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

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.

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 ${\xi }_1^{+}$ and ${\xi }_2^{+}$ stand for two cubic Hermite splines:

$$\begin{array}{c}{\xi }_1^{+}\left(x\right)=:\left(1-3x^2-2x^3\right){\mathcal{X}}_{\left[-1,0\right)}\left(x\right)+\left(1-3x^2+2x^3\right){\mathcal{X}}_{\left[0,1\right)}\left(x\right),\\ {\xi }_2^{+}\left(x\right)=:\left(x+2x^2+x^3\right){\mathcal{X}}_{\left[-1,0\right)}\left(x\right)+\left(x-2x^2+x^3\right){\mathcal{X}}_{\left[0,1\right)}\left(x\right).\end{array}$$

Similarly, the quadratic Hermite splines are defined as

$${\xi }_1^{-}\left(x\right)=:\left(-6x-6x^2\right){\mathcal{X}}_{\left[-1,0\right]},{\xi }_2^{-}\left(x\right)=:\left(1+4x+3x^2\right){\mathcal{X}}_{\left[-1,0\right]}\left(x\right)+\left(1-4x+3x^2\right){\mathcal{X}}_{\left[0,1\right)}\left(x\right).$$

Let $Z_j^0=:\left\{0,1,...,2^j\right\},Z_j^1=:\left\{1,2,...,2^j\right\},Z_j^2=:\left\{0,1,...,2^j-1\right\}$, and $Z_j^3=:\left\{1,2,...,2^j-1\right\}$. For each j ≥ j₀, define the scaling functions on [0,1]:

$$\begin{array}{c}{\xi }_{m;j,k}^{\text{Δ},+}=:{\xi }_m^{+}\left(2^{j}x-k\right){\mathcal{X}}_{\left[0,1\right]},k\in Z_j^0;\\ {\xi }_{1;j,k}^{\text{Δ},-}=:{\xi }_1^{-}\left(2^{j}x-k\right){\mathcal{X}}_{\left[0,1\right]}={\xi }_1^{-}\left(2^{j}x-k\right),k\in Z_j^1;{\xi }_{2;j,k}^{\text{Δ}-}=:{\xi }_2^{-}\left(2^{j}x-k\right){\mathcal{X}}_{\left[0,1\right]},k\in Z_j^0.\end{array}$$

Let $V_j^{\text{Δ},+}=:\text{span}\left\{{\xi }_{1;j,k}^{\text{Δ},+},{\xi }_{2;j,k}^{\text{Δ},+}:k\in Z_j^0\right\},V_j^{\text{Δ},-}=:\text{span}\left\{{\xi }_{1;j,k_1}^{\text{Δ},-},{\xi }_{2;j,k_2}^{\text{Δ},-}:k_1\in Z_j^1,k_2\in Z_j^0\right\}$, then $\left\{V_j^{\text{Δ},+}\right\}$ and $\left\{V_j^{\text{Δ},-}\right\}$ are two MRAs on L²([0,1]) [8]. The corresponding duals ${\tilde{\xi }}_{m;j,k}^{\text{Δ},\pm }$ are given in the sense of distributions:

$$\begin{array}{c}{\tilde{\xi }}_1^{+}=:{\delta }_0,⟨f,{\tilde{\xi }}_{1;j,k}^{\text{Δ},+}⟩=f\left(2^{-j}k\right),k\in Z_j^0;{\tilde{\xi }}_2^{+}=:-{{\delta }^{\prime }}_0,⟨f,{\tilde{\xi }}_{2;j,k}^{\text{Δ},+}⟩=2^{-j}{f}^{\prime }\left(2^{-j}k\right),k\in Z_j^0;\\ {\tilde{\xi }}_1^{-}=:{\mathcal{X}}_{\left[-1,0\right]},⟨f,{\tilde{\xi }}_{1;j,k}^{\text{Δ},-}⟩=2^j\underset{2^{-j}\left(k-1\right)}{\overset{2^{-j}k}{\int }}f\left(x\right)dx,k\in Z_j^1;{\tilde{\xi }}_2^{-}=:{\delta }_0,⟨f,{\tilde{\xi }}_{2;j,k}^{\text{Δ},-}⟩=f\left(2^{-j}k\right),k\in Z_j^0.\end{array}$$

The inperpolating multi-wavelets ${\eta }_{m;j,k}^{\text{Δ},\pm }$ on [0,1] as well as the wavelet spaces are defined by

$${\eta }_{m;j,k}^{\text{Δ},\pm }\left(x\right)=:{\eta }_{m;j,k}^{\pm }\left(x\right){\mathcal{X}}_{\left[0,1\right]}={\eta }_{m;j,k}^{\pm }\left(x\right),k\in Z_j^2;W_j^{\text{Δ},\pm }=:\text{span}\left\{{\eta }_{m;j,k}^{\text{Δ},\pm }:m=1,2,k\in Z_j^2\right\}$$

with ${\eta }_m^{+}=:{\xi }_m^{+}\left(2\cdot -1\right),m=1,2;{\eta }_1^{-}=:{\xi }_1^{-}\left(2\cdot -1\right)-{\xi }_1^{-}\left(2\cdot -2\right);{\eta }_2^{-}=:{\xi }_2^{-}\left(2\cdot -1\right)$. Here and after, h _j,k (·) = h(2^j· -k). The corresponding duals are given by

$$\begin{array}{c}{\tilde{\eta }}_1^{+}=:{\delta }_{\frac{1}{2}}-\frac{1}{2}{\delta }_0-\frac{1}{2}{\delta }_1+\frac{1}{8}{{\delta }^{\prime }}_0-\frac{1}{8}{{\delta }^{\prime }}_1,{\tilde{\eta }}_2^{+}=:\frac{3}{4}{\delta }_0-\frac{3}{4}{\delta }_1-\frac{1}{2}{{\delta }^{\prime }}_{\frac{1}{2}}-\frac{1}{8}{{\delta }^{\prime }}_0-\frac{1}{8}{{\delta }^{\prime }}_1,\\ {\tilde{\eta }}_1^{-}=:{\mathcal{X}}_{\left[0,\frac{1}{2}\right]}-{\mathcal{X}}_{\left[\frac{1}{2},1\right]}-\frac{1}{4}{\delta }_0+\frac{1}{4}{\delta }_1,{\tilde{\eta }}_2^{-}=:{\delta }_{\frac{1}{2}}+\frac{1}{4}{\delta }_0+\frac{1}{4}{\delta }_1-\frac{3}{2}{\mathcal{X}}_{\left[0,1\right]}\end{array}$$

and $⟨f,{\tilde{\eta }}_{m;j,k}^{\text{Δ},\pm }⟩=⟨f\left(\frac{\cdot +k}{2^j}\right),{\tilde{\eta }}_m^{\pm }⟩$. Moreover, there is the following differential relations

$$\begin{array}{c}\frac{d}{dx}{\xi }_{1;j,k}^{\text{Δ},+}\left(x\right)=2^j\left({\xi }_{1;j,k}^{\text{Δ},-}\left(x\right)-{\xi }_{1;j,k}^{\text{Δ},-}\left(x-2^{-j}\right)\right),k\in Z_j^3;\frac{d}{dx}{\xi }_{1;j,0}^{\text{Δ},+}\left(x\right)=-2^j{\xi }_{1;j,1}^{\text{Δ},-}\left(x\right),\\ \frac{d}{dx}{\xi }_{1;j,2^j}^{\text{Δ},+}=2^j{\xi }_{1;j,2^j}^{\text{Δ},-};\frac{d}{dx}{\xi }_{2;j,k}^{\text{Δ},+}=2^j{\xi }_{2;j,k}^{\text{Δ},-},k\in Z_j^0;\frac{d}{dx}{\eta }_{m;j,k}^{\text{Δ},+}=2^{j+1}{\eta }_{m;j,k}^{\text{Δ},-},k\in Z_j^2.\\ \frac{d}{dx}{\tilde{\xi }}_{1;j,k}^{\text{Δ},-}=-2^j\left(\frac{d}{dx}{\tilde{\xi }}_{1;j,k}^{\text{Δ},+}-\frac{d}{dx}{\tilde{\xi }}_{1;j,k-1}^{\text{Δ},+}\right),\frac{d}{dx}{\tilde{\xi }}_{2;j,k}^{\text{Δ},-}=-2^j\frac{d}{dx}{\tilde{\xi }}_{2;j,k}^{\text{Δ},+}.\end{array}$$

(1.1)

2 Curl-free wavelets on the cube

For $\vec{u}\left(x,y,z\right)={\left(u_1,u_2,u_3\right)}^T$, the 3D curl-operator is defined as

$$\text{curl}\vec{u}={\left({\partial }_2{u}_3-{\partial }_3{u}_2,{\partial }_3{u}_1-{\partial }_1{u}_3,{\partial }_1{u}_2-{\partial }_2{u}_1\right)}^T.$$

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

$${\phi }_m^I\left(x_1,x_2,x_3\right)=:{\displaystyle \prod _{v=1}^3}{\xi }_{m_v,v}^I\left(x_v\right),m={\left(m_1,m_2,m_3\right)}^T\in {\left\{1,2\right\}}^3$$

with ${\xi }_{\mu ,v}^I=\left\{\begin{array}{c}\hfill {\xi }_{\mu }^{+},v\in I\hfill \\ \hfill {\xi }_{{\mu }^{}}^{-},v\notin I.\hfill \end{array}\right.$

The corresponding wavelets are

$${\psi }_{e,m}^I\left(x_1,x_2,x_3\right)=:{\displaystyle \prod _{v=1}^3}{\vartheta }_{e_v,m_v,v}^I\left(x_v\right),e\in E_3^{*},m={\left(m_1,m_2,m_3\right)}^T\in {\left\{1,2\right\}}^3.$$

Here and after, $E_3^{*}$ denotes the non-zero apexes of the unite cube and ${\vartheta }_{\ell ,\mu ,v}^I=\left\{\begin{array}{c}\hfill {\xi }_{\mu ,v}^I,\ell =0,\hfill \\ \hfill {\eta }_{\mu ,v}^I,\ell =1.\hfill \end{array}\right.$

Let ${\phi }_{m;j,k}^{\text{Δ},I}=:{\phi }_{m;j,k}^I{\mathcal{X}}_{{\left[0,1\right]}^3}$, which is the tensor product of corresponding interpolatory scaling functions on the interval. The corresponding duals are given similarly. Furthermore, define

$${\vec{\phi }}_{m,i;j,k}^{\text{Δ}}=:{\phi }_{m;j,k}^{\text{Δ},I_0\backslash \left\{i\right\}}{\delta }_i,{\vec{V}}_j^{\text{Δ}}=:\text{span}\left\{{\vec{\phi }}_{m,i;j,k}^{\text{Δ}}:m\in {\left\{1,2\right\}}^3,k\in {\nabla }_j,1\le i\le 3\right\}$$

and the projection operators:

$${\overrightarrow{\text{Λ}}}_j^{\text{Δ}}=:{\text{Λ}}_j^{\text{Δ},\left\{2,3\right\}}{\delta }_1+{\text{Λ}}_j^{\text{Δ},\left\{1,3\right\}}{\delta }_2+{\text{Λ}}_j^{\text{Δ},\left\{1,2\right\}}{\delta }_3,{\overrightarrow{\text{Λ}}}_j^{\text{Δ},*}=:{\text{Λ}}_j^{\text{Δ},\left\{1\right\}}{\delta }_1+{\text{Λ}}_j^{\text{Δ},\left\{2\right\}}{\delta }_2+{\text{Λ}}_j^{\text{Δ},\left\{3\right\}}{\delta }_3.$$

Lemma 2.1[8]. For smooth functions $f:{\left[0,1\right]}^3\to R,\frac{\partial }{\partial x_i}{\text{Λ}}_j^{\text{Δ},I}f={\text{Λ}}_j^{\text{Δ},I\backslash \left\{i\right\}}\left(\frac{\partial f}{\partial x_i}\right),i\in I.$.

Proposition 2.1. For $\vec{f}\in \vec{C}\left(\text{curl};{\left[0,1\right]}^3\right)=:\left\{\vec{\upsilon }\in {\left(C\left({\left[0,1\right]}^3\right)\right)}^3:\text{curl}\vec{\upsilon }\in {\left(C\left({\left[0,1\right]}^3\right)\right)}^3\right\}$, there has $\text{curl}\left({\overrightarrow{\text{Λ}}}_j^{\text{Δ}}\vec{f}\right)={\overrightarrow{\text{Λ}}}_j^{\text{Δ},*}\left(\text{curl}\vec{f}\right)$.

Proof. Note that Lemma 2.1, then

$$\begin{array}{cc}\hfill {\overrightarrow{\text{Λ}}}_j^{\text{Δ},*}\left(\text{curl}\vec{f}\right)& ={\text{Λ}}_j^{\text{Δ},\left\{1\right\}}\left(\frac{\partial f_3}{\partial x_2}-\frac{\partial f_2}{\partial x_3}\right){\delta }_1+{\text{Λ}}_j^{\text{Δ},\left\{2\right\}}\left(\frac{\partial f_1}{\partial x_3}-\frac{\partial f_3}{\partial x_1}\right){\delta }_2+{\text{Λ}}_j^{\text{Δ},\left\{3\right\}}\left(\frac{\partial f_2}{\partial x_1}-\frac{\partial f_1}{\partial x_2}\right){\delta }_3\hfill \\ =\left(\frac{\partial }{\partial x_2}{\text{Λ}}_j^{\text{Δ},\left\{1,2\right\}}{f}_3-\frac{\partial }{\partial x_3}{\text{Δ}}_j^{\text{Δ},\left\{1,3\right\}}{f}_2\right){\delta }_1+\left(\frac{\partial }{\partial x_3}{\text{Λ}}_j^{\text{Δ},\left\{2,3\right\}}{f}_1-\frac{\partial }{\partial x_1}{\text{Λ}}_j^{\text{Δ},\left\{1,2\right\}}{f}_3\right){\delta }_2\hfill \\ +\left(\frac{\partial }{\partial x_1}{\text{Λ}}_j^{\text{Δ},\left\{1,3\right\}}{f}_2-\frac{\partial }{\partial x_2}{\text{Λ}}_j^{\text{Δ},\left\{2,3\right\}}{f}_1\right){\delta }_3,\hfill \end{array}$$

which is $\text{curl}\left({\overrightarrow{\text{Λ}}}_j^{\text{Δ}}\vec{f}\right)$ by definition.

Proposition 2.1 is important, because it tells us that ${\overrightarrow{\text{Λ}}}_j^{\text{Δ}}$ keeps curl-free property. In general, vector-valued wavelets and wavelet spaces are given, respectively, by

$${\vec{\psi }}_{e,m,i;j,k}^{\text{Δ}}=:{\psi }_{e,m;j,k}^{\text{Δ},I_0\backslash \left\{i\right\}}{\delta }_i,{\vec{W}}_j^{\text{Δ}}=:\text{span}\left\{{\vec{\psi }}_{e,m,i;j,k}^{\text{Δ}}:e\in E_3^{*},m\in {\left\{1,2\right\}}^3,1\le i\le 3,k\in {\text{∇}}_j^0\right\}.$$

For $e\in E_3^{*}$ and m ∈ {1, 2}³, we define

$${\vec{\psi }}_{e,m;j,k}^{\text{Δ},c}=:2^{-j}\text{grad}{\psi }_{e,m;j,k}^{\text{Δ},I_0}=2^{-j}\frac{\partial }{\partial x_1}{\psi }_{e,m;j,k}^{\text{Δ},I_0}{\delta }_1+2^{-j}\frac{\partial }{\partial x_2}{\psi }_{e,m;j,k}^{\text{Δ},I_0}{\delta }_2+2^{-j}\frac{\partial }{\partial x_3}{\psi }_{e,m;j,k}^{\text{Δ},I_0}{\delta }_3,k\in {\nabla }_j^1.$$

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

To give a decomposition for ${\vec{W}}_j^{\text{Δ}}$, take

$${\vec{\psi }}_{e,m,i;j,k}^{\text{Δ},\text{non}}=:{\psi }_{e,m;j,k}^{\text{Δ},I_0\backslash \left\{i\right\}}{\delta }_i,k\in {\text{∇}}_j^2$$

for i ∈ I₀\{i _e }. Here, we choose i _e such that $e_{i_e}=1$.

Proposition 2.2. The vector-valued function system $\left\{{\vec{\psi }}_{e,m;j,k_1}^{\text{Δ},c},{\vec{\psi }}_{e,m,i;j,k_2}^{\text{Δ},\text{non}},e\in E_3^{*},m\in {\left\{1,2\right\}}^3,i\ne i_e,k_1\in {\nabla }_j^1,k_2\in {\nabla }_j^2\right\}$ is complete in ${\vec{W}}_j^{\text{Δ}}$.

Proof. It is sufficient to show the statement for j = 0. Let $\vec{f}\in {\vec{W}}_0^{\text{Δ}}$ satisfy

$$\left(\vec{f},{\vec{\psi }}_{e,m;0,k}^{\text{Δ},c}\right)=\left(\overrightarrow{f},{\vec{\psi }}_{e,m,i;0,k}^{\text{Δ},\text{non}}\right)=0$$

for all $e\in E_3^{*},m\in {\left\{1,2\right\}}^3,k\in {\nabla }_j^1\cup {\nabla }_j^2\},1\le i\le 3$and i ≠ i _e . Here, the inner product is in L²([0,1]³). Without loss of generality, one assumes i _e = 1. Then, $\left(\overrightarrow{f},{\vec{\psi }}_{e,m,i;0,k}^{\text{Δ},\text{non}}\right)=0$ leads to

$$\left(f_2,{\psi }_{e,m;0,k}^{\text{Δ},I_0\backslash \left\{2\right\}}\right)=\left(f_3,{\psi }_{e,m;0,k}^{\text{Δ},I_0\backslash \left\{3\right\}}\right)=0.$$

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

$$\left(f_2,\frac{\partial }{\partial x_2}{\psi }_{e,m;0,k}^{\text{Δ},I_0}\right)=\left(f_3,\frac{\partial }{\partial x_3}{\psi }_{e,m;0,k}^{\text{Δ},I_0}\right)=0.$$

Moreover, $\left(\overrightarrow{f},{\vec{\psi }}_{e,m;0,k}^{\text{Δ},c}\right)=0$ reduces to $\left(f_1,\frac{\partial }{\partial x_1}{\psi }_{e,m;0,k}^{\text{Δ},I_0}\right)=0$. Now, it follows that $\left(f_1,{\psi }_{e,m;0,k}^{\text{Δ},I_0\backslash \left\{1\right\}}\right)=0$ from i _e = 1. Finally, $\left(\overrightarrow{f},{\vec{\psi }}_{e,m,i;0,k}^{\text{Δ}}\right)=0$ and $\vec{f}=\vec{0}$ follows from the definition of ${\vec{W}}_0^{\text{Δ}}$.

To give the bi-orthogonal decomposition, we define

$${\overrightarrow{\stackrel{̃}{\psi }}}_{e,m;j,k}^{\text{Δ},c}=\frac{1}{2}{\tilde{\psi }}_{e,m;j,k}^{\text{Δ},I_0\backslash \left\{i_e\right\}}{\delta }_{i_e}.$$

(2.1)

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

$$\text{curl}{\vec{\psi }}_{e,m,i;j,k}^{\text{Δ},\text{non}}=\text{curl}{\psi }_{e,m;j,k}^{\text{Δ},I_0\backslash \left\{i\right\}}{\delta }_i=2^{j+1}{\epsilon }_1{\psi }_{e,m;j,k}^{\text{Δ},I_0\backslash \left\{i,i_e\right\}}{\delta }_{i^{\prime }}+{\epsilon }_2\frac{\partial }{\partial x_{i^{\prime }}}{\psi }_{e,m;j,k}^{\text{Δ},I_0\backslash \left\{i\right\}}{\delta }_{i_e}$$

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

$${\overrightarrow{\stackrel{̃}{\psi }}}_{e,m,i;j,k}^{\text{Δ},\text{non}}=:\frac{1}{2^{j+1}}\text{curl}{\epsilon }_1{\tilde{\psi }}_{e,m;j,k}^{\text{Δ},I_0\backslash \left\{i,i_e\right\}}{\delta }_{i^{\prime }}.$$

(2.2)

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

Proposition 2.3. The set $\left\{{\vec{\psi }}_{e,m;j,k_1}^{\text{Δ},c},{\vec{\psi }}_{e,m,i;j,k_2}^{\text{Δ},\text{non}},e\in E_3^{*},m\in {\left\{1,2\right\}}^3,i\ne i_e,j\ge j_0,k_1\in {\nabla }_j^1,k_2\in {\nabla }_j^2\right\}$ 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) $⟨{\vec{\psi }}_{e^{\prime },m^{\prime },i;j^{\prime },k^{\prime }}^{\text{Δ},\text{non}},{\overrightarrow{\stackrel{̃}{\psi }}}_{e,m;j,k}^{\text{Δ},c}⟩=0$;

(ii) $⟨{\vec{\psi }}_{e,m;j,k}^{\text{Δ},c},{\overrightarrow{\stackrel{̃}{\psi }}}_{e^{\prime },m^{\prime },i;j^{\prime },k^{\prime }}^{\text{Δ},\text{non}}⟩=0$;

(iii) $⟨{\vec{\psi }}_{e,m;j,k}^{\text{Δ},c},{\overrightarrow{\stackrel{̃}{\psi }}}_{e^{\prime },m^{\prime };j^{\prime },k^{\prime }}^{\text{Δ},c}⟩={\delta }_{e,e^{\prime }}{\delta }_{m,m^{\prime }}{\delta }_{j,j^{\prime }}{\delta }_{k,k^{\prime }}$;

(iv) $⟨{\vec{\psi }}_{e_1,m_1,i_1;j_1,k_1}^{\text{Δ},\text{non}},{\overrightarrow{\stackrel{̃}{\psi }}}_{e_2,m_2,i_2;j_2,k_2}^{\text{Δ},\text{non}}⟩={\delta }_{e_1,e_2}{\delta }_{m_1,m_2}{\delta }_{i_1,i_2}{\delta }_{j_1,j_2}{\delta }_{k_1,k_2}$.

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 ${\psi }_{e^{\prime },m^{\prime };j^{\prime },k^{\prime }}^{\text{Δ},I_0\backslash \left\{i\right\}}$ and ${\tilde{\psi }}_{e,m;j,k}^{\text{Δ},I_0\backslash \left\{i\right\}}$.

Note that $⟨\vec{f},\text{curl}\vec{g}⟩=⟨\text{curl}\vec{f},\vec{g}⟩$. Then (ii) follows from curl⋅grad = 0. Furthermore,

$$⟨{\vec{\psi }}_{e,m;j,k}^{\text{Δ},c},{\overrightarrow{\stackrel{̃}{\psi }}}_{e^{\prime },m^{\prime };j^{\prime },k^{\prime }}^{\text{Δ},c}⟩=\frac{1}{2}\times 2^{-j}⟨\frac{\partial }{\partial x_{i_{e^{\prime }}}}{\psi }_{e,m;j,k}^{\text{Δ},I_0},{\tilde{\psi }}_{e^{\prime },m^{\prime };j^{\prime },k^{\prime }}^{\text{Δ},I_0\backslash \left\{i_{e^{\prime }}\right\}}⟩=-\frac{1}{2}\times 2^{-j}⟨{\psi }_{e,m;j,k}^{\text{Δ},I_0},\frac{\partial }{\partial x_{i_{e^{\prime }}}}{\tilde{\psi }}_{e^{\prime },m^{\prime };j^{\prime },k^{\prime }}^{\text{Δ},I_0\backslash \left\{i_{e^{\prime }}\right\}}⟩.$$

Then by the fact $\frac{d}{dx}{\tilde{\eta }}_m^{-}=-2{\tilde{\eta }}_m^{+}$ and the bi-orthogonality of ${\psi }_{e,m}^{\text{Δ},I_0},{\tilde{\psi }}_{e,m}^{\text{Δ},I_0}$, one obtains

$$⟨{\vec{\psi }}_{e,m;j,k}^{\text{Δ},c},{\overrightarrow{\stackrel{̃}{\psi }}}_{e^{\prime },m^{\prime };j^{\prime },k^{\prime }}^{\text{Δ},c}⟩=2^{j^{\prime }-j}⟨{\psi }_{e,m;j,k}^{\text{Δ},I_0},{\tilde{\psi }}_{e^{\prime },m^{\prime };j^{\prime },k^{\prime }}^{\text{Δ},I_0}⟩={\delta }_{e,e^{\prime }}{\delta }_{m,m^{\prime }}{\delta }_{j,j^{\prime }}{\delta }_{k,k^{\prime }}.$$

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

$$\begin{array}{c}{2}^{j_1-j_2}⟨2{\epsilon }_1{\psi }_{e_1,m_1;j_1,k_1}^{\text{Δ},I_0\backslash \left\{i_1,i_{e_1}\right\}}{\delta }_{{i^{\prime }}_1}+2^{-j_1}{\epsilon }_2\frac{\partial }{\partial x_{{i^{\prime }}_1}}{\psi }_{e_1,m_1;j_1,k_1}^{\text{Δ},I_0\backslash \left\{i_1\right\}}{\delta }_{i_{e_1}},\frac{1}{2}{\epsilon }_1{\tilde{\psi }}_{e_2,m_2;j_2,k_2}^{\text{Δ},I_0\backslash \left\{i_2,i_{e_2}\right\}}{\delta }_{{i^{\prime }}_2}⟩\\ ={\delta }_{e_1,e_2}{\delta }_{m_1,m_2}{\delta }_{i_1,i_2}{\delta }_{j_1,j_2}{\delta }_{k_1,k_2}.\end{array}$$

(2.3)

It is easily proved, when e₁ = e₂: In fact, since $i_{e_1}=i_{e_2}$, one can assume i₁ = i₂ and $i_1^{\prime }=i_2^{\prime }$, because $i_1=i_2^{\prime }$ leads to (2.3) obviously. In that case, the left-hand side of (2.3) reduces to $2{\epsilon }_1\cdot \frac{{\epsilon }_1}{2}\cdot 2^{j_1-j_2}⟨{\psi }_{e_1,m_1;j_1,k_1}^{I\backslash \left\{i_1,i_{e_1}\right\}},{\tilde{\psi }}_{e_2,m_2;j_2,k_2}^{I\backslash \left\{i_2,i_{e_2}\right\}}⟩={\delta }_{m_1,m_2}{\delta }_{j_1,j_2}{\delta }_{k_1,k_2}$, which is the desired. To the end, it is sufficient to prove that for e₁ ≠ e₂, that is

$$⟨2{\epsilon }_1{\psi }_{e_1,m_1;j_1,k_1}^{\text{Δ},I_0\backslash \left\{i_1,i_{e_1}\right\}}{\delta }_{{i^{\prime }}_1}+2^{-j_1}{\epsilon }_2\frac{\partial }{\partial x_{{i^{\prime }}_1}}{\psi }_{e_1,m_1;j_1,k_1}^{\text{Δ},I_0\backslash \left\{i_1\right\}}{\delta }_{i_{e_1}},\frac{1}{2}{\epsilon }_1{\tilde{\psi }}_{e_2,m_2;j_2,k_2}^{\text{Δ},I_0\backslash \left\{i_2,i_{e_2}\right\}}{\delta }_{{i^{\prime }}_2}⟩=0.$$

(2.4)

Note that $i_2^{\prime }\in \left\{i_1,i_1^{\prime },i_{e_1}\right\}$. Then the conclusion is obvious when $i_2^{\prime }=i_1$. When $i_2^{\prime }=i_1^{\prime }$, then $\left\{i_1,i_{e_1}\right\}=\left\{i_2,i_{e_2}\right\}$ and the left-hand side of (2.4) reduces to $2{\epsilon }_1\cdot \frac{{\epsilon }_1}{2}⟨{\psi }_{e_1,m_1;j_1,k_1}^{\text{Δ},I_0\backslash \left\{i_1,i_{e_1}\right\}},{\tilde{\psi }}_{e_2,m_2;j_2,k_2}^{\text{Δ},I_0\backslash \left\{i_2,i_{e_2}\right\}}⟩=0$. Hence one only need to show (2.4), when $i_2^{\prime }=i_{e_1}$. However, (2.4) becomes

$$⟨\frac{\partial }{\partial x_{{i^{\prime }}_1}}{\psi }_{e_1,m_1;j_1,k_1}^{\text{Δ},I_0\backslash \left\{i_1\right\}},{\tilde{\psi }}_{e_2,m_2;j_2,k_2}^{\text{Δ},I_0\backslash \left\{i_2,i_{e_2}\right\}}⟩=0.$$

(2.5)

in that case. Since $\left\{i_1,i_1^{\prime },i_{e_1}\right\}=\left\{i_2,i_2^{\prime },i_{e_2}\right\}=I$, two cases should be considered: $i_2=i_1,i_2^{\prime }=i_{e_1},i_{e_2}=i_1^{\prime }$ and $i_2=i_1^{\prime },i_2^{\prime }=i_{e_1},i_{e_2}=i_1$. Using $\frac{d}{dx}{\tilde{\eta }}_m^{-}=-2{\tilde{\eta }}_m^{+}$, the left-hand side of (2.5) is

$$-⟨{\psi }_{e_1,m_1,j_1,k_1}^{\text{Δ},I_0\backslash \left\{i_1\right\}},\frac{\partial }{\partial x_{{i^{\prime }}_1}}{\tilde{\psi }}_{e_2,m_2,j_2,k_2}^{\text{Δ},I_0\backslash \left\{i_2,i_{e_2}\right\}}⟩=2^{j_2+1}⟨{\psi }_{e_1,m_1;j_1,k_1}^{\text{Δ},I_0\backslash \left\{i_1\right\}},{\tilde{\psi }}_{e_2,m_2;j_2,k_2}^{\text{Δ},I_0\backslash \left\{i_2\right\}}⟩=0$$

in the first case; In the second one, the left-hand side of (2.5) becomes $⟨\frac{\partial }{\partial x_{i_2}}{\psi }_{e_1,m_1;j_1,k_1}^{\text{Δ},I_0\backslash \left\{i_1\right\}}{\tilde{\psi }}_{e_2,m_2;j_2,k_2}^{\text{Δ},I_0\backslash \left\{i_2,i_2\right\}}⟩$. According to the differential relation (1.1), $\frac{\partial }{\partial x_{i_2}}{\psi }_{e_1,m_1;j_1,k_1}^{\text{Δ},I_0\backslash \left\{i_1\right\}}$ is a linear combination of ${\psi }_{e_1,m_1;j_1,k_1}^{\text{Δ},I_0\backslash \left\{i_1,i_2\right\}}$. By the bi-orthogonality of ${\psi }_{e_1,m_1;j_1,k_1}^{\text{Δ},I_0\backslash \left\{i_1,i_2\right\}}$ and ${\tilde{\psi }}_{e_2,m_2;j_2,k_2}^{\text{Δ},I_0\backslash \left\{i_1,i_2\right\}}$, 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 $B_q^s\left(L^p\left(\Omega \right)\right)$ is the set of all f ∈ L^p(Ω) such that

$${\left|f\right|}_{B_q^s\left(L^p\left(\Omega \right)\right)}=:{∥\left\{2^{sj}{\omega }_m\left(f,2^{-j},L^p\left(\Omega \right)\right)\right\}∥}_{{\ell }^q}<+\infty$$

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

$${∥f∥}_{B_q^s\left(L^p\left(\Omega \right)\right)}=:{∥f∥}_{L^p\left(\Omega \right)}+{\left|f\right|}_{B_q^s\left(L^p\left(\Omega \right)\right)}.$$

Our Besov space is defined as

$${\hat{B}}_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)=:\left\{\vec{f}\in {\left(B_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)\right)}^3:\frac{\partial }{\partial x_j}{f}_i\in B_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right),i=1,2,3;j\ne i\right\}$$

with the norm

$${∥\vec{f}∥}_{{\hat{B}}_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)}=:{\displaystyle \sum _{i=1}^3}{∥f_i∥}_{B_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)}+{\displaystyle \sum _{i=1}^3}\sum _{\begin{array}{c}1\le j\le 3\\ j\ne i\end{array}}{∥\frac{\partial f_i}{\partial x_j}∥}_{B_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)}.$$

Clearly, $\text{curl}\vec{f}\in {\left(B_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)\right)}^3$, when $\vec{f}\in {\hat{B}}_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)$.

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

Lemma 3.1. If $f\left(x\right),g\left(x\right)\in B_q^s\left(L^p\left(R\right)\right)$, then $f\left(x_1\right)g\left(x_2\right)\in B_q^s\left(L^p\left(R^2\right)\right)$.

For $\alpha ={\left({\alpha }_j\right)}_{j\ge j_0}$ and α _j = (α _j,k ) _k , define

$${∥\alpha ∥}_{{\ell }_{p,q}^s}=:{∥\left(2^{j\left(s-\frac{n}{p}\right)}{∥{\alpha }_j∥}_{{\ell }^p}\right)∥}_{{\ell }^q}.$$

Lemma 3.2[8]. If $\varphi \in B_{\infty }^{\sigma }\left(L^p\left(R^n\right)\right)$ is compactly supported, 0 < p, q ≤ ∞ and 0 < s < σ, then

$$\begin{array}{c}{∥\sum _{k\in {\nabla }_j}{\beta }_k\varphi \left(2^j\cdot -k\right)∥}_{B_q^s\left(L^p\left({\left[0,1\right]}^n\right)\right)}\lesssim 2^{\left(s-\frac{n}{p}\right)j}{∥\beta ∥}_{{\ell }^p},\\ {∥\sum _{j\ge j_0}\sum _{k\in {\nabla }_j}{\alpha }_{j,k}\varphi \left(2^j\cdot -k\right)∥}_{B_q^s\left(L^p\left({\left[0,1\right]}^n\right)\right)}\lesssim {∥\alpha ∥}_{{\ell }_{p,q}^s},\end{array}$$

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

Theorem 3.1. Let ${\vec{\phi }}_{m,i;j,k}^{\text{Δ}},{\vec{\psi }}_{e,m;j,k}^{\text{Δ},c}$, and ${\vec{\psi }}_{e,m,i;j,k}^{\text{Δ},\text{non}}$ be defined in Section 2. If 0 $<s<1+\frac{1}{p}$ and 0 < p, q ≤ ∞, then one has

$$\begin{array}{c}{∥{\displaystyle \sum _{m,i}}{\displaystyle \sum _{k\in {\nabla }_{j_0}}}{\beta }_{m,i,k}{\vec{\phi }}_{m,i;j_0,k}^{\text{Δ}}+{\displaystyle \sum _{j\ge j_0}}{\displaystyle \sum _{e,m,i\ne i_e}}{\displaystyle \sum _{k\in {\nabla }_j^1}}{\alpha }_{e,m;j,k}^c{\vec{\psi }}_{e,m;j,k}^{\text{Δ},c}+{\displaystyle \sum _{j\ge j_0}}{\displaystyle \sum _{e,m}}{\displaystyle \sum _{k\in {\nabla }_j^2}}{\alpha }_{e,m,i;j,k}^{\text{non}}{\vec{\psi }}_{e,m,i;j,k}^{non}∥}_{{\hat{B}}_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)}\\ \lesssim {\displaystyle \sum _{m,i}}{∥{\beta }_{m,i}∥}_{{\ell }^p}+{\displaystyle \sum _{e,m}}\left({\displaystyle \sum _{i\ne i_e}}{∥{\alpha }_{e,m,i}^{\text{non}}∥}_{{\ell }_{p,q}^{s+1}}+{∥{\alpha }_{e,m}^c∥}_{{\ell }_{p,q}^{s+1}}\right).\end{array}$$

Proof. It is enough to prove the following inequality:

• (i) ${∥{\displaystyle \sum _{k\in {\nabla }_{j_0}}}{\beta }_{m,i,k}{\vec{\phi }}_{m,i;j_0,k}^{\text{Δ}}∥}_{{\hat{B}}_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)}\lesssim {∥{\beta }_{m,i}∥}_{\ell p}$;

• (ii) ${∥{\displaystyle \sum _{j\ge j_0}}{\displaystyle \sum _{k\in {\nabla }_j^1}}{\alpha }_{e,m;j,k}^c{\vec{\psi }}_{e,m;j,k}^{\text{Δ},c}∥}_{{\hat{B}}_q^s(L^p\left({\left[0,1\right]}^3\right)}\lesssim {∥{\alpha }_{e,m}^c∥}_{{\ell }_{p,q}^{s+1}}$;

• (iii) ${∥{\displaystyle \sum _{j\ge j_0}}{\displaystyle \sum _{k\in {\nabla }_j^2}}{\alpha }_{e,m,i;j,k}^{\text{non}}{\vec{\psi }}_{e,m,i;j,k}^{non}∥}_{{\hat{B}}_q^s(L^p\left({\left[0,1\right]}^3\right)}\lesssim {∥{\alpha }_{e,m,i}^{\text{non}}∥}_{{\ell }_{p,q}^{s+1}}\left(i\ne i_e\right)$.

Let $\vec{h}=:{\displaystyle \sum _{k\in {\nabla }_{j_0}}}{\beta }_{m,i,k}{\vec{\phi }}_{m,i;j_0,k}^{\text{Δ}}$ and h _ν be the ν th component of $\vec{h}$. Then, for μ ≠ i,

$$h_i={\displaystyle \sum _{k\in {\nabla }_{j_0}}}{\beta }_{m,i;k}{\phi }_m^{I_0\backslash \left\{i\right\}}\left(2^{j_0}x-k\right),\frac{\partial }{\partial x_{\mu }}{h}_i={\displaystyle \sum _{k\in {\nabla }_{j_0}}}{\beta }_{m,i;k}{2}^{j_0}\left(\frac{\partial }{\partial x_{\mu }}{\phi }_m^{I_0\backslash \left\{i\right\}}\right)\left(2^{j_0}x-k\right).$$

Since ${\xi }_m^{+}\in B_{\infty }^{2+\frac{1}{p}}\left(L^p\left(R\right)\right)\subseteq B_{\infty }^{1+\frac{1}{p}}\left(L^p\left(R\right)\right)$ and ${\xi }_m^{-}\in B_{\infty }^{1+\frac{1}{p}}\left(L^p\left(R\right)\right)$, then both ${\phi }_m^{I\backslash \left\{i\right\}}$ and $\frac{\partial }{\partial x_{\mu }}{\phi }_m^{I\backslash \left\{i\right\}}$ are in the Besov space $B_{\infty }^{1+\frac{1}{p}}\left(L^p\left(R^3\right)\right)$, due to Lemma 3.1. Moreover, Lemma 3.2 implies ${∥h_i∥}_{B_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)}\lesssim {∥{\beta }_{m,i}∥}_{{\ell }^p}$ and ${∥\frac{\partial }{\partial x_{\mu }}{h}_i∥}_{B_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)}\lesssim {∥{\beta }_{m,i}∥}_{{\ell }^p}$. Note that h _ν = 0 for ν ≠ i. Finally, the first inequality follows from the definition.

Let $\vec{g}=:{\displaystyle \sum _{j\ge j_0}}{\displaystyle \sum _{k\in {\nabla }_j^1}}{\alpha }_{e,m;j,k}^c{\vec{\psi }}_{e,m;j,k}^{\text{Δ},c}$ and g _ν be the ν th component of $\vec{g}\left(1\le v\le 3\right)$. Then

$$\begin{array}{c}{g}_v={\displaystyle \sum _{j\ge j_0}}{\displaystyle \sum _{k\in {\nabla }_j^1}}{\alpha }_{e,m;j,k}^c\left(\frac{\partial }{\partial x_v}{\psi }_{e,m}^{I_0}\right)\left(2^{j}x-k\right),\\ \frac{\partial }{\partial x_{\mu }}{g}_v={\displaystyle \sum _{j\ge j_0}}{\displaystyle \sum _{k\in {\nabla }_j^1}}{2}^j{\alpha }_{e,m;j,k}^c\left(\frac{{\partial }^2}{\partial x_{\mu }\partial x_v}{\psi }_{e,m}^{I_0}\right)\left(2^{j}x-k\right).\end{array}$$

Similar to the above, $\frac{\partial }{\partial x_v}{\psi }_{e,m}^{I_0},\frac{{\partial }^2}{\partial x_v\partial x_{\mu }}{\psi }_{e,m}^{I_0}\in B_{\infty }^{1+\frac{1}{p}}\left(L^p\left(R^3\right)\right)$. According to Lemma 3.2,

$${∥g_v∥}_{B_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)}\lesssim {∥{\alpha }_{e,m}∥}_{{\ell }_{p,q}^s}\lesssim {∥{\alpha }_{e,m}∥}_{{\ell }_{p,q}^{s+1}}\text{and}{∥\frac{\partial }{\partial x_{\mu }}{g}_v∥}_{B_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)}\lesssim {∥{\alpha }_{e,m}∥}_{{\ell }_{p,q}^{s+1}}.$$

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

Let $W_{\tau }^{\mu }\left(D\right)$ denotes the Sobolev space with regularity exponent μ and domain D. Moreover,

$$E_d\left(f,W_{\tau }^{\mu }(D)\right)=:\underset{P\in {{{\displaystyle \prod }}^{}}_{d-1}}{\text{inf}}{\parallel f-P\parallel }_{W_{\tau }^{\mu }}{(D)}_{}.$$

Furthermore, let σ _j,k = 2^-j([0,1]³ + k) for $k\in {\left(Z_j^2\right)}^3$, the boundary cases are: when there is only one k _i = 2^j(1 ≤ i ≤ 3), σ _j,k is 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, ${{\sigma }_j}_{,\left(2^j,2^j,2^j\right)}=:2^{-j}\left(\left[2^{j-1},2^j\right]\otimes \left[2^{j-1},2^j\right]\otimes \left[2^{j-1},2^j\right]\right)$.

Lemma 3.3[8]. Let $\frac{n}{p}-\frac{n}{\tau }+\mu <s<d,s>0,\mu \in N_0,0<p,q\le \infty ,1\le \tau \le \infty ,j_0\in N_0$.

Then

$$\Vert (2^{j(s-\frac{n}{p}+\frac{n}{\tau }-\mu )}\Vert {\left(E_d\left(f,W_{\tau }^{\mu }\left({\sigma }_{j,k}\right)\right)\right)}_{k\in Z^n}{\Vert {{\ell }^p)}_{j\ge j_0}\Vert }_{{\ell }^q}\lesssim {\left|f\right|}_{B_q^s(L^p({[0,1]}^n))}.$$

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

Lemma 3.4. The following relations hold:

$$\begin{array}{c}⟨f,{\tilde{\eta }}_{1;j,k}^{\text{Δ},+}⟩=f\left(2^{-j}\left(k+\frac{1}{2}\right)\right)-\frac{1}{2}f\left(2^{-j}k\right)-\frac{1}{2}\left(2^{-j}\left(k+1\right)\right)-\frac{1}{8}\cdot 2^{-j}{f}^{\prime }\left(2^{-j}k\right)+\frac{1}{8}\cdot 2^{-j}{f}^{\prime }\left(2^{-j}\left(k+1\right)\right);\\ ⟨f,{\tilde{\eta }}_{2;j,k}^{\text{Δ},+}⟩=2^{-\left(j+1\right)}{f}^{\prime }\left(2^{-j}\left(k+\frac{1}{2}\right)\right)-\frac{3}{4}f\left(2^{-j}\left(k+1\right)\right)+\frac{3}{4}f\left(2^{-j}k\right)+\frac{1}{4}\cdot 2^{-\left(j+1\right)}\left[f^{\prime }\left(2^{-j}k\right)+f^{\prime }\left(2^{-j}\left(k+1\right)\right)\right];\\ ⟨f,{\tilde{\eta }}_{1;j,k}^{\text{Δ},-}⟩=2^j{\displaystyle \underset{2^{-j}k}{\overset{2^{-j}\left(k+\frac{1}{2}\right)}{\int }}}f\left(t\right)dt-2^j{\displaystyle \underset{2^{-j}\left(k+\frac{1}{2}\right)}{\overset{2^{-j}\left(k+1\right)}{\int }}}f\left(t\right)dt-\frac{1}{4}f\left(2^{-j}k\right)+\frac{1}{4}f\left(2^{-j}\left(k+1\right)\right);\\ ⟨f,{\tilde{\eta }}_{2;j,k}^{\text{Δ},-}⟩=-\frac{3}{2}\cdot 2^j{\displaystyle \underset{2^{-j}k}{\overset{2^{-j}\left(k+1\right)}{\int }}}f\left(t\right)dt+\frac{1}{4}f\left(2^{-j}k\right)+\frac{1}{4}f\left(2^{-j}\left(k+1\right)\right)+f\left(2^{-j}\left(k+\frac{1}{2}\right)\right).\end{array}$$

Theorem 3.2. Let $1+\frac{3}{p}<s<3$ and 0 < p, q ≤ ∞. Then ${\beta }_{m,i}={\left(⟨\vec{f},{\overrightarrow{\stackrel{̃}{\phi }}}_{m,i;j_0,k}^{\text{Δ}}⟩\right)}_{k\in {\nabla }_{j_0},}{\alpha }_{e,m}^c={\left(⟨\vec{f},{\overrightarrow{\stackrel{̃}{\psi }}}_{e,m;j,k}^{\text{Δ},c}⟩\right)}_{j\ge j_0,k\in {\nabla }_j^1,}{\alpha }_{e,m,i}^{\text{non}}={\left(⟨\vec{f},{\overrightarrow{\stackrel{̃}{\psi }}}_{e,m,i;j,k}^{\text{Δ},\text{non}}⟩\right)}_{j\ge j_0,k\in {\nabla }_j^2,}{j}_0\ge 1,i\ne i_e$ satisfy

$${\displaystyle \sum _{m,i}}{∥{\beta }_{m,i}∥}_{{\ell }^p}+{\displaystyle \sum _{e,m}}\left({\displaystyle \sum _{i\ne i_e}}{∥{\alpha }_{e,m,i}^{\text{non}}∥}_{{\ell }_{p,q}^{s+1}}+{∥{\alpha }_{e,m}^c∥}_{{\ell }_{p,q}^{s+1}}\right)\lesssim {∥\vec{f}∥}_{{\hat{B}}_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)}.$$

Proof. One only need to show the following inequality:

• (i) ${∥{\beta }_{m,i}∥}_{{\ell }^p}\lesssim {∥\vec{f}∥}_{{\hat{B}}_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)}$;

• (ii) ${∥{\alpha }_{e,m,i}^{\text{non}}∥}_{{\ell }_{p,q}^{s+1}}\lesssim {∥\vec{f}∥}_{{\hat{B}}_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)}$;

• (iii) ${∥{\alpha }_{e,m}^c∥}_{{\ell }_{p,q}^{s+1}}\lesssim {∥\vec{f}∥}_{{\hat{B}}_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)}$.

Note that ${\beta }_{m,i;k}=⟨\vec{f},{\overrightarrow{\stackrel{̃}{\phi }}}_{m,i;j_0,k}^{\text{Δ}}⟩=⟨f_i,{\tilde{\phi }}_{m;j_0,k}^{\text{Δ},I_0\backslash \left\{i\right\}}⟩$. Then one assumes i = 1 without loss of generality and proves first

$${∥{\beta }_{m,1}∥}_{{\ell }^p}={∥{\left(⟨f_1,{\tilde{\phi }}_{m;j_0,k}^{\text{Δ},\left\{2,3\right\}}⟩\right)}_{k\in {{\nabla }^{\prime }}_{j_0}}∥}_{{\ell }^p}\lesssim {∥\vec{f}∥}_{{\hat{B}}_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)}.$$

By the embedding property, $B_q^s\left(L^p\left(D\right)\right)\subseteq W_{\tau }^{\mu }\left(D\right)$ for $s>\frac{1}{p}-\frac{1}{\tau }+\mu$, and

$${∥f∥}_{W_{\tau }^{\mu }\left(D\right)}\lesssim {∥f∥}_{B_q^s\left(L^p\left(D\right)\right)}.$$

(3.1)

When $m_2=m_3=1,\left|⟨f_1,{\tilde{\phi }}_{m;j_0,k}^{\text{Δ},\left\{2,3\right\}}⟩\right|\le {∥f_1∥}_{L^{\infty }\left({\sigma }_{j_0,k-{\delta }_1}\right)}$ for m₁ = 1 or ${∥f_1∥}_{L^{\infty }\left({\sigma }_{j_0,k}\right)}$ for m₁ = 2.

Using (3.1), one obtains

$${∥{\beta }_{m,1}∥}_{{\ell }^p}\le \left\{\begin{array}{c}{\left({\displaystyle \sum _{k\in {{\nabla }^{\prime }}_{j_0}}}{∥f_1∥}_{L^{\infty }\left({\sigma }_{j_0,k-{\delta }_1}\right)}^p\right)}^{\frac{1}{p}},m_1=1\hfill \\ \lesssim {∥f_1∥}_{B_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)\lesssim }{∥\vec{f}∥}_{{\hat{B}}_q^s\left(L^p\left(\left[0,1\right]\right)\right)}\hfill \\ {\left({\displaystyle \sum _{k\in {{\nabla }^{\prime }}_{j_0}}}{∥f_1∥}_{L^{\infty }\left({\sigma }_{j_0,k}\right)}^p\right)}^{\frac{1}{p}},m_1=2\hfill \end{array}.\right.$$

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

$$\left|⟨f_1,{\tilde{\phi }}_{m;j_0,k}^{\text{Δ},\left\{2,3\right\}}⟩\right|=2^{-j_0}\left|⟨\frac{\partial f_1}{\partial x_3},{\tilde{\phi }}_{m;j_0,k}^{\text{Δ},\left\{2\right\}}⟩\right|\le \left\{\begin{array}{c}\hfill 2^{-j_0}{∥\frac{\partial f_1}{\partial x_3}∥}_{W_{\infty }^1\left({\sigma }_{j_0,k-{\delta }_1}\right)},m_1=1\hfill \\ \hfill 2^{-j_0}{∥\frac{\partial f_1}{\partial x_3}∥}_{W_{\infty }^1\left({\sigma }_{j_0,k}\right)},m_1=2.\hfill \end{array}\right.$$

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

For $h\in B_q^s\left(L^p\left({\left[0,1\right]}^3\right)\right)$, we first claim that there are only the following two cases:

• (a) $\left|⟨h,{\tilde{\psi }}_{e,m;j,k}^{\text{Δ},\{i^{}}⟩\right|$