\overline{\mathrm{e}^{iw/2}\sum \limits _{l\in \mathbb {Z}} {\hat{S}^\phi (w+2l\pi )(-1)^l} } } }\right. \nonumber \\&\quad \left. {+\sum \limits _{m\in \mathbb {Z}} {a_{2m-1} \mathrm{e}^{-mwi}\overline{\sum \limits _{l\in \mathbb {Z}} {\hat{S}^\phi (w+2l\pi )} } } }\right) . \end{aligned}$$

(67)

Since $\{S^\phi (x-k)\}_k$ is the interpolatory basis of $V_0$, (39) indicates that $\sum \nolimits _{l\in \mathbb {Z}} {\hat{S}^\phi (w+2l\pi )} =1$. Hence it follows from (67) that

$$\begin{aligned} A_f =\sum \limits _{k\in \mathbb {Z}} {F_k \mathrm{e}^{-ikw}} \left( {\sum \limits _{m\in \mathbb {Z}} {a_{2m-1} \mathrm{e}^{-mwi}} } {+\sum \limits _{m\in \mathbb {Z}} {a_{2m} \mathrm{e}^{-mwi}\mathrm{e}^{-iw/2}\overline{\sum \limits _{l\in \mathbb {Z}} {\hat{S}^\phi (w+2l\pi )(-1)^l} } } }\right) .\nonumber \\ \end{aligned}$$

(68)

It follows from (58) and Theorem 2.4 in [18] that

$$\begin{aligned} \sum \limits _{m\in \mathbb {Z}} {a_{2m} \mathrm{e}^{-iwm}}&=\frac{1}{2^j}\sum \limits _{m\in \mathbb {Z}} {\int \limits _{-\infty }^{+\infty } {\bar{S}^\phi (x)S^\phi (x-2m)dx\times \mathrm{e}^{-iwm}} }\nonumber \\&=\frac{1}{2^{j+1}}\frac{1}{2\pi }\int \limits _{-\infty }^{+\infty } {\vert \hat{S}^\phi (u/2)\vert ^2\sum \limits _{m\in \mathbb {Z}} {\mathrm{e}^{-ium}\times \mathrm{e}^{-iwm}} du} \nonumber \\&=\frac{1}{2^{j+1}}\int \limits _{-\infty }^{+\infty } {\vert \hat{S}^\phi (u/2)\vert ^2\sum \limits _{m\in \mathbb {Z}} {\delta (u+w-2m\pi )} du}. \end{aligned}$$

(69)

Since $\vert \hat{S}^\phi (u/2)\vert ^2$ is even in $u$, it follows from (69) that

$$\begin{aligned} \sum \limits _{m\in \mathbb {Z}} {a_{2m} \mathrm{e}^{-mwi}} =\frac{1}{2^{j+1}}\sum \limits _{k\in \mathbb {Z}} {\left| {\hat{S}^\phi \left( \frac{w+2k\pi }{2}\right) } \right| } ^2. \end{aligned}$$

(70)

Similarly, we have

$$\begin{aligned} \sum \limits _{m\in \mathbb {Z}} {a_{2m-1} \mathrm{e}^{-mwi}} =\frac{1}{2^{j+1}}\mathrm{e}^{-\frac{iw}{2}}\sum \limits _{k\in \mathbb {Z}} {(-1)^k\left| {\hat{S}^\phi \left( \frac{w+2k\pi }{2}\right) } \right| } ^2. \end{aligned}$$

(71)

Inserting (70) and (71) into (68) yields

$$\begin{aligned} A_f&=\frac{\mathrm{e}^{-iw/2}}{2^{j+1}}\sum \limits _{k\in \mathbb {Z}} {F_k \mathrm{e}^{-ikw}} \left( {\sum \limits _{m\in \mathbb {Z}} {(-1)^m\left| {\hat{S}^\phi \left( \frac{w+2m\pi }{2}\right) } \right| ^2} } \right. \nonumber \\&\quad \left. {+\sum \limits _{m\in \mathbb {Z}} {\left| {\hat{S}^\phi (\frac{w+2m\pi }{2})} \right| ^2\overline{\Big ( {\sum \limits _{l\in \mathbb {Z}} {\hat{S}^\phi (w+2l\pi )(-1)^l} }\Big )} } }\right) . \end{aligned}$$

(72)

Cutting summation (72) into two parts, one for $m=2n$ and the other for $m=2n-1$, yields

$$\begin{aligned} \begin{array}{c} A_f {=}\frac{\mathrm{e}^{-iw/2}}{2^{j{+}1}}\sum \limits _{k\in \mathbb {Z}} {F_k \mathrm{e}^{-ikw}} {\times } \left( {\sum \limits _{n\in \mathbb {Z}} {\left| {\hat{S}^\phi (w/2{+}2\pi n)} \right| ^2\overline{\left( {1{+}\sum \limits _{l\in \mathbb {Z}} {(-1)^l\hat{S}^\phi (w+2l\pi )} }\right) } } } \right. \\ \left. {-\sum \limits _{n\in \mathbb {Z}} {\left| {\hat{S}^\phi (w/2+2\pi n+\pi )} \right| ^2\overline{\Big ( {1-\sum \limits _{l\in \mathbb {Z}} {(-1)^l\hat{S}^\phi (w+2l\pi )} }\Big )} } }\right) . \\ \end{array}\nonumber \\ \end{aligned}$$

(73)

Since it follows from (39) that $\sum \limits _{l\in \mathbb {Z}} {\hat{S}^\phi (w+2l\pi )} =\sum \limits _{l\in \mathbb {Z}} {\hat{S}^\phi (w+4l\pi )} +\sum \limits _{l\in \mathbb {Z}} \hat{S}^\phi (w+ 4l\pi +2\pi ) =1$, Eq. (73) gives

$$\begin{aligned} \begin{array}{c} A_f =\frac{\mathrm{e}^{-iw/2}}{2^j}\sum \limits _{k\in \mathbb {Z}} {F_k \mathrm{e}^{-ikw}} \times \left( {\sum \limits _{n\in \mathbb {Z}} {\left| {\hat{S}^\phi (w/2+2\pi n)} \right| ^2\overline{\sum \limits _{l\in \mathbb {Z}} {\hat{S}^\phi (w+4l\pi )} } } } \right. \\ \left. {-\sum \limits _{n\in \mathbb {Z}} {\left| {\hat{S}^\phi (w/2+2\pi n+\pi )} \right| ^2\overline{\sum \limits _{l\in \mathbb {Z}} {\hat{S}^\phi (w+4l\pi +2\pi )} } } }\right) . \\ \end{array} \end{aligned}$$

(74)

Similarly

$$\begin{aligned} \begin{array}{c} B_f =\frac{1}{2^j}\sum \limits _{k\in \mathbb {Z}} {f_s \left( \frac{2k}{2^j}\right) \mathrm{e}^{-ikw}} \times \left( {\sum \limits _{n\in \mathbb {Z}} {\left| {\hat{S}^\phi (w/2+2\pi n)} \right| ^2\overline{\left( {\sum \limits _{l\in \mathbb {Z}} {\hat{S}^\phi (w+4l\pi )} }\right) } } } \right. \\ \left. {+\sum \limits _{n\in \mathbb {Z}} {\left| {\hat{S}^\phi (w/2+2\pi n+\pi )} \right| ^2\overline{\Big ( {\sum \limits _{l\in \mathbb {Z}} {\hat{S}^\phi (w+4l\pi +2\pi )} }\Big )} } }\right) . \\ \end{array} \end{aligned}$$

(75)

Since $\sum \nolimits _{k\in \mathbb {Z}} {\left| {\hat{S}^\phi (w/2+2\pi k)} \right| ^2} \in L^2[-\pi ,\pi ]$ by the frame theory [19] and $\sum \nolimits _{n\in \mathbb {Z}} {\hat{S}^\phi (w+4n\pi )} \in L^2[-2\pi ,2\pi ]$, we have

$$\begin{aligned} \tilde{Q}_s (w)&= -\mathrm{e}^{-iw/2}E_s (w+2\pi )\overline{P_s (w+2\pi )} \nonumber \\&= -\mathrm{e}^{-iw/2}\sum \limits _{k\in \mathbb {Z}} {\left| {\hat{S}^\phi (\frac{w}{2}+\pi +2\pi k)} \right| } ^2\overline{\sum \limits _{n\in \mathbb {Z}} {\hat{S}^\phi \left( w+2\pi +4n\pi \right) } }\nonumber \\&= \frac{1}{2}\sum \limits _{k\in \mathbb {Z}} {\tilde{q}_k \mathrm{e}^{-iwk/2}}, \end{aligned}$$

(76)

with $\{\tilde{q}_k \}_{k\in \mathbb {Z}} \in l^2(\mathbb {Z})$, and

$$\begin{aligned} \hat{\psi }_Q (w)=\tilde{Q}_s (w)\hat{S}^\phi (w/2), \end{aligned}$$

(77)

where $P_s (w)$ and $E_s (w)$ are respectively defined in (3) and (7).

Since $P_s (w)$ is a two-scale symbol corresponding to the scaling function $S^\phi (2^jx)$, the sequence $\{\psi _Q (x-k)\}_{k\in \mathbb {Z}}$ forms a Riesz basis of $W_0$, where the Fourier transform of $\psi _Q (x)$ is $\hat{\psi }_Q (w)$ (see Eq. 5.6.13 in [7]).

Inserting (76) respectively into (74) and (75) yields

$$\begin{aligned} \left\{ {\begin{array}{l} A_f =\frac{1}{2^j}\sum \limits _{k\in \mathbb {Z}} {F_k \mathrm{e}^{-ikw}} \times ( {\tilde{Q}_s (w+2\pi )} {+\tilde{Q}_s (w)})=\frac{1}{2^j}\sum \limits _{k\in \mathbb {Z}} F_k \mathrm{e}^{(-iw/2)\times 2k} \\ \qquad \times \sum \limits _{l\in \mathbb {Z}} {\tilde{q}_{2l} \mathrm{e}^{(-iw/2)\times 2l}} \\ B_f =\frac{1}{2^j\mathrm{e}^{-iw/2}}\sum \limits _{k\in \mathbb {Z}} {f_s \left( \frac{2k}{2^j}\right) \mathrm{e}^{-ikw}} \times ( {\tilde{Q}_s (w+2\pi )} {-\tilde{Q}_s (w)})\\ \qquad =-\frac{1}{2^j}\sum \limits _{k\in \mathbb {Z}} {f_s \left( \frac{2k}{2^j}\right) \mathrm{e}^{(-iw/2)\times }{2k}} \times \sum \limits _{l\in \mathbb {Z}} {\tilde{q}_{2l+1} \mathrm{e}^{(-iw/2)\times 2l}} \\ \end{array}} \right. .\nonumber \\ \end{aligned}$$

(78)

Substituting (78) into (64) gives

$$\begin{aligned}&\sum \limits _{k\in \mathbb {Z}} f_s \left( \frac{2k}{2^j}\right) \mathrm{e}^{(-iw/2)\times {2k-1}} \times \sum \limits _{l\in \mathbb {Z}} \tilde{q}_{2l-1} \mathrm{e}^{(-iw/2)\times {2l-1}} -\sum \limits _{k\in \mathbb {Z}} F_k \mathrm{e}^{(-iw/2)\times {2k}}\nonumber \\&\quad \times \sum \limits _{l\in \mathbb {Z}} {\tilde{q}_{2l} \mathrm{e}^{(-iw/2)\times 2l}} =0. \end{aligned}$$

(79)

Let $\tilde{\psi }_Q (x)$ be the dual of $\psi _Q (x)$ in $W_0$. Consider the “decomposition relation” of scaling function and wavelet, it follows from (3), (76) and Theorem 5.16 in [7] that

$$\begin{aligned} \left\{ {\begin{array}{l} \hat{\tilde{S}}^\phi \left( \frac{w}{2}\right) =\sum \limits _{k\in \mathbb {Z}} {\left( {S^\phi \left( \frac{2k}{2}\right) \mathrm{e}^{(-iw/2)\times 2k}\hat{\tilde{S}}^\phi (w)+\tilde{q}_{2k} \mathrm{e}^{(-iw/2)\times 2k}\hat{\tilde{\psi }}_Q (w)}\right) } \\ \hat{\tilde{S}}^\phi \left( \frac{w}{2}\right) =\sum \limits _{k\in \mathbb {Z}} {\left( {S^\phi \left( \frac{2k-1}{2}\right) \mathrm{e}^{(-iw/2)\times 2k-1}\hat{\tilde{S}}^\phi (w)+\tilde{q}_{2k-1} \mathrm{e}^{(-iw/2)\times 2k-1}\hat{\tilde{\psi }}_Q (w)}\right) } \\ \end{array}} \right. ,\nonumber \\ \end{aligned}$$

(80)

where $\hat{\tilde{S}}^\phi (w)$ is the Fourier transform of $\tilde{S}^\phi (x)$ and $\hat{\tilde{\psi }}_Q (w)$ is the Fourier transform of $\tilde{\psi }_Q (x)$.

Multiplying the two identities in (80) by $\sum \limits _{k\in \mathbb {Z}} {(-F_k )\mathrm{e}^{(-iw/2)\times 2k}}$ and $\sum \limits _{k\in \mathbb {Z}} f_s (\frac{2k}{2^j}) \mathrm{e}^{(-iw/2)\times 2k-1}$ consecutively,

(81)

By (81) and (79)

$$\begin{aligned}&-\sum \limits _{k\in \mathbb {Z}} {F_k \mathrm{e}^{(-iw/2)\times 2k}} \hat{\tilde{S}}^\phi \left( \frac{w}{2}\right) +\sum \limits _{k\in \mathbb {Z}} {f_s \left( \frac{2k}{2^j}\right) \mathrm{e}^{(-iw/2)\times 2k-1}} \hat{\tilde{S}}^\phi \left( \frac{w}{2}\right) \nonumber \\&\quad = \left( {-\sum \limits _{k\in \mathbb {Z}} {F_k \mathrm{e}^{(-iw/2)\times 2k}} \sum \limits _{l\in \mathbb {Z}} {S^\phi \left( \frac{2l}{2}\right) \mathrm{e}^{(-iw/2)\times 2l}} } +\sum \limits _{k\in \mathbb {Z}} {f_s \left( \frac{2k}{2^j}\right) \mathrm{e}^{(-iw/2)\times 2k-1}}\nonumber \right. \\&\left. \qquad \qquad \times \sum \limits _{l\in \mathbb {Z}} {S^\phi \left( \frac{2l-1}{2}\right) \mathrm{e}^{(-iw/2)\times 2l-1}} \right) \hat{\tilde{S}}^\phi (w). \end{aligned}$$

(82)

Taking the inverse Fourier transform on both sides in (82),

$$\begin{aligned} \begin{aligned} f_c (x)&=-\sum \limits _{k\in \mathbb {Z}} {F_k \tilde{S}^\phi (2x-2k)} +\sum \limits _{k\in \mathbb {Z}} {f_s (2k/2^j)\tilde{S}^\phi (2x-2k+1)} \\&=\sum \limits _{m\in \mathbb {Z}} \left( \sum \limits _{n\in \mathbb {Z}} {\left( {f_s \left( \frac{2n}{2^j}\right) S^\phi \left( \frac{2m-2n+1}{2}\right) -F_n S^\phi \left( \frac{2m-2n}{2}\right) }\right) }\right. \\&\quad \left. \times \tilde{S}^\phi (x-m)\right) . \\ \end{aligned} \end{aligned}$$

(83)

Obviously, $f_c (x)$ is an element of $V_0$ since it can be represented as a linear combination of $\{\tilde{S}^\phi (x-m)\}_{m\in \mathbb {Z}}$ in (83). Comparing (83) to (9), we know that $\{-F_k \}_{k\in \mathbb {Z}}$ and $\{f_s (2k/2^j)\}_{k\in \mathbb {Z}}$ respectively correspond to $\{c_k^0 \}_{k\in \mathbb {Z}}$ and $\{d_k^0 \}_{k\in \mathbb {Z}}$. By (6), it follows from Lemma 1 that the coefficients $\{f_s (2k/2^j)\}_{k\in \mathbb {Z}}$ for which (83) holds is unique for any given sequence $\{-F_k \}_{k\in \mathbb {Z}}$. Hence, (83) implies $\{F_k \}_{k\in \mathbb {Z}}$ uniquely determines $\{f_s (2k/2^j)\}_{k\in \mathbb {Z}}$ in (56) for (57) to hold.

Since $\{f_s (2k/2^j)\}_{k\in \mathbb {Z}}$ and $\{F_k \}_{k\in \mathbb {Z}}$ determine a unique element of $V_j$ in (55), Eqs. (57) and (83) imply that the sequence $\{F_k \}_{k\in \mathbb {Z}}$ determines a unique element of $V_j$ whose projection on $V_{j-1}$ vanishes, i.e., $\{F_k \}_{k\in \mathbb {Z}}$ determines a unique element of $W_{j-1}$. This proves Theorem 1. $\square$

Appendix 3: Proof of Lemma 2

Proof

It follows from (17) and (18) that $\psi (x)$ is an element of $V_1 \subset L^2(\mathbb {R})$. By Parseval’s identity,

$$\begin{aligned} \displaystyle \int \limits _{-\infty }^{+\infty } {\psi (x)\overline{\phi (x-n)} dx}&= \frac{1}{2\pi }\int \limits _{-\infty }^{+\infty } {\hat{\psi }(w)\overline{\hat{\phi }(w)} \mathrm{e}^{iwn}dw}\nonumber \\&= \frac{1}{2\pi }\sum \limits _{k\in Z} {\int \limits _{-\pi +2k\pi }^{+\pi +2k\pi } {\hat{\psi }(w)\overline{\hat{\phi }(w)} \mathrm{e}^{iwn}dw} } \\&= \frac{1}{2\pi }\int \limits _{-\pi }^\pi {\sum \limits _{k\in \mathbb {Z}} {\hat{\psi }(w+2k\pi )\overline{\hat{\phi }(w+2k\pi )} } \mathrm{e}^{iwn}dw} , \nonumber \end{aligned}$$

(84)

for $n\in \mathbb {Z}$. Since $W_0$ is orthogonal to $V_0$ and $W_0 \oplus V_0 =V_1$, Eq. (84) implies

$$\begin{aligned} \int _{-\pi }^\pi {\sum \limits _{k\in \mathbb {Z}} {\hat{\psi }(w+2k\pi )\overline{\hat{\phi }(w+2k\pi )} } \mathrm{e}^{iwn}dw} =0 \end{aligned}$$

(85)

if and only if $\psi (x)$ is an element of $W_0$ where $\oplus$ denotes a direct sum of two vector spaces.

Since $\{\mathrm{e}^{iwn}\}_n$ is a complete orthonormal basis of $L^2[-\pi ,\pi ]$, Eq. (85) holds if and only if

$$\begin{aligned} \sum \limits _{k\in \mathbb {Z}} {\hat{\psi }(w+2k\pi )\overline{\hat{\phi }(w+2k\pi )} } =0 \end{aligned}$$

(86)

almost everywhere.

From (42) and (17), Eqs. (85) and (86) imply

$$\begin{aligned} \sum \limits _{k\in \mathbb {Z}} {\hat{\psi }(w+2k\pi )\overline{\hat{\phi }(w+2k\pi )} }&=\sum \limits _{k\in \mathbb {Z}} {\hat{\psi }(w+4k\pi )\overline{\hat{\phi }(w+4k\pi )}}\nonumber \\&\quad +\sum \limits _{k\in \mathbb {Z}} {\hat{\psi }(w+2\pi +4k\pi )\overline{\hat{\phi }(w+2\pi +4k\pi )} } \nonumber \\&=\sum \limits _{k\in \mathbb {Z}} {Q_\psi ( w)\hat{\phi }(\frac{w}{2}+2k\pi )\overline{P_\phi (w)\hat{\phi }(\frac{w}{2}+2k\pi )} }\nonumber \\&\quad +\sum \limits _{k\in \mathbb {Z}}Q_\psi ( {w+2\pi })\hat{\phi }(\frac{w}{2}+2k\pi +\pi )\nonumber \\&\quad \times \overline{P_\phi (w+2\pi )\hat{\phi }(\frac{w}{2}+2k\pi +\pi )} \nonumber \\&=Q_\psi ( w)\sum \limits _{k\in \mathbb {Z}} {\left| {\hat{\phi }(\frac{w}{2}{+}2k\pi )} \right| ^2} \overline{P_\phi ( w)} {+}Q_\psi ( {w+2\pi }) \nonumber \\&\quad \times \sum \limits _{k\in \mathbb {Z}} {\left| {\hat{\phi }(\frac{w}{2}+2k\pi +\pi )} \right| ^2} \overline{P_\phi ( {w+2\pi })} =0 \end{aligned}$$

(87)

holds if and only if $\psi (x)$ is an element of $W_0$. It follows from (87) and (20) that (19) holds if and only if $\psi (x)$ is an element of $W_0$. $\square$

Appendix 4: Proof of Theorem 2

Proof

It follows from (11) and [17] that $\hat{S}^\psi (w)$ can be expressed as

$$\begin{aligned} \hat{S}^\psi (w)=\frac{\mathrm{e}^{-iw/2}\hat{\psi }(w)}{\sum \nolimits _k {(-1)^k\hat{\psi }(w+2k\pi )} } \end{aligned}$$

(88)

where $\hat{\psi }(w)$ is the Fourier transform of a wavelet $\psi (x)$.

Now, we relate the expressions of $\hat{S}^\psi (w)$ and $Q_s (w)$ to that of $S^\phi (x)$ instead of $\psi (x)$ in (88).

By (88) and (14)

$$\begin{aligned} \begin{aligned}&Q_s (w)/\mathrm{e}^{-iw/2}-Q_s (w+2\pi )/\mathrm{e}^{-iw/2}\\&\quad =\frac{1}{\mathrm{e}^{-iw/2}}\left( \sum \limits _{k\in \mathbb {Z}} {\hat{S}^\psi (w+4k\pi )} -\sum \limits _{k\in \mathbb {Z}} {\hat{S}^\psi (u+2\pi +4k\pi )} \right) \\&\quad =\sum \limits _{k\in \mathbb {Z}} {\hat{S}^\psi (w+2k\pi )\mathrm{e}^{i(w+2k\pi )/2}} =1. \\ \end{aligned} \end{aligned}$$

(89)

By (89)

$$\begin{aligned}&Q_s (w)( {\overline{P_s (w+2\pi )} E_s (w\!+\!2\pi )\!+\!E_s (w)\overline{P_s (w)} }) \!=\!Q_s (w)\overline{P_s (w+2\pi )} E_s (w+2\pi ) \nonumber \\&\quad +Q_s (w)E_s (w)\overline{P_s (w)} \nonumber \\&=\mathrm{e}^{-iw/2}\overline{P_s (w+2\pi )} E_s (w+2\pi )+Q_s (w+2\pi )\overline{P_s (w+2\pi )} E_s (w+2\pi ) \nonumber \\&\quad +Q_s (w)E_s (w)\overline{P_s (w)}. \end{aligned}$$

(90)

Since $P_s (w)$ in (3) and $Q_s (w)$ in (14) form a pair of reconstruction filters in an MRA $\{V_j \}_{j\in \mathbb {Z}}$, it follows from Lemma 2 and (90) that

$$\begin{aligned} Q_s (w)( {\overline{P_s (w\!+\!2\pi )} E_s (w+2\pi )\!+\!E_s (w)\overline{P_s (w)} })\!=\!\mathrm{e}^{-iw/2}\overline{P_s (w+2\pi )} E_s (w\!+\!2\pi ).\nonumber \\ \end{aligned}$$

(91)

On the other hand, applying (89) to (16) yields

$$\begin{aligned} \Delta _{P_s ,Q_s } =P_s (w)Q_s (w)-P_s (w+2\pi )Q_s (w)-\mathrm{e}^{-iw/2}P_s (w). \end{aligned}$$

(92)

Consider the formula

$$\begin{aligned} \Delta _{P_s ,Q_s } (E_s (w+2\pi )\overline{P_s (w+2\pi )} +E_s (w)\overline{P_s (w)} ). \end{aligned}$$

(93)

Inserting (92) into (93) yields

$$\begin{aligned}&\Delta _{P_s ,Q_s } ( {E_s (w+2\pi )\overline{P_s (w+2\pi )} +E_s (w)\overline{P_s (w)} }) \nonumber \\&=\left( {P_s (w)-P_s (w+2\pi )}) Q_s (w)( {E_s (w+2\pi )\overline{P_s (w+2\pi )} +E_s (w)\overline{P_s (w)} }) \right. \nonumber \\&\left. \quad -\,\mathrm{e}^{-iw/2}P_s (w)( {E_s (w+2\pi )\overline{P_s (w+2\pi )} +E_s (w)\overline{P_s (w)} }\right) . \end{aligned}$$

(94)

Applying (91) to (94) yields

$$\begin{aligned}&\Delta _{P_s ,Q_s } ( {E_s (w+2\pi )\overline{P_s (w+2\pi )} +E_s (w)\overline{P_s (w)} })\nonumber \\&=\left( {P_s (w)-P_s (w+2\pi )})\mathrm{e}^{-iw/2}\overline{P_s (w+2\pi )} E_s (w+2\pi )\right. \nonumber \\&\left. \quad -\,\mathrm{e}^{-iw/2}P_s (w)( {E_s (w+2\pi )\overline{P_s (w+2\pi )} +E_s (w)\overline{P_s (w)} }\right) \nonumber \\&=-\,\mathrm{e}^{-iw/2}( {E_s (w+2\pi )\left| {P_s (w+2\pi )} \right| ^2+E_s (w)\left| {P_s (w)} \right| ^2}). \end{aligned}$$

(95)

It follows from (2) and (7) that

$$\begin{aligned}&E_s (w+2\pi )\left| {P_s (w+2\pi )} \right| ^2+E_s (w)\left| {P_s (w)} \right| ^2\nonumber \\&\quad =\sum \limits _{k=-\infty }^{+\infty } {\left| {\hat{S}^\phi ( {\frac{w}{2}+2k\pi +\pi })P_s (w+2\pi )} \right| } ^2+\sum \limits _{k=-\infty }^{+\infty } {\left| {\hat{S}^\phi ( {\frac{w}{2}+2k\pi })P_s (w)} \right| } ^2 \nonumber \\&\quad =\sum \limits _{k=-\infty }^{+\infty } {\left| {\hat{S}^\phi (w+2k\pi )} \right| ^2} =E_s (2w). \end{aligned}$$

(96)

Inserting (96) into (95), we have

$$\begin{aligned} -\mathrm{e}^{-iw/2}E_s (2w)=\Delta _{P_s ,Q_s } ( {E_s (w+2\pi )\overline{P_s (w+2\pi )} +E_s (w)\overline{P_s (w)} }). \end{aligned}$$

(97)

Since $( {P_s (w),Q_s (w)})$ forms a pair of reconstruction filters, (15) holds. Hence, by (97)

$$\begin{aligned} E_s (w+2\pi )\overline{P_s (w+2\pi )} +E_s (w)\overline{P_s (w)} =-\mathrm{e}^{-iw/2}E_s (2w)/\Delta _{P_s ,Q_s } . \end{aligned}$$

(98)

Since the sequence $\{S^\phi (x-k)\}_k$ is a frame, there should exist two constants $A_{Es}$ and $B_{Es}$ with $0<A_{Es} \le B_{Es} <+\infty$ such that

$$\begin{aligned} A_{Es} \le E_s (2w)\le B_{Es} \end{aligned}$$

(99)

(Theorem 9 in [19]). Applying (15) and (99) to (98) yields

$$\begin{aligned} A_{Es}^2 /B_\Delta&\le \left| {E_s (w+2\pi )\overline{P_s (w+2\pi )} +E_s (w)\overline{P_s (w)} } \right| ^2 \nonumber \\&=\left| {E_s (2w)/\Delta _{P_s ,Q_s } } \right| ^2\le B_{Es}^2 /A_\Delta , \end{aligned}$$

(100)

which implies that the two-scale condition holds.

Equation (100) implies

$$\begin{aligned} \frac{1}{\overline{P_s (w+2\pi )} E_s (w+2\pi )+E_s (w)\overline{P_s (w)} }\in L^2[-\pi ,\pi ]. \end{aligned}$$

(101)

Hence, it follows from (91) and (101) that

$$\begin{aligned} Q_s (w)=\frac{\mathrm{e}^{-iw/2}E_s (w+2\pi )\overline{P_s (w+2\pi )} }{\overline{P_s (w+2\pi )} E_s (w+2\pi )+E_s (w)\overline{P_s (w)} }, \end{aligned}$$

(102)

which verifies Eq. (21). $\square$