# Approximation of Bivariate Functions via Smooth Extensions

Authors: Zhihua Zhang
Journal: The Scientific World Journal
 (2014)

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## Abstract

For a smooth bivariate function defined on a general domain with arbitrary shape, it is
difficult to do Fourier approximation or wavelet approximation. In order to solve these problems, in this paper,
we give an extension of the bivariate function on a general domain with arbitrary shape to a smooth, periodic
function in the whole space or to a smooth, compactly supported function in the whole space. These smooth
extensions have simple and clear representations which are determined by this bivariate function and some
polynomials. After that, we expand the smooth, periodic function into a Fourier series or a periodic wavelet
series or we expand the smooth, compactly supported function into a wavelet series. Since our extensions are
smooth, the obtained Fourier coefficients or wavelet coefficients decay very fast. Since our extension tools are
polynomials, the moment theorem shows that a lot of wavelet coefficients vanish. From this, with the help of
well-known approximation theorems, using our extension methods, the Fourier approximation and the wavelet
approximation of the bivariate function on the general domain with small error are obtained.


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## Body

## 1. Introduction

In the recent several decades, various approximation tools have been widely developed [1–14]. For example, a smooth periodic function can be approximated by trigonometric polynomials; a square-integrable smooth function can be expanded into a wavelet series and be approximated by partial sum of the wavelet series; and a smooth function on a cube can be approximated well by polynomials. However, for a smooth function on a general domain with arbitrary shape, even if it is infinitely many time differentiable, it is difficult to do Fourier approximation or wavelet approximation. In this paper, we will extend a function on general domain with arbitrary shape to a smooth, periodic function in the whole space or to a smooth, compactly supported function in the whole space. After that, it will be easy to do Fourier approximation or wavelet approximation. For the higher-dimensional case, the method of smooth extensions is similar to that in the two-dimensional case, but the representations of smooth extensions will be too complicated. Therefore, in this paper, we mainly consider the smooth extension of a bivariate function on a planar domain. By the way, for the one-dimensional case, since the bounded domain is reduced to a closed interval, the smooth extension can be regarded as a corollary of the two-dimensional case.This paper is organized as follows. In Section2, we state the main theorems on the smooth extension of the function on the general domain and their applications. In Sections 3 and 4, we give a general method of smooth extensions and complete the proofs of the main theorems. In Section 5, we use our extension method to discuss two important special cases of smooth extensions.Throughout this paper, we denoteT=[0,1]2 and the interior of T by To and always assume that Ω is a simply connected domain. We say that f∈Cq(Ω) if the derivatives (∂i+jf/∂xi∂yj) are continuous on Ω for 0≤i+j≤q. We say that f∈C∞(Ω) if all derivatives (∂i+jf/∂xi∂yj) are continuous on Ω for i,j∈ℤ+. We say that a function h(x,y) is a γ-periodic function if h(x+γk,y+γl)=h(x,y)((x,y)∈T;k,l∈ℤ), where γ is an integer. We appoint that 0!=1 and the notation [α] is the integral part of the real number α.


## 2. Main Theorems and Applications

In this section, we state the main results of smooth extensions and their applications in Fourier analysis and wavelet analysis.


### 2.1. Main Theorems

Our main theorems are stated as follows.Theorem 1.
Letf∈C∞(Ω), where Ω⊂To and the boundary ∂Ω is a piecewise infinitely many time smooth curve. Then for any r∈ℤ+ there is a function F∈Cr(T) such that (i)
F(x,y)=f(x,y)((x,y)∈Ω);(ii)
(∂i+jF/∂xi∂yj)(x,y)=0 on the boundary ∂T for 0≤i+j≤r;(iii)
on the complementT∖Ω,F(x,y) can be expressed locally in the forms
(1)∑j=0Lξj(x)yj‍,or∑j=0Lηj(y)xj‍,or∑i,j=0Lcijxiyj‍,where L is a positive integer and each coefficient cij is constant.Theorem 2.
Letf∈C∞(Ω), where Ω is stated as in Theorem 1. Then, for any r∈ℤ+, there exists a 1-periodic function Fp∈Cr(ℝ2) such that Fp(x,y)=f(x,y)((x,y)∈Ω).Theorem 3.
Letf∈C∞(Ω), where Ω is stated as in Theorem 1. Then, for any r∈ℤ+, there exists a function Fc∈Cr(ℝ2) with compact support T such that Fc(x,y)=f(x,y)((x,y)∈Ω).In Sections3 and 4, we give constructive proofs of Theorems 1–3. In these three theorems, we assume that f∈C∞(Ω). If f∈Cq(Ω) (q is a nonnegative integer), by using the similar method of arguments of Theorems 1–3, we also can obtain the corresponding results.


### 2.2. Applications

Here we show some applications of these theorems.


#### 2.2.1. Approximation by Polynomials

LetF be the smooth extension of f from Ω to T which is stated as in Theorem 1. Then F∈Cr(T) and F=f on Ω. By ΔN, denote the set of all bivariate polynomials in the form ∑n1,n2=-NNcn1,n2xn1yn2. Then
(2)inf⁡P∈ΔN⁡∥f-P∥Lp(Ω)≤inf⁡P∈ΔN⁡∥F-P∥Lp(T),
where ∥·∥Lp(D) is the norm of the space Lp(D). The right-hand side of formula (2) is the best approximation of the extension F in ΔN. By (2), we know that the approximation problem of f by polynomials on a domain Ω is reduced to the well-known approximation problem of its smooth extension F by polynomials on the square T [4, 10].


#### 2.2.2. Fourier Analysis

(i)Approximation by Trigonometric Polynomials. Let Fp be the smooth periodic extension of f as in Theorem 2. Then Fp∈Cr(ℝ2) and Fp=f on Ω. By the well-known results [5, 10], we know that the smooth periodic function Fp can be approximated by bivariate trigonometric polynomials very well. Its approximation error can be estimated by the modulus of continuity of its r time derivatives.ByΔN*, denote the set of all bivariate trigonometric polynomials in the form
(3)∑n1,n2=-NNcn1,n2*e2πi(n1x+n2y).
By Theorem 2, we have
(4)inf⁡P*∈ΔN*⁡∥f-P*∥Lp′(Ω)≤min⁡P*∈ΔN*⁡∥Fp-P*∥Lp′(T).
From this and Theorem 2, we see that the approximation problem of f on Ω by trigonometric polynomials is reduced to a well-known approximation problem of smooth periodic functions [5, 7, 10].(ii)Fourier Series. We expand Fp into a Fourier series [9]
(5)Fp(x,y)=∑(n1,n2)∈ℤ2τn1,n2e2πi(n1x+n2y),
where τn1,n2=∫TFp(x,y)e-2πi(n1x+n2y)dxdy. By Theorem 2, we obtain that, for (x,y)∈Ω,
(6)f(x,y)=∑(n1,n2)∈ℤ2τn1,n2e2πi(n1x+n2y).
Denote the partial sum
(7)sn1,n2(x,y)=∑k1=0n1∑k2=0n2τk1,k2e2πi(k1x+k2y).
Then we have
(8)∥f(x,y)-sn1,n2(x,y)∥Lp′(Ω)≤∥Fp(x,y)-sn1,n2(x,y)∥Lp′(T).
Since the smooth periodic function Fp can be approximated well by the partial sum of its Fourier series [5, 7, 10], from this inequality, we see that we have constructed a trigonometric polynomial sn1,n2(x,y) which can approximate to f on Ω very well.(iii)Odd (Even) Periodic Extension. Let F be the smooth extension of f from Ω to T which is stated in Theorem 1. Define Fo on [-1,1]2 by
(9)Fo(x,y)={F(x,y),(x,y)∈[0,1]2,-F(-x,y),(x,y)∈[-1,0]×[0,1],F(-x,-y),(x,y)∈[-1,0]2,-F(x,-y),(x,y)∈[0,1]×[-1,0].
Then Fo is an odd function. By Theorem 1, we have Fo∈Cr([-1,1]2) and (∂i+jFo/∂xi∂yj)(x,y)=0 on ∂([-1,1]2) for 0≤i+j≤r. Again, doing a 2-periodic extension, we obtain a 2-periodic odd function Fpo and Fpo∈Cr(ℝ2). By the well-known results [5, 7, 10], Fpo can be approximated by sine polynomials very well. Moreover, Fpo can be expanded into the Fourier sine series; that is,
(10)Fpo(x,y)=∑n1=1∞∑n2=1∞αn1,n2sin⁡(πn1x)sin⁡(πn2y),
where the coefficients αn1,n2=4∫TFpo(x,y)sin(πn1x)sin(πn2y)dxdy [9]. Considering the approximation of Fpo by the partial sum, the Fejer sum, and the Vallee-Poussin sum [7, 14] of the Fourier sine series of Fpo, we will obtain the approximation of the original function f on Ω by sine polynomials.DefineFe on [-1,1]2 as follows:
(11)Fe(x,y)={F(x,y),(x,y)∈[0,1]2,F(-x,y),(x,y)∈[-1,0]×[0,1],F(-x,-y),(x,y)∈[-1,0]2,F(x,-y),(x,y)∈[0,1]×[-1,0].
Then Fe is an even function on [-1,1]2. By Theorem 1, Fe∈Cr([-1,1]2) and (∂i+jFe/∂xi∂yj)(x,y)=0 on ∂([-1,1]2) for 0≤i+j≤r. Again, doing a 2-periodic extension, we obtain a 2-periodic even function Fpe and Fpe∈Cr(ℝ2). By the well-known result [5, 10], Fpe can be approximated by cosine polynomials very well. Moreover, Fpe can be expanded into the Fourier cosine series. Considering the partial sum, the Fejer sum, and the Vallee-Poussin sum [5, 7, 14] of the Fourier cosine series of Fpe, we will obtain the approximation of the original function f on Ω by cosine polynomials.


#### 2.2.3. Wavelet Analysis

(i)Periodic Wavelet Series. Let Fp∈Cr(ℝ2) be stated in Theorem 2. Let {ψμ}13 be a bivariate smooth wavelet [2]. Then, under a mild condition, the families
(12)Ψper:={1}⋃‍{ψμ,m,nper,μ=1,2,3;m∈ℤ+;jjjjjjjjjjjjjjjψμ,m,npern=(n1,n2),n1,n2=0,…,2m-1},whereψμ,m,nper=∑l∈ℤ2ψμ,m,n(·+l),ψμ,m,n=2mψμ(2m·-n)
are a periodic wavelet basis. We expand Fp into a periodic wavelet series [2]
(13)Fp=d0,0+∑μ=13∑m=0∞∑n1,n2=02m-1dμ,m,nψμ,m,nper.
From this, we can realize the wavelet approximation of f on Ω, for example, if r=2, its partial sum
(14)s2M(Fp)=d0,0+∑μ=13∑m=0M-1∑n1,n2=02m-1dμ,m,nψμ,m,nper
satisfies ∥Fp-s2M(Fp)∥L2(T)=O(2-2M). From this and Fp(x,y)=f(x,y)((x,y)∈Ω), we will obtain an estimate of wavelet approximation for a smooth function f on the domain Ω.(ii)Wavelet Approximation. Let Fc be the smooth function with a compact support as in Theorem 3. Let ψ be a univariate Daubechies wavelet and ϕ be the corresponding scaling function [2]. Denoting
(15)ψ1(x,y)=ϕ(x)ψ(y),ψ2(x,y)=ψ(x)ϕ(y),gggggggggψ3(x,y)=ψ(x)ψ(y),
then {ψμ(x,y)}μ=13 is a smooth tensor product wavelet. We expand Fc into the wavelet series
(16)Fc(x,y)=∑μ=13∑m∈ℤ∑n∈ℤ2cμ,m,nψμ,m,n(x,y),
where ψμ,m,n=2mψμ(2m·-n) and the wavelet coefficients
(17)cμ,m,n=∫ℝ2Fc(x,y)ψ¯μ,m,n(x,y)dxdy=∫TF(x,y)ψ¯μ,m,n(x,y)dxdy.‍
Since Fc is a smooth function, the wavelet coefficients cμ,m,n decay fast.On the other hand, sinceFc(x,y)=0, (x,y)∈ℝ2∖T, a lot of wavelet coefficients vanish. In fact, when m0∈ℤ and n0∈ℤ2 satisfy supp⁡ψμ,m0,n0⊂(ℝ2∖T), we have cμ,m0,n0=0. Besides, by condition (iii) in Theorem 1, we know that F is univariate or bivariate polynomials on T∖Ω. By the moment theorem [2], we know that more wavelet coefficients vanish.For example, letm*∈ℤ and n*=(n1*,n2*)∈ℤ2 satisfy supp⁡ψm*,n2*⊂[0,α*], where α*=inf⁡{g(x),x1≤x≤x2}. Then we have
(18)c1,m*,n*=2m(∫0x1‍+∫x1x2‍+∫x21‍)ϕ¯m*,n1*(x)c1,m*,n*=×(∫0α*ψ¯m*,n2*(y)F(x,y)dy)dxc1,m*,1=:I1+I2+I3.
By Lemma 8, we know that
(19)F(x,y)=∑j=0Lξj(x)yj,(x,y)∈E1,
where E1={(x,y):x1≤x≤x2,0≤y≤g(x)} and g(x)≥α*(x1≤x≤x2). So
(20)I2=∫x1x2ϕ¯m*,n1*(x)(∫0α*ψ¯m*,n2*(y)(∑j=0Lξj(x)yj)dy‍)dx.
If the Daubechies wavelet ψ chosen by us is L time smooth, then, by using the moment theorem and supp⁡ψm*,n2*⊂[0,α*], we have
(21)∫0α*ψ¯m*,n2*(y)yjdy‍=∫ℝψ¯m*,n2*(y)yjdy=0,kkkl(0≤j≤2r+1).
So I2=0. Similarly, since F(x,y) is bivariate polynomials on rectangles H1 and H3 (see Lemma 11), we have I1=I3=0. Furthermore, by (18), we get c1,m*,n*=0.Therefore, the partial sum of the wavelet series (16) can approximate to Fc very well and few wavelet coefficients can reconstruct Fc. Since Fc=f on Ω, the partial sum of the wavelet series (16) can approximate to the original function f on the domain Ω very well.


## 2.1. Main Theorems

Our main theorems are stated as follows.Theorem 1.
Letf∈C∞(Ω), where Ω⊂To and the boundary ∂Ω is a piecewise infinitely many time smooth curve. Then for any r∈ℤ+ there is a function F∈Cr(T) such that (i)
F(x,y)=f(x,y)((x,y)∈Ω);(ii)
(∂i+jF/∂xi∂yj)(x,y)=0 on the boundary ∂T for 0≤i+j≤r;(iii)
on the complementT∖Ω,F(x,y) can be expressed locally in the forms
(1)∑j=0Lξj(x)yj‍,or∑j=0Lηj(y)xj‍,or∑i,j=0Lcijxiyj‍,where L is a positive integer and each coefficient cij is constant.Theorem 2.
Letf∈C∞(Ω), where Ω is stated as in Theorem 1. Then, for any r∈ℤ+, there exists a 1-periodic function Fp∈Cr(ℝ2) such that Fp(x,y)=f(x,y)((x,y)∈Ω).Theorem 3.
Letf∈C∞(Ω), where Ω is stated as in Theorem 1. Then, for any r∈ℤ+, there exists a function Fc∈Cr(ℝ2) with compact support T such that Fc(x,y)=f(x,y)((x,y)∈Ω).In Sections3 and 4, we give constructive proofs of Theorems 1–3. In these three theorems, we assume that f∈C∞(Ω). If f∈Cq(Ω) (q is a nonnegative integer), by using the similar method of arguments of Theorems 1–3, we also can obtain the corresponding results.


## 2.2. Applications

Here we show some applications of these theorems.


### 2.2.1. Approximation by Polynomials

LetF be the smooth extension of f from Ω to T which is stated as in Theorem 1. Then F∈Cr(T) and F=f on Ω. By ΔN, denote the set of all bivariate polynomials in the form ∑n1,n2=-NNcn1,n2xn1yn2. Then
(2)inf⁡P∈ΔN⁡∥f-P∥Lp(Ω)≤inf⁡P∈ΔN⁡∥F-P∥Lp(T),
where ∥·∥Lp(D) is the norm of the space Lp(D). The right-hand side of formula (2) is the best approximation of the extension F in ΔN. By (2), we know that the approximation problem of f by polynomials on a domain Ω is reduced to the well-known approximation problem of its smooth extension F by polynomials on the square T [4, 10].


### 2.2.2. Fourier Analysis

(i)Approximation by Trigonometric Polynomials. Let Fp be the smooth periodic extension of f as in Theorem 2. Then Fp∈Cr(ℝ2) and Fp=f on Ω. By the well-known results [5, 10], we know that the smooth periodic function Fp can be approximated by bivariate trigonometric polynomials very well. Its approximation error can be estimated by the modulus of continuity of its r time derivatives.ByΔN*, denote the set of all bivariate trigonometric polynomials in the form
(3)∑n1,n2=-NNcn1,n2*e2πi(n1x+n2y).
By Theorem 2, we have
(4)inf⁡P*∈ΔN*⁡∥f-P*∥Lp′(Ω)≤min⁡P*∈ΔN*⁡∥Fp-P*∥Lp′(T).
From this and Theorem 2, we see that the approximation problem of f on Ω by trigonometric polynomials is reduced to a well-known approximation problem of smooth periodic functions [5, 7, 10].(ii)Fourier Series. We expand Fp into a Fourier series [9]
(5)Fp(x,y)=∑(n1,n2)∈ℤ2τn1,n2e2πi(n1x+n2y),
where τn1,n2=∫TFp(x,y)e-2πi(n1x+n2y)dxdy. By Theorem 2, we obtain that, for (x,y)∈Ω,
(6)f(x,y)=∑(n1,n2)∈ℤ2τn1,n2e2πi(n1x+n2y).
Denote the partial sum
(7)sn1,n2(x,y)=∑k1=0n1∑k2=0n2τk1,k2e2πi(k1x+k2y).
Then we have
(8)∥f(x,y)-sn1,n2(x,y)∥Lp′(Ω)≤∥Fp(x,y)-sn1,n2(x,y)∥Lp′(T).
Since the smooth periodic function Fp can be approximated well by the partial sum of its Fourier series [5, 7, 10], from this inequality, we see that we have constructed a trigonometric polynomial sn1,n2(x,y) which can approximate to f on Ω very well.(iii)Odd (Even) Periodic Extension. Let F be the smooth extension of f from Ω to T which is stated in Theorem 1. Define Fo on [-1,1]2 by
(9)Fo(x,y)={F(x,y),(x,y)∈[0,1]2,-F(-x,y),(x,y)∈[-1,0]×[0,1],F(-x,-y),(x,y)∈[-1,0]2,-F(x,-y),(x,y)∈[0,1]×[-1,0].
Then Fo is an odd function. By Theorem 1, we have Fo∈Cr([-1,1]2) and (∂i+jFo/∂xi∂yj)(x,y)=0 on ∂([-1,1]2) for 0≤i+j≤r. Again, doing a 2-periodic extension, we obtain a 2-periodic odd function Fpo and Fpo∈Cr(ℝ2). By the well-known results [5, 7, 10], Fpo can be approximated by sine polynomials very well. Moreover, Fpo can be expanded into the Fourier sine series; that is,
(10)Fpo(x,y)=∑n1=1∞∑n2=1∞αn1,n2sin⁡(πn1x)sin⁡(πn2y),
where the coefficients αn1,n2=4∫TFpo(x,y)sin(πn1x)sin(πn2y)dxdy [9]. Considering the approximation of Fpo by the partial sum, the Fejer sum, and the Vallee-Poussin sum [7, 14] of the Fourier sine series of Fpo, we will obtain the approximation of the original function f on Ω by sine polynomials.DefineFe on [-1,1]2 as follows:
(11)Fe(x,y)={F(x,y),(x,y)∈[0,1]2,F(-x,y),(x,y)∈[-1,0]×[0,1],F(-x,-y),(x,y)∈[-1,0]2,F(x,-y),(x,y)∈[0,1]×[-1,0].
Then Fe is an even function on [-1,1]2. By Theorem 1, Fe∈Cr([-1,1]2) and (∂i+jFe/∂xi∂yj)(x,y)=0 on ∂([-1,1]2) for 0≤i+j≤r. Again, doing a 2-periodic extension, we obtain a 2-periodic even function Fpe and Fpe∈Cr(ℝ2). By the well-known result [5, 10], Fpe can be approximated by cosine polynomials very well. Moreover, Fpe can be expanded into the Fourier cosine series. Considering the partial sum, the Fejer sum, and the Vallee-Poussin sum [5, 7, 14] of the Fourier cosine series of Fpe, we will obtain the approximation of the original function f on Ω by cosine polynomials.


### 2.2.3. Wavelet Analysis

(i)Periodic Wavelet Series. Let Fp∈Cr(ℝ2) be stated in Theorem 2. Let {ψμ}13 be a bivariate smooth wavelet [2]. Then, under a mild condition, the families
(12)Ψper:={1}⋃‍{ψμ,m,nper,μ=1,2,3;m∈ℤ+;jjjjjjjjjjjjjjjψμ,m,npern=(n1,n2),n1,n2=0,…,2m-1},whereψμ,m,nper=∑l∈ℤ2ψμ,m,n(·+l),ψμ,m,n=2mψμ(2m·-n)
are a periodic wavelet basis. We expand Fp into a periodic wavelet series [2]
(13)Fp=d0,0+∑μ=13∑m=0∞∑n1,n2=02m-1dμ,m,nψμ,m,nper.
From this, we can realize the wavelet approximation of f on Ω, for example, if r=2, its partial sum
(14)s2M(Fp)=d0,0+∑μ=13∑m=0M-1∑n1,n2=02m-1dμ,m,nψμ,m,nper
satisfies ∥Fp-s2M(Fp)∥L2(T)=O(2-2M). From this and Fp(x,y)=f(x,y)((x,y)∈Ω), we will obtain an estimate of wavelet approximation for a smooth function f on the domain Ω.(ii)Wavelet Approximation. Let Fc be the smooth function with a compact support as in Theorem 3. Let ψ be a univariate Daubechies wavelet and ϕ be the corresponding scaling function [2]. Denoting
(15)ψ1(x,y)=ϕ(x)ψ(y),ψ2(x,y)=ψ(x)ϕ(y),gggggggggψ3(x,y)=ψ(x)ψ(y),
then {ψμ(x,y)}μ=13 is a smooth tensor product wavelet. We expand Fc into the wavelet series
(16)Fc(x,y)=∑μ=13∑m∈ℤ∑n∈ℤ2cμ,m,nψμ,m,n(x,y),
where ψμ,m,n=2mψμ(2m·-n) and the wavelet coefficients
(17)cμ,m,n=∫ℝ2Fc(x,y)ψ¯μ,m,n(x,y)dxdy=∫TF(x,y)ψ¯μ,m,n(x,y)dxdy.‍
Since Fc is a smooth function, the wavelet coefficients cμ,m,n decay fast.On the other hand, sinceFc(x,y)=0, (x,y)∈ℝ2∖T, a lot of wavelet coefficients vanish. In fact, when m0∈ℤ and n0∈ℤ2 satisfy supp⁡ψμ,m0,n0⊂(ℝ2∖T), we have cμ,m0,n0=0. Besides, by condition (iii) in Theorem 1, we know that F is univariate or bivariate polynomials on T∖Ω. By the moment theorem [2], we know that more wavelet coefficients vanish.For example, letm*∈ℤ and n*=(n1*,n2*)∈ℤ2 satisfy supp⁡ψm*,n2*⊂[0,α*], where α*=inf⁡{g(x),x1≤x≤x2}. Then we have
(18)c1,m*,n*=2m(∫0x1‍+∫x1x2‍+∫x21‍)ϕ¯m*,n1*(x)c1,m*,n*=×(∫0α*ψ¯m*,n2*(y)F(x,y)dy)dxc1,m*,1=:I1+I2+I3.
By Lemma 8, we know that
(19)F(x,y)=∑j=0Lξj(x)yj,(x,y)∈E1,
where E1={(x,y):x1≤x≤x2,0≤y≤g(x)} and g(x)≥α*(x1≤x≤x2). So
(20)I2=∫x1x2ϕ¯m*,n1*(x)(∫0α*ψ¯m*,n2*(y)(∑j=0Lξj(x)yj)dy‍)dx.
If the Daubechies wavelet ψ chosen by us is L time smooth, then, by using the moment theorem and supp⁡ψm*,n2*⊂[0,α*], we have
(21)∫0α*ψ¯m*,n2*(y)yjdy‍=∫ℝψ¯m*,n2*(y)yjdy=0,kkkl(0≤j≤2r+1).
So I2=0. Similarly, since F(x,y) is bivariate polynomials on rectangles H1 and H3 (see Lemma 11), we have I1=I3=0. Furthermore, by (18), we get c1,m*,n*=0.Therefore, the partial sum of the wavelet series (16) can approximate to Fc very well and few wavelet coefficients can reconstruct Fc. Since Fc=f on Ω, the partial sum of the wavelet series (16) can approximate to the original function f on the domain Ω very well.


## 2.2.1. Approximation by Polynomials

LetF be the smooth extension of f from Ω to T which is stated as in Theorem 1. Then F∈Cr(T) and F=f on Ω. By ΔN, denote the set of all bivariate polynomials in the form ∑n1,n2=-NNcn1,n2xn1yn2. Then
(2)inf⁡P∈ΔN⁡∥f-P∥Lp(Ω)≤inf⁡P∈ΔN⁡∥F-P∥Lp(T),
where ∥·∥Lp(D) is the norm of the space Lp(D). The right-hand side of formula (2) is the best approximation of the extension F in ΔN. By (2), we know that the approximation problem of f by polynomials on a domain Ω is reduced to the well-known approximation problem of its smooth extension F by polynomials on the square T [4, 10].


## 2.2.2. Fourier Analysis

(i)Approximation by Trigonometric Polynomials. Let Fp be the smooth periodic extension of f as in Theorem 2. Then Fp∈Cr(ℝ2) and Fp=f on Ω. By the well-known results [5, 10], we know that the smooth periodic function Fp can be approximated by bivariate trigonometric polynomials very well. Its approximation error can be estimated by the modulus of continuity of its r time derivatives.ByΔN*, denote the set of all bivariate trigonometric polynomials in the form
(3)∑n1,n2=-NNcn1,n2*e2πi(n1x+n2y).
By Theorem 2, we have
(4)inf⁡P*∈ΔN*⁡∥f-P*∥Lp′(Ω)≤min⁡P*∈ΔN*⁡∥Fp-P*∥Lp′(T).
From this and Theorem 2, we see that the approximation problem of f on Ω by trigonometric polynomials is reduced to a well-known approximation problem of smooth periodic functions [5, 7, 10].(ii)Fourier Series. We expand Fp into a Fourier series [9]
(5)Fp(x,y)=∑(n1,n2)∈ℤ2τn1,n2e2πi(n1x+n2y),
where τn1,n2=∫TFp(x,y)e-2πi(n1x+n2y)dxdy. By Theorem 2, we obtain that, for (x,y)∈Ω,
(6)f(x,y)=∑(n1,n2)∈ℤ2τn1,n2e2πi(n1x+n2y).
Denote the partial sum
(7)sn1,n2(x,y)=∑k1=0n1∑k2=0n2τk1,k2e2πi(k1x+k2y).
Then we have
(8)∥f(x,y)-sn1,n2(x,y)∥Lp′(Ω)≤∥Fp(x,y)-sn1,n2(x,y)∥Lp′(T).
Since the smooth periodic function Fp can be approximated well by the partial sum of its Fourier series [5, 7, 10], from this inequality, we see that we have constructed a trigonometric polynomial sn1,n2(x,y) which can approximate to f on Ω very well.(iii)Odd (Even) Periodic Extension. Let F be the smooth extension of f from Ω to T which is stated in Theorem 1. Define Fo on [-1,1]2 by
(9)Fo(x,y)={F(x,y),(x,y)∈[0,1]2,-F(-x,y),(x,y)∈[-1,0]×[0,1],F(-x,-y),(x,y)∈[-1,0]2,-F(x,-y),(x,y)∈[0,1]×[-1,0].
Then Fo is an odd function. By Theorem 1, we have Fo∈Cr([-1,1]2) and (∂i+jFo/∂xi∂yj)(x,y)=0 on ∂([-1,1]2) for 0≤i+j≤r. Again, doing a 2-periodic extension, we obtain a 2-periodic odd function Fpo and Fpo∈Cr(ℝ2). By the well-known results [5, 7, 10], Fpo can be approximated by sine polynomials very well. Moreover, Fpo can be expanded into the Fourier sine series; that is,
(10)Fpo(x,y)=∑n1=1∞∑n2=1∞αn1,n2sin⁡(πn1x)sin⁡(πn2y),
where the coefficients αn1,n2=4∫TFpo(x,y)sin(πn1x)sin(πn2y)dxdy [9]. Considering the approximation of Fpo by the partial sum, the Fejer sum, and the Vallee-Poussin sum [7, 14] of the Fourier sine series of Fpo, we will obtain the approximation of the original function f on Ω by sine polynomials.DefineFe on [-1,1]2 as follows:
(11)Fe(x,y)={F(x,y),(x,y)∈[0,1]2,F(-x,y),(x,y)∈[-1,0]×[0,1],F(-x,-y),(x,y)∈[-1,0]2,F(x,-y),(x,y)∈[0,1]×[-1,0].
Then Fe is an even function on [-1,1]2. By Theorem 1, Fe∈Cr([-1,1]2) and (∂i+jFe/∂xi∂yj)(x,y)=0 on ∂([-1,1]2) for 0≤i+j≤r. Again, doing a 2-periodic extension, we obtain a 2-periodic even function Fpe and Fpe∈Cr(ℝ2). By the well-known result [5, 10], Fpe can be approximated by cosine polynomials very well. Moreover, Fpe can be expanded into the Fourier cosine series. Considering the partial sum, the Fejer sum, and the Vallee-Poussin sum [5, 7, 14] of the Fourier cosine series of Fpe, we will obtain the approximation of the original function f on Ω by cosine polynomials.


## 2.2.3. Wavelet Analysis

(i)Periodic Wavelet Series. Let Fp∈Cr(ℝ2) be stated in Theorem 2. Let {ψμ}13 be a bivariate smooth wavelet [2]. Then, under a mild condition, the families
(12)Ψper:={1}⋃‍{ψμ,m,nper,μ=1,2,3;m∈ℤ+;jjjjjjjjjjjjjjjψμ,m,npern=(n1,n2),n1,n2=0,…,2m-1},whereψμ,m,nper=∑l∈ℤ2ψμ,m,n(·+l),ψμ,m,n=2mψμ(2m·-n)
are a periodic wavelet basis. We expand Fp into a periodic wavelet series [2]
(13)Fp=d0,0+∑μ=13∑m=0∞∑n1,n2=02m-1dμ,m,nψμ,m,nper.
From this, we can realize the wavelet approximation of f on Ω, for example, if r=2, its partial sum
(14)s2M(Fp)=d0,0+∑μ=13∑m=0M-1∑n1,n2=02m-1dμ,m,nψμ,m,nper
satisfies ∥Fp-s2M(Fp)∥L2(T)=O(2-2M). From this and Fp(x,y)=f(x,y)((x,y)∈Ω), we will obtain an estimate of wavelet approximation for a smooth function f on the domain Ω.(ii)Wavelet Approximation. Let Fc be the smooth function with a compact support as in Theorem 3. Let ψ be a univariate Daubechies wavelet and ϕ be the corresponding scaling function [2]. Denoting
(15)ψ1(x,y)=ϕ(x)ψ(y),ψ2(x,y)=ψ(x)ϕ(y),gggggggggψ3(x,y)=ψ(x)ψ(y),
then {ψμ(x,y)}μ=13 is a smooth tensor product wavelet. We expand Fc into the wavelet series
(16)Fc(x,y)=∑μ=13∑m∈ℤ∑n∈ℤ2cμ,m,nψμ,m,n(x,y),
where ψμ,m,n=2mψμ(2m·-n) and the wavelet coefficients
(17)cμ,m,n=∫ℝ2Fc(x,y)ψ¯μ,m,n(x,y)dxdy=∫TF(x,y)ψ¯μ,m,n(x,y)dxdy.‍
Since Fc is a smooth function, the wavelet coefficients cμ,m,n decay fast.On the other hand, sinceFc(x,y)=0, (x,y)∈ℝ2∖T, a lot of wavelet coefficients vanish. In fact, when m0∈ℤ and n0∈ℤ2 satisfy supp⁡ψμ,m0,n0⊂(ℝ2∖T), we have cμ,m0,n0=0. Besides, by condition (iii) in Theorem 1, we know that F is univariate or bivariate polynomials on T∖Ω. By the moment theorem [2], we know that more wavelet coefficients vanish.For example, letm*∈ℤ and n*=(n1*,n2*)∈ℤ2 satisfy supp⁡ψm*,n2*⊂[0,α*], where α*=inf⁡{g(x),x1≤x≤x2}. Then we have
(18)c1,m*,n*=2m(∫0x1‍+∫x1x2‍+∫x21‍)ϕ¯m*,n1*(x)c1,m*,n*=×(∫0α*ψ¯m*,n2*(y)F(x,y)dy)dxc1,m*,1=:I1+I2+I3.
By Lemma 8, we know that
(19)F(x,y)=∑j=0Lξj(x)yj,(x,y)∈E1,
where E1={(x,y):x1≤x≤x2,0≤y≤g(x)} and g(x)≥α*(x1≤x≤x2). So
(20)I2=∫x1x2ϕ¯m*,n1*(x)(∫0α*ψ¯m*,n2*(y)(∑j=0Lξj(x)yj)dy‍)dx.
If the Daubechies wavelet ψ chosen by us is L time smooth, then, by using the moment theorem and supp⁡ψm*,n2*⊂[0,α*], we have
(21)∫0α*ψ¯m*,n2*(y)yjdy‍=∫ℝψ¯m*,n2*(y)yjdy=0,kkkl(0≤j≤2r+1).
So I2=0. Similarly, since F(x,y) is bivariate polynomials on rectangles H1 and H3 (see Lemma 11), we have I1=I3=0. Furthermore, by (18), we get c1,m*,n*=0.Therefore, the partial sum of the wavelet series (16) can approximate to Fc very well and few wavelet coefficients can reconstruct Fc. Since Fc=f on Ω, the partial sum of the wavelet series (16) can approximate to the original function f on the domain Ω very well.


## 3. Proofs of the Main Theorems

We first give a partition of the complementT∖Ω.


### 3.1. Partition of the Complement of the DomainΩ in T

SinceΩ⊂To and ∂Ω is a piecewise infinitely many time smooth curve, without loss of generality, we can divide the complement T∖Ω into some rectangles and some trapezoids with a curved side. For convenience of representation, we assume that we can choose four point (xν,yν)∈∂Ω(ν=1,2,3,4) such that T∖Ω can be divided into the four rectangles
(22)H1=[0,x1]×[0,y1],H2=[x2,1]×[0,y2],H3=[x3,1]×[y3,1],H4=[0,x4]×[y4,1]
and four trapezoids with a curved side
(23)E1={(x,y);x1≤x≤x2,0≤y≤g(x)},E2={(x,y);h(y)≤x≤1,y2≤y≤y3},E3={(x,y);x4≤x≤x3,g*(x)≤y≤1},E4={(x,y);0≤x≤h*(y),y1≤y≤y4},
where g∈C∞([x1,x2]),h∈C∞([y2,y3]),g*∈C∞([x4,x3]), and h*∈C∞([y1,y4]) and
(24)0<g(x)<1(x1≤x≤x2),0<h(y)<1(y2≤y≤y3),0<g*(x)<1(x4≤x≤x3),0<h*(y)<1(y1≤y≤y4).
From this, we know that T can be expressed into a disjoint union as follows:
(25)T=Ω⋃(⋃14Eν)‍⋃(⋃14Hν),
where each Eν is a trapezoid with a curved side and each Hν is a rectangle (see Figure 1).Figure 1
Partition of the complement of the domainΩ.In Sections3.2 and 3.3 we will extend f to each Eν and continue to extend to each Hν such that the obtained extension F satisfies the conditions of Theorem 1.


### 3.2. Smooth Extension to Each TrapezoidEν with a Curved Side

By (23), the trapezoid E1 with a curved side y=g(x)(x1≤x≤x2) is represented as
(26)E1={(x,y):x1≤x≤x2,0≤y≤g(x)}.
We define two sequences of functions {ak,1(x,y)}0∞ and {bk,1(x,y)}0∞ as follows:
(27)a0,1(x,y)=yg(x),b0,1(x,y)=y-g(x)-g(x),fffak,1(x,y)=(y-g(x))kk!(yg(x))k+1,bk,1(x,y)=ykk!(y-g(x)-g(x))k+1,k=1,2,….
By (27), we deduce that for x1≤x≤x2,
(28)hhhhhh∂lak,1∂yl(x,g(x))=0(0≤l≤k-1),∂kak,1∂yk(x,g(x))=1;(∂lak,1)∂yl(x,0)=0(0≤l≤k),hhhhhh∂lbk,1∂yl(x,g(x))=0(0≤l≤k);∂lbk,1∂yl(x,0)=0(0≤l≤k-1),(∂kbk,1)∂yk(x,0)=1.OnE1, we define a sequence of functions {S1(k)(x,y)}0∞ by induction.Let(29)S1(0)(x,y)=f(x,g(x))a0,1(x,y)ggg.(x1≤x≤x2,0≤y≤g(x)).
Then, by (27),
(30)S1(0)(x,0)=0,S1(0)(x,g(x))=f(x,g(x)),hhhhhhhhhhhhhhhhhhhhhhhhll(x1≤x≤x2).Let(31)S1(1)(x,y)=S1(0)(x,y)S1(1)(x,y)=+a1,1(x,y)(∂f∂y(x,g(x))-∂S1(0)∂y(x,g(x)))S1(1)(x,y)=-b1,1(x,y)∂S1(0)∂y(x,0)fffffffffffffffff.(x1≤x≤x2,0≤y≤g(x)).
Then, by (27)–(30), we obtain that, for x1≤x≤x2,
(32)S1(1)(x,g(x))=f(x,g(x)),∂S1(1)∂y(x,g(x))=∂f∂y(x,g(x)),S1(1)(x,0)=0,∂S1(1)∂y(x,0)=0.
In general, let
(33)S1(k)(x,y)=S1(k-1)(x,y)+ak,1(x,y)S1(k)(x,y)=×(∂kf∂yk(x,g(x))-∂kS1(k-1)∂yk(x,g(x)))S1(k)(x,y)=-bk,1(x,y)∂kS1(k-1)∂yk(x,0)S1(k)(x,y)ggggggggg=(x1≤x≤x2,0≤y≤g(x)).Lemma 4.
For anyk∈ℤ+, one has S1(k)∈C∞(E1) and
(34)S1(k)(x,y)=∑j=02k+1ζj,1(x)yj‍,(x,y)∈E1.Proof.
Sincef∈C∞(Ω) and g∈C∞([x1,x2]), and g(x)>0(x1≤x≤x2), by the above construction, we know that S1(k)∈C∞(E1) for any k=0,1,….
Fork=0, since
(35)S1(0)(x,y)=f(x,g(x))a0,1(x,y)=f(x,g(x))g(x)y,
(34) holds. We assume that (34) holds for k=l-1; that is,
(36)S1(l-1)(x,y)=∑j=02l-1ζj,1(l-1)(x)yj.
This implies that
(37)∂lS1(l-1)∂yl(x,g(x))=∑j=l2l-1j!(j-l)!ζj,1(l-1)(x)(g(x))j-l,∂lS1(l-1)∂yl(x,0)=l!ζl,1(l-1)(x).
Again, notice that al,1(x,y) and bl,1(x,y) are polynomials of y whose degrees are both 2l+1. From this and (33), it follows that (34) holds for k=l. By induction, (34) holds for all k. Lemma 4 is proved.Below we compute derivatives(∂lS1(k)/∂yl)(x,y)(0≤l≤k) on the curved side Γ1={(x,g(x)):x1≤x≤x2} and the bottom side Δ1={(x,0):x1≤x≤x2} of E1.Lemma 5.
LetS1(k)(x,y) be stated as above. For any k∈ℤ+, one has
(38)∂lS1(k)∂yl(x,g(x))=∂lf∂yl(x,g(x)),∂lS1(k)∂yl(x,0)=0(x1≤x≤x2,0≤l≤k).Proof.
By (30), We have known that, for k=0, (38) holds.
Now we assume that (38) holds for k-1.
Forx1≤x≤x2, by (33), we have
(39)∂lS1(k)∂yl(x,g(x))=∂lS1(k-1)∂yl(x,g(x))+∂lak,1∂yl(x,g(x))×(∂kf∂yk(x,g(x))-∂kS1(k-1)∂yk(x,g(x)))-∂lbk,1∂yl(x,g(x))∂kS1(k-1)∂yk(x,0),kkkkkkkkkkkkkkkhhhhl.k(0≤l≤k).
For l=0,1,…,k-1, by the assumption of induction, we have
(40)∂lS1(k-1)∂yl(x,g(x))=∂lf∂yl(x,g(x)).
By (28), we have (∂lak,1/∂yl)(x,g(x))=0,(∂lbk,1/∂yl)(x,g(x))=0. So we get
(41)∂lS1(k)∂yl(x,g(x))=∂lf∂yl(x,g(x)).
For l=k, note that (∂kak,1/∂yk)(x,g(x))=1 and (∂kbk,1/∂yk)(x,g(x))=0. By (39), we get
(42)∂kS1(k)∂yk(x,g(x))=∂kS1(k-1)∂yk(x,g(x))+(∂kf∂yk(x,g(x))-∂kS1(k-1)∂yk(x,g(x)))=∂kf∂yk(x,g(x)).
The first formula of (38) holds for k.
By (33), we have
(43)∂lS1(k)∂yl(x,0)=∂lS1(k-1)∂yl(x,0)+∂lak,1∂yl(x,0)×(∂kf∂yk(x,g(x))-∂kS1(k-1)∂yk(x,g(x)))-∂lbk,1∂yl(x,0)∂kS1(k-1)∂yk(x,0),hhhhhhhhhhhl(x1≤x≤x2,0≤l≤k).
For l=0,…,k-1, by the assumption of induction and (28), we have (∂lS1(k-1)/∂yl)(x,0)=0 and (∂lak,1/∂yl)(x,0)=(∂lbk,1/∂yl)(x,0)=0. So
(44)∂lS1(k)∂yl(x,0)=0.
For l=k, since (∂kak,1/∂yk)(x,0)=0,(∂kbk,1/∂yk)(x,0)=1, by (43), we have
(45)∂kS1(k)∂yk(x,0)=∂kS1(k-1)∂yk(x,0)-∂kS1(k-1)∂yk(x,0)=0.
The second formula of (38) holds. By induction, (38) holds for all k. From this, we get Lemma 5.Now we compute the mixed derivatives ofS1(k)(x,y) on the curved side Γ1 and bottom side Δ1 of E1.Lemma 6.
LetΓ1 and Δ1 be the curved side and the bottom side of E1, respectively. Then, for k∈ℤ+, (i)
(∂i+jS1(k)/∂xi∂yj)(x,y)=(∂i+jf/∂xi∂yj)(x,y)((x,y)∈Γ1),(ii)
(∂i+jS1(k)/∂xi∂yj)(x,y)=0((x,y)∈Δ1),where 0≤i+j≤k.Proof.
Letx1≤x≤x2. Then we have
(46)ddx(∂l-1f∂yl-1(x,g(x)))=∂lf∂x∂yl-1(x,g(x))+∂lf∂yl(x,g(x))g′(x),(l≥1).
By the Newton-Leibniz formula, we have
(47)∂l-1f∂yl-1(x,g(x))=∂l-1f∂yl-1(x1,g(x1))+∫x1x(∂lf∂x∂yl-1(t,g(t))+∂lf∂yl(t,g(t))g′(t))dt.
Similarly, replacing f by S1(k) in this formula, we have
(48)∂l-1S1(k)∂yl-1(x,g(x))=∂l-1S1(k)∂yl-1(x1,g(x1))+∫x1x(∂lS1(k)∂x∂yl-1(t,g(t))+∂lS1(k)∂yl(t,g(t))g′(t))dt,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhl.(l≥1).
From this and Lemma 5, it follows that, for any x1≤x≤x2, we have
(49)∫x1x∂lS1(k)∂x∂yl-1(t,g(t))dt‍=∫x1x∂lf∂x∂yl-1(t,g(t))dt‍(1≤l≤k).
Finding derivatives on the both sides of this formula, we get
(50)∂lS1(k)∂x∂yl-1(x,g(x))=∂lf∂x∂yl-1(x,g(x))(1≤l≤k).
Now we start from the equality(51)ddx(∂l-1f∂x∂yl-2(x,g(x)))=∂lf∂x2∂yl-2(x,g(x))+∂lf∂x∂yl-1(x,g(x))g′(x),hhhhhhhhhhhhhhhhh(l≥2).
Similar to the argument from (46) to (50), we get
(52)∂lS1(k)∂x2∂yl-2(x,g(x))=∂lf∂x2∂yl-2(x,g(x))(2≤l≤k).
Continuing this procedure, we deduce that (i) holds for 0<i+j≤k. Letting l=0 in Lemma 5, we have S1(k)(x,g(x))=f(x,g(x)); that is, (i) holds for i=j=0. So we get (i).
By Lemma5, (∂jS1(k)/∂yj)(x,0)=0(0≤j≤k). From this and S1(k)∈C∞(E1), we have
(53)∂i+jS1(k)∂xi∂yj(x,0)=0(0≤i+j≤k),
so (ii) holds. Lemma 6 is proved.From this, we get the following.Lemma 7.
For any positive integerr, denote lr=r(r+1)(r+2)(r+3). Let
(54)F(x,y)={S1(lr)(x,y),(x,y)∈E1,f(x,y),(x,y)∈Ω.
Then (i) F∈Clr(Ω⋃‍E1) and F(x,y)=f(x,y)((x,y)∈Ω); (ii) (∂i+jF/∂xi∂yj)(x,y)=0((x,y)∈(E1⋂‍∂T), 0≤i+j≤lr).Proof.
By the assumptionf∈C∞(Ω), Lemma 4: S1(k)∈C∞(E1), and Lemma 6(i):
(55)∂i+jS1(k)∂xi∂yj(x,y)=∂i+jf∂xi∂yj(x,y)hh..((x,y)∈Γ1,0≤i+j≤k),
where Γ1=Ω⋂‍E1, we get (i). By Lemma 6(ii) and E1⋂‍∂T=Δ1, we get (ii). Lemma 7 is proved.Forν=2,3,4, by using a similar method, we define Sν(k)(x,y) on the each trapezoid Eν with a curve side. The representations of Sν(k)(x,y) are stated in Section 4.1.Lemma 8.
For anyν=1,2,3,4, let
(56)F(x,y)={Sν(lr)(x,y),(x,y)∈Eν,f(x,y),(x,y)∈Ω,
where lr=r(r+1)(r+2)(r+3). Then, for ν=1,2,3,4, one has the following: (i)
F∈Clr(Ω⋃‍Eν);(ii)
(∂i+jF/∂xi∂yj)(x,y)=0,(x,y)∈(Eν⋂‍∂T) for 0≤i+j≤lr;(iii)
F(x,y) can be expressed in the form:
(57)F(x,y)=∑j=02lr+1ζj,1(x)yj,(x,y)∈E1,F(x,y)=∑j=02lr+1ζj,2(y)xj,(x,y)∈E2,F(x,y)=∑j=02lr+1ζj,3(x)yj,(x,y)∈E3,F(x,y)=∑j=02lr+1ζj,4(y)xj,(x,y)∈E4.Proof.
By Lemma7, we have
(58)F∈Clr(Ω⋃‍E1),∂i+jF∂xi∂yj(x,y)=0,fff.((x,y)∈(E1⋂‍∂T),0≤i+j≤lr).
Similar to the argument of Lemma 7, for ν=2,3,4, we have
(59)F∈Clr(Ω⋃‍Eν),∂i+jF∂xi∂yj(x,y)=0,ff.((x,y)∈(Eν⋂‍∂T),0≤i+j≤lr).
From this, we get (i) and (ii).
The proof of (iii) is similar to the argument of Lemma4(iii). Lemma 8 is proved.


### 3.3. Smooth Extension to Each RectangleHν

We have completed the smooth extension off to each trapezoid Eν with a curved side. In this subsection we complete the smooth extension of the obtained function F to each rectangle Hν. First we consider the smooth extension of F to H1. We divide this procedure in two steps.Step 1.
In Lemma8, we know that F(x,y)=S4lr(x,y) on E4. Now we construct the smooth extension of S4(lr)(x,y) from E4 to H1, where S4(lr)(x,y) is stated in Section 4.2 and lr=r(r+1)(r+2)(r+3).
Let(60)gggαk,11(y)=(y-y1)kk!(yy1)k+1,βk,11(y)=ykk!(y-y1-y1)k+1,k=0,1,…,
and let
(61)M1(0)(x,y)=S4(lr)(x,y1)α0,11(y),M1(k)(x,y)=M1(k-1)(x,y)+αk,11(y)hhhhhhhhhh×(∂kS4(lr)∂yk(x,y1)-∂kM1(k-1)∂yk(x,y1))M1(k)(x,y)=-βk,11(y)∂kM1(k-1)∂yk(x,0),gggggggggggggggggl.k=1,2,…,τr((x,y)∈H1),
where τr=r(r+2).Lemma 9.
Let{J1,l}14 be four sides of the rectangle H1:(62)J1,1={(x,y1),0≤x≤x1},J1,2={(0,y),0≤y≤y1},J1,3={(x,0),0≤x≤x1},J1,4={(x1,y),0≤y≤y1}.
Then one has the following(i)
M1(τr)(x,y)=∑i,j=02lr+1di,j(1)xiyj‍, where di,j(1) is a constant;(ii)
(∂i+jM1(τr)/∂xi∂yj)(x,y)=(∂i+jS4(lr)/∂xi∂yj)(x,y)((x,y)∈J1,1);(iii)
(∂i+jM1(τr)/∂xi∂yj)(x,y)=0((x,y)∈J1,2);(iv)
(∂i+jM1(τr)/∂xi∂yj)(x,y)=0((x,y)∈J1,3),where 0≤i+j≤τr.Proof.
By Lemma8(iii), we have
(63)S4(lr)(x,y)=∑j=02lr+1ζj,4(y)xj‍.
So (∂kS4(lr)/∂yk)(x,y1) is a polynomial of degree 2lr+1 with respect to x. Since ατr,11(y) and βτr,11(y) are both polynomials of degree 2τr+1, (i) follows from (61).
Similar to the argument of Lemma6, we get (ii) and (iv).
Since(0,y1)∈(E4⋂∂T), by Lemma 7, we have
(64)∂i+jS4(lr)∂xi∂yj(0,y1)=∂i+jF∂xi∂yj(0,y1)=0,(0≤i+j≤lr).
By the definition of M1(0) and (64), we have
(65)∂i+jM1(0)∂xi∂yj(0,y)=∂iS4(lr)∂xi(0,y1)djα0,11dyj(y)=0,hhhhl.(0≤i+j≤lr,y∈ℝ).
We assume that
(66)∂i+jM1(k-1)∂xi∂yj(0,y)=0,(0≤i+j≤lr-12k(k-1)).
By (61), we get
(67)∂i+jM1(k)∂xi∂yj(0,y)=∂i+jM1(k-1)∂xi∂yj(0,y)+djαk,11dyj(y)×(∂k+iS4(lr)∂xi∂yk(0,y1)-∂k+iM1(k-1)∂xi∂yk(0,y1))-djβk,11dyj(y)∂k+iM1(k-1)∂xi∂yk(0,0).
for 0≤i+j≤lr-(1/2)k(k+1), we have 0≤i+j≤lr-(1/2)k(k-1) and 0≤i+k≤lr-(1/2)k(k-1). Again, by the assumption of induction, we get
(68)∂i+jM1(k-1)∂xi∂yj(0,y)=0,∂k+iM1(k-1)∂xi∂yk(0,y1)=0.
By (64), we have (∂k+iS4(lr)/∂xi∂yk)(0,y1)=0. From this and (67), we get
(69)hhhhhhhhh∂i+jM1(k)∂xi∂yj(0,y)=0,(0≤y≤y1,0≤i+j≤lr-k2(k+1)).
Taking k=τr, we have
(70)gggggggg∂i+jM1(τr)∂xi∂yj(0,y)=0,(0≤y≤y1,0≤i+j≤lr-τr2(τr+1)).
Since lr-(τr/2)(τr+1)=(lr/2)≥τr, we get (iii). Lemma 9 is proved.Step 2.
In Lemma8, we know that F(x,y)=S1lr(x,y) on E1. We consider the difference S1(lr)(x,y)-M1(τr)(x,y). Obviously, it is infinitely many time differentiable on E1 since M1(τr)(x,y) is a polynomial. Now we construct its smooth extension from E1 to the rectangle H1 as follows. Let
(71)gggαk,14(x)=(x-x1)kk!(xx1)k+1,βk,14(x)=xkk!(x-x1-x1)k+1,k=0,1,…,
and let
(72)N1(0)(x,y)=(S1(lr)(x1,y)-M1(τr)(x1,y))α0,14(x),N1(k)(x,y)=N1(k-1)(x,y)+αk,14(x)N1(k)(x,y)=×(∂k(S1(lr)-M1(τr))∂xk(x1,y)N1(k)(x,y)==.=∂k(S1(lr)-M1(τr))∂xk-∂kN1(k-1)∂xk(x1,y))gggggggggg-βk,14(x)∂kN1(k-1)∂xk(0,y),ggggggk=0,1,…,r,((x,y)∈H1).
From this, we obtain the following.Lemma 10.
N
1
(
r
)
(
x
,
y
) possesses the following properties: (i)
(∂i+jN1(r)/∂xi∂yj)(x,y)=(∂i+jS1(lr)/∂xi∂yj)(x,y)-(∂i+jM1(τr)/∂xi∂yj)(x,y) on J1,4;(ii)
(∂i+jN1(r)/∂xi∂yj)(x,y)=0 on J1,2;(iii)
(∂i+jN1(r)/∂xi∂yj)(x,y)=0 on J1,1;(iv)
(∂i+jN1(r)/∂xi∂yj)(x,y)=0 on J1,3, where 0≤i+j≤r and {J1,ν}14 are stated in (62);(v)
N1(r)(x,y)=∑i,j=02lr+1τi,j(1)xiyj, where τi,j(1) is a constant.Proof.
The arguments similar to Lemma9(ii) and (iv) give the conclusions (i) and (ii) of this theorem. Now we prove (iii) and (iv).
By Lemma6(i) and Lemma 9(ii), as well as lr≥τr, we get that, for 0≤i+j≤τr,
(73)∂i+jS1(lr)∂xi∂yj(x1,y1)=∂i+jf∂xi∂yj(x1,y1)=∂i+jS4(lr)∂xi∂yj(x1,y1)=∂i+jM1(τr)∂xi∂yj(x1,y1).
So we have
(74)∂i+jN1(0)∂xi∂yj(x,y1)=∂j(S1(lr)-M1(τr))∂yj(x1,y1)diα0,14dxi(x)=0,hhhhhhh(0≤x≤x1,0≤i+j≤τr).
Now we assume that(75)ggggggg∂i+jN1(k-1)∂xi∂yj(x,y1)=0,(0≤x≤x1,0≤i+j≤τr-k2(k-1)).
By (72) and (73),
(76)∂i+jN1(k)∂xi∂yj(x,y1)=∂i+jN1(k-1)∂xi∂yj(x,y1)+diαk,14dxi(x)∂i+jN1(k)∂xi∂yj(x,y1)=×(∂k+j(S1(lr)-M1(τr))∂xk∂yj(x1,y1)∂i+jN1(k)∂xi∂yjggg(x,y1)=∂k+j(S1(lr)-M1(τr))∂xk∂yj-∂k+jN1(k-1)∂xk∂yj(x1,y1))∂i+jN1(k)∂xi∂yj(x,y1)=-diβk,14dxi(x)∂k+jN1(k-1)∂xk∂yj(0,y1)=0,∂i+jN1(k)∂xi∂yj.(0≤x≤x1,0≤i+j≤τr-k(k+1)2).
By induction, we get
(77)hhhhhhhhh∂i+jN1(k)∂xi∂yj(x,y1)=0,(0≤x≤x1,0≤i+j≤τr-k(k+1)2).
From this and τr-(1/2)r(r+1)≥r, we get (iii). By Lemma 6(ii) and Lemma 9(iii), we get that
(78)∂i+jS1(lr)∂xi∂yj(x,0)=0,(0≤i+j≤lr),∂i+jM1(τr)∂xi∂yj(x,0)=0,(0≤i+j≤τr).
From this and (72), by using an argument similar to the proof of (iii), we get (iv).
By Lemma8(iii) and Lemma 9(i), we deduce that (S1(lr)-M1(τr))(x1,y) is a polynomial of degree 2lr+1 with respect to y. From this and (72), we get (v). Lemma 10 is proved.By Lemmas9 and 10, we obtain that for 0≤i+j≤r,
(79)∂i+j(M1(τr)+N1(r))∂xi∂yj(x,y)=∂i+jS4(lr)∂xi∂yj(x,y)on⁡J1,1,(80)∂i+j(M1(τr)+N1(r))∂xi∂yj(x,y)=0on⁡J1,2⋃‍J1,3,(81)∂i+j(M1(τr)+N1(r))∂xi∂yj(x,y)=∂i+jS1(lr)∂xi∂yj(x,y)on⁡J1,4.Lemma 11.
Let(82)F(x,y)={f(x,y),(x,y)∈Ω,S1(lr)(x,y),(x,y)∈E1,S4(lr)(x,y),(x,y)∈E4,M1(τr)(x,y)+N1(r)(x,y),(x,y)∈H1.
Then one has(i)
F∈Cr(Ω⋃E1⋃E4⋃H1),(ii)
(∂i+jF/∂xi∂yj)(x,y)=0,(x,y)∈((E1⋃E4⋃H1)⋂∂T) for 0≤i+j≤r;(iii)
F(x,y)=∑i,j=02lr+1cij(1)xiyj((x,y)∈H1), where each cij(1) is constant.Proof.
By Lemma7, we have F∈Cr(Ω⋃E1⋃E4). Since S1(lr)∈Cr(E1),
(83)M1(τr)+N1(r)∈Cr(H1),E1⋂‍H1=J1,4,
by (81), we deduce that F∈Cr(E1⋃H1). Since S4(lr)∈Cr(E4),
(84)M1(τr)+N1(r)∈Cr(H1),E4⋂‍H1=J1,1,
by (79), we deduce that F∈Cr(H1⋃E4). So we get (i).
By Lemma8(ii),
(85)∂i+jF∂xi∂yj(x,y)=0,(x,y)∈((E1⋃‍E4)⋂‍∂T).
Since H1⋂∂T=J1,2⋃J1,3, by (80), we deduce that
(86)∂i+jF∂xi∂yj(x,y)=0,(x,y)∈(H1⋂‍∂T).
So we get (ii).
From Lemma9(i), Lemma 10(v), and F(x,y)=M1(τr)(x,y)+N1(r)(x,y)((x,y)∈H1), we get (iii). Lemma 11 is proved.Forν=2,3,4, by using a similar method, we define F(x,y)=Mν(τr)(x,y)+Nν(r)(x,y)((x,y)∈Hν), where representations of Mν(τr)(x,y) and Nν(r)(x,y) see Section 4.2.


### 3.4. The Proofs of the Theorems

Proof of Theorem1.
Let(87)F(x,y)={f(x,y),(x,y)∈Ω,Sν(lr)(x,y),(x,y)∈Eν(ν=1,2,3,4),Mν(τr)(x,y)+Nν(r)(x,y),(x,y)∈Hν(ν=1,2,3,4).
By (25), F has been defined on the unit square T. The argument similar to Lemma 11(i)-(ii) shows that
(88)F∈Cr(Ω⋃‍E1⋃‍E4⋃‍H1);F∈Cr(Ω⋃‍E1⋃‍E2⋃‍H2);F∈Cr(Ω⋃‍E2⋃‍E3⋃‍H3);F∈Cr(Ω⋃‍E3⋃‍E4⋃‍H4);
and for 0≤i+j≤r,
(89)∂i+jF∂xi∂yj(x,y)=0,(x,y)∈((E1⋃‍E4⋃‍H1)⋂‍∂T);∂i+jF∂xi∂yj(x,y)=0,(x,y)∈((E1⋃‍E2⋃‍H2)⋂‍∂T);∂i+jF∂xi∂yj(x,y)=0,(x,y)∈((E2⋃‍E3⋃‍H3)⋂‍∂T);∂i+jF∂xi∂yj(x,y)=0,(x,y)∈((E3⋃‍E4⋃‍H4)⋂‍∂T).
From this and Ω⋂∂T=∅, by (25), we have F∈Cr(T) and (∂i+jF/∂xi∂yj)(x,y)=0((x,y)∈∂T,0≤i+j≤r). So we get (i) and (ii).
Similar to the argument of Lemma11(iii), we get
(90)F(x,y)=∑i,j=02lr+1cij(ν)xiyj‍,(x,y)∈Hν(ν=1,2,3,4),
where each cij(ν) is a constant. From this and Lemma 8(iii), we know that, on T∖Ω, F(x,y) can be expressed locally in the form
(91)∑j=02lr+1ξj(x)yjor∑j=02lr+1ηj(x)xjor∑i,j=02lr+1cijxiyj;
(iii) holds. We have completed the proof of Theorem 1.The representation ofF satisfying the conditions of Theorem 1 is given in Section 4.Proof of Theorem2.
LetF be the smooth extension of f from Ω to T which is stated as in Theorem 1. Define Fp by
(92)Fp(x+k,y+l)=F(x,y)((x,y)∈T;k,l∈ℤ).
Then Fp is a 1-periodic function of ℝ2. By Theorem 1, we know that Fp∈Cr(T) and
(93)∂i+jFp∂xi∂yj(x,y)=0((x,y)∈∂T;0≤i+j≤r).
Let Tn1,n2=[n1,n1+1]×[n2,n2+1](n1,n2∈ℤ). Since Fp is 1-periodic function, we have Fp∈Cr(Tn1,n2) and for any n1,n2∈ℤ,
(94)∂i+jFp∂xi∂yj(x,y)=0((x,y)∈∂Tn1,n2;0≤i+j≤r).
Noticing that ℝ2=⋃n1,n2∈ℤTn1,n2, we have Fp∈Cr(ℝ2). By (92) and Theorem 1(i), we get
(95)Fp(x,y)=F(x,y)=f(x,y)((x,y)∈Ω).
Theorem 2 is proved.Proof of Theorem3.
LetF be the smooth extension of f from Ω to T which is stated as in Theorem 1. Define Fc by
(96)Fc(x,y)={F(x,y),(x,y)∈T,0,(x,y)∈ℝ2∖T.
From Theorem 1(ii), we have
(97)∂i+jFc∂xi∂yj(x,y)=0((x,y)∈∂T;0≤i+j≤r).
From this and (96), we get Fc(x,y)∈Cr(ℝ2). By (96) and Theorem 1(i), we get
(98)Fc(x,y)=F(x,y)=f(x,y)((x,y)∈Ω).
Theorem 3 is proved.


## 3.1. Partition of the Complement of the DomainΩ in T

SinceΩ⊂To and ∂Ω is a piecewise infinitely many time smooth curve, without loss of generality, we can divide the complement T∖Ω into some rectangles and some trapezoids with a curved side. For convenience of representation, we assume that we can choose four point (xν,yν)∈∂Ω(ν=1,2,3,4) such that T∖Ω can be divided into the four rectangles
(22)H1=[0,x1]×[0,y1],H2=[x2,1]×[0,y2],H3=[x3,1]×[y3,1],H4=[0,x4]×[y4,1]
and four trapezoids with a curved side
(23)E1={(x,y);x1≤x≤x2,0≤y≤g(x)},E2={(x,y);h(y)≤x≤1,y2≤y≤y3},E3={(x,y);x4≤x≤x3,g*(x)≤y≤1},E4={(x,y);0≤x≤h*(y),y1≤y≤y4},
where g∈C∞([x1,x2]),h∈C∞([y2,y3]),g*∈C∞([x4,x3]), and h*∈C∞([y1,y4]) and
(24)0<g(x)<1(x1≤x≤x2),0<h(y)<1(y2≤y≤y3),0<g*(x)<1(x4≤x≤x3),0<h*(y)<1(y1≤y≤y4).
From this, we know that T can be expressed into a disjoint union as follows:
(25)T=Ω⋃(⋃14Eν)‍⋃(⋃14Hν),
where each Eν is a trapezoid with a curved side and each Hν is a rectangle (see Figure 1).Figure 1
Partition of the complement of the domainΩ.In Sections3.2 and 3.3 we will extend f to each Eν and continue to extend to each Hν such that the obtained extension F satisfies the conditions of Theorem 1.


## 3.2. Smooth Extension to Each TrapezoidEν with a Curved Side

By (23), the trapezoid E1 with a curved side y=g(x)(x1≤x≤x2) is represented as
(26)E1={(x,y):x1≤x≤x2,0≤y≤g(x)}.
We define two sequences of functions {ak,1(x,y)}0∞ and {bk,1(x,y)}0∞ as follows:
(27)a0,1(x,y)=yg(x),b0,1(x,y)=y-g(x)-g(x),fffak,1(x,y)=(y-g(x))kk!(yg(x))k+1,bk,1(x,y)=ykk!(y-g(x)-g(x))k+1,k=1,2,….
By (27), we deduce that for x1≤x≤x2,
(28)hhhhhh∂lak,1∂yl(x,g(x))=0(0≤l≤k-1),∂kak,1∂yk(x,g(x))=1;(∂lak,1)∂yl(x,0)=0(0≤l≤k),hhhhhh∂lbk,1∂yl(x,g(x))=0(0≤l≤k);∂lbk,1∂yl(x,0)=0(0≤l≤k-1),(∂kbk,1)∂yk(x,0)=1.OnE1, we define a sequence of functions {S1(k)(x,y)}0∞ by induction.Let(29)S1(0)(x,y)=f(x,g(x))a0,1(x,y)ggg.(x1≤x≤x2,0≤y≤g(x)).
Then, by (27),
(30)S1(0)(x,0)=0,S1(0)(x,g(x))=f(x,g(x)),hhhhhhhhhhhhhhhhhhhhhhhhll(x1≤x≤x2).Let(31)S1(1)(x,y)=S1(0)(x,y)S1(1)(x,y)=+a1,1(x,y)(∂f∂y(x,g(x))-∂S1(0)∂y(x,g(x)))S1(1)(x,y)=-b1,1(x,y)∂S1(0)∂y(x,0)fffffffffffffffff.(x1≤x≤x2,0≤y≤g(x)).
Then, by (27)–(30), we obtain that, for x1≤x≤x2,
(32)S1(1)(x,g(x))=f(x,g(x)),∂S1(1)∂y(x,g(x))=∂f∂y(x,g(x)),S1(1)(x,0)=0,∂S1(1)∂y(x,0)=0.
In general, let
(33)S1(k)(x,y)=S1(k-1)(x,y)+ak,1(x,y)S1(k)(x,y)=×(∂kf∂yk(x,g(x))-∂kS1(k-1)∂yk(x,g(x)))S1(k)(x,y)=-bk,1(x,y)∂kS1(k-1)∂yk(x,0)S1(k)(x,y)ggggggggg=(x1≤x≤x2,0≤y≤g(x)).Lemma 4.
For anyk∈ℤ+, one has S1(k)∈C∞(E1) and
(34)S1(k)(x,y)=∑j=02k+1ζj,1(x)yj‍,(x,y)∈E1.Proof.
Sincef∈C∞(Ω) and g∈C∞([x1,x2]), and g(x)>0(x1≤x≤x2), by the above construction, we know that S1(k)∈C∞(E1) for any k=0,1,….
Fork=0, since
(35)S1(0)(x,y)=f(x,g(x))a0,1(x,y)=f(x,g(x))g(x)y,
(34) holds. We assume that (34) holds for k=l-1; that is,
(36)S1(l-1)(x,y)=∑j=02l-1ζj,1(l-1)(x)yj.
This implies that
(37)∂lS1(l-1)∂yl(x,g(x))=∑j=l2l-1j!(j-l)!ζj,1(l-1)(x)(g(x))j-l,∂lS1(l-1)∂yl(x,0)=l!ζl,1(l-1)(x).
Again, notice that al,1(x,y) and bl,1(x,y) are polynomials of y whose degrees are both 2l+1. From this and (33), it follows that (34) holds for k=l. By induction, (34) holds for all k. Lemma 4 is proved.Below we compute derivatives(∂lS1(k)/∂yl)(x,y)(0≤l≤k) on the curved side Γ1={(x,g(x)):x1≤x≤x2} and the bottom side Δ1={(x,0):x1≤x≤x2} of E1.Lemma 5.
LetS1(k)(x,y) be stated as above. For any k∈ℤ+, one has
(38)∂lS1(k)∂yl(x,g(x))=∂lf∂yl(x,g(x)),∂lS1(k)∂yl(x,0)=0(x1≤x≤x2,0≤l≤k).Proof.
By (30), We have known that, for k=0, (38) holds.
Now we assume that (38) holds for k-1.
Forx1≤x≤x2, by (33), we have
(39)∂lS1(k)∂yl(x,g(x))=∂lS1(k-1)∂yl(x,g(x))+∂lak,1∂yl(x,g(x))×(∂kf∂yk(x,g(x))-∂kS1(k-1)∂yk(x,g(x)))-∂lbk,1∂yl(x,g(x))∂kS1(k-1)∂yk(x,0),kkkkkkkkkkkkkkkhhhhl.k(0≤l≤k).
For l=0,1,…,k-1, by the assumption of induction, we have
(40)∂lS1(k-1)∂yl(x,g(x))=∂lf∂yl(x,g(x)).
By (28), we have (∂lak,1/∂yl)(x,g(x))=0,(∂lbk,1/∂yl)(x,g(x))=0. So we get
(41)∂lS1(k)∂yl(x,g(x))=∂lf∂yl(x,g(x)).
For l=k, note that (∂kak,1/∂yk)(x,g(x))=1 and (∂kbk,1/∂yk)(x,g(x))=0. By (39), we get
(42)∂kS1(k)∂yk(x,g(x))=∂kS1(k-1)∂yk(x,g(x))+(∂kf∂yk(x,g(x))-∂kS1(k-1)∂yk(x,g(x)))=∂kf∂yk(x,g(x)).
The first formula of (38) holds for k.
By (33), we have
(43)∂lS1(k)∂yl(x,0)=∂lS1(k-1)∂yl(x,0)+∂lak,1∂yl(x,0)×(∂kf∂yk(x,g(x))-∂kS1(k-1)∂yk(x,g(x)))-∂lbk,1∂yl(x,0)∂kS1(k-1)∂yk(x,0),hhhhhhhhhhhl(x1≤x≤x2,0≤l≤k).
For l=0,…,k-1, by the assumption of induction and (28), we have (∂lS1(k-1)/∂yl)(x,0)=0 and (∂lak,1/∂yl)(x,0)=(∂lbk,1/∂yl)(x,0)=0. So
(44)∂lS1(k)∂yl(x,0)=0.
For l=k, since (∂kak,1/∂yk)(x,0)=0,(∂kbk,1/∂yk)(x,0)=1, by (43), we have
(45)∂kS1(k)∂yk(x,0)=∂kS1(k-1)∂yk(x,0)-∂kS1(k-1)∂yk(x,0)=0.
The second formula of (38) holds. By induction, (38) holds for all k. From this, we get Lemma 5.Now we compute the mixed derivatives ofS1(k)(x,y) on the curved side Γ1 and bottom side Δ1 of E1.Lemma 6.
LetΓ1 and Δ1 be the curved side and the bottom side of E1, respectively. Then, for k∈ℤ+, (i)
(∂i+jS1(k)/∂xi∂yj)(x,y)=(∂i+jf/∂xi∂yj)(x,y)((x,y)∈Γ1),(ii)
(∂i+jS1(k)/∂xi∂yj)(x,y)=0((x,y)∈Δ1),where 0≤i+j≤k.Proof.
Letx1≤x≤x2. Then we have
(46)ddx(∂l-1f∂yl-1(x,g(x)))=∂lf∂x∂yl-1(x,g(x))+∂lf∂yl(x,g(x))g′(x),(l≥1).
By the Newton-Leibniz formula, we have
(47)∂l-1f∂yl-1(x,g(x))=∂l-1f∂yl-1(x1,g(x1))+∫x1x(∂lf∂x∂yl-1(t,g(t))+∂lf∂yl(t,g(t))g′(t))dt.
Similarly, replacing f by S1(k) in this formula, we have
(48)∂l-1S1(k)∂yl-1(x,g(x))=∂l-1S1(k)∂yl-1(x1,g(x1))+∫x1x(∂lS1(k)∂x∂yl-1(t,g(t))+∂lS1(k)∂yl(t,g(t))g′(t))dt,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhl.(l≥1).
From this and Lemma 5, it follows that, for any x1≤x≤x2, we have
(49)∫x1x∂lS1(k)∂x∂yl-1(t,g(t))dt‍=∫x1x∂lf∂x∂yl-1(t,g(t))dt‍(1≤l≤k).
Finding derivatives on the both sides of this formula, we get
(50)∂lS1(k)∂x∂yl-1(x,g(x))=∂lf∂x∂yl-1(x,g(x))(1≤l≤k).
Now we start from the equality(51)ddx(∂l-1f∂x∂yl-2(x,g(x)))=∂lf∂x2∂yl-2(x,g(x))+∂lf∂x∂yl-1(x,g(x))g′(x),hhhhhhhhhhhhhhhhh(l≥2).
Similar to the argument from (46) to (50), we get
(52)∂lS1(k)∂x2∂yl-2(x,g(x))=∂lf∂x2∂yl-2(x,g(x))(2≤l≤k).
Continuing this procedure, we deduce that (i) holds for 0<i+j≤k. Letting l=0 in Lemma 5, we have S1(k)(x,g(x))=f(x,g(x)); that is, (i) holds for i=j=0. So we get (i).
By Lemma5, (∂jS1(k)/∂yj)(x,0)=0(0≤j≤k). From this and S1(k)∈C∞(E1), we have
(53)∂i+jS1(k)∂xi∂yj(x,0)=0(0≤i+j≤k),
so (ii) holds. Lemma 6 is proved.From this, we get the following.Lemma 7.
For any positive integerr, denote lr=r(r+1)(r+2)(r+3). Let
(54)F(x,y)={S1(lr)(x,y),(x,y)∈E1,f(x,y),(x,y)∈Ω.
Then (i) F∈Clr(Ω⋃‍E1) and F(x,y)=f(x,y)((x,y)∈Ω); (ii) (∂i+jF/∂xi∂yj)(x,y)=0((x,y)∈(E1⋂‍∂T), 0≤i+j≤lr).Proof.
By the assumptionf∈C∞(Ω), Lemma 4: S1(k)∈C∞(E1), and Lemma 6(i):
(55)∂i+jS1(k)∂xi∂yj(x,y)=∂i+jf∂xi∂yj(x,y)hh..((x,y)∈Γ1,0≤i+j≤k),
where Γ1=Ω⋂‍E1, we get (i). By Lemma 6(ii) and E1⋂‍∂T=Δ1, we get (ii). Lemma 7 is proved.Forν=2,3,4, by using a similar method, we define Sν(k)(x,y) on the each trapezoid Eν with a curve side. The representations of Sν(k)(x,y) are stated in Section 4.1.Lemma 8.
For anyν=1,2,3,4, let
(56)F(x,y)={Sν(lr)(x,y),(x,y)∈Eν,f(x,y),(x,y)∈Ω,
where lr=r(r+1)(r+2)(r+3). Then, for ν=1,2,3,4, one has the following: (i)
F∈Clr(Ω⋃‍Eν);(ii)
(∂i+jF/∂xi∂yj)(x,y)=0,(x,y)∈(Eν⋂‍∂T) for 0≤i+j≤lr;(iii)
F(x,y) can be expressed in the form:
(57)F(x,y)=∑j=02lr+1ζj,1(x)yj,(x,y)∈E1,F(x,y)=∑j=02lr+1ζj,2(y)xj,(x,y)∈E2,F(x,y)=∑j=02lr+1ζj,3(x)yj,(x,y)∈E3,F(x,y)=∑j=02lr+1ζj,4(y)xj,(x,y)∈E4.Proof.
By Lemma7, we have
(58)F∈Clr(Ω⋃‍E1),∂i+jF∂xi∂yj(x,y)=0,fff.((x,y)∈(E1⋂‍∂T),0≤i+j≤lr).
Similar to the argument of Lemma 7, for ν=2,3,4, we have
(59)F∈Clr(Ω⋃‍Eν),∂i+jF∂xi∂yj(x,y)=0,ff.((x,y)∈(Eν⋂‍∂T),0≤i+j≤lr).
From this, we get (i) and (ii).
The proof of (iii) is similar to the argument of Lemma4(iii). Lemma 8 is proved.


## 3.3. Smooth Extension to Each RectangleHν

We have completed the smooth extension off to each trapezoid Eν with a curved side. In this subsection we complete the smooth extension of the obtained function F to each rectangle Hν. First we consider the smooth extension of F to H1. We divide this procedure in two steps.Step 1.
In Lemma8, we know that F(x,y)=S4lr(x,y) on E4. Now we construct the smooth extension of S4(lr)(x,y) from E4 to H1, where S4(lr)(x,y) is stated in Section 4.2 and lr=r(r+1)(r+2)(r+3).
Let(60)gggαk,11(y)=(y-y1)kk!(yy1)k+1,βk,11(y)=ykk!(y-y1-y1)k+1,k=0,1,…,
and let
(61)M1(0)(x,y)=S4(lr)(x,y1)α0,11(y),M1(k)(x,y)=M1(k-1)(x,y)+αk,11(y)hhhhhhhhhh×(∂kS4(lr)∂yk(x,y1)-∂kM1(k-1)∂yk(x,y1))M1(k)(x,y)=-βk,11(y)∂kM1(k-1)∂yk(x,0),gggggggggggggggggl.k=1,2,…,τr((x,y)∈H1),
where τr=r(r+2).Lemma 9.
Let{J1,l}14 be four sides of the rectangle H1:(62)J1,1={(x,y1),0≤x≤x1},J1,2={(0,y),0≤y≤y1},J1,3={(x,0),0≤x≤x1},J1,4={(x1,y),0≤y≤y1}.
Then one has the following(i)
M1(τr)(x,y)=∑i,j=02lr+1di,j(1)xiyj‍, where di,j(1) is a constant;(ii)
(∂i+jM1(τr)/∂xi∂yj)(x,y)=(∂i+jS4(lr)/∂xi∂yj)(x,y)((x,y)∈J1,1);(iii)
(∂i+jM1(τr)/∂xi∂yj)(x,y)=0((x,y)∈J1,2);(iv)
(∂i+jM1(τr)/∂xi∂yj)(x,y)=0((x,y)∈J1,3),where 0≤i+j≤τr.Proof.
By Lemma8(iii), we have
(63)S4(lr)(x,y)=∑j=02lr+1ζj,4(y)xj‍.
So (∂kS4(lr)/∂yk)(x,y1) is a polynomial of degree 2lr+1 with respect to x. Since ατr,11(y) and βτr,11(y) are both polynomials of degree 2τr+1, (i) follows from (61).
Similar to the argument of Lemma6, we get (ii) and (iv).
Since(0,y1)∈(E4⋂∂T), by Lemma 7, we have
(64)∂i+jS4(lr)∂xi∂yj(0,y1)=∂i+jF∂xi∂yj(0,y1)=0,(0≤i+j≤lr).
By the definition of M1(0) and (64), we have
(65)∂i+jM1(0)∂xi∂yj(0,y)=∂iS4(lr)∂xi(0,y1)djα0,11dyj(y)=0,hhhhl.(0≤i+j≤lr,y∈ℝ).
We assume that
(66)∂i+jM1(k-1)∂xi∂yj(0,y)=0,(0≤i+j≤lr-12k(k-1)).
By (61), we get
(67)∂i+jM1(k)∂xi∂yj(0,y)=∂i+jM1(k-1)∂xi∂yj(0,y)+djαk,11dyj(y)×(∂k+iS4(lr)∂xi∂yk(0,y1)-∂k+iM1(k-1)∂xi∂yk(0,y1))-djβk,11dyj(y)∂k+iM1(k-1)∂xi∂yk(0,0).
for 0≤i+j≤lr-(1/2)k(k+1), we have 0≤i+j≤lr-(1/2)k(k-1) and 0≤i+k≤lr-(1/2)k(k-1). Again, by the assumption of induction, we get
(68)∂i+jM1(k-1)∂xi∂yj(0,y)=0,∂k+iM1(k-1)∂xi∂yk(0,y1)=0.
By (64), we have (∂k+iS4(lr)/∂xi∂yk)(0,y1)=0. From this and (67), we get
(69)hhhhhhhhh∂i+jM1(k)∂xi∂yj(0,y)=0,(0≤y≤y1,0≤i+j≤lr-k2(k+1)).
Taking k=τr, we have
(70)gggggggg∂i+jM1(τr)∂xi∂yj(0,y)=0,(0≤y≤y1,0≤i+j≤lr-τr2(τr+1)).
Since lr-(τr/2)(τr+1)=(lr/2)≥τr, we get (iii). Lemma 9 is proved.Step 2.
In Lemma8, we know that F(x,y)=S1lr(x,y) on E1. We consider the difference S1(lr)(x,y)-M1(τr)(x,y). Obviously, it is infinitely many time differentiable on E1 since M1(τr)(x,y) is a polynomial. Now we construct its smooth extension from E1 to the rectangle H1 as follows. Let
(71)gggαk,14(x)=(x-x1)kk!(xx1)k+1,βk,14(x)=xkk!(x-x1-x1)k+1,k=0,1,…,
and let
(72)N1(0)(x,y)=(S1(lr)(x1,y)-M1(τr)(x1,y))α0,14(x),N1(k)(x,y)=N1(k-1)(x,y)+αk,14(x)N1(k)(x,y)=×(∂k(S1(lr)-M1(τr))∂xk(x1,y)N1(k)(x,y)==.=∂k(S1(lr)-M1(τr))∂xk-∂kN1(k-1)∂xk(x1,y))gggggggggg-βk,14(x)∂kN1(k-1)∂xk(0,y),ggggggk=0,1,…,r,((x,y)∈H1).
From this, we obtain the following.Lemma 10.
N
1
(
r
)
(
x
,
y
) possesses the following properties: (i)
(∂i+jN1(r)/∂xi∂yj)(x,y)=(∂i+jS1(lr)/∂xi∂yj)(x,y)-(∂i+jM1(τr)/∂xi∂yj)(x,y) on J1,4;(ii)
(∂i+jN1(r)/∂xi∂yj)(x,y)=0 on J1,2;(iii)
(∂i+jN1(r)/∂xi∂yj)(x,y)=0 on J1,1;(iv)
(∂i+jN1(r)/∂xi∂yj)(x,y)=0 on J1,3, where 0≤i+j≤r and {J1,ν}14 are stated in (62);(v)
N1(r)(x,y)=∑i,j=02lr+1τi,j(1)xiyj, where τi,j(1) is a constant.Proof.
The arguments similar to Lemma9(ii) and (iv) give the conclusions (i) and (ii) of this theorem. Now we prove (iii) and (iv).
By Lemma6(i) and Lemma 9(ii), as well as lr≥τr, we get that, for 0≤i+j≤τr,
(73)∂i+jS1(lr)∂xi∂yj(x1,y1)=∂i+jf∂xi∂yj(x1,y1)=∂i+jS4(lr)∂xi∂yj(x1,y1)=∂i+jM1(τr)∂xi∂yj(x1,y1).
So we have
(74)∂i+jN1(0)∂xi∂yj(x,y1)=∂j(S1(lr)-M1(τr))∂yj(x1,y1)diα0,14dxi(x)=0,hhhhhhh(0≤x≤x1,0≤i+j≤τr).
Now we assume that(75)ggggggg∂i+jN1(k-1)∂xi∂yj(x,y1)=0,(0≤x≤x1,0≤i+j≤τr-k2(k-1)).
By (72) and (73),
(76)∂i+jN1(k)∂xi∂yj(x,y1)=∂i+jN1(k-1)∂xi∂yj(x,y1)+diαk,14dxi(x)∂i+jN1(k)∂xi∂yj(x,y1)=×(∂k+j(S1(lr)-M1(τr))∂xk∂yj(x1,y1)∂i+jN1(k)∂xi∂yjggg(x,y1)=∂k+j(S1(lr)-M1(τr))∂xk∂yj-∂k+jN1(k-1)∂xk∂yj(x1,y1))∂i+jN1(k)∂xi∂yj(x,y1)=-diβk,14dxi(x)∂k+jN1(k-1)∂xk∂yj(0,y1)=0,∂i+jN1(k)∂xi∂yj.(0≤x≤x1,0≤i+j≤τr-k(k+1)2).
By induction, we get
(77)hhhhhhhhh∂i+jN1(k)∂xi∂yj(x,y1)=0,(0≤x≤x1,0≤i+j≤τr-k(k+1)2).
From this and τr-(1/2)r(r+1)≥r, we get (iii). By Lemma 6(ii) and Lemma 9(iii), we get that
(78)∂i+jS1(lr)∂xi∂yj(x,0)=0,(0≤i+j≤lr),∂i+jM1(τr)∂xi∂yj(x,0)=0,(0≤i+j≤τr).
From this and (72), by using an argument similar to the proof of (iii), we get (iv).
By Lemma8(iii) and Lemma 9(i), we deduce that (S1(lr)-M1(τr))(x1,y) is a polynomial of degree 2lr+1 with respect to y. From this and (72), we get (v). Lemma 10 is proved.By Lemmas9 and 10, we obtain that for 0≤i+j≤r,
(79)∂i+j(M1(τr)+N1(r))∂xi∂yj(x,y)=∂i+jS4(lr)∂xi∂yj(x,y)on⁡J1,1,(80)∂i+j(M1(τr)+N1(r))∂xi∂yj(x,y)=0on⁡J1,2⋃‍J1,3,(81)∂i+j(M1(τr)+N1(r))∂xi∂yj(x,y)=∂i+jS1(lr)∂xi∂yj(x,y)on⁡J1,4.Lemma 11.
Let(82)F(x,y)={f(x,y),(x,y)∈Ω,S1(lr)(x,y),(x,y)∈E1,S4(lr)(x,y),(x,y)∈E4,M1(τr)(x,y)+N1(r)(x,y),(x,y)∈H1.
Then one has(i)
F∈Cr(Ω⋃E1⋃E4⋃H1),(ii)
(∂i+jF/∂xi∂yj)(x,y)=0,(x,y)∈((E1⋃E4⋃H1)⋂∂T) for 0≤i+j≤r;(iii)
F(x,y)=∑i,j=02lr+1cij(1)xiyj((x,y)∈H1), where each cij(1) is constant.Proof.
By Lemma7, we have F∈Cr(Ω⋃E1⋃E4). Since S1(lr)∈Cr(E1),
(83)M1(τr)+N1(r)∈Cr(H1),E1⋂‍H1=J1,4,
by (81), we deduce that F∈Cr(E1⋃H1). Since S4(lr)∈Cr(E4),
(84)M1(τr)+N1(r)∈Cr(H1),E4⋂‍H1=J1,1,
by (79), we deduce that F∈Cr(H1⋃E4). So we get (i).
By Lemma8(ii),
(85)∂i+jF∂xi∂yj(x,y)=0,(x,y)∈((E1⋃‍E4)⋂‍∂T).
Since H1⋂∂T=J1,2⋃J1,3, by (80), we deduce that
(86)∂i+jF∂xi∂yj(x,y)=0,(x,y)∈(H1⋂‍∂T).
So we get (ii).
From Lemma9(i), Lemma 10(v), and F(x,y)=M1(τr)(x,y)+N1(r)(x,y)((x,y)∈H1), we get (iii). Lemma 11 is proved.Forν=2,3,4, by using a similar method, we define F(x,y)=Mν(τr)(x,y)+Nν(r)(x,y)((x,y)∈Hν), where representations of Mν(τr)(x,y) and Nν(r)(x,y) see Section 4.2.


## 3.4. The Proofs of the Theorems

Proof of Theorem1.
Let(87)F(x,y)={f(x,y),(x,y)∈Ω,Sν(lr)(x,y),(x,y)∈Eν(ν=1,2,3,4),Mν(τr)(x,y)+Nν(r)(x,y),(x,y)∈Hν(ν=1,2,3,4).
By (25), F has been defined on the unit square T. The argument similar to Lemma 11(i)-(ii) shows that
(88)F∈Cr(Ω⋃‍E1⋃‍E4⋃‍H1);F∈Cr(Ω⋃‍E1⋃‍E2⋃‍H2);F∈Cr(Ω⋃‍E2⋃‍E3⋃‍H3);F∈Cr(Ω⋃‍E3⋃‍E4⋃‍H4);
and for 0≤i+j≤r,
(89)∂i+jF∂xi∂yj(x,y)=0,(x,y)∈((E1⋃‍E4⋃‍H1)⋂‍∂T);∂i+jF∂xi∂yj(x,y)=0,(x,y)∈((E1⋃‍E2⋃‍H2)⋂‍∂T);∂i+jF∂xi∂yj(x,y)=0,(x,y)∈((E2⋃‍E3⋃‍H3)⋂‍∂T);∂i+jF∂xi∂yj(x,y)=0,(x,y)∈((E3⋃‍E4⋃‍H4)⋂‍∂T).
From this and Ω⋂∂T=∅, by (25), we have F∈Cr(T) and (∂i+jF/∂xi∂yj)(x,y)=0((x,y)∈∂T,0≤i+j≤r). So we get (i) and (ii).
Similar to the argument of Lemma11(iii), we get
(90)F(x,y)=∑i,j=02lr+1cij(ν)xiyj‍,(x,y)∈Hν(ν=1,2,3,4),
where each cij(ν) is a constant. From this and Lemma 8(iii), we know that, on T∖Ω, F(x,y) can be expressed locally in the form
(91)∑j=02lr+1ξj(x)yjor∑j=02lr+1ηj(x)xjor∑i,j=02lr+1cijxiyj;
(iii) holds. We have completed the proof of Theorem 1.The representation ofF satisfying the conditions of Theorem 1 is given in Section 4.Proof of Theorem2.
LetF be the smooth extension of f from Ω to T which is stated as in Theorem 1. Define Fp by
(92)Fp(x+k,y+l)=F(x,y)((x,y)∈T;k,l∈ℤ).
Then Fp is a 1-periodic function of ℝ2. By Theorem 1, we know that Fp∈Cr(T) and
(93)∂i+jFp∂xi∂yj(x,y)=0((x,y)∈∂T;0≤i+j≤r).
Let Tn1,n2=[n1,n1+1]×[n2,n2+1](n1,n2∈ℤ). Since Fp is 1-periodic function, we have Fp∈Cr(Tn1,n2) and for any n1,n2∈ℤ,
(94)∂i+jFp∂xi∂yj(x,y)=0((x,y)∈∂Tn1,n2;0≤i+j≤r).
Noticing that ℝ2=⋃n1,n2∈ℤTn1,n2, we have Fp∈Cr(ℝ2). By (92) and Theorem 1(i), we get
(95)Fp(x,y)=F(x,y)=f(x,y)((x,y)∈Ω).
Theorem 2 is proved.Proof of Theorem3.
LetF be the smooth extension of f from Ω to T which is stated as in Theorem 1. Define Fc by
(96)Fc(x,y)={F(x,y),(x,y)∈T,0,(x,y)∈ℝ2∖T.
From Theorem 1(ii), we have
(97)∂i+jFc∂xi∂yj(x,y)=0((x,y)∈∂T;0≤i+j≤r).
From this and (96), we get Fc(x,y)∈Cr(ℝ2). By (96) and Theorem 1(i), we get
(98)Fc(x,y)=F(x,y)=f(x,y)((x,y)∈Ω).
Theorem 3 is proved.


## 4. Representation of the ExtensionF Satisfying Theorem 1

Letf and Ω be stated as in Theorem 1 and let Ω be divided as in Section 3.1. The representation of F satisfying conditions of Theorem 1 is as follows:
(99)F(x,y)={f(x,y),(x,y)∈Ω,Sν(lr)(x,y),(x,y)∈Eν(ν=1,2,3,4),Mν(τr)(x,y)+Nν(r)(x,y),(x,y)∈Hν(ν=1,2,3,4),
where
(100)T=Ω⋃‍(⋃14‍Eν)⋃‍(⋃14‍Hν)
and the rectangles {Hν}14 and the trapezoids {Eν}14 with a curved side are stated in (22) and (23) and lr=r(r+1)(r+2)(r+3) and τr=r(r+2).Below we write out the representations of{Sν(k)(x,y)}14, {Mν(k)(x,y)}14, and {Nν(k)(x,y)}14.


### 4.1. The Construction of EachSν(k)(x,y)

(i) Denote(101)hhhhak,1(x,y)=(y-g(x))kk!(yg(x))k+1,bk,1(x,y)=ykk!(y-g(x)-g(x))k+1,k=0,1,….
Define S1(k)(x,y) by induction as follows:
(102)S1(0)(x,y)=f(x,g(x))a0,1(x,y),S1(k)(x,y)=S1(k-1)(x,y)+ak,1(x,y)S1(k)(x,y)=×(∂kf∂yk(x,g(x))-∂kS1(k-1)∂yk(x,g(x)))S1(k)(x,y)=-bk,1(x,y)∂kS1(k-1)∂yk(x,0),hhhhhhhhhhhhhhhhh.k=1,2,…,((x,y)∈E1).(ii) Denote(103)hhhhak,2(x,y)=(x-h(y))kk!(1-x1-h(y))k+1,bk,2(x,y)=(x-1)kk!(h(y)-xh(y)-1)k+1,k=0,1,….
Define S2(k)(x,y) by induction as follows:
(104)S2(0)(x,y)=f(h(y),y)a0,2(x,y),S2(k)(x,y)=S2(k-1)(x,y)+ak,2(x,y)S2(k)(x,y)=×(∂kf∂xk(h(y),y)-∂kS2(k-1)∂xk(h(y),y))S2(k)(x,y)=-bk,2(x,y)∂kS2(k-1)∂xk(1,y),hhhhhhhhhhhhhhhhk=1,2,…,((x,y)∈E2).(iii) Denote(105)ak,3(x,y)=(y-g*(x))kk!(1-y1-g*(x))k+1,bk,3(x,y)=(y-1)kk!(g*(x)-yg*(x)-1)k+1,k=0,1,….
Define S3(k)(x,y) by induction as follows:
(106)S3(0)(x,y)=f(x,g*(x))a0,3(x,y),S3(k)(x,y)=S3(k-1)(x,y)+ak,3(x,y)S3(k)(x,y)=×(∂kf∂yk(x,g*(x))-∂kS3(k-1)∂yk(x,g*(x)))S3(k)(x,y)=-bk,3(x,y)∂kS3(k-1)∂yk(x,1),jjjjjjjhhhhhhhhhhjjjjjjjjjjk=1,2,…((x,y)∈E3).(iv) Denote(107)ak,4(x,y)=(x-h*(y))kk!(xh*(y))k+1,bk,4(x,y)=xkk!(x-h*(y)-h*(y))k+1,k=0,1,….
Define S4(k)(x,y) by induction as follows:
(108)S4(0)(x,y)=f(h*(y),y)a0,4(x,y),S4(k)(x,y)=S4(k-1)(x,y)+ak,4(x,y)S4(k)(x,y)=×(∂kf∂xk(h*(y),y)-∂kS4(k-1)∂xk(h*(y),y))hhhhhhhhh-bk,4(x,y)∂kS4(k-1)∂xk(0,y),hhhhhhhhhhhhhhhhkkhk=1,2,…((x,y)∈E4).


### 4.2. The Constructions of EachMν(k)(x,y) and Nν(k)(x,y)

(i) Denote(109)αk,11(y)=(y-y1)kk!(yy1)k+1,βk,11(y)=ykk!(y-y1-y1)k+1,k=0,1,….
Define M1(k)(x,y) by induction as follows:
(110)M1(0)(x,y)=S4(lr)(x,y1)α0,11(y),M1(k)(x,y)=M1(k-1)(x,y)+αk,11(y)M1(k)(x,y)=×(∂kS4(lr)∂yk(x,y1)-∂kM1(k-1)∂yk(x,y1))M1(k)(x,y)=-βk,11(y)∂kM1(k-1)∂yk(x,0),hhhhhhhhhhhhhhhk=1,2,…,((x,y)∈H1).Denote(111)αk,14(x)=(x-x1)kk!(xx1)k+1,βk,14(x)=xkk!(x-x1-x1)k+1,k=0,1,….
Define N1(k)(x,y) by induction as follows:
(112)N1(0)(x,y)=(S1(lr)(x1,y)-M1(2r)(x1,y))α0,14(x)N1(k)(x,y)=N1(k-1)(x,y)+αk,14(x)N1(k)(x,y)=×(∂k(S1(4r)-M1(2r))∂xk(x1,y)kkkkkkkkk.kkk.∂k(S1(4r)-M1(2r))∂xk-∂kN1(k-1)∂xk(x1,y))N1(k)(x,y)=-βk,14(x)∂kN1(k-1)∂xk(0,y),hhhhhhhk=1,2,…((x,y)∈H1).(ii) Denote(113)αk,21(x)=(x-x2)kk!(1-x1-x2)k+1,βk,21(x)=(x-1)kk!(x2-xx2-1)k+1,k=0,1,….
Define M2(τr)(x,y) by induction as follows:
(114)M2(0)(x,y)=S1(lr)(x2,y)α0,21(x),M2(k)(x,y)=M2(k-1)(x,y)+αk,21(x)M2(k)(x,y)=×(∂kS1(lr)∂xk(x2,y)M2(k)(x,y)hhh=∂kS1(lr)∂xk-∂kM2(k-1)∂xk(x2,y))M2(k)(x,y)=-βk,21(x)∂kM2(k-1)∂xk(1,y),ggggggggggglk=1,2,…((x,y)∈H2).Denote(115)αk,22(y)=(y-y2)kk!(yy2)k+1,βk,22(y)=ykk!(y-y2-y2)k+1,k=0,1,….
Define N2(r)(x,y) by induction as follows:
(116)N2(0)(x,y)=(S2(lr)-M2(τr))(x,y2)α0,22(y),N2(k)(x,y)=N2(k-1)(x,y)+αk,22(y)N2(k)(x,y)=×(∂k(S2(lr)-M2(τr))∂yk(x,y2)N2(k)(x,y)=hhhh.-∂kN2(k-1)∂yk(x,y2))N2(k)(x,y)=-βk,22(y)∂kN2(k-1)∂yk(x,0),hhhhhhhk=1,2,…((x,y)∈H2).(iii) Denote(117)αk,31(y)=(y-y3)kk!(1-y1-y3)k+1,βk,31(y)=(y-1)kk!(y3-yy3-1)k+1,k=0,1,….
Define M3(k)(x,y) by induction as follows:
(118)M3(0)(x,y)=S2(lr)(x,y3)α0,31(y).M3(k)(x,y)=M3(k-1)(x,y)+αk,31(y)M3(k)(x,y)=×(∂kS2(lr)∂yk(x,y3)-∂kM3(k-1)∂yk(x,y3))M3(k)(x,y)=-βk,31(y)∂kM3(k-1)∂yk(x,1),ggggghhhhhhhgggggk=1,2,…((x,y)∈H3).Denote(119)αk,32(x)=(x-x3)kk!(1-x1-x3)k+1,βk,32(x)=(x-1)kk!(x3-xx3-1)k+1,k=0,1,….
Define N3(k)(x,y) by induction as follows:
(120)N3(0)(x,y)=(S3(lr)-M3(τr))(x3,y)α0,32(x),N3(k)(x,y)=N3(k-1)(x,y)+αk,32(x)N3(k)(x,y)=×(∂k(S3(lr)-M3(τr))∂yk(x3,y)N3(k)(x,y)×=m∂k(S3(lr)-M3(τr))∂yk-∂kN3(k-1)∂xk(x3,y))N3(k)(x,y)=-βk,32(x)∂kN3(k-1)∂xk(1,y),hhhhhhhk=1,2,…((x,y)∈H3).(iv) Denote(121)αk,41(x)=(x-x4)kk!(xx4)k+1,βk,41(x)=(x-x4)kk!(x-x4)k+1,k=0,1,….
Define M4(k)(x,y) by induction as follows:
(122)M4(0)(x,y)=S3(lr)(x4,y)α0,41(x),M4(k)(x,y)=M4(k-1)(x,y)+αk,41(x)M4(k)(x,y)=×(∂kS3(lr)∂xk(x4,y)-∂kM4(k-1)∂xk(x4,y))M4(k)(x,y)=-βk,41(x,y)∂kM4(k-1)∂xk(0,y),hhhhhhhhhhhhhhhhk=1,2,…((x,y)∈H4).Denote(123)αk,42(y)=(y-y4)kk!(1-y1-y4)k+1,βk,42(y)=(y-1)kk!(y4-yy4-1)k+1,k=0,1,….
Define N4(k)(x,y) by induction as follows:
(124)N4(0)(x,y)=(S4(lr)-M4(τr))(x,y4)α0,42(y),N4(k)(x,y)=N4(k-1)(x,y)+αk,42(y)N4(k)(x,y)=×(∂k(S4(lr)-M4(τr))∂yk(x,y4)N4(k)(x,y)=hh.,.-∂kN4(k-1)∂yk(x,y4))N4(k)(x,y)=-βk,42(x,y)∂kN4(k-1)∂xk(x,1),hhhhhhhh.k=1,2,…((x,y)∈H4).


## 4.1. The Construction of EachSν(k)(x,y)

(i) Denote(101)hhhhak,1(x,y)=(y-g(x))kk!(yg(x))k+1,bk,1(x,y)=ykk!(y-g(x)-g(x))k+1,k=0,1,….
Define S1(k)(x,y) by induction as follows:
(102)S1(0)(x,y)=f(x,g(x))a0,1(x,y),S1(k)(x,y)=S1(k-1)(x,y)+ak,1(x,y)S1(k)(x,y)=×(∂kf∂yk(x,g(x))-∂kS1(k-1)∂yk(x,g(x)))S1(k)(x,y)=-bk,1(x,y)∂kS1(k-1)∂yk(x,0),hhhhhhhhhhhhhhhhh.k=1,2,…,((x,y)∈E1).(ii) Denote(103)hhhhak,2(x,y)=(x-h(y))kk!(1-x1-h(y))k+1,bk,2(x,y)=(x-1)kk!(h(y)-xh(y)-1)k+1,k=0,1,….
Define S2(k)(x,y) by induction as follows:
(104)S2(0)(x,y)=f(h(y),y)a0,2(x,y),S2(k)(x,y)=S2(k-1)(x,y)+ak,2(x,y)S2(k)(x,y)=×(∂kf∂xk(h(y),y)-∂kS2(k-1)∂xk(h(y),y))S2(k)(x,y)=-bk,2(x,y)∂kS2(k-1)∂xk(1,y),hhhhhhhhhhhhhhhhk=1,2,…,((x,y)∈E2).(iii) Denote(105)ak,3(x,y)=(y-g*(x))kk!(1-y1-g*(x))k+1,bk,3(x,y)=(y-1)kk!(g*(x)-yg*(x)-1)k+1,k=0,1,….
Define S3(k)(x,y) by induction as follows:
(106)S3(0)(x,y)=f(x,g*(x))a0,3(x,y),S3(k)(x,y)=S3(k-1)(x,y)+ak,3(x,y)S3(k)(x,y)=×(∂kf∂yk(x,g*(x))-∂kS3(k-1)∂yk(x,g*(x)))S3(k)(x,y)=-bk,3(x,y)∂kS3(k-1)∂yk(x,1),jjjjjjjhhhhhhhhhhjjjjjjjjjjk=1,2,…((x,y)∈E3).(iv) Denote(107)ak,4(x,y)=(x-h*(y))kk!(xh*(y))k+1,bk,4(x,y)=xkk!(x-h*(y)-h*(y))k+1,k=0,1,….
Define S4(k)(x,y) by induction as follows:
(108)S4(0)(x,y)=f(h*(y),y)a0,4(x,y),S4(k)(x,y)=S4(k-1)(x,y)+ak,4(x,y)S4(k)(x,y)=×(∂kf∂xk(h*(y),y)-∂kS4(k-1)∂xk(h*(y),y))hhhhhhhhh-bk,4(x,y)∂kS4(k-1)∂xk(0,y),hhhhhhhhhhhhhhhhkkhk=1,2,…((x,y)∈E4).


## 4.2. The Constructions of EachMν(k)(x,y) and Nν(k)(x,y)

(i) Denote(109)αk,11(y)=(y-y1)kk!(yy1)k+1,βk,11(y)=ykk!(y-y1-y1)k+1,k=0,1,….
Define M1(k)(x,y) by induction as follows:
(110)M1(0)(x,y)=S4(lr)(x,y1)α0,11(y),M1(k)(x,y)=M1(k-1)(x,y)+αk,11(y)M1(k)(x,y)=×(∂kS4(lr)∂yk(x,y1)-∂kM1(k-1)∂yk(x,y1))M1(k)(x,y)=-βk,11(y)∂kM1(k-1)∂yk(x,0),hhhhhhhhhhhhhhhk=1,2,…,((x,y)∈H1).Denote(111)αk,14(x)=(x-x1)kk!(xx1)k+1,βk,14(x)=xkk!(x-x1-x1)k+1,k=0,1,….
Define N1(k)(x,y) by induction as follows:
(112)N1(0)(x,y)=(S1(lr)(x1,y)-M1(2r)(x1,y))α0,14(x)N1(k)(x,y)=N1(k-1)(x,y)+αk,14(x)N1(k)(x,y)=×(∂k(S1(4r)-M1(2r))∂xk(x1,y)kkkkkkkkk.kkk.∂k(S1(4r)-M1(2r))∂xk-∂kN1(k-1)∂xk(x1,y))N1(k)(x,y)=-βk,14(x)∂kN1(k-1)∂xk(0,y),hhhhhhhk=1,2,…((x,y)∈H1).(ii) Denote(113)αk,21(x)=(x-x2)kk!(1-x1-x2)k+1,βk,21(x)=(x-1)kk!(x2-xx2-1)k+1,k=0,1,….
Define M2(τr)(x,y) by induction as follows:
(114)M2(0)(x,y)=S1(lr)(x2,y)α0,21(x),M2(k)(x,y)=M2(k-1)(x,y)+αk,21(x)M2(k)(x,y)=×(∂kS1(lr)∂xk(x2,y)M2(k)(x,y)hhh=∂kS1(lr)∂xk-∂kM2(k-1)∂xk(x2,y))M2(k)(x,y)=-βk,21(x)∂kM2(k-1)∂xk(1,y),ggggggggggglk=1,2,…((x,y)∈H2).Denote(115)αk,22(y)=(y-y2)kk!(yy2)k+1,βk,22(y)=ykk!(y-y2-y2)k+1,k=0,1,….
Define N2(r)(x,y) by induction as follows:
(116)N2(0)(x,y)=(S2(lr)-M2(τr))(x,y2)α0,22(y),N2(k)(x,y)=N2(k-1)(x,y)+αk,22(y)N2(k)(x,y)=×(∂k(S2(lr)-M2(τr))∂yk(x,y2)N2(k)(x,y)=hhhh.-∂kN2(k-1)∂yk(x,y2))N2(k)(x,y)=-βk,22(y)∂kN2(k-1)∂yk(x,0),hhhhhhhk=1,2,…((x,y)∈H2).(iii) Denote(117)αk,31(y)=(y-y3)kk!(1-y1-y3)k+1,βk,31(y)=(y-1)kk!(y3-yy3-1)k+1,k=0,1,….
Define M3(k)(x,y) by induction as follows:
(118)M3(0)(x,y)=S2(lr)(x,y3)α0,31(y).M3(k)(x,y)=M3(k-1)(x,y)+αk,31(y)M3(k)(x,y)=×(∂kS2(lr)∂yk(x,y3)-∂kM3(k-1)∂yk(x,y3))M3(k)(x,y)=-βk,31(y)∂kM3(k-1)∂yk(x,1),ggggghhhhhhhgggggk=1,2,…((x,y)∈H3).Denote(119)αk,32(x)=(x-x3)kk!(1-x1-x3)k+1,βk,32(x)=(x-1)kk!(x3-xx3-1)k+1,k=0,1,….
Define N3(k)(x,y) by induction as follows:
(120)N3(0)(x,y)=(S3(lr)-M3(τr))(x3,y)α0,32(x),N3(k)(x,y)=N3(k-1)(x,y)+αk,32(x)N3(k)(x,y)=×(∂k(S3(lr)-M3(τr))∂yk(x3,y)N3(k)(x,y)×=m∂k(S3(lr)-M3(τr))∂yk-∂kN3(k-1)∂xk(x3,y))N3(k)(x,y)=-βk,32(x)∂kN3(k-1)∂xk(1,y),hhhhhhhk=1,2,…((x,y)∈H3).(iv) Denote(121)αk,41(x)=(x-x4)kk!(xx4)k+1,βk,41(x)=(x-x4)kk!(x-x4)k+1,k=0,1,….
Define M4(k)(x,y) by induction as follows:
(122)M4(0)(x,y)=S3(lr)(x4,y)α0,41(x),M4(k)(x,y)=M4(k-1)(x,y)+αk,41(x)M4(k)(x,y)=×(∂kS3(lr)∂xk(x4,y)-∂kM4(k-1)∂xk(x4,y))M4(k)(x,y)=-βk,41(x,y)∂kM4(k-1)∂xk(0,y),hhhhhhhhhhhhhhhhk=1,2,…((x,y)∈H4).Denote(123)αk,42(y)=(y-y4)kk!(1-y1-y4)k+1,βk,42(y)=(y-1)kk!(y4-yy4-1)k+1,k=0,1,….
Define N4(k)(x,y) by induction as follows:
(124)N4(0)(x,y)=(S4(lr)-M4(τr))(x,y4)α0,42(y),N4(k)(x,y)=N4(k-1)(x,y)+αk,42(y)N4(k)(x,y)=×(∂k(S4(lr)-M4(τr))∂yk(x,y4)N4(k)(x,y)=hh.,.-∂kN4(k-1)∂yk(x,y4))N4(k)(x,y)=-βk,42(x,y)∂kN4(k-1)∂xk(x,1),hhhhhhhh.k=1,2,…((x,y)∈H4).


## 5. Corollaries

By using the extension method given in Section3, we discuss the two important special cases.


### 5.1. Smooth Extensions of Functions on a Kind of Domains

LetΩ be a trapezoid with two curved sides:
(125)Ω={(x,y):x1≤x≤x2,η(x)≤y≤ξ(x)},
where L1<η(x)<ξ(x)<L2(x1≤x≤x2),η,ξ∈Cm([x1,x2]). Denote the rectangle D=[x1,x2]×[L1,L2]. Then D=G1⋃‍Ω⋃‍G2, where G1 and G2 are both trapezoids with a curved side:
(126)G1={(x,y):x1≤x≤x2,L1≤y≤η(x)},G2={(x,y):x1≤x≤x2,ξ(x)≤y≤L2}.Suppose thatf∈Cq(Ω) (q is a nonnegative integer). We will smoothly extend f from Ω to the trapezoids G1 and G2 with a curved side, respectively, as in Section 3.2, such that the extension function F is smooth on the rectangle D. Moreover, we will give a precise formula. It shows that the index of smoothness of F depends on not only smoothness of f but also smoothness of η,ξ.Denotea0,1(x,y)=(y-L1)/(η(x)-L1) and
(127)ak,1(x,y)=(y-η(x))kk!(y-L1η(x)-L1)k+1,gggggggk=1,2,…(x1≤x≤x2,y∈ℝ).
We define {S1(k)(x,y)} on G1 as follows. Let
(128)S1(0)(x,y)=f(x,η(x))a1,0(x,y)((x,y)∈G1),
and let k0 be the maximal integer satisfying 1+2+⋯+k0≤q. For k=1,2,…,k0, we define
(129)S1(k)(x,y)=S1(k-1)(x,y)+a1,k(x,y)×(∂kf∂yk(x,η(x))-∂kS1(k-1)∂yk(x,η(x))).
Then S1(k)∈Cλk(G1), where λk=min⁡{q-1-2-⋯-k,m}.Denotea0,2(x,y)=(L2-y)/(L2-ξ(x)) and
(130)ak,2(x,y)=(y-ξ(x))kk!(L2-yL2-ξ(x))k+1,.hhhhk=1,2,…(x1≤x≤x2,y∈ℝ).
We define {S2(k)(x,y)} on G2 as follows. Let
(131)S2(0)(x,y)=f(x,ξ(x))a0,2(x,y)((x,y)∈G2).
For k=1,2,…,k0, define
(132)S2(k)(x,y)=S2(k-1)(x,y)+ak,2(x,y)×(∂kf∂yk(x,ξ(x))-∂kS2(k-1)∂yk(x,ξ(x))),hhhhhhhhhhhhhhhhhh((x,y)∈G2).
Then S2(k)∈Cλk(G2), where λk is stated as above.An argument similar to Lemmas5 and 6 shows that, for 0≤k≤k0 and 0≤i+j≤min⁡{k,λk},
(133)∂i+jS1(k)∂xi∂yj(x,η(x))=∂i+jf∂xi∂yj(x,η(x)),∂i+jS2(k)∂xi∂yj(x,ξ(x))=∂i+jf∂xi∂yj(x,ξ(x)),hhhhhhhhhhhhhhl(x1≤x≤x2).
A direct calculation shows that the number
(134)τ(q,m)=min⁡{[2q+94-32],m}
is the maximal value of integers k satisfying k≤λk, where [·] expresses the integral part. So τ(q,m)≤λτ(q,m).By (133), we get that, for 0≤i+j≤τ(q,m),
(135)∂i+jS1(τ(q,m))∂xi∂yj(x,η(x))=∂i+jf∂xi∂yj(x,ξ(x)),∂i+jS2(τ(q,m))∂xi∂yj(x,ξ(x))=∂i+jf∂xi∂yj(x,ξ(x)),hhhhhhhhhhhhhhhhhl(x1≤x≤x2).
Note that
(136)S1(τ(q,m))∈Cλτ(q,m)(G1),S2(τ(q,m))∈Cλ(q,m)(G2),jjjjjjjjjjjjjjjτ(q,m)≤λτ(q,m)≤q,
and the assumption f∈Cq(Ω). Now we define a function on D by
(137)Fq,m(x,y)={f(x,y),(x,y)∈Ω,S1(τ(q,m))(x,y),(x,y)∈G1,S2(τ(q,m))(x,y),(x,y)∈G2.
From this and (135), we have Fq,m∈Cτ(q,m)(D). This implies the following theorem.Theorem 12.
Let the domainΩ and the rectangle D be stated as above. If f∈Cq(Ω), then the function Fq,m(x,y), defined in (137), is a smooth extension of f from Ω to D and Fq,m∈Cτ(q,m)(D), where τ(q,m) is stated in (134).Especially, forq=0 and m≥0, we have τ(q,m)=0, and so F0,m∈C(D); for q=2 and m≥1, we have τ(q,m)=1, and so F2,1∈C1(D); for q=5 and m≥2, we have τ(q,m)=2, and so F5,2∈C2(D).


### 5.2. Smooth Extensions of Univariate Functions on Closed Intervals

Letf∈Cq([x1,x2]) and [x1,x2]⊂(0,1). In order to extend smoothly f from [x1,x2] to [0,x1], we construct two polynomials
(138)a0(k)(x)=(x-x1)kk!(xx1)k+1,b0(k)(x)=xkk!(x1-xx1)k+1,k=0,1,….
Define S0(0)(x)=f(x1)(x/x1) and for k=1,…,q,
(139)S0(k)(x)=S0(k-1)(x)-a0(k)(x)(f(k)(x1)-(S0(k-1))(k)(x1))-b0(k)(x)(S0(k-1))(k)(0)(0≤x≤x1).
Then S0(q)(x) is a polynomial of degree ≤2q+1.Similar to the proof of Lemma5, we get
(140)(S0(q))(k)(0)=0,(S0(q))(k)(x1)=f(k)(x1),hhhhhhhhhhhhhhhhhhhhhhhhhk=0,1,…,q.
It is also easy to check directly them.Again extend smoothlyf from [x1,x2] to [x2,1], we construct two polynomials
(141)a1(k)(x)=(x-x2)kk!(1-x1-x2)k+1,b1(k)(x)=(x-1)kk!(x2-xx2-1)k+1,k=0,1,….
Define S1(0)(x)=f(x2)((1-x)/(1-x2)) and for k=1,…,q,
(142)S1(k)(x)=S1(k-1)(x)-a1(k)(x)(f(k)(x2)-(S1(k-1))(k)(x2))-b1(k)(x)(S1(k-1))(k)(1)(x2≤x≤1).
Then S1(q)(x) is a polynomial of degree ≤2q+1.Similar to the proof of Lemma5, we get
(143)gggg(S1(q))(k)(x2)=f(k)(x2),(S1(q))(k)(1)=0(k=0,1,…,q).Therefore, we obtain the smooth extensionF from [x1,x2] to [0,1] by
(144)F(x)={f(x),x∈[x1,x2],S0(q)(x),x∈[0,x1],S1(q)(x),x∈[x2,1],
where S0(q)(x) and S1(q)(x) are polynomials of degree 2q+1 defined as above, and F∈Cq([0,1]) and F(l)(0)=F(l)(1)=0(l=0,1,...,q). From this, we get the following.Theorem 13.
Letf∈Cq([x1,x2]) and [x1,x2]⊂(0,1). Then there exists a function F∈Cq([0,1]) satisfying F(x)=f(x)(x1≤x≤x2) and F(l)(0)=F(l)(1)=0(l=0,1,…,q).Letf∈Cq([x1,x2]) and [x1,x2]⊂(0,1), and let F be the smooth extension of f from [x1,x2] to [0,1] which is stated as in Theorem 12. Let Fp be the 1-periodic extension satisfying Fp(x+n)=F(x)(0≤x≤1,n∈ℤ). Then Fp∈Cq(ℝ) and Fp(x)=f(x)(x∈[x1,x2]). We expand F(x) into the Fourier series which converges fast. From this, we get trigonometric approximation of f∈Cq([x1,x2]). We also may do odd extension or even extension of F from [0,1] to [-1,1], and then doing periodic extension, we get the odd periodic extension Fpo∈Cq(ℝ) or the even periodic extension Fpe∈Cq(ℝ). We expand Fpo or Fpe into the sine series and the cosine series, respectively. From this, we get the sine polynomial approximation and the cosine polynomial approximation of f on [x1,x2]. For F∈Cq(x)(x∈[0,1]), we pad zero in the outside of [0,1] and then the obtained function Fc∈Cq(ℝ). We expand Fc into a wavelet series which converges fast. By the moment theorem, a lot of wavelet coefficients are equal to zero. From this, we get wavelet approximation of f∈Cq([x1,x2]).


## 5.1. Smooth Extensions of Functions on a Kind of Domains

LetΩ be a trapezoid with two curved sides:
(125)Ω={(x,y):x1≤x≤x2,η(x)≤y≤ξ(x)},
where L1<η(x)<ξ(x)<L2(x1≤x≤x2),η,ξ∈Cm([x1,x2]). Denote the rectangle D=[x1,x2]×[L1,L2]. Then D=G1⋃‍Ω⋃‍G2, where G1 and G2 are both trapezoids with a curved side:
(126)G1={(x,y):x1≤x≤x2,L1≤y≤η(x)},G2={(x,y):x1≤x≤x2,ξ(x)≤y≤L2}.Suppose thatf∈Cq(Ω) (q is a nonnegative integer). We will smoothly extend f from Ω to the trapezoids G1 and G2 with a curved side, respectively, as in Section 3.2, such that the extension function F is smooth on the rectangle D. Moreover, we will give a precise formula. It shows that the index of smoothness of F depends on not only smoothness of f but also smoothness of η,ξ.Denotea0,1(x,y)=(y-L1)/(η(x)-L1) and
(127)ak,1(x,y)=(y-η(x))kk!(y-L1η(x)-L1)k+1,gggggggk=1,2,…(x1≤x≤x2,y∈ℝ).
We define {S1(k)(x,y)} on G1 as follows. Let
(128)S1(0)(x,y)=f(x,η(x))a1,0(x,y)((x,y)∈G1),
and let k0 be the maximal integer satisfying 1+2+⋯+k0≤q. For k=1,2,…,k0, we define
(129)S1(k)(x,y)=S1(k-1)(x,y)+a1,k(x,y)×(∂kf∂yk(x,η(x))-∂kS1(k-1)∂yk(x,η(x))).
Then S1(k)∈Cλk(G1), where λk=min⁡{q-1-2-⋯-k,m}.Denotea0,2(x,y)=(L2-y)/(L2-ξ(x)) and
(130)ak,2(x,y)=(y-ξ(x))kk!(L2-yL2-ξ(x))k+1,.hhhhk=1,2,…(x1≤x≤x2,y∈ℝ).
We define {S2(k)(x,y)} on G2 as follows. Let
(131)S2(0)(x,y)=f(x,ξ(x))a0,2(x,y)((x,y)∈G2).
For k=1,2,…,k0, define
(132)S2(k)(x,y)=S2(k-1)(x,y)+ak,2(x,y)×(∂kf∂yk(x,ξ(x))-∂kS2(k-1)∂yk(x,ξ(x))),hhhhhhhhhhhhhhhhhh((x,y)∈G2).
Then S2(k)∈Cλk(G2), where λk is stated as above.An argument similar to Lemmas5 and 6 shows that, for 0≤k≤k0 and 0≤i+j≤min⁡{k,λk},
(133)∂i+jS1(k)∂xi∂yj(x,η(x))=∂i+jf∂xi∂yj(x,η(x)),∂i+jS2(k)∂xi∂yj(x,ξ(x))=∂i+jf∂xi∂yj(x,ξ(x)),hhhhhhhhhhhhhhl(x1≤x≤x2).
A direct calculation shows that the number
(134)τ(q,m)=min⁡{[2q+94-32],m}
is the maximal value of integers k satisfying k≤λk, where [·] expresses the integral part. So τ(q,m)≤λτ(q,m).By (133), we get that, for 0≤i+j≤τ(q,m),
(135)∂i+jS1(τ(q,m))∂xi∂yj(x,η(x))=∂i+jf∂xi∂yj(x,ξ(x)),∂i+jS2(τ(q,m))∂xi∂yj(x,ξ(x))=∂i+jf∂xi∂yj(x,ξ(x)),hhhhhhhhhhhhhhhhhl(x1≤x≤x2).
Note that
(136)S1(τ(q,m))∈Cλτ(q,m)(G1),S2(τ(q,m))∈Cλ(q,m)(G2),jjjjjjjjjjjjjjjτ(q,m)≤λτ(q,m)≤q,
and the assumption f∈Cq(Ω). Now we define a function on D by
(137)Fq,m(x,y)={f(x,y),(x,y)∈Ω,S1(τ(q,m))(x,y),(x,y)∈G1,S2(τ(q,m))(x,y),(x,y)∈G2.
From this and (135), we have Fq,m∈Cτ(q,m)(D). This implies the following theorem.Theorem 12.
Let the domainΩ and the rectangle D be stated as above. If f∈Cq(Ω), then the function Fq,m(x,y), defined in (137), is a smooth extension of f from Ω to D and Fq,m∈Cτ(q,m)(D), where τ(q,m) is stated in (134).Especially, forq=0 and m≥0, we have τ(q,m)=0, and so F0,m∈C(D); for q=2 and m≥1, we have τ(q,m)=1, and so F2,1∈C1(D); for q=5 and m≥2, we have τ(q,m)=2, and so F5,2∈C2(D).


## 5.2. Smooth Extensions of Univariate Functions on Closed Intervals

Letf∈Cq([x1,x2]) and [x1,x2]⊂(0,1). In order to extend smoothly f from [x1,x2] to [0,x1], we construct two polynomials
(138)a0(k)(x)=(x-x1)kk!(xx1)k+1,b0(k)(x)=xkk!(x1-xx1)k+1,k=0,1,….
Define S0(0)(x)=f(x1)(x/x1) and for k=1,…,q,
(139)S0(k)(x)=S0(k-1)(x)-a0(k)(x)(f(k)(x1)-(S0(k-1))(k)(x1))-b0(k)(x)(S0(k-1))(k)(0)(0≤x≤x1).
Then S0(q)(x) is a polynomial of degree ≤2q+1.Similar to the proof of Lemma5, we get
(140)(S0(q))(k)(0)=0,(S0(q))(k)(x1)=f(k)(x1),hhhhhhhhhhhhhhhhhhhhhhhhhk=0,1,…,q.
It is also easy to check directly them.Again extend smoothlyf from [x1,x2] to [x2,1], we construct two polynomials
(141)a1(k)(x)=(x-x2)kk!(1-x1-x2)k+1,b1(k)(x)=(x-1)kk!(x2-xx2-1)k+1,k=0,1,….
Define S1(0)(x)=f(x2)((1-x)/(1-x2)) and for k=1,…,q,
(142)S1(k)(x)=S1(k-1)(x)-a1(k)(x)(f(k)(x2)-(S1(k-1))(k)(x2))-b1(k)(x)(S1(k-1))(k)(1)(x2≤x≤1).
Then S1(q)(x) is a polynomial of degree ≤2q+1.Similar to the proof of Lemma5, we get
(143)gggg(S1(q))(k)(x2)=f(k)(x2),(S1(q))(k)(1)=0(k=0,1,…,q).Therefore, we obtain the smooth extensionF from [x1,x2] to [0,1] by
(144)F(x)={f(x),x∈[x1,x2],S0(q)(x),x∈[0,x1],S1(q)(x),x∈[x2,1],
where S0(q)(x) and S1(q)(x) are polynomials of degree 2q+1 defined as above, and F∈Cq([0,1]) and F(l)(0)=F(l)(1)=0(l=0,1,...,q). From this, we get the following.Theorem 13.
Letf∈Cq([x1,x2]) and [x1,x2]⊂(0,1). Then there exists a function F∈Cq([0,1]) satisfying F(x)=f(x)(x1≤x≤x2) and F(l)(0)=F(l)(1)=0(l=0,1,…,q).Letf∈Cq([x1,x2]) and [x1,x2]⊂(0,1), and let F be the smooth extension of f from [x1,x2] to [0,1] which is stated as in Theorem 12. Let Fp be the 1-periodic extension satisfying Fp(x+n)=F(x)(0≤x≤1,n∈ℤ). Then Fp∈Cq(ℝ) and Fp(x)=f(x)(x∈[x1,x2]). We expand F(x) into the Fourier series which converges fast. From this, we get trigonometric approximation of f∈Cq([x1,x2]). We also may do odd extension or even extension of F from [0,1] to [-1,1], and then doing periodic extension, we get the odd periodic extension Fpo∈Cq(ℝ) or the even periodic extension Fpe∈Cq(ℝ). We expand Fpo or Fpe into the sine series and the cosine series, respectively. From this, we get the sine polynomial approximation and the cosine polynomial approximation of f on [x1,x2]. For F∈Cq(x)(x∈[0,1]), we pad zero in the outside of [0,1] and then the obtained function Fc∈Cq(ℝ). We expand Fc into a wavelet series which converges fast. By the moment theorem, a lot of wavelet coefficients are equal to zero. From this, we get wavelet approximation of f∈Cq([x1,x2]).



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*Source: 102062-2014-02-10.xml*