Ergodic Schrödinger operators and its relatedtopics

Shinichi KOTANI

Nanjing University April-May, 2019

Abstract

In the �rst half of this lecture is to provide a basic knowledge forergodic one dimensional Schrödinger operators which will be needed forthe study of the KdV equation starting from ergodic initial data. In thecourse of the lecture history of this area and open problems will be alsomentioned. In the second half of the lecture the construction of solutionsto the KdV equation with ergodic initial data will be given.

1 Introduction

Schrödinger operators with ergodic potentials are physical models to describethe movement of electrons in matters containing impurities, which was proposedby physicist P.W. Anderson [1] in 1958.Ergodic potential8>>>><>>>>: : set, F : �-�eld on , probability measure � on (;F)Rd 3 x =) Tx : ! F-measurable transformation satisfying Tx+y = TxTyergodicity: � (TxAA) = 0 for any x 2 Rd =) �(A) = 0 or 1V : bounded and real valued F-measurable function on ergodic potential: q! (x) = Q (Tx!) x 2 Rd

Schrödinger operator with ergodic potential

L! = ��+ q!, � = @2x1 + � � �+ @2xd

Discrete model8>>>><>>>>: : set, F : �-�eld on , probability measure � on (;F)Zd 3 x =) Tx : ! F-measurable transformation satisfying Tx+y = TxTyergodicity: � (TxAA) = 0 for any x 2 Zd =) �(A) = 0 or 1Q : bounded and real valued F-measurable function on ergodic potential: q! (x) = Q (Tx!) , x 2 Zd

L! = ��d + q!, �d is discrete Laplacian

Example 1 Periodic potential which corresponds to a complete crystal� = Rd=Zd, � Lebesgue measure on Rd=ZdTx! = x+ ! on Rd=Zd

1

Example 2 Quasi-periodic potential� = RD=ZD, � Lebesgue measure on RD=ZDTx! = Ax+ ! on Rd=Zd, A: D � d matrix which is rationally independent

Example 3 Random potential� = C

�Rd�, � a probability measure on C

�Rd�

(Tx!) (�) = ! (�+ x) on C�Rd�

Example of �: For a positive de�nite real valued continuous function k (x) onRd Z

f (! (x1) ; ! (x2) ; � � � ; ! (xn))� (d!)

= (2��)�n=2

ZRnf (y1; y2; � � � ; yn) exp

��12

���1y;y

��dy1 � � � dyn

with� = (k (xi � xj))1�i;j�n n� n matrix:

This � is called Gaussian ergodic ensemble on C�Rd�.

We have similar examples also for discrete potentials.

Since a solution ut(x) to Schrödinger equation

i@tut = L!ut, u0 = f

gives a probability jut(x)j2 dx for a particle to stay near x at time t. Therefore,if for a bounded set B � Rd

(i) limt!1

ZB

jut(x)j2 dx > 0

=) the particle remains in B with positive probability

=) localization � point spectrum

(ii) limt!1

ZB

jut(x)j2 dx = 0

=) the particle escapes from B:

=) delocalization � absolutely continuous spectrum:

The issues we have in mind for ergodic Schrödinger operators are

(a) How does the property of ergodic potentials have an e¤ect on the spectrumof the Schrödinger operators ?(b) How is the dimension of the space related to the spectrum ?

The present status of the above issues is as follows.

(A) Random potentials and dimension dependence(1) The �rst rigorous result was obtained by [13] in 1977 for random potentialsin 1 dimension. They showed the localization (which is called now Andersonlocalization) for any random potentials how small it is if the space dimension isone.

2

(2) In general dimension [10] succeeded in 1983 to show the Anderson localiza-tion for random potentials with big coupling constants or near the edges of thespectrum. Their method is called multi-scale analysis.(3) However, physicist�s conjecture that if the space dimension is larger or equalto 3, then some delocalization region appears in the middle of the spectrum hasnot been veri�ed yet. This problem is supposed to be the most interesting anddi¢ cult problem in this area.

(B) Dependence of potentials randomness or complexityFor periodic potentials it is known that the spectrum is always purely absolutelycontinuous and no localization occurs. Then, the problem is to investigate whatwill happen if the randomness or the complexity of the potentials increases.This problem has been studied intensively especially in one dimension.(4) In one dimension for general ergodic potentials it was shown that the vanish-ing set of the Lyapunov exponent coincides with the absolutely continuous (ac)spectrum (Ishii-Pastur-Kotani). And [17] proved that on the ac spectrum there�ectionless property holds, which has made it possible to show the localizationunder very small randomness.(5) It has been known that almost periodic potentials indicate various aspects ofthe spectrum, especially for the almost Mathieu operator a complete knowledgeof the spectrum was obtained.

(C) Related problems(6) Spacing of the eigenvalues for random Schrödinger operators has been in-vestigated. The results indicate that if the potential is random enough, thenwe have Poissonian feature in the structure of the point spectrum in any spacedimension, and if the randomness decreases, then in one dimension the structureof the point spectrum is described by di¤erent probability distributions.(7) The transition from localization to delocalization in higher dimension is triedto show from the point of view of block random matrices.

An overview for higher dimensional case, especially for multi-scale analysis andblock random matrices can be found in [8], and for one dimensional case [6],[16] and [19] are good references.

3

Contents

1. Introduction2. Weyl functions2.1 Weyl disk and boundary classi�cation2.2 Shift operation and its property2.3 Expansion of Weyl functions when z !12.4 Metric on potentials

3. Remling�s theorem3.1 Estimate of transfer matrices3.2 Re�ectionless property on a.c. spectrum

4. Ergodic Schrödinger operators in one dimension4.1 IDS and Lyapunov exponent4.1.1 Integrated density of states (IDS)4.1.2 Lyapunov exponent4.1.3 Floquet exponent4.2 Moment estimate of Lyapunov exponent

5. KdV equation5.1 Construction of KdV �ow on re�ectionless potentials5.2 Extension of KdV �ow5.3 Ergodic initial data5.4 Some questions on solutions with ergodic initial data

6. Related equations

4

2 Weyl functions

One dimensional Schrödinger operators have functions which are called Weyl(or Weyl-Titchmarsh) functions �rst introduced by Weyl in 1910 and studiedintensively by Titchmarsh later. The Weyl functions correspond to the poten-tials in a one-to-one way and they are good quantities which characterize thespectrum. The Floquet exponents for periodic potentials are invariants of theKdV equation and play a crucial role in solving the equation and the scatteringdata for decaying potentials can replace the role of the Floquet exponents. TheWeyl functions can be de�ned for almost every potentials, however, they are notinvariants for the KdV equation at least explicitly. In later section we expressHirota-Sato�s tau functions by the Weyl functions, which enables us to obtaina general solution to the KdV equation.

2.1 Weyl disk and boundary classi�cation

Let q be a real valued function on R which is integrable on each �nite interval,and the Schrödinger operator with potential q is de�ned by

Lq = �@2x + q.

The precise de�nition of Lq as a self-adjoint operator on L2 (R) will be statedlater in this section.The �rst important notion is the classi�cation of the boundaries �1, which

was initiated by H. Weyl in 1910. The boundaries are classi�ed by the dimen-sions of the spaces

Wz =�u 2 L2 (R+) ; � u00 + qu� zu = 0

. (1)

Since the equation in Wz is of second order, we know dimWz � 2. However,the fact dimWz � 1, equivalently Wz 6= f0g is not trivial. To obtain Wz 6= f0gWeyl introduced a disk.Let 'z; z be linearly independent solutions to

�u00 + qu� zu = 0; ' (0) = 0 (0) = 1; '0 (0) = (0) = 0, (2)

where z is a complex number which will work as a spectral parameter. Forz 2 C+ and a > 0 set

Da (z) =

�m 2 C+;

Z a

0

j'z (x) +m z (x)j2dx � Imm

Im z

�.

The inequality in the de�nition of Da (z) can be written in a form of

A+mC +mC +Bmm � 0,

which is equivalent to ����m+ C

B

����2 � ����CB����2 � A

B.

A little bit of calculation using 'z (a) 0z (a)� '0z (a) z (a) = 1 shows8>>><>>>:

C

B='z (a)

0z (a)� '0z (a) z (a)

z (a) 0z (a)� 0z (a) z (a)����CB

����2 � A

B=��� z (a) 0z (a)� 0z (a) z (a)����2

,

5

which implies that Da (z) forms a non-degenerate disk in C+ with center andradius as follows:8>><>>:

the center = �'z (a) 0z (a)� '0z (a) z (a)

z (a) 0z (a)� 0z (a) z (a)

the radius =��� z (a) 0z (a)� 0z (a) z (a)����1

. (3)

We call Da (z) as Weyl disk. From the de�nition Da (z) � Db (z) holds ifa < b, hence

D1 (z) =\a>0

Da (z)

turns to be a closed disk or one point, which is equivalently stated as follows:Let Wz be a subspace of L2 (R+) de�ned in (1):

(LD) D1 (z) is a disk if and only if dimWz = 2.(LP) D1 (z) degenerates to one point if and only if dimWz = 1.

Therefore, automatically Wz 6= f0g has been proved. To know that the prop-erties (LD) and (LP) occur simultaneously with respect to z 2 C+ we de�ne aVolterra integral operator Vz by

Vzf (x) =

Z x

0

('z (x) z(y)� 'z (y) z(x)) f (y) dy.

Then, we immediately obtain

u = Vzf () �u00 + qu� zu = f , u(0) = u0(0) = 0,

and ('z2 = (I � (z2 � z1)Vz1)

�1'z1

z2 = (I � (z2 � z1)Vz1)�1 z1

for any z1, z2 2 C. Therefore, if 'z1 , z1 2 L2 (R+) are valid, then

(I � (z2 � z1)Vz1)�1

turns out to be a bounded operator in L2 (R+), hence 'z2 , z2 2 L2 (R+) hold,

which implies that the properties (LD) and (LP) are equivalent to the followingstatements:

(LD) dimWz = 2 holds for any z 2 C.(LP) dimWz = 1 holds for any z 2 CnR.

(4)

The boundary +1 is called as of limit circle type if (LD) holds and of limitpoint type if (LP) occurs. The boundary �1 has also a similar classi�cation.A simple su¢ cient condition for q under which Lq satis�es (LP) is known.

Lemma 4 If q is bounded on R+, then the boundary 1 is of limit point type.

Proof. If q is bounded, u 2Wz implies u, u00 2 L2 (R+). Then, in the identities

1

2

d

dxju(x)j2 = Re

�u0(0)u (0) +

Z x

0

u00(y)u(y)dy

�+

Z x

0

ju0(y)j2 dy

6

the �rst term of the right hand side remains bounded as x ! 1. Hence,ifR10ju0(y)j2 dy = 1, then ju(x)j2 should be increasing from some positive

number a on, which contradicts u 2 L2 (R+). Therefore, we have u0 2 L2 (R+).Let u1, u2 2Wz. Then, their Wronskian should be a constant, that is

u01(x)u2(x)� u1(x)u02(x) = c.

Since u1, u01, u2, u02 2 L2 (R+) are valid, c should be 0, which implies dimWz =

1.

A more general condition for (LP) is

"There exists c > 0 such that q(x) � �cx2 for every su¢ ciently large x."

It is known that if q(x) = �cx� with some c > 0, � > 2, then the boundary 1is of limit circle type.Suppose the boundaries �1 are of limit point type for Lq, then it is known

that Lq can be de�ned as a self-adjoint operator on L2 (R) uniquely withoutimposing any boundary condition at �1. Actually, if q is bounded, then themultiplication operator q� de�nes a bounded self-adjoint operator on L2 (R).Since the Laplacian part �@2x is unitarily equivalent to the multiplication j�j

2

on L2 (R) via Fourier transform, �@2x is clearly self-adjoint, hence Lq realizedas their sum turns to be self-adjoint. We consider only the case when theboundaries �1 are of limit point type in what follows. In this case, for anyz 2 C+ there exists a unique m+ (z) 2 D1 (z) satisfyingZ 1

0

j'z (x) +m+ (z) z (x)j2dx � Imm+ (z)

Im z. (5)

Due to �'z (a) = z (a) 2 Da(z) (observe z (a) 6= 0), we know

� lima!1

'z (a)

z (a)= m+ (z)

for any z 2 C+, which shows analyticity of m+ (z) on C+ and Imm+ (z) > 0.Set

f+ (x; z) = 'z (x) +m+ (z) z (x) .

Then, f+ (x; z) satis�es(�f 00+ (x; z) + q(x)f+ (x; z) = zf+ (x; z)

f+ (0; z) = 1, f+ (x; z) 2 L2 (R+). (6)

The above argument shows that there exists a unique solution to the equation(6) and f 0+ (0; z) = m+(z) holds. These m�(z) are called theWeyl functions(or Weyl-Titchmarsh functions) and will play a crucial role in the lectures.

2.2 Shift operation and its properties

Since we treat ergodic potentials later, we de�ne the shift operation �x for q by

(�xq) (�) = q (�+ x) .

We denote m�, f� by m� (z; q), f� (x; z; q) respectively if it is necessary toindicate the dependence of the quantities on q explicitly.

7

Lemma 5 m� (z; q) satisfy(f� (x; z; q) = exp

��R x0m� (z; �yq) dy

��@xm� (z; �xq) = q(x)� z �m� (z; �xq)

2 . (7)

Proof. First note f� (�+ y; z; q) 2 L2 (R�) and

�f 00� (x+ y; z; q) + q (x+ y) f� (x+ y; z; q) = zf� (x+ y; z; q) .

Then, the uniqueness of the solutions implies

f� (x; z; �yq) =f� (x+ y; z; q)

f� (y; z; q),

which yields

m� (z; �yq) = �f 0� (0; z; �yq) = �f 0� (y; z; q)

f� (y; z; q). (8)

Therefore, we have the �rst identity of (7). The Ricatti equations in (7) canbe obtained from (7) also, since f� satisfy the second order linear di¤erentialequations.

This argument using the shift operation was employed by [15] very e¤ectivelyin the study of Schrödinger operators with almost periodic potentials.Another representation of the shifted Weyl functions can be obtained by

transfer matrices. Through f'z (x; q) ; z (x; q)g introduced in (2) de�ne

Tq (x; z) =

� 0z (x; q) �'0z (x; q)� z (x; q) 'z (x; q)

�2 SL (2;C) . (9)

Since �f+ (x; z; q) = 'z (x; q) +m+ (z; q) z (x; q)f� (x; z; q) = 'z (x; q)�m� (z; q) z (x; q)

holds, from (8)

m� (z; �xq) = �'0z (x; q)�m� (z; q)

0z (x; q)

'z (x; q)�m� (z; q) z (x; q)

= Tq (x; z) �m� (z; q) (10)

is valid, where the action of SL (2;C) on C[f1g is de�ned by

T � z = t11z + t12t21z + t22

for T =�t11 t12t21 t22

�. (11)

Lemma 6 Tq (x; z) satis�es a cocycle property

Tq (x+ y; z) = T�yq (x; z)Tq (y; z) , (12)

and if x � 0Tq (x; z) � w 2 C+ if w 2 C+. (13)

8

Proof. Let Hq(x; z) =

�0 z � q (x)�1 0

�. Then, we have

d

dxTq (x; z) =

� 00z (x; q) �'00z (x; q)� 0z (x; q) '0z (x; q)

�= Hq(x; z)Tq (x; z) .

The matrixT (x) = Tq (x+ y; z)Tq (y; z)

�1

satis�es

d

dxT (x) =

d

dxTq (x+ y; z)Tq (y; z)

�1= H�yq(x; z)T (x)

with T (0) = I, hence the uniqueness of the equation shows

T (x) = T�yq (x; z) ,

which is (12).Generally T 2 SL (2;C) maps C+ to C+ if (JTw; Tw) > 0 holds for any

w 2 C+, where

J =

�0 i�i 0

�and w =

�w1

�.

A simple calculation

d

dx(JTq (x; z)w; Tq (x; z)w) = ((JH +H�J)Tq (x; z)w; Tq (x; z)w)

=

��0 00 2 Im z

�Tq (x; z)w; Tq (x; z)w

�,

shows

(JTq (x; z)w; Tq (x; z)w)

= 2 Imw + 2 Im z

Z x

0

��0 00 1

�Tq (y; z)w; Tq (y; z)w

�= 2 Imw + 2 Im z

Z x

0

j'z (y)� w z (y)j2dy > 0 (14)

for z, w 2 C+, hence we have (13).

A holomorphic function on C+ with positive imaginary part is called asHer-glotz function, and Weyl functions m� are Herglotz functions. Any Herglotzfunction m has a representation

m (z) = �+ �z +

Z 1

�1

�1

�� z ��

1 + �2

�� (d�)

with � 2 R, � � 0 and a measure � on R satisfyingZ 1

�1

1

1 + �2� (d�) <1.

We call this measure � as the spectral measure of the Herglotz function m.Since the Weyl functions m� are of Herglotz, their spectral measures �� are

9

called spectral measures or spectral functions for L�q which are de�nedas restriction of Lq on L2 (R�) with Dirichlet boundary condition at x = 0.Especially supp �� coincide with the spectra of the restricted L�q .The Green functions g+z (x; y), gz(x; y) for the operators L

+q , Lq, which

are the integral kernels of�L+q � z

��1, (Lq � z)�1 are given respectively by8>><>>:

g+z (x; y) = g+z (y; x) = z (x) f+(y; z)

Wrons ( z; f+)= z (x) f+(y; z)

gz(x; y) = gz(y; x) =f+ (x; z) f�(y; z)

Wrons (f+; f�)= �f+(x; z)f�(y; z)

m+(z) +m�(z)

(15)

for 0 < x � y and x � y respectively.

2.3 Expansion of Weyl functions when z !1In the discussion of the KdV equation we use an asymptotic expansion of m�(z)at z =1. Although the expansion we need in future is more delicate than thealready known one, we present here the usual expansion, which was studied bymany authors.Atkinson observed the following fact in 1981. For the simplicity of notation

we use k =p�z. If z 2 C+, then Re k > 0 and Im k < 0. Since the radius of

Da(z) has been computed in (3), �'z (a) = z (a), m+(z) 2 Da(z) imply����m+(z) +'z (a)

z (a)

���� � 2��� z (a) 0z (a)� 0z (a) z (a)��� . (16)

Therefore, to obtain an asymptotic behavior of m+(z) as z !1, we have onlyto know the asymptotic behavior of 'z (a), z (a) for a �xed a > 0. To this endwe consider the following integral equation

U(x; k) = ekx +

Z x

0

sinh k (x� y)k

q (y)U(y; k)dy, (17)

which determines the 'z, z:

'z (x) =U(x; k) + U(x;�k)

2, z (x) =

U(x; k)� U(x;�k)2k

.

U(x; k) can be solved by an iteration:8<: Un (x) =

Z x

0

sinh k (x� y)k

q (y)Un�1 (y) dy for n � 1, U0 (x) = ekx

U(x; k) =P1

n=0 Un (x)

.

Since��1� e�2k(x�y)�� � 2 if x� y > 0 and Re k > 0, we have��e�kxUn(x)�� � j2kj�1 Z x

0

���1� e�2k(x�y)��� jq (y)j ��e�kyUn�1 (y)�� dy� jkj�1

Z x

0

jq (y)j��e�kyUn�1 (y)�� dy,10

and8>><>>:��e�kxUn(x)�� � 1

n!

�jkj�1Q(x)

�n��e�kxU 0n(x)�� � Z x

0

����1 + e�2k(x�y)2

���� jq (y)j ��e�kyUn�1 (y)�� dy � jkjn!

�jkj�1Q(x)

�nwith Q(x) =

R x0jq (y)j dy, hence8<:

��U(x; k)� ekx�� �P1n=1 jUn(x)j �

��ekx�� �exp�jkj�1Q(x)�� 1���U 0(x; k)� kekx�� �P1n=1 jU 0n(x)j �

��ekx�� jkj�exp�jkj�1Q(x)�� 1� .

Therefore, we have8>><>>:j'z (x)� cosh kxj = O

�k�1ekx

�j'0z (x)� k sinh kxj = O

�ekx��� z (x)� k�1 sinh kx�� = O�k�2ekx

��� 0z (x)� cosh kx�� = O�k�1ekx

� , (18)

which implies

z (a) 0z (a)� 0z (a) z (a) = 2i Im

�k�1 sinh ka cosh ka

�+O

�k�2e2aRe k

�= 2�1e2aRe ki Im k�1 +O

�k�2e2aRe k

�.

Therefore, if z ! 1 in a sector f� < arg z < � � �g, which is equivalent tok !1 in the sector f��=2 + �=2 < arg k < ��=2g for some � 2 (0; �), then

2��� z (a) 0z (a)� 0z (a) z (a)��� �4e�2aRe k

jIm k�1j ,

which means the exponentially fast decay of the right hand side of (16). Thisobservation makes it possible to reduce the asymptotic problem of m+(z) tothat of �'z (a) = z (a), and this is possible essentially by getting more preciseestimate of (18) if we assume the smoothness of q (see [3], [9]). Their result is

m+(z) = �kn+1Xj=0

cjk�j +O

�k�n�1

�, (19)

if q 2 Cn[0; a] for some a > 0. To obtain a concrete expression for the coe¢ cientsfcjgj�0 we de�ne fcj(x)gj�0 by

m+(z; �xq) = �kn+1Xj=0

cj(x)k�j +O

�k�n�1

�.

Substituting this expansion to the Ricatti equation (11) yields

�kn+1Xj=0

c0j(x)k�j +

0@k n+1Xj=0

cj(x)k�j

1A2

= q(x) + k2 +O�k�n�1

�,

11

we have 8<: c20 = 1, � c00 + 2c0c1 = 0, � c01 + c21 + 2c0c2 = q

�c0j�1 + cj +1

2

Pj�1`=1 c`cj�` = 0 for j � 3

,

which is equivalent to8<: c0 = 1, c1 = 0, c2 = q=2

cj = c0j�1 �1

2

Pj�1`=1 c`cj�` for j � 3

. (20)

Then, fcjgj�0 in (19) is obtained by setting cj = cj(0). This formula wasobtained by [27], [12]. The other Weyl functionm�(z) can be obtained similarly,since m�(z) is nothing but m+(z) with potential bq(x) = q(�x). Let bcj(x) =(�1)j cj(�x). Then, bc0(x) = 1, bc1(x) = 0, bc2(x) = bq(x)=2 and for j � 3

bcj(x) = (�1)j cj(�x) = (�1)j c0j�1 (�x)� (�1)j 12j�1X`=1

c` (�x) cj�` (�x)

= bc0j�1(x)� 12j�1X`=1

bc`bcj�` (x) ,hence fbcj(x)gj�0 becomes the fcj(x)gj�0 for the potential bq. Therefore, wehave

m�(z) = �kn+1Xj=0

(�1)j cjk�j +O�k�n�1

�.

Lemma 7 Suppose q 2 Cn [�a; a] for some a > 0. Then, for any � 2 (0; �)asymptotic expansions(

m+(z) = �k �Pn

j=1 cj+1k�j +O

�k�n�1

�m�(z) = �k �

Pnj=1 (�1)

j+1cj+1k

�j +O�k�n�1

�as z = �k2 !1 in a sector f� < arg z < � � �g hold, where cj is a real numbercj = cj(0) with fcj(x)gj�0 determined by (20). Each cj is a polynomial of�q(0); q0(0); � � � ; q(j�2)(0)

if j � 2.

Later we require q so that this expansion remains valid in a domain D ofC+ whose boundary @D approaches to R+ at +1.

2.4 Metric on potentials

For a �xed M > 0 let

QM = fq; q is real valued and satis�es jq (x)j �M on Rg ,

and de�ne a metric on QM by

d (q1, q2) =1Xn=1

2�n�����Z 1

�1(q1 (x)� q2 (x)) fn(x)dx

���� ^ 1� ,12

where ffngn�1 are continuous functions on R with compact supports such thatffngn�1 are dense in each C([�a; a]) the space of continuous functions on [�a; a]with sup-norm. The convergence in QM is equivalent to the convergence ofZ 1

�1q (x) f(x)dx

for any continuous function f with compact support, and QM is compact withthis metric.

Lemma 8 'z (x; q), z (x; q) are continuous on C � QM , and m� (z; q) arecontinuous on C+ �QM .

Proof. Suppose qn ! q in QM . Generally 'z (x; q) can be obtained by

'z (x; q) = 1 +

Z x

0

(x� y) (q(y)� z)'z (y; q) dy =1Xk=0

�k (x; q) ,

where

�0 (x; q) = 1 and �k (x; q) =Z x

0

(x� y) (q(y)� z)�k�1 (y; q) dy for k � 1.

Then, inductively one can show the convergence �k (x; qn) ! �k (x; q) uni-formly on each compact interval, which leads us to the convergence 'z (x; qn)!'z (x; q). Similarly we have z (x; qn) ! z (x; q), which shows the �rst halfof the statement. Since the radius r (a; q) and the center o (a; q) of the Weyldisk Da (z; q) are described by 'z (x; q), z (x; q), the Weyl disks Da (z; qn) ofqn converge to Da (z; q) for each z 2 C+. For any � > 0 there exists a > 0 suchthat r (a; q) � �. Then, Da (z; qn) � Da (z; q) for this a if n is su¢ ciently large,and

m+ (z; qn) 2 Da (z; q) =) jm+ (z; qn)�m+ (z; q)j � 2r (a; q) � 2�,

which shows the convergence m+ (z; qn) ! m+ (z; q). The convergence ofm� (z; qn) ! m� (z; q) is similar. It is easily seen that this convergence isuniform on each compact set of C+, which completes the proof.

3 Remling�s theorem

Remling [23], [24] revealed the relationship between the absolutely continuousspectrum for one dimensional Schrödinger operators and the re�ectionless prop-erty of the Weyl functions m�. A potential q or corresponding Schrödingeroperator Lq is called re�ectionless on A 2 B (R) if

m+ (� + i0; q) = �m� (� + i0; q) for a.e. � 2 A (21)

holds.

13

3.1 Estimate of transfer matrices

On C+ de�ne a pseudo metric by

(z1; z2) =jz1 � z2jpIm z1

pIm z2

.

Then it is known that

(F (z1) ; F (z2)) � (z1; z2) (22)

holds for any Herglotz function F . Set

T+ = fT 2 SL (2;C) ; T maps C+ into C+:g

and for T 2 T+ de�ne a norm by

� (T ) = supz2C+

2 Im z

(JTz; Tz),

which is not greater than 1 due to (22). Obviously we have inequalities( (T � z1; T � z2) � � (T ) (z1; z2) for any z1, z2 2 C+ and T 2 T+� (T2T1) � � (T2) � (T1) for T1, T2 2 T+:

(23)

For later purpose we have to estimate � (Tq (x; z)) for the transfer matrix Tq (x; z)of (9).

Lemma 9 Suppose jq (x)j � M on R. Then, for any compact set K in C+there exists a positive � such that � < 1 and

� (Tq (n; z)) � �n for any z 2 K and n � 1.

Proof. The space of potentials QM is compact and m� (z; q) is continuouson C+ � QM due to Lemma 8. The continuity of f'z (y; q) ; z (y; q)g on thecompact set K �QM implies that the positive quadratic functions8>><>>:

Q1 (z; w) =

Z 1

0

j'z (y; q)� w z (y; q)j2dy

Q2 (z; w) =

Z 1

0

jw'z (y; q)� z (y; q)j2dy

of w satisfy

infjwj�1

(z;q)2K�QM

Q1 (z; w) jwj�2 = infjwj�1

(z;q)2K�QM

Q2 (z; w) > 0,

hence there exists a constant c > 0 such that Q1 (z; w) � c jwj2 holds for anyw 2 C. Then, (14) shows

(JTq (1; z)w; Tq (1; z)w) = 2 Imw + 2 Im zQ1 (z; w) � 2 Imw + 2c jwj2 Im z,

hence2 Imw

(JTq (1; z)w; Tq (1; z)w)� 2 Imw

2 Imw + 2c jwj2 Im z,

14

which impliessup

z2K; q2QM

� (Tq (1; z)) � �

with

� � supz2K; w2C+

2 Imw

2 Imw + 2c jwj2 Im z< 1.

From the cocycle property of Tq(x; z) we have

Tq (n; z) = T�n�1q (1; z)T�n�2q (1; z) � � �T�1q (1; z)Tq (1; z)

and from (12)

� (Tq (n; z)) �n�1Yk=0

� (T�kq (1; z)) � �n

is valid.

3.2 Re�ectionless property on a.c. spectrum

We follow the argument made by Breimesser-Pearson and Remling. Let H bethe set of all Herglotz functions on C+, namely

H = fm : C+ ! C+; m is holomorphicg :

De�ne for z = x+ iy

!z (S) = Im

�1

�

ZS

1

t� z dt�=1

�

ZS

y

(t� x)2 + y2dt.

!z (S) is related with (z1; z2). Let Ft (z) = (t� z)�1 2 H. Then, (22) shows

(Ft (z1) ; Ft (z2)) � (z1; z2) ,

hence

j!z1 (S)� !z2 (S)j =

���� 1�ZS

Im (Ft(z1)� Ft(z2))���� dt

� 1

�

Z 1

�1jFt(z1)� Ft(z2)j dt

� (z1; z2)1

�

Z 1

�1

pImFt (z1)

pImFt (z2)dt

� (z1; z2)

�1

�

Z 1

�1ImFt (z1) dt

1

�

Z 1

�1ImFt (z1) dt

�1=2holds. Since

R1�1 ImFt (z) dt=� = 1, we have

j!z1 (S)� !z2 (S)j � (z1; z2) for z1, z2 2 C+. (24)

Pearson [21], [22] introduced the following notion of convergence.

15

De�nition 10 A sequence of Herglotz functions fmng converges to m 2 H invalue distribution if

limn!1

ZA

!mn(x) (S) dx =

ZA

!m(x) (S) dx (25)

for all Borel sets A, S of R such that jAj <1.

The following two lemmas were used by Breimesser-Pearson [4], [5]. The�rst lemma rephrases the convergence on C+ by that on R.

Lemma 11 For a sequence fmng of H and m 2 H, the following statementsare equivalent.(1) mn ! m compact uniformly on C+:(2) mn ! m in value distribution.(3) (25) holds for any bounded intervals A, S.

The second one transfers the statement on R to that on C+.

Lemma 12 Let A 2 B (R) with jAj <1. Then

limy#0

supm2M;S2B(R)

����ZA

!m(x+iy) (S) dx�ZA

!m(x) (S) dx

���� = 0:We need the following result which is a stronger statement of the limit point

condition of the boundary +1.

Lemma 13 Let K be a compact subset of C+. Then

limx!1

(Tq (x; z) � w1; Tq (x; z) � w2) = 0;

uniformly in z 2 K, w1, w2 2 C+.

Proof. Note �rst the set

K1 =�Tq (1; z) � w; z 2 K, w 2 C+

de�nes a compact set of C+. Due to (12)

Tq (x; z) � w = T�1q (x� 1; z)Tq (1; z) � w

holds, we have only to show

limx!1

(T�1q (x; z) � w1; T�1q (x; z) � w2) = 0, (26)

uniformly in z 2 K, w1, w2 2 K1. From Lemma 9 we have

� (Tq (n; z)) � �n for any q 2 QM and n � 1:

Hence, for n � x � n+ 1 (9) and the cocycle property imply

� (Tq (x; z)) = ��T�x�nq (n; z)Tq (x� n; z)

�� �n,

16

which shows (26).

To state the theorem we introduce a re�ection operation R on QM by

(Rq) (x) = q (�x) .

Then, obviously identities(m+ (z;Rq) = m� (z; q) , 'z (x;Rq) = 'z (�x; q) , z (x;Rq) = � z (�x; q)�TRq (�x; �) � w = Tq (x; �) � (�w) for any x, � 2 R and w 2 C+

(27)hold. Let �+ac be the absolutely continuous spectrum of L+q (see (15)): Ourtheorem is

Theorem 14 (Remling) Let q 2 QM and eq = limn!1 �xnq for a sequence fxngtending to +1. Then, eq is re�ectionless on �+ac, that is

m+ (� + i0; eq) = �m� (� + i0; eq) for a.e. � 2 �+ac. (28)

Proof. All the necessary things are already prepared and all we have to do isto follow Breimesser-Pearson and Remling. [23] noted that the property (28) isequivalent to

!m�(�+i0;eq) (S) = !m�(�+i0;Req) (�S) for a.e. � 2 �+ac

for any S 2 B (R). Fix � > 0 and A � �+ac with positive Lebesgue measure.There exists a �nite decomposition fAjg0�j�N of A and mj 2 C+ such that

A = A0 [A1 [A2 [ � � � [AN disjoint with jA0j � �,

and (m� (�;Rq) ;mj) � � for 8� 2 Aj .

Here we have written f (�) = f (� + i0) for simplicity. Since TRq (�xn; �) 2SL (2;R) for � 2 R, we have

(TRq (�xn; �) �m� (�;Rq) ; TRq (�xn; �) �mj) = (m� (�;Rq) ;mj) :

Applying the inequality (24) we see�����ZAj

!TRq(�xn;�)�m�(�;Rq) (S) d� �ZAj

!TRq(�xn;�)�mj(S) d�

����� � � jAj j . (29)

Now we try to deform the second integral admitting negligible errors. Theidentity !z (S) = !�z (�S) and (27) imply

!TRq(�xn;�)�mj(S) = !�TRq(�xn;�)�mj

(�S) = !Tq(xn;�)�(�mj) (�S) . (30)

On the other hand, Lemma 12 implies that there exists y > 0 such that for anym 2M �����

ZAj

!m(�+iy) (�S) d� �ZAj

!m(�) (�S) d������ � � jAj j

17

holds. Applying this estimate to m (z) = Tq (xn; z) � (�mj) and Tq (xn; z) �m� (z; q) yields8<:���RAj

!Tq(xn;�)�(�mj) (�S) d� �RAj!Tq(xn;�+iy)�(�mj) (�S) d�

��� � � jAj j���RAj!Tq(xn;�)�m�(�;q) (�S) d� �

RAj!Tq(xn;�+iy)�m�(�+iy;q) (�S) d�

��� � � jAj j.

Then, the estimate of Lemma 13

(Tq (xn; � + iy) �m� (� + iy; q) ; Tq (xn; � + iy) � (�mj)) � �

and (24) imply�����ZAj

!Tq(xn;�+iy)�(�mj) (�S) d� �ZAj

!Tq(xn;�+iy)�m�(�+iy;q) (�S) d������ � � jAj j .

Consequently, we have�����ZAj

!Tq(xn;�+iy)�(�mj) (�S) d� �ZAj

!Tq(xn;�)�m�(�;q) (�S) d������ � 3� jAj j :

Hence, Breimesser-Pearson�s result in this case is obtained from (29), (30),namely����Z

A

!TRq(�xn;�)�m�(�;Rq) (S) d� �ZA

!Tq(xn;�)�m�(�;q) (�S) d����� � 4� jAj+ �;

for every su¢ ciently large n. Since m� (z; q) are continuous with respect to q(TRq (�xn; z) �m� (z;Rq) = m� (z; ��xnRq) = m+ (z; �xnq)! m+ (z; eq)Tq (xn; z) �m� (z; q) = m� (z; �xnq)! m� (z; eq)

holds. ThereforeZA

!m+(�+i0;eq) (S) d� =ZA

!m�(�+i0;eq) (�S) d� =ZA

!�m�(�+i0;eq) (S) d�is valid. Since A is an arbitrary subset of �+ac, we have !m+(�+i0;eq) (S) =!�m�(�+i0;eq) (S) for a.e. � 2 �+ac, which concludes (28).Remark 15 If we denote the absolutely continuous spectrum of L�q and Lq by��ac and �ac respectively, then it is known that

�+ac [ ��ac = �acholds. Suppose the spectrum �+ of L+q coincides with [0;1) and purely ab-solutely continuous there. Then, for any sequence xn !1 such that �xnq ! eqto some eq 2 QM the above theorem says eq is re�ectionless on [0;1). Since thespectrum of L+�xnq are contained in [0;1), hence the spectrum of Leq should bein [0;1), which together with the re�ectionless property on [0;1) implies thateq = 0 identically. Therefore, �xq ! 0 as x!1, which indicates some decayingcondition for the original potential q. Since it is known that if a potential qsatis�es Z 1

0

jq (x)j dx <1,

then L+q has purely absolutely continuous spectrum on [0;1), Remling�s theo-rem asserts that the converse statement is somewhat valid. , one can say theabsolutely continuous spectrum is a rare event.

18

4 Ergodic Schrödinger operators in one dimen-sion

In this section we introduce two important quantities IDS and Lyapunov ex-ponent in the theory of one dimensional ergodic Schrödinger operators. Here,instead of providing its detailed information we only present results which arerelated to the study of the KdV equation. The detail can be found in [6], [16],[19].Throughout the section we assume

jq! (x)j �M

for any x 2 R and ! 2 .

4.1 IDS and Lyapunov exponent

There are two quantities which describes the spectral property of ergodic Schrödingeroperators L!.

4.1.1 Integrated density of states (IDS)

There are two ways of its de�nition. One is constructive and physical. For ` > 0

letn�`;!k

ok�1

be the eigen-values of L! restricted on L2 ([�`; `]) with Dirichletboundary condition at �`, ` and set

N!` (�) =

1

2`#nk � 1; �`;!k � �

o.

Since we are assuming the potential q! is bounded by M , the eigen-values �`;!k

satisfy �`;!k � �M , hence N!` (�) is �nite non-decreasing function on R such

that N!` (�) = 0 for � < �M . The ergodicity of q! assures the existence of the

limit of N!` (�) as `!1 for a.e. ! 2 , and is denoted by N (�), namely

N (�) = lim`!1

N!` (�) .

This quantity is independent of ! due to the ergodicity and is called as inte-grated density of states (IDS) for L!.This quantity can be identi�ed with the expectation of the spectral measure

of L! as follows. Since the Green function g!z (0; 0) = ��m!+ (z) +m

!� (z)

��1(see (15)) is a Herglotz function, there exists a measure �! such that

g!z (0; 0) =

Z 1

�1

1

�� z �! (d�) .

Due to the fact L! � �M the Green function g!z (0; 0) is analytic on Cn[�M;1)and the asymptotics of Lemma 7 implies g!z (0; 0) behaves like

�2p�z��1

whenz !1 suitably, hence the measure �! is supported on [�M;1) and satis�esZ 1

0

1

�+ 1�! (d�) <1.

19

Then, it is known that

N (�) = E�! ((�1; �]) =Z

�! ((�1; �])� (d!)

holds. Therefore, an identity

E (g!z (0; 0)) =Z 1

�1

1

�� z dN (�)

is valid.The following property of N (�) is known.

Theorem 16 N (�) is continuous and non-decreasing on R and satis�es

�! (the spectrum of L!) = supp dN (�) .

The IDS can be de�ned for higher dimensional ergodic Schrödinger operatorsalso and the same property holds except for the continuity.

4.1.2 Lyapunov exponent

There is another important quantity which describes the spectral property moreintimately. Let T! (x; z) be the transfer matrix of (9), namely T! (x; z) is theunique solution to

d

dxT! (x; z) =

�0 z � q! (x)�1 0

�T! (x; z) , T! (0; z) = I.

Then, the cocycle property (12) implies

kT! (x+ y; z)k � T�y! (x; z) kT! (y; z)k ,

hence the subadditive ergode theory shows the existence of the limit of x�1 log kT! (x; z)kas x!1, and we denote it by (z), namely

(z) = limx!1

1

xlog kT! (x; z)k .

This quantity also is independent of ! and is called as Lyapunov exponent:The operator norm k�k for matrices of SL (2;C) is greater or equal to 1, hencekT! (x; z)k � 1 and we have (z) � 0. It should be noted that the limit of(�x)�1 log kT! (x; z)k for x < 0 also the same quantity. For A 2 B (R) we use anotation

Aess = f� 2 R; jA \ (��+ �; �+ �)j > 0 for any � > 0g ,

where j�j denotes the Lebesgue measure of �. Set8<: Z = f� 2 R; (�) = 0g

Zs =�� 2 Z : lim�#0

N (�+ �)�N (�� �)2�

=1�

and letdN = dNac + dNs

be the decomposition of the measure dN into the absolutely continuous partand the singular part.

20

Theorem 17 Let �!ac and �!s be the absolutely continuous spectrum and the

singular spectrum of L! respectively. Then��!ac = Zess�!s = (Zs \ supp dNs) [ ((RnZ) \ supp dN)

holds for a.e. ! 2 . Moreover, we have the re�ectionless property on Z,namely

m!+ (�+ i0) = �m!

� (�+ i0) for a.e. � 2 Z

holds for a.e. ! 2 .

Corollary 18 The spectrum of L! consists only of absolutely continuous partif and only if

dNs = 0 and dN (RnZ) = 0.

Remark 19 Theorem 1 is closely related with Theorem 2, namely, if we regardthe shift operation as a dynamical system, then Theorem 1 states the limitingbehavior of the dynamical system on the absolutely continuous spectrum, andTheorem 2 mentions the property of the equilibrium state.

4.1.3 Floquet exponent

The Floquet exponent was introduced by Johnson-Moser [15] as a generalizationof that of periodic di¤erential equations. From the formula (7) we have

f!+ (x; z; q) = exp

�Z x

0

m�y!+ (z) dy

�,

hence the asymptotic behavior of f!+ is governed by E�m!+ (z)

�which is denoted

by w (z), namelyw(z) = E

�m!+ (z)

�.

This w(z) is called as Floquet exponent and is known to be equal to E�m!� (z)

�also. The importance of w(z) is in its relationship with N (�) and (�), namelywe have

Lemma 20 (Thouless formula) For � 2 R it holds that( (�) = �Rew(�+ i0)N (�) =

1

�Imw(�+ i0)

.

The Ricatti equation (7) yields beautiful identities8><>:w0(z) = �E

��m!+ (z) +m

!� (z)

��1�= E (g!z (0; 0))

�2Rew(z) = E�

Im z

Imm!� (z)

� . (31)

To measure the magnitude of re�ectionlessness we de�ne

R(z) =m+(z) +m�(z)

m+(z) +m�(z), (32)

21

and call it as re�ection coe¢ cient. This quantity was �rst considered byGesztesy-Simon, Rybkin and others as a generalization of the conventional re-�ection coe¢ cient de�ned for decaying potentials. (31) implies another identity

�Rew(z)Im z

� Imw0(z) = 1

4E��

1

Imm!+ (z)

+1

Imm!� (z)

�jR! (z)j2

�, (33)

which was crucial to prove Theorem 17. Set

� (z) = �Rew(z)Im z

� Imw0(z) > 0. (34)

The following lemma will be used for the KdV equation.

Lemma 21 For z 2 C+ we have

E (jR! (z)j) �p2� (z) Imw(z).

Proof. Applying the Schwarz inequality to (33) yields

E (jR! (z)j) � 2

vuutE � 1

Imm!+ (z)

+1

Imm!� (z)

��1!p� (z)

�qE�Imm!

+ (z) + Imm!� (z)

�p� (z) =

p2� (z) Imw(z).

4.2 Moment estimate of Lyapunov exponent

For later purpose we estimate the moments of Lyapunov exponent by the deriv-atives of potentials q!, which was given by [18].(31) implies that the Floquet exponent w satis�es Imw(z), �Rew(z) > 0

for z 2 C+, hence a product �w1w2 of two Floquet exponents w1, w2 gives aHerglotz function. Taking the simple Floquet exponent w0(z) = �

p�z of the

constant potential 0, we have a Herglotz functionp�zw(z).

Lemma 22 Suppose q! (x) 2 C2nb (R) for some n � 0. ThenZ 1

0

�np� (�) d� <1 (35)

is valid.

Proof. (20) and Lemma 7 imply

m!+ (z) +m

!� (z)

= �2p�z �

2n+1Xj=1

c!j+1

�1 + (�1)j+1

�p�z�j +O

�z�n�1

�= �2

p�z � 2

X1�j�2n+1j: odd

c!j+1p�z�j +O

�z�n�1

�,

22

where c!j is a polynomial ofnq!(0); q

0!(0); � � � ; q

(j�2)! (0)

o, hence

w (z) = �p�z �

p�z

n+1Xj=1

a2j (�z)�j + o�z�n�1

�(36)

with aj = E�c!j�. On the other hand, since

p�zw (z) is a Herglotz function

and

Imp� (�+ i0)w (�+ i0) =

��p��N (�) if � < 0p� (�) if � > 0

holds, we have

p�zw (z) = �+ �z +

1

�

Z 0

�M

�p��N (�)�� z d�

+1

�

Z 1

0

�1

�� z ��

�2 + 1

�p� (�) d�

with � 2 R and � � 0. This combined with (36) implies � = 1 andZ 1

0

1

�+ 1

p� (�) d� <1,

hence we have

w (z) = �p�z � ap

�z+

1p�z

Z 0

�M

p��N (�)�� z d�+

1

�p�z

Z 1

0

p� (�)

�� z d�

(37)Substituting this expansion into (37) yields

1

�

Z 1

0

p� (�)

�� z d� =nXj=1

bj (�z)�j +O�z�n�1

�as z ! �1 with some other bj 2 R. Then, inductively we can show the�niteness of the n-th moment.

5 KdV equation

The KdV equation is a equation describing motion of waves on shallow water,and it is known that it has in�nitely many invariants, which makes it possible tosolve it completely for periodic or decaying initial data. However, if the initialdata are neither periodic nor decaying, we cannot use the invariants e¤ectivelyand only several special cases are known to be solvable. In this section we givean approach to this problem by extending the Sato�s method of Grassmannmanifold, since this method admits solutions with periodic or decaying initialdata in a uni�ed way, although they are in a sense �nite dimensional due to itsalgebraic nature.

23

5.1 Construction of KdV �ow on re�ectionless potentials

In this section we state results of [20] without proofs, which will be necessaryin the later section for the construction of solutions to the KdV equation withnon-periodic and non-decaying initial data.The prototype of the tau function was introduced by Hirota [14], and later

Sato [25] rewrote it by determinant and used it as a central notion in his theory.Segal-Wilson [26] developed Sato�s algebraic theory on a Grassmann manifoldGr(2) consisting of subspaces W of H = L2 (jzj =

pr) satisfying the two condi-

tions

(i) f 2W =) z2f 2W(ii) p+ :W ! H+ =

nu 2 H; u(z) =

Pn�0 unz

nois bijective.

Let AW be the unique operator from H+ to H��= H?

+

�satisfying

W = ff 2 H; f = f+ +AW f+ with f+ 2 H+g ,

whose existence is guaranteed by the condition (ii). Since H+ is generated bymultiplying z2 consecutively to the two elements f1; zg, the above two propertiesof W suggest us to introduce two elements of H�

'W (z) = (AW 1) (z) , W (z) = (AW z) (z) .

Then, without di¢ culty we see

W =z2m (1 + 'W (z)) + z

2n (z + W (z))�m;n�0 ,

where h�i denotes the closure of all linear combinations of � in H. By these twofunctions de�ne

mW (z) =z + W (z)

1 + 'W (z)+ a1 (W )

where a1 (W ) is the �rst coe¢ cients of the expansion

'W (z) =1Xn=1

an(W )z�n.

a1 (W ) is added so that we have

mW (z) = z +O�z�1�

as z !1.

Let p+ be the orthogonal projection from H to H+, and set�� =

�g; g = eh h is holomophic on Cn (Is [ iIs) for an s <

pr

�real =�g; g = eh h is real valued on R

.

For W 2 Gr(2), g 2 � the tau function �W (g) is de�ned by

�W (g) = det�I + g�1p+gAW

�,

which is the crucial tool for the theory. Let

�W (g) = exp

1

2�i

Zj�j=

pr

h0 (�) log (1 + 'W (�)) d�

!for g = eh

24

and de�ne�mW

(g) = �W (g) =�W (g),

whereq� (z) =

�1� z��1

��1 2 � if j�j >pr.

The tau function satis�es the following basic properties.

Lemma 23 (1) W 2 Gr(2) if and only if �W (g) 6= 0.(2) �W (g) satis�es the cocycle property

�W (g1g2) = �W (g1) �g1W (g2) if �W (g1) 6= 0.

(3) �W (g) is decomposed as

�W (g) = �W (g)�mW(g) ,

where �mW(g) depends only on mW , namely if W1, W2 2 Gr(2) satisfy mW1 =

mW2, then �mW1

(g) = �mW2(g).

Now we specify W s corresponding to potentials. For a bounded potentialq 2 QM let m� be its Weyl functions and de�ne

m =

��m+

��z2

�if Re z > 0

m���z2

�if Re z < 0

.

Then, m is holomorphic on Cn�h�pM;

pMi[ iR

�and satis�es

Imm(z)

Im z> 0 for z 2 Cn

�h�pM;

pMi[ iR

�.

If m� are re�ectionless on (�1;1) for �1 > 0, then m is holomorphic on

Cn�h�pM;

pMi[ i��p�1;p�1��. Keeping this in mind, we set

Mreflr =

8>><>>:m;(i) m is analytic on Cn (Ir [ iIr)(ii) m(z) = z +O

�z�1�as z !1

(iii) m(z) = m(z), Imm(z)= Im z � 0(iv) m (x) � m (�x) if x > r

9>>=>>; ,where Ir = [�r; r]. (iv) is equivalent to

m+(�x) +m�(�x) < 0 if x > r,

which follows from sp Lq � [�M;1) and the identity

� (m+(�x) +m�(�x))�1 = g�x (0; 0) > 0.

Set

Wm =�f ; f (z) = '

�z2�+m(z)

�z2�with ', 2 H+ (jzj = r)

.

Then, without di¢ culty we see Wm 2 Gr(2) and 'Wm= 0, Wm

= m�z, hencemW = m and

�m (g) = �mW(g) = �Wm (g) =�Wm

(g) = �Wm (g) ,

25

since �Wm(g) = 1.

One can express this �m (g) by a Fredholm determinant of some integraloperator. In the sequel for a function f on a domain D in C satisfying �z 2 Dif z 2 D we de�ne

fe(z) =�f�pz�+ f

��pz��=2, fo(z) =

�f�pz�� f

��pz��=�2pz�,

wherepz is de�ned as a holomorphic function on Cn(�1; 0] satisfying

p1 = 1.

For m 2Mreflprwe de�ne the tau function �m (g) on �real. Let C, C 0 be simple

smooth closed curves surrounding the interval [�r; r] which are directed anti-clockwise.

Cr

C'

­r

f.1

Let � be an analytic function such that �e, �o are analytic on a simply connecteddomain containing C 0 and set

em(z) = m(z)� � (z) .

De�ne an integral kernel Ng;m (z; �) and its integral operator Nm(g) by8>><>>:Ng;m (z; �) =

1

2�i

ZC0

bgo ��0� (g em)e (�) + bge ��0� (g em)o (�)��0 � z

� ��� �0

� mo

��0��1

d�0

(Nm(g)f) (z) =1

2�i

ZC

Ng;m (z; �) f (�) d�,

(38)where bg (z) = g(z)�1. The operator Nm(g) de�nes a trace class operator onL2 (C). Then, we have

Lemma 24 �m (g) is independent of C, C 0 and �, and satis�es(1) �m (g) = det (I +Nm(g)) for g 2 � .(2) �m (g) > 0 holds for g 2 �real.

The potential q can be recovered from m by

q(x) = �2@2x log �m (ex) with ex(z) = exz 2 �real if x 2 R. (39)

SetQreflr =

�q; q(x) = �2@2x log �m (ex) with m 2Mrefl

r

.

Then, we have a �ow fK(g)gg2�real on Qreflprsatisfying8>><>>:

(K (etz) q) (x) = q (x+ t)�K�e�4tz

3�q�(x) = q (x; t) a solution to the KdV equation�

K�etz

2n+1�q�(x) = qn (x; t) a solution to the n-th KdV equation

.

The cocycle property (2) of Lemma 23 can be interpreted by �m(g) as follows.

26

Lemma 25 Suppose g1, g2 2 � and �m (g1) 6= 0. Then, we have

�m (g1g2) = �m (g1) �mg1Wm(g2) exp

1

2�i

Zj�j=r

h02 (�) log�m (g1q�)

�m (g1)d�

!(40)

with g2 = eh2 .

In the proof we use an identity

�W (q�) = 1 + 'W (�) . (41)

One can show also

Lemma 26 Suppose m 2 Mreflprand g 2 �real. Then, mgWm

2 Mreflprholds,

which derives an action g �m = mgWmonMreflp

rby �real.

The resulting space Qreflprof potentials consists of meromorphic functions

on C with no poles on R, and it contains a certain kind of quasi-periodic func-tions as well as rapidly decaying functions, however a direct characterization ofQreflp

ris not known. From the spectral theoretic point of view the associated

Schrödinger operator Lq has purely absolutely continuous spectrum on (r;1)with re�ectionless property and on (�r; r) the spectrum can be arbitrary. Ournext task is to remove the absolute continuity of the spectrum on (r;1) byletting r !1 in the expression of �m (g) = det (I +Nm(g)) of Lemma 24.

5.2 Extension of KdV �ow

In this section we extend the � -function with the help of Lemma 24, and de�nethe KdV �ow on a wider class of potentials q.

5.2.1 Potential class

What we have in mind about the curves C, C 0 is illustrated in the �gure below.

C

C'

−λ b b'0

f.2

The precise shapes of the curves C, C 0 are determined depending on g 2 � sothat ge, go are bounded on the curves, which will be necessary for the operatorNm(g) to be a trace class, or more generally Hilbert-Schmidt class operator.Suppose the curve C is parametrized as

x+ i (�x)��

27

for large �x. For instance if g(z) = etz3

(t 2 R), the exponent of ge behaves asx! �1

z3=2 =�x+ i (�x)��

�3=2= �i (�x)3=2 � 3

2(�x)��+1=2 + o

�jxj��+1=2

�,

hence, if � � 1=2, then g remains bounded on C. Mor generally if h(z) = zn

with odd n, we have to choose � so that � � n=2 � 1. It should be remarkedthat h has to be an odd polynomial with real coe¢ cients. Otherwise g mightbe unbounded on C whichever curve C we choose.The boundedness of ge, go is not su¢ cient in order that the Nm(g) may be a

trace class operator, and the terms me � �e, mo � �o should decay as x! �1,which is the reason why we have attached � to m in the de�nition of Nm(g).Hereafter we �x �0 < 0 and assume that the Schrödinger operator Lq has

its spectrum � (Lq) such that

� (Lq) � [�0;1).

For a �xed b > �0 let ! be a smooth function on (�1; b] satisfying

! (x) > 0 for any x 2 (�1; b) and ! (b) = 0.

For this ! de�ne the curve C! on C by

C! \ C+ = fx+ i! (x) ; x � bg ,

and on C� it is de�ned by taking the complex conjugate of C! on C+. Itsdirection is anti-clockwise. We assume ! satis�es

limx!�1

(�x)n=2�1 ! (x) 2 (0;1) (42)

for an n � 1, which means that ! depends on n. The interior domain D+! and

the exterior domain D�! of C! are de�ned by�

D+! = fz 2 C; jIm zj < !(Re z), Re z < bg

D�! = Cn (D+

! [ C!), (43)

which is illustrated below.

C−λ b0

Dω

Dω

ω

+−

f.3

Moreover, set

H�D�!

�= L2 (C!) -closure of all rational functions with no poles in D�

! .(44)

28

For N > 0 de�ne a space �Nn as

�Nn =

�u 2 H

�D�!

�; 9f 2 H

�D+!

�s.t.

ZC!

jzj2N ju(z)� f(z)j2 jdzj <1�;

(45)

where n comes from (42). For u 2 H (D�! ) and f 2 H (D+

! ) the Cauchy theoremimplies 8><>:

u (z) =1

2�i

RC!

u(�)

z � �d�

0 =1

2�i

RC!

f(�)

z � �d�, if z 2 D�

! ;

hence

u (z) =1

2�i

ZC!

u(�)� f(�)z � � d� (46)

holds. For u 2 �Nn set

kukn;N = inff2H(D+

! )

sZC!

jzj2N ju(z)� f(z)j2 jdzj. (47)

Suppose kukn;N = 0. Then, u = 0 follows from (46) easily. The other propertiesof norms are clear, hence k�kn;N de�nes a norm in �Nn . It should be remarkedthat �Nn depends on !, namely, even if !1, !2 satisfy

limx!�1

(�x)n=2�1 !1 (x) = limx!�1

(�x)n=2�1 !2 (x) ,

we might have H�D�!1

�6= H

�D�!2

�, hence �Nn might be di¤erent for !1, !2.

However, this would not cause any problem for the de�nition of the extendedtau-function.Let C = C! and C 0 = Cc! with c > 1. One can assume D+

! � D+c!,

D�! � D�

c!. If N � 1 is an integer, for u 2 �Nn and b0 > b (46) an identity

1

z � � =N�1Xk=0

(z � b0)�k�1 (�� b0)k + (z � b0)�N (�� b0)N

z � �

implies

u(z) =N�1Xk=0

ak (z � b0)�k�1 +(z � b0)�N

2�i

ZC

(�� b0)N (u (�)� f (�))z � � d� (48)

with

ak =1

2�i

ZC

(�� b0)k (u (�)� f (�)) d�. (49)

Lemma 27 There exists a constant c1 such that�����u (z)�N�1Xk=0

ak (z � b0)�k�1����� � c1 jzj�N+(n�2)=4 kukn;N

holds for any u 2 �Nn and z 2 C 0.

29

Proof. First note that there exists a constant c0 > 0 such thatZC

1

jz � �j2jd�j � c0 jIm zj�1 (50)

holds for any z 2 C 0, sinceinf�2C

jz � �j � (c� 1) jIm zj

and C is very close to R at x = �1. Therefore, we have from (46) and (50)�����u (z)�N�1Xk=0

ak (z � b0)�k�1�����

� jz � b0j�N

2�

ZC

j�� b0jN ju (�)� f (�)jjz � �j jd�j

� c1 jzj�NsZ

C

1

jz � �j2jd�j

sZC

j�j2N ju (�)� f (�)j2 jd�j

� c0c2 jzj�N jIm zj�1=2sZ

C

j�j2N ju (�)� f (�)j2 jd�j

with

c2 =1

2�max�2C

������ b0�

����maxz2C0

���� z

z � b0

���� ,which completes the proof, since jIm zj � const: jzj�n=2+1 if z 2 C 0.Now, we can introduce a potential class by its Weyl functions on which the

KdV �ow is constructed. Let m� be the Weyl functions of the Schrödingeroperator Lq with potential q and set

m(z) =

��m+(�z2) Re z > 0m�(�z2) Re z < 0

andMN

n =�m; me, mo � 1 2 �Nn

,

for n � 1, N > 0.

5.2.2 Extension of �-function

Set a class of g by

�n =�g = eh; h is real odd polynomial with degree � n, and h(0) = 0

for an odd natural number n. The argument below also works if we multiplyrational functions to g, however that makes the notations and the argumentssubtle, so we use this de�nition here. In order to de�ne the � -function form 2 MN

n , g 2 �n set C = C!, C 0 = Cc! with c > 1. The integral kernelNg;m (z; �) on L2(C) and the integral operator on L2(C) are de�ned by8>><>>:

Ng;m (z; �) =1

2�i

ZC0

bgo ��0� (g em)e (�) + bge ��0� (g em)o (�)��0 � z

� ��� �0

� mo

��0��1

d�0

Nm(g)f(z) =1

2�i

ZC

Ng;m (z; �) f (�) d�

(51)

30

as we did in (38). The number N will be speci�ed later, and assume to besu¢ ciently large for the time being. Since me, mo � 1 2 �Nn , there exist f1,f2 2 H (D+

! ) such thatZC

jzj2N jme(z)� f1(z)j2 jdzj <1,ZC

jzj2N jmo(z)� 1� f2(z)j2 jdzj <1

are satis�ed, hence we set eme = me � f1, emo = mo � 1 � f2 with these f1, f2.The estimate of the kernel Ng;m (z; �) is fundamental for the main result. Let

jgjC = maxz2Cjg (z)j .

Since

mo (z) = �m+ (�z) +m� (�z)

2pz

and m� are of Herglotz, we see mo (z) 6= 0 on Cn(�1;��0], hence this togetherwith Lemma 27 implies

min�02C0

��mo

��0��� > 0,

if N > (n� 2) =4.

Lemma 28 Suppose N > (n� 2) =4. Then, there exists a constant c2 such that

jNg;m (z; �)j � c2 jgjC jbgjC0

�min�02C0

��mo

��0�����1 (jeme (�)j+ jemo (�)j) (jzj j�j)(n�2)=4

holds for z, � 2 C.

Proof. From (51)

jNg;m (z; �)j � c3 jgjC jbgjC0

�min�02C0

��mo

��0�����1 Z

C0

jeme (�)j+ jemo (�)j���0 � z�� ���� �0�� d�0

(52)follows. An estimate similar to (50) holds, namelyZ

C0

���0 � z���2 ��d�0�� � c4 jIm zj�1 � c5 jzjn=2�1

if z 2 C, hence applying Schwarz inequality to (52) concludes the proof.Since su¢ cient conditions for trace class operators are involved, we replace

the det by det2 which is de�ned for Hilbert-Schmidt class operators, namely

det2 (I +A) = e�trA det(I +A).

Moreover, the kernelNg;m (z; �) does not decay with respect to z, hence it shouldbe changed so that the resulting operator Nm(g) turns to be of Hilbert-Schmidtclass. Fortunately we have the following property of determinants:

det(I +A) = det(I +H�1AH)

for any operator H. So for a suitable M > 0 set

H(z) = zM ,

31

and replace Ng;m (z; �) with a kernel

eNg;m (z; �) = H(z)�1Ng;m (z; �)H(�).

The necessary su¢ cient condition for the new operator eNm (g) to be a Hilbert-Schmidt class operator isZ

C2

��� eNg;m (z; �)���2 jdzj jd�j <1. (53)

We replace the part trA in the de�nition of det2 by the integralZC

eNg;m (z; z) dz = ZC

Ng;m (z; z) dz (54)

regardless of the tracability of eNm (g). Then we require for Ng;m to satisfyZC

jNg;m (z; z)j dz <1. (55)

If the kernel eNg;m (z; �) satis�es (53) and (55), the ��function is de�ned by�m(g) = exp

�ZC

Ng;m (z; z) dz

�det2

�I + eNm (g)� . (56)

This determinant and trace was used by [29]. A su¢ cient condition for (53)follows from the estimate (52) and it is

N �M + (n� 2) =4 and (n� 2) =2� 2M < �1. (57)

M can be chosen freely if it satis�es n=4 < M � N�(n� 2) =4, which is possibleif N > (n� 1) =2. Since

(jeme (�)j+ jemo (�)j) j�j(n�2)=2 = j�jN (jeme (�)j+ jemo (�)j) j�j�N+(n�2)=2

(55) is satis�ed if �2N + n � 2 < �1, which is equivalent to N > (n� 1) =2.Consequently we have

Lemma 29 Assume N > (n� 1) =2. Then, for M satisfying n=4 < M �N � (n� 2) =4 (53) and (55) are satis�ed, and the � -function �m(g) can bede�ned by (56) for g 2 �n, m 2MN

n . Moreover, so de�ned �m(g) is continuousonMN

n .

Proof. We have only to show the continuity, which require us to prove thecontinuity of the Hilbert-Schmidt norm of Nm(g) and the integral (54) withrespect to m. We prove only the continuity of min�02C0

��mo

��0���, since the

other quantities are easy to treat. Let m1, m2 2 MNn . Applying Lemma 27 to

m1;o �m2;o 2 �Nn yields�����m1;o (z)�m2;o (z)�N�1Xk=0

ak (z � b0)�k�1����� � c1 jzj�N+(n�2)=4 km1;o �m2;okn;N

(58)

32

with

ak =1

2�i

ZC

(�� b0)k (m1;o (�)�m2;o (�)� f (�)) d�,

where f(z) = f1(z)�f2(z) 2 H (D+! ) and fj 2 H (D+

! ) are such that �N (mj;o � fj) 2

L2 (C!). Noting

jakj � c6

ZC

����k (m1;o (�)�m2;o (�)� f (�))��� jd�j

� c6

sZC

����k�N ���2 jd�jsZC

����N (m1;o (�)�m2;o (�)� f (�))���2 jd�j

for k � N � 1, we easily see from (58) that for z 2 C 0

jm1;o (z)�m2;o (z)j � c7

�jzj�1 + jzj�N+(n�2)=4

�km1;o �m2;okn;N ,

which leads us to the continuity of min�02C0��mo

��0���.

We have to be careful about the continuity of �m(g) with respect to g 2 �n,since the sup-metric of �n on C is not suitable. We will reconsider this continuitylater in proving the positivity of �m(g).Although we have de�ned a new � -function by (56), we have to identify it

with the original � -function if m 2Mreflpr.

Lemma 30 The � -function �m(g) de�ned by (56) for g 2 �n coincides with the� -function de�ned by Nm(g) in (38) if m 2Mreflp

r:

Proof. Take an m 2 Mreflpr. For simplicity we use det in (56) instead of

det2. We divide the curve C de�ning eNm (g) into C1, C2, where C1 containsthe interval [�r; r] and is bounded, and C2 is the rest of the curve C extendingto �1. Notice that me, mo have singularities only on [�r; r], hence C2 hasno singularities of me, mo in its inside. And let N1 and N2 be the operatorsobtained by restricting eNm (g) on C1 and C2 respectively. Then, the analyticityof me, mo in the inside of C2 implies

N2N1 = N22 = 0.

Then, we have

det�I + eNm (g)� = det (I +N1 +N2) = det(I +N1),

by which we can identify the two � -functions if m 2Mreflpr.

6 Appendix

6.1 Herglotz functions

A holomorphic function m on C+ is call ed a Herglotz function if it satis�es

Imm(z) > 0 for any z 2 C+.

33

If m is a Herglotz function, Imm (z) is a positive harmonic function on C+,hence it is known that there exists a measure � on R and � � 0 such that

Imm (z) = �y +

Z 1

�1

y

(�� x)2 + y2� (d�)

with z = x+ iy. The measure � should satisfyZ 1

�1

1

1 + �2� (d�) <1.

Since an identity

�y +

Z 1

�1

y

(�� x)2 + y2� (d�) = Im

��z +

Z 1

�1

�1

�� z ��

�2 + 1

�� (d�)

�holds, ther exists a real � such that

m (z) = �+ �z +

Z 1

�1

�1

�� z ��

�2 + 1

�� (d�) . (59)

This is the representation theorem of a Herglotz function. The measure � iscalled a spectral measure of m. The following facts are known for Herglotzfunctions.

Lemma 31 (1) lim�#0R baImm (�+ i�) d� = �� ((a; b)) if � (fag) = � (fbg) =

0.(2) lim�#0 Imm (�+ i�) = ��0 (�) for a.e. � 2 R, where �0 (�) d� is the ab-solutely continuous part of �.(3) m (�+ i0) = lim�#0m (�+ i�) exists �nitely for a.e. � 2 R and satis-�es m (�+ i0) 6= 0. More generally if two Herglotz functions m1, m2 satisfym1 (�+ i0) = m2 (�+ i0) on some A with positive Lebesgue measure, then itholds that m1 = m2 identically.

6.2 Conformal maps from Cn(�1; 0] to D�

Although Riemann mapping theorem says that every simply connected domainon C can be an image of a conformal map on C+, sometimes a quantitativeestimate of it is necessary. In this section we provide a model of conformal mapfrom Cn(�1; 0] to D�

! of (43).A conformal map on C+ is easily obtained if Im 0 (z) has a de�nite sign

on C+. A simple such example is 1 (z) =pz, and a more general conformal

map in this framework can be constructed for an integer k � 1 by an integral

k (z) =k � 1=2k + 1=2

pz +

Z 1

0

pz + t (1 + t)

�k�3=2dt.

This k satis�es

Re k (z) > 0, Im k (z) > 0, Re 0k (z) > 0, Im

0k (z) < 0 (60)

for z 2 C+, hence k maps C+ to k (C+) (� C+), and �k (z) = k (z)2 maps

C+ to �k (C+) (� C+) conformaly. Since k (z) takes real values on [0;1),

34

k (z) and �k (z) can be extended as conformal maps from Cn(�1; 0] onto adomain in fRe z > 0g and a domain in C respectively. Set8><>:

ak = 2R 10s2�1� s2

�k�1ds =

p�� (k)

2� (k + 3=2)

bk = 2ak

�2a2kk (k + 1=2)

2= (k � 1=2)2 + 1

��1=2 .

The image �k (Cn(�1; 0]) is a type of D! in 5.2.1.

Lemma 32 The image �k (Cn(�1; 0]) is descrived as follows:

�k (Cn(�1; 0]) = Cn�z 2 C; jIm zj � ! (Re z) , Re z � a2k

(61)

with positive smooth function ! (x) on (�1; a2k) such that

! (x) =

(2ak (�x)�k+1=2

�1 +O

�x�1

��as x! �1

bk�a2k � x

�1=2 �1 +O

�a2k � x

��as x! a2k � 0

. (62)

Moreover, �k takes a form of

�k (z) = z + f1(z) + z�k+1=2f2(z) (63)

with some real rational functions f1, f2 (that is, fj (z) = fj (z) for j = 1, 2)satisfying (

f1(1) =�k2 � 1=4

��1f2(1) = 2 (�1)k ak

. (64)

Conversely, ��1k (w) has an expression

��1k (w) = w + g1 (w) + w�k+1=2g2(w) (65)

with g1, g2 holomorphic in a neighborhood of 1. Moreover, it holds that

g1 (1) = ��k2 � 1=4

��1, g2 (1) = �2 (�1)k ak: (66)

Proof. Setting s =p(z + t) = (1 + t), we have

k (z) =k � 1=2k + 1=2

pz + 2 (z � 1)�k

Z pz

1

s2�s2 � 1

�k�1ds.

Since the integralR z0s2�s2 � 1

�k�1ds is an odd polynomial of degree 2k + 1,

the integral

p (z) =k � 1=2k + 1=2

(z � 1)k + 2pz�1Z p

z

0

s2�s2 � 1

�k�1ds (67)

de�nes a polynomial of degree k, and

k (z) = (z � 1)�k �p

zp (z)� p (1)�

(68)

holds. It should be noted thatpzp (z)� p (1) has zero of degree k at z = 1, so

k (z) has no singulality at z = 1. Set

s(x) = Re k (x+ i0) , t (x) = Im k (x+ i0)

35

for x 2 R. Then, (68) implies

s(x) =

(�p(1) (x� 1)�k for x < 0(x� 1)�k (

pxp (x)� p (1)) for x � 0

,

t(x) =

�(x� 1)�k

p�xp (x) for x < 0

0 for x � 0 ,

and their asymptotics are

s(x) =

�ak�1 + kx+O

�x2��

as x! �0ak (�x)�k

�1 + kx�1 +O

�x�2

��as x! �1

t (x) =

8<:k � 1=2k + 1=2

p�x (1 +O (x)) as x! �0

p�x�1 +O

�x�1

��as x! �1

, (69)

where we have used8>><>>:p (1) = (�1)k�1 ak, p(0) =

k � 1=2k + 1=2

(�1)k

p(z) = zk ��k � 2

4k2 � 1

�zk�1 + � � �

. (70)

From (68)�k (z) = k (z)

2= z + f1(z) + z

�k+1=2f2(z)

follows, which yields (63) with(f1(z) = (z � 1)�2k

�p (1)

2+ zp(z)2

�� z

f2(z) = �2 (z � 1)�2k p (1) p (z) zk,

and 8>><>>:Re�k (x+ i0) = x+ f1(x) + f2(x)�

�x�k+1=2 if x > 00 if x < 0

Im�k (x+ i0) =

�0 if x > 0p�xx�kf2(x) if x < 0

(71)

is valid, hence (70) shows8<: Re�k (x+ i0) = x+�k2 � 1=4

��1+O

�(�x)�1

�Im�k (x+ i0) = 2ak (�x)

�k+1=2+O

�(�x)�k�1=2

� (72)

as x! �1, and8>>><>>>:Re�k (x+ i0) = a2k +

2ka2k +

�k � 1=2k + 1=2

�2!x+O

�x2�

Im�k (x+ i0) = 2k � 1=2k + 1=2

akp�x+O

�(�x)3=2

� (73)

as x ! �0. Since Re�k (x+ i0) = s(x)2 � t (x)2, Im�k (x+ i0) = 2s (x) t (x),

(60) implies�Re�k (x+ i0) is increasing and moving from�1 to 1Im�k (x+ i0) > 0 on (�1; 0) and 0 on [0;1) . (74)

36

Therefore, ! can be de�ned by an equation

! (Re�k (x+ i0)) = Im�k (x+ i0) .

due to (74), and (71), (73), (74) show ! (x) satis�es (62).We use (68) to show (65). Set # (z) = z2. # is a conformal map from

fRe z > 0g to Cn(�1; 0] and de�ne e k (s) = k (# (s)). Then the function

F (s) = e k (s)� s= �p (1)

�s2 � 1

��k+ s

��s2 � 1

��kp�s2�� 1�.

is a rational function whose poles only at s = �1 and has expansion

F (s) = c1s�1 + c2s

�3 + � � �+ cks�2k+1 + ck+1s�2k+

at s =1 with c1 =�2k2 � 1=2

��1and ck+1 = �p(1), namely the �rst coe¢ cient

of even order starts from 2k. We consider an equation for a given t:

s+ F (s) = t (75)

and �nd a solution of a form

s = t+G(t).

Since the even coe¢ cients of the power series of F vanish up to 2 (k � 1), Lemma33 shows that there exists uniquely such G that G is holomorphic near t = 1and the odd coe¢ cients of G vanishes up to 2 (k � 1). e k (z) is one-to-one onfjzj > r1g and its inverse is given by w + G(w) on fjwj > r2g. Since �k (z) =# ( k (z)) is a conformal map from Cn(�1; 0] to �k (Cn(�1; 0]), its inverse isgiven by ��1k (w) =

�#e �1#�1� (w) for w 2 �m (Cn(�1; 0]). Let

G (t) = Ge�t2�+ tGo(t

2).

Then, we have�#e �1#�1� (w)=�pw +Ge (w) +

pwGo(w)

�2= w +Ge (w)

2+ w

�(Go(w) + 1)

2 � 1�+ 2pwGe (w) (Go(w) + 1) .

Since G has the vanishing even coe¢ cients up to 2 (k � 1), wkGe (w) is holo-morphic near w =1. Therefore, setting�

g1(w) = Ge (w)2+ wGo(w) (Go(w) + 2)

g2(w) = 2wkGe (w) (Go(w) + 1)

,

we have��1k (w) = w + g1(w) + w

�k+1=2g2(w)

with some g1, g2 holomorphic in a neighborhood of 1 satisfying

g1(1) = ��k2 � 1=4

��1, g2 (1) = �2 (�1)k ak,

which completes the proof.

37

Lemma 33 Let F be a power series of s�1 given by F (s) =P1

j=1 ajs�j and

assume it has the positive radius of convergence and consider an equation:

t = s+ F (s). (76)

(i) This equation is uniquely solvable if��t�1�� is su¢ ciently small and it has a

form:s = t+G(t)

with a convergent power series of t�1 given by

G(t) =1Xj=1

xjt�j. (77)

(ii) xn is determined from fajgnj=1 for each n � 1. The �rst three coe¢ cintsare

x1 = �a1, x2 = �a2, x3 = �a21 � a3.

(iii) Suppose F (s) has a form

F (s) =

kXj=1

a2j�1s�2j+1 +

1Xj=2k

ajs�j (78)

for an k � 1. Then, the coe¢ cients xj of G(t) vanish for even j up to 2 (k � 1).Moreover, if a2j 6= 0, then x2j = �a2j.

Proof. Replacing s by s�1 and t by t�1 we see the equation (76) is equivalentto

t =s

1 + sF (s�1). (79)

The condition on F implies

t(0) = 0,dt

ds(0) = 1,

d2t

ds2(0) = 0,

hence the complex function theory shows the existence of the solution s(t) of(79) in a neighborhood of 0 satisfying

s(0) = 0,ds

dt(0) = 1,

d2s

dt2(0) = 0,

which implies the existence of G of the form of (77). One can show inductivelythat the coe¢ cient xn is determined from fajgnj=1. To show (iii) one can assumeaj = 0 for every j � 2k owing to (ii). The relation between F , G is rewritten as

F (s) +G (F (s) + s) = 0.

If we de�ne bf(s) = �f (�s), then the above equation turns tobF (s) + bG� bF (s) + s� = 0.

38

Therefore, if bF (s) = F (s), the uniqueness implies bG (s) = G(s), which showsthe �rst part of (iii). To show the second part we note that if8><>:

F (s) =Pk�1

j=1 ajs�j + aks

�k � F1(s) + aks�k

G(s) =Pk�1

j=1 xjs�j + xks

�k +P1

j=k+1 xjs�j

� G1(s) + xks�k +G2(s)

,

and with some bm

F1(s) +G1 (s+ F1(s)) = bks�k +O

�s�k�1

�(80)

holds, which is veli�ed by induction, then the identity

F1(s) + aks�k +G1(s+ F1(s) + aks

�k) + xk�s+ F1(s) + aks

�k��k+G2(s+ F1(s) + aks

�k) = 0

together with

G1(s+ F1(s) + aks�k) = G1(s+ F1(s)) +O

�s�k�2

�implies xk = �ak � bk. Since, if k is even and a2j = 0 for any j � k=2, then(ii) implies xk = 0, and hence bk = 0. However, clearly bk is determined fromfajg1�j�k�1, hence bk = 0 is valid if a2j = 0 for any j � k=2 � 1 regardless ofthe value ak. Consequently we have xk = �ak if k is even and a2j = 0 for anyj � k=2� 1 holds.

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