Postulation of a Union 𝑋⊂ℙ^𝑟, 𝑟≥4, of a Given Zero-Dimensional Scheme and Several General Lines

Abstract

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

1. Introduction

A scheme𝑋⊂ℙ^𝑟is said to have maximal rank if for all integers 𝑡>0, the restriction map𝐻⁰(𝒪_ℙ^𝑟(𝑡))→𝐻⁰(𝑋,𝒪_𝑋(𝑡))is either injective or surjective, that is, if eitherℎ⁰(ℐ_𝑋(𝑡))=0orℎ¹(ℐ_𝑋(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⊂ℙ^𝑟of𝑡general 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⊂ℙ^𝑟of𝑡disjoint 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⊂ℙ^𝑟,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,𝒪_𝑇∪𝑍(𝑥))=(𝑥+1)𝑡+𝑧for all𝑥≥0. To study the Hilbert function of a general union of𝑍and𝑡disjoint lines, we need the following integers𝑒_{𝑧,𝑟,𝑘}and𝑓_{𝑧,𝑟,𝑘},𝑘≥0, defined by the following relations: 𝑧+(𝑘+1)𝑒_{𝑧,𝑟,𝑘}+𝑓_{𝑧,𝑟,𝑘}=𝑟𝑘+𝑟,0≤𝑓_{𝑧,𝑟,𝑘}≤𝑘. (1) The scheme𝑇∪𝑍has maximal rank if and only ifℎ¹(ℐ_𝑇∪𝑍(𝑘))=0for all integers𝑘with𝑡≤𝑒_{𝑧,𝑟,𝑘}andℎ⁰(ℐ_𝑇∪𝑍(𝑘))=0for the minimal integer𝑘≥𝑐such that either𝑒_{𝑧,𝑟,𝑘}>𝑡or𝑒_{𝑧,𝑟,𝑘}=𝑡and𝑓_{𝑧,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⊂ℙ^𝑟,𝑟≥4, and an integer𝑐>0. Set𝑧∶=deg(𝑍). Assume𝑧≤(_𝑟^𝑟+𝑐). If𝑟=4, assume𝑐≥8. Let𝛿₁be any nonnegative integer such thatℎ⁰(ℐ_𝑍∪𝑋(𝑐))≤𝑓_{𝑧,𝑟,𝑐}+𝛿₁for a general𝑋∈𝐿(𝑟,𝑒_{𝑧,𝑟,𝑐}). Let𝛿₂be any nonnegative integer such thatℎ⁰(ℐ_𝑍∪𝑋(𝑐+1))≤𝑓_{𝑧,𝑟,𝑐+1}+𝛿₂for a general𝑋∈𝐿(𝑟,𝑒_{𝑧,𝑟,𝑐+1}). Set𝛿∶=max{𝛿₁,𝛿₂}.
Fix an integer𝑦>𝑒_{𝑧,𝑟,𝑐+1}, and let𝑌be a general element of𝐿(𝑟,𝑦).(a)For all integers𝑘≥𝑐, eitherℎ⁰(ℐ_𝑍∪𝑌(𝑘))≤𝛿orℎ¹(ℐ_𝑍∪𝑌(𝑘))≤𝛿.(b)Assume𝑓_{𝑧,𝑟,𝑐}≤𝑓_{𝑧,𝑟,𝑐+1}. Then for all integers𝑘≥𝑐, eitherℎ⁰(ℐ_𝑍∪𝑌(𝑘))≤𝛿₁orℎ¹(ℐ_𝑍∪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 integers 𝑟, 𝑧, and  𝑐 such that𝑟≥4and a zero-dimensional scheme𝑍⊂ℙ^𝑟such thatdeg(𝑍)=𝑧andℎ¹(ℐ_𝑍(𝑐))=0. Set𝜏∶=(_𝑟^𝑟+𝑐)−𝑧. If either𝑟≥5or𝑟=4and𝑐≥10, then set𝜏^′∶=𝜏. If𝑟=4and𝑐≤9, then set𝜏^′∶=𝜏+10−𝑐. Fix any integer𝑡≥(^{𝑟+𝑐+𝜏′}_𝑟⁺¹−𝑧)/(𝑐+𝜏^′+1). Let𝑋⊂ℙ^𝑟be a general union of𝑍and𝑡lines. Then𝑋has maximal rank.

If𝑧=(_𝑟^𝑟+𝑐), then we have the following result.

Proposition 3. Fix integers𝑟≥4and𝑐>0. Let𝑍⊂ℙ^𝑟be a zero-dimensional scheme such that deg(𝑍)=(_𝑟^𝑟+𝑐)andℎ⁰(ℐ_𝑍(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 integers𝑒_{𝑧,𝑟,𝑘,𝑥}and𝑓_{𝑧,𝑟,𝑘,𝑥}by the relations 𝑧+(𝑘+1)𝑒_{𝑧,𝑟,𝑘,𝑥}+𝑓_{𝑧,𝑟,𝑘,𝑥}=𝑟𝑟+𝑘+𝑥,0≤𝑓_{𝑧,𝑟,𝑘,𝑥}≤𝑘. (2)

Notice that𝑒_{𝑧,𝑟,𝑘}=𝑒_{𝑧,𝑟,𝑘,0}and𝑓_{𝑧,𝑟,𝑘}=𝑓_{𝑧,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≤_𝑘^𝑟+𝑘, that is, the minimal integer𝑘>0(resp.,𝑘>𝑐) such that𝑡≤𝑒_{𝑧,𝑟,𝑘}.

Taking the equation in (2) minus the same equation for the integer𝑘^′∶=𝑘−1, we get 𝑒_{𝑧,𝑟,𝑘−1,𝑥}+(𝑘+1)𝑒_{𝑧,𝑟,𝑘,𝑥}−𝑒_{𝑧,𝑟,𝑘−1,𝑥}+𝑓_{𝑧,𝑟,𝑘,𝑥}−𝑓_{𝑧,𝑟,𝑘−1,𝑥}=.𝑘+𝑟−1𝑟−1 (3)

Take 𝑍, 𝑧, 𝑐, 𝛿, and 𝛿₁ as in Theorem 1. For each integer𝑘≥𝑐consider the following assertion𝐴_𝑘:𝐴_𝑘:we haveℎ⁰(ℐ_𝑍∪𝐴(𝑘))≤𝑓_{𝑧,𝑟,𝑘}+𝛿for a general𝐴∈𝐿(𝑟,𝑒_{𝑧,𝑟,𝑘}).

Take any𝐴∈𝐿(𝑟,𝑒_{𝑧,𝑟,𝑘})such that𝑍∩𝐴=∅. Sinceℎ⁰(𝑍∪𝐴,𝒪_𝑍∪𝐴(𝑘))=deg(𝑍)+(𝑘+1)deg(𝐴), we haveℎ⁰(ℐ_𝑍∪𝐴(𝑘))=ℎ¹(ℐ_𝑍∪𝐴(𝑘))+𝑓_{𝑧,𝑟,𝑘}. Hence𝐴_𝑘is true if and only ifℎ¹(ℐ_𝑍∪𝐴(𝑘))≤𝛿for a general𝐴∈𝐿(𝑟,𝑒_{𝑧,𝑟,𝑘}).

Call𝐴^′_𝑘, the same assertion with𝛿₁instead of𝛿.

Remark 4. By the definition of the integers𝛿₁and𝛿, the assertions𝐴_𝑐,𝐴^′_𝑐, and𝐴_𝑐+1are true.

For any hyperplane𝐻⊂ℙ^𝑟, and any closed subscheme𝑊⊂ℙ^𝑟, letRes_𝐻(𝑊)be the closed subscheme ofℙ^𝑟withℐ_𝑊∶ℐ_𝐻as its ideal sheaf. If𝑊=𝑍⊔𝑌with𝑍∩𝐻=∅and 𝑌 is a reduced scheme, thenRes_𝐻(𝑊)=𝑍⊔𝑌^′, where𝑌^′is the union of the irreducible components of𝑌not contained in𝐻.

Lemma 5. Fix a hyperplane𝐻⊂ℙ^𝑛,𝑛≥3, an integer𝑘>0and a closed subscheme𝑊⊂ℙ^𝑛. Ifℎ⁰(ℐ_𝑊(𝑘))>ℎ⁰(ℐ_{𝑅𝑒𝑠𝐻(𝑊)}(𝑘−1)), thenℎ⁰(ℐ_{𝑊∪{𝑃}}(𝑘))=ℎ⁰(ℐ_𝑊(𝑘))−1for a general𝑃∈𝐻.

Proof. The assumption implies that𝐻is not contained in the base scheme of|ℐ_𝑊(𝑘)|. The long cohomology exact sequence associated to the following exact sequence 0⟶ℐ_{Res𝐻(𝑊)}(𝑘−1)⟶ℐ_𝑊(𝑘)⟶ℐ_{𝑊∩𝐻,𝐻}(𝑘)⟶0 (4) gives the following inequalities: (1)ℎ⁰(ℐ_𝑊(𝑘))≤ℎ⁰(ℐ_{Res𝐻(𝑊)}(𝑘−1))+ℎ⁰(𝐻,ℐ_{𝑊∩𝐻,𝐻}(𝑘));(2)ℎ¹(ℐ_𝑊(𝑘))≤ℎ¹(ℐ_{Res𝐻(𝑊)}(𝑘−1))+ℎ¹(𝐻,ℐ_{𝑊∩𝐻,𝐻}(𝑘)).
We haveRes_𝐻(𝑊∪{𝑃})=Res_𝐻(𝑊)andℎ⁰(𝐻,ℐ_{(𝑊∩𝐻)∪{𝑃}}(𝑘))=ℎ⁰(𝐻,ℐ_𝑊∩𝐻(𝑘))−1, because a general𝑃∈𝐻is not in the base locus of|ℐ_𝑊(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^𝑛+𝑘. Let𝑋⊂ℙ^𝑛be a general union of𝑒lines and𝑓reducible conics. Thenℎ¹(ℐ_𝑋(𝑘))=0.

Proof. For each𝑃∈Sing(𝑋), let𝐶_𝑃⊂𝑋 be the connected component of𝑋containing𝑃. Write𝑋=𝑌⊔⋃_{𝑃∈Sing(𝑋)}𝐶_𝑃. For each𝑃∈Sing(𝑋), let𝑁_𝑃⊆ℙ^𝑛be a general3-dimensional linear space containing𝐶_𝑃. Let𝐸_𝑃⊂𝑁_𝑃be the sundial with𝐶_𝑃as its support. Set𝑋^′∶=𝑌⊔⋃_{𝑃∈Sing(𝑋)}𝐸_𝑃. Since𝑋and each𝑁_𝑃are general,𝑋^′is a general union of𝑒lines and𝑓sundials. Hence𝑋^′has maximal rank [3]. Since(𝑘+1)(𝑒+2𝑓)≤_𝑛^𝑛+𝑘, we getℎ¹(ℐ_𝑋′(𝑘))=0. Fix𝑃∈Sing(𝑋). Let𝜂be the nilpotent sheaf of𝐸_𝑃. We have an exact sequence 0⟶𝜂⟶𝒪_𝐸𝑃(𝑘)⟶𝒪_𝐶𝑃(𝑘)⟶0. (5) Since𝜂is supported by a single point,𝑃, we haveℎ¹(𝐸_𝑃,𝜂)=0. Hence (5) gives the surjectivity of the restriction map𝐻⁰(𝐸_𝑃,𝒪_𝐸𝑃(𝑘))→𝐻⁰(𝐶_𝑃,𝒪_𝐶𝑃(𝑘)). Looking at all the connected components of𝑋and𝑋^′, we get the surjectivity of the restriction map𝐻⁰(𝑋^′,𝒪_𝑋′(𝑘))→𝐻⁰(𝑋,𝒪_𝑋(𝑘)). Henceℎ¹(ℐ_𝑋(𝑘))≤ℎ¹(ℐ_𝑋′(𝑘))=0.

Lemma 7. 𝐴_𝑘is true for all integers𝑘≥𝑐.

Proof. By Remark 4 we may assume𝑘≥𝑐+2. We use induction on𝑘. Since𝐴_𝑐+1is true, we may assume that𝐴_𝑘−1is true. Fix a hyperplane𝐻⊂ℙ^𝑟such that𝐻∩𝑍=∅. Fix a general𝑌∈𝐿(𝑟,𝑒_{𝑧,𝑟,𝑘−1}). We have𝑌∩𝑍=∅,𝐻∩𝑌is a general union of𝑒_{𝑧,𝑟,𝑘−1}points of𝐻, andℎ⁰(ℐ_𝑍∪𝑌(𝑘))≤𝑓_{𝑧,𝑟,𝑘−1}+𝛿.
(i) First assume𝑓_{𝑧,𝑟,𝑘−1}≤𝑓_{𝑧,𝑟,𝑘}. Let𝐸⊂𝐻be a general union of𝑒_{𝑧,𝑟,𝑘}−𝑒_{𝑧,𝑟,𝑘−1}lines; this is possible, because𝑒_{𝑧,𝑟,𝑘}−𝑒_{𝑧,𝑟,𝑘−1}≥0(Lemma 10). By (3), we have𝑒_{𝑧,𝑟,𝑘−1}+(𝑘+1)(𝑒_{𝑧,𝑟,𝑘}−𝑒_{𝑧,𝑟,𝑘−1})+𝑓_{𝑧,𝑟,𝑘}−𝑓_{𝑧,𝑟,𝑘−1}=^{𝑘+𝑟−1}_𝑟−1. Since𝐸has maximal rank in𝐻[1] and𝑌∩𝐻is general in𝐻, we haveℎ⁰(𝐻,ℐ_{𝑌∩𝐻∪𝐹}(𝑘))=𝑓_{𝑧,𝑟,𝑘}−𝑓_{𝑧,𝑟,𝑘−1}. Castelnuovo’s sequence givesℎ⁰(ℐ_{𝑍∪𝑌∪𝐸}(𝑘))≤𝑓_{𝑧,𝑟,𝑘}+𝛿, proving𝐴_𝑘in this case.
(ii) Now assume𝑓_{𝑧,𝑟,𝑘−1}>𝑓_{𝑧,𝑟,𝑘}. In this case we have𝑒_{𝑧,𝑟,𝑘}−𝑒_{𝑧,𝑟,𝑘−1}≥2(𝑓_{𝑧,𝑟,𝑘−1}−𝑓_{𝑧,𝑟,𝑘})(Lemma 13). Let𝐹₁⊂𝐻be a general union of𝑓_{𝑧,𝑟,𝑘−1}−𝑓_{𝑧,𝑟,𝑘}sundials of𝐻and𝑒_{𝑧,𝑟,𝑘}−𝑒_{𝑧,𝑟,𝑘−1}−2(𝑓_{𝑧,𝑟,𝑘−1}−𝑓_{𝑧,𝑟,𝑘})lines. By [2],𝐹₁has maximal rank in𝐻. Lemma 10 for the integer𝑘^′∶=𝑘−1gives𝑒_{𝑧,𝑟,𝑘−1}≥𝑘−1. Hence𝑒_{𝑧,𝑟,𝑘−1}≥𝑓_{𝑧,𝑟,𝑘−1}−𝑓_{𝑧,𝑟,𝑘}. Hence (3) givesℎ¹(𝐻,ℐ_𝐹1(𝑘))=0. Set𝐹∶=(𝐹₁)_redand𝑆∶=Sing(𝐹). Let𝑈be the union of the degree one connected components of𝐹. Let𝐹₂⊂ℙ^𝑟be 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 of𝑒_{𝑧,𝑟,𝑘}lines, the semicontinuity theorem for cohomology ([5, Theorem III.12.8]) implies that it is sufficient to prove thatℎ⁰(ℐ_{𝑍∪𝑌∪𝐹2}(𝑘))≤𝑓_{𝑧,𝑟,𝑘}+𝛿. We haveRes_𝐻(𝑍∪𝑌∪𝐹₂)=𝑍∪𝑌∪𝑆. The set𝑆is a general union of𝑓_{𝑧,𝑟,𝑘−1}−𝑓_{𝑧,𝑟,𝑘−1}points of𝑆. If𝑆were general points ofℙ^𝑟, then obviouslyℎ⁰(ℐ_{𝑍∪𝑌∪𝑆}(𝑘))=max{0,ℎ⁰(ℐ_𝑍∪𝑌(𝑘))−#(𝑆)}≤𝑓_{𝑧,𝑟,𝑘}+𝛿. To apply𝑓_{𝑧,𝑟,𝑘−1}−𝑓_{𝑧,𝑟,𝑘−1}times Lemma 5 and get the inequalityℎ⁰(ℐ_{𝑍∪𝑌∪𝑆}(𝑘−1))≤𝑓_{𝑧,𝑟,𝑘}+𝛿, we need to check the restriction on the residual scheme in its statement. Assume that this lemma cannot be applied and that𝐴_𝑘fails in this case; that is,ℎ⁰(ℐ_{𝑍∪𝑌∪𝑆}(𝑘−1))>𝑓_{𝑧,𝑟,𝑘}+𝛿. We would getℎ⁰(ℐ_𝑍∪𝑌(𝑘−2))>𝑓_{𝑧,𝑟,𝑘}+𝛿. If𝑘−2=𝑐, we get ℎ⁰(ℐ_{𝑍∪𝑌′}(𝑐)>𝛿) for a general𝑌^′∈𝐿(𝑟,𝑡)(contrary to the definition of𝛿), because𝑒_{𝑧,𝑟,𝑘−1}≥0by Lemma 10. Now assume𝑘≥𝑐+3, and let𝑌₁⊂𝑌a union of𝑒_{𝑧,𝑟,𝑘−2}lines. By𝐴_𝑘−2, we haveℎ⁰(ℐ_𝑍∪𝑌1(𝑘−2))≤𝑓_{𝑧,𝑟,𝑘−2}+𝛿. Any two points ofℙ^𝑟are contained in a line. Henceℎ⁰(ℐ_𝑍∪𝑌(𝑘−2))≤max{0,𝑓_{𝑧,𝑟,𝑘−2}+𝛿−2(deg(𝑌)−deg(𝑌₁))}. Lemma 10 gives2(deg(𝑌)−deg(𝑌₁))≥2(𝑘−1)≥𝑓_{𝑧,𝑟,𝑘−2}. Hence ℎ⁰(ℐ_{𝑍∪𝑌∪𝑆}(𝑘)) = max{0,ℎ⁰(ℐ_𝑍∪𝑌(𝑘))−#(𝑆)}≤𝑓_{𝑧,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 0⟶𝜂⟶𝒪_𝐹1(𝑘)⟶𝒪_𝐹(𝑘)⟶0 (6) gives the surjectivity of the restriction map 𝐻⁰(𝐹₁,𝒪_𝐹1(𝑘))→𝐻⁰(𝐹,𝒪_𝐹(𝑘)). Hence ℎ¹(𝐻,ℐ_𝐹(𝑘))≤ℎ¹(𝐻,ℐ_𝐹1(𝑘))=0. Hence ℎ⁰(𝐻,ℐ_𝐹(𝑘))=^{𝑘+𝑟−1}_𝑟−1−(𝑘+1)(𝑒_{𝑧,𝑟,𝑘}−𝑒_{𝑧,𝑟,𝑘}) + 𝑓_{𝑧,𝑟,𝑘}−𝑓_{𝑧,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(ℐ_{𝑍∪𝑌∪𝐹2}(𝑘))≤𝑓_{𝑧,𝑟,𝑘}+𝛿, concluding the proof of𝐴_𝑘.

Lemma 8. Assume𝑓_{𝑧,𝑟,𝑐}≤𝑓_{𝑧,𝑟,𝑐+1}. Then𝐴^′_𝑘is 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 have𝑒_{𝑧,𝑟,𝑐+1}≥𝑒_{𝑧,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>𝑒_{𝑧,𝑟,𝑘}. Let𝑌be a general element of𝐿(𝑟,𝑡). Thenℎ⁰(ℐ_𝑍∪𝑌(𝑘))≤𝛿.

Proof. It is sufficient to do the case𝑡=𝑒_{𝑧,𝑟,𝑘}+1. Fix a hyperplane𝐻⊂ℙ^𝑟. First assume𝑘≥𝑐+2. Take a general𝐴∈𝐿(𝑟,𝑒_{𝑧,𝑟,𝑘−1}). We have𝐴∩𝑍=∅.𝐴_𝑘−1givesℎ⁰(ℐ_𝐴∪𝑍(𝑘−1))=𝑓_{𝑧,𝑟,𝑘−1}. Let𝐹₁⊂ℙ^𝑟be a general union of𝑒_{𝑧,𝑟,𝑘}+1−2𝑓_{𝑧,𝑟,𝑘−1}lines of𝐻and𝑓_{𝑧,𝑟,𝑘−1}sundials whose support in contained in𝐻. We conclude as in step (ii) of the proof of Lemma 7.

Proof of Theorem 1. First assume𝑧+(𝑘+1)𝑦≥_𝑟^𝑟+𝑘. Since𝑘≥𝑐andℎ⁰(ℐ_𝑍∪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>𝑒_{𝑧,𝑟,𝑘}or𝑦=𝑒_{𝑧,𝑟,𝑘}and𝑓_{𝑧,𝑟,𝑘}, because𝑒_{𝑧,𝑟,𝑢}is a strictly increasing function of𝑢(Lemma 10). Henceℎ⁰(ℐ_𝑍∪𝑌(𝑐))≤𝛿either by Lemma 9 (case𝑓_{𝑧,𝑟,𝑘}>0) or by𝐴_𝑘(case𝑓_{𝑧,𝑟,𝑘}=0).
Now assume𝑧+(𝑘+1)𝑦≤_𝑟^𝑟+𝑘 (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(ℐ_𝑍∪𝐴(𝑘))≤𝛿. By (2) and the assumption𝑦>𝑒_{𝑧,𝑟,𝑐+1}, we have𝑘≥𝑐+2and𝑦≤𝑒_{𝑧,𝑟,𝑘}. Fix a general𝑊⊂𝐿(𝑟,𝑒_{𝑧,𝑟,𝑘}). Let𝑈⊆𝑊be the union of𝑦of the lines of𝑊. By𝐴_𝑘, we haveℎ¹(ℐ_𝑍∪𝑊(𝑘))≤𝛿. Since𝑍∪𝑈is a union of some of the connected components of𝑍∪𝑊, we haveℎ¹(ℐ_𝑍∪𝑈(𝑘))≤𝛿.
Now we check part (b). Since𝑓_{𝑧,𝑟,𝑐}≤𝑓_{𝑧,𝑟,𝑐+1}, Lemma 8 gives𝐴^′_𝑘for all𝑘>𝑐. Then we get Lemma 9 with𝛿₁instead of𝛿. Then we continue as in the proof of part (a).

Proof of Theorem 2. Fix a hyperplane𝐻⊂ℙ^𝑟such that𝐻∩𝑍=∅. If either𝑟≥5or𝑟=4and𝑐≥10for each integer𝑘≥𝑐, set𝑥_𝑘=max{0,−𝜏+𝑘−𝑐−1}. If𝑟=4and𝑐≤9, set𝑥_𝑘∶=−𝜏. Hence 𝑥_𝑐=𝑥_𝑐+1=−𝜏. If𝑟=4and𝑐≤9, we have𝑥_𝑘=−𝜏for all𝑘≤10. Consider the following statement 𝐴_𝑘^′′,𝑘≥𝑐:𝐴_𝑘^′′:we haveℎ⁰(ℐ_𝑍∪𝐴(𝑘))=𝑓_{𝑧,𝑟,𝑘,𝑥𝑘}for a general𝐴∈𝐿(𝑟,𝑒_{𝑧,𝑟,𝑘,𝑥𝑘}).
Take any𝐴∈𝐿(𝑟,𝑒_{𝑧,𝑟,𝑘,𝑥𝑘})such that𝑍∩𝐴=∅. We haveℎ⁰(ℐ_𝑍∪𝐴(𝑘))=ℎ¹(ℐ_𝑍∪𝐴(𝑘))+𝑓_{𝑧,𝑟,𝑘,𝑥𝑘}. Hence𝐴_𝑘^′′is true if and only ifℎ¹(ℐ_𝑍∪𝐴(𝑘))=0for a general𝐴∈𝐿(𝑟,𝑒_{𝑧,𝑟,𝑘,𝑥𝑘}).(a)In this step we prove𝐴_𝑐^′′and𝐴^′′_𝑐+1.𝐴_𝑐^′′is true, because we assumed thatℎ¹(ℐ_𝑍(𝑐))=0. To check𝐴^′′_𝑐+1, we first notice that𝑥_𝑐=𝑥_𝑐+1, and hence we may apply (3) for the integer𝑥∶=𝑥_𝑐. Since𝑓_{𝑧,𝑟,𝑐,𝑥𝑐}=0, we have𝑓_{𝑧,𝑟,𝑐+1,𝑥𝑐+1}≥𝑓_{𝑧,𝑟,𝑐,𝑥𝑐}. Since𝑒_{𝑧,𝑟,𝑐,𝑥𝑐}=0, Lemma 15 gives𝑒_{𝑧,𝑟,𝑐+1,𝑥𝑐}≥𝑒_{𝑧,𝑟,𝑐,𝑥𝑐}. Hence the construction in step (i) of the proof of Lemma 7 works verbatim.(b)In this step we prove𝐴_𝑘^′′for all𝑘≥𝑐. Since the cases𝑘=𝑐,𝑐+1are true by step (a), we may assume that𝑘≥𝑐+2and that𝐴^′′_𝑘−1is true. Taking the difference of (2) with𝑘^′∶=𝑘−1and𝑥∶=𝑥^′and integers𝑘,𝑥, we get 𝑒_{𝑧,𝑟,𝑘−1,𝑥}+(𝑘+1)𝑒_{𝑧,𝑟,𝑘,𝑥}−𝑒_{𝑧,𝑟,𝑘−1,𝑥′}+𝑓_{𝑧,𝑟,𝑘,𝑥}−𝑓_{𝑧,𝑟,𝑘−1,𝑥′}+𝑥−𝑥^′=.𝑘+𝑟−1𝑟−1 (7) We apply (7) with𝑥^′=𝑥_𝑘−1and𝑥=𝑥_𝑘. 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, if𝑥_𝑘−1<0and𝑥_𝑘=𝑥_𝑘−1+1. Fix a general𝑌∈𝐿(𝑟,𝑒_{𝑧,𝑟,𝑘−1,𝑥𝑘−1)}. We have𝑍∩𝑌=∅,ℎ¹(ℐ_𝑍∪𝑌(𝑘−1))=0,ℎ⁰(ℐ_𝑍∪𝑌(𝑘−1))=𝑓_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}−𝑥_𝑘−1, and𝑌∩𝐻is a general union of𝑒_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}points of𝐻. Sinceℎ¹(ℐ_𝑍∪𝑌(𝑘−1))=0,ℎ⁰(ℐ_𝑍∪𝑌(𝑘−1))=𝑓_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}−𝑥_𝑘−1, we haveℎ⁰(ℐ_{𝑍∪𝑌∪{𝑃}}(𝑘−1))=𝑓_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}−𝑥_𝑘−1−1 for a general𝑃∈ℙ^𝑟(and even a general𝑃∈𝐻by Lemma 5 if𝑓_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}−𝑥_𝑘−1−1≥0). This is always the case if𝑥_𝑘−1<0, that is, if𝑥_𝑘−1≠𝑥_𝑘.(b.1)First assume𝑓_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}≤𝑓_{𝑧,𝑟,𝑘,𝑥𝑘}. Since0≤𝑥_𝑘−𝑥_𝑘−1≤1, Lemma 15 gives𝑒_{𝑧,𝑟,𝑘,𝑥𝑘}≥𝑒_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}+2. Let𝐸₁⊂𝐻be a general union of𝑒_{𝑧,𝑟,𝑘,𝑥𝑘}−𝑒_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}−2lines. Take a general reducible conic𝑇⊂𝐻. Let𝑇^′⊂ℙ^𝑟be a general sundial ofℙ^𝑟, 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(ℐ_{𝑍∪𝑌∪𝐸1∪𝑇′}(𝑘))=0. Castelnuovo’s sequence gives ℎ¹(ℐ_{𝑍∪𝑌∪𝐸}(𝑘))=0. Lemma 14 gives 𝑒_{𝑧,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 assume𝑓_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}>𝑓_{𝑧,𝑟,𝑘,𝑥𝑘}. Since0≤𝑥_𝑘−𝑥_𝑘−1≤1, Lemma 16 gives𝑒_{𝑧,𝑟,𝑘,𝑥𝑘}−𝑒_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}≥2(𝑓_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}−𝑓_{𝑧,𝑟,𝑘,𝑥𝑘})+2. We repeat step (ii) of the proof of Lemma 7 taking inside𝐻instead of𝐹a general union𝐹of𝑒_{𝑧,𝑟,𝑘,𝑥𝑘}−𝑒_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}−2(𝑓_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}−𝑓_{𝑧,𝑟,𝑘,𝑥𝑘}+1)lines,𝑓_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}−𝑓_{𝑧,𝑟,𝑘,𝑥𝑘}+1reducible conics. Then we take the general sundials (in𝐻and inℙ^𝑟) with these conics as their supports. To apply Lemma 6, we need𝑒_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}≥𝑓_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}−𝑓_{𝑧,𝑟,𝑘,𝑥𝑘}+1. This inequality is true by Lemma 15.(c)Fix an integer 𝑡≥(^{𝑟+𝑐+𝜏′}_𝑟⁺¹−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≤_𝑟^𝑘+𝑟. Since 𝑡≥(^{𝑟+𝑐+𝜏′}_𝑟⁺¹−𝑧)/(𝑐+𝜏^′+1), either 𝑘≥𝑐+𝜏^′+1 or 𝑘=𝑐+𝜏^′ and 𝑓_{𝑧,𝑐,𝑟,𝑥𝑘}=0. To prove Theorem 2 for the integer 𝑡 it is sufficient to prove ℎ¹(ℐ_𝑍∪𝑌(𝑘))=0 and ℎ⁰(ℐ_𝑍∪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(ℐ_𝑍∪𝐴(𝑘))=0 and ℎ⁰(ℐ_{𝑍∪𝑌𝐵}(𝑘−1))=0. Since 𝑡≥(^{𝑟+𝑐+𝜏′}_𝑟⁺¹−𝑧)/(𝑐+𝜏^′+1), we have 𝑘≥𝜏^′+1. Hence 𝑥_𝑘=𝑥_𝑘−1=0 Since 𝑥_𝑘=0 the existence of 𝐴 is proved as.

3. Numerical Lemmas

Lemma 10. Assume𝑟≥4. Fix an integer𝑐>0. One has𝑒_{𝑧,𝑟,𝑘}≥𝑒_{𝑧,𝑟,𝑘−1}+𝑘for all𝑘≥𝑐+2.

Proof. Assume𝑒_{𝑧,𝑟,𝑘}≤𝑒_{𝑧,𝑟,𝑘−1}+𝑘−1. From (3), we get 𝑒_{𝑧,𝑟,𝑘−1}+(𝑘+1)(𝑘−1)+𝑓_{𝑧,𝑟,𝑘}−𝑓_{𝑧,𝑟,𝑘−1}≥.𝑟+𝑘−1𝑟−1 (8) From (2) for the integer𝑘^′∶=𝑘−1, we get 𝑒_{𝑧,𝑟,𝑘−1}=^{𝑟𝑟+𝑘−1}−𝑧−𝑓_{𝑧,𝑟,𝑘−1}𝑘. (9) Since𝑓_{𝑧,𝑟,𝑘}≤𝑘and𝑘^{𝑟+𝑘−1}_𝑟−1=𝑟_𝑟^{𝑟+𝑘−1}, from (11) and (12), we get 𝑘𝑘²+𝑘−1−(𝑘−1)𝑓_{𝑧,𝑟,𝑘−1}≥()𝑟𝑟−1𝑟+𝑘−1+𝑧. (10) Set𝜇(𝑟,𝑘)∶=(𝑟−1)_𝑟^{𝑟+𝑘−1}−𝑘(𝑘²+𝑘−1). Since𝑓_{𝑧,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^{𝑟+𝑘−1}−^𝑟+𝑘_𝑟+1=^𝑟+𝑘_𝑟+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. Then𝑒_{𝑧,𝑐,𝑘,𝑥}≥𝑒_{𝑧,𝑐,𝑘−1,𝑥}+𝑘for all𝑘≥2.

Proof. Assume𝑒_{𝑧,𝑐,𝑘,𝑥}≤𝑒_{𝑧,𝑐,𝑘−1,𝑥}+𝑘−1. From (3), we get 𝑒_{𝑧,𝑐,𝑘−1,𝑥}+(𝑘+1)(𝑘−1)+𝑓_{𝑧,𝑐,𝑘,𝑥}−𝑓_{𝑧,𝑐,𝑘−1,𝑥}≥.𝑟+𝑘−1𝑟−1 (11) From (2) for the integer𝑘^′∶=𝑘−1, we get 𝑒_{𝑧,𝑐,𝑘−1,𝑥}=^{𝑟𝑟+𝑘−1}+𝑥−𝑧−𝑓_{𝑧,𝑐,𝑘−1,𝑥}𝑘. (12) Since𝑓_{𝑧,𝑐,𝑘,𝑥}≤𝑘,(𝑐+1)𝑡≤(_𝑟^𝑟+𝑐),𝑘^{𝑟+𝑘−1}_𝑟−1=𝑟_𝑟^{𝑟+𝑘−1},𝑧≥0, and𝑥≤0, from (11) and (12), we get 𝑘(𝑘+1)(𝑘−1)−(𝑘−1)𝑓_{𝑧,𝑐,𝑘−1}≥()𝑟𝑟−1.𝑟+𝑘−1 (13) Set𝜙(𝑟,𝑘)∶=(𝑟−1)_𝑟^{𝑟+𝑘−1}−𝑘(𝑘+1)(𝑘−1). Since𝑓_{𝑧,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^{𝑟+𝑘−1}−^𝑟+𝑘_𝑟+1=^𝑟+𝑘_𝑟+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≤(_𝑐^𝑟+𝑐). One has𝑒_{𝑧,𝑟,𝑘}≥𝑒_{𝑧,𝑟,𝑘−1}for all𝑘>𝑐.

Proof. By Lemma 10, it would be sufficient to do the case𝑘=𝑐+1. Assume𝑒_{𝑧,𝑟,𝑘}≤𝑒_{𝑧,𝑟,𝑘−1}−1. From (3) and (12), we get 𝑟()−()𝑓𝑟+𝑘−1−𝑘𝑘+1𝑘−1_{𝑧,𝑟,𝑘−1}≥𝑘𝑟+𝑘−1𝑟−1 (14) which is obviously false.

Lemma 13. Assume𝑟≥4,𝑘>𝑐>0, and𝑓_{𝑧,𝑟,𝑘}<𝑓_{𝑧,𝑟,𝑘−1}. If𝑟=4, assume𝑘≥9. Then𝑒_{𝑧,𝑟,𝑘}−𝑒_{𝑧,𝑟,𝑘−1}≥2(𝑓_{𝑧,𝑟,𝑘−1}−𝑓_{𝑧,𝑟,𝑘}).

Proof. Assume𝑒_{𝑧,𝑟,𝑘}−𝑒_{𝑧,𝑟,𝑘−1}≤2(𝑓_{𝑧,𝑟,𝑘−1}−𝑓_{𝑧,𝑟,𝑘})−1. From (3), we get 𝑒_{𝑧,𝑟,𝑘−1}+(2𝑘+1)𝑓_{𝑧,𝑟,𝑘−1}−𝑓_{𝑧,𝑟,𝑘}−1−𝑘≥.𝑟+𝑘−1𝑟−1 (15) Since𝑒_{𝑧,𝑟,𝑘−1}≤[_𝑟^{𝑘+𝑟−1}−𝑧−𝑓_{𝑧,𝑟,𝑘−1}]/𝑘,𝑘^{𝑟+𝑘−1}_𝑟−1=𝑟_𝑟^{𝑟+𝑘−1}, and𝑓_{𝑧,𝑟,𝑘}≥0, from (15), we get 2𝑘²+𝑘−1𝑓_{𝑧,𝑟,𝑘−1}()≥()𝑟−𝑘𝑘+1𝑟−1𝑟+𝑘−1+𝑧. (16) Since𝑓_{𝑧,𝑟,𝑘−1}≤𝑘, we get ()≥()𝑟2𝑘𝑘+1)(𝑘−1𝑟−1𝑘+𝑟−1+𝑧. (17) Set𝛼(𝑟,𝑘)∶=(𝑟−1)_𝑟^{𝑘+𝑟−1}−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−2}_𝑟−1−_𝑟^{𝑘+𝑟−2}=_𝑟^{𝑘+𝑟−2}[(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≤(_𝑐^𝑟+𝑐), 𝑥^′<0, and𝑥=𝑥^′+1. One has𝑒_{𝑧,𝑟,𝑘,𝑥}≥𝑒_{𝑧,𝑟,𝑘−1,𝑥′}+2.

Proof. Assume𝑒_{𝑧,𝑟,𝑘,𝑥}≤𝑒_{𝑧,𝑟,𝑘−1,𝑥′}+1. By (7), we get 𝑒_{𝑧,𝑟,𝑘−1,𝑥′}+𝑓_{𝑧,𝑟,𝑘,𝑥}−𝑓_{𝑧,𝑟,𝑘−1}+1≥.𝑟+𝑘−1𝑟−1 (18) We have𝑓_{𝑧,𝑟,𝑘,𝑥}−𝑓_{𝑧,𝑟,𝑘−1}≤𝑘. Since𝑥^′≤0, we have𝑒_{𝑧,𝑟,𝑘−1,𝑥′}≤𝑒_{𝑧,𝑟,𝑘−1}≤_𝑟^{𝑟+𝑘−1}/𝑘. Hence 𝑟(),𝑟+𝑘−1+𝑘2𝑘+2≥𝑘𝑟+𝑘−1𝑟−1 (19) Since𝑘^{𝑟+𝑘−1}_𝑟−1=𝑟_𝑟^{𝑟+𝑘−1},𝑟≥4, and𝑘≥2, we get a contradiction.

Lemma 15. Take the setup of Theorem 2. For each integer𝑦>𝑐with𝑥_𝑦<0, one has𝑒_{𝑧,𝑟,𝑦,𝑥𝑦}≥2𝑦.

Proof. By Lemma 14, it is sufficient to do the case𝑦=𝑐+1. In this case, we have𝑥_𝑦=(_𝑟^𝑟+𝑐)−𝑧. Assume𝑒_{𝑧,𝑟,𝑐+1,𝑥𝑦}≤2𝑐+1. Since𝑓_{𝑧,𝑟,𝑐+1,𝑥𝑦}≤𝑐+1, we get(𝑐+2)(2𝑐+1)+𝑐+1≥_𝑟^𝑟+𝑐+1. 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 Theorem 2. Assume𝑟≥4,𝑘>𝑐>0,𝑥_𝑘−1<0, and𝑓_{𝑧,𝑟,𝑘,𝑥𝑘}<𝑓_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}. Then𝑒_{𝑧,𝑟,𝑘,𝑥𝑘}−𝑒_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}≥2(𝑓_{𝑧,𝑟,𝑘−1}−𝑓_{𝑧,𝑟,𝑘})+2.

Proof. Assume𝑒_{𝑧,𝑟,𝑘,𝑥𝑘}−𝑒_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}≤2(𝑓_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}−𝑓_{𝑧,𝑟,𝑘,𝑥𝑘})+1. Since𝑥_𝑘−1<0, we have𝑒_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}<_𝑟^{𝑟+𝑘−1}/𝑘.
From (7) with𝑥_𝑘=𝑥_𝑘−1+1, we get ^𝑟𝑟𝑘^+𝑘−1+(2𝑘+1)𝑓_{𝑧,𝑟,𝑘−1,𝑥𝑘−1}−𝑓_{𝑧,𝑟,𝑘,𝑥𝑘}+()>.𝑘+1−1𝑟+𝑘−1𝑟−1 (20) Since𝑘^{𝑟+𝑘−1}_𝑟−1=𝑟_𝑟^{𝑟+𝑘−1}, we get 2𝑘²()>()𝑟𝑘+1𝑟−1.𝑟+𝑘−1 (21) This inequality is false if either𝑟=4and𝑘≥11or𝑟≥5and𝑘≥2.

Acknowledgment

The author was partially supported by MIUR and GNSAGA of INdAM (Italy).