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Sampling for the V-line Transform with Vertex in a Circle

Duy N. Nguyen111High School for the Gifted, Ho Chi Minh City, Vietnam. Email: [email protected].   and Linh V. Nguyen222University of Idaho, 875 Perimeter Dr, Moscow, ID 83844, USA. Email: [email protected].
Abstract

In this paper, we consider a special V-line transform. It integrates a given function ff over the V-lines whose centers are on a circle centered at the origin and the symmetric axes pass through the origin. We derive two sampling scheme of the transform: the standard and interlaced ones. We prove the an error estimate for the schemes, which is explicitly expressed in term of ff.

1 Introduction

The V-line transform arises in Compton camera imaging. Let us first recall the classical setup single-photon emission computed tomography SPECT, which is a nuclear medicine tomographic imaging technique using gamma rays. In SPECT, weakly radioactive tracers are given to the patient. The radioactive tracers can be detected through the emission of gamma ray photons revealing the information about biochemical processes. Then one uses a gamma camera to record photons that enter the detector surface perpendicularly. That is, the camera measures the integrals of the tracer distribution over straight lines that are orthogonal to its surface, see Fig 1.

Refer to caption
Figure 1: Gamma camera for SPECT. Only orthogonal photons measured in detector, other is removed.

This technique removes most photons and only a few photons are recorded. Therefore, a new type of camera for SPECT, which makes use of Compton scattering, was proposed by Everett [6] and Singh [27]. It uses electronic collimation as an alternative to mechanical collimation, which provides both high efficiency and multiple projections of the object. The camera consists of two plane gamma detectors positioned one behind the other. An emitted photon undergoes Compton scattering in the first detector surface D1D_{1} and is absorbed by the second detector surface D2D_{2}. In each detector surface, the position and the energy of the photon are measured. The scattering angle at D1D_{1} is determined via the Compton scattering formula cosψ=1mc2(E1E2)E1E2\cos\psi=1-\dfrac{mc^{2}(E_{1}–E_{2})}{E_{1}E_{2}}, where mm is the electron mass, cc the speed of light, E1E_{1} the photon energy at D1D_{1}, and E2E_{2} the energy of the photon measured at D2D_{2}. So a photon observed at x1D1x_{1}\in D_{1} and x2D2x_{2}\in D_{2} with the energy E1,E2E_{1},E_{2} respectively must have been emitted on the surface of the circular cone, whose vertex is at x1x_{1}, central axis points from x2x_{2} to x1x_{1}, and the half-opening angle is given by ψ\psi (see Fig. 2).

Refer to caption
Figure 2: Compton camera

As a result, a Compton camera gives us the integrals of emission distribution on conical surfaces whose vertices are on the D1D_{1}. The mathematical problem of Compton camera imaging is to reconstruct the emission distribution from such integrals.

There are quite a few works on the mathematical problems of Compton camera. Specially in the two dimensional space, the cone become V-line and there exist some inversion formulas (e.g. [4, 23, 30]). In the three dimensional space, the space of conical surfaces whose vertices are on a detector surface is a five dimensional manifold. One is in the situation of redundant data. Taking advantage of such redundancy was the topic of several works (see, e.g.,[16]). One, however, may wish to restrict themselves into the lower dimensional data. In this article, we are interested in the sampling theory for Compton camera imaging in two dimensional space (i.e., the VV-line transform).

Refer to caption
Figure 3: Our setups: the V-line transform over all V-lines whose vertices are on the circle of radius rr centered at the origin and whose symmetric axis pass through the origin.

Namely, let ff be a bb-essentially band-limited function supported in the circle of radius r01r_{0}\leq 1 centered at the origin. We consider the V-line transform V(f)V(f) of ff on all the V-lines whose vertex is on the circle of radius r>1r>1 centered at the origin and the symmetric axis passing through the origin, see Fig. 3.

Definition 1.1.

Let ff be a compactly supported function in D1(0)D_{1}(0). The V-line transform Vf(φ,ψ)Vf\left(\varphi,\psi\right) of ff is defined by

Vf:[0,2π)×(0;π)\displaystyle Vf:\left[0,2\pi\right)\times\left(0;\pi\right) \displaystyle\longrightarrow ,\displaystyle\mathbb{R},
(φ,ψ)\displaystyle\left(\varphi,\psi\right) \displaystyle\longmapsto σ=±10+f(rθ(φ)tθ(φ+σψ))𝑑t.\displaystyle\sum\limits_{\sigma=\pm 1}{\int\limits_{0}^{+\infty}{f\left(r\theta\left(\varphi\right)-t\theta\left(\varphi+\sigma\psi\right)\right)dt}}.

We consider the problem of recovering function VfVf from its discrete measures Vf(φk,αm)Vf\left(\varphi_{k},\alpha_{m}\right). That is, we consider efficient sampling schemes to recover the V-line transform. Using Shannon sampling series and classical Fourier analysis, several authors [7, 8, 10, 18] study the reconstruction of the standard Radon and circular Radon transforms from its discrete measurement. More recently, Stefanov [28] employed semi-classical analysis to study the same problem for generalized Radon transform.

In this work, we introduce two sampling schemes. The first one is the standard sampling scheme, which is (φ,ψ)=(φk,ψm)(\varphi,\psi)=(\varphi_{k},\psi_{m}), 0kNφ1,0mNψ10\leq k\leq N_{\varphi}-1,0\leq m\leq N_{\psi}-1. Here, φk\varphi_{k} and ψm\psi_{m} are evenly spaced in their domains. By restricting on the cones that intersect with the support of ff, we only consider

φk=\displaystyle\varphi_{k}= k2πNφ,for0kNφ1,\displaystyle\dfrac{k2\pi}{N_{\varphi}},\hskip 28.45274pt\text{for}\hskip 28.45274pt0\leq k\leq N_{\varphi}-1,
ψm=\displaystyle\psi_{m}= m2πNψ,for0m<Nψarcsin(r0/r)2π.\displaystyle\dfrac{m2\pi}{N_{\psi}},\hskip 28.45274pt\text{for}\hskip 28.45274pt0\leq m<\dfrac{N_{\psi}\arcsin(r_{0}/r)}{2\pi}.

To well recover gg from the discrete data we need

Nφ2πrb,Nψ4rb.\displaystyle N_{\varphi}\geq 2\pi rb,\quad N_{\psi}\geq 4rb.

The total number M0M_{0} of needed samples has to satisfy

M04(rb)2arcsin(r0/r).M_{0}\geq 4(rb)^{2}\arcsin(r_{0}/r).

The second, more efficient, sampling scheme is given by

φk=\displaystyle\varphi_{k}= k2πNφfor0kNφ1,\displaystyle\dfrac{k2\pi}{N_{\varphi}}\hskip 28.45274pt\text{for}\hskip 28.45274pt0\leq k\leq N_{\varphi}-1,
ψk,m=\displaystyle\psi_{k,m}= πmNψfor k, m are parity and0m<Nψarcsin(r0/r)π,\displaystyle\dfrac{\pi m}{N_{\psi}}\hskip 28.45274pt\text{for k, m are parity and}\hskip 28.45274pt0\leq m<\dfrac{N_{\psi}\arcsin(r_{0}/r)}{\pi},

with the sampling conditions

Nφ2πrb,Nψ3rb.\displaystyle N_{\varphi}\geq 2\pi rb,\quad N_{\psi}\geq 3rb.

The total number of samples satisfies

M03(rb)2arcsin(r0/r),M_{0}\geq 3(rb)^{2}\arcsin(r_{0}/r),

which is three quarters of the standard scheme.

The article is organized as follows. In Section 2, we introduce some preliminaries.

2 Preliminaries

The 2D Radon transform. We recall the 2D Radon transform that maps a function ff on 2\mathbb{R}^{2} into the set of its integrals over the lines on 2\mathbb{R}^{2}. If θ(φ)=(cosφ,sinφ)S1\theta(\varphi)=\left(\cos\varphi,\sin\varphi\right)\in S^{1} is a unit vector and ss is a real number. The Radon transform Rf(φ,s)Rf(\varphi,s) is defined as

Rf(φ,s)=xθ=sf(x)𝑑l=+f(scosφtsinφ;ssinφ+tcosφ)𝑑t.Rf(\varphi,s)=\int_{x\cdot\theta=s}f(x)\,dl=\int^{+\infty}_{-\infty}{f\left(s\cos\varphi-t\sin\varphi;s\sin\varphi+t\cos\varphi\right)dt}.

In this notation, Rf(φ,s)Rf(\varphi,s) is the integral of ff over the line with a normal direction θ\theta and of the distance ss from the origin. Let Rφf()=Rf(φ,)R_{\varphi}f(\cdot)=Rf\left(\varphi,\cdot\right), a useful relation between the 1D Fourier transform of RφfR_{\varphi}f and the 2D Fourier transform of ff is as follows

(Rφf)(σ)=(2π)1/2f^(σθ).\left(R_{\varphi}f\right)^{\wedge}(\sigma)=(2\pi)^{1/2}\hat{f}\left(\sigma\cdot\theta\right).

Here, the Fourier transform of a function gg defined on n\mathbb{R}^{n} is given by

g^(ξ)=1(2π)n/2ng(x)eixξ𝑑x.\hat{g}(\xi)=\dfrac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}{g(x)e^{-ix\cdot\xi}dx}.

The inversion formula of 2D Radon transform is

f(x,y)=14π20π+sRf(φ,s)sxcosφysinφ𝑑s𝑑φ.f(x,y)=\dfrac{-1}{4\pi^{2}}\int\limits_{0}^{\pi}{\int\limits_{-\infty}^{+\infty}{\dfrac{\frac{\partial}{\partial s}Rf(\varphi,s)}{s-x\cos\varphi-y\sin\varphi}ds}d\varphi}.

One can see the following relationship between V-line transform and Radon transform

Vf(φ,ψ)=Rf(φ+ψπ2,rsinψ)+Rf(φψπ2,rsinψ).Vf\left(\varphi,\psi\right)=Rf\left(\varphi+\psi-\frac{\pi}{2},r\sin\psi\right)+Rf\left(\varphi-\psi-\frac{\pi}{2},-r\sin\psi\right).

Sampling of periodic functions. Let gg be a function in n\mathbb{R}^{n} that is periodic with respect to nn vectors p1,p2,,pnp_{1},p_{2},...,p_{n}. If the matrix P=(p1,p2,,pn)P=\left(p_{1},p_{2},\cdots,p_{n}\right) is nonsingular then gg is called a P-periodic function. One denotes LP=PnL_{P}=P\mathbb{Z}^{n} and its reciprocal lattice LP=L2πPTL^{\perp}_{P}=L_{2\pi P^{-T}}. We define the discrete Fourier transform of gg be the following function on LPL^{\perp}_{P}:

g^(ξ)=|det(P)|1P[0,1)ng(x)eixξ𝑑x,ξLP.\hat{g}(\xi)=\left|\det(P)\right|^{-1}\int\limits_{P[0,1)^{n}}{g(x)e^{-ix\xi}dx},\hskip 28.45274pt\xi\in L^{\perp}_{P}.

Let us note that gg can be recovered from its Fourier coefficients by the series [29]

g(x)=ξLPg^(ξ)eixξ.g(x)=\sum\limits_{\xi\in L^{\perp}_{P}}{\hat{g}(\xi)e^{ix\xi}}.

Let W be a real nonsingular n×nn\times n - matrix, our goal is study the sampling of gg on LWL_{W}. For this to make, we assume LPLWL_{P}\subset L_{W}. This case is implied if and only if P=WMP=WM with an integer matrix MM. Hence, LWLPL^{\perp}_{W}\subset L^{\perp}_{P}. Our study relies heavily on Poisson summation formula for a P-periodic function gg, see [8]. It reads

yLW/LPg(x+y)=|det(P)det(W)|ξLWg^(ξ)eixξ.\sum\limits_{y\in L_{W}/L_{P}}{g(x+y)}=\left|\dfrac{det(P)}{det(W)}\right|\sum\limits_{\xi\in L_{W}^{\perp}}{\hat{g}(\xi)e^{ix\xi}}.

Here, LW/LPL_{W}/L_{P} is the quotient space whose elements are of the form y+LPy+L_{P}. In particular, using the Poisson summation formula one obtains (see [7, 11])

Theorem 2.1.

Suppose gC(n)g\in C^{\infty}\left(\mathbb{R}^{n}\right) is a P - periodic function. Let KLPK\subset L_{P}^{\perp} be a finite set such that its translates K+ηK+\eta, ηLW\eta\in L_{W}^{\perp} are disjoint, and χK\chi_{K} denotes the characteristic function of K, i.e, χK(ξ)=1\chi_{K}(\xi)=1 if ξK\xi\in K and χK(ξ)=0\chi_{K}(\xi)=0 otherwise. We define the sampling series

SW,Kg(x):=|detWdetP|vLW/LPχK~(xv)g(v).S_{W,K}g\left(x\right):=\left|\dfrac{detW}{detP}\right|\sum\limits_{v\in L_{W}/L_{P}}{\widetilde{\chi_{K}}\left(x-v\right)g(v)}.

Then,

SW,KggL2LP\K|g^(ξ)|𝑑ξ.\left\|{S_{W,K}g-g}\right\|_{L^{\infty}}\leq 2\int\limits_{L_{P}^{\perp}\backslash K}{\left|\widehat{g}(\xi)\right|d\xi}.

Sampling of the V-Line transform. Because of supp(f)D1(0)supp(f)\subset D_{1}(0), so that Vf(φ,ψ)=0Vf(\varphi,\psi)=0 with ψ[π2,π)\psi\in\left[\dfrac{\pi}{2},\pi\right). Consequently, VfVf can expand to an even function in ψ\psi variable and 2π2\pi-periodic in both variables. We will make use of the two dimension form of Theorem 2.1 to recover the V-line transform. For fC(D1(0))f\in{{C}^{\infty}}\left({{D}_{1}}\left(0\right)\right), we denote g(φ,ψ)=Vf(φ,ψ)g\left(\varphi,\psi\right)=Vf\left(\varphi,\psi\right). Since, gg is 2π2\pi-periodic in each variables, then the periodic matrix of gg is P=2πI2×2P=2\pi I_{2\times 2} and L𝒫=2L_{\mathcal{P}}^{\perp}=\mathbb{Z}^{2}. We chose matrix WW which satisfies the condition L𝒫L𝒲L_{\mathcal{P}}\subset L_{\mathcal{W}} as

W=2π(1/P0N/(PQ)1/Q)W=2\pi\left(\begin{array}[]{cc}1/P&0\\ N/(PQ)&1/Q\end{array}\right)

where P,Q,NP,Q,N are three integers such that P,Q>0P,Q>0 and 0N<P0\leq N<P. Therefor,

LW/LP={(sj,tjl):sj=j2πP,tjl=(l+Nj/P)2πQ,j=0,,P1,l=0,,Q1}.L_{W}/L_{P}=\{(s_{j},t_{jl}):s_{j}=\frac{j2\pi}{P},t_{jl}=\frac{(l+Nj/P)2\pi}{Q},j=0,...,P-1,l=0,...,Q-1\}.

The sampling theorem for this setting follows as

Theorem 2.2.

Suppose fC0(D1(0))f\in C^{\infty}_{0}\left(D_{1}(0)\right) and g(φ,ψ)=Vf(φ,ψ)g\left(\varphi,\psi\right)=Vf\left(\varphi,\psi\right). Let K2K\subset\mathbb{Z}^{2} be a finite set such that its translates K+ηK+\eta, ηLW\eta\in L_{W}^{\perp} are disjoint. For x[0,2π)×[0,2π)x\in[0,2\pi)\times[0,2\pi), the sampling series is

SW,Kg(x):=1PQvLW/LPχK~(xv)g(v).S_{W,K}g\left(x\right):=\dfrac{1}{PQ}\sum\limits_{v\in L_{W}/L_{P}}{\widetilde{\chi_{K}}\left(x-v\right)g(v)}.

Then,

SW,KggL22\K|g^(ξ)|dξ.\left\|{S_{W,K}g-g}\right\|_{L^{\infty}}\leq 2\sum_{\mathbb{Z}^{2}\backslash K}{\left|\widehat{g}(\xi)\right|d\xi}.

3 The main results

Our goal is to use the Theorem 2.2 to propose some sampling conditions and derive a corresponding sampling error estimate. For this purpose, we assume f^(ξ)\hat{f}(\xi) be negligible for |ξ|>b|\xi|>b, in the sense that the integral ϵd(f,b):=|ξ|>b|ξ|d|f^(ξ)|𝑑ξ\epsilon_{d}(f,b):=\int_{|\xi|>b}{|\xi|^{d}\,|\hat{f}(\xi)|\,d\xi} is small for all real number dd. Such a function ff is called essentially b-band-limited. Here is the main result of this article:

Theorem 3.1.

Let fC0(D)f\in C_{0}^{\infty}(D) be essentially b-band-limited (b>1)(b>1), and g(φ,ψ)=Vf(φ,ψ)g\left(\varphi,\psi\right)=Vf(\varphi,\psi). Let ϑ<1\vartheta<1 be such that 2ϑ2<r2-\vartheta^{2}<r. We define the set

K={(k,m):|k|<rbϑ;max{|k+m|,|km|}<rb}\displaystyle K=\left\{(k,m):|k|<\dfrac{rb}{\vartheta};\max\left\{|k+m|,|k-m|\right\}<rb\right\}

in 2\mathbb{R}^{2}. Let 𝒲\mathcal{W} be a real non-singular 2×22\times 2 matrix such that the sets K+2π(𝒲1)l,l2K+2\pi\left(\mathcal{W}^{-1}\right)l,\,l\in\mathbb{Z}^{2} are mutually disjoint.
Let

η1(ϑ,γ)=(3(1ϑ2)3/2+9(1ϑ2)31γ)η(ϑ,γ),\displaystyle\eta_{1}(\vartheta,\gamma)=\left(\frac{3}{(1-\vartheta^{2})^{3/2}}+\frac{9}{(1-\vartheta^{2})^{3}}\frac{1}{\gamma}\right)\eta(\vartheta,\gamma),
η2(ϑ,γ)=(9(1ϑ2)3+54(1ϑ2)9/21γ)η(ϑ,γ),\displaystyle\eta_{2}(\vartheta,\gamma)=\left(\frac{9}{(1-\vartheta^{2})^{3}}+\frac{54}{(1-\vartheta^{2})^{9/2}}\frac{1}{\gamma}\right)\eta(\vartheta,\gamma),
η3(ϑ,γ)=γη1(ϑ,γ)+η2(ϑ,γ),\displaystyle\eta_{3}(\vartheta,\gamma)=\gamma\eta_{1}(\vartheta,\gamma)+\eta_{2}(\vartheta,\gamma),

where η(ϑ,γ)=γϑexp(γ3(1ϑ2)3/2)\eta\left(\vartheta,\gamma\right)=\gamma\vartheta\exp\left(-\frac{\gamma}{3}(1-\vartheta^{2})^{3/2}\right). Then for bb being big enough

SW,KggL12πη(ϑ,rb)fL1(2)+4r2πϑ3(2b+1)ϵ1(f,b),\displaystyle\left\|{S_{W,K}g-g}\right\|_{L^{\infty}}\leq\dfrac{12}{\pi}\eta^{*}\left(\vartheta,rb\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}+\dfrac{4r^{2}}{\pi\vartheta^{3}}\left(2b+1\right)\epsilon_{1}(f,b),

where η(ϑ,γ)=max{2bϑ2η1(ϑ,γϑ);1rη2(ϑ,γϑ2);ϑr2r¯2η3(ϑ,γϑ2)}\eta^{*}\left(\vartheta,\gamma\right)=\max\left\{\dfrac{2b}{\vartheta^{2}}\eta_{1}\left(\vartheta,\dfrac{\gamma}{\vartheta}\right);\dfrac{1}{r}\eta_{2}\left(\vartheta,\dfrac{\gamma}{\vartheta^{2}}\right);\dfrac{\vartheta}{r^{2}-\bar{r}^{2}}\eta_{3}\left(\vartheta,\dfrac{\gamma}{\vartheta^{2}}\right)\right\}.

In the rest of this article, we present the proof of this theorem. We then discuss two related sampling schemes.

3.1 Proof of the main result

We first derive some useful property of Fourier coefficient of V-line transform (see [7]).

Lemma 3.2.

Suppose fC0(D1(0))f\in C^{\infty}_{0}\left(D_{1}(0)\right), g(φ,ψ)=Vf(φ,ψ)g\left(\varphi,\psi\right)=Vf\left(\varphi,\psi\right). Let g^k,m\widehat{g}_{k,m} be the Fourier coefficient of gg. Then

g^k,m=ik2π02πf^(σθ(α))[Jkm(σ)+Jk+m(σ)]eikα𝑑α𝑑σ,\widehat{g}_{k,m}=\frac{{{i}^{k}}}{2\pi}\int\limits_{\mathbb{R}}{\int\limits_{0}^{2\pi}{\widehat{f}\left(\sigma\theta\left(\alpha\right)\right)\left[{{J}_{k-m}}\left(\sigma\right)+{{J}_{k+m}}\left(\sigma\right)\right]{{e}^{-ik\alpha}}d\alpha d\sigma}},

with Jk(x)J_{k}(x) is Bessel function of first kind.

Proof.

We have

g^k,m\displaystyle{\widehat{g}}_{k,m} =\displaystyle= 14π202πππg(φ,ψ)ei(kφ+mψ)𝑑φ𝑑ψ\displaystyle\frac{1}{4{{\pi}^{2}}}\int\limits_{0}^{2\pi}{\int\limits_{-\pi}^{\pi}{g\left(\varphi,\psi\right){{e}^{-i\left(k\varphi+m\psi\right)}}}d\varphi d\psi}
=\displaystyle= 14π202ππ2π2[Rf(φ+ψπ2,rθ(φ+ψπ2)θ(φ))+Rf(φψπ2,rθ(φψπ2)θ(φ))]×\displaystyle\frac{1}{{4\pi}^{2}}\int\limits_{0}^{2\pi}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\left[Rf\left(\varphi+\psi-\frac{\pi}{2},r\theta\left(\varphi+\psi-\frac{\pi}{2}\right)\theta\left(\varphi\right)\right)+Rf\left(\varphi-\psi-\frac{\pi}{2},r\theta\left(\varphi-\psi-\frac{\pi}{2}\right)\theta\left(\varphi\right)\right)\right]\times}
ei(kφ+mψ)dφdψ\displaystyle{e}^{-i\left(k\varphi+m\psi\right)}d\varphi d\psi
=\displaystyle= I1+I2,\displaystyle I_{1}+I_{2},

where

I1,2=02ππ2π2Rf(φ+ψπ2,rθ(φ±ψπ2)θ(φ))eimψ𝑑φ𝑑ψ.I_{1,2}=\int\limits_{0}^{2\pi}{\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{Rf\left(\varphi+\psi-\frac{\pi}{2},r\theta\left(\varphi\pm\psi-\frac{\pi}{2}\right)\theta\left(\varphi\right)\right){{e}^{-im\psi}d\varphi}d\psi}}.

Changing variable α=φ+ψπ2\alpha=\varphi+\psi-\frac{\pi}{2}, the integral inside of I1I_{1} becomes

π2π2Rf(φ+ψπ2,rθ(φ+ψπ2)θ(φ))\displaystyle\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{Rf\left(\varphi+\psi-\frac{\pi}{2},r\theta\left(\varphi+\psi-\frac{\pi}{2}\right)\theta\left(\varphi\right)\right)} eimψdψ\displaystyle{{e}^{-im\psi}}d\psi
=\displaystyle= (i)kφπφRf[α,rθ(α)θ(φ)]eimαeimφ𝑑α\displaystyle{\left(-i\right)}^{k}\int\limits_{\varphi-\pi}^{\varphi}{Rf\left[\alpha,r\theta\left(\alpha\right)\theta\left(\varphi\right)\right]{{e}^{-im\alpha}}{{e}^{im\varphi}}d\alpha}
=\displaystyle= (i)k2eimφ02πRf[α,rθ(α)θ(φ)]eimα𝑑α\displaystyle\frac{{{\left(-i\right)}^{k}}}{2}{{e}^{im\varphi}}\int\limits_{0}^{2\pi}{Rf\left[\alpha,r\theta\left(\alpha\right)\theta\left(\varphi\right)\right]{{e}^{-im\alpha}}d\alpha}
=\displaystyle= (i)k22πeimφ02π((Rf)(^α,σ)eirθ(α)θ(φ)σ𝑑σ)eimα𝑑α\displaystyle\frac{{{\left(-i\right)}^{k}}}{2\sqrt{2\pi}}{{e}^{im\varphi}}\int\limits_{0}^{2\pi}{\left(\int\limits_{\mathbb{R}}{(Rf\widehat{)(}\alpha,\sigma){{e}^{ir\theta\left(\alpha\right)\theta\left(\varphi\right)\sigma}}d\sigma}\right)}{{e}^{-im\alpha}}d\alpha
=\displaystyle= (i)k2eimφ02πf^(σθ(α))eimα+irσcos(φα)𝑑σ𝑑α.\displaystyle\frac{{{\left(-i\right)}^{k}}}{2}{{e}^{im\varphi}}\int\limits_{0}^{2\pi}{\int\limits_{\mathbb{R}}{\widehat{f}\left(\sigma\theta\left(\alpha\right)\right){{e}^{-im\alpha+ir\sigma\cos\left(\varphi-\alpha\right)}}d\sigma d\alpha}}.

Hence,

I1=\displaystyle{{I}_{1}}= (i)k8π202πf^(σθ(α))02πei(km)φeimα+irσcos(φα)𝑑φ𝑑σ𝑑α\displaystyle\frac{{{\left(-i\right)}^{k}}}{8{{\pi}^{2}}}\int\limits_{0}^{2\pi}{\int\limits_{\mathbb{R}}{\widehat{f}\left(\sigma\theta\left(\alpha\right)\right)}}\int\limits_{0}^{2\pi}{{{e}^{-i\left(k-m\right)\varphi}}{{e}^{-im\alpha+ir\sigma\cos\left(\varphi-\alpha\right)}}d\varphi d\sigma d\alpha}
=\displaystyle= (i)k8π202πf^(σθ(α))eikα(02πei(km)(φα)+irσcos(φα)𝑑φ)𝑑α𝑑σ.\displaystyle\frac{{{\left(-i\right)}^{k}}}{8{{\pi}^{2}}}\int\limits_{0}^{2\pi}{\int\limits_{\mathbb{R}}{\widehat{f}\left(\sigma\theta\left(\alpha\right)\right){{e}^{-ik\alpha}}\left(\int\limits_{0}^{2\pi}{{{e}^{-i\left(k-m\right)\left(\varphi-\alpha\right)+ir\sigma\cos\left(\varphi-\alpha\right)}}d\varphi}\right)d\alpha d\sigma}}.

Noting that

Jk(x)=ik2π02πeixcosφikφ𝑑φ,J_{k}(x)=\dfrac{i^{-k}}{2\pi}\int_{0}^{2\pi}{e^{ix\cos\varphi-ik\varphi}d\varphi},

we obtain

I1=ik4π02πf^(σθ(α))eikαJkm(rσ)𝑑α𝑑σ.\displaystyle I_{1}=\frac{{{i}^{k}}}{4\pi}\int\limits_{0}^{2\pi}{\int\limits_{\mathbb{R}}{\widehat{f}\left(\sigma\theta\left(\alpha\right)\right){{e}^{-ik\alpha}}}}{{J}_{k-m}}\left(r\sigma\right)d\alpha d\sigma. (1)

Similarly,

I2=\displaystyle I_{2}= ik4π02πf^(σθ(α))eikαJk+m(rσ)𝑑α𝑑σ.\displaystyle\frac{i^{k}}{4\pi}\int\limits_{0}^{2\pi}{\int\limits_{\mathbb{R}}{\widehat{f}\left(\sigma\theta\left(\alpha\right)\right){{e}^{-ik\alpha}}{{J}_{k+m}}\left(r\sigma\right)d\alpha d\sigma}}. (2)

Combining (1) and (2), we conclude

g^k,m=ik4π02πf^(σθ(α))[Jkm(rσ)+Jk+m(rσ)]eikα𝑑α𝑑σ.\hat{g}_{k,m}=\frac{{{i}^{k}}}{4\pi}\int\limits_{0}^{2\pi}{\int\limits_{\mathbb{R}}{\widehat{f}\left(\sigma\theta\left(\alpha\right)\right)\left[{{J}_{k-m}}\left(r\sigma\right)+{{J}_{k+m}}\left(r\sigma\right)\right]{{e}^{-ik\alpha}}d\alpha d\sigma}}.

In order to estimate g^m,k\hat{g}_{m,k} we will need the following result

Lemma 3.3.

Suppose fC0(D1(0))f\in C^{\infty}_{0}\left(D_{1}(0)\right), g(φ,ψ)=Vf(φ,ψ)g\left(\varphi,\psi\right)=Vf\left(\varphi,\psi\right). Let g^k,m\widehat{g}_{k,m} be the Fourier coefficient of gg. Then

|02πf^(σθ(α))Jl(σ)eikα𝑑α𝑑σ|D1(0)|f(x)||σσJl(rσ)Jk(σ|x|)𝑑σ|𝑑x+2ϵ1(f,σ),\Big{|}\int\limits_{\mathbb{R}}\int\limits_{0}^{2\pi}{\widehat{f}\left(\sigma\theta\left(\alpha\right)\right){{J}_{l}}\left(\sigma\right){{e}^{-ik\alpha}}d\alpha d\sigma}\Big{|}\leq\int_{D_{1}(0)}{|f(x)|\left|\int^{\sigma}_{-\sigma}{J_{l}(r\sigma)J_{k}(\sigma|x|)d\sigma}\right|dx}+2\epsilon_{-1}(f,\sigma),
Proof.

Since the Bessel function is bounded by 11,

|02πf^(σθ(α))Jl(σ)eikα𝑑α𝑑σ|\displaystyle\Big{|}\int\limits_{\mathbb{R}}\int\limits_{0}^{2\pi}{\widehat{f}\left(\sigma\theta\left(\alpha\right)\right){{J}_{l}}\left(\sigma\right){{e}^{-ik\alpha}}d\alpha d\sigma}\Big{|} 14π|σσ02πf^(σθ)eikαJl(rσ)𝑑α𝑑σ|+12πϵ1(f,σ).\displaystyle\leq\dfrac{1}{4\pi}\left|\int^{\sigma}_{-\sigma}\int^{2\pi}_{0}\hat{f}(\sigma\theta)e^{-ik\alpha}J_{l}(r\sigma)d\alpha d\sigma\right|+\dfrac{1}{2\pi}\epsilon_{-1}(f,\sigma).

Expressing f^(σθ)\hat{f}(\sigma\theta) by its definition, we obtain

02πf^(σθ(α))eikα𝑑α=\displaystyle\int\limits_{0}^{2\pi}{\widehat{f}\left(\sigma\theta\left(\alpha\right)\right){{e}^{-ik\alpha}}d\alpha}= 12π02πD1(0)f(x)eiσθxeikα𝑑x𝑑α\displaystyle\dfrac{1}{2\pi}\int^{2\pi}_{0}\int_{D_{1}(0)}f(x)e^{-i\sigma\theta x}e^{-ik\alpha}dxd\alpha
=\displaystyle= 12πD1(0)f(x)02πei(α|x|cos(αψ)+k(αψ))eikψ𝑑α𝑑x\displaystyle\dfrac{1}{2\pi}\int_{D_{1}(0)}f(x)\int^{2\pi}_{0}e^{-i\left(\alpha|x|\cos(\alpha-\psi)+k(\alpha-\psi)\right)}e^{-ik\psi}d\alpha dx
=\displaystyle= D1(0)f(x)Jk(σ|x|)𝑑x.\displaystyle\int_{D_{1}(0)}f(x)J_{k}(-\sigma|x|)dx.

This gives us the desired in equality.

The following inequality in [26]

Jv(vs)svexp(v(1s2)1/2))(1+(1s2))1/2,v0,0<s1,J_{v}(vs)\leq\dfrac{s^{v}\exp\left(v(1-s^{2})^{1/2}\right))}{\left(1+(1-s^{2})\right)^{1/2}},\hskip 28.45274ptv\geq 0,~{}0<s\leq 1,

yields

Jv(vs)es3(1s2)3/2,v0, 0<s1.J_{v}(vs)\leq e^{-\frac{s}{3}(1-s^{2})^{3/2}},\hskip 28.45274ptv\geq 0,\,0<s\leq 1.

We obtain

sup|x|1ϑkϑk|Jk(σ|x|)|𝑑σ\displaystyle\sup\limits_{|x|\leq 1}\int\limits_{-\vartheta k}^{\vartheta k}{\left|J_{k}(\sigma|x|)\right|d\sigma} 2kϑsup|x|101Jk(kσϑ|x|)𝑑σ2kϑ01eγ3(1|ϑσ|2)3/2𝑑σ2η(ϑ,γ),\displaystyle\leq 2k\vartheta\sup\limits_{|x|\leq 1}\int\limits_{0}^{1}J_{k}(k\sigma\vartheta|x|)d\sigma\leq 2k\vartheta\int\limits_{0}^{1}e^{-\frac{\gamma}{3}(1-|\vartheta\sigma|^{2})^{3/2}}d\sigma\leq 2\eta(\vartheta,\gamma), (3)

where

η(ϑ,γ)=kϑexp(γ3(1ϑ2)3/2).\eta\left(\vartheta,\gamma\right)=k\vartheta\exp\left(-\frac{\gamma}{3}(1-\vartheta^{2})^{3/2}\right).
Lemma 3.4.

Assume that |k|<ϑ1|l||k|<\vartheta^{-1}|l|. Choosing σ=ϑ|l|\sigma=\vartheta|l|, then the following holds.

D1(0)|f(x)||σσJl(rσ)Jk(σ|x|)𝑑σ|𝑑x2η(ϑ,|l|)fL1(2).\displaystyle\int_{D_{1}(0)}{|f(x)|\left|\int^{\sigma}_{-\sigma}{J_{l}(r\sigma)J_{k}(\sigma|x|)d\sigma}\right|dx}\leq 2\eta\left(\vartheta,|l|\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}.
Proof.

Indeed, using the fact that |Jk(s)|1|J_{k}(s)|\leq 1 for all s1s\geq 1, we obtain

D1(0)|f(x)||σσJl(rσ)Jk(σ|x|)𝑑σ|𝑑x\displaystyle\int_{D_{1}(0)}{|f(x)|\left|\int^{\sigma}_{-\sigma}{J_{l}(r\sigma)J_{k}(\sigma|x|)d\sigma}\right|dx}\leq (|σ|ϑ|l|/r|Jl(rσ)|𝑑σ)fL1(2)\displaystyle\left(\int_{|\sigma|\leq\vartheta|l|/r}\left|J_{l}(r\sigma)\right|d\sigma\right)\,\|f\|_{L^{1}(\mathbb{R}^{2})}
\displaystyle\leq (1r|σ|ϑ|l||Jl(σ)|𝑑σ)fL1(2)\displaystyle\left(\dfrac{1}{r}\int_{|\sigma|\leq\vartheta|l|}\left|J_{l}(\sigma)\right|d\sigma\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}
(3)\displaystyle\stackrel{{\scriptstyle(\ref{E:eta1})}}{{\leq}} 2rη(ϑ,|l|)fL1(2).\displaystyle\dfrac{2}{r}\eta\left(\vartheta,|l|\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}.

We will also need the following result

Lemma 3.5.

Suppose fC0(D1(0))f\in C^{\infty}_{0}\left(D_{1}(0)\right) and r¯:=2ϑ2<r\bar{r}:=2-\vartheta^{2}<r, then for |k|ϑ1|l||k|\geq\vartheta^{-1}|l|,

|02πf^(σθ(α))Jl(σ)eikα𝑑α𝑑σ|12π(r2r¯2)kϑη(k,ϑ)fL1(2).\Big{|}\int\limits_{\mathbb{R}}\int\limits_{0}^{2\pi}{\widehat{f}\left(\sigma\theta\left(\alpha\right)\right){{J}_{l}}\left(\sigma\right){{e}^{-ik\alpha}}d\alpha d\sigma}\Big{|}\leq\dfrac{1}{2\pi(r^{2}-\bar{r}^{2})k\vartheta}\eta(k,\vartheta)\|f\|_{L^{1}(\mathbb{R}^{2})}.
Proof.

From [13], for any positive real number ϵ\epsilon such that ρ:=cosh2ϵr\rho:=\sqrt{\cosh 2\epsilon}\leq r, we have

|02πf^(σθ(α))Jl(σ)eikα𝑑α𝑑σ|\displaystyle\Big{|}\int\limits_{\mathbb{R}}\int\limits_{0}^{2\pi}{\widehat{f}\left(\sigma\theta\left(\alpha\right)\right){{J}_{l}}\left(\sigma\right){{e}^{-ik\alpha}}d\alpha d\sigma}\Big{|}\leq I(f)2πexp(ϵ|k|+δr|l|)\displaystyle\dfrac{I(f)}{2\pi}\exp\left(-\epsilon|k|+\frac{\delta}{r}|l|\right)

where δ:=0ϵcosh2t𝑑t\delta:=\int_{0}^{\epsilon}{\sqrt{\cosh 2t}\,dt} and

I(f):=D1(0)|f(x,y)|r2r¯2(x2+y2)𝑑x𝑑y.I(f):=\int_{D_{1}(0)}{\dfrac{\left|f(x,y)\right|}{r^{2}-\bar{r}^{2}(x^{2}+y^{2})}dxdy}.

Simple calculation give for for all ϵ<1\epsilon<1,

δ=0ϵcosh2t𝑑tϵcosh2ϵ<ϵ(1+ϵ).\delta=\int_{0}^{\epsilon}{\sqrt{\cosh 2t}\,dt}\leq\epsilon\sqrt{\cosh 2\epsilon}<\epsilon(1+\epsilon).

Choosing ϵ=(1ϑ2)\epsilon=\left(1-\vartheta^{2}\right) we obtain

ϵδϑrϵ(11+ϵrϑ)ϵ(1ϑ)ϵ(11ϵ)13ϵ3/2=(1ϑ2)3/2\epsilon-\dfrac{\delta\vartheta}{r}\geq\epsilon(1-\frac{1+\epsilon}{r}\vartheta)\geq\epsilon(1-\vartheta)\geq\epsilon(1-\sqrt{1-\epsilon})\geq\frac{1}{3}\epsilon^{3/2}=\left(1-\vartheta^{2}\right)^{3/2}

Therefore, for |k|ϑ1|l||k|\geq\vartheta^{-1}|l|,

exp(ϵ|k|+δr|l|)exp(|k|3(1ϑ2)3/2).\exp\left(-\epsilon|k|+\frac{\delta}{r}|l|\right)\leq\exp\left(-\frac{|k|}{3}(1-\vartheta^{2})^{3/2}\right).

Moreover, since ρ=cosh(2ϵ)<1+ϵ=r¯\rho=\cosh(2\epsilon)<1+\epsilon=\bar{r},

I(f):=D1(0)|f(x,y)|r2r¯2(x2+y2)𝑑x𝑑y1r2r¯2fL1(2).I(f):=\int_{D_{1}(0)}{\dfrac{\left|f(x,y)\right|}{r^{2}-\bar{r}^{2}(x^{2}+y^{2})}dxdy}\leq\frac{1}{r^{2}-\bar{r}^{2}}\|f\|_{L_{1}(\mathbb{R}^{2})}.

The above two inequalities finish our proof. ∎

We are now ready to prove Theorem 3.1.

Proof of Theorem 3.1.

Due to Theorem 2.2 for the solution, suffices to prove that

22\K|g^k,m|12πη(ϑ,rb)fL1(2)+4r2πϑ3(2b+1)ϵ1(f,b).2\sum_{\mathbb{Z}^{2}\backslash K}\left|\widehat{g}_{k,m}\right|\leq\dfrac{12}{\pi}\eta^{*}\left(\vartheta,rb\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}+\dfrac{4r^{2}}{\pi\vartheta^{3}}\left(2b+1\right)\epsilon_{1}(f,b).

Indeed, from Lemma 3.2, we obtain

|g^k,m|12π|02πf^(σθ(α))Jkm(σ)eikα𝑑α𝑑σ|+12π|02πf^(σθ(α))Jk+m(σ)eikα𝑑α𝑑σ|.|\widehat{g}_{k,m}|\leq\frac{1}{2\pi}\bigg{|}\int\limits_{\mathbb{R}}\int\limits_{0}^{2\pi}\widehat{f}(\sigma\theta\left(\alpha\right)){{J}_{k-m}}\left(\sigma\right)\,e^{-ik\alpha}\,d\alpha d\sigma\bigg{|}+\frac{1}{2\pi}\bigg{|}\int\limits_{\mathbb{R}}\int\limits_{0}^{2\pi}\widehat{f}(\sigma\theta\left(\alpha\right))J_{k+m}\left(\sigma\right)\,e^{-ik\alpha}\,d\alpha d\sigma\bigg{|}.

Therefore,

2\K|g^k,m|S1+S2,\displaystyle\sum_{\mathbb{Z}^{2}\backslash K}\left|\widehat{g}_{k,m}\right|\leq S_{1}+S_{2},

where

S1,2=2\K12π|02πf^(σθ(α))Jkm(σ)eikα𝑑α𝑑σ|.\displaystyle S_{1,2}=\sum_{\mathbb{Z}^{2}\backslash K}\frac{1}{2\pi}\bigg{|}\int\limits_{\mathbb{R}}\int\limits_{0}^{2\pi}\widehat{f}(\sigma\theta\left(\alpha\right)){{J}_{k\mp m}}\left(\sigma\right)\,e^{-ik\alpha}\,d\alpha d\sigma\bigg{|}.

Let us denote

η¯1(ϑ,γ)=m>kη(ϑ,m),η¯2(ϑ,γ)=m>kη¯1(ϑ,m),η¯3(ϑ,γ)=m>kmη(ϑ,m),\displaystyle\bar{\eta}_{1}\left(\vartheta,\gamma\right)=\sum\limits_{m>k}{\eta\left(\vartheta,m\right)},\quad\bar{\eta}_{2}\left(\vartheta,\gamma\right)=\sum\limits_{m>k}{\bar{\eta}_{1}\left(\vartheta,m\right)},\quad\bar{\eta}_{3}\left(\vartheta,\gamma\right)=\sum\limits_{m>k}{m\eta\left(\vartheta,m\right)},

We note that η(ϑ,γ)\eta\left(\vartheta,\gamma\right) exponentially decays as kk\to\infty and

m>kmdη(ϑ,m)ksdη(ϑ,s)=kϑsd+1exp(s3(1ϑ2)3/2)𝑑s,\sum\limits_{m>k}{m^{d}\eta\left(\vartheta,m\right)}\leq\int\limits_{k}^{\infty}{s^{d}\eta\left(\vartheta,s\right)}=\int_{k}^{\infty}{\vartheta s^{d+1}\exp\left(-\frac{s}{3}(1-\vartheta^{2})^{3/2}\right)ds},

for d1d\geq-1 and kk big enough. Direct calculations then show

η¯1(ϑ,γ)\displaystyle\bar{\eta}_{1}\left(\vartheta,\gamma\right) (3(1ϑ2)3/2+9(1ϑ2)31γ)η(ϑ,γ)=η1(ϑ,γ),\displaystyle\leq\left(\frac{3}{(1-\vartheta^{2})^{3/2}}+\frac{9}{(1-\vartheta^{2})^{3}}\frac{1}{\gamma}\right)\eta(\vartheta,\gamma)=\eta_{1}(\vartheta,\gamma), (4)
η¯2(ϑ,γ)\displaystyle\bar{\eta}_{2}\left(\vartheta,\gamma\right) (9(1ϑ2)3+54(1ϑ2)9/21γ)η(ϑ,γ)=η2(ϑ,γ),\displaystyle\leq\left(\frac{9}{(1-\vartheta^{2})^{3}}+\frac{54}{(1-\vartheta^{2})^{9/2}}\frac{1}{\gamma}\right)\eta(\vartheta,\gamma)=\eta_{2}(\vartheta,\gamma), (5)
η¯3(ϑ,γ)\displaystyle\bar{\eta}_{3}\left(\vartheta,\gamma\right) γη¯1(ϑ,γ)+η¯2(ϑ,γ)=η3(ϑ,γ).\displaystyle\leq\gamma\bar{\eta}_{1}(\vartheta,\gamma)+\bar{\eta}_{2}(\vartheta,\gamma)=\eta_{3}(\vartheta,\gamma). (6)
Refer to caption
Figure 4: The set K for ϑ=56,b=5,r=32\vartheta=\frac{5}{6},b=5,r=\frac{3}{2}.

PART 1: Estimate S1S_{1}

We decompose S1=S11+S12+S13S_{1}=S_{11}+S_{12}+S_{13} where each SijS_{ij} is the sum ranging over the region Σij\Sigma_{ij}, where

Σ11:\displaystyle\Sigma_{11}: =\displaystyle= {(k,m)2:|k|<rbϑ2;|km|rbϑ},\displaystyle\left\{(k,m)\in\mathbb{Z}^{2}:|k|<\dfrac{rb}{\vartheta^{2}};|k-m|\geq\dfrac{rb}{\vartheta}\right\},
Σ12:\displaystyle\Sigma_{12}: =\displaystyle= {(k,m)2:|k|rbϑ2;|k|>|km|ϑ},\displaystyle\left\{(k,m)\in\mathbb{Z}^{2}:|k|\geq\dfrac{rb}{\vartheta^{2}};|k|>\dfrac{|k-m|}{\vartheta}\right\},
Σ13:\displaystyle\Sigma_{13}: =\displaystyle= {(k,m)2:|k|rbϑ2;|k||km|ϑ}.\displaystyle\left\{(k,m)\in\mathbb{Z}^{2}:|k|\geq\dfrac{rb}{\vartheta^{2}};|k|\leq\dfrac{|k-m|}{\vartheta}\right\}.
Refer to caption
Figure 5: The parts of set 2\K\mathbb{Z}^{2}\backslash K when we calculate I1I_{1}.
  • For the sum S11S_{11}, we note that |km|ϑ|k||k-m|\geq\vartheta|k| in Σ11\Sigma_{11}. Using Lemma 3.3 and Lemma 3.4, we obtain

    Σ11\displaystyle\Sigma_{11}\leq |k|rbϑ2|km|rbϑ(12rπη(ϑ,|km|)fL1(2)+12πϵ1(f,ϑ|km|r))\displaystyle\sum_{|k|\leq\dfrac{rb}{\vartheta^{2}}}\;\sum_{|k-m|\geq\dfrac{rb}{\vartheta}}\left(\dfrac{1}{2r\pi}\eta\left(\vartheta,|k-m|\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}+\dfrac{1}{2\pi}\epsilon_{-1}\left(f,\dfrac{\vartheta|k-m|}{r}\right)\right)
    \displaystyle\leq |k|rbϑ2lrbϑ(1rπη(ϑ,l)fL1(2)+1πϵ1(f,ϑlr))\displaystyle\sum_{|k|\leq\dfrac{rb}{\vartheta^{2}}}\;\sum_{l\geq\dfrac{rb}{\vartheta}}\left(\dfrac{1}{r\pi}\eta\left(\vartheta,l\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}+\dfrac{1}{\pi}\epsilon_{-1}\left(f,\dfrac{\vartheta l}{r}\right)\right)
    (4)\displaystyle\stackrel{{\scriptstyle(\ref{eq4})}}{{\leq}} 2rbϑ2(1πrη1(ϑ,rbϑ)fL1(2)+rπϑϵ0(f,b))\displaystyle\dfrac{2rb}{\vartheta^{2}}\left(\dfrac{1}{\pi r}\eta_{1}\left(\vartheta,\dfrac{rb}{\vartheta}\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}+\dfrac{r}{\pi\vartheta}\epsilon_{0}(f,b)\right)
    \displaystyle\leq 2bπϑ2η1(ϑ,rbϑ)fL1(2)+2r2bπϑ3ϵ0(f,b),\displaystyle\dfrac{2b}{\pi\vartheta^{2}}\eta_{1}\left(\vartheta,\dfrac{rb}{\vartheta}\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}+\dfrac{2r^{2}b}{\pi\vartheta^{3}}\epsilon_{0}(f,b),

    where we have used, 0<μ0<\mu and b>1b>1 (see, e.g., [7])

    lb/μϵd(f,μl)1μϵd+1(f,b),\sum\limits_{l\geq b/\mu}{\epsilon_{d}\left(f,\mu l\right)\leq\dfrac{1}{\mu}\epsilon_{d+1}\left(f,b\right)}, (7)

    for μ=ϑ/r\mu=\vartheta/r.

  • For the sum S12S_{12}, we notice that |km|<ϑ|k||k-m|<\vartheta|k|. Using Lemma 3.5,

    S12\displaystyle S_{12}\leq |k|rbϑ2ϑ|k|2π(r2r¯2)η(ϑ,|k|)fL1(2)ϑπ(r2r¯2)lrbϑ2lη(ϑ,l)fL1(2)\displaystyle\sum_{|k|\geq\dfrac{rb}{\vartheta^{2}}}\dfrac{\vartheta|k|}{2\pi\left(r^{2}-\bar{r}^{2}\right)}\eta\left(\vartheta,|k|\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}\leq\dfrac{\vartheta}{\pi\left(r^{2}-\bar{r}^{2}\right)}\sum_{l\geq\dfrac{rb}{\vartheta^{2}}}l\,\eta\left(\vartheta,l\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}
    (6)\displaystyle\stackrel{{\scriptstyle(\ref{eq6})}}{{\leq}} ϑπ(r2r¯2)η3(ϑ,rbϑ2)fL1(2).\displaystyle\dfrac{\vartheta}{\pi\left(r^{2}-\bar{r}^{2}\right)}\eta_{3}\left(\vartheta,\dfrac{rb}{\vartheta^{2}}\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}.
  • In Σ13\Sigma_{13}, we get the sum for the mm first. Hence, from Lemma 3.4

    Σ13\displaystyle\Sigma_{13}\leq |k|rbϑ2|km|ϑ|k|(12πrη(ϑ,|km|)fL1(2)+12πϵ1(f,ϑ|km|r))\displaystyle\sum_{|k|\geq\dfrac{rb}{\vartheta^{2}}}\;\sum_{|k-m|\geq\vartheta|k|}\left(\dfrac{1}{2\pi r}\eta\left(\vartheta,|k-m|\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}+\dfrac{1}{2\pi}\epsilon_{-1}\left(f,\dfrac{\vartheta|k-m|}{r}\right)\right)
    (4),(7)\displaystyle\stackrel{{\scriptstyle(\ref{eq4}),(\ref{eq7})}}{{\leq}} |k|rbϑ2(12πrη1(ϑ,ϑ|k|)fL1(2)+r2πϑϵ0(f,ϑ2|k|r))\displaystyle\sum_{|k|\geq\dfrac{rb}{\vartheta^{2}}}\left(\dfrac{1}{2\pi r}\eta_{1}\left(\vartheta,\vartheta|k|\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}+\dfrac{r}{2\pi\vartheta}\epsilon_{0}\left(f,\dfrac{\vartheta^{2}|k|}{r}\right)\right)
    \displaystyle\leq lrbϑ2(1πrη1(ϑ,l)fL1(2)+rπϑϵ0(f,ϑ2lr))\displaystyle\sum_{l\geq\dfrac{rb}{\vartheta^{2}}}\left(\dfrac{1}{\pi r}\eta_{1}\left(\vartheta,l\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}+\dfrac{r}{\pi\vartheta}\epsilon_{0}\left(f,\dfrac{\vartheta^{2}l}{r}\right)\right)
    (5),(7)\displaystyle\stackrel{{\scriptstyle(\ref{eq5}),(\ref{eq7})}}{{\leq}} 1πrη2(ϑ,rbϑ2)fL1(2)+r2πϑ3ϵ1(f,b).\displaystyle\dfrac{1}{\pi r}\eta_{2}\left(\vartheta,\dfrac{rb}{\vartheta^{2}}\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}+\dfrac{r^{2}}{\pi\vartheta^{3}}\epsilon_{1}(f,b).

PART 2: Estimate S2S_{2}

Similarly, we consider the sums S21,S22,S23S_{21},S_{22},S_{23} over the regions

Σ21:|k|<rbϑ2;|k+m|rbϑΣ22:|k|rbϑ2;|k|>|k+m|ϑΣ23:|k|rbϑ2;|k||k+m|ϑ.\displaystyle\begin{matrix}&\Sigma_{21}:&|k|<\dfrac{rb}{\vartheta^{2}}&;&|k+m|\geq\dfrac{rb}{\vartheta}\\ &\Sigma_{22}:&|k|\geq\dfrac{rb}{\vartheta^{2}}&;&|k|>\dfrac{|k+m|}{\vartheta}\\ &\Sigma_{23}:&|k|\geq\dfrac{rb}{\vartheta^{2}}&;&|k|\leq\dfrac{|k+m|}{\vartheta}.\\ \end{matrix}
Refer to caption
Figure 6: The parts of set 2\K\mathbb{Z}^{2}\backslash K when we calculate I2I_{2}.

Then,

S21\displaystyle S_{21}\leq 2bπϑ2η1(ϑ,rbϑ)fL1(2)+2r2bπϑ3ϵ0(f,b),\displaystyle\dfrac{2b}{\pi\vartheta^{2}}\eta_{1}\left(\vartheta,\dfrac{rb}{\vartheta}\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}+\dfrac{2r^{2}b}{\pi\vartheta^{3}}\epsilon_{0}(f,b),
S22\displaystyle S_{22}\leq ϑπ(r2r¯2)η3(ϑ,rbϑ2)fL1(2),\displaystyle\dfrac{\vartheta}{\pi\left(r^{2}-\bar{r}^{2}\right)}\eta_{3}\left(\vartheta,\dfrac{rb}{\vartheta^{2}}\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})},
S23\displaystyle S_{23}\leq 1πrη2(ϑ,rbϑ2)fL1(2)+r2πϑ3ϵ1(f,b).\displaystyle\dfrac{1}{\pi r}\eta_{2}\left(\vartheta,\dfrac{rb}{\vartheta^{2}}\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}+\dfrac{r^{2}}{\pi\vartheta^{3}}\epsilon_{1}(f,b).

FINISHING THE PROOF

Combining the estimates in PART 1 and PART 2, we obtain

2(k,m)K|g^(k,m)|12πη(ϑ,rb)fL1(2)+4r2πϑ3(2b+1)ϵ1(f,b),\displaystyle 2\sum_{(k,m)\notin K}|\hat{g}(k,m)|\leq\dfrac{12}{\pi}\eta^{*}\left(\vartheta,rb\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}+\dfrac{4r^{2}}{\pi\vartheta^{3}}\left(2b+1\right)\epsilon_{1}(f,b),

where η(ϑ,rb)=max{2bϑ2η1(ϑ,rbϑ);1rη2(ϑ,rbϑ2);ϑr2r¯2η3(ϑ,rbϑ2)}\eta^{*}\left(\vartheta,rb\right)=\max\left\{\dfrac{2b}{\vartheta^{2}}\eta_{1}\left(\vartheta,\dfrac{rb}{\vartheta}\right);\dfrac{1}{r}\eta_{2}\left(\vartheta,\dfrac{rb}{\vartheta^{2}}\right);\dfrac{\vartheta}{r^{2}-\bar{r}^{2}}\eta_{3}\left(\vartheta,\dfrac{rb}{\vartheta^{2}}\right)\right\}
According to Theorem 2.2, we conclude

SW,KggL12πη(ϑ,rb)fL1(2)+4r2πϑ3(2b+1)ϵ1(f,b).\displaystyle\left\|{S_{W,K}g-g}\right\|_{L^{\infty}}\leq\dfrac{12}{\pi}\eta^{*}\left(\vartheta,rb\right)\left\|f\right\|_{L^{1}(\mathbb{R}^{2})}+\dfrac{4r^{2}}{\pi\vartheta^{3}}\left(2b+1\right)\epsilon_{1}(f,b).

3.2 Sampling schemes for g(φ,ψ)g(\varphi,\psi)

In this section, we consider two schemes that satisfy the conditions in Theorem 3.1. The first one is the standard scheme, where the sampling locations is the Cartesian product. The second one, more efficient, is an interlaced scheme.

Standard Sampling Scheme

Refer to caption
Figure 7: Geometry of the standard sampling scheme.

For the standard sampling, we chose 2πWT2\pi W^{-T} such that the translates K+2πWTmK+2\pi W^{-T}m are disjoint for any m2m\in\mathbb{Z}^{2}. From Figure 7 we have a choice

2πWT=(2rbϑ2002rbϑ(1+1ϑ))2\pi W^{-T}=\left(\begin{array}[]{cc}\dfrac{2rb}{\vartheta^{2}}&0\\ 0&\dfrac{2rb}{\vartheta}\left(1+\dfrac{1}{\vartheta}\right)\end{array}\right)

So that

W=(πϑ2/rb00πϑ2rb(1+ϑ))=:(2π/Nφ002π/Nψ)W=\left(\begin{array}[]{cc}\pi\vartheta^{2}/rb&0\\ 0&\dfrac{\pi\vartheta^{2}}{rb\left(1+\vartheta\right)}\end{array}\right)=:\left(\begin{array}[]{cc}2\pi/N_{\varphi}&0\\ 0&2\pi/N_{\psi}\end{array}\right)

We assume that NφN_{\varphi} and NψN_{\psi} are integers, otherwise we replace them by [Nφ][N_{\varphi}] and [Nψ][N_{\psi}].
Assume that ff is supported in |x|<r01|x|<r_{0}\leq 1, so mm can be restricted to |m|<Nψarcsin(r0/r)2π|m|<\dfrac{N_{\psi}\arcsin(r_{0}/r)}{2\pi}. Because the function g(φ,ψ)g\left(\varphi,\psi\right) is even in ψ\psi so we only chose ψ0\psi\geq 0. This yields the standard detector system gk,m=g(φk,ψm)g_{k,m}=g\left(\varphi_{k},\psi_{m}\right), where

φk=\displaystyle\varphi_{k}= k2πNφ,for0kNφ1\displaystyle\dfrac{k2\pi}{N_{\varphi}},\hskip 28.45274pt\text{for}\hskip 28.45274pt0\leq k\leq N_{\varphi}-1
ψm=\displaystyle\psi_{m}= m2πNψ,for0m<Nψarcsin(r0/r)2π.\displaystyle\dfrac{m2\pi}{N_{\psi}},\hskip 28.45274pt\text{for}\hskip 28.45274pt0\leq m<\dfrac{N_{\psi}\arcsin(r_{0}/r)}{2\pi}.

The sampling conditions in Theorem 3.1 are satisfied if

Nφ\displaystyle N_{\varphi}\geq 2πrbϑ2,Nψ2rb(1+ϑ)ϑ2.\displaystyle\dfrac{2\pi rb}{\vartheta^{2}},\quad N_{\psi}\geq\dfrac{2rb\left(1+\vartheta\right)}{\vartheta^{2}}.

Efficient Sampling Scheme

Refer to caption
Figure 8: Geometry of efficient sampling scheme.

Again, we need to chose 2πWT2\pi W^{-T} such that the translates K+2πWTmK+2\pi W^{-T}m are disjoint for any m2m\in\mathbb{Z}^{2}. Making the following choice (see Fig. 8)

2πWT=(2rbϑ2rbϑ20rbϑ(2+1ϑ)),2\pi W^{-T}=\left(\begin{array}[]{cc}\dfrac{2rb}{\vartheta^{2}}&-\dfrac{rb}{\vartheta^{2}}\\ 0&\dfrac{rb}{\vartheta}\left(2+\dfrac{1}{\vartheta}\right)\end{array}\right),

We obtain

W=(πϑ2rb0πϑ2rb(1+2ϑ)2πϑ2rb(1+2ϑ))=(2π/Nφ0π/Nψ2π/Nψ)W=\left(\begin{array}[]{cc}\dfrac{\pi\vartheta^{2}}{rb}&0\\ \dfrac{\pi\vartheta^{2}}{rb\left(1+2\vartheta\right)}&\dfrac{2\pi\vartheta^{2}}{rb\left(1+2\vartheta\right)}\end{array}\right)=\left(\begin{array}[]{cc}2\pi/N_{\varphi}&0\\ \pi/N_{\psi}&2\pi/N_{\psi}\end{array}\right)

We obtain the interlaced sampling scheme (φk,ψk,m)\left(\varphi_{k},\psi_{k,m}\right) by

φk=k2πNφ,ψk,m=π(k+2m)Nψ.\displaystyle\varphi_{k}=\dfrac{k2\pi}{N_{\varphi}},\quad\psi_{k,m}=\dfrac{\pi(k+2m)}{N_{\psi}}.

Let l=k+2ml=k+2m then we get the efficient sampling scheme (φk,ψkl)\left(\varphi_{k},\psi_{kl}\right) as

φk=\displaystyle\varphi_{k}= k2πNφfor0kNφ1\displaystyle\dfrac{k2\pi}{N_{\varphi}}\hskip 28.45274pt\text{for}\hskip 28.45274pt0\leq k\leq N_{\varphi}-1
αk,l=\displaystyle\alpha_{k,l}= πlNψfor k, l are parity and0l<Nψarcsin(r0/r)π\displaystyle\dfrac{\pi l}{N_{\psi}}\hskip 28.45274pt\text{for k, l are parity and}\hskip 28.45274pt0\leq l<\dfrac{N_{\psi}\arcsin(r_{0}/r)}{\pi}

which the sampling conditions are

Nφ2πrbϑ2,Nψrb(1+2ϑ)ϑ2.\displaystyle N_{\varphi}\geq\dfrac{2\pi rb}{\vartheta^{2}},\quad N_{\psi}\geq\dfrac{rb\left(1+2\vartheta\right)}{\vartheta^{2}}.

Taking the limit ϑ1\vartheta\rightarrow 1 we obtain

Nφ2πrb,Nψ3rb.\displaystyle N_{\varphi}\geq 2\pi rb,\quad N_{\psi}\geq 3rb.

Acknowlegement

Linh Nguyen’s research is partially supported by the NSF grants, DMS 1212125 and DMS 1616904.

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