research article on a property of a three-dimensional...

Hindawi Publishing CorporationJournal of Discrete MathematicsVolume 2013 Article ID 797249 3 pageshttpdxdoiorg1011552013797249

Research ArticleOn a Property of a Three-Dimensional Matrix

David Blokh

C D Technologies Ltd Israel

Correspondence should be addressed to David Blokh david blokh012netil

Received 11 June 2013 Accepted 24 September 2013

Academic Editor Hong J Lai

Copyright copy 2013 David BlokhThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Let 119878119899be the symmetrical group acting on the set 1 2 119899 and 119909 119910 isin 119878

119899 Consider the set 119882 = (119894 119909(119894) 119910(119894)) | 1 le 119894 le 119899

|119894 minus 119909(119894)| gt 1 or |119894 minus 119910(119894)| gt 1 or |119909(119894) minus 119910(119894)| gt 1The main result of this paper is the following theorem If the number of119882 setentries is more than [1198993] then there exist entries (119894

1 119909(1198941) 119910(1198941)) (1198942 119909(1198942) 119910(1198942)) (1198943 119909(1198943) 119910(1198943)) isin 119882 such that |119894

1minus 119909(1198942)| le 1

|1198941minus119910(1198943)| le 1 and |119909(119894

2) minus 119910(119894

3)| le 1 The application of this theorem to the three-dimensional assignment problem is considered

1 Introduction

Let 1198773119899be the set of 119899 times 119899 times 119899-matrices over the field of real

numbersThree-dimensional matrix not only is an interesting

mathematical object [1ndash3] but also has applications in manyfields such as theoretical physics [4] and operational research[5 6]

Let 119878119899be the symmetrical group acting on the set

1 2 119899 119909 119910 isin 119878119899 119886119894119895119896 isin 1198773

119899and

119866 = 119886119894119895119896|1003816100381610038161003816119894 minus 119895

1003816100381610038161003816 le 1 and |119894 minus 119896| le 1 and1003816100381610038161003816119895 minus 119896

1003816100381610038161003816 le 1

119867 = 119886119894119895119896|1003816100381610038161003816119894 minus 119895

1003816100381610038161003816 gt 1 or |119894 minus 119896| gt 1 or1003816100381610038161003816119895 minus 119896

1003816100381610038161003816 gt 1

119879 = 119886119894119909(119894)119910(119894)

| 1 le 119894 le 119899

(1)

The main result of this paper is the following theorem

Theorem 1 If 119886119894119895119896 isin 119877

3

119899 119909 119910 isin 119878

119899and the number of 119879 cap

119867 set entries is more than [1198993] then there exist entries 119886119894111989511198961

119886119894211989521198962

and 119886119894311989531198963

isin 119879 cap 119867 such that the entry 119886119894111989521198963

isin 119866

We give another formulation of Theorem 1Consider the set119882 = (119894 119909 (119894) 119910 (119894)) | 1 le 119894 le 119899 |119894 minus 119909 (119894)|

gt 1 or1003816100381610038161003816119894 minus 119910 (119894)

1003816100381610038161003816 gt 1 or1003816100381610038161003816119909 (119894) minus 119910 (119894)

1003816100381610038161003816 gt 1

(2)

Theorem 2 If the number of 119882 set entries is more than[1198993] then there exist entries (119894

1 119909(1198941) 119910(1198941)) (1198942 119909(1198942) 119910(1198942))

(1198943 119909(1198943) 119910(1198943)) isin 119882 such that |119894

1minus 119909(1198942)| le 1 |119894

1minus 119910(1198943)| le 1

and |119909(1198942) minus 119910(119894

3)| le 1

2 Proof of Theorem 1

We prove the theorem by contradiction The set of matrix119886119894119895119896 isin 119877

3

119899entries with one index fixed and the two others

having values from 1 to 119899 will be called a layer We denote alayer by 119871119903

119904 where 119903 indicates the location of a fixed index and

119904 indicates its value For example 1198712119904= 119886119894119904119896| 119894 119896 = 1 119899

Furthermore entries from119866will be called basic entries from119867 will be called nonbasic 119879 will be termed a trajectory thelayer containing a basic trajectory entry will be termed a basiclayer and the layer containing a nonbasic trajectory entrywillbe termed a nonbasic layer

If 1198711119904is a nonbasic layer then one layer in the pair

1198712

119904minus1 1198713119904minus1

is basicSuppose to the contrary that layers 1198711

119904 1198712119904minus1

and 1198713119904minus1

arenonbasic Let 119886

11990411989511198961

1198861198942119904minus1119896

2

and 11988611989431198953119904minus1

be the trajectoryentries of 1198711

119904 1198712119904minus1

1198713119904minus1

layers Replace 11988611990411989511198961

1198861198942119904minus1119896

2

and11988611989431198953119904minus1

with 119886119904119904minus1119904minus1

119886119894211989511198961

and 119886119894311989531198962

The nonbasic tra-jectory entries 119886

11990411989511198961

1198861198942119904minus1119896

2

and 11988611989431198953119904minus1

are replaced withentries 119886

119904119904minus1119904minus1 119886119894211989511198961

and 119886119894311989531198962

among which there is abasic one

Similar assertions may be proved for the 1198711119904layer and

the following layer pairs 1198712119904minus1 1198713

119904 1198712119904 1198713

119904minus1 1198712119904 1198713

119904 1198712119904 1198713

119904+1

1198712

119904+1 1198713

119904 and 1198712

119904+1 1198713

119904+1

Three nonbasic layers may not be consecutive

2 Journal of Discrete Mathematics

Suppose that layers 1198711119904minus1

1198711119904 and 1198711

119904+1are nonbasic One

of the layers 1198712119904or 1198713119904is basic In these layers the basic

entries may be 119886119904minus1119904119904minus1

119886119904minus1119904119904

119886119904119904119904minus1

119886119904119904119904 119886119904119904119904+1

119886119904+1119904119904

and 119886

119904+1119904119904+1for the layer 1198712

119904 and 119886

119904minus1119904minus1119904 119886119904minus1119904119904

119886119904119904minus1119904

119886119904119904119904 119886119904+1119904119904

119886119904119904+1119904

119886119904+1119904+1119904

for the layer 1198713119904 However this

contradicts the assumption that the layers 1198711119904minus1

1198711119904 and 1198711

119904+1

are nonbasicThe fact that the two first and the two last layersmay not be nonbasic is proved in a similar way

All assertions given below represent conditions thatprevent replacing nonbasic trajectory entries with entries thatinclude a basic one

Consider the sequence of layers 11987111 1198711

2 119871

1

119904minus1 1198711

119904

1198711

119904+1 119871

1

119899minus1 and 1198711

119899 Given below are possible arrange-

ments of layers in this sequence A basic layer 1198711119904is denoted by

1 a nonbasic layer 1198711119904is denoted by 0

Consider

(1) 11100111 1100111 1110011 110011

(2) 111010111 01111010 01011110 001111010 010111100 0101111010

(3) 0111100 00111100 0011110

(4) 0111 1110 10111 11101 011110

We prove the first arrangement of item 1 Suppose thatthe two nonbasic layers are followed only by two basic layersthat is 00110 Let there be layers 1198711

119904 1198711119904+1

1198711119904+2

1198711119904+3

1198711

119904+4 Consider layer pairs1198712

119904+11198713119904+1

1198712119904+1

1198713119904+2

1198712119904+2

1198713119904+1

1198712119904+2

and 1198713

119904+2 At least one layer of each pair is basic These four

pairs of layers contain at least two basic entries and theseentries are located in layers 1198711

119904+2and 1198711

119904+3 The pair 1198712

119904+3 1198713119904+3

includes a basic layer The first coordinate of the basic layerentry may be 119904 + 2 119904 + 3 or 119904 + 4 But the 1198711

119904+2and 1198711

119904+3

layers are occupied while the 1198711119904+4

layer is nonbasic Thereis a contradiction

The other arrangements are proved in a similar wayThus nonbasic layers may not be arranged closer than

those in the above variants But these variants do not allow forthe composition of a combination containingmore than [1198993]nonbasic layers Hence it follows that if a trajectory includesmore than [1198993] nonbasic entries then one of the variants isviolated and nonbasic trajectory entries can be replaced by aset of entries containing a basic entry

3 Application to the Three-DimensionalAssignment Problem

The three-dimensional assignment problem (AP3) is animportant combinatorial optimization problem It is suf-ficient to note that the particular case of AP3 the 3-dimensional matching problem is one of the six main NP-hard problems [7] The formal AP3 statement is as followsfor a matrix 119886

119894119895119896 isin 119877

3

119899 find permutations 119909 119910 isin 119878

119899such

that sum119899119894=1119886119894119909(119894)119910(119894)

is maximized

In this paper the AP3 is considered for a special class of119899 times 119899 times 119899-matrices 119880(119899) A matrix

10038171003817100381710038171003817119886119894119895119896

10038171003817100381710038171003817isin 119880 (119899) if 119886

119894119895119896ge 0

min119886119894119895119896isin119866

119886119894119895119896ge 119862 sdot max119886119894119895119896isin119867

119886119894119895119896 119862 ge 3

(3)

One of AP3 interpretations is the following There are 119899employees and two job sets of 119899 jobs each If the 119894th employeeperforms the 119895th job of the first set and the 119896th job of the sec-ond set then the effect equals 119886

119894119895119896 It is required to distribute

the jobs among the employees in such away so as tomaximizethe total effect

Let us describe the situation that will lead to the AP3 formatrices from 119880(119899) As a rule the employees are orderedby qualification while the jobs are ordered by complexityA higher effect is reached when a more qualified employeeperforms a more complex job and we may arrive at the AP3for matrices from 119880(119899)

The particular cases of AP3 [5 8ndash11] the nonpolynomialexact algorithms for the AP3 [6 12] and heuristics for theAP3 [13 14] were considered

The NP-hard particular cases of the traveling salesmanproblem with sets of matrices whose structures are similar tothose ofmatrices from119880(119899) have been considered previously[15 16]

Theorem 3 The AP3 for matrices from 119880(119899) is NP-hard

Proof Let 119887119894119895119896 isin 1198773

119899 119887119894119895119896ge 0 and numbers119860119861 are such that

119860 = 3 sdotmax 119887119894119895119896 119861 = 119899119860

Consider 4119899 times 4119899 times 4119899-matrix 119886119894119895119896 with nonnegative

entries

1198864119904minus34119904minus24119904minus2

= 119861 119904 = 1 119899


= 119861 119904 = 1 119899

119886411990441199044119904

= 119861 119904 = 1 119899

119886119903119906V = 119860 119903 119906 V = 1 119899 if |119903 minus 119906| le 1

|119903 minus V| le 1 |119906 minus V| le 1 119886119903119906V = 119861


= 119887119894119895119896

119894 119895 119896 = 1 119899

119886119894119895119896= 0 in other cases

(4)

The matrix 119886119894119895119896 isin 119880(4119899)

All entries that are equal to 119861 belong to the optimal solu-tion of AP3 for 119886

119894119895119896matrix This signifies that if the optimal

solution of AP3 for 119886119894119895119896 matrix is known then the optimal

solution of AP3 for 119887119894119895119896matrix is known as well Therefore

the AP3 for arbitrary matrices is polynomially reducible tothe AP3 for matrices from 119880(119899)

Corollary 4 If 119886119894119895119896 isin 119880(119899) then a set T exists on which the

AP3 optimum is reached and the number of 119879 cap 119867 set entriesis no more than [1198993]

Indeed if for optimal T of matrix 119886119894119895119896 isin 119880(119899) the number

of 119879 cap 119867 set entries is more than [1198993] then according to

Journal of Discrete Mathematics 3

Theorem 1 it is possible to make such replacement of entriessuch that the sum of T set entries will not change and thenumber of 119879 cap 119867 set entries is reduced

Theorem 5 If 119886119894119895119896 isin 119880(119899) then the AP3 optimum for a

1198861015840

119894119895119896 matrix where 1198861015840

119894119895119896= 119886119894119895119896

for 1198861015840119894119895119896isin 119866 and 1198861015840

119894119895119896= 0 for

1198861015840

119894119895119896isin 119867 represents an approximation for the AP3 optimum of

the 119886119894119895119896matrix with a relative error not exceeding 1(3C)

Proof Let 119886119894119895119896 isin 119880(119899) 119898

1is the AP3 optimum for 119886

119894119895119896

119891 119892 isin 119878119899are such that119898

1= sum119899

119894=1119886119894119891(119894)119892(119894)

and the number of119886119894119891(119894)119892(119894)

| 119894 = 1 119899 cap 119867 set entries is no more than [1198993]1198982is the sum of 119886

119894119891(119894)119892(119894)| 119894 = 1 119899 cap 119867 set entries and

1198983= 1198981minus 11989821198981015840 is the AP3 optimum for 1198861015840

119894119895119896

Note that1198983= sum119899

119894=11198861015840

119894119891(119894)119892(119894)and1198981015840 ge 119898

3

Insofar as 119886119894119895119896ge 1198861015840

119894119895119896 119894 119895 119896 = 1 119899 then 119898

1ge 1198981015840

Hence1198983le 1198981015840

le 1198981and (119898

1minus 1198981015840

)1198981le (1198981minus 1198983)1198981=

11989821198981

Since1198982le (13)119899sdotmax

119886119894119895119896isin119867119886119894119895119896

and1198981ge 119899sdotmin

119886119894119895119896isin119866119886119894119895119896

then 11989821198981le 1(3119862) Hence (119898

1minus 1198981015840

)1198981le 1(3119862) and

the relative error of1198981015840 taken as the AP3 approximate solutionfor the 119886

119894119895119896matrix is no more than 1(3119862)

Remark 6 An exact119874(119899)-algorithmofAP3 solution has beenconstructed [9] for 119899 times 119899 times 119899-matrices 119886

119894119895119896 such that 119886

119894119895119896ge

0 for 119886119894119895119896isin 119866 and 119886

119894119895119896= 0 for 119886

119894119895119896isin 119867 using dynamical

programming

Acknowledgments

The author is grateful to Professor Y Dinitz and Professor GGutin for their attention to this work and valuable commentsAlso the author is grateful to the anonymous referee of thisjournal for the valuable comments and suggestions

References

[1] R A Brualdi and J Csima ldquoOn the plane term rank of a threedimensional matrixrdquo Proceedings of the AmericanMathematicalSociety vol 54 no 1 pp 471ndash473 1976

[2] J Csima ldquoOn the plane term rank of three dimensional matri-cesrdquo Discrete Mathematics vol 28 no 2 pp 147ndash152 1979

[3] D K Kim Y A Kim and K Park ldquoGeneralizations of suffixarrays to multi-dimensional matricesrdquo Theoretical ComputerScience vol 302 no 1ndash3 pp 401ndash416 2003

[4] Y Kawamura ldquoCubic matrices generalized spin algebra anduncertainty relationrdquo Progress ofTheoretical Physics vol 110 no3 pp 579ndash587 2003

[5] R E Burkard R Rudolf and G J Woeginger ldquoThree-dimen-sional axial assignment problems with decomposable cost coef-ficientsrdquoDiscrete Applied Mathematics vol 65 no 1ndash3 pp 123ndash139 1996

[6] W P Pierskalla ldquoThe multidimensional assignment problemrdquoOperations Research vol 16 no 2 pp 422ndash431 1968

[7] M R Garey and D C Johnson Computer and Intractability AGuide To the Theory of NP-Completeness W H Freeman SanFrancisco Calif USA 1979

[8] W W Bein P Brucker J K Park and P K Pathak ldquoA Mongeproperty for the d-dimensional transportation problemrdquo Dis-crete Applied Mathematics vol 58 no 2 pp 97ndash109 1995

[9] D A Blokh ldquoAssignment problem for three-index Jacobimatri-cesrdquoAutomation and Remote Control vol 50 no 5 pp 672ndash6771989

[10] Y Crama and F C R Spieksma ldquoApproximation algorithms forthree-dimensional assignment problems with triangle inequal-itiesrdquo European Journal of Operational Research vol 60 no 3pp 273ndash279 1992

[11] F C R Spieksma and G J Woeginger ldquoGeometric three-dimensional assignment problemsrdquo European Journal of Oper-ational Research vol 91 no 3 pp 611ndash618 1996

[12] E Balas and M J Saltzman ldquoAn algorithm for the three-indexassignment problemrdquo Operations Research vol 39 no 1 pp150ndash161 1991

[13] G Gutin B Goldengorin and J Huang ldquoWorst case analysisof Max-Regret greedy and other heuristics for multidimen-sional assignment and traveling salesman problemsrdquo Journal ofHeuristics vol 14 no 2 pp 169ndash181 2008

[14] B-J Kim W L Hightower P M Hahn Y-R Zhu and L SunldquoLower bounds for the axial three-index assignment problemrdquoEuropean Journal of Operational Research vol 202 no 3 pp654ndash668 2010

[15] D Blokh and G Gutin ldquoMaximizing traveling salesman prob-lem for special matricesrdquo Discrete Applied Mathematics vol 56no 1 pp 83ndash86 1995

[16] D Blokh and E Levner ldquoAn approximation algorithm withperformance guarantees for the maximum traveling salesmanproblemon specialmatricesrdquoDiscrete AppliedMathematics vol119 no 1-2 pp 139ndash148 2002

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Stochastic AnalysisInternational Journal of

2 Journal of Discrete Mathematics

Suppose that layers 1198711119904minus1

1198711119904 and 1198711

119904+1are nonbasic One

of the layers 1198712119904or 1198713119904is basic In these layers the basic

entries may be 119886119904minus1119904119904minus1

119886119904minus1119904119904

119886119904119904119904minus1

119886119904119904119904 119886119904119904119904+1

119886119904+1119904119904

and 119886

119904+1119904119904+1for the layer 1198712

119904 and 119886

119904minus1119904minus1119904 119886119904minus1119904119904

119886119904119904minus1119904

119886119904119904119904 119886119904+1119904119904

119886119904119904+1119904

119886119904+1119904+1119904

for the layer 1198713119904 However this

contradicts the assumption that the layers 1198711119904minus1

1198711119904 and 1198711

119904+1

are nonbasicThe fact that the two first and the two last layersmay not be nonbasic is proved in a similar way

All assertions given below represent conditions thatprevent replacing nonbasic trajectory entries with entries thatinclude a basic one

Consider the sequence of layers 11987111 1198711

2 119871

1

119904minus1 1198711

119904

1198711

119904+1 119871

1

119899minus1 and 1198711

119899 Given below are possible arrange-

ments of layers in this sequence A basic layer 1198711119904is denoted by

1 a nonbasic layer 1198711119904is denoted by 0

Consider

(1) 11100111 1100111 1110011 110011

(2) 111010111 01111010 01011110 001111010 010111100 0101111010

(3) 0111100 00111100 0011110

(4) 0111 1110 10111 11101 011110

We prove the first arrangement of item 1 Suppose thatthe two nonbasic layers are followed only by two basic layersthat is 00110 Let there be layers 1198711

119904 1198711119904+1

1198711119904+2

1198711119904+3

1198711

119904+4 Consider layer pairs1198712

119904+11198713119904+1

1198712119904+1

1198713119904+2

1198712119904+2

1198713119904+1

1198712119904+2

and 1198713

119904+2 At least one layer of each pair is basic These four

pairs of layers contain at least two basic entries and theseentries are located in layers 1198711

119904+2and 1198711

119904+3 The pair 1198712

119904+3 1198713119904+3

includes a basic layer The first coordinate of the basic layerentry may be 119904 + 2 119904 + 3 or 119904 + 4 But the 1198711

119904+2and 1198711

119904+3

layers are occupied while the 1198711119904+4

layer is nonbasic Thereis a contradiction

The other arrangements are proved in a similar wayThus nonbasic layers may not be arranged closer than

those in the above variants But these variants do not allow forthe composition of a combination containingmore than [1198993]nonbasic layers Hence it follows that if a trajectory includesmore than [1198993] nonbasic entries then one of the variants isviolated and nonbasic trajectory entries can be replaced by aset of entries containing a basic entry

3 Application to the Three-DimensionalAssignment Problem

The three-dimensional assignment problem (AP3) is animportant combinatorial optimization problem It is suf-ficient to note that the particular case of AP3 the 3-dimensional matching problem is one of the six main NP-hard problems [7] The formal AP3 statement is as followsfor a matrix 119886

119894119895119896 isin 119877

3

119899 find permutations 119909 119910 isin 119878

119899such

that sum119899119894=1119886119894119909(119894)119910(119894)

is maximized

In this paper the AP3 is considered for a special class of119899 times 119899 times 119899-matrices 119880(119899) A matrix

10038171003817100381710038171003817119886119894119895119896

10038171003817100381710038171003817isin 119880 (119899) if 119886

119894119895119896ge 0

min119886119894119895119896isin119866

119886119894119895119896ge 119862 sdot max119886119894119895119896isin119867

119886119894119895119896 119862 ge 3

(3)

One of AP3 interpretations is the following There are 119899employees and two job sets of 119899 jobs each If the 119894th employeeperforms the 119895th job of the first set and the 119896th job of the sec-ond set then the effect equals 119886

119894119895119896 It is required to distribute

the jobs among the employees in such away so as tomaximizethe total effect

Let us describe the situation that will lead to the AP3 formatrices from 119880(119899) As a rule the employees are orderedby qualification while the jobs are ordered by complexityA higher effect is reached when a more qualified employeeperforms a more complex job and we may arrive at the AP3for matrices from 119880(119899)

The particular cases of AP3 [5 8ndash11] the nonpolynomialexact algorithms for the AP3 [6 12] and heuristics for theAP3 [13 14] were considered

The NP-hard particular cases of the traveling salesmanproblem with sets of matrices whose structures are similar tothose ofmatrices from119880(119899) have been considered previously[15 16]

Theorem 3 The AP3 for matrices from 119880(119899) is NP-hard

Proof Let 119887119894119895119896 isin 1198773

119899 119887119894119895119896ge 0 and numbers119860119861 are such that

119860 = 3 sdotmax 119887119894119895119896 119861 = 119899119860

Consider 4119899 times 4119899 times 4119899-matrix 119886119894119895119896 with nonnegative

entries


= 119861 119904 = 1 119899


= 119861 119904 = 1 119899

119886411990441199044119904

= 119861 119904 = 1 119899

119886119903119906V = 119860 119903 119906 V = 1 119899 if |119903 minus 119906| le 1

|119903 minus V| le 1 |119906 minus V| le 1 119886119903119906V = 119861


= 119887119894119895119896

119894 119895 119896 = 1 119899

119886119894119895119896= 0 in other cases

(4)

The matrix 119886119894119895119896 isin 119880(4119899)

All entries that are equal to 119861 belong to the optimal solu-tion of AP3 for 119886

119894119895119896matrix This signifies that if the optimal

solution of AP3 for 119886119894119895119896 matrix is known then the optimal

solution of AP3 for 119887119894119895119896matrix is known as well Therefore

the AP3 for arbitrary matrices is polynomially reducible tothe AP3 for matrices from 119880(119899)

Corollary 4 If 119886119894119895119896 isin 119880(119899) then a set T exists on which the

AP3 optimum is reached and the number of 119879 cap 119867 set entriesis no more than [1198993]

Indeed if for optimal T of matrix 119886119894119895119896 isin 119880(119899) the number

of 119879 cap 119867 set entries is more than [1198993] then according to




1198861015840

119894119895119896 matrix where 1198861015840

119894119895119896= 119886119894119895119896

for 1198861015840119894119895119896isin 119866 and 1198861015840

119894119895119896= 0 for

1198861015840



Proof Let 119886119894119895119896 isin 119880(119899) 119898


119894119895119896


1= sum119899

119894=1119886119894119891(119894)119892(119894)

and the number of119886119894119891(119894)119892(119894)




119894119895119896


119894=11198861015840

119894119891(119894)119892(119894)and1198981015840 ge 119898

3

Insofar as 119886119894119895119896ge 1198861015840

119894119895119896 119894 119895 119896 = 1 119899 then 119898

1ge 1198981015840

Hence1198983le 1198981015840

le 1198981and (119898

1minus 1198981015840

)1198981le (1198981minus 1198983)1198981=

11989821198981


119886119894119895119896isin119867119886119894119895119896


119886119894119895119896isin119866119886119894119895119896

then 11989821198981le 1(3119862) Hence (119898

1minus 1198981015840

)1198981le 1(3119862) and




119894119895119896 such that 119886

119894119895119896ge

0 for 119886119894119895119896isin 119866 and 119886

119894119895119896= 0 for 119886


programming

Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












1198861015840

119894119895119896 matrix where 1198861015840

119894119895119896= 119886119894119895119896

for 1198861015840119894119895119896isin 119866 and 1198861015840

119894119895119896= 0 for

1198861015840



Proof Let 119886119894119895119896 isin 119880(119899) 119898


119894119895119896


1= sum119899

119894=1119886119894119891(119894)119892(119894)

and the number of119886119894119891(119894)119892(119894)




119894119895119896


119894=11198861015840

119894119891(119894)119892(119894)and1198981015840 ge 119898

3

Insofar as 119886119894119895119896ge 1198861015840

119894119895119896 119894 119895 119896 = 1 119899 then 119898

1ge 1198981015840

Hence1198983le 1198981015840

le 1198981and (119898

1minus 1198981015840

)1198981le (1198981minus 1198983)1198981=

11989821198981


119886119894119895119896isin119867119886119894119895119896


119886119894119895119896isin119866119886119894119895119896

then 11989821198981le 1(3119862) Hence (119898

1minus 1198981015840

)1198981le 1(3119862) and




119894119895119896 such that 119886

119894119895119896ge

0 for 119886119894119895119896isin 119866 and 119886

119894119895119896= 0 for 119886


programming

Acknowledgments


References
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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