bất đẳng thức carleman, các hệ quả và mở rộng

12

CHUaNG 2

CAC CONG CT)
Trang chuang nay, chung toi lieU mQt s6 dinh Iy, b6 d~ va cac h~ qua dn
thi€t cho vi~c danh gia cac d<,tiuQng hlnh hQc d6i vdi cac lOp ham F va G.
l

2.1 nilt diing thuc Carleman, cae h~ qua va md r{)ng
n6 d~ 2.1: (nilt diing thuc Carleman)
Gia

stt

w = fez)

A={zl(O<)r
la mQt PBHBG

don

di~p

hlnh

vanh

leu mQt mi~n nhi lien D khong chua di€m

bien trang C1 va bien ngoai C2 san cho

Izl

khan
00

vdi

= R tuong ling vdi C2. GQi S la di~n

tich (trong) cua t~p md do C2 ban bQc, s la di~n tich (ngoai) cua t~p dong do
ban bQc. Khi do, ta co:

(2.1)

S2(~JS.

D~ng thuc xay fa khi va chi khif(z) = az +b vdi a,b la hling s6 va a:;t:O.

~

w=f(z)

A

Or

R

z

w

Hlnh2.1
Chung minh: Xem [4, tr. 212].

C1


13

H~ qua 2.1: (Dinh nghia modun mi~n nhi lien)
Gia sa mi~n nhi lien D qua cac PBHBG / va 1; l~n hiQt bie'n leu hai hlnh
vanhkhan H={wlrthl
R = RI
r
1J

(2.2)

Ti' s6 nay duQc gQi la m6dun cua mi~n nhi lien D va duQcky hi~u la
mod(D) .
Chung minh:

/

1;

~

~
D

HI

0

R

O~

z

R]

WI

ffinh 2.2
Xet PBHBG j:j;-I mi~n HI leu mi~n H, rhea b6 d~ 2.1, ta co

ffR'

hay

~

[ ~'

J ffr'

R
-2- RI
r
1J

(2.2a)

TucJngrtf, ta xet PBHBG 1;0/-] mi~n H ten mi~n H], rhea b6 d~ 2.1, ta co
2
2

1rRl 2

hay

R]
-2-.
1J

R

Tli (2.2a) va (2.2b), suy fa (2.2) .

r

R

[ -; )

2
1r1J

(2.2b)


14

H~ qua 2.2: (Tinh ba't bie'n cua modun mi~n nhf lien)
Ne'u mi~n nhi lien A eo cae thanh phffn bien kh6ng thocii boa thanh mQt
di6m du<;1e ie'n baa giae don di~p len mi~n nhi lien B thl
b

= mod(B) .

mod(A)

(2.3)

Chung minh:

~

f

g

h~

~

~

HI

O~

0

Rj

Rz

w

Hinh 2.3
GQi f la PBHBG don di~p mi~n A len mi~n B . X6t g la PBHBG don di~p
mi~n A !en hlnh vanh khan HI
Bien

={sh < Isl< RI}

va h la PBHBG don di~p mi~n

hlnh vanh khan Hz = {tlrz < ItI< Rz}.

Thea h~ qua 2.1, ta eo:
mod(A) = RI va mod(B) = Rz .
~

rz
D~t cp= hf

thl cp PBHBG don di~p mi~n A len hlnh vanh khan Hz.
la

Theo h~ qua 2.1, ta eo:
mod(A) = Rz va Rj - Rz
rz
lj -- rz

V~y ta eo mod(A) = mod(B)We(2.3) .


15

H~ qua 2.3: (Tinh don di~u cua modun mi~n nhi lien)
Trang m~t ph&ng z cha hai mi~n nhi lien A va B vdi modun tu'dng ung
la R va R] , co Hnh cha't A c B va A ngan cach hai thanh philo bien cua B.
r

lj

Khi do, ta co:
R
-~-. Rl
r

(2.4 )

1j

D&ng thuc xay ra khi va chi khi A = B.
Chung minh:
w=f(z)
~

R

Hinh 2.4
VI mod(B) = RI Den t6n t!;liPBHBG ddn di~p f mi~n BIen
lj

khan

11= {wi'i
nhi lien

A vdi

hlnh vanh

mi~n A trd thanh mi~n

mod(A) = R co bien trong la C] va bien ngaai la Cz saD cha C]
r

baa quanh ha~c trung vdi

!wi

= 1j va Iwl = Rl baa quanh ha~c trung vdi Cz. GQi S

la di~n rich (trong) cua t~p md da Cz baa bQc, s la di~n rich (ngaai) cua t~p
dong da C1baa bQc.


16

Khi do, ta co:
s~m/
S 5, 1rR)2 .

va

(2.4a)
(2.4b)

Vi mod(A) = R nen t6n t(;liPBHBG don di~p g mi~n A leu hlnh vanh
r
khan D = {t Ir

5, It 1 5,

R}.

Ap dl,mgb6 d~ 2.1 cho PBHBG w= g-) (t), ta co:

(2.4c)

s~(:Js,

trong do d~ng thuc xay ra khi va chI khi g-) (t) =at +b vOi a, b la hang s6, a;t: o.
Tuc A la hlnh vanh khan.
Ti'icac k€t qua tren, ta co:

(~ J

< ~(:J.

Ti'i do suy ra (2.4).
f)~ng thuc (j (2.4) xay ra khi va chI khi cac d~ng thuc (j (2.4a), (2.4b) va (2.4c)

cungxayra,tucA=B

hayA=B-

B6 d~ 2.2: (Md r{)ng bilt diing thuc Carleman bdi Thao[12, tr. 521])
Gia

su

w = fez)

la

PBHKABG

mQt

hlnh

vanh

A = {zl (0 <) r < Izi< R (< oo)} leu mQt mi~n nhi lien D khong chua di€m
bien trong c) va bien ngoai C2 sao cho

Izi

khan
00

vdi

= R tu'ong ling vdi C2. GQi S la di~n

tich (trong) cua mi~n do C2 baa bQc, s la di~n tich (ngoai) cua t~p dong do C)
baa bQc. Khi do, ta co:
2

s~(~)K

s.

(2.5)


17

D~ng thuc xay ra khi va chI khi fez) = alzr~.-I+b vdi a,b la h~ng s6 va a ~ o.
Chung minh: Xem [12, tr. 521], [17, tr. 13-14].

2.2 Md r{)ngcae bitt diing thuc Grotzsch va Kiihnau
B6 d~ 2.3: (Bitt diing thuc Grotzsch 1)

Gia sa w = fez) la PBHBG ddn di~p hlnh vanh khan H = {zl(O~)r leu mi~n nhi lien D vdi bien ngoai Iwl= 1va bien trong c, sao cho mi~n gidi h~n

boi c luau chua hlnh troll Iwl,

?

rang tren

C

co p

2bi

diem

Wk

= Me P ,( k = O,1,...,p-1).

Khi do, ta co:
(2.6)

MsT(p,r,s),
trong do d~ng thuc xay ra khi va chi khi f = fo
vanh

khan

H={zl(O<)r
D

= {wi s

Lj

= {w s:>!wI :>/,argw

< Iwl < 1} , (0

c~t

s s < r < 1) bi

la mQt PBHBG

leu
boi

mi~n
p

ddn di~p hlnh

lien

nhi

2

do~n

thang

~ 2;j },(o:> s < / <1),j ~ 1,2,.--,p.

Chung minh: Xem [6, tr. 372] hay [19, tr. 18 -20].
B6 d~ 2.4: (Md r{)ng bitt diing thuc Grotzsch 1 bdi Thao[13, tr. 63])
Gia sa A la hlnh vanh khan R-

hi

c~t cung troll d6ng tam 0 sao cho A trung vdi chinh no boi phep quay z = ze P .
GQi f la PBHKABG mi~n A leu mi~n B n~m trong 0 < Iwl< 1 sao cho dliong
troll Iz 1= R tlidng ling vdi bien trong C giOi h~n mQt t~p dong chua g6c tQa dQ,
du'ong troll Iz 1= 1 tu'dng ling bien ngoai C cua B . Hdn nii'a gia thi€t B trling
2~i

vdi chinh no boi phep quay;:;:' = we P .

18

Khi do, ta co:
(2.7)

M~T(p,R*,m),
vdi M = max{lwl,WEc}, m =min{lwl, WEc}, 0 ~ m ~M < 1.
Ding thuc xay ra khi va ChI khi w=f(z)=ah(u),lal=l,u=bzlzlt-I,lbl=l,

h

la PBHBG don di<$phlnh vanh khan R* < lul< 1 ten mien nhi lie~ sao cho
lul = ltu'ong ling vdi bien ngoai C = {wllw\ =I},

trong

c~

con

lul

= R*

tu'ong ling vdi bien

{~I+ m}u{ ~m,;H,; M,argw~2;j,j ~1,...,+

Chung minh: Xem [13, tr. 63] hay [19, tr. 33 - 35].

Nho phep bien d6i z = Q va W = ml , b6 de 2.4 trd thanh
z
w
H~ qua 2.4:

Giasa A lahlnhvanhkhan

Q
pn,(p =1,2,...;n=0,1,2,...)
2/T'

cung troll d6ng tam 0 sao cho A trung vdi chinh no bdi phep quay;

sa

= ze . Gia
P

f la PBHKABG mien A ten mien B nam trong 0 < Iwl < 00 sao cho Iz I = Q

tu'ong ling vdi bien trong C] baa g6c tQa dQ, du'ong troll Iz1= R tu'ong ling bien
ngoai

C 2 cua B. Hon nua gia thiet B trung vdi chinh no bdi phep quay
2/T'
P

w=we

.

Khi do, ta co:
m2 2::

;1 *, !!!L

T [ p, ( R )

'

M2 J

vdi M2 = max{lwl, WE C2}, mj = min{lwl, WE Cj },j =1,2.

(2.8)


19

B~ng thuc xay fa khi va chi khi w = f(z) = ah(u),Ial= I,u = bzlzlt-),Ibl= 1, h la
PBHBG hlnh vanh khan Qt < lul< Rt len mi~n nhi lien E saD cho lul = Qt tu'ong

voi

va

ling

voi

Gia sa w = fez) la PBHBG don di~p hlnh vanh khan A = {zl(O<)r

< Izi< I}

ling

c,

~

{wI1wi

~

c] = {wllwl= m)}

M,} u{ wllwl

~

m, ,; w,; M"argw

lul=Rt

~ 2;j

tu'ong

,j ~ 1,...,P}.

Chung minh: Xem [13, tr. 64] hay [19, IT.35 - 36].
Be}d~ 2.5: (Ba't diing thuc Grotzsch 2)

len mi~n nhi lien B n~m trong hlnh troll
Iwl

= 1 va bien trong c) saDcho

Izl

Iwl

= 1, c6 bien ngoai C2 Ia du'ong troll

= 1 tu'ong ling

voi C2.

Khi d6, du'ong kinh D cua c) thoa
D::; Do

(2.9)

= 2T(2,r,O),

trong d6 D = Dokhi va chi khi c) la do~n th~ng nh~n w = 0 lam trung di~m.
Chung minh: Xem [8, tr. 220].
Be}d~ 2.6: (Ba't diing thuc Grotzsch 2 md rQng)

Gia sa w=f(z)

la PBHKABG hlnh vanh khan A={zl(O<)r
mi~n nhi lien B c6 bien ngoai C2 va bien trong c) saD cho
C2. B~t M

= max {IwllWE

Izl

= R tu'ong ling voi

C2} . Khi d6, du'ong kinh D cua c) thoa

(2.10)
D';Do ~2MTH~rol
trong d6 D = Do khi va chi khi w= fo (;)

voi; = azlzl-t-] ,Ial = 1 va fo

la PBHBG

don di~p hlnh vanh khan A=FI r-t < 1;1 < R-t} len hlnh troll Iwl< M bi ciit dQc
do~n th~ng nh~n w = 0 lam trung di~m saD cho 1;1= R-t tu'ong ling voi Iwl= M .


20

Chung minh:

.:. Tru'ong hQp 1: K = 1, C2 trung voi du'ong tron Iwl = M

w=j(z)

~
B

R

Bo

(]

M

0

w

------

M

w

~

w=fo(z)
Hinh 2.5

Chi dn thljc hi~n cac phep co dan

; = ~ va ; =

;

, d~ dang du'a tru'ong

h<,Jp
nay ve tru'ong hQp cua b6 de 2.5 voi mien A thay bdi
B thay bdi l3 nQi tie'p trong hih tron

1;1

A= {;I ~

< 1;1 I} va
<

< 1. Trd ve cac bie'n z va w ta thu du'Qc

(2.10) vdi K = 1 cling ke't lu?n cho D = Do'
.:. Tru'ong hQp 2 : K = 1, C2 Ia bien ngoai ba't kl cua B

GQi l3 la mien nhi lien chua B co bien ngoai la

!wi

= M, bien trong Hi C].

Do tinh don di~u cua m6dun mien nhi lien (xem h~ qua 2.3) , ta co:
mod(B) ~ mod(B).

Theo h~ qua 2.2, ta co:
mod(B) = R.
r

M~t khac, gia sa l3 co modun
mod(l3) = ~ .
r


21

V~y
r r
-
Thea tinh chfft don di~u (1.17) cua ham ph\) T(p,r,s), ta co:

(2.lOa)

T[ 2,~,O)~T(2, ~,O}

~
A

B

,,'

,/

Or

R

O

I

..

'
I

I
I
I

C

I

\
\

"M

.

:

Cz,'
I

1
\

\'

'''

'-------- - --

,,

,

,

I

B
M

CIG

w

Hinh 2.6

Ap d\)ng tfu'ong hQp 1, ta co:

DQMT(2,

~,o

J

Ke't hQp voi (2. lOa), suy fa

D:; Do= 2MT( 2, ~, 0).
Tuc (2.10) vOi K = 1.


22

.:. Tru'ong h<;lp3 : K ~ 1, C2 la bien ngoai ba't ld cua B

~
A
,/ '

Or

R

"

:

I

I
r

~,

B

'-\

G

\M
I

I
I

C

\

'-\

J

,

"""'

~

c2,',

"

~~----

u~g(w)
I
"

'\

l»I

O~
u
Hinh 2.7
Mi~n nhi lien B co th~ bie"n baa giac ddn di~p bdi u = g (w) len hlnh vanh
khan

BI = {ulo < fJ < lul< Rj} sao cho C2 tu'dng ung voi lul = RI'
Ap dl;mg tru'ong h<;lp2 cho PBHBO . w = g -I (u) hlnh vanh khan B( leu mi~n

B , ta co:
(2. lOb)
DSlMT(

2,~,0).

M~t khac, hlnh vanh khan Bj co th~ xem la anh cua hlnh vanh khan A qua
phep bie"nhlnh h<;lp
cua PBHKABO f

voi PBHBO g, tuc qua PBHKABO gof.

Do do, theo (1.2), ta co:
~

-<

RJ -

-

r

(R )

*
.

Hdn nii'a, theo Hnh cha't ddn di~u (1.17) cua ham phl,l T(p,r,s), ta co:


24

B6 d~ 2.8: (Ba't diing thuc Kiihnau md rQng)

Trang m~t ph&ng z cho mQt hlnh v~lllhkhan A = {zl(0 <)r < Izl< R}. G<;>i
w = f( z) Ia PBHKABG bie'n mi€n A Ien mi€n nhi lien B co bien ngoai C va

bien trang c sao cho

Izl

= R tlidng ling vdi C. G<;>i Ia di~n rich (trong) cua mi€n
S

do bien ngoai C bao b<;>c D la Quang kinh cua bien trong c.
va
Khi do, ta co:
S In(1D s; 1/

-i(

(2

)

,

,

.

VOl

D&ng thuc xay ra khi va chi khi

r t
(= T 1,( R ) ,0.

(2.13)

)

(

f(z)=fo(~)=bln(1-(~)+c,lbl=1
In(l- ( )

-1..

~=a~zK-'

RR
11

,lal=1.

Chung minh:

w=f(z)

----------.

de

R
AOr

z

w

s=g(w)
B,
1

Q'i
s

~
Hlnh 2.9

1

vdi


25

D§u lien, bi€n baa giac ddn di~p mi€n B boi s = g( w) leu hlnh vanh khan
BJ

= { sl 0

< fj < Isl < I} .

San do, th\lc hi~n PBHBG u = h( s) hlnh vanh khan BJleu mi€n nhi lien B2
gioi h~n boi du'ong troll lul=1 va nh£it c~t L(t)={uIOsaD cho Isl= 1 tu'dng ung voi /u/= 1. Theo dinh nghla ham ph1,1ta co fj = R"(I,t,O)
hay t = T(l,fj,O).
Ap d1,1ng d€ 2.7 cho ph6p bi€n hlnh hQp g-Joh-Jmi€n B2 leu mi€n B, ta co:
b6
Sln(1-t2)
-1(

D ~ ~I
M~t khac,

BJ

".

, VOl t = T (1,fj,0) .

(2.13a)

co th6 xem la anh cua A qua PBHKABG f.g la hQp cua

PBHKABG f va PBHBG g.
Do do, ta co:

~~(~r.
Theo (1.17) v€ tinh ddn di~u cua ham ph1,1T(p,r,s),

ta suy ra
(2.13b)

T(l,~,O)';TH~r

,0).

K€t hQp (2.13a) va (2.13b) ta co (2.13) voi phat bi6u v€ tru'ong hQp d£ng thuc 8
2.3 Ba't diing thuc theo Iy thuye't de)dai ctfc tri
Ly thuy€t dQ dai c\lc tri b~t ngu6n tu mQt s6 cac ba"td£ng thuc lien h~
giii'a modun cua mQt tu giac hay mi€n nhi lien, di~n tich mi€n do va dQdai ng~n
nha"tcua du'ong cong thuQc mQthQ du'ong trai trong mi€n do tinh theo mQt dQ do
ba"tky du'QcAhlfors va Beurling[l] d€ xu'ong nam 1950 dfftro thanh cong C1;1
huu
hi~u d6 giai nhi€u bai loan t6i u'utrong Iy thuy€t hlnh hQc ham bi€n phuc.


26

Trang m~t ph~ng z = x+iy, cho tu giac cong Q co cac dinh lfin Iu'Qtla A, B,
C va D. Qua PBHBG don dit%pw=f(z)=u+iv,
Q'={w=u+ivIO~u~a,O~v~b}

Q du'Qcbi€n ten hlnh chu nh~t

co dinh tu'dng ling lfin Iu'QtIa A', B', C', D' sao

cho A'B' = a; B'C' = b.
GQi r Ia hQ cac du'ong cong r n6i hai canh d6i dit%n va CD cua tugiac
AB

cong Q, <Dla hQ cac ham de>do p=p(Z)~O,ZEQ

sao cho dit%ntich cua tu giac

cong Q theo de> p la huu h~n, nghia Ia
do
Sp(Q)= Hp2(z}iS<+oo.

(2.14)

Q

De>dai cua cac du'ong cong r theo de>do p du'Qctinh b~ng cong thuc

lp(r)= Jp(z)ldzl(~+oo),rEr,pE
(2.15)

Sp(Q)~al~ vdi lp=inflp(r),
b
yer

(2.16)

y

B6 d~ 2.9:
Vdi cac ky hit%u
nhu'tren, ta co:

d~ng thuc xay ra khi va chi khi p(z) = kif (z)l,z EQ,k = canst.
Chung minh:

Ta co
Sp(Q)= Hp2(z}iS= Hp2(z)dxdy
Q

2

dudv

= [fp (Z)jf'(zt
b p2(Z)

=

Q

at b

(Z)

1;/

= II !If'(z)12 dvJu
b

du

2dvJdv ~=J Jdv

0 olf'(z)1
0

~

p2

J

0

1

b p2(Z)

J

~
bo olf'(z)1

b

2dvJdv
0

}

u

-

27

<: 7;

A JI;'~~I dv

du (Do

J'

ap d\lngBDT tich

phiin Schwarzt cho hai ham

p(z) va 1 trendOc;ln[O,b])
If'(z)1
2

1oJ
=b dlJp(z)ldzl J du

(ru Ia nghich anh cua doc;ln
th~ng u = canst,

~

0 ~ u ~ a, 0 ~ v ~ b ).
1a

=b 0
JI~(ru)du ~ a 12
b p'

Ding

!hac

a (2.15) xaY fa

khi

va chi khi

II

I~ gifta

hai ham

I;'~;)I va I la hang

sf), d6ng thai Ip(ru)=lp voi mQi ru,(O~u~a)<=>
p(z)=kl/(z)l,k=const,zEQ
vi khi d6 Ip(ru)= Jp(z)ldzl=k JI/(z)lldzl=kb=lp voimQi uE[O,a]
Yu

Yu

,

.

B6 d~ 2.10:

Trang m~t ph~ng w cho mQt tu giac cong Bo c6 hai cc;lnhn~m tren hai
duang troll Iwl=cva Iwl=d,Oc,

=
Gia sa z = g(w) la PBHKABG mi6n Bo len mi6n .40cua m~t ph~ng z .
Ta d~t Cr = g(Cr) , 0 < c ~ r ~ d < +00.
Hon

1p

mIa,

gia

( Cr) = f p(z) I dz I~
c,

sa

00,

p = p(z) ~ 0

c ~ r ~ d

duQc

xac

dinh

trong

.40 saD

va Sp (.40)= Hp2 (z)dxdy < oo,Z= x+iy
A

theo nghla Lebesgue. Ngoai ra Ip( Cr) ~ I~,c ~ r ~ d.

cho

t6n tc;li


28

Khi do, ta co:
1

2d

Sp(Ao)~ K(l~)

dr
J
crO(r)"

(2.17)

Chung minh:
GQi dS la vi phan cua Sp (Ao) tu'dng ling voi [r,r +dr] c [c,d] , tuc dS xa'p
"
Xl dt theo dQ do p(z) cua anh mi€n D = Bo n{wlr < Iwl< r +dr} bdi z = g( w). Do
dr(> 0) ra't be va O(r) kha tich tren [c,d]co th€ thay D bdi

15= Bon{wlr giac cong 15n~m tren

Iwl

=r

Ham t = Inw bie"nmi€n 15 len hlnh chii'nh~t voi cac q.nh
r + dr
r

In-=

dr
,"
.

dr 'n
In 1+~va,!,,!; r.) Vlv(;J.ymo d uncuatuglaccong
(
r )
r
(
?

/

/

-

'

D Ia

dr
dr
mod(15) = O(r) = r.O(r) .

Theo [ 3, tr. 19], ta co:

.
0 2
1 dr 2 1 dr
(Ip ) , trong d0 Ip = Ip Cx VOl r < x < r + dr
dS ~ K rO (r ) Ip ~ K rO (r )
~

/

( )

/

La'y tich phan hai ve"tren [c,d] ta du'cjc(2.17) .
2.4 Cae b6 d~ khae
B6 d~ 2.11: (Bie'n hai du'ong troD l~ch tam thanh hai du'ong troD d6ng tam)
Ne"u A la mi€n nhi lien gioi h(;J.n di hai du'ong troll Izi= 1 va Iz- hi= lj voi
b
0 < h < 1, 0 < lj < (1- h) du'cjcbie"n baa giac ddn di~p len hlnh vanh khan r < Iwl< 1
thl

r = r(r),h)= 1- h2 +r)2 -~(1-

h2 - r/ Y -4h2r)2
2r)
.

(2.18)


29

Truong

hQp A la mien

0

nhi lien

gioi h~n bdi

Izl= r2 va

Iz- hi = r) voi

thi

r = r(rl'r2,h) = r22
-h2 +r/ -~(r22 -h2 -r/Y
2r)r2
Chung minh: Xem [18, tr. 20

(2.19)

-4h2r12

- 22].

B6 d~ 2.12: ("D~o ham" cua ham ngtiqc cho PBHKABG)
Voi cae ki hi~u d phfin 1.2, giii sa W = f (z)la PBHKABG cua mien chua
z=O voi f(O)=O va m'(O,f»O.
f)~t g = I-I , ta co:
I

(2.20)

m '(0,f) = M* (o,gfX,
I

(2.21)

M'(O,f) = m*(o,gfX.
Chung minh:

Lfty R>O du be , d~t CR ={zllzl=R}va

C~ =/(CR),

r6 rang

t6n

t~i

WIE C~ va z) E C R saD cho

m(R,f)=lw)I=lf(z))I=r,

r>O.

f)~t Lr ={wllwl=r} va Lr =g(Lr)
Vi Lrn~mtrong

Izl~R,taco

M(r,g)=lg(w))I=lzII=R.

Dodo
I

'

"
m,(0 1) =1 m(R,f)
1m
r->O
-I
RK

= 1"
1m

r

r->O

M (r,g

-

)

)K

. M(r,g) -X
=1
1m
r->O [
rK ]

Tudng tlf, lfty R>O du be , d~t CR={zllzl=R}va
t~i W2E C~ va Z2 E CR saD cho

M(R,f)=lw21=lf(z2)I=r,

r>O.

=M *(O,g.)-t

C~ =f(CR),

r6 rang t6n


30

B~ t L, = {wllwl r} va I, = g (L, ) .
=
VI

Izl R
=

nflm trong t~p dong gioi h(;ln bdi I" ta co:

m(r,g) =lg(wz)1=lz21 R.
=
I

Dodo

M'(O,f)=limM(R;f)=lim
,~O

RK

r

-'- =lim
,~O m(r,g)K
,~O

[

m(r;g)
r

]

=m*(O,gft.

-K

H~ qua 2.5:
Cho

K = 1,

ta

co

m'(O,f) = If'(o)1

va

M* (O,g) =lg'(O)I.

m'(O,f) = M* (O,gft trd thanh cong thilc quell thuQc If'(O)1=lg'(OfI.

Luc

do

(2.22)

2.5 Cae daub gia eho lop ham F
BS xay d1!ng cac danh gia cho lOp ham G ta c~n cac danh gia duoi day
cho lOp ham F , tilc lOp ham nguQccua lOp ham G .

Dfnh Iy 2.1:
Duoi cac ky hi~u va giii thie"t d ph~n 1.2, voi mQi f EF, z EA,z *-0,z *- 00 ,
0 < R < 00 , ta co:

(2.23)

S'(O,f)~l,
PSI

(2.24)

~(l-S'(O,f))1Z"Rt,
2

2

S'(O,f)1Z"RK ~ S(R,f)

~ 1Z"RK,

(2.25)

1

~ RK ,

(2.26)

M(R,f) ~ Rt ~S'(O,f),

(2.27)

m(R,f)

m(R,f)

~ 4-; m'(O,f)Rt
1

M(R,f)~4P

(2.28)

-'-

RK,
.l

D(R,f)

,

-'-

~ 2AP RK ,

(2.29)

(2.30)


31

4-im'(O,J)lzlt ~IJ(z)I~4ilzlt,

4-im'(O,J)Rt ~c(R,J)~d(R,J)~4i

(2.31)
Rt .(2.32)

M6i d£ng thuc tu (2.23) d€n (2.21) xay fa khi va chi khi J(z) = azlzlt-l voi lal= 1.
Chung minh: Xem [19, tr. 54 - 56].