# It turns out from 6 the euler relation and the

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It turns out from (6), the Eulerrelation, and the inequalityd0<0 thatft0,Δ0(x0) = 0,which contradicts the non-degeneracycondition. Therefore,Jnegationslash=.ForjJ, we can writeλj(s) =cjsmj+ higher-order terms ins,wherecjnegationslash= 0 andmjQ.PutJ1:={jJ:gj|CInegationslash≡0}.IfJ1=,then∂gj∂xi(φ(s))0for alliIand alljJ.We deduce from the condition (a5) that∂ft(s)∂xi(φ(s))0for alliI.Consequently,∂ft0,Δ0∂xi(x0) = 0for alliI.It follows from (6), the Euler relation, and the inequalityd0<0 thatft0,Δ0(x0) = 0,whichcontradicts the non-degeneracy condition.HenceJ1negationslash=.For eachjJ1, letdjbe the13
minimal value of the linear functioniIαiqionRIΓ(gj) and Δjbe the maximal face ofRIΓ(gj) where this linear function takes its minimum value. We can writegj(φ(s)) =gj,Δj(x0)sdj+ higher-order terms ins,wheregj,Δjis the face function associated withgjand Δj.By the condition (a4), thengj,Δj(x0)=0for alljJ1.(7)On the other hand, foriIandjJ1,∂gj∂xi(φ(s)) =∂gj,Δj∂xi(x0)sdjqi+ higher-order terms ins.Fori /IandjJ1, the functiongj,Δjdoes not depend on the variablexi, and hence,∂gj,Δj∂xi(x0) = 0.(8)The condition (a5) implies that for alliI,∂ft0,Δ0∂xi(x0)sd0qi+· · ·+summationdisplayjJ2cj∂gj,Δj∂xi(x0)sqi+· · ·= 0,(9)where:= minjJ1(mj+dj),J2:={jJ1:=mj+dj}and the dots stand for thehigher-order terms ins. There are three cases to be considered.Case 1:ℓ > d0. By (6) and (9), we have∂ft0,Δ0∂xi(x0) = 0fori= 1, . . . , n.This, together with the Euler relation, implies thatd0ft0,Δ0= 0.Hence,ft0,Δ0(x0) = 0 because ofd0<0.This contradicts the non-degeneracy condition.Case 2:=d0. We deduce from (6), (8) and (9) that∂ft0,Δ0∂xi(x0) +summationdisplayjJ2cj∂gj,Δj∂xi(x0) = 0fori= 1, . . . , n.14
Consequently,0=nsummationdisplayi=1qix0i∂ft0,Δ0∂xi(x0) +nsummationdisplayi=1summationdisplayjJ2cjqix0i∂gj,Δj∂xi(x0)=nsummationdisplayi=1qix0i∂ft0,Δ0∂xi(x0) +summationdisplayjJ2cjnsummationdisplayi=1qix0i∂gj,Δj∂xi(x0)=d0ft0,Δ0(x0) +summationdisplayjJ2cjdjgj,Δj(x0)=d0ft0,Δ0(x0),where the last equation follows from (7). Sinced0<0,we getft0,Δ0(x0) = 0,which contra-dicts the non-degeneracy condition.Case 3:ℓ < d0. By (8) and (9), we obtainsummationdisplayjJ2cj∂gj,Δj∂xi(x0) = 0fori= 1, . . . , n.This fact and (7) combined give a contradiction with the non-degeneracy condition.squareLemma 4.2(Boundedness of singularities at infinity).There exists a real numberr >0such thatΣ(ft|S)Drfor allt[0,1].Proof.Suppose the assertion of the lemma is false. By the Curve Selection Lemma at infinity(see [25] or [17]), we can find a nonempty setI⊂ {1, . . . , n}withft|CInegationslash≡0,a (possiblyempty) setJ⊂ {j∈ {1, . . . , p}:gj|CInegationslash≡0},a vectorqRnwith miniIqi<0 and0 =d(q,Γ(ft|CI)), and analytic curvesφ: (0, ǫ)(C)I,t: (0, ǫ)[0,1],andλj: (0, ǫ)C, jJ,such that the following conditions hold(a1)bardblφ(s)bardbl → ∞ass0;(a2)t(s)t0[0,1] ass0;(a3)ft(s),Δ0(φ(s))→ ∞ass0;(a4)gj,Δj(φ(s)) = 0 for alljJand alls(0, ǫ);(a5)ft(s),Δ0(φ(s)) +jJλj(s)gj,Δj(φ(s)) = 0 for alls(0, ǫ),where Δ0:= Δ(q,Γ(ft|CI) and Δj:= Δ(q,Γ(gj|CI)) forjJ.

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