A priori estimates for the Hill and Dirac operators

0.21 MB·English

by  Evgeny Korotyaev

A priori estimates for the Hill and Dirac operators
Evgeny Korotyaev
∗
February 2, 2008
7
0
0 Abstract
2
Consider the Hill operator Ty = y + q (t)y in L2(R), where q L2(0,1) is a 1-
′′ ′
n − ∈
periodic real potential. The spectrum of T is is absolutely continuous and consists of
a
J bandsseparatedbygapsγ ,n > 1withlength γ > 0. Weobtain aprioriestimates of
n n
| |
6 the gap lengths, effective masses, action variables for the KDV. For example, if µ are
±n
1 theeffective massesassociated withthegapγ = (λ ,λ+),then µ +µ+ 6 C γ 2n 4
n −n n | −n n| | n| −
] for some constant C = C(q) and any n > 1. In order prove these results we use the
P
analysisofaconformalmappingcorrespondingtoquasimomentumoftheHilloperator.
S
That makes possible to reformulate the problems for the differential operator as the
.
h
problems of the conformal mapping theory. Then the proof is based on the analysis of
t
a the conformal mapping and the identities. Moreover, we obtain the similar estimates
m
for the Dirac operator.
[

1
v 1 Introduction and main results
7
3
4 Consider the self-adjoint periodic weighted operator T in L2(R,w2(t)dt), given by
w
1
0 1
7 Twf = w−2(w2f′)′ = f′′ 2pf′, w(t) = eR0xp(s)ds, p H = p L2(T), p(t)dt = 0 .
− − − ∈ { ∈ }
0 Z0
/
h The operator T is well studied, see [K8], [K9], [Kr] and references therein. Define the
w
at unitary transformation U : L2(R,w2dt) L2(R,dt) as multiplication by w. Then T is
w
m unitarily equivalent to the Hill operator T→= UT U 1 given by
w −
:
v d d d2
i T = +p p = +q +q > 0, q = p(t)+p2(t) p 2, (1.1)
X − dt dt − −dxt ′ 0 ′ ′ −k k
(cid:18) (cid:19)(cid:18) (cid:19)
r
a where q = p 2 = 1p2(t)dt and q is a 1-periodic potential (distribution). We mention
0 k k 0 ′
following papers devoted to singular potentials [KM], [K7], [NS], [SS]. We define the Riccati
R
type mapping R : H H by
→
x 1 1
q = R(p) = p(x)+ (p2(t) p 2)dt+ t p2(t)dt, p H. (1.2)
−k k − 2 ∈
Z0 Z0 (cid:18) (cid:19)
∗
Institutfu¨rMathematik,HumboldtUniversit¨atzuBerlin,RudowerChaussee25,12489,Berlin,Germany,
e-mail: [email protected]
1
Recall that the Riccati map R : H H is a real analytic isomorphism between H and H,
→
see [K7]. Thus, if p L2(T), then T corresponds to the Hill operator T with L2-potential.
′ w
∈
We can not introduce the standard fundamental solutions for the operator T, since the
perturbation q is very strong. But the operator T = U 1TU has the standard fundamental
′ w −
solutions ϕ(t,λ),ϑ(t,λ), which satisfy the equation y 2py = λy, λ C and the con-
′′ ′
− − ∈
ditions ϕ(0,λ) = ϑ(0,λ) = 1, ϕ(0,λ) = ϑ(0,λ) = 0. Here and below we use the notation
′ ′
u = ∂u. Introduce the Lyapunov function ∆(λ) = ϕ′(1,λ)+ϑ(1,λ).
′ ∂t 2
It is well known, that the spectrum σ(T) = σ(T ) of T is absolutely continuous and
w
consistsofintervalsσ = [λ+ ,λ ],whereλ+ < λ 6 λ+, n > 1andλ+ = 0. Theintervals
n n 1 −n n 1 −n n 0
σ andσ areseparatedby−thegapγ = (λ−,λ+). Ifagapdegenerates, thatisγ = ,then
n n+1 n −n n n ∅
the corresponding segments σ and σ merge. Note that ∆(λ ) = ( 1)n, n > 1, and there
n n+1 ±n −
exists aunique pointλ [λ ,λ+]suchthat∆(λ ) = 0. Thesequence λ+ < λ 6 λ+ < ...
n ∈ −n n ′ n 0 −1 1
is the spectrum of equation y + qy = zy with the 2-periodic boundary conditions, i.e.
′′
−
y(t+2) = y(t),t R. If λ = λ+ for some n, then this number λ is the double eigenvalue
∈ −n n ±n
of this equation with the 2-periodic boundary conditions. The lowest eigenvalue λ+ is always
0
simple, ∆(λ+) = 1, and the corresponding eigenfunction is 1-periodic. The eigenfunctions,
0
corresponding to the eigenvalue λ , are 1-periodic, when n is even and they are antiperiodic,
±n
i.e. y(t+1) = y(t), t R, when n is odd.
− ∈
We introduce a new parameter (momentum) z = x+iy = √λ,√1 = 1, and momentum
gaps
g = (z ,z+), z = λ > 0, g = g , n > 1, g = .
n n− n n± ±n −n − n 0 ∅
We define a quasimomentum k(z) p= u(z)+iv(z) = arccos∆(z2), z = C g, g = gn,
∈ Z \ ∪
see [MO]. The function k(z) is a conformal mapping from onto a quasimomentum domain
Z
= C n ZΓn, where the slit Γn = [πn+ihn,πn ihn], h0 = 0, andhn = h n > 0 isdefined
Kby the\e∪qua∈tion ∆(λ ) = ( 1)ncoshh . The qua−simomentum k satisfies k(−z) = z +o(1) as
n n
−
z . With each edge of the gap γ = , we associate the effective mass µ+,µ by
| | → ∞ n 6 ∅ 0 ±n
(k πn)2
λ = z2 = λ + − (1+o(1)) as λ λ . (1.3)
±n 2µ → ±n
±n
Introduce the real Hilbert spaces ℓ2 = f = (f ) , f R, f 2 = (2πn)2pf2 <
p n ∞1 n ∈ k kp n>1 n
(cid:26)
P
, p > 0 and let ℓ2 ℓ2. For the sake of the reader, we briefly recall the results existing
∞ 0 ≡
(cid:27)
in the literature about the a priori estimates. Marchenko and Ostrovki [MO1-2] obtained the
estimates: q 6 C(1+h ) h , h 6 C q exp( A q ), for some absolute constants
1 + 1 1 1 1
k k k k k k k k k k
A,C, where h = sup h and q = q +q . These estimates are not sharp since they used
+ n>1 n 1 0 ′
the Bernstein inequality. Using the harmonic measure argument Garnett and Trubowitz
[GT] obtained γ 6 (4+ h ) h .
1 1
k k k k k k
Various estimates for the Hill operator with a potential q L2(0,1) were obtained in
′
∈
the author’s paper [K5-6] and in [KK1], [K3] for the Dirac operator. In the paper [K3], [K5],
[K6] the following sharp estimates for the case q L2(0,1) were obtained:
′
∈
1 1
kγk 6 6kq′k(1+kq′k)3, kq′k 6 4kγk(1+kγk)3, γ = (|γn|)∞1 ,
2
2 h 1 6 π q′ (1+ q′ )31, q′ 6 3 h 1(6+h+)12, h+ = suphn,
k k k k k k k k k k
n>1
and for the case q L2(0,1) [K8]:
∈
q 6 2 h (1+4 h ), h 6 3 q (1+2 q ), q H,
k k k k k k k k k k k k ∈
q 6 48π2 γ (1+ γ )3, γ 6 2 q (1+2 q )3.
1 1 1
k k k k− k k− k k− k k k k
It is well known that v(z) = Imk(z) > vn(z) = |(z −zn−)(z −zn+)|21,z ∈ gn, see [KK1].
Define a function V (z) = v(z) 1 > 0,z g , which is analytic in the domain C C
n vn(z) − ∈ n + ∪ − ∪
(z+ ,z ), see [KK1]. Let M = max V (z),M = max V (z),..
n 1 n−+1 n z gn n n′ z gn n′
− ∈ ∈
Theorem 1.1. Let real q L2(0,1). Then for each n > 1 the following estimates hold
∈
g 2 3 γ 2 z +z+ λ +λ+
λ0 | n| λ 6 | n| M , where z0 = n− n , λ0 = −n n , (1.4)
| n − 4 − n| 8z0 n′ n 2 n 2
n
γ (1+V (z )) γ 3M 2
h | n| n n 6 | n| n′ , (1.5)
| n − 4z0 | 16z03
n n
1
g 2 M γ µ 2
h h 6 | n| M 1+ n′ | n|| ±n| , where h = g z µ , (1.6)
| n − ±n| 2 n′ 4 ±n | n| n± ±n
(cid:18) p (cid:19) (cid:12) (cid:12)
(cid:12) (cid:12)
g 2 (cid:12) (cid:12)
|µ+n +µ−n| 6 4|z+nz| (1+Mn)(1+2zn0Mn′), (cid:12) (cid:12) (1.7)
n n−
g 2 M 2h2
µ µ 6 | n| (1+M )2 1+4z M 1+ n′ γ µ , where µ = n. (1.8)
| ±n ∓ n| 8z0z n n± n′ 4 | n|| ±n| n γ
n n± (cid:20) (cid:18) (cid:19)(cid:21) | n|
p
Remark. 1) Using Theorem 1.1 and estimates
(3π)2
γ 6 (4πn)2(µ+ µ ), h2 6 γ µ , (1.9)
| n| n − −n n 2 | n|| ±n|
from [K4], we obtain the estimates in terms of effective masses.
2) Recall that z = πn 1ei2πnxq(x)dx + ℓd(n), d > 1 as n , see [K8]. This
n± ± | 0 | → ∞
yields asymptotic estimate ρ > π ε as n , for any ε > 0.
n
R − → ∞
Define the action variables A = 4 xv(x)dx > 0, n > 1 for the KDV equation [FM].
n π gn
Theorem 1.2. Let real q L2(0,1) Rand let β = √A > 0,n > 1. Then the following
n n
∈
estimates hold
γ 2 g 4
A | n| (1+V (z0)) 6 | n| (z0M +M ), (1.10)
n − 8z0 n n 25 n n′′ n′
(cid:12) n (cid:12)
(cid:12) (cid:12)
(cid:12)(cid:12)|An − |γn2|hn| 6 |g2n5|4(z(cid:12)(cid:12)n0Mn′′ +6zn0Mn′2 +Mn′), (1.11)
1
γ g m 2 g 3 g
A | n| | n| ±n 6 | n| z0M + | n|M + g M , (1.12)
| n − 2 2 | 4 n n′ 8 n′ | n| n′′
(cid:12) (cid:12) (cid:18) (cid:19)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12) 3
(1+V (z0)) g 3
β γ n n 6 | n| (z0M +M ), (1.13)
n −| n|s 8z0 25 2z0 n n′′ n′
(cid:12) n (cid:12) n
(cid:12) (cid:12)
(cid:12)(cid:12)β |γn|hn 6 |gn|3(cid:12)(cid:12) (z0Mp+6z0M 2 +M ), (1.14)
n − 2 25 2z0 n n′′ n n′ n′
(cid:12) r (cid:12) n
(cid:12) (cid:12)
β (cid:12)(cid:12) |γn|||gn|m±n|(cid:12)(cid:12)14 6 p|gn|2 z0M + |gn|M + g M . (1.15)
| n −r 2 214 | 4 2zn0(cid:18) n n′ 8 n′ | n| n′′(cid:19)
Consider the Dirac operator T actinpg in the space L2(R) L2(R) and given by
D
⊕
d q q 0 1
T = +q (t), q = 1 2 ,
D J dt D D q q J ≡ 1 0
2 1
(cid:18) − (cid:19) (cid:18)− (cid:19)
where q ,q L2(0,1) are real 1-periodic functions of t R. The spectrum of T is purely
1 2 D
∈ ∈
az+bsolu<telzy c6ontzin+u,o(usseean[LdSσ] (fTorD)de=ta∪ilsn)∈.ZsTn,hwe hinetreervsanls=s[z,n+s−1,zn−a]rearseepsapreactterdalbbyanthdes agnadp
n 1 n− n n n+1
g −= (z ,z+) with length g > 0. If a gap g = , then the corresponding spectral
n n− n | n| n ∅
bands s ,s merge. Introduce the 2 2-matrix valued fundamental solution ψ = ψ(t,z)
n n+1
×
satisfying Jψ + q ψ = zψ,ψ(0,z) = I , z C, where I is the 2 2 identity matrix.
′ D 2 2
∈ ×
Define the Lyapunov function ∆(z) = 1Trψ(1,z). Note that ∆(z ) = ( 1)n,n Z, and
2 n± − ∈
the function ∆(z) has a zero z in a ”closed gap” [z ,z+] (see e.g. [LS]). The sequence
′ n n− n
..z+ < z 6 z+ < ... is the spectrum of equation Jy + q y = zy,y = (y ,y ) , with the
0 1− 1 ′ D 1 2 ⊤
2-periodic boundary conditions, i.e. y(t + 2) = y(t),t R. If z = z+ for some n, then
∈ n− n
z is the double eigenvalue of the equation Jy + q y = zy with the 2-periodic boundary
n± ′ zs
conditions. The eigenfunctions, corresponding to the eigenvalue z , are 1-periodic, when n
n±
is even and they are antiperiodic, i.e. y(t+1) = y(t), t R, when n is odd.
− ∈
For each q there exists a unique conformal mapping (the quasimomentum) k :
D
Z →
= C Γ ,Γ = [πn ih ,πn+ih ], given by (see [K6])
n n n n
K \∪ −
cosk(z) = ∆(z), z = C g, g = g , k(z) = z +o(1) as z ,
n
∈ Z \ ∪ | | → ∞
where the height h > 0 is defined by the equation coshh = ( 1)n∆(z ) > 1. With each
n n n
−
edge of the gap g = , we associate the effective mass m by
n 6 ∅ ±n
(k(z) πn)2
z = z + − (1+o(1)) as z z . (1.16)
n± 2m → n±
±n
Theorem 1.3. Let real q L2(0,1). Then for each n Z the following estimates hold
D
∈ ∈
g 2
z z0 6 | n| M , (1.17)
| n − n| 4 n′
g 3
2h g (1+V (z ) 6 | n| M 2, (1.18)
| n −| n| n n | 8 n′
4
g 2 M g m
h h 6 | n| M 1+ n′ 2 g m , where h = | n|| ±n|, (1.19)
| n − ±n| 2 n′ 8 | n|| ±n| ±n 2
(cid:18) (cid:19) r
p
m+ +m 6 g 2(1+M )M , (1.20)
| n −n| | n| n n′
M 2h2
m m 6 g 2(1+M )M 1+ n′ 2 g m , where m = n. (1.21)
| ±n ∓ n| | n| n n′ 8 | n|| ±n| n g
(cid:18) (cid:19) | n|
p
Remark. Using Theorem 1.3 and estimates
g
| n| 6 h 6 π 2 g m 6 2π m , h2 6 2 g m+ m , g 6 2 m . (1.22)
2 n | n|| ±n| | ±n| n | n| n| −n| | n| | ±n|
p p
from [KK1], we obtain the estimates in terms of effective masses.
Recall that the action variables for the NLS equation are given by a = 1 v(x)dx, n
Z. n π gn ∈
R
Theorem 1.4. Let real q L2(0,1) and let b = √a > 0,n Z. Then the following
D n n
∈ ∈
estimates hold
g 4
a ( g 2/8)(1+V (z0)) 6 2| n| M , (1.23)
| n − | n| n n | 44 n′′
g h g 4
a | n| n 6 | n| (M +6M 2), (1.24)
| n − 4 | 27 n′′ n′
g 3/2 g 3 g 4
|an − | n2|52 |m±n|21| 6 |1n6| Mn′ + |2n7| Mn′′, (1.25)
√2 g 3
b 2 3/2 g 1+V (z0) 6 | n| M , (1.26)
| n − − | n| n n | 43 n′′
p
√2 g 3
b g h /2 6 | n| (M +6M 2), (1.27)
| n − | n| n | 27 n′′ n′
|bn −p|g2n45|43|m±n|41| 6 (|gn8|2Mn′√+2|g8n2|3Mn′′). (1.28)
Notethat Theorem 1.4 isused to study electrostatic inverse problems onthe plane[KK3].
Define integrals Q = 1 xpv(x)dx,p Z. Recall the identities for the Hill operator
p π R ∈
Rq p 2 1 1
Q = 0 = k k , Q = (q2 +q (t)2)dt, (1.29)
0 2 2 2 8 0 ′
Z0
(recall that λ+ = 0 for the Hill operator), and for the Dirac operator
0
q 2 1 1 1 1
Q = k Dk = (q2(t)+q2(t))dt, Q = ((q 2 +q 2)+(q2 +q2)2)dt. (1.30)
0 2 2 1 2 2 8 1′ 2′ 1 2
Z0 Z0
5
Theorem 1.5. Let real q L2(0,1) or q L2(0,1). Let ρ = dist(g ,g g ), n Z. Then
D n n n
∈ ∈ \ ∈
for each n Z the following estimates hold
∈
2 √8Q M M
M 6 suph 6 0, M 6 n, M 6 n, (1.31)
n πρn s Z s πρn n′ ρn n′′ ρ2n
∈
π2
M 6 Q , where ρ = inf ρ , (1.32)
n 3ρ2 0 n Z n
n Z ∈
X∈
1 5
6 e5√Q0/2. (1.33)
ρ 2
If q L2(0,1) or q L2(0,1), then
′ ∈ D′ ∈
2 Q
V (x) 6 Q + 2 , x g . (1.34)
n x z+z 0 ρ2 ∈ n
| | | n n−|(cid:18) n(cid:19)
Corollary 1.6. Let real q L2(0p,1). Then the following estimates hold
D
∈
g h 3 π2 √8Q
|an − | n4| n|31 6 Q30 47ρ4 1+ πρ 0 , (1.35)
(cid:18)n Z (cid:19) (cid:18) (cid:19)(cid:18) (cid:19)
X∈
g h g4 π2Q √8Q
a | n| n 6 + 0 1+6 0 , g = sup g . (1.36)
| n − 4 | 27 3ρ4 πρ + | n|
n Z (cid:18) (cid:19) n
X∈
In order to prove these results we consider the quasimomentum of the Hill (or the Dirac)
operator as a conformal mapping. That makes possible to reformulate the problem for the
differential operator as a problem of the conformal mapping theory (see [KK1], [K1]-[K8]
and [MO1]). We use the Poisson integral for the domain C C g and some a priori
+ n
∪ − ∪
estimates from [KK1]. The results of this paper are used in [KK3].
2 Estimates for conformal mappings
Consider a conformal mapping z : C with asymptotics z(iv) = iv(1 + o(1)) as
+ +
K →
v , where z(k),k = u + iv K. The domain = C for some sequence
+ +
→ ∞ ∈ K ∩ K
(hn)n Z ℓ∞,hn > 0 and the domain = C n ZΓn, where Γn = [un ihn,un +ihn] and
u ,n∈ Z∈is strongly increasing sequenKce of re\a∪l n∈umbers such that u − as n .
n n
∈ → ±∞ → ±∞
The difference of any two such mappings equals a real constant. Thus the imaginary part
y(k) = Imz(k) is unique. We call such mapping z(k) the comb mapping. Define the inverse
mapping k( ) : C . It is clear that k(z),z = x+iy C has the continuous extension
+ + +
· → K ∈
into C . We define ”gaps” g , ”bands” s and the ”spectrum” s of the comb mapping by:
+ n n
gn = (zn−,zn+) = (z(un −0),z(un +0)), sn = [zn+−1,zn−], s = ∪n∈Zsn.
It is well known that the set s can not be the spectrum of two different comb mappings [Le].
The function u(z) = Rek(z) is strongly increasing on each band s and u(z) = u for all
n n
6
z [z ,z+], n Z; the function v(z) = Imk(z) equals zero on each band s and is strongly
∈ n− n ∈ n
convex on each gap g = and has the maximum at some point z given by v(z ) = h .
n n n n
6 ∅
If the gap is empty we set z = z . The function z( ) has an analytic extension (by the
n n± ·
symmetry) from the domain onto the domain and z( ) : z( ) = = C g is
K+ K · K → K Z \∪ n
a conformal mapping. For each gap g = (z ,z+) = we define the effective masses m by
n n− n 6 ∅ ±n
(k u )2
z(k) z = − n +O((k u )3) as z z . (2.1)
− n± 2m − n → n±
±n
If g = 0, then we set m = 0. It is clear that if g > 0, then m > 0. These and others
| n| ±n | n| ± ±n
properties of the comb mappings it is possible to find in the papers of Levin [Le].
We emphasize that the comb mapping k(z) with u = πn,n Z corresponds to the
n
∈
quasimomentum for the Dirac operator with periodic coefficients. If the comb mapping
k(z) is symmetric, k( z) = k(z),z , with u = πn,n Z, then it corresponds to
n
− − ∈ Z ∈
the quasimomentum for the Hill operator. Furthermore, the comb mapping is the rotation
number for the Schr¨odinger operator with some almost periodic potential (see [JM]).
For some (hn)n Z ℓ∞R we fix some gap G = ( c,c),c > 0. The function v = Imk(z)
∈ ∈ −
satisfies
1 v(t)dt
v(x) = r(x) 1+V(x) , V(x) = , r(x) = c2 x2 21, x ( c,c), (2.2)
π t x r(t) | − | ∈ −
ZR\g | − |
(cid:0) (cid:1)
see [KK1]. Recall that v (x) < 0 for all x G. The function v(x),x G has a unique
′′
∈ ∈
maximum h at some point z ( c,c) given by
0
∈ −
h = v(z ) = r(z )(1+V ) > c, V = V(z ). (2.3)
0 0 0 0 0
Lemma 2.1. Let a comformal mapping k : have a gap G = ( c,c),c > 0 for some
Z → K −
(hn)n∈Z ∈ ℓ∞ and V0 = V(z0), V0′ = V′(z0),..,r0 = r(z0),ν = −1/v′′(z0) > 0. Then
r2V
z = 0 0′ , where r 6 c, (2.4)
0 0
1+V
0
r3V 2
c r 6 0 0′ , (2.5)
| − 0| 2(1+V )2
0
r3V 2
h c(1+V ) 6 0 0′ , (2.6)
0
| − | 2(1+V )
0
c2
ν 6 hc2((3+c)V 2 +V ). (2.7)
|h − | 0′ 0′′
Proof. We have v (z ) = 0 and 0 = v (x) = r (x)(1 + V(x)) + r(x)V (x) at x = z and
′ 0 ′ ′ ′ 0
(r2) = 2x, which yield (2.4). Substituting (2.4) into the identity z2 = c2 r2, where
r =′ r(z−) we obtain c2 r2 = r04V0′2 , which implies (2.5). 0 − 0
0 0 − 0 (1+V0)2
7
Estimate (2.5) together with the identity h = r (1+V ) yield
0 0
r3V 2
h c(1+V ) = (1+V ) r c 6 0 0′ ,
0 0 0
| − | | − | 2(1+V )
0
which gives (2.6). Differentiating the function v (x)r(x) at z we obtain
′ 0
′
r v (z ) = (rv ) = x(1+V)+r2V = (1+V ) 3z V +r2V ,
0 ′′ 0 ′ ′|z=z0 − ′ |z=z0 − 0 − 0 0′ 0 n′′
(cid:18) (cid:19)
which gives r v (z ) = (1+V ) 3z V +r2V . Then the definition ν = 1/v (z ) implies
0 ′′ 0 − 0 − 0 0′ 0 0′′ n − ′′ 0
r = ν(1+V +3V z r2V ), which yields
0 0 0′ 0 − 0 0′′
r r2 3V z r2V r4V 2
0 = 0 = ν +R, R = ν 0′ 0 − 0 0′′, c2 r2 = 0 0′ . (2.8)
1+V h 1+V − 0 (1+V )2
0 0 0
Thus ch2 = ν +R1,R1 = R+ c2−hr02. Thus |R| 6 νr02(3V0′2 +V0′′) and c2−hr02 6 (1r+03VV00′2)2, which
yield (2.7), where the estimate ν 6 h from [KK4] was used.
Recall the identity from [KK1]
m = c(1+V( c)2 > c. (2.9)
±
± ±
We obtain estimates of the action. Let M = max V(x),M = max V (x),...
x G ′ x G ′
∈ ∈
Lemma 2.2. Let a comformal mapping k : have a gap G = ( c,c),c > 0 for some
Z → K −
(hn)n∈Z ∈ ℓ∞. Then the actions a = π1 Gv(x)dx and A = π4 Gxv(x)dx and b = |a|21 > 0
satisfy
R R
c2 c4
a (1+V(0)) 6 M , (2.10)
′′
| − 2 | 8
c4
A 6 M , (2.11)
′
| | 2
ch c4
a 6 (M +6M 2), (2.12)
′′ ′
| − 2 | 8
c c3 c
a c m 6 (M + M ), (2.13)
′ ′′
| − 2 | ±|| 2 4
p c3
b c (1+V(0))/2 6 M , (2.14)
′′
| − | 8√2
p
c3
b ch/2 6 (M +6M 2), (2.15)
′′ ′
| − | 8√2
p
c3/4 c2 c
1
b m 4 6 (M′ + M′′). (2.16)
| − √2| ±| | 2√2 4
8
Proof. Using the Taylor formula V(x) = V(0)+V (0)x+ xV (s)(x s)ds, we obtain
′ 0 ′′ −
R x
πa = v(x)dx = r(x)(1+V(x))dx = a +a , a = r(x)dx V (s)(x s)ds,
(1) (2) (2) ′′
−
ZG ZG ZG Z0
c2
a = r(x)(1+V(0)+V (0)x)dx = π (1+V(0)).
(1) ′
2
ZG
Consider a . The function V (x) > 0 for all x G, then for some x G we get
(2) ′′ 1
∈ ∈
x x c4
a = r(x)dx V (s)(x s)ds = V (x ) r(x)dx (x s)ds = πV (x )
(2) ′′ − ′′ 1 − ′′ 1 24
ZG Z0 ZG Z0
x
which yields (2.10). Using V(x) = V(0)+ V (s)ds and xr(x)(1+V(0))dx = 0, we have
0 ′ G
4 R 4 R x
A = xv(x)dx = xr(x)dx V (s)ds,
′
π π
ZG ZG Z0
4 x 4M c4M
A 6 x r(x) V (s)ds dx 6 ′ x2r(x)dx = ′,
′
| | π | | | | π 2
ZG Z0 ZG
which yields (2.11). Using (2.3) we have the following identity
h = c(1+V(0))+(h c(1+V ))+cz V (x ), where V = V(z ),
0 0 ′ 1 0 0
−
for some x G, which yields
1
∈
hc c2 c(h c(1+V )) c2z
0 0
a = a (1+V(0)) + − + V (x ).
′ 1
− 2 − 2 2 2
(cid:18) (cid:19)
Using (2.10), (2.6), (2.4) we obtain
hc c4 c4M 2 c4M 2 c4
a 6 M + ′ + ′ < (M +6M 2),
′′ ′′ ′
| − 2 | 23 4 2 8
which yields (2.12). Using (2.9) we obtain
c c2 c4 c3
a √cm 6 a (c2/2)(1+V(0)) + V(0) V( c) 6 M + M ,
′′ ′
| − 2 ±| | − | 2 | − ± | 8 2
which yields (2.13). Using (2.10) we obtain b c (1+V(0))/2 6 c4 M′′ , and the
| − | 8 b+2−21c
estimate a = b2 > c2/2 yields (2.14). Using (2.12) wepobtain b ch/2 6| c4 M′′+|6M′2, and
| − | 8 b+√ch/2
| |
the estimate a = b2 > c2/2,h > c yields (2.15). Due to (2.13) wephave
c3/4 c3 (M + cM )
b m 14 6 ′ 4 ′′ ,
| − √2| ±| | 2 (b+ c3/4 m 41)
√2 | ±|
and m > c,a > c2 gives (2.16).
| ±| 2
9
Lemma 2.3. Let a comformal mapping k : have a gap G = ( c,c),c > 0 for some
Z → K −
(hn)n Z ℓ∞. Then
∈ ∈
V 2√cm
h h 6 2c2 M + 0′ ± , where h = c m , (2.17)
| − ±| ′ 4(1+V )2 ± | ±|
(cid:18) 0 (cid:19)
p
m+ +m 6 4c2(1+M)M , (2.18)
− ′
| |
V 2√cm h2
m m 6 4c2(1+M) M + 0′ ± , where m = . (2.19)
± ′
| ∓ | 4(1+V )2 c
(cid:18) 0 (cid:19)
Proof. Let V = V(z ) and V = V( c). Identities (2.9), (2.3) give
0 0
± ±
h h = r (1+V ) c(1+V ) 6 r V V + r c (1+V ).
± 0 0 0 0 0
| − | | − ± | | − ±| | − | ±
Then estimates (2.5), (2.9) give r c (1+ V ) 6 c2V0′2√c|m±|, and we also have V(z )
| 0 − | ± 2(1+V0)2 | 0 −
V( c) 6 2cM . Combining two last estimates we obtain (2.17).
′
± |
We prove (2.18). Identity (2.9) gives
m+ +m = c(2+V +V ) V V 6 4c2(1+M)M ,
− + + ′
| | − | − −|
which yields (2.18). We prove (2.19). Using (2.9), (2.3) we have
(h+)2 h2 h+ h h+ +h V 2√cm
m+ m = | − | = | − || | 6 h+ +h 2c M + 0′ ± ,
′
| − | c c | | 4(1+V )2
(cid:18) 0 (cid:19)
and h+h+ 6 2c(1+M) yields (2.19).
We rewrite (2.2) for any gap g = (z ,z+) by
n n− n
1 v(t)dt
v(x) = vn(x)(1+Vn(x)), Vn(x) = π v (t) t x , vn(x) = |(x−zn−)(x−zn+)|21, (2.20)
Zg\gn n | − |
x g . We have the following identities and estimates
n
∈
1 v(t)dt 2 v(t)dt
V (x) = , V (x) = > 0, (2.21)
n′ π (t x) t x v (t) n′′ π t x 3v (t)
Zg\gn − | − | n Zg\gn | − | n
V (x) 2V (x)
V (x) 6 n , 0 < V (x) 6 n , (2.22)
| n′ | ρ n′′ ρ2
n n
for all x g . Define the numbers M = max V (x),M = max V (x), . Let
Q = 1 ∈x2pvn(x)dx,p > 0. We prove Thneorem 1.5xf∈ogrnmnore genn′eral case.x∈gn n′
p π R
LemmaR2.4. Let u = infn Z(un+1 un) > 0 and let Q0 < . Then
∗ ∈ − ∞
2 √8Q M M
M 6 suph 6 0, M 6 n, M 6 n, (2.23)
n πρn s Z s πρn n′ ρn n′′ ρ2n
∈
10