A simple pro of for monotonicit y of en trop y in the
quadratic family
Masato TSUJI I
⁄
February 2,1999
Abstract
W e giv e a simple pro offor monotonicit y oftop ological en trop y inthe
quadratic family .W e ua sp ectral prop ert y ofa Ruelle operator.
1In tro duction
In this paper,we giv e a simple pro ofofthefollo wing theorem,due to Sulliv an,Douady-Hubbardand Milnor([6]),on the dynamicsof real
quadaratic maps Q t (x )=t ¡x 2
;R !R .Theorem 1The top olo g ic a lentr o py h top (Q t )of Q t is monotoneincr e as-ingwith r esp e cttothep ar ameter t .Inthepro ofgiv eninction 2,we reduce theorem 1tothefollo wingclaim on beha vior oftheorbit ofthecritical poin t.Theorem 2Iftheorbit o fthecritic alp oint 0isp erio dicwith primep e-rio d n for Q t (i.e.Q n t (0)=0and Q i t (0)6
=0for 0<i<n ),then @
@s
(Q n s (0))j s =t D Q t (Q t (0))
>0
wher e D denote hesp ac e c o or dinate x .
Then we pro ve this claim by makinguofa sp ectral prop ert y ofa Ruelle
operator.Inction 3,we pro ve thefollo winganalogous r esult inCollet-Eckmann ca.
Theorem 3If Q t satisfles Col let-Eckmann c ondition,
that is,lim inf n !1
n
p j D Q n t (
Q t (0))j >1;(1)
thenwe have
lim
n !1
@
@s
(Q n s (0))j s =t D Q n ¡1
t (Q t (0))
>0:(2)
⁄Departmen t ofMathematics,H okkaidoUniv ersit y ,Sapporo,060-0810,J AP AN,
e -mail:tsujii@math.sci.hokudai.ac.jp
1
Theorem3,together witha version o fJak obson’s t heorem in[8,theorem 1],yields
Corollary4If1=4<t<2is a p ar ameter for which Q t satisfles(1)and
lim
n!1
n¡1log j Q n t(0)j=0;(3)
then t is a density p oint ofthe t S ofp ar ameters s for which Q s satisfles (1)and(3).More over,
lim †!+0log Leb([t¡†;t+†]nS)
log†
=2
wher e Leb denotes t heLeb esgue measur e.
Anotheraim ofthis paperisto clarify thecorresp ondence bet ween monotonicit
y ofthequadaratic family a nd that ofthefamily o ften tmaps
T t(x)=1¡t j x j.As we shall sho w in thelast ction,we can pro ve monotonicit
最大的草鱼y ofthefamily often t maps inquite analogous m anner.
Ac knowledgmen t:The author thanks Duncan Sandsfor his suggesting theorem3.
2Proofoftheorem1and2 Letusremembersomedeflnitions
andfacts f romthe kneading t heory[1
,6]. Let f:R!R be a unimo dalmap with0its turning poin ,a con tin u ousmap that is strictly
monotoneincreasing
on theleft o f0and strictly
monotonedecreasing on therigh t of0.The kneading in varian t
of f isan inflnite s equence K(f)=(e1;e2;¢¢¢)ofsymbols L;C and R,
deflnedby
e k=8
<
:
L;if f k(0)<0;
C;if f k(0)=0;
R;if f k(0)>0:
Forquences ofsymbols L;C and R,we deflnetheso-called
signed lexi-cographical
order`inthefollo
wingmanner.First we deflne L`C`R for single symbols.Fort wo difieren t quences
I=(e1;e2;e3;¢¢¢;e n;¢¢¢);
I0=(e01;e02;e03;¢¢¢;e0n;¢¢¢);
withthesame length,l et n be thesmallest num berfor whic h e n6=e0n. When thenum berofthesymbol R in f e1;e2;¢¢¢;e n¡1g is ev en,we deflne I`I0if and onlyif e n`e0n.On theother hand,when thenum berof thesymbol R in f e1;e2;¢¢¢;e n¡1g is odd,we deflne I`I0if and only if e n´e0n.Ob viously t his isa total order on thespace ofquences o fa flxed(flnite o rinflnite)length.
The top ological
en trop y ofa unimo dal map dep ends only on its knead-ingin varian t,and thedep endence is monotonewithresp e ct totheorder above.Thustheorem1follo
ws from
Theorem5K(Q t)ismonotoneincr e asing with r esp e ctto t.
2
But this follo ws from theorem2.In fact,l et K n (Q t )be thetruncated quence ofthekneading in varian t K (Q t )oflength n .Then K n (Q t )is
lo cally c onstan t withresp e ctto t unless Q i
t (0)=0for some 0<i •n .
When Q i
t (0)=0forsome 0<i •n ,thederiv ativ e D Q i ¡1t (Q t (0))is positiv e if and onlyif thenum berofthesymbol R in K i ¡1(Q t )is ev en.From thedeflnition ofthesigned lexicographical order,t heorem 2implies that K n (Q t )isincreasing ateac h ofsuc h parameters.Hence K n (Q t )is monotoneincreasing withresp e ctto t .Since this holds for ev ery n ,the kneading in varian t K (Q t )is monotoneincreasing.
Now we pro ve theorem2.Assume Q n t (0)=0and Q i
t (0)6=0for 0<i <n .Let Q :C !C be thecomplexextension of Q t .A Ruelle operator R is deflnedby
R (ˆ)(z )=
X
Q (y )=z
ˆ(y )
for functions ˆon thecomplex plane.Let ´k (z )=1(z ¡Q k (0))and let
S be thelinear space offunctions spanned by ´k ,k =1;2;¢¢¢;n .W e ha ve R (´k )=
X
y =§
p t ¡z
1
4y 2(y ¡Q k (0))=¡2Q k (0)4y 2((Q k (0))
2¡t +z )=¡(Q k (0))2
2y 2Q k (0)((Q k (0))
2¡t +z )=1
2Q k (0)t ¡Q k +1(0)(z ¡t )(z ¡Q k +1(0))=
´k +1
D Q (Q k (0))¡´1D Q (Q k (0))
(4)
for 1•k <n ,and R (´n )=0.Hence theRuelle operator R prerv es thespace S .The repren tation m atrix for R j S withresp e ct tothebasis f ´k g n
k =1is oftheform
0B B B B B @
¡
1
D Q (Q (0))
¡
1
D Q (Q 2(0))
¡
1
D Q (Q 3(0))
¡
1
D Q (Q 4(0))
¢¢¢
1
¢¢¢01
D Q (Q 2(0))
0¢¢¢001
30¢¢¢...
...悲哀反义词
...
...
.
..
1
C C
C C C A :(5)
W e flnd
det (Id ¡R j S )=det 0
B B
B B B @
1+1
下列表述正确的有D Q (Q (0))
1
D Q (Q 2(0))
1
D Q (Q 3(0))
¢¢¢
¡1
D Q (Q (0))10¢¢¢0¡
1
D Q (Q 2(0))
1¢¢¢00¡
1
D Q (Q 3(0))
¢¢¢......
...
...
1
C C C C C A =
n ¡1X k =0k Y j =1
1=n ¡1X k =0
1=@@s
(Q n s (0))j s =t D Q t (Q t (0))
:(6)
Since theen tries ofthematrix (5)is real,theorem 2follo ws from
3
Lemma 6The sp e ctr a lr adius of R j S issmal
ler than1.pr o of.Letusconsider thePerron-F rob enius operator
P (ˆ)(x )=
X
Q (y )=x
ˆ(y )
j D Q (y )j
2:Ob viously w e ha ve j R k (ˆ)(x )j•P k (j ˆj )(x )for k >0.Tak e a large disk
D suc h that Q ¡1(D )‰D ,and put A =D ¡Q ¡1
(D ).Then
k ˆk =
Z
A
j ˆjj d z j
2
is a norm on theflnite-dimensional vector space S .W e ha ve,for ˆ2S ,
1X k =0
Z A
P k (j ˆj )j dz j 2
=
1X k =0
Z Q ¡k (A )
j ˆjj d z j 2
•
函数求导Z
D
j ˆjj d z j 2
<1:
Thisimplies k R k (ˆ)k !0and,hence,thelemma.
2
W e flnished thepro ofoftheorem1and 2.Here we giv e a remark
on therelation bet weentheabove pro ofand that in[6].In theoriginal pro ofin[6],theyconsider theso-call
ed Thurston map thatacts on the Teic h m uller s pace offlnitely-p oin tedsphere.The poin t ofthepro ofis thatThurston map isa con traction withresp e ctto Teic h m uller m etric.The space S inourpro ofabove corresp ondstothecotangen tspace ofthe Teic h m uller s pace ata flxedpoin t,and theoperator R corresp ondstothe action ofThurston map on thecotangen t space.(cf.[3],[7])Henceour pro ofis a lo cal version o fthat in[6]ina n.W e sho wed that thelo cal con traction wanough.
telnet安装The Ruelle operator R inourpro ofis in teresting initlf.In[4]and [5],Levin,S odinand Yuditski studied t heprop erties of R inmore general tting,and ha ve obtained f orm ulae similar to(6).
3Proofoftheorem 3
The pro ofoftheorem3issimilar to thatfor theorem2except for the poin t that thespace offunction we consider isinflnite-dimensional.Let usassume that Q t satisfles Collet-Ec kmann condition (1)andc hoo r >1suc h that lim inf n !1
n p
j D Q n t (
Q t (0))j >r :Belo w we assumethatthecritical poin t isnot pre-p e rio dic.The ca
when thecritical poin tis pre-p e rio dic is treated attheendofthis ction.Letusconsider a Banac h space
S =
(
(a k )1
k =1
flflflflfl1X
k =1
j a k j¢j D Q k ¡1
t
(Q t (0))j ¢r ¡k <1)4
witha norm jjj (a k )1
k =1jjj =P
1
k =1
j a k j ¢j D Q k ¡1t (Q t (0))j ¢r ¡k .Let ´k (z )=1=(z ¡Q k t (0)).F oreac h (a k )1
k =12S ,we consider
a holomorphic function “((a k )1
k =1
)(z )=1X k =1
a k ´k (z )
deflnedon C n R .
Lemma 7“isan inje ctive m ap fr o m S tothet ofholomorphic func-tions deflne d on C n R .
pr o of.Assume that(a k )1
k =12S isnotzero.Tak e a n 6
=0and m suc h that P k>m j
a k j <j a n j .Then,for y 2R ¡f 0g ,flfl“((a k )1k =1)(Q n t
(0)+y p flfl‚j a n j y
¡X k>m
j a k j y
¡X
k •m;k 6=n
j a k j
j Q t (0)¡Q t (0)
j Since thelast termisboundedas y !0,we ethat“((a k )1
k =1)isnot constan t 0.2
W e consider theRuelle operator R as inthelast ction.From (4)and thedeflnition ofthenorm jjj ¢jjj ,itiasyto ethat R prerv es theimage“(S )and induce a con tin u ouslinear map ':S !S suc h that “–'=R –“.Remark that 'is oftheform(5).
Lemma 8The sp e ctrumof 'c ontaine d in fj z j >r ¡1g isdiscr eteand thec orr esp onding eigensp ac e isflnite-dimensional.pr o of.Let …1:S !S be thepro jection to theflrst componen t,that
is,…1((a k )1k =1)=(a 1;0;0;¢¢¢).W e ha ve k (id ¡…1)–'k •r ¡1
fromthe deflnition ofthenorm.Since theoperator …1–'has one dimensional range,thelemma follo ws fromthegeneral s p ectral t heory .(See[2],pp 709.)2As inthelast ction,we tak e ts D and A on thecomplexplane.Then theoperator 'con tracts t henorm that is induced fromthenorm
k ˆk =
Z
A ¡R
j ˆjj d z j
2
on “(S ),through “.Hence thediscrete sp ectrumintheabove lemma
should be con tained intheunit disk fj z j <1g .Therefore Prop osition 9The sp e ctr a lr adius of 'issmal ler than 1.
Let …n :S !S be thepro jection totheflrst n componen ts.W e appro xi-mate 'by a quence o foperators 'n =…n –'–…n .Lemma 10Ther e exist N >0,M >0and 0<…<1such that jjj 'k n jjj
<M …k
for any n >N and k >0.
5
pr o of.Let B n ='n +(Id ¡…n )–'.Since 'm n =…n –B m
n ,we ha
ve jjj 'm n jjj •jjj B m n jjj :For a =(a k )1k =12S ,we ha
ve jjj '(a )¡B n (a )jjj =jjj …1–'–(Id ¡…n )(
a )jjj •P k>n j a k
=D Q t (Q k t (0))j ¢r ¡1
•P
k>n flflfla k D Q k ¡1t (Q t (0))¢r ¡k ¢r k ¡1D Q k t (Q t (0))fl
flfl•jjj a jjj ¢sup k>n
n flflflr
k ¡1D Q k t
(Q t (0))fl
flflo :Hence jjj '¡B n jjj !0as n !1.Now thelemma follo ws from
lim
m !1
m
p
jjj 'm jjj <1;
whic h is equiv a len t totheconclusion ofprop osition 9.([2,pp.567])
2
By calculation,we e
det (Id ¡'n )=
@
@s
(Q n +1s (0))j s =t D Q t (Q t (0))
白居易的代表作
:(7)
On theother hand,for su–cien t ly large
n ,we ha ve det (Id ¡'n )=exp(Trace log ('n ))=exp ˆ
¡
1X k =1
k ¡1Trace ('k n )!
:
Notice that the p -th diagonal e lemen t ofthematrix 'k n is
zero for p >k becau ofthesp ecial f ormof'.Hencewe ha ve
j Trace ('k n )j•k jjj 'k n jjj
<kM …k for n ‚N ,and obtain
lim
n !1
@
如何看自己的五行@s
(Q n s (0))j s =t D Q n ¡1
t
(Q t (0))‚exp
ˆ
1
X k =1
¡M …k
!
>0:
W e flnished thepro ofoftheorem 3intheca when thecritical poin t is notpre-p e rio dic.When thecritical poin t is pre-p e rio dic,thespace S of functions spanned by ´k ,k =1;2;¢¢¢,is flnite-dimensional.W e cantak e
unique 0<k <‘suc h that Q ‘t (0)=Q k
t (0)and that ´i ,1•i<‘,forma
basis ofthespace
S .By calculation,we e det (Id ¡R j S )=lim
北京道观n !1
@
@s
(Q n s (0))j s =t D Q t (Q t (0))
¢
1¡1
D Q t (Q k t
(0))¶
:From theCollet-Ec kmann condition (1),all theperio dicpoin t of Q t is
hyperb olic r ep elling.Hence j D Q ‘¡k t (Q k
t (0))j is smaller than1.By arguing asinthelast s ection,we e that thesp ectral radius o f R j S is smaller than 1and obtain (2).W e flnished thepro of.
6
