UNIFORM CO-LIPSCHITZ MAPPINGS OF THE PLANE
OLGA MALEVA Received 29 October 2003
We give a sharp estimate on the cardinality of point preimages of a uniform co-Lipschitz mapping on the plane. We also give a necessary and sufficient condition for a ball non- collapsing Lipschitz function to have a point with infinite preimage.
1. Introduction
Consider a mapping f :X→Y between two normed spacesXandY. The function Ωf(d)= sup
x−xX≤d
f(x)−f(x)Y (1.1)
is called the modulus of (uniform) continuity of f. The mapping f is said to be uni- formly continuous ifΩf(d)→0 asd↓0. In this case the modulus of continuity is a sub- additive monotone continuous function. The definition ofΩf implies that f(Br(x))⊂ BΩf(r)(f(x)). (ByBρ(y) andBρ(y) we denote, respectively, the open and the closed ball of radiusρ, centered aty.)
One important class of uniformly continuous mappings is the class of Lipschitz map- pings, that is, those satisfyingΩf(d)≤Ldfor some positiveL. The least suchLis called the Lipschitz constant of the mapping f.
In a similar way, couniformly continuous mappings are defined as those satisfying fBr(x)⊃Bω(r)
f(x), r >0, (1.2)
whereω(r) is a function of the radiusrindependent of the pointx, such thatω(r)>0 for r >0. A particular case is a co-Lipschitz mapping which satisfies
fBr(x)⊃Bcrf(x). (1.3)
We call the best (the largest) such constantcthe co-Lipschitz constant of the mapping f. (Note that in some papers, in particular [3], the co-Lipschitz constant is the reciprocal of ourc.)
A mapping f is called a uniform quotient if it is both uniform and couniform; f is called a Lipschitz quotient if it is Lipschitz and co-Lipschitz.
Copyright©2005 Hindawi Publishing Corporation Abstract and Applied Analysis 2005:5 (2005) 543–562 DOI:10.1155/AAA.2005.543
In Section2, we generalize results which were obtained in [4,5]. We show that for a uniform co-Lipschitz mapping of the plane, the cardinality of the preimage of a point may be estimated in terms of the characteristic constants of the mapping, that is, its co-Lipschitz and weak Lipschitz constants, namely, the maximum number of points that a mapping f can glue together does not exceedL∗f/c. The weak Lipschitz constantL∗f of a uniform mapping f is defined in the following way:
L∗f = lim
d→+∞
Ωf(d)
d <+∞ (1.4)
(see [6] for a discussion of this constant). If f is a Lipschitz mapping,L∗f does not exceed its Lipschitz constant (it can, however, be strictly less than the Lipschitz constant). If the mapping is uniform with weak Lipschitz constantL∗f and isc-co-Lipschitz, thenc≤L∗f.
In this paper, we will use the notion of the index (also called winding number) of a closed curve around a point not on the curve, and the notion ofn-dimensional Hausdorff measure:
Ᏼn(A)=sup
δ>0
inf ∞
j=1
diamCj 2
n
|A⊂∞
j=1
Cj, diamCj≤δ (1.5)
(cf. [1, 2.8.15]). Of course, the diameter in this definition is with respect to the metric given by the norm. Note thatᏴnis so normalized that the measure of the unit ball is equal to 1.
We also settle a special case of the volume ratio problem: if f is a Lipschitz quotient mapping of the plane, then for any measurable setA,λ2(f(A))/λ2(A) is bounded from below by a positive constant depending only on the Lipschitz and co-Lipschitz constants of f.
In Section3, we deal with so-called ball noncollapsing mappings (see [2,5]).
A mapping f is called ball noncollapsing, if the f-image of a ball of radiusralways contains a ball of radiusCr, whereCis a positive constant. The largest suchCis called the BNC constant of the mapping. The difference between BNC and co-Lipschitz mappings is that the ball of radiusCrcontained in f(Br(x)) need not be centered at f(x); the class of BNC mappings is actually strictly wider than that of co-Lipschitz mappings.
In [5], we proved that iff :R2→R2isL-Lipschitz andC-BNC withC/L >1/2, then f is one-to-one. The same statement can be easily shown to be true forf :R1→R1(such a mapping has to be co-Lipschitz by [5, Lemma 4], and therefore, is monotone).
However, whenC/L≤1/2, the mapping is not necessarily one-to-one (considerf(x)=
|x|onR1). In the present paper, we prove that for any pair of positive constants (C,L) the following are equivalent:
(i) there exists f :R1→R1,C-BNC andL-Lipschitz, and a pointx∈R1such that f−1(x) is infinite;
(ii)C/L <1/3.
2. Uniform co-Lipschitz mappings of the plane
In the first part of the paper, we show that under a uniform co-Lipschitz mapping, a point may have up toL∗f/cpreimages. Since for a Lipschitz mapping its weak Lipschitz constant does not exceed its Lipschitz constant, we see that Theorem2.1generalizes our previous result presented in [5, Theorem 1] (which holds for Lipschitz mappings withLinstead of L∗f).
Theorem2.1. Let · be any norm onR2. Iff: (R2, · )→(R2, · )is uniform co-Lip- schitz, withcbeing its co-Lipschitz constant andL∗f its weak Lipschitz constant, and
maxx∈R2#f−1(x)=n, (2.1)
thenc/L∗f ≤1/n.
This theorem immediately yields the existence of the following scale.
Corollary2.2. There is a scale of numbers
0<···< C2(n)<···< C2(1)=C2<1 (2.2) withC(2n)=1/(n+ 1)such that for any norm · of the plane, and any uniform co-Lipschitz mapping f: (R2, · )→(R2, · ), the conditionc/L∗f > C(2n)implies#f−1(x)≤nfor any x∈R2.
Remark 2.3. Once we have such a scale, a natural question is whether the 1/nbounds are precise. In the case of · being the Euclidean norm, the “winding mapping”φn(reiθ)= reniθhas weak Lipschitz constantnand co-Lipschitz constant 1, so the ratio of constants L∗φn/c is equal to the maximum cardinality of a point preimage, which isn.
An analogue of the winding mapping can be constructed for arbitrary norm · . One can define the argument, arg·(y), of any nonzero point y, and then setψn(rx)=r y, wherer≥0, andyis a point on∂B1such that arg·(y)=narg·(x) (see [5, Section 3]
for the description of this construction). In the situation when the unit ball is a regular polygon (or, of course, its affine equivalent), the weak Lipschitz constant ofψnis then shown to be equal ton, the co-Lipschitz constant ofψnis 1, so againL∗ψn/c=n.
We have not yet worked out this example for other norms, so despite the feeling that the estimate is sharp for any given norm (i.e., there exists a mapping f with a maximum ofnpoint preimages and the ratio of constantsL∗f/c equal ton), this question remains open.
Proof of Theorem2.1. Without loss of generality we may assume that f(0)=0. By [3]
there exist a homeomorphismh:R2→R2and a polynomialP(z) of one complex vari- able, such that
f =P◦h. (2.3)
Clearly, degP=maxx∈R2#f−1(x)=n. If n=1, then the statement is obvious. Assume n≥2.
Assume thatc/L∗f >1/n, then by rescaling without loss of generality we may assume thatL∗f <1 andc >1/n. Fix any
c1∈1 n,c
. (2.4)
Let
ε=c1−c
c , (2.5)
thenc1=c(1−ε).
Changinghby a transformation of the formh→ah+b, we may assume thath(0)=0 and the leading coefficientanofP(z) is 1. ThenP(0)= f(0)=0 andP(z) has the form zn+an−1zn−1+···+a1z.
We considerR2as the complex plane, and use the notation|x|for the absolute value of the complex numberx, which is the same as the Euclidean norm ofx∈R2.
Let{z1=0,z2,...,zk}be the set of preimages of zero underf, denote M=max
1≤i≤kzi. (2.6)
Lemma2.4. If f is such as described in the hypothesis of Theorem2.1, andc1is as in (2.4), then there exists anRsuch that for anyxwithx ≥R,f(x) ≥c1x.
Before stating the next lemma, we recall the notation Ind0γfor the index around the origin of a closed curveγ.
Lemma2.5. There existsd >1such that for anyρ > d,
Ind0f∂Bρ(0)=Ind0Ph∂Bρ(0)=n. (2.7) Lemma2.6. IfΓ: [0, 1]→R2is a closed curve withΓ(t) ≥rfor allt∈[0, 1]andInd0Γ= n, then the length ofΓin the sense of the 1-dimensional HausdorffmeasureᏴ1is at least
nᏴ1
∂Br(0). (2.8)
For the proof of Lemma2.4 see [5, Lemma 1] and [4, Lemma 1], for the proof of Lemma2.5see [5, Lemma 2], and for the proof of Lemma2.6see [5, Lemma 3].
Now we return to the proof of Theorem2.1.
ConsiderS=∂B1—the unit sphere in the norm · , as a closed, central symmetric curve in (R2,| · |). For eachεwe denote byP(ε) a polygon inscribed inSwith the follow- ing property: the Euclidean length of each arc between two adjacent vertices ofP(ε) is less thanε.
We assumed in the beginning of the proof thatL∗f <1, therefore there existsd0such thatΩf(d)< dfor alld≥d0.
Letabe the constant of equivalence between the norm · and the Euclidean norm
| · |, that is,
a−1|x| ≤ x ≤a|x| ∀x. (2.9)
Letε0=min{c1/2a,d0/a}and assumeε < ε0. Note that · -lengths of the sides of the polygonP(ε) are less than or equal toaε, which is less thand0. Let(ε) be the smallest · -length among the lengths of sides ofP(ε).
Consider the rescaling ofP(ε) by a factor ofd0/(ε), and denote the new polygon by A1A2···Am(of course, the verticesAi’s and their total numbermdepend onε). For each ione hasAi−Ai+1 ≥d0, hencef Ai−f Ai+1|| ≤Ωf(Ai−Ai+1)<Ai−Ai+1. Thus theᏴ1-lengthᏸof the broken linef(A1)f(A2)···f(Am) does not exceed theᏴ1-length of the broken lineA1A2···Am, which is not more thanᏴ1(∂Bd0/(ε))=(d0/(ε))Ᏼ1(∂B1).
Now we are going to estimate ᏸfrom below. For this purpose, we first prove that Ind0f A1f A2···f Am=nfor sufficiently smallε.
By Lemmas2.4and2.5, there existsRsuch that f(x)> c1xfor allx such that x ≥R, and Ind0f(∂Br)=nfor allr≥R.
Since(ε)≤aε(all sides of the polygonP(ε) were of · -length less than or equal to aε), one has(ε)→0 asε→0, and sod0/(ε)→+∞asε→0. Take suchε1< ε0 so that d0/(ε)> Rfor any positiveε < ε1and fix someε∈(0,ε1).
Note that the · -distance betweenAiand any point on the arcᏭi,i+1=AiAi+1of the · -sphere of radiusd0/(ε) centered at zero is less than or equal toaε·d0/(ε) (this can be easily demonstrated using triangle inequality), soᏭi,i+1⊂Baεd0/(ε)(Ai). Thus
fᏭi,i+1
⊂BΩf(aεd0/(ε))
f Ai
. (2.10)
Sinceaεd0/(ε)≥d0, the choice ofd0yields Ωf
aεd0
(ε)
<aεd0
(ε), (2.11)
and therefore
BΩf(aεd0/(ε))
f Ai⊂Baεd0/(ε)
f Ai. (2.12)
Denote the latter ball by B. Note that by the choice ofε0, the radius ofB is less than (c1/2)d0/(ε), and at the same timef Ai> c1Ai =c1d0/(ε), soB0. Therefore,
n=Ind0f ∂Bd0/(ε)=Ind0f A1f A2···f Am, (2.13) since if we replace f(Ꮽi,i+1) by the segment [f Ai,f Ai+1], the total index does not change.
Now we estimate the · -distance of each segment Ii=[f Ai,f Ai+1] to zero. The length ofIi, as we already showed is less thanAi−Ai+1 ≤aεd0/(ε), and both its ends have norm at leastc1d0/(ε). Thus dist(Ii, 0)≥(1−aε/c1)c1d0/(ε).
Hence by Lemma2.6,ᏸ≥n·(1−aε/c1)c1(d0/(ε))Ᏼ1(∂B1). Thus, we get the follow- ing inequality:
d0
(ε)Ᏼ1
∂B1
≥n·
1−aε c1
c1 d0
(ε)Ᏼ1
∂B1
, (2.14)
or, equivalently,
1≥nc1
1−aε
c1
, (2.15)
which does not hold for sufficiently smallε, since the right-hand side tends tonc1>1 as ε→0.
We arrived at this contradiction because we assumed in the beginning thatc >1/n, which enabled us to choosec > c1>1/n. Thus,c≤1/n, and the theorem is proved.
As a corollary, we obtain an interesting result: a Lipschitz quotient mapping does not collapse areas.
Theorem2.7. If f:R2→R2 is an L-Lipschitz andc-co-Lipschitz mapping with respect to the Euclidean norm, then for any measurable setA⊂R2of finite positive2-dimensional Lebesgue measureλ2(A),
c3 L ≤
λ2 f(A)
λ2(A) . (2.16)
Proof. LetJ f be the Jacobian of f (which is defined almost everywhere onR2since f is Lipschitz). By the coarea formula (see [1, 3.211]),
A
J fdx=
R2#A∩f−1(y)dy. (2.17) LetNbe the maximum cardinality of a point preimage under f. By Theorem2.1,Ndoes not exceedL/c. Since #(A∩f−1(y))=0 is equivalent toy∈ f(A), we conclude
R2#A∩f−1(y)dy≤Nλ2
f(A)≤ L
c
λ2
f(A). (2.18)
But obviously|J f| ≥c2 almost everywhere, thus the left-hand side of (2.17) is at least
c2λ2(A). Thus,λ2(f(A))≥(c3/L)λ2(A).
Note that if f is a Lipschitz quotient (Lipschitz and co-Lipschitz) and (2.16) holds for every measurable setA, then the preimage of each point is finite. In fact, any estimate
λ2
f(A)≥δλ2(A) (2.19)
for all measurableAwould imply that f−1(y) is finite for everyy∈R2.
Indeed, assume there is a point ywithN preimagesx1,x2,...,xN. Consider a radius rso small that the ballsBr(xi) are disjoint. Denote byAthe disjoint union ofBr(xi) for 1≤i≤N. Since f isL-Lipschitz, the image f(A) is a subset of BLr(y). Together with (2.19), this implies (Lr)2≥δNr2, which is equivalent toN≤L2/δ.
This means that if we find an independent way to prove the estimate (2.19) for any Lipschitz quotient mapping of the plane, this will imply the finiteness of point preim- ages and the regularity of the mapping. If this independent way of proving (2.19) works for higher dimensions, we will immediately get quasiregularity (in the sense of [7]) of Lipschitz quotient mappings fromRnto itself, which is now a challenging open question.
3. Ball noncollapsing functions
Theorem3.1. The following statements are equivalent:
(i)there exists f :R1→R1,C-BNC andL-Lipschitz, and a pointx∈R1 such that f−1(x)is infinite;
(ii)C/L <1/3.
Proof. LetC/L <1/3. Without loss of generality we may assume thatL=1 (the general case is obtained by rescaling). Then we find anA >1, such thatC=(1−A−2)/(3−A−2).
For an interval I=[a,b] in R1 define the “hat function”hI(x) by (b−a)/2− |x− (a+b)/2|. Now let the mappingζA:R1→R1be defined by
ζA(x)=
x ifx≤0,
(−1)kh[A−k,A−k+1](x) ifA−k≤x≤A−k+1,k >0 integer,
x−1 ifx >1.
(3.1)
Obviously,ζAis a 1-Lipschitz function (the simplest explanation is that its graph consists of line segments which form an angle of 45◦with thex-axis).
We will check now thatζAis BNC with constant (1−A−2)/(3−A−2).
We reformulate this as the following lemma. Denote by|I|the length of an intervalI inR1.
Lemma3.2. For any nonempty intervalIinR1,
|ζA(I)|
|I| ≥1−A−2
3−A−2 =C. (3.2)
Proof. Let
I1=I∩
−A−1
2A , 1 +A−1 2A2
, I2=I∩
− ∞,−A−1 2A
, I3=I∩
1 +A−1 2A2 ,∞
.
(3.3)
Since max[0,1]ζA=(A−1)/2A2, and min[0,1]ζA= −(A−1)/2A, we conclude that fori=j the intersection ofζA-images of two intervalsIiandIjis an empty set:
ζA>A−1
2A2 on
1 +A−1 2A2 ,∞
,
−A−1
2A ≤ζA≤A−1
2A2 on
−A−1
2A , 1 +A−1 2A2
, ζA<−A−1
2A on
− ∞,−A−1 2A
.
(3.4)
It is clear that|ζA(Ij)| = |Ij|forj=2, 3. Thus, it is enough to show that|ζA(I1)| ≥C|I1|. We will assume|I1| =0, otherwise this inequality is obvious.
LetI1=[a1,b1] andJ=I1∩[0, 1]=[a,b].
In what follows, we use the fact that if [s,t] is contained in an interval of the form [A−(k+1),A−k], then|ζA([s,t])| ≥1/2|[s,t]|by the property of the hat functionh[A−(k+1),A−k]. Case 1. If bothaandbare contained in the same interval of the form [A−(k+1),A−k], then
|ζA([a,b])| ≥(1/2)|[a,b]|.
IfJ=I1, this finishes the proof since 1/2> C.
IfJ=I1, then the only possibility is thatb1> b=1 anda1=a∈[1/A, 1]. ThenζA is negative on [a,b] and is positive on [b,b1], so|ζA(I1)|=|ζA([a,b])|+(b1−b)≥(1/2)|I1|>
C|I1|.
Case 2. Ifaandbare contained in two adjacent intervals of the form [A−(k+1),A−k], then sinceζAhas different signs on those two intervals, we conclude thatζA(J)≥1/2|J|> C|J|. IfJ=I1, then necessarilyb1> b=1 anda1=a∈[1/A2, 1/A]. Letx=1/A−aandy= b1−1. By the construction of the intervalI1,y≤(A−1)/2A2=max[A−2,A−1]ζA. Hence
[maxa1,b1]ζA=min A−1
2A2 , max{x,y}
. (3.5)
Denote this number byα. Then ζAI1=1
2
1− 1 A
+α≥1 2
1− 1
A
+x+y
=1
2I1> CI1 ifα=max{x,y}, ζA
I1=1 2
1− 1 A
+A−1 2A2 =
1 2−
1 2A2 ≥
1 3+ 1
6A− 1 2A2
=1 3
1 +A−1 2A2 −
1 A2
≥1
3I1> CI1 otherwise
(3.6)
(sinceI1⊂[1/A2, 1 + (A−1)/2A2],|I1| ≤1 + (A−1)/2A2−1/A2).
Case 3. IfJoverlaps with three intervals of the form [A−(k+1),A−k], then it must contain one of them. Assume
1
An+3≤a≤ 1 An+2< 1
An+1 ≤b≤ 1
An. (3.7)
Ifb=1, thenn=0, soa∈[1/A3, 1/A2]. Then ζA(J)=
−1 2
A−1 A ,1
2 A−1
A2
, (3.8)
therefore ζA(I1)=ζA(J) (even ifI1=J; the right-hand side is simply the image of the whole interval [−(A−1)/2A, 1 + (A−1)/2A2] containingI1), and
ζAI1=1 2
A−1
A +A−1 A2
=A2−1
2A2 . (3.9)
At the same time
I1≤1 +A−1 2A2 −
−A−1 2A
=3A2−1
2A2 , (3.10)
so,|ζA(I1)| ≥(A2−1)/(3A2−1)· |I1|.
If, however,b <1, thenI1=J. Without loss of generality we may assume thatζA>0 on the interval (1/An+2, 1/An+1), which is contained inJ=I1=[a,b].
Letx=1/An+2−aandy=b−1/An+1. It is clear that max[a,b]ζA=(A−1)/2An+2. (1) Ify≥(A−1)/2An+1, then min[a,b]ζA= −(A−1)/2An+1, and so
ζAI1= A−1
2An+2+ A−1 2An+1=
A2−1
2An+2 (3.11)
and |I1| ≤1/An. We see that |ζA(I1)|/|I1| ≥(A2−1)/2A2 ≥(A2−1)/(3A2−1) (since A >1).
(2) Ifx≤y <(A−1)/2An+1, then min[a,b]ζA= −y, therefore ζA
I1= A−1
2An+2+y≥1 2
A−1 An+2 +x+y
=1
2I1> CI1. (3.12) (3) Ify < x≤(A−1)/2An+3we get min[a,b]ζA= −x, and
ζAI1= A−1
2An+2+x≥1 2
A−1 An+2 +x+y
=1
2|I1|. (3.13) (4) Ify <(A−1)/2An+1,y <x, andx >(A−1)/2An+3, then min[a,b]ζA=−(A−1)/2An+3. In this case
ζAI1= A−1
2An+2+ A−1 2An+3, I1=A−1
An+2 +x+y.
(3.14)
We check that
ζA
I1≥1 3
A−1 An+2 +x+y
. (3.15)
Since y < x≤(A−1)/An+3, we conclude that the right-hand side is not greater than (A−1)/3An+3(A+ 2). The left-hand side is equal to (A−1)/2An+3(A+ 1). So it is enough to check that (A+ 2)/3≤(A+ 1)/2, which is true forA >1.
Case 4. IfJ=[a,b] overlaps with at least four intervals of the form [A−(k+1),A−k], then we consider four possibilities.
(4A)a=0 andb=1. In this case|I1| ≤1 + (A−1)/2A+ (A−1)/2A2=(3A2−1)/2A2 and |ζA(I1)| =(A−1)/2A+ (A−1)/2A2=(A2−1)/2A2. So, |ζA(I1)|/|I1| ≥(A2−1)/
(3A2−1).
(4B)a >0 andb=1. In this caseζA(J)=ζA[a, 1]=ζA[0, 1], since the minimum ofζA
on [a, 1] is attained at the point (1/2)(1/A+ 1) and the maximum—at (1/2) (1/A2+ 1/A).
So,|ζA(I1)| = |ζA[0,b1]| ≥(A2−1)/(3A2−1)|[0,b1]| ≥(A2−1)/(3A2−1)|I1|.
(4C)a >0 andb <1. This case follows from the next one, (4D), in the same way as (4B) follows from (4A): we use thatI1=Jand|ζA(I1)| =ζA|[0,b]|.
(4D)a=0 andb <1. AssumeζA(b)≤0 andb∈[1/An+1, 1/An]. (The caseζA(b)>0 will be treated later in (4D-V) and (4D-VI).) Then
maxJ ζA=max
I1 ζA=1 2
1 An+1−
1 An+2
. (3.16)
Denotex= −a1andy=b−1/An+1.
(4D-I)x≥(A−1)/2An+1. Then minI1ζA= −x. So, ζA
I1=x+ A−1
2An+2, (3.17)
and|I1| ≤x+ 1/An.
In order to prove|ζA(I1)| ≥C|I1|, it suffices to show that x+ A−1
2An+2≥ A2−1 3A2−1
x+ 1
An
. (3.18)
This is equivalent to x
1− A2−1 3A2−1
≥ A2−1 3A2−1·
1 An−
A−1
2An+2. (3.19)
Rewriting this inequality, we get x 2A2
3A2−1 ≥
A−1 3A2−12An+2
2A3−A2+ 1. (3.20)
Sincex≥(A−1)/2An+1, the left-hand side is at least ((A−1)/2An+1)·(2A2/(3A2−1)), so it is enough to prove that
2A2≥ 1 A
2A3−A2+ 1, (3.21)
which is true forA >1.
(4D-II)x <(A−1)/2An+1,y≥(A−1)/2An+1. Then minI1ζA= −(A−1)/2An+1. There- fore,
ζAI1= A−1
2An+2+ A−1 2An+1=
A2−1
2An+2, (3.22)
and|I1| ≤x+ 1/An. We want to show that A2−1
2An+2 ≥ A2−1 3A2−1
x+ 1
An
, (3.23)
which is equivalent to
1 2An+2 ≥
1 3A2−1
x+ 1
An
. (3.24)
Sincex <(A−1)/2An+1, the right-hand side is less than or equal to 1
3A2−1· 3A−1
2An+1 = 1 2An+2
3A2−A 3A2−1
≤ 1
2An+2, (3.25)
sinceA >1.
(4D-III)x,y <(A−1)/2An+1, and max{x,y}≥(A−1)/2An+3. Then minI1ζA=−max{x, y}, so|ζA(I1)| =max{x,y}+ (A−1)/2An+2and|I1| =x+y+ 1/An+1.
We want to show that
max{x,y}+ A−1 2An+2 ≥
A2−1 3A2−1
x+y+ 1 An+1
. (3.26)
Replacingx+yby 2 max{x,y}, we get a stronger inequality max{x,y}
1−2 A2−1 3A2−1
≥ A2−1 3A2−1·
1 An+1−
A−1
2An+2, (3.27) which is equivalent to
max{x,y} A2+ 1 3A2−1≥
A−1 (3A2−1)2An+2
2A2+ 2A−3A2+ 1. (3.28) Since max{x,y} ≥(A−1)/2An+3, it is enough to check that
A−1 2An+3·
A2+ 1≥ A−1 2An+2
−A2+ 2A+ 1. (3.29)
The latter inequality is equivalent toA2+ 1≥A(−A2+ 2A+ 1), which is the same asA3− A2≥A−1, true forA >1.
(4D-IV) max{x,y}<(A−1)/2An+3. Then minI1ζA= −(A−1)/2An+3, so |ζA(I1)| = (A−1)/2An+3+ (A−1)/2An+2and
|I1| =x+y+ 1
An+1 ≤2·1 2
A−1 An+3 + 1
An+1. (3.30)
The inequality |ζA(I1)| ≥(A2−1)/(3A2−1)|I1| then becomes a particular case of (4D-III) (x=y=(A−1)/2An+3).
This finishes the proof of (4D) under the assumptionζA(b)≤0.
Now assumeζA(b)>0.
In order to determine the maximum and minimum values ofζAon the intervalI1, we must comparexwith (A−1)/2An+2, andywith (A−1)/2An+1.
(4D-V) y≥(A−1)/2An+1. Then one has minI1ζA= −max{x, (A−1)/2An+2} and maxI1ζA=(A−1)/2An+1. So|ζA(I1)| =max{x, (A−1)/2An+2}+ (A−1)/2An+1, and
I1≤x+ 1 An≤max
x,A−1
2An+2
+ 1
An. (3.31)
It is enough to prove that
α+ A−1 2An+1≥
A2−1 3A2−1
α+ 1
An
, (3.32)