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Journal of Fourier Analysis and Applications

, Volume 24, Issue 6, pp 1681–1683 | Cite as

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

  • A. San AntolínEmail author
  • R. A. Zalik
Erratum
  • 426 Downloads

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. \(\square \)

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© Springer Science+Business Media New York 2017

Authors and Affiliations

  1. 1.Departamento de MatemáticasUniversidad de AlicanteAlicanteSpain
  2. 2.Department of Mathematics and StatisticsAuburn UniversityAuburnUSA

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