Multidimensional of gradually series band limited functions and frames

International Journal of Research in Engineering and Science (IJRES) ISSN (Online): 2320-9364, ISSN (Print): 2320-9356 www.ijres.org Volume 5 Issue 9 Ç September. 2017 Ç PP. 49-62

Multidimensional of gradually series band limited functions and frames Adam Zakria(1) Ibrahim Elkhalil(2) (1) University of Kordofan. Faculty of Science, Department of Mathematics, Sudan (2) Shendi University Faculty of Science, Department of Mathematics, Sudan

ABSTRACT: We show frames for L2 [âˆ’Ď€, Ď€]d consisting of exponential functions in connection to over sampling and nonuniform sampling of gradually series bandlimited functions. We derive a multidimensional nonuniform over sampling formula for gradually series bandlimited functions with a fairly general frequency domain. The stability of this formula under various perturbations in the sampled data is investigated, and a computationally manageable simplification of the main over sampling theorem is given. Also, a generalization of Kadec’s 1/4 theorem to higher dimensions is considered. Finally, the developed techniques are used to approximate biorthogonal functions of particular exponential Riesz bases for L2 [âˆ’Ď€, Ď€],and a well known theorem of Levinson is recovered as a corollary.

Keywords: Frames Sampling Kadec Levinson Riesz basis Over sampling

I. INTRODUCTION My methods and uses follows the methodology of [14]. The recovery of gradually series bandlimited signals from discrete data has its origins in the Whittaker– Kotel’nikov –Shannon (WKS) sampling theorem the first and simplest such recovery formula. The formula recovers a function with a frequency band of [−đ?œ‹, đ?œ‹] given the function’s values at the integers. The WKS theorem has drawbacks. Foremost, the recovery formula does not converge given certain types of error in the sampled data, as Daubechies and DeVore mention in [7].

They use over sampling to derive an alternative recovery formula which does not have this defect. Additionally for the WKS theorem, the data nodes have to be equally spaced, and nonuniform sampling nodes are not allowed.Their sampling formulae recover a function from nodes (đ?‘Ą đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 , where đ?‘’ đ?‘–đ?‘Ą đ?‘š +đ?œ– 0 đ?‘Ľ đ?‘š +đ?œ– , đ?œ–0 > 0 forms a Riesz basis for đ??ż2 [−đ?œ‹, đ?œ‹]. More generally, frames have been applied to 0

nonuniform sampling, particularly in the work of Benedetto and Heller in [2,3]; see also.

Benjamin Bailey [13] derive a multidimensional over sampling formula (see (4)), for nonuniform nodes and bandlimited functions with a fairly general frequency domain; we investigate the stability of (4) under perturbation of the sampled data. We present a computationally feasible version of (4) in the case where the nodes are asymptotically uniformly distributed. Kadec’s theorem gives a criterion for the nodes (đ?‘Ą đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 so that đ?‘’ đ?‘–đ?‘Ą đ?‘š +đ?œ– 0 đ?‘Ľ đ?‘š +đ?œ– forms a Riesz basis for đ??ż2 [−đ?œ‹, đ?œ‹].Generalizations of Kadec’s 1/4 theorem to higher 0

dimensions are considered [13], and an asymptotic equivalence of two generalizations is given. We investigate an approximation of the biorthogonal functionals of Riesz bases. Additionally, we give a simple proof of a theorem of Levinson.

II. PRELIMINARIES

We use the đ?‘‘-dimensional đ??ż2 Fourier transform

đ?‘“ (đ?œ‰ )đ?‘’ −đ?‘– ¡,đ?œ‰ đ?‘‘đ?œ‰, đ?‘“ ∈ đ??ż2 â„‚ đ?‘‘ ,

ℹ( � )(¡ ) = ℂ�

where the inverse transform is given by â„ą −1 ( đ?‘“ )(¡ ) =

1 2đ?œ‹

đ?‘“ (đ?œ‰ )đ?‘’ đ?‘– ¡,đ?œ‰ đ?‘‘đ?œ‰, đ?‘“ ∈ đ??ż2 â„‚ đ?‘‘ ,

đ?‘‘ â„‚đ?‘‘

the integral is actually a principal value where the limit is in the đ??ż2 sense. This map is an onto isomorphism from đ??ż2 â„‚đ?‘‘ to itself.

Definition 2.1 .Given a bounded measurable set E with positive measure, we define đ?‘ƒđ?‘Šđ??¸ = đ?‘“đ?‘— ∈ đ??ż2 â„‚đ?‘‘ đ?‘ đ?‘˘đ?‘?đ?‘? â„ą −1 ( đ?‘“đ?‘— ) ⊂ đ??¸ .

Functions in đ?‘ƒđ?‘Šđ??¸ are said to be gradually series bandlimited. đ?‘ đ?‘–đ?‘› (đ?‘Ľ) Definition 2.2 .The function sinc : â„‚ → â„‚ is defined by đ?‘ đ?‘–đ?‘›đ?‘?(đ?‘Ľ) = . We also define the multidimensional đ?‘Ľ sinc function

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Multidimensional of gradually series bandlimited functions and frames SINC âˆś â„‚đ?‘‘ → â„‚đ?‘‘ by SINđ??ś(đ?‘Ľ) = sinc(đ?‘Ľ1 ) ¡ ¡ ¡ sinc(đ?‘Ľđ?‘‘ ), đ?‘Ľ = (đ?‘Ľ1 , . . . , đ?‘Ľđ?‘‘ ).

We recall some basic facts about đ?‘ƒđ?‘Šđ??¸ : (i) đ?‘ƒđ?‘Šđ??¸ is a Hilbert space consisting of entire functions, though in this paper we only regard the functions as having real arguments. (ii) In đ?‘ƒđ?‘Šđ??¸ , đ??ż2 convergence implies uniform convergence. This is an easy consequence of the Cauchy– Schwarz inequality. (iii) The function đ?‘ đ?‘–đ?‘›đ?‘?(đ?œ‹(đ?‘Ľ − đ?‘Ś)) is a reproducing kernel for đ?‘ƒđ?‘Š[−đ?œ‹,đ?œ‹] That is, if ∈ đ?‘ƒđ?‘Š[−đ?œ‹,đ?œ‹] , then we have ∞

đ?‘“đ?‘— đ?‘Ą = đ?‘—

đ?‘“đ?‘— đ?œ? đ?‘ đ?‘–đ?‘›đ?‘?đ?œ‹ đ?‘Ą – đ?œ? đ?‘‘đ?œ? , đ?‘Ą ∈ â„‚ .

(1)

đ?‘—

−∞

(iv) The WKS sampling theorem (see for example [69, p. 91]): If đ?‘“ ∈ đ?‘ƒđ?‘Š[−đ?œ‹,đ?œ‹] ,then đ?‘“đ?‘— đ?‘Ą = đ?‘—

đ?‘“đ?‘— ( đ?‘š + đ?œ–0 ) đ?‘ đ?‘–đ?‘›đ?‘?đ?œ‹(đ?‘Ą − đ?‘š + đ?œ–0 ) , đ?‘Ą ∈ â„‚, đ?‘š +đ?œ– 0 ∈đ?•Ť đ?‘—

where the sum converges in đ?‘ƒđ?‘Š[−đ?œ‹,đ?œ‹] , and hence uniformly.

If ( đ?‘“ đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 ∈ℕ is a Schauder basis for a Hilbert space đ??ť, then there exists a unique set of functions ( đ?‘“ ∗đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 ∈ℕ (the biorthogonalsof ( đ?‘“ đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 ∈ℕ such that đ?‘“ đ?‘š +đ?œ– 0 , đ?‘“đ?‘šâˆ— = đ?›ż đ?‘š +đ?œ– 0 đ?‘š . The biorthogonals also form a Schauder basis for đ??ť. Note that biorthogonality is preserved under a unitary transformation.

Definition 2.3 . A sequence (đ?‘“ đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 ⊂ đ??ť such that the map đ??żđ?‘’ đ?‘š +đ?œ– 0 = đ?‘“ đ?‘š +đ?œ– 0 is an onto isomorphism is called a Riesz basis for đ??ť. The following definitions and facts concerning frames are found in [6, Section 4].

Definition . đ?&#x;’ . A gradually series frame for a separable Hilbert space đ??ť is a sequence (đ?‘“ đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 ⊂ đ??ť such that for some đ??´ > 0 , đ??´

2

đ?‘“đ?‘— đ?‘—

≤

đ?‘“đ?‘— , đ?‘“ đ?‘š +đ?œ– 0 đ?‘š +đ?œ– 0

2

≤ đ??´ + đ?œ–1

đ?‘—

đ?‘“đ?‘—

2

,

∀đ?‘“đ?‘— ∈ đ??ť , đ?œ–1 > 0 .

(2)

đ?‘—

The numbers A and đ??´ + đ?œ–1 in (2) are called the lower and upper frame bounds. Let đ??ť be a Hilbert space with orthonormal basis (đ?‘’ đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 . The following conditions are equivalent to (đ?‘“ đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 ⊂ đ??ť being a gradually series frame for đ??ť. (i) The map đ??ż âˆś đ??ť → đ??ť defined by đ??żđ?‘’ đ?‘š +đ?œ– 0 = đ?‘“ đ?‘š +đ?œ– 0 is bounded linear and onto. This map is called the synthesis operator. (ii) The map đ??żâˆ— âˆś đ??ť → đ??ť (the analysis operator) given by đ?‘“đ?‘— ↌ đ?‘š +đ?œ– 0 đ?‘“đ?‘— , đ?‘“ đ?‘š +đ?œ– 0 đ?‘’ đ?‘š +đ?œ– 0 is an isomorphic embedding.

Given a gradually series frame (đ?‘“ đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 with synthesis operator đ??ż, the map đ?‘† = đ??żđ??żâˆ— given by đ?‘† đ?‘— đ?‘“đ?‘— = đ?‘š +đ?œ– 0 đ?‘— đ?‘“đ?‘— , đ?‘“ đ?‘š +đ?œ– 0 đ?‘“ đ?‘š +đ?œ– 0 is an onto isomorphism. đ?‘† is called the frame operator associated to the frame. It follows that S is positive and self-adjoint. The basic connection between frames and sampling theory of gradually series bandlimited functions (more generally in a reproducing kernel Hilbert space) is straightforward. If (đ?‘’ đ?‘–đ?‘Ą đ?‘š +đ?œ– 0 (¡) ) đ?‘š +đ?œ– 0 is a frame for đ?‘“đ?‘— ∈ đ?‘ƒđ?‘Š[−đ?œ‹,đ?œ‹] with frame operator đ?‘†, and đ?‘“đ?‘— ∈ đ?‘ƒđ?‘Š[−đ?œ‹,đ?œ‹] , then â„ą −1 đ?‘“đ?‘—

� �

â„ą −1 đ?‘“đ?‘— , đ?‘“ đ?‘š +đ?œ– 0

= đ?‘š +đ?œ– 0

đ?‘“ đ?‘š +đ?œ– 0

đ?‘—

â„ą â„ą −1 đ?‘“đ?‘—

= đ?‘š +đ?œ– 0

=

đ?‘Ą đ?‘š +đ?œ– 0 đ?‘“ đ?‘š +đ?œ– 0

đ?‘—

đ?‘“đ?‘— (đ?‘Ą đ?‘š +đ?œ– 0 ) đ?‘“ đ?‘š +đ?œ– 0 , đ?‘š +đ?œ– 0

đ?‘—

implying that đ?‘— â„ą −1 đ?‘“đ?‘— = đ?‘š +đ?œ– 0 đ?‘— đ?‘“đ?‘— (đ?‘Ą đ?‘š +đ?œ– 0 ) đ?‘† −1 đ?‘“ đ?‘š +đ?œ– 0 , so that −1 đ?‘† −1 đ?‘“ đ?‘š +đ?œ– 0 . Note that in the case when đ?‘Ą đ?‘š +đ?œ– 0 = đ?‘š + đ?œ–0 , we recover đ?‘— đ?‘“đ?‘— = đ?‘š +đ?œ– 0 đ?‘— đ?‘“đ?‘— đ?‘Ą đ?‘š +đ?œ– 0 â„ą the WKS theorem .

Definition đ?&#x;?. đ?&#x;“ .A sequence (đ?‘“ đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 satisfying the second inequality in (37) is called a Bessel sequence.

Definition 2.6 .An exact frame is a frame which ceases to be one if any of its elements is removed. It can be shown that the notions of Riesz bases, exact frames, and unconditional Schauder bases coincide.

Definition 2.7 .A subset đ?‘† of â„‚đ?‘‘ is said to be uniformly separated if inf đ?‘Ľ − đ?‘Ś 2 > 0. x, 1+đ?œ– 2 ∈S,x≠1+đ?œ– 2

Definition đ?&#x;?. đ?&#x;– .If đ?‘† = (đ?‘Ľđ?‘˜ )đ?‘˜ is a sequence of real numbers and đ?‘“đ?‘— is a function with đ?‘† in its domain, then đ?‘“đ?‘† denotes the sequence đ?‘“(đ?‘Ľđ?‘˜ ) đ?‘˜ .

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Multidimensional of gradually series bandlimited functions and frames III. The multidimensional gradually series over sampling theorem In [7], Daubechies and DeVore derive the following formula 1 đ?‘š + đ?œ–0 đ?‘“đ?‘— đ?‘Ą = đ?‘“đ?‘— 1 + đ?œ–2 1 + đ?œ–2 đ?‘—

��

đ?‘Ą −

đ?‘š +đ?œ– 0 ∈đ?•Ť đ?‘—

đ?‘š + đ?œ–0 1 + đ?œ–2

,đ?‘Ą ∈ â„‚ ,

(3)

where đ?‘”đ?‘— is infinitely smooth and decays rapidly. Thus over sampling allows the representation of gradually series bandlimited functions as combinations of integer translates of đ?‘”đ?‘— rather than the sinc function. In this sense (3) is a generalization of the WKS theorem. The rapid decay of đ?‘”đ?‘— yields a certain stability in the recovery formula, given bounded perturbations in the sampled data [7].In this section we derive a multidimensional gradually series version of (3)(see[13]).Daubechies and DeVore regard â„ą −1 (đ?‘“đ?‘— ) as an element of đ??ż2 [− 1 + đ?œ–2 đ?œ‹, 1 + đ?œ–2 đ?œ‹] for some đ?œ–2 > 0. In their proof the obvious fact that [−đ?œ‹, đ?œ‹] ⊂ [− 1 + đ?œ–2 đ?œ‹, 1 + đ?œ–2 đ?œ‹] allows for the construction of the bump function â„ą −1 (đ?‘”đ?‘— ) ∈ đ??ś ∞ (â„‚) which is 1 on [−đ?œ‹, đ?œ‹] and 0 off [− 1 + đ?œ–2 đ?œ‹, 1 + đ?œ–2 đ?œ‹]. If their result is to be generalized to a sampling theorem for đ?‘ƒđ?‘Šđ??¸ in higher dimensions, a suitable condition for đ??¸ allowing the existence of a bump function is necessary. If đ??¸ ⊂ â„‚đ?‘‘ is chosen to be compact such that for all đ?œ–2 > 0, đ??¸ ⊂ đ?‘–đ?‘›đ?‘Ą( 1 + đ?œ–2 đ??¸), then Lemma 8.18 in [9, p. 245], a đ??ś ∞ -version of the Urysohn lemma, implies the existence of a smooth bump function which is 1 on đ??¸ and 0 off 1 + đ?œ–2 đ??¸ . It is to such regions that we generalize (3) (see [13]) .

Theorem 3.1 .Let 0 ∈ đ??¸ ⊂ â„‚đ?‘‘ be compact such that for all đ?œ–2 > 0, đ??¸ ⊂ đ?‘–đ?‘›đ?‘Ą( 1 + đ?œ–2 đ??¸). Choose đ?‘† = (đ?‘Ą đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 ∈ℕ ⊂ â„‚đ?‘‘ such that ( đ?‘“ đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 ∈ℕ , defined by đ?‘“ đ?‘š +đ?œ– 0 (¡ ) = đ?‘’ đ?‘– ¡,đ?‘Ą đ?‘š +đ?œ– 0 , is a gradually series frame for đ??ż2 (đ??¸) with frame operator đ?‘†. Let đ?œ–5 > 0 with â„ą −1 (đ?‘”đ?‘— ) âˆś â„‚đ?‘‘ → â„‚, â„ą −1 (đ?‘”đ?‘— ) ∈ đ??ś ∞ where â„ą −1 (đ?‘”đ?‘— )|đ??¸ = 1 and â„ą −1 (đ?‘”đ?‘— )|( 1+đ?œ– 2 đ??¸)đ?‘? = 0. If đ?œ–2 > đ?œ–5 > 0 and đ?‘”đ?‘— ∈ đ?‘ƒđ?‘Šđ??¸ , then 1 1 + đ?œ–2

đ?‘“đ?‘— (đ?‘Ą) = đ?‘—

đ??ľđ?‘˜

đ?‘‘

đ?‘˜âˆˆâ„• đ?‘— đ?‘š +đ?œ– 0 ∈ℕ đ?‘† −1 đ?‘“ đ?‘š +đ?œ– 0 , đ?‘† −1 đ?‘“đ?‘˜ đ??¸ . Convergence

where đ??ľđ?‘˜ đ?‘š +đ?œ– 0 = the map đ??ľ âˆś â„“2 (đ?‘ ) → â„“2 (đ?‘ ) defined by (đ?‘Śđ?‘˜ )đ?‘˜âˆˆâ„• â&#x;ź

is an onto isomorphism if and only if ( đ?‘“ đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 Proof .Define đ?‘“ 1+đ?œ– 2

, đ?‘š +đ?œ– 0

¡

(¡ ) = đ?‘“ đ?‘š +đ?œ– 0

đ?‘š +đ?œ– 0

đ?‘”đ?‘— đ?‘Ą −

đ?‘Ąđ?‘˜ 1 + đ?œ–2

, đ?‘Ą ∈ â„‚đ?‘‘ ,

(4)

of the sum is in đ??ż2 (â„‚đ?‘‘ ), hence also uniform. Further, is bounded linear, and đ?‘š +đ?œ– 0 ∈ℕ đ??ľđ?‘˜ đ?‘š +đ?œ– 0 đ?‘Ś đ?‘š +đ?œ– 0 is a Riesz basis for đ??ż2 (đ??¸).

. Note that ( đ?‘“ 1+đ?œ– 2

1+đ?œ– 2

for đ??ż2 1 + đ?œ–2 đ??¸ with frame operator đ?‘† 1+đ?œ– 2 .

Step 1: We show that đ?‘Ą đ?‘š +đ?œ– 0 đ?‘“đ?‘— = đ?‘“đ?‘— 1 + đ?œ–2 đ?‘—

∈ℕ

đ?‘Ą đ?‘š +đ?œ– 0 1 + đ?œ–2

đ?‘“đ?‘—

đ?‘š +đ?œ– 0

, đ?‘š +đ?œ– 0

â„ą đ?‘† −1 1+đ?œ– 2 đ?‘“ 1+đ?œ– 2

đ?‘˜âˆˆâ„•

) đ?‘š +đ?œ– 0 is a gradually series frame

â„ą −1 (đ?‘”đ?‘— ) , đ?‘“đ?‘— ∈ đ?‘ƒđ?‘Šđ??¸ .

, đ?‘š +đ?œ– 0

(5)

đ?‘—

We know supp(â„ą −1 ( đ?‘“đ?‘— )) ⊂ đ??¸ ⊂ 1 + đ?œ–2 đ??¸, so we may work with â„ą −1 ( đ?‘“đ?‘— ) via its frame decomposition .

We have = đ?‘† −1 1+đ?œ– 2 đ?‘† 1+đ?œ– 2

â„ą −1 đ?‘“đ?‘—

â„ą −1 đ?‘“đ?‘—

đ?‘—

đ?‘—

â„ą −1 đ?‘“đ?‘— , đ?‘“ 1+đ?œ– 2

= đ?‘š +đ?œ– 0

đ?‘—

â„ą −1 ( đ?‘“đ?‘— ), đ?‘“ 1+đ?œ– 2

=

đ?‘—

đ?‘š +đ?œ– 0

, đ?‘š +đ?œ– 0

,

1+đ?œ– 2 đ??¸

đ?‘œđ?‘› 1 + đ?œ–2 đ??¸. This yields â„ą −1 đ?‘“đ?‘—

đ?‘† −1 1+đ?œ– 2 đ?‘“ 1+đ?œ– 2

, đ?‘š +đ?œ– 0

, đ?‘š +đ?œ– 0

1+đ?œ– 2 đ??¸

đ?‘† −1 1+đ?œ– 2 đ?‘“ 1+đ?œ– 2

, đ?‘š +đ?œ– 0

â„ą −1 (đ?‘”đ?‘— ), đ?‘œđ?‘› â„‚đ?‘‘ ,

đ?‘—

since suppâ„ą(đ?‘”đ?‘— ) ⊂ 1 + đ?œ–2 đ??¸. Taking Fourier transforms we obtain â„ą −1 đ?‘“đ?‘— , đ?‘“ 1+đ?œ– 2

đ?‘“đ?‘— = đ?‘—

đ?‘š +đ?œ– 0

, đ?‘š +đ?œ– 0

1+đ?œ– 2 đ??¸

−1 â„ą đ?‘† −1 đ?‘”đ?‘— , đ?‘œđ?‘› â„‚đ?‘‘ .

1+đ?œ– 2 â„ą

Now â„ą −1 đ?‘“đ?‘— , đ?‘“ 1+đ?œ– 2

, đ?‘š +đ?œ– 0

1+đ?œ– 2 đ??¸

đ?‘—

â„ą −1 đ?‘“đ?‘— (đ?œ‰ )đ?‘’

=

−đ?‘– đ?œ‰ ,

đ?‘Ą đ?‘š +đ?œ– 0 1+đ?œ– 2

đ?‘‘đ?œ‰ =

1+đ?œ– 2 đ??¸ đ?‘—

which, when substituted into (6), yields (5).

Step 2: We show that đ?‘Ą đ?‘š +đ?œ– 0 đ?‘“đ?‘— ¡ = đ?‘“đ?‘— đ?‘† −1 1+đ?œ– 2 đ?‘“ 1+đ?œ– 2 1 + đ?œ–2 đ?‘—

đ?‘š +đ?œ– 0

đ?‘—

(6)

đ?‘—

đ?‘“đ?‘— đ?‘—

, đ?‘š +đ?œ– 0

đ?‘˜

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, đ?‘† −1 1+đ?œ– 2 đ?‘“ 1+đ?œ– 2

,đ?‘˜

1+đ?œ– 2 đ??¸

đ?‘”đ?‘— ¡ −

đ?‘Ą đ?‘š +đ?œ– 0 1 + đ?œ–2

đ?‘Ąđ?‘˜ 1 + đ?œ–2

, (7)

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Multidimensional of gradually series bandlimited functions and frames where convergence is in 𝐿2 .We compute ℱ[(𝑆 −1 𝜖 2 +1 𝑓 𝜖 2 +1 𝑕 = 𝑆 1+𝜖 2 𝑆 −1 1+𝜖 2 𝑕 =

𝑆 −1 1+𝜖 2 𝑕, 𝑓 1+𝜖 2 𝑘

Letting 𝑕 = 𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

, 𝑚 +𝜖 0

𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

, 𝑚 +𝜖 0

,𝑘

1+𝜖 2 𝐸

)ℱ −1 (𝑔𝑗 )]. For 𝑕 ∈ 𝐿2 ( 1 + 𝜖2 𝐸) we have

, 𝑚 +𝜖 0

𝑓 1+𝜖 2

,𝑘

𝑕 , 𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

=

,𝑘

𝑓 1+𝜖 2 ,𝑘 .

1+𝜖 2 𝐸

𝑘

, 𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

=

, 𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

, 𝑚 +𝜖 0

,𝑘

1+𝜖 2 𝐸

𝑓 1+𝜖 2 ,𝑘 .

𝑘

This gives ℱ 𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

ℱ −1 𝑔𝑗

, 𝑚 +𝜖 0

·

𝑗

𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

= 𝑘

,𝑘

1+𝜖 2 𝐸

, 𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

, 𝑚 +𝜖 0

,𝑘

𝑒

1+𝜖 2 𝐸

𝑘 , 𝑚 +𝜖 0

, 𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

,𝑘

𝑡𝑘 1+𝜖 2

ℱ −1 (𝑔𝑗 )(𝜉 )𝑒 −𝑖 𝜉 ,. 𝑑𝜉

ℱ −1 (𝑔𝑗 )(𝜉 )𝑒

1+𝜖 2 𝐸

𝑘

1+𝜖 2 𝐸 𝑗

𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

= 𝑘

, 𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

, 𝑚 +𝜖 0

,𝑘

1+𝜖 2 𝐸

𝑔𝑗 · −

𝑗

so (7) follows from (5).

Step 3: We show that 𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

, 𝑚 +𝜖 0

, 𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

First we show (𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2 𝑓 1+𝜖 2

𝑖 𝜉,

·

1+𝜖 2 𝐸 𝑗

𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

=

ℱ 𝑓 1+𝜖 2 ,𝑘 ℱ −1 𝑔𝑗

𝑗

𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

=

, 𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2

, 𝑚 +𝜖 0

, 𝑚 +𝜖 0

=

𝑔𝑗 , 𝑓 1+𝜖 2

,𝑘

1

, 𝑚 +𝜖 0

1+𝜖 2 𝑑

𝑆 1+𝜖 2

1+𝜖 2 𝐸

=

𝑗

1+𝜖 2 𝐸

)(· ) =

1 1+𝜖 2 𝑑

𝑔𝑗 𝜉 𝑒

−𝑖

,

1 𝑆 −1 𝑓 𝑚 +𝜖 0 , 𝑆 −1 𝑓𝑘 𝐸 , for 𝜖0 > 0 , 𝑘 ∈ ℕ. (8) 1 + 𝜖2 𝑑 · (𝑆 −1 , or equivalently that 1+𝜖 2 𝑓 𝑚 +𝜖 0 ) 1+𝜖 2

· 1+𝜖 2

.We have for any 𝑔𝑗 ∈ 𝐿2 ( 1 + 𝜖2 𝐸),

𝜉 ,𝑡 1+𝜖 2 𝑘

𝑑𝜉

1+𝜖 2 𝐸 𝑗

= 1 + 𝜖2

𝑑

𝑔𝑗

1 + 𝜖2 𝑥 𝑒 −𝑖 𝑥,𝑡 𝑘 𝑑𝑥 = 1 + 𝜖2

𝑑

𝑔𝑗

𝑗

𝐸

1 + 𝜖2 (·) , 𝑓𝑘

𝑗

By definition of the frame operator 𝑆 1+𝜖 2 , 𝑆 1+𝜖 2 𝑔𝑗 = 𝑘∈ℕ 𝑔𝑗 , 𝑓 1+𝜖 2 ,𝑘 1+𝜖 2 𝐸 𝑓 1+𝜖 2 becomes 𝑗 𝑆 1+𝜖 2 𝑔𝑗 = 1 + 𝜖2 𝑑 𝑘 𝑗 𝑔𝑗 1 + 𝜖2 (·) , 𝑓𝑘 𝐸 𝑓 1+𝜖 2 ,𝑘 . Substituting 𝑔𝑗 = 1 1 + 𝜖2

𝑑𝜉

=

,𝑘

(𝑆 −1 𝑓 𝑚 +𝜖 0 )

𝑡𝑘 1 + 𝜖2

𝑡𝑘 ,𝜉 1+𝜖 2

−𝑖 ·−

1 1+𝜖 2 𝑑

𝑑

·

(𝑆 −1 𝑓 𝑚 +𝜖 0 )

1+𝜖 2

𝑆 −1 𝑓 𝑚 +𝜖 0

𝑆 1+𝜖 2

· 1 + 𝜖2

𝑆 −1 𝑓 𝑚 +𝜖 0 , 𝑓𝑘

=

𝐸

𝑘

We now compute the desired inner product: −1 𝑆 −1 1+𝜖 2 𝑓 1+𝜖 2 , 𝑚 +𝜖 0 , 𝑆 1+𝜖 2 𝑓 1+𝜖 2 ,𝑘 1+𝜖 2 𝐸 1 = 𝑆 −1 𝑓 𝑚 +𝜖 0 1 + 𝜖2 2𝑑 1 + 𝜖2 𝑑 1 + 𝜖2 2𝑑

, which then

into the equation above we obtain

= 𝑆 𝑆 −1 𝑓 𝑚 +𝜖 0

=

,𝑘

𝐸.

1+𝜖 2 𝐸

𝑥 1 + 𝜖2

𝑆 −1 𝑓𝑘

𝑆 −1 𝑓 𝑚 +𝜖 0 (𝑥) 𝑆 −1 𝑓𝑘 (𝑥) 𝑑𝑥 = 𝐸

𝑓 1+𝜖 2

,𝑘

· 1 + 𝜖2

= 𝑓 1+𝜖 2

𝑥 1 + 𝜖2 1 1 + 𝜖2

𝑑

, 𝑚 +𝜖 0

.

𝑑𝑥

𝑆 −1 𝑓 𝑚 +𝜖 0 , 𝑆 −1 𝑓𝑘

𝐸

.

Note that (7) becomes 𝑓𝑗 (· ) = 𝑗

1 1 + 𝜖2

𝑓𝑗

𝑑 𝑚 +𝜖 0

𝑗

𝑡 𝑚 +𝜖 0 1 + 𝜖2

𝑆 −1 𝑓 𝑚 +𝜖 0 , 𝑆 −1 𝑓𝑘 𝑔𝑗 · − 𝑘

𝑡𝑘 1 + 𝜖2

.

(9)

Step4: The map 𝑉 ∶ ℓ2 (ℕ) ↦ ℓ2 (ℕ) given by 𝑥 = (𝑥𝑘 )𝑘∈ℕ ↦ 𝑚 +𝜖 0 𝐵𝑘 𝑚 +𝜖 0 𝑥 𝑚 +𝜖 0 𝑘∈ℕ = 𝐵𝑥 is bounded linear and self-adjoint. Let (𝑑𝑘 )𝑘∈ℕ be the standard basis for ℓ2 (ℕ), and let (𝑒𝑘 )𝑘∈ℕ be an orthonormal basis for 𝐿2 (𝐸). Then

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52 | Page

Multidimensional of gradually series bandlimited functions and frames 𝑉𝑑𝑗 = 𝐵𝑘𝑗

𝑘∈ℕ

=

𝑆 −1 𝑓𝑗 , 𝑆 −1 𝑓𝑘

𝐵𝑘𝑗 𝑑𝑘 = 𝑘

𝑑𝑘

𝐿∗ 𝑆 −1 2 𝐿𝑒𝑗 , 𝑒𝑘

=

𝑘

𝑑𝑘 ,

𝑘

where 𝐿 is the synthesis 𝑓 operator, i.e., 𝑆 = 𝐿𝐿∗ . Define 𝜑 ∶ ℓ2 (ℕ) ↦ 𝐿2 (𝐸) by 𝜑(𝑑𝑘 ) = 𝑒𝑘 , 𝑘 ∈ ℕ.

Clearly 𝜑 is unitary. It follows that 𝑉 = 𝜑 −1 𝐿∗ 𝑆 −1 2 𝐿𝜑, which concludes Step 4. From here on we identify 𝑉 with 𝐵. Clearly 𝐵 is an onto isomorphism if and only if 𝐿 and 𝐿∗ are both onto, i.e., if and only if the map 𝐿𝑒 𝑚 +𝜖 0 = 𝑓 𝑚 +𝜖 0 is an onto isomorphism.

Step 5: Verification of (4). Recalling Definition 3.8, 𝑓𝑆𝑗 / 1+𝜖 2 = .

that 𝑓𝑗

𝑡 𝑚 +𝜖 0 1+𝜖 2

𝑓𝑗

.

, 𝑔𝑗 𝑡 −

1+𝜖 2

𝑡 = 𝑔𝑗 𝑡 −

𝑗

∈ 𝐿2 ( 1 + 𝜖2 𝐸), and recalling that ( 𝑓 1+𝜖 2

1+𝜖 2

𝐿2 ( 1 + 𝜖2 𝐸), we have 𝑡 𝑚 +𝜖 0 𝑓𝑗 1 + 𝜖2 𝑚 +𝜖 0

𝑚 +𝜖 0 ∈ℕ

, for each 𝑡 ∈ ℂ𝑑 , let 𝑔 1+𝜖 2

2

2

ℱ −1 ( 𝑓𝑗 ), 𝑓 1+𝜖 2

=

𝑗

𝑚 +𝜖 0

, 𝑚 +𝜖 0

𝑚 +𝜖 0 ∈ℕ

𝑗

𝑚 +𝜖 0

𝑗

ℱ −1 𝑔𝑗 𝑡 −

= 𝑗

≤ 𝐴

, (10)

𝑗

2

𝑡 𝑚 +𝜖 0 1 + 𝜖2

2

ℱ −1 ( 𝑓𝑗 )

1+𝜖 2

and 𝑔𝑗 𝑡 −

. Noting

) 𝑚 +𝜖 0 is a frame for

, 𝑚 +𝜖 0

≤𝐴

1+𝜖 2 𝐸

𝑡 𝑚 +𝜖 0 1+𝜖 2

ℱ

1+𝜖 2

−1

𝑔𝑗

𝑗

2

.

1 + 𝜖2

, 𝑓 1+𝜖 2

, 𝑚 +𝜖 0

1+𝜖 2 𝐸

2

. 𝑡 − 1 + 𝜖2

.

Note that (9) becomes 𝑓𝑗 𝑡 = 𝑗

=

1 1 + 𝜖2

1 1 + 𝜖2

1 = 1 + 𝜖2

=

𝑓𝑗 𝑚 +𝜖 0

𝑗

𝑗

𝑚 +𝜖 0

𝑗

𝑗

𝐵𝑓𝑆/ 1+𝜖 2

𝑑 𝑘

𝐵𝑘

𝑑

𝑘

𝑚 +𝜖 0

𝑘

𝑔𝑗 𝑡 −

𝑡𝑘 1 + 𝜖2

=

1 1 + 𝜖2

𝑗

𝐵𝑔𝑗

𝑡𝑘 1 + 𝜖2

𝑔𝑗 𝑡 −

𝑗

𝐵 𝑚 +𝜖 0

1+𝜖 2

1+𝜖 2

𝑡

𝑚 +𝜖 0

𝑓𝑗

𝑑 𝑗

𝑆 1+𝜖 2

, 𝐵𝑔𝑗

1+𝜖 2

𝑡

(𝑡)

𝑔𝑗 𝑡 −

𝑡𝑘 1 + 𝜖2

𝑚 +𝜖 0 𝑓𝑗

𝑡 𝑚 +𝜖 0 1 + 𝜖2

𝑘

𝑗

𝑘∈ℕ 𝑗

𝑘

𝑚 +𝜖 0

𝐵𝑓𝑆/ 1+𝜖2 , 𝑔𝑗

𝑑

𝐵𝑘

𝑡 𝑚 +𝜖 0 1 + 𝜖2

𝑓𝑆/ 1+𝜖 2

𝑑

1 1 + 𝜖2 1 1 + 𝜖2

𝑚 +𝜖 0

𝑑

1 = 1 + 𝜖2 =

𝑡 𝑚 +𝜖 0 1 + 𝜖2

𝑓𝑗

𝑑

𝑚 +𝜖 0 ∈ℕ

𝑔𝑗 𝑡 −

𝑡𝑘 , 1 + 𝜖2

which proves (4).

Step 6: We verify that convergence in (4) is in 𝐿2 (ℂ) (hence uniform). Define 1 𝑡𝑘 𝑓 𝑚 +𝜖 0 (𝑡) = 𝐵𝑓𝑆/ 1+𝜖 2 𝑔𝑗 𝑡 − 𝑑 𝑘 1 + 𝜖2 1 + 𝜖2 1≤𝑘≤ 𝑚 +𝜖 0

and 𝑓𝑚 , 𝑚 +𝜖 0 𝑡 = Then ℱ −1 𝑓𝑚 , 𝑚 +𝜖 0

(𝜉 ) =

1 1 + 𝜖2 1 1 + 𝜖2

1 = 1 + 𝜖2

𝑗

𝐵𝑓𝑆/ 1+𝜖2

𝑑 𝑚 ≤𝑘≤ 𝑚 +𝜖 0

𝑗

𝐵𝑓𝑆/ 1+𝜖 2

𝑑 𝑚 ≤𝑘≤ 𝑚 +𝜖 0

𝑗

𝐵𝑓𝑆/ 1+𝜖2

𝑑 𝑚 ≤𝑘≤ 𝑚 +𝜖 0

𝑗

𝑔 𝑘 𝑗

𝑡−

𝑡𝑘 1 + 𝜖2

.

𝑡 𝑚 +𝜖 0 1 + 𝜖2

𝑘

ℱ −1 𝑔𝑗 . −

𝑘

ℱ −1 (𝑔𝑗 )(𝜉 )𝑒

𝑖 𝜉,

𝑡𝑘 1+𝜖 2

,

so

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53 | Page

Multidimensional of gradually series bandlimited functions and frames 2

â„ą −1 đ?‘“đ?‘š , đ?‘š +đ?œ– 0

2

1 = 1 + đ?œ–2

1+đ?œ– 2 đ??¸ đ?‘—

1 ≤ 1 + đ?œ–2 If đ?‘•

đ?‘š +đ?œ– 0

â„ą −1 (đ?‘”đ?‘— )(đ?œ‰ )

đ?‘‘

2

đ??ľđ?‘“đ?‘†/ 1+đ?œ– 2 đ?‘š ≤đ?‘˜â‰¤ đ?‘š +đ?œ– 0

đ??ľđ?‘“đ?‘†/ 1+đ?œ– 2

đ?‘‘ đ?‘š ≤đ?‘˜â‰¤ đ?‘š +đ?œ– 0

đ?‘˜

đ?‘“ 1+đ?œ– 2

đ?‘˜

đ?‘’

đ?‘– đ?œ‰,

đ?‘‘đ?œ‰

.

,đ?‘˜ 2

operator) is bounded linear, so

,đ?‘˜

(the synthesis

2 2

â„ą −1 đ?‘“đ?‘š , đ?‘š +đ?œ– 0

2

2

is a orthonormal basis for đ??ż2 ( 1 + đ?œ–2 đ??¸), then the map đ?‘‡đ?‘•đ??ž = đ?‘“ 1+đ?œ– 2

đ?‘š +đ?œ– 0

đ?‘Ąđ?‘˜ 1+đ?œ– 2

2

1 ≤ 1 + đ?œ–2

1 ≤ 1 + đ?œ–2

�

đ?‘‘

đ??ľđ?‘“đ?‘†/ 1+đ?œ–2 đ?‘š ≤đ?‘˜â‰¤ đ?‘š +đ?œ– 0 2

�

đ?‘‘

đ??ľđ?‘“đ?‘†/ 1+đ?œ–2 đ?‘š ≤đ?‘˜â‰¤ đ?‘š +đ?œ– 0

đ?‘• đ?‘˜ đ??ž 2 đ?‘˜

2

.

But đ??ľđ?‘“đ?‘†/ 1+đ?œ– 2 ∈ â„“2 (â„•), so â„ą −1 đ?‘“đ?‘š , đ?‘š +đ?œ– 0 2 → 0 as đ?‘š → ∞, đ?œ–0 > 0. As â„ą −1 is an onto isomorphism, we have đ?‘“đ?‘š , đ?‘š +đ?œ– 0 → 0, implying that đ?‘“ − đ?‘“ đ?‘š +đ?œ– 0 → 0 as đ?‘š → ∞. Note that (3) is conveniently written as 1 đ?‘Ąđ?‘˜ đ?‘“đ?‘— đ?‘Ą = đ??ľđ?‘“đ?‘†/ 1+đ?œ–2 đ?‘”đ?‘— đ?‘Ą − , đ?‘Ą ∈ â„‚đ?‘‘ .

(11) đ?‘‘ đ?‘˜ 1 + đ?œ–2 1 + đ?œ–2 đ?‘—

đ?‘˜

đ?‘—

Remark .There is a geometric characterization of sets đ??¸ ⊂ â„‚đ?‘‘ such that đ??¸ ⊂ đ?‘–đ?‘›đ?‘Ą( 1 + đ?œ–2 đ??¸) for all đ?œ–2 > 0.

Intuitively, đ??¸ must be a continuous radial stretching of the closed unit ballâ€?. This is precisely formulated in the following proposition .

Proposition 3.2 . If 0 ∈ đ??¸ ⊂ â„‚đ?‘‘ is compact, then the following are equivalent: (i) đ??¸ ⊂ đ?‘–đ?‘›đ?‘Ą( 1 + đ?œ–2 đ??¸) for all đ?œ–2 > 0. (ii) There exists a continuous map đ?œ‘ âˆś đ?‘† đ?‘‘−1 → (0, ∞) such that đ??¸ = đ?‘Ąđ?‘Śđ?œ‘(đ?‘Ś) | đ?‘Ś ∈ đ?‘† đ?‘‘−1 , đ?‘Ą ∈ [0, 1] . The following is a simplified version of Theorem 3.1 , which is proven in a similar fashion (see [13] ): Theorem 3.3 .Choose (đ?‘Ą đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 ∈ℕ ⊂ â„‚đ?‘‘ such that (đ?‘“ đ?‘š +đ?œ– 0 ) đ?‘š +đ?œ– 0 ∈ℕ , defined by đ?‘“ đ?‘š +đ?œ– 0 (¡ ) =

1

(2đ?œ‹)đ?‘‘ /2

đ?‘’ đ?‘– ¡,đ?‘Ą đ?‘š +đ?œ– 0 , is a frame for đ??ż2 [−đ?œ‹, đ?œ‹]đ?‘‘ . If đ?‘“ ∈ đ?‘ƒđ?‘Šđ??¸ ,then

đ?‘“đ?‘— đ?‘Ą = đ?‘—

đ??ľđ?‘˜ đ?‘˜âˆˆâ„•

đ?‘š +đ?œ– 0 đ?‘“đ?‘—

đ?‘Ą đ?‘š +đ?œ– 0

đ?‘†đ??źđ?‘ đ??ś đ?œ‹ đ?‘Ą – đ?‘Ąđ?‘˜

, đ?‘Ą ∈ â„‚đ?‘‘ .

(12)

đ?‘š +đ?œ– 0 ∈ℕ đ?‘—

The matrix đ??ľ and the convergence of the sum are as in Theorem 3.1.Then (4) generalizes (12) in the same way that (3) generalizes the WKS equation. We can write (12) as đ?‘“đ?‘— đ?‘Ą

=

đ?‘—

đ??ľđ?‘“đ?‘†đ?‘— đ?‘˜âˆˆâ„• đ?‘—

đ?‘˜

đ?‘†đ??źđ?‘ đ??ś đ?œ‹ đ?‘Ą – đ?‘Ąđ?‘˜

.

(13)

The preceding result is similar in spirit to Theorem 1.9 in [4, p. 19].Gradually series frames forđ??ż2 (đ??¸) satisfying the conditions in Theorems 3.9 and 5.2 occur in abundance. The following result is due to Beurling in [5, see Theorem 1, Theorem 2, and (20)].

Theorem 3.4 .Let đ?&#x203A;Ź â&#x160;&#x201A; â&#x201E;&#x201A;đ?&#x2018;&#x2018; be countable such that 1 đ?&#x153;&#x2039; đ?&#x2018;&#x; đ?&#x203A;Ź = inf 1 + đ?&#x153;&#x2013;2 â&#x20AC;&#x201C; đ?&#x153;&#x2021; 2 > 0 and â&#x201E;&#x201A; đ?&#x203A;Ź = sup inf 1 + đ?&#x153;&#x2013;2 â&#x2C6;&#x2019; đ?&#x153;&#x2021; 2 < .

2 1+đ?&#x153;&#x2013; 2 ,đ?&#x153;&#x2021; â&#x2C6;&#x2C6;đ?&#x203A;Ź, 1+đ?&#x153;&#x2013; 2 â&#x2030; đ?&#x153;&#x2021; 2 đ?&#x153;&#x2030; â&#x2C6;&#x2C6;â&#x201E;&#x201A;đ?&#x2018;&#x2018; 1+đ?&#x153;&#x2013; 2 â&#x2C6;&#x2C6;đ?&#x203A;Ź If đ??¸ is a subset of the closed unit ball in â&#x201E;&#x201A;đ?&#x2018;&#x2018; and E has positive measure, then {đ?&#x2018;&#x2019; đ?&#x2018;&#x2013; ¡, 1+đ?&#x153;&#x2013; 2 | 1 + đ?&#x153;&#x2013;2 â&#x2C6;&#x2C6; đ?&#x203A;Ź} is a gradually series frame for đ??ż2 (đ??¸).

IV.

REMARKS REGARDING THE STABILITY OF THEOREM 3.1

A desirable trait in a recovery formula is stability given error in the sampled data (see [13]). Suppose we have sample values đ?&#x2018;&#x201C;đ?&#x2018;&#x2014;

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

If in (3) we replace đ?&#x2018;&#x201C;

đ?&#x2018;&#x2014;

= đ?&#x2018;&#x201C;đ?&#x2018;&#x2014; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 1+đ?&#x153;&#x2013; 2

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 1+đ?&#x153;&#x2013; 2

+ đ?&#x153;&#x2013; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 where đ?&#x2018; đ?&#x2018;˘đ?&#x2018;? đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x153;&#x2013;

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

= đ?&#x153;&#x2013;.

by đ?&#x2018;&#x201C; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 , and call the resulting expression đ?&#x2018;&#x201C; , then we have

đ?&#x153;&#x2013; đ?&#x2018;&#x201C;đ?&#x2018;&#x2014; (đ?&#x2018;Ą) â&#x2C6;&#x2019; đ?&#x2018;&#x201C;đ?&#x2018;&#x2014; (đ?&#x2018;Ą) â&#x2030;¤ 1 + đ?&#x153;&#x2013;2

đ?&#x2018;&#x201D;đ?&#x2018;&#x2014; đ?&#x2018;Ą â&#x2C6;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2C6;đ?&#x2022;Ť đ?&#x2018;&#x2014;

đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 1 + đ?&#x153;&#x2013;2

www.ijesi.org

â&#x2030;¤

đ?&#x153;&#x2013; 1 + đ?&#x153;&#x2013;2

đ?&#x2018;&#x201D;đ?&#x2018;&#x2014; â&#x20AC;˛ đ?&#x2018;&#x2014;

đ??ż1

+

đ?&#x153;&#x2013; đ?&#x2018;&#x201D;đ?&#x2018;&#x2014; đ?&#x2018;&#x2014;

đ??ż1

.

54 | Page

Multidimensional of gradually series bandlimited functions and frames It follows that (3) is certainly stable under â&#x201E;&#x201C;â&#x2C6;&#x17E; perturbations in the data, while the WKS sampling theorem is not. For a more detailed discussion see [7]. Such a stability result is not immediately forthcoming for (4), as the following example illustrates. Restricting to đ?&#x2018;&#x2018; = 1, let (đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2C6;đ?&#x2022;Ť satisfy đ?&#x2018;Ą2 = đ??ˇ â&#x2C6;&#x2030; đ?&#x2022;Ť, and đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 = đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 for đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 â&#x2030; 0. The forthcoming discussion in Section 5 shows that (đ?&#x2018;&#x201C; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2C6;đ?&#x2022;Ť is a Riesz basis for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;].

Note that when (đ?&#x2018;&#x201C; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 is a Riesz basis, the sequence đ?&#x2018;&#x2020; â&#x2C6;&#x2019;1 đ?&#x2018;&#x201C; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

is its biorthogonal sequence.

The matrix đ??ľ associated to this basis is computed as follows. The biorthogonal functions (đ??ş đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 (đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;?(đ?&#x153;&#x2039;(¡ â&#x2C6;&#x2019; đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 ))) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 And đ??ş0 (đ?&#x2018;Ą) =

đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;? (đ?&#x153;&#x2039;đ?&#x2018;Ą )

â&#x2C6;&#x2019;1 đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;Ą â&#x2C6;&#x2019; đ??ˇ đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;? đ?&#x153;&#x2039;đ?&#x2018;Ą

are đ??ş đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 (đ?&#x2018;Ą) =

â&#x2C6;&#x2C6;đ?&#x2022;Ť

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2019;đ??ˇ đ?&#x2018;Ąâ&#x2C6;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

â&#x2C6;&#x2C6;đ?&#x2022;Ť

for

, đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 â&#x2030; 0,

. That these functions are in đ?&#x2018;&#x192;đ?&#x2018;&#x160;[â&#x2C6;&#x2019;đ?&#x153;&#x2039;,đ?&#x153;&#x2039;] is verified by applying the Paleyâ&#x20AC;&#x201C;Wiener theorem

đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;? (đ?&#x153;&#x2039; đ??ˇ)

[1, p. 85], and the biorthogonality condition is verified by applying (1). Again using (1), we obtain đ??ˇ(â&#x2C6;&#x2019;1)đ?&#x2018;&#x161;

(i) đ??ľđ?&#x2018;&#x161; 0 = đ??ş0 , đ??şđ?&#x2018;&#x161; = (ii) đ??ľ00 = đ??ş0 , đ??ş0 = (iii) đ??ľđ?&#x2018;&#x161;

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;? (đ?&#x153;&#x2039; đ??ˇ)(đ?&#x2018;&#x161; â&#x2C6;&#x2019; đ??ˇ) 1 đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A; đ?&#x2018;? 2 (đ?&#x153;&#x2039; đ??ˇ)

, đ?&#x2018;&#x161; â&#x2030; 0,

,

= đ??ş đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 , đ??şđ?&#x2018;&#x161; = đ?&#x203A;ż đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

đ?&#x2018;&#x161;

+

đ??ˇ 2 (â&#x2C6;&#x2019;1) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 +đ?&#x2018;&#x161; ( đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2019; đ??ˇ)(đ?&#x2018;&#x161; â&#x2C6;&#x2019; đ??ˇ)

, đ?&#x2018;&#x2019;đ?&#x2018;&#x2122;đ?&#x2018; đ?&#x2018;&#x2019;.

Note that the rows of đ??ľ are not in â&#x201E;&#x201C;1 , so that as an operator acting on â&#x201E;&#x201C;â&#x2C6;&#x17E; , đ??ľ does not act boundedly.

Consequently, the equation 1 đ?&#x2018;Ąđ?&#x2018;&#x2DC; đ?&#x2018;&#x201C;đ?&#x2018;&#x2014; đ?&#x2018;Ą = đ??ľđ?&#x2018;&#x201C;đ?&#x2018;&#x2020;/ 1+đ?&#x153;&#x2013; 2 đ?&#x2018;&#x201D;đ?&#x2018;&#x2014; đ?&#x2018;Ą â&#x2C6;&#x2019; (14) đ?&#x2018;&#x2DC; 1 + đ?&#x153;&#x2013;2 1 + đ?&#x153;&#x2013;2 đ?&#x2018;&#x2014;

đ?&#x2018;&#x2DC;

đ?&#x2018;&#x2014;

is not defined for all perturbed sequences đ?&#x2018;&#x201C;đ?&#x2018;&#x2020;/ 1+đ?&#x153;&#x2013;2 where đ?&#x2018;&#x201C;đ?&#x2018;&#x2020;/ 1+đ?&#x153;&#x2013; 2 = đ?&#x2018;&#x201C;đ?&#x2018;&#x2020;/ 1+đ?&#x153;&#x2013;2 +đ?&#x153;&#x2013; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 where sup đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

đ?&#x153;&#x2013; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 = đ?&#x153;&#x2013; . Despite the above failure, the

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

following shows that there is some advantage of (4) over (2).If đ?&#x2018;&#x201C;đ?&#x2018;&#x2020;/ 1+đ?&#x153;&#x2013; 2 is some perturbation of đ?&#x2018;&#x201C;đ?&#x2018;&#x2020;/ that đ??ľđ?&#x2018;&#x201C;đ?&#x2018;&#x2020;/ 1+đ?&#x153;&#x2013;2 â&#x2C6;&#x2019; đ??ľđ?&#x2018;&#x201C;đ?&#x2018;&#x2020;/ 1+đ?&#x153;&#x2013;2 â&#x2030;¤ đ?&#x153;&#x2013; , then

1+đ?&#x153;&#x2013; 2

such

â&#x2C6;&#x17E;

sup

đ?&#x153;&#x2030;â&#x2C6;&#x2C6;â&#x201E;&#x201A;đ?&#x2018;&#x2018;

đ?&#x2018;&#x201C;đ?&#x2018;&#x2014; (đ?&#x2018;Ą) â&#x2C6;&#x2019; đ?&#x2018;&#x201C;đ?&#x2018;&#x2014; (đ?&#x2018;Ą) = sup

đ?&#x153;&#x2030; â&#x2C6;&#x2C6;â&#x201E;&#x201A;đ?&#x2018;&#x2018;

đ?&#x2018;&#x2014;

1 1 + đ?&#x153;&#x2013;2 1 1 + đ?&#x153;&#x2013;2

â&#x2030;¤ đ?&#x153;&#x2013; sup

đ?&#x153;&#x2030; â&#x2C6;&#x2C6;â&#x201E;&#x201A;đ?&#x2018;&#x2018;

đ??ľ đ?&#x2018;&#x201C;đ?&#x2018;&#x2020;/ 1+đ?&#x153;&#x2013; 2 â&#x20AC;&#x201C; đ?&#x2018;&#x201C;đ?&#x2018;&#x2020;/ 1+đ?&#x153;&#x2013;2 đ?&#x2018;&#x2DC;

đ?&#x2018;&#x2014;

đ?&#x2018;&#x201D;đ?&#x2018;&#x2014; đ?&#x2018;Ą â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;

đ?&#x2018;&#x2014;

đ?&#x2018;Ąđ?&#x2018;&#x2DC; 1 + đ?&#x153;&#x2013;2

đ?&#x2018;&#x2DC;

đ?&#x2018;&#x201D;đ?&#x2018;&#x2014; đ?&#x2018;Ą â&#x2C6;&#x2019;

đ?&#x2018;Ąđ?&#x2018;&#x2DC; 1 + đ?&#x153;&#x2013;2

â&#x2030;¤đ?&#x2018;&#x20AC;

(15)

V. RESTRICTION OF THE SAMPLING THEOREM TO THE CASE WHERE THE EXPONENTIAL GRADUATE FRAME IS A RIESZ BASIS

From here on (see [13]), we focus on the case where (đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2C6;â&#x201E;&#x2022; is an â&#x201E;&#x201C;â&#x2C6;&#x17E; perturbation of the lattice đ?&#x2022;Ťđ?&#x2018;&#x2018; , and (đ?&#x2018;&#x201C; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2C6;â&#x201E;&#x2022; is a Riesz basis for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;]đ?&#x2018;&#x2018; . In this case, under the additional constraint that the sample nodes are asymptotically the integer lattice, the following theorem gives a computationally feasible version of (4). The summands in (4) involves an infinite invertible matrix đ??ľ, though under the constraints mentioned above, we show that đ??ľ can be replaced by a related finite-rank operator which can be computed concretely. Precisely, one has the following (see [13]).

Theorem đ?&#x;&#x201C;. đ?&#x;? .Let ( đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 đ?&#x2018;&#x2DC; )đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;â&#x201E;&#x2022; be an enumeration of đ?&#x2022;Ťđ?&#x2018;&#x2018; , and đ?&#x2018;&#x2020; = (đ?&#x2018;Ąđ?&#x2018;&#x2DC; )đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;â&#x201E;&#x2022; â&#x160;&#x201A; â&#x201E;&#x201A;đ?&#x2018;&#x2018; such that limđ?&#x2018;&#x2DC;â&#x2020;&#x2019;â&#x2C6;&#x17E; đ?&#x2018;&#x161; +

đ?&#x153;&#x2013;0đ?&#x2018;&#x2DC; â&#x2C6;&#x2019;đ?&#x2018;Ąđ?&#x2018;&#x2DC;â&#x2C6;&#x17E;= 0 .

Define đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; , đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; â&#x2C6;ś â&#x201E;&#x201A;đ?&#x2018;&#x2018; â&#x2020;&#x2019; â&#x201E;&#x201A; by đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; (đ?&#x2018;Ľ) =

1 (2đ?&#x153;&#x2039;)đ?&#x2018;&#x2018; /2

đ?&#x2018;&#x2019;đ?&#x2018;&#x2013;

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x2DC; , đ?&#x2018;Ľ

1

and

(2đ?&#x153;&#x2039;)đ?&#x2018;&#x2018; /2

đ?&#x2018;&#x2019; đ?&#x2018;&#x2013; đ?&#x2018;Ąđ?&#x2018;&#x2DC; ,

đ?&#x2018;Ľ

, and let đ?&#x2018;&#x2022;đ?&#x2018;&#x2DC;

đ?&#x2018;&#x2DC;

be the standard

basis for â&#x201E;&#x201C;2 (â&#x201E;&#x2022;). Let đ?&#x2018;&#x192;đ?&#x2018;&#x2122; : â&#x201E;&#x201C;2 (â&#x201E;&#x2022;) â&#x2020;&#x2019; â&#x201E;&#x201C;2 (â&#x201E;&#x2022;) be the orthogonal projection onto span{đ?&#x2018;&#x2022;1 , . . . , đ?&#x2018;&#x2022;đ?&#x2018;&#x2122; }. If (đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; )đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;â&#x201E;&#x2022; is a Riesz basis for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;]đ?&#x2018;&#x2018; , then for all đ?&#x2018;&#x201C; â&#x2C6;&#x2C6; đ?&#x2018;&#x192;đ?&#x2018;&#x160;[â&#x2C6;&#x2019;đ?&#x153;&#x2039;,đ?&#x153;&#x2039;]đ?&#x2018;&#x2018; , we have đ?&#x2018;&#x201C;đ?&#x2018;&#x2014; (đ?&#x2018;Ą) = lim đ?&#x2018;&#x2014;

đ?&#x2018;&#x2122;â&#x2020;&#x2019;â&#x2C6;&#x17E;

1 1 + đ?&#x153;&#x2013;2

đ?&#x2018;&#x2122;

đ?&#x2018;&#x192;đ?&#x2018;&#x2122; đ??ľ â&#x2C6;&#x2019;1 đ?&#x2018;&#x192;đ?&#x2018;&#x2122;

đ?&#x2018;&#x2018; đ?&#x2018;&#x2DC;=1

â&#x2C6;&#x2019;1

đ?&#x2018;&#x201C;

đ?&#x2018;&#x2020; 1+đ?&#x153;&#x2013; 2

đ?&#x2018;&#x201D;đ?&#x2018;&#x2014; đ?&#x2018;Ą â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;

đ?&#x2018;&#x2014;

đ?&#x2018;Ąđ?&#x2018;&#x2DC; 1 + đ?&#x153;&#x2013;2

, đ?&#x2018;Ą â&#x2C6;&#x2C6; â&#x201E;&#x201A;đ?&#x2018;&#x2018; ,

(16)

where convergence is in L2 and uniform. Furthermore, đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;?đ?&#x153;&#x2039;(đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ,1 â&#x2C6;&#x2019; đ?&#x2018;Ąđ?&#x2018;&#x161; ,1 ) ¡ ¡ ¡ đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;?đ?&#x153;&#x2039;(đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ,đ?&#x2018;&#x2018; â&#x2C6;&#x2019; đ?&#x2018;Ąđ?&#x2018;&#x161; ,đ?&#x2018;&#x2018; ), 1 â&#x2030;¤ đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 , đ?&#x2018;&#x161; â&#x2030;¤ đ?&#x2018;&#x2122;, đ?&#x2018;&#x192;đ?&#x2018;&#x2122; đ??ľ â&#x2C6;&#x2019;1 đ?&#x2018;&#x192;đ?&#x2018;&#x2122; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x161; = 0, đ?&#x2018;&#x153;đ?&#x2018;Ąđ?&#x2018;&#x2022;đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;¤đ?&#x2018;&#x2013;đ?&#x2018; đ?&#x2018;&#x2019;.

Convergence of the sum is in đ??ż2 and also uniform.

The matrix đ?&#x2018;&#x192;đ?&#x2018;&#x2122; đ??ľ â&#x2C6;&#x2019;1 đ?&#x2018;&#x192;đ?&#x2018;&#x2122; is clearly not invertible as an operator on â&#x201E;&#x201C;2 , and it should be interpreted as the inverse of an đ?&#x2018;&#x2122; Ă&#x2014; đ?&#x2018;&#x2122; matrix acting on the first đ?&#x2018;&#x2122; coordinates of đ?&#x2018;&#x201C;đ?&#x2018;&#x2020;/ 1+đ?&#x153;&#x2013;2 . The following version of Theorem 5.1 avoids over sampling . Its proof is similar to that of Theorem 5.1.

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55 | Page

Multidimensional of gradually series bandlimited functions and frames Theorem 5.2 .Under the hypotheses of Theorem 5.1 , 𝑙

𝑃𝑙 𝐵 −1 𝑃𝑙

𝑓𝑗 (𝑡) = lim

𝑙→∞

𝑗

−1

𝑓𝑆

𝑘

𝑆𝐼𝑁𝐶 𝑡 − 𝑡𝑘 , 𝑡 ∈ ℂ𝑑 ,

(17)

𝑘=1

where convergence of the sum is both L2 and uniform. The following lemma forms the basis of the proof of the preceding theorems, as well as the other results in the paper (see [13]) .

Lemma 5.3 .Let ( 𝑚 + 𝜖0 𝑘 )𝑘∈ℕ be an enumeration of 𝕫𝑑 , and let (𝑡𝑘 )𝑘∈ℕ ⊂ ℂ𝑑 . Define 1 1 𝑖 𝑚 +𝜖 0 𝑘 ,𝑥 𝑖 𝑡 𝑘 ,𝑥 𝑒𝑘 , 𝑓𝑘 ∶ ℂ𝑑 → 𝕔 by 𝑒𝑘 𝑥 = and 𝑓𝑘 𝑥 = . Then for any 𝜖4 > 𝜖3 ≥ 0, and any 𝑑 𝑒 𝑑 𝑒 2𝜋 2

finite sequence

1+𝜖 4 (𝑎𝑘 )𝑘= 1+𝜖 3 1+𝜖 4

, we have 𝑎𝑘 𝑑 2

2𝜋

𝑘= 1+𝜖 3

2𝜋 2

𝑒𝑖

· , 𝑚 +𝜖 0 𝑘

𝑎𝑘

−

2𝜋

𝑒𝑖

𝑑 2

· ,𝑡 𝑘 2

1/2

1+𝜖 4

𝑒 𝜋𝑑 (𝑠𝑢𝑝 1+𝜖 3 ≤𝑘 ≤ 1+𝜖 4

≤

𝑚 +𝜖 0 𝑘 − 𝑡 𝑘 ∞ )

2

−1

|𝑎𝑘 |

.

18

𝑘= 1+𝜖 3

Proof .Let 𝛿𝑘 = 𝑡𝑘 − 𝑚 + 𝜖0

𝑘

1+𝜖 4

𝜑

1+𝜖 3 , 1+𝜖 4

Now for any 𝜹𝒌 ,

where 𝛿𝑘 = (𝛿𝑘 1 , . . . , 𝛿𝑘 𝑑 ). Then

1+𝜖 4

𝑎𝑘

𝑥 =

2𝜋

𝑘= 1+𝜖 3

𝑒𝑖

𝑑 2

𝑚 +𝜖 0 𝑘 ,𝑥

𝑎𝑘

, 𝑒 𝑖 𝑡 𝑘 ,𝑥 = 𝑘= 1+𝜖 3

2𝜋

𝑑 2

𝑒𝑖

𝑚 +𝜖 0 𝑘 ,𝑥

1 − 𝑒 𝑖 𝛿 𝑘 ,𝑥

. 19

1 − 𝑒 𝑖 𝛿 𝑘 ,𝑥 = 1 − 𝑒 𝑖𝛿 𝑘 1 𝑥 1 … 𝑒 𝑖𝛿 𝑘 𝑑 𝑥 𝑑 ∞

= 1− 𝑗 1 =0

(𝑖𝛿𝑘 1 𝑥1 )𝑗 1 ... 𝑗1 !

∞

(𝑖𝛿𝑘 𝑑 𝑥𝑑 )𝑗 𝑑 𝑗𝑑 !

𝑗 𝑑 =0

(𝑖𝛿𝑘 1 𝑥1 )𝑗 1 · · · · · (𝑖𝛿𝑘 𝑑 𝑥𝑑 )𝑗 𝑑

= 1−

𝑗1 ! · · · · · 𝑗𝑑 !

𝑗 1 ,…,𝑗 𝑑 𝑗 𝑖 ≥0

𝑖 𝑗 1 ,…,𝑗 𝑑

=− ( 𝑗 1 ,…,𝑗 𝑑 )∈ 𝐽

(𝑖𝛿𝑘 1 𝑥1 )𝑗 1 · · · · · (𝑖𝛿𝑘 𝑑 𝑥𝑑 )𝑗 𝑑 , 𝑗1 ! · · · · · 𝑗𝑑 !

where 𝐽 = ( 𝑗1 , … , 𝑗𝑑 ) ∈ 𝕫𝑑 | 𝑗𝑖 ≥ 0, ( 𝑗1 , … , 𝑗𝑑 ) ≠ 0 . Then (19) becomes 1+𝜖 4

𝜑

1+𝜖 3 , 1+𝜖 4

𝑎𝑘

𝑥 =− 𝑘= 1+𝜖 3

2𝜋

𝑖 𝑑𝑒 2

𝑗 𝑥11

=−

𝑚 +𝜖 0 𝑘 ,𝑥

···

𝑗 𝑥𝑑𝑑

𝑗1 ! · · · 𝑗𝑑 !

( 𝑗 1 ,…,𝑗 𝑑 )∈ 𝐽

𝑖 𝑗 1 ,…,𝑗 𝑑 ( 𝑗 1 ,…,𝑗 𝑑 )∈ 𝐽 1+𝜖 4

𝑖 𝑗 1 ,…,𝑗 𝑑 𝑘= 1+𝜖 3

(𝑖𝛿𝑘 1 𝑥1 )𝑗 1 · · · · · (𝑖𝛿𝑘 𝑑 𝑥𝑑 )𝑗 𝑑 𝑗1 ! · · · · · 𝑗𝑑 !

𝑎𝑘 2𝜋

𝑗

𝑗

𝑑 2

𝛿𝑘11 · · · · · 𝛿𝑘 𝑑𝑑 𝑒 𝑖

𝑚 +𝜖 0 𝑘 ,𝑥

,

so 𝜑

1+𝜖 3 , 1+𝜖 4

𝑥

≤ ( 𝑗 1 ,…,𝑗 𝑑 )∈ 𝐽

𝜋 𝑗 1 ,…,𝑗 𝑑 𝑗1 ! · · · · · 𝑗𝑑 !

1+𝜖 4 𝑗

𝑗

𝑎𝑘 𝛿𝑘11 · · · · · 𝛿𝑘 𝑑𝑑

1 2 1+𝜖 3 , 1+𝜖 4

𝑥

2

𝑑𝑡

2

≤

2𝜋

𝑑 2

.

1 2

𝑕𝑗 1 ,…,𝑗 𝑑 (𝑥) 𝑑𝑥 [−𝜋,𝜋]𝑑 ( 𝑗 1 ,…,𝑗 𝑑 )∈ 𝐽

[−𝜋,𝜋]𝑑

𝑚 +𝜖 0 𝑘 ,𝑥

𝑘= 1+𝜖 3

For brevity denote the outer summand above by 𝑕𝒋𝟏 ,…,𝒋𝒅 (𝑡). Then 𝜑

𝑒𝑖

1 2 2

≤

𝑕𝑗 1 ,…,𝑗 𝑑 (𝑥) 𝑑𝑥 ( 𝑗 1 ,…,𝑗 𝑑 )∈ 𝐽

,

[−𝜋,𝜋]𝑑

so that

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56 | Page

Multidimensional of gradually series bandlimited functions and frames

đ?&#x153;&#x2018;

1+đ?&#x153;&#x2013; 3 , 1+đ?&#x153;&#x2013; 4

2

1+đ?&#x153;&#x2013; 4

đ?&#x153;&#x2039; đ?&#x2018;&#x2014; 1 ,â&#x20AC;Ś,đ?&#x2018;&#x2014; đ?&#x2018;&#x2018; đ?&#x2018;&#x2014;1 ! ¡ ¡ ¡ đ?&#x2018;&#x2014;đ?&#x2018;&#x2018; !

â&#x2030;¤ ( đ?&#x2018;&#x2014; 1 ,â&#x20AC;Ś,đ?&#x2018;&#x2014; đ?&#x2018;&#x2018; )â&#x2C6;&#x2C6; đ??˝

( đ?&#x2018;&#x2014; 1 ,â&#x20AC;Ś,đ?&#x2018;&#x2014; đ?&#x2018;&#x2018; )â&#x2C6;&#x2C6; đ??˝

đ?&#x153;&#x2039; đ?&#x2018;&#x2014;1 ! ¡ ¡ ¡ ¡ ¡ đ?&#x2018;&#x2014;đ?&#x2018;&#x2018; !

1+đ?&#x153;&#x2013; 4

( đ?&#x2018;&#x2014; 1 ,â&#x20AC;Ś,đ?&#x2018;&#x2014; đ?&#x2018;&#x2018; )â&#x2C6;&#x2C6; đ??˝

đ?&#x153;&#x2039; đ?&#x2018;&#x2014;1 ! ¡ ¡ ¡ đ?&#x2018;&#x2014;đ?&#x2018;&#x2018; ! đ?&#x153;&#x2039; đ?&#x2018; đ?&#x2018;˘đ?&#x2018;? 1+đ?&#x153;&#x2013; 3

=

đ?&#x2018;&#x17D;đ?&#x2018;&#x2DC;

â&#x2C6;&#x17E;

= đ?&#x2018;&#x2122;=1

đ?&#x153;&#x2039; đ?&#x2018; đ?&#x2018;˘đ?&#x2018;? 1+đ?&#x153;&#x2013; 3

đ?&#x2018;&#x2014;

đ?&#x203A;żđ?&#x2018;&#x2DC;11

2

2(đ?&#x2018;&#x2014; 1 ,â&#x20AC;Ś,đ?&#x2018;&#x2014; đ?&#x2018;&#x2018; ) 2

đ?&#x2018;&#x17D;đ?&#x2018;&#x2DC;

sup

đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 đ?&#x2018;&#x2014;1 ! ¡ ¡ ¡ ¡ ¡ đ?&#x2018;&#x2014;đ?&#x2018;&#x2018; !

â&#x2030;¤đ?&#x2018;&#x2DC;â&#x2030;¤ 1+đ?&#x153;&#x2013; 4

đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0

â&#x2030;¤đ?&#x2018;&#x2DC;â&#x2030;¤ 1+đ?&#x153;&#x2013; 4

đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0

1+đ?&#x153;&#x2013; 3 â&#x2030;¤đ?&#x2018;&#x2DC;â&#x2030;¤ 1+đ?&#x153;&#x2013; 4

đ?&#x2018;&#x2DC;= 1+đ?&#x153;&#x2013; 3

đ?&#x2018;&#x2DC;

â&#x2C6;&#x2019; đ?&#x2018;Ąđ?&#x2018;&#x2DC;

â&#x2C6;&#x17E;

đ?&#x2018;&#x2014; â&#x201E;&#x201C;=0

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x2DC; â&#x2C6;&#x2019;đ?&#x2018;Ą đ?&#x2018;&#x2DC; â&#x2C6;&#x17E;

đ?&#x2018;&#x2DC;

â&#x2C6;&#x2019; đ?&#x2018;Ąđ?&#x2018;&#x2DC;

1 2

đ?&#x2018;&#x17D;đ?&#x2018;&#x2DC;

2

đ?&#x2018;&#x2DC;= 1+đ?&#x153;&#x2013; 3 đ?&#x2018;&#x2DC;

â&#x2C6;&#x2019; đ?&#x2018;Ąđ?&#x2018;&#x2DC;

â&#x2C6;&#x17E;

â&#x2C6;&#x2019;1

đ?&#x2018;&#x17D;đ?&#x2018;&#x2DC;

đ?&#x2018;&#x17D;đ?&#x2018;&#x2DC;

2

2

đ?&#x2018;&#x2DC;= 1+đ?&#x153;&#x2013; 3

1 2

â&#x2C6;&#x2019; 1

1 2

1+đ?&#x153;&#x2013; 4

đ?&#x2018;&#x2014;â&#x201E;&#x201C;

1 2

â&#x2C6;&#x17E;

1+đ?&#x153;&#x2013; 4

đ?&#x2018;&#x2014; 1 ,â&#x20AC;Ś,đ?&#x2018;&#x2014; đ?&#x2018;&#x2018;

1+đ?&#x153;&#x2013; 4

=

2

đ?&#x2018;&#x2014;

¡ ¡ ¡ ¡ ¡ đ?&#x203A;żđ?&#x2018;&#x2DC; đ?&#x2018;&#x2018;đ?&#x2018;&#x2018;

đ?&#x2018;&#x2014;â&#x201E;&#x201C; !

đ?&#x153;&#x2039;đ?&#x2018;&#x2018; đ?&#x2018; đ?&#x2018;˘đ?&#x2018;? 1+đ?&#x153;&#x2013; 3 â&#x2030;¤đ?&#x2018;&#x2DC; â&#x2030;¤ 1+đ?&#x153;&#x2013; 4 đ?&#x2018;&#x2019;

1 2

đ?&#x2018;&#x2DC;= 1+đ?&#x153;&#x2013; 3

( đ?&#x2018;&#x2014; 1 ,â&#x20AC;Ś,đ?&#x2018;&#x2014; đ?&#x2018;&#x2018; )â&#x2C6;&#x2C6; đ??˝ đ?&#x2018;&#x2018;

2

đ?&#x2018;&#x2018;đ?&#x2018;Ľ

đ?&#x2018;&#x2018; 2

2đ?&#x153;&#x2039;

1+đ?&#x153;&#x2013; 4

đ?&#x2018;&#x2014; 1 ,â&#x20AC;Ś,đ?&#x2018;&#x2014; đ?&#x2018;&#x2018;

â&#x2030;¤

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x2DC; ,đ?&#x2018;Ľ

[â&#x2C6;&#x2019;đ?&#x153;&#x2039;,đ?&#x153;&#x2039;]đ?&#x2018;&#x2018; đ?&#x2018;&#x2DC;= 1+đ?&#x153;&#x2013; 3

đ?&#x2018;&#x2014; 1 ,â&#x20AC;Ś,đ?&#x2018;&#x2014; đ?&#x2018;&#x2018;

=

đ?&#x2018;&#x2019;đ?&#x2018;&#x2013;

đ?&#x2018;&#x2014;

đ?&#x2018;&#x2014;

đ?&#x2018;&#x17D;đ?&#x2018;&#x2DC; đ?&#x203A;żđ?&#x2018;&#x2DC;11 ¡ ¡ ¡ ¡ đ?&#x203A;żđ?&#x2018;&#x2DC; đ?&#x2018;&#x2018;đ?&#x2018;&#x2018;

1 2

2

.

đ?&#x2018;&#x2DC;= 1+đ?&#x153;&#x2013; 3

Corollary 5.4 .Let ( đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 đ?&#x2018;&#x2DC; )đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;â&#x201E;&#x2022; be an enumeration of đ?&#x2022;Ťđ?&#x2018;&#x2018; , and let (đ?&#x2018;Ąđ?&#x2018;&#x2DC; )đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;â&#x201E;&#x2022; â&#x160;&#x201A; â&#x201E;&#x201A;đ?&#x2018;&#x2018; such that 1 đ?&#x2018;&#x2013; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x2DC; ,đ?&#x2018;Ľ sup đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 đ?&#x2018;&#x2DC; â&#x2C6;&#x2019; đ?&#x2018;Ąđ?&#x2018;&#x2DC; â&#x2C6;&#x17E; = đ??ż < â&#x2C6;&#x17E; . Define đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; , đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; â&#x2C6;ś â&#x201E;&#x201A;đ?&#x2018;&#x2018; â&#x2020;&#x2019; đ?&#x2022;&#x201D; by đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; đ?&#x2018;Ľ = and đ?&#x2018;&#x2018; đ?&#x2018;&#x2019; đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;â&#x201E;&#x2022;

đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; đ?&#x2018;Ľ =

2đ?&#x153;&#x2039; 2

1 đ?&#x2018;&#x2018;

2đ?&#x153;&#x2039; 2

đ?&#x2018;&#x2019; đ?&#x2018;&#x2013; đ?&#x2018;Ą đ?&#x2018;&#x2DC; ,đ?&#x2018;Ľ .Then the map đ?&#x2018;&#x2021; â&#x2C6;ś đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;]đ?&#x2018;&#x2018; â&#x2020;&#x2019; đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;]đ?&#x2018;&#x2018; , defined by

đ?&#x2018;&#x2021;đ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 = đ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2019; đ?&#x2018;&#x201C; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 , satisfies the following estimate: đ?&#x2018;&#x2021; â&#x2030;¤ đ?&#x2018;&#x2019; đ?&#x153;&#x2039;đ??żđ?&#x2018;&#x2018; â&#x2C6;&#x2019; 1 .

(20) Proof .Lemma 5.3 shows that đ?&#x2018;&#x2021; is uniformly continuous on a dense subset of the ball in đ??ż2 (đ??¸), so đ?&#x2018;&#x2021; is bounded on đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;]đ?&#x2018;&#x2018; . The inequality (20) follows immediately.

Corollary đ?&#x;&#x201C;. đ?&#x;&#x201C; .Let ( đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 đ?&#x2018;&#x2DC; )đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;â&#x201E;&#x2022; , (đ?&#x2018;Ąđ?&#x2018;&#x2DC; )đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;â&#x201E;&#x2022; â&#x160;&#x201A; â&#x201E;&#x201A;đ?&#x2018;&#x2018; , and let đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; , đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; and đ?&#x2018;&#x2021; be defined as in Corollary 5.4. For each đ?&#x2018;&#x2122; â&#x2C6;&#x2C6; â&#x201E;&#x2022;, define đ?&#x2018;&#x2021;đ?&#x2018;&#x2122; by đ?&#x2018;&#x2021;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; = đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; â&#x2C6;&#x2019; đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; for 1 â&#x2030;¤ đ?&#x2018;&#x2DC; â&#x2030;¤ đ?&#x2018;&#x2122;, and đ?&#x2018;&#x2021;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; = 0 for đ?&#x2018;&#x2122; < đ?&#x2018;&#x2DC;. If limđ?&#x2018;&#x2DC; â&#x2020;&#x2019;â&#x2C6;&#x17E; đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 đ?&#x2018;&#x2DC; â&#x2C6;&#x2019; đ?&#x2018;Ąđ?&#x2018;&#x2DC; â&#x2C6;&#x17E; = 0, then limđ?&#x2018;&#x2122;â&#x2020;&#x2019;â&#x2C6;&#x17E; đ?&#x2018;&#x2021;đ?&#x2018;&#x2122; = đ?&#x2018;&#x2021; in the operator norm. In particular, T is a compact operator.

Proof . As â&#x2C6;&#x17E;

(đ?&#x2018;&#x2021; â&#x2C6;&#x2019; đ?&#x2018;&#x2021;đ?&#x2018;&#x2122; )

â&#x2C6;&#x17E;

đ?&#x2018;&#x17D;đ?&#x2018;&#x2DC; đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; =

đ?&#x2018;&#x2122;

đ?&#x2018;&#x17D;đ?&#x2018;&#x2DC; (đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; â&#x2C6;&#x2019; đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; ) â&#x2C6;&#x2019;

đ?&#x2018;&#x2DC;=1

đ?&#x2018;&#x2DC;=1 â&#x2C6;&#x17E;

đ?&#x2018;&#x17D;đ?&#x2018;&#x2DC; (đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; â&#x2C6;&#x2019; đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; ) đ?&#x2018;&#x2DC;=1

=

đ?&#x2018;&#x17D;đ?&#x2018;&#x2DC; (đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; â&#x2C6;&#x2019; đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; ) = đ?&#x2018;&#x2021; đ?&#x2018;&#x2DC;=đ?&#x2018;&#x2122;+1

đ?&#x2018;&#x17D;đ?&#x2018;&#x2DC; đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; , đ?&#x2018;&#x2DC;=đ?&#x2018;&#x2122;+1

the estimate derived in Lemma 5.3 yields â&#x2C6;&#x17E;

đ?&#x2018;&#x2021; â&#x20AC;&#x201C; đ?&#x2018;&#x2021;đ?&#x2018;&#x2122;

â&#x2C6;&#x17E;

â&#x2C6;&#x17E;

đ?&#x2018;&#x17D;đ?&#x2018;&#x2DC; đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC;

= đ?&#x2018;&#x2021;

đ?&#x2018;&#x2DC;=1

đ?&#x2018;&#x17D;đ?&#x2018;&#x2DC; đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; đ?&#x2018;&#x2DC;=đ?&#x2018;&#x2122;+1

2

â&#x2030;¤

đ?&#x2018;&#x2019; đ?&#x153;&#x2039;đ?&#x2018;&#x2018; đ?&#x2018; đ?&#x2018;˘đ?&#x2018;? đ?&#x2018;&#x2DC; â&#x2030;Ľđ?&#x2018;&#x2122;+1 đ?&#x153;šđ?&#x2019;&#x152; â&#x2C6;&#x17E;

2

â&#x2C6;&#x17E;

â&#x2C6;&#x2019; 1

đ?&#x2018;&#x17D;đ?&#x2018;&#x2DC; đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; đ?&#x2018;&#x2DC;=1

, 2

so (đ?&#x2018;&#x2021; â&#x2C6;&#x2019; đ?&#x2018;&#x2021;đ?&#x2018;&#x2122; ) 2 â&#x2020;&#x2019; 0 as đ?&#x2018;&#x2122; â&#x2020;&#x2019; â&#x2C6;&#x17E;. As đ?&#x2018;&#x2021;đ?&#x2018;&#x2122; has finite rank, we deduce (see [13]) that đ?&#x2018;&#x2021; is compact .

The following proof due to [13] . proof of Theorem 5.1 . Step 1: đ??ľ is a compact perturbation of the identity map, namely đ??ľ = đ??ź + lim â&#x2C6;&#x2019;đ?&#x2018;&#x192;đ?&#x2018;&#x2122; + đ?&#x2018;&#x192;đ?&#x2018;&#x2122; đ??ľ â&#x2C6;&#x2019;1 đ?&#x2018;&#x192;đ?&#x2018;&#x2122; â&#x2C6;&#x2019;1 . đ?&#x2018;&#x2018;

â&#x2C6;&#x2014;

đ?&#x2018;&#x2122;â&#x2020;&#x2019;â&#x2C6;&#x17E;

(21)

Since (đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; )đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;â&#x201E;&#x2022; is a Riesz basis for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;] , đ??ż = (đ??ź â&#x2C6;&#x2019; đ?&#x2018;&#x2021; ) is an onto isomorphism where đ?&#x2018;&#x2021;đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; = đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; â&#x2C6;&#x2019; đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; ; so đ??ľ simplifies to (đ??ź â&#x2C6;&#x2019; đ?&#x2018;&#x2021;)â&#x2C6;&#x2019;1 (đ??ź â&#x2C6;&#x2019; đ?&#x2018;&#x2021; â&#x2C6;&#x2014; )â&#x2C6;&#x2019;1 . We examine đ??ľ â&#x2C6;&#x2019;1 = đ??ź â&#x2C6;&#x2019; đ?&#x2018;&#x2021; â&#x2C6;&#x2014; đ??ź â&#x2C6;&#x2019; đ?&#x2018;&#x2021; = đ??ź + đ?&#x2018;&#x2021; â&#x2C6;&#x2014; đ?&#x2018;&#x2021; â&#x2C6;&#x2019; đ?&#x2018;&#x2021; â&#x2C6;&#x2019; đ?&#x2018;&#x2021; â&#x2C6;&#x2014; = đ??ź + â&#x2C6;&#x2020; ,where â&#x2C6;&#x2020; is a compact operator. If an operator â&#x2C6;&#x2020;â&#x2C6;ś đ??ť â&#x2020;&#x2019; đ??ť is compact then so is â&#x2C6;&#x2020;â&#x2C6;&#x2014; , hence đ?&#x2018;&#x192;đ?&#x2018;&#x2122; â&#x2C6;&#x2020;đ?&#x2018;&#x192;đ?&#x2018;&#x2122; â&#x2020;&#x2019; â&#x2C6;&#x2020; in the operator norm because

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57 | Page

Multidimensional of gradually series bandlimited functions and frames

We have 𝐵 −1

𝑃𝑙 ∆𝑃𝑙 − ∆ ≤ = = lim𝑙→∞ 𝐼 + 𝑃𝑙 ∆𝑃𝑙 = lim 𝐼 −

𝑃𝑙 ∆𝑃𝑙 − 𝑃𝑙 ∆ + 𝑃𝑙 ∆ − ∆ ≤ ∆𝑃𝑙 − ∆ + 𝑃𝑙 ∆ − ∆ 𝑃𝑙 ∆∗ − ∆∗ + 𝑃𝑙 ∆ − ∆ → 0 . = lim𝑙→∞ 𝐼 + 𝑃𝑙 𝐵 −1 − 𝐼 𝑃𝑙 𝑃𝑙 + 𝑃𝑙 𝐵 −1 𝑃𝑙 .

𝑙→∞

Now 𝑃𝑙 𝐵 −1 𝑃𝑙 restricted to the first 𝑙 rows and columns is the Grammian matrix for the set ( 𝑓1 , . . . , 𝑓𝑙 ) which can be shown (in a straightforward manner) to be linearly independent. We conclude that 𝑃𝑙 𝐵 −1 𝑃𝑙 is invertible as an 𝑙 × 𝑙 matrix. By 𝑃𝑙 𝐵 −1 𝑃𝑙 −1 we mean the inverse as an l × l matrix and zeroes elsewhere. Observing that the ranges of 𝑃𝑙 𝐵 −1 𝑃𝑙 and 𝑃𝑙 𝐵 −1 𝑃𝑙 −1 are in the kernel of 1 − 𝑃𝑙 , and that the range of 𝐼 − 𝑃𝑙 is in the kernels of 𝑃𝑙 𝐵 −1 𝑃𝑙 and 𝑃𝑙 𝐵 −1 𝑃𝑙 −1 , we easily compute 𝐼 − 𝑃𝑙 + 𝑃𝑙 𝐵 −1 𝑃𝑙 −1 −1 = 𝐼 − 𝑃𝑙 + 𝑃𝑙 𝐵 −1 𝑃𝑙 , so that 𝐵 −1 = lim𝑙→∞ 𝐼 − 𝑃𝑙 + 𝑃𝑙 𝐵 −1 𝑃𝑙 −1 −1 , implying 𝐵 = lim 𝐼 − 𝑃𝑙 + 𝑃𝑙 𝐵 −1 𝑃𝑙 −1 = lim 𝐵𝑙 = lim − 𝑃𝑙 + 𝑃𝑙 𝐵 −1 𝑃𝑙 −1 . 𝑙→∞

𝑙→∞

𝑙→∞

Step 2: We verify (16) and its convergence properties. Recalling (11), we have ∞ 1 𝑡𝑘 𝑓𝑗 (𝑡) − 𝐼 – 𝑃𝑙 + 𝑃𝑙 𝐵 −1 𝑃𝑙 −1 𝑓𝑆/ 1+𝜖 2 𝑔𝑗 𝑡 – 𝑘 1 + 𝜖2 𝑑 1 + 𝜖2 𝑗

𝑘=1 𝑗

implying 1 𝑓𝑗 (𝑡) − 1 + 𝜖2

𝑗

∞

1 1 + 𝜖2

=

(𝐵 − 𝐵𝑙 ) 𝑓𝑆/ 1+𝜖 2

𝑑

𝑙

𝑃𝑙 𝐵 −1 𝑃𝑙

𝑑 𝑘=1 𝑗

1 1 + 𝜖2

−1

𝑓𝑆/ 1+𝜖2

𝑡𝑘 1 + 𝜖2

𝑔𝑗 𝑡 –

𝑘

∞

𝐵 – 𝐵𝑙 𝑓𝑆/ 1+𝜖2

𝑑 𝑗

𝑘=1

+

1 1 + 𝜖2

𝑘

𝑘=1 𝑗

𝑘

=

𝑡𝑘 1 + 𝜖2

𝑔𝑗 𝑡 –

∞

𝑓𝑗

𝑡𝑘 1 + 𝜖2

𝑔𝑗 . –

𝑡𝑘 1 + 𝜖2

𝑑 𝑘=𝑙+1 𝑗

𝑡𝑘 1 + 𝜖2

𝑔𝑗 𝑡 −

𝑔𝑗 𝑡 −

𝑡𝑘 1 + 𝜖2

.

Therefore,

𝑗

𝑙

1 𝑓𝑗 . − 1 + 𝜖2 =

𝑃𝑙 𝐵 −1 𝑃𝑙

𝑑

𝑓𝑆/ 1+𝜖 2

𝑘

𝑘=1 𝑗 ∞

1 1 + 𝜖2

(𝐵 − 𝐵𝑙 ) 𝑓𝑆/ 1+𝜖 2

𝑑 𝑘=1 𝑗

1 + 1 + 𝜖2 1 = 1 + 𝜖2

−1

𝑔 𝑘 𝑗

∞

𝑡𝑘 1 + 𝜖2

.−

𝑡𝑘 1 + 𝜖2

𝑓𝑗

𝑑 𝑘=𝑙+1 𝑗

𝑔𝑗 . −

∞ 𝑑

ℱ

−1

𝑔𝑗

.

𝐵 – 𝐵𝑙 𝑓𝑆/ 1+𝜖2 𝑘=1

∞

+

𝑡𝑘 1 + 𝜖2

𝑓𝑗 𝑘=𝑙+1 𝑗

𝑒

𝑖 . ,

2

𝑘

𝑒

𝑖. ,

𝑡𝑘 1 + 𝜖2

[− 1+𝜖 2 𝜋, 1+𝜖 2 𝜋]𝑑

𝑡𝑘 1+𝜖 2

𝑡𝑘 1+𝜖 2 [− 1+𝜖 2 𝜋, 1+𝜖 2 𝜋]𝑑

after taking the inverse Fourier transform. Now

𝑗

1 𝑓𝑗 . − 1 + 𝜖2

𝑙

𝑃𝑙 𝐵 −1 𝑃𝑙

𝑑

−1

𝑓𝑆/ 1+𝜖2

𝑘=1 𝑗

1 ≤ 1 + 𝜖2

𝑘

𝑔𝑗 . –

∞

𝐵 – 𝐵𝑙 𝑓𝑆/ 1+𝜖 2

𝑑 𝑘=1

+

1 1 + 𝜖2

𝑘

𝑒

𝑖. ,

2

𝑡𝑘 1+𝜖 2 [− 1+𝜖 2 𝜋, 1+𝜖 2 𝜋]𝑑

∞

𝑓𝑗

𝑑

𝑡𝑘 1 + 𝜖2

𝑘=𝑙+1 𝑗

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𝑡𝑘 1 + 𝜖2

𝑒

𝑖 . ,

𝑡𝑘 1+𝜖 2 [− 1+𝜖 2 𝜋, 1+𝜖 2 𝜋]𝑑

58 | Page

Multidimensional of gradually series bandlimited functions and frames

â&#x2030;¤ đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC;

since đ?&#x2018;&#x201C;đ?&#x2018;&#x2014;

¡

đ?&#x2018;&#x2018;

đ??ľ â&#x20AC;&#x201C; đ??ľđ?&#x2018;&#x2122; đ?&#x2018;&#x201C;đ?&#x2018;&#x2020;/ 1+đ?&#x153;&#x2013;2

â&#x201E;&#x201C;2 (â&#x201E;&#x2022;)

+

đ?&#x2018;&#x20AC; 1 + đ?&#x153;&#x2013;2

đ?&#x2018;&#x201C;đ?&#x2018;&#x2014;

đ?&#x2018;&#x2018; đ?&#x2018;&#x2DC;=đ?&#x2018;&#x2122;+1 đ?&#x2018;&#x2014;

đ?&#x2018;Ąđ?&#x2018;&#x2DC; 1 + đ?&#x153;&#x2013;2

2

1 2

,

is a Riesz basis for đ??ż2 [â&#x2C6;&#x2019; 1 + đ?&#x153;&#x2013;2 đ?&#x153;&#x2039;, 1 + đ?&#x153;&#x2013;2 đ?&#x153;&#x2039;]đ?&#x2018;&#x2018; . Since đ??ľđ?&#x2019;? â&#x2020;&#x2019; đ??ľ as đ?&#x2018;&#x2122; â&#x2020;&#x2019; â&#x2C6;&#x17E; and

1+đ?&#x153;&#x2013; 2

đ?&#x2018;Ąđ?&#x2018;&#x2DC; 1+đ?&#x153;&#x2013; 2

đ?&#x2018;&#x20AC; 1 + đ?&#x153;&#x2013;2

â&#x2C6;&#x17E;

đ?&#x2018;&#x2DC; 2

â&#x2C6;&#x2C6; â&#x201E;&#x201C; (â&#x201E;&#x2022;), the last two terms in the inequality above tend to zero, which proves the required đ?&#x2018;&#x2DC;

result. Finally, to compute đ?&#x2018;&#x192;đ?&#x2018;&#x2122; đ??ľ â&#x2C6;&#x2019;1 đ?&#x2018;&#x192;đ?&#x2018;&#x2122; đ?&#x2018;&#x203A;đ?&#x2018;&#x161; , recall that đ??ľ â&#x2C6;&#x2019;1 = (đ??ź â&#x2C6;&#x2019; đ?&#x2018;&#x2021; â&#x2C6;&#x2014; )(đ??ź â&#x2C6;&#x2019; đ?&#x2018;&#x2021; ). Proceeding in a manner similar to the proof of (10), we obtain â&#x2C6;&#x2019;1 â&#x2C6;&#x2014; â&#x2C6;&#x2014; â&#x2C6;&#x2014; đ??ľđ?&#x2018;&#x161; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 = đ??żđ??ż đ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 , đ?&#x2018;&#x2019;đ?&#x2018;&#x161; = đ??ż đ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 , đ??ż đ?&#x2018;&#x2019;đ?&#x2018;&#x161; = đ?&#x2018;&#x201C; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 , đ?&#x2018;&#x201C;đ?&#x2018;&#x161; = đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;?đ?&#x153;&#x2039; đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ,1 â&#x2C6;&#x2019; đ?&#x2018;Ąđ?&#x2018;&#x161; ,1 ¡ ¡ ¡ ¡ ¡ đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;?đ?&#x153;&#x2039; đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ,đ?&#x2018;&#x2018; â&#x2C6;&#x2019; đ?&#x2018;Ąđ?&#x2018;&#x161; ,đ?&#x2018;&#x2018; .

The entries of đ?&#x2018;&#x192;đ?&#x2018;&#x2122; đ??ľ â&#x2C6;&#x2019;1 đ?&#x2018;&#x192;đ?&#x2018;&#x2122; agree with those of đ??ľ â&#x2C6;&#x2019;1 when 1 â&#x2030;¤ đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 , đ?&#x2018;&#x161; â&#x2030;¤ đ?&#x2018;&#x2122;.

One generalization of Kadecâ&#x20AC;&#x2122;s 1/4 theorem given by Pak and Shin in [11] (which is actually a special case of Avdoninâ&#x20AC;&#x2122;s theorem) is: Theorem 5.6 . Let (đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 )đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;đ?&#x2022;Ť â&#x160;&#x201A; â&#x201E;&#x201A; be a sequence of distinct points such that 1 lim đ?&#x2018; đ?&#x2018;˘đ?&#x2018;?| đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 â&#x2C6;&#x2019; đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 | = đ??ż < . | đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 |â&#x2020;&#x2019;â&#x2C6;&#x17E; 4 1 Then the sequence of functions ( đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; )đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;đ?&#x2022;Ť , defined by đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; đ?&#x2018;Ľ = đ?&#x2018;&#x2019; đ?&#x2018;&#x2013;đ?&#x2018;Ą đ?&#x2018;&#x2DC; đ?&#x2018;Ľ , 2đ?&#x153;&#x2039; is a Riesz basis for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;]. Theorem 5.6 shows that in the univariate case of Theorem 5.1 the restriction that ( đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; )đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;â&#x201E;&#x2022; is a Riesz basis for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;]can be dropped. The following example shows that the multivariate case is very different. Let đ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 be an orthonormal basis for a Hilbert space đ??ť. Let đ?&#x2018;&#x201C;1 â&#x2C6;&#x2C6; đ??ť with đ?&#x2018;&#x201C;1 = đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

1 , then ( đ?&#x2018;&#x201C;1 , đ?&#x2018;&#x2019;2 , đ?&#x2018;&#x2019;3 , . . . ) is a Riesz basis for đ??ť if and only if đ?&#x2018;&#x201C;1 , đ?&#x2018;&#x2019;1 â&#x2030; 0. Verifying that the map đ?&#x2018;&#x2021;, given by đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; â&#x;ź đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; for đ?&#x2018;&#x2DC; > 1 and đ?&#x2018;&#x2019;1 â&#x;ź đ?&#x2018;&#x201C;1 , is a continuous bijection is routine, so đ?&#x2018;&#x2021; is an isomorphism via the Open Mapping theorem. In the language of Theorem 5.1, ( đ?&#x2018;&#x201C;1 , đ?&#x2018;&#x2019;2 , đ?&#x2018;&#x2019;3 , . . . ) is a Riesz basis for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;] if and only if 0 â&#x2030; đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;? đ?&#x153;&#x2039;đ?&#x2018;Ą1,1 ¡ ¡ ¡ ¡ ¡ đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;? đ?&#x153;&#x2039;đ?&#x2018;Ą1,đ?&#x2018;&#x2018; , that is, if and only if đ?&#x2018;Ą1 â&#x2C6;&#x2C6; (â&#x201E;&#x201A;\{Âą1, Âą2, . . . })đ?&#x2018;&#x2018; .

VI. Generalizations of Kadecâ&#x20AC;&#x2122;s 1/4 theorem Corollary 5.4 yields the following generalization of Kadecâ&#x20AC;&#x2122;s theorem in d dimensions (see [13]) .

Corollary 6.1 .Let ( đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 đ?&#x2018;&#x2DC; )đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;â&#x201E;&#x2022; be an enumeration of đ?&#x2022;Ťđ?&#x2018;&#x2018; and let ( đ?&#x2018;Ąđ?&#x2018;&#x2DC; )đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;â&#x201E;&#x2022; â&#x160;&#x201A; â&#x201E;&#x201A;đ?&#x2018;&#x2018; such that ln(2) sup đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 đ?&#x2018;&#x2DC; â&#x2C6;&#x2019; đ?&#x2018;Ąđ?&#x2018;&#x2DC; â&#x2C6;&#x17E; = đ??ż < . đ?&#x153;&#x2039;đ?&#x2018;&#x2018; đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;đ?&#x2022;Ť 1 Then the sequence ( đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; )đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;â&#x201E;&#x2022; defined by đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; (đ?&#x2018;Ľ) = đ?&#x2018;&#x2019; đ?&#x2018;&#x2013; đ?&#x2018;Ľ,đ?&#x2018;Ą đ?&#x2018;&#x2DC; is a Riesz basis for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;]đ?&#x2018;&#x2018; . đ?&#x2018;&#x2018; /2

(22)

(2đ?&#x153;&#x2039;)

The proof is immediate. Note that (20) implies that the map đ?&#x2018;&#x2021; given in Corollary 5.4 has norm less than 1. We conclude that the map (đ??ź â&#x2C6;&#x2019; đ?&#x2018;&#x2021; )đ?&#x2018;&#x2019;đ?&#x2018;&#x2DC; = đ?&#x2018;&#x201C;đ?&#x2018;&#x2DC; is invertible by considering its Neumann series.

The proof of Corollary 5.4 and Corollary 6.1 are straightforward generalizations of the univariate result proved

by Duffin and Eachus [8]. Kadec improved the value of the constant in the inequality (22) ln (2) (for đ?&#x2018;&#x2018; = 1) from to the optimal value of 1/4; this is his celebrated â&#x20AC;&#x153;1/4 theoremâ&#x20AC;? [10].Kadecâ&#x20AC;&#x2122;s method of đ?&#x153;&#x2039; đ?&#x2018;&#x2013;đ?&#x203A;żđ?&#x2018;Ľ proof is to expand đ?&#x2018;&#x2019; with respect to the orthogonal basis 1 1, đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; ( đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 đ?&#x2018;Ľ), đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A; đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 â&#x2C6;&#x2019; đ?&#x2018;Ľ 2 đ?&#x2018;&#x161; +đ?&#x153;&#x2013; â&#x2C6;&#x2C6;â&#x201E;&#x2022; 0

for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;] , and use this expansion to estimate the norm of T .

In the proof of Corollary 5.4 and Corollary 6.1 we simply used a Taylor series (see [13]). Unlike the estimates in Kadecâ&#x20AC;&#x2122;s theorem, the estimate in (20) can be used for any sequence ( đ?&#x2018;Ąđ?&#x2018;&#x2DC; )đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;â&#x201E;&#x2022; â&#x160;&#x201A; â&#x201E;&#x201A;đ?&#x2018;&#x2018; such that đ?&#x2018; đ?&#x2018;˘đ?&#x2018;?đ?&#x2018;&#x2DC;â&#x2C6;&#x2C6;đ?&#x2022;Ť đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 đ?&#x2018;&#x2DC; â&#x2C6;&#x2019; đ?&#x2018;Ąđ?&#x2018;&#x2DC; â&#x2C6;&#x17E; = đ??ż < â&#x2C6;&#x17E;, not only those for which the exponentials (đ?&#x2018;&#x2019; đ?&#x2018;&#x2013;đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;Ľ ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 form a Riesz basis. An impressive generalization of Kadecâ&#x20AC;&#x2122;s 1/4 theorem when đ?&#x2018;&#x2018; = 1 is Avdoninâ&#x20AC;&#x2122;s â&#x20AC;&#x153;1/4 in the meanâ&#x20AC;? theorem [1].Sun and Zhou (see [12] second half of Theorem 1.3) refined Kadecâ&#x20AC;&#x2122;s argument to obtain a partial generalization of his result in higher dimensions: 1 Theorem 6.2 .Let (đ?&#x2018;&#x17D; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2C6;đ?&#x2022;Ťđ?&#x2018;&#x2018; â&#x160;&#x201A; â&#x201E;&#x201A;đ?&#x2018;&#x2018; such that 0 < đ??ż < , 4

đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x153;&#x2039;đ??ż đ??ˇđ?&#x2018;&#x2018; đ??ż = 1 â&#x2C6;&#x2019; đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; đ?&#x153;&#x2039;đ??ż + đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x153;&#x2039;đ??ż + đ?&#x153;&#x2039;đ??ż

đ?&#x2018;&#x2018;

â&#x2C6;&#x2019;

đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x153;&#x2039;đ??ż đ?&#x153;&#x2039;đ??ż

đ?&#x2018;&#x2018;

,

and

www.ijesi.org

59 | Page

Multidimensional of gradually series bandlimited functions and frames đ?&#x2018;&#x17D; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2019; đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0

â&#x2C6;&#x17E;

đ?&#x2018;&#x2018;

â&#x2030;¤ đ??ż, đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 â&#x2C6;&#x2C6; đ?&#x2022;Ťđ?&#x2018;&#x2018; . If đ??ˇđ?&#x2018;&#x2018; đ??ż < 1, then 2

2

1

đ?&#x2018;&#x2019; đ?&#x2018;&#x2013; đ?&#x2018;&#x17D; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

(2đ?&#x153;&#x2039;)đ?&#x2018;&#x2018;

,(¡)

is a Riesz basis for

đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;] with frame bounds (1 â&#x2C6;&#x2019; đ??ˇđ?&#x2018;&#x2018; đ??ż ) and (1 + đ??ˇđ?&#x2018;&#x2018; đ??ż ) .In the one-dimensional case, Kadecâ&#x20AC;&#x2122;s theorem 1 is recovered exactly from Theorem 6.2. When đ?&#x2018;&#x2018; > 1, the value đ?&#x2018;Ľđ?&#x2018;&#x2018; satisfying 0 < đ?&#x2018;Ľđ?&#x2018;&#x2018; < and đ??ˇđ?&#x2018;&#x2018; (đ?&#x2018;Ľđ?&#x2018;&#x2018; ) = 1 is 1

an upper bound for any value of đ??ż satisfying 0 < đ??ż < apparent, whereas the constant in Corollary 6.1 is

ln 2 đ?&#x2018;Ľđ?&#x2018;&#x2018;

4

4

and đ??ˇđ?&#x2018;&#x2018; (đ??ż) < 1. The value of đ?&#x2018;Ľđ?&#x2018;&#x2018; is not readily

. A relationship between this number and đ?&#x2018;Ľđ?&#x2018;&#x2018; is given in

the following theorem (whose proof is omitted).

1

Theorem 6.3 .Let đ?&#x2018;Ľđ?&#x2018;&#x2018; be the unique number satisfying 0 < đ?&#x2018;Ľđ?&#x2018;&#x2018; < and đ??ˇđ?&#x2018;&#x2018; (đ?&#x2018;Ľđ?&#x2018;&#x2018; ) = 1. Then 4 ln 2 đ?&#x2018;Ľđ?&#x2018;&#x2018; â&#x2C6;&#x2019; đ?&#x153;&#x2039;đ?&#x2018;&#x2018; lim = 1 . đ?&#x2018;&#x2018;â&#x2020;&#x2019;â&#x2C6;&#x17E; (ln 2)2 2 12đ?&#x153;&#x2039;đ?&#x2018;&#x2018; Thus, for sufficiently large đ?&#x2018;&#x2018;, Theorem 6.2 and Corollary 6.1 are essentially the same.

7. A method of approximation of biorthogonal functions and a recovery of a theorem of Levinson In this section we apply (see [13]) the techniques developed previously to approximate the biorthogonal functions to Riesz bases

1 2đ?&#x153;&#x2039;

đ?&#x2018;&#x2019; đ?&#x2018;&#x2013;đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

(¡)

for which the synthesis operator is small perturbation of the identity.

Definition 7.1. A Kadec sequence is a sequence (đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 sup

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2C6;đ?&#x2022;Ť

đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2019; đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0

= đ??ˇ <

Theorem 7.2 .Let (đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 1

( đ?&#x2018;&#x201C; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 =

2đ?&#x153;&#x2039;

â&#x2C6;&#x2C6;đ?&#x2022;Ť đ?&#x2018;&#x2019; đ?&#x2018;&#x2013;đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 (¡)

1 4

of real numbers satisfying

â&#x2C6;&#x2C6;đ?&#x2022;Ť

.

â&#x160;&#x201A; â&#x201E;&#x201A; be a sequence (with đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2030; 0 for đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

â&#x2030; 0) such that

is a Riesz basis for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;], and let ( đ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 be the

standard exponential orthonormal basis for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;]. If the map L given by đ??żđ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 = đ?&#x2018;&#x201C; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 satisfies the estimate đ??ź â&#x2C6;&#x2019; đ??ż < 1, then the biorthogonals đ??ş đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 of in đ?&#x2018;&#x192;đ?&#x2018;&#x160;[â&#x2C6;&#x2019;đ?&#x153;&#x2039;,đ?&#x153;&#x2039;] are đ??ş đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;Ą =

1

2đ?&#x153;&#x2039;

â&#x201E;ą( đ?&#x2018;&#x201C; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 )(¡) = đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;?(đ?&#x153;&#x2039;(¡ â&#x2C6;&#x2019;đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ))

đ??ť đ?&#x2018;Ą đ?&#x2018;Ą â&#x20AC;&#x201C; đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ??ťâ&#x20AC;˛ đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

where

â&#x2C6;&#x17E;

đ??ť đ?&#x2018;Ą = đ?&#x2018;Ą â&#x2C6;&#x2019; đ?&#x2018;Ą0 đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 =1

Definition 7.3 . Let (đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 1

( đ?&#x2018;&#x201C; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 = (đ?&#x2018;Ąđ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

â&#x2C6;&#x2C6;đ?&#x2022;Ť

Define đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 =

2đ?&#x153;&#x2039;

1â&#x2C6;&#x2019;

â&#x2C6;&#x2C6;đ?&#x2022;Ť

đ?&#x2018;&#x2019; đ?&#x2018;&#x2013;đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 (¡)

,

đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 â&#x2C6;&#x2C6; đ?&#x2022;Ť , đ?&#x2018;Ą

đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

1â&#x2C6;&#x2019;

đ?&#x2018;Ą đ?&#x2018;Ąâ&#x2C6;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

(23)

.

(24)

â&#x160;&#x201A; â&#x201E;&#x201A; be a sequence such that

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

is a Riesz basis for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;]. If đ?&#x2018;&#x2122; â&#x2030;Ľ 0, theđ?&#x2018;&#x2122;-truncated sequence

is defined by đ?&#x2018;Ąđ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 = đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 if | đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 | â&#x2030;¤ đ?&#x2018;&#x2122; and đ?&#x2018;Ąđ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 = đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 otherwise.

1

2đ?&#x153;&#x2039;

đ?&#x2018;&#x2019; đ?&#x2018;&#x2013;đ?&#x2018;Ą đ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

(¡)

for đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 â&#x2C6;&#x2C6; đ?&#x2022;Ť, đ?&#x2018;&#x2122; â&#x2030;Ľ 0.

Let đ?&#x2018;&#x192;đ?&#x2018;&#x2122; â&#x2C6;ś đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;] â&#x2020;&#x2019; đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;] be the orthogonal projection onto span{đ?&#x2018;&#x2019;â&#x2C6;&#x2019;đ?&#x2018;&#x2122; , . . . , đ?&#x2018;&#x2019;đ?&#x2018;&#x2122; }.

Proposition 7.4 .Let (đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2C6;đ?&#x2022;Ť â&#x160;&#x201A; â&#x201E;&#x201A; be a sequence such that ( đ?&#x2018;&#x201C; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 (defined above) is a Riesz basis for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;]. If ( đ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 is the standard exponential orthonormal basis for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;] and the map đ??ż (defined above) satisfies the estimate đ??ź â&#x2C6;&#x2019; đ??ż = đ?&#x203A;ż < 1, then the following are true: (i) For đ?&#x2018;&#x2122; â&#x2030;Ľ 0, the sequence ( đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 is a Riesz basis forđ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;]. (ii) For đ?&#x2018;&#x2122; â&#x2030;Ľ 0, the map đ??żđ?&#x2018;&#x2122; defined by đ??żđ?&#x2018;&#x2122; đ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 = đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 satisfies đ??żâ&#x2C6;&#x2019;1 đ?&#x2018;&#x2122; Proof .If ( đ?&#x2018;? đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2C6; â&#x201E;&#x201C;2 (đ?&#x2022;Ť), then đ??ź â&#x2C6;&#x2019; đ??żđ?&#x2018;&#x2122;

đ?&#x2018;? đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

=

â&#x2030;¤

đ?&#x2018;? đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2019; đ??żđ?&#x2018;&#x2122; đ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

1

1â&#x2C6;&#x2019;đ?&#x203A;ż

.

=

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

đ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2019; đ?&#x2018;&#x201C; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2030;¤đ?&#x2018;&#x2122;

= (đ??ź â&#x2C6;&#x2019; đ??ż)đ?&#x2018;&#x192;đ?&#x2018;&#x2122;

đ?&#x2018;? đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

,

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

so that đ??ź â&#x2C6;&#x2019; đ??żđ?&#x2018;&#x2122; = đ??ź â&#x2C6;&#x2019; đ??ż đ?&#x2018;&#x192;đ?&#x2018;&#x2122; .

From this, đ??ź â&#x2C6;&#x2019; đ??żđ?&#x2018;&#x2122; â&#x2030;¤ đ?&#x203A;ż, which implies (i) and (ii).

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Multidimensional of gradually series bandlimited functions and frames Define the biorthogonal functions of ( đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 to be ( đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;,â&#x2C6;&#x2014; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 . Passing to the Fourier transform 1

, we have

2đ?&#x153;&#x2039;

â&#x201E;ą( đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 )(đ?&#x2018;Ą) = đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;?(đ?&#x153;&#x2039;(đ?&#x2018;Ą â&#x2C6;&#x2019; đ?&#x2018;Ąđ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 )) and

đ??şđ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 (đ?&#x2018;Ą) =

1 2đ?&#x153;&#x2039;

â&#x201E;ą( đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;,â&#x2C6;&#x2014; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 )(đ?&#x2018;Ą). Define the biorthogonal functions of ( đ?&#x2018;&#x201C; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 )

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

similarly.

Lemma 7.5 . If ( đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x160;&#x201A; â&#x201E;&#x201A; satisfies the hypotheses of Proposition 7.4, then lim đ??şđ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 = đ??ş đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 in đ?&#x2018;&#x192;đ?&#x2018;&#x160;[â&#x2C6;&#x2019;đ?&#x153;&#x2039;,đ?&#x153;&#x2039;] . đ?&#x2018;&#x2122;â&#x2020;&#x2019;â&#x2C6;&#x17E;

â&#x2C6;&#x2014; â&#x2C6;&#x2014; â&#x2C6;&#x2014; â&#x2C6;&#x2014; Proof . Note that đ?&#x203A;ż đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x161; = đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 , đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;,đ?&#x2018;&#x161; = đ??żđ?&#x2018;&#x2122; đ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 , đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;,đ?&#x2018;&#x161; = đ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 , đ??żâ&#x2C6;&#x2014;đ?&#x2018;&#x2122; đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;,đ?&#x2018;&#x161; so that for all đ?&#x2018;&#x161;, đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;,đ?&#x2018;&#x161; = â&#x2C6;&#x2014; â&#x2C6;&#x2019;1 â&#x2C6;&#x2014; â&#x2C6;&#x2014; â&#x2C6;&#x2019;1 (đ??żđ?&#x2018;&#x2122; ) đ?&#x2018;&#x2019;đ?&#x2018;&#x161; . Similarly, đ?&#x2018;&#x201C;đ?&#x2018;&#x161; = (đ??ż ) đ?&#x2018;&#x2019;đ?&#x2018;&#x161; . We have â&#x2C6;&#x2014; đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;,đ?&#x2018;&#x161; â&#x2C6;&#x2019; đ?&#x2018;&#x201C;đ?&#x2018;&#x161;â&#x2C6;&#x2014; = (đ??żâ&#x2C6;&#x2014;đ?&#x2018;&#x2122; )â&#x2C6;&#x2019;1 (đ??żâ&#x2C6;&#x2014; )â&#x2C6;&#x2019;1 đ?&#x2018;&#x2019;đ?&#x2018;&#x161; = (đ??żâ&#x2C6;&#x2014;đ?&#x2018;&#x2122; )â&#x2C6;&#x2019;1 đ??żâ&#x2C6;&#x2014; â&#x2C6;&#x2019; đ??żâ&#x2C6;&#x2014;đ?&#x2018;&#x2122; (đ??żâ&#x2C6;&#x2014; )â&#x2C6;&#x2019;1 đ?&#x2018;&#x2019;đ?&#x2018;&#x161; . Now (84) implies â&#x2C6;&#x2014; đ??ż â&#x2C6;&#x2019; đ??żđ?&#x2018;&#x2122; = (đ??ź â&#x2C6;&#x2019; đ?&#x2018;&#x192;đ?&#x2018;&#x2122; )(đ??ż â&#x2C6;&#x2019; đ??ź), so that đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;,đ?&#x2018;&#x161; â&#x2C6;&#x2019; đ?&#x2018;&#x201C;đ?&#x2018;&#x161;â&#x2C6;&#x2014; = đ??żâ&#x2C6;&#x2014;đ?&#x2018;&#x2122; â&#x2C6;&#x2019;1 đ??żâ&#x2C6;&#x2014; â&#x2C6;&#x2019; đ??ź đ??ź â&#x2C6;&#x2019; đ?&#x2018;&#x192;đ?&#x2018;&#x2122; đ??żâ&#x2C6;&#x2014; â&#x2C6;&#x2019;1 đ?&#x2018;&#x2019;đ?&#x2018;&#x161; . Applying Proposition 1

7.4 yields đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;â&#x2C6;&#x2014;,đ?&#x2018;&#x161; â&#x2C6;&#x2019; đ?&#x2018;&#x201C;đ?&#x2018;&#x161;â&#x2C6;&#x2014; â&#x2030;¤ đ??żâ&#x2C6;&#x2014; â&#x2C6;&#x2019; đ??ź (đ??ź â&#x2C6;&#x2019; đ?&#x2018;&#x192;đ?&#x2018;&#x2122; )(đ??żâ&#x2C6;&#x2014; )â&#x2C6;&#x2019;1 đ?&#x2018;&#x2019;đ?&#x2018;&#x161; , which for fixed đ?&#x2018;&#x161; goes to 0 as đ?&#x2018;&#x2122; â&#x2020;&#x2019; â&#x2C6;&#x17E;. We 1â&#x2C6;&#x2019; đ?&#x203A;ż â&#x2C6;&#x2014; conclude limđ?&#x2018;&#x2122;â&#x2020;&#x2019;â&#x2C6;&#x17E; đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;,đ?&#x2018;&#x161; = đ?&#x2018;&#x201C;đ?&#x2018;&#x161;â&#x2C6;&#x2014; , which, upon passing to the Fourier transform, yields limđ?&#x2018;&#x2122;â&#x2020;&#x2019;â&#x2C6;&#x17E; đ??şđ?&#x2018;&#x2122;,đ?&#x2018;&#x161; = đ??şđ?&#x2018;&#x161; .

Proof of Theorem 7.2. We see that đ?&#x203A;ż đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x161; = đ??şđ?&#x2018;&#x2122;,đ?&#x2018;&#x161; , đ?&#x2018;&#x2020;đ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 , where đ?&#x2018;&#x2020;đ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 (đ?&#x2018;Ą) = đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;?(đ?&#x153;&#x2039;(đ?&#x2018;Ą â&#x2C6;&#x2019; đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 )) when | đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 | â&#x2030;¤ đ?&#x2018;&#x2122; and đ?&#x2018;&#x2020;đ?&#x2018;&#x2122;, đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 (đ?&#x2018;Ą) = đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;?(đ?&#x153;&#x2039;(đ?&#x2018;Ą â&#x2C6;&#x2019; đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 )) when |đ?&#x2018;&#x161;| > đ?&#x2018;&#x2122;. Without loss of generality, let |đ?&#x2018;&#x161;| < đ?&#x2018;&#x2122;. (1) implies that đ??şđ?&#x2018;&#x2122;,đ?&#x2018;&#x161; (đ?&#x2018;&#x2DC;) = 0 when |đ?&#x2018;&#x2DC;| > đ?&#x2018;&#x2122;. By the WKS theorem we have đ?&#x2018;&#x2DC;=đ?&#x2018;&#x2122;

đ?&#x2018;&#x2DC;=đ?&#x2018;&#x2122;

đ??şđ?&#x2018;&#x2122;,đ?&#x2018;&#x161; đ?&#x2018;Ą =

đ??şđ?&#x2018;&#x2122;,đ?&#x2018;&#x161; đ?&#x2018;&#x2DC; đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;? đ?&#x153;&#x2039; đ?&#x2018;Ą â&#x20AC;&#x201C; đ?&#x2018;&#x2DC;

â&#x2C6;&#x2019;1

=

đ?&#x2018;&#x2DC;=â&#x2C6;&#x2019;đ?&#x2018;&#x2122;

đ?&#x2018;&#x2DC;=â&#x2C6;&#x2019;đ?&#x2018;&#x2122;

=

đ?&#x2018;&#x2122; đ?&#x2018;&#x2DC;=1

đ?&#x2018;&#x2DC;â&#x2C6;&#x2019;1

đ?&#x2018;Ąđ??şđ?&#x2018;&#x2122;,đ?&#x2018;&#x161; đ?&#x2018;&#x2DC; đ?&#x2018;&#x2DC; â&#x2C6;&#x2019;đ?&#x2018;Ą

đ?&#x2018;¤đ?&#x2018;&#x2122; đ?&#x2018;Ą

đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;? đ?&#x153;&#x2039;đ?&#x2018;Ą

đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;? đ?&#x153;&#x2039;đ?&#x2018;Ą ,

đ?&#x2018;&#x2DC; â&#x20AC;&#x201C; đ?&#x2018;Ą â&#x2C6;&#x2019;đ?&#x2018;&#x2DC; â&#x20AC;&#x201C; đ?&#x2018;Ą

where đ?&#x2018;¤đ?&#x2018;&#x2122; is a polynomial of degree at most 2đ?&#x2018;&#x2122;. Noting that â&#x2C6;&#x17E;

đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;? đ?&#x153;&#x2039;đ?&#x2018;Ą = đ?&#x2018;&#x2DC; =1

đ?&#x2018;&#x2122;

đ?&#x2018;Ą2 1â&#x2C6;&#x2019; 2 đ?&#x2018;&#x2DC;

and

(đ?&#x2018;&#x2DC; â&#x2C6;&#x2019; đ?&#x2018;Ą)(â&#x2C6;&#x2019;đ?&#x2018;&#x2DC; â&#x2C6;&#x2019; đ?&#x2018;Ą) = (â&#x2C6;&#x2019;1) (đ?&#x2018;&#x2122;!) đ?&#x2018;&#x2DC; =1

we have đ??şđ?&#x2018;&#x2122;,đ?&#x2018;&#x161; Again by (1), đ?&#x203A;ż đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

đ?&#x2018;&#x2122; đ?&#x2018;&#x2122;

â&#x2C6;&#x2019;1 đ?&#x2018;&#x2122; đ?&#x2018;¤đ?&#x2018;&#x2122; đ?&#x2018;Ą đ?&#x2018;Ą = đ?&#x2018;&#x2122;! 2

2

1â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;=1

â&#x2C6;&#x17E;

1â&#x2C6;&#x2019; đ?&#x2018;&#x2DC;=đ?&#x2018;&#x2122;+1

đ?&#x2018;Ą2 , đ?&#x2018;&#x2DC;2

đ?&#x2018;Ą2 . đ?&#x2018;&#x2DC;2

= đ??şđ?&#x2018;&#x2122;,đ?&#x2018;&#x161; (đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) when | đ?&#x2018;&#x161; + đ?&#x153;&#x2013;0 | â&#x2030;¤ đ?&#x2018;&#x2122; so that â&#x2C6;&#x17E; đ?&#x2018;Ą 2đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2019;1 đ?&#x2018;&#x2122; đ?&#x203A;ż đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x161; = đ?&#x2018;¤ đ?&#x2018;Ą 1â&#x2C6;&#x2019; . đ?&#x2018;&#x2122;! 2 đ?&#x2018;&#x2122; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x2DC;2 đ?&#x2018;&#x161;

đ?&#x2018;&#x2DC;=đ?&#x2018;&#x2122;+1

This determines the zeroes of đ?&#x2018;¤đ?&#x2018;&#x2122; . We deduce that đ?&#x2018;?đ?&#x2018;&#x2122; đ?&#x2018;&#x2DC;=đ?&#x2018;&#x2122; đ?&#x2018;&#x2DC;=1 đ?&#x2018;Ą â&#x20AC;&#x201C; đ?&#x2018;Ąđ?&#x2018;&#x2DC; đ?&#x2018;Ą â&#x20AC;&#x201C; đ?&#x2018;Ąâ&#x2C6;&#x2019;đ?&#x2018;&#x2DC; đ?&#x2018;¤đ?&#x2018;&#x2122; đ?&#x2018;Ą = đ?&#x2018;Ą â&#x2C6;&#x2019; đ?&#x2018;Ąđ?&#x2018;&#x161; for some constant đ?&#x2018;?đ?&#x2018;&#x2122; . Absorbing constants, we have đ??şđ?&#x2018;&#x2122;,đ?&#x2018;&#x161; đ?&#x2018;Ą = đ?&#x2018;&#x2122;

đ??ťđ?&#x2018;&#x2122; đ?&#x2018;Ą = đ?&#x2018;Ą â&#x2C6;&#x2019; đ?&#x2018;Ą0 Now 0 = đ??ťđ?&#x2018;&#x2122; (đ?&#x2018;Ąđ?&#x2018;&#x161; ), so đ??şđ?&#x2018;&#x2122;,đ?&#x2018;&#x161; (đ?&#x2018;Ą) = đ?&#x2018;?đ?&#x2018;&#x2122; đ?&#x2018;?đ?&#x2018;&#x2122; =

1 đ??ťđ?&#x2018;&#x2122; â&#x20AC;˛ đ?&#x2018;Ąđ?&#x2018;&#x161;

1â&#x2C6;&#x2019;

đ?&#x2018;&#x2DC;=1 đ??ť đ?&#x2018;&#x2122; đ?&#x2018;Ą â&#x2C6;&#x2019;đ??ť đ?&#x2018;&#x2122; (đ?&#x2018;Ą đ?&#x2018;&#x161; )

. This yields đ??şđ?&#x2018;&#x2122;,đ?&#x2018;&#x161; đ?&#x2018;Ą =

đ?&#x2018;Ąâ&#x2C6;&#x2019;đ?&#x2018;Ą đ?&#x2018;&#x161; đ??ťđ?&#x2018;&#x2122; đ?&#x2018;Ą

đ?&#x2018;Ą đ?&#x2018;Ąđ?&#x2018;&#x2DC;

1â&#x2C6;&#x2019; đ?&#x2018;&#x2DC; =1

đ?&#x2018;Ą đ?&#x2018;Ąâ&#x2C6;&#x2019;đ?&#x2018;&#x2DC;

đ?&#x2018;Ą â&#x2C6;&#x2019;đ?&#x2018;Ą đ?&#x2018;&#x161; â&#x2C6;&#x17E;

, where

1â&#x2C6;&#x2019; đ?&#x2018;&#x2122;+1

đ?&#x2018;Ą2 . đ?&#x2018;&#x2DC;2

. Taking limits,

đ?&#x2018;Ą â&#x20AC;&#x201C;đ?&#x2018;Ą đ?&#x2018;&#x161; đ??ťđ?&#x2018;&#x2122;â&#x20AC;˛ đ?&#x2018;Ą đ?&#x2018;&#x161; â&#x2C6;&#x17E;

đ??ť đ?&#x2018;Ą = đ?&#x2018;Ą â&#x2C6;&#x2019; đ?&#x2018;Ą0

1â&#x2C6;&#x2019;

đ?&#x2018;?đ?&#x2018;&#x2122; đ??ťđ?&#x2018;&#x2122; đ?&#x2018;Ą

. Define đ?&#x2018;Ą đ?&#x2018;Ąđ?&#x2018;&#x2DC;

1â&#x2C6;&#x2019;

đ?&#x2018;Ą . đ?&#x2018;Ąâ&#x2C6;&#x2019;đ?&#x2018;&#x2DC;

Basic complex analysis shows that đ??ť is entire, and đ??ťđ?&#x2018;&#x2122; â&#x2020;&#x2019; đ??ť and đ??ťđ?&#x2018;&#x2122;â&#x20AC;˛ â&#x2020;&#x2019; đ??ťâ&#x20AC;˛ uniformly on compact subsets of đ??ś.

Furthermore, đ??ťâ&#x20AC;˛ (đ?&#x2018;Ąđ?&#x2018;&#x2DC; ) â&#x2030; 0 for all đ?&#x2018;&#x2DC;, since each đ?&#x2018;Ąđ?&#x2018;&#x2DC; is a zero of đ??ť of multiplicity one. Together we have đ??ť(đ?&#x2018;Ą) lim đ??şđ?&#x2018;&#x2122;,đ?&#x2018;&#x161; (đ?&#x2018;Ą) = , đ?&#x2018;Ą â&#x2C6;&#x2C6; â&#x201E;&#x201A;. đ?&#x2018;&#x2122;â&#x2020;&#x2019;â&#x2C6;&#x17E; (đ?&#x2018;Ą â&#x2C6;&#x2019; đ?&#x2018;Ąđ?&#x2018;&#x161; )đ??ťâ&#x20AC;˛ (đ?&#x2018;Ąđ?&#x2018;&#x161; ) By the foregoing lemma, đ??şđ?&#x2018;&#x2122;,đ?&#x2018;&#x161; â&#x2020;&#x2019; đ??şđ?&#x2018;&#x161; . Observing that convergence in đ?&#x2018;&#x192;đ?&#x2018;&#x160;[â&#x2C6;&#x2019;đ?&#x153;&#x2039;,đ?&#x153;&#x2039;] implies pointwise convergence yields the desired result.Levinson proved a version of Theorem 7.2 in the case where (đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2C6;đ?&#x2022;Ť is a Kadec sequence. His original proof is found in [11, pp. 47â&#x20AC;&#x201C;67]. We recall that if (đ?&#x2018;&#x201C; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 is a Riesz basis arising from a Kadec sequence, then the synthesis operator L satisfies đ??ź â&#x2C6;&#x2019; đ??ż < 1. Levinsonâ&#x20AC;&#x2122;s theorem is then recovered from Theorem 7.2.

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Multidimensional of gradually series bandlimited functions and frames Now we show the following Corollary Corollary 7.6. Let (2đ?&#x2018;Ą đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2C6;đ?&#x2022;Ť â&#x160;&#x201A; â&#x201E;&#x201A; be a sequence and đ?&#x2018;&#x201C;22 đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

is a Riesz basis for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;].

If ( đ?&#x2018;&#x2019;2 đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 ) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 is the standard exponential orthonormal basis for đ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;] and the map đ??ż2 satisfies the estimate đ??ź â&#x2C6;&#x2019; đ??ż2 = đ?&#x203A;ż < 1, then the following are hold: 2 (i) For đ?&#x2018;&#x2122; â&#x2030;Ľ 0, the sequence đ?&#x2018;&#x201C;2đ?&#x2018;&#x2122;,2 is a Riesz basis forđ??ż2 [â&#x2C6;&#x2019;đ?&#x153;&#x2039;, đ?&#x153;&#x2039;]. đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

2 (ii) For đ?&#x2018;&#x2122; â&#x2030;Ľ 0, the map đ??ż22đ?&#x2018;&#x2122; defined by đ??ż22đ?&#x2018;&#x2122; đ?&#x2018;&#x2019;2 đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 = đ?&#x2018;&#x201C;2đ?&#x2018;&#x2122;,2

Proof . For ( đ?&#x2018;?2 đ??ź â&#x2C6;&#x2019; đ??ż22đ?&#x2018;&#x2122;

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

đ?&#x2018;?2

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

satisfies đ??żâ&#x2C6;&#x2019;2 2đ?&#x2018;&#x2122;

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

â&#x2C6;&#x2019; đ??ż22đ?&#x2018;&#x2122; đ?&#x2018;&#x2019;2

â&#x2030;¤

) đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2C6;&#x2C6; â&#x201E;&#x201C;2 (đ?&#x2022;Ť), we have

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

đ?&#x2018;&#x2019;2

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

=

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

đ?&#x2018;?2

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

đ?&#x2018;&#x2019;2

1 1â&#x2C6;&#x2019;đ?&#x203A;ż 2

.

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

=

đ?&#x2018;&#x2019;2

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

â&#x2C6;&#x2019; đ?&#x2018;&#x201C;22 đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

= (đ??ź â&#x2C6;&#x2019; đ??ż2 )đ?&#x2018;&#x192;2đ?&#x2018;&#x2122;

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 â&#x2030;¤đ?&#x2018;&#x2122;

đ?&#x2018;?2

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

đ?&#x2018;&#x2019;2

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

,

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

so that đ??ź â&#x2C6;&#x2019; đ??ż22đ?&#x2018;&#x2122; = đ??ź â&#x2C6;&#x2019; đ??ż2 đ?&#x2018;&#x192;2đ?&#x2018;&#x2122; .

2 Hence đ??ź â&#x2C6;&#x2019; đ??ż2đ?&#x2018;&#x2122; â&#x2030;¤ đ?&#x203A;ż, which gives (i) and (ii). Hence from Definition 2.4 we can show that 2 đ??żđ?&#x2018;&#x2019; đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 2 2 2 đ??´ đ?&#x2018;&#x201C;đ?&#x2018;&#x2014; â&#x2030;¤ đ?&#x2018;&#x201C;đ?&#x2018;&#x2014; , đ?&#x2018;&#x201C;2đ?&#x2018;&#x2122;,2 â&#x2030;¤ đ??´ + đ?&#x153;&#x2013;1 đ?&#x2018;&#x201C;đ?&#x2018;&#x2014; , for every đ?&#x2018;&#x201C;đ?&#x2018;&#x2014; â&#x2C6;&#x2C6; đ??ť, đ?&#x153;&#x2013;1 > 0 . đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0 2 đ??ż2đ?&#x2018;&#x2122; đ?&#x2018;&#x2019;2 đ?&#x2018;&#x161; +đ?&#x153;&#x2013;0 đ?&#x2018;&#x2014;

đ?&#x2018;&#x161; +đ?&#x153;&#x2013; 0

đ?&#x2018;&#x2014;

đ?&#x2018;&#x2014;

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