ResearchArticle Regularity of Dual Gabor Windowskoasas.kaist.ac.kr/bitstream/10203/223390/1/000325289000001.pdf · 2 AbstractandAppliedAnalysis scaling we obtain the function𝑔(𝑥)

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 747268 8 pageshttpdxdoiorg1011552013747268

Research ArticleRegularity of Dual Gabor Windows

Ole Christensen1 Hong Oh Kim2 and Rae Young Kim3

1 Department of Applied Mathematics and Computer Science Technical University of Denmark Building 3032800 Lyngby Denmark

2Department of Mathematical Sciences KAIST 373-1 Guseong-Dong Yuseong-Gu Daejeon 305-701 Republic of Korea3 Department of Mathematics Yeungnam University 214-1 Dae-Dong Gyeongsan-SiGyeongsangbuk-Do 712-749 Republic of Korea

Correspondence should be addressed to Rae Young Kim rykimynuackr

Received 29 May 2013 Accepted 25 August 2013

Academic Editor Anna Mercaldo

Copyright copy 2013 Ole Christensen et alThis is an open access article distributed under theCreative CommonsAttribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We present a construction of dual windows associated with Gabor frames with compactly supported windows The size of thesupport of the dual windows is comparable to that of the given window Under certain conditions we prove that there exist dualwindows with higher regularity than the canonical dual window On the other hand there are cases where no differentiable dualwindow exists even in the overcomplete case As a special case of our results we show that there exists a common smooth dualwindow for an interesting class of Gabor frames In particular for any value of 119870 isin N there is a smooth function ℎ whichsimultaneously is a dual window for all B-spline generated Gabor frames 119864

119898119887119879119899119861119873(1199092)

119898119899isinN for B-splines 119861119873

of order119873 = 1 2119870 + 1 with a fixed and sufficiently small value of 119887

1 Introduction

A frame 119891119896 in a separable Hilbert space H leads to

expansions of arbitrary elements 119891 isin H in a similar fashionas the well-known orthonormal bases More precisely thereexists at least one so-called dual frame that is a frame ℎ

119896

such that

119891 = sum⟨119891 ℎ119896⟩ 119891119896 forall119891 isin H (1)

Unless 119891119896 is a basis the dual ℎ

119896 is not unique This makes

it natural to search for duals with special prescribed prop-erties In this paper we will consider Gabor frames withtranslation parameter 119886 = 1 that is frames for 1198712(R)that for a certain fixed parameter 119887 gt 0 and a fixedfunction 119892 isin 119871

2

(R) have the form 119864119898119887119879119899119892119898119899isinZ =

1198902120587119894119898119887119909

119892(119909 minus 119899)119898119899isinZ The function 119892 is called the win-

dow function We will construct dual frames of the form119864119898119887119879119899ℎ119898119899isinZ = 119890

2120587119894119898119887119909

ℎ(119909 minus 119899)119898119899isinZ for a suitable func-

tion ℎ isin 1198712(R) to be called the dual windowIt is known that a frame 119864

119898119887119879119899119892119898119899isinZ will be over-

complete if 119887 lt 1 with the redundancy increasing when 119887

decreasesThus the dualwindow is not uniqueWewill inves-tigate whether this freedom can be used to find dual windowswith higher regularity than the canonical dual and compa-rable size of the support We will present cases where this ispossible and other cases where it is not As a special case ofour results we show that there are certain classes of inter-esting frames that have the same dual window For examplefor any value of 119870 isin N there is a smooth function ℎ whichsimultaneously is a dual window for all B-spline generatedGabor frames 119864

119898119887119879119899119861119873(1199092)

119898119899isinN for B-splines119861119873 of order119873 = 1 2119870 + 1 with a fixed and sufficiently small value of119887

Just to give the reader an impression of the results tocome consider the B-spline119861

2The function119861

2is continuous

but not differentiable at the points 0 plusmn1 The Gabor system1198641198981198871198791198991198612119898119899isinZ is a frame for all sufficiently small values of

119887 gt 0 As 119887 tends to zero the redundancy of the frame1198641198981198871198791198991198612119898119899isinZ increases meaning that we get more and

more freedom in the choice of a dual window ℎ Howeverwe will show that it is impossible to find a differentiable dualwindow ℎ supported on supp119861

2= [minus1 1] regardless of the

considered 119887 gt 0 On the other hand by a seemingly innocent

2 Abstract and Applied Analysis

scaling we obtain the function 119892(119909) = 1198612(1199092) Again

119864119898119887119879119899119892119898119899isinZ is a frame for all sufficiently small values of

119887 gt 0 But in contrast with the situation for 1198612 we will show

that one can find infinitely often differentiable dual windowsℎ that are supported on supp119892 = [minus2 2] for any value of119887 isin ]0 12] These examples illustrate that the question ofdifferentiability of the dual windows is a nontrivial issue Theexamples will be derived as a consequence of more generalresults see Example 9

The results will be based on a construction of dual win-dows as presented in Section 2 In Section 3 we consider aparticular casewhere it is possible to obtain smooth dual win-dows regardless of the regularity of the given window 119892Thisis much more than one can hope for in the general case Ageneral approach to the question of differentiability of thedual windows is given in Section 5 Since the necessary con-ditions are quite involved and not very intuitive we first inSection 4 state a version for the case of windows 119892 that aresupported in [minus1 1] For this case we can provide concreteexamples demonstrating that the desired conclusions mightfail if any of the constraints is removed

Note that a complementary approach to duality for Gaborframes that also dealswith the issue of regularity is consideredby Laugesen [1] and Kim [2] For more information about thetheory for Gabor analysis and its applications see [3ndash5]

2 Construction of Dual Frames

In the literature various characterizations of the pairs of dualGabor frames are available For general frames Li gave a char-acterizationin [6] which in the special case of Gabor framesalso lead to a class of dual Gabor frames later in [7] it wasshown that these duals actually characterize all duals In orderto start our analysis we need the duality conditions for Gaborframes by Ron and Shen [8] and Janssen [9] respectively

Lemma 1 Two Bessel sequences 119864119898119887119879119899119886119892119898119899isinZ and

119864119898119887119879119899119886ℎ119898119899isinZ form dual frames for 1198712(R) if and only if

sum

119896isinZ

119892(119909 +119899

119887+ 119896119886)ℎ (119909 + 119896119886) = 119887120575

1198990 ae 119909 isin [0 119886] (2)

We will use the following to apply Lemma 1

Lemma 2 Let 119866 be a real-valued bounded function andassume that for some constant 119860 gt 0

10038161003816100381610038161003816119866 (119909 minus 1)

10038161003816100381610038161003816+10038161003816100381610038161003816119866 (119909)

10038161003816100381610038161003816ge 119860 119909 isin [0 1] (3)

Then there exists a real-valued bounded function withsupp H sube supp119866 cap [minus1 1] such that

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin [0 1] (4)

Proof Consider 119909 isin [0 1]We will define (119909) and (119909 minus 1)simultaneously In case |119866(119909 minus 1)| ge 1198602 put (119909) = 0 and(119909minus1) = 1119866(119909minus1) On the other hand if |119866(119909minus1)| lt 1198602we know that |119866(119909)| ge 1198602 In this case put (119909 minus 1) =

0 and (119909) = 1119866(119909) Clearly we can take = 0 outside[minus1 1]

Wewill nowpresent a general result about the existence offrames with a dual window of a special form Note that incontrast with most results from the literature we do not needto assume that the integer translates of the window functionform a partition of unity

Associated to a function 119892 with support on an interval[minus(2119870 + 1) 2119870 + 1] 119870 isin N we will in the rest of the paperuse the function

119866 (119909) = sum

119896isinZ

119892 (119909 + 2119896) 119909 isin R (5)

Due to the periodicity of 119866 we are mainly interested in 119909 isin

[minus1 1] Note that by the compact support of 119892

119866 (119909) =

119870

sum

119896=minus119870

119892 (119909 + 2119896) 119909 isin [minus1 1] (6)

Theorem 3 Let 119870 isin N cup 0 and let 119887 isin ]0 1(4119870 + 2)]Let 119892 be a real-valued bounded function with supp119892 sube

[minus(2119870 + 1) 2119870 + 1] for which1003816100381610038161003816100381610038161003816100381610038161003816

sum

119899isinZ

119892 (119909 + 119899)

1003816100381610038161003816100381610038161003816100381610038161003816

ge 119860 119909 isin [0 1] (7)

for a constant 119860 gt 0 Then the following hold

(i) The function 119866 in (5) satisfies the conditions inLemma 2

(ii) Take as in Lemma 2 and let

ℎ (119909) = 119887

119870

sum

119896=minus119870

1198792119896 (119909) (8)

Then 119864119898119887119879119899119892119898119899isinZ and 119864

119898119887119879119899ℎ119898119899isinZ form dual frames

for 1198712(R) and ℎ is supported in [minus(2119870 + 1) 2119870 + 1]

Proof To prove (i) we note that by (7) and the definition of119866 for 119909 isin [0 1]

10038161003816100381610038161003816119866 (119909 minus 1)

10038161003816100381610038161003816+10038161003816100381610038161003816119866 (119909)

10038161003816100381610038161003816=

1003816100381610038161003816100381610038161003816100381610038161003816

sum

119896isinZ

119892 (119909 minus 1 + 2119896)

1003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816

sum

119896isinZ

119892 (119909 + 2119896)

1003816100381610038161003816100381610038161003816100381610038161003816

ge

1003816100381610038161003816100381610038161003816100381610038161003816

sum

119896isinZ

119892 (119909 + 119896)

1003816100381610038161003816100381610038161003816100381610038161003816

ge 119860

(9)

Thus119866 satisfies the condition (3) in Lemma 2 Also it is clearthat 119866 is bounded real-valued and 2-periodicTherefore wecan choose a function which is supported in [minus1 1] andsuch that

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin [0 1] (10)

Define ℎ as in (8) In order to prove (ii) we will applyLemma 1 By assumption the function119892has compact supportand is bounded by the construction the function ℎ sharesthese properties It follows that 119864

119898119887119879119899119892119898119899isinZ and

119864119898119887119879119899ℎ119898119899isinZ are Bessel sequences In order to verify that

Abstract and Applied Analysis 3

these sequences form dual frames we need to check that for119909 isin [0 1]

sum

119896isinZ

119892(119909 +119899

119887+ 119896)ℎ (119909 + 119896) = 119887120575

1198990 ae 119909 isin [0 1] (11)

By assumption and construction 119892 and ℎ have support in[minus(2119870 + 1) 2119870 + 1] thus (11) is satisfied for 119899 = 0 whenever1119887 ge 4119870 + 2 that is if 119887 isin ]0 1(4119870 + 2)] For 119899 = 0 andusing the compact support of 119892 we need to check that

2119870

sum

119896=minus2119870minus1

119892 (119909 + 119896) ℎ (119909 + 119896) = 119887 119909 isin [0 1] (12)

For each 119896 isin minus119870 minus119870 + 1 119870 if 119909 isin [0 1] then

119909 minus 1 + 2119896 isin [2119896 minus 1 2119896] 119909 + 2119896 isin [2119896 2119896 + 1] (13)

Thus we have

ℎ (119909 minus 1 + 2119896) = 119887 (119909 minus 1) ℎ (119909 + 2119896) = 119887 (119909)

(14)

This together with (6) and (10) implies for 119909 isin [0 1] that

2119870

sum


119892 (119909 + 119896) ℎ (119909 + 119896)

=

119870

sum

119896=minus119870

119892 (119909 minus 1 + 2119896) ℎ (119909 minus 1 + 2119896)

+

119870

sum

119896=minus119870

119892 (119909 + 2119896) ℎ (119909 + 2119896)

= 119887

119870

sum

119896=minus119870

119892 (119909 minus 1 + 2119896) (119909 minus 1)

+ 119887

119870

sum

119896=minus119870

119892 (119909 + 2119896) (119909)

= 119887 (119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909)) = 119887

(15)

Hence (12) holds Therefore 119864119898119887119879119899119892119898119899isinZ and

119864119898119887119879119899ℎ119898119899isinZ form dual frames for 1198712(R)

3 Smooth Dual Windows for a Class ofGabor Frames

Before we start the general analysis of the dual windows inTheorem 3 we will consider a particular case where we canconstruct smooth compactly supported dual windows re-gardless of the regularity of the window itself

Theorem 4 Let 119870 isin N cup 0 and let 119887 isin ]0 1(4119870 + 2)] Let119892 be a real-valued bounded function with supp119892 sube [minus(2119870 +

1) 2119870 + 1] for which

sum

119899isinZ

119892 (119909 + 2119899) = 1 119909 isin [minus1 1] (16)

Let 119891 R rarr R be any bounded function for which

119891 (119909) = 0 119909 le 0 119891 (119909) = 1 119909 ge 1 (17)

Define the function by

(119909) =

1

2119891 (2 (119909 + 1)) minus1 le 119909 lt minus

1

2

1 minus1

2119891 (minus2119909) minus

1

2le 119909 lt 0

1 minus1

2119891 (2119909) 0 le 119909 lt

1

2

1

2119891 (2 (1 minus 119909))

1

2le 119909 lt 1

0 otherwise

(18)

and let

ℎ (119909) = 119887

119870

sum

119896=minus119870

1198792119896 (119909) (19)

Then the following holds

(i) ℎ is a symmetric function with supp ℎ sube [minus(2119870 + 1)

2119870 + 1](ii) 119864

119898119887119879119899119892119898119899isinZ and 119864

119898119887119879119899ℎ119898119899isinZ are dual frames for

1198712

(R)(iii) If 119891 is chosen to be smooth then the function in (18)

is smooth and consequently the dual window ℎ in (19)is smooth as well

Proof By the assumption (16) we have

119866 (119909) = 1 119909 isin [minus1 1] (20)

Take as in (18) For 119909 isin ]0 12[

(119909 minus 1) + (119909) =119891 (2119909)

2+ (1 minus

119891 (2119909)

2) = 1 (21)

for 119909 isin ]12 1[

(119909 minus 1) + (119909)

= (1 minus119891 (minus2 (119909 minus 1))

2) +

119891 (2 (1 minus 119909))

2= 1

(22)

That is (119909 minus 1) + (119909) = 1 119909 isin [0 1] Combining this with(20) yields

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909)

= (119909 minus 1) + (119909) = 1 119909 isin [0 1]

(23)

By Theorem 3 ℎ is a dual window of 119892 By construction ℎ isa symmetric function with supp ℎ sube [minus(2119870+ 1) 2119870+ 1] Theresult in (iii) follows by direct investigation of the derivativesat minus1 minus12 0 12 1


An example of a smooth function 119891 satisfying condition(17) is (see [10 page 36])

119891 (119909) =

exp[minusexp [ 119909

(1 minus 119909)] minus 1

minus1

] 0 lt 119909 lt 1

0 119909 le 0

1 119909 ge 1

(24)

As noted in the introduction the possibility of construct-ing a smooth dual window is a significant improvementcompared to the use of the canonical dual window whichmight not even be continuous We return to this point inExample 9

Another interesting feature of the construction inTheorem 4 is that the dual window ℎ in (19) is independentof the window 119892 that generates the frame 119864

119898119887119879119899119892119898119899isinZ In

other words we can construct a window ℎ that generates adual frame for all the frames 119864

119898119887119879119899119892119898119899isinZ satisfying the

conditions inTheorem 4 for fixed values of119870 and 119887

Corollary 5 Let119870 isin N and let 119887 isin ]0 1(4119870 + 2)] Considera bounded real-valued function 120601 that is supported on [minus119870119870]and satisfies the partition of unity condition

sum

119899isinZ

120601 (119909 minus 119899) = 1 119909 isin R (25)

Then the function 119892(119909) = 120601(1199092) generates a Gabor frame119864119898119887119879119899119892119898119899isinZ Choosing as in (18) the function ℎ in (19) is

a dual window of 119892

Proof By the choice of the function 120601 supp119892 sube [minus2119870 2119870] sube[minus2119870 minus 1 2119870 + 1] and

sum

119899isinZ

119892 (119909 minus 2119899) = sum

119899isinZ

120601(119909

2minus 119899) = 1 (26)

Choose as in (18) ByTheorem 4 the function ℎ defined byℎ(119909) = 119887sum

119870

119896=minus1198701198792119896(119909) is a dual window of 119892

It is known that the partition of unity condition (25)is satisfied for a large class of functions for example anyscaling function for a multiresolution analysis As a concreteexample recall that the centered B-splines 119861

119873 119873 isin N are

given inductively by 1198611= 120594[minus1212]

119861119873+1

= 119861119873lowast1198611 Any

B-spline satisfies the partition of unity condition and the B-spline 119861

119873has support on the interval [minus11987321198732] Thus for

each fixed value of 119870 isin N the function ℎ in (19) is a dualwindow for each of the B-splines 119861

119873119873 = 1 2119870 and a

fixed choice of 119887 le 1(4119870 + 2)

4 Regularity of the Dual Windowsif supp119892 sube [minus1 1]

Based onTheorem 3 we now aim at a general analysis of therelationship between the regularity of a window 119892 and theassociated dual windows ℎwith comparable support size We

will exhibit cases where the smoothness can be increased andother cases where these dual windows cannot have highersmoothness than thewindow itselfThe general version of ourresult to be stated in Theorem 10 is quite complicated andthe role of the conditions is not intuitively clearTherefore wewill first present inTheorem 6 the corresponding version forwindows 119892 that are supported in [minus1 1] An advantage of thisapproach is that for each of the requirements inTheorem 6wecan provide an example showing that the desired conclusionmight fail if the condition is removed

Given a function 119892 that is supported in [minus1 1] let

119885 = 119909 isin [minus1 1] 119892 (119909) = 0

119864 = 119909 isin [minus1 1] 119892 is not differentiable at 119909 (27)

Theorem 6 Let 119892 be a real-valued bounded function withsupp119892 sube [minus1 1] Assume that for some constant 119860 gt 0

1003816100381610038161003816119892 (119909 minus 1)1003816100381610038161003816

2

+1003816100381610038161003816119892 (119909)

1003816100381610038161003816

2

ge 119860 119909 isin [0 1] (28)

Then the following assertions hold

(1) If 119892 is not differentiable at 0 then there does not exist adifferentiable dual window ℎ with supp ℎ sube [minus1 1] forany 119887 gt 0

(2) Assume that119892 is differentiable at 0 Assume further thatthe set 119885 is a finite union of intervals and points notcontaining 0 that the set 119864 is finite and that

(a) 119864 cap (119864 + 1) = 0(b) 119885 cap (119864 plusmn 1) = 0

Then for any 119887 isin ]0 12] there exists a differentiabledual window ℎ with supp ℎ sube supp119892

Note that (28) is a necessary condition for 119864119898119887119879119899119892119898119899isinZ

being a frame thus it is not a restriction in this context Tosupport the conditions in Theorem 6 we will now provide aseries of examples where just one of these conditions breaksdown and the conclusion in Theorem 6 fails We first give anexample where condition (a) is not satisfied

Example 7 Consider the function

119892 (119909) =

2119909 + 2 119909 isin [minus1 minus1

2]

1 119909 isin [minus1

21

2]

minus2119909 + 2 119909 isin [1

2 1]

0 119909 notin [minus1 1]

(29)

Then 119870 = 0 119885 = plusmn1 and 119864 = plusmn12 plusmn1 so 119864 cap (119864 + 1) =12 119885cap (119864 plusmn 1) = 0 Hence119892 satisfies condition (b) but not(a) in Theorem 6 Now we will show that 119892 does not have adifferentiable dual of any form Suppose that there exists sucha dual ℎ By the duality condition we obtain

119892 (119909 minus 1) ℎ (119909 minus 1) + 119892 (119909) ℎ (119909) = 119887 119909 isin [0 1] (30)


Letting 119863minus119892 and 119863

+119892 denote the left respectively and right

derivatives of 119892 we note that

119892(minus1

2) = 119892(

1

2) = 1 119863

minus119892(minus

1

2) = 2

119863+119892(minus

1

2) = 0 119863

minus119892(

1

2) = 0 119863

+119892(

1

2) = minus2

(31)

Taking the left and right derivatives of (30) at 119909 = 12 thisimplies that

119863ℎ(minus1

2) + 119863ℎ(

1

2) = minus2ℎ (minus

1

2)

119863ℎ (minus1

2) + 119863ℎ(

1

2) = 2ℎ (

1

2)

(32)

But by conditions (30) and (31) we have ℎ(minus12)+ℎ(12) = 119887This is a contradiction Thus a differentiable dual windowdoes not exist

Note that the conclusion in Example 7 is even strongerthan what we asked for no dual window at all can be differ-entiable regardless of its form and support size In the nextexample condition (b) in Theorem 6 is not satisfied and theconclusion breaks down

Example 8 Consider

119892 (119909) =

5

2119909 +

3

2 119909 isin [minus

3

5 minus1

5]


51

5]

minus5

2119909 +

3

2 119909 isin [

1

53

5]

0 119909 notin [minus3

53

5]

(33)

Easy considerations as in Example 7 show that 119892 satisfiescondition (a) but not (b) in Theorem 6 and that 119892 does nothave a differentiable dual of any form We leave the details tothe interested reader

Let us now provide the details for the example mentionedin Section 1

Example 9 Consider the B-spline 1198612 which is continuous

but not differentiable at the points 0 plusmn1 It is an easyconsequence of the results in the literature (see eg [11]) thattheGabor system 119864

1198981198871198791198991198612119898119899isinZ is a frame for all sufficiently

small values of 119887 gt 0 As 119887 tends to zero the redundancyof the frame 119864

1198981198871198791198991198612119898119899isinZ increases meaning that we get

more andmore freedom in the choice of ℎ However since 1198612

is nondifferentiable at 119909 = 0 Theorem 6 implies that none ofthe dual windows supported on [minus1 1] are differentiable forany 119887 isin ]0 12]

On the other hand consider the scaled B-spline 119892(119909) =1198612(1199092) Again 119864


small values of 119887 gt 0 The requirements in Theorem 4 aresatisfied for 119887 isin ]0 12] implying that infinitely often differ-entiable dual windows exists

5 Regularity of the Dual Windows inthe General Case

We will now present the general version of Theorem 6 Themain difference between the results is that the conditions inthe general version are stated in terms of the function 119866 in(5) rather than 119892 itself The proof is in the appendix Given acompactly supported function 119892 R rarr C define 119866 as in (5)and let

119885 = 119909 isin [minus1 1] 119866 (119909) = 0

119864 = 119909 isin [minus1 1] 119866120594[minus11]

is not differentiable at 119909 (34)

Theorem 10 Let119870 isin N cup 0 and let 119887 isin ]0 1(4119870 + 2)] Let119892 be a real-valued bounded function with supp119892 sube [minus(2119870 +

1) 2119870 + 1] Define 119866 as in (5) Then the following assertionshold

(1) If 119866 is not differentiable at 0 then there does not exist adifferentiable dual window ℎ defined as in (8)Assume that for some constant 119860 gt 0

10038161003816100381610038161003816119866 (119909 minus 1)

10038161003816100381610038161003816+10038161003816100381610038161003816119866 (119909)

10038161003816100381610038161003816ge 119860 119909 isin [0 1] (35)

and that the set 119885 is a finite union of intervals and points notcontaining 0

(2) Assume that119866 is differentiable at 0 that the set of points119864 is finite and that

(a) 119864 cap (119864 minus 1) = 0(b) 119866(119909) = 0 119909 isin (119864 minus 1) cup (119864 + 1)

Then 119864119898119887119879119899119892119898119899isinZ is a frame for 1198712(R) and there exists

a differentiable dual window ℎ of the form (8)

Note that the conditions inTheorem 10 (2) are void if119866 isdifferentiable that is the standing assumptions alone implythe existence of a differentiable dual window

For nonnegative functions the conditions inTheorem 10can be formulated in an easier way where we again referdirectly to properties of the function 119892 rather than 119866

Corollary 11 Let119870 isin Ncup0 and let 119887 isin ]0 1(4119870+2)] Let 119892be a nonnegative bounded function with supp119892 sube [minus(2119870 + 1)

2119870 + 1] Assume that for some constant 119860 gt 0

2119870

sum


119892 (119909 + 119896) ge 119860 119909 isin [0 1] (36)

Assume that the set 119885 of zeros of 119892 on [minus(2119870 + 1) 2119870 + 1]

is a finite union of intervals and points that 119892 is differentiableexcept on a finite set 119864 of points and that the sets 119864 and 119885satisfy the following conditions

(a) 0 notin (⋂119870119896=minus119870

(119885 minus 2119896)) cup (⋃119870

119896=minus119870(119864 minus 2119896))

(b) (⋃119870119896=minus119870

(119864 minus 2119896)) cap (⋃119870

119896=minus119870(119864 minus 2119896 + 1)) = 0

(c) (⋂119870119896=minus119870

(119885 minus 2119896)) cap (⋃119870

119896=minus119870(119864 minus 2119896 plusmn 1)) = 0


Then 119864119898119887119879119899119892119898119899isinZ is a frame for 1198712(R) and there exists a

differentiable dual window ℎ supported on [minus(2119870+1) 2119870+1]

Proof We check condition (2) inTheorem 10 Let

119866 (119909) = sum

119896isinZ

119892 (119909 + 2119896) (37)

By an argument similar to the one at the beginning of theproof of Theorem 3 (36) together with (6) implies that

10038161003816100381610038161003816119866 (119909 minus 1)

10038161003816100381610038161003816+10038161003816100381610038161003816119866 (119909)

10038161003816100381610038161003816ge 119860 119909 isin [0 1] (38)

Since 119892 is nonnegative the zeros of 119866 restricted to [minus1 1] are

119885 = (

119870

⋂

119896=minus119870

(119885 minus 2119896)) cap [minus1 1] (39)

A direct calculation shows that if 119892 is differentiable except onthe set 119864 then 119866 is differentiable outside the set ⋃119870

119896=minus119870(119864 minus

2119896) Let

119864 = (

119870

⋃

119896=minus119870

(119864 minus 2119896)) cap [minus1 1] (40)

Then the sets 119864 and 119885 satisfy the conditions (a) and (b) of(2) in Theorem 10 Hence there exists a differentiable dualwindow ℎ

Example 12 Let

119892 (119909) = (5

2119909 + 2)120594

[minus45minus25](119909)

+ 120594[minus2525]

(119909) + (minus5

2119909 + 2)120594

[2545](119909)

(41)

A direct calculation shows that

119864 = plusmn2

5 plusmn4

5 119864 + 1 =

1

53

57

59

5

119864 minus 1 = minus9

5 minus7

5 minus3

5 minus1

5

119885 = [minus1 minus4

5] cup [

4

5 1]

(42)

Thus the conditions (a)ndash(c) in Corollary 11 are satisfiedHence for any 119887 isin ]0 12] there exists a differentiable dualwindow ℎ

Appendix

Proof of Theorem 10

The full proof of Theorem 10 is notationally complicated Wewill therefore formulate the proof for a function 119892 for which

(i) 119870 = 0 in (6) that is 119866(119909) = 119892(119909) 119909 isin [minus1 1]

(ii) 119866 is differentiable except at one point that is 119864 = 119888(iii) the zeroset of 119866 within ] minus 1 1[ consists of just one

interval [119886 119887] (containing the degenerate case of justone point 119886 = 119887 as a special case) that is 119885 = minus1 cup

[119886 119887] cup 1

We leave the obvious modifications to the general case to thereader

We use the following abbreviation

120595 (119886minus

) = lim119909rarr119886

minus

120595 (119909) 120595 (119886+

) = lim119909rarr119886

+

120595 (119909) (A1)

Proof of Theorem 10 (1) We will prove the contrapositiveresult so suppose that a dual window ℎ defined as in (8) with119870 = 0 that is

ℎ (119909) = 119887 (119909) (A2)

with supp sube [minus1 1] is differentiable on R Then is alsodifferentiable on R Then we have

(minus1) = (1) = 119863 (minus1) = 119863 (1) = 0 (A3)

The duality condition can for 119899 = 0 be written as

119892 (119909 minus 1) ℎ (119909 minus 1) + 119892 (119909) ℎ (119909) = 119887 119909 isin [0 1] (A4)

or in terms of 119866 and

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin [0 1] (A5)

It follows from (A3) that

119866 (0) (0) = 1 (A6)

Putting this into (A5) we have

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) ( (119909) minus (0))

+ (0) (119866 (119909) minus 119866 (0)) = 0 119909 isin ]0 1[

(A7)

or by dividing with 119909

119866 (119909 minus 1) (119909 minus 1)

119909+ 119866 (119909) (

(119909) minus (0)

119909)

+ (0) (119866 (119909) minus 119866 (0)

119909) = 0 119909 isin ]0 1[

(A8)

This implies that

119866 (119909) minus 119866 (0)

119909

= minus

119866 (119909 minus 1) ( (119909 minus 1) 119909) + 119866 (119909) (( (119909) minus (0)) 119909)

(0)

119909 isin ]0 1[

(A9)


Using (A3) it now follows that

119863+119866 (0) = minus

119866 (0+

)119863+ (0) + 119866 ((minus1)

+

)119863+ (minus1)

(0)

= minus119866 (0+

)119863 (0)

(0)

(A10)

Similarly

119863minus119866 (0) = minus

119866 (0minus

)119863 (0)

(0)

(A11)

The conditions (A3) and (A4) and the continuity of showthat 119866 is continuous at 119909 = 0 Hence 119863

+119866(0) = 119863

minus119866(0)

which implies that 119866 is differentiable at 0(2) We construct a real-valued differentiable function withsupp sube [minus1 1] so that

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin [0 1]

(A12)

Let

119885 = minus1 cup [119886 119887] cup 1 119864 = 119888 (A13)

There are various scenarios concerning the location of thepoints 119886 119887 and 119888 They are treated in a similar way and wewill assume that minus1 lt 119886 le 119887 lt 0 lt 119888 lt 1 By the condition (2)inTheorem 10 one of the following cases occurs

(1) 119887 lt 119888 minus 1 lt 0(2) minus1 lt 119888 minus 1 lt 119886

The cases are similar so we only consider the case (1) that is

minus1 lt 119886 le 119887 lt 119888 minus 1 lt 0 lt 119888 lt 1 (A14)

Let

= 119885 cup 119864 = minus1 cup [119886 119887] cup 119888 cup 1

= ( cap [minus1 0]) cup (( minus 1) cap [minus1 0])

= (minus1 cup [119886 119887]) cup (119888 minus 1 cup 0)

(A15)

We define (119909) on [minus1 1] as follows first we define ℎ(119909)on cup ( + 1) by

(119909)

=

0 119909 isin cap [minus1 0] = minus1 cup [119886 119887]

1

119866 (119909)

119909 isin ( minus 1) cap [minus1 0] = 119888 minus 1 cup 0

1

119866 (119909)

119909 isin ( + 1) cap [0 1] = 0 cup [119886 + 1 119887 + 1]

0 119909 isin cap [0 1] = 119888 cup 1

(A16)

which is well defined by (A14) Then

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin + 1

(A17)

We now extend (119909) on [minus1 1](cup(+1)) = ([minus1 0]) cup ([0 1] ( + 1)) choose (119909) on [minus1 0] so that is differentiable on [minus1 0] and

119863+ (minus1) = 119863

minus (119886) = 119863

+ (119887) = 0 (A18)

119863 (119888 minus 1) = 119863(1

119866

) (119888 minus 1)

119863minus (0) = 119863(

1

119866

) (0)

(A19)

Then (A16) and (A18)-(A19) imply that is differentiableon ] minus 1 0[ and that

119863+ (minus1) = 0 119863

minus (0) = 119863(

1

119866

) (0) (A20)

We then define (119909) on [0 1] ( + 1) by

(119909) =1 minus 119866 (119909 minus 1) (119909 minus 1)

119866 (119909)

(A21)

which is well defined since

cap [0 1] = 119888 1 sube ( + 1) (A22)

This implies that

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1

119909 isin [0 1] ( + 1)

(A23)

Recall that 119866(119909 minus 1) 119866(119909) and (119909 minus 1) are differentiable on[0 1] ( + 1) By (A21) the same is the case for (119909) Sincesupp sube [minus1 1] it remains to show that is differentiableon + 1 = 0cup [119886+1 119887+1]cup 119888 cup 1 and that119863

minus(1) = 0

Note that 119866 is differentiable on 0 cup [119886 + 1 119887 + 1] Thus by(A16) and (A20) it suffices to show that

119863+ (0) = 119863(

1

119866

) (0) (A24)

119863minus (119886 + 1) = 119863(

1

119866

) (119886 + 1)

119863+ (119887 + 1) = 119863(

1

119866

) (119887 + 1)

(A25)

119863minus (119888) = 119863

+ (119888) 119863

minus (1) = 0 (A26)

We first show (A24) From (A16) we get

(minus1) = 0 119866 (0) (0) = 1 (A27)


Putting this into (A23) we have for 119909 isin [0 1] ( + 1)

119866 (119909 minus 1) ( (119909 minus 1) minus (minus1))

+119866 (119909) ( (119909) minus (0)) + (0) (119866 (119909) minus 119866 (0)) = 0

(A28)


119866 (119909 minus 1) ( (119909 minus 1) minus (minus1)

119909) + 119866 (119909) (

(119909) minus (0)

119909)

+ (0) (119866 (119909) minus 119866 (0)

119909) = 0

(A29)

This implies that

(119909) minus (0)

119909

= minus (119866 (119909 minus 1) (( (119909 minus 1) minus (minus1)) 119909)

+ (0) ((119866 (119909) minus 119866 (0)) 119909))

times (119866 (119909))minus1

119909 isin [0 1] ( + 1)

(A30)

Using (A18) it follows that

119863+ (0) = minus

119866 ((minus1)+

)119863+ (minus1) + (0)119863

+119866 (0)

119866 (0+)

= minus (0)119863

+119866 (0)

119866 (0+)

(A31)

Since 119866(119909) is differentiable at 0 notin we have 119866(0+) = 119866(0)119863+119866(0) = 119863119866(0) This together with (A27) implies that

119863+ (0) = minus

(0)119863119866 (0)

119866 (0)

= minus119863119866 (0)

1198662 (0)

= 119863(1

119866

) (0)

(A32)

This proves that (A24) holdsThe results in (A25) and (A26)can be shown in a similar way

Acknowledgments

The authors would like to thank the reviewers for their usefulcomments and suggestions This research was supported byBasic Science Research Program through the National Re-search Foundation of Korea (NRF) funded by the Ministryof Education (2010-0007614)

References

[1] R S Laugesen ldquoGabor dual spline windowsrdquoApplied and Com-putational Harmonic Analysis vol 27 no 2 pp 180ndash194 2009

[2] I Kim ldquoGabor frames in one dimension with trigonometricspline dual windowsrdquo Preprint 2012

[3] H G Feichtinger and T Strohmer Gabor Analysis and Algo-rithms Theory and Applications Birkhauser Boston MassUSA 1998

[4] H G Feichtinger and T Strohmer Advances in Gabor AnalysisBirkhauser Boston Mass USA 2002

[5] K Grochenig Foundations of Time-Frequency Analysis Bir-khauser Boston Mass USA 2000

[6] S Li ldquoOn general frame decompositionsrdquoNumerical FunctionalAnalysis and Optimization vol 16 no 9-10 pp 1181ndash1191 1995

[7] O Christensen ldquoFrames and generalized shift-invariant sys-temsrdquo in Operator Theory Advances and Applications vol 164pp 193ndash209 Birkhauser Basel Switzerland 2006 Proceedingsof Conference in Vaxjo June 2004

[8] A Ron and Z Shen ldquoWeyl-Heisenberg frames and Riesz basesin 1198712(R119889)rdquo Duke Mathematical Journal vol 89 no 2 pp 237ndash282 1997

[9] A J E M Janssen ldquoThe duality condition forWeyl-Heisenbergframesrdquo in Gabor Analysis Theory and Applications H GFeichtinger and T Strohmer Eds pp 33ndash84 BirkhauserBoston Mass USA 1998

[10] E Lieb and M Loss Analysis vol 14 of Graduate Studies inMathematics American Mathematical Society Providence RIUSA 2nd edition 2001

[11] O Christensen Frames and Bases An Introductory CourseBirkhauser Boston Mass USA 2007

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Volume 2013

International Journal of

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Journal of Function Spaces and Applications

International Journal of Mathematics and Mathematical Sciences


ISRN Geometry


Discrete Dynamics in Nature and Society



Volume 2013

Advances in

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ISRN Algebra


ProbabilityandStatistics

Journal of


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Journal ofApplied Mathematics


Advances in

DecisionSciences



Stochastic AnalysisInternational Journal of

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawi Publishing Corporation httpwwwhindawicom Volume 2013

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Differential EquationsInternational Journal of

Volume 2013


scaling we obtain the function 119892(119909) = 1198612(1199092) Again

119864119898119887119879119899119892119898119899isinZ is a frame for all sufficiently small values of

119887 gt 0 But in contrast with the situation for 1198612 we will show

that one can find infinitely often differentiable dual windowsℎ that are supported on supp119892 = [minus2 2] for any value of119887 isin ]0 12] These examples illustrate that the question ofdifferentiability of the dual windows is a nontrivial issue Theexamples will be derived as a consequence of more generalresults see Example 9

The results will be based on a construction of dual win-dows as presented in Section 2 In Section 3 we consider aparticular casewhere it is possible to obtain smooth dual win-dows regardless of the regularity of the given window 119892Thisis much more than one can hope for in the general case Ageneral approach to the question of differentiability of thedual windows is given in Section 5 Since the necessary con-ditions are quite involved and not very intuitive we first inSection 4 state a version for the case of windows 119892 that aresupported in [minus1 1] For this case we can provide concreteexamples demonstrating that the desired conclusions mightfail if any of the constraints is removed

Note that a complementary approach to duality for Gaborframes that also dealswith the issue of regularity is consideredby Laugesen [1] and Kim [2] For more information about thetheory for Gabor analysis and its applications see [3ndash5]

2 Construction of Dual Frames

In the literature various characterizations of the pairs of dualGabor frames are available For general frames Li gave a char-acterizationin [6] which in the special case of Gabor framesalso lead to a class of dual Gabor frames later in [7] it wasshown that these duals actually characterize all duals In orderto start our analysis we need the duality conditions for Gaborframes by Ron and Shen [8] and Janssen [9] respectively

Lemma 1 Two Bessel sequences 119864119898119887119879119899119886119892119898119899isinZ and

119864119898119887119879119899119886ℎ119898119899isinZ form dual frames for 1198712(R) if and only if

sum

119896isinZ

119892(119909 +119899

119887+ 119896119886)ℎ (119909 + 119896119886) = 119887120575

1198990 ae 119909 isin [0 119886] (2)

We will use the following to apply Lemma 1

Lemma 2 Let 119866 be a real-valued bounded function andassume that for some constant 119860 gt 0

10038161003816100381610038161003816119866 (119909 minus 1)

10038161003816100381610038161003816+10038161003816100381610038161003816119866 (119909)

10038161003816100381610038161003816ge 119860 119909 isin [0 1] (3)

Then there exists a real-valued bounded function withsupp H sube supp119866 cap [minus1 1] such that

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin [0 1] (4)

Proof Consider 119909 isin [0 1]We will define (119909) and (119909 minus 1)simultaneously In case |119866(119909 minus 1)| ge 1198602 put (119909) = 0 and(119909minus1) = 1119866(119909minus1) On the other hand if |119866(119909minus1)| lt 1198602we know that |119866(119909)| ge 1198602 In this case put (119909 minus 1) =

0 and (119909) = 1119866(119909) Clearly we can take = 0 outside[minus1 1]

Wewill nowpresent a general result about the existence offrames with a dual window of a special form Note that incontrast with most results from the literature we do not needto assume that the integer translates of the window functionform a partition of unity

Associated to a function 119892 with support on an interval[minus(2119870 + 1) 2119870 + 1] 119870 isin N we will in the rest of the paperuse the function

119866 (119909) = sum

119896isinZ

119892 (119909 + 2119896) 119909 isin R (5)

Due to the periodicity of 119866 we are mainly interested in 119909 isin

[minus1 1] Note that by the compact support of 119892

119866 (119909) =

119870

sum

119896=minus119870

119892 (119909 + 2119896) 119909 isin [minus1 1] (6)

Theorem 3 Let 119870 isin N cup 0 and let 119887 isin ]0 1(4119870 + 2)]Let 119892 be a real-valued bounded function with supp119892 sube

[minus(2119870 + 1) 2119870 + 1] for which1003816100381610038161003816100381610038161003816100381610038161003816

sum

119899isinZ

119892 (119909 + 119899)

1003816100381610038161003816100381610038161003816100381610038161003816

ge 119860 119909 isin [0 1] (7)

for a constant 119860 gt 0 Then the following hold

(i) The function 119866 in (5) satisfies the conditions inLemma 2

(ii) Take as in Lemma 2 and let

ℎ (119909) = 119887

119870

sum

119896=minus119870

1198792119896 (119909) (8)

Then 119864119898119887119879119899119892119898119899isinZ and 119864

119898119887119879119899ℎ119898119899isinZ form dual frames

for 1198712(R) and ℎ is supported in [minus(2119870 + 1) 2119870 + 1]

Proof To prove (i) we note that by (7) and the definition of119866 for 119909 isin [0 1]

10038161003816100381610038161003816119866 (119909 minus 1)

10038161003816100381610038161003816+10038161003816100381610038161003816119866 (119909)

10038161003816100381610038161003816=

1003816100381610038161003816100381610038161003816100381610038161003816

sum

119896isinZ

119892 (119909 minus 1 + 2119896)

1003816100381610038161003816100381610038161003816100381610038161003816

+

1003816100381610038161003816100381610038161003816100381610038161003816

sum

119896isinZ

119892 (119909 + 2119896)

1003816100381610038161003816100381610038161003816100381610038161003816

ge

1003816100381610038161003816100381610038161003816100381610038161003816

sum

119896isinZ

119892 (119909 + 119896)

1003816100381610038161003816100381610038161003816100381610038161003816

ge 119860

(9)

Thus119866 satisfies the condition (3) in Lemma 2 Also it is clearthat 119866 is bounded real-valued and 2-periodicTherefore wecan choose a function which is supported in [minus1 1] andsuch that

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin [0 1] (10)

Define ℎ as in (8) In order to prove (ii) we will applyLemma 1 By assumption the function119892has compact supportand is bounded by the construction the function ℎ sharesthese properties It follows that 119864

119898119887119879119899119892119898119899isinZ and

119864119898119887119879119899ℎ119898119899isinZ are Bessel sequences In order to verify that



sum

119896isinZ

119892(119909 +119899

119887+ 119896)ℎ (119909 + 119896) = 119887120575

1198990 ae 119909 isin [0 1] (11)


2119870

sum


119892 (119909 + 119896) ℎ (119909 + 119896) = 119887 119909 isin [0 1] (12)



Thus we have

ℎ (119909 minus 1 + 2119896) = 119887 (119909 minus 1) ℎ (119909 + 2119896) = 119887 (119909)

(14)


2119870

sum


119892 (119909 + 119896) ℎ (119909 + 119896)

=

119870

sum

119896=minus119870

119892 (119909 minus 1 + 2119896) ℎ (119909 minus 1 + 2119896)

+

119870

sum

119896=minus119870

119892 (119909 + 2119896) ℎ (119909 + 2119896)

= 119887

119870

sum

119896=minus119870

119892 (119909 minus 1 + 2119896) (119909 minus 1)

+ 119887

119870

sum

119896=minus119870

119892 (119909 + 2119896) (119909)

= 119887 (119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909)) = 119887

(15)






1) 2119870 + 1] for which

sum

119899isinZ

119892 (119909 + 2119899) = 1 119909 isin [minus1 1] (16)


119891 (119909) = 0 119909 le 0 119891 (119909) = 1 119909 ge 1 (17)


(119909) =

1

2119891 (2 (119909 + 1)) minus1 le 119909 lt minus

1

2

1 minus1

2119891 (minus2119909) minus

1

2le 119909 lt 0

1 minus1

2119891 (2119909) 0 le 119909 lt

1

2

1

2119891 (2 (1 minus 119909))

1

2le 119909 lt 1

0 otherwise

(18)

and let

ℎ (119909) = 119887

119870

sum

119896=minus119870

1198792119896 (119909) (19)



2119870 + 1](ii) 119864

119898119887119879119899119892119898119899isinZ and 119864


1198712




119866 (119909) = 1 119909 isin [minus1 1] (20)


(119909 minus 1) + (119909) =119891 (2119909)

2+ (1 minus

119891 (2119909)

2) = 1 (21)

for 119909 isin ]12 1[

(119909 minus 1) + (119909)


2) +

119891 (2 (1 minus 119909))

2= 1

(22)


119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909)

= (119909 minus 1) + (119909) = 1 119909 isin [0 1]

(23)




119891 (119909) =



minus1

] 0 lt 119909 lt 1

0 119909 le 0

1 119909 ge 1

(24)



119898119887119879119899119892119898119899isinZ In





sum

119899isinZ

120601 (119909 minus 119899) = 1 119909 isin R (25)




sum

119899isinZ

119892 (119909 minus 2119899) = sum

119899isinZ

120601(119909

2minus 119899) = 1 (26)


119870



119873 119873 isin N are


119861119873+1

= 119861119873lowast1198611 Any




119873119873 = 1 2119870 and a






119885 = 119909 isin [minus1 1] 119892 (119909) = 0



1003816100381610038161003816119892 (119909 minus 1)1003816100381610038161003816

2

+1003816100381610038161003816119892 (119909)

1003816100381610038161003816

2

ge 119860 119909 isin [0 1] (28)




(a) 119864 cap (119864 + 1) = 0(b) 119885 cap (119864 plusmn 1) = 0





119892 (119909) =

2119909 + 2 119909 isin [minus1 minus1

2]


21

2]

minus2119909 + 2 119909 isin [1

2 1]


(29)







119892(minus1

2) = 119892(

1

2) = 1 119863

minus119892(minus

1

2) = 2

119863+119892(minus

1

2) = 0 119863

minus119892(

1

2) = 0 119863

+119892(

1

2) = minus2

(31)


119863ℎ(minus1

2) + 119863ℎ(

1


1

2)

119863ℎ (minus1

2) + 119863ℎ(

1

2) = 2ℎ (

1

2)

(32)



Example 8 Consider

119892 (119909) =

5

2119909 +

3


3

5 minus1

5]


51

5]

minus5

2119909 +

3

2 119909 isin [

1

53

5]


53

5]

(33)















119885 = 119909 isin [minus1 1] 119866 (119909) = 0

119864 = 119909 isin [minus1 1] 119866120594[minus11]





10038161003816100381610038161003816119866 (119909 minus 1)

10038161003816100381610038161003816+10038161003816100381610038161003816119866 (119909)

10038161003816100381610038161003816ge 119860 119909 isin [0 1] (35)










2119870

sum


119892 (119909 + 119896) ge 119860 119909 isin [0 1] (36)



(a) 0 notin (⋂119870119896=minus119870

(119885 minus 2119896)) cup (⋃119870

119896=minus119870(119864 minus 2119896))

(b) (⋃119870119896=minus119870

(119864 minus 2119896)) cap (⋃119870

119896=minus119870(119864 minus 2119896 + 1)) = 0

(c) (⋂119870119896=minus119870

(119885 minus 2119896)) cap (⋃119870






119866 (119909) = sum

119896isinZ

119892 (119909 + 2119896) (37)


10038161003816100381610038161003816119866 (119909 minus 1)

10038161003816100381610038161003816+10038161003816100381610038161003816119866 (119909)

10038161003816100381610038161003816ge 119860 119909 isin [0 1] (38)


119885 = (

119870

⋂

119896=minus119870

(119885 minus 2119896)) cap [minus1 1] (39)


119896=minus119870(119864 minus

2119896) Let

119864 = (

119870

⋃

119896=minus119870

(119864 minus 2119896)) cap [minus1 1] (40)


Example 12 Let

119892 (119909) = (5

2119909 + 2)120594


+ 120594[minus2525]

(119909) + (minus5

2119909 + 2)120594

[2545](119909)

(41)


119864 = plusmn2

5 plusmn4

5 119864 + 1 =

1

53

57

59

5


5 minus7

5 minus3

5 minus1

5


5] cup [

4

5 1]

(42)


Appendix

Proof of Theorem 10





[119886 119887] cup 1



120595 (119886minus

) = lim119909rarr119886

minus

120595 (119909) 120595 (119886+

) = lim119909rarr119886

+

120595 (119909) (A1)


ℎ (119909) = 119887 (119909) (A2)


(minus1) = (1) = 119863 (minus1) = 119863 (1) = 0 (A3)






119866 (0) (0) = 1 (A6)



+ (0) (119866 (119909) minus 119866 (0)) = 0 119909 isin ]0 1[

(A7)


119866 (119909 minus 1) (119909 minus 1)

119909+ 119866 (119909) (

(119909) minus (0)

119909)

+ (0) (119866 (119909) minus 119866 (0)

119909) = 0 119909 isin ]0 1[

(A8)

This implies that

119866 (119909) minus 119866 (0)

119909

= minus


(0)

119909 isin ]0 1[

(A9)



119863+119866 (0) = minus

119866 (0+

)119863+ (0) + 119866 ((minus1)

+

)119863+ (minus1)

(0)

= minus119866 (0+

)119863 (0)

(0)

(A10)

Similarly

119863minus119866 (0) = minus

119866 (0minus

)119863 (0)

(0)

(A11)


+119866(0) = 119863

minus119866(0)


119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin [0 1]

(A12)

Let

119885 = minus1 cup [119886 119887] cup 1 119864 = 119888 (A13)





Let




(A15)


(119909)

=


1

119866 (119909)


1

119866 (119909)

119909 isin ( + 1) cap [0 1] = 0 cup [119886 + 1 119887 + 1]

0 119909 isin cap [0 1] = 119888 cup 1

(A16)


119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin + 1

(A17)


119863+ (minus1) = 119863

minus (119886) = 119863

+ (119887) = 0 (A18)

119863 (119888 minus 1) = 119863(1

119866

) (119888 minus 1)

119863minus (0) = 119863(

1

119866

) (0)

(A19)


119863+ (minus1) = 0 119863

minus (0) = 119863(

1

119866

) (0) (A20)



119866 (119909)

(A21)


cap [0 1] = 119888 1 sube ( + 1) (A22)

This implies that

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1

119909 isin [0 1] ( + 1)

(A23)


minus(1) = 0


119863+ (0) = 119863(

1

119866

) (0) (A24)

119863minus (119886 + 1) = 119863(

1

119866

) (119886 + 1)

119863+ (119887 + 1) = 119863(

1

119866

) (119887 + 1)

(A25)

119863minus (119888) = 119863

+ (119888) 119863

minus (1) = 0 (A26)


(minus1) = 0 119866 (0) (0) = 1 (A27)




+119866 (119909) ( (119909) minus (0)) + (0) (119866 (119909) minus 119866 (0)) = 0

(A28)



119909) + 119866 (119909) (

(119909) minus (0)

119909)

+ (0) (119866 (119909) minus 119866 (0)

119909) = 0

(A29)

This implies that

(119909) minus (0)

119909


+ (0) ((119866 (119909) minus 119866 (0)) 119909))

times (119866 (119909))minus1

119909 isin [0 1] ( + 1)

(A30)


119863+ (0) = minus

119866 ((minus1)+

)119863+ (minus1) + (0)119863

+119866 (0)

119866 (0+)

= minus (0)119863

+119866 (0)

119866 (0+)

(A31)


119863+ (0) = minus

(0)119863119866 (0)

119866 (0)

= minus119863119866 (0)

1198662 (0)

= 119863(1

119866

) (0)

(A32)


Acknowledgments


References













OperationsResearch

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ISRN Geometry





Volume 2013

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Volume 2013



sum

119896isinZ

119892(119909 +119899

119887+ 119896)ℎ (119909 + 119896) = 119887120575

1198990 ae 119909 isin [0 1] (11)


2119870

sum


119892 (119909 + 119896) ℎ (119909 + 119896) = 119887 119909 isin [0 1] (12)



Thus we have

ℎ (119909 minus 1 + 2119896) = 119887 (119909 minus 1) ℎ (119909 + 2119896) = 119887 (119909)

(14)


2119870

sum


119892 (119909 + 119896) ℎ (119909 + 119896)

=

119870

sum

119896=minus119870

119892 (119909 minus 1 + 2119896) ℎ (119909 minus 1 + 2119896)

+

119870

sum

119896=minus119870

119892 (119909 + 2119896) ℎ (119909 + 2119896)

= 119887

119870

sum

119896=minus119870

119892 (119909 minus 1 + 2119896) (119909 minus 1)

+ 119887

119870

sum

119896=minus119870

119892 (119909 + 2119896) (119909)

= 119887 (119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909)) = 119887

(15)






1) 2119870 + 1] for which

sum

119899isinZ

119892 (119909 + 2119899) = 1 119909 isin [minus1 1] (16)


119891 (119909) = 0 119909 le 0 119891 (119909) = 1 119909 ge 1 (17)


(119909) =

1

2119891 (2 (119909 + 1)) minus1 le 119909 lt minus

1

2

1 minus1

2119891 (minus2119909) minus

1

2le 119909 lt 0

1 minus1

2119891 (2119909) 0 le 119909 lt

1

2

1

2119891 (2 (1 minus 119909))

1

2le 119909 lt 1

0 otherwise

(18)

and let

ℎ (119909) = 119887

119870

sum

119896=minus119870

1198792119896 (119909) (19)



2119870 + 1](ii) 119864

119898119887119879119899119892119898119899isinZ and 119864


1198712




119866 (119909) = 1 119909 isin [minus1 1] (20)


(119909 minus 1) + (119909) =119891 (2119909)

2+ (1 minus

119891 (2119909)

2) = 1 (21)

for 119909 isin ]12 1[

(119909 minus 1) + (119909)


2) +

119891 (2 (1 minus 119909))

2= 1

(22)


119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909)

= (119909 minus 1) + (119909) = 1 119909 isin [0 1]

(23)




119891 (119909) =



minus1

] 0 lt 119909 lt 1

0 119909 le 0

1 119909 ge 1

(24)



119898119887119879119899119892119898119899isinZ In





sum

119899isinZ

120601 (119909 minus 119899) = 1 119909 isin R (25)




sum

119899isinZ

119892 (119909 minus 2119899) = sum

119899isinZ

120601(119909

2minus 119899) = 1 (26)


119870



119873 119873 isin N are


119861119873+1

= 119861119873lowast1198611 Any




119873119873 = 1 2119870 and a






119885 = 119909 isin [minus1 1] 119892 (119909) = 0



1003816100381610038161003816119892 (119909 minus 1)1003816100381610038161003816

2

+1003816100381610038161003816119892 (119909)

1003816100381610038161003816

2

ge 119860 119909 isin [0 1] (28)




(a) 119864 cap (119864 + 1) = 0(b) 119885 cap (119864 plusmn 1) = 0





119892 (119909) =

2119909 + 2 119909 isin [minus1 minus1

2]


21

2]

minus2119909 + 2 119909 isin [1

2 1]


(29)







119892(minus1

2) = 119892(

1

2) = 1 119863

minus119892(minus

1

2) = 2

119863+119892(minus

1

2) = 0 119863

minus119892(

1

2) = 0 119863

+119892(

1

2) = minus2

(31)


119863ℎ(minus1

2) + 119863ℎ(

1


1

2)

119863ℎ (minus1

2) + 119863ℎ(

1

2) = 2ℎ (

1

2)

(32)



Example 8 Consider

119892 (119909) =

5

2119909 +

3


3

5 minus1

5]


51

5]

minus5

2119909 +

3

2 119909 isin [

1

53

5]


53

5]

(33)















119885 = 119909 isin [minus1 1] 119866 (119909) = 0

119864 = 119909 isin [minus1 1] 119866120594[minus11]





10038161003816100381610038161003816119866 (119909 minus 1)

10038161003816100381610038161003816+10038161003816100381610038161003816119866 (119909)

10038161003816100381610038161003816ge 119860 119909 isin [0 1] (35)










2119870

sum


119892 (119909 + 119896) ge 119860 119909 isin [0 1] (36)



(a) 0 notin (⋂119870119896=minus119870

(119885 minus 2119896)) cup (⋃119870

119896=minus119870(119864 minus 2119896))

(b) (⋃119870119896=minus119870

(119864 minus 2119896)) cap (⋃119870

119896=minus119870(119864 minus 2119896 + 1)) = 0

(c) (⋂119870119896=minus119870

(119885 minus 2119896)) cap (⋃119870






119866 (119909) = sum

119896isinZ

119892 (119909 + 2119896) (37)


10038161003816100381610038161003816119866 (119909 minus 1)

10038161003816100381610038161003816+10038161003816100381610038161003816119866 (119909)

10038161003816100381610038161003816ge 119860 119909 isin [0 1] (38)


119885 = (

119870

⋂

119896=minus119870

(119885 minus 2119896)) cap [minus1 1] (39)


119896=minus119870(119864 minus

2119896) Let

119864 = (

119870

⋃

119896=minus119870

(119864 minus 2119896)) cap [minus1 1] (40)


Example 12 Let

119892 (119909) = (5

2119909 + 2)120594


+ 120594[minus2525]

(119909) + (minus5

2119909 + 2)120594

[2545](119909)

(41)


119864 = plusmn2

5 plusmn4

5 119864 + 1 =

1

53

57

59

5


5 minus7

5 minus3

5 minus1

5


5] cup [

4

5 1]

(42)


Appendix

Proof of Theorem 10





[119886 119887] cup 1



120595 (119886minus

) = lim119909rarr119886

minus

120595 (119909) 120595 (119886+

) = lim119909rarr119886

+

120595 (119909) (A1)


ℎ (119909) = 119887 (119909) (A2)


(minus1) = (1) = 119863 (minus1) = 119863 (1) = 0 (A3)






119866 (0) (0) = 1 (A6)



+ (0) (119866 (119909) minus 119866 (0)) = 0 119909 isin ]0 1[

(A7)


119866 (119909 minus 1) (119909 minus 1)

119909+ 119866 (119909) (

(119909) minus (0)

119909)

+ (0) (119866 (119909) minus 119866 (0)

119909) = 0 119909 isin ]0 1[

(A8)

This implies that

119866 (119909) minus 119866 (0)

119909

= minus


(0)

119909 isin ]0 1[

(A9)



119863+119866 (0) = minus

119866 (0+

)119863+ (0) + 119866 ((minus1)

+

)119863+ (minus1)

(0)

= minus119866 (0+

)119863 (0)

(0)

(A10)

Similarly

119863minus119866 (0) = minus

119866 (0minus

)119863 (0)

(0)

(A11)


+119866(0) = 119863

minus119866(0)


119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin [0 1]

(A12)

Let

119885 = minus1 cup [119886 119887] cup 1 119864 = 119888 (A13)





Let




(A15)


(119909)

=


1

119866 (119909)


1

119866 (119909)

119909 isin ( + 1) cap [0 1] = 0 cup [119886 + 1 119887 + 1]

0 119909 isin cap [0 1] = 119888 cup 1

(A16)


119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin + 1

(A17)


119863+ (minus1) = 119863

minus (119886) = 119863

+ (119887) = 0 (A18)

119863 (119888 minus 1) = 119863(1

119866

) (119888 minus 1)

119863minus (0) = 119863(

1

119866

) (0)

(A19)


119863+ (minus1) = 0 119863

minus (0) = 119863(

1

119866

) (0) (A20)



119866 (119909)

(A21)


cap [0 1] = 119888 1 sube ( + 1) (A22)

This implies that

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1

119909 isin [0 1] ( + 1)

(A23)


minus(1) = 0


119863+ (0) = 119863(

1

119866

) (0) (A24)

119863minus (119886 + 1) = 119863(

1

119866

) (119886 + 1)

119863+ (119887 + 1) = 119863(

1

119866

) (119887 + 1)

(A25)

119863minus (119888) = 119863

+ (119888) 119863

minus (1) = 0 (A26)


(minus1) = 0 119866 (0) (0) = 1 (A27)




+119866 (119909) ( (119909) minus (0)) + (0) (119866 (119909) minus 119866 (0)) = 0

(A28)



119909) + 119866 (119909) (

(119909) minus (0)

119909)

+ (0) (119866 (119909) minus 119866 (0)

119909) = 0

(A29)

This implies that

(119909) minus (0)

119909


+ (0) ((119866 (119909) minus 119866 (0)) 119909))

times (119866 (119909))minus1

119909 isin [0 1] ( + 1)

(A30)


119863+ (0) = minus

119866 ((minus1)+

)119863+ (minus1) + (0)119863

+119866 (0)

119866 (0+)

= minus (0)119863

+119866 (0)

119866 (0+)

(A31)


119863+ (0) = minus

(0)119863119866 (0)

119866 (0)

= minus119863119866 (0)

1198662 (0)

= 119863(1

119866

) (0)

(A32)


Acknowledgments


References













OperationsResearch

Advances in








Volume 2013


Combinatorics





ISRN Geometry





Volume 2013

Advances in


ISRN Algebra



Journal of






Advances in

DecisionSciences










Volume 2013



119891 (119909) =



minus1

] 0 lt 119909 lt 1

0 119909 le 0

1 119909 ge 1

(24)



119898119887119879119899119892119898119899isinZ In





sum

119899isinZ

120601 (119909 minus 119899) = 1 119909 isin R (25)




sum

119899isinZ

119892 (119909 minus 2119899) = sum

119899isinZ

120601(119909

2minus 119899) = 1 (26)


119870



119873 119873 isin N are


119861119873+1

= 119861119873lowast1198611 Any




119873119873 = 1 2119870 and a






119885 = 119909 isin [minus1 1] 119892 (119909) = 0



1003816100381610038161003816119892 (119909 minus 1)1003816100381610038161003816

2

+1003816100381610038161003816119892 (119909)

1003816100381610038161003816

2

ge 119860 119909 isin [0 1] (28)




(a) 119864 cap (119864 + 1) = 0(b) 119885 cap (119864 plusmn 1) = 0





119892 (119909) =

2119909 + 2 119909 isin [minus1 minus1

2]


21

2]

minus2119909 + 2 119909 isin [1

2 1]


(29)







119892(minus1

2) = 119892(

1

2) = 1 119863

minus119892(minus

1

2) = 2

119863+119892(minus

1

2) = 0 119863

minus119892(

1

2) = 0 119863

+119892(

1

2) = minus2

(31)


119863ℎ(minus1

2) + 119863ℎ(

1


1

2)

119863ℎ (minus1

2) + 119863ℎ(

1

2) = 2ℎ (

1

2)

(32)



Example 8 Consider

119892 (119909) =

5

2119909 +

3


3

5 minus1

5]


51

5]

minus5

2119909 +

3

2 119909 isin [

1

53

5]


53

5]

(33)















119885 = 119909 isin [minus1 1] 119866 (119909) = 0

119864 = 119909 isin [minus1 1] 119866120594[minus11]





10038161003816100381610038161003816119866 (119909 minus 1)

10038161003816100381610038161003816+10038161003816100381610038161003816119866 (119909)

10038161003816100381610038161003816ge 119860 119909 isin [0 1] (35)










2119870

sum


119892 (119909 + 119896) ge 119860 119909 isin [0 1] (36)



(a) 0 notin (⋂119870119896=minus119870

(119885 minus 2119896)) cup (⋃119870

119896=minus119870(119864 minus 2119896))

(b) (⋃119870119896=minus119870

(119864 minus 2119896)) cap (⋃119870

119896=minus119870(119864 minus 2119896 + 1)) = 0

(c) (⋂119870119896=minus119870

(119885 minus 2119896)) cap (⋃119870






119866 (119909) = sum

119896isinZ

119892 (119909 + 2119896) (37)


10038161003816100381610038161003816119866 (119909 minus 1)

10038161003816100381610038161003816+10038161003816100381610038161003816119866 (119909)

10038161003816100381610038161003816ge 119860 119909 isin [0 1] (38)


119885 = (

119870

⋂

119896=minus119870

(119885 minus 2119896)) cap [minus1 1] (39)


119896=minus119870(119864 minus

2119896) Let

119864 = (

119870

⋃

119896=minus119870

(119864 minus 2119896)) cap [minus1 1] (40)


Example 12 Let

119892 (119909) = (5

2119909 + 2)120594


+ 120594[minus2525]

(119909) + (minus5

2119909 + 2)120594

[2545](119909)

(41)


119864 = plusmn2

5 plusmn4

5 119864 + 1 =

1

53

57

59

5


5 minus7

5 minus3

5 minus1

5


5] cup [

4

5 1]

(42)


Appendix

Proof of Theorem 10





[119886 119887] cup 1



120595 (119886minus

) = lim119909rarr119886

minus

120595 (119909) 120595 (119886+

) = lim119909rarr119886

+

120595 (119909) (A1)


ℎ (119909) = 119887 (119909) (A2)


(minus1) = (1) = 119863 (minus1) = 119863 (1) = 0 (A3)






119866 (0) (0) = 1 (A6)



+ (0) (119866 (119909) minus 119866 (0)) = 0 119909 isin ]0 1[

(A7)


119866 (119909 minus 1) (119909 minus 1)

119909+ 119866 (119909) (

(119909) minus (0)

119909)

+ (0) (119866 (119909) minus 119866 (0)

119909) = 0 119909 isin ]0 1[

(A8)

This implies that

119866 (119909) minus 119866 (0)

119909

= minus


(0)

119909 isin ]0 1[

(A9)



119863+119866 (0) = minus

119866 (0+

)119863+ (0) + 119866 ((minus1)

+

)119863+ (minus1)

(0)

= minus119866 (0+

)119863 (0)

(0)

(A10)

Similarly

119863minus119866 (0) = minus

119866 (0minus

)119863 (0)

(0)

(A11)


+119866(0) = 119863

minus119866(0)


119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin [0 1]

(A12)

Let

119885 = minus1 cup [119886 119887] cup 1 119864 = 119888 (A13)





Let




(A15)


(119909)

=


1

119866 (119909)


1

119866 (119909)

119909 isin ( + 1) cap [0 1] = 0 cup [119886 + 1 119887 + 1]

0 119909 isin cap [0 1] = 119888 cup 1

(A16)


119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin + 1

(A17)


119863+ (minus1) = 119863

minus (119886) = 119863

+ (119887) = 0 (A18)

119863 (119888 minus 1) = 119863(1

119866

) (119888 minus 1)

119863minus (0) = 119863(

1

119866

) (0)

(A19)


119863+ (minus1) = 0 119863

minus (0) = 119863(

1

119866

) (0) (A20)



119866 (119909)

(A21)


cap [0 1] = 119888 1 sube ( + 1) (A22)

This implies that

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1

119909 isin [0 1] ( + 1)

(A23)


minus(1) = 0


119863+ (0) = 119863(

1

119866

) (0) (A24)

119863minus (119886 + 1) = 119863(

1

119866

) (119886 + 1)

119863+ (119887 + 1) = 119863(

1

119866

) (119887 + 1)

(A25)

119863minus (119888) = 119863

+ (119888) 119863

minus (1) = 0 (A26)


(minus1) = 0 119866 (0) (0) = 1 (A27)




+119866 (119909) ( (119909) minus (0)) + (0) (119866 (119909) minus 119866 (0)) = 0

(A28)



119909) + 119866 (119909) (

(119909) minus (0)

119909)

+ (0) (119866 (119909) minus 119866 (0)

119909) = 0

(A29)

This implies that

(119909) minus (0)

119909


+ (0) ((119866 (119909) minus 119866 (0)) 119909))

times (119866 (119909))minus1

119909 isin [0 1] ( + 1)

(A30)


119863+ (0) = minus

119866 ((minus1)+

)119863+ (minus1) + (0)119863

+119866 (0)

119866 (0+)

= minus (0)119863

+119866 (0)

119866 (0+)

(A31)


119863+ (0) = minus

(0)119863119866 (0)

119866 (0)

= minus119863119866 (0)

1198662 (0)

= 119863(1

119866

) (0)

(A32)


Acknowledgments


References













OperationsResearch

Advances in








Volume 2013


Combinatorics





ISRN Geometry





Volume 2013

Advances in


ISRN Algebra



Journal of






Advances in

DecisionSciences










Volume 2013





119892(minus1

2) = 119892(

1

2) = 1 119863

minus119892(minus

1

2) = 2

119863+119892(minus

1

2) = 0 119863

minus119892(

1

2) = 0 119863

+119892(

1

2) = minus2

(31)


119863ℎ(minus1

2) + 119863ℎ(

1


1

2)

119863ℎ (minus1

2) + 119863ℎ(

1

2) = 2ℎ (

1

2)

(32)



Example 8 Consider

119892 (119909) =

5

2119909 +

3


3

5 minus1

5]


51

5]

minus5

2119909 +

3

2 119909 isin [

1

53

5]


53

5]

(33)















119885 = 119909 isin [minus1 1] 119866 (119909) = 0

119864 = 119909 isin [minus1 1] 119866120594[minus11]





10038161003816100381610038161003816119866 (119909 minus 1)

10038161003816100381610038161003816+10038161003816100381610038161003816119866 (119909)

10038161003816100381610038161003816ge 119860 119909 isin [0 1] (35)










2119870

sum


119892 (119909 + 119896) ge 119860 119909 isin [0 1] (36)



(a) 0 notin (⋂119870119896=minus119870

(119885 minus 2119896)) cup (⋃119870

119896=minus119870(119864 minus 2119896))

(b) (⋃119870119896=minus119870

(119864 minus 2119896)) cap (⋃119870

119896=minus119870(119864 minus 2119896 + 1)) = 0

(c) (⋂119870119896=minus119870

(119885 minus 2119896)) cap (⋃119870






119866 (119909) = sum

119896isinZ

119892 (119909 + 2119896) (37)


10038161003816100381610038161003816119866 (119909 minus 1)

10038161003816100381610038161003816+10038161003816100381610038161003816119866 (119909)

10038161003816100381610038161003816ge 119860 119909 isin [0 1] (38)


119885 = (

119870

⋂

119896=minus119870

(119885 minus 2119896)) cap [minus1 1] (39)


119896=minus119870(119864 minus

2119896) Let

119864 = (

119870

⋃

119896=minus119870

(119864 minus 2119896)) cap [minus1 1] (40)


Example 12 Let

119892 (119909) = (5

2119909 + 2)120594


+ 120594[minus2525]

(119909) + (minus5

2119909 + 2)120594

[2545](119909)

(41)


119864 = plusmn2

5 plusmn4

5 119864 + 1 =

1

53

57

59

5


5 minus7

5 minus3

5 minus1

5


5] cup [

4

5 1]

(42)


Appendix

Proof of Theorem 10





[119886 119887] cup 1



120595 (119886minus

) = lim119909rarr119886

minus

120595 (119909) 120595 (119886+

) = lim119909rarr119886

+

120595 (119909) (A1)


ℎ (119909) = 119887 (119909) (A2)


(minus1) = (1) = 119863 (minus1) = 119863 (1) = 0 (A3)






119866 (0) (0) = 1 (A6)



+ (0) (119866 (119909) minus 119866 (0)) = 0 119909 isin ]0 1[

(A7)


119866 (119909 minus 1) (119909 minus 1)

119909+ 119866 (119909) (

(119909) minus (0)

119909)

+ (0) (119866 (119909) minus 119866 (0)

119909) = 0 119909 isin ]0 1[

(A8)

This implies that

119866 (119909) minus 119866 (0)

119909

= minus


(0)

119909 isin ]0 1[

(A9)



119863+119866 (0) = minus

119866 (0+

)119863+ (0) + 119866 ((minus1)

+

)119863+ (minus1)

(0)

= minus119866 (0+

)119863 (0)

(0)

(A10)

Similarly

119863minus119866 (0) = minus

119866 (0minus

)119863 (0)

(0)

(A11)


+119866(0) = 119863

minus119866(0)


119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin [0 1]

(A12)

Let

119885 = minus1 cup [119886 119887] cup 1 119864 = 119888 (A13)





Let




(A15)


(119909)

=


1

119866 (119909)


1

119866 (119909)

119909 isin ( + 1) cap [0 1] = 0 cup [119886 + 1 119887 + 1]

0 119909 isin cap [0 1] = 119888 cup 1

(A16)


119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin + 1

(A17)


119863+ (minus1) = 119863

minus (119886) = 119863

+ (119887) = 0 (A18)

119863 (119888 minus 1) = 119863(1

119866

) (119888 minus 1)

119863minus (0) = 119863(

1

119866

) (0)

(A19)


119863+ (minus1) = 0 119863

minus (0) = 119863(

1

119866

) (0) (A20)



119866 (119909)

(A21)


cap [0 1] = 119888 1 sube ( + 1) (A22)

This implies that

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1

119909 isin [0 1] ( + 1)

(A23)


minus(1) = 0


119863+ (0) = 119863(

1

119866

) (0) (A24)

119863minus (119886 + 1) = 119863(

1

119866

) (119886 + 1)

119863+ (119887 + 1) = 119863(

1

119866

) (119887 + 1)

(A25)

119863minus (119888) = 119863

+ (119888) 119863

minus (1) = 0 (A26)


(minus1) = 0 119866 (0) (0) = 1 (A27)




+119866 (119909) ( (119909) minus (0)) + (0) (119866 (119909) minus 119866 (0)) = 0

(A28)



119909) + 119866 (119909) (

(119909) minus (0)

119909)

+ (0) (119866 (119909) minus 119866 (0)

119909) = 0

(A29)

This implies that

(119909) minus (0)

119909


+ (0) ((119866 (119909) minus 119866 (0)) 119909))

times (119866 (119909))minus1

119909 isin [0 1] ( + 1)

(A30)


119863+ (0) = minus

119866 ((minus1)+

)119863+ (minus1) + (0)119863

+119866 (0)

119866 (0+)

= minus (0)119863

+119866 (0)

119866 (0+)

(A31)


119863+ (0) = minus

(0)119863119866 (0)

119866 (0)

= minus119863119866 (0)

1198662 (0)

= 119863(1

119866

) (0)

(A32)


Acknowledgments


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119866 (119909) = sum

119896isinZ

119892 (119909 + 2119896) (37)


10038161003816100381610038161003816119866 (119909 minus 1)

10038161003816100381610038161003816+10038161003816100381610038161003816119866 (119909)

10038161003816100381610038161003816ge 119860 119909 isin [0 1] (38)


119885 = (

119870

⋂

119896=minus119870

(119885 minus 2119896)) cap [minus1 1] (39)


119896=minus119870(119864 minus

2119896) Let

119864 = (

119870

⋃

119896=minus119870

(119864 minus 2119896)) cap [minus1 1] (40)


Example 12 Let

119892 (119909) = (5

2119909 + 2)120594


+ 120594[minus2525]

(119909) + (minus5

2119909 + 2)120594

[2545](119909)

(41)


119864 = plusmn2

5 plusmn4

5 119864 + 1 =

1

53

57

59

5


5 minus7

5 minus3

5 minus1

5


5] cup [

4

5 1]

(42)


Appendix

Proof of Theorem 10





[119886 119887] cup 1



120595 (119886minus

) = lim119909rarr119886

minus

120595 (119909) 120595 (119886+

) = lim119909rarr119886

+

120595 (119909) (A1)


ℎ (119909) = 119887 (119909) (A2)


(minus1) = (1) = 119863 (minus1) = 119863 (1) = 0 (A3)






119866 (0) (0) = 1 (A6)



+ (0) (119866 (119909) minus 119866 (0)) = 0 119909 isin ]0 1[

(A7)


119866 (119909 minus 1) (119909 minus 1)

119909+ 119866 (119909) (

(119909) minus (0)

119909)

+ (0) (119866 (119909) minus 119866 (0)

119909) = 0 119909 isin ]0 1[

(A8)

This implies that

119866 (119909) minus 119866 (0)

119909

= minus


(0)

119909 isin ]0 1[

(A9)



119863+119866 (0) = minus

119866 (0+

)119863+ (0) + 119866 ((minus1)

+

)119863+ (minus1)

(0)

= minus119866 (0+

)119863 (0)

(0)

(A10)

Similarly

119863minus119866 (0) = minus

119866 (0minus

)119863 (0)

(0)

(A11)


+119866(0) = 119863

minus119866(0)


119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin [0 1]

(A12)

Let

119885 = minus1 cup [119886 119887] cup 1 119864 = 119888 (A13)





Let




(A15)


(119909)

=


1

119866 (119909)


1

119866 (119909)

119909 isin ( + 1) cap [0 1] = 0 cup [119886 + 1 119887 + 1]

0 119909 isin cap [0 1] = 119888 cup 1

(A16)


119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin + 1

(A17)


119863+ (minus1) = 119863

minus (119886) = 119863

+ (119887) = 0 (A18)

119863 (119888 minus 1) = 119863(1

119866

) (119888 minus 1)

119863minus (0) = 119863(

1

119866

) (0)

(A19)


119863+ (minus1) = 0 119863

minus (0) = 119863(

1

119866

) (0) (A20)



119866 (119909)

(A21)


cap [0 1] = 119888 1 sube ( + 1) (A22)

This implies that

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1

119909 isin [0 1] ( + 1)

(A23)


minus(1) = 0


119863+ (0) = 119863(

1

119866

) (0) (A24)

119863minus (119886 + 1) = 119863(

1

119866

) (119886 + 1)

119863+ (119887 + 1) = 119863(

1

119866

) (119887 + 1)

(A25)

119863minus (119888) = 119863

+ (119888) 119863

minus (1) = 0 (A26)


(minus1) = 0 119866 (0) (0) = 1 (A27)




+119866 (119909) ( (119909) minus (0)) + (0) (119866 (119909) minus 119866 (0)) = 0

(A28)



119909) + 119866 (119909) (

(119909) minus (0)

119909)

+ (0) (119866 (119909) minus 119866 (0)

119909) = 0

(A29)

This implies that

(119909) minus (0)

119909


+ (0) ((119866 (119909) minus 119866 (0)) 119909))

times (119866 (119909))minus1

119909 isin [0 1] ( + 1)

(A30)


119863+ (0) = minus

119866 ((minus1)+

)119863+ (minus1) + (0)119863

+119866 (0)

119866 (0+)

= minus (0)119863

+119866 (0)

119866 (0+)

(A31)


119863+ (0) = minus

(0)119863119866 (0)

119866 (0)

= minus119863119866 (0)

1198662 (0)

= 119863(1

119866

) (0)

(A32)


Acknowledgments


References













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Volume 2013


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Volume 2013



119863+119866 (0) = minus

119866 (0+

)119863+ (0) + 119866 ((minus1)

+

)119863+ (minus1)

(0)

= minus119866 (0+

)119863 (0)

(0)

(A10)

Similarly

119863minus119866 (0) = minus

119866 (0minus

)119863 (0)

(0)

(A11)


+119866(0) = 119863

minus119866(0)


119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin [0 1]

(A12)

Let

119885 = minus1 cup [119886 119887] cup 1 119864 = 119888 (A13)





Let




(A15)


(119909)

=


1

119866 (119909)


1

119866 (119909)

119909 isin ( + 1) cap [0 1] = 0 cup [119886 + 1 119887 + 1]

0 119909 isin cap [0 1] = 119888 cup 1

(A16)


119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1 119909 isin + 1

(A17)


119863+ (minus1) = 119863

minus (119886) = 119863

+ (119887) = 0 (A18)

119863 (119888 minus 1) = 119863(1

119866

) (119888 minus 1)

119863minus (0) = 119863(

1

119866

) (0)

(A19)


119863+ (minus1) = 0 119863

minus (0) = 119863(

1

119866

) (0) (A20)



119866 (119909)

(A21)


cap [0 1] = 119888 1 sube ( + 1) (A22)

This implies that

119866 (119909 minus 1) (119909 minus 1) + 119866 (119909) (119909) = 1

119909 isin [0 1] ( + 1)

(A23)


minus(1) = 0


119863+ (0) = 119863(

1

119866

) (0) (A24)

119863minus (119886 + 1) = 119863(

1

119866

) (119886 + 1)

119863+ (119887 + 1) = 119863(

1

119866

) (119887 + 1)

(A25)

119863minus (119888) = 119863

+ (119888) 119863

minus (1) = 0 (A26)


(minus1) = 0 119866 (0) (0) = 1 (A27)




+119866 (119909) ( (119909) minus (0)) + (0) (119866 (119909) minus 119866 (0)) = 0

(A28)



119909) + 119866 (119909) (

(119909) minus (0)

119909)

+ (0) (119866 (119909) minus 119866 (0)

119909) = 0

(A29)

This implies that

(119909) minus (0)

119909


+ (0) ((119866 (119909) minus 119866 (0)) 119909))

times (119866 (119909))minus1

119909 isin [0 1] ( + 1)

(A30)


119863+ (0) = minus

119866 ((minus1)+

)119863+ (minus1) + (0)119863

+119866 (0)

119866 (0+)

= minus (0)119863

+119866 (0)

119866 (0+)

(A31)


119863+ (0) = minus

(0)119863119866 (0)

119866 (0)

= minus119863119866 (0)

1198662 (0)

= 119863(1

119866

) (0)

(A32)


Acknowledgments


References













OperationsResearch

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Volume 2013


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Volume 2013




+119866 (119909) ( (119909) minus (0)) + (0) (119866 (119909) minus 119866 (0)) = 0

(A28)



119909) + 119866 (119909) (

(119909) minus (0)

119909)

+ (0) (119866 (119909) minus 119866 (0)

119909) = 0

(A29)

This implies that

(119909) minus (0)

119909


+ (0) ((119866 (119909) minus 119866 (0)) 119909))

times (119866 (119909))minus1

119909 isin [0 1] ( + 1)

(A30)


119863+ (0) = minus

119866 ((minus1)+

)119863+ (minus1) + (0)119863

+119866 (0)

119866 (0+)

= minus (0)119863

+119866 (0)

119866 (0+)

(A31)


119863+ (0) = minus

(0)119863119866 (0)

119866 (0)

= minus119863119866 (0)

1198662 (0)

= 119863(1

119866

) (0)

(A32)


Acknowledgments


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OperationsResearch

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Volume 2013


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Volume 2013

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