• 検索結果がありません。

UNIFORM CO-LIPSCHITZ MAPPINGS OF THE PLANE

N/A
N/A
Protected

Academic year: 2022

シェア "UNIFORM CO-LIPSCHITZ MAPPINGS OF THE PLANE"

Copied!
20
0
0

読み込み中.... (全文を見る)

全文

(1)

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 :XY between two normed spacesXandY. The function Ωf(d)= sup

xxXd

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 asd0. In this case the modulus of continuity is a sub- additive monotone continuous function. The definition ofΩf implies that f(Br(x)) Bf(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

(2)

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 exceedLf/c. The weak Lipschitz constantLf of a uniform mapping f is defined in the following way:

Lf = lim

d+

f(d)

d <+ (1.4)

(see [6] for a discussion of this constant). If f is a Lipschitz mapping,Lf does not exceed its Lipschitz constant (it can, however, be strictly less than the Lipschitz constant). If the mapping is uniform with weak Lipschitz constantLf and isc-co-Lipschitz, thencLf.

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:

n(A)=sup

δ>0

inf

j=1

diamCj 2

n

|A

j=1

Cj, diamCjδ (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 :R2R2isL-Lipschitz andC-BNC withC/L >1/2, then f is one-to-one. The same statement can be easily shown to be true forf :R1R1(such a mapping has to be co-Lipschitz by [5, Lemma 4], and therefore, is monotone).

However, whenC/L1/2, 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 :R1R1,C-BNC andL-Lipschitz, and a pointxR1such that f1(x) is infinite;

(ii)C/L <1/3.

(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 toLf/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 Lf).

Theorem2.1. Let · be any norm onR2. Iff: (R2, · )(R2, · )is uniform co-Lip- schitz, withcbeing its co-Lipschitz constant andLf its weak Lipschitz constant, and

maxx∈R2#f1(x)=n, (2.1)

thenc/Lf 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/Lf > C(2n)implies#f1(x)nfor any xR2.

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(re)= reniθhas weak Lipschitz constantnand co-Lipschitz constant 1, so the ratio of constants Lφn/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, wherer0, 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 againLψn/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 constantsLf/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:R2R2and a polynomialP(z) of one complex vari- able, such that

f =Ph. (2.3)

Clearly, degP=maxx∈R2#f1(x)=n. If n=1, then the statement is obvious. Assume n2.

(4)

Assume thatc/Lf >1/n, then by rescaling without loss of generality we may assume thatLf <1 andc >1/n. Fix any

c11 n,c

. (2.4)

Let

ε=c1c

c , (2.5)

thenc1=c(1ε).

Changinghby a transformation of the formhah+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+an1zn1+···+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 ofxR2.

Let{z1=0,z2,...,zk}be the set of preimages of zero underf, denote M=max

1ikzi. (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 anyxwithxR,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 Hausdorffmeasure1is 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 thatLf <1, therefore there existsd0such thatΩf(d)< dfor alldd0.

Letabe the constant of equivalence between the norm · and the Euclidean norm

| · |, that is,

a1|x| ≤ xa|x| ∀x. (2.9)

(5)

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 hasAiAi+1d0, hencef Aif Ai+1|| ≤f(AiAi+1)<AiAi+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))1(∂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 xR, and Ind0f(∂Br)=nfor allrR.

Since(ε)(all sides of the polygonP(ε) were of · -length less than or equal to aε), one has(ε)0 asε0, and sod0/(ε)+asε0. 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+1Baεd0/(ε)(Ai). Thus

fi,i+1

Bf(aεd0/(ε))

f Ai

. (2.10)

Sinceaεd0/(ε)d0, the choice ofd0yields Ωf

aεd0

(ε)

<aεd0

(ε), (2.11)

and therefore

Bf(aεd0/(ε))

f AiBaεd0/(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 thanAiAi+1aεd0/(ε), and both its ends have norm at leastc1d0/(ε). Thus dist(Ii, 0)(1aε/c1)c1d0/(ε).

Hence by Lemma2.6,ᏸn·(1aε/c1)c1(d0/(ε))Ᏼ1(∂B1). Thus, we get the follow- ing inequality:

d0

(ε)1

∂B1

n·

1 c1

c1 d0

(ε)1

∂B1

, (2.14)

(6)

or, equivalently,

1nc1

1

c1

, (2.15)

which does not hold for sufficiently smallε, since the right-hand side tends tonc1>1 as ε0.

We arrived at this contradiction because we assumed in the beginning thatc >1/n, which enabled us to choosec > c1>1/n. Thus,c1/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:R2R2 is an L-Lipschitz andc-co-Lipschitz mapping with respect to the Euclidean norm, then for any measurable setAR2of 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#Af1(y)dy. (2.17) LetNbe the maximum cardinality of a point preimage under f. By Theorem2.1,Ndoes not exceedL/c. Since #(Af1(y))=0 is equivalent toy f(A), we conclude

R2#Af1(y)dy2

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 f1(y) is finite for everyyR2.

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 1iN. Since f isL-Lipschitz, the image f(A) is a subset of BLr(y). Together with (2.19), this implies (Lr)2δNr2, which is equivalent toNL2/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.

(7)

3. Ball noncollapsing functions

Theorem3.1. The following statements are equivalent:

(i)there exists f :R1R1,C-BNC andL-Lipschitz, and a pointxR1 such that f1(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=(1A2)/(3A2).

For an interval I=[a,b] in R1 define the “hat function”hI(x) by (ba)/2− |x (a+b)/2|. Now let the mappingζA:R1R1be defined by

ζA(x)=

x ifx0,

(1)kh[Ak,Ak+1](x) ifAkxAk+1,k >0 integer,

x1 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 45with thex-axis).

We will check now thatζAis BNC with constant (1A2)/(3A2).

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| 1A2

3A2 =C. (3.2)

Proof. Let

I1=I

A1

2A , 1 +A1 2A2

, I2=I

− ∞,A1 2A

, I3=I

1 +A1 2A2 ,

.

(3.3)

Since max[0,1]ζA=(A1)/2A2, and min[0,1]ζA= −(A1)/2A, we conclude that fori=j the intersection ofζA-images of two intervalsIiandIjis an empty set:

ζA>A1

2A2 on

1 +A1 2A2 ,

,

A1

2A ζAA1

2A2 on

A1

2A , 1 +A1 2A2

, ζA<A1

2A on

− ∞,A1 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.

(8)

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),Ak], then|ζA([s,t])| ≥1/2|[s,t]|by the property of the hat functionh[A(k+1),Ak]. Case 1. If bothaandbare contained in the same interval of the form [A(k+1),Ak], 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])|+(b1b)(1/2)|I1|>

C|I1|.

Case 2. Ifaandbare contained in two adjacent intervals of the form [A(k+1),Ak], 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/Aaandy= b11. By the construction of the intervalI1,y(A1)/2A2=max[A2,A1]ζA. Hence

[maxa1,b1]ζA=min A1

2A2 , max{x,y}

. (3.5)

Denote this number byα. Then ζAI1=1

2

1 1 A

+α1 2

1 1

A

+x+y

=1

2I1> CI1 ifα=max{x,y}, ζA

I1=1 2

1 1 A

+A1 2A2 =

1 2

1 2A2

1 3+ 1

6A 1 2A2

=1 3

1 +A1 2A2

1 A2

1

3I1> CI1 otherwise

(3.6)

(sinceI1[1/A2, 1 + (A1)/2A2],|I1| ≤1 + (A1)/2A21/A2).

Case 3. IfJoverlaps with three intervals of the form [A(k+1),Ak], then it must contain one of them. Assume

1

An+3a 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

A1 A ,1

2 A1

A2

, (3.8)

therefore ζA(I1)=ζA(J) (even ifI1=J; the right-hand side is simply the image of the whole interval [(A1)/2A, 1 + (A1)/2A2] containingI1), and

ζAI1=1 2

A1

A +A1 A2

=A21

2A2 . (3.9)

(9)

At the same time

I11 +A1 2A2

A1 2A

=3A21

2A2 , (3.10)

so,|ζA(I1)| ≥(A21)/(3A21)· |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+2aandy=b1/An+1. It is clear that max[a,b]ζA=(A1)/2An+2. (1) Ify(A1)/2An+1, then min[a,b]ζA= −(A1)/2An+1, and so

ζAI1= A1

2An+2+ A1 2An+1=

A21

2An+2 (3.11)

and |I1| ≤1/An. We see that |ζA(I1)|/|I1| ≥(A21)/2A2 (A21)/(3A21) (since A >1).

(2) Ifxy <(A1)/2An+1, then min[a,b]ζA= −y, therefore ζA

I1= A1

2An+2+y1 2

A1 An+2 +x+y

=1

2I1> CI1. (3.12) (3) Ify < x(A1)/2An+3we get min[a,b]ζA= −x, and

ζAI1= A1

2An+2+x1 2

A1 An+2 +x+y

=1

2|I1|. (3.13) (4) Ify <(A1)/2An+1,y <x, andx >(A1)/2An+3, then min[a,b]ζA=−(A1)/2An+3. In this case

ζAI1= A1

2An+2+ A1 2An+3, I1=A1

An+2 +x+y.

(3.14)

We check that

ζA

I11 3

A1 An+2 +x+y

. (3.15)

Since y < x(A1)/An+3, we conclude that the right-hand side is not greater than (A1)/3An+3(A+ 2). The left-hand side is equal to (A1)/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),Ak], then we consider four possibilities.

(4A)a=0 andb=1. In this case|I1| ≤1 + (A1)/2A+ (A1)/2A2=(3A21)/2A2 and |ζA(I1)| =(A1)/2A+ (A1)/2A2=(A21)/2A2. So, |ζA(I1)|/|I1| ≥(A21)/

(3A21).

(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]| ≥(A21)/(3A21)|[0,b1]| ≥(A21)/(3A21)|I1|.

(10)

(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=b1/An+1.

(4D-I)x(A1)/2An+1. Then minI1ζA= −x. So, ζA

I1=x+ A1

2An+2, (3.17)

and|I1| ≤x+ 1/An.

In order to prove|ζA(I1)| ≥C|I1|, it suffices to show that x+ A1

2An+2 A21 3A21

x+ 1

An

. (3.18)

This is equivalent to x

1 A21 3A21

A21 3A21·

1 An

A1

2An+2. (3.19)

Rewriting this inequality, we get x 2A2

3A21

A1 3A212An+2

2A3A2+ 1. (3.20)

Sincex(A1)/2An+1, the left-hand side is at least ((A1)/2An+1)·(2A2/(3A21)), so it is enough to prove that

2A2 1 A

2A3A2+ 1, (3.21)

which is true forA >1.

(4D-II)x <(A1)/2An+1,y(A1)/2An+1. Then minI1ζA= −(A1)/2An+1. There- fore,

ζAI1= A1

2An+2+ A1 2An+1=

A21

2An+2, (3.22)

and|I1| ≤x+ 1/An. We want to show that A21

2An+2 A21 3A21

x+ 1

An

, (3.23)

which is equivalent to

1 2An+2

1 3A21

x+ 1

An

. (3.24)

(11)

Sincex <(A1)/2An+1, the right-hand side is less than or equal to 1

3A21· 3A1

2An+1 = 1 2An+2

3A2A 3A21

1

2An+2, (3.25)

sinceA >1.

(4D-III)x,y <(A1)/2An+1, and max{x,y}≥(A1)/2An+3. Then minI1ζA=−max{x, y}, so|ζA(I1)| =max{x,y}+ (A1)/2An+2and|I1| =x+y+ 1/An+1.

We want to show that

max{x,y}+ A1 2An+2

A21 3A21

x+y+ 1 An+1

. (3.26)

Replacingx+yby 2 max{x,y}, we get a stronger inequality max{x,y}

12 A21 3A21

A21 3A21·

1 An+1

A1

2An+2, (3.27) which is equivalent to

max{x,y} A2+ 1 3A21

A1 (3A21)2An+2

2A2+ 2A3A2+ 1. (3.28) Since max{x,y} ≥(A1)/2An+3, it is enough to check that

A1 2An+3·

A2+ 1 A1 2An+2

A2+ 2A+ 1. (3.29)

The latter inequality is equivalent toA2+ 1A(A2+ 2A+ 1), which is the same asA3 A2A1, true forA >1.

(4D-IV) max{x,y}<(A1)/2An+3. Then minI1ζA= −(A1)/2An+3, so |ζA(I1)| = (A1)/2An+3+ (A1)/2An+2and

|I1| =x+y+ 1

An+1 2·1 2

A1 An+3 + 1

An+1. (3.30)

The inequality |ζA(I1)| ≥(A21)/(3A21)|I1| then becomes a particular case of (4D-III) (x=y=(A1)/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 (A1)/2An+2, andywith (A1)/2An+1.

(4D-V) y(A1)/2An+1. Then one has minI1ζA= −max{x, (A1)/2An+2} and maxI1ζA=(A1)/2An+1. So|ζA(I1)| =max{x, (A1)/2An+2}+ (A1)/2An+1, and

I1x+ 1 Anmax

x,A1

2An+2

+ 1

An. (3.31)

It is enough to prove that

α+ A1 2An+1

A21 3A21

α+ 1

An

, (3.32)

参照

関連したドキュメント

Finally, we give an example to show how the generalized zeta function can be applied to graphs to distinguish non-isomorphic graphs with the same Ihara-Selberg zeta

The inclusion of the cell shedding mechanism leads to modification of the boundary conditions employed in the model of Ward and King (199910) and it will be

Incidentally, it is worth pointing out that an infinite discrete object (such as N) cannot have a weak uniformity since a compact space cannot contain an infinite (uniformly)

It is suggested by our method that most of the quadratic algebras for all St¨ ackel equivalence classes of 3D second order quantum superintegrable systems on conformally flat

Answering a question of de la Harpe and Bridson in the Kourovka Notebook, we build the explicit embeddings of the additive group of rational numbers Q in a finitely generated group

It turns out that the symbol which is defined in a probabilistic way coincides with the analytic (in the sense of pseudo-differential operators) symbol for the class of Feller

Then it follows immediately from a suitable version of “Hensel’s Lemma” [cf., e.g., the argument of [4], Lemma 2.1] that S may be obtained, as the notation suggests, as the m A

In our previous paper [Ban1], we explicitly calculated the p-adic polylogarithm sheaf on the projective line minus three points, and calculated its specializa- tions to the d-th