Đánh giá lớp phép biến hình Á bảo giác lên vành khăn bị cắt theo các cung tròn đối xứng quay 7

28

Lu(m van Th(lc sj Toan h(Jc - Tntclng Thu(1n

Chuang 5
cA C DANH GIA LOP HAM G
Trang chu'ang nay, chung toi danh gia cae d£;lilu'cjngd~c tru'ng cha mi~n
chuffn cling nhu' moduli cua cae lOp ham G. Vi~c danh gia ban kfnh
Q (g), 9 E G, dong m9t vai tro quail trQng trang vi~c danh gia cae d£;li
lu'cjngkhac, VI the' chung Wi b~t dftu voi danh gia nay.

5.1

Danh gia ban kinh Q(g)

Dinh Iy 5.1. VcJicae giG thief va ky hi£lu iJ ehLtdng 2, V9 E G, ta e6
K

S~~)

(1+

r'

(5.1)

< Q(g) < /i-*,

trong d6 dang thue trai xdy ra khi B = Bo va g(w) = alwIK-1w,w E
1

1

B, lal = 1 va dang thue phdi xdy ra khi B = Bo va g(w) = blwlK-W,w E
B, Ibl= 1.
Chung minh. Ap d\lng b6 d~ 4.1 cha PBHKABG ngu'cjcf
ta co
2

32

>

81

>

81

(Q~g)r+ps (R~9)r
2

(Q~g))K
+ps

C2.~ psrIf

9 E G,

2

(doR(g) < 1).

Tli day, suy fa c~n du'oi cua Q(g)

Q(g) >

= g-l,

=

(1+

S~~)rIf.

(5.2)


Khi B = Bo va g(w) = alwIK-1w, la! = 1 thl d~ng thuc Kayfa.
M~H khac, ne'u f.1 > 1, ap d\lng [13,dinh 19 1] cho PBHKABG 9 E G

ta nh~n dllc;1c

2

7r12

>

7rQ2 (g) {l K .

Suy fa

1

Q(g) <

{l-K.

Tli day, nhd [14,c6ng thuc 2.5], ta co d~ng thuc Kay fa khi B
-k-1

= Bo

~

va g(w) = blwl w, Ibl= 1, wEB.
Ne'u {l = 1 thl daub gia c~n tfen Ia hi~n nhien va d~ng thuc kh6ng
th~ Kay fa.
0
H~ qua 5.1. VI 8(B) <

7r - 81 - p8, 81
(5.1) dzt(fc vief dztcJi d(lng
P8

(

Q(g) > qK 1 - -:;

>

7rq2, cg,n dztcJicua Q(g) trong

-if

)

(g E G),

(5.3)

suy ra

Q(g) >
Deing thac (5.3) ho(ic (5.4) xdy ra

{:}

(5.4)

qK.

B = Bo va g(w) = alwIK-1w, w E

B, lal = 1.

5.2

Danh gia c~n dtioi ban kinh R(g)
K

Vi R(g) > Q(g), \/g E G, tli dinh 19 5.1, ta co R(g) > (1 + S~~))-2.
M~t khac, ta con co dinh 19 sail:

Dinh Iy 5.2. V6i caegid thief va ky hi?u iJ chztdng 2, \/9 E G va 8 > 0,
ta co cae danh gia:
1f

R(g) >
R(g) >

(

1- 8(B)

)

82

( 82 -

p8

29

2

81{l KI

)

(5.5)
j

If

.

(5.6)


Cluing minh. Thea (5.2), voi 8 >
1
82

a
1

1<

+ p8 ( R(g) )

> 81 ( Q(g))

-*

1<

>R

(5.7)

(81+ p8).

Suy fa

R --k
82

81+ p8

= 82 - 8(B), ta co (5.5).

M~it khac, tli (5.7) va (5.1), ta co
2

82

> 81

(1'*)K

1

1<

+ p8 ( R(g) )

.

D

Tli day suy fa (5.6).

5.3

Cae danh gia khae eho Q(g), R(g) va Ig(w)1

Blob Iy 5.3. VJi cae gia thitt va ky hi£1u(J chZlcJng2, 'l/g E G, ta co
-x

Q(g)

<

1

J-L

RK(p, Iwl,q) < Ig(w)1< RK (p, 1~I,q) - RK (p, -K
1;I,q
Q (g)

<

f-l

,

(5.8)

,

(5.9)

)

(qK <)RK(p,d,q) < R(g) < RK (p,~,q) - RK (p,~,q)
RK(p,d,q)RK (p,~,q) < Q(g).

(5.10)

Cluing minh. Ap d\lng bfft d~ng thuc phai cua (4.20) cha PBHKABG
w = j(z), z E A, ham ngu'Qccua z = g(w), wEB, voi 9 E G, ta co
Iwl < t voi t = T [v, Ig(w) I*",q] .
Do do, theo dinh nghla cua hai ham sf{ ph\! T(p, r, 8) va R(p, t, 8),
E G va Vw E B, ta co
1

Ig(w)IK = R(p, t, q)
30


va rhea tinh don di~u (3.11) cua ham R(p, t, s), ta co
R(p, t, q) > R(p, Iwl, q).
Tli do suy ra c~n duoi cua Ig(w) I trong (5.8).
M~t khac, ta nh~n duQc tli ba't d~ng thlic trai cua (4.20), \:/g E G va
\:/w E B,

Iw I

>

;

v6it'

=T

[p,

C~~i I) k , q]

.

Tli do tu'dng t1/ nhu' tren, \:/gE G va \:/w E B, ta co
.

-k

Q(g)

q

I

( Ig(w)1 )

= R(p, t, q) > R ( p,~,

q) ,

Ke't hcjp voi (5.1), suy ra cac c~n tren cua Ig(w) i trong (5.8).
Tu'ong t1/, nho ba't d~ng thlic (4.22), ta co th~ chi ra cac ba't d~ng thlic
(5.9).
f)anh gia c~n du'oi (5.10) d6i voi Q(g) duQc suy ra tr1/c tie'p t11(5.9).
D
H~ qua 5.2. Tit (3.24), ta nh(m du(lc cac danh gia ddn gilln.
v fii cac gill thief va ky hi~u iJ chuang 2, \:/gE G, ta co

4-*lwIK < Ig(w)1 < 4{fQ(g) C~I) K < 4*p-k
K

K

K

(~)

4-P dK < R(g) < 4P Q(g)

4-~

(c )

4 _2f,'

CJ

K

K

<

C~I) K,
K

1

4P jL-K

( ~)

'

< Q(g) < 1
R(g)

(5.11)
(5.12)
(5.13)

,

(q:r< Q(g).

(5.14)

Tit (3.12) va (3.16), ta thfly rling cac h~ slf chi phl:l thuQc vaa K va p
trang (5.11)-(5.14) la tot nh{{t.
Chzl y 1. Truong hcjp cac thanh ph~n bien o-jthoai hoa thanh p di~m
roi r(;lc,hi~n nhien danh gia (5.11) vftn con dung ne'u ta thay cac da'u <
bdi <. Nhu' v~y, bAng cach thac tri~n lien t\lC ham z

31

= 9 (w)

t(;lip di~m


bien dfi nell, ta thffy (5.11) v~n con dung cho PBHKABG d6i xung quay
p lfin z = g(w) mi~n nhi lien B nQi tie'p tfong hlnh vanh khan q < Iwl < 1
leD hlnh vanh khan Q < Izi < 1.
Chu j 2. (5.14) co th~ s~c hon (5.4) khi q -+ 0, C -+ 0 vdi di~u ki~n
d = canst va" -q = canst.
c
H~ qua 5.3. TruiJng hC;pC1, C2 va cac CJjlan ll1c;tla cac dl1(Jngtran
Iwl = Qt, Iwl = 1 va cac nhat cdt tren duiJngtran Iwl = R', ta co cac
danh gia sau
~

,1
Q'K

< Q < Q K,

(5.15)

,1

.
RK

(5.16)

R" Q')
(p,R', Q') < R < RK (p,
Q ~,

Trang (5.15) dling thac traixay ra {::}B = Bo, g(w) = alwIK-1w, w E
1 1
B, lal = 1 va dling thac phai xay ra {::}B = Bo, g(w) = blwlK- w, w E
B, Ibl = 1.
Chang minh. Th~t v~y, khi do q = M1 = Q', C = d = R', J.t = J"
S(B) = 7r(1 - Q'2), 81 = 7rQ'2, 8 = O.
Do do, ap d1;lngdinh 19 5.1, ta nh~n du'Qc(5.15), cling vdi di~u ki~n
xay fa ding thuc.
D
Tu (5.9) va (5.15), nh~n du'Qc(5.16).
H~ qua 5.4. Ktt hC;p(5.2) vai (5.9), ta tim l(Ii c(ln dual cila Q(g) co thi
sdc h(Jn (5.1) nhu sau
1
82

>

81

81

( Q(g) )

1<

1

+ p8 ( R(g) )

( Q(g) )
1

>

1<

1<

K

q
-k -k
p,~, q J.t
.

( (

+ p8 R

) )

Suy ra

-~
82 - p8.R2 (p,~, q) J.t~

Q(g) >

(

81
32

)

.


H~ qua 5.5. Tit (5.9) va (5.10), ta nh(m dLt(le,nhiJ (3.12) va (3.17), cae
danh gia sau day ddi vai Ide d{j h{ji I¥ ctla R(g), ~i~i va Q(g) lrong eae

tntiJng h(lp gifJi h(ln

K1f2

1- R(g)

<

1-

RK(p,d,O)

~ K[l-

R(p,d,O)]

~

1
8
2p n p(l-d)

(5.17)

khi d -+ 1, tlle la R(g) -+ 1 khi d -+ 1.
CJ 0
1 - Q(9) < 1 - RK
R(g)
P, c'

(

~

khi ~

-;

I, tlic fa

~i: i

-;

!{

[ 1-

1 khi ~

R (p, ~,O )]

~

C

~

~

(5.18)

1.

-;

K [(1- R(p, d, 0)) + (1 - R (p,~, 0))]

!{ 1f2
K 1f2
8

2p In p(l-d)

+

8

(5.19)

2p In p(l-~)

~ -+

1.
e
FJanh giG (5.19) n6i r2ingQ(g) ddn tdi 1 ntu d

5.4

K1f28
2p In p(l-~)

< 1-RK(p,d,0)RK(p,~,0)

1-Q(g)

khi d -+ 1 va

)

-+

1 va :l. -+ 1.
c

Danh gia g6c md j3(g)

R5 rang ta luan co 0 < {3(g) < 21f, 9 E G, tuy nhien ta mu6n co danh gia
p
t6t hdn trong nhfi'ng tru'ong hejp nao do. Mu6n v~y ta dung phu'dng phap
dQ d~li-di~n tich hay con gQi Ia dQ dai Qtc tri do Ahlfors va Beurling [1]
d~ xu'ong nam 1950, giup giiH quye't nhi~u bai toan t6i u'u trong PBHBG.
Md rQng phu'dng phap do cho PBHKABG, ta co b6 d~ sau:
B6 d~ 5.1. Trong m(lt phdng z eho hlnh ehil nh~t
D = {z = x + iyl 0 < x < a, 0 < y < b}.

33


Gia sa ham so' W = j(z) th1!c hi~n mQt PBHKABG hlnh chTl nh(lt D
ZenmQt ta giac Gong H cila m(it phdng W saD cho cac dlnh 0, a, a + ib
va ib cila D ztm Zufft tu(jng ang V(ji cac dlnh WI, W2, W3 va W4 cila H.
GQi r ZahQ cac cung r trong H noi cc;mhWIW2wJi c(;mhW3W4cila H.

GiGsa co ham dQdo p = p(w)
0 < Ip(r) =

>0

lien tf:lC trong H saD cho

lp,dW,<

00,

V"{

E

r

va
0 < SetH) =

JJ HP2dudv <

00,

W =

u +iv.

lJ(it
l p = inf l p ( r
fEr

).

Khi do, ta co
Sp(H)

>

1 a 2
K blpo

(5.20)

Ddng thac (j (5.20) co thl xay ra.
W4
W3

a + ib

ib

WI
D

0

,Dx

x

a

W2

Hinh 5.1: PBHKABG hlnh chu nh?t D Jen tu giac cong H.

Chang minh. *Tru'dng h
=1

D~t

5x=Dn{zl~z=x}varx=j(5x),
Thea giii thie't, ta co

Oa

Sp(H)

= JJ D p21f'(zWdxdy
34

=

J0 dx 1. p21f'(z)12Idyl.


Theo ba't d~ng thuc Schwarz I, ta nh~n du'QC\Ix E (0, a)

L pV(zWldyll,

Idyl >

(L pl!'(z)lIdyl)
2,

va do foxIdyl= b > 0 Denco

l
Do do, d€

p21f'(zWldvl

y 1x E f,

> ~

2

ta co

a

SetH)

(l plf'(zJ[[dvl)
.
2

a

i J0 (l plf'(z)lIdYI)
dx= i J0 (1.

>

2

pldwl) dx

a
12

J

> y;lp dx =

a2
y;lp.

0

*Tru'ong hQp K > 1.

Xet T/= h(w) la PBHBG tu giac H leD hlnh chii'nh~t
D' = {'TJ= S + it 0 < s < a', 0 < t < b'}
I

sao cho cac dlnh WI, W2, W3 va W4 cua H l~n hiQt tu'ong umg vdi cac

dlnh 0, a', a' + ib' va ib' cua D'.

Ap dvng chung minh tren cho anh x~ ngu'Qch-I, ta co
a'
Sp(H) > bll~.
M~t khac, anh x~ h 0 f la PBHKABG hlnh chii'nh~t D leD hlnh chii'
nh~t D' Den co

a'
1a
->-b' - K b.

IBilt dAng thuc co d~ng

!
!
U g(x)h(x)dx ) ,

trong do giii thie't g(x), h(x) lien t\lc teen do~n [Xl,X2]va dAng thuc xiiy fa khi va chi khi g(x)
X E [Xl, X2], C = canst.

35

= Ch(x),


Tli do co (5.20).
f)~ng thuc (j (5,20) co th€ xay fa, ch~ng h~n khi H tIling voi D',
a'
'

1a 2
'- a
1a '2
'b'
l
K
Th " " kh d
1
(b )
)
(
P W = va b =
b"
~t v~y,
1 0 KbP = K b
=a =
'

Sp(H).

D

B6 d~ 5.2. Trang m(it phang z eho miin
E = {zl rl < Izi < r2, 'PI < arg z < 'P2}.
Gia sit ham sa W = j(z) thl;tehita giac Gong H cila m(it phang w sao cho cae dlnh Zl = rl ei'P2, Z2 =
ei'Pl, Z3 = r2eiWI, 'W2, W3va W4cila H. GQi r fa hQ cae cung "( trong H m5'i C(mh WIW2
vdi qmh W3W4cila H.
Vai cae ky hi

Sp(H) >

~ 'P2In--r:ll~,

(5.21)

rI

Dang thric (j (5,21) co thi xay ra.

Z4

~

! r)

1'9,

WI

A'f)

1/'?2
01"\

/67

r2

H

Z3

W4

W,
'"'II{>
W2

Z2

i In r2
rl

~
~6x

0

-x

-!.pi + i In r2
rl

~ -\PI

Hinh 5,2: PBHKABG mi~n E Jen tti' giac cong H.

36

Chang m inh. Qua phep bie"nd6i

= x + iff = rp2+ i In~rl =

z

(rp2

- rp)+ i In ~, z = reirp,
rl

mi~n E se bie"nthanh hlnh chITnh~t

z x + iY 0 < x < - 0 < y < In ~: } ,
- . r2
,r2
1
tfong d 0 cac d m h Zl = 0 , Z2 = rp2-rpl, Z3 = rp2-rpl +1, 1n - va Z4 = 1,nD

'

=

,

{ =

'P2

I

'p"

,,-

?

.

rl
rl
Iftn htCjt tu'ong ling vdi Zl, Z2, Z3 va Z4, va cac cling Arp,6x, 1rptu'ong ling
vdi nhau nhu' tfong hlnh 5.2.
Tli b6 d~ 5.1, suy fa (5.21).
D
Dinh Iy 5.4. V cJieae giG thilt va ky hi0 < q < M1 < C < d < m2 < M2 = 1, ta co eae bat dang thue:
21r- p{3I 2 m2 + p{3 I 2 ~ + p{3 I 2
1 nM
RnM
IndIn 1 In 1 In Q
Q
R
21r- p{3 In2 m2 +
In (4~ qCd)

M1

In2 ~

p{3
In (4~~)

M1

trong do
Spa (B)

=

+

fi

p{3

m2

< K s (B)
Po'

In2 m2 < K2S (B).

In (4~~)

d

dudv
BU

2

+V

va

(5.22)

(5.23)

Po

.
2'

W=U+'lV.

Dang thac (j (5.22) co thl xay ra.
Chang minh. f)~t

Arp= A n {z arg z
I

= rp}

va vdi m6i j (j = 0, ...,p - 1), d~t

Qlj = {z
Q2j = {z

Q3j = {z

QQ <

Izi

-a+(2j+1)7r

p


< R, ex+ (2j - 1)7r < arg z <
P

-0:

p}

,
+ (2j + 1)7r
P
}

R < Izi < 1, ex+ (2j - 1)7r < argz < -a + (2j + 1)7r .
P
P}
37

,


1

Hinh 5.3: PBHKABG mi~n chuffn A len mi~n B ling vdi (p = 2).

ngu<;1cf

=

0 < lp(1k) =

1

g-1 bie'n mi~n A leu mi~n B, trong do Qlj
bie'n thanh H1j co mQt c~nh tren C1 va mQt c~nh tren c2, Q2j bie'n thanh
H2j co mQt qmh n~m tren C1 va mQt c~nh n~m tren O"j,Q3j bie'n thanh
H3j co ffiQtqlllh nam tren (J"jva mQt c~nh nam tren c2.
Gia su fk lftn 1u'<;1t
la hQ cac cling 1k trong Hkj (k = 1,2,3) lftn hi<;1t
n6i C1 vdi C2, n6i C1 voi O"j,n6i O"jvoi C2 va gia sa co p = p(w) > 0
lien t\1Ctrong B sao cho
PBHKABG

pldwl

< 00,

'V1k

E fk (k

= 1,2,3)

"fk

va
0 < Sp(Hkj) =

jf

p2dudv <

(k = 1,2,3), w = u + iv.

00

Hkj

£)~t

lkp = "fkEfk
inf Ip(1k) (k = 1,2,3).

D~ dang tha"y rang
f1::)

U f(A

f1

=

'\pEfi

U

A
-£1+(2j+ 1)~::;
Tit do, ap dvng b6 d.; 5.2 cho mi.;n Q'j, yiJi p( w) =

1 20: 2
!{ ---yllp < Sp(H1j).
InQ
vdi

m2

hp = inf lp(11) =
"fiEfi

J

fvh

38

-1
1

W

1dwl
1

m2 ,
= In M
1

I~I' wEB, co


tuc Hi
In- 1
Q

VI 2po; =

2'if

(5.24)

< pK Sp(HI),

2'if - p{3 In2 m2

M1

- p{3.

1

Tu'dng ttf, d6i vdi mi~n Q2j, vdi p(w) = Iwl' wEB,
f2 ~

U

co

U

f(A'P) voi r2 =

A.pEfz

1'1"

(X+(2j-l)~:::;'P:::;-(X+(2j+l)~

2
-1

'if

(P

!(

-0;

R

) Z2 < Sp(H2j).
-

2p

In Q

vdi

c

.

l2p

= bEG
Inf

1

J _w

lp(12) =

I

C

idwi
I

= In M 1 '

Ml

tile IiI

p{3

In-R
Q

In2 ~ < pKSp(H2j).
M1-

Tu'dng tif, d6i vdi mii\n Q3j, vdi p( w)

f3 ~

U

r3 =

f(A'P) vdi

A
= I~I' WEB, co
U

1'1"

(X+(2j-l)~:::;'P:::;-(X+(2j+l)~

2
-1
!(

'if

(P

-0;

1

)

In R

vdi

= I'3Ef3
inf lp(13) =

-< Sp(H3j).

Z2
3p

mz
l3p

(5.25)

Jd
39

-l
I

w

2,

lldwi = In md


tuc Ia
PP

2m2

(5.26)

d < pKSp(H3j).

~In

InR

Tli (5.24), (5.25) va (5.26), VOlchu

9

pK(Sp(H1j) + Sp(H2j) + Sp(H3j)) = KSpo(B)
ta nh~n duqc (5.22).
Danh gia (5.23) duqc Suy fa tu (5.22) va h~ qua 5.2.
Ne'u B = Bo, tuc ml = M1 = q, C = d = r, m2 = M2 = 1 va ne'u
1 1
1
Z = g(w) = IwIK- w, tuc Izl = IwlK hay Iwl = IzIK, q = QK, r = RK,
thl
K

".

ve tral (5 .22)

-

21r -

In
-

21r -

PP 1n2 -1 + - PP 1n2 r- + - PP 1n 2 -1

~

In R

Q

PP1 2 ~

In -1
Q

-

q

Q

q

In

PP 1 2 RK

~

r

R

PP 1 2 ~

n QK + In -R n QK + In -1 n RK
Q
R
1
R
1

= I{2(21r - pp) In Q + I{2p{3In Q + K2p{3ln R
1
1
1
= 21rK2 In Q = 21rK In QK = 21r!{In q'

. 5
ve phal ( .22) = K
K

?

dudv

J U +v
q2

2

27f 1

-

K

JJ
0

27f

~d:2de

q

= K

1

J deJ
0

q

~~= 27rKln~.

V~y, ta co dAng thuc (j (5.22).

0

Chl1j 3. Cac bftt dAng thuc (5.22) va (5.23) v~n con dung khi m2 <
ho~c C < M1 ho~c m2 < d ne'u ta Iftn Iuqt d~t In

M1

:~ = 0 ho~cIn;1 = 0

ho~c In :2 = 0, d6ng nghIa VOlvi<$cta khong d~ 9 de'n s1;id6ng g6p cua
di~n Hch cae t~p con cua B khong thoa gia thie't dinh 195.4.
40


H~ qua 5.6. Vlii cae gid thief va ky hi~u iJ chu(Jflg 2, V9 E G, wEB va
0 < q < M1 < C < d < m2 < M2 = 1, kef h(Jp (5.23) vlii bift dang
27r
thac hiln nhien 0 < {3(g) < -, ta co danh gia cho {3(g), Vg E G:
p
2

27r
0, 1-

max

p

(

Chu

K 2 In -1 In 415dc
q 2

In

[

(

~
M1

27r

q

)] }

< {3 (g) < -.p

y 4. Khi m2 = 1, M1 = q, ta xet
J(2In.! In 4~s..

27r

1-

-

( )

q

P

qd

In2 rn2
M1

[

]

K2InlIn

27r

=

-

1-

P
[

-

4js..
qd

In2 1q

)]

(

27r

p

q

(

1 - K21n 4~q'J)

[

Inl.q

]

2

K2 In

D~t C =
saG cho

( )
411

In-q1

s..

qd ,

trong tru'onghQpd = canst, cho q ---+ 0, C ---+ 0

~ = canst,thl C ---+ O. V~y 27r > {3(g) > 27r(1 - C), tile c~n

q
p
p
du'dicua {3(g)trong h~ qua 5.6 Ia ffiQtdaub gia s~c, it ra cho tru'onghQp
da neUe

41