Journal of Fourier Analysis and Applications

, Volume 24, Issue 6, pp 1681–1683

# Erratum to: Some Smooth Compactly Supported Tight Wavelet Frames with Vanishing Moments

• A. San Antolín
• R. A. Zalik
Erratum

## 1 Erratum to: J Fourier Anal Appl DOI 10.1007/s00041-015-9442-x

The line between the displayed formulas (16) and (17) was copied incorrectly from [41, Theorem 1]. It should read as follows: “Suppose that there exist trigonometric polynomials $${\widetilde{P}}_1({\mathbf {t}}), \ldots , {\widetilde{P}}_M(\mathbf{t})$$ such that”. In addition, in the proof of Lemma 3 we overlooked to prove that the functions $${\widetilde{P}}_{n,m}^{(j)}(\mathbf{t})$$ are $${\mathbb {Z}}^n$$-periodic. This makes it necessary to reformulate Lemma 3. The statement and proof of Theorem 3 remain the same, but we wish to emphasize that the polynomials $$L_{0}(A^T{\mathbf {t}})$$ and $$L_{1}(A^T{\mathbf {t}})$$ are generated by the algorithm described in Theorem E.

### Lemma 3

Let $$\Omega := \{0,1/2\}^d\setminus \varvec{\Gamma }_{A^T}$$, let $$u_{n,m}(t)$$ and $$h_{n,m}(t)$$ be trigonometric polynomials that satisfy (19), let $$P_{n,m}(\mathbf {t})$$ be defined by (11), let $${\mathbf{u}} \in {\mathbb {Z}}^d$$ be such that $$r_1(A) \cdot {\mathbf{u}} =1/2$$, let $$K=2^d-2$$, and let $$\rho : \Omega \rightarrow \{ d+1,\ldots , K + d \}$$ be a bijection. Let
• $${\widetilde{P}}_{n,m}^{(j)}(A^T{\mathbf {t}}) := h_{n,m}( t_j)\prod _{s={j+1}}^d u_{n,m}( t_s), \ j=1,\ldots , d - 1$$,

• $${\widetilde{P}}_{n,m}^{(d)}(A^T{\mathbf {t}}) := h_{n,m}( t_d),$$

and
\begin{aligned} {\widetilde{P}}_{n,m}^{(\rho ({\mathbf {r}})) }(A^T{\mathbf {t}}):= & {} \frac{1}{2} \big [ (P_{n,m}(\mathbf {t}+\mathbf {r})+ P_{n,m}(\mathbf {t}+\mathbf {r}+ \mathbf {r}_1(A) )) \\&+\, e^{i 2\pi {\mathbf {t}} \cdot {\mathbf{u}}} (P_{n,m}(\mathbf {t}+\mathbf {r})- P_{n,m}(\mathbf {t}+\mathbf {r}+ \mathbf {r}_1(A))) \big ], \qquad \mathbf {r}\in \Omega , \end{aligned}
then $${\widetilde{P}}_{n,m}^{(j)}({\mathbf {t}})$$, $$j=1\ldots , K+d$$, are trigonometric polynomials and
\begin{aligned} \sum _{\mathbf {r} \in \varvec{\Gamma }_{A^T} } | P_{n,m}({\mathbf {t}}+ \mathbf {r})|^2+ \sum _{j=1}^{K+ d } |{\widetilde{P}}_{n,m}^{(j)}(A^T{\mathbf {t}})|^2 = 1. \end{aligned}
(20)

### Proof

We start by showing that the $${\widetilde{P}}_{n,m}^{(j)}({\mathbf {t}})$$ are $${\mathbb {Z}}^d$$-periodic polynomials. Assume first that $$1 \le j \le d-1$$. Since $$g_{n,2m}(t ) + g_{n,2m}(t + 1/2 )$$ has period 1 / 2 we readily see that also the polynomials $$h_{n,m}(t)$$ and $$u_{n,m}(t)$$ have period 1 / 2. This in turn implies that $$P_{n,m}(A^T{\mathbf {t}})$$ is $$(1/2){\mathbb {Z}}^d$$-periodic. It will therefore suffice to show that if $$\mathbf {k} \in {\mathbb {R}}^d$$ and $$\mathbf {x} = (A^T)^{-1} \mathbf {k}$$, then $$\mathbf {x} \in (1/2){\mathbb {Z}}^d$$. Since the determinant of $$A^T$$ equals $$\pm 2$$ and the columns of $$A^T$$ are in $${\mathbb {Z}}^d$$ this readily follows by an application of Cramer’s rule.

From the definition it is also obvious that $${\widetilde{P}}_{n,m}^{(d)}({\mathbf {t}})$$ is $${\mathbb {Z}}^d$$-periodic.

We now establish the $${\mathbb {Z}}^d$$-periodicity of the functions $${\widetilde{P}}_{n,m}^{(\rho ({\mathbf {r}})) }({\mathbf {t}})$$. Let $$k \in {\mathbb {Z}}^d$$. If $$\mathbf {k}= A^T(\mathbf {k}_1)$$ for some $$\mathbf {k}_1 \in {\mathbb {Z}}^d$$, then the $${\mathbb {Z}}^d$$-periodicity of the polynomials $$P_{n,m}({\mathbf {t}})$$ readily imply that $${\widetilde{P}}_{n,m}^{(j)}({\mathbf {t}} + \mathbf {k}) = {\widetilde{P}}_{n,m}^{(j)}({\mathbf {t}})$$. On the other hand, if $$\mathbf {k}= A^T(\mathbf {r}_1(A)+ \mathbf {k}_2)$$ for some $$\mathbf {k}_2 \in {\mathbb {Z}}^d$$, the assertion follows by observing that $$2 \mathbf {r}_1(A) \in {\mathbb {Z}}^n$$ and $$e^{i 2\pi ({\mathbf {t}} + \mathbf {r}_1(A)) \cdot {\mathbf{u}} } = - e^{i 2\pi {\mathbf {t}} \cdot {\mathbf{u}} }$$.

Let $$\varvec{\Gamma }=\varvec{\Gamma }_{A^T}$$. We claim that for $$\mathbf {r}\in \Omega$$ there exists an unique $$\widetilde{\mathbf {r}} \in \Omega$$, $$\widetilde{\mathbf {r}} \ne \mathbf {r}$$, such that $$\mathbf {r}+ \mathbf {r}_1(A) + \mathbf {k}_3 = \widetilde{\mathbf {r}}$$ for some $$\mathbf {k}_3 \in {\mathbb {Z}}^d$$. Let us verify this assertion. Since $$\mathbf {r}+ \mathbf {r}_1(A) \in \{ 0,{\frac{1}{2}}, 1 \}^d$$, there exists an unique $$\mathbf {k}_3 \in {\mathbb {Z}}^d$$ such that $$\mathbf {r}+ \mathbf {r}_1(A) + \mathbf {k}_3 \in \{ 0,{\frac{1}{2}}\}^d$$. Let $$\widetilde{\mathbf {r}} := \mathbf {r}+ \mathbf {r}_1(A) + \mathbf {k}_3$$. We need to show that $$\widetilde{\mathbf {r}}$$ is neither $$(0,\ldots , 0)$$ nor $$\mathbf {r}_1(A)$$ nor r. If $$\widetilde{\mathbf {r}} = (0,\ldots , 0)$$ then $$\mathbf {r}+ \mathbf {r}_1(A) \in \{ 0, 1 \}^d$$. This implies that $$\mathbf {r}= \mathbf {r}_1(A)$$, which contradicts the hypothesis that $$\mathbf {r}\in \Omega$$. In similar fashion we can see that $$\widetilde{\mathbf {r}}$$ is neither $$\mathbf {r}_1(A)$$ nor $$\mathbf {r}$$.

Conversely, there exists an unique $${\mathbf {k}}_4 \in {\mathbb {Z}}^d$$ such that $$\widetilde{\mathbf {r}}+ \mathbf {r}_1(A) + {\mathbf {k}}_4 = \mathbf {r}$$. Indeed, repeating the preceding argument we conclude that there exists an unique $${\mathbf {k}}_5 \in {\mathbb {Z}}^d$$ such that $$\widetilde{\mathbf {r}}+ \mathbf {r}_1(A) + {\mathbf {k}}_5 \in \{ 0, {\frac{1}{2}}\}^d$$. Let $${\mathbf {k}}_4 := {\mathbf {k}}_5$$. Since $$\widetilde{\mathbf {r}}= \mathbf {r}+ \mathbf {r}_1(A) + {\mathbf {k}}_3$$, it follows that $$\mathbf {r}+ 2 \mathbf {r}_1(A) + {\mathbf {k}}_3 + {\mathbf {k}}_4 \in \{ 0, {\frac{1}{2}}\}^d.$$ Bearing in mind that $$2 \mathbf {r}_1(A) \in {\mathbb {Z}}^d$$ and $$\mathbf {r}\in \Omega$$, we have $$2 \mathbf {r}_1(A) + {\mathbf {k}}_3 + {\mathbf {k}}_4 = \mathbf{0}$$. Thus
\begin{aligned} \widetilde{\mathbf {r}}+ \mathbf {r}_1(A) + {\mathbf {k}}_4= \mathbf {r}+ 2\mathbf {r}_1(A) + {\mathbf {k}}_3 + {\mathbf {k}}_4 = \mathbf {r}. \end{aligned}
We have therefore shown that there exist two disjoint sets $$\Omega _1, \Omega _2 \subset \Omega$$, such that $$\Omega = \Omega _1 \cup \Omega _2$$ and for any $$\mathbf {r}\in \Omega _1$$ there exists an unique $$\widetilde{\mathbf {r}} \in \Omega _2$$ such that $$\widetilde{\mathbf {r}} = \mathbf {r}+ \mathbf {r}_1(A) + \mathbf {k}$$ and $$\mathbf {r}= \widetilde{\mathbf {r}}+ \mathbf {r}_1(A) + {{\mathbf m}}$$ for some $$\mathbf {k}, {{\mathbf m}} \in {\mathbb {Z}}^d$$. Since, moreover, $${\widetilde{P}}_{n,m}^{(\rho ({\mathbf {r}})) }(A^T{\mathbf {t}})$$ and $${\widetilde{P}}_{n,m}^{(\rho (\widetilde{\mathbf {r}})) }(A^T{\mathbf {t}})$$ are complex conjugates of each other, we readily see that
\begin{aligned} |{\widetilde{P}}_{n,m}^{(\rho ({\mathbf {r}})) }(A^T{\mathbf {t}}) |^2 + |{\widetilde{P}}_{n,m}^{(\rho (\widetilde{\mathbf {r}})) }(A^T{\mathbf {t}}) |^2 = |P_{n,m}(\mathbf {t}+\mathbf {r})|^2+ |P_{n,m}(\mathbf {t}+\mathbf {r}+ \mathbf {r}_1(A) )|^2. \end{aligned}
Therefore
\begin{aligned}&\sum _{j=d+1}^{K+d} |{\widetilde{P}}_{n,m}^{(j)}(A^T{\mathbf {t}}) |^2 \\&\quad = \sum _{\mathbf {r}\in \Omega } |{\widetilde{P}}_{n,m}^{(\rho ({\mathbf {r}})) }(A^T{\mathbf {t}}) |^2 = \sum _{\mathbf {r}\in \Omega _1} |{\widetilde{P}}_{n,m}^{(\rho ({\mathbf {r}})) }(A^T{\mathbf {t}}) |^2 + \sum _{\widetilde{\mathbf {r}} \in \Omega _2} |{\widetilde{P}}_{n,m}^{(\rho ({\widetilde{\mathbf {r}}})) }(A^T{\mathbf {t}}) |^2 \\&\quad = \sum _{\mathbf {r}\in \Omega } |P_{n,m}(\mathbf {t}+\mathbf {r})|^2. \end{aligned}
The remainder of the proof is a repetition of the argument used in the original version of this lemma. $$\square$$