Interpolatory curl-free wavelets on bounded domains and characterization of Besov spaces

Yingchun Jiang

doi:10.1186/1029-242X-2012-68

Research

Open Access

Interpolatory curl-free wavelets on bounded domains and characterization of Besov spaces

Yingchun Jiang¹Email author

Journal of Inequalities and Applications20122012:68

https://doi.org/10.1186/1029-242X-2012-68

Received: 27 December 2011

Accepted: 23 March 2012

Published: 23 March 2012

Abstract

Based on interpolatory Hermite splines on rectangular domains, the interpolatory curl-free wavelets and its duals are first constructed. Then we use it to characterize a class of vector-valued Besov spaces. Finally, the stability of wavelets that we constructed are studied.

MR(2000) Subject Classification: 42C15; 42C40.

Keywords

interpolatorycurl-freewaveletsbounded domainsBesov spaces

1 Introduction

Due to its potential use in many physical problems, like the simulation of incompressible fluids or in electromagnetism, curl-free wavelet bases have been advocated in several articles and most of the study focus on the cases of R² and R³[1–4]. However, it is reasonable to study the corresponding wavelet bases on bounded domains because of some practical use. At the same time, the stability and the characterization of function spaces are also necessary in some applications, such as the adaptive wavelet methods. In recent years, divergence-free and curl-free wavelets on bounded domains begin to be studied [5–8]. In particular, [8] use the truncation method to obtain interpolatory spline wavelets on rectangular domains from [3]. Inspired by this, we mainly study the interpolatory 3D curl-free wavelet bases on the cube and its applications for characterizing the vector-valued Besov spaces.

In Section 2, we first give the construction of interpolatory curl-free wavelets and its duals on the cube. The characterization of a class of vector-valued Besov spaces are given in part 3. Finally, we also study the stability of the corresponding curl-free wavelets.

Now, we begin with some notations and formulae, which will be used later on. Let

ξ_{1}^{+}

and

ξ_{2}^{+}

stand for two cubic Hermite splines:

\begin{gathered} ξ_{1}^{+} (x) = : (1 - 3 x^{2} - 2 x^{3}) X_{[- 1, 0)} (x) + (1 - 3 x^{2} + 2 x^{3}) X_{[0, 1)} (x), \\ ξ_{2}^{+} (x) = : (x + 2 x^{2} + x^{3}) X_{[- 1, 0)} (x) + (x - 2 x^{2} + x^{3}) X_{[0, 1)} (x) . \end{gathered}

Similarly, the quadratic Hermite splines are defined as

ξ_{1}^{-} (x) = : (- 6 x - 6 x^{2}) X_{[- 1, 0]}, ξ_{2}^{-} (x) = : (1 + 4 x + 3 x^{2}) X_{[- 1, 0]} (x) + (1 - 4 x + 3 x^{2}) X_{[0, 1)} (x) .

Let

Z_{j}^{0} = : {0, 1, . . ., 2^{j}}, Z_{j}^{1} = : {1, 2, . . ., 2^{j}}, Z_{j}^{2} = : {0, 1, . . ., 2^{j} - 1}

, and

Z_{j}^{3} = : {1, 2, . . ., 2^{j} - 1}

. For each j ≥ j₀, define the scaling functions on [0,1]:

\begin{gathered} ξ_{m; j, k}^{Δ, +} = : ξ_{m}^{+} (2^{j} x - k) X_{[0, 1]}, k \in Z_{j}^{0}; \\ ξ_{1; j, k}^{Δ, -} = : ξ_{1}^{-} (2^{j} x - k) X_{[0, 1]} = ξ_{1}^{-} (2^{j} x - k), k \in Z_{j}^{1}; ξ_{2; j, k}^{Δ -} = : ξ_{2}^{-} (2^{j} x - k) X_{[0, 1]}, k \in Z_{j}^{0} . \end{gathered}

Let

V_{j}^{Δ, +} = : span \{ξ_{1; j, k}^{Δ, +}, ξ_{2; j, k}^{Δ, +} : k \in Z_{j}^{0}\}, V_{j}^{Δ, -} = : span \{ξ_{1; j, k_{1}}^{Δ, -}, ξ_{2; j, k_{2}}^{Δ, -} : k_{1} \in Z_{j}^{1}, k_{2} \in Z_{j}^{0}\}

, then

{V_{j}^{Δ, +}}

and

{V_{j}^{Δ, -}}

are two MRAs on L²([0,1]) [8]. The corresponding duals

{\tilde{ξ}}_{m; j, k}^{Δ, \pm}

are given in the sense of distributions:

\begin{gathered} {\tilde{ξ}}_{1}^{+} = : δ_{0}, ⟨f, {\tilde{ξ}}_{1; j, k}^{Δ, +}⟩ = f (2^{- j} k), k \in Z_{j}^{0}; {\tilde{ξ}}_{2}^{+} = : - {δ^{'}}_{0}, ⟨f, {\tilde{ξ}}_{2; j, k}^{Δ, +}⟩ = 2^{- j} f^{'} (2^{- j} k), k \in Z_{j}^{0}; \\ {\tilde{ξ}}_{1}^{-} = : X_{[- 1, 0]}, ⟨f, {\tilde{ξ}}_{1; j, k}^{Δ, -}⟩ = 2^{j} \int_{2^{- j} (k - 1)}^{2^{- j} k} f (x) d x, k \in Z_{j}^{1}; {\tilde{ξ}}_{2}^{-} = : δ_{0}, ⟨f, {\tilde{ξ}}_{2; j, k}^{Δ, -}⟩ = f (2^{- j} k), k \in Z_{j}^{0} . \end{gathered}

The inperpolating multi-wavelets

η_{m; j, k}^{Δ, \pm}

on [0,1] as well as the wavelet spaces are defined by

η_{m; j, k}^{Δ, \pm} (x) = : η_{m; j, k}^{\pm} (x) X_{[0, 1]} = η_{m; j, k}^{\pm} (x), k \in Z_{j}^{2}; W_{j}^{Δ, \pm} = : span \{η_{m; j, k}^{Δ, \pm} : m = 1, 2, k \in Z_{j}^{2}\}

with

η_{m}^{+} = : ξ_{m}^{+} (2 \cdot - 1), m = 1, 2; η_{1}^{-} = : ξ_{1}^{-} (2 \cdot - 1) - ξ_{1}^{-} (2 \cdot - 2); η_{2}^{-} = : ξ_{2}^{-} (2 \cdot - 1)

. Here and after, h_j,k(·) = h(2^j· -k). The corresponding duals are given by

\begin{gathered} {\tilde{η}}_{1}^{+} = : δ_{\frac{1}{2}} - \frac{1}{2} δ_{0} - \frac{1}{2} δ_{1} + \frac{1}{8} {δ^{'}}_{0} - \frac{1}{8} {δ^{'}}_{1}, {\tilde{η}}_{2}^{+} = : \frac{3}{4} δ_{0} - \frac{3}{4} δ_{1} - \frac{1}{2} {δ^{'}}_{\frac{1}{2}} - \frac{1}{8} {δ^{'}}_{0} - \frac{1}{8} {δ^{'}}_{1}, \\ {\tilde{η}}_{1}^{-} = : X_{[0, \frac{1}{2}]} - X_{[\frac{1}{2}, 1]} - \frac{1}{4} δ_{0} + \frac{1}{4} δ_{1}, {\tilde{η}}_{2}^{-} = : δ_{\frac{1}{2}} + \frac{1}{4} δ_{0} + \frac{1}{4} δ_{1} - \frac{3}{2} X_{[0, 1]} \end{gathered}

and

⟨f, {\tilde{η}}_{m; j, k}^{Δ, \pm}⟩ = ⟨f (\frac{\cdot + k}{2^{j}}), {\tilde{η}}_{m}^{\pm}⟩

. Moreover, there is the following differential relations

\begin{gathered} \frac{d}{d x} ξ_{1; j, k}^{Δ, +} (x) = 2^{j} (ξ_{1; j, k}^{Δ, -} (x) - ξ_{1; j, k}^{Δ, -} (x - 2^{- j})), k \in Z_{j}^{3}; \frac{d}{d x} ξ_{1; j, 0}^{Δ, +} (x) = - 2^{j} ξ_{1; j, 1}^{Δ, -} (x), \\ \frac{d}{d x} ξ_{1; j, 2^{j}}^{Δ, +} = 2^{j} ξ_{1; j, 2^{j}}^{Δ, -}; \frac{d}{d x} ξ_{2; j, k}^{Δ, +} = 2^{j} ξ_{2; j, k}^{Δ, -}, k \in Z_{j}^{0}; \frac{d}{d x} η_{m; j, k}^{Δ, +} = 2^{j + 1} η_{m; j, k}^{Δ, -}, k \in Z_{j}^{2} . \\ \frac{d}{d x} {\tilde{ξ}}_{1; j, k}^{Δ, -} = - 2^{j} (\frac{d}{d x} {\tilde{ξ}}_{1; j, k}^{Δ, +} - \frac{d}{d x} {\tilde{ξ}}_{1; j, k - 1}^{Δ, +}), \frac{d}{d x} {\tilde{ξ}}_{2; j, k}^{Δ, -} = - 2^{j} \frac{d}{d x} {\tilde{ξ}}_{2; j, k}^{Δ, +} . \end{gathered}

(1.1)

2 Curl-free wavelets on the cube

For

\vec{u} (x, y, z) = {(u_{1}, u_{2}, u_{3})}^{T}

, the 3D curl-operator is defined as

curl \vec{u} = {(\partial_{2} u_{3} - \partial_{3} u_{2}, \partial_{3} u_{1} - \partial_{1} u_{3}, \partial_{1} u_{2} - \partial_{2} u_{1})}^{T} .

Let I ⊆ {1, 2, 3} =: I₀, define scaling functions

φ_{m}^{I} (x_{1}, x_{2}, x_{3}) = : \prod_{v = 1}^{3} ξ_{m_{v}, v}^{I} (x_{v}), m = {(m_{1}, m_{2}, m_{3})}^{T} \in {1, 2}^{3}

with $ξ_{μ, v}^{I} = \{\begin{matrix} ξ_{μ}^{+}, v \in I \\ ξ_{μ^{}}^{-}, v \notin I . \end{matrix}$

The corresponding wavelets are

ψ_{e, m}^{I} (x_{1}, x_{2}, x_{3}) = : \prod_{v = 1}^{3} ϑ_{e_{v}, m_{v}, v}^{I} (x_{v}), e \in E_{3}^{*}, m = {(m_{1}, m_{2}, m_{3})}^{T} \in {1, 2}^{3} .

Here and after, $E_{3}^{*}$ denotes the non-zero apexes of the unite cube and $ϑ_{ℓ, μ, v}^{I} = \{\begin{matrix} ξ_{μ, v}^{I}, ℓ = 0, \\ η_{μ, v}^{I}, ℓ = 1 . \end{matrix}$

Let

φ_{m; j, k}^{Δ, I} = : φ_{m; j, k}^{I} X_{{[0, 1]}^{3}}

, which is the tensor product of corresponding interpolatory scaling functions on the interval. The corresponding duals are given similarly. Furthermore, define

{\vec{φ}}_{m, i; j, k}^{Δ} = : φ_{m; j, k}^{Δ, I_{0} \ {i}} δ_{i}, {\vec{V}}_{j}^{Δ} = : span \{{\vec{φ}}_{m, i; j, k}^{Δ} : m \in {1, 2}^{3}, k \in \nabla_{j}, 1 \leq i \leq 3\}

and the projection operators:

{\vec{Λ}}_{j}^{Δ} = : Λ_{j}^{Δ, {2, 3}} δ_{1} + Λ_{j}^{Δ, {1, 3}} δ_{2} + Λ_{j}^{Δ, {1, 2}} δ_{3}, {\vec{Λ}}_{j}^{Δ, *} = : Λ_{j}^{Δ, {1}} δ_{1} + Λ_{j}^{Δ, {2}} δ_{2} + Λ_{j}^{Δ, {3}} δ_{3} .

Lemma 2.1[8]. For smooth functions $f : {[0, 1]}^{3} \to R, \frac{\partial}{\partial x_{i}} Λ_{j}^{Δ, I} f = Λ_{j}^{Δ, I \ {i}} (\frac{\partial f}{\partial x_{i}}), i \in I .$ .

Proposition 2.1. For $\vec{f} \in \vec{C} (curl; {[0, 1]}^{3}) = : {\vec{υ} \in {(C ({[0, 1]}^{3}))}^{3} : curl \vec{υ} \in {(C ({[0, 1]}^{3}))}^{3}}$ , there has $curl ({\vec{Λ}}_{j}^{Δ} \vec{f}) = {\vec{Λ}}_{j}^{Δ, *} (curl \vec{f})$ .

Proof. Note that Lemma 2.1, then

\begin{aligned} {\vec{Λ}}_{j}^{Δ, *} (curl \vec{f}) & = Λ_{j}^{Δ, {1}} (\frac{\partial f_{3}}{\partial x_{2}} - \frac{\partial f_{2}}{\partial x_{3}}) δ_{1} + Λ_{j}^{Δ, {2}} (\frac{\partial f_{1}}{\partial x_{3}} - \frac{\partial f_{3}}{\partial x_{1}}) δ_{2} + Λ_{j}^{Δ, {3}} (\frac{\partial f_{2}}{\partial x_{1}} - \frac{\partial f_{1}}{\partial x_{2}}) δ_{3} \\ = (\frac{\partial}{\partial x_{2}} Λ_{j}^{Δ, {1, 2}} f_{3} - \frac{\partial}{\partial x_{3}} Δ_{j}^{Δ, {1, 3}} f_{2}) δ_{1} + (\frac{\partial}{\partial x_{3}} Λ_{j}^{Δ, {2, 3}} f_{1} - \frac{\partial}{\partial x_{1}} Λ_{j}^{Δ, {1, 2}} f_{3}) δ_{2} \\ + (\frac{\partial}{\partial x_{1}} Λ_{j}^{Δ, {1, 3}} f_{2} - \frac{\partial}{\partial x_{2}} Λ_{j}^{Δ, {2, 3}} f_{1}) δ_{3}, \end{aligned}

which is $curl ({\vec{Λ}}_{j}^{Δ} \vec{f})$ by definition.

Proposition 2.1 is important, because it tells us that

{\vec{Λ}}_{j}^{Δ}

keeps curl-free property. In general, vector-valued wavelets and wavelet spaces are given, respectively, by

{\vec{ψ}}_{e, m, i; j, k}^{Δ} = : ψ_{e, m; j, k}^{Δ, I_{0} \ {i}} δ_{i}, {\vec{W}}_{j}^{Δ} = : span \{{\vec{ψ}}_{e, m, i; j, k}^{Δ} : e \in E_{3}^{*}, m \in {1, 2}^{3}, 1 \leq i \leq 3, k \in ∇_{j}^{0}\} .

For

e \in E_{3}^{*}

and m ∈ {1, 2}³, we define

{\vec{ψ}}_{e, m; j, k}^{Δ, c} = : 2^{- j} grad ψ_{e, m; j, k}^{Δ, I_{0}} = 2^{- j} \frac{\partial}{\partial x_{1}} ψ_{e, m; j, k}^{Δ, I_{0}} δ_{1} + 2^{- j} \frac{\partial}{\partial x_{2}} ψ_{e, m; j, k}^{Δ, I_{0}} δ_{2} + 2^{- j} \frac{\partial}{\partial x_{3}} ψ_{e, m; j, k}^{Δ, I_{0}} δ_{3}, k \in \nabla_{j}^{1} .

Clearly, $curl ({\vec{ψ}}_{e, m; j, k}^{Δ, c}) = \vec{0}$ .

To give a decomposition for

{\vec{W}}_{j}^{Δ}

, take

{\vec{ψ}}_{e, m, i; j, k}^{Δ, non} = : ψ_{e, m; j, k}^{Δ, I_{0} \ {i}} δ_{i}, k \in ∇_{j}^{2}

for i ∈ I₀\{i_e}. Here, we choose i_esuch that $e_{i_{e}} = 1$ .

Proposition 2.2. The vector-valued function system ${{\vec{ψ}}_{e, m; j, k_{1}}^{Δ, c}, {\vec{ψ}}_{e, m, i; j, k_{2}}^{Δ, non}, e \in E_{3}^{*}, m \in {1, 2}^{3}, i \neq i_{e}, k_{1} \in \nabla_{j}^{1}, k_{2} \in \nabla_{j}^{2}}$ is complete in ${\vec{W}}_{j}^{Δ}$ .

Proof. It is sufficient to show the statement for j = 0. Let

\vec{f} \in {\vec{W}}_{0}^{Δ}

satisfy

(\vec{f}, {\vec{ψ}}_{e, m; 0, k}^{Δ, c}) = (\vec{f}, {\vec{ψ}}_{e, m, i; 0, k}^{Δ, non}) = 0

for all

e \in E_{3}^{*}, m \in {1, 2}^{3}, k \in \nabla_{j}^{1} \cup \nabla_{j}^{2}}, 1 \leq i \leq 3

and i ≠ i_e. Here, the inner product is in L²([0,1]³). Without loss of generality, one assumes i_e= 1. Then,

(\vec{f}, {\vec{ψ}}_{e, m, i; 0, k}^{Δ, non}) = 0

leads to

(f_{2}, ψ_{e, m; 0, k}^{Δ, I_{0} \ {2}}) = (f_{3}, ψ_{e, m; 0, k}^{Δ, I_{0} \ {3}}) = 0 .

By the definition of ψ^Iand differential relations (1.1), one knows

(f_{2}, \frac{\partial}{\partial x_{2}} ψ_{e, m; 0, k}^{Δ, I_{0}}) = (f_{3}, \frac{\partial}{\partial x_{3}} ψ_{e, m; 0, k}^{Δ, I_{0}}) = 0 .

Moreover, $(\vec{f}, {\vec{ψ}}_{e, m; 0, k}^{Δ, c}) = 0$ reduces to $(f_{1}, \frac{\partial}{\partial x_{1}} ψ_{e, m; 0, k}^{Δ, I_{0}}) = 0$ . Now, it follows that $(f_{1}, ψ_{e, m; 0, k}^{Δ, I_{0} \ {1}}) = 0$ from i_e= 1. Finally, $(\vec{f}, {\vec{ψ}}_{e, m, i; 0, k}^{Δ}) = 0$ and $\vec{f} = \vec{0}$ follows from the definition of ${\vec{W}}_{0}^{Δ}$ .

To give the bi-orthogonal decomposition, we define

{\vec{\tilde{ψ}}}_{e, m; j, k}^{Δ, c} = \frac{1}{2} {\tilde{ψ}}_{e, m; j, k}^{Δ, I_{0} \ {i_{e}}} δ_{i_{e}} .

(2.1)

Assume I = {i, i_e, i'}, then

curl {\vec{ψ}}_{e, m, i; j, k}^{Δ, non} = curl ψ_{e, m; j, k}^{Δ, I_{0} \ {i}} δ_{i} = 2^{j + 1} ε_{1} ψ_{e, m; j, k}^{Δ, I_{0} \ {i, i_{e}}} δ_{i^{'}} + ε_{2} \frac{\partial}{\partial x_{i^{'}}} ψ_{e, m; j, k}^{Δ, I_{0} \ {i}} δ_{i_{e}}

with |ε₁| = |ε₂| = 1 and ε₁ε₂ = -1. Now, define

{\vec{\tilde{ψ}}}_{e, m, i; j, k}^{Δ, non} = : \frac{1}{2^{j + 1}} curl ε_{1} {\tilde{ψ}}_{e, m; j, k}^{Δ, I_{0} \ {i, i_{e}}} δ_{i^{'}} .

(2.2)

Here, the derivatives are meant in the sense of distributions. Now, we state the main result:

Proposition 2.3. The set ${{\vec{ψ}}_{e, m; j, k_{1}}^{Δ, c}, {\vec{ψ}}_{e, m, i; j, k_{2}}^{Δ, non}, e \in E_{3}^{*}, m \in {1, 2}^{3}, i \neq i_{e}, j \geq j_{0}, k_{1} \in \nabla_{j}^{1}, k_{2} \in \nabla_{j}^{2}}$ is a bi-orthogonal wavelet basis of L²([0,1]³)³ with duals defined in (2.1) and (2.2).

Proof. According to Proposition 2.2, one only need show

(i) $⟨{\vec{ψ}}_{e^{'}, m^{'}, i; j^{'}, k^{'}}^{Δ, non}, {\vec{\tilde{ψ}}}_{e, m; j, k}^{Δ, c}⟩ = 0$ ;

(ii) $⟨{\vec{ψ}}_{e, m; j, k}^{Δ, c}, {\vec{\tilde{ψ}}}_{e^{'}, m^{'}, i; j^{'}, k^{'}}^{Δ, non}⟩ = 0$ ;

(iii) $⟨{\vec{ψ}}_{e, m; j, k}^{Δ, c}, {\vec{\tilde{ψ}}}_{e^{'}, m^{'}; j^{'}, k^{'}}^{Δ, c}⟩ = δ_{e, e^{'}} δ_{m, m^{'}} δ_{j, j^{'}} δ_{k, k^{'}}$ ;

(iv) $⟨{\vec{ψ}}_{e_{1}, m_{1}, i_{1}; j_{1}, k_{1}}^{Δ, non}, {\vec{\tilde{ψ}}}_{e_{2}, m_{2}, i_{2}; j_{2}, k_{2}}^{Δ, non}⟩ = δ_{e_{1}, e_{2}} δ_{m_{1}, m_{2}} δ_{i_{1}, i_{2}} δ_{j_{1}, j_{2}} δ_{k_{1}, k_{2}}$ .

The identity (i) holds obviously for i ≠ i_e. For i = i_e, since i ≠ i_e', then i_e≠ i_e', which means e ≠ e'. Finally, the result (i) follows from the bi-orthogonality of $ψ_{e^{'}, m^{'}; j^{'}, k^{'}}^{Δ, I_{0} \ {i}}$ and ${\tilde{ψ}}_{e, m; j, k}^{Δ, I_{0} \ {i}}$ .

Note that

⟨\vec{f}, curl \vec{g}⟩ = ⟨curl \vec{f}, \vec{g}⟩

. Then (ii) follows from curl⋅grad = 0. Furthermore,

⟨{\vec{ψ}}_{e, m; j, k}^{Δ, c}, {\vec{\tilde{ψ}}}_{e^{'}, m^{'}; j^{'}, k^{'}}^{Δ, c}⟩ = \frac{1}{2} \times 2^{- j} ⟨\frac{\partial}{\partial x_{i_{e^{'}}}} ψ_{e, m; j, k}^{Δ, I_{0}}, {\tilde{ψ}}_{e^{'}, m^{'}; j^{'}, k^{'}}^{Δ, I_{0} \ {i_{e^{'}}}}⟩ = - \frac{1}{2} \times 2^{- j} ⟨ψ_{e, m; j, k}^{Δ, I_{0}}, \frac{\partial}{\partial x_{i_{e^{'}}}} {\tilde{ψ}}_{e^{'}, m^{'}; j^{'}, k^{'}}^{Δ, I_{0} \ {i_{e^{'}}}}⟩ .

Then by the fact

\frac{d}{d x} {\tilde{η}}_{m}^{-} = - 2 {\tilde{η}}_{m}^{+}

and the bi-orthogonality of

ψ_{e, m}^{Δ, I_{0}}, {\tilde{ψ}}_{e, m}^{Δ, I_{0}}

, one obtains

⟨{\vec{ψ}}_{e, m; j, k}^{Δ, c}, {\vec{\tilde{ψ}}}_{e^{'}, m^{'}; j^{'}, k^{'}}^{Δ, c}⟩ = 2^{j^{'} - j} ⟨ψ_{e, m; j, k}^{Δ, I_{0}}, {\tilde{ψ}}_{e^{'}, m^{'}; j^{'}, k^{'}}^{Δ, I_{0}}⟩ = δ_{e, e^{'}} δ_{m, m^{'}} δ_{j, j^{'}} δ_{k, k^{'}} .

Now, it remains to prove (iv), which is equivalent to

\begin{gathered} 2^{j_{1} - j_{2}} ⟨2 ε_{1} ψ_{e_{1}, m_{1}; j_{1}, k_{1}}^{Δ, I_{0} \ {i_{1}, i_{e_{1}}}} δ_{{i^{'}}_{1}} + 2^{- j_{1}} ε_{2} \frac{\partial}{\partial x_{{i^{'}}_{1}}} ψ_{e_{1}, m_{1}; j_{1}, k_{1}}^{Δ, I_{0} \ {i_{1}}} δ_{i_{e_{1}}}, \frac{1}{2} ε_{1} {\tilde{ψ}}_{e_{2}, m_{2}; j_{2}, k_{2}}^{Δ, I_{0} \ {i_{2}, i_{e_{2}}}} δ_{{i^{'}}_{2}}⟩ \\ = δ_{e_{1}, e_{2}} δ_{m_{1}, m_{2}} δ_{i_{1}, i_{2}} δ_{j_{1}, j_{2}} δ_{k_{1}, k_{2}} . \end{gathered}

(2.3)

It is easily proved, when e₁ = e₂: In fact, since

i_{e_{1}} = i_{e_{2}}

, one can assume i₁ = i₂ and

i_{1}^{'} = i_{2}^{'}

, because

i_{1} = i_{2}^{'}

leads to (2.3) obviously. In that case, the left-hand side of (2.3) reduces to

2 ε_{1} \cdot \frac{ε_{1}}{2} \cdot 2^{j_{1} - j_{2}} ⟨ψ_{e_{1}, m_{1}; j_{1}, k_{1}}^{I \ {i_{1}, i_{e_{1}}}}, {\tilde{ψ}}_{e_{2}, m_{2}; j_{2}, k_{2}}^{I \ {i_{2}, i_{e_{2}}}}⟩ = δ_{m_{1}, m_{2}} δ_{j_{1}, j_{2}} δ_{k_{1}, k_{2}}

, which is the desired. To the end, it is sufficient to prove that for e₁ ≠ e₂, that is

⟨2 ε_{1} ψ_{e_{1}, m_{1}; j_{1}, k_{1}}^{Δ, I_{0} \ {i_{1}, i_{e_{1}}}} δ_{{i^{'}}_{1}} + 2^{- j_{1}} ε_{2} \frac{\partial}{\partial x_{{i^{'}}_{1}}} ψ_{e_{1}, m_{1}; j_{1}, k_{1}}^{Δ, I_{0} \ {i_{1}}} δ_{i_{e_{1}}}, \frac{1}{2} ε_{1} {\tilde{ψ}}_{e_{2}, m_{2}; j_{2}, k_{2}}^{Δ, I_{0} \ {i_{2}, i_{e_{2}}}} δ_{{i^{'}}_{2}}⟩ = 0 .

(2.4)

Note that

i_{2}^{'} \in {i_{1}, i_{1}^{'}, i_{e_{1}}}

. Then the conclusion is obvious when

i_{2}^{'} = i_{1}

. When

i_{2}^{'} = i_{1}^{'}

, then

{i_{1}, i_{e_{1}}} = {i_{2}, i_{e_{2}}}

and the left-hand side of (2.4) reduces to

2 ε_{1} \cdot \frac{ε_{1}}{2} ⟨ψ_{e_{1}, m_{1}; j_{1}, k_{1}}^{Δ, I_{0} \ {i_{1}, i_{e_{1}}}}, {\tilde{ψ}}_{e_{2}, m_{2}; j_{2}, k_{2}}^{Δ, I_{0} \ {i_{2}, i_{e_{2}}}}⟩ = 0

. Hence one only need to show (2.4), when

i_{2}^{'} = i_{e_{1}}

. However, (2.4) becomes

⟨\frac{\partial}{\partial x_{{i^{'}}_{1}}} ψ_{e_{1}, m_{1}; j_{1}, k_{1}}^{Δ, I_{0} \ {i_{1}}}, {\tilde{ψ}}_{e_{2}, m_{2}; j_{2}, k_{2}}^{Δ, I_{0} \ {i_{2}, i_{e_{2}}}}⟩ = 0 .

(2.5)

in that case. Since

{i_{1}, i_{1}^{'}, i_{e_{1}}} = {i_{2}, i_{2}^{'}, i_{e_{2}}} = I

, two cases should be considered:

i_{2} = i_{1}, i_{2}^{'} = i_{e_{1}}, i_{e_{2}} = i_{1}^{'}

and

i_{2} = i_{1}^{'}, i_{2}^{'} = i_{e_{1}}, i_{e_{2}} = i_{1}

. Using

\frac{d}{d x} {\tilde{η}}_{m}^{-} = - 2 {\tilde{η}}_{m}^{+}

, the left-hand side of (2.5) is

- ⟨ψ_{e_{1}, m_{1}, j_{1}, k_{1}}^{Δ, I_{0} \ {i_{1}}}, \frac{\partial}{\partial x_{{i^{'}}_{1}}} {\tilde{ψ}}_{e_{2}, m_{2}, j_{2}, k_{2}}^{Δ, I_{0} \ {i_{2}, i_{e_{2}}}}⟩ = 2^{j_{2} + 1} ⟨ψ_{e_{1}, m_{1}; j_{1}, k_{1}}^{Δ, I_{0} \ {i_{1}}}, {\tilde{ψ}}_{e_{2}, m_{2}; j_{2}, k_{2}}^{Δ, I_{0} \ {i_{2}}}⟩ = 0

in the first case; In the second one, the left-hand side of (2.5) becomes $⟨\frac{\partial}{\partial x_{i_{2}}} ψ_{e_{1}, m_{1}; j_{1}, k_{1}}^{Δ, I_{0} \ {i_{1}}} {\tilde{ψ}}_{e_{2}, m_{2}; j_{2}, k_{2}}^{Δ, I_{0} \ {i_{2}, i_{2}}}⟩$ . According to the differential relation (1.1), $\frac{\partial}{\partial x_{i_{2}}} ψ_{e_{1}, m_{1}; j_{1}, k_{1}}^{Δ, I_{0} \ {i_{1}}}$ is a linear combination of $ψ_{e_{1}, m_{1}; j_{1}, k_{1}}^{Δ, I_{0} \ {i_{1}, i_{2}}}$ . By the bi-orthogonality of $ψ_{e_{1}, m_{1}; j_{1}, k_{1}}^{Δ, I_{0} \ {i_{1}, i_{2}}}$ and ${\tilde{ψ}}_{e_{2}, m_{2}; j_{2}, k_{2}}^{Δ, I_{0} \ {i_{1}, i_{2}}}$ , one receives the desired conclusion.

3 Characterization for Besov spaces

We shall characterize a class of vector-valued Besov spaces in this section. For 0 < p, q ≤ ∞ and s > 0, the Besov space

B_{q}^{s} (L^{p} (Ω))

is the set of all f ∈ L^p(Ω) such that

{|f|}_{B_{q}^{s} (L^{p} (Ω))} = : {∥{2^{s j} ω_{m} (f, 2^{- j}, L^{p} (Ω))}∥}_{ℓ^{q}} < + \infty

with m = [s] + 1 and ω_m(f, 2^-j, L^p(Ω)) the classical m-order modulus of smoothness. The corresponding norm is defined by

{∥f∥}_{B_{q}^{s} (L^{p} (Ω))} = : {∥f∥}_{L^{p} (Ω)} + {|f|}_{B_{q}^{s} (L^{p} (Ω))} .

Our Besov space is defined as

{\hat{B}}_{q}^{s} (L^{p} ({[0, 1]}^{3})) = : {\vec{f} \in {(B_{q}^{s} (L^{p} ({[0, 1]}^{3})))}^{3} : \frac{\partial}{\partial x_{j}} f_{i} \in B_{q}^{s} (L^{p} ({[0, 1]}^{3})), i = 1, 2, 3; j \neq i}

with the norm

{∥\vec{f}∥}_{{\hat{B}}_{q}^{s} (L^{p} ({[0, 1]}^{3}))} = : \sum_{i = 1}^{3} {∥f_{i}∥}_{B_{q}^{s} (L^{p} ({[0, 1]}^{3}))} + \sum_{i = 1}^{3} \sum_{\begin{array}{c} 1 \leq j \leq 3 \\ j \neq i \end{array}} {∥\frac{\partial f_{i}}{\partial x_{j}}∥}_{B_{q}^{s} (L^{p} ({[0, 1]}^{3}))} .

Clearly, $curl \vec{f} \in {(B_{q}^{s} (L^{p} ({[0, 1]}^{3})))}^{3}$ , when $\vec{f} \in {\hat{B}}_{q}^{s} (L^{p} ({[0, 1]}^{3}))$ .

The following lemma is easily proved by the definition of modulus of smoothness:

Lemma 3.1. If $f (x), g (x) \in B_{q}^{s} (L^{p} (R))$ , then $f (x_{1}) g (x_{2}) \in B_{q}^{s} (L^{p} (R^{2}))$ .

For

α = {(α_{j})}_{j \geq j_{0}}

and α_j= (α_j,k)_k, define

{∥α∥}_{ℓ_{p, q}^{s}} = : {∥(2^{j (s - \frac{n}{p})} {∥α_{j}∥}_{ℓ^{p}})∥}_{ℓ^{q}} .

Lemma 3.2[8]. If

ϕ \in B_{\infty}^{σ} (L^{p} (R^{n}))

is compactly supported, 0 < p, q ≤ ∞ and 0 < s < σ, then

\begin{gathered} {∥\sum_{k \in \nabla_{j}} β_{k} ϕ (2^{j} \cdot - k)∥}_{B_{q}^{s} (L^{p} ({[0, 1]}^{n}))} ≲ 2^{(s - \frac{n}{p}) j} {∥β∥}_{ℓ^{p}}, \\ {∥\sum_{j \geq j_{0}} \sum_{k \in \nabla_{j}} α_{j, k} ϕ (2^{j} \cdot - k)∥}_{B_{q}^{s} (L^{p} ({[0, 1]}^{n}))} ≲ {∥α∥}_{ℓ_{p, q}^{s}}, \end{gathered}

where ∇_j=: {k : suppϕ(2^j⋅ -k) ⊆ [0,1] ⁿ}.

Theorem 3.1. Let

{\vec{φ}}_{m, i; j, k}^{Δ}, {\vec{ψ}}_{e, m; j, k}^{Δ, c}

, and

{\vec{ψ}}_{e, m, i; j, k}^{Δ, non}

be defined in Section 2. If 0

< s < 1 + \frac{1}{p}

and 0 < p, q ≤ ∞, then one has

\begin{gathered} {∥\sum_{m, i} \sum_{k \in \nabla_{j_{0}}} β_{m, i, k} {\vec{φ}}_{m, i; j_{0}, k}^{Δ} + \sum_{j \geq j_{0}} \sum_{e, m, i \neq i_{e}} \sum_{k \in \nabla_{j}^{1}} α_{e, m; j, k}^{c} {\vec{ψ}}_{e, m; j, k}^{Δ, c} + \sum_{j \geq j_{0}} \sum_{e, m} \sum_{k \in \nabla_{j}^{2}} α_{e, m, i; j, k}^{non} {\vec{ψ}}_{e, m, i; j, k}^{n o n}∥}_{{\hat{B}}_{q}^{s} (L^{p} ({[0, 1]}^{3}))} \\ ≲ \sum_{m, i} {∥β_{m, i}∥}_{ℓ^{p}} + \sum_{e, m} (\sum_{i \neq i_{e}} {∥α_{e, m, i}^{non}∥}_{ℓ_{p, q}^{s + 1}} + {∥α_{e, m}^{c}∥}_{ℓ_{p, q}^{s + 1}}) . \end{gathered}

Proof. It is enough to prove the following inequality:

(i)

${∥\sum_{k \in \nabla_{j_{0}}} β_{m, i, k} {\vec{φ}}_{m, i; j_{0}, k}^{Δ}∥}_{{\hat{B}}_{q}^{s} (L^{p} ({[0, 1]}^{3}))} ≲ {∥β_{m, i}∥}_{ℓ p}$ ;
(ii)

${∥\sum_{j \geq j_{0}} \sum_{k \in \nabla_{j}^{1}} α_{e, m; j, k}^{c} {\vec{ψ}}_{e, m; j, k}^{Δ, c}∥}_{{\hat{B}}_{q}^{s} (L^{p} ({[0, 1]}^{3})} ≲ {∥α_{e, m}^{c}∥}_{ℓ_{p, q}^{s + 1}}$ ;
(iii)

${∥\sum_{j \geq j_{0}} \sum_{k \in \nabla_{j}^{2}} α_{e, m, i; j, k}^{non} {\vec{ψ}}_{e, m, i; j, k}^{n o n}∥}_{{\hat{B}}_{q}^{s} (L^{p} ({[0, 1]}^{3})} ≲ {∥α_{e, m, i}^{non}∥}_{ℓ_{p, q}^{s + 1}} (i \neq i_{e})$ .

Let

\vec{h} = : \sum_{k \in \nabla_{j_{0}}} β_{m, i, k} {\vec{φ}}_{m, i; j_{0}, k}^{Δ}

and h_νbe the ν th component of

\vec{h}

. Then, for μ ≠ i,

h_{i} = \sum_{k \in \nabla_{j_{0}}} β_{m, i; k} φ_{m}^{I_{0} \ {i}} (2^{j_{0}} x - k), \frac{\partial}{\partial x_{μ}} h_{i} = \sum_{k \in \nabla_{j_{0}}} β_{m, i; k} 2^{j_{0}} (\frac{\partial}{\partial x_{μ}} φ_{m}^{I_{0} \ {i}}) (2^{j_{0}} x - k) .

Since $ξ_{m}^{+} \in B_{\infty}^{2 + \frac{1}{p}} (L^{p} (R)) \subseteq B_{\infty}^{1 + \frac{1}{p}} (L^{p} (R))$ and $ξ_{m}^{-} \in B_{\infty}^{1 + \frac{1}{p}} (L^{p} (R))$ , then both $φ_{m}^{I \ {i}}$ and $\frac{\partial}{\partial x_{μ}} φ_{m}^{I \ {i}}$ are in the Besov space $B_{\infty}^{1 + \frac{1}{p}} (L^{p} (R^{3}))$ , due to Lemma 3.1. Moreover, Lemma 3.2 implies ${∥h_{i}∥}_{B_{q}^{s} (L^{p} ({[0, 1]}^{3}))} ≲ {∥β_{m, i}∥}_{ℓ^{p}}$ and ${∥\frac{\partial}{\partial x_{μ}} h_{i}∥}_{B_{q}^{s} (L^{p} ({[0, 1]}^{3}))} ≲ {∥β_{m, i}∥}_{ℓ^{p}}$ . Note that h_ν= 0 for ν ≠ i. Finally, the first inequality follows from the definition.

Let

\vec{g} = : \sum_{j \geq j_{0}} \sum_{k \in \nabla_{j}^{1}} α_{e, m; j, k}^{c} {\vec{ψ}}_{e, m; j, k}^{Δ, c}

and g_νbe the ν th component of

\vec{g} (1 \leq v \leq 3)

. Then

\begin{gathered} g_{v} = \sum_{j \geq j_{0}} \sum_{k \in \nabla_{j}^{1}} α_{e, m; j, k}^{c} (\frac{\partial}{\partial x_{v}} ψ_{e, m}^{I_{0}}) (2^{j} x - k), \\ \frac{\partial}{\partial x_{μ}} g_{v} = \sum_{j \geq j_{0}} \sum_{k \in \nabla_{j}^{1}} 2^{j} α_{e, m; j, k}^{c} (\frac{\partial^{2}}{\partial x_{μ} \partial x_{v}} ψ_{e, m}^{I_{0}}) (2^{j} x - k) . \end{gathered}

Similar to the above,

\frac{\partial}{\partial x_{v}} ψ_{e, m}^{I_{0}}, \frac{\partial^{2}}{\partial x_{v} \partial x_{μ}} ψ_{e, m}^{I_{0}} \in B_{\infty}^{1 + \frac{1}{p}} (L^{p} (R^{3}))

. According to Lemma 3.2,

{∥g_{v}∥}_{B_{q}^{s} (L^{p} ({[0, 1]}^{3}))} ≲ {∥α_{e, m}∥}_{ℓ_{p, q}^{s}} ≲ {∥α_{e, m}∥}_{ℓ_{p, q}^{s + 1}} and {∥\frac{\partial}{\partial x_{μ}} g_{v}∥}_{B_{q}^{s} (L^{p} ({[0, 1]}^{3}))} ≲ {∥α_{e, m}∥}_{ℓ_{p, q}^{s + 1}} .

Finally, one receives the second inequality and the last one follows analogously.

Let

W_{τ}^{μ} (D)

denotes the Sobolev space with regularity exponent μ and domain D. Moreover,

E_{d} (f, W_{τ}^{μ} (D)) = : inf_{P \in {\prod^{}}_{d - 1}} {∥ f - P ∥}_{W_{τ}^{μ}} {(D)}_{} .

Furthermore, let σ_j,k= 2^-j([0,1]³ + k) for $k \in {(Z_{j}^{2})}^{3}$ , the boundary cases are: when there is only one k_i= 2^j(1 ≤ i ≤ 3), σ_j,kis defined as replacing the i th position of 2^-j([k₁, k₁ + 1] ⊗ [k₂, k₂ +1] ⊗ [k₃, k₃ +1]) by [2^j-1, 2^j]; when k_i= k_i'= 2ⁱfor i, i' ∈ {1, 2, 3}, both the positions i and i' are replaced by [2^j-1, 2^j]; finally, ${σ_{j}}_{, (2^{j}, 2^{j}, 2^{j})} = : 2^{- j} ([2^{j - 1}, 2^{j}] \otimes [2^{j - 1}, 2^{j}] \otimes [2^{j - 1}, 2^{j}])$ .

Lemma 3.3[8]. Let $\frac{n}{p} - \frac{n}{τ} + μ < s < d, s > 0, μ \in N_{0}, 0 < p, q \leq \infty, 1 \leq τ \leq \infty, j_{0} \in N_{0}$ .

Then

‖ (2^{j (s - \frac{n}{p} + \frac{n}{τ} - μ)} ‖ {(E_{d} (f, W_{τ}^{μ} (σ_{j, k})))}_{k \in Z^{n}} {‖ {ℓ^{p})}_{j \geq j_{0}} ‖}_{ℓ^{q}} ≲ {| f |}_{B_{q}^{s} (L^{p} ({[0, 1]}^{n}))} .

The following lemma can be easily proved, but it is important for proving Theorem 3.2:

Lemma 3.4. The following relations hold:

\begin{gathered} ⟨f, {\tilde{η}}_{1; j, k}^{Δ, +}⟩ = f (2^{- j} (k + \frac{1}{2})) - \frac{1}{2} f (2^{- j} k) - \frac{1}{2} (2^{- j} (k + 1)) - \frac{1}{8} \cdot 2^{- j} f^{'} (2^{- j} k) + \frac{1}{8} \cdot 2^{- j} f^{'} (2^{- j} (k + 1)); \\ ⟨f, {\tilde{η}}_{2; j, k}^{Δ, +}⟩ = 2^{- (j + 1)} f^{'} (2^{- j} (k + \frac{1}{2})) - \frac{3}{4} f (2^{- j} (k + 1)) + \frac{3}{4} f (2^{- j} k) + \frac{1}{4} \cdot 2^{- (j + 1)} [f^{'} (2^{- j} k) + f^{'} (2^{- j} (k + 1))]; \\ ⟨f, {\tilde{η}}_{1; j, k}^{Δ, -}⟩ = 2^{j} \int_{2^{- j} k}^{2^{- j} (k + \frac{1}{2})} f (t) d t - 2^{j} \int_{2^{- j} (k + \frac{1}{2})}^{2^{- j} (k + 1)} f (t) d t - \frac{1}{4} f (2^{- j} k) + \frac{1}{4} f (2^{- j} (k + 1)); \\ ⟨f, {\tilde{η}}_{2; j, k}^{Δ, -}⟩ = - \frac{3}{2} \cdot 2^{j} \int_{2^{- j} k}^{2^{- j} (k + 1)} f (t) d t + \frac{1}{4} f (2^{- j} k) + \frac{1}{4} f (2^{- j} (k + 1)) + f (2^{- j} (k + \frac{1}{2})) . \end{gathered}

Theorem 3.2. Let

1 + \frac{3}{p} < s < 3

and 0 < p, q ≤ ∞. Then

β_{m, i} = {(⟨\vec{f}, {\vec{\tilde{φ}}}_{m, i; j_{0}, k}^{Δ}⟩)}_{k \in \nabla_{j_{0}},} α_{e, m}^{c} = {(⟨\vec{f}, {\vec{\tilde{ψ}}}_{e, m; j, k}^{Δ, c}⟩)}_{j \geq j_{0}, k \in \nabla_{j}^{1},} α_{e, m, i}^{non} = {(⟨\vec{f}, {\vec{\tilde{ψ}}}_{e, m, i; j, k}^{Δ, non}⟩)}_{j \geq j_{0}, k \in \nabla_{j}^{2},} j_{0} \geq 1, i \neq i_{e}

satisfy

\sum_{m, i} {∥β_{m, i}∥}_{ℓ^{p}} + \sum_{e, m} (\sum_{i \neq i_{e}} {∥α_{e, m, i}^{non}∥}_{ℓ_{p, q}^{s + 1}} + {∥α_{e, m}^{c}∥}_{ℓ_{p, q}^{s + 1}}) ≲ {∥\vec{f}∥}_{{\hat{B}}_{q}^{s} (L^{p} ({[0, 1]}^{3}))} .

Proof. One only need to show the following inequality:

(i)

${∥β_{m, i}∥}_{ℓ^{p}} ≲ {∥\vec{f}∥}_{{\hat{B}}_{q}^{s} (L^{p} ({[0, 1]}^{3}))}$ ;
(ii)

${∥α_{e, m, i}^{non}∥}_{ℓ_{p, q}^{s + 1}} ≲ {∥\vec{f}∥}_{{\hat{B}}_{q}^{s} (L^{p} ({[0, 1]}^{3}))}$ ;
(iii)

${∥α_{e, m}^{c}∥}_{ℓ_{p, q}^{s + 1}} ≲ {∥\vec{f}∥}_{{\hat{B}}_{q}^{s} (L^{p} ({[0, 1]}^{3}))}$ .

Note that

β_{m, i; k} = ⟨\vec{f}, {\vec{\tilde{φ}}}_{m, i; j_{0}, k}^{Δ}⟩ = ⟨f_{i}, {\tilde{φ}}_{m; j_{0}, k}^{Δ, I_{0} \ {i}}⟩

. Then one assumes i = 1 without loss of generality and proves first

{∥β_{m, 1}∥}_{ℓ^{p}} = {∥{(⟨f_{1}, {\tilde{φ}}_{m; j_{0}, k}^{Δ, {2, 3}}⟩)}_{k \in {\nabla^{'}}_{j_{0}}}∥}_{ℓ^{p}} ≲ {∥\vec{f}∥}_{{\hat{B}}_{q}^{s} (L^{p} ({[0, 1]}^{3}))} .

By the embedding property,

B_{q}^{s} (L^{p} (D)) \subseteq W_{τ}^{μ} (D)

for

s > \frac{1}{p} - \frac{1}{τ} + μ

, and

{∥f∥}_{W_{τ}^{μ} (D)} ≲ {∥f∥}_{B_{q}^{s} (L^{p} (D))} .

(3.1)

When $m_{2} = m_{3} = 1, |⟨f_{1}, {\tilde{φ}}_{m; j_{0}, k}^{Δ, {2, 3}}⟩| \leq {∥f_{1}∥}_{L^{\infty} (σ_{j_{0}, k - δ_{1}})}$ for m₁ = 1 or ${∥f_{1}∥}_{L^{\infty} (σ_{j_{0}, k})}$ for m₁ = 2.

Using (3.1), one obtains

{∥β_{m, 1}∥}_{ℓ^{p}} \leq \{\begin{matrix} {(\sum_{k \in {\nabla^{'}}_{j_{0}}} {∥f_{1}∥}_{L^{\infty} (σ_{j_{0}, k - δ_{1}})}^{p})}^{\frac{1}{p}}, m_{1} = 1 \\ ≲ {∥f_{1}∥}_{B_{q}^{s} (L^{p} ({[0, 1]}^{3})) ≲} {∥\vec{f}∥}_{{\hat{B}}_{q}^{s} (L^{p} ([0, 1]))} \\ {(\sum_{k \in {\nabla^{'}}_{j_{0}}} {∥f_{1}∥}_{L^{\infty} (σ_{j_{0}, k})}^{p})}^{\frac{1}{p}}, m_{1} = 2 \end{matrix} .

When m₃ = 2 (similarly for m₂ = 2), we obtain

|⟨f_{1}, {\tilde{φ}}_{m; j_{0}, k}^{Δ, {2, 3}}⟩| = 2^{- j_{0}} |⟨\frac{\partial f_{1}}{\partial x_{3}}, {\tilde{φ}}_{m; j_{0}, k}^{Δ, {2}}⟩| \leq \{\begin{matrix} 2^{- j_{0}} {∥\frac{\partial f_{1}}{\partial x_{3}}∥}_{W_{\infty}^{1} (σ_{j_{0}, k - δ_{1}})}, m_{1} = 1 \\ 2^{- j_{0}} {∥\frac{\partial f_{1}}{\partial x_{3}}∥}_{W_{\infty}^{1} (σ_{j_{0}, k})}, m_{1} = 2 . \end{matrix}

Note that $1 + \frac{1}{p} < s < 3$ , then the same arguments as above lead to (i).

For

h \in B_{q}^{s} (L^{p} ({[0, 1]}^{3}))

, we first claim that there are only the following two cases:

(a)

$|⟨h, {\tilde{ψ}}_{e, m; j, k}^{Δ, {i^{'}}}⟩| ≲ 2^{- j} E_{3} (h, W_{\infty}^{1} (σ_{j, k}))$ or $2^{- j} E_{3} (h, W_{\infty}^{1} (σ_{j, k - δ_{i}}))$ ;
(b)

$|⟨h, {\tilde{ψ}}_{e, m; j, k}^{Δ, {i^{'}}}⟩| ≲ E_{3} (h, L^{\infty} (σ_{j, k}))$ or $E_{3} (h, L^{\infty} (σ_{j, k - δ_{i}}))$ .

In fact, by the vanishing moment property of the dual wavelets, that is,

{\tilde{η}}_{m; j, k}^{+} (P_{1}) = 0, P_{1} \in \prod_{3}; {\tilde{η}}_{m; j, k}^{-} (P_{2}) = 0, P_{2} \in \prod_{2} .

Then

⟨h, {\tilde{ψ}}_{e, m; j, k}^{Δ, {i^{'}}}⟩ = ⟨h - P, {\tilde{ψ}}_{e, m; j, k}^{Δ, {i^{'}}}⟩

for each P ∈ Π₂. Hence, if e_i'= 1 or e_i'= 0 but m_i'= 2, the differential relation (1.1) implies that

|⟨h, {\tilde{ψ}}_{e, m; j, k}^{Δ, {i^{'}}}⟩| ≲ 2^{- j} |⟨\frac{\partial}{\partial x_{i^{'}}} (h - P), {\tilde{ψ}}_{e, m; j, k}^{Δ, 0̸}⟩| ≲ \{\begin{matrix} 2^{- j} {∥h - P∥}_{W_{\infty}^{1} (σ_{j}, k - δ_{i})}, & e_{i} = 0, m_{i} = 1; \\ 2^{- j} {∥h - P∥}_{W_{\infty}^{1} (σ_{j}, k),} & o . w . \end{matrix}

Moreover, the (a) part follows from the definition of

E_{3} (h, W_{\infty}^{1} (D))

; If e_i'= 0 and m_i'= 1,

|⟨h, {\tilde{ψ}}_{e, m; j, k}^{Δ, {i^{'}}}⟩| = |⟨h - P, {\tilde{ψ}}_{e, m; j, k}^{Δ, {i^{'}}}⟩| ≲ \{\begin{matrix} {∥h - P∥}_{L^{\infty} (σ_{j, k - δ_{i}})}, & e_{i} = 0, m_{i} = 1; \\ {∥h - P∥}_{L^{\infty} (σ_{j, k})}, & o . w . \end{matrix}

and the (b) part follows.

Now, one is ready to estimate

α_{e, m, i; j, k}^{non}

and

α_{e, m; j, k}^{c}

. By the definition of

{\vec{\tilde{ψ}}}_{e, m, i}^{Δ, non}

, one knows

\begin{gathered} || \end{gathered}