Shearlet approximations to the inverse of a family of linear operators

Abstract

The Radon transform plays an important role in applied mathematics. It is a fundamental problem to reconstruct images from noisy observations of Radon data. Compared with traditional methods, Colona, Easley and etc. apply shearlets to deal with the inverse problem of the Radon transform and receive more effective reconstruction. This paper extends their work to a class of linear operators, which contains Radon, Bessel and Riesz fractional integration transforms as special examples.

MSC:42C15, 42C40.

1 Introduction and preliminary

The Radon transform is an important tool in medical imaging. Although $f\in L^1({\mathbb{R}}^2)$ can be recovered analytically from the Radon data $Rf(\theta ,t)$, the solution is unstable and those data are corrupted by some noise in practice [1]. In order to recover the object f stably and control the amplification of noise in the reconstruction, many methods of regularization were introduced including the Fourier method, singular value decomposition, etc. [2]. However, those methods produced a blurred version of the original one.

Curvelets and shearlets were then proposed, which proved to be efficient in dealing with edges [3–7]. In 2002, Candés and Donoho applied curvelets [5] to the inverse problem

$$Y=Rf+\epsilon W,$$

(1.1)

where the recovered function f is compactly supported and twice continuously differentiable away from a smooth edge; W denotes a Wiener sheet; ε is a noisy level. Because curvelets have complicated structure, Colonna, Easley, etc. used shearlets to deal with the problem (1.1) in 2010 and received an effective reconstructive algorithm [8].

Note that the Bessel transform and the Riesz fractional integration transform arise in many scientific areas ranging from physical chemistry to extragalactic astronomy. Then this paper considers a more general problem,

$$Y=Kf+\epsilon W,$$

(1.2)

where K stands for a linear operator mapping the Hilbert space $L^2({\mathbb{R}}^2)$ to another Hilbert space Y and satisfies

$${\left(K^{\ast }Kf\right)}^{\wedge }(\xi )={(b+{|\xi |}^2)}^{-\alpha }\hat{f}(\xi )$$

(1.3)

with $b>0$, $\alpha >0$ ($K^{\ast }$ is the conjugate operator of K). Here and in what follows, $\hat{f}$ denotes the Fourier transform of f. The next section shows that Radon, Bessel and Riesz fractional integration transforms satisfy the condition (1.3).

The current paper is organized as follows. Section 2 presents three examples for (1.3) and several lemmas. An approximation result is proved in the last section, which contains Theorem 4.2 of [8] as a special case.

At the end of this section, we introduce some basic knowledge of shearlets, which will be used in our discussions. The Fourier transform of a function $f\in L^1({\mathbb{R}}^2)\cap L^2({\mathbb{R}}^2)$ is defined by

$$\hat{f}(\xi )={\int }_{{\mathbb{R}}^2}f(x)e^{-2\pi ix\cdot \xi }dx.$$

The classical method extends that definition to $L^2({\mathbb{R}}^2)$ functions.

There exist many different constructions for discrete shearlets. We introduce the construction [8] by taking two functions ${\psi }_1,{\psi }_2$ of one variable such that ${\hat{\psi }}_1,{\hat{\psi }}_2\in C^{\mathrm{\infty }}(\mathbb{R})$ with their supports $supp{\hat{\psi }}_1\subset [-\frac{1}{2},-\frac{1}{16}]\cup [\frac{1}{16},\frac{1}{2}]$, $supp{\hat{\psi }}_2\subset [-1,1]$ and

$$\sum _{j\ge 0}{\left|\widehat{{\psi }_1}\left(2^{-2j}\omega \right)\right|}^2=1(|\omega |\ge \frac{1}{8}),\sum _{l=-2^j}^{2^j-1}{\left|\widehat{{\psi }_2}(2^j\omega -l)\right|}^2=1(|\omega |\le 1).$$

Here, $C^{\mathrm{\infty }}({\mathbb{R}}^n)$ stands for infinitely many times differentiable functions on the Euclidean space ${\mathbb{R}}^n$. Then two shearlet functions ${\psi }^{(0)}$, ${\psi }^{(1)}$ are defined by

$${\hat{\psi }}^{(0)}(\xi ):={\hat{\psi }}_1({\xi }_1){\hat{\psi }}_2\left(\frac{{\xi }_2}{{\xi }_1}\right)\text{and}{\hat{\psi }}^{(1)}(\xi ):=\widehat{{\psi }_1}({\xi }_2)\widehat{{\psi }_2}\left(\frac{{\xi }_1}{{\xi }_2}\right)$$

respectively.

To introduce discrete shearlets, we need two shear matrices

$$B_0=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right),B_1=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$$

and two dilation matrices

$$A_0=\left(\begin{array}{cc}4& 0\\ 0& 2\end{array}\right),A_1=\left(\begin{array}{cc}2& 0\\ 0& 4\end{array}\right).$$

Define discrete shearlets ${\psi }_{j,l,k}^{(d)}(x):=2^{\frac{3}{2}j}{\psi }^{(d)}(B_d^l{A}_d^{j}x-k)$ for $j\ge 0$, $-2^j\le l\le 2^j-1$, $k\in {\mathbb{Z}}^2$ and $d=0,1$. Then there exists $\hat{\phi }\in C_0^{\mathrm{\infty }}({\mathbb{R}}^2)$ such that

$$\{{\phi }_{j_0,k}(x),{\psi }_{j,l,k}^{(d)}(x),j\ge j_0\ge 0,-2^j\le l\le 2^j-1,k\in {\mathbb{Z}}^2,d=0,1\}$$

forms a Parseval frame of $L^2({\mathbb{R}}^2)$, where ${\phi }_{j_0,k}(x):=2^{j_0}\phi (2^{j_0}x-k)$. More precisely, for $f\in L^2({\mathbb{R}}^2)$,

$$f(x)=\sum _{k\in {\mathbb{Z}}^2}〈f,{\phi }_{j_0,k}〉{\phi }_{j_0,k}(x)+\sum _{d=0}^1\sum _{j\ge j_0}\sum _{l=-2^j}^{2^j-1}\sum _{k\in {\mathbb{Z}}^2}〈f,{\psi }_{j,l,k}^{(d)}〉{\psi }_{j,l,k}^{(d)}(x)$$

holds in $L^2({\mathbb{R}}^2)$. It should be pointed out that ${\psi }_{j,l,k}^{(d)}(x)$ are modified for $l=-2^j$ and $2^j-1$, as seen in [8].

2 Examples and lemmas

In this section, we provide three important examples of a linear operator K satisfying ${(K^{\ast }Kf)}^{\wedge }(\xi )={(b+{|\xi |}^2)}^{-\alpha }\hat{f}(\xi )$ and present some lemmas which will be used in the next section. To introduce the first one, define a subspace of $L^2({\mathbb{R}}^2)$,

$$D\left({\mathbb{R}}^2\right):=\{f\in L^1\left({\mathbb{R}}^2\right),f\text{ is bounded}\}\subseteq \{f\in L^2\left({\mathbb{R}}^2\right),{\int }_{{\mathbb{R}}^2}{|\xi |}^{-1}{|\hat{f}(\xi )|}^{2}d\xi <+\mathrm{\infty }\}$$

and a Hilbert space

$$L^2([0,\pi )\times \mathbb{R}):=\{f(\theta ,t),{\int }_0^{\pi }{\int }_{\mathbb{R}}{|f(\theta ,t)|}^{2}dtd\theta <+\mathrm{\infty }\}$$

with the inner product $〈f,g〉:={\int }_0^{\pi }{\int }_{\mathbb{R}}f(\theta ,t)\overline{g(\theta ,t)}dtd\theta$.

Example 2.1 Let $L_{\theta ,t}:=\{(x,y),xcos\theta +ysin\theta =t\}\subseteq {\mathbb{R}}^2$ and $ds(x,y)$ be the Euclidean measure on the line $L_{\theta ,t}$. Then the classical Radon transform R: $D(\mathbb{R})\to L^2([0,\pi )\times \mathbb{R})$ defined by

$$Rf(\theta ,t)={\int }_{L_{\theta ,t}}f(x,y)ds(x,y)$$

satisfies ${(R^{\ast }Rf)}^{\wedge }(\xi )={|\xi |}^{-1}\hat{f}(\xi )$.

Proof By the definition of $D({\mathbb{R}}^2)$, ${\int }_{{\mathbb{R}}^2}{|\xi |}^{-1}\hat{f}(\xi )\overline{\hat{g}(\xi )}d\xi <+\mathrm{\infty }$ for $f,g\in D({\mathbb{R}}^2)$. It is easy to see that ${\int }_0^{2\pi }d\theta {\int }_0^{+\mathrm{\infty }}\hat{f}(\omega cos\theta ,\omega sin\theta )\overline{\hat{g}(\omega cos\theta ,\omega sin\theta )}d\omega ={\int }_0^{\pi }d\theta {\int }_{\mathbb{R}}\hat{f}(\omega cos\theta ,\omega sin\theta )\times \overline{\hat{g}(\omega cos\theta ,\omega sin\theta )}d\omega$. This with the Fourier slice theorem ([1, 9]) and the Plancherel formula leads to

$$\begin{array}{rcl}{\int }_{{\mathbf{R}}^2}{|\xi |}^{-1}\hat{f}(\xi )\overline{\hat{g}(\xi )}d\xi & =& {\int }_0^{2\pi }d\theta {\int }_0^{+\mathrm{\infty }}\hat{f}(\omega cos\theta ,\omega sin\theta )\overline{\hat{g}(\omega cos\theta ,\omega sin\theta )}d\omega \\ =& {\int }_0^{\pi }d\theta {\int }_{\mathbb{R}}{(R_{\theta }f)}^{\wedge }(\omega )\overline{{(R_{\theta }g)}^{\wedge }(\omega )}d\omega \\ =& {\int }_0^{\pi }d\theta {\int }_{\mathbb{R}}Rf(\theta ,t)\overline{Rg(\theta ,t)}dt=〈Rf,Rg〉,\end{array}$$

where $R_{\theta }f(t):=Rf(\theta ,t)$. Moreover, $〈{(R^{\ast }Rf)}^{\wedge },\hat{g}〉=〈R^{\ast }Rf,g〉=〈Rf,Rg〉=〈{|\xi |}^{-1}\hat{f}(\xi ),\hat{g}(\xi )〉$ for each $g\in D({\mathbb{R}}^2)$.

Because $D({\mathbb{R}}^2)$ is dense in $L^2({\mathbb{R}}^2)$, one receives the desired conclusion ${(R^{\ast }Rf)}^{\wedge }(\xi )={|\xi |}^{-1}\hat{f}(\xi )$. Here, $R^{\ast }Rf\in L^2({\mathbb{R}}^2)$ for $f\in D({\mathbb{R}}^2)$. In fact, $R^{\ast }Rf=4\pi I_{1}f$ by [10], where $I_1$ is the Riesz fractional integration transform defined by

$$I_1(f)(x):=C_{\alpha }{\int }_{{\mathbb{R}}^2}\frac{f(x-y)}{|y|}dy$$

with some normalizing constant $C_{\alpha }$ [11]. We rewrite $I_1(f)(x)={\int }_{|y|\le r}\frac{f(x-y)}{|y|}dy+{\int }_{|y|>r}\frac{f(x-y)}{|y|}dy=:J_1+J_2$. Let $h(y)=\frac{1}{|y|}{1}_{B(0,1)}(y)$, $h_r(y)=\frac{1}{r^2}h(\frac{y}{r})$, where $B(0,1)$ stands for the unit ball of ${\mathbb{R}}^2$ and $1_A$ represents an indicator function on the set A. Then $J_1={\int }_{|y|\le r}\frac{f(x-y)}{|y|}dy=r{\int }_{|y|\le r}{h}_r(y)f(x-y)dy\le rMf(x)$ by Theorem 9 of reference [[12], p.59], where Mf is the Hardy-Littlewood maximal function of f.

On the other hand, the Holder inequality implies

$$J_2\le {\parallel f\parallel }_{\frac{p}{p-1}}{\left({\int }_{|y|>r}\frac{1}{{|y|}^p}dy\right)}^{\frac{1}{p}}\le r^{2-p}{\parallel f\parallel }_{\frac{p}{p-1}}$$

with $p>3$. Take $r={[Mf(x)]}^{-\frac{1}{p-1}}$, one gets $I_1(f)(x)\le {[Mf(x)]}^{\frac{p-2}{p-1}}(1+{\parallel f\parallel }_{\frac{p}{p-1}})$ and ${\parallel I_1(f)\parallel }_2\le (1+{\parallel f\parallel }_{\frac{p}{p-1}}){\parallel Mf\parallel }_{\frac{2(p-2)}{p-1}}^{\frac{p-2}{p-1}}\lesssim (1+{\parallel f\parallel }_{\frac{p}{p-1}}){\parallel f\parallel }_{\frac{2(p-2)}{p-1}}^{\frac{p-2}{p-1}}<+\mathrm{\infty }$, since $f\in L^{\frac{p}{p-1}}({\mathbb{R}}^2)\cap L^{\frac{2(p-2)}{p-1}}({\mathbb{R}}^2)$ due to the assumption $f\in D({\mathbb{R}}^2)$ and $\frac{2(p-2)}{p-1}>1$, $\frac{p}{p-1}>1$.

In order to introduce the next example, we use $f\ast g$ to denote the convolution of f and g. □

Example 2.2 The Bessel operator $B_{\alpha }:L^2({\mathbb{R}}^2)\to L^2({\mathbb{R}}^2)$ defined by $B_{\alpha }f=b_{\alpha }\ast f$ with ${\hat{b}}_{\alpha }(\xi )={(1+{|\xi |}^2)}^{-\frac{\alpha }{2}}$ and $\alpha >0$ satisfies

$${\left(B_{\alpha }^{\ast }{B}_{\alpha }f\right)}^{\wedge }(\xi )={(1+{|\xi |}^2)}^{-\alpha }\hat{f}(\xi ).$$

Proof It is known that $b_{\alpha }(x)\in L^1({\mathbb{R}}^2)$ for $\alpha >0$ [11]. Hence, ${(B_{\alpha }f)}^{\wedge }(\xi )={\hat{b}}_{\alpha }(\xi )\hat{f}(\xi )={(1+{|\xi |}^2)}^{-\frac{\alpha }{2}}\hat{f}(\xi )$. For $f,g\in L^2({\mathbb{R}}^2)$, $〈{(B_{\alpha }^{\ast }{B}_{\alpha }f)}^{\wedge },\hat{g}〉=〈B_{\alpha }^{\ast }{B}_{\alpha }f,g〉=〈B_{\alpha }f,B_{\alpha }g〉=〈{(B_{\alpha }f)}^{\wedge },{(B_{\alpha }g)}^{\wedge }〉=〈{(1+{|\xi |}^2)}^{-\alpha }\hat{f}(\xi ),\hat{g}(\xi )〉$. Thus,

$${\left(B_{\alpha }^{\ast }{B}_{\alpha }f\right)}^{\wedge }(\xi )={(1+{|\xi |}^2)}^{-\alpha }\hat{f}(\xi ).$$

 □

To introduce the Riesz fractional integration transform, we define

$$D=\{f\in L^2\left({\mathbb{R}}^2\right),f\text{ has compact support}\}\subseteq L^1\left({\mathbb{R}}^2\right)\cap L^2\left({\mathbb{R}}^2\right).$$

Then $D\subseteq L^s({\mathbb{R}}^2)$ ($1\le s\le 2$). For $f\in D$ and $0<\alpha <1$, the Riesz fractional integration transform is defined by

$$I_{\alpha }(f)(x):=C_{\alpha }{\int }_{{\mathbb{R}}^2}\frac{f(y)}{{|x-y|}^{2-\alpha }}dy\in L^2\left({\mathbb{R}}^2\right),$$

(2.1)

where $C_{\alpha }$ is the normalizing constant [11]. In order to show ${(I_{\alpha }^{\ast }{I}_{\alpha }f)}^{\wedge }(\xi )={|\xi |}^{-2\alpha }\hat{f}(\xi )$ for $f\in D$ and $0<\alpha <1/2$, we need a lemma ([11], Lemma 2.15).

Lemma 2.1 Let $S({\mathbb{R}}^2)$ be the Schwartz space and $\mathrm{\Psi }=\{\psi \in S({\mathbb{R}}^2),\frac{{\partial }^{\beta }}{\partial x^{\beta }}\psi (0)=0,\beta \in {\mathbb{Z}}^{+}\times {\mathbb{Z}}^{+}\}$ with ${\mathbb{Z}}^{+}$ being the non-negative integer set. Define $\mathrm{\Phi }:=\{\phi =\hat{\psi },\psi \in \mathrm{\Psi }\}$. Then with $\alpha >0$,

$${(I_{\alpha }f)}^{\wedge }(\xi )={|\xi |}^{-\alpha }\hat{f}(\xi )$$

holds for each $f\in \mathrm{\Phi }$.

Example 2.3 The transform $I_{\alpha }$ defined by (2.1) satisfies ${(I_{\alpha }^{\ast }{I}_{\alpha }f)}^{\wedge }(\xi )={|\xi |}^{-2\alpha }\hat{f}(\xi )$ for $f\in D$ and $0<\alpha <\frac{1}{2}$.

Proof As proved in Examples 2.1, 2.2, it is sufficient to show that for $f\in D$,

$${(I_{\alpha }f)}^{\wedge }(\xi )={|\xi |}^{-\alpha }\hat{f}(\xi ).$$

(2.2)

One proves (2.2) firstly for $f\in C_0^{\mathrm{\infty }}({\mathbb{R}}^2)$. Take $\mu (r)\in C^{\mathrm{\infty }}([0,\mathrm{\infty }))$ with $0\le \mu (r)\le 1$ and

$$\mu (r)=\{\begin{array}{cc}1,\hfill & r\ge 2;\hfill \\ 0,\hfill & 0\le r\le 1.\hfill \end{array}$$

Define ${\psi }_N(\xi ):=\mu (N|\xi |)\hat{f}(\xi )$. Then ${\psi }_N(\xi )\in \mathrm{\Psi }$ and $f_N(x):={\check{\psi }}_N(x)={\hat{\psi }}_N(-x)\in \mathrm{\Phi }$. By Lemma 2.1,

$${(I_{\alpha }{f}_N)}^{\wedge }(\xi )={|\xi |}^{-\alpha }{\hat{f}}_N(\xi ).$$

(2.3)

Let $k(x)$ be the inverse Fourier transform of the function $1-\mu (|x|)$ and $k_N(x):=\frac{1}{N^2}k(\frac{x}{N})$. Then ${\int }_{{\mathbb{R}}^2}k(x)dx=1$ and $f_N(x)=f(x)-k_N\ast f(x)$. Moreover, the classical approximation theorem [11] tells

$$\underset{N\to \mathrm{\infty }}{lim}{\parallel f_N-f\parallel }_p=0$$

for $p>1$. On the other hand, ${\parallel I_{\alpha }{f}_N-I_{\alpha }f\parallel }_2={\parallel I_{\alpha }(f_N-f)\parallel }_2\le C{\parallel f_N-f\parallel }_{\frac{2}{1+\alpha }}$ due to Theorem 16 [[12], p.69]. Hence, ${lim}_{N\to \mathrm{\infty }}{\parallel {(I_{\alpha }{f}_N)}^{\wedge }(\xi )-{(I_{\alpha }f)}^{\wedge }(\xi )\parallel }_2={lim}_{N\to \mathrm{\infty }}{\parallel I_{\alpha }{f}_N-I_{\alpha }f\parallel }_2=0$. That is,

$$\underset{N\to \mathrm{\infty }}{lim}{(I_{\alpha }{f}_N)}^{\wedge }(\xi )={(I_{\alpha }f)}^{\wedge }(\xi )$$

(2.4)

in $L^2({\mathbb{R}}^2)$ sense. Note that ${\parallel {|\xi |}^{-\alpha }{\hat{f}}_N(\xi )-{|\xi |}^{-\alpha }\hat{f}(\xi )\parallel }_2^2={\int }_{{\mathbb{R}}^2}{|\xi |}^{-2\alpha }{|\hat{f}(\xi )|}^2{[1-\mu (N|\xi |)]}^{2}d\xi$; ${|\xi |}^{-2\alpha }{|\hat{f}(\xi )|}^2\in L^1({\mathbb{R}}^2)$ with $0<\alpha <\frac{1}{2}$ and ${lim}_{N\to \mathrm{\infty }}[1-\mu (N|\xi |)]=0$. Then

$$\underset{N\to \mathrm{\infty }}{lim}{\parallel {|\xi |}^{-\alpha }{\hat{f}}_N(\xi )-{|\xi |}^{-\alpha }\hat{f}(\xi )\parallel }_2=0$$

(2.5)

thanks to the Lebesgue dominated convergence theorem, which means ${lim}_{N\to \mathrm{\infty }}{|\xi |}^{-\alpha }\times {\hat{f}}_N(\xi )={|\xi |}^{-\alpha }\hat{f}(\xi )$ in $L^2({\mathbb{R}}^2)$ sense. This with (2.3), (2.4) shows (2.2) for $f\in C_0^{\mathrm{\infty }}({\mathbb{R}}^2)$.

In order to show (2.2) for $f\in D$, one can find $g\in C_0^{\mathrm{\infty }}({\mathbb{R}}^2)$ such that ${\int }_{{\mathbb{R}}^2}g(x)dx=1$ and ${lim}_{N\to \mathrm{\infty }}{\parallel f\ast g_N-f\parallel }_p=0$ ($p\ge 1$) by Theorem 4.2.1 in [13], where $g_N(\cdot )=N^{2}g(N\cdot )$. Since $f\ast g_N\in C_0^{\mathrm{\infty }}({\mathbb{R}}^2)$, the above proved fact says

$${(I_{\alpha }(f\ast g_N))}^{\wedge }(\xi )={|\xi |}^{-\alpha }{(f\ast g_N)}^{\wedge }(\xi ).$$

(2.6)

The same arguments as (2.4) and (2.5) show that ${lim}_{N\to \mathrm{\infty }}{(I_{\alpha }(f\ast g_N))}^{\wedge }(\xi )={(I_{\alpha }f)}^{\wedge }(\xi )$ and ${lim}_{N\to \mathrm{\infty }}{|\xi |}^{-\alpha }{(f\ast g_N)}^{\wedge }(\xi )={|\xi |}^{-\alpha }\hat{f}(\xi )$. Hence,

$${(I_{\alpha }f)}^{\wedge }(\xi )={|\xi |}^{-\alpha }\hat{f}(\xi ).$$

This completes the proof of (2.2) for $f\in D$. □

Next, we prove a lemma which will be used in the next section. For convenience, here and in what follows, we define $\mathcal{M}=N\cup M$ with $N={\mathbb{Z}}^2$, $M:=\{\mu =(j,l,k,d):j\ge j_0,-2^j\le l\le 2^j-1,k\in {\mathbb{Z}}^2,d=0,1\}$. Then the shearlet system (introduced in Section 1) can be represented as $\{s_{\mu }:\mu \in \mathcal{M}\}$, where $s_{\mu }={\psi }_{\mu }={\psi }_{j,l,k}^{(d)}$ if $\mu \in M$, and $s_{\mu }={\phi }_{\mu }={\phi }_{j_0,k}$ if $\mu \in N$.

Lemma 2.2 Let K satisfy ${(K^{\ast }Kf)}^{\wedge }(\xi )={(b+{|\xi |}^2)}^{-\alpha }\hat{f}(\xi )$ and $\{s_{\mu },\mu \in \mathcal{M}\}$ be shearlets introduced in the first section. Define ${\hat{\sigma }}_{\mu }(\xi )={(b+{|\xi |}^2)}^{\alpha }{\hat{s}}_{\mu }(\xi )$ and $U_{\mu }:=2^{-2\alpha j}K{\sigma }_{\mu }$. Then $\parallel U_{\mu }\parallel \le C$ and for $\mu \in \mathcal{M}$,

$$〈f,s_{\mu }〉=2^{2\alpha j}〈Kf,U_{\mu }〉.$$

Proof By the Plancherel formula and the assumption ${\hat{\sigma }}_{\mu }(\xi )={(b+{|\xi |}^2)}^{\alpha }{\hat{s}}_{\mu }(\xi )$, one knows that $〈f,s_{\mu }〉=〈\hat{f},{\hat{s}}_{\mu }〉=〈\hat{f}(\xi ),{(b+{|\xi |}^2)}^{-\alpha }{\hat{\sigma }}_{\mu }(\xi )〉$. Moreover,

$$〈f,s_{\mu }〉=〈\hat{f}(\xi ),{\left(K^{\ast }K{\sigma }_{\mu }\right)}^{\wedge }(\xi )〉=〈f,K^{\ast }K{\sigma }_{\mu }〉=〈Kf,K{\sigma }_{\mu }〉=2^{2\alpha j}〈Kf,U_{\mu }〉$$

due to ${(K^{\ast }Kf)}^{\wedge }(\xi )={(b+{|\xi |}^2)}^{-\alpha }\hat{f}(\xi )$ and $U_{\mu }:=2^{-2\alpha j}K{\sigma }_{\mu }$.

Next, one shows $\parallel U_{\mu }\parallel \le C$. Note that ${\parallel K{\sigma }_{\mu }\parallel }^2=〈K{\sigma }_{\mu },K{\sigma }_{\mu }〉=〈K^{\ast }K{\sigma }_{\mu },{\sigma }_{\mu }〉=〈{(K^{\ast }K{\sigma }_{\mu })}^{\wedge },{\hat{\sigma }}_{\mu }〉$, ${(K^{\ast }K{\sigma }_{\mu })}^{\wedge }={(b+{|\xi |}^2)}^{-\alpha }{\hat{\sigma }}_{\mu }(\xi )$ and ${\hat{\sigma }}_{\mu }(\xi )={(b+{|\xi |}^2)}^{\alpha }{\hat{s}}_{\mu }(\xi )$. Then ${\parallel K{\sigma }_{\mu }\parallel }^2=〈{\hat{s}}_{\mu }(\xi ),{(b+{|\xi |}^2)}^{\alpha }{\hat{s}}_{\mu }(\xi )〉$ and

$${\parallel U_{\mu }\parallel }^2=2^{-4\alpha j}{\parallel K{\sigma }_{\mu }\parallel }^2=2^{-4\alpha j}{\int }_{{\mathbb{R}}^2}{(b+{|\xi |}^2)}^{\alpha }{|{\hat{s}}_{\mu }(\xi )|}^{2}d\xi .$$

Because $supp{\hat{s}}_{\mu }\subseteq C_j:={[-2^{2j-1},2^{2j-1}]}^2\setminus {[-2^{2j-4},2^{2j-4}]}^2$, one receives ${\parallel U_{\mu }\parallel }^2=2^{-4\alpha j}{\int }_{C_j}{(b+{|\xi |}^2)}^{\alpha }{|{\hat{s}}_{\mu }(\xi )|}^{2}d\xi \le C$. This completes the proof of Lemma 2.2. □

At the end of this section, we introduce two theorems which are important for our discussions. As in [8], we use ${\mathit{STAR}}^2(A)$ to denote all sets $B\subseteq {[0,1]}^2$ with $C^2$ boundary ∂B given by

$$\beta (\theta )=\left(\begin{array}{c}\rho (\theta )cos\theta \\ \rho (\theta )sin\theta \end{array}\right)$$

in a polar coordinate system. Here, $\rho (\theta )\le {\rho }_0<1$ and $|{\rho }^{″}(\theta )|\le A$. Define ${\epsilon }^2(A):=\{f=f_0+f_1{X}_B,B\in {\mathit{STAR}}^2(A)\}$, where $f_0,f_1\in C_0^2({[0,1]}^2)$ are compactly supported on ${[0,1]}^2$. Let $c_{\mu }:=〈f,s_{\mu }〉$, $M_j:=\{(j,l,k,d),|k|\le 2^{2j+1},-2^j\le l\le 2^j-1,d=0,1\}$ and

$$R(j,\epsilon )=\{\mu \in M_j:|c_{\mu }|>\epsilon \}.$$

Then with $\mathrm{♯}R(j,\epsilon )$ standing for the cardinality of $R(j,\epsilon )$, the following conclusion holds [8].

Theorem 2.3 For $f\in {\epsilon }^2(A)$, $\mathrm{♯}R(j,\epsilon )\le C{\epsilon }^{-\frac{2}{3}}$ and

$$\sum _{\mu \in M_j}{|c_{\mu }|}^2\le C{2}^{-2j}.$$

Theorem 2.4 [14]

Let $X\sim N(u,1)$ and $t=\sqrt{2log({\eta }^{-1})}$ with $0<\eta \le \frac{1}{2}$. Then

$$E{|T_s(X,t)-u|}^2=[2log\left({\eta }^{-1}\right)+1](\eta +min\{u^2,1\}),$$

where $N(u,1)$ denotes the normal distribution with mean u and variance 1, while $T_s(y,t):=sgn(y){(|y|-t)}_{+}$ is the soft thresholding function.

3 Main theorem

In this section, we give an approximation result, which extends the result [[8], Theorem 4.2] from the Radon transform to a family of linear operators. To do that, we introduce a set $\mathcal{N}(\epsilon )$ of significant shearlet coefficients as follows. Let

$$s_1=\frac{1}{\frac{9}{2}+6\alpha }{log}_2\left({\epsilon }^{-1}\right),s_2=\frac{1}{\frac{3}{2}+2\alpha }{log}_2\left({\epsilon }^{-1}\right),$$

and $j_0=\lceil s_1\rceil$, $j_1=\lceil s_2\rceil$. Define $\mathcal{N}(\epsilon ):=M(\epsilon )\cup N(\epsilon )\subseteq \mathcal{M}$, where

Consider the model $Y=Kf+\epsilon W$ with ${(K^{\ast }Kf)}^{\wedge }(\xi )={(b+{|\xi |}^2)}^{-\alpha }\hat{f}(\xi )$. Lemma 2.2 tells that $y_{\mu }:=2^{2\alpha j}〈Y,U_{\mu }〉=〈f,s_{\mu }〉+\epsilon 2^{2\alpha j}{n}_{\mu }$, where $n_{\mu }$ is Gaussian noise with zero mean and bounded variance ${\sigma }_{\mu }^2={\parallel U_{\mu }\parallel }^2\le C$ [15]. Let $c_{\mu }=〈f,s_{\mu }〉$ and $\tilde{f}={\sum }_{\mu \in \mathcal{N}(\epsilon )}{\tilde{c}}_{\mu }{s}_{\mu }$ with

$${\tilde{c}}_{\mu }=\{\begin{array}{cc}{T}_s(y_{\mu },\epsilon \sqrt{2log(\mathrm{♯}\mathcal{N}(\epsilon ))}{2}^{2\alpha j}{\sigma }_{\mu }),\hfill & \mu \in \mathcal{N}(\epsilon );\hfill \\ 0,\hfill & \text{otherwise},\hfill \end{array}$$

where $T_s(y,t)$ is the soft thresholding function. Then the following result holds.

Theorem 3.1 Let $f\in {\epsilon }^2(A)$ be the solution to $Y=Kf+\epsilon W$ with ${(K^{\ast }Kf)}^{\wedge }(\xi )={(b+{|\xi |}^2)}^{-\alpha }\hat{f}(\xi )$ and $\tilde{f}$ be defined as above. Then

$$\underset{f\in {\epsilon }^2(A)}{sup}E{\parallel \tilde{f}-f\parallel }^2\le Clog\left({\epsilon }^{-1}\right){\epsilon }^{\frac{2}{\frac{3}{2}+2\alpha }}(\epsilon \to 0).$$

Here and in what follows, E stands for the expectation operator.

Proof Since $\{s_{\mu },\mu \in \mathcal{M}\}$ is a Parseval frame, $f={\sum }_{\mu \in \mathcal{M}}{c}_{\mu }{s}_{\mu }$ and $\tilde{f}={\sum }_{\mu \in \mathcal{N}(\epsilon )}{\tilde{c}}_{\mu }{s}_{\mu }$, $\tilde{f}-f={\sum }_{\mu \in \mathcal{M}}({\tilde{c}}_{\mu }-c_{\mu })s_{\mu }$. Moreover, ${\parallel \tilde{f}-f\parallel }^2={\sum }_{\mu \in \mathcal{M}}{|{\tilde{c}}_{\mu }-c_{\mu }|}^2$ and

$$E{\parallel \tilde{f}-f\parallel }^2=\sum _{\mu \in \mathcal{N}(\epsilon )}E{|\widetilde{c_{\mu }}-c_{\mu }|}^2+\sum _{\mu \in \mathcal{N}{(\epsilon )}^C}{|c_{\mu }|}^2.$$

(3.1)

In order to estimate ${\sum }_{\mu \in N{(\epsilon )}^C}{|c_{\mu }|}^2$, one observes ${\sum }_{\mu \in M_j}{|c_{\mu }|}^2\le C{2}^{-2j}$ due to Theorem 2.3. Then ${\sum }_{j>j_1}{\sum }_{\mu \in M_j}{|c_{\mu }|}^2\le C{\sum }_{j>j_1}{2}^{-2j}\le C{2}^{-2j_1}$. By $2^{j_1}\lesssim {\epsilon }^{-\frac{1}{\frac{3}{2}+2\alpha }}$,

$$\sum _{j>j_1}\sum _{\mu \in M_j}{|c_{\mu }|}^2\lesssim {\epsilon }^{\frac{2}{\frac{3}{2}+2\alpha }}.$$

(3.2)

(Here and in what follows, $A\lesssim B$ denotes $A\le CB$ for some constant $C>0$).

Next, one considers $c_{\mu }$ for $j_0\le j\le j_1$ and $|k|\ge 2^{2j+1}$. Note that $|{\psi }^{(d)}(x)|\le C_m{(1+|x|)}^{-m}$ ($d=0,1$, $m=1,2,\dots$). Then $|{\psi }_{j,l,k}^{(d)}(x)|\le C_m{2}^{\frac{3}{2}j}{(1+|B_d^l{A}_d^{j}x-k|)}^{-m}$. Since $f\in {\epsilon }^2(A)$, $suppf\subset Q_0:={[0,1]}^2$ and

$$\left|〈f,{\psi }_{j,l,k}^{(d)}〉\right|\le C_m{2}^{\frac{3}{2}j}{\parallel f\parallel }_{\mathrm{\infty }}{\int }_{Q_0}{(1+|B_d^l{A}_d^{j}x-k|)}^{-m}dx.$$

On the other hand, $|B_d^l{A}_d^{j}x|\le \parallel B_d^l{A}_d^j\parallel |x|\le 2^{2j}|x|\le \sqrt{2}{2}^{2j}$ for $x\in Q_0$. Hence, ${(1+|B_d^l{A}_d^{j}x-k|)}^{-m}\le {(1+|k|-|B_d^l{A}_d^{j}x|)}^{-m}\le {(|k|-\sqrt{2}{2}^{2j})}^{-m}$ for $|k|\ge 2^{2j+1}$. Moreover, ${\sum }_{|k|\ge 2^{2j+1}}{|c_{\mu }|}^2\le 2^{3j}{\sum }_{|k|\ge 2^{2j+1}}{(|k|-\sqrt{2}{2}^{2j})}^{-2m}=$ $2^{3j}{\sum }_{n=1}^{\mathrm{\infty }}{\sum }_{2^{2j+n}\le |k|\le 2^{2j+n+1}}{2}^{-4mj}{(2^n-\sqrt{2})}^{-2m}\lesssim$ $2^{3j}{2}^{-4mj}\times {\sum }_{n=1}^{\mathrm{\infty }}{2}^{2(2j+n+1)}{(2^n-\sqrt{2})}^{-2m}\lesssim 2^{3j}{2}^{-2j(2m-2)}$, since m can be chosen big enough. Therefore,

$$\sum _{j=j_0}^{j_1}\sum _{l=-2^j}^{2^j-1}\sum _{|k|\ge 2^{2j+1}}{|c_{\mu }|}^2\le C_m\sum _{j=j_0}^{\mathrm{\infty }}{2}^{8j}{2}^{-4mj}\lesssim 2^{-8j_0}\le 2^{-6j_0}\lesssim {\epsilon }^{\frac{2}{\frac{3}{2}+2\alpha }}$$

(3.3)

due to the choice of $j_0$. The similar (even simpler) arguments show ${\sum }_{|k|\ge 2^{2j_0+1}}{|〈f,{\phi }_{j_0,k}〉|}^2\lesssim {\epsilon }^{\frac{2}{\frac{3}{2}+2\alpha }}$ with ${\phi }_{j_0,k}(x)=2^{j_0}\phi (2^{j_0}x-k)$. This with (3.2) and (3.3) leads to

$$\sum _{\mu \in \mathcal{N}{(\epsilon )}^C}{|c_{\mu }|}^2\lesssim {\epsilon }^{\frac{2}{\frac{3}{2}+2\alpha }}.$$

(3.4)

Finally, one estimates ${\sum }_{\mu \in \mathcal{N}(\epsilon )}E{|\widetilde{c_{\mu }}-c_{\mu }|}^2$. By the definition of $y_{\mu }$, ${\epsilon }^{-1}{2}^{-2\alpha j}{\sigma }_{\mu }^{-1}{y}_{\mu }\sim N({\epsilon }^{-1}{2}^{-2\alpha j}{\sigma }_{\mu }^{-1}{c}_{\mu },1)$. Applying Theorem 2.4 with ${\eta }^{-1}=\mathrm{♯}\mathcal{N}(\epsilon )$, one obtains that

Hence, $E{|T_s[y_{\mu },\epsilon 2^{2\alpha j}{\sigma }_{\mu }\sqrt{2log(\mathrm{♯}\mathcal{N}(\epsilon ))}]-c_{\mu }|}^2\lesssim [2log(\mathrm{♯}\mathcal{N}(\epsilon ))+1][{\epsilon }^2\frac{2^{4\alpha j}{\sigma }_{\mu }^2}{\mathrm{♯}\mathcal{N}(\epsilon )}+min\{c_{\mu }^2,{\epsilon }^2{2}^{4\alpha j}{\sigma }_{\mu }^2\}]$. By ${\tilde{c}}_{\mu }=T_s[y_{\mu },\epsilon 2^{2\alpha j}{\sigma }_{\mu }\sqrt{2log(\mathrm{♯}\mathcal{N}(\epsilon ))}]$ for $\mu \in \mathcal{N}(\epsilon )$, one knows that

(3.5)

Note that $\mathcal{N}(\epsilon )\cap M_j\subset \{(j,l,k,d):|k|\le 2^{2j+1},|l|\le 2^j\}$. Then $\mathrm{♯}\mathcal{N}(\epsilon )\le C{\sum }_{j\le j_1}{2}^{5j}\lesssim 2^{5j_1}\lesssim {\epsilon }^{-\frac{5}{\frac{3}{2}+2\alpha }}$, and $log(\mathrm{♯}\mathcal{N}(\epsilon ))\lesssim \frac{10}{\frac{3}{2}+2\alpha }log({\epsilon }^{-1})\lesssim log({\epsilon }^{-1})$. Since $\{{\sigma }_{\mu }:\mu \in \mathcal{M}\}$ is uniformly bounded, $[2log(\mathrm{♯}\mathcal{N}(\epsilon ))+1]{\epsilon }^2{\sum }_{\mu \in \mathcal{N}(\epsilon )}\frac{2^{4\alpha j}{\sigma }_{\mu }^2}{\mathrm{♯}\mathcal{N}(\epsilon )}\lesssim log({\epsilon }^{-1}){\epsilon }^2{2}^{4\alpha j_1}$. This with the choice of $2^{j_1}$ shows that

$$[2log(\mathrm{♯}\mathcal{N}(\epsilon ))+1]{\epsilon }^2\sum _{\mu \in \mathcal{N}(\epsilon )}\frac{2^{4\alpha j}{\sigma }_{\mu }^2}{\mathrm{♯}\mathcal{N}(\epsilon )}\lesssim log\left({\epsilon }^{-1}\right){\epsilon }^{\frac{3}{\frac{3}{2}+2\alpha }}\lesssim log\left({\epsilon }^{-1}\right){\epsilon }^{\frac{2}{\frac{3}{2}+2\alpha }}.$$

(3.6)

It remains to estimate $[2log(\mathrm{♯}\mathcal{N}(\epsilon ))+1]{\sum }_{\mu \in \mathcal{N}(\epsilon )}min\{c_{\mu }^2,{\epsilon }^2{2}^{4\alpha j}{\sigma }_{\mu }^2\}$. Clearly,

$$\sum _{\mu \in \mathcal{N}(\epsilon )}min\{c_{\mu }^2,{\epsilon }^2{2}^{4\alpha j}\}=\sum _{\{\mu \in \mathcal{N}(\epsilon ):|c_{\mu }|\ge 2^{2\alpha j}\epsilon \}}{2}^{4\alpha j}{\epsilon }^2+\sum _{\{\mu \in \mathcal{N}(\epsilon ):|c_{\mu }|<2^{2\alpha j}\epsilon \}}{|c_{\mu }|}^2.$$

(3.7)

By Theorem 2.3, ${\sum }_{\{\mu \in M_j:|c_{\mu }|\ge 2^{2\alpha j}\epsilon \}}{2}^{4\alpha j}{\epsilon }^2\lesssim {(2^{2\alpha j}\epsilon )}^{-\frac{2}{3}}{2}^{4\alpha j}{\epsilon }^2\lesssim 2^{\frac{8}{3}\alpha j}{\epsilon }^{\frac{4}{3}}$. Hence,

$$\sum _{\{\mu \in \mathcal{N}(\epsilon ):|c_{\mu }|\ge 2^{2\alpha j}\epsilon \}}{2}^{4\alpha j}{\epsilon }^2=\sum _{j=j_0}^{j_1}\sum _{\{\mu \in M_j:|c_{\mu }|\ge 2^{2\alpha j}\epsilon \}}{2}^{4\alpha j}{\epsilon }^2\lesssim 2^{\frac{8}{3}\alpha j_1}{\epsilon }^{\frac{4}{3}}.$$

(3.8)

On the other hand, ${\sum }_{\{\mu \in \mathcal{N}(\epsilon ):|c_{\mu }|<2^{2\alpha j}\epsilon \}}{|c_{\mu }|}^2={\sum }_{j=j_0}^{j_1}{\sum }_{n=0}^{\mathrm{\infty }}{\sum }_{\{2^{2\alpha j-n-1}\epsilon <|c_{\mu }|\le 2^{2\alpha j-n}\epsilon \}}{|c_{\mu }|}^2$. According to Theorem 2.3, $\mathrm{♯}R(j,2^{2\alpha j-n-1}\epsilon )\lesssim 2^{-\frac{2}{3}(2\alpha j-n-1)}{\epsilon }^{-\frac{2}{3}}$ and

$$\sum _{\{2^{2\alpha j-n-1}\epsilon <|c_{\mu }|\le 2^{2\alpha j-n}\epsilon \}}{|c_{\mu }|}^2\lesssim 2^{-\frac{2}{3}(2\alpha j-n-1)}{\epsilon }^{-\frac{2}{3}}{2}^{2(2\alpha j-n)}{\epsilon }^2\lesssim 2^{\frac{8}{3}\alpha j}{2}^{-\frac{4}{3}n}{2}^{\frac{2}{3}}{\epsilon }^{\frac{4}{3}}.$$

Therefore,

$$\sum _{\{\mu \in \mathcal{N}(\epsilon ):|c_{\mu }|<2^{2\alpha j}\epsilon \}}{|c_{\mu }|}^2\le \sum _{j=j_0}^{j_1}\sum _{n=0}^{\mathrm{\infty }}{2}^{\frac{8}{3}\alpha j}{2}^{-\frac{4}{3}n}{\epsilon }^{\frac{4}{3}}\le 2^{\frac{8}{3}\alpha j_1}{\epsilon }^{\frac{4}{3}}.$$

Combining this with (3.7) and (3.8), one knows that ${\sum }_{\mu \in \mathcal{N}(\epsilon )}min\{c_{\mu }^2,{\epsilon }^2{2}^{4\alpha j}\}\lesssim 2^{\frac{8}{3}\alpha j_1}{\epsilon }^{\frac{4}{3}}$. Furthermore,

$$\sum _{\mu \in \mathcal{N}(\epsilon )}min\{c_{\mu }^2,{\epsilon }^2{2}^{4\alpha j}\}\lesssim {\epsilon }^{\frac{2}{2\alpha +\frac{3}{2}}}$$

(3.9)

thanks to $2^{j_1}\lesssim {\epsilon }^{-\frac{1}{2\alpha +\frac{3}{2}}}$. Now, it follows from (3.5), (3.6) and (3.9) that

$$\sum _{\mu \in \mathcal{N}(\epsilon )}E{|\widetilde{c_{\mu }}-c_{\mu }|}^2\lesssim log\left({\epsilon }^{-1}\right){\epsilon }^{\frac{2}{2\alpha +\frac{3}{2}}}.$$

This with (3.1) and (3.4) leads to the desired conclusion ${sup}_{f\in {\epsilon }^2(A)}E{\parallel \tilde{f}-f\parallel }^2\le Clog({\epsilon }^{-1})\times {\epsilon }^{\frac{2}{\frac{3}{2}+2\alpha }}$. The proof is completed. □