# Postulation of a UnionX⊂ℙr, r≥4, of a Given Zero-Dimensional Scheme and Several General Lines

Authors: E. Ballico
Journal: ISRN Geometry
 (2013)

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## Abstract

We study the Hilbert function of the unionX of a given zero-dimensional scheme Z⊂ℙr, r≥4, and t general lines of ℙr, proving, for example, that if t≫deg(Z), then X has maximal rank.


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## Body

## 1. Introduction

A schemeX⊂ℙris said to have maximal rank if for all integers t>0, the restriction mapH0(𝒪ℙr(t))→H0(X,𝒪X(t))is either injective or surjective, that is, if eitherh0(ℐX(t))=0orh1(ℐX(t))=0, that is, ifXimposes the “expected” number of conditions to the vector space of the homogeneous degreetpolynomials inr+1variables. Hartshorne and Hirschowitz proved that for allt>0andr≥3, a general unionX⊂ℙroftgeneral lines has maximal rank [1]. Carlini et al. considered several cases in which we allow unions of linear spaces with certain multiplicities (for zero-dimensional subspaces these are general unions of fat points) [2–4]. LetL(r,t)denote the set of all unionsX⊂ℙroftdisjoint lines. In this paper (following [4]) we consider the case in whichXis a disjoint union of a given zero-dimensional schemeZand a generalY∈L(r,t).Fix a zero-dimensional schemeZ⊂ℙr,r≥4, and integersc>0andt≥0. Setz:=deg(Z). In this note we study the Hilbert function of a general union ofZandtdisjoint lines. We assume almost nothing onZ. We only prescribe the integerz:=deg(Z)and an integerc>0. Take a generalT∈L(r,t). SinceZis fixed, whileTis general, we haveZ∩T=∅, and henceh0(T∪Z,𝒪T∪Z(x))=(x+1)t+zfor allx≥0. To study the Hilbert function of a general union ofZandtdisjoint lines, we need the following integersez,r,kandfz,r,k,k≥0, defined by the following relations:
(1)z+(k+1)ez,r,k+fz,r,k=(k+rr),0≤fz,r,k≤k.
The schemeT∪Zhas maximal rank if and only ifh1(ℐT∪Z(k))=0for all integerskwitht≤ez,r,kandh0(ℐT∪Z(k))=0for the minimal integerk≥csuch that eitherez,r,k>torez,r,k=tandfz,r,k=0.We first prove the following result, which says that if we may handle general unions ofZand general lines with respect to homogeneous polynomials in two consecutive degrees (not small with respect to the integerz:=deg(Z)), then these information propagates to higher degree polynomials.Theorem 1.
Fix a zero-dimensional schemeZ⊂ℙr,r≥4, and an integerc>0. Setz:=deg(Z). Assumez≤(r+cr). Ifr=4, assumec≥8. Letδ1be any nonnegative integer such thath0(ℐZ∪X(c))≤fz,r,c+δ1for a generalX∈L(r,ez,r,c). Letδ2be any nonnegative integer such thath0(ℐZ∪X(c+1))≤fz,r,c+1+δ2for a generalX∈L(r,ez,r,c+1). Setδ:=max{δ1,δ2}.
Fix an integery>ez,r,c+1, and letYbe a general element ofL(r,y).(a)
For all integersk≥c, eitherh0(ℐZ∪Y(k))≤δorh1(ℐZ∪Y(k))≤δ.(b)
Assumefz,r,c≤fz,r,c+1. Then for all integersk≥c, eitherh0(ℐZ∪Y(k))≤δ1orh1(ℐZ∪Y(k))≤δ1.Then we go to a case in which we only assume something about the integerz=deg(Z)and prove the following result.Theorem 2.
Fix positive integersr, z, and  c such thatr≥4and a zero-dimensional schemeZ⊂ℙrsuch thatdeg(Z)=zandh1(ℐZ(c))=0. Setτ:=(r+cr)-z. If eitherr≥5orr=4andc≥10, then setτ′:=τ. Ifr=4andc≤9, then setτ′:=τ+10-c. Fix any integert≥((r+c+τ′+1r)-z)/(c+τ′+1). LetX⊂ℙrbe a general union ofZandtlines. ThenXhas maximal rank.Ifz=(r+cr), then we have the following result.Proposition 3.
Fix integersr≥4andc>0. LetZ⊂ℙrbe a zero-dimensional scheme such that deg(Z)=(r+cr)andh0(ℐZ(c))=0. Then for each integert≥0, a general union ofZandtlines has maximal rank.In the caseZ=(c+1)P, we get the caser≥4of [4]. We omit the proof of Proposition 3, because the proof of [4] works verbatim (e.g., quote [4, Lemma5.1] instead of Lemmas 10, 12, and 13).


## 2. The Proofs

For all integersz≥0,r≥4,c≥0,x∈ℤ, andk≥0, define the integersez,r,k,xandfz,r,k,xby the relations
(2)z+(k+1)ez,r,k,x+fz,r,k,x=(r+kr)+x,0≤fz,r,k,x≤k.Notice thatez,r,k=ez,r,k,0andfz,r,k=fz,r,k,0. For anyt≥0,c>0, the critical value of the triple(z,r,t)(resp.,(z,r,c)) is the minimal integerk>0(resp.,k>c) such thatz+(k+1)t≤(r+kk), that is, the minimal integerk>0(resp.,k>c) such thatt≤ez,r,k.Taking the equation in (2) minus the same equation for the integerk′:=k-1, we get
(3)ez,r,k-1,x+(k+1)(ez,r,k,x-ez,r,k-1,x)+fz,r,k,x-fz,r,k-1,x=(k+r-1r-1).TakeZ, z, c, δ, and δ1 as in Theorem 1. For each integerk≥cconsider the following assertionAk:Ak:
we haveh0(ℐZ∪A(k))≤fz,r,k+δfor a generalA∈L(r,ez,r,k).Take anyA∈L(r,ez,r,k)such thatZ∩A=∅. Sinceh0(Z∪A,𝒪Z∪A(k))=deg(Z)+(k+1)deg(A), we haveh0(ℐZ∪A(k))=h1(ℐZ∪A(k))+fz,r,k. HenceAkis true if and only ifh1(ℐZ∪A(k))≤δfor a generalA∈L(r,ez,r,k).CallAk′, the same assertion withδ1instead ofδ.Remark 4.
By the definition of the integersδ1andδ, the assertionsAc,Ac′, andAc+1are true.For any hyperplaneH⊂ℙr, and any closed subschemeW⊂ℙr, letResH(W)be the closed subscheme ofℙrwithℐW:ℐHas its ideal sheaf. IfW=Z⊔YwithZ∩H=∅and Y is a reduced scheme, thenResH(W)=Z⊔Y′, whereY′is the union of the irreducible components ofYnot contained inH.Lemma 5.
Fix a hyperplaneH⊂ℙn,n≥3, an integerk>0and a closed subschemeW⊂ℙn. Ifh0(ℐW(k))>h0(ℐResH(W)(k-1)), thenh0(ℐW∪{P}(k))=h0(ℐW(k))-1for a generalP∈H.Proof.
The assumption implies thatHis not contained in the base scheme of|ℐW(k)|. The long cohomology exact sequence associated to the following exact sequence
(4)0⟶ℐResH(W)(k-1)⟶ℐW(k)⟶ℐW∩H,H(k)⟶0
gives the following inequalities: (1)
h0(ℐW(k))≤h0(ℐResH(W)(k-1))+h0(H,ℐW∩H,H(k));(2)
h1(ℐW(k))≤h1(ℐResH(W)(k-1))+h1(H,ℐW∩H,H(k)).
We haveResH(W∪{P})=ResH(W)andh0(H,ℐ(W∩H)∪{P}(k))=h0(H,ℐW∩H(k))-1, because a generalP∈His not in the base locus of|ℐW(k)|. Apply (4) first toWand then toW∪{P}.As in [2–4], we will call Casteluonovo’s sequence for any of the inequalities in the proof of Lemma 5.Lemma 6.
Fix integersn≥3,k>0,e≥0, andf≥0such that(k+1)(e+2f)≤(n+kn). LetX⊂ℙnbe a general union ofelines andfreducible conics. Thenh1(ℐX(k))=0.Proof.
For eachP∈Sing(X), letCP⊂X be the connected component ofXcontainingP. WriteX=Y⊔⋃P∈Sing(X)‍CP. For eachP∈Sing(X), letNP⊆ℙnbe a general3-dimensional linear space containingCP. LetEP⊂NPbe the sundial withCPas its support. SetX′:=Y⊔⋃P∈Sing(X)‍EP. SinceXand eachNPare general,X′is a general union ofelines andfsundials. HenceX′has maximal rank [3]. Since(k+1)(e+2f)≤(n+kn), we geth1(ℐX′(k))=0. FixP∈Sing(X). Letηbe the nilpotent sheaf ofEP. We have an exact sequence
(5)0⟶η⟶𝒪EP(k)⟶𝒪CP(k)⟶0.
Sinceηis supported by a single point,P, we haveh1(EP,η)=0. Hence (5) gives the surjectivity of the restriction mapH0(EP,𝒪EP(k))→H0(CP,𝒪CP(k)). Looking at all the connected components ofXandX′, we get the surjectivity of the restriction mapH0(X′,𝒪X′(k))→H0(X,𝒪X(k)). Henceh1(ℐX(k))≤h1(ℐX′(k))=0.Lemma 7.
A
kis true for all integersk≥c.Proof.
By Remark4 we may assumek≥c+2. We use induction onk. SinceAc+1is true, we may assume thatAk-1is true. Fix a hyperplaneH⊂ℙrsuch thatH∩Z=∅. Fix a generalY∈L(r,ez,r,k-1). We haveY∩Z=∅,H∩Yis a general union ofez,r,k-1points ofH, andh0(ℐZ∪Y(k))≤fz,r,k-1+δ.
(i) First assumefz,r,k-1≤fz,r,k. LetE⊂Hbe a general union ofez,r,k-ez,r,k-1lines; this is possible, becauseez,r,k-ez,r,k-1≥0(Lemma 10). By (3), we haveez,r,k-1+(k+1)(ez,r,k-ez,r,k-1)+fz,r,k-fz,r,k-1=(k+r-1r-1). SinceEhas maximal rank inH[1] andY∩His general inH, we haveh0(H,ℐY∩H∪F(k))=fz,r,k-fz,r,k-1. Castelnuovo’s sequence givesh0(ℐZ∪Y∪E(k))≤fz,r,k+δ, provingAkin this case.
(ii) Now assumefz,r,k-1>fz,r,k. In this case we haveez,r,k-ez,r,k-1≥2(fz,r,k-1-fz,r,k)(Lemma 13). LetF1⊂Hbe a general union offz,r,k-1-fz,r,ksundials ofHandez,r,k-ez,r,k-1-2(fz,r,k-1-fz,r,k)lines. By [2],F1has maximal rank inH. Lemma 10 for the integerk′:=k-1givesez,r,k-1≥k-1. Henceez,r,k-1≥fz,r,k-1-fz,r,k. Hence (3) givesh1(H,ℐF1(k))=0. SetF:=(F1)redandS:=Sing(F). LetUbe the union of the degree one connected components ofF. LetF2⊂ℙrbe a general union of disjoint lines and sundials withFhas its reduction. Since(Y∪F2)∩Z=∅andY∪F2is a degeneration of a family of disjoint unions ofez,r,klines, the semicontinuity theorem for cohomology ([5, Theorem III.12.8]) implies that it is sufficient to prove thath0(ℐZ∪Y∪F2(k))≤fz,r,k+δ. We haveResH(Z∪Y∪F2)=Z∪Y∪S. The setSis a general union offz,r,k-1-fz,r,k-1points ofS. IfSwere general points ofℙr, then obviouslyh0(ℐZ∪Y∪S(k))=max{0,h0(ℐZ∪Y(k))-#(S)}≤fz,r,k+δ. To applyfz,r,k-1-fz,r,k-1times Lemma 5 and get the inequalityh0(ℐZ∪Y∪S(k-1))≤fz,r,k+δ, we need to check the restriction on the residual scheme in its statement. Assume that this lemma cannot be applied and thatAkfails in this case; that is,h0(ℐZ∪Y∪S(k-1))>fz,r,k+δ. We would geth0(ℐZ∪Y(k-2))>fz,r,k+δ. Ifk-2=c, we get h0(ℐZ∪Y′(c)>δ) for a generalY′∈L(r,t)(contrary to the definition ofδ), becauseez,r,k-1≥0by Lemma 10. Now assumek≥c+3, and letY1⊂Ya union ofez,r,k-2lines. ByAk-2, we haveh0(ℐZ∪Y1(k-2))≤fz,r,k-2+δ. Any two points ofℙrare contained in a line. Henceh0(ℐZ∪Y(k-2))≤max{0,fz,r,k-2+δ-2(deg(Y)-deg(Y1))}. Lemma 10 gives2(deg(Y)-deg(Y1))≥2(k-1)≥fz,r,k-2. Hence h0(ℐZ∪Y∪S(k)) = max{0,h0(ℐZ∪Y(k))-#(S)}≤fz,r,k+δ.
We haveH∩(Z∪Y∪F2)=(Y∩H)∪F. Letηbe the nilpotent sheaf ofF1. Sinceηis supported by the finite setS, we haveh1(F1,η⊗L)=0for each line bundleLonF1. The exact sequence
(6)0⟶η⟶𝒪F1(k)⟶𝒪F(k)⟶0
gives the surjectivity of the restriction map H0(F1,𝒪F1(k))→H0(F,𝒪F(k)). Hence h1(H,ℐF(k))≤h1(H,ℐF1(k))=0. Hence h0(H,ℐF(k))=(k+r-1r-1)-(k+1)(ez,r,k-ez,r,k) + fz,r,k-fz,r,k-1. SinceY∩His formed by general points ofH, (3) gives h0(H,ℐ(Z∪Y∪F2)∩H(k))=0. Castelnuovo’s sequence gives h0(ℐZ∪Y∪F2(k))≤fz,r,k+δ, concluding the proof ofAk.Lemma 8.
Assumefz,r,c≤fz,r,c+1. ThenAk′is true for allk≥c.Proof.
A
c
′ is true (Remark 4). Now assumek=c+1. To copy step (i) of the proof of Lemma 7, it is sufficient to haveez,r,c+1≥ez,r,c. Use Lemma 12. The casek≥c+2is done as in the proof of Lemma 7.Lemma 9.
FixZas in Theorem 1 and integersk>candt>ez,r,k. LetYbe a general element ofL(r,t). Thenh0(ℐZ∪Y(k))≤δ.Proof.
It is sufficient to do the caset=ez,r,k+1. Fix a hyperplaneH⊂ℙr. First assumek≥c+2. Take a generalA∈L(r,ez,r,k-1). We haveA∩Z=∅.Ak-1givesh0(ℐA∪Z(k-1))=fz,r,k-1. LetF1⊂ℙrbe a general union ofez,r,k+1-2fz,r,k-1lines ofHandfz,r,k-1sundials whose support in contained inH. We conclude as in step (ii) of the proof of Lemma 7.Proof of Theorem1.
First assumez+(k+1)y≥(r+kr). Sincek≥candh0(ℐZ∪Y(c))≤δfor a generalY∈L(r,t)andy≥t, we may assumek>c. In this range the assumption is satisfied if and only if eithery>ez,r,kory=ez,r,kandfz,r,k, becauseez,r,uis a strictly increasing function ofu(Lemma 10). Henceh0(ℐZ∪Y(c))≤δeither by Lemma 9 (casefz,r,k>0) or byAk(casefz,r,k=0).
Now assumez+(k+1)y≤(r+kr) (we allow the equality). By the semicontinuity theorem for cohomology ([5, Theorem III.12.8]), it is sufficient to findA∈L(r,y)such thatA∩Z=∅and,h1(ℐZ∪A(k))≤δ. By (2) and the assumptiony>ez,r,c+1, we havek≥c+2andy≤ez,r,k. Fix a generalW⊂L(r,ez,r,k). LetU⊆Wbe the union ofyof the lines ofW. ByAk, we haveh1(ℐZ∪W(k))≤δ. SinceZ∪Uis a union of some of the connected components ofZ∪W, we haveh1(ℐZ∪U(k))≤δ.
Now we check part (b). Sincefz,r,c≤fz,r,c+1, Lemma 8 givesAk′for allk>c. Then we get Lemma 9 withδ1instead ofδ. Then we continue as in the proof of part (a).Proof of Theorem2.
Fix a hyperplaneH⊂ℙrsuch thatH∩Z=∅. If eitherr≥5orr=4andc≥10for each integerk≥c, setxk=max{0,-τ+k-c-1}. Ifr=4andc≤9, setxk:=-τ. Hence xc=xc+1=-τ. Ifr=4andc≤9, we havexk=-τfor allk≤10. Consider the following statement Ak′′,k≥c:Ak′′:
we haveh0(ℐZ∪A(k))=fz,r,k,xkfor a generalA∈L(r,ez,r,k,xk).
Take anyA∈L(r,ez,r,k,xk)such thatZ∩A=∅. We haveh0(ℐZ∪A(k))=h1(ℐZ∪A(k))+fz,r,k,xk. HenceAk′′is true if and only ifh1(ℐZ∪A(k))=0for a generalA∈L(r,ez,r,k,xk).(a)
In this step we proveAc′′andAc+1′′.Ac′′is true, because we assumed thath1(ℐZ(c))=0. To checkAc+1′′, we first notice thatxc=xc+1, and hence we may apply (3) for the integerx:=xc. Sincefz,r,c,xc=0, we havefz,r,c+1,xc+1≥fz,r,c,xc. Sinceez,r,c,xc=0, Lemma 15 givesez,r,c+1,xc≥ez,r,c,xc. Hence the construction in step (i) of the proof of Lemma 7 works verbatim.(b)
In this step we proveAk′′for allk≥c. Since the casesk=c,c+1are true by step (a), we may assume thatk≥c+2and thatAk-1′′is true. Taking the difference of (2) withk′:=k-1andx:=x′and integersk,x, we get
(7)ez,r,k-1,x+(k+1)(ez,r,k,x-ez,r,k-1,x′)+fz,r,k,x-fz,r,k-1,x′+x-x′=(k+r-1r-1).
We apply (7) withx′=xk-1andx=xk. Hence eitherx′=x(casek≥τ+2+c) orx′=x-1. In both cases, we have0≤x-x′≤1. Ifx=x′, then we may apply the numerical Lemmas 10, 12, and 13 used in the proof of Theorem 1. We need different numerical lemmas ifx≠x′, that is, ifxk-1<0andxk=xk-1+1.Fix a generalY∈L(r,ez,r,k-1,xk-1). We haveZ∩Y=∅,h1(ℐZ∪Y(k-1))=0,h0(ℐZ∪Y(k-1))=fz,r,k-1,xk-1-xk-1, andY∩His a general union ofez,r,k-1,xk-1points ofH. Sinceh1(ℐZ∪Y(k-1))=0,h0(ℐZ∪Y(k-1))=fz,r,k-1,xk-1-xk-1, we haveh0(ℐZ∪Y∪{P}(k-1))=fz,r,k-1,xk-1-xk-1-1 for a generalP∈ℙr(and even a generalP∈Hby Lemma 5 iffz,r,k-1,xk-1-xk-1-1≥0). This is always the case ifxk-1<0, that is, ifxk-1≠xk.(b.1)
First assumefz,r,k-1,xk-1≤fz,r,k,xk. Since0≤xk-xk-1≤1, Lemma 15 givesez,r,k,xk≥ez,r,k-1,xk-1+2. LetE1⊂Hbe a general union ofez,r,k,xk-ez,r,k-1,xk-1-2lines. Take a general reducible conicT⊂H. LetT′⊂ℙrbe a general sundial ofℙr, and letT′′⊂Hbe a general sundial ofHwithTas its support. SetF:=E1∪T. SinceZ∪Y∪E1∪T′is a specialization of a union ofZanddeg(Y∪E1∪T′)lines, it is sufficient to prove that h1(ℐZ∪Y∪E1∪T′(k))=0. Castelnuovo’s sequence gives h1(ℐZ∪Y∪E(k))=0. Lemma 14 gives ez,r,k-1>0. By (7) to apply Lemma 5, we need that Y∩H has as least one point; this is the reason why this proof does not work if k=c+1.(b.2)
Now assumefz,r,k-1,xk-1>fz,r,k,xk. Since0≤xk-xk-1≤1, Lemma 16 givesez,r,k,xk-ez,r,k-1,xk-1≥2(fz,r,k-1,xk-1-fz,r,k,xk)+2. We repeat step (ii) of the proof of Lemma 7 taking insideHinstead ofFa general unionF~ofez,r,k,xk-ez,r,k-1,xk-1-2(fz,r,k-1,xk-1-fz,r,k,xk+1)lines,fz,r,k-1,xk-1-fz,r,k,xk+1reducible conics. Then we take the general sundials (inHand inℙr) with these conics as their supports. To apply Lemma 6, we needez,r,k-1,xk-1≥fz,r,k-1,xk-1-fz,r,k,xk+1. This inequality is true by Lemma 15.(c)
Fix an integert≥((r+c+τ′+1r)-z)/(c+τ′+1). Let k be the critical value for the triple (r,c,t), i.e. the minimal positive integer such that z+(k+1)t≤(k+rr). Since t≥((r+c+τ′+1r)-z)/(c+τ′+1), either k≥c+τ′+1 or k=c+τ′ and fz,c,r,xk=0. To prove Theorem 2 for the integer t it is sufficient to prove h1(ℐZ∪Y(k))=0 and h0(ℐZ∪Y(k-1))=0 for a general Y∈L(r,t). Since L(r,t) is irreducible, the semicontinuity theorem for cohomology says that to prove Theorem 2 for the integer t it is sufficient to prove the existence of A∈L(r,t) and B∈L(r,t) such that A∩Z=B∩Z=∅, h1(ℐZ∪A(k))=0 and h0(ℐZ∪YB(k-1))=0. Since t≥((r+c+τ′+1r)-z)/(c+τ′+1), we have k≥τ′+1. Hence xk=xk-1=0 Since xk=0 the existence of A is proved as.


## 3. Numerical Lemmas

Lemma 10.
Assumer≥4. Fix an integerc>0. One hasez,r,k≥ez,r,k-1+kfor allk≥c+2.Proof.
Assumeez,r,k≤ez,r,k-1+k-1. From (3), we get
(8)ez,r,k-1+(k+1)(k-1)+fz,r,k-fz,r,k-1≥(r+k-1r-1).
From (2) for the integerk′:=k-1, we get
(9)ez,r,k-1=((r+k-1r)-z-fz,r,k-1)k.
Sincefz,r,k≤kandk(r+k-1r-1)=r(r+k-1r), from (11) and (12), we get
(10)k(k2+k-1)-(k-1)fz,r,k-1≥(r-1)(r+k-1r)+z.
Setμ(r,k):=(r-1)(r+k-1r)-k(k2+k-1). Sincefz,r,k-1≥0andz>0, to get a contradiction, it is sufficient to prove thatμ(r,k)≥0. First assumer=4. We have8μ(4,k)=(k+3)(k+2)(k+1)(k-1)-8k(k2+k-1). Obviously8μ(4,k)≥0ifk≥6. We haveμ(4,5)=210-5·29,μ(4,4)=105-76>0, and μ(4,3)=45-33>0.
We haveμ(r+1,k)-μ(r,k)=k(r+k-1r)-(r+kr+1)=(r+kr+1){[k(r+1)/(r+k)]-1}>0. By induction onr, we get the lemma for allk≥5.Lemma 11.
Fix integersr≥4,c>0,z≥0, andx≤0. Thenez,c,k,x≥ez,c,k-1,x+kfor allk≥2.Proof.
Assumeez,c,k,x≤ez,c,k-1,x+k-1. From (3), we get
(11)ez,c,k-1,x+(k+1)(k-1)+fz,c,k,x-fz,c,k-1,x≥(r+k-1r-1).
From (2) for the integerk′:=k-1, we get
(12)ez,c,k-1,x=((r+k-1r)+x-z-fz,c,k-1,x)k.
Sincefz,c,k,x≤k,(c+1)t≤(r+cr),k(r+k-1r-1)=r(r+k-1r),z≥0, andx≤0, from (11) and (12), we get
(13)k(k+1)(k-1)-(k-1)fz,c,k-1≥(r-1)(r+k-1r).
Setϕ(r,k):=(r-1)(r+k-1r)-k(k+1)(k-1). Sincefz,c,k-1≥0, to get a contradiction it is sufficient to prove thatϕ(r,k)>0. First assumer=4. We have8ϕ(4,k)=(k+3)(k+2)(k+1)k-8k(k+1)(k-1)=(k+1)k(k2-3k+8). Henceϕ(4,k)>0for allk≥2. Now assumer>4. We haveϕ(r+1,k)-ϕ(r,k)=k(r+k-1r)-(r+kr+1)=(r+kr+1){[k(r+1)/(r+k)]-1}>0. By induction onr, we get the lemma.Lemma 12.
Assumer≥4. Fix integersc,zsuch thatc>0and0≤z≤(r+cc). One hasez,r,k≥ez,r,k-1for allk>c.Proof.
By Lemma10, it would be sufficient to do the casek=c+1. Assumeez,r,k≤ez,r,k-1-1. From (3) and (12), we get
(14)(r+k-1r)-k(k+1)-(k-1)fz,r,k-1≥k(r+k-1r-1)
which is obviously false.Lemma 13.
Assumer≥4,k>c>0, andfz,r,k<fz,r,k-1. Ifr=4, assumek≥9. Thenez,r,k-ez,r,k-1≥2(fz,r,k-1-fz,r,k).Proof.
Assumeez,r,k-ez,r,k-1≤2(fz,r,k-1-fz,r,k)-1. From (3), we get
(15)ez,r,k-1+(2k+1)(fz,r,k-1-fz,r,k)-1-k≥(r+k-1r-1).
Sinceez,r,k-1≤[(k+r-1r)-z-fz,r,k-1]/k,k(r+k-1r-1)=r(r+k-1r), andfz,r,k≥0, from (15), we get
(16)(2k2+k-1)fz,r,k-1-k(k+1)≥(r-1)(r+k-1r)+z.
Sincefz,r,k-1≤k, we get
(17)2k(k+1)(k-1)≥(r-1)(k+r-1r)+z.
Setα(r,k):=(r-1)(k+r-1r)-k(2k2-2). Sincez>0, to get a contradiction and hence to prove the lemma, it is sufficient to prove thatα(r,k)>0. First assumer=4. We have 8α(4,k)=(k+3)(k+2)(k+1)k-8k(2k2-2)=k(k+1)[(k+3)(k+2)-16k+16]. Hence α(4,k)≥0if and only ifk≥9.
Now assumer>4. We haveα(r,k)-α(r-1,k)=k(r+k-2r-1)-(k+r-2r)=(k+r-2r)[(rk/(k-1)-1]>0for allk≥2. We haveα(5,3)=3·35-21-48>0, andα(5,k)is an increasing function ofk. Henceα(r,k)≥0if eitherr≥5ork≥9.Lemma 14.
Assumer≥4. Fix integers k, c, z, x, and x′such thatk>c>0, 0≤z≤(r+cc), x′<0, andx=x′+1. One hasez,r,k,x≥ez,r,k-1,x′+2.Proof.
Assumeez,r,k,x≤ez,r,k-1,x′+1. By (7), we get
(18)ez,r,k-1,x′+fz,r,k,x-fz,r,k-1+1≥(r+k-1r-1).
We havefz,r,k,x-fz,r,k-1≤k. Sincex′≤0, we haveez,r,k-1,x′≤ez,r,k-1≤(r+k-1r)/k. Hence
(19)(r+k-1r)+k(2k+2)≥k(r+k-1r-1),
Sincek(r+k-1r-1)=r(r+k-1r),r≥4, andk≥2, we get a contradiction.Lemma 15.
Take the setup of Theorem2. For each integery>cwithxy<0, one hasez,r,y,xy≥2y.Proof.
By Lemma14, it is sufficient to do the casey=c+1. In this case, we havexy=(r+cr)-z. Assumeez,r,c+1,xy≤2c+1. Sincefz,r,c+1,xy≤c+1, we get(c+2)(2c+1)+c+1≥(r+c+1r). The right hand side of this inequality is an increasing function ofr. Forr=4, this inequality fails for allc≥2, because24(2c2+6c+3)<(c+5)(c+4)(c+3)(c+2)for allc≥2.Lemma 16.
Take the setup of Theorem2. Assumer≥4,k>c>0,xk-1<0, andfz,r,k,xk<fz,r,k-1,xk-1. Thenez,r,k,xk-ez,r,k-1,xk-1≥2(fz,r,k-1-fz,r,k)+2.Proof.
Assumeez,r,k,xk-ez,r,k-1,xk-1≤2(fz,r,k-1,xk-1-fz,r,k,xk)+1. Sincexk-1<0, we haveez,r,k-1,xk-1<(r+k-1r)/k.
From (7) withxk=xk-1+1, we get
(20)(r+k-1r)k+(2k+1)(fz,r,k-1,xk-1-fz,r,k,xk)+(k+1)-1>(r+k-1r-1).
Sincek(r+k-1r-1)=r(r+k-1r), we get
(21)2k2(k+1)>(r-1)(r+k-1r).
This inequality is false if eitherr=4andk≥11orr≥5andk≥2.



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*Source: 391721-2013-12-12.xml*