Research Article The Largest Laplacian Spectral Radius of...

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 563067 13 pageshttpdxdoiorg1011552013563067

Research ArticleThe Largest Laplacian Spectral Radius of UnicyclicGraphs with Fixed Diameter

Haixia Zhang

Department of Mathematics Taiyuan University of Science and Technology Taiyuan 030024 China

Correspondence should be addressed to Haixia Zhang zhanghaixiasshotmailcom

Received 27 September 2013 Accepted 11 November 2013

Academic Editor Baolin Wang

Copyright copy 2013 Haixia ZhangThis 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

We identify graphs with the maximal Laplacian spectral radius among all unicyclic graphs with 119899 vertices and diameter 119889

1 Introduction

Following [1] let 119866 = (119881(119866) 119864(119866)) be a simple undirectedgraph on 119899 vertices and119898 edges (so 119899 = |119881(119866)| is its order and119898 = |119864(119866)| is its size) For V isin 119881(119866) 119889

119866(V) or 119889(V) denotes

the degree of V and 119873(V) denotes the set of all neighbors ofvertex V A pendant vertex is a vertex of degree 1 and a pendantedge is an edge incident with a pendant vertex Let 119875119881(119866) =V 119889119866(V) = 1 For two vertices 119906 and V (119906 = V) the distance

between 119906 and V is the number of edges in the shortest pathjoining 119906 and V The diameter of a graph is the maximumdistance between any two vertices of 119866 Let 119875 = V0V1 V119904(119904 ge 1) be a path of 119866 with 119889(V1) = sdot sdot sdot = 119889(V119904minus1) = 2 (unless119904 = 1) If 119889(V0) 119889(V119904) ge 3 then we call 119875 an internal path of119866 if 119889(V0) ge 3 and 119889(V119904) = 1 then we call 119875 a pendant pathof 119866 if the subgraph induced by 119881(119875) in 119866 is 119875 itself thatis 119866[119881(119875)] = 119875 then we call 119875 an induced path Obviouslythe shortest path between any two distinct vertices of 119866 is aninduced path We will use 119866 minus V 119866 minus 119906V to denote the graphobtained from 119866 by deleting a vertex V isin 119881(119866) or an edge119906V isin 119864(119866) respectively (this notation is naturally extended ifmore than one vertex or edge is deleted)

Denote by119862119899and119875119899the cycle and the pathwith 119899 vertices

respectively We call 119866 a unicyclic graph if 119898 = 119899 where 119899 isthe number of vertices and119898 is the number of edges We willuseC119889

119899to denote the sets of all unicyclic graphswith 119899 vertices

and diameter 119889 Let 119896119899be a graph of order 119899 obtained from

the cycle1198624 by attaching 119899minus119889minus2 pendant edges and a path oflength 119889minus119896minus1 at one vertex of the cycle and a path of length

119896 minus 1 to another nonadjacent vertex of the cycle respectivelywhere 0 le 119896 minus 1 le 119889 minus 119896 minus 1

Let 119871(119866) = 119863(119866) minus 119860(119866) be the Laplacian matrix where119863(119866) is the diagonalmatrix and119860(119866) is the adjacencymatrixThe matrix 119871(119866) is real symmetric and positive semidefinitethe eigenvalues of 119871(119866) can be arranged as 120583

1(119866) ge sdot sdot sdot ge

120583119899(119866) = 0 where the largest eigenvalue 120583

1(119866) is called the

Laplacian spectral radius of 119866The investigation on the Laplacian spectral radius of

graphs is an important topic in the theory of graph spectraRecently the problem concerning graphs with maximalLaplacian spectral radius of a given class of graphs hasbeen studied extensively Li et al [2] determined thosegraphs which maximized Laplacian spectral radius amongall bipartite graphs with (edge-) connectivity at most 119896 andcharacterized graphs of order 119899 with 119896 cut-edges havingLaplacian spectral radius equal to 119899 X L Zhang and H PZhang [3] studied the largest Laplacian spectral radius ofthe bipartite graphs with 119899 vertices and 119896 cut edges and thebicyclic bipartite graphs respectively The Laplacian spectralradius of unicyclic graphs has been studied by many authors(see [4ndash6]) Liu et al [7] determined the graphs with thelargest Laplacian spectral radii among all unicyclic graphsand bicyclic graphs with 119899 vertices and 119896 pendant verticesHua et al [8] determined extremal graphs with maximalLaplacian spectral radius among all unicyclic graphs withgiven order and given pendant vertices number

In 2007 Liu et al [9] determined graphswith themaximalspectral radius among all unicyclic graphs with 119899 vertices

2 Journal of Applied Mathematics

and diameter 119889 In 2012 He and Li [6] identified graphswith themaximal signless Laplacian spectral radius among allunicyclic graphs with 119899 vertices of diameter 119889 Next Guo [4]considered the Laplacian spectral radius of unicyclic graphswith fixed diameter and proposed Conjecture 1

In this paper we prove the conjecture as Theorem 1

Theorem 1 Let 119866 be a graph in C119889119899 3 le 119889 le 119899 minus 2 Consider

the following

(i) If 119889 is odd 1205831(119866) le 120583

1(lfloor1198892rfloor

119899) and equality holds if

and only if 119866 cong lfloor1198892rfloor

119899

(ii) If 119889 is even and 119899 minus 119889 minus 2 = 0 1 1205831(119866) le 120583


119899)

and equality holds if and only if 119866 cong lfloor1198892rfloor

119899

(iii) If 119889 is even and 119899 minus 119889 minus 2 ge 2 1205831(119866) le 1205831(

lfloor1198892rfloorminus1

119899)

and equality holds if and only if 119866 cong lfloor1198892rfloorminus1

119899

The rest of this paper is organized as follows In Section 2we present some notations and lemmas which will be usedlater on In Section 3 we determine graphs with the largestLaplacian spectral radius among all unicyclic graphs with 119899vertices and diameter 119889

2 Lemmas

In this section we list some lemmas which will be used toprove our main results

Lemma 2 (see [10]) Suppose that 119906 V are two distinct verticesof a connected graph 119866 Let 119866

119905be the graph obtained from 119866

by attaching 119905 new paths VV1198941V1198942 V

119894119902119894(119894 = 1 2 119905) at V

Let 119883 = (1199091 1199092 119909119899)119879 where 119909119894 corresponds to the vertex

V119894 (1 le 119894 le 119899) be a unit eigenvector of 119866119905 corresponding to1205831(119866119905) ge 4 Let

119866119906= 119866119905minus VV11minus VV21minus sdot sdot sdot minus VV

1199051

+ 119906V11+ 119906V21+ sdot sdot sdot + 119906V

1199051

(1)

If |119909119906| ge |119909V| then 1205831(119866119906) ge 120583

1(119866119905) Further if |119909

119906| gt |119909V|

then 1205831(119866119906) gt 1205831(119866119905)

Lemma 3 (see [10]) Let 119906V be a pendant edge of a connectedgraph 119866 with 119899 ge 2 vertices and let V be a pendant vertexLet 119866

1 1198662 119866

119896(119896 ge 2) be 119896 disjoint connected graphs

and let V119894be a vertex of 119866

119894(119894 = 1 2 119896) Let 1198661015840 be the

graph obtained by adding 119896 new edges VV1 VV2 VV

119896among

1198661198661 1198662 119866

119896 Let

119866lowast= 1198661015840minus VV1minus VV2minus sdot sdot sdot minus VV

119896

+ 119906V1 + 119906V2 + sdot sdot sdot + 119906V119896(2)

(i) If 119899 = 2 then 1205831(119866lowast) = 1205831(1198661015840)

(ii) If 119899 ge 3 then 1205831(119866lowast) ge 120583

1(1198661015840) with equality if and

only if either1205831(119866lowast) = 1205831(119866) or there exists some 119894 (1 le

119894 le 119896) such that 1205831(119866lowast) = 1205831(119866119894)

Let V be a vertex of a connected graph 119866 with at least twovertices Let 119866

119896119897(119897 ge 119896 ge 1) be the graph obtained from 119866

by attaching two new paths 119875 V(= V0)V1V2 V

119896and 119876

V(= V0)11990611199062 119906

119897of length 119896 and 119897 respectively at V where

1199061 1199062 119906

119897and V

1 V2 V

119896are distinct new vertices Let

119866119896minus1119897+1

= 119866119896119897minus V119896minus1

V119896+ 119906119897V119896

Lemma 4 (see [11]) Let 119866 be a connected graph on 119899 ge

2 vertices and V be a vertex of 119866 Let 119866119896119897 be the graphdefined as previously mentioned If 119897 ge 119896 ge 1 then1205831(119866119896minus1119897+1) le 1205831(119866119896119897) with equality if and only if there existsa unit eigenvector of 119866119896119897 corresponding to 1205831(119866119896119897) taking thevalue 0 on vertex V

Lemma 5 (see [1]) Let 1198661015840 be a graph obtained by deleting anedge from the graph 119866 Then 120583119894(119866) ge 120583119894(119866

1015840) ge 120583119894+1(119866) 119894 =

1 119899 minus 1

Let 1198781198943be a graph obtained from the cycle 1198623 by attaching

119894 pendant edges at one vertex of the cycle 1198623

Lemma 6 (see [5]) Let 119866 be a unicyclic graph on 119899 verticesthen 120583

1(119866) ge 120583

1(119862119899) when 119899 = 4 the equality holds if and only

if119866 cong 119862119899 when 119899 = 4 the equality holds if and only if119866 cong 119862

4

119866 cong 1198781

3

Lemma 7 Let 119866 be a connected graph with at least one edgeletΔ(119866) be its maximal degree and let 119889119894 be the degree of vertexV119894 and119898119894 = sumV119895isin119873(V119894)

119889119895119889119894 then

(i) 1205831(119866) ge Δ(119866) + 1 the equality holds if and only if

Δ(119866) = 119899 minus 1 [12]

(ii) 1205831(119866) le max119889

119894+ 119898119894| V119894isin 119881(119866) the equality

holds if and only if119866 is regular or semiregular bipartitegraph [13]

Let 119871V119894(119866) be the principal submatrix obtained from119871(119866) by deleting the corresponding row and column of V

119894

Generally let119871119878(119866) be the principal submatrix obtained from

119871(119866) by deleting the corresponding rows and columns of allvertices of 119878 For any square matrix 119861 denote by Φ(119861) =

Φ(119861 119909) = det(119909119868 minus 119861) the characteristic polynomial of 119861In particular if 119861 = 119871(119866) we write Φ(119871(119866)) by Φ(119866) forconvenience If 119866 = 119906 then suppose thatΦ(119871119906(119866)) = 1

Let 119866 = 1198661119906 V1198662 be the graph obtained by joining thevertex 119906 of the graph 1198661 to the vertex of V of the graph 1198662 byan edge We call 119866 a connected sum of 1198661 at 119906 and 1198662 at V

Lemma 8 Let 1198661 and 1198662 be two graphs If Φ(1198661) gt Φ(1198662)

for 119909 ge 1205831(1198662) then 1205831(1198661) lt 1205831(1198662) (In general let 119891(119909)and 119892(119909) be polynomials with positive leading coefficients If119891(119909) gt 119892(119909) for 119909 ge 120583

1(119892(119909)) then 120583

1(119891(119909)) lt 120583

1(119892(119909))

where 1205831(119892(119909)) and 120583

1(119891(119909)) are the largest roots of 119892(119909) = 0

and 119891(119909) = 0 resp)

Proof If 1205831(1198661) ge 120583

1(1198662) then Φ(119866

1)|119909=1205831(1198661)

= 0Φ(1198662)|119909=1205831(1198661)

ge 0 a contradiction

Journal of Applied Mathematics 3

Lemma 9 (see [14]) Let 119866 = 1198661119906 V119866

2be a connected sum

of 1198661at 119906 and 119866

2at V then

Φ (119866) = Φ (1198661)Φ (119866

2) minus Φ (119866

1)Φ (119871V (1198662))

minus Φ (1198662)Φ (119871119906 (1198661))

(3)

Lemma 10 (see [14]) Let 119866 be a connected graph with 119899

vertices which consists of a subgraph119867 and 119899minus |119881(119867)| distinctpendant edges (not in119867) attaching to a vertex V in119867 Then

Φ (119866) = (119909 minus 1)119899minus|119881(119867)|

Φ (119867)

minus (119899 minus |119881 (119867)|) 119909(119909 minus 1)119899minus|119881(119867)|minus1

Φ(119871V (119867))

(4)

Lemma 11 (see [15]) Let 119863119899 (119899 ge 1) be the matrix obtained

from 119871(119875119899+2) by deleting the rows and columns corresponding

to two pendant vertices of 119875119899+2

suppose that Φ(1198630) = 1

Φ(119863minus119899) = 0 then

(i) xΦ(119863119899minus1) = Φ(119875119899)(ii) Φ(119863

119899+1) = (119909 minus 2)Φ(119863

119899) minus Φ(119863

119899minus1)

(iii) Φ(119863119898+1

)Φ(119863119899)minusΦ(119863

119898)Φ(119863119899+1) = Φ(119863

119898)Φ(119863119899minus1)minus

Φ(119863119898minus1)Φ(119863119899) (119899119898 ge 1)(iv) Φ(119862119899) = Φ(119863119899) minus Φ(119863119899minus2) + 2(minus1)

119899+1

From Lemma 11(i) all eigenvalues of 119863119899

are 2 +

2 cos(119894120587(119899 + 1)) where 1 le 119894 le 119899 Other characterizationsofΦ(119863119899) can be shown below

Lemma 12 Let 119863119899 (119899 ge 1) be the matrix as above Consider

the following

(i) If 119899 ge 1 then Φ(119863119899) gt Φ(119863

119899minus1) when 119909 ge 4

(ii) If 119899 ge 1 then Φ(119863119899) gt 2Φ(119863

119899minus1) when 119909 ge 5

(iii) Φ(119863119898)Φ(119863119899) minus Φ(119863

119898minus1)Φ(119863119899+1) = Φ(119863

119899minus119898) where

0 le 119898 le 119899

Proof From Lemma 11(ii) it is easy to prove (i) and (ii) byintroduction on 119899

By Lemma 11(iii) we have

Φ(119863119898)Φ (119863119899) minus Φ (119863119898minus1)Φ (119863119899+1)

= Φ (119863119898minus1

)Φ (119863119899minus1) minus Φ (119863

119898minus2)Φ (119863

119899)

= Φ (119863119898minus2

)Φ (119863119899minus2) minus Φ (119863

119898minus3)Φ (119863

119899minus1)

= sdot sdot sdot = Φ (1198630)Φ (119863

119899minus119898) minus Φ (119863

minus1)Φ (119863

119899minus119898+1)

= Φ (119863119899minus119898

)

(5)

as desired

Lemma 13 Suppose that 119906 V are two adjacent vertices of thecycle 119862119902 where 119902 is even Let 119867119896119897 (119897 ge 119896 ge 1) be thegraph obtained from 119862

119902by attaching two new paths 119875 V(=

V0) V1V2 V

119896and 119876 119906(= 119906

0) 11990611199062 119906

119897of length 119896 and

119897 at V and 119906 respectively where 1199061 1199062 119906

119897and V1 V2 V

119896

are distinct new vertices Let 119867119896minus1119897+1

= 119867119896119897minus V119896minus1

V119896+ 119906119897V119896

Then 1205831(119867119896minus1119897+1

) lt 1205831(119867119896119897)

Proof Using Lemma 9 we have

Φ(119867119896minus1119897+1

) minus Φ (119867119896119897)

= 119909 [Φ (119871V119896minus1 (119867119896minus1119897)) minus Φ (119871119906119897 (119867119896minus1119897))]

= 119909 [(119909 minus 1)Φ (119871V119896minus1 119906119897 (119867119896minus1119897)) minus Φ (119871V119896minus1 119906119897119906119897minus1 (119867119896minus1119897))]

minus [(119909 minus 1)Φ (119871119906119897 V119896minus1 (119867119896minus1119897))

minusΦ (119871119906119897 V119896minus1V119896minus2 (119867119896minus1119897))]

= sdot sdot sdot = 119909 [119887119896119897 (119909) minus 119886119896119897 (

119909)]

(6)

where119886119896119897 (119909) = [Φ (119862119902

) minus 2Φ (119863119902minus1) + Φ (119863

119902minus2)]Φ (119863

119897minus119896)

minus [Φ (119863119902minus1) minus Φ (119863

119902minus2)]Φ (119863

119897minus119896minus1)

119887119896119897 (119909) = [Φ (119863119902minus1

) minus Φ (119863119902minus2)]Φ (119863

119897minus119896+1)

minus Φ (119863119902minus2)Φ (119863119897minus119896)

(7)

From Lemmas 11(ii) and 11(iv) (6) becomes

Φ(119867119896minus1119897+1

) minus Φ (119867119896119897)

= 119909Φ (119863119897minus119896) [(119909 minus 2)Φ (119863119902minus2

) minus 2Φ (119863119902minus3) + 2]

(8)

which is greater than 0 when 119909 ge 4 by Lemma 12(i) And1205831(119867119896119897) gt 4 follows from Lemma 7(i) Thus 1205831(119867119896minus1119897+1) lt1205831(119867119896119897) holds by Lemma 8

For 119866 isin C119889119899 we have 119899 ge 3 and 1 le 119889 le 119899 minus 2 If 119889 = 1

then 119866 cong 1198623 If 119889 = 2 then 119866 cong 119862

4 119866 cong 119862

5or 119866 cong 119878

119899minus3

3 By

Lemma 7 1205831(119878119899minus3

3) has the largest Laplacian spectral radius

Therefore in the following we assume that 3 le 119889 le 119899minus2Let 119867

0be the unicyclic graph of order 119889 + 2 shown

in Figure 1 Let 1198670(1199012 119901

119889 119901119889+2) be a graph of order 119899

obtained from 1198670by attaching 119901

119894pendant vertices to each

V119894isin 119881(119867

0) V1 V119889+1 respectively where 119901

119889+2= 0 when

119896 = 1 or 119896 = 119889 Denote that

H119889

119899= 119867

0(1199012 119901

119889 119901119889+2)

119889

sum

119894=2

119901119894+ 119901119889+2

= 119899 minus 119889 minus 2

H119889

119899= 1198670(0 0 119901

119894 0 0) = 119867

0(119901119894) 119901119894= 119899 minus 119889 minus 2

(9)

Lemma 14 Let 119866 isinH119889119899 Then there is a graph 119866lowast isinH

119889

119899such

that 1205831(119866lowast) ge 1205831(119866)

Proof Let 119866 isinH119889119899 H119889

119899and let 119883 = (1199091 1199092 119909119899)

119879

be a unit eigenvector of 1205831(119866) where 119909

119894corresponds to the

vertex V119894 (1 le 119894 le 119899) Let 119905 = |119901119894 119901119894 = 0| Then 119905 ge 2Let 119901119894 119901119895 = 0 119894 lt 119895 Assume without loss of generality that|119909119894| ge |119909119895| Let119873(V119895)⋂119875119881(119866) = 1199061 1199062 119906119901119895 Let

119866lowast

1= 119866 minus V

1198951199061minus sdot sdot sdot minus V

119895119906119901119895+ V1198941199061+ sdot sdot sdot + V

119894119906119901119895 (10)


d+2

d+1k+21 kminus1 k k+1

middot middot middotmiddot middot middot

H0

(a)

d+1k+21 kminus1 k k+1 k+3

U0

d+2


(b)

Figure 1

By Lemma 2 we have 1205831(119866lowast) ge 120583

1(119866) Note that 119866

1isin H119889

119899

for 119905 = 2 and 1198661isinH119889119899 H119889

119899for 119905 gt 2 If 119905 gt 2 then we

will use119866lowast1to repeat the above step until the cardinality of 119901

119894

being nonzero is only one So we have 119866lowast2 119866lowast

3 119866

lowast

119905minus1and

1205831(119866lowast

2) le 1205831(119866lowast

3) le sdot sdot sdot le 120583

1(119866lowast

119905minus1) Note that 119866lowast

119905minus1isin H119889

119899 and

hence the lemma holds

Lemma 15 For any 119866 isin H119889

119899 1205831(119866) le 120583

1(1198670(119901119896+1)) where

0 le 119896minus1 le 119889minus119896 the equality holds if and only if119866 cong 1198670(119901119896+1)

Proof Suppose that 119899minus119889minus2 = 119905 If 119905 = 0 the result is obviousIf 119905 ge 1 by Lemma 7 we have

1205831 (119866) lt max max 119889

119894+ 119898119894| V119894isin 119881 (119866) | 119866 isinH

119889

119899

119894 notin 119889 + 2 119896 119896 + 1

= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (1198670 (119901119896+1)) + 1

lt 1205831(1198670(119901119896+1))

(11)

Case 1 119896 minus 1 lt 119889 minus 119896When 119894 = 119889 + 2 let119873(V

119894)⋂119875119881(119866) = 119906

1 1199062 119906

119905 Let

119866lowast=

1198670(119901119894) minus V1198941199061minus sdot sdot sdot minus V

119894119906119905+ V1198961199061+ sdot sdot sdot + V

119896119906119905

1003816100381610038161003816119909119896

1003816100381610038161003816ge1003816100381610038161003816119909119894

1003816100381610038161003816

1198670(119901119894) minus V119896minus1

V119896+ V119896minus1

V119894

1003816100381610038161003816119909119896

1003816100381610038161003816lt1003816100381610038161003816119909119894

1003816100381610038161003816

(12)

Then in all cases 119866lowast cong 1198670(119901119896) Thus by Lemma 21205831(1198670(119901119889+2)) le 1205831(1198670(119901119896))

Next we show that Φ(1198670(119901119896)) gt Φ(1198670(119901119896+1)) for 119909 ge

1205831(1198670(119901119896+1)) Because 119867

0minus V119896cong 119875119896minus1

⋃119875119889minus119896+2

we canget Φ(119867

0) in which the rows and columns correspond to

vertices as the ordering V1 V

119896minus1 V119889+2 V119896+1 V119896+2 V

119889+1

Furthermore let 11986411= [119890119894119895] be a square matrix of order 119896minus1

where 119890119896minus1119896minus1

= 1 and 119890119894119895= 0whenever 119894 = 119896minus1 and 119895 = 119896minus1

let 119865119896119896= [119891119894119895] be a square matrix of order 119889 minus 119896 + 2 where

119891119896119896= 1 and 119891

119894119895= 0 whenever 119894 = 119896 and 119895 = 119896 Then

119871V119896 (1198670) = (

119871 (119875119896minus1) + 11986411

00 119871 (119875

119889minus119896+2) + 11986511+ 11986522

) (13)

Hence

Φ(119871V119896 (1198670))

= Φ (119871 (119875119896minus1) + 11986411)Φ (119871 (119875

119889minus119896+2) + 11986511+ 11986522)

(14)

In order to simplify the notation we denoteΦ(119871(119875119896minus1) +11986411)

and Φ(119871(119875119889minus119896+2

) + 11986511+ 11986522) by 119891

119896minus11(119909) and 119891

119889minus119896+22(119909)

respectively Similarly

Φ(119871V119896+1 (1198670))

= Φ (119871 (119875119889minus119896) + 11986411)Φ (119871 (119875

119896+1) + 11986511+ 11986522)

= 119891119889minus1198961 (

119909) 119891119896+12 (119909)

(15)

In general by Lemma 11(ii) we have

11989111989911(119909) = Φ (1198631198991

) + Φ (1198631198991minus1)

11989111989922

(119909) = (119909 minus 1) (119909 minus 3)Φ (1198631198992minus2) minus (2119909 minus 3)Φ (1198631198992minus3

)

(16)

Hence by Lemma 12(iii)

Φ(119871V119896+1 (1198670)) minus Φ (119871V119896 (1198670))

= 119891119889minus1198961 (

119909) 119891119896+12 (119909) minus 119891119896minus11 (

119909) 119891119889minus119896+22 (119909)

= (119909 minus 1) (119909 minus 3)

times [Φ (119863119896minus1)Φ (119863

119889minus119896minus1) minus Φ (119863

119896minus2)Φ (119863

119889minus119896)]

+ (2119909 minus 3) [Φ (119863119896minus1)Φ (119863


119896minus2)Φ (119863

119889minus119896)]

= 119909 (119909 minus 2)Φ (119863119889minus2119896)

(17)

From (17) and Lemma 10

Φ(1198670(119901119896)) minus Φ (119867

0(119901119896+1))

= 119905 (119909 minus 2) (119909 minus 1)119905minus11199092Φ(119863119889minus2119896

) gt 0

(18)

holds for 119909 ge 1205831(1198670(119901119896+1))

Thus 1205831(1198670(119901119896)) lt 1205831(1198670(119901119896+1)) follows from Lemma 8

Case 2 119896 minus 1 = 119889 minus 119896First note that119867

0(119901119896) cong 119867

0(119901119896+1)

Next since1198670minus V119889+2

= 119875119889+1

we can deriveΦ(119871V119889+2(1198670))

in which the rows and columns correspond to vertices as theordering V

1 V

119896minus1 V119896 V119896+1 V

119889+1 Then

Φ(119871V119889+2 (1198670))

= [Φ (119863119896minus1) + Φ (119863

119896minus2)]

times [(1199092minus 5119909 + 5)Φ (119863

119896minus1) minus (2119909 minus 5)Φ (119863119896minus1

)]

minus [Φ (119863119896minus2) + Φ (119863

119896minus3)]

times [(119909 minus 2)Φ (119863119896minus1) minus 2Φ (119863

119896minus2)]

(19)


Combining Lemma 10 with (15) and (19) we get

Φ(1198670 (119901119889+2)) minus Φ (1198670 (119901119896+1))

= 119905(119909 minus 1)119905minus1119909Φ (119863

119896minus2) [(119909 minus 2)Φ (119863119896minus1

) minus 2Φ (119863119896minus2)]

(20)which is greater than 0 when 119909 ge 120583

1(1198670(119901119896+1)) by

Lemma 12(i) Thus by Lemma 8 1205831(1198670(119901119889+2)) lt

1205831(1198670(119901119896+1)) holds Hence the proof is completed

Let Δ119896119899be a graph of order 119899 obtained from a triangle by

attaching 119899minus119889minus2 pendant edges and a path of length 119889minus119896 atone vertex of the triangle and a path of length 119896minus1 to anothervertex of the triangle respectively where 0 le 119896 minus 1 le 119889 minus 119896

Lemma 16 1205831(Δ119896minus1119899 ) lt 1205831(Δ119896

119899) where 2 le 119896 le lceil1198892rceil

Proof Suppose that 119899 minus 119889 minus 2 = 119905 by Lemma 7(i) 1205831(Δ119896

119899) gt

119905 + 4 Let119867lowast = 1198670minus V1 by Lemma 9

Φ(Δ119896minus1

119889+2) minus Φ (Δ

119896

119889+2)

= 119909 [Φ (119871V2 (119867lowast)) minus Φ (119871V119889+1 (119867

lowast))]

= 119909 [(119909 minus 1)Φ (119871V2 V119889+1 (119867lowast)) minus Φ (119871V2 V119889+1V119889 (119867

lowast))

minus (119909 minus 1)Φ (119871V119889+1 V2 (119867lowast)) + Φ (119871V119889+1V2 V3 (119867

lowast))]

= sdot sdot sdot = 119909 [119892119889119896 (

119909) minus 119891119889119896 (119909)]

(21)where

119891119889119896 (

119909) = (1199093minus 81199092+ 18119909 minus 8)Φ (119863

119889minus2119896+1)

minus (1199092minus 5119909 + 5)Φ (119863

119889minus2119896)

119892119889119896 (119909) = (119909

2minus 5119909 + 5)Φ (119863119889minus2119896+2)

minus (119909 minus 2)Φ (119863119889minus2119896+1)

(22)

By Lemmas 10 and 11(ii) and (15) and (21)

Φ(Δ119896minus1

119899) minus Φ (Δ

119896

119899)

= (119909 minus 1)119905[Φ (Δ

119896minus1

119889+2) minus Φ (Δ

119896

119889+2)]

minus 119905119909(119909 minus 1)119905minus1[Φ (119871V119896 (Δ

119896minus1

119889+2)) minus Φ (119871V119896+1 (Δ

119896

119889+2))]

= (119909 minus 1)1199051199092(119909 minus 4)Φ (119863119889minus2119896+1) minus 119905119909(119909 minus 1)

119905minus1

times [119891119889minus119896+11 (

119909) 1198911198962 (119909) minus 119891119889minus1198961 (

119909) 119891119896+12 (119909)]

= 1199092(119909 minus 1)

119905minus1

times [(119909 minus 1) (119909 minus 119905 minus 4)Φ (119863119889minus2119896+1) + 119905 (119909 minus 2)Φ (119863119889minus2119896

)]

gt 0

(23)

for 119909 ge 1205831(Δ119896

119899)

So 1205831(Δ119896minus1

119899) lt 1205831(Δ

119896

119899) follows from Lemma 8

In view of Lemma 16 the next corollary is obvious

Corollary 17 1205831(Δ119896

119899) le 1205831(Δlceil1198892rceil

119899) where 1 le 119896 le lceil1198892rceil the

equality holds if and only if Δ119896119899cong Δlceil1198892rceil

119899

Let 1198800be the unicyclic graph of order 119889 + 2 shown

in Figure 1 Let 1198800(1199012 119901




V119894isin 119881(119880

0) V1 V119889+1 respectively Denote

U119889

119899= 1198800(1199012 119901

119889 119901119889+2)

119889

sum

119894=2

119901119894+ 119901119889+2

= 119899 minus 119889 minus 2

U119889

119899= 1198800(0 0 119901

119894 0 0) = 119880

0(119901119894) 119901119894= 119899 minus 119889 minus 2

(24)

Lemma 18 Let 119866 isinU119889119899 Then there is a graph 119866lowast isin U

119889

119899such

that 1205831(119866lowast) ge 1205831(119866)

Proof The proof is similar to that of Lemma 14

Lemma 19 For any 119866 isin U119889

119899 1205831(119866) le 120583

1(1198800(119901119896+2)) where

0 le 119896 minus 1 le 119889 minus 119896 minus 1 the equality holds if and only if 119866 cong

1198800(119901119896+2)

Proof Suppose that 119899 minus 119889 minus 2 = 119905 If 119905 = 0 the result is trivialIf 119905 ge 1 by Lemma 7 we have

1205831 (119866) lt max max 119889

119894+ 119898119894| V119894isin 119881 (119866) | 119866 isin U

119889

119899

119894 notin 119889 + 2 119896 119896 + 1 119896 + 2

= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (1198800(119901119896+2)) + 1

lt 1205831 (1198800 (119901119896+2))

(25)

Case 1 119894 = 119896If 119896 minus 1 = 119889 minus 119896 minus 1 119880

0(119901119896) cong 1198800(119901119896+2)

If 119896 minus 1 lt 119889 minus 119896 minus 1 we can obtain Φ(119871V119896+2(1198800)) in whichthe rows and columns correspond to vertices as the orderingV119889+2 V119896+1 V119896 V

1 V119896+3 V

119889+1 Then by Lemma 11(ii)

Φ(119871V119896+2 (1198800))

= (119909 minus 2) [(1199092minus 4119909 + 2)Φ (119863

119896minus1) minus 2 (119909 minus 1)Φ (119863119896minus2

)]

times [Φ (119863119889minus119896minus1

) + Φ (119863119889minus119896minus2

)]

(26)

Similarly

Φ(119871V119896 (1198800))

= (119909 minus 2) [(1199092minus 4119909 + 2)Φ (119863

119889minus119896minus1) minus 2 (119909 minus 1)Φ (119863119889minus119896minus2

)]

times [Φ (119863119896minus1) + Φ (119863

119896minus2)]

(27)


Combining the two equations above with Lemmas 10 and12(iii) we get

Φ(1198800(119901119896)) minus Φ (119880

0(119901119896+2))

= 119905119909(119909 minus 1)119905minus1[Φ (119871V119896+2 (1198800)) minus Φ (119871V119896 (1198800))]

= 1199051199092(119909 minus 2)

2(119909 minus 1)

119905minus1Φ(119863119889minus2119896minus1

) gt 0

(28)

for 119909 ge 1205831(1198800(119901119896+2))

From Lemma 8 1205831(1198800(119901119896)) lt 120583

1(1198800(119901119896+2)) holds

Case 2 119894 = 119896 + 1 119889 + 2From Lemma 7 when 119905 ge 2 we have

1205831 (1198800 (119901119896+1))

lt max 119889119894+ 119898119894| V119894isin 119881 (119880

0(119901119896+1))

= 119905 + 2 +

3 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (1198800 (119901119896+2)) + 1

lt 1205831(1198800(119901119896+2))

(29)

For 119905 = 1 we can obtain Φ(119871V119896+1(1198800)) in which therows and columns correspond to vertices as the orderingV1 V119896minus1 V119896 V119889+2 V119896+2 V119889+1 Then

Φ(119871V119896+1 (1198800)) = [Φ (119863119896minus1) + Φ (119863119896minus2)] 119891 (119909)

minus [Φ (119863119896minus2) + Φ (119863

119896minus3)] 119892 (119909)

= [(119909 minus 2) 119892 (119909) minus ℎ (119909)]Φ (119863119896minus1)

minus [2119892 (119909) + ℎ (119909)]Φ (119863119896minus2)

(30)

where

119891 (119909) = (1199093minus 71199092+ 14119909 minus 7)Φ (119863119889minus119896minus1)

minus (21199092minus 9119909 + 7)Φ (119863


119892 (119909) = (119909 minus 1) (119909 minus 3)Φ (119863119889minus119896minus1) minus (2119909 minus 3)Φ (119863119889minus119896minus2)

ℎ (119909) = (119909 minus 3) 119892 (119909) minus 119891 (119909)

= (119909 minus 2)Φ (119863119889minus119896minus1) minus 2Φ (119863


(31)

By (26) and (30)

Φ(119871V119896+2 (1198800)) minus Φ (119871V119896+1 (1198800))

= 119909 (119909 minus 1)Φ (119863119889minus119896minus2) [(119909 minus 3)Φ (119863119896minus1

) minus 2Φ (119863119896minus2)]

minus Φ (119863119889minus119896minus1)Φ (119863119896minus2)

(32)

From Lemmas 11(ii) and 12(i) when 119909 ge 5

(119909 minus 1)Φ (119863119889minus119896minus2) minus Φ (119863


= Φ (119863119889minus119896minus2

) + Φ (119863119889minus119896minus3

) gt 0

(119909 minus 3)Φ (119863119896minus1) minus 3Φ (119863

119896minus2)

=

(1199092minus 5119909 + 3)Φ(119863119896minus2)minus (119909 minus 3)Φ(119863119896minus3) gt 0 if 119896 ge 2

119909 minus 3 gt 0 if 119896 = 1(33)

hold (since (1199092 minus5119909+3)minus (119909minus3) = 1199092 minus6119909+6 gt 0 for 119909 ge 5)By Lemma 9 and (32) when 119909 ge 5

Φ(1198800(119901119896+1)) minus Φ (119880

0(119901119896+2))

= 119909 [Φ (119871V119896+2 (1198800)) minus Φ (119871V119896+1 (1198800))] gt 0(34)

From Lemma 8 1205831(1198800(119901119896+1)) lt 1205831(1198800(119901119896+2)) holds

Hence we complete the proof

Let 119896119899be a graph of order 119899 obtained from the cycle 119862

4

by attaching 119899 minus 119889 minus 2 pendant edges and a path of length119889minus119896minus1 at one vertex of the cycle and a path of length 119896minus1 toanother nonadjacent vertex of the cycle respectively where0 le 119896 minus 1 le 119889 minus 119896 minus 1

Lemma 20 Let 119896119899be a graph defined as above then

(i) if 119896 minus 1 lt 119889 minus 119896 minus 1 then 1205831(119896minus1

119899) lt 120583

1(119896

119899) where

2 le 119896 le lfloor1198892rfloor(ii) if 119896 minus 1 = 119889 minus 119896 minus 1(ie 119896 = lfloor1198892rfloor) then

(a) when 119899 minus 119889 minus 2 = 0 or 1 1205831(119896minus1119899 ) lt 1205831(119896

119899)

(b) when 119899 minus 119889 minus 2 ge 2 1205831(119896

119899) lt 1205831(

119896minus1

119899)

Proof Suppose that 119899minus119889minus2 = 119905 by Lemma 7(i)1205831(119896

119899) gt 119905+4

holds Let 119880lowast = 1198800 minus V1 by Lemma 9

Φ(119896minus1

119889+2) minus Φ (

119896

119889+2)


lowast))]

= 119909 [Φ (119871V119889+1 V2 V3 (119880lowast)) minus Φ (119871V2 V119889+1V119889 (119880

lowast))]

= sdot sdot sdot = 119909 [119904119889119896 (

119909) minus 119905119889119896 (119909)]

(35)

where

119904119889119896 (

119909) = (119909 minus 2) [(119909 minus 1) (119909 minus 4)Φ (119863119889minus2119896+1)

minus (119909 minus 2)Φ (119863119889minus2119896)]

119905119889119896 (

119909) = (119909 minus 2) [(119909 minus 3) (1199092minus 5119909 + 2)Φ (119863

119889minus2119896)

minus (119909 minus 1) (119909 minus 4)Φ (119863119889minus2119896minus1) ]

(36)



Φ(119896minus1

119899) minus Φ (

119896

119899)

= (119909 minus 1)119905[Φ (

119896minus1

119889+2) minus Φ (

119896

119889+2)]

minus 119905119909(119909 minus 1)119905minus1[Φ (119871V119896+1 (

119896minus1

119889+2)) minus Φ (119871V119896+2 (

119896

119889+2))]

= 1199092(119909 minus 1)

119905minus1(119909 minus 2) 119906 (119909)

(37)

where 119906(119909) = (1199092minus(119905+5)119909+4)Φ(119863119889minus2119896)+119905(119909minus2)Φ(119863119889minus2119896minus1)When 119896 minus 1 lt 119889 minus 119896 minus 1 by Lemma 11(ii) we get

119906 (119909) = (119909 minus 2) (1199092minus (119905 + 5) 119909 + 119905 + 4)Φ (119863119889minus2119896minus1

)

minus (1199092minus (119905 + 5) 119909 + 4)Φ (119863119889minus2119896minus2

)

(38)

Denote by 119890(119905) the largest root of 1199092 minus (119905 + 5)119909 + 4 = 0 Then

1199092minus (119905 + 5) 119909 + 4 le 0 if 119909 isin (119905 + 4 119890 (119905)]

1199092minus (119905 + 5) 119909 + 119905 + 4

gt 1199092minus (119905 + 5) 119909 + 4 gt 0 if 119909 gt 119890 (119905)

(39)

Hence by Lemma 12(i) (37) is greater than 0 when 119909 ge

1205831(119896

119899) So 1205831(

119896minus1

119899) lt 1205831(

119896


When 119896 minus 1 = 119889 minus 119896 minus 1 (37) becomes

Φ(119896minus1

119899) minus Φ (

119896

119899)

= 1199092(119909 minus 1)

119905minus1(119909 minus 2) [(119909

2minus (119905 + 5) 119909 + 4)]

(40)

If 119905 = 1 let 27and 1

7be two graphs with 119889 = 4 Through

Maple 15 the largest root of 1199092minus6119909+4 = 0 is e(1) = 52363 (upto four decimal places) which is less than 120583

1(2

7) = 53145

and 1205831(1

7) = 53008 From Lemma 5 120583

1(2

7) le 1205831(119896

119899) and

1205831(1

7) le 1205831(

119896minus1

119899) hold when 119899 ge 7 So (40) is greater than

0 for 119909 ge 1205831(119896

119899) By Lemma 8 120583

1(119896minus1

119899) lt 1205831(119896

119899) holds

If 119905 ge 2 by Lemmas 11(ii) and 12(iii)

119905 (119909) minus [(1199092minus (119905 + 5) 119909 + 4)]

times [(119909 minus 2)Φ2(119863119898) minus (119905 + 4)Φ (119863119898

)Φ (119863119898minus1

)]

= 2119905 (119909 minus 1)Φ2(119863119898) minus [(119905

2+ 3119905 + 4) 119909 minus 4]

times Φ (119863119898)Φ (119863

119898minus1) + 2 (119905 + 2) (119909 minus 1)Φ

2(119863119898minus1

)

= [21199051199092minus (1199052+ 9119905 + 4) 119909 + (4119905 + 4)]Φ (119863119898)Φ (119863119898minus1)

minus 2119905 (119909 minus 1)Φ (119863119898)Φ (119863119898minus2)

+ 2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

= [21199051199092minus (1199052+ 9119905 + 4) 119909 + (4119905 + 4)]Φ (119863119898

)Φ (119863119898minus1

)

+ 4 (119909 minus 1)Φ2(119863119898minus1

) + 2119905 (119909 minus 1)

(41)

d+3 d+2

d+11 kminus1 vk k+1


W0

(a)

d+2d+3



k

d+3

(b)

Figure 2

where

119905 (119909) = [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]

times Φ2(119863119898) minus [(119905 + 4) 119909

2minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863

119898minus1) + 2 (119905 + 2) (119909 minus 1)Φ

2(119863119898minus1

)

(42)

(one may refer to (52) in Lemma 24)By Lemma 12(ii) (119909minus2)Φ2(119863

119898)minus(119905+4)Φ(119863

119898)Φ(119863119898minus1

) gt

0 when 119909 ge 119905 + 4 And by derivative 21199051199092 minus (1199052 + 9119905 + 4)119909 +(4119905 + 4) gt 0 when 119909 ge 119905 + 4

Thus (41) is greater than 0 for 119909 ge 119890(119905) gt 119905 + 4 FromLemma 8 120583

1(119896

119899) lt 119890(119905) holds Put 119909 = 120583

1(119896

119899) into (40)

whose right side is less than 0 So 1205831(119896

119899) lt 1205831(

119896minus1

119899) We

complete the proof

Form Lemma 20 the below corollary holds

Corollary 21 When 1 le 119896 le lfloor1198892rfloor

(i) if 119889 is odd then 1205831(119896

119899) le 1205831(lfloor1198892rfloor

119899) the equality holds

if and only if 119896119899cong lfloor1198892rfloor

119899

(ii) if 119889 is even then

(a) when 119899 minus 119889 minus 2 = 0 1 1205831(119896

119899) le 1205831(lfloor1198892rfloor

119899) the

equality holds if and only if 119896119899cong lfloor1198892rfloor

119899


119899) le 1205831(lfloor1198892rfloorminus1

119899) the

equality holds if and only if 119896119899cong lfloor1198892rfloorminus1

119899

Let 1198820be the unicyclic graph of order 119889 + 3 shown in

Figure 2 Let1198820(1199012 119901119889 119901119889+2 119901119889+3) be a graph of order 119899obtained from 1198820

by attaching 119901119894pendant vertices to each

V119894 isin 119881(1198670) V1 V119889+1 respectively When 119896 = 2 119901119889+2 and119901119889+3 = 0 Denote that

W119889

119899= 119882

0(1199012 119901

119889 119901119889+2 119901119889+3)

119889

sum

119894=2

119901119894 + 119901119889+2 + 119901119889+3 = 119899 minus 119889 minus 3

W119889

119899= 1198820(0 0 119901

119894 0 0)

= 1198820(119901119894) 119901119894= 119899 minus 119889 minus 3

(43)


Lemma 22 Let119866 isinW119889119899 Then there is a graph119866lowast isinW

119889

119899such

that 1205831(119866lowast) ge 1205831(119866)


Lemma 23 For any119866 isinW119889

119899 1205831(119866) le 1205831(1198820(119901119896)) where 1 le

119896minus1 le 119889minus119896+1 the equality holds if and only if119866 cong 1198820(119901119896+1)


1205831 (119866)

lt max max 119889119894+ 119898119894| V119894isin 119881 (119866) | 119866 isinW

119889

119899 119894 = 119896

= 119905 + 2 +

2 + 4 + 119905

119905 + 2

lt 119905 + 5 = Δ (1198820(119901119896)) + 1

lt 1205831 (1198820 (119901119896))

(44)

Hence the lemma holds

3 Main Results and Their Proofs

In this section we first show that 1205831(nablalfloor1198892rfloor+1

119899) lt 1205831(Δ

lceil1198892rceil

119899) lt


119899)

Lemma 24 1205831(Δlceil1198892rceil

119899) lt 1205831(lfloor1198892rfloor

119899)

Proof Suppose that 119899 minus 119889 minus 2 = 119905 by Lemma 7 119905 + 4 lt

1205831(Δlceil1198892rceil

119899) 1205831(lfloor1198892rfloor

119899) lt 119905 + 5 holds We distinguish the

following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)Let Δ119898+1

119899and 119898

119899be two graphs on the left of Figure 3 If

119905 = 0 denote Δ119898+1119899

and 119898119899by 1198661 and 1198662 respectively Let

1198671 = 1198661 minus V119898+3 minus sdot sdot sdot minus V119889+1 and1198672 = 1198662 minus V119898+3 minus sdot sdot sdot minus V119889+1By Lemma 9

Φ(1198661) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198671)

minus Φ (119875119898)Φ (119871V119898+2 (1198671))

Φ (1198662) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198672)

minus Φ (119875119898)Φ (119871V119898+2 (1198672))

(45)

where

Φ(1198671) = 119909 (119909 minus 3) [(119909 minus 3)Φ (119863119898

) minus 2Φ (119863119898minus1

)]

Φ (119871V119898+2 (1198671))

= (119909 minus 1) (119909 minus 3)Φ (119863119898) minus (2119909 minus 3)Φ (119863119898minus1

)

Φ (1198672)

= 119909 (119909 minus 2)

times [(119909 minus 2) (119909 minus 4)Φ (119863119898minus1) minus 2 (119909 minus 3)Φ (119863119898minus2

)]

Φ (119871V119898+2 (1198672))

= (119909 minus 2) [(1199092minus 4119909 + 2)Φ (119863119898minus1) minus 2 (119909 minus 1)Φ (119863119898minus2)]

(46)

Note thatΦ(119875119898) = 119909Φ(119863

119898minus1) andΦ(119871V119898+2(119875119898)) = Φ(119863119898minus1)+

Φ(119863119898minus2

)Combining the equations above with Lemma 10 we get

Φ(Δ119898+1

119899)

= (119909 minus 1)119905minus1[(119909 minus 1)Φ (1198661

) minus 119905119909Φ (119871V119898+2 (1198661))]

= 119909(119909 minus 1)119905minus1(119909 minus 1) (119909 minus 3) (119909 minus 3 minus 119905)Φ

2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (6119905 + 12)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 3) 119909 minus 3 (119905 + 1)]Φ2(119863119898minus1

)

(47)

Φ(119898

119899)


) minus 119905119909Φ (119871V119898+2 (1198662))]

= 119909 (119909 minus 2) (119909 minus 1)119905minus1

times (119909 minus 1) [1199093minus (119905 + 8) 119909

2+ (4119905 + 18) 119909 minus (2119905 + 10)]

times Φ2(119863119898minus1)

minus [31199093minus (3119905 + 19) 119909

2+ 8 (119905 + 4) 119909 minus 4 (119905 + 4)]

times Φ (119863119898minus1

)Φ (119863119898minus2

) + 2 (119909 minus 1) (119909 minus 3 minus 119905)

times Φ2(119863119898minus1

)

(48)

Hence by Lemmas 11(ii) and 12(i) when1205831(119898+1

119899) le 119909 lt 119905+5

Φ(Δ119898+1

119899) minus Φ (

119898

119899)

= 119909(119909 minus 1)119905[119889 (119909)Φ

2(119863119898minus1

) + 119890 (119909)Φ (119863119898minus1)Φ (119863

119898minus2)

minus119903 (119909)Φ2(119863119898minus2

)]

gt 119909(119909 minus 1)119905[119889 (119909) + 119890 (119909) minus 119903 (119909)]Φ

2(119863119898minus2

) ge 0

(49)


where

119889 (119909) = minus (119909 minus 1) (119909 minus 119905 minus 5) gt 0

119890 (119909) = (1199092minus 2119909 + 2) (119909 minus 119905 minus 4) gt 0

119903 (119909) = (119909 minus 1) (119909 minus 119905 minus 3) gt 0

119889 (119909) + 119890 (119909) minus 119903 (119909) = (119909 minus 2)2(119909 minus 119905 minus 4) gt 0

(50)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1

119899) holds by Lemma 8

Case 2 119889 = 2119898 + 2 (119898 ge 1)LetΔ119898+1119899

and119898+1119899

be two graphs on the right of Figure 3If 119905 = 0 denoteΔ119898+1

119899and119898+1

119899by119866lowast1and119866lowast

2 respectively By

similar computations as Case 1 we have

Φ(Δ119898+1

119899)


lowast

1) minus 119905119909Φ (119871V119898+2 (119866

lowast

1))]

= 119909(119909 minus 1)119905minus1(119909 minus 1)

2(119909 minus 3) (119909 minus 4 minus 119905)Φ

2(119863119898)

minus (119909 minus 1) [31199092minus (3119905 + 16) 119909 + (6119905 + 18)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 4) 119909 + 3 (119905 + 2)]Φ

2(119863119898minus1)

(51)

Φ(119898+1

119899)


lowast

2) minus 119905119909Φ (119871V119898+3 (119866

lowast

2))]


times [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]Φ

2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863119898minus1)

+2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

(52)

Hence by Lemmas 11(ii) 12(i) and 12(iii) when 119909 ge 119905 + 4

Φ(Δ119898+1

119899) minus Φ (

119898+1

119899)

= 119909(119909 minus 1)119905

times minus [(119905 + 1) 1199092minus (3119905 + 5) 119909 + (119905 + 4)]Φ

2(119863119898)

+ [(119905 + 1) 1199093minus (5119905 + 5) 119909

2+ (7119905 + 10) 119909 minus (2119905 + 6)]

times Φ (119863119898)Φ (119863

119898minus1)

minus [(2119905 + 2) 1199092minus (4119905 + 4) 119909 + (119905 + 2)]Φ

2(119863119898minus1

)

1middot middot middotmiddot middot middot

middot middot middot

2m+3

m+2

m+3m+1m 2m+2

m+1

n

(a)



m+2

m+3m+1m

2m+4

2m+3

m+1

n

(b)



2m+3

m+2

m+3m+1m 2m+2

⋄m

n

(c)

middot middot middot1



m+3

m+1 m+2m

2m+4

2m+3

⋄m+1

n

(d)

1


2m+3

m+2

m+1m

2m+4

2m+2


m+1

n

(e)

1


2m+4

m+3m+2m+1

2m+5

2m+3


m+2

n

(f)

Figure 3

= 119909(119909 minus 1)119905

times [119901 (119909)Φ (119863119898minus1) minus 119902 (119909)Φ (119863119898minus2

)]Φ (119863119898minus1

) minus 119908 (119909)

ge 119909(119909 minus 1)119905Φ(119863119898minus1

)

times [(119901 (119909) minus 119908 (119909))Φ (119863119898minus1) minus 119902 (119909)Φ (119863119898minus2

)] gt 0

(53)

where

119901 (119909) = 21199093minus (119905 + 9) 119909

2+ (119905 + 9) 119909 minus 2 gt 0

119902 (119909) = 2(119909 minus 1)2gt 0

119908 (119909) = (119905 + 1) 1199092minus (3119905 + 5) 119909 + (119905 + 4) gt 0

119901 (119909) minus 119908 (119909) minus 119902 (119909)

= 21199093minus (2119905 + 12) 119909

2+ (4119905 + 18) 119909 minus (119905 + 8) gt 0

(54)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1


Let nabla119896119899be a graph of order 119899 obtained from a triangle by

attaching 119899 minus 119889 minus 3 pendant edges a path of length 119896 minus 1 anda path of length 119889 minus 119896 + 1 at one vertex of the triangle where1 le 119896 minus 1 le 119889 minus 119896 + 1

Lemma 25 1205831(nablalfloor1198892rfloor+1119899) lt 1205831(Δ

lceil1198892rceil

119899)

Proof Suppose that 119899 minus 119889 minus 3 = 119905 by Lemma 7(i)1205831(nablalfloor1198892rfloor+1

119899) 1205831(Δlceil1198892rceil

119899) gt 119905 + 5 holds We distinguish the

following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)


Letnabla119898+1119899

be a graph on the left of Figure 3 If 119905 = 0 denotenabla119898+1

119899by 1198663 Let119867

3= 1198663minus V1minus sdot sdot sdot minus V

119898 By Lemma 9

Φ(1198663) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]Φ (1198673)

minus Φ (119875119898)Φ (119871V119898+1 (1198673))

(55)

where

Φ(1198673) = 119909 (119909 minus 3)

times [(119909 minus 1) (119909 minus 4)Φ (119863119898) minus (119909 minus 3)Φ (119863119898minus1

)]

Φ (119871V119898+1 (1198673)) = (119909 minus 1) (119909 minus 3)

times [(119909 minus 1)Φ (119863119898) minus Φ (119863

119898minus1)]

(56)

By Lemma 10 we get

Φ(nabla119898+1

119899)


) minus 119905119909Φ (119871V119898+1 (1198663))]

= 119909 (119909 minus 3) (119909 minus 1)119905

times (119909 minus 1) (119909 minus 119905 minus 4)Φ2(119863119898) minus [(119905 + 4) 119909 minus (2119905 + 6)]

times Φ (119863119898)Φ (119863

119898minus1) + (119905 + 2)Φ

2(119863119898minus1

)

(57)

By (47) we have

Φ(Δ119898+1

119899) = 119909(119909 minus 1)

119905

times (119909 minus 1) (119909 minus 3) (119909 minus 119905 minus 4)Φ2(119863119898)

minus [(119905 + 5) 1199092minus (6119905 + 22) 119909 + (6119905 + 18)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 5) 119909 minus 3 (119905 + 2)]Φ2(119863119898minus1

)

(58)

Hence

Φ(nabla119898+1

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)

times [(119909 minus 119905 minus 4)Φ (119863119898) minus (119905 + 3)Φ (119863119898minus1

)]

(59)

When119898 = 1 (59) = 1199092(119909 minus 1)119905+1(119909 minus 119905 minus 5) gt 0 holds for119909 gt 119905 + 5 So 120583

1(nabla119898+1

119899) lt 1205831(Δ119898+1


When 119898 ge 2 1205831(nabla119898+1

119899) lt 1205831(Δ119898+1

119899) also holds (the proof

will be given in Case 2 of the lemma)

Case 2 119889 = 2119898 + 2 (119898 ge 1)

Let nabla119898+2119899

be a graph on the right of Figure 3 By similarcomputations as in Case 1 we have

Φ(nabla119898+2

119899) = 119909 (119909 minus 3) (119909 minus 1)

119905

times (119909 minus 1)2(119909 minus 119905 minus 5)Φ

2(119863119898)

minus 2 (119909 minus 1) (119909 minus 119905 minus 4)Φ (119863119898)Φ (119863

119898minus1)

+ (119909 minus 119905 minus 3)Φ2(119863119898minus1

)

(60)

By (51) we get

Φ(Δ119898+1

119899)



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 5) 119909 + (3119905 + 9)]Φ

2(119863119898minus1

)

(61)

Hence

Φ(nabla119898+2

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)

times [(119909 minus 119905 minus 4)Φ (119863119898+1) minus (119905 + 3)Φ (119863119898

)]

(62)

Let 119905119896(119909) = (119909 minus 119905 minus 4)Φ(119863

119896) minus (119905 + 3)Φ(119863

119896minus1) and the

largest root of 119905119896(119909) = 0 is denoted by 120583

1(119905119896(119909)) where 119896 ge 0

We first show that 1205831(119905119896(119909)) is strictly increasing

We use the induction on 119896 Clearly 1205831(1199050(119909)) = 119905 + 4 lt

1205831(1199051(119909)) = 119905 + 5 holds Generally assume that 120583

1(119905119896minus1(119909)) lt

1205831(119905119896(119909)) then by Lemmas 11(ii) and 12(iii)

119905119896+1 (

119909) = (119909 minus 2) 119905119896 (119909) minus 119905119896minus1 (

119909) (63)

1199052

119896(119909) minus 119905119896+1 (

119909) 119905119896minus1 (119909)

= minus [(119905 + 2) 1199092minus (119905 + 2) (119905 + 5) 119909 minus 1] = minusV (119909)

(64)

Put 119909 = 1205831(119905119896(119909)) into (63) whose right side is less than 0 So1205831(119905119896(119909)) lt 1205831(119905119896+1(119909))

Furthermore 1205831(119905119896(119909)) has the upper boundDenote that the largest root of V(119909) = 0 is1205831(V(119909)) If there

exists some119898 such that 1205831(119905119898(119909)) ge 1205831(V(119909)) we substitute 119896with 119898 and put 119909 = 1205831(119905119898(119909)) into (64) So the right side ofit is less than and equal to 0 and the left side is greater than 0a contradiction

Let 11986610158403= nabla119898+2

119899minus V1minus sdot sdot sdot minus V

119898minus1minus V119898+5

minus sdot sdot sdot minus V2119898+3

and1198661015840

1= Δ119898+1





By (60) and (62)

Φ(1198661015840

3) = 119909(119909 minus 1)

119905(119909 minus 3) (119909

2minus 3119909 + 1) 119903 (119909)

Φ (1198661015840

3) minus Φ (119866

1015840

1)

= 1199092(119909 minus 1)

119905[1199093minus (119905 + 8) 119909

2+ (3119905 + 16) 119909 minus (119905 + 6)]

= 1199092(119909 minus 1)

119905(119903 (119909) + 1)

(65)

where 119903(119909) = 1199093 minus (119905 + 8)1199092 + (3119905 + 16)119909 minus (119905 + 7)By Lemma 7(i) 120583

1(1198661015840

3) is the largest root of 119903(119909) = 0 If

1205831(1198661015840

3) ge 1205831(119866

1015840

1) then [Φ(1198661015840

3)minusΦ(119866

1015840

1)]|119909=1205831(119866

1015840

3) le 0 and 119903(119909)+

1|119909=1205831(119866

1015840

3)gt 0 a contradiction So 120583

1(1198661015840

3) lt 1205831(1198661015840

1)

By derivative when 119909 gt 119905 + 5

119909V (119909) minus (119905 + 2) 119903 (119909)

= 3 (119905 + 2) 1199092minus (31199052+ 22119905 + 33) 119909 + (119905 + 2) (119905 + 7) gt 0

(66)

From Lemma 8 1205831(V(119909)) lt 120583

1(1198661015840

3) holds Furthermore

1205831(1198661015840

3) le 1205831(nabla119898+2

119899) and 120583

1(1198661015840

1) le 1205831(Δ119898+1

119899) hold by Lemma 5

Hence (62) is greater than 0 for 119909 ge 1205831(Δ119898+1

119899) From

Lemma 8 we get 1205831(nabla119898+2

119899) lt 1205831(Δ119898+1

119899) as desired

Let ◻119896119889+3

be the unicyclic graph of order 119889 + 3 shown inFigure 2

Lemma 26 1205831(◻lceil1198892rceil

119889+3) lt 1205831(Δlceil1198892rceil

119889+3)

Proof Note that by Lemma 7(i) 1205831(Δlceil1198892rceil

119889+3) gt 5

Case 1 119889 = 2119898 + 1(119898 ge 1)By Lemma 9 we have

Φ(◻119898+1

2119898+4) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]

times Φ (1198674) minus Φ (119875

119898)Φ (119871V119898+1 (1198674))

= 119909 [(119909 minus 2)2(119909 minus 4)Φ

2(119863119898) minus 4 (119909 minus 2) (119909 minus 3)

times Φ (119863119898)Φ (119863

119898minus1) + (3119909 minus 8)Φ

2(119863119898minus1

)]

(67)

where1198674= ◻119898+1

2119898+4minus V1minus sdot sdot sdot minus V

119898

By (47) we get

Φ(Δ119898+1

2119898+4)

= 119909 [(119909 minus 1) (119909 minus 3) (119909 minus 4)Φ2(119863119898) minus (5119909

2minus 22119909 + 18)

times Φ (119863119898)Φ (119863

119898minus1) + (5119909 minus 6)Φ

2(119863119898minus1

)]

(68)

Hence by Lemmas 11(ii) 12(ii) and 12(iii) when 119909 ge 5

Φ(◻119898+1

2119898+4) minus Φ (Δ

119898+1

2119898+4)

= 119909 [(119909 minus 4)Φ2(119863119898) + (119909

2minus 2119909 minus 6)Φ (119863

119898)Φ (119863

119898minus1)

minus2 (119909 + 1)Φ2(119863119898minus1

)]

= 119909Φ (119863119898minus1

) [2 (1199092minus 4119909 + 1)Φ (119863

119898) minus (3119909 minus 2)Φ (119863119898minus1

)]

+ 119909 (119909 minus 4)

gt 119909Φ2(119863119898minus1

) (41199092minus 19119909 + 6) + 119909 (119909 minus 4) gt 0

(69)

So 1205831(◻119898+1

2119898+4) lt 1205831(Δ119898+1

2119898+4) follows from Lemma 8

Case 2 119889 = 2119898 + 2 (119898 ge 1)By a similar proof as of Case 1 120583

1(◻119898+1

2119898+5) lt 120583

1(Δ119898+1

2119898+5)

holds

Next we give the proof of Theorem 1 which is the mostimportant result

Proof of Theorem 1 Let 119866 isin C119889119899and 119883 = (119909

1 1199092 119909

119899)119879



vertex V119894(1 le 119894 le 119899)

Choose119866 isin C119889119899such that the Laplacian spectral radius of

119866 is as large as possible Then by Lemma 6 we can assumethat 119866 =119862119899 Let 119875119889+1 = V1V2 V119889+1 be the induced path oflength 119889 and let 119862119902 be the only cycle in 119866 Since 119866 =119862119899 wehave min119889(V1) 119889(V119889+1) = 1 say 119889(V1) = 1 We first showsome claims

Claim 1 119881(119862119902)⋂119881(119875

119889+1) = 0

Proof of Claim 1Otherwise since119866 is connected there existsan only path 119875 = V

119894V119896V119896+1 V

119897minus1V119897connecting 119862

119902and 119875119889+1

where V

119894isin 119881(119862

119902) V119897isin 119881(119875

119889+1)and V

119896 V

119897minus1isin 119881(119866)

(119881(119862119902)⋃119881(119875119889+1

))For each 119895 let 119879

119895be a rooted tree (with 119903

119895as its root)

attached at V119895(119896 le 119895 le 119897 minus 1) where the order of 119879

119895is

119899119895 We assume that all trees but 119879

119894 are kept fixed while 119879

119894

(along with its root) can be changed Suppose that 119879119894= 119878119899119894

Let V be a vertex belonging to 119879119894 chosen so that 119889(V) gt 2

and that 119889(V 119903119894) (the distance between V and 119903119894) is the largestBy Lemma 4 (applied in the reverse direction) the Laplacianspectral radius is increased when any hanging path at V isreplaced by a hanging star (namely edges of a hanging pathnow become the hanging edges at V) If the same is repeatedfor other hanging paths at V we get one star attached at V(its central vertex is identified with V) whose size is equal tothe sum of the lengths of the aforementioned paths Let 119908 bea vertex in 119879

119894 adjacent to V and belonging to the (unique)

path between 119903119894and V By Lemma 3 the Laplacian spectral

radius is increased when all hanging edges at V become thehanging edges at 119908 Note also that 119889(119908 119903

119894) = 119889(V 119903

119894) minus 1 By

repeating the same procedure (for any other vertex as V) wearrive at 1198661 where the rooted tree 119879119894 becomes a star 119878119899119894 sothat 120583

1(1198661) ge 1205831(119866)

By the same way to other rooted trees we arrive at 1198662

where every rooted tree 119879119895 cong 119878119899119895

(119896 le 119895 le 119897minus1) and 1205831(1198662) ge1205831(1198661) From 119866

2 applying Lemma 2 we arrive at 119866

3 where

only a rooted tree 119878119899119896+sdotsdotsdot+119899119897minus1

is attached at some vertex V119898isin

V119896 V

119897minus1 So 120583

1(1198663) ge 1205831(1198662)

From 1198663 by Lemma 3 the Laplacian spectral radius

increased when all vertices adjacent to V119897become adjacent


vertices to V119897minus1

By repeating the same procedure (for anyother vertex of 119875 minus V

119894 as V119897) we arrive at 1198664 where119881(119862119902)⋂119881(119875

119889+1) = 0 and 120583

1(1198664) ge 1205831(1198663)

Hence we have 1205831(1198664) ge 1205831(119866) a contradiction

By Claim 1 119881(119862119902)⋂119881(119875

119889+1) = 0 Denote that

119862119902

= V119896V119896+1 V

119897minus1V119897V119889+2

V119889+3 V

119904V119896(119904 ge 119889 + 2)

where V119896 V119896+1 V

119897minus1 V119897 = 119881(119862

119902)⋂119881(119875

119889+1) and

V119889+2 V119889+3 V

119904 = 119881(119862

119902) 119881(119875

119889+1)

Claim 2 119889(V) = 1 for V isin 119881(119866) (119881(119862119902)⋃119881(119875119889+1))

Proof of Claim 2 Consider other rooted trees attached at119881(119862119902) and 119881(119875119889+1) respectively By a similar proof as Claim1 (the procedure until 119866

3) we can get

119894and 120583

1(119894) ge

1205831(119866) where only a rooted tree is attached at V

119894(V119894isin

119881(119862119902)⋃119881(119875

119889+1) V

1 V119889+1) a contradiction

Claim 3 119897 = 119896 + 1 and 119904 = 119889 + 2

Proof of Claim 3 Denote that 119860 = V | V isin 119881(119866)

(119881(119862119902)⋃119881(119875119889+1

) = V119904+1 V

119899 and |119860| = 119899 minus 119904 = 119905 ge 0

Case 1 119897 = 119896 By Lemma 7

1205831(119894) le max max 119889V119894 + 119898V119894 | V119894 isin 119881 (119894) | 119894 = 119896

= 119905 + 2 +

2 + 4 + 119905

119905 + 2

le 119905 + 5 = Δ (119896) + 1

lt 1205831 (119896) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

= 119905 + 4 +

2 + 2 + 2 + 2 + 119905

119905 + 4

le 119905 + 6 le 119899 minus 119889 + 2

= Δ (nabla119896

119899) + 1 lt 120583

1(nabla119896

119899)

(70)

for 119904 ge 119889 + 4 (|119862119902| ge 4)

When 119904 = 119889 + 3 (|119862119902| = 3)

119894cong 1198820(119901119894) By Lemma 23

1205831(119894) lt 1205831(nabla119896

119899) for any 119894 = 119896 where

119896cong nabla119896

119899

Hence by Lemma 4 nablalfloor1198892rfloor+1119899

has the larger Laplacianspectral radius Furthermore from Lemmas 24 and 251205831(lfloor1198892rfloor

119899) gt 1205831(nablalfloor1198892rfloor+1

119899) holds

Case 2 119897 = 119896 + 1

Subcase 21 119905 ge 1 By Lemma 7

1205831 (119894)

le max max 119889V119894 + 119898V119894 | V119894 isin 119881 (119894) 119894 = 119896 119896 + 1

= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (119896) + 1 (Δ (

119896+1) + 1)

lt 1205831(119896) (1205831(119896+1)) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

le 119905 + 3 +

2 + 3 + 119905

119905 + 3

lt 119905 + 5 le 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(71)

for 119904 ge 119889 + 3 (|119862119902| ge 4)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) By Lemma 15

1205831(119894) lt 1205831(Δ119896

119899) for any 119894 = 119896 + 1 where

119896+1cong Δ119896

119899

Subcase 22 119905 = 0 By Lemma 7

1205831 (119894) le max max 119889V119894 + 119898V119894 | V119894 isin 119881 (119894) 119894 = 1 119889 + 1

= 3 +

2 + 3 + 2

3

lt 6 lt 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(72)

for 119904 ge 119889 + 4 (|119862119902| ge 5)

When 119904 = 119889 + 3 (|119862119902| = 4) by Lemma 13 120583

1(◻119896

119889+3) lt

1205831(◻lceil1198892rceil

119889+3) Furthermore from Lemma 26 1205831(Δ

lceil1198892rceil

119889+3) gt

1205831(◻lceil1198892rceil

119889+3)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) cong Δ119896

119899

Hence in view of Subcases 21 and 22 byCorollary 17 andLemma 24 120583


119899) gt 1205831(Δlceil1198892rceil

119899) holds

Case 3 119897 ge 119896 + 2 By Lemma 7

1205831(119894)


= 119905 + 3 +

2 + 2 + 2 + 119905

119905 + 3

le 119905 + 5 le 119899 minus 119889 + 2 = Δ (119896

119899) + 1

lt 1205831 (119896

119899)

(73)

for 119904 ge 119889 + 3 (|119862119902| ge 5)

When 119904 = 119889 + 2 (|119862119902| = 4)

119894cong 1198800(119901119894) by Lemma 19

1205831(119894) lt 1205831(

119896

119899) for any 119894 = 119896 + 2 where 119896+2 cong

119896

119899

Hence in view of Cases 1 2 and 3 and Corollary 21 if 119889is odd lfloor1198892rfloor

119899has the largest Laplacian spectral radius if 119889 is

even and 119899 minus 119889 minus 2 = 0 1 lfloor1198892rfloor119899

has the largest Laplacianspectral radius If 119889 is even and 119899 minus 119889 minus 2 ge 2 lfloor1198892rfloorminus1

119899has the

largest Laplacian spectral radius a contradictionBy Claims 1 2 and 3Theorem 1 follows immediately

Acknowledgments

The author would like to express sincere gratitude to thereferees for a very careful reading of the paper and for alltheir insightful comments and valuable suggestions whichled to improving this paper The research of the author issupported by the National Natural Science Foundation ofChina (nos 11371078 and 61303020) and the National NaturalScience Foundation of Shanxi Province (no 2012011019-2)


References

[1] D Cvetkovic M Doob and H Sachs Spectra of GraphsAcademic Press New York NY USA 1976

[2] J X Li W C Shiu and W C Han ldquoThe Laplacian spectralradius of some graphsrdquo Linear Algebra and Its Applications vol431 no 1-2 pp 99ndash103 2009

[3] X L Zhang and H P Zhang ldquoThe Laplacian spectral radius ofsome bipartite graphsrdquo Linear Algebra and Its Applications vol428 no 7 pp 1610ndash1619 2008

[4] S G Guo ldquoOn the Laplacian spectral radius of unicyclic graphswith fixed diameterrdquoArs Combinatoria C vol 6 pp 47ndash58 2012

[5] S G Guo ldquoThe largest eigenvalues of Laplacian matrices ofunicyclic graphsrdquo Applied Mathematics A vol 16 no 2 pp 131ndash135 2001 (Chinese)

[6] S S He and S C Li ldquoOn the signless Laplacian index ofunicyclic graphs with fixed diameterrdquo Linear Algebra and ItsApplications vol 436 no 1 pp 252ndash261 2012

[7] M H Liu X Z Tan and B L Liu ldquoThe (signless) Laplacianspectral radius of unicyclic and bicyclic graphs with n verticesand k pendant verticesrdquo Czechoslovak Mathematical Journalvol 60 no 3 pp 849ndash867 2010

[8] F L Hua G H Yu and A Ilic ldquoThe Laplacian spectral radiusfor unicyclic graphs with given independence numberrdquo LinearAlgebra and Its Applications vol 433 no 5 pp 934ndash944 2010

[9] HQ LiuM Lu andF Tian ldquoOn the spectral radius of unicyclicgraphs with fixed diameterrdquo Linear Algebra and Its Applicationsvol 420 no 2-3 pp 449ndash457 2007

[10] J M Guo ldquoThe Laplacian spectral radius of a graph underperturbationrdquo Computers and Mathematics with Applicationsvol 54 no 5 pp 709ndash720 2007

[11] JM Guo ldquoThe effect on the Laplacian spectral radius of a graphby adding or grafting edgesrdquo Linear Algebra and Its Applicationsvol 413 no 1 pp 59ndash71 2006

[12] R Grone and R Merris ldquoThe Laplacian spectrum of a graphrdquoSIAM Journal on DiscreteMathematics vol 7 no 2 pp 221ndash2291994

[13] R Merris ldquoA note on Laplacian graph eigenvaluesrdquo LinearAlgebra and Its Applications vol 285 no 1ndash3 pp 33ndash35 1998

[14] C X He J Y Shao and J He ldquoOn the Laplacian spectral radiiof bicyclic graphsrdquo Discrete Mathematics vol 308 no 24 pp5981ndash5995 2008

[15] J M Guo ldquoA conjecture on the algebraic connectivity ofconnected graphs with fixed girthrdquo Discrete Mathematics vol308 no 23 pp 5702ndash5711 2008

Submit your manuscripts athttpwwwhindawicom

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Mathematical PhysicsAdvances in

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OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


and diameter 119889 In 2012 He and Li [6] identified graphswith themaximal signless Laplacian spectral radius among allunicyclic graphs with 119899 vertices of diameter 119889 Next Guo [4]considered the Laplacian spectral radius of unicyclic graphswith fixed diameter and proposed Conjecture 1

In this paper we prove the conjecture as Theorem 1

Theorem 1 Let 119866 be a graph in C119889119899 3 le 119889 le 119899 minus 2 Consider

the following

(i) If 119889 is odd 1205831(119866) le 120583


119899) and equality holds if

and only if 119866 cong lfloor1198892rfloor

119899

(ii) If 119889 is even and 119899 minus 119889 minus 2 = 0 1 1205831(119866) le 120583


119899)

and equality holds if and only if 119866 cong lfloor1198892rfloor

119899

(iii) If 119889 is even and 119899 minus 119889 minus 2 ge 2 1205831(119866) le 1205831(

lfloor1198892rfloorminus1

119899)

and equality holds if and only if 119866 cong lfloor1198892rfloorminus1

119899

The rest of this paper is organized as follows In Section 2we present some notations and lemmas which will be usedlater on In Section 3 we determine graphs with the largestLaplacian spectral radius among all unicyclic graphs with 119899vertices and diameter 119889

2 Lemmas

In this section we list some lemmas which will be used toprove our main results

Lemma 2 (see [10]) Suppose that 119906 V are two distinct verticesof a connected graph 119866 Let 119866

119905be the graph obtained from 119866

by attaching 119905 new paths VV1198941V1198942 V

119894119902119894(119894 = 1 2 119905) at V

Let 119883 = (1199091 1199092 119909119899)119879 where 119909119894 corresponds to the vertex

V119894 (1 le 119894 le 119899) be a unit eigenvector of 119866119905 corresponding to1205831(119866119905) ge 4 Let

119866119906= 119866119905minus VV11minus VV21minus sdot sdot sdot minus VV

1199051

+ 119906V11+ 119906V21+ sdot sdot sdot + 119906V

1199051

(1)

If |119909119906| ge |119909V| then 1205831(119866119906) ge 120583

1(119866119905) Further if |119909

119906| gt |119909V|

then 1205831(119866119906) gt 1205831(119866119905)

Lemma 3 (see [10]) Let 119906V be a pendant edge of a connectedgraph 119866 with 119899 ge 2 vertices and let V be a pendant vertexLet 119866

1 1198662 119866

119896(119896 ge 2) be 119896 disjoint connected graphs

and let V119894be a vertex of 119866

119894(119894 = 1 2 119896) Let 1198661015840 be the

graph obtained by adding 119896 new edges VV1 VV2 VV

119896among

1198661198661 1198662 119866

119896 Let

119866lowast= 1198661015840minus VV1minus VV2minus sdot sdot sdot minus VV

119896

+ 119906V1 + 119906V2 + sdot sdot sdot + 119906V119896(2)

(i) If 119899 = 2 then 1205831(119866lowast) = 1205831(1198661015840)

(ii) If 119899 ge 3 then 1205831(119866lowast) ge 120583

1(1198661015840) with equality if and

only if either1205831(119866lowast) = 1205831(119866) or there exists some 119894 (1 le

119894 le 119896) such that 1205831(119866lowast) = 1205831(119866119894)

Let V be a vertex of a connected graph 119866 with at least twovertices Let 119866

119896119897(119897 ge 119896 ge 1) be the graph obtained from 119866

by attaching two new paths 119875 V(= V0)V1V2 V

119896and 119876

V(= V0)11990611199062 119906

119897of length 119896 and 119897 respectively at V where

1199061 1199062 119906

119897and V

1 V2 V

119896are distinct new vertices Let

119866119896minus1119897+1

= 119866119896119897minus V119896minus1

V119896+ 119906119897V119896

Lemma 4 (see [11]) Let 119866 be a connected graph on 119899 ge

2 vertices and V be a vertex of 119866 Let 119866119896119897 be the graphdefined as previously mentioned If 119897 ge 119896 ge 1 then1205831(119866119896minus1119897+1) le 1205831(119866119896119897) with equality if and only if there existsa unit eigenvector of 119866119896119897 corresponding to 1205831(119866119896119897) taking thevalue 0 on vertex V

Lemma 5 (see [1]) Let 1198661015840 be a graph obtained by deleting anedge from the graph 119866 Then 120583119894(119866) ge 120583119894(119866

1015840) ge 120583119894+1(119866) 119894 =

1 119899 minus 1

Let 1198781198943be a graph obtained from the cycle 1198623 by attaching

119894 pendant edges at one vertex of the cycle 1198623

Lemma 6 (see [5]) Let 119866 be a unicyclic graph on 119899 verticesthen 120583

1(119866) ge 120583

1(119862119899) when 119899 = 4 the equality holds if and only

if119866 cong 119862119899 when 119899 = 4 the equality holds if and only if119866 cong 119862

4

119866 cong 1198781

3

Lemma 7 Let 119866 be a connected graph with at least one edgeletΔ(119866) be its maximal degree and let 119889119894 be the degree of vertexV119894 and119898119894 = sumV119895isin119873(V119894)

119889119895119889119894 then

(i) 1205831(119866) ge Δ(119866) + 1 the equality holds if and only if

Δ(119866) = 119899 minus 1 [12]

(ii) 1205831(119866) le max119889

119894+ 119898119894| V119894isin 119881(119866) the equality

holds if and only if119866 is regular or semiregular bipartitegraph [13]

Let 119871V119894(119866) be the principal submatrix obtained from119871(119866) by deleting the corresponding row and column of V

119894

Generally let119871119878(119866) be the principal submatrix obtained from

119871(119866) by deleting the corresponding rows and columns of allvertices of 119878 For any square matrix 119861 denote by Φ(119861) =

Φ(119861 119909) = det(119909119868 minus 119861) the characteristic polynomial of 119861In particular if 119861 = 119871(119866) we write Φ(119871(119866)) by Φ(119866) forconvenience If 119866 = 119906 then suppose thatΦ(119871119906(119866)) = 1

Let 119866 = 1198661119906 V1198662 be the graph obtained by joining thevertex 119906 of the graph 1198661 to the vertex of V of the graph 1198662 byan edge We call 119866 a connected sum of 1198661 at 119906 and 1198662 at V

Lemma 8 Let 1198661 and 1198662 be two graphs If Φ(1198661) gt Φ(1198662)

for 119909 ge 1205831(1198662) then 1205831(1198661) lt 1205831(1198662) (In general let 119891(119909)and 119892(119909) be polynomials with positive leading coefficients If119891(119909) gt 119892(119909) for 119909 ge 120583

1(119892(119909)) then 120583

1(119891(119909)) lt 120583

1(119892(119909))

where 1205831(119892(119909)) and 120583

1(119891(119909)) are the largest roots of 119892(119909) = 0

and 119891(119909) = 0 resp)

Proof If 1205831(1198661) ge 120583

1(1198662) then Φ(119866

1)|119909=1205831(1198661)

= 0Φ(1198662)|119909=1205831(1198661)

ge 0 a contradiction


Lemma 9 (see [14]) Let 119866 = 1198661119906 V119866

2be a connected sum

of 1198661at 119906 and 119866

2at V then

Φ (119866) = Φ (1198661)Φ (119866

2) minus Φ (119866

1)Φ (119871V (1198662))

minus Φ (1198662)Φ (119871119906 (1198661))

(3)



Φ (119866) = (119909 minus 1)119899minus|119881(119867)|

Φ (119867)


Φ(119871V (119867))

(4)





Φ(119863minus119899) = 0 then

(i) xΦ(119863119899minus1) = Φ(119875119899)(ii) Φ(119863

119899+1) = (119909 minus 2)Φ(119863

119899) minus Φ(119863

119899minus1)

(iii) Φ(119863119898+1

)Φ(119863119899)minusΦ(119863

119898)Φ(119863119899+1) = Φ(119863

119898)Φ(119863119899minus1)minus


119899+1


are 2 +



the following

(i) If 119899 ge 1 then Φ(119863119899) gt Φ(119863

119899minus1) when 119909 ge 4

(ii) If 119899 ge 1 then Φ(119863119899) gt 2Φ(119863

119899minus1) when 119909 ge 5

(iii) Φ(119863119898)Φ(119863119899) minus Φ(119863

119898minus1)Φ(119863119899+1) = Φ(119863


0 le 119898 le 119899



Φ(119863119898)Φ (119863119899) minus Φ (119863119898minus1)Φ (119863119899+1)

= Φ (119863119898minus1

)Φ (119863119899minus1) minus Φ (119863

119898minus2)Φ (119863

119899)

= Φ (119863119898minus2

)Φ (119863119899minus2) minus Φ (119863

119898minus3)Φ (119863

119899minus1)

= sdot sdot sdot = Φ (1198630)Φ (119863

119899minus119898) minus Φ (119863

minus1)Φ (119863

119899minus119898+1)

= Φ (119863119899minus119898

)

(5)

as desired



V0) V1V2 V

119896and 119876 119906(= 119906

0) 11990611199062 119906



119897and V1 V2 V

119896


= 119867119896119897minus V119896minus1

V119896+ 119906119897V119896

Then 1205831(119867119896minus1119897+1

) lt 1205831(119867119896119897)


Φ(119867119896minus1119897+1

) minus Φ (119867119896119897)






119909)]

(6)

where119886119896119897 (119909) = [Φ (119862119902

) minus 2Φ (119863119902minus1) + Φ (119863

119902minus2)]Φ (119863

119897minus119896)


119902minus2)]Φ (119863


119887119896119897 (119909) = [Φ (119863119902minus1

) minus Φ (119863119902minus2)]Φ (119863

119897minus119896+1)


(7)


Φ(119867119896minus1119897+1

) minus Φ (119867119896119897)


) minus 2Φ (119863119902minus3) + 2]

(8)




4 119866 cong 119862

5or 119866 cong 119878

119899minus3

3 By

Lemma 7 1205831(119878119899minus3




in Figure 1 Let 1198670(1199012 119901




V119894isin 119881(119867


119889+2= 0 when

119896 = 1 or 119896 = 119889 Denote that

H119889

119899= 119867

0(1199012 119901

119889 119901119889+2)

119889

sum

119894=2

119901119894+ 119901119889+2

= 119899 minus 119889 minus 2

H119889

119899= 1198670(0 0 119901

119894 0 0) = 119867

0(119901119894) 119901119894= 119899 minus 119889 minus 2

(9)


119889

119899such

that 1205831(119866lowast) ge 1205831(119866)

Proof Let 119866 isinH119889119899 H119889

119899and let 119883 = (1199091 1199092 119909119899)

119879




119866lowast

1= 119866 minus V


119895119906119901119895+ V1198941199061+ sdot sdot sdot + V

119894119906119901119895 (10)


d+2



H0

(a)


U0

d+2


(b)

Figure 1


1(119866) Note that 119866

1isin H119889

119899

for 119905 = 2 and 1198661isinH119889119899 H119889



119894


3 119866

lowast

119905minus1and

1205831(119866lowast

2) le 1205831(119866lowast


1(119866lowast



119899 and



119899 1205831(119866) le 120583

1(1198670(119901119896+1)) where



1205831 (119866) lt max max 119889

119894+ 119898119894| V119894isin 119881 (119866) | 119866 isinH

119889

119899

119894 notin 119889 + 2 119896 119896 + 1

= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (1198670 (119901119896+1)) + 1

lt 1205831(1198670(119901119896+1))

(11)


119894)⋂119875119881(119866) = 119906

1 1199062 119906

119905 Let

119866lowast=


119894119906119905+ V1198961199061+ sdot sdot sdot + V

119896119906119905

1003816100381610038161003816119909119896

1003816100381610038161003816ge1003816100381610038161003816119909119894

1003816100381610038161003816

1198670(119901119894) minus V119896minus1

V119896+ V119896minus1

V119894

1003816100381610038161003816119909119896

1003816100381610038161003816lt1003816100381610038161003816119909119894

1003816100381610038161003816

(12)



1205831(1198670(119901119896+1)) Because 119867


⋃119875119889minus119896+2

we canget Φ(119867



119896minus1 V119889+2 V119896+1 V119896+2 V

119889+1





119891119896119896= 1 and 119891


119871V119896 (1198670) = (

119871 (119875119896minus1) + 11986411

00 119871 (119875

119889minus119896+2) + 11986511+ 11986522

) (13)

Hence

Φ(119871V119896 (1198670))

= Φ (119871 (119875119896minus1) + 11986411)Φ (119871 (119875

119889minus119896+2) + 11986511+ 11986522)

(14)


and Φ(119871(119875119889minus119896+2

) + 11986511+ 11986522) by 119891

119896minus11(119909) and 119891

119889minus119896+22(119909)


Φ(119871V119896+1 (1198670))

= Φ (119871 (119875119889minus119896) + 11986411)Φ (119871 (119875

119896+1) + 11986511+ 11986522)

= 119891119889minus1198961 (

119909) 119891119896+12 (119909)

(15)


11989111989911(119909) = Φ (1198631198991

) + Φ (1198631198991minus1)

11989111989922


)

(16)


Φ(119871V119896+1 (1198670)) minus Φ (119871V119896 (1198670))

= 119891119889minus1198961 (

119909) 119891119896+12 (119909) minus 119891119896minus11 (

119909) 119891119889minus119896+22 (119909)

= (119909 minus 1) (119909 minus 3)

times [Φ (119863119896minus1)Φ (119863


119896minus2)Φ (119863

119889minus119896)]

+ (2119909 minus 3) [Φ (119863119896minus1)Φ (119863


119896minus2)Φ (119863

119889minus119896)]

= 119909 (119909 minus 2)Φ (119863119889minus2119896)

(17)


Φ(1198670(119901119896)) minus Φ (119867

0(119901119896+1))


) gt 0

(18)

holds for 119909 ge 1205831(1198670(119901119896+1))



0(119901119896) cong 119867

0(119901119896+1)


= 119875119889+1

we can deriveΦ(119871V119889+2(1198670))


1 V

119896minus1 V119896 V119896+1 V

119889+1 Then

Φ(119871V119889+2 (1198670))

= [Φ (119863119896minus1) + Φ (119863

119896minus2)]

times [(1199092minus 5119909 + 5)Φ (119863


)]

minus [Φ (119863119896minus2) + Φ (119863

119896minus3)]


119896minus2)]

(19)



Φ(1198670 (119901119889+2)) minus Φ (1198670 (119901119896+1))

= 119905(119909 minus 1)119905minus1119909Φ (119863


) minus 2Φ (119863119896minus2)]


1(1198670(119901119896+1)) by





Lemma 16 1205831(Δ119896minus1119899 ) lt 1205831(Δ119896



119899) gt


Φ(Δ119896minus1

119889+2) minus Φ (Δ

119896

119889+2)


lowast))]


lowast))


lowast))]

= sdot sdot sdot = 119909 [119892119889119896 (

119909) minus 119891119889119896 (119909)]

(21)where

119891119889119896 (

119909) = (1199093minus 81199092+ 18119909 minus 8)Φ (119863

119889minus2119896+1)

minus (1199092minus 5119909 + 5)Φ (119863

119889minus2119896)

119892119889119896 (119909) = (119909

2minus 5119909 + 5)Φ (119863119889minus2119896+2)


(22)


Φ(Δ119896minus1

119899) minus Φ (Δ

119896

119899)

= (119909 minus 1)119905[Φ (Δ

119896minus1

119889+2) minus Φ (Δ

119896

119889+2)]


119896minus1

119889+2)) minus Φ (119871V119896+1 (Δ

119896

119889+2))]


119905minus1

times [119891119889minus119896+11 (

119909) 1198911198962 (119909) minus 119891119889minus1198961 (

119909) 119891119896+12 (119909)]

= 1199092(119909 minus 1)

119905minus1


)]

gt 0

(23)

for 119909 ge 1205831(Δ119896

119899)

So 1205831(Δ119896minus1

119899) lt 1205831(Δ

119896



Corollary 17 1205831(Δ119896

119899) le 1205831(Δlceil1198892rceil



119899


in Figure 1 Let 1198800(1199012 119901




V119894isin 119881(119880


U119889

119899= 1198800(1199012 119901

119889 119901119889+2)

119889

sum

119894=2

119901119894+ 119901119889+2

= 119899 minus 119889 minus 2

U119889

119899= 1198800(0 0 119901

119894 0 0) = 119880

0(119901119894) 119901119894= 119899 minus 119889 minus 2

(24)


119889

119899such

that 1205831(119866lowast) ge 1205831(119866)



119899 1205831(119866) le 120583

1(1198800(119901119896+2)) where


1198800(119901119896+2)


1205831 (119866) lt max max 119889

119894+ 119898119894| V119894isin 119881 (119866) | 119866 isin U

119889

119899

119894 notin 119889 + 2 119896 119896 + 1 119896 + 2

= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (1198800(119901119896+2)) + 1

lt 1205831 (1198800 (119901119896+2))

(25)


0(119901119896) cong 1198800(119901119896+2)


1 V119896+3 V


Φ(119871V119896+2 (1198800))

= (119909 minus 2) [(1199092minus 4119909 + 2)Φ (119863


)]


) + Φ (119863119889minus119896minus2

)]

(26)

Similarly

Φ(119871V119896 (1198800))

= (119909 minus 2) [(1199092minus 4119909 + 2)Φ (119863


)]

times [Φ (119863119896minus1) + Φ (119863

119896minus2)]

(27)



Φ(1198800(119901119896)) minus Φ (119880

0(119901119896+2))


= 1199051199092(119909 minus 2)

2(119909 minus 1)


) gt 0

(28)

for 119909 ge 1205831(1198800(119901119896+2))

From Lemma 8 1205831(1198800(119901119896)) lt 120583

1(1198800(119901119896+2)) holds


1205831 (1198800 (119901119896+1))

lt max 119889119894+ 119898119894| V119894isin 119881 (119880

0(119901119896+1))

= 119905 + 2 +

3 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (1198800 (119901119896+2)) + 1

lt 1205831(1198800(119901119896+2))

(29)


Φ(119871V119896+1 (1198800)) = [Φ (119863119896minus1) + Φ (119863119896minus2)] 119891 (119909)

minus [Φ (119863119896minus2) + Φ (119863

119896minus3)] 119892 (119909)


minus [2119892 (119909) + ℎ (119909)]Φ (119863119896minus2)

(30)

where


minus (21199092minus 9119909 + 7)Φ (119863



ℎ (119909) = (119909 minus 3) 119892 (119909) minus 119891 (119909)



(31)

By (26) and (30)

Φ(119871V119896+2 (1198800)) minus Φ (119871V119896+1 (1198800))


) minus 2Φ (119863119896minus2)]


(32)




= Φ (119863119889minus119896minus2

) + Φ (119863119889minus119896minus3

) gt 0


119896minus2)

=


119909 minus 3 gt 0 if 119896 = 1(33)


Φ(1198800(119901119896+1)) minus Φ (119880

0(119901119896+2))

= 119909 [Φ (119871V119896+2 (1198800)) minus Φ (119871V119896+1 (1198800))] gt 0(34)




4




119899) lt 120583

1(119896

119899) where



119899)


119899) lt 1205831(

119896minus1

119899)


119899) gt 119905+4


Φ(119896minus1

119889+2) minus Φ (

119896

119889+2)


lowast))]


lowast))]

= sdot sdot sdot = 119909 [119904119889119896 (

119909) minus 119905119889119896 (119909)]

(35)

where

119904119889119896 (



119905119889119896 (


119889minus2119896)


(36)



Φ(119896minus1

119899) minus Φ (

119896

119899)

= (119909 minus 1)119905[Φ (

119896minus1

119889+2) minus Φ (

119896

119889+2)]


119896minus1

119889+2)) minus Φ (119871V119896+2 (

119896

119889+2))]

= 1199092(119909 minus 1)

119905minus1(119909 minus 2) 119906 (119909)

(37)



)


)

(38)


1199092minus (119905 + 5) 119909 + 4 le 0 if 119909 isin (119905 + 4 119890 (119905)]

1199092minus (119905 + 5) 119909 + 119905 + 4

gt 1199092minus (119905 + 5) 119909 + 4 gt 0 if 119909 gt 119890 (119905)

(39)


1205831(119896

119899) So 1205831(

119896minus1

119899) lt 1205831(

119896



Φ(119896minus1

119899) minus Φ (

119896

119899)

= 1199092(119909 minus 1)

119905minus1(119909 minus 2) [(119909

2minus (119905 + 5) 119909 + 4)]

(40)

If 119905 = 1 let 27and 1



1(2

7) = 53145

and 1205831(1

7) = 53008 From Lemma 5 120583

1(2

7) le 1205831(119896

119899) and

1205831(1

7) le 1205831(

119896minus1


0 for 119909 ge 1205831(119896

119899) By Lemma 8 120583

1(119896minus1

119899) lt 1205831(119896

119899) holds


119905 (119909) minus [(1199092minus (119905 + 5) 119909 + 4)]


)Φ (119863119898minus1

)]

= 2119905 (119909 minus 1)Φ2(119863119898) minus [(119905

2+ 3119905 + 4) 119909 minus 4]

times Φ (119863119898)Φ (119863

119898minus1) + 2 (119905 + 2) (119909 minus 1)Φ

2(119863119898minus1

)

= [21199051199092minus (1199052+ 9119905 + 4) 119909 + (4119905 + 4)]Φ (119863119898)Φ (119863119898minus1)


+ 2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

= [21199051199092minus (1199052+ 9119905 + 4) 119909 + (4119905 + 4)]Φ (119863119898

)Φ (119863119898minus1

)

+ 4 (119909 minus 1)Φ2(119863119898minus1

) + 2119905 (119909 minus 1)

(41)

d+3 d+2

d+11 kminus1 vk k+1


W0

(a)

d+2d+3



k

d+3

(b)

Figure 2

where

119905 (119909) = [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]

times Φ2(119863119898) minus [(119905 + 4) 119909

2minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863

119898minus1) + 2 (119905 + 2) (119909 minus 1)Φ

2(119863119898minus1

)

(42)


119898)minus(119905+4)Φ(119863

119898)Φ(119863119898minus1

) gt



1(119896

119899) lt 119890(119905) holds Put 119909 = 120583

1(119896

119899) into (40)


119899) lt 1205831(

119896minus1

119899) We

complete the proof



(i) if 119889 is odd then 1205831(119896

119899) le 1205831(lfloor1198892rfloor



119899



119899) le 1205831(lfloor1198892rfloor

119899) the


119899



119899) the


119899





W119889

119899= 119882

0(1199012 119901

119889 119901119889+2 119901119889+3)

119889

sum

119894=2

119901119894 + 119901119889+2 + 119901119889+3 = 119899 minus 119889 minus 3

W119889

119899= 1198820(0 0 119901

119894 0 0)

= 1198820(119901119894) 119901119894= 119899 minus 119889 minus 3

(43)



119889

119899such

that 1205831(119866lowast) ge 1205831(119866)



119899 1205831(119866) le 1205831(1198820(119901119896)) where 1 le



1205831 (119866)


119889

119899 119894 = 119896

= 119905 + 2 +

2 + 4 + 119905

119905 + 2

lt 119905 + 5 = Δ (1198820(119901119896)) + 1

lt 1205831 (1198820 (119901119896))

(44)




119899) lt 1205831(Δ

lceil1198892rceil

119899) lt


119899)


119899) lt 1205831(lfloor1198892rfloor

119899)



119899) 1205831(lfloor1198892rfloor


following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)Let Δ119898+1

119899and 119898


119905 = 0 denote Δ119898+1119899



Φ(1198661) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198671)

minus Φ (119875119898)Φ (119871V119898+2 (1198671))

Φ (1198662) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198672)

minus Φ (119875119898)Φ (119871V119898+2 (1198672))

(45)

where

Φ(1198671) = 119909 (119909 minus 3) [(119909 minus 3)Φ (119863119898

) minus 2Φ (119863119898minus1

)]

Φ (119871V119898+2 (1198671))


)

Φ (1198672)

= 119909 (119909 minus 2)


)]

Φ (119871V119898+2 (1198672))


(46)

Note thatΦ(119875119898) = 119909Φ(119863


Φ(119863119898minus2


Φ(Δ119898+1

119899)


) minus 119905119909Φ (119871V119898+2 (1198661))]


2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (6119905 + 12)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 3) 119909 minus 3 (119905 + 1)]Φ2(119863119898minus1

)

(47)

Φ(119898

119899)


) minus 119905119909Φ (119871V119898+2 (1198662))]



2+ (4119905 + 18) 119909 minus (2119905 + 10)]

times Φ2(119863119898minus1)

minus [31199093minus (3119905 + 19) 119909

2+ 8 (119905 + 4) 119909 minus 4 (119905 + 4)]


)Φ (119863119898minus2



)

(48)


119899) le 119909 lt 119905+5

Φ(Δ119898+1

119899) minus Φ (

119898

119899)

= 119909(119909 minus 1)119905[119889 (119909)Φ

2(119863119898minus1

) + 119890 (119909)Φ (119863119898minus1)Φ (119863

119898minus2)

minus119903 (119909)Φ2(119863119898minus2

)]

gt 119909(119909 minus 1)119905[119889 (119909) + 119890 (119909) minus 119903 (119909)]Φ

2(119863119898minus2

) ge 0

(49)


where





(50)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1


Case 2 119889 = 2119898 + 2 (119898 ge 1)LetΔ119898+1119899

and119898+1119899


119899and119898+1


2 respectively By


Φ(Δ119898+1

119899)


lowast

1) minus 119905119909Φ (119871V119898+2 (119866

lowast

1))]



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 4) 119909 + 3 (119905 + 2)]Φ

2(119863119898minus1)

(51)

Φ(119898+1

119899)


lowast

2) minus 119905119909Φ (119871V119898+3 (119866

lowast

2))]


times [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]Φ

2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863119898minus1)

+2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

(52)


Φ(Δ119898+1

119899) minus Φ (

119898+1

119899)

= 119909(119909 minus 1)119905


2(119863119898)

+ [(119905 + 1) 1199093minus (5119905 + 5) 119909

2+ (7119905 + 10) 119909 minus (2119905 + 6)]

times Φ (119863119898)Φ (119863

119898minus1)

minus [(2119905 + 2) 1199092minus (4119905 + 4) 119909 + (119905 + 2)]Φ

2(119863119898minus1

)



2m+3

m+2

m+3m+1m 2m+2

m+1

n

(a)



m+2

m+3m+1m

2m+4

2m+3

m+1

n

(b)



2m+3

m+2

m+3m+1m 2m+2

⋄m

n

(c)




m+3

m+1 m+2m

2m+4

2m+3

⋄m+1

n

(d)

1


2m+3

m+2

m+1m

2m+4

2m+2


m+1

n

(e)

1


2m+4

m+3m+2m+1

2m+5

2m+3


m+2

n

(f)

Figure 3

= 119909(119909 minus 1)119905


)]Φ (119863119898minus1

) minus 119908 (119909)

ge 119909(119909 minus 1)119905Φ(119863119898minus1

)


)] gt 0

(53)

where

119901 (119909) = 21199093minus (119905 + 9) 119909

2+ (119905 + 9) 119909 minus 2 gt 0

119902 (119909) = 2(119909 minus 1)2gt 0

119908 (119909) = (119905 + 1) 1199092minus (3119905 + 5) 119909 + (119905 + 4) gt 0

119901 (119909) minus 119908 (119909) minus 119902 (119909)

= 21199093minus (2119905 + 12) 119909

2+ (4119905 + 18) 119909 minus (119905 + 8) gt 0

(54)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1





lceil1198892rceil

119899)


119899) 1205831(Δlceil1198892rceil


following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)


Letnabla119898+1119899


119899by 1198663 Let119867


119898 By Lemma 9

Φ(1198663) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]Φ (1198673)

minus Φ (119875119898)Φ (119871V119898+1 (1198673))

(55)

where

Φ(1198673) = 119909 (119909 minus 3)


)]

Φ (119871V119898+1 (1198673)) = (119909 minus 1) (119909 minus 3)


119898minus1)]

(56)

By Lemma 10 we get

Φ(nabla119898+1

119899)


) minus 119905119909Φ (119871V119898+1 (1198663))]

= 119909 (119909 minus 3) (119909 minus 1)119905


times Φ (119863119898)Φ (119863

119898minus1) + (119905 + 2)Φ

2(119863119898minus1

)

(57)

By (47) we have

Φ(Δ119898+1

119899) = 119909(119909 minus 1)

119905


minus [(119905 + 5) 1199092minus (6119905 + 22) 119909 + (6119905 + 18)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 5) 119909 minus 3 (119905 + 2)]Φ2(119863119898minus1

)

(58)

Hence

Φ(nabla119898+1

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(59)


1(nabla119898+1

119899) lt 1205831(Δ119898+1


When 119898 ge 2 1205831(nabla119898+1

119899) lt 1205831(Δ119898+1



Case 2 119889 = 2119898 + 2 (119898 ge 1)

Let nabla119898+2119899


Φ(nabla119898+2

119899) = 119909 (119909 minus 3) (119909 minus 1)

119905


2(119863119898)


119898minus1)


)

(60)

By (51) we get

Φ(Δ119898+1

119899)



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 5) 119909 + (3119905 + 9)]Φ

2(119863119898minus1

)

(61)

Hence

Φ(nabla119898+2

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(62)

Let 119905119896(119909) = (119909 minus 119905 minus 4)Φ(119863

119896) minus (119905 + 3)Φ(119863



1(119905119896(119909)) where 119896 ge 0




1(119905119896minus1(119909)) lt


119905119896+1 (


119909) (63)

1199052

119896(119909) minus 119905119896+1 (

119909) 119905119896minus1 (119909)


(64)




Let 11986610158403= nabla119898+2




and1198661015840

1= Δ119898+1





By (60) and (62)

Φ(1198661015840

3) = 119909(119909 minus 1)

119905(119909 minus 3) (119909

2minus 3119909 + 1) 119903 (119909)

Φ (1198661015840

3) minus Φ (119866

1015840

1)

= 1199092(119909 minus 1)

119905[1199093minus (119905 + 8) 119909

2+ (3119905 + 16) 119909 minus (119905 + 6)]

= 1199092(119909 minus 1)

119905(119903 (119909) + 1)

(65)


1(1198661015840


1205831(1198661015840

3) ge 1205831(119866

1015840

1) then [Φ(1198661015840

3)minusΦ(119866

1015840

1)]|119909=1205831(119866

1015840

3) le 0 and 119903(119909)+

1|119909=1205831(119866

1015840


1(1198661015840

3) lt 1205831(1198661015840

1)


119909V (119909) minus (119905 + 2) 119903 (119909)

= 3 (119905 + 2) 1199092minus (31199052+ 22119905 + 33) 119909 + (119905 + 2) (119905 + 7) gt 0

(66)

From Lemma 8 1205831(V(119909)) lt 120583

1(1198661015840


1205831(1198661015840

3) le 1205831(nabla119898+2

119899) and 120583

1(1198661015840

1) le 1205831(Δ119898+1



119899) From


119899) lt 1205831(Δ119898+1

119899) as desired

Let ◻119896119889+3



119889+3) lt 1205831(Δlceil1198892rceil

119889+3)


119889+3) gt 5


Φ(◻119898+1

2119898+4) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]

times Φ (1198674) minus Φ (119875

119898)Φ (119871V119898+1 (1198674))

= 119909 [(119909 minus 2)2(119909 minus 4)Φ


times Φ (119863119898)Φ (119863

119898minus1) + (3119909 minus 8)Φ

2(119863119898minus1

)]

(67)

where1198674= ◻119898+1


119898

By (47) we get

Φ(Δ119898+1

2119898+4)


2minus 22119909 + 18)

times Φ (119863119898)Φ (119863

119898minus1) + (5119909 minus 6)Φ

2(119863119898minus1

)]

(68)


Φ(◻119898+1

2119898+4) minus Φ (Δ

119898+1

2119898+4)

= 119909 [(119909 minus 4)Φ2(119863119898) + (119909

2minus 2119909 minus 6)Φ (119863

119898)Φ (119863

119898minus1)

minus2 (119909 + 1)Φ2(119863119898minus1

)]

= 119909Φ (119863119898minus1

) [2 (1199092minus 4119909 + 1)Φ (119863


)]

+ 119909 (119909 minus 4)

gt 119909Φ2(119863119898minus1

) (41199092minus 19119909 + 6) + 119909 (119909 minus 4) gt 0

(69)

So 1205831(◻119898+1

2119898+4) lt 1205831(Δ119898+1



1(◻119898+1

2119898+5) lt 120583

1(Δ119898+1

2119898+5)

holds



1 1199092 119909

119899)119879



vertex V119894(1 le 119894 le 119899)



Claim 1 119881(119862119902)⋂119881(119875

119889+1) = 0


119894V119896V119896+1 V


119902and 119875119889+1

where V

119894isin 119881(119862

119902) V119897isin 119881(119875

119889+1)and V

119896 V

119897minus1isin 119881(119866)

(119881(119862119902)⋃119881(119875119889+1

))For each 119895 let 119879


119895as its root)


119895is



119894







119894) = 119889(V 119903

119894) minus 1 By


1(1198661) ge 1205831(119866)





3 where



V119896 V

119897minus1 So 120583

1(1198663) ge 1205831(1198662)







119889+1) = 0 and 120583

1(1198664) ge 1205831(1198663)


By Claim 1 119881(119862119902)⋂119881(119875


119862119902

= V119896V119896+1 V

119897minus1V119897V119889+2

V119889+3 V

119904V119896(119904 ge 119889 + 2)

where V119896 V119896+1 V

119897minus1 V119897 = 119881(119862

119902)⋂119881(119875

119889+1) and

V119889+2 V119889+3 V

119904 = 119881(119862

119902) 119881(119875

119889+1)

Claim 2 119889(V) = 1 for V isin 119881(119866) (119881(119862119902)⋃119881(119875119889+1))


3) we can get

119894and 120583

1(119894) ge


119894(V119894isin

119881(119862119902)⋃119881(119875

119889+1) V


Claim 3 119897 = 119896 + 1 and 119904 = 119889 + 2


(119881(119862119902)⋃119881(119875119889+1

) = V119904+1 V

119899 and |119860| = 119899 minus 119904 = 119905 ge 0

Case 1 119897 = 119896 By Lemma 7


= 119905 + 2 +

2 + 4 + 119905

119905 + 2

le 119905 + 5 = Δ (119896) + 1

lt 1205831 (119896) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

= 119905 + 4 +

2 + 2 + 2 + 2 + 119905

119905 + 4

le 119905 + 6 le 119899 minus 119889 + 2

= Δ (nabla119896

119899) + 1 lt 120583

1(nabla119896

119899)

(70)

for 119904 ge 119889 + 4 (|119862119902| ge 4)

When 119904 = 119889 + 3 (|119862119902| = 3)

119894cong 1198820(119901119894) By Lemma 23

1205831(119894) lt 1205831(nabla119896

119899) for any 119894 = 119896 where


119899




119899) holds

Case 2 119897 = 119896 + 1


1205831 (119894)


= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (119896) + 1 (Δ (

119896+1) + 1)

lt 1205831(119896) (1205831(119896+1)) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

le 119905 + 3 +

2 + 3 + 119905

119905 + 3

lt 119905 + 5 le 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(71)

for 119904 ge 119889 + 3 (|119862119902| ge 4)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) By Lemma 15

1205831(119894) lt 1205831(Δ119896

119899) for any 119894 = 119896 + 1 where

119896+1cong Δ119896

119899



= 3 +

2 + 3 + 2

3

lt 6 lt 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(72)

for 119904 ge 119889 + 4 (|119862119902| ge 5)

When 119904 = 119889 + 3 (|119862119902| = 4) by Lemma 13 120583

1(◻119896

119889+3) lt

1205831(◻lceil1198892rceil


lceil1198892rceil

119889+3) gt

1205831(◻lceil1198892rceil

119889+3)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) cong Δ119896

119899



119899) gt 1205831(Δlceil1198892rceil

119899) holds

Case 3 119897 ge 119896 + 2 By Lemma 7

1205831(119894)


= 119905 + 3 +

2 + 2 + 2 + 119905

119905 + 3

le 119905 + 5 le 119899 minus 119889 + 2 = Δ (119896

119899) + 1

lt 1205831 (119896

119899)

(73)

for 119904 ge 119889 + 3 (|119862119902| ge 5)

When 119904 = 119889 + 2 (|119862119902| = 4)

119894cong 1198800(119901119894) by Lemma 19

1205831(119894) lt 1205831(

119896


119896

119899





119899has the


Acknowledgments



References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Lemma 9 (see [14]) Let 119866 = 1198661119906 V119866

2be a connected sum

of 1198661at 119906 and 119866

2at V then

Φ (119866) = Φ (1198661)Φ (119866

2) minus Φ (119866

1)Φ (119871V (1198662))

minus Φ (1198662)Φ (119871119906 (1198661))

(3)



Φ (119866) = (119909 minus 1)119899minus|119881(119867)|

Φ (119867)


Φ(119871V (119867))

(4)





Φ(119863minus119899) = 0 then

(i) xΦ(119863119899minus1) = Φ(119875119899)(ii) Φ(119863

119899+1) = (119909 minus 2)Φ(119863

119899) minus Φ(119863

119899minus1)

(iii) Φ(119863119898+1

)Φ(119863119899)minusΦ(119863

119898)Φ(119863119899+1) = Φ(119863

119898)Φ(119863119899minus1)minus


119899+1


are 2 +



the following

(i) If 119899 ge 1 then Φ(119863119899) gt Φ(119863

119899minus1) when 119909 ge 4

(ii) If 119899 ge 1 then Φ(119863119899) gt 2Φ(119863

119899minus1) when 119909 ge 5

(iii) Φ(119863119898)Φ(119863119899) minus Φ(119863

119898minus1)Φ(119863119899+1) = Φ(119863


0 le 119898 le 119899



Φ(119863119898)Φ (119863119899) minus Φ (119863119898minus1)Φ (119863119899+1)

= Φ (119863119898minus1

)Φ (119863119899minus1) minus Φ (119863

119898minus2)Φ (119863

119899)

= Φ (119863119898minus2

)Φ (119863119899minus2) minus Φ (119863

119898minus3)Φ (119863

119899minus1)

= sdot sdot sdot = Φ (1198630)Φ (119863

119899minus119898) minus Φ (119863

minus1)Φ (119863

119899minus119898+1)

= Φ (119863119899minus119898

)

(5)

as desired



V0) V1V2 V

119896and 119876 119906(= 119906

0) 11990611199062 119906



119897and V1 V2 V

119896


= 119867119896119897minus V119896minus1

V119896+ 119906119897V119896

Then 1205831(119867119896minus1119897+1

) lt 1205831(119867119896119897)


Φ(119867119896minus1119897+1

) minus Φ (119867119896119897)






119909)]

(6)

where119886119896119897 (119909) = [Φ (119862119902

) minus 2Φ (119863119902minus1) + Φ (119863

119902minus2)]Φ (119863

119897minus119896)


119902minus2)]Φ (119863


119887119896119897 (119909) = [Φ (119863119902minus1

) minus Φ (119863119902minus2)]Φ (119863

119897minus119896+1)


(7)


Φ(119867119896minus1119897+1

) minus Φ (119867119896119897)


) minus 2Φ (119863119902minus3) + 2]

(8)




4 119866 cong 119862

5or 119866 cong 119878

119899minus3

3 By

Lemma 7 1205831(119878119899minus3




in Figure 1 Let 1198670(1199012 119901




V119894isin 119881(119867


119889+2= 0 when

119896 = 1 or 119896 = 119889 Denote that

H119889

119899= 119867

0(1199012 119901

119889 119901119889+2)

119889

sum

119894=2

119901119894+ 119901119889+2

= 119899 minus 119889 minus 2

H119889

119899= 1198670(0 0 119901

119894 0 0) = 119867

0(119901119894) 119901119894= 119899 minus 119889 minus 2

(9)


119889

119899such

that 1205831(119866lowast) ge 1205831(119866)

Proof Let 119866 isinH119889119899 H119889

119899and let 119883 = (1199091 1199092 119909119899)

119879




119866lowast

1= 119866 minus V


119895119906119901119895+ V1198941199061+ sdot sdot sdot + V

119894119906119901119895 (10)


d+2



H0

(a)


U0

d+2


(b)

Figure 1


1(119866) Note that 119866

1isin H119889

119899

for 119905 = 2 and 1198661isinH119889119899 H119889



119894


3 119866

lowast

119905minus1and

1205831(119866lowast

2) le 1205831(119866lowast


1(119866lowast



119899 and



119899 1205831(119866) le 120583

1(1198670(119901119896+1)) where



1205831 (119866) lt max max 119889

119894+ 119898119894| V119894isin 119881 (119866) | 119866 isinH

119889

119899

119894 notin 119889 + 2 119896 119896 + 1

= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (1198670 (119901119896+1)) + 1

lt 1205831(1198670(119901119896+1))

(11)


119894)⋂119875119881(119866) = 119906

1 1199062 119906

119905 Let

119866lowast=


119894119906119905+ V1198961199061+ sdot sdot sdot + V

119896119906119905

1003816100381610038161003816119909119896

1003816100381610038161003816ge1003816100381610038161003816119909119894

1003816100381610038161003816

1198670(119901119894) minus V119896minus1

V119896+ V119896minus1

V119894

1003816100381610038161003816119909119896

1003816100381610038161003816lt1003816100381610038161003816119909119894

1003816100381610038161003816

(12)



1205831(1198670(119901119896+1)) Because 119867


⋃119875119889minus119896+2

we canget Φ(119867



119896minus1 V119889+2 V119896+1 V119896+2 V

119889+1





119891119896119896= 1 and 119891


119871V119896 (1198670) = (

119871 (119875119896minus1) + 11986411

00 119871 (119875

119889minus119896+2) + 11986511+ 11986522

) (13)

Hence

Φ(119871V119896 (1198670))

= Φ (119871 (119875119896minus1) + 11986411)Φ (119871 (119875

119889minus119896+2) + 11986511+ 11986522)

(14)


and Φ(119871(119875119889minus119896+2

) + 11986511+ 11986522) by 119891

119896minus11(119909) and 119891

119889minus119896+22(119909)


Φ(119871V119896+1 (1198670))

= Φ (119871 (119875119889minus119896) + 11986411)Φ (119871 (119875

119896+1) + 11986511+ 11986522)

= 119891119889minus1198961 (

119909) 119891119896+12 (119909)

(15)


11989111989911(119909) = Φ (1198631198991

) + Φ (1198631198991minus1)

11989111989922


)

(16)


Φ(119871V119896+1 (1198670)) minus Φ (119871V119896 (1198670))

= 119891119889minus1198961 (

119909) 119891119896+12 (119909) minus 119891119896minus11 (

119909) 119891119889minus119896+22 (119909)

= (119909 minus 1) (119909 minus 3)

times [Φ (119863119896minus1)Φ (119863


119896minus2)Φ (119863

119889minus119896)]

+ (2119909 minus 3) [Φ (119863119896minus1)Φ (119863


119896minus2)Φ (119863

119889minus119896)]

= 119909 (119909 minus 2)Φ (119863119889minus2119896)

(17)


Φ(1198670(119901119896)) minus Φ (119867

0(119901119896+1))


) gt 0

(18)

holds for 119909 ge 1205831(1198670(119901119896+1))



0(119901119896) cong 119867

0(119901119896+1)


= 119875119889+1

we can deriveΦ(119871V119889+2(1198670))


1 V

119896minus1 V119896 V119896+1 V

119889+1 Then

Φ(119871V119889+2 (1198670))

= [Φ (119863119896minus1) + Φ (119863

119896minus2)]

times [(1199092minus 5119909 + 5)Φ (119863


)]

minus [Φ (119863119896minus2) + Φ (119863

119896minus3)]


119896minus2)]

(19)



Φ(1198670 (119901119889+2)) minus Φ (1198670 (119901119896+1))

= 119905(119909 minus 1)119905minus1119909Φ (119863


) minus 2Φ (119863119896minus2)]


1(1198670(119901119896+1)) by





Lemma 16 1205831(Δ119896minus1119899 ) lt 1205831(Δ119896



119899) gt


Φ(Δ119896minus1

119889+2) minus Φ (Δ

119896

119889+2)


lowast))]


lowast))


lowast))]

= sdot sdot sdot = 119909 [119892119889119896 (

119909) minus 119891119889119896 (119909)]

(21)where

119891119889119896 (

119909) = (1199093minus 81199092+ 18119909 minus 8)Φ (119863

119889minus2119896+1)

minus (1199092minus 5119909 + 5)Φ (119863

119889minus2119896)

119892119889119896 (119909) = (119909

2minus 5119909 + 5)Φ (119863119889minus2119896+2)


(22)


Φ(Δ119896minus1

119899) minus Φ (Δ

119896

119899)

= (119909 minus 1)119905[Φ (Δ

119896minus1

119889+2) minus Φ (Δ

119896

119889+2)]


119896minus1

119889+2)) minus Φ (119871V119896+1 (Δ

119896

119889+2))]


119905minus1

times [119891119889minus119896+11 (

119909) 1198911198962 (119909) minus 119891119889minus1198961 (

119909) 119891119896+12 (119909)]

= 1199092(119909 minus 1)

119905minus1


)]

gt 0

(23)

for 119909 ge 1205831(Δ119896

119899)

So 1205831(Δ119896minus1

119899) lt 1205831(Δ

119896



Corollary 17 1205831(Δ119896

119899) le 1205831(Δlceil1198892rceil



119899


in Figure 1 Let 1198800(1199012 119901




V119894isin 119881(119880


U119889

119899= 1198800(1199012 119901

119889 119901119889+2)

119889

sum

119894=2

119901119894+ 119901119889+2

= 119899 minus 119889 minus 2

U119889

119899= 1198800(0 0 119901

119894 0 0) = 119880

0(119901119894) 119901119894= 119899 minus 119889 minus 2

(24)


119889

119899such

that 1205831(119866lowast) ge 1205831(119866)



119899 1205831(119866) le 120583

1(1198800(119901119896+2)) where


1198800(119901119896+2)


1205831 (119866) lt max max 119889

119894+ 119898119894| V119894isin 119881 (119866) | 119866 isin U

119889

119899

119894 notin 119889 + 2 119896 119896 + 1 119896 + 2

= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (1198800(119901119896+2)) + 1

lt 1205831 (1198800 (119901119896+2))

(25)


0(119901119896) cong 1198800(119901119896+2)


1 V119896+3 V


Φ(119871V119896+2 (1198800))

= (119909 minus 2) [(1199092minus 4119909 + 2)Φ (119863


)]


) + Φ (119863119889minus119896minus2

)]

(26)

Similarly

Φ(119871V119896 (1198800))

= (119909 minus 2) [(1199092minus 4119909 + 2)Φ (119863


)]

times [Φ (119863119896minus1) + Φ (119863

119896minus2)]

(27)



Φ(1198800(119901119896)) minus Φ (119880

0(119901119896+2))


= 1199051199092(119909 minus 2)

2(119909 minus 1)


) gt 0

(28)

for 119909 ge 1205831(1198800(119901119896+2))

From Lemma 8 1205831(1198800(119901119896)) lt 120583

1(1198800(119901119896+2)) holds


1205831 (1198800 (119901119896+1))

lt max 119889119894+ 119898119894| V119894isin 119881 (119880

0(119901119896+1))

= 119905 + 2 +

3 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (1198800 (119901119896+2)) + 1

lt 1205831(1198800(119901119896+2))

(29)


Φ(119871V119896+1 (1198800)) = [Φ (119863119896minus1) + Φ (119863119896minus2)] 119891 (119909)

minus [Φ (119863119896minus2) + Φ (119863

119896minus3)] 119892 (119909)


minus [2119892 (119909) + ℎ (119909)]Φ (119863119896minus2)

(30)

where


minus (21199092minus 9119909 + 7)Φ (119863



ℎ (119909) = (119909 minus 3) 119892 (119909) minus 119891 (119909)



(31)

By (26) and (30)

Φ(119871V119896+2 (1198800)) minus Φ (119871V119896+1 (1198800))


) minus 2Φ (119863119896minus2)]


(32)




= Φ (119863119889minus119896minus2

) + Φ (119863119889minus119896minus3

) gt 0


119896minus2)

=


119909 minus 3 gt 0 if 119896 = 1(33)


Φ(1198800(119901119896+1)) minus Φ (119880

0(119901119896+2))

= 119909 [Φ (119871V119896+2 (1198800)) minus Φ (119871V119896+1 (1198800))] gt 0(34)




4




119899) lt 120583

1(119896

119899) where



119899)


119899) lt 1205831(

119896minus1

119899)


119899) gt 119905+4


Φ(119896minus1

119889+2) minus Φ (

119896

119889+2)


lowast))]


lowast))]

= sdot sdot sdot = 119909 [119904119889119896 (

119909) minus 119905119889119896 (119909)]

(35)

where

119904119889119896 (



119905119889119896 (


119889minus2119896)


(36)



Φ(119896minus1

119899) minus Φ (

119896

119899)

= (119909 minus 1)119905[Φ (

119896minus1

119889+2) minus Φ (

119896

119889+2)]


119896minus1

119889+2)) minus Φ (119871V119896+2 (

119896

119889+2))]

= 1199092(119909 minus 1)

119905minus1(119909 minus 2) 119906 (119909)

(37)



)


)

(38)


1199092minus (119905 + 5) 119909 + 4 le 0 if 119909 isin (119905 + 4 119890 (119905)]

1199092minus (119905 + 5) 119909 + 119905 + 4

gt 1199092minus (119905 + 5) 119909 + 4 gt 0 if 119909 gt 119890 (119905)

(39)


1205831(119896

119899) So 1205831(

119896minus1

119899) lt 1205831(

119896



Φ(119896minus1

119899) minus Φ (

119896

119899)

= 1199092(119909 minus 1)

119905minus1(119909 minus 2) [(119909

2minus (119905 + 5) 119909 + 4)]

(40)

If 119905 = 1 let 27and 1



1(2

7) = 53145

and 1205831(1

7) = 53008 From Lemma 5 120583

1(2

7) le 1205831(119896

119899) and

1205831(1

7) le 1205831(

119896minus1


0 for 119909 ge 1205831(119896

119899) By Lemma 8 120583

1(119896minus1

119899) lt 1205831(119896

119899) holds


119905 (119909) minus [(1199092minus (119905 + 5) 119909 + 4)]


)Φ (119863119898minus1

)]

= 2119905 (119909 minus 1)Φ2(119863119898) minus [(119905

2+ 3119905 + 4) 119909 minus 4]

times Φ (119863119898)Φ (119863

119898minus1) + 2 (119905 + 2) (119909 minus 1)Φ

2(119863119898minus1

)

= [21199051199092minus (1199052+ 9119905 + 4) 119909 + (4119905 + 4)]Φ (119863119898)Φ (119863119898minus1)


+ 2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

= [21199051199092minus (1199052+ 9119905 + 4) 119909 + (4119905 + 4)]Φ (119863119898

)Φ (119863119898minus1

)

+ 4 (119909 minus 1)Φ2(119863119898minus1

) + 2119905 (119909 minus 1)

(41)

d+3 d+2

d+11 kminus1 vk k+1


W0

(a)

d+2d+3



k

d+3

(b)

Figure 2

where

119905 (119909) = [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]

times Φ2(119863119898) minus [(119905 + 4) 119909

2minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863

119898minus1) + 2 (119905 + 2) (119909 minus 1)Φ

2(119863119898minus1

)

(42)


119898)minus(119905+4)Φ(119863

119898)Φ(119863119898minus1

) gt



1(119896

119899) lt 119890(119905) holds Put 119909 = 120583

1(119896

119899) into (40)


119899) lt 1205831(

119896minus1

119899) We

complete the proof



(i) if 119889 is odd then 1205831(119896

119899) le 1205831(lfloor1198892rfloor



119899



119899) le 1205831(lfloor1198892rfloor

119899) the


119899



119899) the


119899





W119889

119899= 119882

0(1199012 119901

119889 119901119889+2 119901119889+3)

119889

sum

119894=2

119901119894 + 119901119889+2 + 119901119889+3 = 119899 minus 119889 minus 3

W119889

119899= 1198820(0 0 119901

119894 0 0)

= 1198820(119901119894) 119901119894= 119899 minus 119889 minus 3

(43)



119889

119899such

that 1205831(119866lowast) ge 1205831(119866)



119899 1205831(119866) le 1205831(1198820(119901119896)) where 1 le



1205831 (119866)


119889

119899 119894 = 119896

= 119905 + 2 +

2 + 4 + 119905

119905 + 2

lt 119905 + 5 = Δ (1198820(119901119896)) + 1

lt 1205831 (1198820 (119901119896))

(44)




119899) lt 1205831(Δ

lceil1198892rceil

119899) lt


119899)


119899) lt 1205831(lfloor1198892rfloor

119899)



119899) 1205831(lfloor1198892rfloor


following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)Let Δ119898+1

119899and 119898


119905 = 0 denote Δ119898+1119899



Φ(1198661) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198671)

minus Φ (119875119898)Φ (119871V119898+2 (1198671))

Φ (1198662) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198672)

minus Φ (119875119898)Φ (119871V119898+2 (1198672))

(45)

where

Φ(1198671) = 119909 (119909 minus 3) [(119909 minus 3)Φ (119863119898

) minus 2Φ (119863119898minus1

)]

Φ (119871V119898+2 (1198671))


)

Φ (1198672)

= 119909 (119909 minus 2)


)]

Φ (119871V119898+2 (1198672))


(46)

Note thatΦ(119875119898) = 119909Φ(119863


Φ(119863119898minus2


Φ(Δ119898+1

119899)


) minus 119905119909Φ (119871V119898+2 (1198661))]


2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (6119905 + 12)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 3) 119909 minus 3 (119905 + 1)]Φ2(119863119898minus1

)

(47)

Φ(119898

119899)


) minus 119905119909Φ (119871V119898+2 (1198662))]



2+ (4119905 + 18) 119909 minus (2119905 + 10)]

times Φ2(119863119898minus1)

minus [31199093minus (3119905 + 19) 119909

2+ 8 (119905 + 4) 119909 minus 4 (119905 + 4)]


)Φ (119863119898minus2



)

(48)


119899) le 119909 lt 119905+5

Φ(Δ119898+1

119899) minus Φ (

119898

119899)

= 119909(119909 minus 1)119905[119889 (119909)Φ

2(119863119898minus1

) + 119890 (119909)Φ (119863119898minus1)Φ (119863

119898minus2)

minus119903 (119909)Φ2(119863119898minus2

)]

gt 119909(119909 minus 1)119905[119889 (119909) + 119890 (119909) minus 119903 (119909)]Φ

2(119863119898minus2

) ge 0

(49)


where





(50)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1


Case 2 119889 = 2119898 + 2 (119898 ge 1)LetΔ119898+1119899

and119898+1119899


119899and119898+1


2 respectively By


Φ(Δ119898+1

119899)


lowast

1) minus 119905119909Φ (119871V119898+2 (119866

lowast

1))]



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 4) 119909 + 3 (119905 + 2)]Φ

2(119863119898minus1)

(51)

Φ(119898+1

119899)


lowast

2) minus 119905119909Φ (119871V119898+3 (119866

lowast

2))]


times [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]Φ

2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863119898minus1)

+2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

(52)


Φ(Δ119898+1

119899) minus Φ (

119898+1

119899)

= 119909(119909 minus 1)119905


2(119863119898)

+ [(119905 + 1) 1199093minus (5119905 + 5) 119909

2+ (7119905 + 10) 119909 minus (2119905 + 6)]

times Φ (119863119898)Φ (119863

119898minus1)

minus [(2119905 + 2) 1199092minus (4119905 + 4) 119909 + (119905 + 2)]Φ

2(119863119898minus1

)



2m+3

m+2

m+3m+1m 2m+2

m+1

n

(a)



m+2

m+3m+1m

2m+4

2m+3

m+1

n

(b)



2m+3

m+2

m+3m+1m 2m+2

⋄m

n

(c)




m+3

m+1 m+2m

2m+4

2m+3

⋄m+1

n

(d)

1


2m+3

m+2

m+1m

2m+4

2m+2


m+1

n

(e)

1


2m+4

m+3m+2m+1

2m+5

2m+3


m+2

n

(f)

Figure 3

= 119909(119909 minus 1)119905


)]Φ (119863119898minus1

) minus 119908 (119909)

ge 119909(119909 minus 1)119905Φ(119863119898minus1

)


)] gt 0

(53)

where

119901 (119909) = 21199093minus (119905 + 9) 119909

2+ (119905 + 9) 119909 minus 2 gt 0

119902 (119909) = 2(119909 minus 1)2gt 0

119908 (119909) = (119905 + 1) 1199092minus (3119905 + 5) 119909 + (119905 + 4) gt 0

119901 (119909) minus 119908 (119909) minus 119902 (119909)

= 21199093minus (2119905 + 12) 119909

2+ (4119905 + 18) 119909 minus (119905 + 8) gt 0

(54)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1





lceil1198892rceil

119899)


119899) 1205831(Δlceil1198892rceil


following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)


Letnabla119898+1119899


119899by 1198663 Let119867


119898 By Lemma 9

Φ(1198663) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]Φ (1198673)

minus Φ (119875119898)Φ (119871V119898+1 (1198673))

(55)

where

Φ(1198673) = 119909 (119909 minus 3)


)]

Φ (119871V119898+1 (1198673)) = (119909 minus 1) (119909 minus 3)


119898minus1)]

(56)

By Lemma 10 we get

Φ(nabla119898+1

119899)


) minus 119905119909Φ (119871V119898+1 (1198663))]

= 119909 (119909 minus 3) (119909 minus 1)119905


times Φ (119863119898)Φ (119863

119898minus1) + (119905 + 2)Φ

2(119863119898minus1

)

(57)

By (47) we have

Φ(Δ119898+1

119899) = 119909(119909 minus 1)

119905


minus [(119905 + 5) 1199092minus (6119905 + 22) 119909 + (6119905 + 18)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 5) 119909 minus 3 (119905 + 2)]Φ2(119863119898minus1

)

(58)

Hence

Φ(nabla119898+1

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(59)


1(nabla119898+1

119899) lt 1205831(Δ119898+1


When 119898 ge 2 1205831(nabla119898+1

119899) lt 1205831(Δ119898+1



Case 2 119889 = 2119898 + 2 (119898 ge 1)

Let nabla119898+2119899


Φ(nabla119898+2

119899) = 119909 (119909 minus 3) (119909 minus 1)

119905


2(119863119898)


119898minus1)


)

(60)

By (51) we get

Φ(Δ119898+1

119899)



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 5) 119909 + (3119905 + 9)]Φ

2(119863119898minus1

)

(61)

Hence

Φ(nabla119898+2

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(62)

Let 119905119896(119909) = (119909 minus 119905 minus 4)Φ(119863

119896) minus (119905 + 3)Φ(119863



1(119905119896(119909)) where 119896 ge 0




1(119905119896minus1(119909)) lt


119905119896+1 (


119909) (63)

1199052

119896(119909) minus 119905119896+1 (

119909) 119905119896minus1 (119909)


(64)




Let 11986610158403= nabla119898+2




and1198661015840

1= Δ119898+1





By (60) and (62)

Φ(1198661015840

3) = 119909(119909 minus 1)

119905(119909 minus 3) (119909

2minus 3119909 + 1) 119903 (119909)

Φ (1198661015840

3) minus Φ (119866

1015840

1)

= 1199092(119909 minus 1)

119905[1199093minus (119905 + 8) 119909

2+ (3119905 + 16) 119909 minus (119905 + 6)]

= 1199092(119909 minus 1)

119905(119903 (119909) + 1)

(65)


1(1198661015840


1205831(1198661015840

3) ge 1205831(119866

1015840

1) then [Φ(1198661015840

3)minusΦ(119866

1015840

1)]|119909=1205831(119866

1015840

3) le 0 and 119903(119909)+

1|119909=1205831(119866

1015840


1(1198661015840

3) lt 1205831(1198661015840

1)


119909V (119909) minus (119905 + 2) 119903 (119909)

= 3 (119905 + 2) 1199092minus (31199052+ 22119905 + 33) 119909 + (119905 + 2) (119905 + 7) gt 0

(66)

From Lemma 8 1205831(V(119909)) lt 120583

1(1198661015840


1205831(1198661015840

3) le 1205831(nabla119898+2

119899) and 120583

1(1198661015840

1) le 1205831(Δ119898+1



119899) From


119899) lt 1205831(Δ119898+1

119899) as desired

Let ◻119896119889+3



119889+3) lt 1205831(Δlceil1198892rceil

119889+3)


119889+3) gt 5


Φ(◻119898+1

2119898+4) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]

times Φ (1198674) minus Φ (119875

119898)Φ (119871V119898+1 (1198674))

= 119909 [(119909 minus 2)2(119909 minus 4)Φ


times Φ (119863119898)Φ (119863

119898minus1) + (3119909 minus 8)Φ

2(119863119898minus1

)]

(67)

where1198674= ◻119898+1


119898

By (47) we get

Φ(Δ119898+1

2119898+4)


2minus 22119909 + 18)

times Φ (119863119898)Φ (119863

119898minus1) + (5119909 minus 6)Φ

2(119863119898minus1

)]

(68)


Φ(◻119898+1

2119898+4) minus Φ (Δ

119898+1

2119898+4)

= 119909 [(119909 minus 4)Φ2(119863119898) + (119909

2minus 2119909 minus 6)Φ (119863

119898)Φ (119863

119898minus1)

minus2 (119909 + 1)Φ2(119863119898minus1

)]

= 119909Φ (119863119898minus1

) [2 (1199092minus 4119909 + 1)Φ (119863


)]

+ 119909 (119909 minus 4)

gt 119909Φ2(119863119898minus1

) (41199092minus 19119909 + 6) + 119909 (119909 minus 4) gt 0

(69)

So 1205831(◻119898+1

2119898+4) lt 1205831(Δ119898+1



1(◻119898+1

2119898+5) lt 120583

1(Δ119898+1

2119898+5)

holds



1 1199092 119909

119899)119879



vertex V119894(1 le 119894 le 119899)



Claim 1 119881(119862119902)⋂119881(119875

119889+1) = 0


119894V119896V119896+1 V


119902and 119875119889+1

where V

119894isin 119881(119862

119902) V119897isin 119881(119875

119889+1)and V

119896 V

119897minus1isin 119881(119866)

(119881(119862119902)⋃119881(119875119889+1

))For each 119895 let 119879


119895as its root)


119895is



119894







119894) = 119889(V 119903

119894) minus 1 By


1(1198661) ge 1205831(119866)





3 where



V119896 V

119897minus1 So 120583

1(1198663) ge 1205831(1198662)







119889+1) = 0 and 120583

1(1198664) ge 1205831(1198663)


By Claim 1 119881(119862119902)⋂119881(119875


119862119902

= V119896V119896+1 V

119897minus1V119897V119889+2

V119889+3 V

119904V119896(119904 ge 119889 + 2)

where V119896 V119896+1 V

119897minus1 V119897 = 119881(119862

119902)⋂119881(119875

119889+1) and

V119889+2 V119889+3 V

119904 = 119881(119862

119902) 119881(119875

119889+1)

Claim 2 119889(V) = 1 for V isin 119881(119866) (119881(119862119902)⋃119881(119875119889+1))


3) we can get

119894and 120583

1(119894) ge


119894(V119894isin

119881(119862119902)⋃119881(119875

119889+1) V


Claim 3 119897 = 119896 + 1 and 119904 = 119889 + 2


(119881(119862119902)⋃119881(119875119889+1

) = V119904+1 V

119899 and |119860| = 119899 minus 119904 = 119905 ge 0

Case 1 119897 = 119896 By Lemma 7


= 119905 + 2 +

2 + 4 + 119905

119905 + 2

le 119905 + 5 = Δ (119896) + 1

lt 1205831 (119896) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

= 119905 + 4 +

2 + 2 + 2 + 2 + 119905

119905 + 4

le 119905 + 6 le 119899 minus 119889 + 2

= Δ (nabla119896

119899) + 1 lt 120583

1(nabla119896

119899)

(70)

for 119904 ge 119889 + 4 (|119862119902| ge 4)

When 119904 = 119889 + 3 (|119862119902| = 3)

119894cong 1198820(119901119894) By Lemma 23

1205831(119894) lt 1205831(nabla119896

119899) for any 119894 = 119896 where


119899




119899) holds

Case 2 119897 = 119896 + 1


1205831 (119894)


= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (119896) + 1 (Δ (

119896+1) + 1)

lt 1205831(119896) (1205831(119896+1)) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

le 119905 + 3 +

2 + 3 + 119905

119905 + 3

lt 119905 + 5 le 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(71)

for 119904 ge 119889 + 3 (|119862119902| ge 4)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) By Lemma 15

1205831(119894) lt 1205831(Δ119896

119899) for any 119894 = 119896 + 1 where

119896+1cong Δ119896

119899



= 3 +

2 + 3 + 2

3

lt 6 lt 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(72)

for 119904 ge 119889 + 4 (|119862119902| ge 5)

When 119904 = 119889 + 3 (|119862119902| = 4) by Lemma 13 120583

1(◻119896

119889+3) lt

1205831(◻lceil1198892rceil


lceil1198892rceil

119889+3) gt

1205831(◻lceil1198892rceil

119889+3)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) cong Δ119896

119899



119899) gt 1205831(Δlceil1198892rceil

119899) holds

Case 3 119897 ge 119896 + 2 By Lemma 7

1205831(119894)


= 119905 + 3 +

2 + 2 + 2 + 119905

119905 + 3

le 119905 + 5 le 119899 minus 119889 + 2 = Δ (119896

119899) + 1

lt 1205831 (119896

119899)

(73)

for 119904 ge 119889 + 3 (|119862119902| ge 5)

When 119904 = 119889 + 2 (|119862119902| = 4)

119894cong 1198800(119901119894) by Lemma 19

1205831(119894) lt 1205831(

119896


119896

119899





119899has the


Acknowledgments



References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










d+2



H0

(a)


U0

d+2


(b)

Figure 1


1(119866) Note that 119866

1isin H119889

119899

for 119905 = 2 and 1198661isinH119889119899 H119889



119894


3 119866

lowast

119905minus1and

1205831(119866lowast

2) le 1205831(119866lowast


1(119866lowast



119899 and



119899 1205831(119866) le 120583

1(1198670(119901119896+1)) where



1205831 (119866) lt max max 119889

119894+ 119898119894| V119894isin 119881 (119866) | 119866 isinH

119889

119899

119894 notin 119889 + 2 119896 119896 + 1

= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (1198670 (119901119896+1)) + 1

lt 1205831(1198670(119901119896+1))

(11)


119894)⋂119875119881(119866) = 119906

1 1199062 119906

119905 Let

119866lowast=


119894119906119905+ V1198961199061+ sdot sdot sdot + V

119896119906119905

1003816100381610038161003816119909119896

1003816100381610038161003816ge1003816100381610038161003816119909119894

1003816100381610038161003816

1198670(119901119894) minus V119896minus1

V119896+ V119896minus1

V119894

1003816100381610038161003816119909119896

1003816100381610038161003816lt1003816100381610038161003816119909119894

1003816100381610038161003816

(12)



1205831(1198670(119901119896+1)) Because 119867


⋃119875119889minus119896+2

we canget Φ(119867



119896minus1 V119889+2 V119896+1 V119896+2 V

119889+1





119891119896119896= 1 and 119891


119871V119896 (1198670) = (

119871 (119875119896minus1) + 11986411

00 119871 (119875

119889minus119896+2) + 11986511+ 11986522

) (13)

Hence

Φ(119871V119896 (1198670))

= Φ (119871 (119875119896minus1) + 11986411)Φ (119871 (119875

119889minus119896+2) + 11986511+ 11986522)

(14)


and Φ(119871(119875119889minus119896+2

) + 11986511+ 11986522) by 119891

119896minus11(119909) and 119891

119889minus119896+22(119909)


Φ(119871V119896+1 (1198670))

= Φ (119871 (119875119889minus119896) + 11986411)Φ (119871 (119875

119896+1) + 11986511+ 11986522)

= 119891119889minus1198961 (

119909) 119891119896+12 (119909)

(15)


11989111989911(119909) = Φ (1198631198991

) + Φ (1198631198991minus1)

11989111989922


)

(16)


Φ(119871V119896+1 (1198670)) minus Φ (119871V119896 (1198670))

= 119891119889minus1198961 (

119909) 119891119896+12 (119909) minus 119891119896minus11 (

119909) 119891119889minus119896+22 (119909)

= (119909 minus 1) (119909 minus 3)

times [Φ (119863119896minus1)Φ (119863


119896minus2)Φ (119863

119889minus119896)]

+ (2119909 minus 3) [Φ (119863119896minus1)Φ (119863


119896minus2)Φ (119863

119889minus119896)]

= 119909 (119909 minus 2)Φ (119863119889minus2119896)

(17)


Φ(1198670(119901119896)) minus Φ (119867

0(119901119896+1))


) gt 0

(18)

holds for 119909 ge 1205831(1198670(119901119896+1))



0(119901119896) cong 119867

0(119901119896+1)


= 119875119889+1

we can deriveΦ(119871V119889+2(1198670))


1 V

119896minus1 V119896 V119896+1 V

119889+1 Then

Φ(119871V119889+2 (1198670))

= [Φ (119863119896minus1) + Φ (119863

119896minus2)]

times [(1199092minus 5119909 + 5)Φ (119863


)]

minus [Φ (119863119896minus2) + Φ (119863

119896minus3)]


119896minus2)]

(19)



Φ(1198670 (119901119889+2)) minus Φ (1198670 (119901119896+1))

= 119905(119909 minus 1)119905minus1119909Φ (119863


) minus 2Φ (119863119896minus2)]


1(1198670(119901119896+1)) by





Lemma 16 1205831(Δ119896minus1119899 ) lt 1205831(Δ119896



119899) gt


Φ(Δ119896minus1

119889+2) minus Φ (Δ

119896

119889+2)


lowast))]


lowast))


lowast))]

= sdot sdot sdot = 119909 [119892119889119896 (

119909) minus 119891119889119896 (119909)]

(21)where

119891119889119896 (

119909) = (1199093minus 81199092+ 18119909 minus 8)Φ (119863

119889minus2119896+1)

minus (1199092minus 5119909 + 5)Φ (119863

119889minus2119896)

119892119889119896 (119909) = (119909

2minus 5119909 + 5)Φ (119863119889minus2119896+2)


(22)


Φ(Δ119896minus1

119899) minus Φ (Δ

119896

119899)

= (119909 minus 1)119905[Φ (Δ

119896minus1

119889+2) minus Φ (Δ

119896

119889+2)]


119896minus1

119889+2)) minus Φ (119871V119896+1 (Δ

119896

119889+2))]


119905minus1

times [119891119889minus119896+11 (

119909) 1198911198962 (119909) minus 119891119889minus1198961 (

119909) 119891119896+12 (119909)]

= 1199092(119909 minus 1)

119905minus1


)]

gt 0

(23)

for 119909 ge 1205831(Δ119896

119899)

So 1205831(Δ119896minus1

119899) lt 1205831(Δ

119896



Corollary 17 1205831(Δ119896

119899) le 1205831(Δlceil1198892rceil



119899


in Figure 1 Let 1198800(1199012 119901




V119894isin 119881(119880


U119889

119899= 1198800(1199012 119901

119889 119901119889+2)

119889

sum

119894=2

119901119894+ 119901119889+2

= 119899 minus 119889 minus 2

U119889

119899= 1198800(0 0 119901

119894 0 0) = 119880

0(119901119894) 119901119894= 119899 minus 119889 minus 2

(24)


119889

119899such

that 1205831(119866lowast) ge 1205831(119866)



119899 1205831(119866) le 120583

1(1198800(119901119896+2)) where


1198800(119901119896+2)


1205831 (119866) lt max max 119889

119894+ 119898119894| V119894isin 119881 (119866) | 119866 isin U

119889

119899

119894 notin 119889 + 2 119896 119896 + 1 119896 + 2

= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (1198800(119901119896+2)) + 1

lt 1205831 (1198800 (119901119896+2))

(25)


0(119901119896) cong 1198800(119901119896+2)


1 V119896+3 V


Φ(119871V119896+2 (1198800))

= (119909 minus 2) [(1199092minus 4119909 + 2)Φ (119863


)]


) + Φ (119863119889minus119896minus2

)]

(26)

Similarly

Φ(119871V119896 (1198800))

= (119909 minus 2) [(1199092minus 4119909 + 2)Φ (119863


)]

times [Φ (119863119896minus1) + Φ (119863

119896minus2)]

(27)



Φ(1198800(119901119896)) minus Φ (119880

0(119901119896+2))


= 1199051199092(119909 minus 2)

2(119909 minus 1)


) gt 0

(28)

for 119909 ge 1205831(1198800(119901119896+2))

From Lemma 8 1205831(1198800(119901119896)) lt 120583

1(1198800(119901119896+2)) holds


1205831 (1198800 (119901119896+1))

lt max 119889119894+ 119898119894| V119894isin 119881 (119880

0(119901119896+1))

= 119905 + 2 +

3 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (1198800 (119901119896+2)) + 1

lt 1205831(1198800(119901119896+2))

(29)


Φ(119871V119896+1 (1198800)) = [Φ (119863119896minus1) + Φ (119863119896minus2)] 119891 (119909)

minus [Φ (119863119896minus2) + Φ (119863

119896minus3)] 119892 (119909)


minus [2119892 (119909) + ℎ (119909)]Φ (119863119896minus2)

(30)

where


minus (21199092minus 9119909 + 7)Φ (119863



ℎ (119909) = (119909 minus 3) 119892 (119909) minus 119891 (119909)



(31)

By (26) and (30)

Φ(119871V119896+2 (1198800)) minus Φ (119871V119896+1 (1198800))


) minus 2Φ (119863119896minus2)]


(32)




= Φ (119863119889minus119896minus2

) + Φ (119863119889minus119896minus3

) gt 0


119896minus2)

=


119909 minus 3 gt 0 if 119896 = 1(33)


Φ(1198800(119901119896+1)) minus Φ (119880

0(119901119896+2))

= 119909 [Φ (119871V119896+2 (1198800)) minus Φ (119871V119896+1 (1198800))] gt 0(34)




4




119899) lt 120583

1(119896

119899) where



119899)


119899) lt 1205831(

119896minus1

119899)


119899) gt 119905+4


Φ(119896minus1

119889+2) minus Φ (

119896

119889+2)


lowast))]


lowast))]

= sdot sdot sdot = 119909 [119904119889119896 (

119909) minus 119905119889119896 (119909)]

(35)

where

119904119889119896 (



119905119889119896 (


119889minus2119896)


(36)



Φ(119896minus1

119899) minus Φ (

119896

119899)

= (119909 minus 1)119905[Φ (

119896minus1

119889+2) minus Φ (

119896

119889+2)]


119896minus1

119889+2)) minus Φ (119871V119896+2 (

119896

119889+2))]

= 1199092(119909 minus 1)

119905minus1(119909 minus 2) 119906 (119909)

(37)



)


)

(38)


1199092minus (119905 + 5) 119909 + 4 le 0 if 119909 isin (119905 + 4 119890 (119905)]

1199092minus (119905 + 5) 119909 + 119905 + 4

gt 1199092minus (119905 + 5) 119909 + 4 gt 0 if 119909 gt 119890 (119905)

(39)


1205831(119896

119899) So 1205831(

119896minus1

119899) lt 1205831(

119896



Φ(119896minus1

119899) minus Φ (

119896

119899)

= 1199092(119909 minus 1)

119905minus1(119909 minus 2) [(119909

2minus (119905 + 5) 119909 + 4)]

(40)

If 119905 = 1 let 27and 1



1(2

7) = 53145

and 1205831(1

7) = 53008 From Lemma 5 120583

1(2

7) le 1205831(119896

119899) and

1205831(1

7) le 1205831(

119896minus1


0 for 119909 ge 1205831(119896

119899) By Lemma 8 120583

1(119896minus1

119899) lt 1205831(119896

119899) holds


119905 (119909) minus [(1199092minus (119905 + 5) 119909 + 4)]


)Φ (119863119898minus1

)]

= 2119905 (119909 minus 1)Φ2(119863119898) minus [(119905

2+ 3119905 + 4) 119909 minus 4]

times Φ (119863119898)Φ (119863

119898minus1) + 2 (119905 + 2) (119909 minus 1)Φ

2(119863119898minus1

)

= [21199051199092minus (1199052+ 9119905 + 4) 119909 + (4119905 + 4)]Φ (119863119898)Φ (119863119898minus1)


+ 2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

= [21199051199092minus (1199052+ 9119905 + 4) 119909 + (4119905 + 4)]Φ (119863119898

)Φ (119863119898minus1

)

+ 4 (119909 minus 1)Φ2(119863119898minus1

) + 2119905 (119909 minus 1)

(41)

d+3 d+2

d+11 kminus1 vk k+1


W0

(a)

d+2d+3



k

d+3

(b)

Figure 2

where

119905 (119909) = [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]

times Φ2(119863119898) minus [(119905 + 4) 119909

2minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863

119898minus1) + 2 (119905 + 2) (119909 minus 1)Φ

2(119863119898minus1

)

(42)


119898)minus(119905+4)Φ(119863

119898)Φ(119863119898minus1

) gt



1(119896

119899) lt 119890(119905) holds Put 119909 = 120583

1(119896

119899) into (40)


119899) lt 1205831(

119896minus1

119899) We

complete the proof



(i) if 119889 is odd then 1205831(119896

119899) le 1205831(lfloor1198892rfloor



119899



119899) le 1205831(lfloor1198892rfloor

119899) the


119899



119899) the


119899





W119889

119899= 119882

0(1199012 119901

119889 119901119889+2 119901119889+3)

119889

sum

119894=2

119901119894 + 119901119889+2 + 119901119889+3 = 119899 minus 119889 minus 3

W119889

119899= 1198820(0 0 119901

119894 0 0)

= 1198820(119901119894) 119901119894= 119899 minus 119889 minus 3

(43)



119889

119899such

that 1205831(119866lowast) ge 1205831(119866)



119899 1205831(119866) le 1205831(1198820(119901119896)) where 1 le



1205831 (119866)


119889

119899 119894 = 119896

= 119905 + 2 +

2 + 4 + 119905

119905 + 2

lt 119905 + 5 = Δ (1198820(119901119896)) + 1

lt 1205831 (1198820 (119901119896))

(44)




119899) lt 1205831(Δ

lceil1198892rceil

119899) lt


119899)


119899) lt 1205831(lfloor1198892rfloor

119899)



119899) 1205831(lfloor1198892rfloor


following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)Let Δ119898+1

119899and 119898


119905 = 0 denote Δ119898+1119899



Φ(1198661) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198671)

minus Φ (119875119898)Φ (119871V119898+2 (1198671))

Φ (1198662) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198672)

minus Φ (119875119898)Φ (119871V119898+2 (1198672))

(45)

where

Φ(1198671) = 119909 (119909 minus 3) [(119909 minus 3)Φ (119863119898

) minus 2Φ (119863119898minus1

)]

Φ (119871V119898+2 (1198671))


)

Φ (1198672)

= 119909 (119909 minus 2)


)]

Φ (119871V119898+2 (1198672))


(46)

Note thatΦ(119875119898) = 119909Φ(119863


Φ(119863119898minus2


Φ(Δ119898+1

119899)


) minus 119905119909Φ (119871V119898+2 (1198661))]


2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (6119905 + 12)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 3) 119909 minus 3 (119905 + 1)]Φ2(119863119898minus1

)

(47)

Φ(119898

119899)


) minus 119905119909Φ (119871V119898+2 (1198662))]



2+ (4119905 + 18) 119909 minus (2119905 + 10)]

times Φ2(119863119898minus1)

minus [31199093minus (3119905 + 19) 119909

2+ 8 (119905 + 4) 119909 minus 4 (119905 + 4)]


)Φ (119863119898minus2



)

(48)


119899) le 119909 lt 119905+5

Φ(Δ119898+1

119899) minus Φ (

119898

119899)

= 119909(119909 minus 1)119905[119889 (119909)Φ

2(119863119898minus1

) + 119890 (119909)Φ (119863119898minus1)Φ (119863

119898minus2)

minus119903 (119909)Φ2(119863119898minus2

)]

gt 119909(119909 minus 1)119905[119889 (119909) + 119890 (119909) minus 119903 (119909)]Φ

2(119863119898minus2

) ge 0

(49)


where





(50)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1


Case 2 119889 = 2119898 + 2 (119898 ge 1)LetΔ119898+1119899

and119898+1119899


119899and119898+1


2 respectively By


Φ(Δ119898+1

119899)


lowast

1) minus 119905119909Φ (119871V119898+2 (119866

lowast

1))]



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 4) 119909 + 3 (119905 + 2)]Φ

2(119863119898minus1)

(51)

Φ(119898+1

119899)


lowast

2) minus 119905119909Φ (119871V119898+3 (119866

lowast

2))]


times [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]Φ

2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863119898minus1)

+2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

(52)


Φ(Δ119898+1

119899) minus Φ (

119898+1

119899)

= 119909(119909 minus 1)119905


2(119863119898)

+ [(119905 + 1) 1199093minus (5119905 + 5) 119909

2+ (7119905 + 10) 119909 minus (2119905 + 6)]

times Φ (119863119898)Φ (119863

119898minus1)

minus [(2119905 + 2) 1199092minus (4119905 + 4) 119909 + (119905 + 2)]Φ

2(119863119898minus1

)



2m+3

m+2

m+3m+1m 2m+2

m+1

n

(a)



m+2

m+3m+1m

2m+4

2m+3

m+1

n

(b)



2m+3

m+2

m+3m+1m 2m+2

⋄m

n

(c)




m+3

m+1 m+2m

2m+4

2m+3

⋄m+1

n

(d)

1


2m+3

m+2

m+1m

2m+4

2m+2


m+1

n

(e)

1


2m+4

m+3m+2m+1

2m+5

2m+3


m+2

n

(f)

Figure 3

= 119909(119909 minus 1)119905


)]Φ (119863119898minus1

) minus 119908 (119909)

ge 119909(119909 minus 1)119905Φ(119863119898minus1

)


)] gt 0

(53)

where

119901 (119909) = 21199093minus (119905 + 9) 119909

2+ (119905 + 9) 119909 minus 2 gt 0

119902 (119909) = 2(119909 minus 1)2gt 0

119908 (119909) = (119905 + 1) 1199092minus (3119905 + 5) 119909 + (119905 + 4) gt 0

119901 (119909) minus 119908 (119909) minus 119902 (119909)

= 21199093minus (2119905 + 12) 119909

2+ (4119905 + 18) 119909 minus (119905 + 8) gt 0

(54)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1





lceil1198892rceil

119899)


119899) 1205831(Δlceil1198892rceil


following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)


Letnabla119898+1119899


119899by 1198663 Let119867


119898 By Lemma 9

Φ(1198663) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]Φ (1198673)

minus Φ (119875119898)Φ (119871V119898+1 (1198673))

(55)

where

Φ(1198673) = 119909 (119909 minus 3)


)]

Φ (119871V119898+1 (1198673)) = (119909 minus 1) (119909 minus 3)


119898minus1)]

(56)

By Lemma 10 we get

Φ(nabla119898+1

119899)


) minus 119905119909Φ (119871V119898+1 (1198663))]

= 119909 (119909 minus 3) (119909 minus 1)119905


times Φ (119863119898)Φ (119863

119898minus1) + (119905 + 2)Φ

2(119863119898minus1

)

(57)

By (47) we have

Φ(Δ119898+1

119899) = 119909(119909 minus 1)

119905


minus [(119905 + 5) 1199092minus (6119905 + 22) 119909 + (6119905 + 18)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 5) 119909 minus 3 (119905 + 2)]Φ2(119863119898minus1

)

(58)

Hence

Φ(nabla119898+1

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(59)


1(nabla119898+1

119899) lt 1205831(Δ119898+1


When 119898 ge 2 1205831(nabla119898+1

119899) lt 1205831(Δ119898+1



Case 2 119889 = 2119898 + 2 (119898 ge 1)

Let nabla119898+2119899


Φ(nabla119898+2

119899) = 119909 (119909 minus 3) (119909 minus 1)

119905


2(119863119898)


119898minus1)


)

(60)

By (51) we get

Φ(Δ119898+1

119899)



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 5) 119909 + (3119905 + 9)]Φ

2(119863119898minus1

)

(61)

Hence

Φ(nabla119898+2

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(62)

Let 119905119896(119909) = (119909 minus 119905 minus 4)Φ(119863

119896) minus (119905 + 3)Φ(119863



1(119905119896(119909)) where 119896 ge 0




1(119905119896minus1(119909)) lt


119905119896+1 (


119909) (63)

1199052

119896(119909) minus 119905119896+1 (

119909) 119905119896minus1 (119909)


(64)




Let 11986610158403= nabla119898+2




and1198661015840

1= Δ119898+1





By (60) and (62)

Φ(1198661015840

3) = 119909(119909 minus 1)

119905(119909 minus 3) (119909

2minus 3119909 + 1) 119903 (119909)

Φ (1198661015840

3) minus Φ (119866

1015840

1)

= 1199092(119909 minus 1)

119905[1199093minus (119905 + 8) 119909

2+ (3119905 + 16) 119909 minus (119905 + 6)]

= 1199092(119909 minus 1)

119905(119903 (119909) + 1)

(65)


1(1198661015840


1205831(1198661015840

3) ge 1205831(119866

1015840

1) then [Φ(1198661015840

3)minusΦ(119866

1015840

1)]|119909=1205831(119866

1015840

3) le 0 and 119903(119909)+

1|119909=1205831(119866

1015840


1(1198661015840

3) lt 1205831(1198661015840

1)


119909V (119909) minus (119905 + 2) 119903 (119909)

= 3 (119905 + 2) 1199092minus (31199052+ 22119905 + 33) 119909 + (119905 + 2) (119905 + 7) gt 0

(66)

From Lemma 8 1205831(V(119909)) lt 120583

1(1198661015840


1205831(1198661015840

3) le 1205831(nabla119898+2

119899) and 120583

1(1198661015840

1) le 1205831(Δ119898+1



119899) From


119899) lt 1205831(Δ119898+1

119899) as desired

Let ◻119896119889+3



119889+3) lt 1205831(Δlceil1198892rceil

119889+3)


119889+3) gt 5


Φ(◻119898+1

2119898+4) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]

times Φ (1198674) minus Φ (119875

119898)Φ (119871V119898+1 (1198674))

= 119909 [(119909 minus 2)2(119909 minus 4)Φ


times Φ (119863119898)Φ (119863

119898minus1) + (3119909 minus 8)Φ

2(119863119898minus1

)]

(67)

where1198674= ◻119898+1


119898

By (47) we get

Φ(Δ119898+1

2119898+4)


2minus 22119909 + 18)

times Φ (119863119898)Φ (119863

119898minus1) + (5119909 minus 6)Φ

2(119863119898minus1

)]

(68)


Φ(◻119898+1

2119898+4) minus Φ (Δ

119898+1

2119898+4)

= 119909 [(119909 minus 4)Φ2(119863119898) + (119909

2minus 2119909 minus 6)Φ (119863

119898)Φ (119863

119898minus1)

minus2 (119909 + 1)Φ2(119863119898minus1

)]

= 119909Φ (119863119898minus1

) [2 (1199092minus 4119909 + 1)Φ (119863


)]

+ 119909 (119909 minus 4)

gt 119909Φ2(119863119898minus1

) (41199092minus 19119909 + 6) + 119909 (119909 minus 4) gt 0

(69)

So 1205831(◻119898+1

2119898+4) lt 1205831(Δ119898+1



1(◻119898+1

2119898+5) lt 120583

1(Δ119898+1

2119898+5)

holds



1 1199092 119909

119899)119879



vertex V119894(1 le 119894 le 119899)



Claim 1 119881(119862119902)⋂119881(119875

119889+1) = 0


119894V119896V119896+1 V


119902and 119875119889+1

where V

119894isin 119881(119862

119902) V119897isin 119881(119875

119889+1)and V

119896 V

119897minus1isin 119881(119866)

(119881(119862119902)⋃119881(119875119889+1

))For each 119895 let 119879


119895as its root)


119895is



119894







119894) = 119889(V 119903

119894) minus 1 By


1(1198661) ge 1205831(119866)





3 where



V119896 V

119897minus1 So 120583

1(1198663) ge 1205831(1198662)







119889+1) = 0 and 120583

1(1198664) ge 1205831(1198663)


By Claim 1 119881(119862119902)⋂119881(119875


119862119902

= V119896V119896+1 V

119897minus1V119897V119889+2

V119889+3 V

119904V119896(119904 ge 119889 + 2)

where V119896 V119896+1 V

119897minus1 V119897 = 119881(119862

119902)⋂119881(119875

119889+1) and

V119889+2 V119889+3 V

119904 = 119881(119862

119902) 119881(119875

119889+1)

Claim 2 119889(V) = 1 for V isin 119881(119866) (119881(119862119902)⋃119881(119875119889+1))


3) we can get

119894and 120583

1(119894) ge


119894(V119894isin

119881(119862119902)⋃119881(119875

119889+1) V


Claim 3 119897 = 119896 + 1 and 119904 = 119889 + 2


(119881(119862119902)⋃119881(119875119889+1

) = V119904+1 V

119899 and |119860| = 119899 minus 119904 = 119905 ge 0

Case 1 119897 = 119896 By Lemma 7


= 119905 + 2 +

2 + 4 + 119905

119905 + 2

le 119905 + 5 = Δ (119896) + 1

lt 1205831 (119896) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

= 119905 + 4 +

2 + 2 + 2 + 2 + 119905

119905 + 4

le 119905 + 6 le 119899 minus 119889 + 2

= Δ (nabla119896

119899) + 1 lt 120583

1(nabla119896

119899)

(70)

for 119904 ge 119889 + 4 (|119862119902| ge 4)

When 119904 = 119889 + 3 (|119862119902| = 3)

119894cong 1198820(119901119894) By Lemma 23

1205831(119894) lt 1205831(nabla119896

119899) for any 119894 = 119896 where


119899




119899) holds

Case 2 119897 = 119896 + 1


1205831 (119894)


= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (119896) + 1 (Δ (

119896+1) + 1)

lt 1205831(119896) (1205831(119896+1)) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

le 119905 + 3 +

2 + 3 + 119905

119905 + 3

lt 119905 + 5 le 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(71)

for 119904 ge 119889 + 3 (|119862119902| ge 4)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) By Lemma 15

1205831(119894) lt 1205831(Δ119896

119899) for any 119894 = 119896 + 1 where

119896+1cong Δ119896

119899



= 3 +

2 + 3 + 2

3

lt 6 lt 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(72)

for 119904 ge 119889 + 4 (|119862119902| ge 5)

When 119904 = 119889 + 3 (|119862119902| = 4) by Lemma 13 120583

1(◻119896

119889+3) lt

1205831(◻lceil1198892rceil


lceil1198892rceil

119889+3) gt

1205831(◻lceil1198892rceil

119889+3)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) cong Δ119896

119899



119899) gt 1205831(Δlceil1198892rceil

119899) holds

Case 3 119897 ge 119896 + 2 By Lemma 7

1205831(119894)


= 119905 + 3 +

2 + 2 + 2 + 119905

119905 + 3

le 119905 + 5 le 119899 minus 119889 + 2 = Δ (119896

119899) + 1

lt 1205831 (119896

119899)

(73)

for 119904 ge 119889 + 3 (|119862119902| ge 5)

When 119904 = 119889 + 2 (|119862119902| = 4)

119894cong 1198800(119901119894) by Lemma 19

1205831(119894) lt 1205831(

119896


119896

119899





119899has the


Acknowledgments



References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Φ(1198670 (119901119889+2)) minus Φ (1198670 (119901119896+1))

= 119905(119909 minus 1)119905minus1119909Φ (119863


) minus 2Φ (119863119896minus2)]


1(1198670(119901119896+1)) by





Lemma 16 1205831(Δ119896minus1119899 ) lt 1205831(Δ119896



119899) gt


Φ(Δ119896minus1

119889+2) minus Φ (Δ

119896

119889+2)


lowast))]


lowast))


lowast))]

= sdot sdot sdot = 119909 [119892119889119896 (

119909) minus 119891119889119896 (119909)]

(21)where

119891119889119896 (

119909) = (1199093minus 81199092+ 18119909 minus 8)Φ (119863

119889minus2119896+1)

minus (1199092minus 5119909 + 5)Φ (119863

119889minus2119896)

119892119889119896 (119909) = (119909

2minus 5119909 + 5)Φ (119863119889minus2119896+2)


(22)


Φ(Δ119896minus1

119899) minus Φ (Δ

119896

119899)

= (119909 minus 1)119905[Φ (Δ

119896minus1

119889+2) minus Φ (Δ

119896

119889+2)]


119896minus1

119889+2)) minus Φ (119871V119896+1 (Δ

119896

119889+2))]


119905minus1

times [119891119889minus119896+11 (

119909) 1198911198962 (119909) minus 119891119889minus1198961 (

119909) 119891119896+12 (119909)]

= 1199092(119909 minus 1)

119905minus1


)]

gt 0

(23)

for 119909 ge 1205831(Δ119896

119899)

So 1205831(Δ119896minus1

119899) lt 1205831(Δ

119896



Corollary 17 1205831(Δ119896

119899) le 1205831(Δlceil1198892rceil



119899


in Figure 1 Let 1198800(1199012 119901




V119894isin 119881(119880


U119889

119899= 1198800(1199012 119901

119889 119901119889+2)

119889

sum

119894=2

119901119894+ 119901119889+2

= 119899 minus 119889 minus 2

U119889

119899= 1198800(0 0 119901

119894 0 0) = 119880

0(119901119894) 119901119894= 119899 minus 119889 minus 2

(24)


119889

119899such

that 1205831(119866lowast) ge 1205831(119866)



119899 1205831(119866) le 120583

1(1198800(119901119896+2)) where


1198800(119901119896+2)


1205831 (119866) lt max max 119889

119894+ 119898119894| V119894isin 119881 (119866) | 119866 isin U

119889

119899

119894 notin 119889 + 2 119896 119896 + 1 119896 + 2

= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (1198800(119901119896+2)) + 1

lt 1205831 (1198800 (119901119896+2))

(25)


0(119901119896) cong 1198800(119901119896+2)


1 V119896+3 V


Φ(119871V119896+2 (1198800))

= (119909 minus 2) [(1199092minus 4119909 + 2)Φ (119863


)]


) + Φ (119863119889minus119896minus2

)]

(26)

Similarly

Φ(119871V119896 (1198800))

= (119909 minus 2) [(1199092minus 4119909 + 2)Φ (119863


)]

times [Φ (119863119896minus1) + Φ (119863

119896minus2)]

(27)



Φ(1198800(119901119896)) minus Φ (119880

0(119901119896+2))


= 1199051199092(119909 minus 2)

2(119909 minus 1)


) gt 0

(28)

for 119909 ge 1205831(1198800(119901119896+2))

From Lemma 8 1205831(1198800(119901119896)) lt 120583

1(1198800(119901119896+2)) holds


1205831 (1198800 (119901119896+1))

lt max 119889119894+ 119898119894| V119894isin 119881 (119880

0(119901119896+1))

= 119905 + 2 +

3 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (1198800 (119901119896+2)) + 1

lt 1205831(1198800(119901119896+2))

(29)


Φ(119871V119896+1 (1198800)) = [Φ (119863119896minus1) + Φ (119863119896minus2)] 119891 (119909)

minus [Φ (119863119896minus2) + Φ (119863

119896minus3)] 119892 (119909)


minus [2119892 (119909) + ℎ (119909)]Φ (119863119896minus2)

(30)

where


minus (21199092minus 9119909 + 7)Φ (119863



ℎ (119909) = (119909 minus 3) 119892 (119909) minus 119891 (119909)



(31)

By (26) and (30)

Φ(119871V119896+2 (1198800)) minus Φ (119871V119896+1 (1198800))


) minus 2Φ (119863119896minus2)]


(32)




= Φ (119863119889minus119896minus2

) + Φ (119863119889minus119896minus3

) gt 0


119896minus2)

=


119909 minus 3 gt 0 if 119896 = 1(33)


Φ(1198800(119901119896+1)) minus Φ (119880

0(119901119896+2))

= 119909 [Φ (119871V119896+2 (1198800)) minus Φ (119871V119896+1 (1198800))] gt 0(34)




4




119899) lt 120583

1(119896

119899) where



119899)


119899) lt 1205831(

119896minus1

119899)


119899) gt 119905+4


Φ(119896minus1

119889+2) minus Φ (

119896

119889+2)


lowast))]


lowast))]

= sdot sdot sdot = 119909 [119904119889119896 (

119909) minus 119905119889119896 (119909)]

(35)

where

119904119889119896 (



119905119889119896 (


119889minus2119896)


(36)



Φ(119896minus1

119899) minus Φ (

119896

119899)

= (119909 minus 1)119905[Φ (

119896minus1

119889+2) minus Φ (

119896

119889+2)]


119896minus1

119889+2)) minus Φ (119871V119896+2 (

119896

119889+2))]

= 1199092(119909 minus 1)

119905minus1(119909 minus 2) 119906 (119909)

(37)



)


)

(38)


1199092minus (119905 + 5) 119909 + 4 le 0 if 119909 isin (119905 + 4 119890 (119905)]

1199092minus (119905 + 5) 119909 + 119905 + 4

gt 1199092minus (119905 + 5) 119909 + 4 gt 0 if 119909 gt 119890 (119905)

(39)


1205831(119896

119899) So 1205831(

119896minus1

119899) lt 1205831(

119896



Φ(119896minus1

119899) minus Φ (

119896

119899)

= 1199092(119909 minus 1)

119905minus1(119909 minus 2) [(119909

2minus (119905 + 5) 119909 + 4)]

(40)

If 119905 = 1 let 27and 1



1(2

7) = 53145

and 1205831(1

7) = 53008 From Lemma 5 120583

1(2

7) le 1205831(119896

119899) and

1205831(1

7) le 1205831(

119896minus1


0 for 119909 ge 1205831(119896

119899) By Lemma 8 120583

1(119896minus1

119899) lt 1205831(119896

119899) holds


119905 (119909) minus [(1199092minus (119905 + 5) 119909 + 4)]


)Φ (119863119898minus1

)]

= 2119905 (119909 minus 1)Φ2(119863119898) minus [(119905

2+ 3119905 + 4) 119909 minus 4]

times Φ (119863119898)Φ (119863

119898minus1) + 2 (119905 + 2) (119909 minus 1)Φ

2(119863119898minus1

)

= [21199051199092minus (1199052+ 9119905 + 4) 119909 + (4119905 + 4)]Φ (119863119898)Φ (119863119898minus1)


+ 2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

= [21199051199092minus (1199052+ 9119905 + 4) 119909 + (4119905 + 4)]Φ (119863119898

)Φ (119863119898minus1

)

+ 4 (119909 minus 1)Φ2(119863119898minus1

) + 2119905 (119909 minus 1)

(41)

d+3 d+2

d+11 kminus1 vk k+1


W0

(a)

d+2d+3



k

d+3

(b)

Figure 2

where

119905 (119909) = [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]

times Φ2(119863119898) minus [(119905 + 4) 119909

2minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863

119898minus1) + 2 (119905 + 2) (119909 minus 1)Φ

2(119863119898minus1

)

(42)


119898)minus(119905+4)Φ(119863

119898)Φ(119863119898minus1

) gt



1(119896

119899) lt 119890(119905) holds Put 119909 = 120583

1(119896

119899) into (40)


119899) lt 1205831(

119896minus1

119899) We

complete the proof



(i) if 119889 is odd then 1205831(119896

119899) le 1205831(lfloor1198892rfloor



119899



119899) le 1205831(lfloor1198892rfloor

119899) the


119899



119899) the


119899





W119889

119899= 119882

0(1199012 119901

119889 119901119889+2 119901119889+3)

119889

sum

119894=2

119901119894 + 119901119889+2 + 119901119889+3 = 119899 minus 119889 minus 3

W119889

119899= 1198820(0 0 119901

119894 0 0)

= 1198820(119901119894) 119901119894= 119899 minus 119889 minus 3

(43)



119889

119899such

that 1205831(119866lowast) ge 1205831(119866)



119899 1205831(119866) le 1205831(1198820(119901119896)) where 1 le



1205831 (119866)


119889

119899 119894 = 119896

= 119905 + 2 +

2 + 4 + 119905

119905 + 2

lt 119905 + 5 = Δ (1198820(119901119896)) + 1

lt 1205831 (1198820 (119901119896))

(44)




119899) lt 1205831(Δ

lceil1198892rceil

119899) lt


119899)


119899) lt 1205831(lfloor1198892rfloor

119899)



119899) 1205831(lfloor1198892rfloor


following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)Let Δ119898+1

119899and 119898


119905 = 0 denote Δ119898+1119899



Φ(1198661) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198671)

minus Φ (119875119898)Φ (119871V119898+2 (1198671))

Φ (1198662) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198672)

minus Φ (119875119898)Φ (119871V119898+2 (1198672))

(45)

where

Φ(1198671) = 119909 (119909 minus 3) [(119909 minus 3)Φ (119863119898

) minus 2Φ (119863119898minus1

)]

Φ (119871V119898+2 (1198671))


)

Φ (1198672)

= 119909 (119909 minus 2)


)]

Φ (119871V119898+2 (1198672))


(46)

Note thatΦ(119875119898) = 119909Φ(119863


Φ(119863119898minus2


Φ(Δ119898+1

119899)


) minus 119905119909Φ (119871V119898+2 (1198661))]


2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (6119905 + 12)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 3) 119909 minus 3 (119905 + 1)]Φ2(119863119898minus1

)

(47)

Φ(119898

119899)


) minus 119905119909Φ (119871V119898+2 (1198662))]



2+ (4119905 + 18) 119909 minus (2119905 + 10)]

times Φ2(119863119898minus1)

minus [31199093minus (3119905 + 19) 119909

2+ 8 (119905 + 4) 119909 minus 4 (119905 + 4)]


)Φ (119863119898minus2



)

(48)


119899) le 119909 lt 119905+5

Φ(Δ119898+1

119899) minus Φ (

119898

119899)

= 119909(119909 minus 1)119905[119889 (119909)Φ

2(119863119898minus1

) + 119890 (119909)Φ (119863119898minus1)Φ (119863

119898minus2)

minus119903 (119909)Φ2(119863119898minus2

)]

gt 119909(119909 minus 1)119905[119889 (119909) + 119890 (119909) minus 119903 (119909)]Φ

2(119863119898minus2

) ge 0

(49)


where





(50)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1


Case 2 119889 = 2119898 + 2 (119898 ge 1)LetΔ119898+1119899

and119898+1119899


119899and119898+1


2 respectively By


Φ(Δ119898+1

119899)


lowast

1) minus 119905119909Φ (119871V119898+2 (119866

lowast

1))]



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 4) 119909 + 3 (119905 + 2)]Φ

2(119863119898minus1)

(51)

Φ(119898+1

119899)


lowast

2) minus 119905119909Φ (119871V119898+3 (119866

lowast

2))]


times [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]Φ

2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863119898minus1)

+2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

(52)


Φ(Δ119898+1

119899) minus Φ (

119898+1

119899)

= 119909(119909 minus 1)119905


2(119863119898)

+ [(119905 + 1) 1199093minus (5119905 + 5) 119909

2+ (7119905 + 10) 119909 minus (2119905 + 6)]

times Φ (119863119898)Φ (119863

119898minus1)

minus [(2119905 + 2) 1199092minus (4119905 + 4) 119909 + (119905 + 2)]Φ

2(119863119898minus1

)



2m+3

m+2

m+3m+1m 2m+2

m+1

n

(a)



m+2

m+3m+1m

2m+4

2m+3

m+1

n

(b)



2m+3

m+2

m+3m+1m 2m+2

⋄m

n

(c)




m+3

m+1 m+2m

2m+4

2m+3

⋄m+1

n

(d)

1


2m+3

m+2

m+1m

2m+4

2m+2


m+1

n

(e)

1


2m+4

m+3m+2m+1

2m+5

2m+3


m+2

n

(f)

Figure 3

= 119909(119909 minus 1)119905


)]Φ (119863119898minus1

) minus 119908 (119909)

ge 119909(119909 minus 1)119905Φ(119863119898minus1

)


)] gt 0

(53)

where

119901 (119909) = 21199093minus (119905 + 9) 119909

2+ (119905 + 9) 119909 minus 2 gt 0

119902 (119909) = 2(119909 minus 1)2gt 0

119908 (119909) = (119905 + 1) 1199092minus (3119905 + 5) 119909 + (119905 + 4) gt 0

119901 (119909) minus 119908 (119909) minus 119902 (119909)

= 21199093minus (2119905 + 12) 119909

2+ (4119905 + 18) 119909 minus (119905 + 8) gt 0

(54)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1





lceil1198892rceil

119899)


119899) 1205831(Δlceil1198892rceil


following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)


Letnabla119898+1119899


119899by 1198663 Let119867


119898 By Lemma 9

Φ(1198663) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]Φ (1198673)

minus Φ (119875119898)Φ (119871V119898+1 (1198673))

(55)

where

Φ(1198673) = 119909 (119909 minus 3)


)]

Φ (119871V119898+1 (1198673)) = (119909 minus 1) (119909 minus 3)


119898minus1)]

(56)

By Lemma 10 we get

Φ(nabla119898+1

119899)


) minus 119905119909Φ (119871V119898+1 (1198663))]

= 119909 (119909 minus 3) (119909 minus 1)119905


times Φ (119863119898)Φ (119863

119898minus1) + (119905 + 2)Φ

2(119863119898minus1

)

(57)

By (47) we have

Φ(Δ119898+1

119899) = 119909(119909 minus 1)

119905


minus [(119905 + 5) 1199092minus (6119905 + 22) 119909 + (6119905 + 18)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 5) 119909 minus 3 (119905 + 2)]Φ2(119863119898minus1

)

(58)

Hence

Φ(nabla119898+1

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(59)


1(nabla119898+1

119899) lt 1205831(Δ119898+1


When 119898 ge 2 1205831(nabla119898+1

119899) lt 1205831(Δ119898+1



Case 2 119889 = 2119898 + 2 (119898 ge 1)

Let nabla119898+2119899


Φ(nabla119898+2

119899) = 119909 (119909 minus 3) (119909 minus 1)

119905


2(119863119898)


119898minus1)


)

(60)

By (51) we get

Φ(Δ119898+1

119899)



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 5) 119909 + (3119905 + 9)]Φ

2(119863119898minus1

)

(61)

Hence

Φ(nabla119898+2

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(62)

Let 119905119896(119909) = (119909 minus 119905 minus 4)Φ(119863

119896) minus (119905 + 3)Φ(119863



1(119905119896(119909)) where 119896 ge 0




1(119905119896minus1(119909)) lt


119905119896+1 (


119909) (63)

1199052

119896(119909) minus 119905119896+1 (

119909) 119905119896minus1 (119909)


(64)




Let 11986610158403= nabla119898+2




and1198661015840

1= Δ119898+1





By (60) and (62)

Φ(1198661015840

3) = 119909(119909 minus 1)

119905(119909 minus 3) (119909

2minus 3119909 + 1) 119903 (119909)

Φ (1198661015840

3) minus Φ (119866

1015840

1)

= 1199092(119909 minus 1)

119905[1199093minus (119905 + 8) 119909

2+ (3119905 + 16) 119909 minus (119905 + 6)]

= 1199092(119909 minus 1)

119905(119903 (119909) + 1)

(65)


1(1198661015840


1205831(1198661015840

3) ge 1205831(119866

1015840

1) then [Φ(1198661015840

3)minusΦ(119866

1015840

1)]|119909=1205831(119866

1015840

3) le 0 and 119903(119909)+

1|119909=1205831(119866

1015840


1(1198661015840

3) lt 1205831(1198661015840

1)


119909V (119909) minus (119905 + 2) 119903 (119909)

= 3 (119905 + 2) 1199092minus (31199052+ 22119905 + 33) 119909 + (119905 + 2) (119905 + 7) gt 0

(66)

From Lemma 8 1205831(V(119909)) lt 120583

1(1198661015840


1205831(1198661015840

3) le 1205831(nabla119898+2

119899) and 120583

1(1198661015840

1) le 1205831(Δ119898+1



119899) From


119899) lt 1205831(Δ119898+1

119899) as desired

Let ◻119896119889+3



119889+3) lt 1205831(Δlceil1198892rceil

119889+3)


119889+3) gt 5


Φ(◻119898+1

2119898+4) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]

times Φ (1198674) minus Φ (119875

119898)Φ (119871V119898+1 (1198674))

= 119909 [(119909 minus 2)2(119909 minus 4)Φ


times Φ (119863119898)Φ (119863

119898minus1) + (3119909 minus 8)Φ

2(119863119898minus1

)]

(67)

where1198674= ◻119898+1


119898

By (47) we get

Φ(Δ119898+1

2119898+4)


2minus 22119909 + 18)

times Φ (119863119898)Φ (119863

119898minus1) + (5119909 minus 6)Φ

2(119863119898minus1

)]

(68)


Φ(◻119898+1

2119898+4) minus Φ (Δ

119898+1

2119898+4)

= 119909 [(119909 minus 4)Φ2(119863119898) + (119909

2minus 2119909 minus 6)Φ (119863

119898)Φ (119863

119898minus1)

minus2 (119909 + 1)Φ2(119863119898minus1

)]

= 119909Φ (119863119898minus1

) [2 (1199092minus 4119909 + 1)Φ (119863


)]

+ 119909 (119909 minus 4)

gt 119909Φ2(119863119898minus1

) (41199092minus 19119909 + 6) + 119909 (119909 minus 4) gt 0

(69)

So 1205831(◻119898+1

2119898+4) lt 1205831(Δ119898+1



1(◻119898+1

2119898+5) lt 120583

1(Δ119898+1

2119898+5)

holds



1 1199092 119909

119899)119879



vertex V119894(1 le 119894 le 119899)



Claim 1 119881(119862119902)⋂119881(119875

119889+1) = 0


119894V119896V119896+1 V


119902and 119875119889+1

where V

119894isin 119881(119862

119902) V119897isin 119881(119875

119889+1)and V

119896 V

119897minus1isin 119881(119866)

(119881(119862119902)⋃119881(119875119889+1

))For each 119895 let 119879


119895as its root)


119895is



119894







119894) = 119889(V 119903

119894) minus 1 By


1(1198661) ge 1205831(119866)





3 where



V119896 V

119897minus1 So 120583

1(1198663) ge 1205831(1198662)







119889+1) = 0 and 120583

1(1198664) ge 1205831(1198663)


By Claim 1 119881(119862119902)⋂119881(119875


119862119902

= V119896V119896+1 V

119897minus1V119897V119889+2

V119889+3 V

119904V119896(119904 ge 119889 + 2)

where V119896 V119896+1 V

119897minus1 V119897 = 119881(119862

119902)⋂119881(119875

119889+1) and

V119889+2 V119889+3 V

119904 = 119881(119862

119902) 119881(119875

119889+1)

Claim 2 119889(V) = 1 for V isin 119881(119866) (119881(119862119902)⋃119881(119875119889+1))


3) we can get

119894and 120583

1(119894) ge


119894(V119894isin

119881(119862119902)⋃119881(119875

119889+1) V


Claim 3 119897 = 119896 + 1 and 119904 = 119889 + 2


(119881(119862119902)⋃119881(119875119889+1

) = V119904+1 V

119899 and |119860| = 119899 minus 119904 = 119905 ge 0

Case 1 119897 = 119896 By Lemma 7


= 119905 + 2 +

2 + 4 + 119905

119905 + 2

le 119905 + 5 = Δ (119896) + 1

lt 1205831 (119896) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

= 119905 + 4 +

2 + 2 + 2 + 2 + 119905

119905 + 4

le 119905 + 6 le 119899 minus 119889 + 2

= Δ (nabla119896

119899) + 1 lt 120583

1(nabla119896

119899)

(70)

for 119904 ge 119889 + 4 (|119862119902| ge 4)

When 119904 = 119889 + 3 (|119862119902| = 3)

119894cong 1198820(119901119894) By Lemma 23

1205831(119894) lt 1205831(nabla119896

119899) for any 119894 = 119896 where


119899




119899) holds

Case 2 119897 = 119896 + 1


1205831 (119894)


= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (119896) + 1 (Δ (

119896+1) + 1)

lt 1205831(119896) (1205831(119896+1)) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

le 119905 + 3 +

2 + 3 + 119905

119905 + 3

lt 119905 + 5 le 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(71)

for 119904 ge 119889 + 3 (|119862119902| ge 4)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) By Lemma 15

1205831(119894) lt 1205831(Δ119896

119899) for any 119894 = 119896 + 1 where

119896+1cong Δ119896

119899



= 3 +

2 + 3 + 2

3

lt 6 lt 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(72)

for 119904 ge 119889 + 4 (|119862119902| ge 5)

When 119904 = 119889 + 3 (|119862119902| = 4) by Lemma 13 120583

1(◻119896

119889+3) lt

1205831(◻lceil1198892rceil


lceil1198892rceil

119889+3) gt

1205831(◻lceil1198892rceil

119889+3)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) cong Δ119896

119899



119899) gt 1205831(Δlceil1198892rceil

119899) holds

Case 3 119897 ge 119896 + 2 By Lemma 7

1205831(119894)


= 119905 + 3 +

2 + 2 + 2 + 119905

119905 + 3

le 119905 + 5 le 119899 minus 119889 + 2 = Δ (119896

119899) + 1

lt 1205831 (119896

119899)

(73)

for 119904 ge 119889 + 3 (|119862119902| ge 5)

When 119904 = 119889 + 2 (|119862119902| = 4)

119894cong 1198800(119901119894) by Lemma 19

1205831(119894) lt 1205831(

119896


119896

119899





119899has the


Acknowledgments



References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Φ(1198800(119901119896)) minus Φ (119880

0(119901119896+2))


= 1199051199092(119909 minus 2)

2(119909 minus 1)


) gt 0

(28)

for 119909 ge 1205831(1198800(119901119896+2))

From Lemma 8 1205831(1198800(119901119896)) lt 120583

1(1198800(119901119896+2)) holds


1205831 (1198800 (119901119896+1))

lt max 119889119894+ 119898119894| V119894isin 119881 (119880

0(119901119896+1))

= 119905 + 2 +

3 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (1198800 (119901119896+2)) + 1

lt 1205831(1198800(119901119896+2))

(29)


Φ(119871V119896+1 (1198800)) = [Φ (119863119896minus1) + Φ (119863119896minus2)] 119891 (119909)

minus [Φ (119863119896minus2) + Φ (119863

119896minus3)] 119892 (119909)


minus [2119892 (119909) + ℎ (119909)]Φ (119863119896minus2)

(30)

where


minus (21199092minus 9119909 + 7)Φ (119863



ℎ (119909) = (119909 minus 3) 119892 (119909) minus 119891 (119909)



(31)

By (26) and (30)

Φ(119871V119896+2 (1198800)) minus Φ (119871V119896+1 (1198800))


) minus 2Φ (119863119896minus2)]


(32)




= Φ (119863119889minus119896minus2

) + Φ (119863119889minus119896minus3

) gt 0


119896minus2)

=


119909 minus 3 gt 0 if 119896 = 1(33)


Φ(1198800(119901119896+1)) minus Φ (119880

0(119901119896+2))

= 119909 [Φ (119871V119896+2 (1198800)) minus Φ (119871V119896+1 (1198800))] gt 0(34)




4




119899) lt 120583

1(119896

119899) where



119899)


119899) lt 1205831(

119896minus1

119899)


119899) gt 119905+4


Φ(119896minus1

119889+2) minus Φ (

119896

119889+2)


lowast))]


lowast))]

= sdot sdot sdot = 119909 [119904119889119896 (

119909) minus 119905119889119896 (119909)]

(35)

where

119904119889119896 (



119905119889119896 (


119889minus2119896)


(36)



Φ(119896minus1

119899) minus Φ (

119896

119899)

= (119909 minus 1)119905[Φ (

119896minus1

119889+2) minus Φ (

119896

119889+2)]


119896minus1

119889+2)) minus Φ (119871V119896+2 (

119896

119889+2))]

= 1199092(119909 minus 1)

119905minus1(119909 minus 2) 119906 (119909)

(37)



)


)

(38)


1199092minus (119905 + 5) 119909 + 4 le 0 if 119909 isin (119905 + 4 119890 (119905)]

1199092minus (119905 + 5) 119909 + 119905 + 4

gt 1199092minus (119905 + 5) 119909 + 4 gt 0 if 119909 gt 119890 (119905)

(39)


1205831(119896

119899) So 1205831(

119896minus1

119899) lt 1205831(

119896



Φ(119896minus1

119899) minus Φ (

119896

119899)

= 1199092(119909 minus 1)

119905minus1(119909 minus 2) [(119909

2minus (119905 + 5) 119909 + 4)]

(40)

If 119905 = 1 let 27and 1



1(2

7) = 53145

and 1205831(1

7) = 53008 From Lemma 5 120583

1(2

7) le 1205831(119896

119899) and

1205831(1

7) le 1205831(

119896minus1


0 for 119909 ge 1205831(119896

119899) By Lemma 8 120583

1(119896minus1

119899) lt 1205831(119896

119899) holds


119905 (119909) minus [(1199092minus (119905 + 5) 119909 + 4)]


)Φ (119863119898minus1

)]

= 2119905 (119909 minus 1)Φ2(119863119898) minus [(119905

2+ 3119905 + 4) 119909 minus 4]

times Φ (119863119898)Φ (119863

119898minus1) + 2 (119905 + 2) (119909 minus 1)Φ

2(119863119898minus1

)

= [21199051199092minus (1199052+ 9119905 + 4) 119909 + (4119905 + 4)]Φ (119863119898)Φ (119863119898minus1)


+ 2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

= [21199051199092minus (1199052+ 9119905 + 4) 119909 + (4119905 + 4)]Φ (119863119898

)Φ (119863119898minus1

)

+ 4 (119909 minus 1)Φ2(119863119898minus1

) + 2119905 (119909 minus 1)

(41)

d+3 d+2

d+11 kminus1 vk k+1


W0

(a)

d+2d+3



k

d+3

(b)

Figure 2

where

119905 (119909) = [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]

times Φ2(119863119898) minus [(119905 + 4) 119909

2minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863

119898minus1) + 2 (119905 + 2) (119909 minus 1)Φ

2(119863119898minus1

)

(42)


119898)minus(119905+4)Φ(119863

119898)Φ(119863119898minus1

) gt



1(119896

119899) lt 119890(119905) holds Put 119909 = 120583

1(119896

119899) into (40)


119899) lt 1205831(

119896minus1

119899) We

complete the proof



(i) if 119889 is odd then 1205831(119896

119899) le 1205831(lfloor1198892rfloor



119899



119899) le 1205831(lfloor1198892rfloor

119899) the


119899



119899) the


119899





W119889

119899= 119882

0(1199012 119901

119889 119901119889+2 119901119889+3)

119889

sum

119894=2

119901119894 + 119901119889+2 + 119901119889+3 = 119899 minus 119889 minus 3

W119889

119899= 1198820(0 0 119901

119894 0 0)

= 1198820(119901119894) 119901119894= 119899 minus 119889 minus 3

(43)



119889

119899such

that 1205831(119866lowast) ge 1205831(119866)



119899 1205831(119866) le 1205831(1198820(119901119896)) where 1 le



1205831 (119866)


119889

119899 119894 = 119896

= 119905 + 2 +

2 + 4 + 119905

119905 + 2

lt 119905 + 5 = Δ (1198820(119901119896)) + 1

lt 1205831 (1198820 (119901119896))

(44)




119899) lt 1205831(Δ

lceil1198892rceil

119899) lt


119899)


119899) lt 1205831(lfloor1198892rfloor

119899)



119899) 1205831(lfloor1198892rfloor


following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)Let Δ119898+1

119899and 119898


119905 = 0 denote Δ119898+1119899



Φ(1198661) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198671)

minus Φ (119875119898)Φ (119871V119898+2 (1198671))

Φ (1198662) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198672)

minus Φ (119875119898)Φ (119871V119898+2 (1198672))

(45)

where

Φ(1198671) = 119909 (119909 minus 3) [(119909 minus 3)Φ (119863119898

) minus 2Φ (119863119898minus1

)]

Φ (119871V119898+2 (1198671))


)

Φ (1198672)

= 119909 (119909 minus 2)


)]

Φ (119871V119898+2 (1198672))


(46)

Note thatΦ(119875119898) = 119909Φ(119863


Φ(119863119898minus2


Φ(Δ119898+1

119899)


) minus 119905119909Φ (119871V119898+2 (1198661))]


2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (6119905 + 12)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 3) 119909 minus 3 (119905 + 1)]Φ2(119863119898minus1

)

(47)

Φ(119898

119899)


) minus 119905119909Φ (119871V119898+2 (1198662))]



2+ (4119905 + 18) 119909 minus (2119905 + 10)]

times Φ2(119863119898minus1)

minus [31199093minus (3119905 + 19) 119909

2+ 8 (119905 + 4) 119909 minus 4 (119905 + 4)]


)Φ (119863119898minus2



)

(48)


119899) le 119909 lt 119905+5

Φ(Δ119898+1

119899) minus Φ (

119898

119899)

= 119909(119909 minus 1)119905[119889 (119909)Φ

2(119863119898minus1

) + 119890 (119909)Φ (119863119898minus1)Φ (119863

119898minus2)

minus119903 (119909)Φ2(119863119898minus2

)]

gt 119909(119909 minus 1)119905[119889 (119909) + 119890 (119909) minus 119903 (119909)]Φ

2(119863119898minus2

) ge 0

(49)


where





(50)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1


Case 2 119889 = 2119898 + 2 (119898 ge 1)LetΔ119898+1119899

and119898+1119899


119899and119898+1


2 respectively By


Φ(Δ119898+1

119899)


lowast

1) minus 119905119909Φ (119871V119898+2 (119866

lowast

1))]



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 4) 119909 + 3 (119905 + 2)]Φ

2(119863119898minus1)

(51)

Φ(119898+1

119899)


lowast

2) minus 119905119909Φ (119871V119898+3 (119866

lowast

2))]


times [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]Φ

2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863119898minus1)

+2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

(52)


Φ(Δ119898+1

119899) minus Φ (

119898+1

119899)

= 119909(119909 minus 1)119905


2(119863119898)

+ [(119905 + 1) 1199093minus (5119905 + 5) 119909

2+ (7119905 + 10) 119909 minus (2119905 + 6)]

times Φ (119863119898)Φ (119863

119898minus1)

minus [(2119905 + 2) 1199092minus (4119905 + 4) 119909 + (119905 + 2)]Φ

2(119863119898minus1

)



2m+3

m+2

m+3m+1m 2m+2

m+1

n

(a)



m+2

m+3m+1m

2m+4

2m+3

m+1

n

(b)



2m+3

m+2

m+3m+1m 2m+2

⋄m

n

(c)




m+3

m+1 m+2m

2m+4

2m+3

⋄m+1

n

(d)

1


2m+3

m+2

m+1m

2m+4

2m+2


m+1

n

(e)

1


2m+4

m+3m+2m+1

2m+5

2m+3


m+2

n

(f)

Figure 3

= 119909(119909 minus 1)119905


)]Φ (119863119898minus1

) minus 119908 (119909)

ge 119909(119909 minus 1)119905Φ(119863119898minus1

)


)] gt 0

(53)

where

119901 (119909) = 21199093minus (119905 + 9) 119909

2+ (119905 + 9) 119909 minus 2 gt 0

119902 (119909) = 2(119909 minus 1)2gt 0

119908 (119909) = (119905 + 1) 1199092minus (3119905 + 5) 119909 + (119905 + 4) gt 0

119901 (119909) minus 119908 (119909) minus 119902 (119909)

= 21199093minus (2119905 + 12) 119909

2+ (4119905 + 18) 119909 minus (119905 + 8) gt 0

(54)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1





lceil1198892rceil

119899)


119899) 1205831(Δlceil1198892rceil


following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)


Letnabla119898+1119899


119899by 1198663 Let119867


119898 By Lemma 9

Φ(1198663) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]Φ (1198673)

minus Φ (119875119898)Φ (119871V119898+1 (1198673))

(55)

where

Φ(1198673) = 119909 (119909 minus 3)


)]

Φ (119871V119898+1 (1198673)) = (119909 minus 1) (119909 minus 3)


119898minus1)]

(56)

By Lemma 10 we get

Φ(nabla119898+1

119899)


) minus 119905119909Φ (119871V119898+1 (1198663))]

= 119909 (119909 minus 3) (119909 minus 1)119905


times Φ (119863119898)Φ (119863

119898minus1) + (119905 + 2)Φ

2(119863119898minus1

)

(57)

By (47) we have

Φ(Δ119898+1

119899) = 119909(119909 minus 1)

119905


minus [(119905 + 5) 1199092minus (6119905 + 22) 119909 + (6119905 + 18)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 5) 119909 minus 3 (119905 + 2)]Φ2(119863119898minus1

)

(58)

Hence

Φ(nabla119898+1

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(59)


1(nabla119898+1

119899) lt 1205831(Δ119898+1


When 119898 ge 2 1205831(nabla119898+1

119899) lt 1205831(Δ119898+1



Case 2 119889 = 2119898 + 2 (119898 ge 1)

Let nabla119898+2119899


Φ(nabla119898+2

119899) = 119909 (119909 minus 3) (119909 minus 1)

119905


2(119863119898)


119898minus1)


)

(60)

By (51) we get

Φ(Δ119898+1

119899)



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 5) 119909 + (3119905 + 9)]Φ

2(119863119898minus1

)

(61)

Hence

Φ(nabla119898+2

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(62)

Let 119905119896(119909) = (119909 minus 119905 minus 4)Φ(119863

119896) minus (119905 + 3)Φ(119863



1(119905119896(119909)) where 119896 ge 0




1(119905119896minus1(119909)) lt


119905119896+1 (


119909) (63)

1199052

119896(119909) minus 119905119896+1 (

119909) 119905119896minus1 (119909)


(64)




Let 11986610158403= nabla119898+2




and1198661015840

1= Δ119898+1





By (60) and (62)

Φ(1198661015840

3) = 119909(119909 minus 1)

119905(119909 minus 3) (119909

2minus 3119909 + 1) 119903 (119909)

Φ (1198661015840

3) minus Φ (119866

1015840

1)

= 1199092(119909 minus 1)

119905[1199093minus (119905 + 8) 119909

2+ (3119905 + 16) 119909 minus (119905 + 6)]

= 1199092(119909 minus 1)

119905(119903 (119909) + 1)

(65)


1(1198661015840


1205831(1198661015840

3) ge 1205831(119866

1015840

1) then [Φ(1198661015840

3)minusΦ(119866

1015840

1)]|119909=1205831(119866

1015840

3) le 0 and 119903(119909)+

1|119909=1205831(119866

1015840


1(1198661015840

3) lt 1205831(1198661015840

1)


119909V (119909) minus (119905 + 2) 119903 (119909)

= 3 (119905 + 2) 1199092minus (31199052+ 22119905 + 33) 119909 + (119905 + 2) (119905 + 7) gt 0

(66)

From Lemma 8 1205831(V(119909)) lt 120583

1(1198661015840


1205831(1198661015840

3) le 1205831(nabla119898+2

119899) and 120583

1(1198661015840

1) le 1205831(Δ119898+1



119899) From


119899) lt 1205831(Δ119898+1

119899) as desired

Let ◻119896119889+3



119889+3) lt 1205831(Δlceil1198892rceil

119889+3)


119889+3) gt 5


Φ(◻119898+1

2119898+4) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]

times Φ (1198674) minus Φ (119875

119898)Φ (119871V119898+1 (1198674))

= 119909 [(119909 minus 2)2(119909 minus 4)Φ


times Φ (119863119898)Φ (119863

119898minus1) + (3119909 minus 8)Φ

2(119863119898minus1

)]

(67)

where1198674= ◻119898+1


119898

By (47) we get

Φ(Δ119898+1

2119898+4)


2minus 22119909 + 18)

times Φ (119863119898)Φ (119863

119898minus1) + (5119909 minus 6)Φ

2(119863119898minus1

)]

(68)


Φ(◻119898+1

2119898+4) minus Φ (Δ

119898+1

2119898+4)

= 119909 [(119909 minus 4)Φ2(119863119898) + (119909

2minus 2119909 minus 6)Φ (119863

119898)Φ (119863

119898minus1)

minus2 (119909 + 1)Φ2(119863119898minus1

)]

= 119909Φ (119863119898minus1

) [2 (1199092minus 4119909 + 1)Φ (119863


)]

+ 119909 (119909 minus 4)

gt 119909Φ2(119863119898minus1

) (41199092minus 19119909 + 6) + 119909 (119909 minus 4) gt 0

(69)

So 1205831(◻119898+1

2119898+4) lt 1205831(Δ119898+1



1(◻119898+1

2119898+5) lt 120583

1(Δ119898+1

2119898+5)

holds



1 1199092 119909

119899)119879



vertex V119894(1 le 119894 le 119899)



Claim 1 119881(119862119902)⋂119881(119875

119889+1) = 0


119894V119896V119896+1 V


119902and 119875119889+1

where V

119894isin 119881(119862

119902) V119897isin 119881(119875

119889+1)and V

119896 V

119897minus1isin 119881(119866)

(119881(119862119902)⋃119881(119875119889+1

))For each 119895 let 119879


119895as its root)


119895is



119894







119894) = 119889(V 119903

119894) minus 1 By


1(1198661) ge 1205831(119866)





3 where



V119896 V

119897minus1 So 120583

1(1198663) ge 1205831(1198662)







119889+1) = 0 and 120583

1(1198664) ge 1205831(1198663)


By Claim 1 119881(119862119902)⋂119881(119875


119862119902

= V119896V119896+1 V

119897minus1V119897V119889+2

V119889+3 V

119904V119896(119904 ge 119889 + 2)

where V119896 V119896+1 V

119897minus1 V119897 = 119881(119862

119902)⋂119881(119875

119889+1) and

V119889+2 V119889+3 V

119904 = 119881(119862

119902) 119881(119875

119889+1)

Claim 2 119889(V) = 1 for V isin 119881(119866) (119881(119862119902)⋃119881(119875119889+1))


3) we can get

119894and 120583

1(119894) ge


119894(V119894isin

119881(119862119902)⋃119881(119875

119889+1) V


Claim 3 119897 = 119896 + 1 and 119904 = 119889 + 2


(119881(119862119902)⋃119881(119875119889+1

) = V119904+1 V

119899 and |119860| = 119899 minus 119904 = 119905 ge 0

Case 1 119897 = 119896 By Lemma 7


= 119905 + 2 +

2 + 4 + 119905

119905 + 2

le 119905 + 5 = Δ (119896) + 1

lt 1205831 (119896) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

= 119905 + 4 +

2 + 2 + 2 + 2 + 119905

119905 + 4

le 119905 + 6 le 119899 minus 119889 + 2

= Δ (nabla119896

119899) + 1 lt 120583

1(nabla119896

119899)

(70)

for 119904 ge 119889 + 4 (|119862119902| ge 4)

When 119904 = 119889 + 3 (|119862119902| = 3)

119894cong 1198820(119901119894) By Lemma 23

1205831(119894) lt 1205831(nabla119896

119899) for any 119894 = 119896 where


119899




119899) holds

Case 2 119897 = 119896 + 1


1205831 (119894)


= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (119896) + 1 (Δ (

119896+1) + 1)

lt 1205831(119896) (1205831(119896+1)) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

le 119905 + 3 +

2 + 3 + 119905

119905 + 3

lt 119905 + 5 le 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(71)

for 119904 ge 119889 + 3 (|119862119902| ge 4)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) By Lemma 15

1205831(119894) lt 1205831(Δ119896

119899) for any 119894 = 119896 + 1 where

119896+1cong Δ119896

119899



= 3 +

2 + 3 + 2

3

lt 6 lt 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(72)

for 119904 ge 119889 + 4 (|119862119902| ge 5)

When 119904 = 119889 + 3 (|119862119902| = 4) by Lemma 13 120583

1(◻119896

119889+3) lt

1205831(◻lceil1198892rceil


lceil1198892rceil

119889+3) gt

1205831(◻lceil1198892rceil

119889+3)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) cong Δ119896

119899



119899) gt 1205831(Δlceil1198892rceil

119899) holds

Case 3 119897 ge 119896 + 2 By Lemma 7

1205831(119894)


= 119905 + 3 +

2 + 2 + 2 + 119905

119905 + 3

le 119905 + 5 le 119899 minus 119889 + 2 = Δ (119896

119899) + 1

lt 1205831 (119896

119899)

(73)

for 119904 ge 119889 + 3 (|119862119902| ge 5)

When 119904 = 119889 + 2 (|119862119902| = 4)

119894cong 1198800(119901119894) by Lemma 19

1205831(119894) lt 1205831(

119896


119896

119899





119899has the


Acknowledgments



References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Φ(119896minus1

119899) minus Φ (

119896

119899)

= (119909 minus 1)119905[Φ (

119896minus1

119889+2) minus Φ (

119896

119889+2)]


119896minus1

119889+2)) minus Φ (119871V119896+2 (

119896

119889+2))]

= 1199092(119909 minus 1)

119905minus1(119909 minus 2) 119906 (119909)

(37)



)


)

(38)


1199092minus (119905 + 5) 119909 + 4 le 0 if 119909 isin (119905 + 4 119890 (119905)]

1199092minus (119905 + 5) 119909 + 119905 + 4

gt 1199092minus (119905 + 5) 119909 + 4 gt 0 if 119909 gt 119890 (119905)

(39)


1205831(119896

119899) So 1205831(

119896minus1

119899) lt 1205831(

119896



Φ(119896minus1

119899) minus Φ (

119896

119899)

= 1199092(119909 minus 1)

119905minus1(119909 minus 2) [(119909

2minus (119905 + 5) 119909 + 4)]

(40)

If 119905 = 1 let 27and 1



1(2

7) = 53145

and 1205831(1

7) = 53008 From Lemma 5 120583

1(2

7) le 1205831(119896

119899) and

1205831(1

7) le 1205831(

119896minus1


0 for 119909 ge 1205831(119896

119899) By Lemma 8 120583

1(119896minus1

119899) lt 1205831(119896

119899) holds


119905 (119909) minus [(1199092minus (119905 + 5) 119909 + 4)]


)Φ (119863119898minus1

)]

= 2119905 (119909 minus 1)Φ2(119863119898) minus [(119905

2+ 3119905 + 4) 119909 minus 4]

times Φ (119863119898)Φ (119863

119898minus1) + 2 (119905 + 2) (119909 minus 1)Φ

2(119863119898minus1

)

= [21199051199092minus (1199052+ 9119905 + 4) 119909 + (4119905 + 4)]Φ (119863119898)Φ (119863119898minus1)


+ 2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

= [21199051199092minus (1199052+ 9119905 + 4) 119909 + (4119905 + 4)]Φ (119863119898

)Φ (119863119898minus1

)

+ 4 (119909 minus 1)Φ2(119863119898minus1

) + 2119905 (119909 minus 1)

(41)

d+3 d+2

d+11 kminus1 vk k+1


W0

(a)

d+2d+3



k

d+3

(b)

Figure 2

where

119905 (119909) = [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]

times Φ2(119863119898) minus [(119905 + 4) 119909

2minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863

119898minus1) + 2 (119905 + 2) (119909 minus 1)Φ

2(119863119898minus1

)

(42)


119898)minus(119905+4)Φ(119863

119898)Φ(119863119898minus1

) gt



1(119896

119899) lt 119890(119905) holds Put 119909 = 120583

1(119896

119899) into (40)


119899) lt 1205831(

119896minus1

119899) We

complete the proof



(i) if 119889 is odd then 1205831(119896

119899) le 1205831(lfloor1198892rfloor



119899



119899) le 1205831(lfloor1198892rfloor

119899) the


119899



119899) the


119899





W119889

119899= 119882

0(1199012 119901

119889 119901119889+2 119901119889+3)

119889

sum

119894=2

119901119894 + 119901119889+2 + 119901119889+3 = 119899 minus 119889 minus 3

W119889

119899= 1198820(0 0 119901

119894 0 0)

= 1198820(119901119894) 119901119894= 119899 minus 119889 minus 3

(43)



119889

119899such

that 1205831(119866lowast) ge 1205831(119866)



119899 1205831(119866) le 1205831(1198820(119901119896)) where 1 le



1205831 (119866)


119889

119899 119894 = 119896

= 119905 + 2 +

2 + 4 + 119905

119905 + 2

lt 119905 + 5 = Δ (1198820(119901119896)) + 1

lt 1205831 (1198820 (119901119896))

(44)




119899) lt 1205831(Δ

lceil1198892rceil

119899) lt


119899)


119899) lt 1205831(lfloor1198892rfloor

119899)



119899) 1205831(lfloor1198892rfloor


following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)Let Δ119898+1

119899and 119898


119905 = 0 denote Δ119898+1119899



Φ(1198661) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198671)

minus Φ (119875119898)Φ (119871V119898+2 (1198671))

Φ (1198662) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198672)

minus Φ (119875119898)Φ (119871V119898+2 (1198672))

(45)

where

Φ(1198671) = 119909 (119909 minus 3) [(119909 minus 3)Φ (119863119898

) minus 2Φ (119863119898minus1

)]

Φ (119871V119898+2 (1198671))


)

Φ (1198672)

= 119909 (119909 minus 2)


)]

Φ (119871V119898+2 (1198672))


(46)

Note thatΦ(119875119898) = 119909Φ(119863


Φ(119863119898minus2


Φ(Δ119898+1

119899)


) minus 119905119909Φ (119871V119898+2 (1198661))]


2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (6119905 + 12)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 3) 119909 minus 3 (119905 + 1)]Φ2(119863119898minus1

)

(47)

Φ(119898

119899)


) minus 119905119909Φ (119871V119898+2 (1198662))]



2+ (4119905 + 18) 119909 minus (2119905 + 10)]

times Φ2(119863119898minus1)

minus [31199093minus (3119905 + 19) 119909

2+ 8 (119905 + 4) 119909 minus 4 (119905 + 4)]


)Φ (119863119898minus2



)

(48)


119899) le 119909 lt 119905+5

Φ(Δ119898+1

119899) minus Φ (

119898

119899)

= 119909(119909 minus 1)119905[119889 (119909)Φ

2(119863119898minus1

) + 119890 (119909)Φ (119863119898minus1)Φ (119863

119898minus2)

minus119903 (119909)Φ2(119863119898minus2

)]

gt 119909(119909 minus 1)119905[119889 (119909) + 119890 (119909) minus 119903 (119909)]Φ

2(119863119898minus2

) ge 0

(49)


where





(50)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1


Case 2 119889 = 2119898 + 2 (119898 ge 1)LetΔ119898+1119899

and119898+1119899


119899and119898+1


2 respectively By


Φ(Δ119898+1

119899)


lowast

1) minus 119905119909Φ (119871V119898+2 (119866

lowast

1))]



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 4) 119909 + 3 (119905 + 2)]Φ

2(119863119898minus1)

(51)

Φ(119898+1

119899)


lowast

2) minus 119905119909Φ (119871V119898+3 (119866

lowast

2))]


times [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]Φ

2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863119898minus1)

+2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

(52)


Φ(Δ119898+1

119899) minus Φ (

119898+1

119899)

= 119909(119909 minus 1)119905


2(119863119898)

+ [(119905 + 1) 1199093minus (5119905 + 5) 119909

2+ (7119905 + 10) 119909 minus (2119905 + 6)]

times Φ (119863119898)Φ (119863

119898minus1)

minus [(2119905 + 2) 1199092minus (4119905 + 4) 119909 + (119905 + 2)]Φ

2(119863119898minus1

)



2m+3

m+2

m+3m+1m 2m+2

m+1

n

(a)



m+2

m+3m+1m

2m+4

2m+3

m+1

n

(b)



2m+3

m+2

m+3m+1m 2m+2

⋄m

n

(c)




m+3

m+1 m+2m

2m+4

2m+3

⋄m+1

n

(d)

1


2m+3

m+2

m+1m

2m+4

2m+2


m+1

n

(e)

1


2m+4

m+3m+2m+1

2m+5

2m+3


m+2

n

(f)

Figure 3

= 119909(119909 minus 1)119905


)]Φ (119863119898minus1

) minus 119908 (119909)

ge 119909(119909 minus 1)119905Φ(119863119898minus1

)


)] gt 0

(53)

where

119901 (119909) = 21199093minus (119905 + 9) 119909

2+ (119905 + 9) 119909 minus 2 gt 0

119902 (119909) = 2(119909 minus 1)2gt 0

119908 (119909) = (119905 + 1) 1199092minus (3119905 + 5) 119909 + (119905 + 4) gt 0

119901 (119909) minus 119908 (119909) minus 119902 (119909)

= 21199093minus (2119905 + 12) 119909

2+ (4119905 + 18) 119909 minus (119905 + 8) gt 0

(54)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1





lceil1198892rceil

119899)


119899) 1205831(Δlceil1198892rceil


following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)


Letnabla119898+1119899


119899by 1198663 Let119867


119898 By Lemma 9

Φ(1198663) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]Φ (1198673)

minus Φ (119875119898)Φ (119871V119898+1 (1198673))

(55)

where

Φ(1198673) = 119909 (119909 minus 3)


)]

Φ (119871V119898+1 (1198673)) = (119909 minus 1) (119909 minus 3)


119898minus1)]

(56)

By Lemma 10 we get

Φ(nabla119898+1

119899)


) minus 119905119909Φ (119871V119898+1 (1198663))]

= 119909 (119909 minus 3) (119909 minus 1)119905


times Φ (119863119898)Φ (119863

119898minus1) + (119905 + 2)Φ

2(119863119898minus1

)

(57)

By (47) we have

Φ(Δ119898+1

119899) = 119909(119909 minus 1)

119905


minus [(119905 + 5) 1199092minus (6119905 + 22) 119909 + (6119905 + 18)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 5) 119909 minus 3 (119905 + 2)]Φ2(119863119898minus1

)

(58)

Hence

Φ(nabla119898+1

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(59)


1(nabla119898+1

119899) lt 1205831(Δ119898+1


When 119898 ge 2 1205831(nabla119898+1

119899) lt 1205831(Δ119898+1



Case 2 119889 = 2119898 + 2 (119898 ge 1)

Let nabla119898+2119899


Φ(nabla119898+2

119899) = 119909 (119909 minus 3) (119909 minus 1)

119905


2(119863119898)


119898minus1)


)

(60)

By (51) we get

Φ(Δ119898+1

119899)



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 5) 119909 + (3119905 + 9)]Φ

2(119863119898minus1

)

(61)

Hence

Φ(nabla119898+2

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(62)

Let 119905119896(119909) = (119909 minus 119905 minus 4)Φ(119863

119896) minus (119905 + 3)Φ(119863



1(119905119896(119909)) where 119896 ge 0




1(119905119896minus1(119909)) lt


119905119896+1 (


119909) (63)

1199052

119896(119909) minus 119905119896+1 (

119909) 119905119896minus1 (119909)


(64)




Let 11986610158403= nabla119898+2




and1198661015840

1= Δ119898+1





By (60) and (62)

Φ(1198661015840

3) = 119909(119909 minus 1)

119905(119909 minus 3) (119909

2minus 3119909 + 1) 119903 (119909)

Φ (1198661015840

3) minus Φ (119866

1015840

1)

= 1199092(119909 minus 1)

119905[1199093minus (119905 + 8) 119909

2+ (3119905 + 16) 119909 minus (119905 + 6)]

= 1199092(119909 minus 1)

119905(119903 (119909) + 1)

(65)


1(1198661015840


1205831(1198661015840

3) ge 1205831(119866

1015840

1) then [Φ(1198661015840

3)minusΦ(119866

1015840

1)]|119909=1205831(119866

1015840

3) le 0 and 119903(119909)+

1|119909=1205831(119866

1015840


1(1198661015840

3) lt 1205831(1198661015840

1)


119909V (119909) minus (119905 + 2) 119903 (119909)

= 3 (119905 + 2) 1199092minus (31199052+ 22119905 + 33) 119909 + (119905 + 2) (119905 + 7) gt 0

(66)

From Lemma 8 1205831(V(119909)) lt 120583

1(1198661015840


1205831(1198661015840

3) le 1205831(nabla119898+2

119899) and 120583

1(1198661015840

1) le 1205831(Δ119898+1



119899) From


119899) lt 1205831(Δ119898+1

119899) as desired

Let ◻119896119889+3



119889+3) lt 1205831(Δlceil1198892rceil

119889+3)


119889+3) gt 5


Φ(◻119898+1

2119898+4) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]

times Φ (1198674) minus Φ (119875

119898)Φ (119871V119898+1 (1198674))

= 119909 [(119909 minus 2)2(119909 minus 4)Φ


times Φ (119863119898)Φ (119863

119898minus1) + (3119909 minus 8)Φ

2(119863119898minus1

)]

(67)

where1198674= ◻119898+1


119898

By (47) we get

Φ(Δ119898+1

2119898+4)


2minus 22119909 + 18)

times Φ (119863119898)Φ (119863

119898minus1) + (5119909 minus 6)Φ

2(119863119898minus1

)]

(68)


Φ(◻119898+1

2119898+4) minus Φ (Δ

119898+1

2119898+4)

= 119909 [(119909 minus 4)Φ2(119863119898) + (119909

2minus 2119909 minus 6)Φ (119863

119898)Φ (119863

119898minus1)

minus2 (119909 + 1)Φ2(119863119898minus1

)]

= 119909Φ (119863119898minus1

) [2 (1199092minus 4119909 + 1)Φ (119863


)]

+ 119909 (119909 minus 4)

gt 119909Φ2(119863119898minus1

) (41199092minus 19119909 + 6) + 119909 (119909 minus 4) gt 0

(69)

So 1205831(◻119898+1

2119898+4) lt 1205831(Δ119898+1



1(◻119898+1

2119898+5) lt 120583

1(Δ119898+1

2119898+5)

holds



1 1199092 119909

119899)119879



vertex V119894(1 le 119894 le 119899)



Claim 1 119881(119862119902)⋂119881(119875

119889+1) = 0


119894V119896V119896+1 V


119902and 119875119889+1

where V

119894isin 119881(119862

119902) V119897isin 119881(119875

119889+1)and V

119896 V

119897minus1isin 119881(119866)

(119881(119862119902)⋃119881(119875119889+1

))For each 119895 let 119879


119895as its root)


119895is



119894







119894) = 119889(V 119903

119894) minus 1 By


1(1198661) ge 1205831(119866)





3 where



V119896 V

119897minus1 So 120583

1(1198663) ge 1205831(1198662)







119889+1) = 0 and 120583

1(1198664) ge 1205831(1198663)


By Claim 1 119881(119862119902)⋂119881(119875


119862119902

= V119896V119896+1 V

119897minus1V119897V119889+2

V119889+3 V

119904V119896(119904 ge 119889 + 2)

where V119896 V119896+1 V

119897minus1 V119897 = 119881(119862

119902)⋂119881(119875

119889+1) and

V119889+2 V119889+3 V

119904 = 119881(119862

119902) 119881(119875

119889+1)

Claim 2 119889(V) = 1 for V isin 119881(119866) (119881(119862119902)⋃119881(119875119889+1))


3) we can get

119894and 120583

1(119894) ge


119894(V119894isin

119881(119862119902)⋃119881(119875

119889+1) V


Claim 3 119897 = 119896 + 1 and 119904 = 119889 + 2


(119881(119862119902)⋃119881(119875119889+1

) = V119904+1 V

119899 and |119860| = 119899 minus 119904 = 119905 ge 0

Case 1 119897 = 119896 By Lemma 7


= 119905 + 2 +

2 + 4 + 119905

119905 + 2

le 119905 + 5 = Δ (119896) + 1

lt 1205831 (119896) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

= 119905 + 4 +

2 + 2 + 2 + 2 + 119905

119905 + 4

le 119905 + 6 le 119899 minus 119889 + 2

= Δ (nabla119896

119899) + 1 lt 120583

1(nabla119896

119899)

(70)

for 119904 ge 119889 + 4 (|119862119902| ge 4)

When 119904 = 119889 + 3 (|119862119902| = 3)

119894cong 1198820(119901119894) By Lemma 23

1205831(119894) lt 1205831(nabla119896

119899) for any 119894 = 119896 where


119899




119899) holds

Case 2 119897 = 119896 + 1


1205831 (119894)


= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (119896) + 1 (Δ (

119896+1) + 1)

lt 1205831(119896) (1205831(119896+1)) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

le 119905 + 3 +

2 + 3 + 119905

119905 + 3

lt 119905 + 5 le 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(71)

for 119904 ge 119889 + 3 (|119862119902| ge 4)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) By Lemma 15

1205831(119894) lt 1205831(Δ119896

119899) for any 119894 = 119896 + 1 where

119896+1cong Δ119896

119899



= 3 +

2 + 3 + 2

3

lt 6 lt 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(72)

for 119904 ge 119889 + 4 (|119862119902| ge 5)

When 119904 = 119889 + 3 (|119862119902| = 4) by Lemma 13 120583

1(◻119896

119889+3) lt

1205831(◻lceil1198892rceil


lceil1198892rceil

119889+3) gt

1205831(◻lceil1198892rceil

119889+3)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) cong Δ119896

119899



119899) gt 1205831(Δlceil1198892rceil

119899) holds

Case 3 119897 ge 119896 + 2 By Lemma 7

1205831(119894)


= 119905 + 3 +

2 + 2 + 2 + 119905

119905 + 3

le 119905 + 5 le 119899 minus 119889 + 2 = Δ (119896

119899) + 1

lt 1205831 (119896

119899)

(73)

for 119904 ge 119889 + 3 (|119862119902| ge 5)

When 119904 = 119889 + 2 (|119862119902| = 4)

119894cong 1198800(119901119894) by Lemma 19

1205831(119894) lt 1205831(

119896


119896

119899





119899has the


Acknowledgments



References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119889

119899such

that 1205831(119866lowast) ge 1205831(119866)



119899 1205831(119866) le 1205831(1198820(119901119896)) where 1 le



1205831 (119866)


119889

119899 119894 = 119896

= 119905 + 2 +

2 + 4 + 119905

119905 + 2

lt 119905 + 5 = Δ (1198820(119901119896)) + 1

lt 1205831 (1198820 (119901119896))

(44)




119899) lt 1205831(Δ

lceil1198892rceil

119899) lt


119899)


119899) lt 1205831(lfloor1198892rfloor

119899)



119899) 1205831(lfloor1198892rfloor


following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)Let Δ119898+1

119899and 119898


119905 = 0 denote Δ119898+1119899



Φ(1198661) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198671)

minus Φ (119875119898)Φ (119871V119898+2 (1198671))

Φ (1198662) = [Φ (119875

119898) minus Φ (119871V119898+2 (119875119898))]Φ (1198672)

minus Φ (119875119898)Φ (119871V119898+2 (1198672))

(45)

where

Φ(1198671) = 119909 (119909 minus 3) [(119909 minus 3)Φ (119863119898

) minus 2Φ (119863119898minus1

)]

Φ (119871V119898+2 (1198671))


)

Φ (1198672)

= 119909 (119909 minus 2)


)]

Φ (119871V119898+2 (1198672))


(46)

Note thatΦ(119875119898) = 119909Φ(119863


Φ(119863119898minus2


Φ(Δ119898+1

119899)


) minus 119905119909Φ (119871V119898+2 (1198661))]


2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (6119905 + 12)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 3) 119909 minus 3 (119905 + 1)]Φ2(119863119898minus1

)

(47)

Φ(119898

119899)


) minus 119905119909Φ (119871V119898+2 (1198662))]



2+ (4119905 + 18) 119909 minus (2119905 + 10)]

times Φ2(119863119898minus1)

minus [31199093minus (3119905 + 19) 119909

2+ 8 (119905 + 4) 119909 minus 4 (119905 + 4)]


)Φ (119863119898minus2



)

(48)


119899) le 119909 lt 119905+5

Φ(Δ119898+1

119899) minus Φ (

119898

119899)

= 119909(119909 minus 1)119905[119889 (119909)Φ

2(119863119898minus1

) + 119890 (119909)Φ (119863119898minus1)Φ (119863

119898minus2)

minus119903 (119909)Φ2(119863119898minus2

)]

gt 119909(119909 minus 1)119905[119889 (119909) + 119890 (119909) minus 119903 (119909)]Φ

2(119863119898minus2

) ge 0

(49)


where





(50)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1


Case 2 119889 = 2119898 + 2 (119898 ge 1)LetΔ119898+1119899

and119898+1119899


119899and119898+1


2 respectively By


Φ(Δ119898+1

119899)


lowast

1) minus 119905119909Φ (119871V119898+2 (119866

lowast

1))]



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 4) 119909 + 3 (119905 + 2)]Φ

2(119863119898minus1)

(51)

Φ(119898+1

119899)


lowast

2) minus 119905119909Φ (119871V119898+3 (119866

lowast

2))]


times [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]Φ

2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863119898minus1)

+2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

(52)


Φ(Δ119898+1

119899) minus Φ (

119898+1

119899)

= 119909(119909 minus 1)119905


2(119863119898)

+ [(119905 + 1) 1199093minus (5119905 + 5) 119909

2+ (7119905 + 10) 119909 minus (2119905 + 6)]

times Φ (119863119898)Φ (119863

119898minus1)

minus [(2119905 + 2) 1199092minus (4119905 + 4) 119909 + (119905 + 2)]Φ

2(119863119898minus1

)



2m+3

m+2

m+3m+1m 2m+2

m+1

n

(a)



m+2

m+3m+1m

2m+4

2m+3

m+1

n

(b)



2m+3

m+2

m+3m+1m 2m+2

⋄m

n

(c)




m+3

m+1 m+2m

2m+4

2m+3

⋄m+1

n

(d)

1


2m+3

m+2

m+1m

2m+4

2m+2


m+1

n

(e)

1


2m+4

m+3m+2m+1

2m+5

2m+3


m+2

n

(f)

Figure 3

= 119909(119909 minus 1)119905


)]Φ (119863119898minus1

) minus 119908 (119909)

ge 119909(119909 minus 1)119905Φ(119863119898minus1

)


)] gt 0

(53)

where

119901 (119909) = 21199093minus (119905 + 9) 119909

2+ (119905 + 9) 119909 minus 2 gt 0

119902 (119909) = 2(119909 minus 1)2gt 0

119908 (119909) = (119905 + 1) 1199092minus (3119905 + 5) 119909 + (119905 + 4) gt 0

119901 (119909) minus 119908 (119909) minus 119902 (119909)

= 21199093minus (2119905 + 12) 119909

2+ (4119905 + 18) 119909 minus (119905 + 8) gt 0

(54)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1





lceil1198892rceil

119899)


119899) 1205831(Δlceil1198892rceil


following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)


Letnabla119898+1119899


119899by 1198663 Let119867


119898 By Lemma 9

Φ(1198663) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]Φ (1198673)

minus Φ (119875119898)Φ (119871V119898+1 (1198673))

(55)

where

Φ(1198673) = 119909 (119909 minus 3)


)]

Φ (119871V119898+1 (1198673)) = (119909 minus 1) (119909 minus 3)


119898minus1)]

(56)

By Lemma 10 we get

Φ(nabla119898+1

119899)


) minus 119905119909Φ (119871V119898+1 (1198663))]

= 119909 (119909 minus 3) (119909 minus 1)119905


times Φ (119863119898)Φ (119863

119898minus1) + (119905 + 2)Φ

2(119863119898minus1

)

(57)

By (47) we have

Φ(Δ119898+1

119899) = 119909(119909 minus 1)

119905


minus [(119905 + 5) 1199092minus (6119905 + 22) 119909 + (6119905 + 18)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 5) 119909 minus 3 (119905 + 2)]Φ2(119863119898minus1

)

(58)

Hence

Φ(nabla119898+1

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(59)


1(nabla119898+1

119899) lt 1205831(Δ119898+1


When 119898 ge 2 1205831(nabla119898+1

119899) lt 1205831(Δ119898+1



Case 2 119889 = 2119898 + 2 (119898 ge 1)

Let nabla119898+2119899


Φ(nabla119898+2

119899) = 119909 (119909 minus 3) (119909 minus 1)

119905


2(119863119898)


119898minus1)


)

(60)

By (51) we get

Φ(Δ119898+1

119899)



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 5) 119909 + (3119905 + 9)]Φ

2(119863119898minus1

)

(61)

Hence

Φ(nabla119898+2

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(62)

Let 119905119896(119909) = (119909 minus 119905 minus 4)Φ(119863

119896) minus (119905 + 3)Φ(119863



1(119905119896(119909)) where 119896 ge 0




1(119905119896minus1(119909)) lt


119905119896+1 (


119909) (63)

1199052

119896(119909) minus 119905119896+1 (

119909) 119905119896minus1 (119909)


(64)




Let 11986610158403= nabla119898+2




and1198661015840

1= Δ119898+1





By (60) and (62)

Φ(1198661015840

3) = 119909(119909 minus 1)

119905(119909 minus 3) (119909

2minus 3119909 + 1) 119903 (119909)

Φ (1198661015840

3) minus Φ (119866

1015840

1)

= 1199092(119909 minus 1)

119905[1199093minus (119905 + 8) 119909

2+ (3119905 + 16) 119909 minus (119905 + 6)]

= 1199092(119909 minus 1)

119905(119903 (119909) + 1)

(65)


1(1198661015840


1205831(1198661015840

3) ge 1205831(119866

1015840

1) then [Φ(1198661015840

3)minusΦ(119866

1015840

1)]|119909=1205831(119866

1015840

3) le 0 and 119903(119909)+

1|119909=1205831(119866

1015840


1(1198661015840

3) lt 1205831(1198661015840

1)


119909V (119909) minus (119905 + 2) 119903 (119909)

= 3 (119905 + 2) 1199092minus (31199052+ 22119905 + 33) 119909 + (119905 + 2) (119905 + 7) gt 0

(66)

From Lemma 8 1205831(V(119909)) lt 120583

1(1198661015840


1205831(1198661015840

3) le 1205831(nabla119898+2

119899) and 120583

1(1198661015840

1) le 1205831(Δ119898+1



119899) From


119899) lt 1205831(Δ119898+1

119899) as desired

Let ◻119896119889+3



119889+3) lt 1205831(Δlceil1198892rceil

119889+3)


119889+3) gt 5


Φ(◻119898+1

2119898+4) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]

times Φ (1198674) minus Φ (119875

119898)Φ (119871V119898+1 (1198674))

= 119909 [(119909 minus 2)2(119909 minus 4)Φ


times Φ (119863119898)Φ (119863

119898minus1) + (3119909 minus 8)Φ

2(119863119898minus1

)]

(67)

where1198674= ◻119898+1


119898

By (47) we get

Φ(Δ119898+1

2119898+4)


2minus 22119909 + 18)

times Φ (119863119898)Φ (119863

119898minus1) + (5119909 minus 6)Φ

2(119863119898minus1

)]

(68)


Φ(◻119898+1

2119898+4) minus Φ (Δ

119898+1

2119898+4)

= 119909 [(119909 minus 4)Φ2(119863119898) + (119909

2minus 2119909 minus 6)Φ (119863

119898)Φ (119863

119898minus1)

minus2 (119909 + 1)Φ2(119863119898minus1

)]

= 119909Φ (119863119898minus1

) [2 (1199092minus 4119909 + 1)Φ (119863


)]

+ 119909 (119909 minus 4)

gt 119909Φ2(119863119898minus1

) (41199092minus 19119909 + 6) + 119909 (119909 minus 4) gt 0

(69)

So 1205831(◻119898+1

2119898+4) lt 1205831(Δ119898+1



1(◻119898+1

2119898+5) lt 120583

1(Δ119898+1

2119898+5)

holds



1 1199092 119909

119899)119879



vertex V119894(1 le 119894 le 119899)



Claim 1 119881(119862119902)⋂119881(119875

119889+1) = 0


119894V119896V119896+1 V


119902and 119875119889+1

where V

119894isin 119881(119862

119902) V119897isin 119881(119875

119889+1)and V

119896 V

119897minus1isin 119881(119866)

(119881(119862119902)⋃119881(119875119889+1

))For each 119895 let 119879


119895as its root)


119895is



119894







119894) = 119889(V 119903

119894) minus 1 By


1(1198661) ge 1205831(119866)





3 where



V119896 V

119897minus1 So 120583

1(1198663) ge 1205831(1198662)







119889+1) = 0 and 120583

1(1198664) ge 1205831(1198663)


By Claim 1 119881(119862119902)⋂119881(119875


119862119902

= V119896V119896+1 V

119897minus1V119897V119889+2

V119889+3 V

119904V119896(119904 ge 119889 + 2)

where V119896 V119896+1 V

119897minus1 V119897 = 119881(119862

119902)⋂119881(119875

119889+1) and

V119889+2 V119889+3 V

119904 = 119881(119862

119902) 119881(119875

119889+1)

Claim 2 119889(V) = 1 for V isin 119881(119866) (119881(119862119902)⋃119881(119875119889+1))


3) we can get

119894and 120583

1(119894) ge


119894(V119894isin

119881(119862119902)⋃119881(119875

119889+1) V


Claim 3 119897 = 119896 + 1 and 119904 = 119889 + 2


(119881(119862119902)⋃119881(119875119889+1

) = V119904+1 V

119899 and |119860| = 119899 minus 119904 = 119905 ge 0

Case 1 119897 = 119896 By Lemma 7


= 119905 + 2 +

2 + 4 + 119905

119905 + 2

le 119905 + 5 = Δ (119896) + 1

lt 1205831 (119896) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

= 119905 + 4 +

2 + 2 + 2 + 2 + 119905

119905 + 4

le 119905 + 6 le 119899 minus 119889 + 2

= Δ (nabla119896

119899) + 1 lt 120583

1(nabla119896

119899)

(70)

for 119904 ge 119889 + 4 (|119862119902| ge 4)

When 119904 = 119889 + 3 (|119862119902| = 3)

119894cong 1198820(119901119894) By Lemma 23

1205831(119894) lt 1205831(nabla119896

119899) for any 119894 = 119896 where


119899




119899) holds

Case 2 119897 = 119896 + 1


1205831 (119894)


= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (119896) + 1 (Δ (

119896+1) + 1)

lt 1205831(119896) (1205831(119896+1)) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

le 119905 + 3 +

2 + 3 + 119905

119905 + 3

lt 119905 + 5 le 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(71)

for 119904 ge 119889 + 3 (|119862119902| ge 4)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) By Lemma 15

1205831(119894) lt 1205831(Δ119896

119899) for any 119894 = 119896 + 1 where

119896+1cong Δ119896

119899



= 3 +

2 + 3 + 2

3

lt 6 lt 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(72)

for 119904 ge 119889 + 4 (|119862119902| ge 5)

When 119904 = 119889 + 3 (|119862119902| = 4) by Lemma 13 120583

1(◻119896

119889+3) lt

1205831(◻lceil1198892rceil


lceil1198892rceil

119889+3) gt

1205831(◻lceil1198892rceil

119889+3)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) cong Δ119896

119899



119899) gt 1205831(Δlceil1198892rceil

119899) holds

Case 3 119897 ge 119896 + 2 By Lemma 7

1205831(119894)


= 119905 + 3 +

2 + 2 + 2 + 119905

119905 + 3

le 119905 + 5 le 119899 minus 119889 + 2 = Δ (119896

119899) + 1

lt 1205831 (119896

119899)

(73)

for 119904 ge 119889 + 3 (|119862119902| ge 5)

When 119904 = 119889 + 2 (|119862119902| = 4)

119894cong 1198800(119901119894) by Lemma 19

1205831(119894) lt 1205831(

119896


119896

119899





119899has the


Acknowledgments



References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










where





(50)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1


Case 2 119889 = 2119898 + 2 (119898 ge 1)LetΔ119898+1119899

and119898+1119899


119899and119898+1


2 respectively By


Φ(Δ119898+1

119899)


lowast

1) minus 119905119909Φ (119871V119898+2 (119866

lowast

1))]



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 4) 119909 + 3 (119905 + 2)]Φ

2(119863119898minus1)

(51)

Φ(119898+1

119899)


lowast

2) minus 119905119909Φ (119871V119898+3 (119866

lowast

2))]


times [1199093minus (119905 + 7) 119909

2+ (4119905 + 14) 119909 minus (2119905 + 8)]Φ

2(119863119898)

minus [(119905 + 4) 1199092minus (6119905 + 16) 119909 + (4119905 + 12)]

times Φ (119863119898)Φ (119863119898minus1)

+2 (119905 + 2) (119909 minus 1)Φ2(119863119898minus1

)

(52)


Φ(Δ119898+1

119899) minus Φ (

119898+1

119899)

= 119909(119909 minus 1)119905


2(119863119898)

+ [(119905 + 1) 1199093minus (5119905 + 5) 119909

2+ (7119905 + 10) 119909 minus (2119905 + 6)]

times Φ (119863119898)Φ (119863

119898minus1)

minus [(2119905 + 2) 1199092minus (4119905 + 4) 119909 + (119905 + 2)]Φ

2(119863119898minus1

)



2m+3

m+2

m+3m+1m 2m+2

m+1

n

(a)



m+2

m+3m+1m

2m+4

2m+3

m+1

n

(b)



2m+3

m+2

m+3m+1m 2m+2

⋄m

n

(c)




m+3

m+1 m+2m

2m+4

2m+3

⋄m+1

n

(d)

1


2m+3

m+2

m+1m

2m+4

2m+2


m+1

n

(e)

1


2m+4

m+3m+2m+1

2m+5

2m+3


m+2

n

(f)

Figure 3

= 119909(119909 minus 1)119905


)]Φ (119863119898minus1

) minus 119908 (119909)

ge 119909(119909 minus 1)119905Φ(119863119898minus1

)


)] gt 0

(53)

where

119901 (119909) = 21199093minus (119905 + 9) 119909

2+ (119905 + 9) 119909 minus 2 gt 0

119902 (119909) = 2(119909 minus 1)2gt 0

119908 (119909) = (119905 + 1) 1199092minus (3119905 + 5) 119909 + (119905 + 4) gt 0

119901 (119909) minus 119908 (119909) minus 119902 (119909)

= 21199093minus (2119905 + 12) 119909

2+ (4119905 + 18) 119909 minus (119905 + 8) gt 0

(54)

So 1205831(Δ119898+1

119899) lt 1205831(119898+1





lceil1198892rceil

119899)


119899) 1205831(Δlceil1198892rceil


following two cases

Case 1 119889 = 2119898 + 1 (119898 ge 1)


Letnabla119898+1119899


119899by 1198663 Let119867


119898 By Lemma 9

Φ(1198663) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]Φ (1198673)

minus Φ (119875119898)Φ (119871V119898+1 (1198673))

(55)

where

Φ(1198673) = 119909 (119909 minus 3)


)]

Φ (119871V119898+1 (1198673)) = (119909 minus 1) (119909 minus 3)


119898minus1)]

(56)

By Lemma 10 we get

Φ(nabla119898+1

119899)


) minus 119905119909Φ (119871V119898+1 (1198663))]

= 119909 (119909 minus 3) (119909 minus 1)119905


times Φ (119863119898)Φ (119863

119898minus1) + (119905 + 2)Φ

2(119863119898minus1

)

(57)

By (47) we have

Φ(Δ119898+1

119899) = 119909(119909 minus 1)

119905


minus [(119905 + 5) 1199092minus (6119905 + 22) 119909 + (6119905 + 18)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 5) 119909 minus 3 (119905 + 2)]Φ2(119863119898minus1

)

(58)

Hence

Φ(nabla119898+1

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(59)


1(nabla119898+1

119899) lt 1205831(Δ119898+1


When 119898 ge 2 1205831(nabla119898+1

119899) lt 1205831(Δ119898+1



Case 2 119889 = 2119898 + 2 (119898 ge 1)

Let nabla119898+2119899


Φ(nabla119898+2

119899) = 119909 (119909 minus 3) (119909 minus 1)

119905


2(119863119898)


119898minus1)


)

(60)

By (51) we get

Φ(Δ119898+1

119899)



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 5) 119909 + (3119905 + 9)]Φ

2(119863119898minus1

)

(61)

Hence

Φ(nabla119898+2

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(62)

Let 119905119896(119909) = (119909 minus 119905 minus 4)Φ(119863

119896) minus (119905 + 3)Φ(119863



1(119905119896(119909)) where 119896 ge 0




1(119905119896minus1(119909)) lt


119905119896+1 (


119909) (63)

1199052

119896(119909) minus 119905119896+1 (

119909) 119905119896minus1 (119909)


(64)




Let 11986610158403= nabla119898+2




and1198661015840

1= Δ119898+1





By (60) and (62)

Φ(1198661015840

3) = 119909(119909 minus 1)

119905(119909 minus 3) (119909

2minus 3119909 + 1) 119903 (119909)

Φ (1198661015840

3) minus Φ (119866

1015840

1)

= 1199092(119909 minus 1)

119905[1199093minus (119905 + 8) 119909

2+ (3119905 + 16) 119909 minus (119905 + 6)]

= 1199092(119909 minus 1)

119905(119903 (119909) + 1)

(65)


1(1198661015840


1205831(1198661015840

3) ge 1205831(119866

1015840

1) then [Φ(1198661015840

3)minusΦ(119866

1015840

1)]|119909=1205831(119866

1015840

3) le 0 and 119903(119909)+

1|119909=1205831(119866

1015840


1(1198661015840

3) lt 1205831(1198661015840

1)


119909V (119909) minus (119905 + 2) 119903 (119909)

= 3 (119905 + 2) 1199092minus (31199052+ 22119905 + 33) 119909 + (119905 + 2) (119905 + 7) gt 0

(66)

From Lemma 8 1205831(V(119909)) lt 120583

1(1198661015840


1205831(1198661015840

3) le 1205831(nabla119898+2

119899) and 120583

1(1198661015840

1) le 1205831(Δ119898+1



119899) From


119899) lt 1205831(Δ119898+1

119899) as desired

Let ◻119896119889+3



119889+3) lt 1205831(Δlceil1198892rceil

119889+3)


119889+3) gt 5


Φ(◻119898+1

2119898+4) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]

times Φ (1198674) minus Φ (119875

119898)Φ (119871V119898+1 (1198674))

= 119909 [(119909 minus 2)2(119909 minus 4)Φ


times Φ (119863119898)Φ (119863

119898minus1) + (3119909 minus 8)Φ

2(119863119898minus1

)]

(67)

where1198674= ◻119898+1


119898

By (47) we get

Φ(Δ119898+1

2119898+4)


2minus 22119909 + 18)

times Φ (119863119898)Φ (119863

119898minus1) + (5119909 minus 6)Φ

2(119863119898minus1

)]

(68)


Φ(◻119898+1

2119898+4) minus Φ (Δ

119898+1

2119898+4)

= 119909 [(119909 minus 4)Φ2(119863119898) + (119909

2minus 2119909 minus 6)Φ (119863

119898)Φ (119863

119898minus1)

minus2 (119909 + 1)Φ2(119863119898minus1

)]

= 119909Φ (119863119898minus1

) [2 (1199092minus 4119909 + 1)Φ (119863


)]

+ 119909 (119909 minus 4)

gt 119909Φ2(119863119898minus1

) (41199092minus 19119909 + 6) + 119909 (119909 minus 4) gt 0

(69)

So 1205831(◻119898+1

2119898+4) lt 1205831(Δ119898+1



1(◻119898+1

2119898+5) lt 120583

1(Δ119898+1

2119898+5)

holds



1 1199092 119909

119899)119879



vertex V119894(1 le 119894 le 119899)



Claim 1 119881(119862119902)⋂119881(119875

119889+1) = 0


119894V119896V119896+1 V


119902and 119875119889+1

where V

119894isin 119881(119862

119902) V119897isin 119881(119875

119889+1)and V

119896 V

119897minus1isin 119881(119866)

(119881(119862119902)⋃119881(119875119889+1

))For each 119895 let 119879


119895as its root)


119895is



119894







119894) = 119889(V 119903

119894) minus 1 By


1(1198661) ge 1205831(119866)





3 where



V119896 V

119897minus1 So 120583

1(1198663) ge 1205831(1198662)







119889+1) = 0 and 120583

1(1198664) ge 1205831(1198663)


By Claim 1 119881(119862119902)⋂119881(119875


119862119902

= V119896V119896+1 V

119897minus1V119897V119889+2

V119889+3 V

119904V119896(119904 ge 119889 + 2)

where V119896 V119896+1 V

119897minus1 V119897 = 119881(119862

119902)⋂119881(119875

119889+1) and

V119889+2 V119889+3 V

119904 = 119881(119862

119902) 119881(119875

119889+1)

Claim 2 119889(V) = 1 for V isin 119881(119866) (119881(119862119902)⋃119881(119875119889+1))


3) we can get

119894and 120583

1(119894) ge


119894(V119894isin

119881(119862119902)⋃119881(119875

119889+1) V


Claim 3 119897 = 119896 + 1 and 119904 = 119889 + 2


(119881(119862119902)⋃119881(119875119889+1

) = V119904+1 V

119899 and |119860| = 119899 minus 119904 = 119905 ge 0

Case 1 119897 = 119896 By Lemma 7


= 119905 + 2 +

2 + 4 + 119905

119905 + 2

le 119905 + 5 = Δ (119896) + 1

lt 1205831 (119896) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

= 119905 + 4 +

2 + 2 + 2 + 2 + 119905

119905 + 4

le 119905 + 6 le 119899 minus 119889 + 2

= Δ (nabla119896

119899) + 1 lt 120583

1(nabla119896

119899)

(70)

for 119904 ge 119889 + 4 (|119862119902| ge 4)

When 119904 = 119889 + 3 (|119862119902| = 3)

119894cong 1198820(119901119894) By Lemma 23

1205831(119894) lt 1205831(nabla119896

119899) for any 119894 = 119896 where


119899




119899) holds

Case 2 119897 = 119896 + 1


1205831 (119894)


= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (119896) + 1 (Δ (

119896+1) + 1)

lt 1205831(119896) (1205831(119896+1)) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

le 119905 + 3 +

2 + 3 + 119905

119905 + 3

lt 119905 + 5 le 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(71)

for 119904 ge 119889 + 3 (|119862119902| ge 4)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) By Lemma 15

1205831(119894) lt 1205831(Δ119896

119899) for any 119894 = 119896 + 1 where

119896+1cong Δ119896

119899



= 3 +

2 + 3 + 2

3

lt 6 lt 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(72)

for 119904 ge 119889 + 4 (|119862119902| ge 5)

When 119904 = 119889 + 3 (|119862119902| = 4) by Lemma 13 120583

1(◻119896

119889+3) lt

1205831(◻lceil1198892rceil


lceil1198892rceil

119889+3) gt

1205831(◻lceil1198892rceil

119889+3)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) cong Δ119896

119899



119899) gt 1205831(Δlceil1198892rceil

119899) holds

Case 3 119897 ge 119896 + 2 By Lemma 7

1205831(119894)


= 119905 + 3 +

2 + 2 + 2 + 119905

119905 + 3

le 119905 + 5 le 119899 minus 119889 + 2 = Δ (119896

119899) + 1

lt 1205831 (119896

119899)

(73)

for 119904 ge 119889 + 3 (|119862119902| ge 5)

When 119904 = 119889 + 2 (|119862119902| = 4)

119894cong 1198800(119901119894) by Lemma 19

1205831(119894) lt 1205831(

119896


119896

119899





119899has the


Acknowledgments



References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Letnabla119898+1119899


119899by 1198663 Let119867


119898 By Lemma 9

Φ(1198663) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]Φ (1198673)

minus Φ (119875119898)Φ (119871V119898+1 (1198673))

(55)

where

Φ(1198673) = 119909 (119909 minus 3)


)]

Φ (119871V119898+1 (1198673)) = (119909 minus 1) (119909 minus 3)


119898minus1)]

(56)

By Lemma 10 we get

Φ(nabla119898+1

119899)


) minus 119905119909Φ (119871V119898+1 (1198663))]

= 119909 (119909 minus 3) (119909 minus 1)119905


times Φ (119863119898)Φ (119863

119898minus1) + (119905 + 2)Φ

2(119863119898minus1

)

(57)

By (47) we have

Φ(Δ119898+1

119899) = 119909(119909 minus 1)

119905


minus [(119905 + 5) 1199092minus (6119905 + 22) 119909 + (6119905 + 18)]

times Φ (119863119898)Φ (119863

119898minus1)

+ [(2119905 + 5) 119909 minus 3 (119905 + 2)]Φ2(119863119898minus1

)

(58)

Hence

Φ(nabla119898+1

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(59)


1(nabla119898+1

119899) lt 1205831(Δ119898+1


When 119898 ge 2 1205831(nabla119898+1

119899) lt 1205831(Δ119898+1



Case 2 119889 = 2119898 + 2 (119898 ge 1)

Let nabla119898+2119899


Φ(nabla119898+2

119899) = 119909 (119909 minus 3) (119909 minus 1)

119905


2(119863119898)


119898minus1)


)

(60)

By (51) we get

Φ(Δ119898+1

119899)



2(119863119898)


times Φ (119863119898)Φ (119863

119898minus1)

+ [21199092minus 2 (119905 + 5) 119909 + (3119905 + 9)]Φ

2(119863119898minus1

)

(61)

Hence

Φ(nabla119898+2

119899) minus Φ (Δ

119898+1

119899)

= 1199092(119909 minus 1)

119905Φ(119863119898minus1)


)]

(62)

Let 119905119896(119909) = (119909 minus 119905 minus 4)Φ(119863

119896) minus (119905 + 3)Φ(119863



1(119905119896(119909)) where 119896 ge 0




1(119905119896minus1(119909)) lt


119905119896+1 (


119909) (63)

1199052

119896(119909) minus 119905119896+1 (

119909) 119905119896minus1 (119909)


(64)




Let 11986610158403= nabla119898+2




and1198661015840

1= Δ119898+1





By (60) and (62)

Φ(1198661015840

3) = 119909(119909 minus 1)

119905(119909 minus 3) (119909

2minus 3119909 + 1) 119903 (119909)

Φ (1198661015840

3) minus Φ (119866

1015840

1)

= 1199092(119909 minus 1)

119905[1199093minus (119905 + 8) 119909

2+ (3119905 + 16) 119909 minus (119905 + 6)]

= 1199092(119909 minus 1)

119905(119903 (119909) + 1)

(65)


1(1198661015840


1205831(1198661015840

3) ge 1205831(119866

1015840

1) then [Φ(1198661015840

3)minusΦ(119866

1015840

1)]|119909=1205831(119866

1015840

3) le 0 and 119903(119909)+

1|119909=1205831(119866

1015840


1(1198661015840

3) lt 1205831(1198661015840

1)


119909V (119909) minus (119905 + 2) 119903 (119909)

= 3 (119905 + 2) 1199092minus (31199052+ 22119905 + 33) 119909 + (119905 + 2) (119905 + 7) gt 0

(66)

From Lemma 8 1205831(V(119909)) lt 120583

1(1198661015840


1205831(1198661015840

3) le 1205831(nabla119898+2

119899) and 120583

1(1198661015840

1) le 1205831(Δ119898+1



119899) From


119899) lt 1205831(Δ119898+1

119899) as desired

Let ◻119896119889+3



119889+3) lt 1205831(Δlceil1198892rceil

119889+3)


119889+3) gt 5


Φ(◻119898+1

2119898+4) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]

times Φ (1198674) minus Φ (119875

119898)Φ (119871V119898+1 (1198674))

= 119909 [(119909 minus 2)2(119909 minus 4)Φ


times Φ (119863119898)Φ (119863

119898minus1) + (3119909 minus 8)Φ

2(119863119898minus1

)]

(67)

where1198674= ◻119898+1


119898

By (47) we get

Φ(Δ119898+1

2119898+4)


2minus 22119909 + 18)

times Φ (119863119898)Φ (119863

119898minus1) + (5119909 minus 6)Φ

2(119863119898minus1

)]

(68)


Φ(◻119898+1

2119898+4) minus Φ (Δ

119898+1

2119898+4)

= 119909 [(119909 minus 4)Φ2(119863119898) + (119909

2minus 2119909 minus 6)Φ (119863

119898)Φ (119863

119898minus1)

minus2 (119909 + 1)Φ2(119863119898minus1

)]

= 119909Φ (119863119898minus1

) [2 (1199092minus 4119909 + 1)Φ (119863


)]

+ 119909 (119909 minus 4)

gt 119909Φ2(119863119898minus1

) (41199092minus 19119909 + 6) + 119909 (119909 minus 4) gt 0

(69)

So 1205831(◻119898+1

2119898+4) lt 1205831(Δ119898+1



1(◻119898+1

2119898+5) lt 120583

1(Δ119898+1

2119898+5)

holds



1 1199092 119909

119899)119879



vertex V119894(1 le 119894 le 119899)



Claim 1 119881(119862119902)⋂119881(119875

119889+1) = 0


119894V119896V119896+1 V


119902and 119875119889+1

where V

119894isin 119881(119862

119902) V119897isin 119881(119875

119889+1)and V

119896 V

119897minus1isin 119881(119866)

(119881(119862119902)⋃119881(119875119889+1

))For each 119895 let 119879


119895as its root)


119895is



119894







119894) = 119889(V 119903

119894) minus 1 By


1(1198661) ge 1205831(119866)





3 where



V119896 V

119897minus1 So 120583

1(1198663) ge 1205831(1198662)







119889+1) = 0 and 120583

1(1198664) ge 1205831(1198663)


By Claim 1 119881(119862119902)⋂119881(119875


119862119902

= V119896V119896+1 V

119897minus1V119897V119889+2

V119889+3 V

119904V119896(119904 ge 119889 + 2)

where V119896 V119896+1 V

119897minus1 V119897 = 119881(119862

119902)⋂119881(119875

119889+1) and

V119889+2 V119889+3 V

119904 = 119881(119862

119902) 119881(119875

119889+1)

Claim 2 119889(V) = 1 for V isin 119881(119866) (119881(119862119902)⋃119881(119875119889+1))


3) we can get

119894and 120583

1(119894) ge


119894(V119894isin

119881(119862119902)⋃119881(119875

119889+1) V


Claim 3 119897 = 119896 + 1 and 119904 = 119889 + 2


(119881(119862119902)⋃119881(119875119889+1

) = V119904+1 V

119899 and |119860| = 119899 minus 119904 = 119905 ge 0

Case 1 119897 = 119896 By Lemma 7


= 119905 + 2 +

2 + 4 + 119905

119905 + 2

le 119905 + 5 = Δ (119896) + 1

lt 1205831 (119896) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

= 119905 + 4 +

2 + 2 + 2 + 2 + 119905

119905 + 4

le 119905 + 6 le 119899 minus 119889 + 2

= Δ (nabla119896

119899) + 1 lt 120583

1(nabla119896

119899)

(70)

for 119904 ge 119889 + 4 (|119862119902| ge 4)

When 119904 = 119889 + 3 (|119862119902| = 3)

119894cong 1198820(119901119894) By Lemma 23

1205831(119894) lt 1205831(nabla119896

119899) for any 119894 = 119896 where


119899




119899) holds

Case 2 119897 = 119896 + 1


1205831 (119894)


= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (119896) + 1 (Δ (

119896+1) + 1)

lt 1205831(119896) (1205831(119896+1)) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

le 119905 + 3 +

2 + 3 + 119905

119905 + 3

lt 119905 + 5 le 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(71)

for 119904 ge 119889 + 3 (|119862119902| ge 4)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) By Lemma 15

1205831(119894) lt 1205831(Δ119896

119899) for any 119894 = 119896 + 1 where

119896+1cong Δ119896

119899



= 3 +

2 + 3 + 2

3

lt 6 lt 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(72)

for 119904 ge 119889 + 4 (|119862119902| ge 5)

When 119904 = 119889 + 3 (|119862119902| = 4) by Lemma 13 120583

1(◻119896

119889+3) lt

1205831(◻lceil1198892rceil


lceil1198892rceil

119889+3) gt

1205831(◻lceil1198892rceil

119889+3)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) cong Δ119896

119899



119899) gt 1205831(Δlceil1198892rceil

119899) holds

Case 3 119897 ge 119896 + 2 By Lemma 7

1205831(119894)


= 119905 + 3 +

2 + 2 + 2 + 119905

119905 + 3

le 119905 + 5 le 119899 minus 119889 + 2 = Δ (119896

119899) + 1

lt 1205831 (119896

119899)

(73)

for 119904 ge 119889 + 3 (|119862119902| ge 5)

When 119904 = 119889 + 2 (|119862119902| = 4)

119894cong 1198800(119901119894) by Lemma 19

1205831(119894) lt 1205831(

119896


119896

119899





119899has the


Acknowledgments



References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










By (60) and (62)

Φ(1198661015840

3) = 119909(119909 minus 1)

119905(119909 minus 3) (119909

2minus 3119909 + 1) 119903 (119909)

Φ (1198661015840

3) minus Φ (119866

1015840

1)

= 1199092(119909 minus 1)

119905[1199093minus (119905 + 8) 119909

2+ (3119905 + 16) 119909 minus (119905 + 6)]

= 1199092(119909 minus 1)

119905(119903 (119909) + 1)

(65)


1(1198661015840


1205831(1198661015840

3) ge 1205831(119866

1015840

1) then [Φ(1198661015840

3)minusΦ(119866

1015840

1)]|119909=1205831(119866

1015840

3) le 0 and 119903(119909)+

1|119909=1205831(119866

1015840


1(1198661015840

3) lt 1205831(1198661015840

1)


119909V (119909) minus (119905 + 2) 119903 (119909)

= 3 (119905 + 2) 1199092minus (31199052+ 22119905 + 33) 119909 + (119905 + 2) (119905 + 7) gt 0

(66)

From Lemma 8 1205831(V(119909)) lt 120583

1(1198661015840


1205831(1198661015840

3) le 1205831(nabla119898+2

119899) and 120583

1(1198661015840

1) le 1205831(Δ119898+1



119899) From


119899) lt 1205831(Δ119898+1

119899) as desired

Let ◻119896119889+3



119889+3) lt 1205831(Δlceil1198892rceil

119889+3)


119889+3) gt 5


Φ(◻119898+1

2119898+4) = [Φ (119875119898) minus Φ (119871V119898 (119875119898))]

times Φ (1198674) minus Φ (119875

119898)Φ (119871V119898+1 (1198674))

= 119909 [(119909 minus 2)2(119909 minus 4)Φ


times Φ (119863119898)Φ (119863

119898minus1) + (3119909 minus 8)Φ

2(119863119898minus1

)]

(67)

where1198674= ◻119898+1


119898

By (47) we get

Φ(Δ119898+1

2119898+4)


2minus 22119909 + 18)

times Φ (119863119898)Φ (119863

119898minus1) + (5119909 minus 6)Φ

2(119863119898minus1

)]

(68)


Φ(◻119898+1

2119898+4) minus Φ (Δ

119898+1

2119898+4)

= 119909 [(119909 minus 4)Φ2(119863119898) + (119909

2minus 2119909 minus 6)Φ (119863

119898)Φ (119863

119898minus1)

minus2 (119909 + 1)Φ2(119863119898minus1

)]

= 119909Φ (119863119898minus1

) [2 (1199092minus 4119909 + 1)Φ (119863


)]

+ 119909 (119909 minus 4)

gt 119909Φ2(119863119898minus1

) (41199092minus 19119909 + 6) + 119909 (119909 minus 4) gt 0

(69)

So 1205831(◻119898+1

2119898+4) lt 1205831(Δ119898+1



1(◻119898+1

2119898+5) lt 120583

1(Δ119898+1

2119898+5)

holds



1 1199092 119909

119899)119879



vertex V119894(1 le 119894 le 119899)



Claim 1 119881(119862119902)⋂119881(119875

119889+1) = 0


119894V119896V119896+1 V


119902and 119875119889+1

where V

119894isin 119881(119862

119902) V119897isin 119881(119875

119889+1)and V

119896 V

119897minus1isin 119881(119866)

(119881(119862119902)⋃119881(119875119889+1

))For each 119895 let 119879


119895as its root)


119895is



119894







119894) = 119889(V 119903

119894) minus 1 By


1(1198661) ge 1205831(119866)





3 where



V119896 V

119897minus1 So 120583

1(1198663) ge 1205831(1198662)







119889+1) = 0 and 120583

1(1198664) ge 1205831(1198663)


By Claim 1 119881(119862119902)⋂119881(119875


119862119902

= V119896V119896+1 V

119897minus1V119897V119889+2

V119889+3 V

119904V119896(119904 ge 119889 + 2)

where V119896 V119896+1 V

119897minus1 V119897 = 119881(119862

119902)⋂119881(119875

119889+1) and

V119889+2 V119889+3 V

119904 = 119881(119862

119902) 119881(119875

119889+1)

Claim 2 119889(V) = 1 for V isin 119881(119866) (119881(119862119902)⋃119881(119875119889+1))


3) we can get

119894and 120583

1(119894) ge


119894(V119894isin

119881(119862119902)⋃119881(119875

119889+1) V


Claim 3 119897 = 119896 + 1 and 119904 = 119889 + 2


(119881(119862119902)⋃119881(119875119889+1

) = V119904+1 V

119899 and |119860| = 119899 minus 119904 = 119905 ge 0

Case 1 119897 = 119896 By Lemma 7


= 119905 + 2 +

2 + 4 + 119905

119905 + 2

le 119905 + 5 = Δ (119896) + 1

lt 1205831 (119896) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

= 119905 + 4 +

2 + 2 + 2 + 2 + 119905

119905 + 4

le 119905 + 6 le 119899 minus 119889 + 2

= Δ (nabla119896

119899) + 1 lt 120583

1(nabla119896

119899)

(70)

for 119904 ge 119889 + 4 (|119862119902| ge 4)

When 119904 = 119889 + 3 (|119862119902| = 3)

119894cong 1198820(119901119894) By Lemma 23

1205831(119894) lt 1205831(nabla119896

119899) for any 119894 = 119896 where


119899




119899) holds

Case 2 119897 = 119896 + 1


1205831 (119894)


= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (119896) + 1 (Δ (

119896+1) + 1)

lt 1205831(119896) (1205831(119896+1)) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

le 119905 + 3 +

2 + 3 + 119905

119905 + 3

lt 119905 + 5 le 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(71)

for 119904 ge 119889 + 3 (|119862119902| ge 4)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) By Lemma 15

1205831(119894) lt 1205831(Δ119896

119899) for any 119894 = 119896 + 1 where

119896+1cong Δ119896

119899



= 3 +

2 + 3 + 2

3

lt 6 lt 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(72)

for 119904 ge 119889 + 4 (|119862119902| ge 5)

When 119904 = 119889 + 3 (|119862119902| = 4) by Lemma 13 120583

1(◻119896

119889+3) lt

1205831(◻lceil1198892rceil


lceil1198892rceil

119889+3) gt

1205831(◻lceil1198892rceil

119889+3)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) cong Δ119896

119899



119899) gt 1205831(Δlceil1198892rceil

119899) holds

Case 3 119897 ge 119896 + 2 By Lemma 7

1205831(119894)


= 119905 + 3 +

2 + 2 + 2 + 119905

119905 + 3

le 119905 + 5 le 119899 minus 119889 + 2 = Δ (119896

119899) + 1

lt 1205831 (119896

119899)

(73)

for 119904 ge 119889 + 3 (|119862119902| ge 5)

When 119904 = 119889 + 2 (|119862119902| = 4)

119894cong 1198800(119901119894) by Lemma 19

1205831(119894) lt 1205831(

119896


119896

119899





119899has the


Acknowledgments



References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra













119889+1) = 0 and 120583

1(1198664) ge 1205831(1198663)


By Claim 1 119881(119862119902)⋂119881(119875


119862119902

= V119896V119896+1 V

119897minus1V119897V119889+2

V119889+3 V

119904V119896(119904 ge 119889 + 2)

where V119896 V119896+1 V

119897minus1 V119897 = 119881(119862

119902)⋂119881(119875

119889+1) and

V119889+2 V119889+3 V

119904 = 119881(119862

119902) 119881(119875

119889+1)

Claim 2 119889(V) = 1 for V isin 119881(119866) (119881(119862119902)⋃119881(119875119889+1))


3) we can get

119894and 120583

1(119894) ge


119894(V119894isin

119881(119862119902)⋃119881(119875

119889+1) V


Claim 3 119897 = 119896 + 1 and 119904 = 119889 + 2


(119881(119862119902)⋃119881(119875119889+1

) = V119904+1 V

119899 and |119860| = 119899 minus 119904 = 119905 ge 0

Case 1 119897 = 119896 By Lemma 7


= 119905 + 2 +

2 + 4 + 119905

119905 + 2

le 119905 + 5 = Δ (119896) + 1

lt 1205831 (119896) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

= 119905 + 4 +

2 + 2 + 2 + 2 + 119905

119905 + 4

le 119905 + 6 le 119899 minus 119889 + 2

= Δ (nabla119896

119899) + 1 lt 120583

1(nabla119896

119899)

(70)

for 119904 ge 119889 + 4 (|119862119902| ge 4)

When 119904 = 119889 + 3 (|119862119902| = 3)

119894cong 1198820(119901119894) By Lemma 23

1205831(119894) lt 1205831(nabla119896

119899) for any 119894 = 119896 where


119899




119899) holds

Case 2 119897 = 119896 + 1


1205831 (119894)


= 119905 + 2 +

2 + 3 + 119905

119905 + 2

le 119905 + 4 = Δ (119896) + 1 (Δ (

119896+1) + 1)

lt 1205831(119896) (1205831(119896+1)) lt max 119889V119894 + 119898V119894 | V119894 isin 119881 (119896)

le 119905 + 3 +

2 + 3 + 119905

119905 + 3

lt 119905 + 5 le 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(71)

for 119904 ge 119889 + 3 (|119862119902| ge 4)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) By Lemma 15

1205831(119894) lt 1205831(Δ119896

119899) for any 119894 = 119896 + 1 where

119896+1cong Δ119896

119899



= 3 +

2 + 3 + 2

3

lt 6 lt 119899 minus 119889 + 2 = Δ (Δ119896

119899) + 1

lt 1205831(Δ119896

119899)

(72)

for 119904 ge 119889 + 4 (|119862119902| ge 5)

When 119904 = 119889 + 3 (|119862119902| = 4) by Lemma 13 120583

1(◻119896

119889+3) lt

1205831(◻lceil1198892rceil


lceil1198892rceil

119889+3) gt

1205831(◻lceil1198892rceil

119889+3)

When 119904 = 119889 + 2 (|119862119902| = 3)

119894cong 1198670(119901119894) cong Δ119896

119899



119899) gt 1205831(Δlceil1198892rceil

119899) holds

Case 3 119897 ge 119896 + 2 By Lemma 7

1205831(119894)


= 119905 + 3 +

2 + 2 + 2 + 119905

119905 + 3

le 119905 + 5 le 119899 minus 119889 + 2 = Δ (119896

119899) + 1

lt 1205831 (119896

119899)

(73)

for 119904 ge 119889 + 3 (|119862119902| ge 5)

When 119904 = 119889 + 2 (|119862119902| = 4)

119894cong 1198800(119901119894) by Lemma 19

1205831(119894) lt 1205831(

119896


119896

119899





119899has the


Acknowledgments



References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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