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Sampling Expansions in Reproducing Kernel Hilbert andBanach SpacesDeguang Han a , M. Zuhair Nashed a & Qiyu Sun aa Department of Mathematics , University of Central Florida , Orlando, Florida, USAPublished online: 22 Dec 2009.

To cite this article: Deguang Han , M. Zuhair Nashed & Qiyu Sun (2009) Sampling Expansions in Reproducing Kernel Hilbert andBanach Spaces, Numerical Functional Analysis and Optimization, 30:9-10, 971-987, DOI: 10.1080/01630560903408606

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Numerical Functional Analysis and Optimization, 30(9–10):971–987, 2009Copyright © Taylor & Francis Group, LLCISSN: 0163-0563 print/1532-2467 onlineDOI: 10.1080/01630560903408606

SAMPLING EXPANSIONS IN REPRODUCING KERNELHILBERT AND BANACH SPACES

Deguang Han, M. Zuhair Nashed, and Qiyu Sun

Department of Mathematics, University of Central Florida, Orlando, Florida, USA

� Given a countable subset � of a set �, we investigate the construction of all the reproducingkernel Hilbert (resp. Banach) spaces on � that have � as a sampling (resp. p-sampling) set.Unlike the general p-frames, we prove that every p-sampling set for a reproducing kernel Banachspace yields a reconstruction formula. Some applications are given to demonstrate the generalconstruction.

Keywords Frames; p-frames; Reproducing kernel Banach spaces; Reproducing kernelHilbert spaces; Sampling.

Mathematics Subject Classification Primary 42C15, 46C05, 47B10.

1. INTRODUCTION

The Whittaker–Shannon–Kotel’nikov (WSK) sampling theorem statesthat if a square-integrable function f is bandlimited to [−�, �], so it isrepresentable as

f (t) =∫ �

−�

e−ixt g (x)dx , t ∈ �

for some function g ∈ L2(−�, �), then f can be reconstructed from itssamples, f (k�/�), that are taken at the equally spaced nodes k�/� on thetime axis �. Moreover

‖f ‖2 =∑k∈�

∣∣∣∣f(k��

)∣∣∣∣2

, (1.1)

Received 27 April 2009; revised 15 August 2009; accepted 24 August 2009Address correspondence to Qiyu Sun, Department of Mathematics, University of Central

Florida, Orlando, FL 32816, USA; E-mail: qsun@mail.ucf.edu

971

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972 D. Han et al.

and

f (t) =∞∑

k=−∞f(k��

)sin (�t − k�)(�t − k�)

, t ∈ � (1.2)

where the series is absolutely and uniformly convergent on any compactset of the real line.

The WSK-sampling theorem has lots of engineering applications andhave been generalized in numerous contexts, the reader may refer to thesurvey articles and monographs [7, 14, 16, 23, 24]. It is observed thatthe space of all square-integrable functions bandlimited to [−�, �] is areproducing kernel Hilbert space on the real line. Here a reproducing kernelHilbert space � (RKHS for short) is a (complex) Hilbert space of functionson a domain � such that the evaluation function is continuous, i.e., forany x ∈ � there exists a positive constant Cx such that

|f (x)| ≤ Cx‖f ‖, f ∈ � (1.3)

[5]. Nashed and his collaborators studied sampling theory for reproducingkernel Hilbert space in his series of articles, see [17–20]. In general, areproducing kernel Hilbert space � on a domain � may not admit astable sampling set, that is, there does not exist a countable subset � of �such that

A‖f ‖ ≤( ∑

�∈�|f (�)|2

)1/2

≤ B‖f ‖, f ∈ � ,

where A,B are positive constants. This leads to the natural questionof characterizing all the reproducing kernel Hilbert spaces that have asampling set. In this article we discuss the general construction of all thereproducing kernel Hilbert spaces that have a sampling set, and also wecharacterize all the reproducing kernel Hilbert spaces on a set � that have� as a prescribed sampling set. Our results generalize the correspondingwork related to Riesz bases [20].

The concept of reproducing kernel Hilbert spaces has a naturalgeneralization in Banach spaces. Like the Hilbert space case, thesampling theory for a Banach space involves Banach space frame (forreconstruction) and p-frames (for determination). However, a p-frame ina Banach space does not necessarily yield a reconstruction formula [9].In Section 3, we will prove that for every p-sampling set (correspondingto a p-frame) in a reproducing kernel Banach space automatically yieldsa reconstruction formula. Similar to the reproducing Hilbert space case,for a given set � and a countable subset � ⊂ �, we also have a generalconstruction for all the reproducing kernel Banach spaces of functions on

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Sampling Expansion 973

� with � as a p-sampling set. Some applications will be discussed in thelast section of this article.

2. SAMPLING SETS FOR REPRODUCING KERNELHILBERT SPACES

In this section we describe a general construction for all thereproducing kernel Hilbert spaces which admit a sampling set �. We firstintroduce a few definitions and some preliminary results that will beneeded in proving our main results.

A frame for a separable Hilbert space � is a sequence �xn� in � suchthat there exist two positive constants A,B with the property that

A‖x‖2 ≤∑n

|〈x , xn〉|2 ≤ B‖x‖2, x ∈ � � (2.1)

The optimal constants (maximal for A and minimal for B) are called framebounds. When A = B = 1, �xn� is called a normalized tight frame. A sequence�xn� is called Bessel if we only require the right side inequality of (2.1)holds. In the proof of our main results we also need a concept of stronglydisjoint or orthogonal frames introduced in [6, 12]: Two Bessel sequences �xn�and �yn� are called strongly disjoint if

∑n〈x , xn〉yn = 0 holds for all x ∈ � .

Let �xn� be a frame for a Hilbert space � . The associated analysisoperator T is the bounded linear operator defined by Tx = ∑

n〈x , xn〉en ,where �en� is the standard orthonormal basis for 2(�). It is easy to verifythat the range space T� is closed and T is bounded invertible linearoperator from � to T� . Moreover, T ∗en = xn for each n and T ∗ is calledthe synthesis operator. Let S = T ∗T (this operator is refereed as the frameoperator for �xn�). Then �S−1xn� is the standard dual frame which provides usthe reconstruction formula:

x =∑n

〈x , S−1xn〉xn =∑n

〈x , xn〉S−1xn , x ∈ � , (2.2)

where the convergence is in the Hilbert space norm.Let � be a reproducing kernel Hilbert space of functions on a set �. A

sampling set is a countable subset � of � with the property that there existtwo positive constants A and B such that

A‖f ‖2 ≤∑�∈�

|f (�)|2 ≤ B‖f ‖2, f ∈ � � (2.3)

By the Riesz representation theorem, for any x ∈ � there exists hx ∈ �such that

f (x) = 〈f , hx〉, f ∈ � � (2.4)

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974 D. Han et al.

The function

k(s, t) = 〈hs , ht 〉 (2.5)

is known as the reproducing kernel of the RKHS � . By (2.1)–(2.4),we conclude that the evaluation linear functional sequence �h�, � ∈ �� is aframe for � , and that any function f in � can be reconstructed from itssampling on � by the following reconstruction formula:

f =∑�∈�

f (�)g�, f ∈ � , (2.6)

where �g� : � ∈ �� is a dual frame of �h� : � ∈ ��. The series in (2.6)converges in the Hilbert space norm. It is very desirable for ourconsideration of sampling to have the pointwise and/or uniformconvergence for the reconstruction formula. The following well-knownresult tells us that the norm convergence in a RKHS implies the pointwiseconvergence, and also the uniform convergence if the reproducing kernelis uniformly bounded.

Lemma 2.1. Suppose that � is a reproducing kernel Hilbert space on � withreproducing kernel k(s, t). If fn → f in the Hilbert space norm, then �fn(t)� isconvergent to f (t) for each t ∈ �. Moreover, the convergent is uniform on � if‖k(·, t)‖ is uniformly bounded.

By (2.6) and Lemma 2.1, we obtain a reconstruction formula in areproducing kernel Hilbert space that converges in the Hilbert space normand also pointwise.

Theorem 2.2. Assume that � is a sampling set for a reproducing kernel Hilbertspace � on a set �. Then there exists a frame �g�, � ∈ �� for � such that

f (t) =∑�∈�

f (�)g�(t), f ∈ � , (2.7)

where the convergence is both in norm and pointwise.

Now we introduce a procedure to construct reproducing kernelHilbert spaces that have a sampling set, and later we will show thatevery reproducing kernel Hilbert space that have a sampling set can beconstructed in that procedure. Let � be a set and let � be a countablesubset of �. Assume that �xn� is a frame for a Hilbert space �, and Sn : � →� (or �) is a sequence of functions satisfying the following two conditions:

(H1) �Sn(t)� ∈ 2(�) for every t ∈ �.(H2) �� : � ∈ �� is a frame for 2(�), where � = �Sn(�)�n∈�.

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Sampling Expansion 975

Now we define F by

F (t) =∑n∈�

Sn(t)xn , t ∈ �� (2.8)

The function F (t) is well-defined since �xn� is a frame and∑

n |Sn(t)|2 <∞.Clearly if x = 0 then 〈x , F (t)〉 = 0 for all t ∈ �. Conversely if 〈x , F (t)〉 = 0for all t ∈ �, then the 2-sequence � := (〈xn , x〉)n∈� satisfies

〈�, �〉 =∑n

Sn(�)〈x , xn〉 = 0 for all � ∈ ��

This, together with the assumptions that �� : � ∈ �� is a frame for 2(�)and that �xn� is a frame for �, implies that x = 0. Using the above functionF , we construct a space � of functions on � by

� = �〈x , F (t)〉 : x ∈ ��� (2.9)

Therefore the linear map

� x �−→ 〈x , F (t)〉 ∈ �

is one-to-one and onto. So � becomes a Hilbert space with the innerproduct on � defined by 〈f , g 〉 = 〈x , y〉 if f (t) = 〈x , F (t)〉 and g (t) =〈y, F (t)〉. Moreover the space � is a reproducing kernel Hilbert spacebecause for any t ∈ �, we have

|f (t)| = |〈x , F (t)〉| ≤ ‖F (t)‖‖x‖ = ‖F (t)‖‖f ‖ for all f ∈ � � (2.10)

For the reproducing kernel space � in (2.9), we have the followingresult about its sampling sets:

Theorem 2.3.

(i) � is a sampling set for the reproducing kernel Hilbert space � in (2.9).(ii) Every reproducing kernel Hilbert space admitting a sampling set � can be

constructed in the way described above.

To prove Theorem 2.3, we recall a lemma in [12].

Lemma 2.4. Let �xn� be a frame for a Hilbert space � and T be its associatedanalysis operator. Let �yn� be the standard dual of �xn� and P be the orthogonalprojection from 2(�) onto T�. Then

(i) �Tyn� and �P⊥en� are strongly disjoint.(ii) �Tyn + P⊥en� is a frame for 2(�).

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976 D. Han et al.

Proof of Theorem 2.3. (i) Since �xn� and ��� are frames for � and2(�) respectively, there exist two positive constants A and B such that

A‖x‖2 ≤∑n∈�

|〈x , xn〉|2 ≤ B‖x‖2, x ∈ �,

and

A‖�‖2 ≤∑�∈�

|〈�, �〉|2 ≤ B‖�‖2, � ∈ 2(�)�

Take any f ∈ � , and write f (t) = 〈x , F (t)〉 for x ∈ �. Then

f (�) =∑n∈�

Sn(�)〈x , xn〉 = 〈�, �〉, � ∈ �,

where � = �〈x , xn〉� ∈ 2(�). Therefore

∑�∈�

|f (�)|2 =∑�∈�

|〈�, �〉|2 ≤ B‖�‖2

= B∑n∈�

|〈x , xn〉|2 ≤ B2‖x‖2 = B2‖f ‖2

and similarly,

∑�∈�

|f (�)|2 =∑�∈�

|〈�, �〉|2 ≥ A‖�‖2 ≥ A2‖x‖2 = A2‖f ‖2�

This proves that � is a sampling set for the reproducing kernel Hilbertspace � .

(ii) Now assume that � is a reproducing kernel Hilbert space whichadmits a countable sampling set �. Then there exists two positive constantsA and B such that

A‖f ‖2 ≤∑�∈�

|f (�)|2 ≤ B‖f ‖2, f ∈ � � (2.11)

By (2.11), �x� : � ∈ �� is a frame for � , where x� := h�, the evaluationfunctional at the sampling location �. Let �g� : � ∈ �� be the standard dualframe of �x� : � ∈ ��, and T : � → 2(�) be the analysis operator for �g��.By Lemma 2.4, �Tx�� and �P⊥e�� are strongly disjoint and �Tx� + P⊥e�� isa frame for 2(�), where P is the orthogonal projection of T� and �e�� isthe standard orthonormal basis for 2(�).

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Sampling Expansion 977

Now we define

S�(t) ={g�(t) if t �∈ �,

g�(�′) + 〈P⊥e�, e�′ 〉 if t = �′ ∈ ��

Then S�(t) satisfies the two conditions (H1) and (H2). In fact,

(H1):∑

�∈� |S�(t)|2 < ∞ follows from

∑�∈�

|〈P⊥e�, e�′ 〉|2 = ‖P⊥e�′‖2 < ∞

and ∑�∈�

|g�(t)|2 ≤∑�∈�

|〈g�, ht 〉|2 < ∞

by the frame property of �g��, where ht is the evaluation functionalat the location t ∈ �.

(H2): �� := �S�′(�)��′∈�, � ∈ �� is a frame for 2(�) by Lemma 2.4 and thefact that � = Tx� + P⊥e�, � ∈ �.

Define F (t) = ∑�∈� S�(t)x�, t ∈ �. Then it remains to prove that the

operator

� f �−→ 〈f , F (t)〉 ∈ �

is an identity. Take any f ∈ � . For t �∈ �, we have

〈f , F (t)〉 =∑�∈�

S�(t)〈f , x�〉 =∑�∈�

f (�)g�(t) = f (t), (2.12)

where the last equality follows from the pointwise convergence of thereconstruction formula (2.7). While for t = �′ ∈ �, we obtain

〈f , F (t)〉 =∑�∈�

S�(�′)〈f , x�〉

=∑�∈�

f (�)(g�(�′) + 〈P⊥e�, e�′ 〉)

=∑�∈�

f (�)g�(�′) +∑�∈�

〈f , x�〉〈P⊥e�, e�′ 〉

=∑�∈�

f (�)g�(�′) = f (t), (2.13)

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978 D. Han et al.

where we have used Theorem 2.2 and Lemma 2.4 to obtain the last twoidentities. Combining the two cases in (2.12) and (2.13) leads to

f (t) = 〈f , F (t)〉 for all f ∈ � and t ∈ �,

and hence completes the proof.

As a special case, we call a sampling set for � exact if the correspondingevaluation functional sequence is a Riesz basis for � . In any nonexactsampling case, we always have a over-sampling for the function space. Thefollowing follows immediately from the standard frame theory (cf. [12]):

Proposition 2.5. A sampling set � for a Hilbert space � of functions is exactif and only if �(f (�))�∈� : f ∈ �� = 2(�).

In the general construction described before Proposition 2.5, if wefurther require that both �xn� and ��� be Riesz bases for H and 2(�),respectively, then we have �(f (�)) : f ∈ �� = TTxH = 2(�), where Tx andT are the analysis operator for �x�� and ���, respectively. Thus, � isan exact sampling set for � by Proposition 2.5. Similar to the proofof Theorem 2.3(ii), any reproducing kernel Hilbert space with an exactsampling set can be constructed this way. Thus we have

Theorem 2.6. The following are equivalent:

(i) � is a reproducing kernel Hilbert space with an exact sampling set.(ii) � can be constructed as in the general construction with �xn� and ��� being

Riesz bases for H and 2(�), respectively.

3. SAMPLING SETS FOR REPRODUCING KERNELBANACH SPACES

A reproducing kernel Banach space is a Banach space � of functions on aset � such that the evaluation functions f → f (t) is continuous for eacht ∈ �. A countable subset � of � is called a p-sampling set (p ≥ 1) for aBanach space � if there exist two positive constants A and B such that

A‖f ‖� ≤( ∑

�∈�|f (�)|p

)1/p

≤ B‖f ‖�, f ∈ �� (3.1)

In this section, we investigate p-sampling in reproducing kernel Banachspaces.

Let X be a Banach space and X ∗ be its dual. The dual relation g (x) willbe still denoted by 〈x , g 〉, where x ∈ X and g ∈ X ∗. A countable set �gk� ⊂ X ∗

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Sampling Expansion 979

is called a p-frame if there exist positive constants A and B such that

A‖x‖X ≤( ∑

k∈�|〈x , gk〉|p

)1/p

≤ B‖x‖X , x ∈ X , (3.2)

so, the analysis operator T is defined by

T : X x �−→ �〈x , gk〉� ∈ p(�)

is bounded from both above and below. If �gk� is p-frame for a Banachspace X , and if, in addition, there exists a bounded linear operator R :p → X such that RT = IX , the identity operator on X , then �gk� is calleda Banach frame with respect to the p space. In this case we have thereconstruction formula:

x =∑k∈�

〈x , gk〉g̃k , x ∈ X ,

where g̃k = Rek with �ek , k ∈ �� is the standard basis of p .Unlike the Hilbert space case, the reconstruction operator R does

not necessarily exist for p-frames. It is observed in [9] that there existsa p-frame �gk� for a Banach space X , for which no family �g̃k� ⊂ Xsatisfies that

x =∑�

〈x , gk〉g̃k , x ∈ X �

That is why the existence of a reconstruction operator is often required inthe definition of Banach frames [1, 3, 8, 9, 11]. However, a reconstructionformula is possible for some special p-frames. For instance, it wasproved in [3] that under certain natural conditions on the generators,a translation p-frame for a finitely generated shift-invariant subspace isalways a Banach frame. In general, a p-frame �gk� is a Banach frame if andonly if the range space TX of the analysis operator T is complementedin p [9].

Similarly to sampling in a Hilbert space, p-sampling for a Banach spaceare related to Banach space frame and p-frame in that space. In particular,for a Banach space � of functions on a set �, a countable subset � of � isp-sampling set for the space � if and only if �h�� is p-frame for the space �,where h� is the evaluation functional: x → x(�). In the following result, weshow that a reconstruction formula always exists for the p-frame �h�, � ∈ ��associated with a p-sampling set � on a reproducing kernel Banach space.

Theorem 3.1. Assume that � is a p-sampling set for a reproducing kernelBanach space � on a set �. Then there exists a sequence of functions �S�(t)� on �

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980 D. Han et al.

such that the following are satisfied:

(B1) �S�(t)� ∈ q(�) for every t ∈ �, where 1p + 1

q = 1.(B2) �� : � ∈ �� is a p-frame for T�, where � = �S�′(�) : �′ ∈ �� and T :

� → p(�) is defined by Tf = �f (�′)��′∈�.(B3) f (t) = ∑

�∈� f (�)S�(t), t ∈ �, with the pointwise convergence.

Proof. Since � is a reproducing kernel Banach space, it follows that foreach t ∈ � there exists a bounded linear functional ht on � such thatf (t) = 〈f , ht 〉 for all f ∈ �. Note that the analysis operator T associatedwith the sampling functionals for � is both bounded and bounded below.Therefore T −1 : T� → � is a bounded linear operator, and so (T −1)∗ht isbounded linear functional on T� which is closed subspace of p . Extend(T −1)∗ht to a bounded linear functional, say St , on (p)∗ = q , and writeSt = �S�(t)�. Then, clearly �S�(t)� satisfies (B1). For (B3),

f (t) = 〈f , ht 〉 = 〈T −1Tf , ht 〉 = 〈Tf , (T −1)∗ht 〉= 〈Tf , St 〉 =

∑�∈�

f (�)S�(t), (3.3)

and the convergence in the summation f (t) = ∑�∈� f (�)S�(t) is pointwise.

By (3.3), we have

f (�) =∑�′∈�

f (�′)S�′(�) = 〈Tf , �〉,

and therefore ( ∑�∈�

|〈Tf , �〉|p)1/p

=( ∑

�∈�|f (�)|p

)1/p

�

This together with the definition of the sampling set � implies that

( ∑�∈�

|〈Tf , �〉|p)1/p

≤ B‖f ‖ ≤ B‖T −1‖‖Tf ‖

and ( ∑�∈�

|〈Tf , �〉|p)1/p

≥ A‖f ‖ ≥ A‖T −1‖‖Tf ‖,

where A and B are the two constants defining the p-sampling set �. Hence��� is p-frame for T� from the definition of p-frame, and then (B2) holds.

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Sampling Expansion 981

Now we consider the general construction of reproducing kernelBanach spaces that have � as a p-sampling set. More precisely, we will prove

Theorem 3.2. Let �yk� ⊂ X ∗ be p-frame for a Banach space X and � be acountable subset of �. Suppose that Sk(t), k ∈ �, satisfy

(B1) �Sk(t)� ∈ q(�) for every t ∈ �, where 1p + 1

q = 1.(B2′) �� : � ∈ �� is a p-frame for p(�), where � = �Sk(�)�.

Then

� :={ ∑

k∈�〈x , yk〉Sk(t), x ∈ X

}

is a reproducing kernel Banach space with � as a p-sampling set of �, where thenorm of f ∈ � is defined to be ‖x‖ when f = ∑

k∈�〈x , yk〉Sk(t).Proof. Let A,B and C ,D be the p-frame bounds corresponding to thep-frames �yk� and ���. To see that � is well defined (and so it is Banachspace isometric to X ), we only need to check that if

∑k〈x , yk〉Sk(t) = 0 for

all t , then x = 0. In fact, if∑

k〈x , yk〉Sk(t) = 0 for all t , then∑

k〈x , yk〉Sk(�) =0 for all � ∈ �, which implies that 〈�, �〉 = 0 for each � ∈ �, where � =�〈x , yk〉�. Since ��� is a p-frame for p , it follows that � = 0, and thereforex = 0 since �yk� is a p-frame for X .

Secondly we show that � is a reproducing kernel Banach space. Foreach t ∈ �, we have

|f (t)| ≤( ∑

k

|〈x , yk〉|p)1/p( ∑

k

|Sk(t)|q)1/q

≤ B‖x‖( ∑

k

|Sk(t)|q)1/q

= B( ∑

k

|Sk(t)|q)1/q

‖f ‖,

where 1/p + 1/q = 1. Thus f → f (t) is continuous, and so � is areproducing kernel Banach space.

Finally we prove that � is a p-sampling set. Note that

f (�) =∑k

〈x , yk〉Sk(�) = 〈Tx , �〉

where Tx = �〈x , yk〉�. Thus( ∑�∈�

|f (�)|p)1/p

=( ∑

�∈�|〈Tx , �〉|p

)1/p

≤ D‖Tx‖

= D( ∑

k

|〈x , yk〉|p)1/p

≤ DB‖x‖ = DB‖f ‖,

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982 D. Han et al.

and similarly

( ∑�∈�

|f (�)|p)1/p

≥ CA‖f ‖�

Therefore � is a p-sampling set for �, as claimed.

Remark 3.3. When the sampling set � induces a bounded unconditionalbases, i.e, when T� = p , it is clear from the construction that �S�� ⊂ �(in fact, S� = T −1e�, where e� = �� ∈ p). However, unlike the Hilbert spacesampling case, we don’t know whether �S�� can be always chosen in such away that each S� is in �. Therefore we ask.

Question. Is it true that we can always choose �S�� in such a way thateach S� ∈ �? If so, is the convergence in

f =∑k

f (�)S�(t)

always in the Banach space norm?

4. APPLICATIONS

We give a few applications of Theorem 3.2 in this section.Denote by W (L1) the space of all measurable functions f such that

‖f ‖W (L1) :=∑k∈�d

supx∈k+[0,1)d

|f (x)| < ∞�

For 1, � � � , N in W (L1), we define the shift-invariant space Vp( 1, � � � , N )by

Vp( 1, � � � , N ) :={ N∑

n=1

∑k∈�d

cn(k) n(t − k) : �cn(k)� ∈ p},

and equip Vp( 1, � � � , N ) by usual Lp norm, where 1 ≤ p ≤ ∞. Forsampling and reconstruction of signals in a shift-invariant space, the readermay refer to [2, 4, 21, 23] and references therein.

Theorem 4.1. Let 1 ≤ p ≤ ∞, � be a coutable subset of �d , and 1, � � � , N becontinuous functions in W (L1). Assume that Vp( 1, � � � , N ) is a closed subspaceof Lp. Then

(i) Vp( 1, � � � , N ) is a reproducing kernel Banach space.

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Sampling Expansion 983

(ii) � is a p-sampling set for Vp( 1, � � � , N ) provided that �� : � ∈ �� is ap-frame for p(�d × �N ) with � = ��n(� − k) : (k,n) ∈ �d × �N �, where�N = �1, � � � ,N �.

Proof. (i) By the assumption on the closedeness condition ofVp( 1, � � � , N ), it is shown in [3] that there exists a positive constant Bsuch that

B−1‖f ‖p ≤ inf{‖�cn(k)�‖p : f =

N∑n=1

∑k∈�d

cn(k) n(t − k)}

≤ B‖f ‖Lp (4.1)

holds for all f ∈ Vp( 1, � � � , N ). For every t ∈ �d and

f =N∑

n=1

∑k∈�d

cn(k) n(t − k) ∈ Vp( 1, � � � , N )

we have

|f (t)| =∣∣∣∣

N∑n=1

∑k∈�d

cn(k) n(t − k)∣∣∣∣

≤ ‖�cn(k)�‖∞N∑

n=1

∑k∈�d

| n(t − k)|

≤ ‖�cn(k)�‖p

( N∑n=1

‖ n‖W (L1)

)� (4.2)

Therefore, by (4.1) and (4.2), we obtain

|f (t)| ≤ inf{‖�cn(k)�‖p : f =

N∑n=1

∑k∈�d

cn(k) n(t − k)}

×N∑

n=1

‖ n‖W (L1)

≤ B‖f ‖Lp

( N∑n=1

‖ n‖W (L1)

),

and hence the evaluation functional f → f (t) is a continuous onVp( 1, � � � , N ). This proves that Vp( 1, � � � , N ) is a reproducingkernel Banach space. Moreover, the reproducing kernel kt belongsto V1( 1, � � � , N ) ⊂ Vp( 1, � � � , N )

∗ for every t ∈ �. Recalling fromTheorem 1 in [3] that there exist functions �1, � � � ,�N in V1( 1, � � � , N )

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984 D. Han et al.

with the property that

f (t) =N∑

n=1

∑k∈�d

〈f ,�n(· − k)〉 n(t − k), f ∈ Vp( 1, � � � , N ),

we then have the following expression for the reproducing kernel kt forthe reproducing kernel Banach space Vp( 1, � � � , N ):

kt(s) =N∑

n=1

∑k∈�d

n(t − k)�n(s − k)�

(ii) If f (t) = ∑Nn=1

∑k∈�d cn(k) n(t − k) = ∑N

n=1

∑k∈�d dn(k) n(t − k)

for some �cn(k)�, �dn(k)� ∈ p(�d × �N ), then∑N

n=1

∑k∈�d (cn(k) −

dn(k)) n(t − k) = 0. In particular we have �cn(k) − dn(k)� is orthogonal to� for all � ∈ �. Thus cn(k) = dn(k) for all 1 ≤ n ≤ N and k ∈ �d by thep-frame assumption, which implies that any function f in Vp( 1, � � � , N )

has a unique representation f (t) = ∑Nn=1

∑k∈�d cn(k) n(t − k). By (4.1)

and the above unique representation of a function in Vp( 1, � � � , N ), 1, � � � , N is a p-Riesz basis for Vp( 1, � � � , N ), i.e., there exist positiveconstants A,B > 0 such that

A‖f ‖Lp ≤ ‖�cn(k)�‖p ≤ B‖f ‖Lp (4.3)

whenever f = ∑Nn=1

∑k∈�d cn(k) n(t − k) ∈ Vp( 1, � � � , N ). Let ek,n(m, j) =

�(k,n),(m,j). Then clearly �ek,n : (k,n) ∈ �d × �N � is a p-frame for p(�d × �N ).Let

� :={ N∑

n=1

∑k∈�d

〈x , ek,n〉 n(t − k), x ∈ p(�d × �N )

}

be as defined in the proof of Theorem 3.2. Then we have thatVp( 1, � � � , N ) = � and that the norm ‖f ‖� = ‖x‖p = ‖�〈x , ek,n〉�‖p isequivalent the Lp -norm of Vp( 1, � � � , N ) by (4.3). This together withTheorem 3.2 implies that � is a p-sampling set for Vp( 1, � � � , N ).

Now we consider the second application.

Theorem 4.2. Let � be a real function that satisfies∫��(t)2(1 + |t |)2�dt < ∞ (4.4)

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Sampling Expansion 985

for some � > 1/2, and

the Fourier transform �̂(�) does not vanish for all � ∈ �� (4.5)

Write

��(u, v) =∫��(t − u)�(t − v)dt , (4.6)

and let U := �uj�j∈� be such that

∞> supj �=j ′

|uj − uj ′ | ≥ infj �=j ′

|uj − uj ′ | ≥ � > 0� (4.7)

Define

Xp ={ ∑

j∈�c(j)�(u,uj) : �c(j)� ∈ p(�)

}(4.8)

with ‖f ‖ = ‖�c(j)�‖p and 1 ≤ p ≤ ∞. Then the set U is a p-sampling set for Xp.

Proof. For any u ∈ �, we have∑j

|�(u,uj)|2(1 + |u − uj |)2�

≤∑j

(2�

∫�

|�(t − u)||�(t − uj)|((1 + |t − u|)� + (1 + |t − uj |)�

)dt

)2

≤ 22�+1∑j

( ∫�

|�(t − u)||�(t − uj)|(1 + |t − u|)�dt)2

+ 22�+1∑j

( ∫�

|�(t − u)||�(t − uj)|(1 + |t − uj |)�dt)2

= I + II� (4.9)

Noting that there exists an absolute constant C such that

∑l∈�

( ∫ l+1

l|�(t − u)|2dt

)1/2

≤( ∑

l∈�

∫ l+1

l|�(t − u)|2dt × (1 + |l − u|)2�

)1/2

×(∑

l∈�(1+ |l −u|)−2�

)1/2

≤ C for all u ∈�,

(4.10)

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986 D. Han et al.

and

∑j

( ∫ l+1

l|�(t − uj)|2dt

)1/2

≤( ∑

j

∫ l+1

l|�(t − uj)|2dt × (1 + |l − uj |)2�

)1/2

×( ∑

j

(1 + |l − uj |)−2�

)1/2

≤ C for all l ∈ �, (4.11)

we then have the following estimate for the item I in (4.9):

I ≤∑j

(∑l∈�

(∫ l+1

l|�(t − u)|2(1 + |t − u|)2�dt

)1/2(∫ l+1

l|�(t − uj)|2dt

)1/2)2

≤∑j

( ∑l∈�

∫ l+1

l|�(t − u)|2(1 + |t − u|)2�dt

( ∫ l+1

l|�(t − uj)|2dt

)1/2)

×∑l ′∈�

( ∫ l ′+1

l ′|�(t − uj)|2dt

)1/2

≤ C1 (4.12)

for some positive constant C1 independent of u. Similarly, we have thefollowing estimate for the item II in (4.9):

II ≤ C2 (4.13)

for some positive constant C2 independent of u. Combining (4.12)and (4.13), and recalling �(u, v) = �(v,u), we conclude that

supi

∑j

|��(ui ,uj)|2(1+ |ui −uj |)2� + supj

∑i

|��(ui ,uj)|2(1+ |ui −uj |)2� <∞�

(4.14)

By Theorem 3.2 in [17], the matrix A = [�(ui ,uj)] is an invertible matrixon 2(�). This, together with the estimate (4.14) and the Wiener lemmafor infinite matrices in [22], implies that A−1 = (b(i , j)) satisfies

supi

∑j

|b(i , j)|2(1+ |ui − uj |)2� + supj

∑i

|b(i , j)|2(1+ |ui −uj |)2� <∞�

(4.15)

Therefore �i : i ∈ �� is a frame for q(�), where i = ��(ui ,uj)� by (4.15).From Theorem 3.2, it then follows that �uj� is a sampling set for Xp .

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Sampling Expansion 987

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