Sampling and reconstruction in shift-invariant spaces on $\mathbb {R}^d$

Abstract

Let $\phi \in W(C,\ell ^1)$ such that $\{\tau _n\phi :n\in \mathbb {Z}^d\}$ forms a Riesz basis for $V(\phi )$. It is shown that $\mathbb {Z}^d$ is a stable set of sampling for $V(\phi )$ if and only if $\Phi ^\dagger (x)\ne 0$, for every $x\in \mathbb {T}^d$, where $\Phi ^{\dagger }(x):=\sum _{n\in \mathbb {Z}^d}\phi (n)e^{2\pi in\cdot x},~~ x\in \mathbb {T}^d$. Sampling formulae are provided for reconstructing a function $f\in V(\phi )$ from uniform samples using Zak transform and complex analytic technique. The problem of sampling and reconstruction is discussed in the case of irregular samples also. The theory is illustrated with some examples, and numerical implementation for reconstruction of a function from its nonuniform samples is provided using MATLAB.

1 Introduction

Shift-invariant spaces play an important role in modern analysis because of their rich underlying theory and their applications in various fields such as wavelets, theory of frames, approximation theory, nonuniform sampling problems, and so on. We refer to [3] for more information about shift-invariant spaces.

A sampling problem involves the reconstruction of a function $f$ on $\mathbb {R}^d$ from its samples $f(x_j)$, $j\in \mathbb {Z}^d$. Unless a priori some conditions are assumed on $f$, the problem is not meaningful. In fact, one has to obtain some conditions on a sampling set $X=\{x_j\in \mathbb {R}^d:j\in \mathbb {J}\}$, $\mathbb {J}$, a countable set, to recover a certain class of functions from the sample set $X$ in a stable and unique manner. Since sampling problem deals with real-world applications, it is also important to find a stable numerical algorithm to obtain a reconstruction procedure.

The fundamental Shannon’s sampling theorem states that if $f\in L^2(\mathbb {R})\bigcap B_{[-1/2,1/2]}$, then $f$ can be reconstructed from its samples $\{f(k):k\in \mathbb {Z}\}$ by the formula

$$\begin{aligned} f(x)=\sum _{k\in \mathbb {Z}}f(k)\,\,\mathrm{sinc}(x-k), \end{aligned}$$

where $\mathrm{sinc} y=\dfrac{\sin \pi y}{\pi y}$ and $B_{[-1/2,1/2]}$ denotes the class of functions whose Fourier transform have support in $[-1/2,1/2]$. Here, the sample set being the set of integers is uniform. The reconstruction formula of Shannon’s sampling theorem is equivalent to the fact that $\{e^{2\pi ik\xi }:k\in \mathbb {Z}\}$ forms an orthonormal basis for $L^2(-1/2,1/2)$. This equivalence has been extended to a nonuniform sampled data. In fact, Kadec proved that if $X=\{x_k\in \mathbb {R}:|x_k-k|\le L<1/4,k\in \mathbb {Z}\}$, then the set $\{e^{2\pi ix_k\xi }:k\in \mathbb {Z}\}$ is a Riesz basis for $L^2(-1/2,1/2)$. For a more general sampling set, the sampling condition is stated in Beurling density. See [12]. A sampling theorem for wavelet spaces is provided in [22].

The problem of nonuniform sampling in general shift-invariant spaces was studied by Aldroubi and Grochenig in [3]. In [2], they obtained sufficient conditions for a stable set of sampling for shift-invariant spaces $V(\beta _n)$ resulting out of B-spline functions $\beta _n$. They also proved equivalent conditions for a stable set of sampling. In [3], they have discussed the detailed development of nonuniform sampling problem in shift-invariant spaces. Following this paper, Grochenig and Schwab [9] obtained an efficient numerical algorithm for reconstructing functions from $V(\beta _n)$ on $[a, b]$.

Sun [20] showed that signals in locally finite dimensional shift-invariant space $V$ can be locally reconstructed from its samples on any sampling set with sufficiently large density. In [13], Luo gives estimation of errors of the reconstruction formula for nonuniform sampling in shift-invariant spaces. In fact, it is shown that the accuracy of the reconstruction formula mainly depends on the decay property of the generator and the sampling function. The paper [21] deals with the reconstruction of functions belonging to shift-invariant subspaces of $L^2(\mathbb {R})$ from averaged samples using frames.

If $\phi$ is a continuous function having polynomial decay in time as well as frequency domain such that its Fourier transform is nonvanishing on the unit interval and if $\sum _{k\in \mathbb {Z}}\widehat{\widetilde{\phi }}(t+k)\ne 0$, for every $t\in [0,1]$, where $\widetilde{\phi }(x)=\phi (-x)$, then it is shown in [11] that integers and perturbed version of integers form a stable set of sampling for $V(\phi )$. In [16], it is shown that if $\phi$ is a continuous function having polynomial decay in frequency domain such that its Fourier transform is nonvanishing on the unit interval, then any function in $V(\phi )$ can be locally reconstructed from its nonuniform samples. An application of this local reconstruction procedure for voice system is discussed in [17]. In [5], Atreas obtained a stable reconstruction formula for functions belonging to $V(\phi )$, where $\phi$ has certain decay and nonvanishing property for $\sum _{n\in \mathbb {Z}}\phi (n)e^{2\pi i nx}$. He also established a finite reconstruction formula approximating $f\in V(\phi )$ on bounded interval.

In general, sampling and reconstruction model can be stable with respect to different kinds of small perturbations. In [1], the authors address the problem of describing the sampling model which may arise due to perturbation of generators of the underlying shift-invariant spaces, or due to some uncertainty about the characteristics of the measuring devices or due to the perturbation of the sampling set.

In this paper, we study the problem of sampling and reconstruction in shift-invariant spaces $V(\phi )$ in $L^2(\mathbb {R}^d)$. In Sect. 2, we provide the required notations and background. In Sect. 3, we consider the problem of uniform sampling. Let $\phi \in W(C,\ell ^1)$ such that $\{\tau _n\phi :n\in \mathbb {Z}^d\}$ is a Riesz basis for $V(\phi )$. We show that $\mathbb {Z}^d$ forms a stable set of sampling for $V(\phi )$ if and only if $\Phi ^\dagger (x)\ne 0$ for every $x\in \mathbb {T}^d$, where $\Phi ^{\dagger }(x):=\sum _{n\in \mathbb {Z}^d}\phi (n)e^{2\pi in\cdot x},~~ x\in \mathbb {T}^d$. We provide two type of sampling formula for functions $f\in V(\phi )$ using Zak transform and another using complex analytic technique. We also discuss a verifiable condition for $\Phi ^{\dagger }(x)\ne 0$ for $x\in \mathbb {T}^d$. We illustrate the theory with examples. In Sect. 4, we discuss the reconstruction of a function $f\in V(\phi )$ from perturbed version of uniform samples. In Sect. 5, we address the problem of nonuniform sampling for $V(\phi )$ on $\mathbb {R}^d$. Finally, in Sect. 6, we discuss a reconstruction method for a function $f\in V(\phi )\subset L^2(\mathbb {R}^2)$ from its nonuniform samples and illustrate with a numerical example (digital image), using MATLAB.

2 Notations and background

Definition 2.1

A closed subspace $M$ of $L^2(\mathbb {R}^d)$ is called a shift invariant space if $\tau _{n}\phi \in M$, for every $\phi \in M$ and $n\in \mathbb {Z}^d$, where $\tau _n$ is the translation operator defined by $\tau _n\phi (x)=\phi (x-n)$, for all $x\in \mathbb {R}^d$. For $\phi \in L^2(\mathbb {R}^d)$, $\overline{span\{\tau _n\phi :n\in \mathbb {Z}^d\}}$ is called shift invariant space generated by $\phi$ and will be denoted by $V(\phi )$. Here $\overline{span~X}$ denotes the closure of $span~X$.

Definition 2.2

A sequence of vectors $\{f_n:~n\in \mathbb {Z}^d\}$ in a separable Hilbert space $\mathcal {H}$ is said to be a Riesz basis if $\overline{span\{f_n\}}=\mathcal {H}$ and there exist constants $0< c \le C<\infty$ such that

$$\begin{aligned} c\sum _{n\in \mathbb {Z}^d}|d_n|^2\le \big \Vert \sum _{n\in \mathbb {Z}^d}d_nf_n\big \Vert ^2_{\mathcal {H}}\le C\sum _{n\in \mathbb {Z}^d}|d_n|^2, \end{aligned}$$

(2.1)

for all $(d_n)\in \ell ^2(\mathbb {Z}^d)$. Equivalently, a Riesz basis is an image of an orthonormal basis under a bounded invertible operator.

Remark 2.1

If $\{f_n\}$ is a Riesz basis for $\mathcal {H}$, then the image of $\{f_n\}$ under a bounded invertible operator is also a Riesz basis for $\mathcal {H}$.

It is well known that $\{\tau _n\phi :n\in \mathbb {Z}^d\}$ is a Riesz basis for $V(\phi )$ if and only if

(2.2)

where $G_\phi (\xi ):=\sum _{n\in \mathbb {Z}^d}|\widehat{\phi }(\xi +n)|^2$. It follows from (2.2) that $\Vert G_\phi \Vert _0:={\mathop {\mathrm{essinf}} \limits _{{\xi \in \mathbb {T}}^d}}\,\,G_\phi (\xi )$ and $\Vert G_\phi \Vert _{\infty }:={\mathop {\mathrm{esssup}}\limits _{{\xi \in \mathbb {T}}^d}}\,\, G_\phi (\xi )$ are the optimal Riesz bounds for $\{\tau _n\phi :n\in \mathbb {Z}^d\}$. Further, if $\{\tau _n\phi :n\in \mathbb {Z}^d\}$ is a Riesz basis for $V(\phi )$ with bounds $\Vert G_\phi \Vert _{0}$, $\Vert G_\phi \Vert _{\infty }$, then the dual basis is given by $\{\tau _n\psi :n\in \mathbb {Z}^d\}$, where $\psi$ satisfies the relation $\widehat{\psi }=\widehat{\phi }{G_\phi ^{-1}}$. The optimal Riesz bounds of $\{\tau _n\psi :n\in \mathbb {Z}^d\}$ are $\Vert G_\phi \Vert _{\infty }^{-1}$, and $\Vert G_\phi \Vert _{0}^{-1}$. If $d=1$ and $\phi (x)=\mathrm{sinc} x$, then $V(\phi )$ coincides with the space of bandlimited functions, namely the functions whose Fourier transform has support in $[-1/2,1/2]$.

Recall that a closed subspace $V$ of $L^2(\mathbb {R}^d)$ is said to be a reproducing kernel Hilbert space if there exists a function $K_x\in V$ such that

(2.3)

The function $K(x,y):=K_x(y)=\langle K_x,K_y\rangle$ is called the reproducing kernel of $V$.

Remark 2.2

If $\{\phi _n\}$ is a Riesz basis for $V$, then

$$\begin{aligned} K(x,y)=\sum _n\overline{\phi _n(x)}\widetilde{\phi }_n(y) \end{aligned}$$

(2.4)

is a reproducing kernel for $V$, where $\{\widetilde{\phi }_n\}$ is the dual basis of $\{\phi _n\}$.

Definition 2.3

A sequence of vectors $\{f_n:~n\in \mathbb {Z}^d\}$ in a separable Hilbert space $\mathcal {H}$ is said to be a frame if there exist constants $0< m \le M<\infty$ such that

$$\begin{aligned} m\Vert f\Vert _\mathcal {H}^2\le \sum _{n\in \mathbb {Z}^d}|\langle f,f_n\rangle _\mathcal {H}|^2\le M\Vert f\Vert _\mathcal {H}^2, \end{aligned}$$

(2.5)

for every $f\in \mathcal {H}$.

Clearly $\{f_n\}$ is complete in $\mathcal {H}$. It is easy to see that the elements of a frame must satisfy $\Vert f_n\Vert \le \sqrt{M}$.

Definition 2.4

A set $X=\{x_{n}:n\in \mathbb {Z}^d\}\subset \mathbb {R}^d$ is said to be a stable set of sampling for $V(\phi )$ if there exist constants $r$, $R>0$ such that

$$\begin{aligned} r\Vert f\Vert _{L^2(\mathbb {R}^d)} \le \displaystyle {\Big (\sum _{n\in \mathbb {Z}^d}}|f(x_{n})|^2\Big )^{1/2}\le R\Vert f\Vert _{L^2(\mathbb {R}^d)}, \end{aligned}$$

(2.6)

for every $f\in V(\phi )$.

Definition 2.5

The Wiener amalgam space $W(C,\ell ^1)$ is defined as

$$\begin{aligned} W(C,\ell ^1):=\left\{ f\in C(\mathbb {R}^d):\sum _{n\in \mathbb {Z}^d} \max \limits _{x\in [0,1]^d}|f(x+n)|<\infty \right\} . \end{aligned}$$

(2.7)

If $\phi \in W(C,\ell ^1)$, then each function in $V(\phi )$ is continuous. Hence, $V(\phi )$ is a reproducing kernel Hilbert space. In particular, if $\phi$ is a continuous function having compact support, then $V(\phi )$ is a reproducing kernel Hilbert space. In addition, if $\widehat{\phi }\in W(C,\ell ^1)$, then $\phi$ satisfies the Poisson summation formula.

Definition 2.6

The Zak transform $Z\phi$ of $\phi \in L^2(\mathbb {R}^d)$ is the function on $\mathbb {R}^{2d}$ defined by

$$\begin{aligned} (Z\phi )(x,y):=\sum _{n\in \mathbb {Z}^d}\phi (x-n)e^{2\pi in\cdot y}. \end{aligned}$$

(2.8)

The following properties of the Zak transform are well known. See [10].

$$\begin{aligned} Z\widehat{\phi }(x,y)&= e^{2\pi ix\cdot y}Z\phi (-y,x), a.e.,~~ x,y \in \mathbb {R}^d. \end{aligned}$$

(2.9)

$$\begin{aligned} Z\phi (k,y)&= e^{2\pi ik\cdot y}Z\phi (0,y), k\in \mathbb {Z}^d. \end{aligned}$$

(2.10)

For each $F\in L^2(\mathbb {T}^d)$, we define

$$\begin{aligned} (\mathcal {T}F)(x):= \sum _{n\in \mathbb {Z}^d}\langle F,e^{2\pi in\cdot }\rangle \phi (x+n), ~~x\in \mathbb {R}^d. \end{aligned}$$

(2.11)

Clearly, $\mathcal {T}F\in V(\phi )$. Notice that $(\mathcal {T}e^{2\pi i m\cdot })(x)=\phi (x+m)$. The operator $\mathcal {T}$ is bounded above and below. In fact,

$$\begin{aligned} \Vert \mathcal {T}F\Vert _{L^2(\mathbb {R}^d)}^2=\Vert \sum _{n\in \mathbb {Z}^d}\langle F,e^{2\pi i n\cdot }\rangle \phi (\cdot +n)\Vert _{L^2(\mathbb {R}^d)}^2. \end{aligned}$$

(2.12)

Since $\{\tau _n\phi :n\in \mathbb {Z}^d\}$ forms a Riesz basis for $V(\phi )$, we have,

$$\begin{aligned} \Vert G_\phi \Vert _0\Vert F\Vert _{L^2(\mathbb {T}^d)}^2\le \Vert \mathcal {T}F\Vert _{L^2(\mathbb {R}^d)}^2\le \Vert G_\phi \Vert _{\infty }\Vert F\Vert _{L^2(\mathbb {T}^d)}^2. \end{aligned}$$

(2.13)

Moreover, for each $f\in V(\phi )$, there exists a unique $(a_n)\in \ell ^2(\mathbb {Z}^d)$ such that $f=\sum _{n\in \mathbb {Z}^d}a_n\phi (\cdot +n)$. Since $\{e^{2\pi in \cdot } :n\in \mathbb {Z}^d\}$ forms an orthonormal basis for $L^2(\mathbb {T}^d)$, by Riesz-Fischer’s theorem, there exists a unique $F\in L^2(\mathbb {T}^d)$ such that $\langle F,e^{2\pi in\cdot }\rangle =a_n$, $n\in \mathbb {Z}^d$. For this $F$, we have $\mathcal {T}F=f$. Hence $\mathcal {T}$ is onto. Thus, $\mathcal {T}$ is a bounded invertible linear operator from $L^2(\mathbb {T}^d)$ onto $V(\phi )$. Since $\Vert G_\phi \Vert _0$ and $\Vert G_\phi \Vert _{\infty }$ are the optimal Riesz bounds, we get $\Vert \mathcal {T}\Vert ^2=\Vert G_\phi \Vert _{\infty }$, and $\Vert \mathcal {T}^{-1}\Vert ^2=\Vert G_\phi \Vert ^{-1}_{0}$.

From (2.11), we have,

$$\begin{aligned} (\mathcal {T}F)(x)&= \langle F,\sum _{n\in \mathbb {Z}^d}\overline{\phi (x+n)}e^{2\pi in\cdot }\rangle _{L^2(\mathbb {T}^d)}\nonumber \\&= \langle F,\overline{Z\phi (x,.)}\rangle _{L^2(\mathbb {T}^d)}. \end{aligned}$$

(2.14)

For further information, we refer to [7].

Consider the infinite system

$$\begin{aligned} \sum _{k\in \mathbb {Z}^d}d_k\phi (x_j-k)=f(x_j),~~j\in \mathbb {Z}^d. \end{aligned}$$

(2.15)

Let $U$ be the infinite matrix with entries

$$\begin{aligned} U_{jk}=[\phi (x_j-k)],~~~j,k\in \mathbb {Z}^d. \end{aligned}$$

(2.16)

Then we have

Theorem 2.1

Let $V(\phi )$ be a reproducing kernel Hilbert space such that $\phi$ satisfying (2.2). Then, the following statements are equivalent.

• (1) The set $X=\{x_j:j\in \mathbb {Z}^d\}$ is a stable set of sampling for $V(\phi )$.

• (2) There exist $A$, $B>0$ such that

  $$\begin{aligned} A\Vert D\Vert _{\ell ^2(\mathbb {Z}^d)}^2\le \Vert UD\Vert _{\ell ^2(\mathbb {Z}^d)}^2\le B\Vert D\Vert _{\ell ^2(\mathbb {Z}^d)}^2, \end{aligned}$$

  for every $D\in \ell ^2(\mathbb {Z}^d)$.

• (3) The set of reproducing kernels $\{K_{x_j}:j\in \mathbb {Z}^d\}$ is a frame for $V(\phi )$.

• (4) The set $\{Z\phi (x_j,\cdot ):j\in \mathbb {Z}^d\}$ is a frame for $L^2(\mathbb {T}^d)$.

Proof

The equivalence of $(1)$, $(2)$, and $(3)$ is proved in [3]. From (2.3), and (2.14), we have,

$$\begin{aligned} \sum _{j\in \mathbb {Z}^d}|f(x_j)|^2=\sum _{j\in \mathbb {Z}^d}|\langle f,K_{x_j}\rangle |_{L^2(\mathbb {R}^d)}^2=\sum _{j\in \mathbb {Z}^d}|\langle F,(Z\phi )(x_j,\cdot )\rangle |_{L^2(\mathbb {T}^d)}^2, \end{aligned}$$

where $\mathcal {T}F=f$, showing that $(3)$ and $(4)$ are equivalent. $\square$

Remark 2.3

We also have

$$\begin{aligned} A\Vert G_\phi \Vert _{\infty }^{-1}\Vert f\Vert ^2_2\le \sum _{j\in \mathbb {Z}^d}| f(x_j)|^2 \le B\Vert G_\phi \Vert _{0}^{-1}\Vert f\Vert ^2_2, \end{aligned}$$

(2.17)

which shows that $A\Vert G_\phi \Vert _{\infty }^{-1}$, $B\Vert G_\phi \Vert _{0}^{-1}$ are the sampling bounds of $X$.

Let $A$ be a bounded linear operator on $\ell ^2(\mathbb {Z}^d)$ with the associated matrix given by $A=[a_{rs}]$. Then, $A$ is called a Laurent operator if $a_{r-k,s-k}=a_{r,s}$, for every $r$, $s$, $k$ $\in \mathbb {Z}^d$. Let $m$ be a bounded complex valued measurable function defined on $\mathbb {T}^d$. Define $M:L^2(\mathbb {T}^d)\rightarrow L^2(\mathbb {T}^d)$ by

$$\begin{aligned} (Mf)(x):=m(x)f(x). \end{aligned}$$

Then $M$ is a bounded operator on $L^2(\mathbb {T}^d)$ with $\Vert M\Vert =\Vert m\Vert _\infty$. The matrix of $M$ with respect to the orthonormal basis $\{e_n(\cdot )=e^{2\pi in\cdot }$: $n\in \mathbb {Z}^d\}$ is given by

$$\begin{aligned} M_{rs}=\int \limits _{\mathbb {T}^d}m(x)e^{-2\pi i(r-s)\cdot x}\mathrm{d}x \end{aligned}$$

It follows that the operator $A=\mathcal {F}M\mathcal {F}^{-1}$ is a Laurent operator described by the matrix $[\widehat{m}(r-s)]$, where $\mathcal {F}:L^2(\mathbb {T}^d)\rightarrow \ell ^2(\mathbb {Z}^d)$ is the Fourier transform defined by $\mathcal {F}f=\widehat{f}$, $\widehat{f}(n)=\langle f,e_n\rangle$, $n\in \mathbb {Z}^d$. In this case, we say that $A$ is a Laurent operator defined by the function $m$. The function $m$ is called the symbol of $A$.

Theorem 2.2

([8]) Let A be a Laurent operator defined by the continuous function $m$. Then $A$ is invertible if and only if $m(x)\ne 0$, for every $x\in \mathbb {T}^d$. In this case, $A^{-1}$ will also be a Laurent operator with the symbol $\dfrac{1}{m}$. The matrix of $A^{-1}$ is given by $A^{-1}=\left[ \widehat{\left( \dfrac{1}{m}\right) }(r-s)\right]$.

3 Regular sampling in shift-invariant subspaces

We define, for $\phi \in W(C,\ell ^1)$, the function $\Phi ^\dagger$ given by

$$\begin{aligned} \Phi ^{\dagger }(x):=\sum _{n\in \mathbb {Z}^d}\phi (n)e^{2\pi in\cdot x},~~ x\in \mathbb {T}^d. \end{aligned}$$

(3.1)

Since $\phi \in W(C,\ell ^1)$, the infinite sum in (3.1) converges uniformly on $\mathbb {T}^d$, from which it follows that $\Phi ^\dagger$ is continuous on $\mathbb {T}^d$. If we assume $\phi$ to be both real and even, then $\Phi ^\dagger$ will also be real and even.

Theorem 3.1

Suppose that $\phi \in W(C,\ell ^1)$. Then, the operator $U$ described by the matrix $U_{rs}=\phi (r-s)$, $r$, $s\in \mathbb {Z}^d$ is a Laurent operator on $\ell ^2(\mathbb {Z}^d)$ with the symbol $\Phi ^{\dagger }$. Moreover, the operator $U$ satisfies the inequalities

$$\begin{aligned} \Vert \Phi ^\dagger \Vert _0^2\Vert D\Vert _{\ell ^2(\mathbb {Z}^d)}^2\le \Vert UD\Vert _{\ell ^2(\mathbb {Z}^d)}^2\le \Vert \Phi ^\dagger \Vert _{\infty }^2\Vert D\Vert _{\ell ^2(\mathbb {Z}^d)}^2, \end{aligned}$$

(3.2)

for every $D\in \ell ^2(\mathbb {Z}^d)$, where $\Vert \Phi ^\dagger \Vert _\infty =\max \limits _{\mathbb {T}^d}|\Phi ^\dagger (x)|$ and $\Vert \Phi ^\dagger \Vert _0=\min \limits _{\mathbb {T}^d}|\Phi ^\dagger (x)|$.

Proof

Using Parseval’s identity, we have

$$\begin{aligned} \Vert UD\Vert _{\ell ^2(\mathbb {Z}^d)}^2&= \sum _{m\in \mathbb {Z}^d}|(UD)(m)|^2\nonumber \\&= \int \limits _{\mathbb {T}^d}\big |\sum _{m\in \mathbb {Z}^d}\sum _{n\in \mathbb {Z}^d}\phi (m-n)D_ne^{2\pi im\cdot x}\big |^2\mathrm{d}x \nonumber \\&= \int \limits _{\mathbb {T}^d}\big |\sum _{j\in \mathbb {Z}^d}\sum _{n\in \mathbb {Z}^d}\phi (j)D_ne^{2\pi i(n+j)\cdot x}\big |^2\mathrm{d}x \nonumber \\&= \int \limits _{\mathbb {T}^d}\big |\sum _{j\in \mathbb {Z}^d}\phi (j)e^{2\pi ij\cdot x}\sum _{n\in \mathbb {Z}^d}D_ne^{2\pi in\cdot x}\big |^2\mathrm{d}x \nonumber \\&= \int \limits _{\mathbb {T}^d}|\Phi ^\dagger (x)|^2\big |\sum _{n\in \mathbb {Z}^d}D_ne^{2\pi in\cdot x}\big |^2\mathrm{d}x,\\ \nonumber \end{aligned}$$

(3.3)

for every $D\in \ell ^2(\mathbb {Z}^d)$. Then $U$ satisfies (3.2). We define $Mf(x):=\Phi ^\dagger (x)f(x)$, where $M_{rs}=\widehat{\Phi ^\dagger }(r-s)$. But it can be easily seen that $\widehat{\Phi ^\dagger }(r-s)=\phi (r-s)$, from which it follows that $U=\mathcal {F}M\mathcal {F}^{-1}$ is a Laurent operator on $\ell ^2(\mathbb {Z}^d)$ with the symbol $\Phi ^\dagger$. $\square$

As a consequence, we obtain the following result using theorem 2.1.

Corollary 3.1

Let $\phi \in W(C,\ell ^1)$ and satisfy the condition (2.2). Then $\mathbb {Z}^d$ is a stable set of sampling for $V(\phi )$ if and only if $\Phi ^\dagger (x)\ne 0$ for every $x\in \mathbb {T}^d$.

Example 3.1

We consider a function $\vartheta (\xi )$ defined on the interval $0\le \xi \le 1$ satisfying the following properties:

• (1) $0\le \vartheta (\xi )\le 1$,

• (2) $\vartheta (\xi )+\vartheta (1-\xi )=1$,

• (3) $\vartheta (\xi )$ is a monotonically decreasing function,

• (4) $\vartheta (\xi )=1$, $0\le \xi \le \dfrac{1}{3}$.

The function $\vartheta$ is extended to the real line by setting $\vartheta (\xi )=\vartheta (-\xi )$ for $-1\le \xi \le 0$ and $\vartheta (\xi )=0$ for $|\xi | > 1$. Then the Meyer scaling function is defined as

$$\begin{aligned} \phi (x):=\displaystyle \int \limits _{-1}^{1}\sqrt{\vartheta (\xi )}e^{2\pi i\xi x}\mathrm{d}\xi . \end{aligned}$$

(3.4)

It is well known that $\{\tau _n\phi :n\in \mathbb {Z}^d\}$ forms an orthonormal basis for $V(\phi )$. In addition, if $\sqrt{\vartheta (\xi )}$ is infinitely differentiable, then $\phi \in S(\mathbb {R})$. We refer to [15] for further details.

Since $\phi \in S(\mathbb {R})\subset L^1(\mathbb {R})$, it follows from lemma $6_7$ of [23], $\phi \in W(C,\ell ^1)$. Since $\widehat{\phi }(\xi )=\sqrt{\vartheta (\xi )}$, we also have $\widehat{\phi }\in W(C,\ell ^1)$. By Poisson summation formula,

$$\begin{aligned} \Phi ^{\dagger }(x)&= \sum _{n\in \mathbb {Z}}\widehat{\phi }(x+n)\\&= \sum _{n\in \mathbb {Z}}\sqrt{\vartheta (x+n)}. \end{aligned}$$

If $0\le x\le \dfrac{1}{2}$, then $\vartheta (x)>0$, and if $\dfrac{1}{2}<x<1$, then $\vartheta (x-1)>0$. Therefore, $\Phi ^\dagger (x)\ne 0$ for every $x\in [0,1]$. By corollary 3.1, $\mathbb {Z}$ is a stable sampling set for $V(\phi )$.

3.1 A sampling formula for $f\in V(\phi )$

Let $\phi \in W(C,\ell ^1)$. In addition, let $\phi$ be a real-valued even function such that $\Phi ^\dagger (x)\ne 0$ for every $x\in \mathbb {T}^d$. Define

$$\begin{aligned} \theta (x):=\int \limits _{\mathbb {T}^d}\dfrac{Z\phi (x,\xi )}{Z\phi (0,\xi )}\mathrm{d}\xi , x\in \mathbb {R}^d, \end{aligned}$$

(3.5)

where $Z\phi$ is the Zak transform of $\phi$. Then, $\theta$ is a real-valued continuous even function on $\mathbb {R}^d$. Using (2.10), it follows that

$$\begin{aligned} \theta (n)=\delta _{n0}, ~~\mathrm for ~~ n\in \mathbb {Z}^d. \end{aligned}$$

(3.6)

Using (2.9), for every $x\in \mathbb {R}^d$, we have

$$\begin{aligned} \theta (x)&= \int \limits _{\mathbb {T}^d}\dfrac{Z\widehat{\phi }(\xi ,-x)}{\Phi ^\dagger (\xi )}e^{2\pi i\xi \cdot x}\mathrm{d}\xi \\&= \int \limits _{\mathbb {T}^d}\big (\sum _{k\in \mathbb {Z}^d}\widehat{\phi }(\xi +k)e^{2\pi ik\cdot x}\big )\Phi ^\dagger (\xi )^{-1}e^{2\pi i\xi \cdot x}\mathrm{d}x\\&= \int \limits _{\mathbb {R}^d}\widehat{\phi }(\xi ) \Phi ^\dagger (\xi )^{-1}e^{2\pi i\xi \cdot x}\mathrm{d}x, \end{aligned}$$

from which it follows that

$$\begin{aligned} \widehat{\theta }(\xi )=\widehat{\phi }(\xi )\Phi ^\dagger (\xi )^{-1}. \end{aligned}$$

(3.7)

Now, by applying inverse Fourier transform, we get

$$\begin{aligned} \theta (x)=\sum _{k\in \mathbb {Z}^d}\widehat{\left( \dfrac{1}{\Phi ^\dagger }\right) }(k)\phi (x+k). \end{aligned}$$

(3.8)

Clearly $\Phi ^\dagger (x)=1$ if and only if $\phi (n)=\delta _{n0}$. From (3.8), $\Phi ^\dagger (x)=1$ if and only if $\theta (x)=\phi (x)$.

Since $\phi$ is even, $U$ is self-adjoint. Then, it follows from theorem 3.1 that the Laurent operator $U=[\phi (i-j)]$ is invertible. Further, $U^{-1}=\left[ \widehat{\left( \dfrac{1}{\Phi ^\dagger }\right) }(i-j)\right]$.

If $\{\tau _n\phi :n\in \mathbb {Z}^d\}$ is a Riesz basis for $V(\phi )$, then $\{\tau _n\theta :n\in \mathbb {Z}^d\}$ is the image of $\{\tau _n\phi :n\in \mathbb {Z}^d\}$ under a bounded invertible operator $U^{-1}=\left[ \widehat{\left( \dfrac{1}{\Phi ^\dagger }\right) }(i-j)\right]$. Hence $\{\tau _n\theta :n\in \mathbb {Z}^d\}$ is a Riesz basis for $V(\phi )$.

Theorem 3.2

Let $\phi \in W(C,\ell ^1)$ such that $\{\tau _n\phi :n\in \mathbb {Z}^d\}$ is a Riesz basis for $V(\phi )$. Suppose that $\phi$ is a real-valued even function such that $\Phi ^\dagger (x)\ne 0$ for every $x\in \mathbb {T}^d$. Then for every $f\in V(\phi )$, we have,

$$\begin{aligned} f(x)=\sum _{n\in \mathbb {Z}^d}f(n)\theta (x-n),x\in \mathbb {R}^d, \end{aligned}$$

(3.9)

where $\theta$ is defined as in (3.5).

Although the above theorem has not been mentioned in the literature explicitly, this can be obtained from proposition 7 of [4] using lemma 2.9 of [3] and Wiener’s lemma. However, for the sake of completion, we give the proof.

Proof of theorem 3.2

Let $f\in V(\phi )$. Then there exists a unique sequence $(c_k)\in \ell ^2(\mathbb {Z}^d)$ such that

$$\begin{aligned} f(x)=\sum _{k\in \mathbb {Z}^d}c_k\phi (x-k),x\in \mathbb {R}^d. \end{aligned}$$

Notice that $Uc=\{f(n)\}_{n\in \mathbb {Z}^d}$. Since $U$ is invertible, we have,

$$\begin{aligned} f(x)&= \sum _{k\in \mathbb {Z}^d}\left( U^{-1}(f(n))\right) _k\phi (x-k)\nonumber \\&= \sum _{k\in \mathbb {Z}^d}\sum _{n\in \mathbb {Z}^d}\left( U^{-1}_{kn}f(n)\right) \phi (x-k)\nonumber \\&= \sum _{k\in \mathbb {Z}^d}\sum _{n\in \mathbb {Z}^d}\left[ \widehat{\left( \dfrac{1}{\Phi ^\dagger }\right) }(k-n)\right] f(n)\phi (x-k)\nonumber \\&= \sum _{n\in \mathbb {Z}^d}f(n)\sum _{k\in \mathbb {Z}^d}\left[ \widehat{\left( \dfrac{1}{\Phi ^\dagger }\right) }(k-n)\right] \phi (x-k) \end{aligned}$$

(3.10)

Since $\Phi ^\dagger$ is a real-valued even function, (3.10) becomes

$$\begin{aligned} f(x)&=\sum _{n\in \mathbb {Z}^d}f(n)\sum _{k\in \mathbb {Z}^d}\left[ \widehat{\left( \dfrac{1}{\Phi ^\dagger }\right) }(n-k)\right] \phi (x-k)\\&=\sum _{n\in \mathbb {Z}^d}f(n)\sum _{r\in \mathbb {Z}^d}\left[ \widehat{\left( \dfrac{1}{\Phi ^\dagger }\right) }(r)\right] \phi (x-n+r)\\&=\sum _{n\in \mathbb {Z}^d}f(n)\theta (x-n). \end{aligned}$$

$\square$

Example 3.2

Consider the function $\phi _y(x)=e^{-2\pi y|x|}$, $y>0$. The Fourier transform of $\phi _y$ is given by

$$\begin{aligned} \widehat{\phi _y}(\xi )=\dfrac{1}{\pi }\dfrac{y}{y^2+\xi ^2}. \end{aligned}$$

Moreover $\phi _y$, $\widehat{\phi _y}\in W(C,\ell ^1)$. For this $\phi _y$,

$$\begin{aligned} G_{\phi _y}(x)=\dfrac{1}{\pi ^2}\sum _{n\in \mathbb {Z}}\dfrac{y^2}{(y^2+(x+n)^2)^2} \end{aligned}$$

and by Poisson summation formula,

$$\begin{aligned} \Phi ^\dagger _y(x)&=\dfrac{1}{\pi }\sum _{n\in \mathbb {Z}}\dfrac{y}{y^2+(x+n)^2}. \end{aligned}$$

Clearly, the above two series converge uniformly on $[0,1]$. Then $\Phi ^\dagger _y$ and $G_{\phi _y}$ are continuous functions on $[0,1]$. Further, $\Phi ^\dagger _y(x)\ne 0$ for every $x\in [0,1]$. Since $G_{\phi _y}(x)>0$ for every $x\in [0,1]$, there exist two positive constants $c$, $C$ such that $c\le G_{\phi _y}(x)\le C$, for every $x\in [0,1]$, from which it follows that $\{\tau _n\phi _y:n\in \mathbb {Z}^d\}$ is a Riesz basis for $V(e^{-2\pi y\cdot })$. Thus $\phi _y$ satisfies the hypothesis of theorem 3.2.

Let $r=e^{-2\pi y}$. Clearly $0\le r<1$. Then

$$\begin{aligned} \Phi ^\dagger _y(x)=\sum _{n\in \mathbb {Z}}r^{|n|}e^{2\pi inx} =\dfrac{1-r^2}{1+r^2-re^{-2\pi i x}-re^{2\pi i x}}. \end{aligned}$$

Notice that the only nonzero Fourier coefficients of $\dfrac{1}{\Phi ^\dagger _y}$ are given by

$$\begin{aligned} \widehat{\left( \dfrac{1}{\Phi ^\dagger _y}\right) }(0)=\coth 2\pi y,\quad \quad \widehat{\left( \dfrac{1}{\Phi ^\dagger _y}\right) }(\pm 1)=-\dfrac{1}{2} \mathrm{cosech}2\pi y. \end{aligned}$$

Therefore,

$$\begin{aligned} \theta (x)=\coth 2\pi y~\phi (x)-\dfrac{1}{2} \mathrm{cosech}2\pi y~[\phi (x-1)+\phi (x+1)]. \end{aligned}$$

(3.11)

Thus the sampling formula for $f\in V(\phi _y)$is given by

$$\begin{aligned} f(x)=\sum _{n\in \mathbb {Z}}f(n)[\coth 2\pi y~\phi (x-n)-\dfrac{1}{2} \mathrm{{cosech}}2\pi y~[\phi (x-n-1)+\phi (x-n+1)]].\quad \end{aligned}$$

(3.12)

Example 3.3

We now obtain sampling formula for B-splines. The B-splines $Q_n$ are defined inductively as follows:

$$\begin{aligned} Q_1(x):=\chi _{[0,1]}(x), \end{aligned}$$

$$\begin{aligned} Q_{n+1}(x)= (Q_n*Q_1)(x):=\displaystyle {\int \limits _{-\infty }^{\infty }Q_n(x-t)Q_1(t)dt}, n\ge 1. \end{aligned}$$

Using $Q_n$, the symmetric B-splines are defined by

$$\begin{aligned} \beta _n(x):=Q_n(x+\dfrac{n}{2}), n=1,2,3,\ldots \end{aligned}$$

The following properties of $\beta _n$ are well known:

• $\beta _n(x)=\dfrac{1}{(n-1)!}\sum _{j=0}^{n}(-1)^j \left( {\begin{array}{c}n\\ j\end{array}}\right) \left( x+\dfrac{n}{2}-j\right) _+^{n-1}$, $n\ge 2$, where  $x_+=\max (0,x)$.

• $\beta _n$ is an even function with support $\left[ -\dfrac{n}{2},\dfrac{n}{2}\right]$.

• $\int _{-\infty }^{\infty }\beta _n(y+k)\overline{\beta _n(y)}\mathrm{d}y=\beta _{2n}(k)$.

• $\beta _n(x)> 0$, for $|x|<\dfrac{n}{2}$.

• The system $\{\beta _{n}(\cdot -k):k\in \mathbb {Z}\}$ forms a Riesz basis for $V(\beta _{n})$ with optimal Riesz bounds $\Vert G_{\beta _n}\Vert _0=\dfrac{1}{(2n-1)!}\displaystyle \prod _{k=1}^{n-1}\dfrac{(1+\lambda _k)^2}{|\lambda _k|}$ and $\Vert G_{\beta _n}\Vert _\infty =1$ where $\lambda _k$’s are zeros of the Euler–Frobenius polynomial defined by

  $$\begin{aligned} \Pi _{2n-1}(z)=(2n-1)!z^{n-1}\sum _{r=-n+1}^{n-1}\beta _{2n}(r)z^r. \end{aligned}$$

  (3.13)

Let $\phi (x)=\beta _{2l}(x)$. Then $\Phi ^\dagger (x)=\displaystyle {\sum _{r=-l+1}^{l-1}\beta _{2l}(r)e^{2\pi irx}}\ne 0$, for every $x\in [0,1]$. $($See [19]$)$. Moreover, the sampling formula

$$\begin{aligned} f(x)=\sum _{k\in \mathbb {Z}^d}f(k)\beta _{2l}(x-k) \end{aligned}$$

holds if and only if $l=1$. Indeed, if $l=1$, we have $\beta _2(k)=\delta _{k0}$, for every $k\in \mathbb {Z}$. Conversely, if $l\ge 2$, $l-1$ is a positive integer and $l-1\in (-l,l)$. Therefore, $\beta _{2l}(l-1)>0$ which implies that $\beta _2(k)\ne \delta _{k0}$, for every $k\in \mathbb {Z}$. The Fourier coefficients of $\dfrac{1}{\Phi ^\dagger }$ are given by

$$\begin{aligned} {\widehat{\left( \dfrac{1}{\Phi ^\dagger }\right) }}(m)=(2l-1)!\sum _{\lambda _i\in \mathbb {D}}\alpha _i\lambda _i^{|m|+l-2},~m\in \mathbb {Z}, \end{aligned}$$

(3.14)

where $\mathbb {D}$ denotes the unit disc in $\mathbb {C}$, $\alpha _i=\sum _{\begin{array}{c} r=1 \\ r\ne i \end{array}}^{2l-2} (\lambda _i-\lambda _r)^{-1}$, and $\lambda _i$’s are zeros of the Euler–Frobenius polynomial $\Pi _{2l-1}(z)$. Therefore,

$$\begin{aligned} \theta (x)= (2l-1)!\sum _{m\in \mathbb {Z}}\sum _{\lambda _i\in \mathbb {D}}\alpha _i\lambda _i^{|m|+l-2}\beta _{2l}(x+m). \end{aligned}$$

(3.15)

Thus the sampling formula for $f\in V(\beta _{2l})$ is given by

$$\begin{aligned} f(x)=(2l-1)!\sum _{k\in \mathbb {Z}} \sum _{m\in \mathbb {Z}}\sum _{\lambda _i\in \mathbb {D}}f(k)\alpha _i\lambda _i^{|m|+l-2}\beta _{2l}(x+m-k). \end{aligned}$$

(3.16)

In particular, when $\phi (x)=\beta _4(x)$, the sampling formula for $f\in V(\beta _4)$ is given by

$$\begin{aligned} f(x)=\sqrt{3}\sum _{k\in \mathbb {Z}}\sum _{m\in \mathbb {Z}}f(k)(-2+\sqrt{3})^{|m|}\beta _4(x+m-k). \end{aligned}$$

(3.17)

Example 3.4

Let us take

$$\begin{aligned} F_n(x,y):=\beta _n(x)\beta _n(y), x,y\in \mathbb {R}. \end{aligned}$$

Then the system $\{F_{n}(\cdot -r,\cdot -s):(r,s)\in \mathbb {Z}^2\}$ forms a Riesz basis for $V(F_{n})$ with optimal Riesz bounds $\Vert G_{F_n}\Vert _{0}=\Vert G_{\beta _n}\Vert _0^2$ and $\Vert G_{F_n}\Vert _\infty =1$.

Let $\phi (x,y)=F_{2l}(x,y)$. Then $\Phi ^\dagger (x,y)={\sum _{r=-l+1}^{l-1}\beta _{2l}(r)e^{2\pi i rx}}{\sum _{s=-l+1}^{l-1}\beta _{2l}(s)e^{2\pi isy}}$ $\ne 0$. The Fourier coefficients of $\dfrac{1}{\Phi ^\dagger }$ are given by

$$\begin{aligned} \widehat{\left( \dfrac{1}{\Phi ^\dagger }\right) }(m,n) =((2l-1)!)^2\sum _{\lambda _i\in \mathbb {D}}\sum _{\lambda _j\in \mathbb {D}}\alpha _i\alpha _j\lambda _i^{|m|+l-2}\lambda _j^{|n|+l-2} \end{aligned}$$

(3.18)

Therefore,

$$\begin{aligned} \theta (x,y)=\sum _{m,n\in \mathbb {Z}} \sum _{\lambda _i,\lambda _j\in \mathbb {D}}\alpha _i\alpha _j\lambda _i^{|m|+l-2}\lambda _j^{|n|+l-2}F_{2l}(x+m,y+n). \end{aligned}$$

(3.19)

Thus the sampling formula for $f\in V(F_{2l})$ is given by

$$\begin{aligned} f(x,y)=((2l-1)!)^2\sum _{r,s\in \mathbb {Z}} \sum _{m,n\in \mathbb {Z}} \sum _{\lambda _i,\lambda _j\in \mathbb {D}}f(r,s)\alpha _i\alpha _j\lambda _i^{|m|+l-2}\lambda _j^{|n|+l-2}F_{2l}(x+m-r,y+n-s). \end{aligned}$$

(3.20)

3.2 A verifiable criterion for $\Phi ^\dagger \ne 0$

It is important to know when $\Phi ^\dagger (x)\ne 0$, $x\in [0,1]$. On the one hand, if we take $\sum _{n\ne 0}|\phi (n)|<|\phi (0)|$, then the result is true. On the other hand, we notice that if $\phi$ has compact support, then $\Phi ^\dagger$ turns out to be a trigonometric polynomial. Thus, the question of $\Phi ^\dagger (x)\ne 0$, $x\in [0,1]$ boils down to show that the polynomial has no zeros on the unit circle. Consider a polynomial with complex coefficients

$$\begin{aligned} p(z)=a_0+a_1z+\cdots +a_nz^n,~~ a_n\ne 0. \end{aligned}$$

(3.21)

The conjugate polynomial $p^*$ of $p$ is defined as $p^{*}(z):=z^n\overline{p\left( \dfrac{1}{\overline{z}}\right) }$. A polynomial $p(z)$ of degree $n$ is self-inversive if $p^*(z)=p(z)$. We now assume that the polynomial $p$ has real coefficients. Put

$$\begin{aligned} \Delta _k(p):=\mathrm det [A_k(p)+B_k(p)]\cdot \mathrm det [A_k(p)-B_k(p)],~~ k=1,2,\ldots ,n, \end{aligned}$$

(3.22)

where

$$\begin{aligned} A_k(p):= \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} a_0 &{} a_1 &{} \cdots &{} a_{k-1} \\ 0 &{} a_0 &{} \cdots &{} a_{k-2} \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} a_0 \end{array} \right) ,~~ B_k(p):= \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} a_{n-k+1} &{} \cdots &{} a_{n-1} &{} a_n \\ a_{n-k+2} &{} \cdots &{} a_n &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ a_n &{} 0 &{} \cdots &{} 0 \end{array} \right) . \end{aligned}$$

Let us denote by $\mathrm var (b_1, b_2,\ldots , b_n)$ to be the number of sign changes of the real sequence $\{b_1, b_2,\ldots , b_n\}$. Then the following result is well known:

• Schur–Cohn criterion [14]: If $\Delta _k(p)\ne 0$, $k=1,2,3,\ldots , n$, then $p(z)$ has no zeros on the unit circle and $\mathrm{{var}}(1,\Delta _1(p),\ldots ,\Delta _n(p))$ zeros inside the unit circle.

Clearly one cannot apply the Schur–Cohn criterion for self-inversive polynomials. But the derivative of self-inversive polynomial need not be a self-inversive polynomial. The relationship between the number of zeros of $p'(z)$ and the number of zeros of $p(z)$ is given by the following theorem.

Theorem 3.3

(Schur–Cohn) [18] Let $p(z)$ be a self-inversive polynomial of degree $n$. Suppose that $p(z)$ has exactly $\mu$ zeros on the unit circle and $p'(z)$ has exactly $\nu$ zeros in the closed unit disc. Then $\mu =2(\nu +1)-n$.

Based on these discussions and the observation that $p(z)$ is a self-inversive polynomial if $\phi$ is a real-valued continuous function having compact support, we obtain the following result.

Theorem 3.4

Let $\phi$ be a real-valued continuous even function having support $[-l,l]$, where $l$ is an integer $\ge 2$. Define $p(z)=\sum _{k=0}^{2l-2}\phi (k-l+1)z^{k}$. If $\{\phi (k):k=-l,\cdots , l\}$ are chosen so that $\Delta _k(p')\ne 0$, for every $k=1,2,\ldots ,2l-1$ and $\mathrm{{var}}(1,\Delta _1(p'),\ldots , \Delta _{2l-1}(p'))=l-2$, then $\Phi ^\dagger$ has no zeros on $[0,1]$.

Example 3.5

Consider the function

$$\begin{aligned} \phi (x)= \left\{ \begin{array}{ccc} 4-2|x|&{} \text{ if } |x|\le 1\text{, }\\ 3-|x|&{} \text{ if } 2\le |x|\le 3\text{, }\\ 0 &{}\text{ otherwise. } \end{array} \right. \end{aligned}$$

Clearly, $\phi$ is a real-valued continuous even function having support $[-3,3]$. For this $\phi$, $p(z)=1+2z+4z^2+2z^3+z^4$. Then $p'(z)=2+8z+6z^2+4z^3$ and $\Delta _1(p')=-12$, $\Delta _2(p')=-240$, $\Delta _3(p')=-5120$. Therefore, $\mathrm{{var}}(1,-12,-240,-5120)=1$. This implies that $p(z)$ has no zeros on the unit circle.

3.3 A yet another sampling formula

If $\phi$ is a real-valued continuous function having support $[-l,l]$, where $l$ is an integer $\ge 2$, then we rewrite the sampling formula (3.9) in a different way using complex analytic technique.

Theorem 3.5

Let $\phi \in W(C,\ell ^1)$ such that $\{\tau _n\phi :n\in \mathbb {Z}\}$ is a Riesz basis for $V(\phi )$. Suppose that $\phi$ is a real-valued even function having support $[-l,l]$, where $l$ is an integer $\ge 2$. If $\Phi ^\dagger (x)\ne 0$ for every $x\in \mathbb {T}$, then every $f\in V(\phi )$ can be written as

$$\begin{aligned} f(x)=\sum _{n\in \mathbb {Z}}\sum _{m\in \mathbb {Z}}\sum _{\lambda _i\in \mathbb {D}} f(n) \mathrm{Res}\left( \frac{z^{|m|+l-2}}{\sum _{n=0}^{2l-2}\phi (n-l+1)z^n},\lambda _i\right) \phi (x-m-n), \end{aligned}$$

(3.23)

where $\mathbb {D}$ is the unit disc and $\lambda _i$’s are zeros of the polynomial $\sum _{n=0}^{2l-2}\phi (n-l+1)z^n$ inside $\mathbb {D}$.

Proof

If $\phi$ is a real-valued even continuous function having compact support $[-l,l]$, then $\Phi ^{\dagger }(x)=\sum _{n=-l+1}^{l-1}\phi (n)e^{2\pi in x}$ is a trigonometric polynomial of degree $l-1$. Since $\phi (x)$ is a real-valued even function, the Fourier coefficients of the function $\dfrac{1}{\Phi ^\dagger }$ satisfy the relation $\widehat{\left( \dfrac{1}{\Phi ^\dagger }\right) }(m)=\widehat{\left( \dfrac{1}{\Phi ^\dagger }\right) }(-m)$, $m\in \mathbb {Z}$. Hence,

$$\begin{aligned} \widehat{\left( \dfrac{1}{\Phi ^\dagger }\right) }(m)&= {\int \limits _0^1\frac{e^{2\pi i|m|x}}{\sum _{n=-l+1}^{l-1}\phi (n)e^{2\pi inx}}\mathrm{d}x},\nonumber \\&= {\int \limits _0^1\dfrac{e^{2\pi i|m|x}}{e^{2\pi i(-l+1)}\sum _{n=0}^{2l-2}\phi (n-l+1)e^{2\pi inx}}}\mathrm{d}x,\nonumber \\&= {\int \limits _0^1\frac{e^{2\pi i(|m|+l-1)x}}{\sum _{n=0}^{2l-2}\phi (n-l+1)e^{2\pi inx}}}\mathrm{d}x,\nonumber \\&= \dfrac{1}{2\pi i}\int \limits _{|z|=1}\frac{z^{|m|+l-2}}{\sum _{n=0}^{2l-2}\phi (n-l+1)z^n}\mathrm{d}z \end{aligned}$$

(3.24)

Notice that $|m|+l-2\ge 0$ and $\phi (-l+1)\ne 0$. Consequently, the integrand in (3.24) is in rational polynomial form. Therefore, the Fourier coefficients of $\dfrac{1}{\Phi ^\dagger }$ are given by

$$\begin{aligned} \widehat{\left( \frac{1}{\Phi ^\dagger }\right) }(m) =\sum _{\lambda _i\in \mathbb {D}}\mathrm Res \left( \dfrac{z^{|m|+l-2}}{\sum _{n=0}^{2l-2}\phi (n-l+1)z^n},\lambda _i\right) , \end{aligned}$$

(3.25)

where $\mathbb {D}$ is the unit disc and $\lambda _i$’s are zeros of the polynomial $\sum _{n=0}^{2l-2}\phi (n-l+1)z^n$ inside $\mathbb {D}$. Hence, from theorem 3.4, every $f\in V(\phi )$ can be written as

$$\begin{aligned} f(x)=\sum _{n\in \mathbb {Z}}\sum _{m\in \mathbb {Z}}\sum _{\lambda _i\in \mathbb {D}} f(n)\mathrm Res \left( \dfrac{z^{|m|+l-2}}{\sum _{n=0}^{2l-2}\phi (n-l+1)z^n},\lambda _i\right) \phi (x+m-n). \end{aligned}$$

$\square$

4 Perturbation of uniform sampling in shift-invariant spaces

Let $\{f_n\}$ be a frame in a separable Hilbert space $\mathcal {H}$ with frame bounds $m$, $M$. Consider the frame operator $S:\mathcal {H}\rightarrow \mathcal {H}$ defined by

$$\begin{aligned} Sf:=\sum _{n}\langle f,f_n\rangle f_n. \end{aligned}$$

(4.1)

Then $\{S^{-1}f_n\}$ is a frame with frame bounds $M^{-1}$, $m^{-1}$. This is called the canonical dual frame of $\{f_n\}$. Moreover, if $\{f_n\}$ is a Riesz basis for $\mathcal {H}$, then the Riesz bounds coincide with frame bounds. The following result on perturbation of frames will be used later.

Theorem 4.1

([6]) Let $\{f_n\}$ be a frame for $\mathcal {H}$ with frame bounds $m$, $M$ and let $\{g_n\}$ be a sequence in $\mathcal {H}$. If there exists a constant $R<m$ such that

$$\begin{aligned} \sum _{n}|\langle f_n-g_n,f\rangle |^2\le R\Vert f\Vert ^2, \mathrm for each ~~ f\in \mathcal {H}, \end{aligned}$$

(4.2)

then $\{g_n\}$ is a frame for $\mathcal {H}$ with frame bounds $m\left( 1-\sqrt{\dfrac{R}{m}}\right) ^2$, $M\left( 1+\sqrt{\dfrac{R}{M}}\right) ^2$. If $\{f_n\}$ is a Riesz basis for $\mathcal {H}$, then $\{g_n\}$ is a Riesz basis for $\mathcal {H}$.

Let us consider the infinite matrix $A=\left[ \phi (m-n+\delta _m)-\phi (m-n)\right]$.

Theorem 4.2

Let $\{\tau _n\phi :n\in \mathbb {Z}^d\}$ be a Riesz basis for $V(\phi )$ such that $\Phi ^\dagger (x)\ne 0$ for every $x\in \mathbb {T}^d$. Define

$$\begin{aligned} \Vert A\Vert :=\sum _{m\in \mathbb {Z}^d}(\sum _{n\in \mathbb {Z}^d}|\phi (m-n)-\phi (m-n+\delta _m)|)^2. \end{aligned}$$

(4.3)

If $\Vert A\Vert <\Vert \Phi ^\dagger \Vert _0^2\Vert G_{\phi }\Vert _0\Vert G_{\phi }\Vert _{\infty }^{-1}$, then $\{m+\delta _m:m\in \mathbb {Z}^d\}$ is a stable set of sampling for $V(\phi )$.

Proof

It follows from corollary 3.1 that $\mathbb {Z}^d$ is a stable set of sampling for $V(\phi )$. Using theorem 2.1 and remark 2.3 that $\{K_m(\cdot ):m\in \mathbb {Z}^d\}$ forms a frame for $V(\phi )$ with lower frame bound $\Vert \Phi ^\dagger \Vert _0^2\Vert G_{\phi }\Vert _{\infty }^{-1}$, where

$$\begin{aligned} K(x,y)=\sum _{n\in \mathbb {Z}^d}\overline{\tau _n\phi (x)}\tau _n\psi (y). \end{aligned}$$

Since $\{\tau _n\psi :n\in \mathbb {Z}^d\}$ is the dual Riesz basis of $\{\tau _n\phi :n\in \mathbb {Z}^d\}$, we have, $\Vert \tau _n\phi \Vert _2^2\le \Vert G_{\phi }\Vert _0^{-1}$, for every $n\in \mathbb {Z}^d$.

Consider the infinite sum

$$\begin{aligned}&\sum _{m\in \mathbb {Z}^d}|\langle f,K_m(\cdot )-K_{m+\delta _m}(\cdot )\rangle |^2\nonumber \\&\quad =\sum _{m\in \mathbb {Z}^d}\big |\int \limits _{\mathbb {R}^d}f(y)\overline{[K_m(\cdot )-K_{m+\delta _m}(\cdot )]}~\mathrm{d}y\big |^2\nonumber \\&\quad =\sum _{m\in \mathbb {Z}^d}\big |\int \limits _{\mathbb {R}^d}f(y) \sum _{n\in \mathbb {Z}^d}[\phi (m-n)-\phi (m-n+\delta _m)]\overline{\psi (y-n)}~\mathrm{d}y\big |^2\nonumber \\&\quad \le \sum _{m\in \mathbb {Z}^d}\int \limits _{\mathbb {R}^d}|f(y)|^2~\mathrm{d}y \int \limits _{\mathbb {R}^d}\big |\sum _{n\in \mathbb {Z}^d}[\phi (m-n)-\phi (m-n+\delta _m)]\overline{\psi (y-n)}\big |^2~\mathrm{d}y\nonumber \\&\quad = \Vert f\Vert ^2_2\sum _{m\in \mathbb {Z}^d} \int \limits _{\mathbb {R}^d}\big |\sum _{n\in \mathbb {Z}^d}[\phi (m-n)-\phi (m-n+\delta _m)]\overline{\psi (y-n)}\big |^2~\mathrm{d}y \end{aligned}$$

(4.4)

Using Minkowski’s integral inequality, (4.4) becomes

$$\begin{aligned}&\sum _{m}|\langle f,K_m(\cdot )-K_{m+\delta _m}(\cdot )\rangle |^2\nonumber \\&\quad \le \Vert f\Vert ^2_2\sum _{m\in \mathbb {Z}^d} \big [\sum _{n\in \mathbb {Z}^d}|\phi (m-n)-\phi (m-n+\delta _m)|\big (\int \limits _{\mathbb {R}^d}\overline{\psi (y-n)}|^2\mathrm{d}y\big )^{1/2}\big ]^2\nonumber \\&\quad =\Vert f\Vert ^2_2\sum _{m\in \mathbb {Z}^d}\big [\sum _{n\in \mathbb {Z}^d}|\phi (m-n)-\phi (m-n+\delta _m)|\Vert \tau _n\psi \Vert _2\big ]^2\nonumber \\&\quad \le \Vert f\Vert ^2_2\Vert G_{\phi }\Vert _0^{-1}\sum _{m\in \mathbb {Z}^d}\big [\sum _{n\in \mathbb {Z}^d}|\phi (m-n)-\phi (m-n+\delta _m)|\big ]^2\nonumber \\&\quad =\Vert G_{\phi }\Vert _0^{-1}\Vert A\Vert \Vert f\Vert ^2_2 \end{aligned}$$

(4.5)

If $\Vert A\Vert <\Vert \Phi ^\dagger \Vert _0^2\Vert G_{\phi }\Vert _0\Vert G_{\phi }\Vert _{\infty }^{-1}$, then by theorem 4.1, $\{K_{m+\delta _m}:m\in \mathbb {Z}^d\}$ forms a frame for $V(\phi )$. Using theorem 2.1, $\{m+\delta _m:m\in \mathbb {Z}^d\}$ is a stable set of sampling for $V(\phi )$. $\square$

Example 4.1

Take $\phi (x)=\beta _{2}(x)$. Choose $\delta _m=\dfrac{2}{\pi m}$. Then it is easy to verify that $\Vert A\Vert <1$. Then it follows by the above theorem $\{m+\delta _m:m\in \mathbb {Z}^d\}$ is a stable set of sampling for $V(\beta _2)$.

Suppose that $\phi \in W(C,\ell ^1)$ and $\Phi ^\dagger (x)\ne 0$ for every $x\in \mathbb {T}^d$. Then the operator $U$ described by the matrix $U_{rs}=\phi (r-s)$, $r$, $s\in \mathbb {Z}^d$ is an invertible Laurent operator on $\ell ^2(\mathbb {Z}^d)$ with symbol $\Phi ^{\dagger }$. Let us define the matrix

$$\begin{aligned} U_{X_\delta }=[\phi (x_m-n)], \end{aligned}$$

where $X_\delta =\{x_m=m+\delta _m:\Vert \delta _m\Vert <\delta <1,m\in \mathbb {Z}^d\}$ is a sampling set on $\mathbb {R}^d$ for some $\delta >0$. Define $\omega _\phi :\mathbb {R}^+\rightarrow \mathbb {R}^+$ by

$$\begin{aligned} \omega _\phi (\rho ):=\sum _{m\in \mathbb {Z}^d}\sup \limits _{\Vert x\Vert \le \rho }|\phi (x+m)-\phi (m)|, \end{aligned}$$

(4.6)

and

$$\begin{aligned} \delta _\phi :=\inf \{\rho >0:\omega _\phi (\rho )\ge \Vert \Phi ^\dagger \Vert _0\}. \end{aligned}$$

(4.7)

Here $\Vert x\Vert$ for $x\in \mathbb {R}^d$ can be taken as any norm on $\mathbb {R}^d$. Then $\omega _\phi$ is well defined on $\mathbb {R}^+$. Since $\omega _\phi$ is continuous on $\mathbb {R}^+$, $0<\delta _{\phi }<\infty$.

Let $\phi$, $X_\delta$, $U_{X_\delta }$, and $\delta _\phi$ be as above. Given $\phi$, we say that $U_{X_\delta }$ belongs to the class $\mathcal {F}_{\delta _\phi }$ if the matrix $U_{X_\delta }$ is produced from a sampling set $X_\delta$ such that $0\le \delta <\delta _\phi$.

We notice that for any $D\in \ell ^2(\mathbb {Z}^d)$, we have

$$\begin{aligned} \Vert \Phi ^\dagger \Vert _0^2\Vert D\Vert ^2_{\ell ^2(\mathbb {Z}^d)}\le \Vert UD\Vert ^2_{\ell ^2(\mathbb {Z}^d)}\le \Vert \Phi ^\dagger \Vert _\infty ^2\Vert D\Vert ^2_{\ell ^2(\mathbb {Z}^d)}, \end{aligned}$$

(4.8)

which follows from theorem 3.2. Consequently, we obtain the following result whose proof is similar to that of [5] for one dimension.

Lemma 4.1

If the infinite matrix $U_{X_\delta }$ belongs to the class $\mathcal {F}_{\delta _\phi }$, then there exist $A$, $B>0$ such that

$$\begin{aligned} A\Vert D\Vert ^2_{\ell ^2(\mathbb {Z}^d)}\le \Vert U_{X_\delta }D\Vert ^2_{\ell ^2(\mathbb {Z}^d)}\le B\Vert D\Vert ^2_{\ell ^2(\mathbb {Z}^d)}, \end{aligned}$$

(4.9)

for every $D\in \ell ^2(\mathbb {Z}^d)$, where $A=[\Vert \Phi ^\dagger \Vert _0-\omega _\phi (\delta )]^2$ and $B=[\Vert \Phi ^\dagger \Vert _0+\omega _\phi (\delta )]^2$.

Remark 4.1

From the Lemma 4.1, the range of $U_{X_\delta }$ is closed. Hence the pseudo inverse of $U_{X_\delta }$ exists and is given by $U_{X_\delta }^\dagger =(U_{X_\delta }^*U_{X_\delta })^{-1}U_{X_\delta }^*$.

Now, by Lemma 4.1, $U_{X_\delta }^*U_{X_\delta }$ is invertible for $0\le \delta <\delta _\phi$. Then, by theorem 2.1, the set $X_\delta$ is a stable set of sampling for $V(\phi )$. Now, following remark 4.1, we obtain the following result whose proof is similar to that [5] for one dimension if we replace $U_{X_\delta }^{-1}$ by the pseudo inverse $U_{X_\delta }^{\dagger }$.

Theorem 4.3

Let $\phi \in W(C,\ell ^1)$ such that $\{\tau _n\phi :n\in \mathbb {Z}^d\}$ is a Riesz basis for $V(\phi )$. Suppose that $\Phi ^\dagger (x)\ne 0$ for every $x\in \mathbb {T}^d$. Then for any $0\le \delta <\delta _\phi$, the set $X_\delta$ is a stable set of sampling for $V(\phi )$. Moreover, every $f\in V(\phi )$ is uniquely reconstructed from the sample set $X_\delta$ by the formula

$$\begin{aligned} f(x)=\sum _{n\in \mathbb {Z}^d}f(x_n)\psi _n^{X_\delta }(x), \end{aligned}$$

(4.10)

where $\psi _n^{X_\delta }(x):=\sum _{k\in \mathbb {Z}^d}(U_{X_\delta }^{\dagger })_{kn}\phi (x-k)$.

Example 4.2

Consider the linear spline

$$\begin{aligned} Q_2(x)= \left\{ \begin{array}{c@{\quad }c@{\quad }c} x&{} \text{ if } x\in [0,1]\text{, }\\ 2-x&{} \text{ if } x\in [1,2]\text{, }\\ 0 &{}\text{ otherwise. } \end{array} \right. \end{aligned}$$

For any $0\le \rho \le 1$, we have

$$\begin{aligned} \omega _{Q_2}(\rho )=\sup \limits _{|x|\le \rho }|Q_2(x)| +\sup \limits _{|x|\le \rho }|Q_2(x+1)-1|+\sup \limits _{|x|\le \rho }|Q_2(x+2)| =3\rho . \end{aligned}$$

Then $\delta _{Q_2}=\inf \{\rho \in [0,1]:3\rho \ge 1\}=\dfrac{1}{3}$.

Let $H_2(x,y):=Q_2(x)Q_2(y)$. Then for any $0\le \rho \le 1$, $\omega _{H_2}(\rho )=3\rho ^2$. Here $\Vert (x,y)\Vert =\max \{|x|, |y|\}$. Consequently, $\delta _{H_2}=\inf \{\rho \in [0,1]:3\rho ^2\ge 1\}=\dfrac{1}{\sqrt{3}}$.

5 Irregular sampling in shift-invariant spaces

Definition 5.1

For $\delta >0$, a set $Y=\{y_j:j\in \mathbb {Z}^d\}$ is said to be in a $\delta$-neighborhood of $X=\{x_j:j\in \mathbb {Z}^d\}$ if $\Vert x_j-y_j\Vert <\delta$ for every $j\in \mathbb {Z}^d$.

Let $U_X$ denote the bounded operator on $\ell ^2(\mathbb {Z}^d)$ described by the matrix $[\phi (x_i-j)]$. The proof of the following theorem is similar to [11] in one dimension. However, for the sake of completion, we provide the proof here.

Theorem 5.1

Let $\phi$ be a continuous function having compact support in $\mathbb {R}^d$. Then, for every $\epsilon >0$, there exists $\delta >0$ such that whenever $Y$ is in a $\delta$-neighborhood of $X$, $\Vert U_X-U_Y\Vert <\epsilon$.

Proof

Since $\phi$ is a continuous function having compact, there exists an integer $N\in \mathbb {N}$ such that

$$\begin{aligned} \sum _{j\in \mathbb {Z}^d}|\phi (x_i-j)-\phi (y_i-j)|=\sum _{\Vert j\Vert \le N}|\phi (x_i-j)-\phi (y_i-j)|. \end{aligned}$$

(5.1)

Here $\Vert j\Vert$ denote the Euclidean norm of $j$. Notice that $\phi$ is uniformly continuous in $\mathbb {R}^d$. Therefore, given $\epsilon >0$, there exists $\delta >0$ such that

$$\begin{aligned} |\phi (x)-\phi (y)|<\dfrac{\epsilon }{\#(j:\Vert j\Vert \le N)},~~ \mathrm whenever ~~\Vert x-y\Vert <\delta . \end{aligned}$$

(5.2)

Suppose that $Y$ is in this $\delta$-neighborhood. Then $\Vert x_i-y_i\Vert <\delta$, for every $i\in \mathbb {Z}^d$. Hence

$$\begin{aligned} \sum _{j\in \mathbb {Z}^d}|\phi (x_i-j)-\phi (y_i-j)|<\epsilon ,~~ \mathrm for every ~~ i\in \mathbb {Z}^d. \end{aligned}$$

(5.3)

This implies that

$$\begin{aligned} \sup \limits _{i\in \mathbb {Z}^d}\sum _{j\in \mathbb {Z}^d}|\phi (x_i-j)-\phi (y_i-j)|<\epsilon \end{aligned}$$

(5.4)

Similarly, we can prove that

$$\begin{aligned} \sup \limits _{j\in \mathbb {Z}^d}\sum _{i\in \mathbb {Z}^d}|\phi (x_i-j)-\phi (y_i-j)|<\epsilon \end{aligned}$$

(5.5)

By Schur’s test, we know that for any bounded operator $A$ on $\ell ^2(\mathbb {Z}^d)$ described by a matrix $[a_{ij}]$, we have

$$\begin{aligned} \Vert A\Vert \le \sqrt{\alpha \beta }, \end{aligned}$$

where $\alpha =\sup \limits _{i\in \mathbb {Z}^d}\sum _{j\in \mathbb {Z}^d}|a_{ij}|$, and $\beta =\sup \limits _{j\in \mathbb {Z}^d}\sum _{i\in \mathbb {Z}^d}|a_{ij}|$. Thus using (5.4) and (5.5), we get $\Vert U_X-U_Y\Vert <\epsilon$. $\square$

Corollary 5.1

Suppose that $\phi$ is a continuous function having compact support satisfying the condition (2.2). If $X$ is stable set of sampling for $V(\phi )$, then there exists $\delta >0$ such that whenever $Y$ is in $\delta$-neighborhood of $X$, $Y$ is a stable set of sampling for $V(\phi )$. Moreover, every $f\in V(\phi )$ is uniquely reconstructed from the sample set $\{y_n:n\in \mathbb {Z}^d\}$ by the formula

$$\begin{aligned} f(x)=\sum _{n\in \mathbb {Z}^d}f(y_n)\psi _n^{Y}(x), \end{aligned}$$

(5.6)

where $\psi _n^{Y}(x):=\sum _{k\in \mathbb {Z}^d}(U_{Y}^{-1})_{kn}\phi (x-k)$.

Proof

Since $X$ is a stable set of sampling for $V(\phi )$, $U_X$ is invertible. Take $\epsilon =\dfrac{1}{\Vert U_X^{-1}\Vert }$. By theorem 5.1, there exists $\delta >0$ such that $\Vert U_X-U_Y\Vert <\dfrac{1}{\Vert U_X^{-1}\Vert }$. This implies that $U_Y$ is invertible. Hence $Y$ is a stable set of sampling for $V(\phi )$ whenever $Y$ is in this $\delta$-neighborhood. Further, for every $f\in V(\phi )$, there exists a unique sequence $D=(d_k)\in \ell ^2(\mathbb {Z}^d)$ such that

$$\begin{aligned} f(x)=\sum _{k\in \mathbb {Z}^d}d_k\phi (x-k),x\in \mathbb {R}^d. \end{aligned}$$

Notice that $U_{Y}D=\{f(y_n)\}_{n\in \mathbb {Z}^d}$. Define $L_{Y}(f):=\{f(y_n)\}_{n\in \mathbb {Z}^d}$.

By Lemma 4.1, $U_Y$ is invertible for $0\le \delta <\delta _\phi$. Thus, by theorem 2.1, the set $Y$ is a stable set of sampling for $V(\phi )$. Since the matrix $U_Y$ is invertible, we have,

$$\begin{aligned} f(x)&=\sum _{k\in \mathbb {Z}^d}d_k\phi (x-k)\\&=\sum _{k\in \mathbb {Z}^d}\big [U_Y^{-1}L_{Y}(f)\big ]_k\phi (x-k)\\&=\sum _{k\in \mathbb {Z}^d}\big [\sum _{n\in \mathbb {Z}^d}(U_Y^{-1})_{kn}f(x_n)\big ]\phi (x-k)\\&=\sum _{n\in \mathbb {Z}^d}f(y_n)\sum _{k\in \mathbb {Z}^d}(U_{Y}^{-1})_{kn}\phi (x-k)\\&=\sum _{n\in \mathbb {Z}^d}f(y_n)\psi _n^{Y}(x), \end{aligned}$$

where $\psi _n^{Y}(x)=\sum _{k\in \mathbb {Z}^d}(U_{Y}^{-1})_{kn}\phi (x-k)$. $\square$

Corollary 5.2

Suppose that $\phi$ is a continuous function having compact support in $\mathbb {R}^d$ satisfying the condition (2.2). Let $\Phi ^\dagger (x)\ne 0$ for every $x\in \mathbb {T}^d$. If there exists $\delta >0$ such that $X$ is in a $\delta$-neighborhood of $\mathbb {Z}^d$, then $X$ is a stable set of sampling for $V(\phi )$.

Proof

The proof follows from theorem 3.2 and corollary 5.1. $\square$

6 Implementation and illustration

Let $\mathcal {H}=\bigoplus \limits _{i\in \mathbb {Z}}\mathcal {H}_i$, where $\mathcal {H}_i=\ell ^2(\mathbb {Z})$. Let $(d_{kl})\in \ell ^2(\mathbb {Z}^2)$. For each $k\in \mathbb {Z}$, we define $D_{k}:=(\cdots ,d_{k,-1},d_{k,0},d_{k,1},\cdots )^T$, and $D:=(\cdots ,D_{-1},D_{0},D_{1},\cdots )$. Clearly $\Vert D\Vert _{\mathcal {H}}^2:=\sum _{k\in \mathbb {Z}}\Vert D_{k}\Vert _{\ell ^2(\mathbb {Z})}^2<\infty$.

Consider the linear transformation $\mathbb {U}:\mathcal {H}\rightarrow \mathcal {H}$ described by the block matrix $\mathbb {U}$ with matrix entries $U_{ij}$ by the usual action

$$\begin{aligned} F_i=(\mathbb {U}D)_i=\sum _{j\in \mathbb {Z}}U_{ij}D_j,~~ i\in \mathbb {Z} \end{aligned}$$

(6.1)

on $\mathcal {H}$, where $U_{ij}$ is the linear transformation on $\ell ^2(\mathbb {Z})$ described by the matrix $U_{ij}$ by the usual action on $\ell ^2(\mathbb {Z})$.

We shall abbreviate the system (6.1) by

$$\begin{aligned} F=\mathbb {U}D, \end{aligned}$$

(6.2)

where $F=(\cdots ,F_{-1},F_{0},F_{1},\cdots )^T$, $D=(\cdots ,D_{-1},D_{0},D_{1},\cdots )^T$ and

$$\begin{aligned} \mathbb {U}= \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} &{} \vdots &{} \vdots &{} \vdots &{} \\ \cdots &{} U_{-1,-1} &{} U_{-1,0} &{} U_{-1,1}&{} \cdots \\ \cdots &{} U_{0,-1} &{} U_{0,0} &{} U_{0,1} &{} \cdots \\ \cdots &{} U_{1,-1} &{} U_{1,0}&{} U_{1,1} &{} \cdots \\ &{} \vdots &{} \vdots &{} \vdots &{} \end{array} \right) \end{aligned}$$

Let $\phi$ be a continuous function on $\mathbb {R}^2$ with compact support such that $\mathrm supp ~\phi \subseteq [-S,S]^2$. Then the shift-invariant space

$$\begin{aligned} V(\phi )=\left\{ f(x,y)=\sum _{k\in \mathbb {Z}}\sum _{l\in \mathbb {Z}}d_{kl}\phi (x-k,y-l):(d_{kl})\in \ell ^2(\mathbb {Z}^2)\right\} \end{aligned}$$

is a reproducing kernel Hilbert space. Recall that the set $X=\{(x_m,y_n):(m,n)\in \mathbb {Z}^2\}$ said to be a stable set of sampling for $V(\phi )$ if there exist constants $r$, $R>0$ such that

$$\begin{aligned} r\Vert f\Vert _{L^2(\mathbb {R}^2)} \le \displaystyle \left( \,\sum _{(m,n)\in \mathbb {Z}^2}|f(x_m,y_n)|^2\right) ^{1/2}\le R\Vert f\Vert _{L^2(\mathbb {R}^2)}, \end{aligned}$$

(6.3)

for every $f\in V(\phi )$.

Let $f_{mn}=f(x_m,y_n)$. For each $k\in \mathbb {Z}$, we define $F_{k}:=(\cdots ,f_{k,-1},f_{k,0},f_{k,1},\cdots )^T$, and $F:=(\cdots ,F_{-1},F_{0},F_{1},\cdots )$.

Then (6.3) is equivalent to

$$\begin{aligned} r\Vert f\Vert _{L^2(\mathbb {R}^2)} \le \Vert F\Vert _{\mathcal {H}}^2\le R\Vert f\Vert _{L^2(\mathbb {R}^2)}, \end{aligned}$$

(6.4)

for every $f\in V(\phi )$.

In order to reconstruct a function from its samples $f(x_m,y_n)$, consider the infinite system

$$\begin{aligned} \sum _{(k,l)\in \mathbb {Z}^2}d_{kl}\phi (x_m-k,y_n-l)=f(x_m,y_n),~~(m,n)\in \mathbb {Z}^2. \end{aligned}$$

(6.5)

The elements of $\mathbb {Z}^2$ are arranged in lexicographic order. Let $\mathbb {U}$ be a infinite block matrix with matrix entries

$$\begin{aligned} U_{mk}=[\phi (x_m-k,y_n-l)]_{n,l},~~(m,k)\in \mathbb {Z}^2. \end{aligned}$$

(6.6)

Then (6.5) can be written as

$$\begin{aligned} F=\mathbb {U}D, \end{aligned}$$

(6.7)

where $F:=(\cdots ,F_{-1},F_{0},F_{1},\cdots )^T$, $D:=(\cdots ,D_{-1},D_{0},D_{1},\cdots )^T$. Then we have the following as in the case of theorem 2.1.

Theorem 6.1

Let $\mathcal {H}=\bigoplus \limits _{i\in \mathbb {Z}}\mathcal {H}_i$, where $\mathcal {H}_i=\ell ^2(\mathbb {Z})$. Let $\phi$ be a continuous function having compact support satisfying the condition (2.2). Then, the following are equivalent.

• (1) The set $X=\{(x_m,y_n):(m,n)\in \mathbb {Z}^2\}$ is a stable set of sampling for $V(\phi )$.

• (2) There exist $A$, $B>0$ such that

  $$\begin{aligned} A\Vert D\Vert _{\mathcal {H}}^2\le \Vert \mathbb {U}D\Vert _{\mathcal {H}}^2\le B\Vert D\Vert _{\mathcal {H}}^2, \end{aligned}$$

  for every $D\in \mathcal {H}$.

• (3) The set of reproducing kernels $\{K_{(x_m,y_n)}:(m,n)\in \mathbb {Z}^2\}$ is a frame for $V(\phi )$.

We have the following results as in the case of one dimension. We refer to [9] for details.

Lemma 6.1

The restriction of $f\in V(\phi )$ to the rectangle $[M_1,M_0]\times [N_1,N_0]$ is determined uniquely by the coefficients $(d_{kl})$ for $(k,l)\in [M_1-S,M_0-S]\times [N_1-S,N_0-S]\bigcap \mathbb {Z}^2$.

Proof

Since $\phi (x-k,y-l)=0$ if $|x-k|\ge S$ or $|y-l|\ge S$, we obtain

$$\begin{aligned} M_0-S\le x-S<k<x+S\le M_1+S, \end{aligned}$$

and

$$\begin{aligned} N_0-S\le y-S<k<y+S\le N_1+S. \end{aligned}$$

Consequently, we have,

$$\begin{aligned} f(x,y)&=\sum _{(k,l)\in \mathbb {Z}^2}d_{kl}\phi (x-k,y-l)\\&=\sum _{|x-k|<S}\sum _{|y-l|<S}d_{kl}\phi (x-k,y-l)\\&=\sum _{k=M_0-S+1}^{M_1+S-1}\sum _{l=N_0-S+1}^{N_1-S+1}d_{kl}\phi (x-k,y-l) \end{aligned}$$

$\square$

The above result tells that the exact reconstruction of $f\in V(\phi )$ requires only $(M_1-M_0+2S-1)(N_1-N_0+2S-1)$ unknown coefficients $d_{kl}$ with $(k,l)\in [M_1-S,M_0-S]\times [N_1-S,N_0-S]\bigcap \mathbb {Z}^2$. Hence, we need at least $(M_1-M_0+2S-1)(N_1-N_0+2S-1)$ samples in $[M_1,M_0]\times [N_1,N_0]$ for the coefficients to be determined uniquely.

Lemma 6.2

If $\mathrm supp ~\phi \subseteq [-S,S]^2$, then the block matrix $\mathbb {U}^*\mathbb {U}$ is a band block matrix of bandwidth $2S$.

Proof

By the definition of $\mathbb {U}^*\mathbb {U}$, we have

$$\begin{aligned} (\mathbb {U}^*\mathbb {U})_{kl}&=\sum _{j\in \mathbb {Z}}U_{jk}^*U_{jl}\\&=\sum _{j\in \mathbb {Z}}[\phi (x_j-k,y_n-l)]_{n,l}^*[\phi (x_j-l,y_n-s)]_{n,s}. \end{aligned}$$

Since $\phi (x_j-l,y_n-s)=0$ if $|x_j-k|\ge S$, we find that $(\mathbb {U}^*\mathbb {U})_{kl}$ can be nonzero matrix if both $|x_j-k|<S$ and $|x_j-l|<S$. In other words, $(\mathbb {U}^*\mathbb {U})_{kl}$ is nonzero matrix implies that

$$\begin{aligned} |k-l|\le |k-x_j|+|x_j-l|<2S. \end{aligned}$$

$\square$

In order to present the reconstruction method, we need certain facts about Moore–Penrose inverse.

Definition 6.1

Let $A$ be an $m\times n$ matrix with complex entries. The Moore–Penrose inverse of $A$ is the unique $n\times m$ matrix $A^\dagger$ which satisfies the following conditions:

• (1) $AA^\dagger$ and $A^\dagger A$ are hermitian,

• (2) $AA^\dagger A=A$,

• (3) $A^\dagger A A^\dagger =A^\dagger$.

If $m=n$, and the matrix $A$ is invertible, then $A^\dagger =A^{-1}$. If the columns of $A$ are linearly independent, then $A^\dagger =(A^*A)^{-1}A^*$.

Let $H_1$, $H_2$ be Hilbert spaces over the field of complex numbers. Consider the operator equation

$$\begin{aligned} Ax=b, \end{aligned}$$

(6.8)

where $b\in H_2$ and $A$ is a bounded linear operator from $H_1$ to $H_2$. If the operator $A$ is invertible, then one has the solution $x=A^{-1}b$. But in general, (6.8) may have infinitely many solution $(N(A)\ne 0)$ or may not have any solution at all $(b\notin R(A))$. Here, $N(A)$ denotes the null space of $A$, $R(A)$ denotes the range space of $A$. Even if the equation has no solution, then one can assign a generalized solution $u\in H$ to (6.8) which satisfies the following inequality

$$\begin{aligned} \Vert Au-b\Vert \le \Vert Ax-b\Vert \end{aligned}$$

(6.9)

for every $x\in H_1$. This generalized solution is called a least square solution of equation $Ax=b$. If columns of $A$ are linearly independent, the least square solution is given by $(A^*A)^{-1}A^*b$. In general, the least square solution of $Ax=b$ is given by $A^\dagger b$.

From Lemmas 6.1 and 6.2, to reconstruct a function $f\in V(\phi )$ from its samples in the rectangle $[M_1,M_0]\times [N_1,N_0]$, we have to consider the truncated version of the matrices of $\mathbb {U}$. We shall denote $\mathbb {U}^{M_0,M_1,N_0,N_1}$ as the truncated matrix of $\mathbb {U}$. Now let us consider the system

$$\begin{aligned} \mathbb {U}^{M_0,M_1,N_0,N_1}D=F^{M_0,M_1,N_0,N_1}. \end{aligned}$$

(6.10)

In order to find $D$, we shall convert the block matrix equation form (6.10) into usual matrix form. Then, we obtain the solution $D$ as $D=\mathbb {U}^\dagger F$.

In our numerical implementation, we have used MATLAB. We used the shift-invariant space $V(\phi )$ with

$$\begin{aligned} \phi (x,y)=\beta _2(x/9)\beta _2(y/9), \end{aligned}$$

where $\beta _2(x)$ is the B-spline of order 2. Thus, the support of $\phi$ is $[-9,9]$ and S=2.

In the following, Fig. 1 denotes the original image, Fig. 2 denotes the uniform sampled image, and Fig. 3 is the reconstructed image from the uniform sample 6.2.

Fig. 1

Original image

Fig. 2

Uniform sample

Fig. 3

Reconstructed image with error 41.73 %

Figure 4 denotes the nonuniform sampled image and Fig. 5 is the reconstructed image from the nonuniform sample 6.4.

Fig. 4

Non uniform sample

Fig. 5

Reconstructed image with error 30.71 %